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.
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Lesson Settings & Tools
| | 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.
Which of the graphs corresponds to the table?
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.
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Moving a Point Along a Line
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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
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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.
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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 |
b To calculate Paulina's profit when selling 5 glasses of lemonade, the graph from Part A will be used.
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Writing a Direct Variation
Tadeo is given the following math 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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Writing a Direct Variation Knowing a Point
Tearrik is given the graph of a direct variation and one of its points.
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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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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.
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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.
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
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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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Modeling With Direct Variation
Tiffaniqua is jogging on Saturday morning. As she jogs, Tiffaniqua keeps a constant speed of 8km/h.
Which of these is the graph of the direct variation from Part A?
How many hours would it take her to travel 12 kilometers?
How many hours will it take her to travel 44 kilometers? 
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.
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=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.
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.
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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.
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.
d Since these two values are too large to be seen in the graph, the formula from Part A will be used.
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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 |
Direct Variation and Proportional Relationships
Exercise 3.1
Suppose y varies directly with x. How does the value of y change if the value of x is tripled?
A.& y increases by 3.
B.& y decreases by 3.
C.& y is tripled.
D.& y is decreased by one third.
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Since we know that y varies directly with x, we can represent the relationship with a direct variation equation. y=kx Let's say that the graph of the linear equation passes through the point ( a, b). In this case, the value of b is equal to k a. This can be shown by substituting the point ( a, b) into the equation. y=kx substitute b=k a We want to determine how the value of y changes when x is tripled. To do so, we will substitute x= 3a into the equation y=kx and find the corresponding value for y.
Now we can substitute ka= b to find y.
As we can see, when x is tripled, y is also tripled. This corresponds to option C.
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Direct Variation and Proportional Relationships
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