Grade 9 Unit 2: Direct and Indirect Proportion
1. Rearranging Formulae
Explanation
Rearranging formulae involves changing the subject of a formula. This is a key skill in algebra and is crucial for solving equations involving multiple variables. The subject of a formula is the variable that is being solved for, and rearranging the formula means isolating this variable on one side of the equation. This process often involves inverse operations such as addition, subtraction, multiplication, division, and taking roots.
Steps for Rearranging Formulae
- Identify the subject of the formula: Determine which variable you need to isolate.
- Perform inverse operations: Use operations that will isolate the subject variable. Apply these operations to both sides of the equation to maintain equality.
- Simplify the equation: Combine like terms and simplify the equation to solve for the subject variable.
Example 1: Rearranging the Equation of a Line
Consider the standard form of the equation of a line:
\[ Ax + By = C \]
where \( A \), \( B \), and \( C \) are constants.
Make \( y \) the subject:
1. \( Ax + By = C \)
2. Subtract \( Ax \) from both sides:
\[ By = C – Ax \]
3. Divide both sides by \( B \):
\[ y = \frac{C – Ax}{B} \]
Thus, the rearranged formula is:
\[ y = \frac{C – Ax}{B} \]
Example 2: Rearranging the Formula for the Area of a Triangle
The formula for the area of a triangle is:
\[ A = \frac{1}{2}bh \]
where \( A \) is the area, \( b \) is the base, and \( h \) is the height.
Make \( h \) the subject:
1. \( A = \frac{1}{2}bh \)
2. Multiply both sides by 2:
\[ 2A = bh \]
3. Divide both sides by \( b \):
\[ h = \frac{2A}{b} \]
Thus, the rearranged formula is:
\[ h = \frac{2A}{b} \]
Example 3: Rearranging the Formula for the Volume of a Cylinder
The formula for the volume of a cylinder is:
\[ V = \pi r^2 h \]
where \( V \) is the volume, \( r \) is the radius, and \( h \) is the height.
Make \( h \) the subject:
1. \( V = \pi r^2 h \)
2. Divide both sides by \( \pi r^2 \):
\[ h = \frac{V}{\pi r^2} \]
Thus, the rearranged formula is:
\[ h = \frac{V}{\pi r^2} \]
2. Direct Proportion
Explanation
Direct proportion, also known as direct variation, occurs when two variables increase or decrease in the same ratio. In other words, if one variable doubles, the other doubles; if one is halved, the other is halved. The relationship can be expressed with the equation:
\[ y = kx \]
where \( y \) and \( x \) are the variables, and \( k \) is the constant of proportionality.
Characteristics of Direct Proportion
- The ratio \(\frac{y}{x}\) is constant and equal to \( k \).
- The graph of \( y \) against \( x \) is a straight line that passes through the origin (0, 0).
- The equation can be written in the form \( y = kx \).
Steps to Identify Direct Proportion
- Calculate the ratio: Find the ratio \(\frac{y}{x}\). If it is constant for all pairs of values, the variables are directly proportional.
- Graph the relationship: Plot \( y \) against \( x \). If the points lie on a straight line through the origin, the variables are directly proportional.
Example 1: Identifying Direct Proportion from a Table of Values
Consider the following table of values:
| x | y |
|---|---|
| 1 | 3 |
| 2 | 6 |
| 3 | 9 |
| 4 | 12 |
Calculate the ratio \(\frac{y}{x}\) for each pair of values:
\(\frac{3}{1} = 3\), \(\frac{6}{2} = 3\), \(\frac{9}{3} = 3\), \(\frac{12}{4} = 3\)
Since the ratio is constant, \( y \) is directly proportional to \( x \).
The constant of proportionality \( k \) is 3, so the equation of the relationship is:
\( y = 3x \)
Example 2: Finding the Constant of Proportionality
A car travels at a constant speed. The distance \( d \) it travels is directly proportional to the time \( t \) it travels. If the car travels 150 kilometers in 3 hours, find the constant of proportionality and write the equation for the relationship.
Identify the given values:
\( d = 150 \text{ km}, \quad t = 3 \text{ hours} \)
Use the formula for direct proportion \( d = kt \) to find \( k \):
\( 150 = k \times 3 \implies k = \frac{150}{3} = 50 \)
The constant of proportionality \( k \) is 50. The equation for the relationship is:
\( d = 50t \)
Example 3: Graphing a Direct Proportion
A shop sells apples. The total cost \( C \) is directly proportional to the number of kilograms of apples bought \( n \). If 5 kilograms of apples cost $20, graph the relationship and determine the equation.
Identify the given values:
\( C = 20 \text{ dollars}, \quad n = 5 \text{ kg} \)
Find the constant of proportionality \( k \):
\( 20 = k \times 5 \implies k = \frac{20}{5} = 4 \)
The constant of proportionality \( k \) is 4. The equation for the relationship is:
\( C = 4n \)
3. Inverse (Indirect) Proportion
Explanation
Inverse proportion, also known as inverse variation, occurs when one variable increases while the other decreases in such a way that their product is constant. In other words, if one variable doubles, the other is halved; if one is tripled, the other is reduced to one-third. The relationship can be expressed with the equation:
\[ y = \frac{k}{x} \]
where \( y \) and \( x \) are the variables, and \( k \) is the constant of proportionality.
Characteristics of Inverse Proportion
- The product \( y \times x \) is constant and equal to \( k \).
- The graph of \( y \) against \( x \) is a hyperbola.
- The equation can be written in the form \( y = \frac{k}{x} \).
Steps to Identify Inverse Proportion
- Calculate the product: Find the product \( y \times x \). If it is constant for all pairs of values, the variables are inversely proportional.
- Graph the relationship: Plot \( y \) against \( x \). If the points form a hyperbola, the variables are inversely proportional.
Example 1: Identifying Inverse Proportion from a Table of Values
Consider the following table of values:
| x | y |
|---|---|
| 1 | 12 |
| 2 | 6 |
| 3 | 4 |
| 4 | 3 |
Calculate the product \( y \times x \) for each pair of values:
\( 1 \times 12 = 12 \), \( 2 \times 6 = 12 \), \( 3 \times 4 = 12 \), \( 4 \times 3 = 12 \)
Since the product is constant, \( y \) is inversely proportional to \( x \).
The constant of proportionality \( k \) is 12, so the equation of the relationship is:
\( y = \frac{12}{x} \)
Example 2: Finding the Constant of Proportionality
A worker can complete a task in a certain number of hours. The time \( t \) it takes to complete the task is inversely proportional to the number of workers \( n \). If 4 workers can complete the task in 3 hours, find the constant of proportionality and write the equation for the relationship.
Identify the given values:
\( t = 3 \text{ hours}, \quad n = 4 \text{ workers} \)
Use the formula for inverse proportion \( t = \frac{k}{n} \) to find \( k \):
\( 3 = \frac{k}{4} \implies k = 3 \times 4 = 12 \)
The constant of proportionality \( k \) is 12. The equation for the relationship is:
\( t = \frac{12}{n} \)
Example 3: Graphing an Inverse Proportion
The intensity of light \( I \) is inversely proportional to the square of the distance \( d \) from the source. If the light intensity is 16 units when the distance is 2 meters, graph the relationship and determine the equation.
Identify the given values:
\( I = 16 \text{ units}, \quad d = 2 \text{ meters} \)
Find the constant of proportionality \( k \):
\( I = \frac{k}{d^2} \implies 16 = \frac{k}{2^2} = \frac{k}{4} \implies k = 16 \times 4 = 64 \)
The constant of proportionality \( k \) is 64. The equation for the relationship is:
\( I = \frac{64}{d^2} \)
4. Different Types of Variation
Questions
1. Rearranging Formulae
Rearrange the formula to make \( x \) the subject:
\[ y = 5x + 3 \]
2. Direct Proportion
If \( y \) is directly proportional to \( x \) and \( y = 10 \) when \( x = 2 \), find \( y \) when \( x = 5 \).
3. Inverse (Indirect) Proportion
If \( y \) is inversely proportional to \( x \) and \( y = 4 \) when \( x = 3 \), find \( y \) when \( x = 6 \).
4. Different Types of Variation
The intensity of light \( I \) is inversely proportional to the square of the distance \( d \) from the source. If the intensity is 100 units when the distance is 2 meters, find the constant of proportionality and write the equation for the relationship.
Solutions
1. Rearranging Formulae
Rearrange the formula to make \( x \) the subject:
\[ y = 5x + 3 \]
**Solution:**
- Subtract 3 from both sides: \[ y - 3 = 5x \]
- Divide both sides by 5: \[ x = \frac{y - 3}{5} \]
2. Direct Proportion
If \( y \) is directly proportional to \( x \) and \( y = 10 \) when \( x = 2 \), find \( y \) when \( x = 5 \).
**Solution:**
- Write the proportional relationship: \[ y = kx \]
- Find \( k \) using the given values: \[ 10 = k \times 2 \implies k = 5 \]
- Use \( k \) to find \( y \) when \( x = 5 \): \[ y = 5 \times 5 = 25 \]
3. Inverse (Indirect) Proportion
If \( y \) is inversely proportional to \( x \) and \( y = 4 \) when \( x = 3 \), find \( y \) when \( x = 6 \).
**Solution:**
- Write the proportional relationship: \[ y = \frac{k}{x} \]
- Find \( k \) using the given values: \[ 4 = \frac{k}{3} \implies k = 12 \]
- Use \( k \) to find \( y \) when \( x = 6 \): \[ y = \frac{12}{6} = 2 \]
4. Different Types of Variation
The intensity of light \( I \) is inversely proportional to the square of the distance \( d \) from the source. If the intensity is 100 units when the distance is 2 meters, find the constant of proportionality and write the equation for the relationship.
**Solution:**
- Write the proportional relationship: \[ I = \frac{k}{d^2} \]
- Find \( k \) using the given values: \[ 100 = \frac{k}{2^2} \implies 100 = \frac{k}{4} \implies k = 400 \]
- Write the equation: \[ I = \frac{400}{d^2} \]
Practice Problems
Types of Variation
Non-linear Direct Proportion
Definition: When one quantity increases or decreases in a non-linear direct proportion to another, it is called non-linear direct proportion. The relationship can be expressed with the equation:
\[ y = kx^n \]
where \( y \) and \( x \) are the variables, \( k \) is the constant of proportionality, and \( n \) is a positive real number greater than 1.
Inverse Non-linear Proportion
Definition: When one quantity increases while the other decreases in a non-linear manner such that their product raised to a power is constant, it is called inverse non-linear proportion. The relationship can be expressed with the equation:
\[ y = \frac{k}{x^n} \]
where \( y \) and \( x \) are the variables, \( k \) is the constant of proportionality, and \( n \) is a positive real number greater than 1.
Example 1: Non-linear Direct Proportion
The area \( A \) of a circle is directly proportional to the square of its radius \( r \). If the area is 50.24 square units when the radius is 4 units, find the constant of proportionality and write the equation for the relationship.
Identify the given values:
\( A = 50.24 \text{ square units}, \quad r = 4 \text{ units} \)
Use the formula for non-linear direct proportion \( A = kr^2 \) to find \( k \):
\( 50.24 = k \times 4^2 \implies 50.24 = 16k \implies k = \frac{50.24}{16} = 3.14 \)
The constant of proportionality \( k \) is 3.14. The equation for the relationship is:
\( A = 3.14r^2 \)
Example 2: Inverse Non-linear Proportion
The gravitational force \( F \) between two masses is inversely proportional to the square of the distance \( d \) between them. If the force is 100 units when the distance is 2 meters, find the constant of proportionality and write the equation for the relationship.
Identify the given values:
\( F = 100 \text{ units}, \quad d = 2 \text{ meters} \)
Use the formula for inverse non-linear proportion \( F = \frac{k}{d^2} \) to find \( k \):
\( 100 = \frac{k}{2^2} \implies 100 = \frac{k}{4} \implies k = 100 \times 4 = 400 \)
The constant of proportionality \( k \) is 400. The equation for the relationship is:
\( F = \frac{400}{d^2} \)