A fish tank is initially filled with 400 liters of water containing 1 g/liter of dissolved oxygen. At noon, oxygenated water containing 10g/liter of oxygen flows in at a rate of 5 liters per minute and the well-mixed water is pumped out at a rate of 7 liters per minute.Let A(t) represent the amount of dissolved oxygen in the tank at time t. a) Write the differential equation that represents the problem.b) Solve the differential equation. c) At 1 p.m., what is the amount of dissolve oxygen in the fish tank

Answer:

a. [tex]dA(t)/dt = 50 - \frac{7A(t)}{400} where A(0) = 400[/tex]

b. [tex]A(t) = \frac{20000}{7} + \frac{17200}{7}e^{-7t/400}[/tex]

c. 3.717 kg

Step-by-step explanation:

a) Write the differential equation that represents the problem.

Let A(t) be the amount of dissolved oxygen in the tank at any time, t.

The net flow rate dA(t)/dt = mass flow in - mass flow out

Since 10 g/l of oxygen flows in at a rate of 5 l/min, the mass flow in is 10 g/l × 5 l/min = 50 g/min

Since A(t) is the amount of oxygen present in the tank at time, t, and the volume of the tank is 400 liters. The concentration of oxygen in the tank is thus A(t)/400 g/l.

Also, water is being pumped out at a rate of 7 l/min. So, the mass flow out is thus concentration × flow rate out = A(t)/400 g/l × 7 l/min = 7A(t)/400 g/min

So, dA(t)/dt = mass flow in - mass flow out

dA(t)/dt = 50 - 7A(t)/400 with A(0) = 400 l × 1 g/l = 400 g since the tank initially contains 1 g/l of dissolved oxygen and has a volume of 400 l

So, the differential equation is

dA(t)/dt = 50 - 7A(t)/400  where A(0) = 400 g

[tex]dA(t)/dt = 50 - \frac{7A(t)}{400} where A(0) = 400[/tex]

b) Solve the differential equation

To solve the equation, we use separation of variables, so

dA(t)/dt = 50 - 7A(t)/400  where A(0) = 400 g

dA(t)/(50 - 7A(t)/400) = dt

Integrating both sides, we have

∫dA(t)/(50 - 7A(t)/400) = ∫dt

-7/400 ÷ -7/400∫dA(t)/(50 - 7A(t)/400) = ∫dt

1/ (-7/400)∫-7/400dA(t)/(50 - 7A(t)/400) = ∫dt

(-400/7)㏑(50 - 7A(t)/400) = t + C

㏑(50 - 7A(t)/400) = -7t/400 + (-7/400)C

㏑(50 - 7A(t)/400) = -7t/400 + C'     (C' = (-7/400)C)

taking exponents of both sides, we have

50 - 7A(t)/400 = exp[(-7t/400) + C']

50 - 7A(t)/400 = exp(-7t/400)expC'

[tex]50 - \frac{7A(t)}{400} = e^{-7t/400}e^{C'} \\50 - \frac{7A(t)}{400} = Ae^{-7t/400} A = e^{C'}\\ \frac{7A(t)}{400} = 50 - Ae^{-7t/400} \\A(t) = \frac{400}{7} X 50 - \frac{400}{7} Ae^{-7t/400} \\A(t) = \frac{20000}{7} - \frac{400}{7} Ae^{-7t/400}[/tex]

when t = 0 , A(0) = 400. So,

[tex]A(t) = \frac{20000}{7} - \frac{400}{7} Ae^{-7t/400} \\A(0) = \frac{20000}{7} - \frac{400}{7} Ae^{-7(0)/400}\\A(0) = \frac{20000}{7} - \frac{400}{7} Ae^{0/400}\\A(0) = \frac{20000}{7} - \frac{400}{7} Ae^{0}\\A(0) = \frac{20000}{7} - \frac{400}{7} A\\400 = \frac{20000}{7} - \frac{400}{7} A\\\frac{400}{7} A = 400 - \frac{20000}{7}\\\frac{400}{7} A = \frac{2800}{7} - \frac{20000}{7}\\\frac{400}{7} A = -\frac{17200}{7}\\\\A = -\frac{17200}{7} X \frac{7}{400} \\A = -43[/tex]

So,

[tex]A(t) = \frac{20000}{7} - \frac{400}{7} X -43e^{-7t/400} \\A(t) = \frac{20000}{7} + \frac{17200}{7}e^{-7t/400}[/tex]

c) At 1 p.m., what is the amount of dissolve oxygen in the fish tank.

At 1 p.m, t = 60 min

So, the amount of dissolved oxygen in the fish tank is A(60)

So,

[tex]A(t) = \frac{20000}{7} + \frac{17200}{7}e^{-7t/400}\\A(60) = \frac{20000}{7} + \frac{17200}{7}e^{-7X60/400}\\A(60) = \frac{20000}{7} + \frac{17200}{7}e^{-420/400}\\A(60) = \frac{20000}{7} + \frac{17200}{7}e^{-1.05}\\A(60) = \frac{20000}{7} + \frac{17200}{7} X 0.3499\\A(60) = \frac{20000}{7} + \frac{6018.93}{7} \\A(60) = \frac{26018.93}{7} \\A(60) = 3716.99 g[/tex]

A(60) ≅ 3717 g

A(60) ≅ 3.717 kg

Helppp (7th grade)

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