Transcript Lesson 113 Direct Variation ~ Inverse Variation LESSON PRESENTATION Inverse Variation LESSON PRESENTATION Direct Variation Example 113.1 Example 113.3 Example 113.2 Example 113.4 Example 113.5 Prepared by: David Crockett Math Department.

Lesson 113
Direct Variation
~
Inverse Variation
LESSON PRESENTATION
Inverse Variation
LESSON PRESENTATION
Direct Variation
Example 113.1
Example 113.3
Example 113.2
Example 113.4
Example 113.5
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Direct Variation
In mathematics, two quantities are said to be
proportional if they vary in such a way that one of the
quantities is a constant multiple of the other, or
equivalently if they have a constant ratio.
Proportion also refers to the equality of two ratios.
For example: Suppose at a particular athletic event there
are twice as many boys as girls. We can describe this
relations ship using the equation
B  2G
We can use this equation to find the number of boys at the
event if we know the number of girls. In this case the “2”
is the constant ratio.
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Direct Variation
Given two variables x and y, y is (directly) proportional
to x (y varies directly as x) if there is a non-zero
constant k such that
y  kx
and the constant ratio
k 
y
x
is called the proportionality constant or constant
of proportionality (or constant of variation).
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Direct Variation
Example 113.1 The mass of a substance varies directly as the volume
of the substance. If the mass of 2 liters of the substance is 10
kilograms, what will be the volume of 35 kilograms of the substance?
Solution: 1) We first recognize the words “varies directly” that imply
the relationship
M  kV
2) Read the problem again to find the values of M and V
that can be used to find the value of the constant k.
Use these values to solve for k.
10  k  2 
3) Replace k in the equation with
the value we found.
M  5V
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
k 5
4) Reread the problem again to
find that we are asked to find
the value of V when M is 35.
M  5V
35  5V
7 liters  V
Direct Variation
Example 113.2 The distance traversed by a car traveling at a constant
speed is directly proportional to the time spent traveling. If the car goes
75 kilometers in 5 hours, how far will it go in 7 hours?
D  kT
D  15T
75  k  5 
D  15  7 
15  k
D  105 km
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Direct Variation
Example 113.3 Under certain conditions the pressure of a gas varies
directly as the temperature. When the pressure is 800 pascals, the
temperature is 400 K. What is the temperature when the pressure is 400
pascals.
P  kT
800  k  400 
2k
P  2T
400  2  T
200  T
The temperature is 200 K when the pressure is 400 pascals.
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
Inverse Variation
When a problem states that one variable varies
inversely as the other variable or that the value of one
variable is inversely proportional to the value of the
other variable, an equation of the form
k
V 
W
is implied, where k is the constant of proportionality
and V and K are the variables.
Let’s compare the equations for direct and inverse
variation.
Direct Variation
V  kW
Inverse Variation
V 
k
W
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Inverse Variation
Example 113.4 Under certain conditions, the pressure of a perfect gas
varies inversely as the volume. When the pressure of a quantity of gas
7 pascals, the volume is 75 liters. What would be the volume if the
pressure is increased to 15 pascals?
k
P 
P
525
V
V
k
525
7
15 
V
75
525  k
15V  525
V 
525
15
V  35 liters
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Inverse Variation
inverselyproportional
proportional
Example 113.5 To travel a fixed distance, the rate is inversely
to the time required. When the rate is 60 kilometers per hour, the time
required is 4 hours. What would be the time required for the same distance
if the rate is increased to 80 kilometers per hour?
k
R 
R 
240
T
T
k
240
60 
80 
T
4
240  k
80T  240
T 
240
80
T  3 hours
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Prepared by: David Crockett Math Department