Welcome to the Chapter on Proportion!
Hello! This chapter is where you learn to express relationships between quantities using the powerful language of algebra. Proportion is all about how things change together—whether they rise or fall in tandem, or if one goes up while the other goes down.
Understanding proportion is essential not just for high-level mathematics, but for everyday life, from scaling a recipe to calculating fuel consumption. We will use the proportional symbol (\(\propto\)) and turn these proportional statements into accurate algebraic equations that allow us to calculate unknown values. Let's dive in!
Section 1: Direct Proportion (Linear Variation)
1.1 What is Direct Proportion?
Two quantities are in direct proportion if they increase or decrease together at the same constant rate.
Think of it this way: If you double the input, you double the output.
Example: The cost of bananas. If one banana costs $0.50, two bananas cost $1.00. The cost is directly proportional to the number of bananas.
The Algebraic Form
If quantity \(y\) is directly proportional to quantity \(x\), we write:
$$\text{Proportional relationship: } y \propto x$$
To turn this relationship into an equation, we introduce a value called the Constant of Proportionality, usually denoted by \(k\).
$$\text{Equation: } y = kx$$
Key Term: Constant of Proportionality (\(k\))
This is the fixed number that relates the two quantities. You can find it by dividing \(y\) by \(x\): \(k = \frac{y}{x}\).
Step-by-Step: Solving a Direct Proportion Problem
If \(C\) is directly proportional to \(T\), and \(C = 15\) when \(T = 3\), find \(C\) when \(T = 10\).
- Write the Proportion: $$C \propto T$$
- Form the Equation using \(k\): $$C = kT$$
- Find \(k\): Use the first pair of values (\(C=15, T=3\)). $$15 = k(3) \implies k = \frac{15}{3} = 5$$
- Write the Complete Formula: $$C = 5T$$
- Solve for the Unknown: Use the formula to find \(C\) when \(T = 10\). $$C = 5(10) = 50$$
Key Takeaway: Direct proportion means $y = kx$. Always find $k$ first to establish the unique formula for that problem.
Section 2: Inverse Proportion
2.1 What is Inverse Proportion?
Two quantities are in inverse proportion if one quantity increases while the other quantity decreases, such that their product remains constant.
Example: Imagine digging a trench. If one worker takes 10 hours, two workers will take 5 hours (half the time). The time taken is inversely proportional to the number of workers.
Think of it this way: If you double the input, you halve the output.
The Algebraic Form
If quantity \(y\) is inversely proportional to quantity \(x\), it means \(y\) is directly proportional to the reciprocal of \(x\) (\(\frac{1}{x}\)). We write:
$$\text{Proportional relationship: } y \propto \frac{1}{x}$$
Again, we introduce the constant \(k\) to form the equation:
$$\text{Equation: } y = \frac{k}{x} \quad \text{or equivalently, } xy = k$$
Memory Aid: The word 'Inverse' should remind you of 'Invert' (turn upside down). The variable must be inverted (put in the denominator).
Step-by-Step: Solving an Inverse Proportion Problem
If \(P\) is inversely proportional to \(V\), and \(P = 20\) when \(V = 4\), find \(P\) when \(V = 8\).
- Write the Proportion: $$P \propto \frac{1}{V}$$
- Form the Equation using \(k\): $$P = \frac{k}{V}$$
- Find \(k\): Use the first pair of values (\(P=20, V=4\)). $$20 = \frac{k}{4} \implies k = 20 \times 4 = 80$$
- Write the Complete Formula: $$P = \frac{80}{V}$$
- Solve for the Unknown: Use the formula to find \(P\) when \(V = 8\). $$P = \frac{80}{8} = 10$$
Did you notice? When V doubled (from 4 to 8), P halved (from 20 to 10). This confirms inverse proportion!
Key Takeaway: Inverse proportion means $y = \frac{k}{x}$. The constant $k$ is found by multiplying the pair of values: $k = xy$.
Section 3: Other Types of Variation (Extended Content)
The concept of proportion extends beyond simple linear relationships. The syllabus requires you to handle proportions involving powers and roots:
3.1 Direct Proportion involving Powers and Roots
Sometimes, a quantity is proportional not just to \(x\), but to \(x^2\), \(x^3\), or \(\sqrt{x}\). The method remains the same—just replace \(x\) in the equation \(y=kx\) with the correct function of \(x\).
| Description of Relationship | Proportional Symbol | Algebraic Equation |
|---|---|---|
| \(y\) is proportional to the square of \(x\) | \(y \propto x^2\) | \(y = kx^2\) |
| \(y\) is proportional to the cube of \(x\) | \(y \propto x^3\) | \(y = kx^3\) |
| \(y\) is proportional to the square root of \(x\) | \(y \propto \sqrt{x}\) | \(y = k\sqrt{x}\) |
3.2 Inverse Proportion involving Powers and Roots
If the relationship is inverse, you place the function of \(x\) in the denominator.
| Description of Relationship | Proportional Symbol | Algebraic Equation |
|---|---|---|
| \(y\) is inversely proportional to the square of \(x\) | \(y \propto \frac{1}{x^2}\) | \(y = \frac{k}{x^2}\) |
| \(y\) is inversely proportional to the cube root of \(x\) | \(y \propto \frac{1}{\sqrt[3]{x}}\) | \(y = \frac{k}{\sqrt[3]{x}}\) |
Example: Scaling Area
The area (\(A\)) of a shape is proportional to the square of its radius (\(r\)).
If \(A = 50\) when \(r = 5\), find the formula connecting \(A\) and \(r\).
1. Write the Proportion: \(A \propto r^2\)
2. Form the Equation: \(A = kr^2\)
3. Find \(k\): \(50 = k(5^2) \implies 50 = 25k \implies k = 2\)
4. Complete Formula: \(A = 2r^2\)
Common Mistake to Avoid: Students often forget to apply the square or the square root when finding \(k\). If \(y \propto x^3\), you must substitute \(x^3\) into the equation, not just \(x\).
Section 4: Identifying the Best Variation Model (Extended Skill)
Sometimes, you are given a set of data points and asked to determine the relationship—is it \(y \propto x\), \(y \propto x^2\), or \(y \propto 1/x\)?
You must calculate the constant \(k\) for each possible model. The correct model is the one where \(k\) is consistent (or nearly consistent) for all data pairs.
4.1 Steps to Identify the Correct Model
Suppose you have two variables \(X\) and \(Y\).
Test 1: Is it Direct Proportion (\(Y = kX\))?
Calculate \(k_1 = \frac{Y}{X}\) for every pair of data points. If all $k_1$ values are the same, the model is \(Y \propto X\).
Test 2: Is it Direct Proportion to the Square (\(Y = kX^2\))?
Calculate \(k_2 = \frac{Y}{X^2}\) for every pair of data points. If all $k_2$ values are the same, the model is \(Y \propto X^2\).
Test 3: Is it Inverse Proportion (\(Y = \frac{k}{X}\))?
Calculate \(k_3 = YX\) for every pair of data points. If all $k_3$ values are the same, the model is \(Y \propto \frac{1}{X}\).
Example of Model Identification
Data points: (2, 18), (3, 40.5). What is the relationship?
Let's test the models:
- Test \(Y \propto X\) (\(k = Y/X\)):
Pair 1: \(k = 18 / 2 = 9\)
Pair 2: \(k = 40.5 / 3 = 13.5\)
The values of \(k\) are different. This model is INCORRECT. - Test \(Y \propto X^2\) (\(k = Y/X^2\)):
Pair 1: \(k = 18 / (2^2) = 18 / 4 = 4.5\)
Pair 2: \(k = 40.5 / (3^2) = 40.5 / 9 = 4.5\)
The values of \(k\) are the same! This model is CORRECT.
Conclusion: The relationship is \(Y = 4.5X^2\).
Tip for Struggling Students: Don't panic when you see data tables. Just systematically calculate the potential constant \(k\) for $x$, $x^2$, $1/x$, etc., using the equations $k = y/x$, $k = y/x^2$, or $k = xy$. One of them *must* work!
Section 5: Summary and Key Formula Review
Quick Review Box: The Algebra of Proportion (E2.8)
The core skill in this chapter is always the same: turn the proportional statement into an equation using \(k\), find \(k\), and use the final equation to solve the problem.
| Type of Proportion | Proportional Symbol | Equation (How to find \(k\)) |
|---|---|---|
| \(y\) is directly proportional to \(x\) | \(y \propto x\) | \(y = kx \implies k = y/x\) |
| \(y\) is inversely proportional to \(x\) | \(y \propto 1/x\) | \(y = k/x \implies k = xy\) |
| \(y\) is proportional to \(x^n\) | \(y \propto x^n\) | \(y = kx^n \implies k = y/x^n\) |
| \(y\) is inversely proportional to \(x^n\) | \(y \propto 1/x^n\) | \(y = k/x^n \implies k = yx^n\) |
Final Thought: Proportion problems look complicated because they involve reading carefully to determine the correct relationship ($x$, $x^2$, $1/\sqrt{x}$, etc.). Take your time, set up the relationship correctly with $\propto$, and the rest is straightforward algebraic substitution! You've got this!