A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. Which differential equation does y(t) satisfy (in kg/min)?

Question

A very large tank initially contains 100L of pure water. Starting at time t=0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. Which differential equation does y(t) satisfy (in kg/min)?

BlackTom AI · Accepted Answer

Question restatement: A tank starts with 100 L of pure water. Inflow adds solution with salt concentration 0.8 kg/L at 5 L/min, and outflow drains at 3 L/min, with thorough mixing. Let y(t) be kilograms of salt in the tank after t minutes. We seek the differential equation dy/dt that y(t) satisfies.

Option A: dy/dt = 4
This would imply a constant rate of change of salt, independent of the amount currently in the tank. It ignores the outflow that depends on how much salt is present and the changing volume, so it cannot be correct.

Option B: dy/dt = 4 − 3y/(100+2t)
Correct reasoning is that the inflow salt rate is concentration × inflow rate = (0.8 kg/L) × (5 L/min) = 4 kg/min. The tank volume increases as 100 L plus net inflow of (5−3)=2 L/min, giving V(t) = 100 + 2t. The outflow salt rate is (salt concentration in tank) × (outflow rate) = (y(t)/V(t)) × 3 L/min = 3y/(100+2t). Thus dy/dt = 4 − 3y/(100+2t), which matches this option.

Option C: dy/dt = 4 − 3y
This form would require a constant outflow rate of salt proportional to y with a coefficient of 3, but it ignores the changing volume in the denominator and uses a constant outflow rate. Since the volume is growing with time, the outflow term should involve y divided by (100+2t). Hence this is incorrect.

Option D: dy/dt = 4 − 3y/(100)
This expression fixes the volume at 100 L, which is only true at t = 0. As time progresses, the volume increases due to the higher inflow than outflow, so using a constant 100 L in the denominator misrepresents the actual dynamics. Therefore this option is not correct.

Summary: The correct formulation must incorporate the time-varying tank volume V(t) = 100 + 2t in the outflow term and reflect the constant inflow salt rate of 4 kg/min. With these considerations, dy/dt = 4 − 3y/(100+2t) is the appropriate differential equation.

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