A tank holds 360 gallons of brine containing 2 pounds of salt per gallon. Clear

Question

A tank holds 360 gallons of brine containing 2 pounds of salt per gallon. Clear water flows into the tank at the rate of 10 gallons per minute, and the mixture, kept uniform by stirring, runs out at the same rate. If S(t) is the amount of salt in solution at time t, then the amount of salt in a typical gallon of solution is given by Amount of salt / Amount of fluid = S(t) / 360 Write a differential equation for the rate of change of S(t) using the fact that dS / dt = [rate at which salt enters tank] - [rate at which salt leaves tank] dS / dt =

Explanation / Answer

rate at which salt enters the tank = 0 because no salt is entering rate at which salt leaves the tank = amount of fluid flowing out per minute * S(t)/360 = 10*S(t)/360 = S(t)/36 hence dS/dt = -S(t)/36 S(0) = 2 because initially 2 lb of salt is present per gallon -36dt = dS/S(t) hence integration we get ln(S(t))-ln(S(0)) = -36t hence S(t) = 2e^(-36t) if total salt content is to be calculated then multiply by 360

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