) A tank contains 19 liters of water to start, 4 liters of water flow into the tank while 3 liters of water flow out of the tank per minute. Write a differential for the amount of water A(t) (in liters) in the tank at time t in minutes. =0A(0)= Solve the differential equation: A(t)= Note: use A,A′, etc instead of A(t), dAdt (t) in your answers.
Question
) A tank contains 19 liters of water to start, 4 liters of water flow into the tank while 3 liters of water flow out of the tank per minute. Write a differential for the amount of water A(t) (in liters) in the tank at time t in minutes. =0A(0)= Solve the differential equation: A(t)= Note: use A,A′, etc instead of A(t), dAdt (t) in your answers.
Solution
The rate of change of the amount of water in the tank is given by the difference between the rate at which water flows into the tank and the rate at which water flows out of the tank.
The rate at which water flows into the tank is 4 liters per minute, and the rate at which water flows out of the tank is 3 liters per minute. Therefore, the rate of change of the amount of water in the tank is 4 - 3 = 1 liter per minute.
This gives us the differential equation for the amount of water A(t) in the tank at time t:
dA/dt = 1
The initial condition is given as A(0) = 19 liters.
To solve the differential equation, we integrate both sides with respect to t:
∫dA = ∫dt A(t) = t + C
where C is the constant of integration. We can find the value of C using the initial condition A(0) = 19:
19 = 0 + C C = 19
Therefore, the solution to the differential equation is:
A(t) = t + 19
This means that the amount of water in the tank at time t is t + 19 liters.
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