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Ch. 9 - First-Order Differential Equations
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 9 - First-Order Differential EquationsProblema 9.PE.44b
Capitolo 9, Problema 9.PE.44b

In Exercises 43 and 44, let S represent the pounds of salt in a tank at time t minutes. Set up a differential equation representing the given information and the rate at which S changes. Then solve for S and answer the particular questions.
Pure water flows into a tank at the rate of 4 gal/min, and the well-stirred mixture flows out of the tank at the rate of 5 gal/min. The tank initially holds 200 gal of solution containing 50 pounds of salt.
b. How many pounds of salt are in the tank after 1 minute? after 30 minutes?

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Identify the variables and given information: Let \(S(t)\) be the amount of salt (in pounds) in the tank at time \(t\) minutes. The inflow rate of pure water is 4 gal/min, and the outflow rate of the mixture is 5 gal/min. The initial volume is 200 gallons with 50 pounds of salt.
Determine the volume of solution in the tank at time \(t\): Since water flows in at 4 gal/min and flows out at 5 gal/min, the volume decreases by 1 gal/min. So, the volume at time \(t\) is \(V(t) = 200 - t\) gallons.
Set up the differential equation for the rate of change of salt \(S(t)\): The rate of salt entering is zero because pure water contains no salt. The rate of salt leaving is the concentration of salt in the tank times the outflow rate. The concentration is \(\frac{S(t)}{V(t)}\), so the rate out is \(5 \times \frac{S(t)}{V(t)}\). Therefore, the differential equation is: \[\frac{dS}{dt} = 0 - 5 \times \frac{S(t)}{200 - t} = - \frac{5S}{200 - t}\]
Solve the differential equation: This is a separable equation. Rewrite as \[\frac{dS}{S} = - \frac{5}{200 - t} dt\] Integrate both sides to find \(S(t)\), remembering to include the constant of integration and use the initial condition \(S(0) = 50\) pounds to solve for it.
Use the solution \(S(t)\) to find the amount of salt after 1 minute and after 30 minutes by substituting \(t = 1\) and \(t = 30\) into the expression for \(S(t)\).

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Setting Up a Differential Equation for Mixing Problems

In mixing problems, the rate of change of a substance in a tank is modeled by a differential equation that accounts for inflow and outflow rates. The inflow may add or dilute the substance, while the outflow removes it proportionally to its concentration. Writing this equation requires expressing the rate of change of salt as the difference between salt entering and leaving the tank.
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Classifying Differential Equations

Solving First-Order Linear Differential Equations

The differential equation from mixing problems is typically first-order linear and can be solved using integrating factors or separation of variables. The solution gives the amount of substance as a function of time, allowing prediction of concentration or quantity at any moment. Understanding this method is essential to find explicit formulas for S(t).
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Solving Separable Differential Equations

Interpreting Initial Conditions and Variable Volume

Initial conditions specify the starting amount of salt and volume, crucial for solving the differential equation uniquely. When inflow and outflow rates differ, the tank volume changes over time, affecting concentration calculations. Properly incorporating variable volume ensures accurate modeling of salt concentration and total amount.
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Initial Value Problems
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