Chapter 6: Linear Law

Overview

Linear Law is a powerful technique in Additional Mathematics that transforms non-linear relationships into linear forms for easier analysis. This chapter explores the principles of linear law, focusing on how to convert various non-linear equations into the form Y = mX + c. By mastering these linearization techniques, students can use linear graph analysis methods to solve complex non-linear problems efficiently.

Learning Objectives

After completing this chapter, you will be able to:

• Distinguish between linear and non-linear relationships
• Apply various linearization techniques to non-linear equations
• Plot and interpret linear graphs
• Determine constants and make predictions using linear law
• Apply linear law to solve real-world problems

Key Concepts

6.1 Linear and Non-Linear Relations

Linear Relations

A linear relationship produces a straight-line graph with the form:

$$y = mx + c$$

Where:

• m = gradient (slope)
• c = y-intercept

Characteristics:

• Constant rate of change
• Straight line when plotted
• Simple interpretation

Non-Linear Relations

A non-linear relationship produces a curved graph. Common types include:

• Quadratic: $y = ax^2 + bx + c$
• Exponential: $y = ab^x$
• Power: $y = ax^n$
• Reciprocal: $y = \frac{a}{x} + b$

Characteristics:

• Variable rate of change
• Curved when plotted
• More complex analysis required

Purpose of Linear Law

The main goal of linear law is to transform non-linear relationships into linear form ($Y = mX + c$) so that:

• Linear graph analysis can be used
• Gradient and intercept can be easily determined
• Predictions and extrapolations become more reliable

6.2 Application of Linear Law

Linearization Techniques

1. Quadratic Relationships: $y = ax^2 + b$

   • Transform to: $y = ax^2 + b$
   • Plot $y$ vs $x^2$
   • Gradient = $a$, y-intercept = $b$

2. Power Relationships: $y = ax^n$

   • Apply logarithms: $\log y = \log a + n \log x$
   • Plot $\log y$ vs $\log x$
   • Gradient = $n$, y-intercept = $\log a$

3. Exponential Relationships: $y = ab^x$

   • Apply logarithms: $\log y = \log a + x \log b$
   • Plot $\log y$ vs $x$
   • Gradient = $\log b$, y-intercept = $\log a$

4. Reciprocal Relationships: $y = \frac{a}{x} + b$

   • Rearrange: $xy = a + bx$
   • Plot $xy$ vs $x$
   • Gradient = $b$, y-intercept = $a$

Best Fit Line

The line of best fit is a straight line drawn through a scatter plot that:

• Passes through as many data points as possible
• Has roughly equal numbers of points above and below it
• Minimizes the overall distance from all data points

Important Formulas and Methods

Linearization Formulas

Graph Plotting Methods

Steps for Linear Law Analysis:

1. Identify the type of non-linear relationship
2. Apply appropriate linearization technique
3. Calculate transformed variables
4. Plot transformed data on graph paper
5. Draw line of best fit
6. Determine gradient and intercept
7. Extract original constants
8. Make predictions or solve for required values

Solved Examples

Example 1: Quadratic Relationship

The relationship between variables $x$ and $y$ is given by $y = ax^2 + b$. The following data is obtained:

Find the values of a and b.

Solution:

Transform to linear form: $y = ax^2 + b$ Calculate $x^2$ and plot $y$ vs $x^2$:

Plot $y$ vs $x^2$ and find gradient and intercept.

Using first and last points: Gradient = $\frac{61 - 5}{25 - 1} = \frac{56}{24} = \frac{7}{3} \approx 2.333$ Y-intercept = when $x^2 = 0$, from equation $y = ax^2 + b$, $y = b$

Using point (1,5): $5 = a(1)^2 + b \Rightarrow 5 = a + b$ Using point (25,61): $61 = a(25) + b \Rightarrow 61 = 25a + b$

Subtract first equation from second: $56 = 24a \Rightarrow a = \frac{56}{24} = \frac{7}{3} \approx 2.333$ Then $b = 5 - \frac{7}{3} = \frac{8}{3} \approx 2.667$

Therefore: $a = \frac{7}{3}, b = \frac{8}{3}$

Example 2: Power Relationship

The relationship is $y = ax^n$. Given data:

Find a and n.

Solution:

Transform using logarithms: $\log y = \log a + n \log x$ Calculate $\log x$ and $\log y$:

Plot $\log y$ vs $\log x$ and find gradient and intercept.

Using first and last points: Gradient = $\frac{2.2833 - 1.0792}{0.9031 - 0.3010} = \frac{1.2041}{0.6021} \approx 2.000$

Using point ($\log x = 0.3010, \log y = 1.0792$): $\log y = n \log x + \log a$ $1.0792 = 2(0.3010) + \log a$ $1.0792 = 0.6020 + \log a$ $\log a = 0.4772 \Rightarrow a = 10^{0.4772} \approx 3.000$

Therefore: $a = 3, n = 2$ (relationship is $y = 3x^2$)

Example 3: Exponential Relationship

Given the relationship $y = ab^x$ and data:

Find a and b.

Solution:

Transform using logarithms: $\log y = \log a + x \log b$ Calculate $\log y$:

Plot $\log y$ vs $x$ and find gradient and intercept.

Using first and last points: Gradient = $\frac{1.9031 - 1.0000}{4 - 1} = \frac{0.9031}{3} \approx 0.3010$

Using point ($x=1, \log y=1.0000$): $\log y = (\log b)x + \log a$ $1.0000 = 0.3010(1) + \log a$ $\log a = 1.0000 - 0.3010 = 0.6990 \Rightarrow a = 10^{0.6990} \approx 5.000$

Also, $\log b = 0.3010 \Rightarrow b = 10^{0.3010} \approx 2.000$

Therefore: $a = 5, b = 2$ (relationship is $y = 5 \times 2^x$)

Example 4: Reciprocal Relationship

Given $y = \frac{a}{x} + b$ and data:

Find a and b.

Solution:

Transform: $xy = a + bx$ Calculate $xy$:

Plot $xy$ vs $x$ and find gradient and intercept.

Using first and last points: Gradient = $\frac{28 - 7}{8 - 1} = \frac{21}{7} = 3$

Using point ($x=1, xy=7$): $xy = bx + a$ $7 = 3(1) + a \Rightarrow a = 4$

Therefore: $a = 4, b = 3$ (relationship is $y = \frac{4}{x} + 3$)

Example 5: Practical Application

The growth of bacteria is modeled by $N = N_0 e^{kt}$, where $N$ is the population at time $t$. The following data was obtained:

Find the initial population $N_0$ and the growth rate k.

Solution:

Transform the equation: $N = N_0 e^{kt}$ Take natural logarithms: $\ln N = \ln N_0 + kt$

Calculate $\ln N$:

Plot $\ln N$ vs $t$ and find gradient and intercept.

Using first and last points: Gradient = $\frac{6.1701 - 4.6052}{8 - 0} = \frac{1.5649}{8} \approx 0.1956$

Using point ($t=0, \ln N=4.6052$): $\ln N = kt + \ln N_0$ $4.6052 = 0.1956(0) + \ln N_0 \Rightarrow \ln N_0 = 4.6052 \Rightarrow N_0 \approx 100$

Therefore: $N_0 = 100, k \approx 0.1956$ per hour

Mathematical Derivations

Linearization of Power Functions

Given $y = ax^n$ Take logarithms: $\log y = \log(ax^n) = \log a + \log(x^n) = \log a + n \log x$ This is in the form $Y = mX + c$ where:

• $Y = \log y$
• $m = n$ (gradient)
• $X = \log x$
• $c = \log a$ (y-intercept)

Linearization of Exponential Functions

Given $y = ab^x$ Take logarithms: $\log y = \log(ab^x) = \log a + \log(b^x) = \log a + x \log b$ This is in the form $Y = mX + c$ where:

• $Y = \log y$
• $m = \log b$ (gradient)
• $X = x$
• $c = \log a$ (y-intercept)

Real-World Applications

1. Physics - Ohm's Law

Voltage and current in a resistor: $V = IR$ If we plot $V$ vs $I$, we get a straight line with gradient = $R$ (resistance)

2. Chemistry - Rate Laws

Rate = $k[A]^m[B]^n$ Take logarithms: $\log(\text{rate}) = \log k + m \log[A] + n \log[B]$ Multiple linear regression can determine $m$ and $n$

3. Biology - Enzyme Kinetics

Michaelis-Menten equation: $v = \frac{V_{max}[S]}{K_m + [S]}$ Transform to: $\frac{1}{v} = \frac{K_m}{V_{max}}\frac{1}{[S]} + \frac{1}{V_{max}}$ Plot $\frac{1}{v}$ vs $\frac{1}{[S]}$ to find $V_{max}$ and $K_m$

4. Economics - Demand Functions

Demand often follows power law: $Q = aP^b$ Take logarithms: $\log Q = \log a + b \log P$ Plot $\log Q$ vs $\log P$ to find price elasticity $b$

Complex Problem-Solving Techniques

Problem: Experimental data follows $y = ax^2 + bx + c$. Use linear law to find $a, b, c$.

Solution: This is more complex since there are three parameters. One approach is:

1. Plot $y$ vs $x^2$ (assuming $b$ is small or zero)
2. If not accurate, use finite differences or other methods

Problem: Two variables $x$ and $y$ are related by $y^2 = ax + b$. Find $a$ and $b$ using linear law.

Solution: Transform: $y^2 = ax + b$ Plot $y^2$ vs $x$:

• Gradient = $a$
• Y-intercept = $b$

Problem: The relationship between $x$ and $y$ is given by $y = ax^n + b$. Use logarithms to linearize.

Solution: This is challenging because of the $+b$ term. One approach:

1. Estimate $b$ from data (minimum or asymptotic value)
2. Let $z = y - b$, then $z = ax^n$
3. Take logarithms: $\log z = \log a + n \log x$
4. Plot $\log z$ vs $\log x$ to find $a$ and $n$

Summary Points

• Linear law transforms non-linear relationships into linear forms
• Different types of non-linear relationships require specific linearization techniques
• Plotting transformed variables allows linear graph analysis
• Gradient and y-intercept give the original constants
• Applications span physics, chemistry, biology, and economics
• Careful choice of transformation is crucial for accuracy

Common Mistakes to Avoid

1. Incorrect transformation - Choose the right linearization method for the equation type
2. Calculation errors - Double-check transformed variables and gradients
3. Plotting mistakes - Use appropriate scales and label axes correctly
4. Extrapolation errors - Be careful when extending beyond data range
5. Interpretation errors - Remember what gradient and intercept represent in original terms

SPM Exam Tips

Exam Strategies

1. Identify equation type - Recognize whether it's quadratic, exponential, power, or reciprocal
2. Apply correct transformation - Use the appropriate linearization technique
3. Calculate transformed values carefully - Show all calculations in working
4. Plot accurately - Use proper scales and draw a good line of best fit
5. Extract constants correctly - Relate gradient and intercept back to original equation

Key Exam Topics

• Quadratic relationships ($y = ax^2 + b$) (25% of questions)
• Power relationships ($y = ax^n$) (25% of questions)
• Exponential relationships ($y = ab^x$) (25% of questions)
• Reciprocal relationships ($y = \frac{a}{x} + b$) (15% of questions)
• Practical applications (10% of questions)

Time Management Tips

• Basic linearization: 3-4 minutes
• Plotting and analysis: 5-6 minutes
• Finding constants: 4-5 minutes
• Applications: 6-8 minutes
• Complex problems: 8-10 minutes

Practice Problems

Level 1: Basic Linearization

1. Given $y = ax^2 + b$ and data:

   x	y
   1	4
   2	7
   3	14
   4	25
   Find $a$ and $b$.

2. Given $y = ax^n$ and data:

   x	y
   2	8
   3	27
   4	64
   5	125
   Find $a$ and $n$.

Level 2: Intermediate Problems

3. Given $y = ab^x$ and data:

   x	y
   0	3
   1	6
   2	12
   3	24
   Find $a$ and $b$.

4. Given $y = \frac{a}{x} + b$ and data:

   x	y
   1	8
   2	5
   4	3
   8	2
   Find $a$ and $b$.

Level 3: Applications

5. Physics: The period $T$ of a pendulum is related to length $L$ by $T = 2\pi\sqrt{\frac{L}{g}}$. Given data:

   L (m)	T (s)
   1	2.01
   2	2.84
   3	3.48
   4	4.01
   Find $g$ (acceleration due to gravity).

6. Chemistry: The rate of reaction follows rate = $k[A]^2$. Given data:

   [A] (mol/L)	Rate (mol/L/s)
   0.1	0.01
   0.2	0.04
   0.3	0.09
   0.4	0.16
   Find $k$.

7. Biology: Bacterial growth follows $N = N_0 e^{kt}$. Given data:

   t (hours)	N
   0	500
   1	670
   2	897
   3	1200
   Find $N_0$ and $k$.

Level 4: Complex Problems

8. The relationship is $y = ax^2 + bx$. Find $a$ and $b$ using linear law.

9. Given data following $y = a\sqrt{x} + b$, find $a$ and $b$.

10. Economics: Demand function $Q = aP^b$. Given data:

    Price (RM)	Quantity demanded
    10	1000
    20	354
    30	192
    40	125
    Find the price elasticity of demand ($b$).

Did You Know? 📚

Linear law techniques were essential in the development of scientific computing before electronic computers. Scientists would manually plot transformed data on graph paper to extract parameters from experimental data. The method of least squares, developed by Carl Friedrich Gauss in 1809, provides the mathematical foundation for finding the best fit line, revolutionizing data analysis in all sciences.

Quick Reference Guide

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Linear law is a versatile tool that bridges the gap between complex non-linear relationships and simple linear analysis. Mastering these techniques will enable you to solve a wide range of practical problems in science, engineering, and economics.