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\begin{document}

{\bf Question}

$(*)$ A tank contains a well-stirred solution of 5 kg salt and 500
L of water. Starting at $t=0$, fresh water is poured into the
well-stirred solution at a rate of 4L/min, and the mixture leaves
at the same rate.

\begin{enumerate}

\item What is the differential equation governing the amount $x(t)$ of
salt in the tank at time $t$?

\item How long will it take for the concentration of salt to reach a
level of 0.1\%?

\item The next day the procedure is repeated but the exit pipe has
become partially blocked so that the mixture only leaves at a rate
of 2 L/min. The capacity of the tank is 1000 L, what is the
concentration of the mixture when it first overflows?

\end{enumerate}




\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds V(t)=$volume of water in tank in Litres. $\ds \hspace{0.25in}
V(0)=500$

$\ds S(t)=$mass of solution in tank in kg. $\ds \hspace{0.5in}
S(0)=5$

Now balance the water:

$\ds \begin{array}{ccccc} rate\ of\ change &=& rate\ of\ water\ in
&-& rate\ of\ water\ out\\ of\ volume\\ \frac{dV}{dt} &=& 4&-&4
\end{array}$

$\ds \Rightarrow \frac{dV}{dt}=0 \Rightarrow V(t)=500$

Now balance the salt:

$\ds \begin{array}{ccccc} rate\ of\ change &=& rate\ of\ salt\ in
&-& rate\ of\ salt\ out\\ of\ salt\\ \frac{dS}{dt} &=&
0&-&4\left(\frac{S}{V}\right) \end{array}$

(no salt is coming in and the concentration in the tank is $\ds
\frac{S}{V}$).

$\ds \frac{dS}{dt}=-\frac{4S}{500} \Rightarrow
S=5e^{\left(-\frac{4t}{500}\right)}$

\item[b)]
The concentration of salt gets to 0.1\% of initial value when

$\ds S=0.005$ i.e. $\ds 0.005=5e^{-\frac{4t}{500}} \Rightarrow
t=863 \rm{minutes} \approx 14 \rm{hours}$

\item[c)]
Water balance is:

$\ds \frac{dV}{dt}=4-2 \Rightarrow \frac{dV}{dt}=2 \Rightarrow
V(t)=500+2t$

hence it overflows at $\ds t=250 mins$

Salt balance is:

$\ds \frac{dS}{dt}=-\frac{2S}{V} \Rightarrow
\frac{dS}{dt}=-\frac{2S}{500+2t}$ solve by separation of
variables.

$\ds\int\frac{1}{S}dS=\int\frac{-2}{500+2t}dt \Rightarrow \ln
S=-\ln(500+2t)+A$

$\ds S(0)=5 \rm{\ then\ gives\ } \ln
S=-\ln(500+2t)+\ln(500)+\ln(5)$

$\ds S(t)=5\left(\frac{500}{500+2t}\right)$

concentration
=$\ds\frac{S(t)}{V(t)}=5\left(\frac{500}{500+2t}\right)\left(\frac{1}{500+2t}\right)$

and at t=250 mins (at overflow) $\ds\frac{S}{V}=2.5*10^{-3}kg/L$
\end{itemize}


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