2.12 Functions in modelling: limitations and refinements — Practise with help

Work each one first, closed book. For every model, write down the domain and one assumption before you calculate. At the end, turn the number back into a sentence about the situation, with a unit.

Warm-up: remember

Work out $2^2$ and $(-3)^2$ .
Solve $t^2 - 4 = 0$ .
Complete the square for $t^2 - 6t + 5$ .

Answers

$2^2 = 4$ and $(-3)^2 = 9$ .
$t^2 = 4$ , so $t = 2$ or $t = -2$ .
$t^2 - 6t + 5 = (t - 3)^2 - 4$ .

If question 3 was hard, revisit completing the square: half of $-6$ is $-3$ .

Practise with help

1. A stone is dropped from a cliff. Its distance fallen in metres after $t$ seconds is modelled by $d = 5t^2$ . Work out the distance fallen after $3$ seconds.

Hint

Substitute $t = 3$ into the model.

Next step

Work out $5 \times 3^2$ .

Full working

Substitute $t = 3$ :

d = 5 \times 3^2 = 5 \times 9 = 45.

The stone has fallen $45$ m. The domain is $t \geq 0$ , because a negative time has no meaning here. The model assumes no air resistance.

2. For the model $d = 5t^2$ , work out the time taken to fall $80$ m.

Hint

Set $d = 80$ and solve for $t$ . A time is not negative.

Next step

$5t^2 = 80$ , so $t^2 = 16$ .

Full working

Set the distance to $80$ :

5t^2 = 80, \qquad t^2 = 16.

The solutions are $t = 4$ or $t = -4$ . A fall cannot happen before it starts, so $t \geq 0$ and the answer is $t = 4$ seconds. This is an example of rejecting a solution that is outside the domain.

3. The height of a ball is modelled by $h = 2 + 6t - 5t^2$ . Complete the square and state the greatest height.

Hint

Take the factor $-5$ from the $t$ terms, or complete the square directly and watch the signs.

Next step

$h = -5\left(t^2 - \frac{6}{5}t\right) + 2$ . Complete the square inside the bracket.

Full working

Take out the factor $-5$ :

h = -5\left(t^2 - \frac{6}{5}t\right) + 2.

Complete the square inside the bracket:

t^2 - \frac{6}{5}t = \left(t - \frac{3}{5}\right)^2 - \frac{9}{25}.

Substitute back:

h = -5\left(t - \frac{3}{5}\right)^2 + \frac{9}{5} + 2 = -5\left(t - \frac{3}{5}\right)^2 + \frac{19}{5}.

The greatest height is $\frac{19}{5} = 3.8$ m, reached at $t = 0.6$ s.

4. For the model $h = 2 + 6t - 5t^2$ , find when the ball hits the ground.

Hint

The ground is $h = 0$ . Solve the quadratic and reject the negative root.

Next step

Solve $5t^2 - 6t - 2 = 0$ with the formula. The discriminant is $36 + 40 = 76$ .

Full working

Set $h = 0$ and rearrange:

5t^2 - 6t - 2 = 0.

The formula gives

t = \frac{6 \pm \sqrt{36 + 40}}{10} = \frac{6 \pm \sqrt{76}}{10}.

Now $\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}$ , so $t = \frac{3 \pm \sqrt{19}}{5}$ . Numerically, $t \approx 1.472$ or $t \approx -0.272$ . The negative value is outside the domain, so the ball lands after about $1.47$ s.

5. A model predicts a value of $12.4$ . A measurement gives $12.1$ . With a tolerance of $0.5$ , decide whether the model is acceptable.

Hint

Work out the size of the difference first, then compare it with the tolerance.

Next step

$|12.1 - 12.4| = 0.3$ . Compare $0.3$ with $0.5$ .

Full working

The difference is

|12.1 - 12.4| = 0.3.

This is less than the tolerance $0.5$ , so the prediction is within the agreed error and the model is acceptable at this point. A single fit does not prove the model is right; it only fails to rule it out here.

6. A company models its cost for $n$ items as $C = 80 + 5n + 0.2n^2$ pounds, for whole numbers $n \geq 1$ . Work out the cost of $15$ items, and explain one limitation of the model.

Hint

Substitute $n = 15$ . For the limitation, think about whether the $n^2$ term keeps the cost realistic for very large $n$ .

Next step

$C = 80 + 5 \times 15 + 0.2 \times 15^2$ . Work out each term.

Full working

Substitute $n = 15$ :

C = 80 + 5 \times 15 + 0.2 \times 225 = 80 + 75 + 45 = 200.

The cost is £200. One limitation: the $n^2$ term means the cost grows faster and faster as $n$ increases. For a large enough $n$ this stops being realistic, because a factory does not pay ever-increasing costs per item. The model is meant for the range of $n$ where it was fitted.