1 Example 1 Syrup from a 500 liter container of 10% sugar syrup flows into a candy making machine at 2 liters per minute. The vat is kept full by adding.

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1 Example 1 Syrup from a 500 liter container of 10% sugar syrup flows into a candy making machine at 2 liters per minute. The vat is kept full by adding 5% syrup. When is the concentration of syrup in the vat reduced to 7%? What is the concentration of syrup in the vat in the long run? Solution Let A(t) denote the number of liters of sugar in the vat after t minutes. Sugar enters the vat (.05)(2) liters per minute. Each liter of syrup in the vat contains A/500 liters of sugar. Hence sugar leaves the vat at (A/500)(2) liters per minute. Thus the rate of change of sugar in the vat is A / = Rate In – Rate Out =.10 – A/250. This is a separable differential equation. If.10 – A/250 = 0, then A = 25. However A  25 for this vat because A(0) = (.10)(500) = 50, not 25. Now divide the above equation by.10 – A/250:

2 Integrate this equation with respect to t: To find the value of the constant K use the fact that A(0) = (.10)(500) = 50. Hence Thus K = 25 and

3 The concentration in the vat is 7% when A(t) = (7/100)(500) = 35 liters. This occurs when In the long run: Therefore the concentration of sugar in this 500 liter tank in the long run is 100  (25/500) = 5%.