Further modelling AI HL

IB Mathematics: Applications and Interpretation HL · Unit 2: Functions

Lesson 2.8 · Further Modelling

Half-life, logarithmic, sinusoidal, logistic and piecewise models.

Faith connection

When a widow had little food left, the prophet Elijah listened to her need and trusted God to provide for her household. Scripture says, “The barrel of meal wasted not, neither did the cruse of oil fail” (1 Kings 17:16, KJV). Loving others means paying attention to their circumstances and responding with care. Mathematical models also begin by looking carefully at a situation and its limits. Come to Jesus, our Savior and Lord, and ask him to help you use what you learn to serve others wisely.

Learning objectives

By the end of this lesson, you should be able to:

use exponential models to calculate half-life;
interpret and use natural logarithmic models of the form $f(x)=a+b\ln x$ ;
identify amplitude, period, phase shift and midline in a sinusoidal model written in radians;
interpret the parameters and carrying capacity of a logistic model;
use piecewise functions and choose parameters that make adjoining pieces continuous;
evaluate predictions against the context and the model’s domain.

Which model fits the situation?

Pattern	Possible model	Feature to interpret
Repeated reduction by the same proportion	Exponential decay	Half-life
Continued growth at a slowing rate	Logarithmic	Input must be positive
Regular cycle	Sinusoidal	Amplitude and period
Growth that levels off near a limit	Logistic	Carrying capacity
Different rules for different intervals	Piecewise	Boundary conditions

Real data may not follow any of these patterns exactly. Always check the domain and the assumptions behind a prediction.

1. Exponential decay and half-life

A quantity has a half-life of $H$ time units if half of the amount present remains after every interval of length $H$ . If the initial amount is $A_0$ , then:

$A(t)=A_0\left(\frac12\right)^{t/H}$

An equivalent model is $A(t)=A_0e^{-kt}$ , where $k>0$ and:

$H=\frac{\ln2}{k}$

Worked example · Radioactive decay.

A sample initially contains $96$ mg of a substance with a half-life of $8$ hours. Its mass after $t$ hours is:

$A(t)=96\left(\frac12\right)^{t/8}$

After $24$ hours, three half-lives have passed, giving $A(24)=96(1/2)^3=12$ mg. To find when $15$ mg remain:

$96\left(\frac12\right)^{t/8}=15$

$t=8\frac{\ln(15/96)}{\ln(1/2)}\approx21.4\text{ hours}$

The model approaches zero but never reaches zero at a finite time.

2. Natural logarithmic models

A logarithmic model can have the form:

$f(x)=a+b\ln x,\qquad x>0$

Here $\ln x$ is the natural logarithm. Since $\ln1=0$ , the parameter $a$ is the output when $x=1$ . If $b>0$ , the model increases, but its increase becomes slower as $x$ grows. The line $x=0$ is a vertical asymptote for the unrestricted mathematical function.

Worked example · Learning time.

Suppose the time $T(n)$ , in minutes, needed to complete a task is modelled by $T(n)=a+b\ln n$ , where $n$ is the number of steps and $n\geq1$ . Measurements give $T(1)=4$ and $T(e^2)=10$ .

Since $\ln1=0$ , we obtain $a=4$ . Since $\ln(e^2)=2$ :

$10=4+2b\quad\Longrightarrow\quad b=3$

$T(n)=4+3\ln n$

The model predicts $T(e^3)=4+3(3)=13$ minutes. It should be used only over a range of step counts where the modelling assumptions remain sensible.

3. Sinusoidal models and phase shift

A sinusoidal model describes approximately periodic behaviour:

$f(x)=a\sin\bigl(b(x-c)\bigr)+d$

In this lesson, angles are measured in radians unless a degree symbol is shown. For $b\neq0$ , the amplitude is $|a|$ , the period is $2\pi/|b|$ , and the principal axis or midline is $y=d$ . The parameter $c$ gives a horizontal translation, often called a phase shift. For $b>0$ , increasing $c$ shifts the sine graph right.

The maximum and minimum values of the unrestricted model are $d+|a|$ and $d-|a|$ .

Worked example · Water level.

A water level, in metres, is modelled by:

$W(t)=12+4\sin\left(\frac{\pi}{6}(t-2)\right)$

Here $t$ is measured in hours. The amplitude is $4$ m, the midline is $12$ m, and the period is:

$\frac{2\pi}{\pi/6}=12\text{ hours}$

The sine graph is shifted $2$ hours right. Its highest level is $16$ m and its lowest is $8$ m. The first maximum after $t=2$ occurs when the sine input equals $\pi/2$ :

$\frac{\pi}{6}(t-2)=\frac{\pi}{2}\quad\Longrightarrow\quad t=5\text{ hours}$

4. Logistic models and carrying capacity

Growth often slows as space, food or other resources become limited. A logistic model is:

$P(t)=\frac{L}{1+Ce^{-kt}},\qquad L>0,\ C>0,\ k>0$

The horizontal asymptote $P=L$ is called the carrying capacity. The parameter $k$ controls how rapidly the population changes, while $C$ helps determine its initial value:

$P(0)=\frac{L}{1+C}$

The graph approaches $L$ from below as $t$ increases. A logistic model is useful only if a limiting value is plausible in the context.

Worked example · Population in a confined habitat.

A population is modelled by:

$P(t)=\frac{1000}{1+9e^{-0.4t}}$

Initially, $P(0)=1000/(1+9)=100$ . The carrying capacity is $1000$ . To estimate when the population reaches $800$ :

$800=\frac{1000}{1+9e^{-0.4t}}$

$9e^{-0.4t}=0.25\quad\Longrightarrow\quad t=\frac{\ln36}{0.4}\approx8.96$

The model predicts $800$ individuals after approximately $8.96$ time units. Its predictions depend on the carrying capacity and other conditions remaining reasonably stable.

5. Piecewise models

A piecewise function uses different formulas on different parts of its domain. For any given input, select the rule belonging to that input’s interval.

Worked example · Delivery charges.

A service charges €5 for orders up to and including $3$ kg, then €2 for each additional kilogram. If $m$ is the mass in kilograms, a model is:

$C(m)= \begin{cases} 5,&0\leq m\leq3,\\ 5+2(m-3),&m>3. \end{cases}$

An order of $2$ kg costs €5. An order of $4.5$ kg costs $5+2(1.5)=€8$ . The two pieces meet at a cost of €5 when $m=3$ , so this idealised model has no jump there.

Joining two pieces without a jump

To make two pieces meet at a boundary, calculate what each rule gives as the input approaches the boundary. Set those values equal. Then check which piece supplies the actual value at the boundary.

Worked example · Find a parameter.

Consider:

$f(x)= \begin{cases} 1+x,&0\leq x<2,\\ ax^2+x,&x\geq2. \end{cases}$

The first rule approaches $1+2=3$ as $x$ approaches $2$ from the left. The second rule gives $4a+2$ at $x=2$ . To make the pieces meet:

$4a+2=3\quad\Longrightarrow\quad a=\frac14$

With $a=1/4$ , both pieces meet at $(2,3)$ .

In an exam, describe this as choosing a parameter so the pieces join continuously. A formal definition of continuity is not needed for this lesson.

How to assess a further model

Half-life: Is repeated proportional decay plausible?
Logarithmic: Are all input values positive, and does the situation show slowing growth?
Sinusoidal: Is the behaviour genuinely periodic, with a reasonably stable amplitude and period?
Logistic: Is there evidence for a limiting capacity?
Piecewise: Are the boundaries and charges or rates clearly defined?

For calculator work, use radian mode for sinusoidal formulas in radians. Set a viewing window that covers the contextual domain, and check any numerical solution against the original model.

IB-style Paper 1 · Skill practice

Question 1 · Half-life

A sample contains $120$ mg of a substance at $t=0$ and has a half-life of $6$ hours.

Write a model for its mass $M(t)$ .
Find $M(18)$ .
Find when $30$ mg remain.

Show solution

$M(t)=120(1/2)^{t/6}$ . Since $18$ hours is three half-lives, $M(18)=120/8=15$ mg. The amount $30$ mg is one-quarter of the initial amount, so it remains after two half-lives, or $12$ hours.

Question 2 · Logarithmic parameters

A function has the form $f(x)=a+b\ln x$ for $x>0$ . Given $f(1)=5$ and $f(e^3)=14$ , find $a$ and $b$ . Then solve $f(x)=20$ .

Show solution

Since $\ln1=0$ , $a=5$ . As $\ln(e^3)=3$ , $14=5+3b$ , giving $b=3$ . Thus $f(x)=5+3\ln x$ . Solving $5+3\ln x=20$ gives $\ln x=5$ , hence $x=e^5$ .

Question 3 · Sinusoidal features

For $f(x)=3\sin\left(2(x-\pi/4)\right)+7$ , state the amplitude, period, phase shift, midline and range.

Show solution

The amplitude is $3$ , period is $2\pi/2=\pi$ , phase shift is $\pi/4$ to the right, and midline is $y=7$ . The range is $4\leq f(x)\leq10$ .

Question 4 · Logistic parameters

A population follows $P(t)=\dfrac{600}{1+Ce^{-0.3t}}$ and initially contains $75$ individuals. Find $C$ , and interpret $600$ .

Show solution

$75=600/(1+C)$ , so $1+C=8$ and $C=7$ . The value $600$ is the model’s carrying capacity: the population approaches $600$ as $t$ increases.

Question 5 · Piecewise rule

A printing service charges €10 for up to $40$ pages, followed by €0.15 for each additional page. Let $n$ be a positive whole number of pages. Write a piecewise cost function and find the cost of $68$ pages.

Show solution

$C(n)= \begin{cases} 10,&1\leq n\leq40,\\ 10+0.15(n-40),&n>40. \end{cases}$

For $68$ pages, the additional count is $28$ , so $C(68)=10+0.15(28)=€14.20$ .

Question 6 · Joining two pieces

Find $k$ so that the pieces of the following function meet at $x=4$ :

$g(x)= \begin{cases} 2x+1,&x<4,\\ kx^2-3,&x\geq4. \end{cases}$

Show solution

The first piece approaches $2(4)+1=9$ . At $x=4$ , the second piece gives $16k-3$ . Setting $16k-3=9$ gives $k=3/4$ . Both pieces then meet at $(4,9)$ .

Applications and interpretation

Application 1 · Medical tracer

A medical tracer has an initial activity of $640$ units and a half-life of $5$ hours.

Construct an activity model $A(t)$ .
Estimate when its activity first falls below $100$ units.
Explain why the model does not predict a finite time when activity becomes exactly zero.

Show solution

$A(t)=640(1/2)^{t/5}$ . The threshold occurs when:

$640\left(\frac12\right)^{t/5}=100$

$t=5\frac{\ln(100/640)}{\ln(1/2)}\approx13.4\text{ hours}$

The activity is below $100$ after approximately $13.4$ hours. An exponential decay model approaches zero as time increases but stays positive at every finite time.

Application 2 · Seasonal temperature

The average temperature of a greenhouse, in degrees Celsius, is approximated by $T(t)=18+5\sin\left(\frac{\pi}{6}(t-3)\right)$ , where $t$ is hours after midnight.

State the period, maximum and minimum temperatures.
Find the first time after $t=3$ when the temperature reaches its maximum.
Find $T(6)$ .

Show solution

The period is $2\pi/(\pi/6)=12$ hours. The maximum is $23^\circ$ C and the minimum is $13^\circ$ C. A maximum occurs when $(\pi/6)(t-3)=\pi/2$ , giving $t=6$ . Also, $T(6)=18+5\sin(\pi/2)=23^\circ$ C.

Application 3 · Growth with a limit

A laboratory population is modelled by $N(t)=\dfrac{800}{1+15e^{-0.5t}}$ , where $t$ is in days.

Find the initial population and carrying capacity.
Estimate when the population reaches $400$ .
Explain why an exponential growth model might give unrealistic long-term predictions for this population.

Show solution

$N(0)=800/(1+15)=50$ , and the carrying capacity is $800$ . To find when $N=400$ , solve $1+15e^{-0.5t}=2$ , giving:

$t=2\ln15\approx5.42\text{ days}$

An unrestricted exponential growth model increases without a limiting value. The logistic model reflects the assumption of limited laboratory resources.

Application 4 · A graduated delivery charge

A company charges €8 for parcels weighing at most $2$ kg. Parcels weighing more than $2$ kg and at most $5$ kg incur €2 for each kilogram above $2$ kg, charged proportionally. Above $5$ kg, the company charges €3 per additional kilogram, also proportionally. Let $m$ be the parcel mass in kilograms.

Construct a piecewise model $C(m)$ for $m>0$ .
Check that its pieces meet at $m=2$ and $m=5$ .
Find the charge for a $7.5$ kg parcel.

Show solution

$C(m)= \begin{cases} 8,&0<m\leq2,\\ 8+2(m-2),&2<m\leq5,\\ 14+3(m-5),&m>5. \end{cases}$

At $m=2$ , both adjoining rules give €8. At $m=5$ , the middle rule gives $8+2(3)=€14$ , matching the starting value of the final rule. Therefore $C(7.5)=14+3(2.5)=€21.50$ .

Application 5 · Deciding between models

A protected habitat has a population of $100$ at the start. Over the next four years, counts are $166$ , $251$ , $349$ and $449$ . A student suggests $E(t)=100(1.65)^t$ . Another suggests $L(t)=\dfrac{600}{1+5e^{-0.5t}}$ .

Find each model’s year 4 prediction to the nearest whole number.
Compare the predictions with the observed year 4 count.
Explain why a good year 4 prediction alone is insufficient to establish which model should be used for many more years.

Show solution

$E(4)=100(1.65)^4\approx741$ . The logistic model predicts $L(4)=600/(1+5e^{-2})\approx358$ . The observed count is $449$ , so both models miss it, with the logistic model closer at year 4. A useful judgement must also consider the other observations, measurement uncertainty and whether a carrying capacity of $600$ is supported by the habitat. Neither model should be extrapolated far ahead solely because it performs better at one point.

Paper 1 checklist

Use radians unless a degree symbol is supplied. State positive-input restrictions for logarithmic models. Interpret carrying capacity as a modelled limit. In piecewise questions, check the correct interval and its boundary. Distinguish a prediction from an observation, and explain whether extrapolation is sensible.

Lesson 2.8 · Further Modelling

Faith connection

Learning objectives

Which model fits the situation?

1. Exponential decay and half-life

2. Natural logarithmic models

3. Sinusoidal models and phase shift

4. Logistic models and carrying capacity

5. Piecewise models

Joining two pieces without a jump

How to assess a further model

IB-style Paper 1 · Skill practice

Question 1 · Half-life

Question 2 · Logarithmic parameters

Question 3 · Sinusoidal features

Question 4 · Logistic parameters

Question 5 · Piecewise rule

Question 6 · Joining two pieces

Applications and interpretation

Application 1 · Medical tracer

Application 2 · Seasonal temperature

Application 3 · Growth with a limit

Application 4 · A graduated delivery charge

Application 5 · Deciding between models

Paper 1 checklist

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Further modelling AI HL

Faith connection

Learning objectives

Which model fits the situation?

1. Exponential decay and half-life

2. Natural logarithmic models

3. Sinusoidal models and phase shift

4. Logistic models and carrying capacity

5. Piecewise models

Joining two pieces without a jump

How to assess a further model

IB-style Paper 1 · Skill practice

Question 1 · Half-life

Question 2 · Logarithmic parameters

Question 3 · Sinusoidal features

Question 4 · Logistic parameters

Question 5 · Piecewise rule

Question 6 · Joining two pieces

Applications and interpretation

Application 1 · Medical tracer

Application 2 · Seasonal temperature

Application 3 · Growth with a limit

Application 4 · A graduated delivery charge

Application 5 · Deciding between models

Paper 1 checklist

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