Question

A tank originally contains 100 gal of fresh water. Then water containing .5 Ib of salt water per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at a same rate.After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving ar the same rate. Find the amount of salt in the tank at the end of an additional 10 min.

The answer in the book is 7.42 Ib

how do you work it out?

Answer 1

First model:

\(\frac{d y}{d t}=y_{i n}-y_{o u t}\)

\(\frac{\mathrm{dy}}{\mathrm{dt}}=\left(0.5 \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)-\left(\frac{y}{100} \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)\)

\(\frac{d y}{d t}=1-\frac{y}{50}\)

\(\frac{\mathrm{dy}}{1-\frac{y}{50}}=\mathrm{dt}\)

\(-50 \ln \left|1-\frac{y}{50}\right|=t+C\)

\(1-\frac{y}{50}=e^{-\frac{t}{50}} e^{C_{2}}\)

\(y=50-k e^{-\frac{t}{50}}\)

\(y(0)=0\)

\(0=50-k ; k=50\)

\(y(t)=50\left(1-e^{-\frac{t}{50}}\right)\)

At \(t=10\) minutes:

\(y(10)=9.063 \mathrm{lbs}\)

We need this because it will become our initial condition for the second model.

Second model: There is no salt going in, so:

\(\frac{\mathrm{dy}}{\mathrm{d} t}=y_{\mathrm{in}}-y_{\mathrm{out}}\)

\(\frac{\mathrm{dy}}{\mathrm{dt}}=\left(0 \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)-\left(\frac{y}{100} \frac{\mathrm{lb}}{\mathrm{gal}}\right)\left(2 \frac{\mathrm{gal}}{\mathrm{min}}\right)\)

\(\frac{d y}{d t}=-\frac{y}{50}\)

\(\frac{d y}{y}=-\frac{d t}{50}\)

\(\ln |y|=-\frac{t}{50}+C\)

$$ y=k e^{-\frac{t}{50}} $$

Now be careful here. The origin of this differential equation occurs when \(t=10\) for the OLD differential equation. Which means \(y(10)\) in the OLD MODEL is \(y(0)\) in the NEW MODEL because in the eyes of the new model, the previous 10 minutes didn't even happen because salt was going into the tank.

So we say:

\(y(0)=9.063\)

\(9.063=k\)

\(y(t)=9.063 e^{-\frac{t}{50}}\)

\(y(10)=7.42 \mathrm{lbs}\)