# Thread: Differntial Caculus Salt Problem

1. ## Differntial Caculus Salt Problem

A resevoir contains 700 liters of pure water. Brine contaning 0.02 kg/L of salt enters at a rate of 5 L/min. Another source of brine contanining 0.05 kg/L enters at a rate of 2L/min. The resevoir is well mixed and drains at a rate of 7L/min.How much salt is thier in the resevoir (i) after t min (ii) after 10 min?

2. In our problem, the level of water remains the same, as we have $\displaystyle 5\frac{\mbox{L}}{\mbox{min}}$ and $\displaystyle 2\frac{\mbox{L}}{\mbox{min}}$ flowing in and $\displaystyle 7\frac{\mbox{L}}{\mbox{min}}$ flowing out.

If we call the salt level $\displaystyle S(t)$, then we have

$\displaystyle S(0)=0\,\mbox{kg}.$

To find $\displaystyle \frac{dS}{dt}$, we subtract the rate of salt flowing out from the rate flowing in:

$\displaystyle \frac{dS}{dt}=\frac{dS_{\mbox{\scriptsize{in}}}}{d t}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}.$

The rate flowing in is

\displaystyle \begin{aligned} \frac{dS_{\scriptsize{\mbox{in}}}}{dt}&=0.02\frac{ \mbox{kg}}{\mbox{L}}\cdot 5\frac{\mbox{L}}{\mbox{min}}+0.05\frac{\mbox{kg}}{ \mbox{L}}\cdot 2\frac{\mbox{L}}{\mbox{min}} \\ &=0.1\frac{\mbox{kg}}{\mbox{min}}+0.1\frac{\mbox{k g}}{\mbox{min}}\\ &=0.2\frac{\mbox{kg}}{\mbox{min}}. \end{aligned}

Since the fluid is well-mixed, the rate flowing out will be in proportion to $\displaystyle 7\,\mbox{L}$ as the current salt level is to the entire resevoir. Therefore,

\displaystyle \begin{aligned} \frac{dS_{\scriptsize{\mbox{out}}}}{dt}&=\frac{S}{ 700\,\mbox{L}}\cdot 7\frac{\mbox{L}}{\mbox{min}} \\ &=\frac{S}{100}\,\mbox{min}^{-1}.\\ \end{aligned}

For $\displaystyle \frac{dS}{dt}$, we obtain

\displaystyle \begin{aligned} \frac{dS}{dt}&=\frac{dS_{\mbox{\scriptsize{in}}}}{ dt}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}\\ &=0.2\frac{\mbox{kg}}{\mbox{min}}-\frac{S}{100}\,\mbox{min}^{-1}. \end{aligned}

Now, all that remains is to solve the differential equation using the initial conditions given.