Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 37

Q:
A 200-gallon tank is full of a solution containing 25 pounds of concentrate. Starting at time t = 0, distilled water is admitted to the tank at a rate of 10 gallons per minute, and the well-stirred solution is withdrawn at the same rate. Find the amount of concentrate Q in the solution as a function of t.
A:
At time zero, we start with a concentration of (25 lb)/(200 gal) , which we can reduce to (0.125 lb)/(gal) . This is the first expression in the formula because it is the starting concentration.
Replacing 10 gallons of the solution with 10 gallons of pure (distilled) water is equivalent to reducing the total concentration by 5% (.05) every minute after time 0. That is, at the end of each minute, the solution will be 95% (0.95) as concentrated as the previous one.
This is an example of continuous exponential decay. In other words, the concentration will decrease exponentially as a function of time. Thus, we will use the general formula
Q(t) = Q_0*e^(k*t)
In the above formula, Q(t) is the concentration at t minutes, Q_0 is our initial concentration, k is the rate of growth/decay in Q (0.95), and t is the time in minutes.
Therefore the final formula will look like this:
Q(t) = 0.125 (lb)/(gal)*e^(0.95*t)
where t is the time in minutes.