R13: Equations of Proportion
Interpret equations that describe direct and inverse proportion
📋 Key Concepts
Higher Tier Only: This topic extends proportion to include more complex relationships including powers and roots.
Types of Proportional Relationships
| Relationship | Equation | Form |
|---|---|---|
| y ∝ x | y = kx | Direct proportion |
| y ∝ x² | y = kx² | Proportional to square |
| y ∝ x³ | y = kx³ | Proportional to cube |
| y ∝ √x | y = k√x | Proportional to square root |
| y ∝ 1/x | y = k/x | Inverse proportion |
| y ∝ 1/x² | y = k/x² | Inversely proportional to square |
📝 Identifying the Relationship
Method: Look at the structure of the equation to identify the type of proportion.
Example 1
Identify the type of proportion in each equation:
a) y = 3x²
b) y = 15/x
c) y = 4√x
Solution:
a) y ∝ x² (y is directly proportional to x²)
b) y ∝ 1/x (y is inversely proportional to x)
c) y ∝ √x (y is directly proportional to √x)
📝 Finding the Constant k
Method: Substitute known values and solve for k.
Example 2
y is directly proportional to x². When x = 3, y = 36.
Find the equation.
Solution:
y = kx²
36 = k × 9
k = 4
Equation: y = 4x²
Example 3
y is inversely proportional to x². When x = 4, y = 5.
Find y when x = 10.
Solution:
y = k/x²
5 = k/16
k = 80
y = 80/x²
When x = 10: y = 80/100 = 0.8
📝 Solving Complex Problems
Example 4
The force between two objects is inversely proportional to the square of the distance between them. When distance = 2m, force = 50N. Find the force when distance = 5m.
Solution:
F = k/d²
50 = k/4
k = 200
F = 200/d²
When d = 5: F = 200/25 = 8N
Example 5
The area of a circle is proportional to the square of its radius. If radius = 3, area = 28.27 (to 2 dp). Find the constant k and interpret its meaning.
Solution:
A = kr²
28.27 = k × 9
k = 3.14 (approximately π)
The constant is π - this is the formula A = πr²
📝 Tables and Graphs
Identifying from data: Test whether y/x, y/x², xy, or x²y is constant.
Example 6
x: 1, 2, 3 | y: 6, 24, 54. Find the relationship.
Solution:
Test y/x: 6, 12, 18 (not constant)
Test y/x²: 6, 6, 6 (constant!)
Relationship: y = 6x²
❓ Practice Questions
Q1: y ∝ x³. When x = 2, y = 40. Find the equation.
Q2: y ∝ 1/x². When x = 3, y = 8. Find y when x = 6.
Q3: y ∝ √x. When x = 16, y = 12. Find y when x = 25.
Q4: State whether each shows y ∝ x, y ∝ x² or y ∝ 1/x: x: 2, 4, 6 | y: 8, 32, 72
Q5: The kinetic energy is proportional to the square of speed. When speed = 10, energy = 500. Find energy when speed = 15.
✅ Answers
- y = 5x³
- y = 2 (k = 72)
- y = 15 (k = 3)
- y ∝ x² (y/x² = 2)
- 1125 (E = 5v²)
🎯 Exam Tips
- Identify the form from the equation structure
- Square powers: y = kx², y = kx³
- Inverse square: y = k/x²
- Test data: if y/x² constant, then y ∝ x²
- Always substitute to find k first
🧠 Problem-Solving Strategies
Problem-Solving
Set up equations from proportional relationships: y ∝ x means y = kx, y ∝ x² means y = kx², y ∝ 1/x means y = k/x. Always find k first using given values, then solve for the unknown. Check answers by substituting back. For graphs, y = kx gives a straight line through the origin.
Multi-Step Problem
y is directly proportional to x². When x = 3, y = 36. Find y when x = 5, and find x when y = 100.
Solution: y = kx². 36 = k × 9, so k = 4. When x = 5: y = 4 × 25 = 100. When y = 100: 100 = 4x², x² = 25, x = 5.
⚠️ Common Errors
Watch Out!
1. Wrong: Writing y = kx when y is proportional to x² Correct: y ∝ x² means y = kx², not y = kx. The square applies to the variable, not the constant.
2. Wrong: When y ∝ √x, writing y = kx² Correct: y ∝ √x means y = k√x = kx^(1/2). Square root is x^0.5, not x².
3. Wrong: Finding k by dividing y by x when the relationship is y = kx² Correct: If y = kx² and y = 36 when x = 3, then k = 36/9 = 4 (divide by x², not x).
✍️ 6-Mark Exam Question
Extended Answer
6 marks: The force (F) between two magnets is inversely proportional to the square of the distance (d) between them. When d = 2 cm, F = 50 N. (a) Find F when d = 5 cm. (b) Find d when F = 200 N. (c) What happens to F when the distance is halved? Explain using the formula.
F = k/d². 50 = k/4, so k = 200.
(a) F = 200/25 = 8 N.
(b) 200 = 200/d², so d² = 1, d = 1 cm.
(c) If d halves, d becomes d/2. New F = 200/(d/2)² = 200/(d²/4) = 4 × 200/d² = 4F. The force becomes 4 times greater because the square of half the distance is a quarter.
Mark scheme: M1 for finding k, A1 k = 200, M1 for F when d = 5, A1 8N, M1 for d when F = 200, A1 d = 1cm, A1 explanation with F quadrupling
📊 AO3: Reason & Interpret
Reasoning and Interpretation
The kinetic energy (KE) of an object is directly proportional to the square of its speed: KE = kv². When v = 8 m/s, KE = 200 J.
(a) Find the kinetic energy when v = 12 m/s.
(b) By what factor does KE increase when the speed doubles?
(c) A car driver says "Going 10 mph over the limit isn't much more dangerous because the speed increase is small." Use the proportional relationship to comment on this.
Answers: (a) k = 200/64 = 3.125. KE = 3.125 × 144 = 450 J. (b) If v doubles, KE = k(2v)² = 4kv². KE increases by factor of 4. (c) Since KE ∝ v², even a small speed increase causes a large energy increase. At 30 mph → 40 mph, the energy increases by (40/30)² = 1.78 times. This means crashes at slightly higher speeds are significantly more dangerous, so the driver is wrong.
📝 Exam Questions by Topic
🎬 Video Resources
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