## A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into th

Question

A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped 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.

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2 weeks 2021-09-12T06:02:06+00:00 2 Answers 0

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is …

A'(t) = 6 -6/300·A(t)

Rewriting, we have …

A'(t) +(1/50)A(t) = 6

This has solution …

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve …

A'(t) = -q/50e^-(t/50) = 6 – (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, …

A(0) = 300 +q = 40

q = -260

So, the complete solution is …

A(t) = 300 -260e^(-t/50)

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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.