	17. A tank initially contains 50 litres of a solution of a chemical in water. Pure water is poured into the container at a rate of 4 litres per minute and the solution is drained from the container at a rate of 5 litres per minute. Denote the number of litres of the chemical in the container at time t by x(t).

Use a differential equation to develop a model for x(t).

To develop a differential equation model for \( x(t) \), the amount of chemical in the tank at time \( t \), let's analyze the given problem:

1. The tank initially contains 50 liters of a solution.
2. Pure water is added at a rate of 4 liters per minute.
3. The solution is drained at a rate of 5 liters per minute.

Given that pure water is added, it does not change the amount of chemical in the solution. However, the draining process affects the amount of chemical because both the chemical and the water are being drained together.

Let's denote:
- \( V(t) \): the total volume of the solution in the tank at time \( t \).
- \( x(t) \): the amount of chemical in the tank at time \( t \).

Initially:
\[ V(0) = 50 \text{ liters} \]
\[ x(0) = \text{initial amount of chemical (let's denote it as } x_0) \]

### Volume of Solution \( V(t) \)
The rate of change of the volume of the solution is the difference between the inflow rate and the outflow rate:
\[ \frac{dV}{dt} = \text{inflow rate} - \text{outflow rate} = 4 - 5 = -1 \text{ liters per minute} \]

Thus, the volume of the solution at any time \( t \) is:
\[ V(t) = 50 - t \]

### Amount of Chemical \( x(t) \)
The rate of change of the amount of chemical in the tank is influenced only by the outflow since the inflow is pure water:
\[ \frac{dx}{dt} = \text{outflow rate of chemical} \]

The outflow rate of the chemical depends on its concentration in the solution. The concentration of the chemical at any time \( t \) is:
\[ \text{Concentration} = \frac{x(t)}{V(t)} = \frac{x(t)}{50 - t} \]

The amount of chemical drained per minute is:
\[ \text{Outflow rate of chemical} = 5 \times \left( \frac{x(t)}{50 - t} \right) = \frac{5x(t)}{50 - t} \]

Thus, the differential equation governing \( x(t) \) is:
\[ \frac{dx}{dt} = -\frac{5x(t)}{50 - t} \]

### Initial Condition
\[ x(0) = x_0 \]

In summary, the differential equation model for \( x(t) \) is:
\[ \frac{dx}{dt} = -\frac{5x(t)}{50 - t}, \quad x(0) = x_0 \]

Do you want more details or have any questions?

Here are some questions for further exploration:
1. How do you solve the differential equation \(\frac{dx}{dt} = -\frac{5x(t)}{50 - t}\)?
2. What is the behavior of \( x(t) \) as \( t \) approaches 50 minutes?
3. How does the concentration of the chemical change over time?
4. What would happen if the rates of inflow and outflow were different?
5. How can this model be applied to other similar scenarios?

**Tip:** When solving differential equations, always check the initial conditions and the physical constraints of the problem to ensure your solution makes sense in the given context.

Learn how to model the concentration of a chemical in a draining tank using differential equations. Understand the differential equation dx/dt = -5x(t) / (50 - t) and its application to real-world scenarios. Suitable for advanced high school students and undergraduates.

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