Continuation

June 5, 2019

(Video 8) (Book p. 52)

Mixing problem

Image a tank containing a solution— a mixture of solute and solvent— such as salt dissolved in water. And we want to know

$x(t)=\mathrm{amount\ of\ salt\ in\ tank\ at\ time\ } t$

Suppose we have an opening on top, and a solution with concentration of $c_i$ grams of solute per liter of solution flows into the tank at a constant rate of $r_i$ litters per seconds; there’s another different-sized opening in the bottom, and that solution in the tank flows out at a different constant rate of $r_0$ liters per second. Let $c_0$ be the concentration of solute in the tank, then for a very small time interval $\Delta t$ , the change of amount of salt in the tank is

$\Delta x=\mathrm{grams\ input}-\mathrm{grams\ output}\approx (r_ic_i-r_0c_0)\Delta t$

If we know the volume of the liquid (solution) in the tank at time $t$ , then we can write

$c_0(t)=\frac{x(t)}{V(t)}$

Combine all these

$\frac{dx}{dt}=r_ic_i-\frac{x(t)}{V(t)}r_0$

Writing in standard form

$x'+\frac{r_0}{V(t)}x=r_ic_i$