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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
Capitolo 9, Problema 9.PE.44c

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.


c. When will the tank have exactly 5 pounds of salt and how many gallons of solution will be in the tank?

Guida verificata passo dopo passo
1
Identify the variables and given information: Let \(S(t)\) be the amount of salt (in pounds) in the tank at time \(t\) (in 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 of solution is 200 gallons, and initially, there are 50 pounds of salt.
Express 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. Therefore, 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 proportional to the concentration of salt in the tank times the outflow rate. The concentration is \(\frac{S(t)}{V(t)}\), so the rate out is \(5 \times \frac{S(t)}{V(t)}\). Thus, 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 it as: \[ \frac{dS}{S} = - \frac{5}{200 - t} dt \] Integrate both sides to find \(S(t)\), including the constant of integration determined by the initial condition \(S(0) = 50\).
Use the solution \(S(t)\) to find when the tank has exactly 5 pounds of salt by solving \(S(t) = 5\). Then, find the volume at that time using \(V(t) = 200 - t\) gallons.

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