A fish tank initially contains 20 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 2 liters per minute. The solution is mixed well and drained at 2 liters per minute.  Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c. 

To find the rate of change in the amount of salt \( x(t) \) in the fish tank after \( t \) minutes, we start by setting up a differential equation based on the inflow and outflow of the solution.

### Problem Setup

1. **Volume of the tank**: The tank always contains 20 liters of solution, since the inflow and outflow rates are equal (both are 2 liters per minute).
   
2. **Concentration of salt in the incoming brine**: The concentration of salt in the incoming brine is \( c \) grams per liter (unknown).

3. **Inflow of salt**: Brine flows in at a rate of 2 liters per minute, so the rate at which salt enters the tank is:
   \[
   \text{Rate of salt in} = \text{(flow rate of incoming brine)} \times \text{(concentration of salt in incoming brine)} = 2c \, \text{grams per minute}.
   \]

4. **Concentration of salt in the tank**: At any time \( t \), the amount of salt in the tank is \( x(t) \) grams. Since the tank has a constant volume of 20 liters, the concentration of salt in the tank is:
   \[
   \frac{x(t)}{20} \, \text{grams per liter}.
   \]

5. **Outflow of salt**: Solution flows out at 2 liters per minute, so the rate at which salt leaves the tank is:
   \[
   \text{Rate of salt out} = \text{(flow rate of outgoing solution)} \times \text{(concentration of salt in the tank)} = 2 \times \frac{x(t)}{20} = \frac{x(t)}{10} \, \text{grams per minute}.
   \]

### Differential Equation for \( \frac{dx}{dt} \)

The rate of change of the amount of salt in the tank, \( \frac{dx}{dt} \), is the difference between the rate of salt coming in and the rate of salt going out:

\[
\frac{dx}{dt} = \text{Rate of salt in} - \text{Rate of salt out}.
\]

Substitute the expressions for the rate of salt in and the rate of salt out:

\[
\frac{dx}{dt} = 2c - \frac{x(t)}{10}.
\]

### Formula for the Rate of Change in the Amount of Salt

Thus, the formula for the rate of change in the amount of salt \( x \) in terms of \( x \) and the unknown concentration of incoming brine \( c \) is:

\[
\frac{dx}{dt} = 2c - \frac{x}{10}.
\]

This is the required differential equation representing the rate of change of salt in the tank.

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Would you like me to solve this differential equation, or do you have any other questions?

### Related Questions

1. How do you solve the differential equation \( \frac{dx}{dt} = 2c - \frac{x}{10} \)?
2. What is the steady-state amount of salt in the tank if the concentration \( c \) of the incoming brine is known?
3. How would the differential equation change if the inflow rate was different from the outflow rate?
4. What if the tank initially contained some amount of salt instead of pure water?
5. How does the concentration \( c \) affect the long-term behavior of the amount of salt in the tank?

### Tip

Remember that for problems involving rates of change in mixing tanks, it's helpful to carefully define the inflow and outflow rates and relate them to concentrations to set up the correct differential equation.

A fish tank initially contains 20 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 2 liters per minute. The solution is mixed well and drained at 2 liters per minute. Let x be the amount of salt, in grams, in the fish tank after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x and the unknown concentration of incoming brine c.

Solve a differential equation for a fish tank mixing problem. Learn how to find the rate of change in the amount of salt (dx/dt) based on inflow and outflow of brine. This problem involves setting up and solving a first-order linear differential equation, suitable for Grades 11-12 and college-level students.

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