OCR MEI Further Pure Core 2020 November — Question 16 25 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2020
SessionNovember
Marks25
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFirst order differential equations (integrating factor)
TypeApplied/modelling contexts
DifficultyChallenging +1.2 This is a structured multi-part question on integrating factors with clear scaffolding. Parts (a) and (b) are routine verification and standard integrating factor calculation. Parts (c) and (d) involve solving first-order ODEs with given integrating factors, which is methodical rather than insightful. Part (e) requires simple numerical comparison. While lengthy, each step follows standard techniques without requiring novel problem-solving approaches, making it slightly above average difficulty for Further Maths students.
Spec4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations
16 The population density \(P\), in suitable units, of a certain bacterium at time \(t\) hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is \(A\). \begin{enumerate}[label=(\alph*)] \item One simple model is to assume that the rate of change of population density is directly proportional to \(A - P\).
  1. Formulate a differential equation for this model.
  2. Verify that \(P = A \left( 1 - \mathrm { e } ^ { - k t } \right)\), where \(k\) is a positive constant, satisfies
    An alternative model uses the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \mathrm { Q } ( t )$$ where \(\mathrm { Q } ( t )\) is a function of \(t\).
  3. Find the integrating factor for this differential equation, showing that it can be written in the $$\text { form } \frac { \sqrt { 1 + t ^ { 2 } } } { t } \text {. }$$
  4. Suppose that \(\mathrm { Q } ( t ) = 0\). $$\text { (i) Show that } P = \frac { A t } { \sqrt { 1 + t ^ { 2 } } } \text {. }$$ (ii) Find the time predicted by this model for the population density to reach half its longterm value. Give your answer correct to the nearest minute.
  5. Now suppose that \(\mathrm { Q } ( t ) = \frac { t \mathrm { e } ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }\). $$\text { Show that } \left. P = \frac { A t - t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } } \text {. [You may assume that } \lim _ { t \rightarrow \infty } t e ^ { - t } = 0 . \right]$$ It is found that the long-term value of \(P\) is 10, and \(P\) reaches half this value after 37 minutes.
  6. Determine which of the models proposed in parts (c) and (d) is more consistent with these data.
Question 16:
AnswerMarks Guidance
16(a) (i)
[1]3.3
16(a) (ii)
​​ ​ ​ ​ ​
as t → ∞ , e− ​ kt ​ → 0 so P → A
AnswerMarks
​ ​ ​ ​ ​ ​B1
B1
B1
AnswerMarks
[3]1.1
3.4
AnswerMarks
3.4or by integration by separating
variables
PPMMTT
Y420/01 Mark SchemeNovember 2020
16 (b)
B1 1.1
M1 3.1a attempt at partial fractions
1 = A(1 + t2)​ + (Bt + C)t
​ ​ ​​ ​ ​ ​ ​ ​
t = 0 ⇒ A = 1 M1 2.1 substituting values and/or need to attempt 3 values of t
​ ​ ​
t2 ​: 0 = A + B ⇒ B = −1 equating coeffs (or alternative) for mark
​ ​ ​ ​ ​ ​ ​
t: C = 0
​ ​ ​
A1 2.1
IF
M1 2.1
A1 2.1
combining lns
M1 2.1
NB AG
E1cao 2.2a
[8]
17
AnswerMarks Guidance
16(b) 1 = A(1 + t2)​ + (Bt + C)t
​ ​ ​​ ​ ​ ​ ​ ​
t = 0 ⇒ A = 1
​ ​ ​
t2 ​: 0 = A + B ⇒ B = −1
​ ​ ​ ​ ​ ​ ​
t: C = 0
​ ​ ​
AnswerMarks
IFB1
M1
M1
A1
M1
A1
M1
E1cao
AnswerMarks
[8]1.1
3.1a
2.1
2.1
2.1
2.1
2.1
AnswerMarks
2.2aattempt at partial fractions
substituting values and/or
equating coeffs
combining lns
AnswerMarks
NB AGneed to attempt 3 values of t
(or alternative) for mark
PPMMTT
Y420/01 Mark SchemeNovember 2020
16 (c) (i)
M1 2.1
⇒
A1 1.1
⇒
= k
​ M1 3.3
So k = A
​​ ​
E1 2.2a
⇒
NB AG
[4]
16 (c) (ii)
1 A = At M1 3.4 substituting P = 0.5A and
​ ​
2 √1+t2 squaring to solve for t
​
⇒√(1 + t2)​ = 2t
​​ ​
⇒ 3t2 ​=1
​​
⇒ t = 1/√3 = 35 mins A1 3.2a
​​
[2]
18
AnswerMarks Guidance
16(c) (i)
⇒
= k
​
So k = A
​​ ​
AnswerMarks
⇒M1
A1
M1
E1
AnswerMarks
[4]2.1
1.1
3.3
AnswerMarks Guidance
2.2aNB AG
16(c) (ii)
A =
2 √1+t2
⇒√(1 + t2)​ = 2t
​​ ​
⇒ 3t2 ​=1
​​
⇒ t = 1/√3 = 35 mins
AnswerMarks
​​M1
A1
AnswerMarks
[2]3.4
3.2asubstituting P = 0.5A and
​ ​
squaring to solve for t
​
PPMMTT
Y420/01 Mark SchemeNovember 2020
16 (d)
M1 1.1
A1 1.1
⇒ A1 1.1
= c = long term value of P M1 3.1b
​​ ​
so c = A
​​ ​
E1 2.2a NB AG
⇒
[5]
16 (e) A = 10
​
By first model, when t = 37/60 , P = 5.25
​​ ​ ​
By second model, P = 4.97 B1 3.4 Award for either P seen
​ ​
So 2nd ​model fits better B1 3.5a the other value of P and
​ [2] conclusion
19
AnswerMarks Guidance
16(d) ⇒
= c = long term value of P
​​ ​
so c = A
​​ ​
AnswerMarks
⇒M1
A1
A1
M1
E1
AnswerMarks
[5]1.1
1.1
1.1
3.1b
AnswerMarks Guidance
2.2aNB AG
16(e) A = 10
​
By first model, when t = 37/60 , P = 5.25
​​ ​ ​
By second model, P = 4.97
​ ​
So 2nd ​model fits better
AnswerMarks
​B1
B1
AnswerMarks
[2]3.4
3.5aAward for either P seen
the other value of P and
conclusion
PPMMTT
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Question 16:
16 | (a) | (i) | B1
[1] | 3.3
16 | (a) | (ii) | when t = 0, P = A(1 − 1) = 0
​​ ​ ​ ​ ​
as t → ∞ , e− ​ kt ​ → 0 so P → A
​ ​ ​ ​ ​ ​ | B1
B1
B1
[3] | 1.1
3.4
3.4 | or by integration by separating
variables
PPMMTT
Y420/01 Mark SchemeNovember 2020
16 (b)
B1 1.1
M1 3.1a attempt at partial fractions
1 = A(1 + t2)​ + (Bt + C)t
​ ​ ​​ ​ ​ ​ ​ ​
t = 0 ⇒ A = 1 M1 2.1 substituting values and/or need to attempt 3 values of t
​ ​ ​
t2 ​: 0 = A + B ⇒ B = −1 equating coeffs (or alternative) for mark
​ ​ ​ ​ ​ ​ ​
t: C = 0
​ ​ ​
A1 2.1
IF
M1 2.1
A1 2.1
combining lns
M1 2.1
NB AG
E1cao 2.2a
[8]
17
16 | (b) | 1 = A(1 + t2)​ + (Bt + C)t
​ ​ ​​ ​ ​ ​ ​ ​
t = 0 ⇒ A = 1
​ ​ ​
t2 ​: 0 = A + B ⇒ B = −1
​ ​ ​ ​ ​ ​ ​
t: C = 0
​ ​ ​
IF | B1
M1
M1
A1
M1
A1
M1
E1cao
[8] | 1.1
3.1a
2.1
2.1
2.1
2.1
2.1
2.2a | attempt at partial fractions
substituting values and/or
equating coeffs
combining lns
NB AG | need to attempt 3 values of t
(or alternative) for mark
PPMMTT
Y420/01 Mark SchemeNovember 2020
16 (c) (i)
M1 2.1
⇒
A1 1.1
⇒
= k
​ M1 3.3
So k = A
​​ ​
E1 2.2a
⇒
NB AG
[4]
16 (c) (ii)
1 A = At M1 3.4 substituting P = 0.5A and
​ ​
2 √1+t2 squaring to solve for t
​
⇒√(1 + t2)​ = 2t
​​ ​
⇒ 3t2 ​=1
​​
⇒ t = 1/√3 = 35 mins A1 3.2a
​​
[2]
18
16 | (c) | (i) | ⇒
⇒
= k
​
So k = A
​​ ​
⇒ | M1
A1
M1
E1
[4] | 2.1
1.1
3.3
2.2a | NB AG
16 | (c) | (ii) | 1 At
A =
2 √1+t2
⇒√(1 + t2)​ = 2t
​​ ​
⇒ 3t2 ​=1
​​
⇒ t = 1/√3 = 35 mins
​​ | M1
A1
[2] | 3.4
3.2a | substituting P = 0.5A and
​ ​
squaring to solve for t
​
PPMMTT
Y420/01 Mark SchemeNovember 2020
16 (d)
M1 1.1
A1 1.1
⇒ A1 1.1
= c = long term value of P M1 3.1b
​​ ​
so c = A
​​ ​
E1 2.2a NB AG
⇒
[5]
16 (e) A = 10
​
By first model, when t = 37/60 , P = 5.25
​​ ​ ​
By second model, P = 4.97 B1 3.4 Award for either P seen
​ ​
So 2nd ​model fits better B1 3.5a the other value of P and
​ [2] conclusion
19
16 | (d) | ⇒
= c = long term value of P
​​ ​
so c = A
​​ ​
⇒ | M1
A1
A1
M1
E1
[5] | 1.1
1.1
1.1
3.1b
2.2a | NB AG
16 | (e) | A = 10
​
By first model, when t = 37/60 , P = 5.25
​​ ​ ​
By second model, P = 4.97
​ ​
So 2nd ​model fits better
​ | B1
B1
[2] | 3.4
3.5a | Award for either P seen
the other value of P and
conclusion
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The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
16 The population density $P$, in suitable units, of a certain bacterium at time $t$ hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is $A$.
\begin{enumerate}[label=(\alph*)]
\item One simple model is to assume that the rate of change of population density is directly proportional to $A - P$.
\begin{enumerate}[label=(\roman*)]
\item Formulate a differential equation for this model.
\item Verify that $P = A \left( 1 - \mathrm { e } ^ { - k t } \right)$, where $k$ is a positive constant, satisfies

\begin{itemize}
\end{enumerate}\item this differential equation,
  \item the initial condition,
  \item the long-term condition.
\end{itemize}

An alternative model uses the differential equation

$$\frac { \mathrm { d } P } { \mathrm {~d} t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \mathrm { Q } ( t )$$

where $\mathrm { Q } ( t )$ is a function of $t$.
\item Find the integrating factor for this differential equation, showing that it can be written in the

$$\text { form } \frac { \sqrt { 1 + t ^ { 2 } } } { t } \text {. }$$
\item Suppose that $\mathrm { Q } ( t ) = 0$.

$$\text { (i) Show that } P = \frac { A t } { \sqrt { 1 + t ^ { 2 } } } \text {. }$$

(ii) Find the time predicted by this model for the population density to reach half its longterm value. Give your answer correct to the nearest minute.
\item Now suppose that $\mathrm { Q } ( t ) = \frac { t \mathrm { e } ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }$.

$$\text { Show that } \left. P = \frac { A t - t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } } \text {. [You may assume that } \lim _ { t \rightarrow \infty } t e ^ { - t } = 0 . \right]$$

It is found that the long-term value of $P$ is 10, and $P$ reaches half this value after 37 minutes.
\item Determine which of the models proposed in parts (c) and (d) is more consistent with these data.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2020 Q16 [25]}}
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