OCR MEI C4 — Question 2 19 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeLong-term behaviour analysis
DifficultyStandard +0.3 This is a standard C4 differential equations question with routine verification (part ii), straightforward partial fractions integration (part iii), and simple model comparison. All techniques are textbook exercises requiring no novel insight, though part (iii) involves several steps. Slightly easier than average due to the guided structure and standard methods.
Spec1.06i Exponential growth/decay: in modelling context1.08k Separable differential equations: dy/dx = f(x)g(y)
The growth of a tree is modelled by the differential equation $$10\frac{dh}{dt} = 20 - h,$$ where \(h\) is its height in metres and the time \(t\) is in years. It is assumed that the tree is grown from seed, so that \(h = 0\) when \(t = 0\).
  1. Write down the value of \(h\) for which \(\frac{dh}{dt} = 0\), and interpret this in terms of the growth of the tree. [1]
  2. Verify that \(h = 20(1 - e^{-0.1t})\) satisfies this differential equation and its initial condition. [5]
The alternative differential equation $$200\frac{dh}{dt} = 400 - h^2$$ is proposed to model the growth of the tree. As before, \(h = 0\) when \(t = 0\).
  1. Using partial fractions, show by integration that the solution to the alternative differential equation is $$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}.$$ [9]
  2. What does this solution indicate about the long-term height of the tree? [1]
  3. After a year, the tree has grown to a height of 2 m. Which model fits this information better? [3]
Question 2:
AnswerMarks Guidance
2(i) h = 20, stops growing
[1]AG need interpretation
(ii)h = 20 – 20et/10
dh/dt = 2et/10
20 et/10= 20 – 20(1 – et/10) = 20 – h
=10dh/dt
when t = 0, h = 20(1 – 1 ) = 0
………………………………………………………
OR verifying by integration
dh dt
 
20h 10
ln(20h)0.1tc
h0,t0,cln20
ln(20h)0.1tln20
20h
ln( )0.1t
20
20h20e0.1t
AnswerMarks
h20(1e0.1t)M1A1
M1
A1
B1
M1
A1
B1
M1
A1
AnswerMarks
[5]differentiation (for M1 need ket/10, k const)
oe eg 20 – h = 20 – 20(1 – et/10) = 20et/10
= 10dh/dt (showing sides equivalent)
initial conditions
……………………………………………………………………………..
sep correctly and intending to integrate
correct result (condone omission of c, although no further marks are possible)
condone ln (h – 20) as part of the solution at this stage
constant found from expression of correct form (at any stage) but B0 if say
c = ln (–20) (found using ln (h – 20))
combining logs and anti-logging (correct rules)
correct form (do not award if B0 above)
AnswerMarks Guidance
2(iii) 200 A B
 
(20h)(20h) 20h 20h
 200 = A(20  h) + B(20 + h)
h = 20  200 = 40 B, B = 5
h = 20  200 = 40A , A = 5
200 dh/dt = 400  h2
200
  dhdt
400h2
5 5
 (  )dhdt
20h 20h
 5ln(20 + h) 5ln(20  h) = t + c
When t = 0, h = 0  0 = 0 + c  c = 0
20h
 5ln t
20h
20h
 et/5
20h
 20 + h ===  h)et/5 = 2200 t/5  h et/5
 h + h et/5 = 20 et/5  20
 h(et/5+ 1) = 20(et/5  1)
20(et/51)
 h
et/51
20(1et/5)
 h *
AnswerMarks
1et/5M1
A1
A1
M1
A1
B1
M1
DDMM11
A1
AnswerMarks
[9]cover up, substitution or equating coeffs
separating variables and intending to integrate (condone sign error)
substituting partial fractions
ft their A,B, condone absence of c, Do not allow ln(h-20) for A1.
cao need to show this. c can be found at any stage. NB c = ln(–1) (from ln(h –20))
or similar scores B0.
anti-logging an equation of the correct form . Allow if c = 0 clearly stated
(provided that c = 0) even if B mark is not awarded, but do not allow if c omitted.
Can ft their c.
maki h the subject, dependent on previous mark
NB method marks can be in either order, in which case the dependence is the other
way around.(In which case, 20 + h is divided by 20 – h first to isolate h).
AG must have obtained B1 (for c) in order to obtain final A1.
AnswerMarks Guidance
2(iv) As t  ∞, h  20. So long-term height is 20m.
[1]www
2(v) 1st model h = 20(1  e01) = 1.90..
2nd model h = 20(e1/5 1)/(e1/5 + 1)= 1.99..
AnswerMarks
so 2nd model fits data betterBB11
B1
B1 dep
AnswerMarks
[3]O 1st model h = 2 gives t = 1.05..
2nd model h = 2 gives t = 1.003..
dep previous B1s correct
AnswerMarks Guidance
Quueessttiionanswer Marks 
Question 2:
2 | (i) | h = 20, stops growing | B1
[1] | AG need interpretation
(ii) | h = 20 – 20et/10
dh/dt = 2et/10
20 et/10= 20 – 20(1 – et/10) = 20 – h
=10dh/dt
when t = 0, h = 20(1 – 1 ) = 0
………………………………………………………
OR verifying by integration
dh dt
 
20h 10
ln(20h)0.1tc
h0,t0,cln20
ln(20h)0.1tln20
20h
ln( )0.1t
20
20h20e0.1t
h20(1e0.1t) | M1A1
M1
A1
B1
M1
A1
B1
M1
A1
[5] | differentiation (for M1 need ket/10, k const)
oe eg 20 – h = 20 – 20(1 – et/10) = 20et/10
= 10dh/dt (showing sides equivalent)
initial conditions
……………………………………………………………………………..
sep correctly and intending to integrate
correct result (condone omission of c, although no further marks are possible)
condone ln (h – 20) as part of the solution at this stage
constant found from expression of correct form (at any stage) but B0 if say
c = ln (–20) (found using ln (h – 20))
combining logs and anti-logging (correct rules)
correct form (do not award if B0 above)
2 | (iii) | 200 A B
 
(20h)(20h) 20h 20h
 200 = A(20  h) + B(20 + h)
h = 20  200 = 40 B, B = 5
h = 20  200 = 40A , A = 5
200 dh/dt = 400  h2
200
  dhdt
400h2
5 5
 (  )dhdt
20h 20h
 5ln(20 + h) 5ln(20  h) = t + c
When t = 0, h = 0  0 = 0 + c  c = 0
20h
 5ln t
20h
20h
 et/5
20h
 20 + h ===  h)et/5 = 2200 t/5  h et/5
 h + h et/5 = 20 et/5  20
 h(et/5+ 1) = 20(et/5  1)
20(et/51)
 h
et/51
20(1et/5)
 h *
1et/5 | M1
A1
A1
M1
A1
B1
M1
DDMM11
A1
[9] | cover up, substitution or equating coeffs
separating variables and intending to integrate (condone sign error)
substituting partial fractions
ft their A,B, condone absence of c, Do not allow ln(h-20) for A1.
cao need to show this. c can be found at any stage. NB c = ln(–1) (from ln(h –20))
or similar scores B0.
anti-logging an equation of the correct form . Allow if c = 0 clearly stated
(provided that c = 0) even if B mark is not awarded, but do not allow if c omitted.
Can ft their c.
maki h the subject, dependent on previous mark
NB method marks can be in either order, in which case the dependence is the other
way around.(In which case, 20 + h is divided by 20 – h first to isolate h).
AG must have obtained B1 (for c) in order to obtain final A1.
2 | (iv) | As t  ∞, h  20. So long-term height is 20m. | B1
[1] | www
2 | (v) | 1st model h = 20(1  e01) = 1.90..
2nd model h = 20(e1/5 1)/(e1/5 + 1)= 1.99..
so 2nd model fits data better | BB11
B1
B1 dep
[3] | O 1st model h = 2 gives t = 1.05..
2nd model h = 2 gives t = 1.003..
dep previous B1s correct
Quueessttiion | answer | Marks | Guidance
The growth of a tree is modelled by the differential equation
$$10\frac{dh}{dt} = 20 - h,$$
where $h$ is its height in metres and the time $t$ is in years. It is assumed that the tree is grown from seed, so that $h = 0$ when $t = 0$.

\begin{enumerate}[label=(\roman*)]
\item Write down the value of $h$ for which $\frac{dh}{dt} = 0$, and interpret this in terms of the growth of the tree. [1]

\item Verify that $h = 20(1 - e^{-0.1t})$ satisfies this differential equation and its initial condition. [5]
\end{enumerate}

The alternative differential equation
$$200\frac{dh}{dt} = 400 - h^2$$
is proposed to model the growth of the tree. As before, $h = 0$ when $t = 0$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Using partial fractions, show by integration that the solution to the alternative differential equation is
$$h = \frac{20(1 - e^{-0.2t})}{1 + e^{-0.2t}}.$$ [9]

\item What does this solution indicate about the long-term height of the tree? [1]

\item After a year, the tree has grown to a height of 2 m. Which model fits this information better? [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4  Q2 [19]}}
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