Question

Initially, two large tanks $\mathbf{A}$ and $\mathbf{B}$ each hold $\mathbf{100}$gallons of brine. The well-stirred liquid is pumped between the tanks as shown in Figure 3.R.4. Use the information given in the figure to construct a mathematical model for the number of pounds of salt ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ and ${\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ at time $\mathbf{t}$ in tanks$\mathbf{A}$ and $\mathbf{B}$, respectively.

Short answer

A mathematical model for the number of pounds of salt in tanks A and B as $\left({\mathrm{x}}_1\right)$and $\left({\mathrm{x}}_2\right)$is $\frac{{\mathrm{dx}}_1}{\mathrm{dt}}=14-\frac{2}{25}{\mathrm{x}}_1+\frac{1}{100}{\mathrm{x}}_2$ and $\frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{1}{20}{\mathrm{x}}_1-\frac{1}{20}{\mathrm{x}}_2$.

Step-by-step solution

Obtain the system of differential equations for salt in tanks A and B:

It is given that two tanks A and B of a capacity 100 gallon of liquid pumped in and out at the same rate, then we have well-stirred liquid containing 2lb for each gallon, is pumped into tank A as a rate 7 gal/min as shown in the figure below, and the objective is to obtain the system of differential equations for salt in tanks A and B as the following:

Let the number of pounds of salt in tanks A and B as $\left({\mathrm{x}}_1\right)$and $\left({\mathrm{x}}_2\right)$respectively.

After that, for tank A, obtain the rate of change for salt ${\mathrm{x}}_1$as:

$$\begin{gathered} \frac{{\mathrm{dx}}_1}{\mathrm{dt}}=\mathrm{Input}\mathrm{rate}\mathrm{of}\mathrm{salt}\mathrm{for}\mathrm{tank}\mathrm{A}-\mathrm{Output}\mathrm{rate}\mathrm{of}\mathrm{salt}\mathrm{for}\mathrm{tank}\mathrm{A} \\ \frac{{\mathrm{dx}}_1}{\mathrm{dt}}=\left(7\mathrm{gal}/\min \times 2\mathrm{lb}/\mathrm{gal}+1\mathrm{gal}/\min \times \frac{{\mathrm{x}}_2}{100}\mathrm{lb}/\mathrm{gal}\right)-\left(3\mathrm{gal}/\min \times \frac{{\mathrm{x}}_1}{100}\mathrm{lb}/\mathrm{gal}+5\mathrm{gal}/\min \times \frac{{\mathrm{x}}_1}{100}\mathrm{lb}/\mathrm{gal}\right) \\ \frac{{\mathrm{dx}}_1}{\mathrm{dt}}=14+\frac{{\mathrm{x}}_2}{100}+\frac{3{\mathrm{x}}_1}{100}-\frac{5{\mathrm{x}}_1}{100} \\ \frac{{\mathrm{dx}}_1}{\mathrm{dt}}=14-\frac{8}{100}{\mathrm{x}}_1+\frac{1}{100}{\mathrm{x}}_2 \\ \frac{{\mathrm{dx}}_1}{\mathrm{dt}}=14-\frac{2}{25}{\mathrm{x}}_1+\frac{1}{100}{\mathrm{x}}_2 \end{gathered}$$

Find the rate of change for salt x2

After that, for tank B, obtain the rate of change for salt ${\mathrm{x}}_2$as:

$$\begin{gathered} \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\mathrm{Input}\mathrm{rate}\mathrm{of}\mathrm{salt}\mathrm{for}\mathrm{tank}\mathrm{B}-\mathrm{Output}\mathrm{rate}\mathrm{of}\mathrm{salt}\mathrm{for}\mathrm{tank}\mathrm{B} \\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\left(5\mathrm{gal}/\min \times \frac{1}{100}{\mathrm{x}}_1\mathrm{lb}/\mathrm{gal}\right)-\left(1\mathrm{gal}/\min \times \frac{1}{100}{\mathrm{x}}_2\mathrm{lb}/\mathrm{gal}+4\mathrm{gal}/\min \times \frac{1}{100}{\mathrm{x}}_2\mathrm{lb}/\mathrm{gal}\right) \\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{1}{20}{\mathrm{x}}_1-\frac{1}{100}{\mathrm{x}}_2-\frac{4}{100}{\mathrm{x}}_2 \\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{1}{20}{\mathrm{x}}_1-\frac{5}{20}{\mathrm{x}}_2 \\ \frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{1}{20}{\mathrm{x}}_1-\frac{1}{20}{\mathrm{x}}_2 \end{gathered}$$

Thus, a mathematical model for the number of pounds of salt in tanks A and B as $\left({\mathrm{x}}_1\right)$and $\left({\mathrm{x}}_2\right)$is $\frac{{\mathrm{dx}}_1}{\mathrm{dt}}=14-\frac{2}{25}{\mathrm{x}}_1+\frac{1}{100}{\mathrm{x}}_2$and $\frac{{\mathrm{dx}}_2}{\mathrm{dt}}=\frac{1}{20}{\mathrm{x}}_1-\frac{1}{20}{\mathrm{x}}_2$.