IB Mathematics Applications & Interpretation Topic…

Why HL Modelling Needs More Shapes

The SL models (linear, quadratic, simple exponential and basic sine) fit many situations, but not all.
Populations that grow then level off, quantities that halve at a steady rate, cycles that start off-centre, and rules that switch at thresholds each need a richer function.
HL adds five modelling tools: the exponential with base $e$, the natural-log model, the phase-shifted sinusoid, the logistic curve, and piecewise functions.
For every one of them the exam skill is the same: recognise the shape, pin down the parameters from data, and interpret what they mean in context.

Exponential Models With Base e: Growth, Decay And Half-Life

Continuous growth or decay is modelled with the natural base $e$: $$ N(t) = N_0 e^{kt} $$
$N_0$ is the initial amount (the value at $t=0$), and $k$ is the continuous rate: $k>0$ grows, $k<0$ decays.
The half-life is the time for a decaying quantity to fall to half its value, and it can be derived directly:
1. set $N=\tfrac{1}{2}N_0$, so $\tfrac{1}{2}N_0 = N_0 e^{kt}$, and divide by $N_0$ to get $e^{kt}=\tfrac{1}{2}$,
2. take natural logs: $kt = \ln\tfrac{1}{2} = -\ln 2$, giving a half-life of $t_{1/2}=\dfrac{-\ln 2}{k}$ (positive because $k<0$ for decay).

Example question

A radioactive sample follows $N = 50\,e^{-0.03t}$, with $t$ in hours. Find its half-life, to 3 significant figures.

Solution

Set $N=25$: $25 = 50\,e^{-0.03t}$, so $e^{-0.03t}=0.5$.
Take logs: $$-0.03t = \ln 0.5 = -\ln 2$$
Solve: $$t = \dfrac{\ln 2}{0.03} = 23.1 \text{ hours}$$

Note

A base-$e$ model and a base-$b$ model describe the same curves; $e$ is the natural choice when the rate is continuous.
Simple growth and decay were introduced in SL 2.5, and the same models arise as differential equations in AHL 5.14.

Natural-Logarithm Models a + b ln x

Some quantities rise quickly and then keep climbing, but ever more slowly and with no ceiling; the model $f(x)=a+b\ln x$ captures this.
It is only defined for $x>0$, and unlike the logistic it has no horizontal asymptote, it just grows slower and slower.
$b$ controls how fast it climbs and $a$ shifts it vertically.
Because $\ln x$ is the inverse of $e^x$, a log model is the natural partner of an exponential one (SL 1.5).

Example question

A model has the form $f(x)=a+b\ln x$. Given $f(1)=2$ and $f(e)=5$, find $a$ and $b$.

Solution

At $x=1$: $\ln 1 = 0$, so $f(1)=a=2$.
At $x=e$: $\ln e = 1$, so $a+b=5$.
Therefore $b=3$ and the model is $f(x)=2+3\ln x$.

Sinusoidal Models With A Phase Shift

Periodic data (tides, daylight hours, temperature cycles) uses $$ f(x) = a\sin\big(b(x-c)\big) + d $$
The amplitude and midline come straight from the maximum and minimum:
1. amplitude $a = \dfrac{\text{max}-\text{min}}{2}$, half the peak-to-trough range,
2. principal axis $d = \dfrac{\text{max}+\text{min}}{2}$, the midline the curve oscillates about.
The period $T$ fixes $b$ through $T=\dfrac{2\pi}{b}$, so $b=\dfrac{2\pi}{T}$.
The phase shift $c$ slides the whole wave horizontally so it starts in the right place.

Common Mistake

At HL, trigonometric models use radians (AHL 3.7), so the period is $\dfrac{2\pi}{b}$, never $\dfrac{360}{b}$.
Set your GDC to radian mode before fitting or evaluating a sinusoidal model.

Example question

Tide depth $d$ (metres) varies between a low of $2$ and a high of $8$, with period $12$ hours and the first high tide at $t=5$. Write a model $d = a\sin\big(b(t-c)\big)+d_0$.

Solution

Amplitude: $$a = \dfrac{8-2}{2} = 3$$
Principal axis: $$d_0 = \dfrac{8+2}{2} = 5$$
Period: $$b = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$$
A high tide is where the sine peaks, at argument $\tfrac{\pi}{2}$: $\tfrac{\pi}{6}(5-c)=\tfrac{\pi}{2}$, so $5-c=3$ and $c=2$.
Model: $$d = 5 + 3\sin\!\left(\tfrac{\pi}{6}(t-2)\right)$$

The Logistic Model: Growth That Levels Off At A Ceiling

When growth is fast at first but saturates (limited resources, market saturation, an epidemic), use the logistic model: $$ f(x) = \dfrac{L}{1 + C e^{-kx}} $$
$L$ is the carrying capacity, the horizontal asymptote the curve approaches but never reaches.
The ceiling can be derived: as $x\to\infty$, $e^{-kx}\to 0$, so the denominator $\to 1$ and $f\to L$.
The starting value is $f(0)=\dfrac{L}{1+C}$, and $k$ controls how steeply the curve rises.
The graph is S-shaped, steepest at its point of inflexion where $f=\dfrac{L}{2}$, then it bends over toward $L$.

Common Mistake

$L$ is a limit, not a value the curve reaches: the model gets within any percentage of $L$ but never equals it.
Answer logistic questions as “within $X\%$ of the maximum”, not “reaches the maximum”.

Example question

A population is modelled by $P = \dfrac{1200}{1+5e^{-0.4t}}$, with $t$ in years. State the carrying capacity and the initial population.

Solution

Carrying capacity: as $t\to\infty$, $e^{-0.4t}\to 0$, so $P\to 1200$.
Initial population: at $t=0$, $P=\dfrac{1200}{1+5}=200$.
So the population starts at $200$ and levels off toward $1200$ without reaching it.

Piecewise Models Switch Rules At Thresholds

Some situations follow different rules over different intervals: tax bands, phone tariffs, parking charges, delivery costs.
A piecewise function gives a separate expression on each interval of the domain, and the intervals fit together to cover it.
Always check which interval an input falls in before choosing the rule, and watch the endpoints where the rule changes.

Example question

A delivery cost (dollars) is $C(x)=5$ for $0<x\leq 2$ kg, and $C(x)=5+2(x-2)$ for $x>2$ kg. Find $C(2)$ and $C(5)$.

Solution

$x=2$ lies in the first interval, so $C(2)=5$.
$x=5$ lies in the second interval: $C(5)=5+2(5-2)=5+6=11$.
So a $2$ kg parcel costs $\$5$ and a $5$ kg parcel costs $\$11$.

Exam technique

None of these model equations is in the formula booklet; when a question uses one it is given in the stem, so recognise the form and read off what each parameter means.
Where data is given, fit parameters with GDC regression (exponential, sinusoidal, logistic), then interpret in context and state a reasonable domain.
Keep trigonometric models in radians and get the period right with $b=\dfrac{2\pi}{T}$.
For decay and logistic questions, answer “within $X\%$ of” the limit rather than “equals” it.

The HL Models At A Glance

Each model has a signature shape and a small set of parameters you extract and interpret.

Model	Form	Key parameters
Exponential (base e)	N₀ e^{kt}	N₀ start, k rate, half-life = ln2/\|k\|
Natural log	a + b ln x	x > 0, no ceiling, b sets speed
Sinusoidal	a sin(b(x − c)) + d	a amplitude, d midline, T = 2π/b, c phase shift
Logistic	L / (1 + C e^{−kx})	L carrying capacity, inflection at L/2
Piecewise	different rule per interval	mind which interval and the thresholds

Active recall

How do you find the half-life from $N=N_0 e^{kt}$?
For a sinusoidal model, how are the amplitude and principal axis found from the maximum and minimum?
In the logistic model $\dfrac{L}{1+Ce^{-kx}}$, what does $L$ represent and why is it never reached?
Which of these model formulas appear in the formula booklet?
What must you always check before applying a piecewise rule to a given input?

Why HL Modelling Needs More Shapes

The SL models (linear, quadratic, simple exponential and basic sine) fit many situations, but not all.
Populations that grow then level off, quantities that halve at a steady rate, cycles that start off-centre, and rules that switch at thresholds each need a richer function.
HL adds five modelling tools: the exponential with base $e$, the natural-log model, the phase-shifted sinusoid, the logistic curve, and piecewise functions.
For every one of them the exam skill is the same: recognise the shape, pin down the parameters from data, and interpret what they mean in context.

Exponential Models With Base e: Growth, Decay And Half-Life

Continuous growth or decay is modelled with the natural base $e$: $$ N(t) = N_0 e^{kt} $$
$N_0$ is the initial amount (the value at $t=0$), and $k$ is the continuous rate: $k>0$ grows, $k<0$ decays.
The half-life is the time for a decaying quantity to fall to half its value, and it can be derived directly:
1. set $N=\tfrac{1}{2}N_0$, so $\tfrac{1}{2}N_0 = N_0 e^{kt}$, and divide by $N_0$ to get $e^{kt}=\tfrac{1}{2}$,
2. take natural logs: $kt = \ln\tfrac{1}{2} = -\ln 2$, giving a half-life of $t_{1/2}=\dfrac{-\ln 2}{k}$ (positive because $k<0$ for decay).

Example question

A radioactive sample follows $N = 50\,e^{-0.03t}$, with $t$ in hours. Find its half-life, to 3 significant figures.

Solution

Set $N=25$: $25 = 50\,e^{-0.03t}$, so $e^{-0.03t}=0.5$.
Take logs: $$-0.03t = \ln 0.5 = -\ln 2$$
Solve: $$t = \dfrac{\ln 2}{0.03} = 23.1 \text{ hours}$$

Note

A base-$e$ model and a base-$b$ model describe the same curves; $e$ is the natural choice when the rate is continuous.
Simple growth and decay were introduced in SL 2.5, and the same models arise as differential equations in AHL 5.14.

Natural-Logarithm Models a + b ln x

Some quantities rise quickly and then keep climbing, but ever more slowly and with no ceiling; the model $f(x)=a+b\ln x$ captures this.
It is only defined for $x>0$, and unlike the logistic it has no horizontal asymptote, it just grows slower and slower.
$b$ controls how fast it climbs and $a$ shifts it vertically.
Because $\ln x$ is the inverse of $e^x$, a log model is the natural partner of an exponential one (SL 1.5).

Example question

A model has the form $f(x)=a+b\ln x$. Given $f(1)=2$ and $f(e)=5$, find $a$ and $b$.

Solution

At $x=1$: $\ln 1 = 0$, so $f(1)=a=2$.
At $x=e$: $\ln e = 1$, so $a+b=5$.
Therefore $b=3$ and the model is $f(x)=2+3\ln x$.

Sinusoidal Models With A Phase Shift

Periodic data (tides, daylight hours, temperature cycles) uses $$ f(x) = a\sin\big(b(x-c)\big) + d $$
The amplitude and midline come straight from the maximum and minimum:
1. amplitude $a = \dfrac{\text{max}-\text{min}}{2}$, half the peak-to-trough range,
2. principal axis $d = \dfrac{\text{max}+\text{min}}{2}$, the midline the curve oscillates about.
The period $T$ fixes $b$ through $T=\dfrac{2\pi}{b}$, so $b=\dfrac{2\pi}{T}$.
The phase shift $c$ slides the whole wave horizontally so it starts in the right place.

Common Mistake

At HL, trigonometric models use radians (AHL 3.7), so the period is $\dfrac{2\pi}{b}$, never $\dfrac{360}{b}$.
Set your GDC to radian mode before fitting or evaluating a sinusoidal model.

Example question

Tide depth $d$ (metres) varies between a low of $2$ and a high of $8$, with period $12$ hours and the first high tide at $t=5$. Write a model $d = a\sin\big(b(t-c)\big)+d_0$.

Solution

Amplitude: $$a = \dfrac{8-2}{2} = 3$$
Principal axis: $$d_0 = \dfrac{8+2}{2} = 5$$
Period: $$b = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$$
A high tide is where the sine peaks, at argument $\tfrac{\pi}{2}$: $\tfrac{\pi}{6}(5-c)=\tfrac{\pi}{2}$, so $5-c=3$ and $c=2$.
Model: $$d = 5 + 3\sin\!\left(\tfrac{\pi}{6}(t-2)\right)$$

The Logistic Model: Growth That Levels Off At A Ceiling

When growth is fast at first but saturates (limited resources, market saturation, an epidemic), use the logistic model: $$ f(x) = \dfrac{L}{1 + C e^{-kx}} $$
$L$ is the carrying capacity, the horizontal asymptote the curve approaches but never reaches.
The ceiling can be derived: as $x\to\infty$, $e^{-kx}\to 0$, so the denominator $\to 1$ and $f\to L$.
The starting value is $f(0)=\dfrac{L}{1+C}$, and $k$ controls how steeply the curve rises.
The graph is S-shaped, steepest at its point of inflexion where $f=\dfrac{L}{2}$, then it bends over toward $L$.

Common Mistake

$L$ is a limit, not a value the curve reaches: the model gets within any percentage of $L$ but never equals it.
Answer logistic questions as “within $X\%$ of the maximum”, not “reaches the maximum”.

Example question

A population is modelled by $P = \dfrac{1200}{1+5e^{-0.4t}}$, with $t$ in years. State the carrying capacity and the initial population.

Solution

Carrying capacity: as $t\to\infty$, $e^{-0.4t}\to 0$, so $P\to 1200$.
Initial population: at $t=0$, $P=\dfrac{1200}{1+5}=200$.
So the population starts at $200$ and levels off toward $1200$ without reaching it.

Piecewise Models Switch Rules At Thresholds

Some situations follow different rules over different intervals: tax bands, phone tariffs, parking charges, delivery costs.
A piecewise function gives a separate expression on each interval of the domain, and the intervals fit together to cover it.
Always check which interval an input falls in before choosing the rule, and watch the endpoints where the rule changes.

Example question

A delivery cost (dollars) is $C(x)=5$ for $0<x\leq 2$ kg, and $C(x)=5+2(x-2)$ for $x>2$ kg. Find $C(2)$ and $C(5)$.

Solution

$x=2$ lies in the first interval, so $C(2)=5$.
$x=5$ lies in the second interval: $C(5)=5+2(5-2)=5+6=11$.
So a $2$ kg parcel costs $\$5$ and a $5$ kg parcel costs $\$11$.

Exam technique

None of these model equations is in the formula booklet; when a question uses one it is given in the stem, so recognise the form and read off what each parameter means.
Where data is given, fit parameters with GDC regression (exponential, sinusoidal, logistic), then interpret in context and state a reasonable domain.
Keep trigonometric models in radians and get the period right with $b=\dfrac{2\pi}{T}$.
For decay and logistic questions, answer “within $X\%$ of” the limit rather than “equals” it.

The HL Models At A Glance

Each model has a signature shape and a small set of parameters you extract and interpret.

Model	Form	Key parameters
Exponential (base e)	N₀ e^{kt}	N₀ start, k rate, half-life = ln2/\|k\|
Natural log	a + b ln x	x > 0, no ceiling, b sets speed
Sinusoidal	a sin(b(x − c)) + d	a amplitude, d midline, T = 2π/b, c phase shift
Logistic	L / (1 + C e^{−kx})	L carrying capacity, inflection at L/2
Piecewise	different rule per interval	mind which interval and the thresholds

Active recall

How do you find the half-life from $N=N_0 e^{kt}$?
For a sinusoidal model, how are the amplitude and principal axis found from the maximum and minimum?
In the logistic model $\dfrac{L}{1+Ce^{-kx}}$, what does $L$ represent and why is it never reached?
Which of these model formulas appear in the formula booklet?
What must you always check before applying a piecewise rule to a given input?

IB Mathematics Applications & Interpretation AHL 2.9—HL Modelling Functions Notes

IB Mathematics Applications & Interpretation AHL 2.9—HL Modelling Functions Notes

Why HL Modelling Needs More Shapes

Exponential Models With Base e: Growth, Decay And Half-Life

Natural-Logarithm Models a + b ln x

Sinusoidal Models With A Phase Shift

The Logistic Model: Growth That Levels Off At A Ceiling

Piecewise Models Switch Rules At Thresholds

The HL Models At A Glance

Why HL Modelling Needs More Shapes

Exponential Models With Base e: Growth, Decay And Half-Life

Natural-Logarithm Models a + b ln x

Sinusoidal Models With A Phase Shift

The Logistic Model: Growth That Levels Off At A Ceiling

Piecewise Models Switch Rules At Thresholds

The HL Models At A Glance

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation AHL 2.9—HL Modelling Functions Notes

IB Mathematics Applications & Interpretation AHL 2.9—HL Modelling Functions Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

Why HL Modelling Needs More Shapes

Exponential Models With Base e: Growth, Decay And Half-Life

Natural-Logarithm Models a + b ln x

Sinusoidal Models With A Phase Shift

The Logistic Model: Growth That Levels Off At A Ceiling

Piecewise Models Switch Rules At Thresholds

The HL Models At A Glance

Why HL Modelling Needs More Shapes

Exponential Models With Base e: Growth, Decay And Half-Life

Natural-Logarithm Models a + b ln x

Sinusoidal Models With A Phase Shift

The Logistic Model: Growth That Levels Off At A Ceiling

Piecewise Models Switch Rules At Thresholds

The HL Models At A Glance