| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - N₀)) |
| Difficulty | Standard +0.3 This is a standard C4 differential equations question with clear scaffolding through five parts. Parts (i)-(ii) involve straightforward integration of an exponential and finding a limit. Parts (iii)-(v) follow a typical textbook pattern: partial fractions, separation of variables, and interpretation. While it requires multiple techniques, each step is routine and heavily guided, making it slightly easier than average for A-level. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \int 10e^{-\frac{1}{2}t}\,dt\) | M1 | separate variables and intend to integrate |
| \(= -20e^{-\frac{1}{2}t}+c\) | A1 | \(-20e^{-\frac{1}{2}t}\) |
| when \(t=0\), \(v=0 \Rightarrow 0=-20+c \Rightarrow c=20\) | M1 | finding \(c\) |
| so \(v=20-20e^{-\frac{1}{2}t}\) | A1 cao | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| As \(t\to\infty\), \(e^{-\frac{1}{2}t}\to 0 \Rightarrow v\to 20\) | M1 | |
| So long term speed is \(20\) m s\(^{-1}\) | A1 | ft for their \(c>0\), found [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{(w-4)(w+5)} = \frac{A}{w-4}+\frac{B}{w+5} = \frac{A(w+5)+B(w-4)}{(w-4)(w+5)}\) | M1 | cover up, substitution or equating coefficients |
| \(w=4\): \(1=9A \Rightarrow A=\frac{1}{9}\) | A1 | \(\frac{1}{9}\) |
| \(w=-5\): \(1=-9B \Rightarrow B=-\frac{1}{9}\) | A1 | \(-\frac{1}{9}\) |
| \(\Rightarrow \frac{1}{(w-4)(w+5)} = \frac{1}{9(w-4)}-\frac{1}{9(w+5)}\) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \int\frac{dw}{(w-4)(w+5)} = \int-\frac{1}{2}\,dt\) | M1 | separating variables |
| \(\Rightarrow \int\left[\frac{1}{9(w-4)}-\frac{1}{9(w+5)}\right]dw = \int-\frac{1}{2}\,dt\) | M1 | substituting their partial fractions |
| \(\Rightarrow \frac{1}{9}\ln(w-4)-\frac{1}{9}\ln(w+5) = -\frac{1}{2}t+c\) | A1ft | integrating correctly (condone absence of \(c\)) |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t=0\), \(w=10 \Rightarrow c=\frac{1}{9}\ln\frac{6}{15}=\frac{1}{9}\ln\frac{2}{5}\) | M1 | correctly evaluating \(c\) (at any stage) |
| \(\Rightarrow \ln\frac{w-4}{w+5} = -\frac{9}{2}t+\ln\frac{2}{5}\) | M1 | combining \(\ln\)s (at any stage) |
| \(\Rightarrow \frac{w-4}{w+5} = e^{-\frac{9}{2}t+\ln\frac{2}{5}} = \frac{2}{5}e^{-4.5t}\) | E1 | www [6] |
| Answer | Marks | Guidance |
|---|---|---|
| As \(t\to\infty\), \(e^{-4.5t}\to 0 \Rightarrow w-4\to 0\) | M1 | |
| So long term speed is \(4\) m s\(^{-1}\) | A1 | [2] |
## Question 7(i):
$v = \int 10e^{-\frac{1}{2}t}\,dt$ | M1 | separate variables and intend to integrate
$= -20e^{-\frac{1}{2}t}+c$ | A1 | $-20e^{-\frac{1}{2}t}$
when $t=0$, $v=0 \Rightarrow 0=-20+c \Rightarrow c=20$ | M1 | finding $c$
so $v=20-20e^{-\frac{1}{2}t}$ | A1 cao | [4] |
---
## Question 7(ii):
As $t\to\infty$, $e^{-\frac{1}{2}t}\to 0 \Rightarrow v\to 20$ | M1 |
So long term speed is $20$ m s$^{-1}$ | A1 | ft for their $c>0$, found [2] |
---
## Question 7(iii):
$\frac{1}{(w-4)(w+5)} = \frac{A}{w-4}+\frac{B}{w+5} = \frac{A(w+5)+B(w-4)}{(w-4)(w+5)}$ | M1 | cover up, substitution or equating coefficients
$w=4$: $1=9A \Rightarrow A=\frac{1}{9}$ | A1 | $\frac{1}{9}$
$w=-5$: $1=-9B \Rightarrow B=-\frac{1}{9}$ | A1 | $-\frac{1}{9}$
$\Rightarrow \frac{1}{(w-4)(w+5)} = \frac{1}{9(w-4)}-\frac{1}{9(w+5)}$ | [4] |
---
## Question 7(iv):
$\frac{dw}{dt} = -\frac{1}{2}(w-4)(w+5)$
$\Rightarrow \int\frac{dw}{(w-4)(w+5)} = \int-\frac{1}{2}\,dt$ | M1 | separating variables
$\Rightarrow \int\left[\frac{1}{9(w-4)}-\frac{1}{9(w+5)}\right]dw = \int-\frac{1}{2}\,dt$ | M1 | substituting their partial fractions
$\Rightarrow \frac{1}{9}\ln(w-4)-\frac{1}{9}\ln(w+5) = -\frac{1}{2}t+c$ | A1ft | integrating correctly (condone absence of $c$)
$\Rightarrow \frac{1}{9}\ln\frac{w-4}{w+5} = -\frac{1}{2}t+c$
When $t=0$, $w=10 \Rightarrow c=\frac{1}{9}\ln\frac{6}{15}=\frac{1}{9}\ln\frac{2}{5}$ | M1 | correctly evaluating $c$ (at any stage)
$\Rightarrow \ln\frac{w-4}{w+5} = -\frac{9}{2}t+\ln\frac{2}{5}$ | M1 | combining $\ln$s (at any stage)
$\Rightarrow \frac{w-4}{w+5} = e^{-\frac{9}{2}t+\ln\frac{2}{5}} = \frac{2}{5}e^{-4.5t}$ | E1 | www [6] |
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## Question 7(v):
As $t\to\infty$, $e^{-4.5t}\to 0 \Rightarrow w-4\to 0$ | M1 |
So long term speed is $4$ m s$^{-1}$ | A1 | [2] |7 A skydiver drops from a helicopter. Before she opens her parachute, her speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ after time $t$ seconds is modelled by the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = 10 \mathrm { e } ^ { - \frac { 1 } { 2 } t }$$
When $t = 0 , v = 0$.\\
(i) Find $v$ in terms of $t$.\\
(ii) According to this model, what is the speed of the skydiver in the long term?
She opens her parachute when her speed is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Her speed $t$ seconds after this is $w \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and is modelled by the differential equation
$$\frac { \mathrm { d } w } { \mathrm {~d} t } = - \frac { 1 } { 2 } ( w - 4 ) ( w + 5 )$$
(iii) Express $\frac { 1 } { ( w - 4 ) ( w + 5 ) }$ in partial fractions.\\
(iv) Using this result, show that $\frac { w - 4 } { w + 5 } = 0.4 \mathrm { e } ^ { - 4.5 t }$.\\
(v) According to this model, what is the speed of the skydiver in the long term?
\hfill \mbox{\textit{OCR MEI C4 Q7 [18]}}