A 500-gallon tank initially contains 200 gallons of brine containing 85 pounds of dissolved salt. Brine containing 1 pounds of salt per gallon flows into the tank at the rate of 44 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).
dx/dt=ri*ci-ro*co where x is amount of salt in tank.
but volume=200+(ri-ro)t=200+(44-1)t
= 200-43t
so dx/dt=44*1 -x/(200-43t) check that.
see http://math2.uncc.edu/~sjbirdso/diffyQ/handouts/mixture%20solution.pdf for the rest
To set up a differential equation for the amount of salt in the tank at time t, we need to consider the rate of change of salt in the tank.
Let A(t) represent the amount of salt in the tank at time t. The rate of change of salt in the tank (dA/dt) is given by the difference between the rate of salt flowing in and the rate of salt flowing out of the tank.
Rate of salt flowing in = rate of inflow * concentration of salt in inflow
Rate of salt flowing in = 44 gallons/min * 1 pounds/gallon = 44 pounds/min
Rate of salt flowing out = rate of outflow * concentration of salt in the tank
Rate of salt flowing out = 1 gallon/min * (A(t)/V(t)) pounds/gallon
Since the volume of the tank is given as 500 gallons, we have:
Rate of salt flowing out = 1 gallon/min * (A(t)/500) pounds/gallon
Therefore, we can write the differential equation as:
dA/dt = 44 - (A(t)/500)
To find the amount of salt in the tank when it is full, we need to solve this differential equation and find the value of A(t) when t approaches infinity.
To solve the differential equation, we can separate the variables:
(dA/dt) + (A(t)/500) = 44
We can also rewrite the equation as:
dA/(44 - A/500) = dt
Next, we integrate both sides of the equation:
∫(dA/(44 - A/500)) = ∫dt
This simplifies to:
500 ln(44 - A/500) = t + C
Where C is the constant of integration.
To find the value of C, we can use the initial condition given in the problem:
When t = 0, A(0) = 85 (initial amount of salt)
Substituting these values into the equation, we have:
500 ln(44 - 85/500) = 0 + C
500 ln(22.83) = C
C ≈ 8253.67
Therefore, the equation becomes:
500 ln(44 - A/500) = t + 8253.67
To find the amount of salt in the tank when it is full, we need to find the value of A(t) as t approaches infinity.
As t approaches infinity, the logarithmic term ln(44 - A/500) approaches negative infinity, and hence its argument (44 - A/500) approaches 0.
Therefore, to find the amount of salt A(t) when the tank is full, we set the argument of the logarithmic term to 0:
0 = 44 - A/500
Solving for A, we have:
A = 44 * 500
A = 22,000 pounds
So, when the tank is full, there will be 22,000 pounds of salt in it.
To set up the differential equation for the amount of salt A(t) in the tank at time t, we can consider the rate of change of salt in the tank. The rate of change of salt will be the difference between the rate at which salt flows in and the rate at which it flows out.
Let's call the amount of salt in the tank at time t as A(t), measured in pounds. The rate at which salt flows in is given by 1 pound per gallon, and it flows in at a rate of 44 gallons per minute. So, the rate of salt flowing in at time t is 1 * 44 = 44 pounds per minute.
The rate at which salt flows out is given by the rate at which the mixed solution flows out of the tank, which is 1 gallon per minute. Since the concentration of salt in the tank is A(t) pounds per 500 gallons, the rate of salt flowing out at time t is A(t) * (1/500) pounds per minute.
Therefore, the differential equation for the amount of salt A(t) in the tank at time t is:
dA/dt = 44 - (A(t) / 500)
To solve this differential equation, we need to find the function A(t) and use initial conditions.
To find the amount of salt in the tank when it is full, we can find the limit of A(t) as t approaches infinity. In this case, the tank is full when it reaches its maximum capacity of 500 gallons.
Substituting A(t) = 500 into the differential equation:
dA/dt = 44 - (500/500) = 44 - 1 = 43 pounds per minute
So, the amount of salt in the tank when it is full is:
Lim(A(t)) as t approaches infinity = 43 * t
Since the tank is initially filled with 85 pounds of salt, we need to solve the differential equation using this initial condition:
A(0) = 85
After solving the differential equation with this initial condition, we can find the amount of salt in the tank when it is full by substituting t = infinity into the solution.
I hope this helps!