3. Direct Variation and Proportional Relationships
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Chapter 2
3.
Direct Variation and Proportional Relationships
Direct variation occurs where one variable changes directly in relation to another. In simple terms, if one variable increases, the other does too, maintaining a consistent ratio. Direct variation graphs visualize this relationship, showing a straight line passing through the origin. This connection between variables is termed as proportional.
Recognizing this relationship is crucial for tasks like budgeting, scaling designs, or predicting trends. In real-world scenarios, understanding proportional variables can simplify complex problems, making it easier to make informed decisions.
| Student Learning Objectives: |
|---|
|
| | 10 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Direct Variation and Proportional Relationships
Slide of 10
If two quantities are related and one of them varies, then the other quantity will also vary. Depending on the type of relationship they have, the second quantity will vary in a specific way. This lesson will explore the ideas of direct variation and proportional relationships.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
- Table of Values
- Making a Table of Values
- Linear Relations
- Graphing a Linear Relation Using a Table of Values
Here are a few practice exercises before getting started with this lesson.
a
Consider the table of values.
| x | y=2x+1 |
|---|---|
| 1 | 3 |
| 2 | A |
| 3 | B |
| 4 | 9 |
| 5 | C |
Find the values of A, B, and C.
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b
Consider the table of values.
| x | y=x+1 |
|---|---|
| - 1 | 0 |
| 0 | 1 |
| 1 | 2 |
| 2 | 3 |
| 3 | 4 |
Consider now the following graphs.
Which of the graphs corresponds to the table?
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Explore
Moving a Point Along a Line
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Discussion
Direct Variation
Direct variation, also known as direct proportionality or proportional relationship, occurs when two variables, x and y, have a relationship that forms a linear function passing through the origin.
y=kx
The constant k is the constant of variation. It defines the slope of the line. When k=0, the relationship is not in direct variation. In the example below, the constant of variation is k=1.5.
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Example
Graphing a Direct Variation
Paulina is selling lemonade to save some money for her summer vacation. For each glass of lemonade she sells, Paulina makes a profit of $ 0.50. She models this situation with a direct variation. p= 0.5g Here, p is the profit made when g glasses are sold. Also, 0.5 is the constant of variation.
a
Consider the following graphs.
Which of the graphs corresponds to the given direct variation?
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b
What is Paulina's profit if she sells 5 glasses?
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Hint
a
Draw the graph of the direct variation and compare it with the given graphs.
b
Use the graph from Part A.
Solution
a
To determine which of the four graphs represents the given direct variation, the graph of p=0.5g will be drawn by making a table of values.
| g | 0.5g | p=0.5g |
|---|---|---|
| 0 | 0.5( 0) | 0 |
| 1 | 0.5( 1) | 0.5 |
| 2 | 0.5( 2) | 1 |
| 3 | 0.5( 3) | 1.5 |
| 4 | 0.5( 4) | 2 |
Now, the obtained points in the table will be plotted and connected with a straight line.
The obtained graph is the same as Graph II. Note that in the context of the situation, negative values of g do not make sense, since Paulina cannot sell a negative number of glasses of lemonade.
b
To calculate Paulina's profit when selling 5 glasses of lemonade, the graph from Part A will be used.
The profit for 5 glasses is $2.50. Note that this can be verified using the equation. To do so, 5 will be substituted for g. p=0.5g substitute p&=0.5( 5) p&=2.5 ✓
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Example
Writing a Direct Variation
Tadeo is given the following math homework.
Help Tadeo do his homework!
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Hint
The general form of a direct variation is y=kx, where k is the constant of variation.
Solution
Start by writing the general form of a direct variation.
y=kx
Here, k is the constant of variation. To find its value, x=16 and y=12 will be substituted into the equation.
The constant of variation is 34. With this information, the equation can be written. y=kx substitute y= 3/4x Finally, this equation can be used to find the value of y when x= 24.
When x=24, the value of y is 18.
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Example
Writing a Direct Variation Knowing a Point
Tearrik is given the graph of a direct variation and one of its points.
To complete his homework before going dancing, he wants to find the equation of the direct variation shown in the graph. Help him do this!
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Hint
The equation of a direct variation is y=kx, where k is the constant of variation.
Solution
Start by recalling the general form of a direct variation.
y=kx
Here, k is the constant of variation. To find its value, the point ( 4, 5) will be used. To do this, x= 4 and y= 5 will be substituted into the above equation.
The constant of variation is 1.25. With this information, the formula can be written. y=kx substitute y= 1.25x
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Example
Writing a Direct Variation Knowing Its Graph
This time, Tearrik is given the graph of a direct variation, but none of its points are plotted.
Once again, to complete his homework before going dancing, he wants to find the equation of the direct variation shown in the graph. Help him do this!
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Hint
Use any point on the line.
Solution
Recall the general form of a direct variation.
y=kx
Here, k is the constant of variation. To find its value, any point on the given graph can be used. For simplicity, the point (- 10,8) will be considered.
To find the constant of variation, x= - 10 and y= 8 will be substituted into the general formula.
y=kx
SubstituteII
x= - 10, y= 8
8=k( - 10)
▼
Solve for k
DivEqn
.LHS /(- 10).=.RHS /(- 10).
8/- 10=k
MoveNegDenomToFrac
Put minus sign in front of fraction
- 8/10=k
CalcQuot
Calculate quotient
- 0.8=k
RearrangeEqn
Rearrange equation
k=- 0.8
The constant of variation is - 0.8. With this information, the equation of the direct variation of the given graph can be written. y=kx substitute y= - 0.8x
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Pop Quiz
Finding the Constant of Variation
Find the constant of variation of the direct variation whose graph is given. If the answer is not an integer, write it as a decimal rounded to one decimal place.
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Example
Modeling With Direct Variation
Tiffaniqua is jogging on Saturday morning. As she jogs, Tiffaniqua keeps a constant speed of 8km/h.
a
Let d represent the distance in kilometers and t the time in hours. Write a direct variation in terms of d and t to represent this situation.
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b
Consider the following graphs.
Which of these is the graph of the direct variation from Part A?
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c
How many kilometers would Tiffaniqua travel in 45 minutes?
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d
How many kilometers would Tiffaniqua travel in 6 hours and 45 minutes?
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Hint
a
What is the constant of variation?
b
If Tiffaniqua's speed is 8km/h, then she travels 8 kilometers in one hour.
c
Use the graph from Part B.
d
Use the formula from Part A.
Solution
a
The distance traveled d varies directly with the time t. Since Tiffaniqua's speed is 8km/h, the constant of variation is 8.
d=kt substitute d= 8t
b
The graph of a direct variation passes through the origin. Therefore, the point (0,0) is on the graph of d=8t. To find another point on the line, any value can be substituted for t in the formula. For simplicity, t=1 will be used.
d=8t substitute d&=8( 1) d&= 8 It has been found that the point ( 1, 8) is also on the line. To draw the graph, these two points will be plotted and the line through them will be drawn.
This graph corresponds to Graph III. Note that in the context of the situation, a negative value of t does not make sense, since Tiffaniqua cannot run for a negative amount of hours.
c
To find how many kilometers Tiffaniqua would travel in 45 minutes and how many hours it would take her to travel 12 kilometers, the graph from Part B will be used.
It can be seen that, if Tiffaniqua jogged at a constant speed of 8 kilometers per hour, she would travel 6 kilometers in 45 minutes. Similarly, it would take her one and a half hours to travel 12 kilometers.
d
Since these two values are too large to be seen in the graph, the formula from Part A will be used.
d=8t Recall that d is the distance in kilometers and t the time in hours. To find how many kilometers Tiffaniqua would travel in 6 hours and 45 minutes, 45 minutes need to be expressed in hours. 45min convert 4560h =0.75h Therefore, 6 hours and 45 minutes are 6.75 hours. This value can be substituted for t in the equation. d=8t substitute d&=8( 6.75) d&=54 In 6 hours and 45 minutes, Tiffaniqua would travel 54 kilometers. Finally, to calculate how many hours it would take her to travel 44 kilometers at this speed, 44 will be substituted for d in the formula.
If she kept her constant speed, it would take Tiffaniqua 5.5 hours to travel 44 kilometers.
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Closure
Inverse Variation
Another type of variation is inverse variation. Here, one variable is the quotient of the constant of variation and the other variable, which cannot be zero. y=k/x, x≠ 0 Inverse variation occurs when the product of the variables is constant. y=k/x ⇔ xy=k As in direct variation, the constant of variation cannot be zero. For example, let the constant of variation of an inverse variation be 10. To draw its graph, a table of values will be first made. Only positive values will be considered for the x-variable.
| x | 10/x | y=10/x |
|---|---|---|
| 1 | 10/1 | 10 |
| 2 | 10/2 | 5 |
| 3 | 10/3 | ≈ 3.3 |
| 4 | 10/4 | 2.5 |
| 5 | 10/5 | 2 |
| 6 | 10/6 | ≈ 1.7 |
| 7 | 10/7 | ≈ 1.4 |
| 8 | 10/8 | 1.25 |
| 9 | 10/9 | ≈ 1.1 |
| 10 | 10/10 | 1 |
Next, the points found in the table will be plotted and connected.
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Direct Variation and Proportional Relationships
Exercise 2.1
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If there is a direct variation between the variables x and y, then we can represent this relation with the following equation. y=kx In this form, k ≠ 0. We will rewrite the given equation to isolate y on the left hand side. Let's do it!
Notice that we cannot rewrite the equation in the form y=kx because we have a constant term -2 on the right-hand side. Therefore, the equation does not represent a direct variation.
We will apply the same process to isolate the variable y on the left-hand side of the equation. Let's do it!
Since we can write the equation in the form y=kx, where k=-2 in this case, the given equation does represent a direct variation.
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Hint & Answer
S
Solution
| Time (minutes) | 15 | 30 | 45 | 60 | 75 |
|---|---|---|---|---|---|
| Price (dollars) | 3 | 6 | 9 | 12 | 15 |
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We are asked to write an equation relating the time x at the playground to the price y of playing there. To do so, we will first examine how these variables change accordingly. Time:& 15 +15 30 +15 45 ... Price:& 3 +3 6 +3 9 ... Notice that the price increases by 3 dollars for every 15 minutes of play. This means that there is a constant rate of change between the duration of play and the price. Therefore, we can express the relation between these variables with a direct variation. y= mx In this form, m ≠ 0 and is the constant of variation. To find its value, we can substitute one of the values from the table into the equation. Let's randomly choose the second time and price value, $6 for 30 minutes.
Now that we know the constant of variation, we can write a direct variation equation that relates the time x of play to its price y. y= 0.2x
To find the time the kid spent at the playground if the cost was $ 20, we can use the equation that we found in the previous part. y=0.2x Let's substitute y=20 into the direct variation equation and solve for x.
For $20, the kid played at the playground for 100 minutes.
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S
Solution
A map of Italy is scaled so that a length of 4.5 inches represents 90 miles. How far apart are Bologna and Rome if they are 11.9 inches apart on the map?
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The number of inches on the map between Bologna and Rome varies directly with the number of miles between the cities. We must write a direct variation equation, but we first need to determine the constant of variation. We know that that 4.5 inches represent 90 miles, so we can write the following ratio of miles to inches. 90 miles/4.5 inches We can simplify this value by dividing both numbers by 4.5. When we have 1 inch in the denominator, we can use it as a constant of variation. 90 miles/4.5/4.5 inches/4.5=20 miles/1 inch Now we can write the equation. Let's call the distance in miles between the cities M and the distance in inches i. M=20i To find how far apart the cities are in miles, we can substitute the given number of inches for i into the equation and simplify.
The cities are 238 miles apart.
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Solution
Direct Variation and Proportional Relationships
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