2.12 Functions in modelling: limitations and refinements — Learn

Prerequisites: completing the square and the discriminant from 2.3; solving quadratics from 2.4; inequalities from 2.6; graph transformations from 2.11.

By the end of this section you can: build a function to model a real situation; state the assumptions and the domain; work out and interpret the greatest value and the zeros; use the model to answer a question in context; and refine a model when a measurement disagrees.

1. The modelling cycle

A model is a mathematical description of a real situation. Modelling runs a cycle.

Translate. Read the situation and choose a function that fits it.
Solve. Use the algebra to find what the question asks.
Interpret. Turn the answer back into the situation, with units.
Refine. Compare with data. If the model is wrong, change an assumption and repeat.

Every model makes assumptions that simplify the real world. A good answer states them, because they are also the model's limitations.

2. A quadratic model

A thrown object, moving under gravity, has a height that is a quadratic in time. This is the standard projectile model.

Example 1. A ball is thrown from $1.5$ m above the ground. Its height in metres after $t$ seconds is

h = 1.5 + 4t - 5t^2.

The coefficient of $t^2$ is negative, so the graph is an upside-down parabola: the ball rises and then falls. The constant $1.5$ is the height at $t = 0$ , so the throw starts from a hand $1.5$ m above the ground.

Find the greatest height. Complete the square:

h = -5\left(t - \frac{2}{5}\right)^2 + \frac{23}{10}.

The largest value is $\frac{23}{10} = 2.3$ m, reached at $t = \frac{2}{5} = 0.4$ s.

Find when the ball lands. Solve $h = 0$ :

5t^2 - 4t - 1.5 = 0.

The discriminant is $16 + 30 = 46$ , so

t = \frac{4 \pm \sqrt{46}}{10}.

The two values are about $-0.278$ and $1.078$ . A negative time is outside the situation, so the ball lands after about $1.08$ s. The domain of the model is $0 \le t \le 1.078$ , and the model is not meant to be used outside it.

3. Reading the model in context

Each feature of the graph has a meaning in the situation.

Feature of the graph	What it says about the ball
the $y$ -intercept	the height it was released from
the vertex	the greatest height and when it happens
the positive root	when it lands
the coefficient of $t^2$	how fast the ball falls, from gravity

Example 2. Use the model to find when the ball is $2$ m high.

Set $h = 2$ :

1.5 + 4t - 5t^2 = 2, \qquad 5t^2 - 4t + 0.5 = 0.

The quadratic formula gives

t = \frac{4 \pm \sqrt{6}}{10},

so $t \approx 0.155$ s or $t \approx 0.645$ s. The ball is $2$ m high twice: once on the way up and once on the way down. Two solutions are expected, and both are realistic.

4. Refining the model

A model is tested by comparing its prediction with a measurement. If the numbers disagree, an assumption is changed and the model is tried again.

Example 3. A sensor measures the ball at $2.28$ m when $t = 0.4$ s, but the model predicts $2.3$ m.

The difference is

|2.28 - 2.3| = 0.02 \text{ m}.

Whether this matters depends on the tolerance: the size of error the question allows. With a tolerance of $0.05$ m, a difference of $0.02$ m is acceptable, and the model is not changed. With a tighter tolerance it would be.

Watch out. A model answers only the question it was built for.

Using it outside its domain. The height model stops when the ball lands. Substituting $t = 3$ gives a negative height, which is not a real position; the formula continues past the end of the situation, but the model does not. Always state the domain and check an answer against it.

Forgetting units. $2.3$ is not an answer on its own; $2.3$ metres is. Write the unit into the final sentence.

Mistaking the model for reality. A prediction that fits one measurement is not a proof that the model is right. Re-check: say what the model assumes, and compare more than one prediction with more than one measurement.

Progress check

Return to the chapter problem. A survey measures the arch again. The height at the centre is $4.2$ m, not $4$ m, and the arch still meets the road at $x = \pm 4$ .

Complete the refined model $y = 4.2 - \frac{4.2x^2}{16}$ and show that it is zero at $x = 4$ .
Work out the height of the refined arch at $x = 1.25$ .
The lorry roof is $3.2$ m. Work out the new clearance and compare it with the clearance from the original model.
Write one sentence about whether the extra $0.2$ m at the centre is worth the change in the model.

You should now be able to

translate a situation into a function;
state the assumptions of a model and its domain;
find a greatest value by completing the square and zeros by solving;
check that each answer is inside the domain and give it a unit;
compare a prediction with a measurement using a tolerance;
refine a model when the data disagree, and say what the refinement changes.