Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 60

a) Show that the concentration of salt after $t$ minutes (in grams per liter) is $\displaystyle C(t) = \frac{30t}{200 + t}$. If a tank contains $5000 L$ of pure water. Brine that contains $30 g$ of salt per liter of water is pumped into the tank at a rate of $25 L/min$.

Let $M(t) = \rho ft$, where $M(t) = $ total mass of salt as a function of time

$\rho$ = density of salt (g/L)

$f$ = flow rate of Brine water into the tank (L/min)

$t$ = time (min)

Volume of tank

$V(t) = 5000 L + ft$

So we have,


$
\begin{equation}
\begin{aligned}

C(t) = \frac{M(t)}{V(t)} =& \frac{\rho ft}{5000 L + ft}
\\
\\
=& \frac{\displaystyle \left( 30 \frac{g}{\cancel{L}} \right) \left( 25 \frac{\cancel{L}}{\cancel{min}} \right) (t \cancel{min}) }{\displaystyle 5000 L + \left( 25 \frac{L}{\cancel{min}} (t \cancel{min} \right) }
\\
\\
=& \frac{(30g)(25)(t)}{5000 L + (25 L) (t)}
\\
\\
=& \frac{\cancel{25} (30t) g }{\cancel{25} (200 + t)L}
\\
\\
C(t) =& \frac{30 t}{200 + t} \frac{g}{L}

\end{aligned}
\end{equation}
$


b) Find what happens to the concentration if $t \to \infty$.


$
\begin{equation}
\begin{aligned}

\lim_{x \to \infty} C(t) =& \lim_{t \to \infty} \frac{30t}{200 + t} \frac{g}{L}
\\
\\
=& \lim_{t \to \infty} \frac{\displaystyle \frac{30 \cancel{t}}{\cancel{t}}}{\displaystyle \frac{200 }{t} + \frac{\cancel{t}}{\cancel{t}} } \frac{g}{L}
\\
\\
=& \lim_{t \to \infty} \frac{30}{\displaystyle \frac{200}{t} + 1} \frac{g}{L}
\\
\\
=& \frac{30}{\displaystyle \lim_{x \to \infty} \frac{200}{t} + 1} \frac{g}{L}
\\
\\
=& \frac{30}{0 + 1} \frac{g}{L}
\\
\\
=& 30 g/L

\end{aligned}
\end{equation}
$


The concentration of salt in the tank as $t$ approaches $\infty$, approaches $30 g/L$. This is the same as the concentration of salt in the Brine. Therefore, the Brine overpowers the pure water.