Lesson Explainer: Direct Variation | Nagwa Lesson Explainer: Direct Variation | Nagwa

Lesson Explainer: Direct Variation Mathematics • Third Year of Preparatory School

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

Let’s begin by defining what is meant by direct variation.

Definition: Direct Variation or Direct Proportion

Two variables are said to be in direct variation, or direct proportion, if their ratio is constant.

This type of relationship is often written as π‘¦βˆπ‘₯, which is read as 𝑦 is directly proportional to π‘₯. Since their ratio is constant, we must have that 𝑦π‘₯=π‘š for some constant π‘šβ‰ 0, provided π‘₯β‰ 0, where π‘š is called the constant of variation or constant of proportionality.

Multiplying both sides of the previous equation through by π‘₯, we see that 𝑦=π‘šπ‘₯.

This then allows π‘₯=0, since this gives 𝑦=0.

There are many examples of real-world phenomena that follow a directly proportional relationship. For example, if a body is moving at a constant velocity of 5 m/s, then its distance traveled after 𝑑 seconds is given by 𝑑=5𝑑.

Hence, the distance traveled by a body (moving at a constant velocity) is directly proportional to the amount of time traveled. We can substitute in values of 𝑑 or 𝑑 and then solve the equation for the missing unknown. For example, when 𝑑=10m, we see that 10=5𝑑𝑑=105=2.s

In our first example, we will determine the constant of proportionality from the values of two variables that directly vary.

Example 1: Finding the Constant of Proportionality

If π‘¦βˆπ‘₯ and 𝑦=14 when π‘₯=6, determine the constant of proportionality.

Answer

We recall that saying that π‘¦βˆπ‘₯ means that the ratio between corresponding values of 𝑦 and π‘₯ remains constant, so 𝑦π‘₯=π‘š for some constant π‘šβ‰ 0, where π‘š is called the constant of proportionality. We can substitute 𝑦=14 and π‘₯=6 into this equation to get π‘š=1416=73.

Hence, the constant of proportionality is 73.

Let’s see an example of using this definition to determine the value of an unknown in a directly proportional relationship.

Example 2: Solving Proportion Equations Involving Direct Variation

If π‘¦βˆπ‘₯ and π‘₯=75 when 𝑦=25, find the value of 𝑦 when π‘₯=30.

Answer

We recall that saying that π‘¦βˆπ‘₯ means that the ratio between corresponding values of 𝑦 and π‘₯ remains constant, so 𝑦π‘₯=π‘š for some constant π‘šβ‰ 0, provided π‘₯β‰ 0. In particular, we can rearrange this equation to get 𝑦=π‘šπ‘₯, which then allows π‘₯=0.

We can substitute 𝑦=25 and π‘₯=75 into the first equation to get π‘š=2575=13.

Substituting this value of π‘š into the linear equation gives us 𝑦=13π‘₯.

We can then substitute π‘₯=30 into this equation to find the corresponding value of 𝑦: 𝑦=13Γ—30=10.

Therefore, the value of 𝑦 when π‘₯=30 is 10.

In the previous example, we saw that the corresponding pairs of values of π‘₯ and 𝑦 were in proportion, since 1030=2575=13. This is true for any directly proportional variables, and we can show this from the definition.

If π‘¦βˆπ‘₯ and the variable π‘₯ takes the values π‘₯ and π‘₯ and the corresponding 𝑦-values are π‘¦οŠ§ and π‘¦οŠ¨, then 𝑦=π‘šπ‘₯,𝑦=π‘šπ‘₯, and if we divide the two equations, we get 𝑦𝑦=π‘₯π‘₯.

So, we have that π‘¦οŠ§, π‘¦οŠ¨, π‘₯, and π‘₯ are in proportion.

We can also recall that a linear function is one in the form 𝑦=π‘šπ‘₯+𝑏, which when graphed has a slope of π‘š and a 𝑦-intercept of 𝑏. Hence, if π‘¦βˆπ‘₯, then π‘₯ is a linear function in 𝑦 and its graph is a straight line passing through the origin. It is also worth noting that the same result is true in reverse: if 𝑦=π‘šπ‘₯, then π‘¦βˆπ‘₯. This gives us the following result.

Definition: Direct Variation as a Linear Function

If π‘¦βˆπ‘₯, then 𝑦 is a linear function in π‘₯ and its graph is a straight line that passes through the origin.

Let’s see an example of how to use this property to correctly determine the graph of a directly proportional relationship.

Example 3: Identifying the Graph of Direct Variation

Which of the given graphs represents the direct variation between π‘₯ and 𝑦?

Answer

We recall that saying that π‘¦βˆπ‘₯ means that the ratio between corresponding values of π‘₯ and 𝑦 remains constant, so 𝑦π‘₯=π‘š for some constant π‘šβ‰ 0, provided π‘₯β‰ 0. In particular, we can rearrange this equation to get 𝑦=π‘šπ‘₯, which then allows π‘₯=0.

We then recall that the equation 𝑦=π‘šπ‘₯+𝑏 is a straight line with slope π‘š and 𝑦-intercept 𝑏. So, the graph of any directly proportional relationship is a straight line that has a 𝑦-intercept of 0, meaning it passes through the origin.

Option b is the only straight line passing through the origin, so it is the only graph of a directly proportional relationship.

Since the graph of a straight line passing through the origin represents two variables in direct variation, with the slope as the coefficient of proportionality, we can determine information about the relationship from this graph. Let’s see an example of this.

By considering the graph below, find the coefficient of proportionality between 𝑦 and π‘₯, and determine the value of 𝑦 when π‘₯=4.

First, we know that the coefficient of proportionality is the slope of the line, so we can find the slope of the line by using two points on the line (π‘₯,𝑦) and (π‘₯,𝑦), where π‘š=π‘¦βˆ’π‘¦π‘₯βˆ’π‘₯.

We know that the line passes through the origin (0,0) and we can see in the diagram that the line passes through (2,1). Substituting these points into the formula gives π‘š=1βˆ’02βˆ’0=12.

Hence, 𝑦=12π‘₯.

We can then find the value of 𝑦 when π‘₯=4 in two different ways.

  • We can substitute π‘₯=4 into this equation to get 𝑦=12Γ—4=2.
  • We can find the 𝑦-coordinate of the point on the graph with π‘₯-coordinate 4. We see that this is 2.

It is also worth recalling that graphs and equations are not the only way of representing linear relations. For example, we can represent these as ordered pairs or as entries in a table. Let’s see an example of how to identify a directly proportional relationship from a table.

Example 4: Recognizing Direct Variation from a Table

Which table does not show π‘₯ varying directly with 𝑦?

  1. π‘₯123
    𝑦122436
  2. π‘₯102030
    𝑦246
  3. π‘₯20βˆ’2
    𝑦80βˆ’8
  4. π‘₯531
    𝑦61030
  5. π‘₯246
    𝑦1.534.5

Answer

We recall that saying that π‘₯ varies directly with 𝑦 means that the ratio between corresponding values of π‘₯ and 𝑦 remains constant, so 𝑦π‘₯=π‘š for some constant π‘šβ‰ 0 provided π‘₯β‰ 0. We can allow π‘₯=0 by rewriting the equation as 𝑦=π‘šπ‘₯, which gives π‘₯=0 and 𝑦=0.

Hence, we need to determine which table has ratios of corresponding 𝑦- and π‘₯-values that do not remain constant. We can calculate the ratios for each table separately. We will do this by adding an extra row to each table calculating the ratio of the terms in the same column.

In table A, we see that 121=12, 242=12, and 363=12, giving us the following.

π‘₯123
𝑦122436
𝑦π‘₯121212

Since the ratio of the corresponding 𝑦- and π‘₯-values stays constant, this table represents direct variation with constant of proportionality 12.

In table B, we see that 210=0.2, 420=0.2, and 630=0.2, giving us the following.

π‘₯102030
𝑦246
𝑦π‘₯0.20.20.2

Since the ratio of the corresponding 𝑦- and π‘₯-values stays constant, this table represents direct variation with constant of proportionality 0.2.

In table C, we see that 82=4 and βˆ’8βˆ’2=4; however, we cannot evaluate 00 since we cannot divide by 0. We recall that direct variation between π‘₯ and 𝑦 can be written as the equation 𝑦=π‘šπ‘₯, so when π‘₯=0, we have 𝑦=0. Hence, this still represents direct variation with constant of proportionality 4.

In table D, we see that 65=1.2, 103=3.333…, and 301=30.

π‘₯531
𝑦61030
𝑦π‘₯1.23.33…30

These ratios differ, so this does not represent direct variation.

For due diligence, we can also check table E. We have that 1.52=0.75, 34=0.75, and 4.56=0.75, giving us the following table.

π‘₯246
𝑦1.534.5
𝑦π‘₯0.750.750.75

Since the ratio of the corresponding 𝑦- and π‘₯-values stays constant, this table represents direct variation with constant of proportionality 0.75.

Hence, the relation in table D does not represent direct variation between π‘₯ and 𝑦.

In our next example, we will determine which of a list of equations represents two variables in direct variation.

Example 5: Identifying an Equation of Direct Variation

Which of the following relations represents a direct variation between the two variables π‘₯ and 𝑦?

  1. π‘₯6=3𝑦
  2. 𝑦=π‘₯+2
  3. π‘₯𝑦=6
  4. π‘₯5=𝑦4

Answer

We recall that saying that 𝑦 varies directly with π‘₯ means that the ratio between corresponding values of 𝑦 and π‘₯ remains constant, so 𝑦π‘₯=π‘š for some constant π‘šβ‰ 0, provided π‘₯β‰ 0. We can rewrite this equation as 𝑦=π‘šπ‘₯ to allow π‘₯=0, and we note that any variables in direct variation must satisfy an equation of this form. So, we need to determine which of the given equations can be written in this form.

In relation A, we can multiply the equation through by 6𝑦 to get π‘₯𝑦=18.

We then divide the equation through by π‘₯ to get 𝑦=18π‘₯, which is not in the form 𝑦=π‘šπ‘₯ for a nonzero constant π‘š, so this equation does not represent direct variation.

In relation B, we have a linear function in the form 𝑦=π‘šπ‘₯+𝑏, and we note that linear functions only represent direct variation with 𝑏=0. Since 𝑏=2, this does not represent direct variation.

In relation C, we can divide the equation through by π‘₯ to get 𝑦=6π‘₯.

This is not in the form 𝑦=π‘šπ‘₯ for a nonzero constant π‘š, so this equation does not represent direct variation.

In relation D, we multiply the equation through by 4 to get 𝑦=45π‘₯.

This is now written in the form 𝑦=π‘šπ‘₯, where the constant of variation is 45.

Hence, only the relation in option D represents direct variation.

Let’s now see a few real-world examples of direct variation and how we can use this relationship to determine the value of unknowns.

Example 6: Finding the Constant of Variation from an Equation

The amount of meat required to feed a captive lion is given by the equation 𝑀=9𝑑, where 𝑀 is the weight of the meat in kilograms needed to feed a lion for 𝑑 days. What is the unit rate of this proportional relationship?

Answer

We recall that saying that 𝑀 varies directly with 𝑑 means that the ratio between corresponding values of 𝑀 and 𝑑 remains constant, so 𝑀𝑑=π‘š for some constant π‘šβ‰ 0, provided 𝑑≠0; this constant π‘š is called the constant of proportionality.

In the question we are told that 𝑀=9𝑑, so we can conclude that the ratio of 𝑀 and 𝑑 stays at a constant value of 9. Hence, they are directly proportional and the constant of proportionality is 9.

The question asks for the unit rate of this relationship, and we recall that this is the ratio of two quantities when the second quantity is 1. So, we can find the unit rate by substituting 𝑑=1 into the equation 𝑀𝑑=9, which gives 𝑀1=9/.kgdaykgday

Since, 𝑀 is measured in kilograms and 𝑑 is measured in days, we see that the unit rate is 9 kg/day.

In the previous example, we showed a useful result: the unit rate of a directly proportional relationship is equal to the constant of proportionality, since this is the ratio of any two terms.

In our next example, we will use direct variation to determine the weight of a given object on the moon.

Example 7: Solving Proportion Equations Involving Direct Variation

An object that weighs 120 N on Earth weighs 20 N 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 N.

Answer

We are told in the question that the weight of a body on Earth is directly proportional to its weight on the Moon. We recall that this means that the ratio of any body’s weight on the Moon to its weight on Earth is constant. This can be mathematically written as π‘Šπ‘Š=π‘š,MoonEarth for some constant π‘šβ‰ 0, and the weight on Earth is nonzero.

We can find the value of π‘š by using the first pair of values given in the question π‘Š=20MoonN and π‘Š=120EarthN. Hence, using the equation given above and substituting this pair of values, we find π‘š=20120=16.

Therefore, we have π‘Šπ‘Š=16,MoonEarth which can be also written as π‘Š=16Γ—π‘Š.MoonEarth

We have found that the weight of the given body on the Moon is one-sixth of its weight on Earth. Hence, the weight on the Moon of a body whose weight on Earth is 126 N is π‘Š=16Γ—126=21.MoonN

In our final example, we will find the relationship between two variables π‘₯ and 𝑦, given that 𝑦 varies directly with a given linear function of π‘₯.

Example 8: Writing an Equation between Two Proportional Quantities

Given that π‘¦βˆ(π‘₯+7), where π‘₯=13 when 𝑦=34, find the relation between π‘₯ and 𝑦.

Answer

We recall that π‘¦βˆ(π‘₯+7) means that there exists some nonzero constant π‘š such that 𝑦=π‘š(π‘₯+7).

We can determine the value of π‘š by substituting π‘₯=13 and 𝑦=34 into this equation: 34=π‘š(13+7)34=20π‘š.

We divide the equation through by 20 to get π‘š=3420=1710.

Substituting this value back into the linear equation gives 𝑦=1710(π‘₯+7).

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • Two variables are said to be in direct variation, or direct proportion, if their ratio is constant.
  • Direct variation between two variables 𝑦 and π‘₯ is written as π‘¦βˆπ‘₯. It is mathematically described as 𝑦=π‘šπ‘₯, where π‘š is called the constant of variation or constant of proportionality.
  • Since 𝑦π‘₯=π‘š in a directly proportional relationship, π‘š is also the unit rate.
  • A linear relationship between two variables 𝑦 and π‘₯ is described as 𝑦=π‘šπ‘₯+𝑏; hence, only those linear relationships where 𝑏=0 correspond to direct variation.
  • The graph of direct variation (of a proportional relationship) is a straight line that passes through the origin.

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