Applications in Ch1

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1.3 Applications of Systems of Linear Equations 25

1.3 Applications of Systems of Linear Equations

Set up and solve a system of equations to fit a polynomial functionto a set of data points.

Set up and solve a system of equations to represent a network.

Systems of linear equations arise in a wide variety of applications. In this section youwill look at two applications, and you will see more in subsequent chapters. The firstapplication shows how to fit a polynomial function to a set of data points in the plane.The second application focuses on networks and Kirchhoff’s Laws for electricity.

POLYNOMIAL CURVE FITTINGSuppose points in the -plane

represent a collection of data and you are asked to find a polynomial function of degree

whose graph passes through the specified points. This procedure is called polynomial curvefitting. If all -coordinates of the points are distinct, then there is precisely one polynomialfunction of degree (or less) that fits the points, as shown in Figure 1.4.

To solve for the coefficients of substitute each of the points into thepolynomial function and obtain linear equations in variables

Example 1 demonstrates this procedure with a second-degree polynomial.

Polynomial Curve Fitting

Determine the polynomial whose graph passes through thepoints and

SOLUTION

Substituting 2, and 3 into and equating the results to the respective -valuesproduces the system of linear equations in the variables and shown below.

The solution of this system is

and

so the polynomial function is

Figure 1.5 shows the graph of p .

p x 24 28 x 8 x 2 .

a 2 8a 1 28,a 0 24,

p 3 a 0 a 1 3 a 2 3 2 a 0 3a 1 9a 2 12 p 2 a 0 a 1 2 a 2 2

2

a 0 2a 1 4a 2 0

p 1 a 0 a 1 1 a 2 1 2 a 0 a 1 a 2 4

a 2a 1 ,a 0 , y p x x 1,

3, 12 .2, 0 ,1, 4 , p x a 0 a 1 x a 2 x 2

a 0 a 1 x n a 2 x n2 . . . a n 1 x n

n 1 yn

.

.

.

a 0 a 1 x 2 a 2 x 22 . . . a n 1 x 2n 1 y2

a 0 a 1 x 1 a 2 x 12 . . . a n 1 x 1n 1 y1

a n 1a 0 , a 1 , a 2 , . . . ,nnn p x ,n

nn 1 x

p x a 0 a 1 x a 2 x 2 . . . a n 1 x n 1

n 1

x 1 , y1 , x 2 , y2 , . . . , x n , yn

xyn

Figure 1.4

Polynomial C urve Fittin g

( x 1, y 1)

( x 2, y 2)

( x 3, y 3)

( x n , y n )

y

x

Figure 1.5

x

(1, 4)

(2, 0)

(3, 12)

2

4

6

8

10

12

12 3 4

y

p( x )

Simulation

Explore this concept further withan electronic simulation availableat www.cengagebrain.com .



Polynomial Curve Fitting

Find a polynomial that fits the points

and

SOLUTION

Because you are given five points, choose a fourth-degree polynomial function

Substituting the given points into produces the following system of linearequations.

The solution of these equations is

which means the polynomial function is

Figure 1.6 shows the graph of

The system of linear equations in Example 2 is relatively easy to solve becausethe -values are small. For a set of points with large -values, it is usually best totranslate the values before attempting the curve-fitting procedure. The next exampledemonstrates this approach.

Translating Large -Values Before Curve Fitting

Find a polynomial that fits the points

SOLUTIONBecause the given -values are large, use the translation to obtain

This is the same set of points as in Example 2. So, the polynomial that fits these points is

Letting you have

p x 1 54 x 2008 101

24 x 2008 2 34 x 2008 3 17

24 x 2008 4 .

z x 2008,

1 54 z

10124 z 2 3

4 z3 17

24 z4 .

p z 124 24 30 z 101 z2 18 z 3 17 z 4

2, 10 .1, 4 ,0, 1 ,1, 5 ,2, 3 ,

z 5 , y5 z 4 , y4 z 3 , y3 z 2 , y2 z 1 , y1

z x 2008 x

2010, 10 .2009, 4 ,2008, 1 ,2007, 5 ,2006, 3 ,

x 5 , y5 x 4 , y4 x 3 , y3 x 2 , y2 x 1 , y1

x

x x

p.

124 24 30 x 101 x 2 18 x 3 17 x 4 .

p x 1 3024 x 101

24 x 2 1824 x 3 17

24 x 4

a 41724a 3

1824 ,a 2

10124 ,a 1

3024 ,a 0 1,

a 0 2a 1 4a 2 8a 3 16a 4 10

a 0 a 1 a 2 a 3 a 4 4

a 0 1

a 0 a 1 a 2 a 3 a 4 5

a 0 2a 1 4a 2 8a 3 16a 4 3

p x

p x a 0 a 1 x a 2 x 2 a 3 x 3 a 4 x 4 .

2, 10 .1, 4 ,0, 1 ,1, 5 ,2, 3 ,

26 Chapter 1 Systems of Linear Equations

Figure 1.6

(2, 10)

(1, 4)

(0, 1)

(− 1, 5)

(− 2, 3)

x − 3 − 1 1 2

4

8

10

y

p( x )



An Application of Curve Fitting

Find a polynomial that relates the periods of the three planets that are closest to the Sunto their mean distances from the Sun, as shown in the table. Then test the accuracy of the fit by using the polynomial to calculate the period of Mars. (In the table, the meandistance is given in astronomical units, and the period is given in years.)

SOLUTION

Begin by fitting a quadratic polynomial function

to the points

and

The system of linear equations obtained by substituting these points into is

The approximate solution of the system is

which means that an approximation of the polynomial function is

Using to evaluate the period of Mars produces

years.

Note that the actual period of Mars is 1.881 years. Figure 1.7 compares the estimatewith the actual period graphically.

Figure 1.7

Period(inyears)

Mean distance from the S un(in astronomical units)

x

1.5

2.0

1.0

0.5

0.5 1.0 1.5 2.0

Merc ury (0.387, 0.241)

Ven us

Earth

Mars(1.524, 1.881)

y

(0.723, 0.615)

(1.000, 1.000)

y = p( x )

p 1.524 1.918

p x

p x 0.0634 0.6119 x 0.4515 x 2 .

a 2 0.4515a 1 0.6119,a 0 0.0634,

a 0 a 1 a 2 1.

a 0 0.723 a 1 0.723 2 a 2 0.615

a 0 0.387 a 1 0.387 2 a 2 0.241

p x

1, 1 .0.723, 0.615 ,0.387, 0.241 ,

p x a 0 a 1 x a 2 x 2

Planet Mercury Venus Earth Mars

Mean Distance 0.387 0.723 1.000 1.524

Period 0.241 0.615 1.000 1.881




Planet Mercury Venus Earth Mars

Mean Distance x 0.387 0.723 1.000 1.524

ln x 0.949 0.324 0.0 0.421

Period y 0.241 0.615 1.000 1.881

ln y 1.423 0.486 0.0 0.632

As illustrated in Example 4, a polynomial that fits some of the points in a dataset is not necessarily an accurate model for other points in the data set. Generally, thefarther the other points are from those used to fit the polynomial, the worse the fit. Forinstance, the mean distance of Jupiter from the Sun is 5.203 astronomical units. Using

in Example 4 to approximate the period gives 15.343 years—a poor estimate of Jupiter’s actual period of 11.860 years.

The problem of curve fitting can be difficult. Types of functions other than

polynomial functions may provide better fits. For instance, look again at the curve-fittingproblem in Example 4. Taking the natural logarithms of the given distances and periodsproduces the following results.

Now, fitting a polynomial to the logarithms of the distances and periods produces thelinear relationship

shown in Figure 1.8.

Figure 1.8

From it follows that or In other words, the square of theperiod (in years) of each planet is equal to the cube of its mean distance (in astronomicalunits) from the Sun. Johannes Kepler first discovered this relationship in 1619.

y 2 x 3 . y x 3 2 ,ln y 32 ln x ,

ln y = ln x 32

ln x

Merc ury

Ven us

EarthMars

1

− 1

− 2

2

1 2− 2

ln y

ln y 32 ln x

p x


LINEARALGEBRAAPPLIED

Researchers in Italy studying the acoustical noise levelsfrom vehicular traffic at a busy three-way intersection on acollege campus used a system of linear equations to modelthe traffic flow at the intersection. To help formulate thesystem of equations, “operators” stationed themselves atvarious locations along the intersection and counted thenumbers of vehicles going by. (Source: Acoustical Noise Analysis in Road Intersections: A Case Study, Guarnaccia, Claudio,Recent Advances in Acoustics & Music, Proceedings of the 11thWSEAS International Conference on Acoustics & Music: Theory & Applications, June, 2010)

gemphotography/Shutterstock.com



NETWORK ANALYSISNetworks composed of branches and junctions are used as models in such fieldsas economics, traffic analysis, and electrical engineering. In a network model,you assume that the total flow into a junction is equal to the total flow out of the

junction. For instance, the junction shown in Figure 1.9 has 25 units flowing intoit, so there must be 25 units flowing out of it. You can represent this with the

linear equation

Figure 1.9

Because each junction in a network gives rise to a linear equation, you can analyzethe flow through a network composed of several junctions by solving a system oflinear equations. Example 5 illustrates this procedure.

Analysis of a Network

Set up a system of linear equations to represent the network shown in Figure 1.10. Thensolve the system.

SOLUTIONEach of the network’s five junctions gives rise to a linear equation, as follows.

The augmented matrix for this system is

Gauss-Jordan elimination produces the matrix

From the matrix above, you can see that

and

Letting you have

where is any real number, so this system has infinitely many solutions.t

x 5 t x 4 t 10, x 3 t 10, x 2 t 30, x 1 t 10,

t x 5 ,

x 4 x 5 10. x 3 x 5 10, x 2 x 5 30, x 1 x 5 10,

1

0000

0

1000

0

0100

0

0010

1

1110

10

301010

0

.

10010

10100

01100

01001

00011

2020201010

.

Junction 1

Junction 2

Junction 3

Junction 4Junction 5

x 1

x 1

x 2

x 2 x 3 x 3

x 4

x 4 x 5 x 5

202020

1010

25

x 2

x 1

x 1 x 2 25.


Figure 1.10

10 10

x 1 x 4

x 3 x 2

x 5

1

3

4

2

5

20



In Example 5, suppose you could controlthe amount of flow along the branch labeledUsing the solution of Example 5, you could thencontrol the flow represented by each of the othervariables. For instance, letting wouldreduce the flow of and to zero, as shown inFigure 1.11.

You may be able to see how the type of network analysis demonstrated in Example 5could be used in problems dealing with the flow of traffic through the streets of a cityor the flow of water through an irrigation system.

An electrical network is another type of network where analysis is commonlyapplied. An analysis of such a system uses two properties of electrical networks knownas Kirchhoff’s Laws.

1. All the current flowing into a junction must flow out of it.

2. The sum of the products ( is current and is resistance) around a closed pathis equal to the total voltage in the path.

In an electrical network, current is measured in amperes, or amps resistance ismeasured in ohms and the product of current and resistance is measured involts The symbol represents a battery. The larger vertical bar denoteswhere the current flows out of the terminal. The symbol denotes resistance. Anarrow in the branch indicates the direction of the current.

Analysis of an Electrical Network

Determine the currents and for the electrical network shown in Figure 1.12.

SOLUTIONApplying Kirchhoff’s first law to either junction produces

Junction 1 or Junction 2

and applying Kirchhoff’s second law to the two paths produces

Path 1Path 2

So, you have the following system of three linear equations in the variablesand .

Applying Gauss-Jordan elimination to the augmented matrix

produces the reduced row-echelon form

which means amp, amps, and amp. I 3 1 I 2 2 I 1 1

100

010

001

121

130

122

104

078

2 I 2 4 I 3 8 3 I 1 2 I 2 7

I 1 I 2 I 3 0

I 3 I 2 , I 1 ,

R2 I 2 R3 I 3 2 I 2 4 I 3 8. R1 I 1 R2 I 2 3 I 1 2 I 2 7

I 1 I 3 I 2

I 3 I 2 , I 1 ,

V .,

A ,

R I IR

x 3 x 1t 10

x 5 .


20 00 20

10

10 10

1

3

4

2

5

20

Figure 1.11

REMARKA closed path is a sequenceof branches such that thebeginning point of the firstbranch coincides with the endpoint of the last branch.

Figure 1.12

1 2

Path 1

Path 2

I 1

I 2

I 3

R1 = 3

R2 = 2

R3 = 4 8 V

7 V

Ω

Ω

Ω



Analysis of an Electrical Network

Determine the currents and for the electrical network shown inFigure 1.13.

Figure 1.13

SOLUTION

Applying Kirchhoff’s first law to the four junctions producesJunction 1Junction 2Junction 3Junction 4

and applying Kirchhoff’s second law to the three paths produces

Path 1Path 2Path 3

You now have the following system of seven linear equations in the variablesand .

The augmented matrix for this system is

Using Gauss-Jordan elimination, a graphing utility, or a software program, solve thissystem to obtain

and

meaning amp, amps, amp, amp, amps, and amps. I 6 2 I 5 3 I 4 1 I 3 1 I 2 2 I 1 1

I 6 2 I 5 3, I 4 1, I 3 1, I 2 2, I 1 1,

1100200

1100440

1010010

0101020

0011022

0011004

0000

101714

.

I 5 I 5

2 I 52 I 5

I 6 I 6

4 I 6

0000

101714

I 1 I 1

2 I 1

I 2 I 2

4 I 24 I 2

I 3

I 3

I 3

I 4

I 4

2 I 4

I 6 I 5 , I 4 , I 3 I 2 , I 1 ,

2 I 5 4 I 6 14. 4 I 2 I 3 2 I 4 2 I 5 17

2 I 1 4 I 2 10

I 4 I 6 I 5 I 3 I 6 I 5 I 1 I 4 I 2 I 1 I 3 I 2

1 3

2 4

17 V

10 V 14 V

Path 1 Path 3Path 2

R1 = 2

R2 = 4

R3 = 1

R4 = 2

R5 = 2

R6 = 4 I 2 I 5

I 3

I 4

I 6 I 1

Ω

Ω

Ω Ω

Ω

Ω

I 6 I 5 , I 4 , I 3 , I 2 , I 1 ,


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