Further Mathematics (9665) | Linear graphs

FPP1.2: Study Notes - Reducing Relations to Linear Graphs

Welcome to the chapter on Linear Graphs! If you're thinking, "Wait, why are we doing basic lines in Further Maths?", the answer is simple: We're not just drawing lines, we're using lines to solve complicated, non-linear problems.

This skill—called linearisation—is incredibly powerful. It allows us to take curvy, tricky relationships found in physics, biology, or finance and turn them into simple straight lines. Why? Because straight lines are predictable! If we can plot data and get a straight line, we can easily find the constants governing the entire relationship (like finding \(k\) in \(y = kx^n\)).

Don't worry if this chapter involves logarithms. We will break down exactly why and how they make curves straight!

1. The Core Idea: Transforming Non-Linear Relations

What is Linearisation?

Linearisation is the process of manipulating a non-linear equation so that it matches the familiar form of a straight line:

\[Y = mX + c\]

Here, the key is the capital letters:

• \(Y\) is the transformed dependent variable (might be \(y\), \(\log y\), or \(\frac{1}{y}\)).
• \(X\) is the transformed independent variable (might be \(x\), \(\log x\), or \(\frac{1}{x}\)).
• \(m\) is the gradient (slope) of the straight line.
• \(c\) is the \(Y\)-intercept.

Our job is to figure out the right substitution or mathematical tool (usually logarithms) to turn the given formula into this linear format.

Analogy Alert!

Imagine you have a curvy road map. Linearisation is like putting on a pair of special mathematical glasses that make the curvy road appear perfectly straight. Once it looks straight, measuring the slope and intercept is easy, giving you crucial information about the original curvy road.

2. Key Types of Non-Linear Relations and Their Transformations

You must be able to recognise the structure of the equation and choose the appropriate transformation. Some equations require simple algebraic rearrangement; others require logarithms.

Type A: Simple Substitution Laws

These laws become linear by simple algebraic manipulation or substitution, without needing logarithms.

Example A1: Inverse Law (\(\frac{1}{y} = \frac{1}{x} + k\))

If you encounter a relation like \(\frac{1}{y} = \frac{1}{x} + k\), notice it is already linear in terms of the reciprocal variables.

Step-by-step Transformation:

1. Let \(Y = \frac{1}{y}\).
2. Let \(X = \frac{1}{x}\).

The equation becomes: \[Y = X + k\]

Analysis:

• Gradient \(m = 1\).
• \(Y\)-intercept \(c = k\).

If you plot \(Y\) against \(X\), the slope should be 1, and the intercept will immediately tell you the value of the constant \(k\).

Example A2: Polynomial Law (\(y^2 = ax + b\))

Here, one variable is raised to a power, but the other is not.

Step-by-step Transformation:

1. Let \(Y = y^2\).
2. Let \(X = x\). (The variable that is *not* transformed).

The equation becomes: \[Y = aX + b\]

Analysis:

• Gradient \(m = a\).
• \(Y\)-intercept \(c = b\).

This is straightforward: Plot \(y^2\) on the vertical axis against \(x\) on the horizontal axis.

Type B: Laws Requiring Logarithms

When the variable you want to solve for is in an exponent (like \(b^x\)) or when the original variables are multiplied/divided and raised to powers (like \(ax^n\)), we use logarithms to "pull down" the exponents and separate the terms.

Remember: Logarithms convert multiplication into addition and powers into multiplication. This is exactly what we need to get to the \(Y = mX + c\) addition format!

The syllabus mentions using logarithms to base 10 where appropriate, but any consistent base (including the natural log, \(\ln\)) will work. For simplicity in the working, we will use \(\log\) to represent a consistent logarithm base (like \(\log_{10}\)).

Key Log Rules Review (Quick Refresher)

• \(\log (A \times B) = \log A + \log B\)
• \(\log (\frac{A}{B}) = \log A - \log B\)
• \(\log (A^n) = n \log A\)

Example B1: The Power Law (\(y = ax^n\))

This describes relationships where \(y\) is proportional to a power of \(x\). Here, \(a\) and \(n\) are the unknown constants we want to find.

Step-by-step Transformation:

1. Take logs of both sides:
   \[\log y = \log (ax^n)\]
2. Use the multiplication rule:
   \[\log y = \log a + \log (x^n)\]
3. Use the power rule:
   \[\log y = \log a + n \log x\]
4. Rearrange into \(Y = mX + c\) format:
   \[\log y = n (\log x) + \log a\]

The Linear Equation:

\[Y = nX + \log a\]

Analysis (Crucial!):

• \(Y\) variable: \(\log y\)
• \(X\) variable: \(\log x\)
• Gradient \(m\): \(n\) (The exponent)
• \(Y\)-intercept \(c\): \(\log a\)

Memory Aid (Power Law): To find the *power* (\(n\)), you need a log-log plot (plot \(\log y\) against \(\log x\)).

Example B2: The Exponential Law (\(y = ab^x\))

This describes exponential growth or decay. Here, \(a\) and \(b\) are the unknown constants.

Step-by-step Transformation:

1. Take logs of both sides:
   \[\log y = \log (ab^x)\]
2. Use the multiplication rule:
   \[\log y = \log a + \log (b^x)\]
3. Use the power rule:
   \[\log y = \log a + x (\log b)\]
4. Rearrange into \(Y = mX + c\) format:
   \[\log y = (\log b) x + \log a\]

The Linear Equation:

\[Y = (\log b) X + \log a\]

Analysis (Crucial!):

• \(Y\) variable: \(\log y\)
• \(X\) variable: \(x\) (Notice \(x\) is NOT logged!)
• Gradient \(m\): \(\log b\)
• \(Y\)-intercept \(c\): \(\log a\)

Memory Aid (Exponential Law): If the variable \(x\) is in the *exponent*, you only need to log the \(y\) variable. This is a log-linear plot (plot \(\log y\) against \(x\)).

3. Working with Numerical Data and Estimation

The main purpose of linearisation is using real-world data points \((x, y)\) to find the constants \(a\), \(b\), or \(n\) in the original non-linear equation.

Step-by-Step Procedure for Finding Unknown Constants

Step 1: Identify the Relation and Transform

Look at the given equation (e.g., \(y = ax^n\)). Determine the required linear plot variables \(Y\) and \(X\). (For \(y = ax^n\), we need \(Y = \log y\) and \(X = \log x\)).

Step 2: Transform the Data

Use your original \((x, y)\) data points and calculate the new, transformed data points \((X, Y)\).
Example: If the original data point is \((2, 10)\) and you need a log-log plot, the new point is \((\log 2, \log 10) = (0.301, 1)\).

Step 3: Plot the Linear Graph

Plot the new \((X, Y)\) points. These should form a reasonably straight line. Draw the line of best fit.

Step 4: Calculate the Gradient (\(m\)) and Intercept (\(c\))

Choose two points, \((X_1, Y_1)\) and \((X_2, Y_2)\), that lie on your line of best fit (not necessarily original data points, as data often has experimental error).

Gradient: \[m = \frac{Y_2 - Y_1}{X_2 - X_1}\] Intercept: Find the point where the line crosses the \(Y\)-axis (where \(X=0\)).

Step 5: Relate \(m\) and \(c\) back to the Original Constants

Use the relationships established in Section 2 to find the original constants.

Example (Power Law \(y = ax^n\)):

• We know \(m = n\). So, the gradient you calculated is the value of \(n\).
• We know \(c = \log a\). This means \(a\) is the anti-log of \(c\).
  If using \(\log_{10}\), then \(a = 10^c\).
  If using natural log (\(\ln\)), then \(a = e^c\).

Example Walkthrough: Finding constants for \(y = ab^x\)

Suppose you are given data \((x, y)\) and told the relationship is \(y = ab^x\). You plot \(\log y\) against \(x\) and find the gradient of the resulting straight line is \(m = 0.5\) and the \(Y\)-intercept is \(c = 1.2\).

Goal: Find \(a\) and \(b\).

Recall the linear form: \[\log y = (\log b) x + \log a\]

1. Find \(b\) (from the gradient):
We know \(m = \log b\).
\[0.5 = \log_{10} b\]
Anti-log both sides:
\[b = 10^{0.5} \approx 3.16\]

2. Find \(a\) (from the intercept):
We know \(c = \log a\).
\[1.2 = \log_{10} a\]
Anti-log both sides:
\[a = 10^{1.2} \approx 15.85\]

Conclusion: The original relationship is approximately \(y = 15.85(3.16)^x\).

4. Dealing with More Complex Linearisations

Sometimes the equation requires a combination of log rules and algebraic separation.

Example C1: Logarithmic Base Swapping (Implied in the syllabus)

If you see a law involving \(\log_B y\), you might still want to plot using logs of base 10 (or natural logs) for ease of calculation. You must ensure all terms are handled consistently.

Consider the law: \[\frac{1}{y} = A x + B\] If you let \(Y = \frac{1}{y}\) and \(X = x\), the constants \(A\) and \(B\) are found directly from the gradient and intercept. Sometimes, the original law might involve constants we want to separate from the variables, for example:

\[y = \frac{ax}{x+b}\]

This looks complex, but if we take the reciprocal of both sides:

\[\frac{1}{y} = \frac{x+b}{ax}\] \[\frac{1}{y} = \frac{x}{ax} + \frac{b}{ax}\] \[\frac{1}{y} = \frac{1}{a} + \left(\frac{b}{a}\right) \frac{1}{x}\]

Transformation:

• Let \(Y = \frac{1}{y}\)
• Let \(X = \frac{1}{x}\)

The Linear Equation: \[Y = \left(\frac{b}{a}\right) X + \frac{1}{a}\]

Analysis:

• Gradient \(m = \frac{b}{a}\).
• \(Y\)-intercept \(c = \frac{1}{a}\).

Now you can find \(a\) from the intercept (\(a = \frac{1}{c}\)) and then use that value along with the gradient to find \(b\) (\(b = ma\)).