Question 4 - A-Level Maths

OCR MEI Further Numerical Methods 2024 June — Question 4 10 marks

Exam Board	OCR MEI
Module	Further Numerical Methods (Further Numerical Methods)
Year	2024
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Numerical integration
Type	Trapezium rule applied to real-world data
Difficulty	Moderate -0.5 This is a straightforward application of Newton's forward difference formula with routine calculations. Parts (a)-(c) involve filling a difference table and applying a standard formula, while part (d) requires simple substitution and comparison. The question is slightly easier than average as it's mostly procedural with no novel problem-solving required, though it does test understanding of when polynomial models are appropriate.
Spec	1.04e Sequences: nth term and recurrence relations 4.05a Roots and coefficients: symmetric functions

4 Between 1946 and 2012 the mean monthly maximum temperature of the water surface of a lake in northern England has been recorded by environmental scientists. Some of the data are shown in Table 4.1. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Table 4.1}

Month	May	June	July	August	September
\(t =\) Time in months	0	1	2	3	4
\(T =\) Mean temperature in \({ } ^ { \circ } \mathrm { C }\)	8.8	13.2	15.4	15.4	13.3

\end{table} Table 4.2 shows a difference table for the data. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Table 4.2}

\(t\)	\(T\)	\(\Delta T\)	\(\Delta T ^ { 2 }\)
0	8.8

1	13.2

2	15.4

3	15.4

4	13.3

\end{table}

Complete the copy of the difference table in the Printed Answer Booklet.
Explain why a quadratic model may be appropriate for these data.
Use Newton's forward difference interpolation formula to construct an interpolating polynomial of degree 2 for these data. This polynomial is used to model the relationship between \(T\) and \(t\). Between 1946 and 2012 the mean monthly maximum temperature of the water surface of the lake was recorded as \(8.9 ^ { \circ } \mathrm { C }\) for October and \(7.5 ^ { \circ } \mathrm { C }\) for November.
Determine whether the model is a good fit for the temperatures recorded in October and November. A scientist recorded the mean monthly maximum temperature of the water surface of the lake in 2022. Some of the data are shown in Table 4.3. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 4.3}
Month May June July August September
\(t =\) Time in months 0 1 2 3 4
\(T =\) Mean temperature in \({ } ^ { \circ } \mathrm { C }\) 10.3 14.7 16.9 16.9 14.8
\end{table}
Adapt the polynomial found in part (c) so that it can be used to model the relationship between \(T\) and \(t\) for the data in Table 4.3.

Show mark scheme Show mark scheme source

Question 4:

Answer	Marks	Guidance
4	(a)	t
A1	1.1
1.1	at least 3 first differences correct

all correct

Answer	Marks
0	8.8

4.4

Answer	Marks	Guidance
1	13.2	-2.2

2.2

Answer	Marks	Guidance
2	15.4	-2.2

0

Answer	Marks	Guidance
3	15.4	-2.1

-2.1

Answer	Marks
4	13.3

[2]

Answer	Marks	Guidance
4	(b)	because the second differences are

approximately equal oe

or

because the third differences are 0 and 0.1 and

Answer	Marks	Guidance
0.1 is approximately 0 oe	B1	2.2b

eg approximately constant

eg roughly the same

do not allow

eg roughly equivalent

eg relatively similar

[1]

Answer	Marks	Guidance
4	(c)	𝑡(𝑡−1)

8.8+4.4𝑡−2.2

2!

8.8+5.5𝑡−1.1𝑡2

Answer	Marks
𝑇 = 8.8+5.5𝑡−1.1𝑡2	M1

A1

Answer	Marks
A1	3.3

1.1

Answer	Marks
2.5	FT their differences from part (a) and one sign error in substitution;

allow other variable

two of three terms correct; allow other variable

all three terms correct; allow other variable;

may see eg y in terms of x

fully correct with correct variables used

[4]

Answer	Marks	Guidance
4	(d)	[t = 5,] T = 8.8 (which is close to 8.9) so good fit

for October isw

[t = 6,] T = 2.2 (which is a long way from 7.5),

Answer	Marks
so bad fit for November	B1
B1	3.4
3.2b	FT their quadratic following award of at least M1;

if B0B0 allow SC1 for 8.8 and 2.2 seen

or

allow SC1 for calculation of both values of T found FT their

quadratic following award of at least M1 in part (c)

[2]

Answer	Marks	Guidance
4	(e)	𝑇 = 10.3+5.5𝑡−1.1𝑡2

[1]

Question 4:
4 | (a) | t | T | ∆T | ∆2𝑇 | M1
A1 | 1.1
1.1 | at least 3 first differences correct
all correct
0 | 8.8
4.4
1 | 13.2 | -2.2
2.2
2 | 15.4 | -2.2
0
3 | 15.4 | -2.1
-2.1
4 | 13.3
[2]
4 | (b) | because the second differences are
approximately equal oe
or
because the third differences are 0 and 0.1 and
0.1 is approximately 0 oe | B1 | 2.2b | allow
eg approximately constant
eg roughly the same
do not allow
eg roughly equivalent
eg relatively similar
[1]
4 | (c) | 𝑡(𝑡−1)
8.8+4.4𝑡−2.2
2!
8.8+5.5𝑡−1.1𝑡2
𝑇 = 8.8+5.5𝑡−1.1𝑡2 | M1
A1
A1
A1 | 3.3
1.1
1.1
2.5 | FT their differences from part (a) and one sign error in substitution;
allow other variable
two of three terms correct; allow other variable
all three terms correct; allow other variable;
may see eg y in terms of x
fully correct with correct variables used
[4]
4 | (d) | [t = 5,] T = 8.8 (which is close to 8.9) so good fit
for October isw
[t = 6,] T = 2.2 (which is a long way from 7.5),
so bad fit for November | B1
B1 | 3.4
3.2b | FT their quadratic following award of at least M1;
if B0B0 allow SC1 for 8.8 and 2.2 seen
or
allow SC1 for calculation of both values of T found FT their
quadratic following award of at least M1 in part (c)
[2]
4 | (e) | 𝑇 = 10.3+5.5𝑡−1.1𝑡2 | B1 | 2.2a | FT their quadratic; must be T in terms of t
[1]

Show LaTeX source

4 Between 1946 and 2012 the mean monthly maximum temperature of the water surface of a lake in northern England has been recorded by environmental scientists. Some of the data are shown in Table 4.1.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 4.1}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Month & May & June & July & August & September \\
\hline
$t =$ Time in months & 0 & 1 & 2 & 3 & 4 \\
\hline
$T =$ Mean temperature in ${ } ^ { \circ } \mathrm { C }$ & 8.8 & 13.2 & 15.4 & 15.4 & 13.3 \\
\hline
\end{tabular}
\end{center}
\end{table}

Table 4.2 shows a difference table for the data.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 4.2}
\begin{tabular}{|l|l|l|l|}
\hline
$t$ & $T$ & $\Delta T$ & $\Delta T ^ { 2 }$ \\
\hline
0 & 8.8 &  &  \\
\hline
 &  &  &  \\
\hline
1 & 13.2 &  &  \\
\hline
 &  &  &  \\
\hline
2 & 15.4 &  &  \\
\hline
 &  &  &  \\
\hline
3 & 15.4 &  &  \\
\hline
 &  &  &  \\
\hline
4 & 13.3 &  &  \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Complete the copy of the difference table in the Printed Answer Booklet.
\item Explain why a quadratic model may be appropriate for these data.
\item Use Newton's forward difference interpolation formula to construct an interpolating polynomial of degree 2 for these data.

This polynomial is used to model the relationship between $T$ and $t$. Between 1946 and 2012 the mean monthly maximum temperature of the water surface of the lake was recorded as $8.9 ^ { \circ } \mathrm { C }$ for October and $7.5 ^ { \circ } \mathrm { C }$ for November.
\item Determine whether the model is a good fit for the temperatures recorded in October and November.

A scientist recorded the mean monthly maximum temperature of the water surface of the lake in 2022. Some of the data are shown in Table 4.3.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 4.3}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
Month & May & June & July & August & September \\
\hline
$t =$ Time in months & 0 & 1 & 2 & 3 & 4 \\
\hline
$T =$ Mean temperature in ${ } ^ { \circ } \mathrm { C }$ & 10.3 & 14.7 & 16.9 & 16.9 & 14.8 \\
\hline
\end{tabular}
\end{center}
\end{table}
\item Adapt the polynomial found in part (c) so that it can be used to model the relationship between $T$ and $t$ for the data in Table 4.3.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2024 Q4 [10]}}

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