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Lady bird [3.3K]
3 years ago
13

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

mped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t. A(t) =
Mathematics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

A(t)=240-220e^{-\frac{t}{40}}

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

  • Volume of the tank = 240 liters
  • Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

R_{in}=(concentration of salt in inflow)(input rate of fluid)

R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}

R_{out}=(concentration of salt in outflow)(output rate of fluid)

R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}

Rate of change of the amount of salt in the tank:

\dfrac{dA}{dt}=R_{in}-R_{out}

\dfrac{dA}{dt}=6-\dfrac{A}{40}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}

Taking the integral of both sides

\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}

Recall that when t=0, A(t)=20 (our initial condition)

20=240+Ce^{-\frac{0}{40}}\\20-240=C\\C=-220\\$Therefore, the number A(t) of grams of salt in the tank at time t\\A(t)=240-220e^{-\frac{t}{40}}

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<h3><em>Good</em><em> </em><em>Luck</em><em>!</em></h3>

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