Secondary 4 Additional Mathematics: Linear Law Turns Nonlinear Relationships Into Straight-Line Evidence
Linear-law problems are not really about drawing straight lines. They are about choosing a transformation that converts a nonlinear relationship into the standard form Y = mX + c, then reading the hidden parameters from the gradient and intercept of the transformed graph.
At Secondary 4, this becomes a powerful representation skill. The original variables may follow a power law, exponential law or another rearrangeable relationship. Instead of attacking the parameters directly in the original form, the learner creates new variables that expose a linear structure. The graph then becomes an algebraic measuring instrument.
Linearisation is successful when the transformed axes make the hidden constants appear as gradient and intercept.
The Simple Answer
The target structure is
Y = mX + c.
The transformed variables Y and X may not be the original y and x. Once the equation is rearranged into straight-line form:
- m is the gradient of the transformed graph;
- c is the vertical intercept;
- the original parameters can be recovered by matching them to m and c.
The key is to state clearly what is plotted against what. A mathematically correct transformation is incomplete if the axes are not identified.
Power Law: y = Axn
Take logarithms:
log y = log A + n log x.
Match this to Y = mX + c:
- Y = log y;
- X = log x;
- gradient m = n;
- intercept c = log A.
Therefore a graph of log y against log x is straight if the power model is appropriate. The gradient recovers n directly, and A is recovered from the intercept.
Worked Example 1: Recover a Power Parameter
Suppose a graph of log y against log x has gradient 2.5 and vertical intercept 0.3010 when common logarithms are used. If
y = Axn,
then n = 2.5 and log A = 0.3010. Since 100.3010 is approximately 2, A ≈ 2. The recovered model is approximately
y = 2x2.5.
The graph did not merely show a trend. Its straight-line parameters decoded the nonlinear model.
Exponential Law: y = Aekx
Take natural logarithms:
ln y = ln A + kx.
This already matches straight-line form with
- Y = ln y;
- X = x;
- gradient = k;
- intercept = ln A.
A graph of ln y against x should therefore be straight if the model y = Aekx fits the data.
Worked Example 2: Exponential Parameter Recovery
Suppose the straight-line graph of ln y against x has gradient −0.4 and intercept ln 6. Then the original model is
y = 6e−0.4x.
The negative gradient signals exponential decay. The intercept corresponds to the value at x = 0 because e0 = 1.
Transformations Without Logarithms
Not every linearisation uses logarithms. Suppose
y = a + b/x.
Let X = 1/x and Y = y. Then
Y = bX + a.
A graph of y against 1/x is straight with gradient b and intercept a. The correct transformation depends on the algebraic form of the original relationship.
The Most Important Skill: Match the Entire Straight-Line Form
Students often stop after “taking logs”. That is not enough. The transformed equation must be rearranged until it clearly matches Y = mX + c. Only then can the axes, gradient and intercept be identified without ambiguity.
For example, if
y = A(Bx),
then taking logarithms gives
log y = log A + x log B.
A graph of log y against x has gradient log B and intercept log A. The parameter B is not the gradient itself; it must be recovered by reversing the logarithm.
Gradient and intercept may contain the parameters rather than equal them directly.
Worked Example 3: Parameter Hidden Inside the Gradient
If y = ABx and the graph of log10y against x has gradient 0.4771, then
log10B = 0.4771.
Therefore B ≈ 3. A common error would be to write B = 0.4771, confusing the transformed gradient with the original parameter.
Axis Choice Is Part of the Mathematics
If the transformed relation is ln y = kx + ln A, the horizontal axis is x and the vertical axis is ln y. If the relation is log y = n log x + log A, the horizontal axis is log x, not x. If the relation is y = b(1/x) + a, the horizontal axis is 1/x.
Writing the wrong axes can make a correct transformation unusable. Always state the plotting instruction explicitly.
Worked Example 4: Read Parameters From Two Points on a Linearised Graph
Suppose Y = mX + c is the transformed form and two points on the straight line are (1, 4) and (5, 12). Then
m = (12 − 4)/(5 − 1) = 2.
Substitute into Y = 2X + c using (1,4):
4 = 2 + c → c = 2.
The original parameters then depend on how m and c were defined by the transformation. Never interpret them before returning to the original model mapping.
Choosing a Transformation Backward From Y = mX + c
When a question asks you to show that a relationship can be represented by a straight-line graph, work backward from the target structure:
- decide which part of the original equation can become Y;
- identify a second expression that can become X;
- rearrange or transform until one appears linearly in the other;
- collect the remaining constants into gradient and intercept positions;
- state the axes and parameter meanings.
This prevents random manipulation. Every transformation should move the relationship closer to Y = mX + c.
Linearisation as Model Testing
If transformed data lie approximately on a straight line, that provides evidence that the chosen model family may be appropriate over the observed range. If the points bend systematically away from a line, the model may be inadequate or the transformation may be wrong.
The straight line therefore does more than recover parameters. It also gives a visual diagnostic of model fit. The model remains an approximation of the observed relationship, not an automatic law of nature.
Verification: Rebuild the Original Relationship
After recovering parameters, substitute them back into the original nonlinear model and test one known data point or condition. This catches errors caused by misreading the gradient, using the wrong logarithm base or forgetting to reverse an intercept transformation.
For example, if log A = c, then A must be recovered using the same base as the logarithm. If natural logarithms were used, A = ec. If common logarithms were used, A = 10c.
The Linear-Law Decision Tree
- What is the original model form? Power, exponential, reciprocal or another transformable relation?
- Which operation makes a variable appear linearly? Logarithm, reciprocal, square, root or rearrangement?
- Can the result be written exactly as Y = mX + c?
- What are X and Y? State the axes.
- What do m and c represent in terms of the original parameters?
- What inverse operation recovers the parameters?
- Can the recovered model reproduce a known condition?
Common Secondary 4 Linear-Law Errors
- Taking logarithms but not rearranging into Y = mX + c.
- Plotting the original variables instead of the transformed variables.
- Calling log A the parameter A.
- Calling log B the parameter B.
- Using the wrong logarithm base when reversing the transformation.
- Reading gradient from an imprecise pair of points when better graph information is available.
- Forgetting that transformed intercepts may need exponentiation to recover original constants.
- Recovering parameters but not checking the original model.
- Treating an approximately straight transformed plot as perfect proof that the model is exact.
Transfer Across the A-Math System
Linear-law reasoning connects several A-Math ideas. Logarithms change multiplicative structure into additive structure. Coordinate geometry provides gradient and intercept tools. Function transformations teach how changing the representation changes what information is visible. Modelling gives the parameters meaning. The same learner who can move confidently between these layers is much better equipped for unfamiliar questions.
A Six-Stage Training Sequence
- Match simple transformed equations directly to Y = mX + c.
- Linearise power and exponential relationships using logarithms.
- Linearise reciprocal and rearranged algebraic models without logarithms.
- State transformed axes and parameter meanings precisely.
- Recover parameters from gradients and intercepts, including inverse transformations.
- Verify recovered models against known points and mixed modelling questions.
Checkpoint: Linear-Law Control
- If y = Axn, what should be plotted to obtain a straight line after logarithms?
- What does the gradient represent in that transformed graph?
- If ln y = kx + ln A, what does the intercept represent?
- Why is B not equal to the gradient in log y = x log B + log A?
- What is the best final verification after recovering parameters?
Checkpoint Answers
- Plot log y against log x.
- The exponent n.
- ln A, so A = e raised to that intercept.
- The gradient is log B; B must be recovered by reversing the logarithm.
- Substitute the recovered parameters into the original model and test a known condition or data point.
Wintour House V1.0 Learning Standard
Wintour House V1.0 treats linear law as controlled representation change. CivDJ routing identifies the nonlinear object, selects a transformation that preserves the relationship while exposing a straight-line form, maps transformed gradient and intercept back to the original parameters, and verifies the recovered model in its original representation.
The straight line is not the original problem. It is the representation that makes the hidden constants easy to read.
