SOLUTION: A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.5kg/L is added at a rate of 6L/min. The solution is kept th

Question 1186677: A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.5kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min. Answer the following questions.
1. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. What differential equation does y satisfy? Use the variable y for y(t).
Answer (in kilograms per minute): dydt=
2. How much salt is in the tank after 20 minutes?
Answer (in kilograms):

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve this mixing problem:
**1. Differential Equation:**
The rate of change of salt in the tank (dy/dt) is determined by the difference between the rate of salt entering and the rate of salt leaving.
* **Rate in:** The solution enters at 6 L/min with a concentration of 0.5 kg/L. So, the rate of salt entering is 6 L/min * 0.5 kg/L = 3 kg/min.
* **Rate out:** The solution leaves at 4 L/min. The concentration of salt in the tank at time *t* is y(t) / (100 + (6-4)t) = y(t) / (100 + 2t). So, the rate of salt leaving is 4 L/min * [y(t) / (100 + 2t)] kg/L = (4y) / (100 + 2t) kg/min.
Therefore, the differential equation is:
dy/dt = 3 - (4y) / (100 + 2t)
**2. Amount of Salt after 20 Minutes:**
To find the amount of salt after 20 minutes, we need to solve the differential equation with the initial condition y(0) = 0 (since the tank initially contains pure water). This is a first-order linear differential equation, and it can be solved using an integrating factor.
However, a simpler approach is to use the code I provided earlier, which employs a numerical method to solve the differential equation.
Using the code, the amount of salt in the tank after 20 minutes is approximately 44.49 kg.

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