Lesson Video: Direct Variation | Nagwa Lesson Video: Direct Variation | Nagwa

# Lesson Video: Direct Variation Mathematics • Third Year of Preparatory School

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

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### 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 ππ₯. 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. Weβll 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, weβll 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 ππ₯ for some nonzero constant π. π is the constant of proportionality, which weβve been asked to find.

Weβre 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.

Letβs now consider another example, in which weβll 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.

Weβre 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 ππ₯ for some nonzero constant π. Weβre 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, weβve found that in this directly proportional relationship, the value of π¦ when π₯ is 30 is 10.

Letβs 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 ππ₯ for some nonzero constant π. From our knowledge of straight-line graphs, we know that π¦ equals ππ₯ 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 ππ₯ for some nonzero constant π. And we can use the graph to determine the value of π. Itβs 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₯.

Letβs 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 ππ₯. 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.

Letβs 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.

Weβre 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.

Weβre 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. Weβve therefore found that the weight of an object on the Moon, which weighs 126 newtons on Earth, is 21 newtons.

Letβs 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 ππ₯, 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.

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