(Solved): Consider a water tank with two pipes into the tank and a drain at the bottom of the tank. Assume th ...

Consider a water tank with two pipes into the tank and a drain at the bottom of the tank. Assume that water can enter through pipe A at a rate of 6 liters/minute, enter through pipe B at a rate of 4 liters/minute and exit through the drain at a rate of 8 liters/minute. Assume also that everything in the tank is well-mixed (instantaneously). (a) Suppose that we start with a tank containing 100 liters of pure water. At $t=0$ we open both pipes and the drain. After the process has been running for exactly 10 minutes, a supervisor realizes that the water entering through pipe A is polluted with 0.5 grams of pollutant per liter of water. How much pollutant (in grams) is in the tank at the moment the error is discovered? Hint: Let $y(t)$ be the volume of water in the tank and let $z(t)$ be the mass of the pollutant at time $t$. Write down the equations for $y^{?}(t)$ and $z^{?}(t)$ in terms of $y(t)$ and $z(t)$. Solve for $y(t)$ first then use this to solve for $z(t)$. (b) The supervisor immediately shuts down pipe A upon discovering the error. Pipe B and the drain is allowed to continue to function as before. How long will it take, after pipe A is shut down, for the quantity (in grams) of the pollutant to reach one half of the level in part (a)?