OCR MEI Further Numerical Methods 2023 June — Question 2 8 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2023
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicNumerical integration
TypeTrapezium rule applied to real-world data
DifficultyStandard +0.3 This is a straightforward application of standard numerical methods from Further Maths. Part (a) requires recognizing that Newton forward differences need equal spacing (trivial check). Part (b) is a direct application of Lagrange interpolation formula with 3 pointsβ€”mechanical calculation. Parts (c) and (d) involve substituting values and making basic observations about polynomial behavior. While this is Further Maths content, it's entirely procedural with no problem-solving insight required, making it easier than an average A-level question overall.
Spec4.05a Roots and coefficients: symmetric functions
2 A car tyre has a slow puncture. Initially the tyre is inflated to a pressure of 34.5 psi . The pressure is checked after 3 days and then again after 5 days. The time \(t\) in days and the pressure, \(P\) psi, are shown in the table below. You are given that the pressure in a car tyre is measured in pounds per square inch (psi).
\(t\)035
\(P\)34.529.427.0
The owner of the car believes the relationship between \(P\) and \(t\) may be modelled by a polynomial.
  1. Explain why it is not possible to use Newton's forward difference interpolation method for these data.
  2. Use Lagrange's form of the interpolating polynomial to find an interpolating polynomial of degree 2 for these data. The car owner uses the polynomial found in part (b) to model the relationship between \(P\) and \(t\).
    Subsequently it is found that when \(t = 6 , P = 26.0\) and when \(t = 10 , P = 24.4\).
  3. Determine whether the owner's model is a good fit for these data.
  4. Explain why the model would not be suitable in the long term.
Question 2:
AnswerMarks Guidance
2(a) because the values of t are not evenly spaced
oeB1 1.2
eg because the data are not evenly spaced
eg because the x values are not evenly spaced
[1]
AnswerMarks Guidance
2(b) (π‘‘π‘‘βˆ’0)(π‘‘π‘‘βˆ’3) (π‘‘π‘‘βˆ’0)(π‘‘π‘‘βˆ’5)
(5βˆ’0)(5βˆ’3)Γ—27.0 + (3βˆ’0)(3βˆ’5)Γ—29.4+
(π‘‘π‘‘βˆ’3)(π‘‘π‘‘βˆ’5)
(0βˆ’3)(0βˆ’5)Γ—34.5
2 or
0.1𝑑𝑑 βˆ’2𝑑𝑑+34.5
2
𝑃𝑃2(𝑑𝑑) = 0.1𝑑𝑑 βˆ’2𝑑𝑑+34.5
AnswerMarks
2M1
A1
A1
AnswerMarks
A13.3
1.1
1.1
AnswerMarks
1.1allow sign errors in substitution of x-values and one numerical
error
all substitutions correct
two terms correct
all correct; A0 if other variable used or P omitted; do not
allow f(t)
AnswerMarks Guidance
𝑃𝑃 = 0.1𝑑𝑑 βˆ’2𝑑𝑑+34.5[4]
2(c) t = 6 gives P = 26.1 β‰ˆ 26.0 so good fit
t =10 gives P =24.5 β‰ˆ 24.4 so good fitB1
B13.4
3.4if B0B0 allow SC1 for 26.1 and 24.5 seen
may see one comment for both comparisons
[2]
AnswerMarks Guidance
2(d) model predicts tyre pressures will start to
increase (after t = 10)B1 3.5b
eg
AnswerMarks
[1]as 𝑑𝑑 β†’ ∞,𝑃𝑃 β†’ ∞
2
1.81651017847
1.81159652649
1.81159008685
1.81159008298
[2]
Question 2:
2 | (a) | because the values of t are not evenly spaced
oe | B1 | 1.2 | do not allow
eg because the data are not evenly spaced
eg because the x values are not evenly spaced
[1]
2 | (b) | (π‘‘π‘‘βˆ’0)(π‘‘π‘‘βˆ’3) (π‘‘π‘‘βˆ’0)(π‘‘π‘‘βˆ’5)
(5βˆ’0)(5βˆ’3)Γ—27.0 + (3βˆ’0)(3βˆ’5)Γ—29.4+
(π‘‘π‘‘βˆ’3)(π‘‘π‘‘βˆ’5)
(0βˆ’3)(0βˆ’5)Γ—34.5
2 or
0.1𝑑𝑑 βˆ’2𝑑𝑑+34.5
2
𝑃𝑃2(𝑑𝑑) = 0.1𝑑𝑑 βˆ’2𝑑𝑑+34.5
2 | M1
A1
A1
A1 | 3.3
1.1
1.1
1.1 | allow sign errors in substitution of x-values and one numerical
error
all substitutions correct
two terms correct
all correct; A0 if other variable used or P omitted; do not
allow f(t)
𝑃𝑃 = 0.1𝑑𝑑 βˆ’2𝑑𝑑+34.5 | [4]
2 | (c) | t = 6 gives P = 26.1 β‰ˆ 26.0 so good fit
t =10 gives P =24.5 β‰ˆ 24.4 so good fit | B1
B1 | 3.4
3.4 | if B0B0 allow SC1 for 26.1 and 24.5 seen
may see one comment for both comparisons
[2]
2 | (d) | model predicts tyre pressures will start to
increase (after t = 10) | B1 | 3.5b | allow
eg
[1] | as 𝑑𝑑 β†’ ∞,𝑃𝑃 β†’ ∞
2
1.81651017847
1.81159652649
1.81159008685
1.81159008298
[2]
2 A car tyre has a slow puncture. Initially the tyre is inflated to a pressure of 34.5 psi . The pressure is checked after 3 days and then again after 5 days. The time $t$ in days and the pressure, $P$ psi, are shown in the table below. You are given that the pressure in a car tyre is measured in pounds per square inch (psi).

\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$t$ & 0 & 3 & 5 \\
\hline
$P$ & 34.5 & 29.4 & 27.0 \\
\hline
\end{tabular}
\end{center}

The owner of the car believes the relationship between $P$ and $t$ may be modelled by a polynomial.
\begin{enumerate}[label=(\alph*)]
\item Explain why it is not possible to use Newton's forward difference interpolation method for these data.
\item Use Lagrange's form of the interpolating polynomial to find an interpolating polynomial of degree 2 for these data.

The car owner uses the polynomial found in part (b) to model the relationship between $P$ and $t$.\\
Subsequently it is found that when $t = 6 , P = 26.0$ and when $t = 10 , P = 24.4$.
\item Determine whether the owner's model is a good fit for these data.
\item Explain why the model would not be suitable in the long term.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2023 Q2 [8]}}
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