Math Challenges

Submissions for Problem #6

Problem #6

A tank contains 10 liters of pure water. Saltwater containing 0.5 kilograms of salt per liter enters the tank through pipe A at a rate of 2 liters per minute, while saltwater containing 0.2 kilograms of salt per liter enters through pipe B at a rate of 3 liters per minute. The solution is well mixed and leaves through pipe C at a rate of 5 liters per minute.

i) What amount of salt is in the tank at time t?
ii) How much salt will eventually be in the tank?

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zapwai

Solution: \( \displaylines{\frac{dS}{\differentialD t}=\left(0.5\text{ kg/L}\right)\left(2\text{ L/min}\right)+\left(0.2\text{ kg/L}\right)\left(3\text{ L/min}\right)-\left(\frac{S}{10}\text{ kg/L}\right)\left(5\text{ L/min}\right)\\ \frac{dS}{\differentialD t}=1.6-\frac12S=\frac{16-5S}{10}\\ \frac{dS}{16-5S}=\frac{1}{10}\differentialD t\\ -\frac15\ln\left(16-5S\right)=\frac{1}{10}t+C\Rightarrow\ln\left(16-5S\right)=-\frac{t}{2}+C\\ 16-5S=Ke^{-\frac{t}{2}}\Rightarrow16-Ke^{-\frac{t}{2}}=5S\\ S\left(t\right)=\frac{16}{5}-Ke^{-\frac12t}\\ \text{Initial value was 0 salt (pure water)}\Rightarrow S\left(0\right)=0\Rightarrow K=\frac{16}{5}\\ S\left(t\right)=\frac{16}{5}-\frac{16}{5}e^{-\frac12t}\\ \lim_{t\to\infty}S\left(t\right)=\frac{16}{5}\text{ kg}} \)

Explanation:
In this case the volume is fixed (5 L/min in, 5 L/min out) at 10L. We set up a differential equation for the change in salt per minute and solve it. For part (i) we get a function, and part (ii) we take a limit.