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A tank contains 100L of water. A solution with a salt concentration of 0.8kg/L is added at a rate of 7L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 5L/min. Answer the following questions.
1. If y(t) is the amount of salt (in kilograms) after tt minutes, what is the differential equation for which yy is satisfied? Use the variable y for y(t).
Answer (in kilograms per minute): dy/dt=
2
2. How much salt is in the tank after 50 minutes?
Answer (in kilograms)
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To answer these questions, we need to understand the rate of change of salt concentration in the tank.
1. To determine the differential equation, let's first consider the amount of salt being added to the tank per minute. The salt concentration is 0.8 kg/L, and the solution is being added at a rate of 7 L/min. So, the amount of salt being added per minute is (0.8 kg/L) * (7 L/min) = 5.6 kg/min.
Next, let's consider the amount of salt being drained from the tank per minute. The solution is being drained at a rate of 5 L/min. Therefore, the amount of salt being drained per minute is the salt concentration of 0.8 kg/L multiplied by 5 L/min, which is 4 kg/min.
Now, let's consider the rate of change of salt in the tank. The rate of change of salt in the tank is equal to the rate at which salt is being added minus the rate at which salt is being drained. Therefore, the differential equation for yy in terms of time is:
dy/dt = (rate of salt being added) - (rate of salt being drained)
= 5.6 kg/min - 4 kg/min
= 1.6 kg/min.
So, the differential equation for yy is dy/dt = 1.6 kg/min.
2. To find the amount of salt in the tank after 50 minutes, we can solve the initial value problem using the differential equation. The differential equation dy/dt = 1.6 kg/min represents the rate of change of salt in the tank.
To find the amount of salt in the tank after 50 minutes, we need to integrate the differential equation with respect to time from 0 to 50, and then evaluate the integral.
∫ dy = ∫ 1.6 dt
Integrating both sides gives us:
y = 1.6t + C
To find the constant C, we need to use the initial condition that the initial amount of salt in the tank is 0 since it starts with only water. Therefore, when t = 0, y = 0.
0 = 1.6 * 0 + C
C = 0
So, our equation becomes:
y = 1.6t
To find the amount of salt in the tank after 50 minutes, substitute t = 50 into the equation:
y = 1.6 * 50
y = 80 kg
Therefore, the amount of salt in the tank after 50 minutes is 80 kilograms.