Question

As shown in figure 8.2.6 two large connected mixing tanks A and B initially contain 100 gallons of brine. Liquid is pumped in and out of the tanks as indicated in the figure. the mixture pumped between and out of a tank is assumed to be well-stirred.

a) construct a mathematical mode in the form of a linear system of first-order differential equations for the number of poundsx₁(t)andx₂(t)of salt in tanks A and B is respectively, at time t. write system in matrix form.[Hint: Review section 3.3]

b)Use the eigenvalue method of this section to solve the linear system in part(a) subject${\mathit{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{20}\mathbf{,}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$

c) Use graphing utility or CAS to plot the graph of${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in the same coordinate plane.

d) suppose the system of mixing tanks is to be turned off when the number of pounds of salt in tank b equal that in tank A. Use a root-finding application of a CAS or calculator to approximate that time.

Short answer

Write system in matrix form.

a) $\frac{d{x}_1}{dt}=$Input Rate of Salt − Output Rate of Salt

$X\text{'}=\left(\begin{array}{cc}-\frac{3}{100}& \frac{1}{100}\\ \frac{1}{50}& -\frac{1}{50}\end{array}\right)X$

b)$\det \left(A-\lambda I\right)=0\to \left|\begin{array}{cc}-\frac{3}{100}-\lambda & \frac{1}{100}\\ \frac{1}{50}& -\frac{1}{50}-\lambda \end{array}\right|=0$

$20=\frac{25}{3}-c_2\to c_2=-\frac{35}{3}$

Therefore,

$$\begin{gathered} x\left(t\right)=\frac{25}{3}\left(1 \\ 2\right)e^{-0.01t}-\frac{35}{3}\left(-1 \\ 1\right)e^{-0.04t} \end{gathered}$$

c)

d)

$t=34.321$minutes

Step-by-step solution

definition of: tank A and tank B

Its Input Rate of Salt − Output Rate of Salt.

For tank A

$\frac{d{x}_1}{dt}=$Input Rate of Salt − Output Rate of Salt

$$\begin{gathered} =\left(2\text{gal/min}\right)\left(01\text{b/gal}\right)+\left(1\text{gal/min}\right)\left(\frac{x_2}{100}1\text{b/gal}\right)-\left(2\text{gal/min}\right)\left(\frac{x_1}{100}1\text{b/gal}\right) \\ \left(1\text{gal/min}\right)\left(\frac{x_1}{100}1\text{b/gal}\right) \\ =\frac{x_2}{100}1\text{b/min}-\frac{3x}{100}1\text{b/min} \end{gathered}$$

For tank B

Input Rate of Salt − Output Rate of Salt

$$\begin{gathered} \frac{d{x}_2}{dt}=\left(2\text{gal/min}\right)\left(\frac{x_1}{100}1\text{b/gal}\right)-\left(1\text{gal/min}\right)\left(\frac{x_2}{100}1\text{b/gal}\right)-\left(2\text{gal/min}\right)\frac{x_2}{100}\text{Ib} \\ =\frac{x_1}{50}.1\text{b/min}-\frac{x_2}{50}.1\text{b/min} \end{gathered}$$

$$\begin{gathered} \frac{d{x}_1}{dt}=\frac{3}{100}{x}_1+\frac{1}{100}{x}_2, \\ \frac{d{x}_2}{dt}=\frac{1}{50}{x}_1-\frac{1}{50}{x}_2 \end{gathered}$$

Given, matrix form

$X\text{'}=\left(\begin{array}{cc}-\frac{3}{100}& \frac{1}{100}\\ \frac{1}{50}& -\frac{1}{50}\end{array}\right)X$

$A=\left(\begin{array}{cc}-\frac{3}{100}& \frac{1}{100}\\ \frac{1}{50}& -\frac{1}{50}\end{array}\right)$

b)The eigenvalue method of this section to solve the linear system.

$$\begin{gathered} \det \left(A-\lambda I\right)=0\to \left|\begin{array}{cc}-\frac{3}{100}-\lambda & \frac{1}{100}\\ \frac{1}{50}& -\frac{1}{50}-\lambda \end{array}\right|=0 \\ \left(-\frac{3}{100}-\lambda \right)\left(-\frac{1}{50}-\lambda \right)-\left(\frac{1}{100}\right)\left(\frac{1}{50}\right)=0 \\ \frac{3}{100}+\frac{3}{100}\lambda +\frac{1}{50}\lambda +{\lambda }^2-\frac{1}{500}=0 \\ {\lambda }^2+\frac{1}{20}\lambda +\frac{1}{2500}=0 \\ \left(\lambda +\frac{1}{100}\right)\left(\lambda +\frac{1}{25}\right)=0 \end{gathered}$$

Eigenvalues:

$$\begin{gathered} {\lambda }_1=-\frac{1}{100}=-0.01 \\ {\lambda }_2=-\frac{1}{25}=-0.04 \end{gathered}$$

For ${\lambda }_1=-0.01:$

role="math" localid="1663977548762" $$\begin{gathered} \left(A + 0.01I | 0\right)= \left(-0.3 \\ 0.02-0.02\right. \left.-\left(-0.01\right) \\ -0.01 \right| \left.0 \\ 0\right) \\ =\left(-0.02 \\ 0.02\right. \left. 0.01 \\ - 0.01 \right|\left.0 \\ 0\right) \end{gathered}$$

Apply row operation $R_2+R_1\to R_2$

$$\begin{gathered} =\left(-0.02 \\ 0\right. \left. 0.01 \\ 0 \right|\left.0 \\ 0\right) \end{gathered}$$

$-0.02K_1+0.01K_2=0\to K_1=\frac{1}{2}{K}_2$

$K_2=2$yields $K_1=1$this gives an eigenvector and a corresponding solution vector.

$$\begin{gathered} K_1=\left(1 \\ 2\right),x_1=\left(1 \\ 2\right)e^{-0.01t} \end{gathered}$$

For${\lambda }_2=-0.04:$

role="math" localid="1663978155453" $$\begin{gathered} \left(A+0.04I\left|0\right.\right)=\left(-0.01 \\ 0.02\right. \left. 0.01 \\ - 0.02 \right|\left.0 \\ 0\right) \end{gathered}$$

Row operation:

$0.5R_2+R_1\to R_2$

$$\begin{gathered} =\left(-0.01 \\ 0\right. \left.0.01 \\ 0 \right|\left.0 \\ 0\right) \end{gathered}$$

$0.01K_1+0.01K_2=0\to K_1=-0.01K_2$

$K_2=1,K_1=-1$This gives an eigenvalues solution vector:

$$\begin{gathered} K_2=\left(-1 \\ 1\right),x_2=\left(-1 \\ 1\right)e^{-0.04t} \end{gathered}$$

General solution system is:

$$\begin{gathered} x=c_1\left(1 \\ 2\right)e^{-0.01t}+c_2\left(-1 \\ 1\right)e^{-0.04t} \end{gathered}$$

Apply the initial value condition:

$$\begin{gathered} x\left(0\right)=\left(20 \\ 5\right) \\ \left(20 \\ 5\right)=c_1\left(1 \\ 2\right)+c_2\left(-1 \\ 1\right) \end{gathered}$$

$$\begin{gathered} 20=c_1-c_2\to 1 \\ 5=2c_1+c_2\to 2 \end{gathered}$$

Two equations together yields

$$\begin{gathered} 20+5=c_1+2c_1 \\ 25=3c_1 \\ {c}_1=\frac{25}{3} \end{gathered}$$

Substitute 1 obtain value

$$\begin{gathered} 20=\frac{25}{3}-c_2 \\ {c}_2=-\frac{35}{3} \end{gathered}$$

Therefore,

$$\begin{gathered} x\left(t\right)=\frac{25}{3}\left(1 \\ 2\right)e^{-0.01t}-\frac{35}{3}\left(-1 \\ 1\right)e^{-0.04t} \end{gathered}$$

Plot

The plot is point the value of graphing.

c) Use graphing utility

$x_1\left(t\right)=\frac{25}{3}{e}^{-0.01t}+\frac{35}{3}{e}^{-0.04t}$and $x_2\left(t\right)=\frac{50}{3}{e}^{-0.01t}-\frac{35}{3}{e}^{-0.04t}$

d) system of mixing tanks is to be turned off when the number of pounds of salt in tank b equal that in tank A.

The amount of salt in both tanks are the same when $x_1\left(t\right)$and $x_2\left(t\right)$intersect. From the graph, we can see that they intersect at $\left(8.869,34.321\right)$. Therefore, at time $t=34.321$, the amount of salt in both tanks are the same.

Hence,

a) $X\text{'}=\left(\begin{array}{cc}-\frac{3}{100}& \frac{1}{100}\\ \frac{1}{50}& -\frac{1}{50}\end{array}\right)X$

b) $$\begin{gathered} x\left(t\right)=\frac{25}{3}\left(1 \\ 2\right)e^{-0.01t}-\frac{35}{3}\left(-1 \\ 1\right)e^{-0.04t} \end{gathered}$$

c)

d)$t=34.321\text{minutes}$