| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure with Technology (Further Pure with Technology) |
| Year | 2023 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations with non-standard force laws |
| Difficulty | Standard +0.8 This is a substantial differential equations question requiring separation of variables, partial fractions, and interpretation of logistic growth models, followed by numerical methods analysis. While the techniques are A-level standard, the multi-part structure, need for careful algebraic manipulation in solving the logistic equation, and the numerical methods component with non-standard exponents make this moderately challenging but within reach of strong Further Maths students. |
| Spec | 1.09d Newton-Raphson method4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | (i) |
| Answer | Marks |
|---|---|
| d t | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | Consideration of the signs in at |
| Answer | Marks |
|---|---|
| (ii) | K P e rt |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | B1 | |
| [1] | 1.1a | By CAS |
| (iii) | In both cases P(t)→K as t → | B1 |
| [1] | 1.1 | |
| (iv) | P ( t ) is decreasing if P K and increasing if |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | B1 | |
| [1] | 1.2 | Allow ‘P approaches K from above |
| Answer | Marks | Guidance |
|---|---|---|
| (v) | K is the long term population. | B1 |
| [1] | 3.2a | |
| (b) | (i) | If G1 contains the value of h |
| Answer | Marks |
|---|---|
| and copy down | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 2.5 | Give reasonable BOD on possible |
| Answer | Marks |
|---|---|
| (ii) | Approximations using h = 0.1 to 7 d.p are |
| Answer | Marks | Guidance |
|---|---|---|
| P(3) ≈ 979.5970106 | B1 | |
| [1] | 1.1 | Need all correct to at least 3 s.f. |
| (iii) | Approximations using h = 0.05 to 7 d.p are |
| Answer | Marks | Guidance |
|---|---|---|
| P(3) ≈ 981.0771079 | B1 | |
| [1] | 1.1 | Need all correct to at least 3 s.f. |
| (iv) | There is an increase in the approximations when | |
| using h = 0.05 compared to h = 0.1. | M1 | 1.1 |
| 𝑡 = 1 and 𝑡 = 2). | A suitable annotated |
| Answer | Marks | Guidance |
|---|---|---|
| Therefore, values are likely to be underestimates. | A1 | |
| [2] | 3.2b | Tangent to curve is below the |
Question 4:
4 | (a) | (i) | d P P
= r P 1 −
d t K
d P
= 0 P = 0 o r P = K
d t
Since r > 0, by considering the signs of the
P
factors P and 1− for P > 0,
K
d P
0 0 P K
d t
d P
0 K P
d t | B1
M1
A1
A1
[4] | 1.1a
1.1
1.1
1.1 | Consideration of the signs in at
least one case.
If M0, then SC1 if at least one of
final two conditions correct.
Condone inclusion of 𝑃 < 0 for
final case.
(ii) | K P e rt
P ( t ) = 0
K + P ( e rt − 1 )
0 | B1
[1] | 1.1a | By CAS
(iii) | In both cases P(t)→K as t → | B1
[1] | 1.1
(iv) | P ( t ) is decreasing if P K and increasing if
0
P K .
0 | B1
[1] | 1.2 | Allow ‘P approaches K from above
when 𝑃 > 𝐾 and from below
0
when 𝑃 < 𝐾.
0
(v) | K is the long term population. | B1
[1] | 3.2a
(b) | (i) | If G1 contains the value of h
then
A1 contains 0
B1 contains 1
C1 contains =2*(B1^1.25)*((1-B1/1000)^1.5)
A2 contains =A1+$G$1
B2 contains =B1+$G$1*C1
C2 contains =2*(B2^1.25)*((1-B2/1000)^1.5)
and copy down | B1
B1
B1
B1
[4] | 3.1a
3.1a
3.1a
2.5 | Give reasonable BOD on possible
transcription errors and consider
correct answers to 4(b)(ii),
4(b)(iii), 4(b)(iv) as evidence of
correct formulae in the
spreadsheet.
Allows for h to be varied.
Columns for t and P
n n
dP
Calculation of at each stage.
dt
Could be in separate column.
Formulae for t and P . Should
n+1 n+1
include clear indication, for
example, “copy down”, for how
formulae are generated.
(ii) | Approximations using h = 0.1 to 7 d.p are
P(1) ≈ 9.5679218
P(2) ≈ 421.7281066
P(3) ≈ 979.5970106 | B1
[1] | 1.1 | Need all correct to at least 3 s.f.
(iii) | Approximations using h = 0.05 to 7 d.p are
P(1) ≈ 11.7740227
P(2) ≈ 672.3104086
P(3) ≈ 981.0771079 | B1
[1] | 1.1 | Need all correct to at least 3 s.f.
(iv) | There is an increase in the approximations when
using h = 0.05 compared to h = 0.1. | M1 | 1.1 | The gradient increases (between
𝑡 = 1 and 𝑡 = 2). | A suitable annotated
diagram specific to this case
could receive M1 A1.
A smaller ℎ (generally) gives a better estimate.
Therefore, values are likely to be underestimates. | A1
[2] | 3.2b | Tangent to curve is below the
curve for 𝑡 ≤ 2. Hence values
are likely to be underestimates
(including for 𝑡 = 3 due to the
cumulative effect of continually
underestimating for 𝑡 ≤ 2). o.e.
PMT
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Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.4 In this question you are required to consider the family of differential equations
$$\frac { d P } { d t } = r P \left( 1 - \frac { P } { K } \right) , \quad t \geqslant 0 , \quad P ( t ) \geqslant 0 \left( ^ { * } \right)$$
where $r$ and $K$ are positive constants. This differential equation can be used as a model for the size of a population $P$ as a function of time $t$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the values of $P$ for which
\begin{itemize}
\item $\frac { \mathrm { dP } } { \mathrm { dt } } = 0$
\item $\frac { \mathrm { dP } } { \mathrm { dt } } > 0$
\item $\frac { \mathrm { dP } } { \mathrm { dt } } < 0$
\item Solve the equation (*) subject to the initial condition that $P = P _ { 0 }$ when $t = 0$.
\item Find a property common to your solution in (ii) in the cases $\mathrm { P } _ { 0 } > \mathrm { K }$ and $\mathrm { P } _ { 0 } < \mathrm { K }$.
\item State a feature of your solution in (iii) for the case $\mathrm { P } _ { 0 } > \mathrm { K }$ which is different to the case $P _ { 0 } < K$.
\item Interpret the value $K$ when $P ( t )$ is the size of a population at time $t$.
\item In this question you will explore the limitations of using the Euler method to approximate solutions to the differential equation
\end{itemize}
$$\frac { d P } { d t } = 2 P ^ { 1.25 } \left( 1 - \frac { P } { 1000 } \right) ^ { 1.5 } , t \geqslant 0 , P ( t ) \geqslant 0 ( * * )$$
The diagram shows the tangent field to (**), and a solution in which $P = 1$ when $t = 0$, produced using a much more accurate numerical method.\\
\includegraphics[max width=\textwidth, alt={}, center]{4715d0f0-a860-4189-802f-1d2d019e1115-4_899_1552_1763_319}\\
(i) The Euler method for the solution of the differential equation $f ( t , P ) = \frac { d P } { d t }$ is as follows
$$P _ { n + 1 } = P _ { n } + h f \left( t _ { n } , P _ { n } \right)$$
It is given that $t _ { 0 } = 0$ and $P _ { 0 } = 1$.
\begin{itemize}
\item Construct a spreadsheet to solve (**) using the Euler method so that the value of $h$ can be varied.
\item State the formulae you have used in your spreadsheet.\\
(ii) Use your spreadsheet with $h = 0.1$ to approximate
\item $P ( 1 )$
\item $P ( 2 )$
\item $P ( 3 )$\\
(iii) Use your spreadsheet with $h = 0.05$ to approximate
\item $P ( 1 )$
\item $P ( 2 )$
\item $P ( 3 )$\\
(iv) State, with reasons, whether the estimates to $P ( t )$ given in your spreadsheet are likely to be overestimates or underestimates to the exact values.\\
(v) With reference to the diagram, explain any noticeable feature identified when comparing the approximations given to $P ( 2 )$ in (ii) and (iii).
\end{itemize}
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure with Technology 2023 Q4 [18]}}