(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.7kg/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 =
Question
(1 point) A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.7kg/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 =
Solution
The rate of change of the amount of salt in the tank is given by the difference between the rate at which salt is added and the rate at which salt is removed.
The rate at which salt is added is given by the product of the rate at which the solution is added and the concentration of salt in the solution, which is 6 L/min * 0.7 kg/L = 4.2 kg/min.
The rate at which salt is removed is given by the product of the rate at which the solution is drained and the concentration of salt in the solution in the tank at time t. The concentration of salt in the solution in the tank at time t is y(t) / (100 + 2t), where y(t) is the amount of salt in the tank at time t and 100 + 2t is the volume of the solution in the tank at time t (since the volume of the solution in the tank increases by 2 L/min). So, the rate at which salt is removed is 4 L/min * y(t) / (100 + 2t) = 4y(t) / (100 + 2t) kg/min.
Therefore, the differential equation that y satisfies is:
dy/dt = 4.2 - 4y(t) / (100 + 2t)
So, the answer is dy/dt = 4.2 - 4y / (100 + 2t) kg/min.
Similar Questions
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