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KSSM DLP Form 5 Mathematics › Chapter 1 Variation › 1.3 Combined & Joint Variation
👤 Student (Class 5 STEM)
Stage A

Mission Brief: Combining Direct & Inverse Variations

⏱️ 2–3 mins
Authentic Multi-Variable Challenge

🏛️ The School Hall Chair Logistics Dilemma

Ahead of the SPM Graduation Ceremony, student volunteers arrange chairs in the school hall.

1,000 Chairs Arranged by 5 volunteers in 2.0 hours
→
What If? We double chairs to 2,000 AND double volunteers to 10?

⚡ Multi-Variable Insight: Time taken \(t\) varies directly as chair count \(c\) (more chairs $\to$ more time), but inversely as worker count \(p\) (more volunteers $\to$ less time)! $$ t \propto \frac{c}{p} \iff t = \frac{k c}{p} $$ When both chairs and volunteers double simultaneously, the time required remains exactly 2.0 hours!

DSKP 1.3.1

Determine Relations

Synthesize three or more variables into unified variation relations \(y \propto \frac{x^m}{z^n}\) and equations \(y = \frac{k x^m}{z^n}\).

DSKP 1.3.2

Solve Multi-Variable Problems

Execute calculations using the constant ratio \(k = \frac{y z^n}{x^m}\) across STEM, physics, and industrial applications.

Synthesis TP 5/6

Combined Reasoning

Analyze how simultaneous changes in numerators (direct) and denominators (inverse) govern system outcomes.

📖 Mathematical Symbol Decoder: Combined Variation Rules

Textbook Reference: Form 5 p. 26–27
\( y \propto \frac{x^m}{z^n} \)

Combined Variation: \(y\) varies directly as \(x^m\) and inversely as \(z^n\).

\( y = \frac{k x^m}{z^n} \)

Equation Form: Where \(k\) is a non-zero constant evaluated by \(k = \frac{y z^n}{x^m}\).

Fraction Anchor

Direct / Joint variables: Multiply in the numerator (top).
Inverse variables: Multiply in the denominator (bottom).

Invariant Ratio

For any two states: \(\frac{y_1 z_1^n}{x_1^m} = \frac{y_2 z_2^n}{x_2^m} = k\).

Stage B

Discovery Lab: 3D Dynamic Cylinder & Multi-Variable Studio

⏱️ 5–10 mins

Explore textbook Mind Mobilisation 8 (p. 26). Adjust Volume (\(x\)) and Base Area (\(z\)) independently to observe how cylinder height (\(y = \frac{x}{z}\)) reacts!

🧪 3D Geometric Cylinder Apparatus

Vol = 200 cm³, Area = 50 cm² → Height = 4.0 cm
Ground Reference y = 4.0 cm
200 cm³
50 cm²
Quick Presets:
Volume, \(x\) (Direct) 200 cm³
Base Area, \(z\) (Inverse) 50 cm²
Calculated Height, \(y = \frac{x}{z}\) 4.0 cm
Graph (a): Height \(y\) against Volume \(x\)
Direct: Straight Line
x y O
Holding area \(z\) constant: \(y\) varies directly as \(x\).
Graph (b): Height \(y\) against Base Area \(z\)
Inverse: Hyperbola
z y O
Holding volume \(x\) constant: \(y\) varies inversely as \(z\).

💡 Multi-Variable Conceptual Check:

1. If Volume \(x\) is doubled (\(\times 2\)) AND Base Area \(z\) is doubled (\(\times 2\)) simultaneously, what happens to cylinder height \(y\)?

2. In the formula \(y = \frac{k x^m}{z^n}\), how is the invariant constant \(k\) determined from any given set of coordinates?

Stage C

Interactive Worked Examples: Step-by-Step Formalization

⏱️ 10–15 mins

Master official textbook Examples 22, 23, and 24. Work through the intermediate calculation checkpoints to unlock detailed mathematical reasoning.

Textbook Example 22 (p. 27)

Expressing Equation with Direct Square & Inverse Square Root

It is given that \(y\) varies directly as the square of \(x\) and inversely as the square root of \(z\). If \(y = 8\) when \(x = 4\) and \(z = 36\), express \(y\) in terms of \(x\) and \(z\).

Step 1: Calculate the constant of variation \(k\) from \(8 = \frac{k(4)^2}{\sqrt{36}} = \frac{16k}{6}\):
\(k = \frac{8 \times \sqrt{36}}{4^2} = \frac{8 \times 6}{16} = \)
Textbook Example 23 (p. 27–28)

Determining Multiple Unknowns from Tabular Data

The table shows values of \(P\), \(Q\), and \(R\). It is given that \(P\) varies directly as the cube of \(Q\) and inversely as \(R\).

Quantity Reference Calibration Unknown \(R = x\) Unknown \(Q = y\)
\(P\) 4 3.6 0.081
\(Q\) 2 6 \(y\)
\(R\) 0.6 \(x\) 2.7
1. Establish Formula

$$P = \frac{k Q^3}{R} \implies 4 = \frac{k (2)^3}{0.6} = \frac{8k}{0.6} \implies k = \frac{2.4}{8} = 0.3 = \frac{3}{10}$$

$$\text{Equation: } P = \frac{3 Q^3}{10 R}$$

2. Calculate Unknown \(x\)

When \(P = 3.6, Q = 6, R = x\):

$$3.6 = \frac{3 (6)^3}{10 x} = \frac{3(216)}{10 x} = \frac{64.8}{x} \implies x = \frac{64.8}{3.6} = \mathbf{18}$$

3. Calculate Unknown \(y\)

When \(P = 0.081, Q = y, R = 2.7\):

$$0.081 = \frac{3 y^3}{10(2.7)} = \frac{3 y^3}{27} = \frac{y^3}{9} \implies y^3 = 0.081 \times 9 = 0.729 \implies y = \sqrt[3]{0.729} = \mathbf{0.9}$$

Textbook Example 24 (p. 28–29)

Tyre Ground Pressure & Practical Engineering Insight

The pressure \(p\) (\(\text{N m}^{-2}\)) on the tyre of a wheelbarrow varies directly as total mass \(m\) (kg) and inversely as contact surface area \(l\) (\(\text{m}^2\)). Given \(p = 45\,000\text{ N m}^{-2}\) when \(m = 90\text{ kg}\) and \(l = 0.02\text{ m}^2\).

Calculation for \(m = 120\text{ kg}, l = 0.5\text{ m}^2\)

$$k = \frac{p \cdot l}{m} = \frac{45\,000 \times 0.02}{90} = 10 \implies p = \frac{10m}{l}$$

Substitute new values:

$$p = \frac{10(120)}{0.5} = \frac{1200}{0.5} = \mathbf{2\,400\text{ N m}^{-2}}$$

Stage D

Guided Practice: Scaffolding with Multi-Tier Hints

⏱️ 5–10 mins

Solve each textbook practice problem. If you encounter difficulty, expand the 3-step hint sequence to assist your algebraic setup 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, procedural, and applied mastery of Subtopic 1.3 Combined & Joint Variation.