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A tank contains 210 liters of fluid in which 30 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 3 L/min; the well-mixed solution is pumped out at the same rate. Let the A(t) be the number of grams of salt in the tank at time t. Find the rate at which the number of grams of salt in the tank is changing at time t.

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oladayoademosu

Answer:

[tex]dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex] g/min

Step-by-step explanation:

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

Since water is pumped in at a rate of 3 L/min, and it contains no salt, the concentration in is thus 0 g/L. the mass flow in is thus 0 g/L × 3 L/min = 0 g/min.

Let A(t) be the mass present at any time, t. The concentration at any time ,t is thus A(t)/volume = A(t)/210. Since water flows out at a rate of 3 L/min, the mass flow out is thus, A(t)/210 g/L × 3 L/min = A(t)/70 g/min.

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

dA(t)/dt = 0 g/min - A(t)/70 g/min

dA(t)/dt =  - A(t)/70 g/min

 Since the tank initially contains 30 g of salt, the initial mass of salt in the tank is 30 g. So A(0) = 30

So, the initial value problem is thus  

dA(t)/dt =  - A(t)/70 , A(0) = 30

Separating the variables, we have

dA(t)/A(t) = -dt/70

Integrating both sides, we have

∫dA(t)/A(t) = ∫-dt/70

㏑A(t) = -t/70 + C

taking exponents of both sides, we have

A(t) = exp(-t/70 + C)

A(t) = exp(-t/70)expC

[tex]A(t) = e^{-t/70}e^{C}\\A(t) = Ce^{-t/70} whereC = e^{C}[/tex]

A(0) = 30, So

[tex]A(0) = Ce^{-0/70}\\30 = Ce^{0}\\C = 30[/tex]

[tex]A(t) = 30e^{-t/70}[/tex]

The rate at which the number of grams of salt in the tank is changing at time t is

[tex]dA(t)/dt = \frac{d30e^{-t/70} }{dt} \\dA(t)/dt = -\frac{30e^{-t/70} }{70}\\dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex]

So, the rate at which the number of grams of salt in the tank is changing at time t is

[tex]dA(t)/dt = -\frac{3}{7}e^{-t/70}[/tex] g/min

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