Respuesta :
Answer:
- [tex]y(t) = 0.9 e^{6 \times 10^{-4} \cdot 60} + 0.1 = 0.9 e^{36 \times 10^{-3} } + 0.1[/tex]
- [tex]t^* \approx 3663.7084[/tex] minutes
Step-by-step explanation:
Let [tex]y(t)[/tex] be the amount of chlorine in the tank at time [tex]t[/tex].
Then, its time rate of change, [tex]y'[/tex], 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
[tex]6 \cdot 0.001 \% = 6 \cdot 0.00001 = 6 \times 10^{-5}[/tex]
Now, the outflow is 6 gal of the water in minute. That is [tex]\frac{6}{10,000}[/tex] of the total water content in the pool, hence [tex]6 \times 10^{-4}[/tex] of the chlorine content [tex]y(t)[/tex], that is
[tex]6 \times 10^{-4}y(t)[/tex]
The amount of water in the pool at any moment [tex]t[/tex] is 10,000 gal since the water is withdrawn from the pool at the same rate at which it is added.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
[tex]y(0) = 0.01 \% 10,000 = 10^{-4} \cdot 10^4 = 1[/tex]
Thus, the model is
[tex]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[/tex]
This is linear ODE, with
[tex]P(t ) = 6 \times 10^{-4} \quad \wedge \quad Q(t ) = 6 \times 10^{-5}[/tex]
Hence,
[tex]h = \int P \; dt = \int 6 \times 10^{-4} \, dt = 6 \times 10^{-4}t[/tex]
So, an integrating factor is
[tex]e^h = e^{6 \times 10^{-4}t}[/tex]
and the general solution is
[tex]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}[/tex]
Now, we can use the initial condition to determine the numeric value of the constant [tex]c[/tex]. Substitute [tex]0[/tex] for [tex]t[/tex] and [tex]1[/tex] for [tex]y[/tex] in the last equation
[tex]1 = y(0) = ce^{6 \times 10^{-4}\cdot 0} + 0.1 \implies 1 = c+0.1 \implies c = 0.9[/tex]
So,we obtain that
[tex]y(t) = 0.9 e^{6 \times 10^{-4}t} + 0.1[/tex]
is the amount of chlorine in the pool at any moment [tex]t[/tex].
To determine the amount of chlorine in the pool after 1 hour, we need to substitute 60 minutes for [tex]t[/tex] in the equation above.
[tex]y(t) = 0.9 e^{6 \times 10^{-4} \cdot 60} + 0.1 = 0.9 e^{36 \times 10^{-3} } + 0.1[/tex]
Let [tex]t^*[/tex] be the time when the pool water will be 0.002% chlorine. Then,
[tex]y(t^*) = 0.002 \% = 2 \cdot 10^{-5} \cdot 10^4 = \frac{2}{10} = \frac{1}{5} = 0.2[/tex]
We need to solve this equation for [tex]t^*[/tex].
[tex]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[/tex]
