To solve the problem of determining the mass of salt in the tank over time, we need to set up and solve a differential equation based on the rate of change of salt in the tank.

Problem Restatement

• A brine solution flows into a tank at a constant rate of 6 liters per minute.
• The tank initially holds 500 liters of brine with 0.5 kg of dissolved salt.
• The concentration of salt in the incoming brine is 0.05 kg/liter.
• The mixed solution flows out of the tank at a rate of 3 liters per minute.
• We want to determine the mass of salt in the tank after $t$ minutes, as well as the concentration of salt in the tank at that time.

Step-by-Step Solution

1. Define Variables:

   • $V(t)$ = volume of solution in the tank at time $t$ (in liters)
   • $S(t)$ = mass of salt in the tank at time $t$ (in kg)
   • $C(t)$ = concentration of salt in the tank at time $t$ (in kg/liter)

2. Volume of Solution in the Tank: The volume of solution changes because of the difference between the inflow and outflow rates: $\frac{dV}{dt} = 6 - 3 = 3 \text{ liters/min}$ Since the initial volume is 500 liters: $V(t) = 500 + 3t$

3. Rate of Change of Salt: The rate of change of salt in the tank is given by the difference between the rate of salt entering and the rate of salt leaving the tank: $\frac{dS}{dt} = (\text{rate of salt entering}) - (\text{rate of salt leaving})$

   • Rate of salt entering: $6 \text{ liters/min} \times 0.05 \text{ kg/liter} = 0.3 \text{ kg/min}$
   • Rate of salt leaving: $\frac{S(t)}{V(t)} \times 3 \text{ liters/min}$

   So, $\frac{dS}{dt} = 0.3 - \frac{3S(t)}{500 + 3t}$

4. Solve the Differential Equation: The differential equation can be written as: $\frac{dS}{dt} + \frac{3S}{500 + 3t} = 0.3$

   This is a first-order linear differential equation. To solve it, we can use the integrating factor method: $\mu(t) = e^{\int \frac{3}{500 + 3t} dt} = e^{\ln|500 + 3t|} = 500 + 3t$

   Multiplying through by the integrating factor: $(500 + 3t) \frac{dS}{dt} + 3S = 0.3 (500 + 3t)$

   Simplify and integrate both sides: $\frac{d}{dt} [(500 + 3t) S] = 0.3 (500 + 3t)$

   Integrate with respect to $t$: $(500 + 3t) S = 0.3 \int (500 + 3t) dt$ $(500 + 3t) S = 0.3 \left( 500t + \frac{3t^2}{2} \right) + C$ $(500 + 3t) S = 150t + 0.45t^2 + C$

   Solve for $S(t)$: $S(t) = \frac{150t + 0.45t^2 + C}{500 + 3t}$

   Use the initial condition $S(0) = 0.5$: $0.5 = \frac{C}{500}$ $C = 250$

   So the solution is: $S(t) = \frac{150t + 0.45t^2 + 250}{500 + 3t}$

5. Concentration of Salt in the Tank: $C(t) = \frac{S(t)}{V(t)} = \frac{S(t)}{500 + 3t}$

Final Answers

1. Mass of Salt in the Tank After $t$ Minutes: $S(t) = \frac{150t + 0.45t^2 + 250}{500 + 3t}$

2. Concentration of Salt in the Tank at Time $t$: $C(t) = \frac{S(t)}{500 + 3t}$

Would you like more details or have any questions?

Suggested Questions to Ask:

1. How do we determine the integrating factor for a differential equation?
2. What is the physical interpretation of the initial condition $S(0) = 0.5$?
3. How can we verify the solution to the differential equation?
4. Can you explain the importance of the rate of change in this context?
5. What happens to the concentration of salt as time goes to infinity?

Tip:

When solving differential equations, carefully handle initial conditions to ensure the particular solution accurately reflects the given scenario.