Questions: At time t=0, a tank contains 25 oz of salt dissolved in 50 gallons of water. Then brine containing 40 z of salt per gallon of brine is allowed to enter the tank at a rate of 5 gal / min and the mixed solution is drained from the tank at the same rate. (a) How much salt is in the tank at an arbitrary time? oz. (b) How much salt is in the tank at time 15 min ? oz.

Solution Steps

To solve this problem, we need to set up and solve a differential equation that models the amount of salt in the tank over time.

1. Define the variables:

   • Let \( S(t) \) be the amount of salt in the tank at time \( t \) (in minutes).
   • The rate of salt entering the tank is \( 5 \, \text{gal/min} \times 40 \, \text{oz/gal} = 200 \, \text{oz/min} \).
   • The rate of salt leaving the tank is proportional to the concentration of salt in the tank, which is \( \frac{S(t)}{50} \, \text{oz/gal} \times 5 \, \text{gal/min} = \frac{S(t)}{10} \, \text{oz/min} \).

2. Set up the differential equation:

   • 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.
   • This gives us the differential equation: \( \frac{dS}{dt} = 200 - \frac{S(t)}{10} \).

3. Solve the differential equation:

   • This is a first-order linear differential equation and can be solved using standard techniques.

4. Find the particular solution using the initial condition:

   • At \( t = 0 \), \( S(0) = 25 \).

5. Evaluate the solution at \( t = 15 \) minutes.

We start by defining the differential equation that models the amount of salt in the tank over time. Let \( S(t) \) be the amount of salt in the tank at time \( t \) (in minutes). The rate of salt entering the tank is \( 200 \, \text{oz/min} \), and the rate of salt leaving the tank is \( \frac{S(t)}{10} \, \text{oz/min} \). Therefore, the differential equation is: \[ \frac{dS}{dt} = 200 - \frac{S(t)}{10} \]

To solve this first-order linear differential equation, we find the general solution: \[ S(t) = C_1 e^{-t/10} + 2000 \]

We use the initial condition \( S(0) = 25 \) to find the particular solution. Substituting \( t = 0 \) and \( S(0) = 25 \) into the general solution: \[ 25 = C_1 e^{0} + 2000 \implies 25 = C_1 + 2000 \implies C_1 = 25 - 2000 = -1975 \] Thus, the particular solution is: \[ S(t) = -1975 e^{-t/10} + 2000 \]

To find the amount of salt in the tank at \( t = 15 \) minutes, we substitute \( t = 15 \) into the particular solution: \[ S(15) = -1975 e^{-15/10} + 2000 = -1975 e^{-1.5} + 2000 \] Using the value \( e^{-1.5} \approx 0.2231 \): \[ S(15) \approx -1975 \times 0.2231 + 2000 \approx -440.1225 + 2000 \approx 1559.8775 \]

Final Answer