Chapter 7: Direct Variation
Advanced Math
Section 7.1: Direct Variation, Slope,
and Tangent

When two variable quantities have a
constant ratio, their relationship is called
direct variation.

The constant ratio is called the variation
constant.
Page 359 in your textbook…

For the roof, the height of a post depends on
its horizontal distance from the eaves.

Although the height increases as the
distance decreases, the ratio of these two
quantities is constant.

Ratio of Quantities = variation constant
–
–

Height/Distance = 0.75
Height = 0.75(Distance)
You can either use this equation to find the
height of a post needed to support the roof at
another point.
Sample Question #1

Suppose the
carpenter of the roof
shown on page 359
decides that a post
should be 3 feet
from the eaves.
What should the
height of the post
be?
H= 0.75(D)
H = 0.75(3)
H = 2.25
The post should be
2.25 feet long.
Try this one on your own…
Distance from eaves to
first post is 10/4 or 2.5
feet.
H = 0.75(d)
H = 0.75(2.5)
H = 1.875
The shortest post should
be 1.875 feet long.

Suppose the carpenter
of the roof shown on
page 359 wants only
three support post
between the eaves and
the post at the peak.
Assuming that the
posts are evenly
spaced, how long
should the shortest
support post be?
Slope…

The slope of a line is the ratio of rise to run
for any two points on the line.


Rise = vertical change between two points
Run = horizontal change between two points

Slope = rise/run
Sample Question #2

Find the slope of the
roof shown on page
359. Use the eaves
and the peak as the
two points.
Step One: Draw a Sketch.
Step Two: Find the slope.
The roof rises 7.5 feet
across a distance of 10
feet.
Slope = rise/run
Slope = 7.5/10
Try this one on your own…

Slope = rise/run
Slope = 9/18
Slope = 1/2
Suppose the post
from the horizontal
beam to the peak of
the roof is 9 feet
high and the length
of the beam from
the eaves to the
post is 18 feet.
Tangent…

The tangent ratio of an acute angle
compares the two legs of a right triangle.

Tangent = leg opposite / leg adjacent
Sample Question #3

Look again at the
roof on page 359.
What is the tangent
ratio of the angle
between the roof
and the horizontal
beam?
Tan <A = leg opposite <A / leg adjacent <A
Tan <A = 7.5/10
or 0.75
Try this one on your own…

Tangent <A = BC / AC
Consider the roof on
the board…
–
Tan <A = 6/15
Tan <A = 2/5 or 0.4
What is the tangent
ratio of the angle
between the roof
and the horizontal
beam?
Section 7.2: A Direct Variation Model

Read pages 367-370.

Get with a partner and begin your homework.
Section 7.3: Circumference and Arc
Length
overhead
Section 7.4: Direct Variation with y = kx

You can model direct variation with an equation y =
kx.

Y = kx is called the general form for direct variation.

Y = dependent variable

K = variation constant

X = control variable
Sample #1



Y = kx
336 = k (40)
K = 8.4





Y = 8.4x
Y = 8.4(130)
Y = 1092
In 130 hrs, Cindy would
make $1092.
The amount Cindy earns varies directly with the
number of hours she works. She earns $336 in a 40hour work week. How much will Cindy earn in 130
hours?
Try this one on your own…

Efren set an empty fish tank on a bathroom scale. When he
added 20 quarts of water, the weight increase was 41 ¾
pounds. He plans to have 28 quarts of water in the tank when
he sets the tank up for his fish. If the weight of the water varies
directly with the number of quarts used, how much will the
water in the tank weigh?
About 58 pounds
Direct Variation Graphs

The graph of a direct variation equation
y = kx is a line that passes through the origin.

The variation constant, k, is the slope of the
line.
Sample # 2

Is direct variation a good model for the data
shown in each graph? Explain why or why
not.
Try these on your own…

Is direct variation a good model for the data
shown in each graph? Explain why or why
not.
Sample # 3

The direct variation
graph shown is a
line with slope 3.
Write an equation of
the line.
Try this one on your own…

The direct variation
graph shown is a
line with slope 3/2.
Write an equation
for the graph.
Sample #4

Graph y = -2x.

Try this one on your
own…
–
Graph y = -2/3x.
Section 7.5: Using Dimensional
Analysis

When you cancel units of measurement as if they
are numbers, you are using a problem solving
strategy called dimensional analysis.

A conversion factor is a ratio of two equal
quantities that are measured in different units.
–
Example: 12 inches in 1 foot – 12:1
Sample #1

The distance Milo
travels varies directly
with the number of
hours he drives. Milo
drives 145 miles in 3
hours.
–
–
Identify the control
variable and the
dependent variable.
Express the variation
constant as a rate.

Control/Independent
Variable = time traveled
(hours)

Dependent Variable =
distance traveled
(miles)

145 miles / 3 hours =
48.3 miles per hour
Try this one on your own…

The thickness of a stack of
copier paper varies directly
with the number of sheets in
the stack. A new pack of
copier paper is 2 inches
thick and contains 500
sheets.
–
–
Identify the control variable
and dependent variable.
Express the variation
constant as a rate.

Control/Independent =
thickness of stack of paper

Dependent = number of
sheets

Variation Constant = 2
inches/500 sheets = 0.004
inches per sheet
Sample #2

How many centimeters are
in 3 inches?
–

There are 2.54 centimeters
in 1 inch.
Write a direct variation
equation that helps you
convert inches to
centimeters.
–
X = number of inches
Y = number of centimeters
K = 2.54
–
Equation: y = 2.54x
–
3 inches x 2.54 cm/1 in
–
7.62 centimeters

What is the variation
constant?
–
2.54 centimeters per inch
Try this one on your own…

How many pounds are in
70.5 kg?

155.4525 pounds

Write a direct variation
equation for converting
kilograms to pounds.
–

Y = 2.205X
What is the variation
constant?
–
2.205 pounds per kilograms

Pedro Zepeda weighs 70.5 kg.

There are 2.205 pounds per 1 kilogram.
Sample #3






In 1990 a French train set a speed record of 515
km/h on the Atlantique line. Find its speed in miles
per hour?
1 kilometer = 0.621 miles
515 km / 1 hr = ? / ?
515 km / 1 hr = ? mi / ? km
515 km / 1 hr = 0.621 mi / 1 km
320 miles per hour
Try this one on your own…

A large crude oil pipeline from Canada to the United States has
a flow of 8.3 million gallons per day. How many liters per day
does the pipeline handle?

There are 3.78531 liters per 1 gallon.
31.42 million liters of crude oil per day
Sample #4
rate x conversion
factors = new rate

Convert 30 mi/hr to ft/s


1 mile = 5,280 feet
1 hour = 60 minutes
1 minute = 60 seconds

Rate x Conversion Factors = New Rate


Section 7.6: Areas of Circles and
Sectors

Area Formula
–
–
What is it?
Are they units squared or cubed?
Sample #1

Find the area of a circle whose radius is 3 units.
Round the answer to the nearest square unit.

28.3 units squared

Try this one on your own…
–
–
Find the area of a circle whose radius is 7.5 units. Round
the answer to the nearest square units.
176.7 units squared
Sample #2

In a center-pivot irrigation system, a moving
arm sprinkles water over a circular region.
How long must the arm be to water an area
of 586,000 meters squared?

431.9 = radius
Try this one on your own…

The amount of water a pipe can carry
depends on the area of the opening at the
end. An engineer wants a large sewer pipe to
have an opening of 13 feet squared. What
should the radius of the pipe be? Round the
answer to the nearest tenth of a foot.

2.0 = radius

The region of a circle formed by a central
angle and its arc is called a sector.

area of sector = measure of the central angle
area of circle =
360 degrees
Sample #3

A fixed camera that is part of the TV security system
at a shopping mall parking lot has a range of about
250 feet. The angle of vision of the camera is 100
degrees. Over how large an area can the camera
see when it is in operation?

Steps to Solve
–
–
Draw a Picture
Set up the Equation
Sample #4

About how many times more pizza do you
get when you buy the large size instead of
the small size?
–
–
Large is 16 inches in diameter.
Small is 12 inches in diameter.
Try this one on your own…

The camera is Sample #3 replaced an older
camera that had a viewing angle of 110
degrees but a range of only 175 feet. How
many times greater is the viewing area of the
newer camera than that of the older camera?

1.86 times