500-gallon tank initially contains 200 gallons of brine containing 100 pounds of dissolved salt. Brine containing 2 pounds of salt per gallon flows into the tank at the rate of 4 gallons per minute, and the well-stirred mixture flows out of the tank at the rate of 1 gallon per minute. Set up a differential equation for the amount of salt A(t) in the tank at time t. How much salt is in the tank when it is full? (Round your answer to the 2 decimal places
To set up the differential equation for the amount of salt A(t) in the tank at time t, we need to consider the rate at which salt is flowing into and out of the tank.
Let's say the amount of salt in the tank at time t is A(t) pounds. The rate at which brine flows into the tank is 4 gallons per minute, and the concentration of salt in the incoming brine is 2 pounds per gallon. Therefore, the rate at which salt flows into the tank is 4 * 2 = 8 pounds of salt per minute.
The rate at which the mixture flows out of the tank is 1 gallon per minute. Since the mixture in the tank is well-stirred, the concentration of salt in the outgoing mixture is the same as the concentration in the tank at time t, which is A(t) / V(t), where V(t) is the volume of the mixture in the tank at time t.
The rate at which salt flows out of the tank is therefore 1 * (A(t) / V(t)) pounds of salt per minute.
Now, considering the overall change in the amount of salt in the tank, we can set up the differential equation:
dA/dt = rate of salt inflow - rate of salt outflow
Based on the information above, the differential equation becomes:
dA/dt = 8 - (A(t) / V(t))
To solve this differential equation, we would need additional information about the relationship between the volume V(t) and the time t.
To find out how much salt is in the tank when it is full, we need to determine the maximum capacity of the tank. Given that the tank has a capacity of 500 gallons and initially contains 200 gallons of brine, the maximum volume V(t) = 500 - 200 = 300 gallons.
Using this maximum volume, we can substitute V(t) = 300 into the differential equation:
dA/dt = 8 - (A(t) / 300)
Now, we can solve this differential equation to find the amount of salt A(t) in the tank when it is full. However, since the question asks for the answer rounded to 2 decimal places, we would need to use numerical methods or a computer program to solve the equation and obtain the precise value.
We have
dA/dt = 2*4 - 1/(200+3t)
A(0) = 100
and we want to find A(100)
So, just solve the differential equation, and then plug in t=100, since at a net gain of 3 gal/min, the tank will be full at t=100.