A tank holds 450 gallons of brine containing 2 pounds of salt per gallon. Clear

Question

A tank holds 450 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) / 450 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] is dS / dt = ?S / 45. Solve this differential equation to obtain S(t). [Hint: What is S(0)?] S(t) =

Explanation / Answer

S(0) = 2* because initially 2 lb of salt is present per gallon dS/S = -45dt integrating we get : ln(S(t)) - ln(S(0)) = -45t S(t) = 2e^(-45t) if we want total salt content multiply by 450

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