linear depreciation

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Linear Functions and Mathematical Modeling in the Port of Long Beach

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By the end of this lesson you will:

• Write linear equations using two points or using a point and the slope of the line.

• Write a linear equation to model the depreciation of capital goods.

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The following California Standards will be addressed:

• Algebra 5: Students solve multi-step problems, including word problems, involving linear equations and linear inequalities in one variable and provide justification for each step.

• Algebra 8: Students understand the concepts of parallel lines and perpendicular lines and how their slopes are related. Students are able to find the equation of a line perpendicular to a given line that passes through a given point.

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Real World Applications…

We can use math to calculate the height of the scrap metal pile using the Pythagorean Theorem, a2+b2=c2.

We can calculate the volume of scrap metal that a truck can hold using V = lwh.

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Capital Equipment can be Cranes, Trucks, Railways…

• The Port of Long Beach invests millions of dollars in building infrastructure and purchasing equipment.

• As it ages, infrastructure and equipment loses value, or depreciates.

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Why learn about modeling and linear equations?

• Mathematics is a symbolic language that we use to represent and study the world around us. In this lesson you will use Algebra to model simple depreciation as used in business.

• Using math to study real world problems will provide a better understanding of the uses of math outside of the classroom.

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Think of a car…

• When you buy a car, it loses value in the same way as the equipment and infrastructure at the Port of Long Beach.

• For example, if you bought a 2007 Nissan and were to sell it in one year, you would not be able to sell it for the price you paid — you would sell it for less because it has depreciated in value.

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A little review…

Before we can model using a linear equation, we must remember how to write a linear equation.

We will begin by reviewing slope and two ways to use points to write a linear equation.

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The slope of a line…

• The slope is the rate of change of y with respect to x. Visually we see this as the steepness of the line. (Think of a really steep hill you would have to walk up or down.)

10

Y

X

m=1

Steep

ness

On this graph, with each single unit increase in x, there is a single unit increase in y.

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The larger the slope…

• As the absolute value of the slope becomes larger, the line becomes steeper, moving toward vertical.

Y

X

m=1m

=2

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The smaller the slope…

• As absolute value of the slope becomes smaller, the line becomes flatter, moving toward horizontal.

Y

X

m=1

m= ½

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Tell a neighbor

• The slope represents the of a line.

• As the slope becomes larger, the line becomes .

• True or False: A line with slope of 1/8 will be flatter, moving toward horizontal.

steepness

steeper, moving toward vertical

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A Positive Slope

• A line with a positive slope is drawn up and to the right

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A Negative Slope

• A line with a negative slope is drawn down and to the right.

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Slope of a Line

• If points (x1, y1) and (x2,y2) are two points on a non-vertical line, then the slope of the line is given by the equation:

my y

x x

2 1

2 1

17

Y

X

L1

y2 - y1

x2 - x1

The slope is the change in the y over the change in the x.

m =

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Calculating the Slope

• Find the slope of the line that passes through the point (5, -7) and (2, 4).

my y

x x

2 1

2 1

m

4 7

2 5

( )

3

11

3

11

3

11

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What does the slope tell us?

• The slope of the line that passes through the given points is .

• This line is drawn .

• Is this line steep –going toward vertical, or is it flatter – going toward horizontal?

m 113

down and to the right

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Calculating the Slope

• Find the slope of the line that passes through the point (3,10) and (-3, 8).

m

8 10

3 3

my y

x x

2 1

2 1

m

2

6

m1

3

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What does the slope tell us?

• Discuss the following questions with your neighbor:

• Is the line with this slope, m = 1/3, drawn up or down to the right?

• Is this line steeper – going toward vertical or is it flatter – going toward horizontal?

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The Slope of Vertical Lines

• The slope of a vertical line is undefined. If you select two points on a vertical line and solve for the slope, you will end up with a zero in the denominator.

Y

X

m = undefined

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The Slope of Horizontal Lines

• The slope of a horizontal line is zero. If you select two points on a horizontal line and solve for the slope, you will end up with a zero in the numerator.

Y

X

m = 0

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Parallel Lines

• Two lines are parallel if they have the same slope, m1= m2.

m 1

m 2

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Perpendicular Lines

• Two lines are perpendicular if the slopes of the two lines are negative reciprocals.

m1(m2) = -1

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Parallel, Perpendicular, Neither?

• L1 passes through the points (1,-2) and (4,2), L2 passes through the points (-1,-2)and (3,6). Determine whether lines are parallel, perpendicular or neither.

my y

x x

m

m

m

22 1

2 1

2

2

2

6 2

3 1

8

42

( )

( )

my y

x x

m

m

12 1

2 1

1

1

2 2

4 14

3

( )

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Are they parallel?

• The slope of L1 is 4/3 and the slope of L2 is 2.

• Are these lines parallel?

• Raise your right hand if you think the answer is yes and your left if you think the answer is no.

• No, the slopes are not the same.

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Are they perpendicular?

• The slope of L1 is 4/3 and the slope of L2 is 2.

• Are these lines perpendicular?• Raise your right hand if you think the

answer is yes and your left if you think the answer is no.

• No, their product is not negative 1.

3

82

3

4

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Neither

• Lines L1 and L2 are neither parallel nor perpendicular.

• To have been parallel, the slopes would have to have been equal.

• To have been perpendicular, the slopes would have to have been negative reciprocals (the product of the two slopes will equal -1).

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Parallel, Perpendicular, Neither?

• L1 passes through the points (-2,5) and (4,2), L2 passes through the points (-1,-2)and (3,6). Determine whether lines are parallel, perpendicular or neither. Write your answer on your response board.

my y

x x

m

m

m

12 1

2 1

1

1

1

2 5

4 2

3

61

2

( )

my y

x x

m

m

m

22 1

2 1

2

2

2

6 2

3 1

8

42

( )

( )

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Are they parallel?

• The slope of L1 is -1/2 and the slope of L2 is 2.

• Are these lines parallel? • No.• Why?• The slopes are not the same.

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Are they perpendicular?

• The slope of L1 is -1/2 and the slope of L2 is 2.

• Are these lines perpendicular?

• Yes• Why?• The product of the two slopes in -1.

They are negative reciprocals.

122

1

33

Tell a friend…

• Explain to your neighbor how you determine whether a line is perpendicular or parallel.

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Find the Equation of a Line Using Point-Slope Form.

• We can write the equation of a line if we know two points on the line or a point on the line and the slope of the line.

• We can use the Point–Slope Form which is given by the equation

y – y1 = m(x – x1)

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Given two points…

Find the equation of the line that passes through the points (4, 5) and (6, -1).

Step 1: Find the slope.

my y

x x

2 1

2 1

m

1 5

6 4

m 6

2

m 3

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Example Continued…

Step 2: Using point (4,5) and m= -3, substitute the values into the equation.

y – y1 = m(x – x1).

y – 5=-3(x – 4)y – 5=-3x +12y =-3x + 17

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Find the equation…

Find the equation of the line that passes through the point (2,3) and (-4, -6) using the Point-Slope equation, y – y1 = m(x – x1)

m

m

m

6 3

4 29

63

2

y y m x x

y x

y x

y x

1 1

32

32

32

3 2

3 3

( )

( )

Step 1 Step 2

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Given a point and the slope…

Find the equation of the line that passes through the point (6, -2) and has a slope of 2.

y – (-2) = 2(x – 6)y + 2 = 2x – 12y = 2x - 14

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Tell a friend…

• Explain to your neighbor how you found the equation of the line.

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Practice

1. Write the equation of the line passing through the points (5,1) and (-6, -4).

2. Write the equation of the line with a slope of 0 and passing through the point (6,9).

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Solution to Practice #1

1. Write the equation of the line passing through the points (5,1) and (-6, -4).

m

m

m

4 1

6 55

115

11

y y m x x

y x

y x

y x

y x

y x

1 1

511

511

2511

511

2511

511

2511

1111

511

1411

1 5

1

1

( )

( )

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Solution to Practice #2

2. Write the equation of the line with a slope of 0 and passing through the point (6,9).

y y m x x

y x

y

y

1 1

9 0 6

9 0

9

( )

( )

Is this line vertical or horizontal? Why?

It is horizontal because the slope is 0.

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Finding the equation of a line using slope-intercept form.

• Another way to find the equation of a line is by using the slope intercept form:

y = mx + b• Given two points we can find the slope of

the line. • With the slope and a point we solve for b. • Once we have m and b we substitute the

values in to the equation.

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Slope-Intercept Form

Write the equation of the line that passesthrough point (1,3) and point (4, -6).

my y

x x

m

m

m

( )

( )2 1

2 1

6 3

4 19

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1. Find the slope. 2. Find b

y mx b

b

b

b

3 3 1

3 3

6

( )

3. Substitute

y mx b

y x

3 6

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Slope-Intercept

Using the slope-intercept form the linear equation, find the equation of the line that passes through the point (4,1) and (5,-3).

Write your solutions on your response board.

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Solution

1. Slope

my y

x x

m

m

2 1

2 1

3 1

5 44

2. Find b

y mx b

b

b

b

1 4 4

1 16

17

( )

3. Substitute

y mx b

y x

4 17

Find the equation of the line that passes through the points (4,1) and (5,-3).

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Practice

• Find the equation of the line passing through the given points, use the specified method.

• A) Point-Slope, (4,7) and (-2, -3).

• B) Slope – Intercept for (-4, -6) and (5, 8).

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Solution to A

Point-Slope, (4,7) and (-2, -3).

my y

x x

m

m

m

2 1

2 1

3 7

2 410

65

3

1. Slope 2. Substitute and simplify

y y m x x

y x

y x

y x

y x

y x

y x

1 1

53

53

203

53

203

53

203

33

53

203

213

53

13

7 4

7

7

7

( )

( )

( )

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Solution to B

Use slope – intercept for (-4, -6) and (5, 8).

my y

x x

m

m

2 1

2 1

8 6

5 4

14

9

( )

( )

1. Slope 2. Find b 3. Substitute

y mx b

b

b

b

b

b

b

6 4

6

6

6

149

569

1269

99

569

549

569

29

( )

( )

y mx b

y x

14

929

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Mathematical Modeling

• Mathematics is simply a symbolic language used to study the world around us.

• Mathematical Modeling is the process of formulating real-world situations into the language of mathematics.

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Mathematical Modeling

• Some mathematical models are very precise. For example, finance equations will give you an exact calculation of interest or the future value of an investment.

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Compound Interest

• Principal= 10,000• Interest Rate= 7%• Compounding

Period =12• Years=30

A P

A

A

A

A

rm

m t

( )

, ( )

, ( . )

, ( . )

$81, .

( )( )

. ( )( )

1

10 000 1

10 000 10058

10 000 8116

164 97

0712

12 30

360

If you were to deposit $10,000 at 7% interest for 20 years compounded monthly, it would accumulate to $81,164.97.

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Models for Estimating

Some mathematical models will provide only an estimate.

For example, using an exponential function can predict global warming.

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Simple Depreciation

• We will be writing linear equations for the straight line method of linear depreciation to study the depreciation of capital investments made by the Port of Long Beach.

• Depreciation is the decrease or loss in value of capital due to age, wear or market conditions. In accounting it is the allowance made for a loss in the value of capital.

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The Port of Long Beach

• Located in our own backyard, the Port of Long Beach is the second-busiest port in the United States.

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Port of Long Beach

• Over $100 billion dollars with of cargo passes through the Port each year.

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Port of Long Beach

• Long Beach-generated trade supports 1.4 million jobs throughout the U.S. and generates about $15 billion in annual trade-related wages.

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The Port

• The Port of Long Beach manages the facilities of the Port.

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The Port

• It is responsible for infrastructure.

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The Port

• Its revenue comes from tariffs that shipping companies and importers pay for the cargo received and shipped out of Long Beach.

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The Port of Long Beach

• Toward that end, the Port of Long Beach makes large capital investments in in infrastructure for improvements, expansion and safety. The following examples illustrate some of those investments.

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Linear Depreciation

• In 2002 the Port of Long Beach purchased a ZPMC Crane for Pier T for the amount of $6,811,461.73. The crane is to be depreciated over 15 years with a scrap value of $0.

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Linear Depreciation

• In 2002 the Port of Long Beach purchased a ZPMC Crane for Pier T for the amount of $6,811,461.73. The crane is to be depreciated over 15 years with a scrap value of $0.

• Write an expression that will calculate the value of the crane at the end of year (t).

• What is the value of the crane in 2007?

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Simple Depreciation

• Write an expression that will calculate the value of the crane at the end of year (t).

• We will need two coordinates of the form (time, value).

• Time is the independent variable.• Value is the dependent variable.

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Linear Depreciation

In 2002, (this will be t = 0) the Port of Long Beach purchased a ZPMC Crane for Pier T for the amount of $6,811,461.73. The crane is to be depreciated over 15 years with a scrap value of $0.

(0, $6,811,461.73) (15, $0)&

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Finding the Coordinates

• First coordinate: The crane was worth $6,811,461.73 at the time of purchase (t=0). Our first coordinate is

(0, $6,811,461.73)

• Second coordinate: After 15 years, the crane will have a value of $0, so the second coordinate is

(15, $0)

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Given the coordinates we can solve the problem…

• Using (0, $6,811,461.73) and (15,0), we can find the equation.

m

m

m

( , , . )

( , )

, , .

, .

0 6 811 46173

15 0

6 811 46173

15454 097 45

Find the slope Substitute and Simplify

736,811,461.x454,097.45y

15)(x454,097.450y

)xm(xyy 11

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The linear equation…

The linear equation expressing the cranes value at the end of t years is given by

y = -454,097.45x + 6,811,461.73

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Rate of Depreciation

• Given, y = - 454,097.45x +

6,811,461.73

• the slope is the rate of depreciation or $454,097.45 per year.

• Note the y-intercept is the original value of the crane.

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What is the value of the crane in 2007?

The crane was purchase in 2002, so 2007 would be t=5. To solve, substitute 5 in for x.

y=-454,097.45x+6,811,461.73y=-454,097.45(5)+6,811,461.73y=4,540,974.48

At the end of 2007, the book value of the crane was $4,540,974.48.

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Check for Understanding

In 1984, a tractor/loader was purchased for use at the Port of Long Beach for a price of $29,041.01. The tractor/loader was depreciated using the straight-line method over 8 years. Find the linear equation expressing the tractor’s book value at the end of t years. What is the rate of depreciation? Check your answer with your neighbor.

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Solution

• Your two coordinates are (0,$29,041.01) and (8,$0)

1. Find the slope 2. Find the equation

13.363008

01.041,290

m

m

01.041,2913.3630

)8(13.36300

)( 11

xy

xy

xxmyy

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Rate of Depreciation?

• The linear equation expressing the crane’s value at the end of t years is given by

y = -3,630.13x + 29,041.01

• What is the rate of depreciation?$3,630.13 per year

• What was the value of the crane in 1987? y = -3,630.13x + 29,041.01y = -3,630.13(3)+ 29,041.01y = 18,150.62

The value of the crane was $18,150.62.

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Practice

A truck scale purchase at a cost of $151,999.75 in 1986 has a scrap value of $0 at the end of 10 years. If the straight-line method of depreciation is used,

• A) Find the rate of depreciation.• B) Find the linear equation expressing the

book value of the scale at the end of t years.

• C) Find the book value at the end of 7 years.

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Check Your Work

A) The rate of depreciation is the slope.

B) The linear equation is

C) Find the book value at the end of 7 years.

y x 15 199 98 151 999 75, . , .

m

m

0 151 999 75

10 015 199 98

, .

, .

y = $45,600.03

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Closure

• Write a brief paragraph explaining the method for writing simple depreciation equations. Include an explanation of depreciation.

• Share your paragraph with your neighbor.