Acid Solution Tank Differential Equation: Determining Volume of Acid Over Time

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The discussion focuses on deriving the differential equation for the volume of acid in a tank with a constant inflow and outflow rate. The inflow rate of acid is calculated as 1.2L per minute from a 20% acid solution. The outflow rate is expressed as y(t)/v(t) * 8, where v(t) represents the decreasing volume of liquid in the tank over time. The resulting differential equation is confirmed as dy/dt + (8/(200-2t))y = 1.2. This equation accurately models the volume of acid in the tank over time.
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An acid solution flows at a constant rate of 6L/min into a tank which initially holds 200L of a 0.5% acid solution. The solution in the tank is kept well mixed and flows out of the tank at 8L/min. If the solution entering the tank is 20% acid then determine the volume of acid in the tank after t minutes.

I just want to make sure I have the differential equation right.
let y(t) be the amount of acid solution in the tank.
rate in = 0.2 * 6 = 1.2L acid per min
rate out = y(t)/v(t) * 8, where v(t) = 200 - 2t is the volume of liquid in the tank

So I got

<br /> \frac{dy}{dt} + \frac{8}{200-2t}y=1.2<br />

Can someone please confirm that this is the right equation to be working with?
 
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Yes, that is correct.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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