Study the table and graph of the linear function y = 2x shown below.
The value of y depends directly on the corresponding
value of x. That is, as the value of x increases, the value of y
increases. Note that when x = 0, y = 0. Therefore, the line
passes through the origin. The equation y = 2x is called a direct variation. We say that
y varies directly as x.
A direct variation describes a linear relationship because y
= 0.4x is a linear equation.
Since the value of y depends on the value of x,
y is called the dependent variable. x is called the independent variable.
As stated, a direct variation can be written in the form y
= x. Notice that when x = 0, y = 0. The graph of a
direct variation passes through the origin.
In the equation y = kx, k is called the constant of variation.
Example
Determine whether the equation is a direct variation.
1. y = 4x
Alternative Solutions :
Graph
the equation.
The graph
passes through the origin. So, the equation is a direct variation. The constant of variation is 4.
2. y = x – 1
Alternative Solutions :
Graph
the equation.
The graph
does not pass through the origin. So, the equation is not a direct variation.
Many real-world problems involve direct variation.
Example
Astronomy Link
3. The
weight of an object on Venus varies directly as its weight on Earth. An object
that weighs 80 pounds on Earth weighs 72 pounds on Venus. How much would an
object weigh on Venus if its weight on Earth is 90 pounds?
Alternative Solutions :
First, find the constant of variation. Let x weight on Earth and let y weight on Venus, since the weight on Venus
depends on the weight on Earth.
y = kx Definition of direct variation
72 = k ⋅ 80 Replace y with 72 and x with 80.
Divide each
side by 80.
0.9 = k Simplify.
Next, find the amount of gasoline
needed for a 345-mile trip.
y =
kx Definition
of direct variation
y =
0.9(90)
Replace k with
0.9 and x with 90
y =
81 Simplify.
So, the object would
weigh 81 pounds on Venus.
Direct
variations can be used to solve rate problems.
Example
Travel Link
4. The length of a trip varies directly as the
amount of gasoline used. Suppose 4 gallons of gasoline were used for a 120-mile
trip. At that rate, how many gallons of gasoline are needed for a 345-mile
trip?
Alternative Solutions :
Let l represent the length of the trip and
let g represent the amount of gasoline used for the trip. The statement the
length varies directly as the amount of gasoline translates into the equation l = kg
in the same way as y varies directly as x
translates into y = kx. Find the
value of k.
l = kg
Direct
variation
120 = k ⋅ 4 Replace
l with 120 and g with 4
Divide
each side by 4.
30 = k Simplify.
Next,
find the amount of gasoline needed for a 345-mile trip.
l = kg Direct
variation
345 = 30 g
Replace
l with 345 and k with 30.
Divide each side by 30.
11.5 = g Simplify.
A 345-mile
trip would use 11.5 gallons of gasoline.
Direct variations are also related to proportions. Using the
table on page 264, many proportions can be formed. Two examples are shown.
Three general forms for proportions like these are 
and 
. You can use any of these forms to solve direct variation
problems.
Example
5. Suppose y varies directly as x
and y = 36 when x = 9. Find x when y = 48.
Alternative Solutions :
You can also use direct variation to convert measurements.
Example
Measurement Link
6. If there are 12 inches in 1 foot, how
many inches are in 4.5 feet?
Alternative Solutions :
Sumber
Thanks for reading Direct Variation. Please share...!
