Functions in mathematical modelling

A function model represents how one quantity depends on another. You choose a function whose behaviour matches the situation, define its variables and domain, estimate any parameters, then test whether its predictions are sensible.

For example, if a taxi fare has a fixed charge of £3.20 and costs £1.80 per mile, the fare for a journey of (d) miles is

[ C(d)=3.20+1.80d, \qquad d\geq 0. ]

The formula is only one part of the model. The variables, units, domain and assumptions are equally important.

Prerequisites

You should be comfortable with:

The modelling cycle

Most modelling questions follow the same cycle.

Define the quantities. State what each variable measures and include units.
Simplify reality. Decide which effects matter and which can be ignored.
Choose a function family. Use the context, a graph, or data to decide whether a linear, quadratic, exponential or other model is plausible.
Find the parameters. Use given information or data to determine constants.
Use the model. Calculate, solve, differentiate, integrate, or interpret its graph.
Check the result. Test units, signs, scale and whether the input lies in the model’s domain.
Refine the model. Compare predictions with reality and change assumptions or the function if necessary.

Modelling is not the claim that reality is a formula. It is the deliberate use of a simplified mathematical description for a stated purpose.

Choosing an appropriate function

The form of a function determines its behaviour.

Model	Typical form	Behaviour it can represent
Linear	(y=a+bx)	Constant absolute change per unit of (x)
Quadratic	(y=ax^2+bx+c)	Curvature with one turning point
Power	(y=ax^n)	Scaling laws, areas, volumes and inverse powers
Reciprocal	(y=\dfrac{a}{x})	Inverse proportionality
Exponential	(y=ab^x) or (y=Ae^{kx})	Constant proportional or percentage change
Piecewise	Different rules on different intervals	Thresholds, tariffs and changing regimes

These are clues, not guarantees. A curve that looks quadratic over a short interval may behave very differently outside that interval.

Example 1: building and interpreting a linear model

A water tank contains (120) litres and is filled at a constant rate of (8) litres per minute.

Let (V(t)) be the volume in litres after (t) minutes. The initial value is (120), and the constant rate of change is (8), so

[ V(t)=120+8t. ]

Suppose the tank has capacity (400) litres. The physical domain is not simply (t\geq 0). The model only applies until the tank is full:

[ 120+8t\leq 400 ]

[ 8t\leq 280 ]

[ 0\leq t\leq 35. ]

Therefore a complete model is

[ V(t)=120+8t, \qquad 0\leq t\leq 35. ]

At (t=20),

[ V(20)=120+8(20)=280, ]

so the model predicts (280) litres.

Assumptions and limitations

This model assumes that:

the inflow remains exactly (8) litres per minute
no water leaks or evaporates
the tank does not overflow
time starts when the tank contains exactly (120) litres

The formula predicts (V(50)=520), but that is not a meaningful prediction because (50) lies outside the model’s domain. Substituting into a formula is not enough. You must check whether the substitution is valid in context.

Example 2: finding a quadratic model from information

The height of a ball is modelled by

[ h(t)=at^2+bt+c, ]

where (h) is measured in metres and (t) in seconds. The ball starts (1.5) m above the ground, has height (9.5) m after (1) second, and height (12.5) m after (2) seconds.

Use each observation to form an equation.

At (t=0):

[ h(0)=c=1.5. ]

At (t=1):

[ a+b+1.5=9.5 ]

[ a+b=8. \tag{1} ]

At (t=2):

[ 4a+2b+1.5=12.5 ]

[ 4a+2b=11. \tag{2} ]

Doubling equation (1) gives (2a+2b=16). Subtracting this from equation (2):

[ 2a=-5, ]

so (a=-2.5). Then

[ b=8-(-2.5)=10.5. ]

The model is

[ h(t)=-2.5t^2+10.5t+1.5. ]

Because the coefficient of (t^2) is negative, the graph has a maximum. Completing the square gives

[ h(t)=-2.5(t-2.1)^2+12.525. ]

The model therefore predicts a maximum height of (12.525) m at (t=2.1) s.

To find when the ball reaches the ground, solve (h(t)=0):

[ -2.5t^2+10.5t+1.5=0. ]

Using the quadratic formula gives (t\approx -0.138) or (t\approx 4.338). The negative solution is mathematically valid for the extended curve but not relevant after the ball is released. A reasonable physical domain is therefore

[ 0\leq t\leq 4.338. ]

What the model omits

This simple quadratic treats acceleration as constant and ignores air resistance, wind, spin and the size of the ball. It may be accurate enough over a few seconds, but it is not a universal description of flight.

Example 3: direct and inverse proportion

The energy (E) stored in a stretched spring is modelled as proportional to the square of its extension (x):

[ E\propto x^2. ]

Therefore

[ E=kx^2 ]

for some constant (k). If an extension of (0.04) m stores (0.12) J, then

[ 0.12=k(0.04)^2, ]

[ k=\frac{0.12}{0.0016}=75. ]

The fitted model is

[ E=75x^2. ]

At (x=0.06) m,

[ E=75(0.06)^2=0.27\text{ J}. ]

Notice that increasing (x) from (0.04) to (0.06) multiplies it by (1.5), so energy is multiplied by

[ 1.5^2=2.25. ]

This is not constant absolute change, so a linear model would be inappropriate.

For inverse proportion, (y\propto \frac1x), so (y=\frac{k}{x}). Doubling (x) halves (y). In both cases, the constant of proportionality has units. Here, since (E=kx^2), (k) has units (\text{J m}^{-2}).

Example 4: exponential growth from a percentage rate

A culture contains (800) cells and grows by (18%) per hour. A percentage increase acts on the current quantity, so the hourly multiplier is

[ 1+\frac{18}{100}=1.18. ]

After (t) hours, the model is

[ N(t)=800(1.18)^t, \qquad t\geq 0. ]

After (6) hours,

[ N(6)=800(1.18)^6\approx 2159.64. ]

Since a cell count is discrete, report approximately (2160) cells rather than (2159.64) cells.

To estimate when the culture first exceeds (5000) cells, solve

[ 800(1.18)^t>5000. ]

Taking logarithms:

[ t\log(1.18)>\log(6.25), ]

[ t>\frac{\log(6.25)}{\log(1.18)}\approx 11.08. ]

The continuous model reaches (5000) after about (11.08) hours. If measurements are made only at whole hours, the first recorded value above (5000) occurs at (t=12).

Why indefinite exponential growth is unrealistic

The model assumes a constant percentage growth rate and unlimited resources. Eventually, lack of nutrients, waste accumulation and limited space reduce growth. A more refined model might use a function that approaches a maximum population.

Example 5: a piecewise pricing model

A delivery company charges £5 for parcels up to (2) kg, then £1.50 for each kilogram above (2) kg, up to a maximum mass of (10) kg.

Let (m) be the parcel mass in kilograms. The cost is

[ C(m)= \begin{cases} 5, & 0<m\leq 2,\ 5+1.5(m-2), & 2<m\leq 10. \end{cases} ]

The term (m-2) is essential. It counts only the mass above (2) kg. Using (5+1.5m) would incorrectly charge the additional rate on the whole mass.

For a (6.4) kg parcel,

[ C(6.4)=5+1.5(6.4-2)=£11.60. ]

At the boundary (m=2), the first rule gives (C(2)=5). The second expression would also approach (5), so there is no sudden jump in cost. Real companies may round mass up to a billing unit, in which case a step function would model the tariff more accurately.

Example 6: testing and refining a model

A student models the stopping distance (D), in metres, of a car travelling at speed (v), in metres per second, by

[ D(v)=0.7v+0.05v^2. ]

The linear term can represent thinking distance, while the quadratic term can represent braking distance.

At (v=20):

[ D(20)=0.7(20)+0.05(20)^2=14+20=34\text{ m}. ]

At (v=40):

[ D(40)=0.7(40)+0.05(40)^2=28+80=108\text{ m}. ]

Doubling speed does not double stopping distance because the quadratic term becomes four times as large.

Suppose an observed stopping distance at (20\text{ m s}^{-1}) is (42) m. The residual is

[ \text{residual}=\text{observed}-\text{predicted}=42-34=8\text{ m}. ]

A positive residual means the model underestimates this observation. One residual does not prove that the model is poor. However, a consistent pattern in residuals suggests a structural problem. Possible refinements include changing the coefficients for wet roads, including reaction time explicitly, or introducing variables for tyre condition and road gradient.

Interpolation and extrapolation

Suppose a model is fitted using temperatures between (5^\circ\text{C}) and (30^\circ\text{C}).

Predicting at (18^\circ\text{C}) is interpolation because the input lies within the observed range.
Predicting at (45^\circ\text{C}) is extrapolation because the input lies outside the observed range.

Extrapolation is usually less reliable. A model can fit known data closely while behaving absurdly outside that range. When giving a prediction, ask both:

Is the input inside the mathematical domain of the function?
Is it inside the practical range for which the model was developed?

These are different questions. The function (f(x)=2x+1) is mathematically defined for every real (x), but a model using it may only be valid for (0\leq x\leq 10).

Checking a function model

Use several independent checks.

Units

Every term added together must have the same units. In

[ s(t)=4+3t+2t^2, ]

if (s) is measured in metres and (t) in seconds, then (4) has units m, (3) has units (\text{m s}^{-1}), and (2) has units (\text{m s}^{-2}). The numerical constants hide these units, but they still matter.

Initial value

Evaluate the model at the starting input. For (N(t)=800(1.18)^t),

[ N(0)=800, ]

which agrees with the stated initial population.

Direction and shape

If the quantity should increase, check that the function increases on the relevant domain. If growth should slow, a model whose gradient increases may have the wrong shape.

Boundary behaviour

Check what happens near the endpoints. A reciprocal model may become unbounded as (x\to 0), even when the real quantity cannot.

Scale and rounding

A population cannot be negative, a probability must lie between (0) and (1), and a count usually requires a whole number. Round only at the end unless the context requires values to change in discrete steps.

Common misconceptions

Treating every increasing pattern as linear

Linear growth adds the same amount in equal intervals. Exponential growth multiplies by the same factor. The sequences

[ 10,15,20,25 ]

and

[ 10,15,22.5,33.75 ]

both increase, but only the first has a constant difference. The second has a constant ratio of (1.5).

Forgetting to define variables

Writing (C=3.2+1.8d) without saying what (C) and (d) mean makes the model incomplete. State quantities and units.

Confusing a parameter with a variable

In (f(x)=ax+b), (x) is the input variable. Once a particular model is chosen, (a) and (b) are fixed parameters. Different parameter values produce different members of the same function family.

Ignoring the contextual domain

A negative time, fractional person, or temperature beyond the experiment’s range may be algebraically possible but contextually meaningless.

Claiming that a model is simply correct or incorrect

Models are judged by purpose and accuracy. A simple linear model may be useful for short term prediction even when the underlying process is not exactly linear.

Giving an unqualified calculator answer

Write what the number means, include units, and state whether it is an estimate. For discrete quantities, explain the direction of rounding when it affects the conclusion.

Practice questions

1. Linear model

A printer costs £240 to hire plus £0.035 per page. Write a model for the total cost (C(n)) of printing (n) pages. Find the cost of (12{,}000) pages and state a suitable domain if at most (50{,}000) pages may be printed.

Answer

[ C(n)=240+0.035n, \qquad 0\leq n\leq 50{,}000, ]

where (n) is a whole number. Then

[ C(12{,}000)=240+0.035(12{,}000)=£660. ]

2. Comparing models

Two savings plans are modelled by

[ A(t)=500+75t ]

and

[ B(t)=500(1.08)^t, ]

where (t) is measured in years. Explain the difference between the two growth assumptions and calculate both balances after (5) years.

Answer

(A) adds a constant £75 each year. (B) increases by (8%) of its current balance each year.

[ A(5)=500+75(5)=£875, ]

[ B(5)=500(1.08)^5\approx £734.66. ]

Although (A) is larger after (5) years, exponential growth will eventually overtake linear growth if both models continue unchanged.

3. Model criticism

The mass of a puppy is modelled by (M(t)=3(1.12)^t), where (M) is in kilograms and (t) is age in weeks. Give two reasons why the model should not be used for the puppy’s entire life.

Answer

The model assumes a constant weekly percentage increase and therefore predicts unlimited exponential growth. Real growth slows and eventually stops. It may also be invalid near birth or in adulthood because nutrition, health and biological development alter the growth rate.

4. Quadratic optimisation

The daily profit from selling (x) units is modelled by

[ P(x)=-2x^2+240x-3000, ]

where (0\leq x\leq 100). Find the value of (x) that maximises the modelled profit, and find that profit.

Answer

Complete the square:

[ P(x)=-2(x-60)^2+4200. ]

The maximum occurs at (x=60), which lies in the stated domain. The maximum modelled daily profit is £4200.

Summary

A function model connects an input quantity to an output quantity using a simplified mathematical rule. A strong model answer should:

define variables and units
justify the function family from the situation or data
determine parameters accurately
state a realistic contextual domain
interpret calculations in words
identify assumptions and limitations
distinguish interpolation from extrapolation
check units, boundaries, scale and rounding
refine the model when predictions and observations disagree

The most important habit is to move repeatedly between mathematics and context. Build the formula from the situation, perform the mathematics, then translate the result back and ask whether it makes sense.

What to learn next

Explore composite and inverse functions when models involve several linked processes.
Learn about growth and decay for proportional change over time.
Study linearising data to test power and exponential models.
Apply these ideas to series in modelling and interpreting context.