Question

Three large tanks contain brine, as shown in Figure 3.3.8. Use the information in the figure to construct a mathematical model for the number of pounds of salt \({{\bf{x}}_1}\left( {\bf{t}} \right),{\rm{ }}{{\bf{x}}_2}\left( {\bf{t}} \right),\)and \({x_3}(t)\)at time t in tanks A, B, and C, respectively. Without solving the system. predict limiting values of \({{\bf{x}}_1}\left( {\bf{t}} \right),{\rm{ }}{{\bf{x}}_2}\left( {\bf{t}} \right),\)and \({x_3}(t)\)as \(t \to \infty \)

Short answer

The system of differential equations for salt in tanks\(A,{\rm{ }}B\;\)and\(C\)are\(\frac{{d{x_1}}}{{dt}} = - \frac{1}{{50}}{x_1}\)

\(\frac{{d{x_2}}}{{dt}} = \frac{1}{{50}}{x_1} - \frac{2}{{75}}{x_2}\)

\(\frac{{d{x_3}}}{{dt}} = \frac{2}{{75}}{x_2} - \frac{1}{{25}}{x_3}\)

Step-by-step solution

Define the newton law of differential equation

Differential equations are used in research and engineering to model physical quantities that fluctuate over time.

Define the rate of change \(\frac{{d{x_1}}}{{dt}} = {\rm{ Input rate of salt for tank A}} - {\rm{ Output rate of salt for tank A}}\)

Find the rate of change in salts for A tank

Let the number of pound so salts be\(A,B{\rm{ and }}C{\rm{ as }}\left( {{x_1}} \right),\left( {{x_2}} \right){\rm{ and }}\left( {{x_3}} \right)\)

\(\frac{{d{x_1}}}{{dt}} = {\rm{ Input rate of salt for }}{\mathop{\rm tank}\nolimits} {\rm{A}} - {\rm{ Output rate of salt for }}{\mathop{\rm tank}\nolimits} {\rm{A}}\)

\(\frac{{d{x_1}}}{{dt}} = (4{\rm{gal}}/{\rm{min}} \times 0{\rm{Ib}}/{\rm{gal}}) - \left( {4{\rm{gal}}/{\rm{min}} \times \frac{{{x_1}}}{{200}}{\rm{Ib}}/{\rm{gal}}} \right)\),

\(\frac{{d{x_1}}}{{dt}} = - \frac{1}{{50}}{x_1}\)

Find the rate of change in salts for B tank 

Then\(\frac{{d{x_2}}}{{dt}} = {\rm{ Input rate of salt for }}{\mathop{\rm tank}\nolimits} {\rm{B}} - {\rm{ Output rate of salt for }}{\mathop{\rm tank}\nolimits} {\rm{B}}\)

\(\frac{{d{x_2}}}{{dt}} = \left( {4{\rm{gal}}/{\rm{min}} \times \frac{{{x_1}}}{{200}}{\rm{Ib}}/{\rm{gal}}} \right) - \left( {4{\rm{gal}}/{\rm{min}} \times \frac{{{x_2}}}{{150}}{\rm{Ib}}/{\rm{gal}}} \right)\)

\(\frac{{d{x_2}}}{{dt}} = \frac{1}{{50}}{x_1} - \frac{2}{{75}}{x_2}\)

Find the rate of change in salts for C tank

Again

\(\frac{{d{x_3}}}{{dt}} = {\rm{ Input rate of salt for tank C}} - {\rm{ Output rate of salt for tank C}}\)

\(\frac{{d{x_3}}}{{dt}} = \left( {4{\rm{gal}}/\min \times \frac{{{x_2}}}{{150}}{\rm{Ib}}/{\rm{gal}}} \right) - \left( {4{\rm{gal}}/{\rm{min}} \times \frac{{{x_3}}}{{100}}{\rm{Ib}}/{\rm{gal}}} \right)\)

\(\frac{{d{x_3}}}{{dt}} = \frac{2}{{75}}{x_2} - \frac{1}{{25}}{x_3}\)

\( = 37.8{\rm{ seconds }}\)

The system of differential equations for salt in tanks\(A,{\rm{ }}B\;\)and\(C\)are\(\frac{{d{x_1}}}{{dt}} = - \frac{1}{{50}}{x_1}\)

\(\frac{{d{x_2}}}{{dt}} = \frac{1}{{50}}{x_1} - \frac{2}{{75}}{x_2}\)

\(\frac{{d{x_3}}}{{dt}} = \frac{2}{{75}}{x_2} - \frac{1}{{25}}{x_3}\)