Numerical Methods in Context – OCR FSMQ Additional Maths

1 Why Numerical Methods Matter in Practice

NM9: apply numerical methods in context where appropriate. The specification's own example is "determine the velocity from a displacement–time curve" — a situation where there is no algebraic formula to differentiate, only measurements.

When numerical methods are the only option

You have data, not a formula — measurements from an experiment
The equation cannot be solved algebraically
The function cannot be integrated by any standard method

In real engineering, science and finance, all three are the normal case rather than the exception. Numerical methods are not second-best — for most real problems, they are the method.

The translation table

Gradient of a displacement–time graph → velocity
Gradient of a velocity–time graph → acceleration
Area under a velocity–time graph → distance
Area under a rate of flow graph → total volume

2 Finding a Rate from Data

Worked Example 1 — The specification's own example

A vehicle's displacement is recorded every second. Estimate its velocity at $t = 3$ s and at $t = 6$ s.

$t$ (s)	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$s$ (m)	$0$	$3$	$11$	$24$	$40$	$57$	$73$

①At $t = 3$, use the straddling chord from $t=2$ to $t=4$:

②$v \approx \dfrac{40 - 11}{4 - 2} = \dfrac{29}{2} = 14.5$ m s⁻¹

③At $t = 6$ there is no data beyond, so only a one-sided chord is available:

④$v \approx \dfrac{73 - 57}{6 - 5} = 16$ m s⁻¹

⑤The $t=3$ estimate is the more reliable; the $t=6$ one is biased towards the average speed over the preceding second.

Always say which chord you used and why. Stating that a straddling chord was chosen because data exists on both sides shows the examiner you understand the accuracy issue, not just the arithmetic.

Worked Example 2 — A rate of cooling

A cup of tea cools as follows. Estimate the rate of cooling at $t = 10$ minutes, and comment on how the rate changes.

$t$ (min)	$0$	$5$	$10$	$15$	$20$
$T$ (°C)	$90$	$70$	$56$	$46$	$39$

①Straddling chord from $t=5$ to $t=15$:

②$\dfrac{46 - 70}{15 - 5} = \dfrac{-24}{10} = -2.4$ °C per minute

③The rate is negative because the tea is cooling. The tea is losing about $2.4$ °C every minute at that moment.

④Early on: $\dfrac{70-90}{5} = -4$ °C min⁻¹. Later: $\dfrac{39-46}{5} = -1.4$ °C min⁻¹.

⑤The cooling is slowing down — because the tea is getting closer to room temperature, so the temperature difference driving the heat loss is smaller.

Interpret the sign and the units. "$-2.4$ °C per minute" carries the direction and the scale; "$2.4$" alone carries neither.

3 Finding a Total from a Rate

Worked Example 3 — Distance from speed readings

A cyclist's speed is recorded every $10$ seconds. Estimate the distance travelled in the minute.

$t$ (s)	$0$	$10$	$20$	$30$	$40$	$50$	$60$
$v$ (m s⁻¹)	$4$	$6$	$9$	$11$	$10$	$8$	$5$

①Distance is the area under the speed–time graph. There are $7$ ordinates, so $6$ strips with $h = 10$.

②Ends: $4 + 5 = 9$

③Middles: $6 + 9 + 11 + 10 + 8 = 44$, doubled gives $88$

④Distance $\approx \tfrac12(10)(9 + 88) = 5 \times 97 = 485$ m

⑤Sense check: the average speed looks to be around $8$ m s⁻¹ over $60$ s, which would give roughly $480$ m ✓

Always sense-check a large numerical answer. A rough mental estimate catches a misplaced factor of ten far more reliably than re-reading your arithmetic.

Worked Example 4 — Volume from a flow rate

Water flows into a tank at the rates shown. Estimate the total volume added in $8$ hours, and say whether your estimate is too high or too low.

$t$ (h)	$0$	$2$	$4$	$6$	$8$
Rate (litres/h)	$120$	$100$	$85$	$75$	$70$

①Four strips of width $h = 2$.

②Ends: $120 + 70 = 190$

③Middles: $100 + 85 + 75 = 260$, doubled gives $520$

④Volume $\approx \tfrac12(2)(190 + 520) = 710$ litres

⑤The successive drops are $20, 15, 10, 5$ — shrinking, so the graph bends upwards (it is falling but levelling off).

⑥So the chords lie above the curve, and $710$ litres is an over-estimate.

Reading the curvature from the differences works even without a formula: if the successive changes are shrinking in size, the graph is bending upwards.

4 Solving Equations That Arise in Context

Worked Example 5 — A design problem

An open cylindrical can of radius $r$ cm and height $2r$ cm must hold $500$ cm³. Find $r$ to $2$ decimal places.

①$V = \pi r^2 (2r) = 2\pi r^3 = 500$

②$r^3 = \dfrac{500}{2\pi} = 79.5775$

③Here the cube root can be taken directly: $r = 4.30$ cm — no numerical method needed.

④But suppose the can had a lid and the surface area were fixed at $400$ cm² instead: $2\pi r^2 + 2\pi r(2r) = 6\pi r^2 = 400$, again solvable directly.

⑤The methods become necessary when the equation mixes powers, as in the next example.

Check first whether algebra will do. Reaching for a numerical method when a direct route exists wastes time and invites error.

Worked Example 6 — An equation that needs a numerical method

A box has a square base of side $x$ cm and height $(10 - x)$ cm. Its volume is $200$ cm³. Find $x$.

①$V = x^2(10-x) = 200$

②$10x^2 - x^3 = 200$, so $x^3 - 10x^2 + 200 = 0$.

③Try factors of $200$: $\mathrm{f}(5) = 125 - 250 + 200 = 75$; $\mathrm{f}(8) = 512 - 640 + 200 = 72$; $\mathrm{f}(10) = 200$. None is zero, so there is no neat root.

④$\mathrm{f}(6) = 216 - 360 + 200 = 56$ and $\mathrm{f}(4) = 64 - 160 + 200 = 104$ — both positive.

⑤$\mathrm{f}(7) = 343 - 490 + 200 = 53$, $\mathrm{f}(9) = 729 - 810 + 200 = 119$. The minimum is around $x \approx 6.7$ with value about $52$ — still positive.

⑥So $\mathrm{f}(x) > 0$ for all $0 < x < 10$: there is no solution in the physically sensible range.

⑦The greatest possible volume is at the maximum of $x^2(10-x)$: $\dfrac{\mathrm{d}V}{\mathrm{d}x} = 20x - 3x^2 = 0$ gives $x = \tfrac{20}{3} = 6.67$, and then $V = 44.44 \times 3.33 = 148$ cm³.

⑧Since the maximum volume is only $148$ cm³, a box of $200$ cm³ with these proportions is impossible.

A numerical search that finds no sign change is still an answer. Combined with the maximum from calculus, it proves the target is unreachable — a genuinely useful modelling conclusion.

5 Quick Reference

Gradient

A rate of change in the context's units.

Area

A total accumulated over the interval.

$s$–$t$ gradient

Velocity.

$v$–$t$ area

Distance.

Rate–time area

Total amount.

Straddling chord

Use it wherever data allows.

Units

Always state them, with the sign.

Curvature from data

Shrinking differences $\Rightarrow$ bends upwards.

Sense-check

Estimate roughly before trusting the answer.

Interpret

Answer in context, not just as a number.

6 Practice Questions

Question 1

What does the gradient of a displacement–time graph represent?

▶ Show solution

The velocity, in units of displacement per unit time — for example m s⁻¹.

Question 2

What does the area under a velocity–time graph represent?

▶ Show solution

The distance travelled (or change in displacement), in metres.

Question 3

Displacement readings are $s = 12$ m at $t = 2$ s and $s = 30$ m at $t = 4$ s. Estimate the velocity at $t = 3$ s.

▶ Show solution

Straddling chord: $\dfrac{30-12}{4-2} = \dfrac{18}{2} = 9$ m s⁻¹

Question 4

A tap's flow rate is $5$, $7$, $8$ litres per minute at $t = 0, 1, 2$ minutes. Use the trapezium rule to estimate the volume delivered.

▶ Show solution

Two strips, $h = 1$.

$\approx \tfrac12(1)\Big[(5+8) + 2(7)\Big] = \tfrac12(27) = 13.5$ litres

Question 5

Temperature readings are $80$ °C at $t=0$ and $65$ °C at $t=4$ min. Estimate the average rate of cooling.

▶ Show solution

$\dfrac{65-80}{4} = \dfrac{-15}{4} = -3.75$ °C per minute.

The negative sign shows the temperature is falling.

Question 6

Explain why an estimate at the last data point is less reliable than one in the middle.

▶ Show solution

At the last point there is no data beyond it, so no straddling chord can be formed.

Only a one-sided chord is available, which reflects the average behaviour over the interval before the point rather than at it, so the errors on the two sides cannot cancel.

Question 7

Speeds of $0$, $8$, $14$, $18$ m s⁻¹ are recorded at $t = 0, 5, 10, 15$ s. Estimate the distance travelled.

▶ Show solution

Three strips, $h = 5$.

Ends: $0 + 18 = 18$. Middles: $8 + 14 = 22$, doubled $= 44$.

$\approx \tfrac12(5)(18+44) = 2.5 \times 62 = 155$ m

Question 8

Using the data in Question 7, is your answer an over- or under-estimate? Justify it.

▶ Show solution

The speed increases by $8$, then $6$, then $4$ — the increases are shrinking.

So the graph is rising but levelling off: it bends downwards, and the chords lie below the curve.

The estimate of $155$ m is therefore an under-estimate.

Question 9

Explain how you would improve the estimate in Question 7 if you could take more readings.

▶ Show solution

Record the speed more often — say every $2.5$ seconds instead of every $5$.

That halves the strip width $h$, and since the trapezium rule's error is roughly proportional to $h^2$, the error would fall by a factor of about four.

The extra ordinates let the chords follow the curve's bend much more closely.

Question 10

A rocket's height above the ground is recorded during the first $10$ seconds of flight:

$t$ (s)	$0$	$2$	$4$	$6$	$8$	$10$
$h$ (m)	$0$	$14$	$52$	$112$	$196$	$300$

(a) Estimate the rocket's velocity at $t = 4$ s and at $t = 8$ s. (b) Use these to estimate the acceleration between those times. (c) The mission requires the rocket to pass $250$ m within $10$ seconds. Use a change of sign argument on the data to confirm this happened, and estimate when. (d) Comment on the reliability of your answer to (c).

▶ Show solution

(a) At $t = 4$, straddling chord from $t=2$ to $t=6$:

$v \approx \dfrac{112 - 14}{6-2} = \dfrac{98}{4} = 24.5$ m s⁻¹

At $t = 8$, straddling chord from $t=6$ to $t=10$:

$v \approx \dfrac{300 - 112}{10-6} = \dfrac{188}{4} = 47$ m s⁻¹

(b) Acceleration is the rate of change of velocity:

$a \approx \dfrac{47 - 24.5}{8 - 4} = \dfrac{22.5}{4} = 5.625$ m s⁻²

This is an average over $4 \leqslant t \leqslant 8$, best regarded as the acceleration at the midpoint $t = 6$ s.

(c) Consider $\mathrm{g}(t) = h(t) - 250$, which is zero when the height is exactly $250$ m.

$\mathrm{g}(8) = 196 - 250 = -54$ (negative)

$\mathrm{g}(10) = 300 - 250 = 50$ (positive)

The sign changes, and height varies continuously with time, so the rocket passed $250$ m somewhere between $t=8$ and $t=10$ s.

To estimate when, interpolate linearly between the two readings. The height must rise $54$ m of the $104$ m gained over those $2$ seconds:

$t \approx 8 + 2 \times \dfrac{54}{104} = 8 + 1.04 = \mathbf{9.0}$ s (2 s.f.)

(d) Two limitations are worth noting.

First, the linear interpolation assumes the height rises at a constant rate between $t=8$ and $t=10$. But the rocket is accelerating, so it actually covers less than half the gap in the first half-second and more later — meaning the true crossing is slightly earlier than $9.0$ s.

Second, the data is only recorded every $2$ seconds. Readings every $0.5$ s would allow a much tighter bracket and remove most of the interpolation error.

The conclusion that the rocket did pass $250$ m is nevertheless secure — that rests only on the change of sign, not on the interpolation.

Numerical Methods in Context (NM9) · OCR FSMQ Additional Maths · Created with MathJax