2.12 Functions in modelling: limitations and refinements — Practise

Work through these on your own. For each one, write the model's domain and one assumption before you calculate. Give every final answer with a unit and one sentence of interpretation. When a measurement disagrees, say which model fits and what the fit does not show.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A ball is thrown upwards. Its height in metres after $t$ seconds is modelled by $h = 1.5 + 4t - 5t^2$ . Complete the square and state the greatest height.
For the model $h = 1.5 + 4t - 5t^2$ , solve $h = 0$ and give the realistic landing time. Give both exact roots first.
For the model $h = 1.5 + 4t - 5t^2$ , work out the height at $t = 0$ and explain what the constant $1.5$ means.
For the model $h = 1.5 + 4t - 5t^2$ , find the two times at which the ball is $2$ m high.
A sensor records the readings $(1, 3)$ , $(2, 5)$ , $(3, 8)$ and $(4, 9)$ . Find the least-squares line $y = a + bx$ for these data.
A model predicts a value of $2.3$ . A measurement gives $2.28$ . Decide whether the prediction is within a tolerance of $0.05$ .
The cost in pounds of making $n$ items is modelled by $C = 50 + 12n + 0.5n^2$ . Work out the cost of making $20$ items.
For the cost model $C = 50 + 12n + 0.5n^2$ , find the smallest whole number of items for which the cost is at least £250.
Show that the two forms of the height model $h = 1.5 + 4t - 5t^2$ and $h = -5\left(t - \frac{2}{5}\right)^2 + \frac{23}{10}$ are the same.
A model for the average cost per item is $A = \dfrac{100}{x} + 2x$ pounds, for $x > 0$ . Work out the average cost when $x = 5$ and when $x = 10$ .
A refined model for the ball is $h = 1.6 + 4.1t - 5t^2$ . Work out its height at $t = 0.4$ and compare it with the original model.
A measurement gives $2.28$ m. Model A predicts $2.30$ m and model B predicts $2.44$ m. With a tolerance of $0.05$ , decide which model fits the measurement, and say what the decision does not show.

Answers and concise solutions

1. The completed square and the greatest height

Answer: Complete the square: $h = -5\left(t - \frac{2}{5}\right)^2 + \frac{23}{10}$ . The squared term is subtracted, so $h$ is greatest when $t = \frac{2}{5}$ . The greatest height is $\frac{23}{10} = 2.3$ m. The coefficient of $t^2$ is negative, so the parabola opens downwards and the vertex is a maximum.

2. The roots and the realistic time

Answer: Solve $5t^2 - 4t - 1.5 = 0$ with the formula. The discriminant is $16 + 30 = 46$ , so $t = \frac{4 \pm \sqrt{46}}{10}$ . Numerically these are about $-0.278$ and $1.078$ . A negative time is outside the model, so the ball lands at about $1.08$ s. The domain of the model is $0 \le t \le \frac{4 + \sqrt{46}}{10}$ .

3. The initial height and its meaning

Answer: At $t = 0$ , $h = 1.5$ . The constant term is the height at the instant the model starts, so the ball is released from $1.5$ m above the ground. This is the $y$ -intercept of the graph.

4. The two times

Answer: Set $1.5 + 4t - 5t^2 = 2$ and rearrange: $5t^2 - 4t + 0.5 = 0$ . The discriminant is $16 - 10 = 6$ , so $t = \frac{4 \pm \sqrt{6}}{10}$ . Numerically the two times are about $0.155$ s and $0.645$ s. The ball is at $2$ m once on the way up and once on the way down, so two times is the expected answer.

5. The gradient and intercept

Answer: The least-squares line has gradient $b = \frac{21}{10} = 2.1$ and intercept $a = 1$ , so $y = 1 + 2.1x$ . The gradient is positive, so the reading increases as the input increases. The line is a fit to these four points only, and it says nothing about readings outside them.

6. The decision

Answer: The difference is $|2.28 - 2.3| = 0.02$ , which is less than $0.05$ . So the prediction is within the stated tolerance. A tolerance is part of the claim: without one, a decimal prediction and a measurement can never be compared fairly.

7. The cost

Answer: Substitute $n = 20$ : $C = 50 + 12 \times 20 + 0.5 \times 20^2 = 50 + 240 + 200 = 490$ . The cost is £490. The fixed term $50$ is the set-up cost, and the $n^2$ term makes the cost grow faster as the run gets longer.

8. The number of items

Answer: Set $50 + 12n + 0.5n^2 = 250$ : multiply by $2$ and rearrange to $n^2 + 24n - 400 = 0$ . The formula gives $n = -12 \pm 4\sqrt{34}$ . Only the positive root is realistic: $n = -12 + 4\sqrt{34} \approx 11.32$ . Since $n$ is a whole number of items, the smallest count is $12$ . The negative root is rejected because a number of items cannot be negative.

9. The identity

Answer: Expand the completed-square form: $-5\left(t - \frac{2}{5}\right)^2 = -5\left(t^2 - \frac{4}{5}t + \frac{4}{25}\right) = -5t^2 + 4t - \frac{4}{5}$ . Add $\frac{23}{10}$ : $-5t^2 + 4t - \frac{4}{5} + \frac{23}{10} = -5t^2 + 4t + \frac{3}{2}$ . Since $1.5 = \frac{3}{2}$ , this is the expanded model. The two forms are equal for every $t$ .

10. The two average costs

Answer: At $x = 5$ : $A = \frac{100}{5} + 10 = 20 + 10 = 30$ . At $x = 10$ : $A = \frac{100}{10} + 20 = 10 + 20 = 30$ . Both runs give the same average cost, which suggests the model has a smallest value somewhere between the two. The model is not defined at $x = 0$ , because it would divide by zero, and a run of zero items has no average cost; the domain is $x > 0$ .

11. The refined value

Answer: Substitute $t = 0.4$ : $h = 1.6 + 4.1 \times 0.4 - 5 \times 0.4^2 = 1.6 + 1.64 - 0.8 = 2.44$ m. The original model gives $2.3$ m at the same time. The refined model gives a higher peak, because the constant and the linear coefficient were increased after comparing with measurements.

12. The decision and its limit

Answer: Model A is within the tolerance: $|2.30 - 2.28| = 0.02 < 0.05$ . Model B is not: $|2.44 - 2.28| = 0.16 > 0.05$ . So model A fits this measurement and model B does not. This shows only that A is closer at this one point. One measurement cannot show that a model is correct in general; more data, and a thought about the assumptions, are needed before either model can be trusted.