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KSSM DLP Form 5 Mathematics › Chapter 1 Variation › 1.2 Inverse Variation
👤 Student (Class 5 STEM)
Stage A

Mission Brief: The Principle of Inverse Trade-offs

⏱️ 2–3 mins
Authentic Real-World Dilemma

🚗 The Road Trip Dilemma: Speed vs Travel Time

Imagine your family is traveling from Kuala Lumpur to Ipoh (a distance of 200 km).

50 km/h Slow Crawl Traffic → 4.0 hours
↔
100 km/h Smooth Highway Cruise → 2.0 hours

⚡ Key Discovery: When your speed is doubled ($\times 2$), your travel time is halved ($\div 2$). Their product is invariant: $$ \text{Speed} \times \text{Time} = 50 \times 4.0 = 100 \times 2.0 = 200\text{ km (Constant!)} $$ This is the hallmark of Inverse Variation (*Ubahan Songsang*)!

DSKP 1.2.1

Explain Meaning

Understand that when variable $y$ increases, variable $x$ decreases at the same rate, yielding constant product $xy = k$.

DSKP 1.2.2

Determine Relations

Formulate $y \propto \frac{1}{x^n}$ and equation $y = \frac{k}{x^n}$ where $k$ is the constant of variation.

DSKP 1.2.3

Solve Applications

Apply Equation and Proportion methods to solve SPM & real-world problems (Boyle's Law, Gravity, Worker efficiency).

📖 Mathematical Symbol Decoder

Textbook Reference: Form 5 p. 18–21
\( y \propto \frac{1}{x} \)

Variation Relation: Read as "\(y\) varies inversely as \(x\)" or "\(y\) is inversely proportional to \(x\)".

\( y = \frac{k}{x} \)

Equation Form: Where \(k\) is a non-zero constant known as the constant of variation (\(k = xy\)).

\( y \propto \frac{1}{x^n} \)

General Powers: For \(n = 1, 2, 3, \frac{1}{2}, \frac{1}{3}\), the product is constant: \(k = x^n y\).

Dual Graphs

Graph of \(y\) vs \(x\): Hyperbola curve.
Graph of \(y\) vs \(\frac{1}{x^n}\): Straight line starting from origin \((0, 0)\).

Stage B

Discovery Lab: The Boyle's Law Gas Syringe & Dual Graph Studio

⏱️ 5–10 mins

Manipulate the gas volume inside the sealed cylinder. Observe how gas molecule collisions generate pressure, and discover why the product \(P \times V\) never changes!

🔬 Gas Syringe Apparatus (Boyle's Law)

Volume = 60 cm³, Pressure = 20.0 kPa
CLOSED PUSH 20 40 60 80 100 cm³
60 cm³
Quick Presets:
Measured Pressure, \(P\) 20.00 kPa
Reciprocal, \(\frac{1}{V}\) 0.0167 cm⁻³
Product \(P \times V = k\) 1200.0 (Constant!)
Graph (a): \(P\) against \(V\)
Hyperbola Curve
V P O
As volume \(V\) expands, pressure \(P\) drops asymptotically.
Graph (b): \(P\) against \(\frac{1}{V}\)
Straight Line from Origin
1/V P O
Forms a straight line passing through \((0, 0)\). Gradient \(m = k = 1200\)!

💡 Conceptual Discovery Check:

1. When the gas volume is compressed from \(100\text{ cm}^3\) down to \(50\text{ cm}^3\) (halved), what happens to the gas pressure?

2. Which graphical representation proves that \(P\) varies inversely as \(V\)?

Stage C

Interactive Worked Examples: Step-by-Step Mastery

⏱️ 10–15 mins

Master the official textbook worked examples. Solve the intermediate checkpoints below to reveal full mathematical reasoning and contrast alternative methods.

Textbook Example 16 (p. 21)

Determining Variation from Tabular Data

The time needed \(t\) (minutes) to assemble furniture varies inversely as the number of workers \(x\).

Number of workers, \(x\) 2 3 4 5 6
Time needed, \(t\) (min) 180 120 90 72 60
Product \(xt\) 2(180) = 360 3(120) = 360 4(90) = 360 5(72) = 360 6(60) = 360
Product \(x^2 t\) 4(180) = 720 9(120) = 1080 16(90) = 1440 25(72) = 1800 36(60) = 2160
Interactive Checkpoint: Based on the table calculations above, which statement correctly identifies the relationship?
Textbook Example 18 (p. 22)

Finding the Equation with Square Roots

It is given that \(x = 0.25\) when \(y = 3\). Express \(y\) in terms of \(x\) if \(y\) varies inversely as the square root of \(x\).

Step 1: Calculate the constant of variation \(k\) in \(y = \frac{k}{\sqrt{x}}\):
\(k = y \times \sqrt{x} = 3 \times \sqrt{0.25} = 3 \times 0.5 = \)
Textbook Example 20 (p. 23)

Equation Method vs Proportion Method

It is given that \(p\) varies inversely as \(q\). If \(p = 2\) when \(q = 7\), calculate the value of \(p\) when \(q = 1.6\).

Method 1: Equation Method

  1. Formulate equation: \(p = \frac{k}{q}\)
  2. Find constant \(k\): \(2 = \frac{k}{7} \implies k = 2 \times 7 = 14\)
  3. Write formula: \(p = \frac{14}{q}\)
  4. Substitute \(q = 1.6\): \(p = \frac{14}{1.6} = 8.75\)
Textbook Example 21 (p. 25)

Boyle's Law Physical Application

According to Boyle's Law, gas pressure \(p\) of a trapped gas is inversely proportional to volume \(V\). When \(V = 80\text{ cm}^3\), \(p = 190.25\text{ kPa}\). Calculate the volume of gas in \(\text{cm}^3\) when the pressure is \(121.76\text{ kPa}\).

1. Understand

\(p \propto \frac{1}{V}\). Known: \(V_1 = 80\text{ cm}^3, p_1 = 190.25\text{ kPa}\). Target: find \(V_2\) when \(p_2 = 121.76\text{ kPa}\).

2. Devise & Implement

Constant \(k = p_1 V_1 = 190.25 \times 80 = 15\,220\text{ kPa}\cdot\text{cm}^3\).

Equation: \(p = \frac{15\,220}{V} \implies 121.76 = \frac{15\,220}{V_2}\).

$$V_2 = \frac{15\,220}{121.76} = 125\text{ cm}^3$$

3. Conclusion

When the gas pressure is \(121.76\text{ kPa}\), the gas volume expands to \(125\text{ cm}^3\).

Stage D

Guided Practice: Scaffolding with Multi-Tier Hints

⏱️ 5–10 mins

Solve each textbook practice problem. If you need assistance, unlock multi-tier hints to guide your solution without spoiling the answer!

Stage E

Independent Challenge: Tiered SPM Practice

⏱️ 10–15 mins
Select Difficulty Tier:
Stage F

Exit Ticket & Student Learning Report

⏱️ 3–5 mins

Complete the 3 diagnostic assessment items to prove your conceptual and procedural mastery of Subtopic 1.2 Inverse Variation.