Modelling real-life situations with variables, formulae and graphs, including linear, piecewise linear and exponential growth and decay models, and using the model to make predictions.

What this dot point is asking

The SQA wants you to take a described situation, define sensible variables, and represent it with a formula or a graph. You then use that model to predict values, read information off a graph, and comment on how well the model fits. The standard model families at Higher are linear, piecewise linear, and exponential growth or decay.

Defining variables and building a formula

Modelling starts by naming the quantities. Choose a letter for the input (often time, distance or quantity) and a letter for the output, and state the units. A clear statement such as "let $C$ be the cost in pounds and $m$ the distance in miles" earns communication marks and stops you mixing units later.

A linear model fits any situation with a fixed start and a constant rate: a standing charge plus a unit rate, a tank draining at a steady litres-per-minute, or a wage of a base plus an hourly rate. It has the form $y = mx + c$, where $c$ is the value when $x = 0$ and $m$ is the change in $y$ for each unit increase in $x$.

Piecewise linear models

Many real charges change once a threshold is crossed: an off-peak then peak rate, free units then a charge per unit, or a parking tariff with a flat fee then an hourly rate. A piecewise linear model uses one linear rule below the threshold and a different rule above it.

The second rule must join the first at the threshold so the graph has no jump unless the situation really does jump. In the example above, at $u = 100$ the first rule gives $\pounds 20$ and the second rule gives $20 + 0.35(0) = \pounds 20$, so the pieces meet.

Exponential growth and decay

When a quantity changes by a fixed percentage each period rather than a fixed amount, the model is exponential. A population growing by $8\%$ a year, an investment earning compound interest, or radioactive decay all follow $y = a \times b^{t}$, where $a$ is the starting value and $b$ is the multiplier per period.

A $6\%$ annual fall in value gives $b = 0.94$, so a car worth $\pounds 18\,000$ is modelled by $V = 18000 \times 0.94^{t}$. After $4$ years, $V = 18000 \times 0.94^{4} = 18000 \times 0.7807 = \pounds 14\,053$. The key signal of an exponential model is that successive ratios are constant, whereas a linear model has constant differences.

Reading and using a graph

A graph of the model lets you read values without recalculating. On a linear graph the gradient is the rate and the intercept is the start; on an exponential growth curve the values rise ever more steeply. To predict, read up from the input axis to the curve and across to the output; to solve a target, read across from the output and down. Interpolating inside the data is reliable; extrapolating far beyond it is risky because the model may stop holding.

Evaluating the model

Every model is a simplification, and the SQA rewards a clear comment on its fit and limits.

Try this

Q1. A plumber charges a $\pounds 40$ call-out fee plus $\pounds 35$ per hour. Write a formula for the cost $C$ of a job lasting $h$ hours and find the cost of a $3$ hour job. [2 marks]

• Cue. $C = 40 + 35h$, so for $h = 3$, $C = 40 + 105 = \pounds 145$.

Q2. A sample of $80$ mg of a substance decays by $15\%$ each day. Write a formula for the mass $m$ after $d$ days and find the mass after $5$ days. [3 marks]

• Cue. $b = 0.85$, so $m = 80 \times 0.85^{d}$; after $5$ days $m = 80 \times 0.85^{5} = 35.5$ mg.

Q3. A car park charges $\pounds 2$ for up to $2$ hours, then $\pounds 1.50$ per extra hour. Write the cost for a stay of $t$ hours when $t > 2$. [2 marks]

• Cue. $C = 2 + 1.50(t - 2)$ for $t > 2$; for example a $5$ hour stay costs $2 + 1.50 \times 3 = \pounds 6.50$.