| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| 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 first-order linear differential equation with straightforward separation of variables, followed by routine algebraic manipulation and substitution. While it requires multiple steps across four parts, each individual step uses well-practiced techniques (separating variables, integration, applying initial conditions, finding limits, and solving logarithmic equations). The context is familiar from mechanics, and no novel insight is required—making it slightly easier than average for an A-level Further Maths question. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \dfrac{1}{9.8 - kv}\,dv = \int 1\,dt\) | M1 | Attempt to separate variables and integrate |
| \(-\dfrac{1}{k}\ln(9.8 - kv) = t + c\) | A1 | Correct rearrangement |
| A1 | Correctly integrated — must include \(+c\) | |
| \(9.8 - kv = Ae^{-kt}\) | M1 | Attempt to write in form \(v=\) |
| When \(t=0\), \(v=0\) giving \(0 = \frac{1}{k}(9.8 - Ae^0)\), \(A = 9.8\) | M1 | Using initial conditions leading to value for constant |
| A1 | Correct value for \(A\) (or \(c = -\frac{1}{k}\ln 9.8\)) | |
| Hence \(v = \dfrac{9.8}{k}(1 - e^{-kt})\) | A1 | Must be in form \(v = ...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Graph through \((0,0)\) with correct general shape, horizontal asymptote labelled \(\dfrac{9.8}{k}\) | B1, B1 | Graph through \((0,0)\) with correct general shape; horizontal asymptote labelled \(\frac{9.8}{k}\). Ignore any graph for \(t<0\); any graph involving \(\sqrt{t}\) must be concave |
| Answer | Marks | Guidance |
|---|---|---|
| As \(t \to \infty\), \(v \to \dfrac{9.8}{k} = 7\) | M1 | Recognises the limiting value |
| \(k = 1.4\) | A1 | Any form from correct \(v\). Allow M1 for their \(v\) used, provided it has a non-zero limiting value |
| Alternative: Limiting value when \(\dfrac{dv}{dt} = 0\), \(k = \dfrac{9.8}{7} = 1.4\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(-1.4t = \ln 0.5 \Rightarrow t = 0.495\) | B1 | Allow for 0.50 or awrt 0.495. May substitute in other versions of equation relating \(v\) and \(t\). No FT from incorrect expression for \(v\) |
## Question 15:
**Part (a):**
$\int \dfrac{1}{9.8 - kv}\,dv = \int 1\,dt$ | M1 | Attempt to separate variables and integrate
$-\dfrac{1}{k}\ln(9.8 - kv) = t + c$ | A1 | Correct rearrangement
| A1 | Correctly integrated — must include $+c$
$9.8 - kv = Ae^{-kt}$ | M1 | Attempt to write in form $v=$
When $t=0$, $v=0$ giving $0 = \frac{1}{k}(9.8 - Ae^0)$, $A = 9.8$ | M1 | Using initial conditions leading to value for constant
| A1 | Correct value for $A$ (or $c = -\frac{1}{k}\ln 9.8$)
Hence $v = \dfrac{9.8}{k}(1 - e^{-kt})$ | A1 | Must be in form $v = ...$
**Part (b):**
Graph through $(0,0)$ with correct general shape, horizontal asymptote labelled $\dfrac{9.8}{k}$ | B1, B1 | Graph through $(0,0)$ with correct general shape; horizontal asymptote labelled $\frac{9.8}{k}$. Ignore any graph for $t<0$; any graph involving $\sqrt{t}$ must be concave
**Part (c):**
As $t \to \infty$, $v \to \dfrac{9.8}{k} = 7$ | M1 | Recognises the limiting value
$k = 1.4$ | A1 | Any form from correct $v$. Allow M1 for their $v$ used, provided it has a non-zero limiting value
Alternative: Limiting value when $\dfrac{dv}{dt} = 0$, $k = \dfrac{9.8}{7} = 1.4$ | M1, A1 |
**Part (d):**
$v = 3.5 \Rightarrow 1 - e^{-1.4t} = \dfrac{3.5}{7}$
Hence $-1.4t = \ln 0.5 \Rightarrow t = 0.495$ | B1 | Allow for 0.50 or awrt 0.495. May substitute in other versions of equation relating $v$ and $t$. No FT from incorrect expression for $v$
---15 A model for the motion of a small object falling through a thick fluid can be expressed using the differential equation\\
$\frac { \mathrm { d } v } { \mathrm {~d} t } = 9.8 - k v$,\\
where $v \mathrm {~ms} ^ { - 1 }$ is the velocity after $t \mathrm {~s}$ and $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Given that $v = 0$ when $t = 0$, solve the differential equation to find $v$ in terms of $t$ and $k$.
\item Sketch the graph of $v$ against $t$.
Experiments show that for large values of $t$, the velocity tends to $7 \mathrm {~ms} ^ { - 1 }$.
\item Find the value of $k$.
\item Find the value of $t$ for which $v = 3.5$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q15 [12]}}