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1.10 Applications of Linear Systems: Class Notes

Definitions

#1

The accompanying figure shows a network in which the flow rate and direction of flow in certain branches are known. Find the flow rates and directions of flow in the remaining branches.

A diamond-shaped network with a node at the top, left, right, and bottom. A flow of 50 enters the top node from above. A flow of 30 leaves the left node to the left. A flow of 60 leaves the right node to the right. A flow of 40 enters the bottom node from below. On the branch from the bottom node to the right node, a flow of 50 moves toward the right node. The other three branches (bottom to left, left to top, top to right) have unknown flows.

Solution

Pick a direction for the flow in each unknown branch. Label the nodes $A$ (left), $B$ (top), $C$ (right), and $D$ (bottom), and let $x_1$ be the flow from $D$ to $A$, $x_2$ the flow from $A$ to $B$, and $x_3$ the flow from $B$ to $C$.

The same network with nodes labeled $A$ (left), $B$ (top), $C$ (right), $D$ (bottom). $x_1$ flows from $D$ to $A$, $x_2$ from $A$ to $B$, and $x_3$ from $B$ to $C$.

At each node, rate in = rate out:

$$\begin{array}{ccc}{\displaystyle \text{Node }A:}\hfill & \hfill {\displaystyle x_1}& {\displaystyle =30+x_2}\hfill \\ {\displaystyle \text{Node }B:}\hfill & \hfill {\displaystyle x_2+50}& {\displaystyle =x_3}\hfill \\ {\displaystyle \text{Node }C:}\hfill & \hfill {\displaystyle x_3+50}& {\displaystyle =60}\hfill \\ {\displaystyle \text{Node }D:}\hfill & \hfill {\displaystyle 40}& {\displaystyle =x_1+50}\hfill \end{array}$$

Moving the variables to one side:

$$\begin{array}{cc}\hfill {\displaystyle x_1-x_2}& {\displaystyle =30}\hfill \\ \hfill {\displaystyle x_2-x_3}& {\displaystyle =-50}\hfill \\ \hfill {\displaystyle x_3}& {\displaystyle =10}\hfill \\ \hfill {\displaystyle x_1}& {\displaystyle =-10}\hfill \end{array}$$

Solving: use an appropriate method. (Ideally, practice the methods from this course, but honestly, in this case it’s overkill.)

$$x_3=10\Rightarrow x_2-10=-50\Rightarrow x_2=-40$$

$$x_1=30+x_2=30-40\Rightarrow x_1=-10$$

The value $x_1=-10$ also agrees with the equation at node $D$.

A negative flow means the flow goes the opposite way from the direction we picked. So $x_1=-10$ is a flow of 10 from $A$ to $D$, and $x_2=-40$ is a flow of 40 from $B$ to $A$.

The solved network: 40 flows from $B$ to $A$, 10 flows from $A$ to $D$, and 10 flows from $B$ to $C$.

Answer: 40 from $B$ to $A$, 10 from $A$ to $D$, and 10 from $B$ to $C$.

Tip: Check flow conservation at node $A$: 40 flows in, and 30 + 10 = 40 flows out.

Definitions

Ohm’s Law

Kirchhoff’s Laws (summary)

#6

Analyze the given electrical circuit by finding the unknown currents.

A diamond-shaped circuit with node $A$ at the top and node $B$ at the bottom, joined by a middle wire with a $4\Omega$ resistor and current $I_1$ pointing up, toward $A$. Left side: from $A$, a $6\Omega$ resistor with current $I_2$ pointing away from $A$, then a 1 V battery whose negative terminal is on the upper side and positive terminal on the lower side, down to $B$. Right side: from $A$, a 2 V battery whose positive terminal is on the upper side and negative terminal on the lower side, then a $2\Omega$ resistor with current $I_3$ pointing up, from $B$ toward $A$.

Solution

This circuit contains 3 closed loops: the left loop, the right loop, and the outer loop. We go around each one clockwise.

The same circuit with node $A$ at the top and $B$ at the bottom, and three clockwise loops marked: the left loop (through the $6\Omega$ resistor, the 1 V battery, and the $4\Omega$ resistor), the right loop (through the 2 V battery, the $2\Omega$ resistor, and the $4\Omega$ resistor), and the outer loop (around the outside, skipping the middle wire).

Nodes. Count current into a node as positive and current out of a node as negative:

$$\begin{array}{cc}\hfill {\displaystyle \text{Node }A:}& {\displaystyle I_1+I_3-I_2=0}\hfill \\ \hfill {\displaystyle \text{Node }B:}& {\displaystyle I_2-I_1-I_3=0}\hfill \end{array}$$

These are equivalent equations (each is the other times $-1$), so we only use one of them.

Clockwise right loop. Starting at $A$: going through the 2 V battery from + to − is a drop of 2. Going through the $2\Omega$ resistor against $I_3$ is a rise of $2I_3$. Going up through the $4\Omega$ resistor with $I_1$ is a drop of $4I_1$.

$$\begin{array}{ccc}\hfill {\displaystyle \text{voltage rises}}\hfill & \hfill {\displaystyle \text{voltage drops}}\hfill & {\displaystyle }\hfill \\ \hfill {\displaystyle 2I_3}\hfill & \hfill {\displaystyle -4I_1-2}\hfill & {\displaystyle =0}\hfill \end{array}$$

Clockwise left loop. Starting at $B$: going through the 1 V battery from + to − is a drop of 1. Going through the $6\Omega$ resistor against $I_2$ is a rise of $6I_2$. Going down through the $4\Omega$ resistor against $I_1$ is a rise of $4I_1$.

$$\begin{array}{ccc}\hfill {\displaystyle \text{voltage rises}}\hfill & \hfill {\displaystyle \text{voltage drops}}\hfill & {\displaystyle }\hfill \\ \hfill {\displaystyle 6I_2+4I_1}\hfill & \hfill {\displaystyle -1}\hfill & {\displaystyle =0}\hfill \end{array}$$

Outer loop. $2I_3+6I_2-1-2=0$. This equation is redundant: it is the sum of the left-loop and right-loop equations, so we don’t need it.

System:

$$\begin{array}{cc}\hfill {\displaystyle I_1-I_2+I_3}& {\displaystyle =0(\text{Node }A)}\hfill \\ \hfill {\displaystyle -4I_1+2I_3}& {\displaystyle =2}\hfill \\ \hfill {\displaystyle 4I_1+6I_2}& {\displaystyle =1}\hfill \end{array}$$

In matrix form:

$$\left[\begin{array}{ccc}1& -1& 1\\ -4& 0& 2\\ 4& 6& 0\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\\ I_3\end{array}\right]=\left[\begin{array}{c}0\\ 2\\ 1\end{array}\right]$$

Row reduce the augmented matrix:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_2\to 4R_1+R_2$ and $R_3\to -4R_1+R_3$, then ${\displaystyle R_2\to \frac{1}{2}{R}_2}$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 10}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_3\to 5R_2+R_3$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 11}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]$$

The entries below the diagonal are now zero. Next clear the third column: $R_1\to 11R_1-R_3$ and $R_2\to 11R_2-3R_3$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 11}\hfill & \hfill {\displaystyle -11}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -6}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -22}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -7}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 11}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]$$

$R_1\to 2R_1-R_2$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 22}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -5}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -22}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -7}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 11}\hfill & \hfill {\displaystyle 6}\hfill \end{array}\right]$$

Divide each row by its pivot:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -5/22}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 7/22}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 6/11}\hfill \end{array}\right]$$

Tip: Since $I_1$ is negative, the arrow given for $I_1$ points in the wrong direction: a current of ${\displaystyle \frac{5}{22}}$ A flows down through the $4\Omega$ resistor.

Tip: Check with the outer loop: ${\displaystyle 2\left(\frac{6}{11}\right)+6\left(\frac{7}{22}\right)=\frac{12}{11}+\frac{21}{11}=3=1+2}$.

Theorem 1.10.1 Polynomial Interpolation

#16

The accompanying figure shows the graph of a cubic polynomial. Find the polynomial.

Graph on a grid with $x$ from 0 to 8 and $y$ from 0 to 10. The curve starts at $(0,0)$, passes through $(2,5)$, reaches a peak at $(4,8)$, passes through $(6,3)$, and crosses the $x$-axis just before $x=6.5$.

Solution

The polynomial is a cubic, so it has the form of a general cubic polynomial:

$$p(x)=a{x}^3+b{x}^2+cx+d$$

A cubic has 4 coefficients, so we need 4 points (Theorem 1.10.1 with $n=4$). Read them off the grid:

The same graph with the points $(0,0)$, $(2,5)$, $(4,8)$, and $(6,3)$ marked on the curve.

For our augmented matrix setup, write the polynomial as

$$p(x)=d+cx+b{x}^2+a{x}^3$$

Substitute each point:

$$\begin{array}{cc}{\displaystyle (0,0):}\hfill & {\displaystyle 0=d+0c+0^{2}b+0^{3}a}\hfill \\ {\displaystyle (2,5):}\hfill & {\displaystyle 5=d+2c+4b+8a}\hfill \\ {\displaystyle (4,8):}\hfill & {\displaystyle 8=d+4c+16b+64a}\hfill \\ {\displaystyle (6,3):}\hfill & {\displaystyle 3=d+6c+36b+216a}\hfill \end{array}$$

The first equation gives $d=0$. The other three equations, with the columns in the order $c,b,a$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 5}\hfill \\ \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 16}\hfill & \hfill {\displaystyle 64}\hfill & \hfill {\displaystyle 8}\hfill \\ \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 36}\hfill & \hfill {\displaystyle 216}\hfill & \hfill {\displaystyle 3}\hfill \end{array}\right]$$

Tip: Ordering the columns $c,b,a$ (instead of $a,b,c$) puts the smallest numbers in the first column, which makes the row reduction easier. With the order $a,b,c$, the first column would be 8, 64, 216.

${\displaystyle R_2\to \frac{1}{4}{R}_2}$ and ${\displaystyle R_3\to \frac{1}{3}{R}_3}$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 5}\hfill \\ \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 16}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 12}\hfill & \hfill {\displaystyle 72}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_1\leftrightarrow R_2$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 16}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 8}\hfill & \hfill {\displaystyle 5}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 12}\hfill & \hfill {\displaystyle 72}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_2\to R_2-2R_1$ and $R_3\to R_3-2R_1$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 16}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle -24}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 40}\hfill & \hfill {\displaystyle -3}\hfill \end{array}\right]$$

$R_3\to R_3+R_2$, then ${\displaystyle R_3\to \frac{1}{16}{R}_3}$:

$$\left[\begin{array}{cccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 4}\hfill & \hfill {\displaystyle 16}\hfill & \hfill {\displaystyle 2}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -4}\hfill & \hfill {\displaystyle -24}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -1/8}\hfill \end{array}\right]$$

Back substitute. Row 3 gives ${\displaystyle a=-\frac{1}{8}}$. Row 2:

$${\displaystyle -4b-24(-\frac{1}{8})=1\Rightarrow -4b+3=1\Rightarrow b=\frac{1}{2}}$$

Row 1:

$${\displaystyle c+4\left(\frac{1}{2}\right)+16(-\frac{1}{8})=2\Rightarrow c+2-2=2\Rightarrow c=2}$$

Recall $d=0$.