### Video Transcript

In this video, we will learn how to
describe direct variation between two variables and use this to solve word
problems.

Two variables are said to be in
direct variation or direct proportion if their ratio is constant. This means that if the two
variables π₯ and π¦ are in direct variation, we can say that π¦ over π₯ is equal to
some nonzero constant π, provided π₯ is also nonzero. Multiplying both sides of this
equation by π₯ gives π¦ is equal to ππ₯. We can now include π₯ equal to zero
because it then follows that π¦ also equals zero. The constant π is called the
constant of variation or constant of proportionality.

When two variables are in direct
variation, we represent this using the notation shown, which simply means π¦ is
directly proportional to π₯. A graph of two directly
proportional variables is a straight line passing through the origin, with a slope
equal to the constant of proportionality π. Weβll see more detail on this later
in the video.

There are many real-world examples
of variables that follow a directly proportional relationship. For example, if an object such as a
car is traveling at a constant velocity, then the distance traveled is directly
proportional to the time the object has been traveling for. In our first example, weβll see how
to determine the constant of proportionality in a directly proportional
relationship.

If π¦ is directly proportional
to π₯ and π¦ equals 14 when π₯ equals six, determine the constant of
proportionality.

Recall first that this notation
means π¦ is directly proportional to π₯ or π¦ and π₯ are in direct
variation. This means that the ratio
between π¦ and π₯ is constant, which we can express as π¦ is equal to ππ₯ for
some nonzero constant π. π is the constant of
proportionality, which weβve been asked to find.

Weβre told that π¦ is equal to
14 when π₯ is equal to six, which means we have a pair of values of π₯ and π¦
that we can substitute into this equation to determine the value of π. Doing so gives the equation 14
is equal to six π. We then solve for π by
dividing both sides of the equation by six. This fraction can be simplified
by dividing both the numerator and denominator by two. And so we find that the
constant of proportionality π is equal to seven over three.

Letβs now consider another example,
in which weβll use the definition of direct variation to determine the value of an
unknown in a directly proportional relationship.

If π¦ is directly proportional
to π₯ and π₯ equals 75 when π¦ equals 25, find the value of π¦ when π₯ equals
30.

Weβre told that π¦ is directly
proportional to π₯ or π¦ and π₯ are in direct variation, which means the
relationship between π₯ and π¦ can be expressed as π¦ is equal to ππ₯ for some
nonzero constant π. Weβre also given that π₯ is
equal to 75 when π¦ is equal to 25. And so we have a pair of values
of π₯ and π¦ that we can use to determine the constant of proportionality
π.

Substituting these values gives
25 is equal to π multiplied by 75, or simply 25 equals 75π. To find the value of π, we
need to divide both sides of this equation by 75, which gives π equals 25 over
75. We can simplify this fraction
by dividing both the numerator and denominator by 25. And we find that the constant
of proportionality π is equal to one-third.

We now know that the
relationship between π₯ and π¦ is π¦ equals one-third π₯, or π¦ equals π₯ over
three. To find the value of π¦ when π₯
equals 30, we simply need to substitute this value of π₯ and evaluate. That gives π¦ equals 30 over
three, which is 10. So, by first determining the
constant of proportionality, weβve found that in this directly proportional
relationship, the value of π¦ when π₯ is 30 is 10.

Letβs now consider what the graph
representing a directly proportional relationship might look like.

We know that when two variables π₯
and π¦ are in direct variation, then this can be expressed as π¦ equals ππ₯ for
some nonzero constant π. From our knowledge of straight-line
graphs, we know that π¦ equals ππ₯ plus π is the equation of a straight line with
slope π and π¦-intercept π. Here, the value of π is zero, so
we have a straight line passing through the origin. The graph of a directly
proportional relationship might therefore look like this.

However, is it also possible that
the constant of proportionality π could be negative, in which case the slope of the
line would be negative and the relationship would look like the line drawn in
pink. The important thing is that the
line must be straight, and it must pass through the origin.

We can use this knowledge to
determine information about such a relationship from a graph. For example, suppose we are told
that π¦ is in direct variation with π₯ and given a graph of values of π₯ plotted
against values of π¦. We know that π¦ will be equal to
ππ₯ for some nonzero constant π. And we can use the graph to
determine the value of π. Itβs given by the slope of the
graph, which in this case is two. And so we can deduce that the
constant of proportionality is two and π¦ is equal to two π₯. The value of π is also known as
the unit rate because it represents the change in π¦ corresponding to a change of
one unit in π₯.

Letβs now consider an example in
which we determine which of a set of values of π₯ and π¦ are in direct
variation.

Which table does not show π₯
varying directly with π¦?

We recall first that if π₯
varies directly with π¦, then this means that the ratio between values of π₯ and
π¦ remains constant. So we can say that when π₯ is
nonzero, π¦ over π₯ is equal to some nonzero constant π. We can test whether this is
true for each of the tables given by calculating the ratio between each pair of
π₯- and π¦-values and seeing whether it remains constant.

In table (A), the values of π¦
over π₯ are 12 over one, 24 over two, and 36 over three. Each of these ratios simplifies
to 12. And so for table (A), the value
of π¦ over π₯ is constant, and so π₯ does vary directly with π¦.

In table (B), the ratios are
two over 10, four over 20, and six over 30. Each of these simplifies to
one-fifth, or 0.2. So the values in table (B) do
demonstrate direct variation.

In table (C), the ratios are
eight over two, zero over zero, and negative eight over negative two. The first and last ratios each
simplify to four, but zero over zero is undefined. We therefore need to recall
that direct variation between π₯ and π¦ can also be expressed as π¦ equals
ππ₯. And it then follows that when
π₯ is equal to zero, π¦ is also equal to zero.

The pair of values zero, zero
satisfy any direct variation relationship, which corresponds to the graph of any
directly proportional relationship passing through the origin, where both π₯ and
π¦ are zero. In particular, they satisfy the
same direct variation as the other values in table (C), with π equal to
four. And so table (C) does show π₯
varying directly with π¦.

In table (D), the values of π¦
over π₯ are six over five, 10 over three, and 30 over one. Evaluating these as decimals
gives 1.2, 3.3 recurring, and 30. As these values are not
constant, this means that the values in table (D) are not in direct
variation.

Finally, we calculate the
ratios in table (E). Each of these simplifies to
0.75. So this confirms that the
values in table (E) are in direct variation.

We can therefore conclude that
the only table that does not show π₯ varying directly with π¦ is table (D),
because the ratio of π¦ over π₯ is not constant.

Letβs now consider a real-life
application of direct variation.

An object that weighs 120
newtons on Earth weighs 20 newtons on the Moon. Given that the weight of an
object on Earth is directly proportional to its weight on the Moon, find the
weight of an object on the Moon given that its weight on Earth is 126
newtons.

Weβre told that the weight of
an object on Earth is directly proportional to its weight on the Moon. We can express this as weight
on the Earth is equal to π times the weight on the Moon, where π represents
the constant of proportionality. We want to find the weight of
an object on the Moon given that its weight on Earth is 126 newtons. And to do this, we first need
to determine the value of π.

Weβre told that an object that
weighs 120 newtons on Earth weighs 20 newtons on the Moon. So we can substitute this pair
of values into the direct proportion relationship. Doing so gives 120 is equal to
20π. To solve for π, we divide both
sides of this equation by 20, giving π equals six.

We now know that the weight of
an object on Earth is six times its weight on the Moon. We then substitute 126 for the
weight of this particular object on the Earth to give the equation 126 equals
six times its weight on the Moon. To find the weight of the same
object on the Moon, we divide both sides of the equation by six. Weβve therefore found that the
weight of an object on the Moon, which weighs 126 newtons on Earth, is 21
newtons.

Letβs now summarize the key points
from this video. Two variables are said to be in
direct variation, or direct proportion, if their ratio is constant. We can express this using the
proportionality symbol. The relationship between π₯ and π¦
can also be written as π¦ equals ππ₯, where π is a nonzero constant of
proportionality. The graph of a directly
proportional relationship is a straight line that passes through the origin. The slope of this graph gives the
constant of proportionality π, which is also known as the unit rate. The slope of the graph can also be
negative if the value of π is negative.

Finally, we saw that direct
variation can be used to model a number of real-life phenomena, such as the
relationship between distance and time for objects traveling at a constant velocity
and the weights of objects on Earth and on the Moon.