Mathematics A30
Module 2
Lesson 14
Mathematics A30
Inverse Variation
193
Lesson 14
Inverse Variation
Introduction
In a direct variation, when one variable increases, the other variable increases by the
same proportion. If one variable doubles, the other variable doubles. If one variable
triples, the other variable triples. This concept was studied in Mathematics 10.
This course studies another type of variation, that of inverse variation. A definite
relationship can still be shown between the set of ordered pairs of a proportion that varies
inversely. In this relationship, for example, each time one variable doubles, the other
variable is one half as much. If one variable triples, the other variable decreases by one
third.
Inverse variation is very evident in the field of science; for example in pressure and
volume relationships and in relationships involving force. The concept of inverse variation
will be shown using equations, tables and graphs. These concepts will then be applied to
real world situations.
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Objectives
After completing this lesson, you will be able to
• identify and graph examples of inverse variation taken from real world
situations.
• state the domain and range, along with any restrictions, for the graphs of
inverse variations.
• determine the constant of proportionality of an inverse relation.
• solve problems that involve inverse variation.
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Lesson 14
14.1 Inverse Variation
Two types of variation have already been studied in Mathematics 10. They are direct
variation and partial variation. Some of the same concepts will be used in this lesson.
The table will show you a quick review of the concepts from these two types of variation.
Direct Variation
Partial Variation
Word Statement
x varies directly as y.
x varies partially as y.
Variation Statement
xy
xy
Equation
x
=k
y
where k is the constant of
variation.
y = kx + c
where k is the constant of
variation and c is a fixed
number.
Chart
Diagram
y = kx or
x01234
y01234
y
x01234
y34567
y
x
x
With both of these variations, an increase in one of the variables results in a
corresponding increase in the other variable. It could also be that a decrease in one of the
variables would result in a corresponding decrease in the other variable.
The difference between the two types of variation is that partial variation starts with a
fixed number or amount. The graph of a partial variation relation does not go through the
origin.
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The word inverse means "reversed in direction, position or tendency", or "the direct
opposite".
3
4
•
is the multiplicative inverse of .
4
3
•
If one variable increases, the other variable decreases.
•
 6 is the additive inverse of +6.
Sometimes a change in one quantity brings about an opposite change in another quantity.
As one amount increases, the other decreases.
A variation statement shows that one quantity varies as the inverse of the other quantity.
•
If x is one variable and y is the other variable,
1
.
x 
y
There are many situations in the real world where one variable increases and the other
variable decreases.
•
•
An increase in the age of a car results in a decrease in its value.
An increase in oven temperature results in a decrease in the cooking time.
Example 1
It will take 720 man hours to clean up one section of the Assiniboine River in
Moose Jaw where flood waters did a lot of damage.
a)
Create a table to show how the job can be accomplished by varying the
number of workers.
b)
What observations can you make about the relationship between the
number of workers and the time that each works?
c)
Graph the relation.
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Solution:
a)
Number of Workers (x)
Hours Each Worker Works ( y)
1
720
2
360
3
240
4
180
5
144
6
120
7
103
8
90
9
80
10
72
b)
Observations:
•
As the number of workers increases, the number of hours per worker
decreases.
•
If you multiply the number of workers times the number of hours worked, the
answer will always be 720.
c)
Graph the relation.
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What is the shape of the graph?
Example 2
Road conditions vary when driving between Glaslyn and Wilkie which is a
distance of 120 km.
a)
b)
c)
Create a table of values to show how long the trip would take at
different speeds.
What observations can you make about the relationship between the
speed and the time that it takes to travel between Glaslyn and Wilkie?
Graph the relation.
Solution:
a)
b)
Speed
(km/hr)
Time to Travel
(Hours)
40
3
60
2
70
1.7
80
1.5
90
1.3
100
1.2
110
1.1
120
1
Observations:
•
•
The time taken to travel from Glaslyn to Wilkie decreases as the speed
increases.
If you multiply together the speed and the time to travel, the answer will
always be the same.
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c)
Graph the relation.
t
3
Tim e t o
T ravel
(H our s)
2
1
20 40 60 80 100 120 140
Speed (km /hr )
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Lesson 14
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Activity 14.1
This activity is to be handed in with the assignment for evaluation.
• Find an example in the real world where one of the quantities increases and
this brings about a decrease in the other quantity.
• Explain the situation and include real world data where you can.
• Provide a table of values, and a graph.
• Analyze the information.
x
y
y
x
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In each of the first two examples it was found that the product of the numbers in each
ordered pair is a constant. This is true of all relation in which the variables vary inversely.
This value is called the constant of proportionality.
Example 3
Determine if the following set of ordered pairs is a relation in which one
variable varies inversely with the other variable. If it does, state the constant
of proportionality.
110,
 5,  55,  10,  11,  50, 2, 275, 5, 110, 10, 55
Solution:
Multiply each set of ordered pairs together.
•
•
110   5 = 550
2  275 =550
•
•
 55  10 = 550
5  110 = 550
•
•
11   50 = 550
10  55 = 550
Yes, in this relation the variable varies inversely. The constant of proportionality is 550,
the constant value that is the product of the two variables.
Constant of Proportionality
If x 
1
, then xy = k where k is the constant of proportionality.
y
Inverse Variation Equation
From these examples, the standard form of an inverse variation has been shown.
•
If the variable x varies inversely with the variable y, then
1
x
y
xy  k
and for any two ordered pairs x1 , y1 , x 2 , y2  of the relation
x1 y1  k and x 2 y2  k .
Therefore, x1 y1  x 2 y2 .
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Example 4
Find the missing value in the following variation if x 
x 1 = 8,
y1 = 14, x 2 = ?,
1
.
y
y2 = 4
Solution:
•
•
•
This is an inverse variation.
The equation for a proportion that is an inverse variation is x1 y1  x 2 y2 .
The values of the variables are:
•
x1  8
•
x2  ?
•
y1  14
•
y2  4
Write the equation.
Substitute the known values.
Simplify.
Check the solution.
x1 y1  x 2 y2
814 = x2 4
8 14  = x
2
4
28 = x 2
x1 y1  x 2 y2
814  =? 284 
112 = 112 ()
The value of x 2 is 28.
Example 5
Find the missing value in the proportion if p varies inversely as q.
p1 = 6.8 p2 = 3.0 q1 = 15.6 q2 = ? .
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Solution:
•
•
The equation for a proportion that is an inverse variation is p1 q1  p2 q 2 .
The values of the variables are:
•
p1 = 6.8
•
p2 = 3.0
•
q1 = 15 .6
•
q2 = ?
Write the equation.
Substitute the known values.
Simplify.
p1 q1  p2 q 2
6 .8 15 .6  = 3.0  q2
6 .8 15 .6  =
q2
3 .0
35 .36 = q 2
Check the solution.
x1 y1  x 2 y2
6 .8 15 .6  =? 3 .0 35 .36 
106 .08 = 106 .08 ()
The value of q 2 is 35.36.
The majority of situations in the real world require that the domain only include positive
values. This restricts the graph of the function to occur only in the first quadrant. In these
cases the domain and range are both greater than zero.
Domain:
Range:
{x | x   ; x > 0}
{ y | y  ; y > 0}
y
x
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The values in an inverse variation can include negative values. Here is an example of an
inverse variation and its graph, where the domain is the set of all except zero.
Example 6
The values in the following table show an inverse variation. State the
constant of proportionality and graph the relation.
1
x 
y
x
 12
6
4
3
2
1
1
2
3
4
6
12
y
1
2
3
4
6
 12
12
6
4
3
2
1
Solution:
The constant of proportionality is 12.
xy = 12 for all ordered pairs of x and y.
•
Graph the relation.
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What happens when the constant of proportionality is a negative number?
Which of the following graphs represents the inverse variation xy =  5 ?
Summary of Inverse Variation:
There are five ways of describing a proportion that varies inversely.
1.
Word Statement
The time needed to complete a job (t) varies inversely as the number of workers (w).
2.
Variation Statement
t 
3.
1
w
Equation
wt = k where k is the constant of proportionality.
t1w1  t2w2
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4.
Table
wt = 20
5.
w
1
2
4
5
8
10
t (hrs)
20
10
5
4
2.5
2

20 is the constant of proportionality.
Graph
Exercise 14.1
1.
Write the inverse variation statement for each of the following sentences. State a
letter for each of the variables.
a.
b.
c.
d.
e.
The volume of a gas is inversely proportional to the pressure applied to it.
The frequency that a radio station broadcasts at varies inversely as the
wavelength.
In a rectangle of a given area, the length varies inversely as the width.
The atmospheric pressure is inversely proportional to the altitude above sea
level.
The value of a vehicle is inversely proportional to the age of the vehicle.
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2.
Determine which of the following represent an inverse variation. For those that do,
state the constant of proportionality.
a.
b.
c.
1 , 2 , 2 , 1 ,  4 ,
1  1
 , 6, 
2  3

1, 2 , 2 , 4 , 3 , 6 , 4 , 8 
x
4
8
12
16
y
300
150
100
75
d.
3.
Sketch the graph of the following inverse variations.
a.
b.
c.
d.
xy =
xy =
xy =
xy =
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6
9 x > 0, y > 0
 10 x < 0 , y > 0
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Lesson 14
y
y
x
4.
y
x
y
x
x
For each inverse variation, find the missing value.
a.
b.
c.
d.
1
t
1
v 
p
1
s 
t
1
x 
y
s 
14.2
s1 = 75
s 2 = 100
t1 = 6
t2 = ?
v1 = 450
p1 = 10
v2 = ?
p2 = 15
t 1 = 20
t2 = ?
s1 = 360
s 2 = 800
x 2 = 14
y2 = ?
x1 =
3
4
y1 =
8
27
Applications of Inverse Variation
This section will show you how the principle of inverse variation is applied to situations
that occur in every day life. In each of the examples it is important to be able to read
through the problem and apply each of the 5 ways of analyzing a proportion that varies
inversely.
The examples in this section are only a few of the ones that can be found in the real world.
Gears and Pulleys
When two gears rotate together, the speed varies as the number of teeth in each of the
gears.
•
Gear A has 50 teeth.
•
Gear B has 25 teeth.
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Gear A
50 teeth
25 teeth
Gear B
If gear A makes one revolution, gear B will make two revolutions and will therefore be
going twice as fast.
The speed of gears is usually expressed in revolutions per minute (rpm).
Example 1
Bob is riding his road bicycle. He has the gears set so that there are 20 teeth
on the back gear and 32 teeth on the front gear. With this setting, the speed
of the back gear is 120 rpm. What is the speed of the front gear?
Solution:
Read the problem.
•
•
•
The back gear has 20 teeth and a turning speed of 120 rpm.
The front gear has 32 teeth.
Find the turning speed of the front gear.
Develop a plan.
•
•
•
The speed varies inversely as the number of teeth in each of the gears.
This means that:
1
s  , where s is the speed in rpm and t is the number of teeth.
t
t1s1  t2 s2 , where t1 , s1 represent the front gear and t2 , s2 the back.
Substitute the known values into the equation that is determined by the inverse
variation.
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Carry out the plan.
•
•
•
•
t1
t2
s2
s1
=
=
=
=
32
20
120
?
Write the equation.
Substitute the known values.
Simplify.
Check the solution.
t 1 s1  t 2 s 2
32 s1  20 120 
20 120 
s1 
32
s1  75 rpm
32 75  =? 20 120 
2 400 = 2 400 ()
Write a concluding statement.
The front gear would be turning at a speed of 75 rpm.
If Bob wanted his front gear to go faster, would he shift the gears so that he
increases or decreases the number of teeth in his front gear?
Pulleys operate on the same principle as gears except that the speed is inversely
proportional to the diameter of the pulley.
1
s 
d
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Example 2
Two pulleys work together. The smaller pulley has a diameter of 2 cm and
turns at a speed of 1750 rpm. What size of pulley connected to the small
pulley is needed to create a speed of 560 rpm?
Solution:
Read the problem.
•
•
•
Pulley A has a diameter of 2 cm and a speed of 1750 rpm.
Pulley B has a speed of 560 rpm.
Find the diameter of pulley B.
Develop a plan.
•
•
•
The speed of a pulley is inversely proportional to the diameter.
This means that:
1
s 
d
d 1 s1  d 2 s 2
Substitute the known values into the equation that is determined by the inverse
variation.
Carry out the plan.
•
•
•
•
d1 = 2
d2 = ?
s1 = 1750
s 2 = 560
Write the equation.
Substitute the known values.
Simplify.
Check the solution.
d 1 s1  d 2 s 2
2 1750  = d 2 560 
2 1750  =
d2
560
6 .25 = d 2
2 1750  =? 6 .25 560 
3500 = 3500 ()
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Lesson 14
Write a concluding statement.
The diameter of the larger pulley must be 6.25 cm in order for the speed of that pulley to
be 560 rpm.
Boyle's Law
Boyle's Law states that the volume of a gas is inversely proportional to the pressure
exerted on it provided the temperature stays the same.
1
v 
p
•
•
The volume of a gas is often measured in litres (l).
The most common measurement for pressure is kilopascals or kPa.
Example 3
A 1.34 litre container is full of oxygen. A sheet is covering the container and
the only pressure that is being exerted on the gas is that of the atmosphere
which is 101 kPa. A cylinder is placed over the container and a pressure of
202 kPa is exerted. What will be the volume of the gas?
Solution:
Read the problem.
•
•
•
•
The initial volume is 1.34 litres.
The initial pressure is 101 kPa.
The final pressure is 202 kPa.
Find the final volume.
Develop a plan.
•
•
•
The volume of a gas is inversely proportional to the pressure exerted on it.
This means that:
1
v 
p
v1 p1  v2 p2
Substitute the known values into the equation.
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Carry out the plan.
•
•
•
•
v1 = 1.34
v2 = ?
p1 = 101
p2 = 202
Write the equation.
Substitute the known values.
Simplify.
v1 p1  v2 p2
1.34 101  = v 2 202 
1.34 101 
=v2
202
0 .67 = v 2
Check the solution.
1.34 101  =? 0 .67 202 
135 .34 = 135 .34
()
Write a concluding statement.
The volume of gas in the cylinder would now be 0.67 litres.
•
You may have seen an easier method of doing Example 3. By doubling the pressure,
1
this would have resulted in there being the volume.
2
Sometimes the inverse variation of two variables is in a different proportion.
•
•
As one variable increases, the second variable decreases by a proportion that is
squared.
1
x  2
y
As one variable increases, the second variable decreases by a proportion that is the
square root.
1
x 
y
When this occurs, the variation equation is the same, but the different proportion, either
the square or the square root, must be applied to the variable.
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Variation Statement
x 
x 
Equation
x 1  y1  = x 2  y 2 
1
2
2
y
1
2
x1 y1 = x 2 y2
y
Example 4
1
In the inverse variation x 
x1 = 18
y1 = 4 x 2 = ?
2
y
y2 = 3
, find the missing value.
Solution:
•
•
In this inverse variation, x varies as the square of y.
2
2
The equation for this variation is x 1  y1  = x 2  y 2  .
•
The values of the variables are:
•
x1 = 18
•
y1 = 4
•
x2 = ?
•
y2 = 3
Write the equation.
Substitute the known values.
Simplify.
x 1  y1  = x 2  y 2 
2
2
18 4 2 = x 2 32
1816 = x 2 9
1816 = x
2
9
32 = x 2
Check the solution.
184 2 =? 3232
1816 =? 329
288 = 288 ()
The missing value for x 2 is 32.
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What is the constant of proportionality in this example?
Example 5
In the inverse variation x 
x1 = 6
y1 = 25 x 2 = 3
1
, find the missing value.
y
y2 = ?
Solution:
•
•
•
In this inverse variation, x varies as the square root of y.
The equation for this variation is x1 y1  x 2 y 2 .
The values of the variables are:
•
x1 = 6
•
y1 = 25
•
x2 = 3
•
y2 = ?
Write the equation.
Substitute the known values.
Simplify.
x1 y1 = x 2 y2
6  25 = 3 
6 5  = 3 
30
=
3
10 =
10 2
y2
y2
y2
Square both sides.
y2
= y2
100 = y 2
Check the solution.
6  25 =? 3  100
6 5  =? 3 10 
30 = 30
()
The missing value for y2 is 100.
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Square both sides of an equation to remove a square root from an isolated term.
What is the constant of proportionality in Example 5?
The constant of proportionality is found the same way with inverse variations that have
squared or square root variables.
•
x1 y1 = k
•
x 1 y1 = k
•
x 1 y1 = k
2
Sound
There are many inverse variations associated with sound in terms of the frequency or
pitch of the sound.
Frequency or the pitch of a sound is the number of vibrations in a second and is measured
in hertz (Hz).
With a vibrating string, the frequency of the sound produced is inversely proportional to
the length of the string if the tension remains the same.
•
•
The longer the string, the fewer vibrations.
The shorter the string, the more vibrations.
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Example 6
In an experiment it was found that the frequency of a vibrating string that
was 1.8 metres in length was 200 Hz. What would the length have to be to
give a frequency of 720 Hz if the tension on the string is held constant? (The
unit for frequency is Hz (Hertz) where 1 Hz is 1 cycle per sec.)
Solution:
Read the problem.
•
•
•
•
The two variables are the frequency of a vibrating string and the length of the
string.
These two variables vary inversely.
When the length is 1.8 m, the frequency is 200 Hz.
Find the length when the frequency is 720 Hz.
Develop a plan.
•
•
•
•
•
Let f represent frequency.
Let l represent length.
1
f 
l
The variation equation is f 1 l1  f 2 l 2 .
Substitute the known values into the variation equation.
Carry out the plan.
•
The values of the variables are:
•
f 1 = 200
•
f 2 = 720
•
l 1 = 1.8
•
l2 = ?
Write the equation.
Substitute the known values.
Simplify.
f 1 l1  f 2 l 2
200 1.8  = 720  l 2
200 1.8 
720
= l2
0 .5 = l 2
Check the solution.
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200 1.8  =? 720 0 .5 
360 = 360
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()
Lesson 14
Write a concluding statement.
The length of the string must be 0.5 metres for the frequency to be 720 Hz.
There is also another variation with the frequency of the sound produced by a stringed
instrument.
The frequency of the sound produced is inversely proportional to the square root of the
density of the string.
•
•
•
Frequency is measured in hertz (Hz).
Density of a string is measured in grams per centimetre cubed (g/cm3).
1
f 
D
Example 7
A certain type of guitar string has a density of 1.44 g/cm3 and a frequency of
500 Hz. What is the frequency of a guitar string with a density of 4 g/cm3 if
the tension and length are unchanged?
Solution:
Read the problem.
•
•
•
•
The two variables are the frequency of a vibrating string and the density of the
string.
These two variables vary inversely as the square root.
When the density is 1.44 g/cm3, the frequency is 500 Hz.
Find the frequency when the density is 4 g/cm3.
Develop a plan.
•
•
•
•
•
Let f represent frequency.
Let D represent density.
1
f 
D
The variation equation is f1 D1  f2 D2 .
Substitute the known values into the variation equation.
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Lesson 14
Carry out the plan.
•
The values of the variables are:
•
f 1 = 500
•
f2 = ?
•
D1 = 1.44
•
D2 = 4
Write the equation.
Substitute the known values.
Simplify.
f 1 D1 = f 2 D2
500 1.44 =
500 1.2 =
f2 4
f 2 2 
600
= f2
2
300 = f 2
Check the solution.
500 1.44 =? 300 4
5001.2 =? 3002
600 = 600
()
Write a concluding statement.
The frequency of the guitar string with a density of 4 g/cm3 is 300 Hz.
Another interesting proportion with relation to sound is that the intensity or loudness of a
sound is inversely proportional to the square of the distance that you are away from the
sound.
•
The further away you are, the quieter the sound.
•
The closer you are, the louder the sound.
1
•
I  2
d
•
•
The distance (d) is usually measured in metres (m).
The intensity (I) is measured in decibels (Db).
Mathematics A30
226
Lesson 14
Example 8
At a fireworks display one of the explosives travels 250 metres in the air and
explodes with an intensity of 12 decibels. Damage can be done to the eardrum
at 60 decibels. How close to the ground could the fireworks explode and not
exceed 60 decibels?
Solution:
Read the problem.
•
•
•
•
The two variables are the intensity of sound and the distance from the source.
The intensity varies inversely as the square of the distance from the sound.
When the distance is 250 m, the loudness is 12 Db.
Find the distance when the loudness is 60 Db.
Develop a plan.
•
•
•
•
•
Let d represent distance.
Let I represent intensity.
1
I  2
d
2
2
The variation equation is I 1 d1  I 2 d2 .
Substitute the known values into the variation equation.
Carry out the plan.
•
The values of the variables are:
•
I1 = 12
•
I 2 = 60
•
d1 = 250
•
d2 = ?
Write the equation.
Substitute the known values.
Simplify.
2
I 1 d1  I 2 d2
2
12250 2 = 60 d2 2
12 62 500  = 60 d 2 2
750 000
2
=d2
60
12 500 = d 2
111 .8 = d 2
Mathematics A30
227
Lesson 14
Check the solution.
12 250 2
2
?
= 60 111 .80339 
750 000  750 000 ()
Write a concluding statement.
The fireworks would have to be at least 111.8 m in the air for the loudness to not exceed
60 Db.
What happens when a variation that is inversely proportional to the square is
graphed?
How would you graph the relation determined by the variation in Example 8?
You already know that the graph of an inverse variation is a curved line. In most real
world situations, the values of the variables are positive and therefore the graph occurs
only in the first quadrant.
When you graph a curved line, you need at least three points. More than three points
would give even a better indication of the slope of the curve. In order to graph the relation
determined by the variation in Example 8, it is necessary to find one more ordered pair.
Example 9
Sketch the graph of the relation determined by the variation in Example 8,
1
where I  2 , and two ordered pairs were found to be 12, 250 and
d
60,110.8 .
Solution:
•
•
•
Find a third ordered pair.
Choose an intensity of 30 Db.
2
2
The variation equation is I 1 d1 = I 2 d2 .
Mathematics A30
228
Lesson 14
•
The values of the variables now are:
•
I 1 = 12
•
I 2 = 30
•
d1 = 250
•
d2 = ?
2
Write the equation.
Substitute the known values.
Simplify.
I 1 d1  I 2 d2
2
12 250 2 = 30  d 2 2
12 62 500  = 30  d 2 2
750 000
2
=d2
30
25 000 = d 2
158 .1 = d 2
12 250 2
= 30 158 .1139 
750 000  750 000 ()
Check the solution.
2
?
The third ordered pair is (30, 158.1).
Graph the relation.
d
300
Distance 200
(m)
100
10 20 30 40 50 60
Intensity
(Db)
I
Observations:
•
The graph is the same shape as other inverse variation graphs.
•
With this real world application of variation, the domain and range include only
positive numbers.
Mathematics A30
229
Lesson 14
Exercise 14.2
1.
Fill in the blanks in the table.
Word Statement
Variation Statement
Equation
Height (h) above sea level varies
inversely with the pressure (p).
r 
1
d
2
2
A1 d1 = A 2 d 2
t varies as
2
l
2.
The turning speed of a gear is inversely proportional to the number of teeth in the
gear. The pinion gear of a car has 8 teeth and is turning at a speed of 1925 rpm. It
is meshed with the ring gear which has 35 teeth. How fast are the ring gear and
axel turning?
3.
The volume of a gas is inversely proportional to the pressure applied to it. If
450 cm3 of gas is under 20 kPa of pressure, what will the volume be if the pressure
is increased to 60 kPa?
4.
A radio station is broadcasting on a wavelength of 380 m and has a frequency of 800
kHz. The frequency varies inversely as the wavelength. Find the wavelength of a
radio station which has a frequency of 650 kHz.
Mathematics A30
230
Lesson 14
5.
The resistance of a length of wire to an electric current is inversely proportional to
the square of the wire's diameter. If an 8 mm diameter wire has a resistance of 10
ohms, find the resistance of a 5 mm diameter wire.
6.
The illuminance of a surface (I) varies inversely as the square of the distance (d)
between the source and the screen. An LCD Projector is 5 metres away from a
screen with an illuminance of 200 lumens/m2.
a)
How bright would the screen be if you moved one metre back?
b)
How bright would the screen be if you moved one metre closer?
Mathematics A30
231
Lesson 14
Answers to Exercises
1.
a.
Let v represent volume.
Let p represent pressure.
1
v
p
b.
Let f represent frequency.
Let w represent wavelength.
1
f
w
c.
l
d.
P
e.
v
2.
a.
b.
c.
d.
yes; k = 2
no
yes; k = 1200
yes; k =  8
3.
a.
Exercise 14.1
Mathematics A30
1
w
1
a
1
a
b.
232
Lesson 14
c.
4.
Mathematics A30
d.
a.
s1t1  s2t2
756  100t2 
450
 t2
100
4.5  t2
b.
v1 p1  v2 p2
45010  v2 15
4500
 v2
15
300  v2
c.
t1s1  t2 s2
20360  t2 800
7200
 t2
800
9  t2
d.
x1 y1  x 2 y2
 3  8 
    14 y2
 4  27 
2
 14 y2
9
 1  2 
    y2
 14  9 
1
 y2
63
233
Lesson 14
Exercise 14.2
1.
Word Statement
Variation Statement
Height (h) above sea level varies
inversely with the pressure (p).
h
Equation
1
p
h1 p1  h 2 p2
1
r1 d 1  r2 d 2
r varies inversely as the square
of d.
r 
A varies inversely as the square
of d.
A
1
d2
A1 d1 = A 2 d 2
t
1
t 1 l1  t 2 l 2
t varies inversely as
2.
d
l.
2
2
2
2
2
l
1
t
t1  8
s
s1  1925
t 2  35
s2  ?
t 1 s1  t 2 s 2
8 1925   35 s 2
8 1925   s
2
35
440  s 2
The ring gear is turning at a speed of 440 rpm.
Mathematics A30
234
Lesson 14
3.
1
p
v1  450
v
p1  20
v2  ?
p 2  60
v 1 p1  v 2 p 2
450 20   v 2 60 
450 20   v
2
60
150  v 2
The volume would be 150 cm 3 .
4.
1
w
f 1  800
f 
w 1  380
f 2  650
w2  ?
800 380   650 w 2 
800 380   w
2
650
467 .7  w 2
The wavelength is 467.7 m.
Mathematics A30
235
Lesson 14
5.
1
d2
r1  10
r
d1  8
r2  ?
d2  5
2
r1 d1  r2 d2
2
10 8 2  r2 52
10 64   r2 25
10 64   r
2
25
25.6  r2
The resistance is 25.6 ohms.
6.
a.
1
d2
I 1  200
I
d1  5
I2  ?
 One metre back
d2  6
2
I 1 d1  I 2 d2
2
200 52  I 2 6 2
200 25  I 2 36 
200 25  I
2
36
138.9  I 2
The illuminance of the surface if you moved one metre
back would be 138.9 lumens/m2.
Mathematics A30
236
Lesson 14
b.
One metre closer:
I 1  200
d1  5
I2  ?
 one metre closer
d2  4
2
I 1 d1  I 2 d2
2
200 52  I 2 4 2
200 25   I 2 (16)
200 25   I
2
16
312.5  I 2
The illuminance of the surface if you moved one metre
closer would be 312.5 lumes/m2.
Mathematics A30
237
Lesson 14
Mathematics A30
Module 2
Assignment 14
Mathematics A30
237
Assignment 14
5. Staple the completed barcode
sheet on top of this address sheet
(upper left corner.)
4. Staple this sheet to the
appropriately-numbered
assignment. Use one address
sheet for each assignment.
3. Complete the details
in this address box.
2. Number all the pages and place
them in order.
1. Write your name and address and
the course name and assignment
number in the upper right corner of
the first page of each assignment.
Before submitting your
assignment, please complete
the following procedures:
Postal Code:
City/Town, Province
Street Address or P.O. Box
Name
Country
Print your name and address, with postal code. This address sheet
will be used when mailing back your corrected assignment.
Assignment Number
14
Mark Assigned:
Distance-Learning Teacher’s Name
Course Title
Mathematics A30
Course Number
8404
Student Number
Staple here to the
upper left corner of
your assignment
Assignment 14
Values
(40)
A.
Multiple Choice: Select the correct answer for each of the following and place
a check () beside it.
1.
2.
3.
The domain of y 
1
is ***.
x
____
____
____
a.
b.
c.
x 0 x R
x  0, x  R
xR
____
d.
x  0, x  R
The equation which does not express an inverse variation between the
variables is ***.
k
x4
a1b1  a2b2
a1 b2

a 2 b1
a1 a2

b1 b2
y
____
a.
____
b.
____
c.
____
d.
If x 
1
and x is 3 when y is 13, then the constant of proportionality is
y
***.
Mathematics A30
____
____
a.
b.
____
c.
____
d.
39
1
1
39
1
13
241
Assignment 14
4.
5.
6.
If a 
1
and a is 3 when d is 2, the constant of proportionality is ***.
d3
____
____
____
a.
b.
c.
____
d.
If distance travelled is inversely proportional to the square root of the
time taken, and distance is 13 m, after 9 sec, then the constant of
proportionality is ***.
____
____
____
a.
b.
c.
____
d.
Mathematics A30
39
117
9 13
1
3
If the strength of a beam is inversely proportional to its length, and a
beam 5 m long can support at most 100 kg, the number of kg an 8 m
long beam can support is ***.
____
____
____
____
7.
6
1
24
1
8
a.
b.
c.
d.
500
62.5
160
50
The statement “The amount of current I flowing through a circuit is
inversely proportional to the amount of resistance R in the circuit.” is
written in symbols as ***.
____
a.
____
b.
____
c.
____
d.
K
R
I  KR
I
K
R
I1 R2  I 2 R1
I 
242
Assignment 14
8.
9.
If the area of a rectangle is to remain constant then ***.
____
a.
____
____
____
b.
c.
d.
The most likely graph of the inverse variation y 
____
____
____
____
10.
a.
b.
c.
d.
1
is ***.
x2
A
B
C
D
The graph that best describes the variation statement, “The height of a
tree is inversely proportional to the cube of the diameter.” is ***.
____
____
____
____
Mathematics A30
the area remains unchanged if the length is increased by 1
and the width is decreased by 1
the area increases as the length increases
the length is inversely proportional to the width
the length is directly proportional to the width
a.
b.
c.
d.
A
B
C
D
243
Assignment 14
11.
A 3 m bar is used to lift a stone. A 90 kg person exerts all his weight to
1
one end of the bar which is 2 m from the fulcrum and balances the
2
stone. The weight of the stone is approximately ***.
____
____
____
____
a.
b.
c.
d.
500 kg
450 kg
400 kg
45 kg
For each of the questions from 12 to 17 determine the correct ratio of y to x.
12.
13.
Mathematics A30
7
5
y x
3
4
____
a.
____
b.
____
c.
____
d.
15
28
28
15
5
4
7
3
5
10

7 y 21x
____
a.
____
b.
____
c.
____
d.
2
3
3
5
5
3
3
2
244
Assignment 14
14.
15.
16.
Mathematics A30
3x  2 y 2

5y
3
____
a.
____
b.
____
c.
____
d.
9
16
16
9
9
4
4
9
7 y  15 5x  10

3
2
____
a.
____
b.
____
c.
____
d.
14
15
45
44
44
45
15
14
3x  y
 9
3y  x
____
a.
____
b.
____
c.
____
d.
3
13
1
3
3
13
3
245
Assignment 14
17.
18.
19.
12 y
9x

5  4 y 2  3x
____
a.
____
b.
____
c.
____
d.
An equation which is not equivalent to a1b1  a2b2 is ***.
____
a.
____
b.
____
c.
____
d.
a1 b2

a 2 b1
a1 a2

b2 b1
a1 b1

a 2 b2
a 2 b2  b1 a1
The given graph is that of a ***.
____
____
____
____
Mathematics A30
3
8
8
15
15
8
3
5
a.
b.
c.
d.
direct variation
partial variation
inverse variation
infinite variation
246
Assignment 14
20.
If distance (d) is directly proportional to time (t) and d = 15 km after
1
2 hrs, then after 3 hrs more, d is ***.
2
____
____
____
____
Mathematics A30
a.
b.
c.
d.
18 km
45 km
37.5 km
33 km
247
Assignment 14
Mathematics A30
248
Assignment 14
Answer Part B and Part C in the space provided. Evaluation of your solution
to each problem will be based on the following.
(5)
(5)
B.
•
A correct mathematical method for solving the problem is shown.
•
The final answer is accurate and a check of the answer is shown where
asked for by the question.
•
The solution is written in a style that is clear, logical, well organized,
uses proper terms, and states a conclusion.
1.
If 6 men do a job in 12 days, how long would 18 men take, working at
the same rate?
2.
If y varies inversely as 3x  2 and y = 24 when x = 1, find x when y is
15.
Mathematics A30
249
Assignment 14
(5)
3.
If v varies inversely as the cube of t, and V  297 when t = 2, what is V
when t = 3?
(5)
4.
The weight of a body at or above the earth's surface varies inversely as
the square of the distance from the earth's center. What does a 450 kg
object weigh 500 km away from the earth's surface? A 450 kg object
weighs
4 410 Newtons at the surface of the earth. Newton is a unit of measure
for weight. Use 6 500 km as the radius of the earth in your calculations.
Mathematics A30
250
Assignment 14
(5)
5.
If a uniform bar is to balance on a fulcrum, the ratio of the larger weight
to the smaller weight must equal the ratio of the smaller distance to the
larger distance.
w1
w2
d1
d2
If an 8 kg weight is 7.2 m from the fulcrum, how far from the fulcrum
is a 9 kg weight which balances it?
(5)
6.
Mathematics A30
If the illumination of a book 9 m from a lamp is 150 lumens/m2 find the
illumination of the book 3 m closer to the lamp.
251
Assignment 14
(5)
7.
A 12 cm diameter pulley runs at 240 rpm and runs an 8 cm diameter
pulley. How fast does the 8 cm pulley rotate?
(5)
8.
A gear with 48 teeth makes 5 rpm and webs with a gear having 20 teeth
which in turn webs with a gear having 16 teeth. What is the speed of the
smallest gear?
Mathematics A30
252
Assignment 14
(10)
C.
1.
A study was conducted which measures the appearance of sunburn on
skin after exposure for a fixed length of time to solar radiation of
different wave lengths. The data recorded is given below where the
wavelength is in nanometers (nm) and the sunburn appearance is
measured on a scale from zero to one.
The data suggests that there is an inverse relationship between the
variables since, as the wavelengths increase, the appearance of sunburn
decreases. The problem is to determine the variation equation; i.e., is it
k
k
k
y ,y 2 ,y
, or some other relation?
x
x
x
Do your investigation by showing your calculations in the table. From
your calculations write a conclusion at the end of the table.
w
x
Wave length
w  300
Sunburn
appearance (y)
306
6
1.00
310
10
0.77
315
15
0.60
320
20
0.55
325
0.49
330
0.40
340
0.37
360
0.30
370
0.28
Conclusion:
Mathematics A30
253
Assignment 14
(5)
2.
Attach the completed Activity 14.1 to the assignment.
(5)
3.
(STUDENT JOURNAL)
Write a brief summary of this lesson which would be suitable for review
or study purposes.
100 
Mathematics A30
254
Assignment 14