Mathematics, 14.02.2020 01:16 joceeeeelyn2899

A swimming pool whose volume is 10 comma 000 gal contains water that is 0.01% chlorine. Starting at tequals0, city water containing 0.001% chlorine is pumped into the pool at a rate of 6 gal/min. The pool water flows out at the same rate. What is the percentage of chlorine in the pool after 1 hour? When will the pool water be 0.002% chlorine?

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Answer from: harmonytaylor13

$y(t) = 0.9 e^{6 \times 10^{-4} \cdot 60} + 0.1 = 0.9 e^{36 \times 10^{-3} } + 0.1$ $t^* \approx 3663.7084$ minutes

Step-by-step explanation:

Let y(t) be the amount of chlorine in the tank at time .

Then, its time rate of change, , is the difference between the chlorine inflow and outflow rate (Balance law).

Since 6 gallons of water runs in the pool per minute, containing 0.001 % chlorine, the chlorine inflow rate is

$6 \cdot 0.001 \% = 6 \cdot 0.00001 = 6 \times 10^{-5}$

Now, the outflow is 6 gal of the water in minute. That is $\frac{6}{10,000}$ of the total water content in the pool, hence $6 \times 10^{-4}$ of the chlorine content y(t) , that is

$6 \times 10^{-4}y(t)$

The amount of water in the pool at any moment is 10,000 gal since the water is withdrawn from the pool at the same rate at which it is added.

Initially the pool contained 10 000 gal of water with 0.01 % chlorine, therefore we obtain the initial condition

$y(0) = 0.01 \% 10,000 = 10^{-4} \cdot 10^4 = 1$

Thus, the model is

$y'(t) =6 \times 10^{-5} - 6 \times 10^{-4}, \; y(0) = 1 \iff y' + 6 \times 10^{-4}y = 6 \times 10^{-5}, \; y(0) =1$

This is linear ODE, with

$P(t ) = 6 \times 10^{-4} \quad \wedge \quad Q(t ) = 6 \times 10^{-5}$

Hence,

$h = \int P \; dt = \int 6 \times 10^{-4} \, dt = 6 \times 10^{-4}t$

So, an integrating factor is

$e^h = e^{6 \times 10^{-4}t}$

and the general solution is

$y(t) = e^{-h} \left( c+ \int Qe^h \; dt \right) \\\\\phantom{y(t)} = e^{6 \times 10^{-4}t} \left( c+ \int 6 \times 10^{-5}e^{-6 \times 10^{-4}t} \; dt\right)\\\\\phantom{y(t)} = e^{6 \times 10^{-4}t} \left( c+ \frac{6 \times 10^{-5}}{-6 \times 10^{-4}} e^{-6 \times 10^{-4}t}\right) \\\\\phantom{y(t)} = ce^{6 \times 10^{-4}t} + 10^{-1}$

Now, we can use the initial condition to determine the numeric value of the constant . Substitute for and for in the last equation

$1 = y(0) = ce^{6 \times 10^{-4}\cdot 0} + 0.1 \implies 1 = c+0.1 \implies c = 0.9$

So,we obtain that

$y(t) = 0.9 e^{6 \times 10^{-4}t} + 0.1$

is the amount of chlorine in the pool at any moment .

To determine the amount of chlorine in the pool after 1 hour, we need to substitute 60 minutes for in the equation above.

$y(t) = 0.9 e^{6 \times 10^{-4} \cdot 60} + 0.1 = 0.9 e^{36 \times 10^{-3} } + 0.1$

Let t^* be the time when the pool water will be 0.002% chlorine. Then,

$y(t^*) = 0.002 \% = 2 \cdot 10^{-5} \cdot 10^4 = \frac{2}{10} = \frac{1}{5} = 0.2$

We need to solve this equation for t^* .

$0.2 = 0.9 e^{6 \times 10^{-4}t^* } + 0.1 \implies 0.1 = 0.9 e^{-6 \times 10^{-4}t^* } \\\\\phantom{0.2 = 0.9 e^{6 \times 10^{-4}t^* } + 0.1}\implies 0.111= e^{-6 \times 10^{-4}t^* } \\\\\phantom{0.2= 0.9 e^{-6 \times 10^{-4}t^* } + 0.1} \implies -2.1982 = -6 \times 10^{-4}t^* \\\\\phantom{0.2= 0.9 e^{-6 \times 10^{-4}t^* } + 0.1} \implies t^* \approx 3663.7084$

Answer from: Quest

answer: 180

step-by-step explanation:

12-2=10

20-2=18

18x10=180

Answer from: Quest

step-by-step explanation:

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A swimming pool whose volume is 10 comma 000 gal contains water that is 0.01% chlorine. Starting at...