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AHL 2.9—HL modelling functions

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Overview of HL Modelling Functions

At Higher Level, Math AI extends modelling beyond simple linear and quadratic functions into a rich toolkit of function types that can describe complex, real-world behaviour. This subtopic covers five key model families:

  • Exponential models , rapid growth or decay (radioactive decay, compound interest)
  • Natural logarithmic models , diminishing returns, perceived sensation
  • Sinusoidal models , periodic/cyclical phenomena (tides, temperature, sound)
  • Logistic models , growth with a carrying capacity (populations, disease spread)
  • Piecewise models , behaviour that changes abruptly at defined thresholds

A critical HL skill is not just applying these models, but selecting the most appropriate one given data or context, and fitting parameters using either algebra or a GDC.

Exam Tip

When you see a modelling question, always ask: Is the quantity growing/decaying without bound? Levelling off? Oscillating? Changing rule at a threshold? Your answer points directly to the model family.

Exponential Models and Half-Life

Exponential Model: A function of the form f(x)=a⋅bx (or equivalently f(x)=a⋅ekx) where the rate of change is proportional to the current value. If k>0 the model describes growth; if k<0 it describes decay.

The decay form most commonly used in physics and pharmacology is:

N(t)=N0​⋅e−λt

where:

  • N(t) = quantity remaining at time t
  • N0​ = initial quantity
  • λ = decay constant (always positive)

Half-Life: The time T1/2​ required for a quantity to reduce to exactly half its current value. It is related to the decay constant by:
T1/2​=λln2​

Deriving the half-life formula: Set N(t)=2N0​​:
2N0​​=N0​⋅e−λT1/2​
21​=e−λT1/2​
ln(21​)=−λT1/2​
T1/2​=λln2​

Example

Radioactive decay problem

A sample contains 100 g of a radioactive substance with a half-life of 5 hours. Find the amount remaining after 12 hours.

Step 1: Find λ.
λ=T1/2​ln2​=5ln2​≈0.1386 h−1

Step 2: Write the model.
N(t)=100⋅e−(ln2/5)t

Step 3: Evaluate at t=12.
N(12)=100⋅e−(ln2/5)(12)=100⋅2−12/5≈29.3 g

Interpretation: After 12 hours (2.4 half-lives), just under 30% of the original sample remains.

Common Mistake

A common error is writing λ=T1/2​/ln2 instead of λ=ln2/T1/2​. Always check: a larger λ should mean faster decay (shorter half-life), so λ and T1/2​ must be inversely proportional.

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12 more sections in this topic

← Previous topicAHL 2.8—Transformations of graphs, composite transformationsNext topic →AHL 2.10—Scaling large numbers, log-log graphs
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