Mathematical modelling

A mathematical model represents a real situation using variables, equations, graphs, diagrams or probability distributions. It deliberately ignores some detail so that mathematics can answer a useful question.

For example, the equation

$$d=vt$$

models distance $d$ travelled at constant speed $v$ for time $t$. It is not universally true. It is useful only when the speed is constant, the units are consistent, and $d$ means distance along the route.

The central habit of modelling is therefore not merely to calculate. It is to ask:

1. What do the symbols mean?
2. Which assumptions make the mathematics possible?
3. Is the result sensible in the original situation?

Prerequisites

You should be able to:

• substitute into and rearrange formulae;
• read tables and graphs;
• round numbers appropriately;
• convert units and compound measures;
• distinguish exact values from approximations.

The modelling cycle

A complete modelling solution moves between the real situation and mathematics.

$$\boxed{ \text{real problem} \longrightarrow \text{assumptions and variables} \longrightarrow \text{mathematical model} \longrightarrow \text{solution} \longrightarrow \text{interpretation and evaluation} }$$

If the answer is not accurate enough for its purpose, refine the assumptions or choose a different model and repeat the cycle.

Stage	Questions to ask
Define	What is known? What must be found? What are the units and allowed values?
Simplify	Which effects are important? Which can reasonably be ignored?
Formulate	Which variables and relationships represent the situation?
Solve	Which algebraic, graphical, numerical or statistical method applies?
Interpret	What does the mathematical answer mean in context?
Validate	Does it fit observations, units, constraints and common sense?
Refine	Which assumption or parameter should change?

In an examination, these stages may be compressed into one question. Your written solution should still make them visible.

Define variables precisely

A variable needs a meaning, a unit and often a domain. Compare

$$x=5$$

with

$$t=5\text{ minutes after the tank begins draining}.$$

The second statement can be interpreted and checked. The first cannot.

Suppose $n$ is the number of complete rows of seats. Then

$$n\in\{0,1,2,\ldots\},$$

not merely $n\geq0$. A model may produce $n=14.7$, but the context permits only a whole number. Whether to use $14$ or $15$ depends on the question, not on the usual rounding rule.

Worked example 1: identify the variables and domain

A taxi charges a fixed fee of $\pounds 3.20$ plus $\pounds 1.80$ per mile. Model the fare for a journey of $m$ miles.

Let

$$m=\text{journey length in miles}, \qquad C=\text{fare in pounds}.$$

The model is

$$\boxed{C=3.20+1.80m}, \qquad m\geq0.$$

Here $3.20$ is the fare at $m=0$, and $1.80$ is the rate of change of fare with distance, measured in pounds per mile.

If the meter records each completed tenth of a mile, the actual domain is discrete:

$$m\in\{0,0.1,0.2,\ldots\}.$$

That operational detail changes the model even though the algebraic formula looks the same.

Assumptions are part of the model

An assumption is a statement accepted while building the model. Assumptions reduce complexity, but each one limits when the result can be trusted.

Common assumptions include:

• an object is a particle, so its size and rotation are ignored;
• a string is light and inextensible;
• air resistance is negligible;
• acceleration is constant;
• observations are independent;
• a sample is representative of a population;
• a percentage rate remains constant;
• a relationship seen within the data continues beyond it.

Good assumptions are specific. Write air resistance is negligible, not conditions are normal.

An assumption is not a fact

The phrase model the car as a particle does not claim that the car has no size. It claims that its size is irrelevant to the quantity being studied. That may be reasonable for a 200 km journey and unreasonable when parking in a 4.8 m space.

Worked example 2: expose the hidden assumptions

A runner completes $400$ m in $50$ s. Using

$$v=\frac{d}{t},$$

we find

$$v=\frac{400}{50}=8\text{ m s}^{-1}.$$

This is the runner’s mean speed, not necessarily the speed at every instant. To use $d=vt$ to predict that the runner covers $640$ m in the next $80$ s, we must assume that the same mean speed continues:

$$d=8\times80=640\text{ m}.$$

The prediction may fail because of fatigue, acceleration from rest, bends or changing conditions. The arithmetic can be correct while the model is poor.

Choose the model from the behaviour

Different patterns suggest different mathematical structures.

Behaviour	Typical model	Recognising feature
constant amount added per unit	$y=a+bx$	constant gradient
constant factor per interval	$y=Ab^x$	constant percentage change
quantity proportional to a power	$y=Ax^n$	scaling $x$ scales $y$ predictably
periodic variation	$y=a+b\sin(cx+d)$	repeating cycle
constant acceleration	$v=u+at$	velocity changes linearly with time
random outcome	probability distribution	probabilities describe uncertainty

This table is a starting point, not proof. Several models may fit a small data set. Context and validation decide which is useful.

Worked example 3: linear or exponential?

Two savings plans both start with $\pounds 1000$.

• Plan A adds $\pounds 60$ each year.
• Plan B adds $6\%$ of the current balance each year.

After $n$ years, Plan A is linear:

$$A_n=1000+60n.$$

Plan B is exponential because each balance is multiplied by $1.06$:

$$B_n=1000(1.06)^n.$$

After $10$ years,

$$A_{10}=1000+60(10)=1600,$$

whereas

$$B_{10}=1000(1.06)^{10}=1790.847\ldots.$$

Thus the models predict $\pounds1600$ and $\pounds1790.85$ respectively.

The phrase adds $6\%$ does not mean add $60$ every year. The percentage acts on a changing balance.

Keep units consistent

Every term added in an equation must have the same units. In

$$s=ut+\frac12at^2,$$

if $u$ is in $\text{m s}^{-1}$, $a$ in $\text{m s}^{-2}$ and $t$ in seconds, then

$$[ut]=(\text{m s}^{-1})(\text{s})=\text{m}$$

and

$$[at^2]=(\text{m s}^{-2})(\text{s}^2)=\text{m}.$$

Both terms have units of length, as $s$ must.

Unit analysis can reject an impossible formula, but it cannot prove a formula correct. Both $s=ut+at^2$ and $s=ut+\frac12at^2$ are dimensionally consistent.

Worked example 4: repair inconsistent units

A cyclist travels at $18\text{ km h}^{-1}$ for $40$ seconds. Find the distance travelled under a constant speed assumption.

Convert the speed first:

$$18\text{ km h}^{-1} =18\times\frac{1000}{3600}\text{ m s}^{-1} =5\text{ m s}^{-1}.$$

Then

$$d=vt=5\times40=200\text{ m}.$$

Therefore the model predicts

$$\boxed{d=200\text{ m}}.$$

Multiplying $18$ directly by $40$ combines hours with seconds and has no valid interpretation.

Solve, then return to the context

A mathematical solution is not automatically a contextual answer. Check:

• sign: can the quantity be negative?
• domain: must it be a whole number or lie in a fixed interval?
• scale: is the order of magnitude plausible?
• precision: do the data justify the stated accuracy?
• meaning: which root or solution answers the question?

Worked example 5: reject an inadmissible solution

The height of a ball is modelled by

$$h=1.2+14t-4.9t^2,$$

where $h$ is in metres and $t$ is seconds after release. Find when the model says the ball reaches the ground.

Set $h=0$:

$$1.2+14t-4.9t^2=0,$$

or

$$4.9t^2-14t-1.2=0.$$

Using the quadratic formula,

$$t=\frac{14\pm\sqrt{(-14)^2-4(4.9)(-1.2)}}{2(4.9)} =\frac{14\pm\sqrt{219.52}}{9.8}.$$

This gives

$$t\approx2.940 \qquad\text{or}\qquad t\approx-0.0833.$$

The model begins at $t=0$, so the negative value describes an extrapolation before release and is outside the contextual domain. Hence

$$\boxed{t\approx2.94\text{ s}}.$$

Do not call the negative root wrong. It solves the equation, but it is inadmissible for this question.

Worked example 6: rounding depends on purpose

A minibus holds $16$ passengers. How many minibuses are required for $93$ passengers?

$$\frac{93}{16}=5.8125.$$

Five minibuses are insufficient, so the answer must be rounded up:

$$\boxed{6\text{ minibuses}}.$$

By contrast, if $93$ identical items are packed into complete boxes of $16$, the number of full boxes is

$$\boxed{5},$$

with $13$ items left. The same calculation leads to different rounding because the questions ask for different things.

Validate and evaluate

Validation compares the model with reality or with information not used to construct it. Useful checks include:

1. substitute a known case;
2. compare predictions with fresh observations;
3. inspect residuals, the differences $\text{observed}-\text{predicted}$;
4. test boundary and extreme cases;
5. check whether parameters have sensible signs and sizes;
6. reconsider the assumptions.

Worked example 7: validate against data

Water depth $d$ in a tank is modelled by

$$d=1.80-0.12t,$$

where $d$ is in metres and $t$ is in minutes. At $t=5$, the observed depth is $1.24$ m.

The predicted depth is

$$d=1.80-0.12(5)=1.20\text{ m}.$$

The residual is

$$\text{observed}-\text{predicted}=1.24-1.20=0.04\text{ m}.$$

So the model underestimates the observed depth by $0.04$ m.

Whether this is acceptable depends on purpose. An error of $4$ cm may be harmless for a rough display and unacceptable for an overflow warning system.

The model also predicts $d=0$ when

$$1.80-0.12t=0,$$

so $t=15$ minutes. For $t>15$, it predicts negative depth, which is physically impossible. Its sensible domain is at most

$$0\leq t\leq15.$$

Validity depends on range and purpose

A model is rarely simply valid or invalid. State where it is useful, how accurate it is, and for what purpose. A local approximation may be excellent near observed values and poor far away.

Interpolation and extrapolation

Interpolation predicts within the range of observed data. Extrapolation predicts outside that range.

If temperatures have been recorded for $10\leq t\leq30$ minutes, estimating at $t=18$ is interpolation, while estimating at $t=80$ is extrapolation.

Extrapolation is usually less reliable because:

• the relationship may change;
• a physical limit may be reached;
• omitted effects may become important;
• small parameter errors can produce large long term errors.

For example, a linear population model with positive gradient eventually predicts unlimited growth. Resource limits make that implausible over a sufficiently long period.

Sensitivity and model refinement

A model is sensitive to a parameter if a small change in that parameter produces a large change in the prediction.

Worked example 8: assess sensitivity

A population after $20$ years is modelled by

$$P=5000(1+r)^{20},$$

where $r$ is the annual growth rate as a decimal.

For $r=0.030$,

$$P=5000(1.03)^{20}=9030.56\ldots.$$

For the slightly larger estimate $r=0.032$,

$$P=5000(1.032)^{20}=9386.88\ldots.$$

A change of only $0.2$ percentage points in the annual rate changes the twenty year prediction by about

$$9386.88-9030.56=356.32.$$

This does not make the model useless. It means the rate must be estimated carefully and long term predictions should be reported with appropriate caution.

Possible refinements include:

• allowing a parameter to vary with time;
• replacing a straight line with a curve;
• splitting the domain into different regimes;
• adding an omitted force or constraint;
• collecting more representative data;
• stating a narrower domain of validity.

Each refinement adds complexity. Prefer the simplest model accurate enough for the intended use.

Common misconceptions

The model gave a number, so the answer is reliable

Calculation only shows what follows from the model. Reliability also depends on assumptions, data quality, domain and purpose.

More decimal places make a prediction more accurate

Decimal places show numerical precision, not model accuracy. If a length was measured as $2.4$ m, reporting a prediction as $17.638291$ m usually gives unjustified precision.

A good fit proves cause

An association can support prediction without establishing causation. A third variable or coincidence may explain the pattern. See correlation and regression.

All solutions of the equation answer the problem

An equation may have negative, non integer or out of range solutions. Apply the contextual domain after solving.

An unrealistic assumption makes the whole model worthless

All models simplify. The proper question is whether the assumption causes unacceptable error for the required purpose.

A model answer structure

For an extended modelling question, use this compact structure:

1. Define: Let $t$ be time in seconds and $h$ be height in metres.
2. Assume: Assume constant acceleration and negligible air resistance.
3. Model: write the equation and its domain.
4. Solve: show the mathematics with units.
5. Interpret: select admissible solutions and answer in context.
6. Evaluate: state a limitation and its likely effect.

A useful evaluation connects cause to consequence. For example, ignoring air resistance makes the predicted range too large because drag reduces the horizontal speed.

Merely writing the model is unrealistic gives no mathematical insight.

Self check

1. Variables and units

A tap fills a container at a constant rate of $0.35$ litres per second. The container initially holds $2.0$ litres.

1. Write a model for the volume $V$ after $t$ seconds.
2. State the units of the gradient.
3. Find the predicted volume after $12$ seconds.

Answer

Let $V$ be volume in litres and $t$ be time in seconds. Then

$$V=2.0+0.35t, \qquad t\geq0,$$

until the container becomes full or the flow changes. The gradient has units litres per second. At $t=12$,

$$V=2.0+0.35(12)=6.2\text{ litres}.$$

2. Interpret a solution

A model for the number of crates required gives $n=23.12$. Each crate holds at most the stated capacity. What value of $n$ should be used?

Answer

Use

$$n=24.$$

The number of crates must be a whole number, and $23$ crates would not provide enough capacity.

3. Evaluate an extrapolation

A linear model fitted to the height of a plant during its first six weeks predicts a height of $9.4$ m after five years. Give two reasons to doubt this prediction.

Answer

Five years lies far outside the observed six week range, so this is substantial extrapolation. The plant’s growth rate is unlikely to remain constant because maturity, seasonal conditions, disease and limited resources can change it. The linear relationship should therefore not be assumed to continue for five years.

4. Compare models

A quantity starts at $80$. Model L adds $12$ each period. Model E increases by $12\%$ each period. Find both predictions after $5$ periods.

Answer

For the linear model,

$$L=80+12(5)=140.$$

For the exponential model,

$$E=80(1.12)^5=140.987\ldots\approx141.$$

The similar five period answers do not make the models equivalent. Their predictions separate increasingly over time.

5. Diagnose a model

A model predicts the stopping distance $s$ of a car from $s=0.08v^2$, where $s$ is in metres and $v$ is in metres per second. A student substitutes $v=72$ for a speed of $72\text{ km h}^{-1}$. Explain and correct the error.

Answer

The formula requires metres per second, but $72$ is in kilometres per hour. Convert:

$$72\text{ km h}^{-1}=72\times\frac{1000}{3600}=20\text{ m s}^{-1}.$$

Therefore

$$s=0.08(20)^2=32\text{ m}.$$

Next steps

• Learn how algebraic structure represents context in functions in modelling.
• Build constant rate and trend models with straight lines.
• Model repeated percentage change in exponential growth and decay.
• Apply modelling assumptions to motion in kinematics language and constant acceleration.
• Study data based model fitting in correlation and regression.
• Develop changing rate models in differential equation models.

The transferable principle is simple: solve the mathematics, then judge the solution as a statement about the real situation.