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galben [10]
3 years ago
5

A tank contains 120 liters of fluid in which 50 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

mped into the tank at a rate of 3 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.
Mathematics
1 answer:
gtnhenbr [62]3 years ago
7 0

If A(t) is the amount of salt in the tank at time t, then A(0) = 50 g and

A'(t) = (1 g/L)*(3 L/min) - (A(t)/120 g/L)*(3 L/min)

\implies A'+\dfrac A{40}=3

Multiply both sides by e^{t/40}, so that the left side can be condensed as the derivative of a product:

e^{t/40}A'+\dfrac{e^{t/40}}{40}A=3e^{t/40}

\left(e^{t/40}A\right)'=3e^{t/40}

Integrate both sides and solve for A(t).

e^{t/40}A=120e^{t/40}+C

\implies A(t)=120+Ce^{-t/40}

Given that A(0) = 50 g, we have

50=120+C\implies C=-70

so that the amount of salt in the tank at time t is

\boxed{A(t)=120-70e^{-t/40}}

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