What is 9.5: Numerical methods in context?

A contextual numerical model translates a physical condition, such as a target value or accumulated quantity, into a root, recurrence or numerical integral.

How many marks are 9.5 questions worth?

9.5 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 9.5 appeared in?

We have not yet indexed a verified past-paper appearance for 9.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

9.5

Use numerical methods to solve problems in context.

Draft — not yet indexed

Numerical methods in context

Worked answers and methods for 9.5 on Edexcel A-level Maths 9MA0.

Explanation

A contextual numerical model translates a physical condition, such as a target value or accumulated quantity, into a root, recurrence or numerical integral.
Define the function and units before applying the method, choose an interval or starting value that is meaningful in the context, and state the requested accuracy.
For a threshold $M(t)=C$ , solve $M(t)-C=0$ numerically; for a varying rate $q(t)$ , approximate the accumulated change with $\int q(t)\,dt$ .
A calculator value is not a complete contextual answer: a common error is to omit units, ignore the model's domain or round in a way that makes a safety decision invalid.

Worked example

A culture is modelled by $P(t)=200e^{0.3t}$ , where $t$ is in hours. Use Newton-Raphson on $f(t)=200e^{0.3t}-1000$ with $t_0=5$ to find $t_1$ and $t_2$ . Hence estimate when the culture reaches $1000$ cells.

1.Differentiate to get $f'(t)=60e^{0.3t}$ .
2.Use $t_{n+1}=t_n-\frac{200e^{0.3t_n}-1000}{60e^{0.3t_n}}$ .
3.Substitution of $t_0=5$ gives $t_1=5.385502\ldots$ , and the next substitution gives $t_2=5.364857\ldots$ .
4.The positive time is therefore about $5.365$ hours.

Answer: $t_1=5.38550$ and $t_2=5.36486$ .; The culture reaches $1000$ cells after approximately $5.365$ hours.

Common mistakes

Don't select a root that satisfies the equation but lies outside the interval allowed by the context.
Don't carry out a numerical iteration but give no contextual time, units or check that the root is meaningful.

Exam tip

After numerical iteration, translate the root back into the context and report a sensible accuracy with units.

Worked practice

Tier 1 · Easy

A greenhouse temperature is modelled by

T(t)=18+4t-t^2

degrees Celsius, where

t

is hours after noon. Show that the temperature returns to

20^\circ\mathrm{C}

at a time between

3

and

4

hours after noon.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$T(3)-20=1$ and $T(4)-20=-2$ , so the return time lies between $3$ and $4$ hours after noon.	2
Notes
Define $f(t)=T(t)-20=-2+4t-t^2$ . Then $f(3)=-2+12-9=1$ and $f(4)=-2+16-16=-2$ . The model is continuous and the sign changes, so a solution occurs in $(3,4)$ .

(2 marks)

Tier 2 · Standard

A particle's height is modelled by

h(t)=20t-4.9t^2

metres. Show that the positive time when the particle returns to the ground lies in

(4.0,4.1)

seconds and, by considering the sign of

h(4.05)

, show that it lies in

(4.05,4.10)

seconds.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$h(4.0)=1.6>0$ and $h(4.1)=-0.369<0$ $h(4.05)=0.62775>0$ , so the root lies in $(4.05,4.10)$ .	4
Notes
Calculate $h(4.0)=20(4)-4.9(4^2)=1.6$ and $h(4.1)=20(4.1)-4.9(4.1^2)=-0.369$ . Continuity gives a root in $(4.0,4.1)$ . The midpoint is $4.05$ , and $h(4.05)=20(4.05)-4.9(4.05^2)=0.62775>0$ . The sign change is therefore between $4.05$ and $4.10$ seconds.

(4 marks)

Tier 3 · Hard

Water enters a tank at rates

10,14,18,21,19,15,11

litres per minute at times

0,5,10,15,20,25,30

minutes. The tank initially contains

350

litres and has capacity

800

litres. Use the trapezium rule to estimate whether the tank overflows during the

30

minutes, and estimate any excess at the end.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Estimated inflow $=487.5$ litres and estimated final volume $=837.5$ litres. The model predicts overflow, with $37.5$ litres in excess of capacity at $30$ minutes.	5
Notes
With spacing $h=5$ , the trapezium estimate is $\frac52[10+11+2(14+18+21+19+15)]=\frac52(195)=487.5$ litres. Adding the initial $350$ litres gives $837.5$ litres, which is $837.5-800=37.5$ litres above capacity, so the tank must have overflowed by the end.

(5 marks)

Tier 1 · Easy

A sensor output is modelled by

S(t)=t^3+t

volts for

t\geq0

, where

t

is measured in seconds. Show that the output reaches

20

volts at a time between

2

and

3

seconds.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$S(2)-20=-10$ and $S(3)-20=10$ , so the model reaches $20$ volts for some $t\in(2,3)$ seconds.	2
Notes
Let $f(t)=S(t)-20=t^3+t-20$ . Then $f(2)=8+2-20=-10$ and $f(3)=27+3-20=10$ . The model is continuous, so the change of sign guarantees a solution between $2$ and $3$ seconds.

(2 marks)

Tier 2 · Standard

A curing model predicts that the completion time

t

hours satisfies

t=2.4+0.35\ln(t+1)

. Starting with

t_0=3

, use

t_{n+1}=2.4+0.35\ln(t_n+1)

to find

t_1

t_2

and

t_3

, giving each iterate to

6

decimal places. Hence give the predicted completion time to

3

decimal places.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$t_1=2.885203$ , $t_2=2.875011$ and $t_3=2.874092$ . The predicted completion time is $2.874$ hours.	4
Notes
Successive substitution gives $t_1=2.885203\ldots$ , $t_2=2.875011\ldots$ and $t_3=2.874092\ldots$ . Continuing gives $t_4=2.874009\ldots$ and the values settle to $2.8740007\ldots$ . Therefore the model predicts a completion time of $2.874$ hours to $3$ decimal places.

(4 marks)

Tier 3 · Hard

The volume of liquid in a vessel is modelled by

V(h)=h^3+4h^2

litres when the depth is

h

metres, for

h>0

. Show that the depth at which

V=70

litres lies between

3

m and

4

m. Starting with

h_0=3.2

, apply Newton-Raphson twice, giving each iterate to

6

decimal places, and hence give the required depth to

3

significant figures.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$V(3)-70=-7$ and $V(4)-70=58$ , so the depth lies in $(3,4)$ m. $h_1=3.133807$ , $h_2=3.132719$ , so the required depth is $3.13$ m to $3$ significant figures.	6
Notes
Set $f(h)=h^3+4h^2-70$ . Since $f(3)=27+36-70=-7$ and $f(4)=64+64-70=58$ , continuity gives a root in $(3,4)$ . Also $f'(h)=3h^2+8h$ , so $h_{n+1}=h_n-(h_n^3+4h_n^2-70)/(3h_n^2+8h_n)$ . From $h_0=3.2$ , this gives $h_1=3.1338068\ldots=3.133807$ ( $6$ d.p.) and then $h_2=3.1327194\ldots=3.132719$ ( $6$ d.p.). Thus the required positive depth is $3.13$ m to $3$ significant figures.

(6 marks)

Tier 2 · Standard

A radio tag's reliable range

R

metres is modelled by

R=40+4\cos(R/20)

. Starting with

R_0=45

, use

R_{n+1}=40+4\cos(R_n/20)

to find

R_1

R_5

, giving each iterate to

4

decimal places. Use unrounded values in your working. Hence estimate the reliable range to the nearest

0.1

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$R_1=37.4873$ , $R_2=38.8043$ , $R_3=38.5557$ , $R_4=38.6022$ and $R_5=38.5935$ . The reliable range is approximately $38.6$ m.	4
Notes
Substituting unrounded values successively gives $R_1=37.4873055\ldots$ , $R_2=38.8042885\ldots$ , $R_3=38.5557086\ldots$ , $R_4=38.6021810\ldots$ and $R_5=38.5934763\ldots$ . The alternating values are settling near $38.6$ , so the model gives a reliable range of approximately $38.6$ m to the nearest $0.1$ m.

(4 marks)

Tier 3 · Hard

The distance

D

metres travelled by an automated trolley after

t

seconds is modelled by

D(t)=5t+8\sin(0.5t)

for

0\leq t\leq8

. Show that there is a unique time in

(5,6)

at which

D=30

. Derive a Newton-Raphson recurrence and, starting with

t_0=5.5

, find

t_1

and

t_2

5

decimal places. Use unrounded values in your working. Hence give the time to

3

significant figures.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$D(5)<30<D(6)$ and $D'(t)>0$ , so there is exactly one required time in $(5,6)$ . $t_{n+1}=t_n-\dfrac{5t_n+8\sin(0.5t_n)-30}{5+4\cos(0.5t_n)}$ ; $t_1=5.07531$ and $t_2=5.12236$ . The time is $5.12$ s to $3$ significant figures.	6
Notes
Let $f(t)=5t+8\sin(0.5t)-30$ . Then $f(5)=8\sin2.5-5<0$ and $f(6)=8\sin3>0$ , so continuity gives a root in $(5,6)$ . Also $f'(t)=5+4\cos(0.5t)\geq1>0$ , so the root is unique. Newton-Raphson gives $t_{n+1}=t_n-[5t_n+8\sin(0.5t_n)-30]/[5+4\cos(0.5t_n)]$ . From $t_0=5.5$ , unrounded substitution gives $t_1=5.075305513\ldots$ and $t_2=5.122362632\ldots$ . Thus the estimated time is $5.12$ s to $3$ significant figures.

(6 marks)

Tier 3 · Hard

A drying-process output is modelled by

S(t)=t+2e^{-t}

, where

t

is the time in hours. The process is complete when

S=3.5

. Show that the completion time lies in

(3,4)

and is unique in this interval. Starting with this interval, repeatedly test the midpoint of the current bracket and retain the half containing the sign change; carry out five such steps. State the resulting interval, use its midpoint to estimate the completion time, and give the maximum possible error in this estimate.

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	For $f(t)=t+2e^{-t}-3.5$ , $f(3)<0$ and $f(4)>0$ ; also $f'(t)=1-2e^{-t}>0$ on $(3,4)$ , so the root there is unique. The five tested midpoints are $3.5$ , $3.25$ , $3.375$ , $3.4375$ and $3.40625$ . The resulting interval is $(3.40625,3.4375)$ , so the midpoint estimate is $3.421875$ hours with maximum possible error $0.015625$ hours.	7
Notes
Let $f(t)=t+2e^{-t}-3.5$ . Then $f(3)=-0.400425\ldots<0$ and $f(4)=0.536631\ldots>0$ . Since $f'(t)=1-2e^{-t}>0$ when $3<t<4$ , there is exactly one root in the interval. The midpoint signs are $f(3.5)>0$ , $f(3.25)<0$ , $f(3.375)<0$ , $f(3.4375)>0$ and $f(3.40625)<0$ . The successive brackets therefore end at $(3.40625,3.4375)$ . Its midpoint is $(3.40625+3.4375)/2=3.421875$ , and half the interval width is $(3.4375-3.40625)/2=0.015625$ , which is the maximum possible error.

(7 marks)

Q10

Tier 3 · Hard

10.

An optical instrument has a calibration factor

c>1

satisfying

c+\ln c=4

. Show that there is exactly one possible factor between

2.9

and

3

. By testing suitable decimal values and using changes of sign, first locate

c

in an interval of width

0.01

and then refine this to an interval of width

0.001

. Hence give the calibration factor to

3

significant figures.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	For $f(c)=c+\ln c-4$ , $f(2.9)<0<f(3)$ and $f'(c)=1+1/c>0$ , so there is exactly one possible factor in $(2.9,3)$ . $f(2.92)<0<f(2.93)$ , and then $f(2.926)<0<f(2.927)$ , so $2.926<c<2.927$ . The calibration factor is $2.93$ to $3$ significant figures.	6
Notes
Let $f(c)=c+\ln c-4$ . Then $f(2.9)=-0.035289\ldots$ and $f(3)=0.098612\ldots$ , so continuity gives a root in $(2.9,3)$ . Since $f'(c)=1+1/c>0$ for $c>1$ , this root is unique. Now $f(2.92)=-0.008416\ldots$ and $f(2.93)=0.005002\ldots$ , giving the width- $0.01$ bracket $(2.92,2.93)$ . Refining within it, $f(2.926)=-0.000363\ldots$ and $f(2.927)=0.000978\ldots$ , giving the width- $0.001$ bracket $(2.926,2.927)$ . Every number in this final bracket rounds to $2.93$ to $3$ significant figures.

(6 marks)

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