| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - direct proportionality (dN/dt = kN) |
| Difficulty | Moderate -0.3 This is a standard exponential growth differential equation question that follows a textbook template. Part (i) requires writing dN/dt = kN (basic recall), part (ii) is a routine separation of variables with 6 marks suggesting full working is expected but the method is standard, parts (iii) and (iv) involve straightforward substitution and logarithm manipulation. While it requires multiple techniques across 12 marks total, each step follows predictable patterns with no novel problem-solving or geometric insight required, making it slightly easier than the average A-level question. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks |
|---|---|
| \(\frac{dN}{dt} = kN\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{N} \, dN = \int k \, dt\) | M1 | |
| \(\ln | N | = kt + c\) |
| \(t = 0, N = N_0 \Rightarrow \ln | N_0 | = c\) |
| \(\ln | N | = kt + \ln |
| \(\ln \left | \frac{N}{N_0}\right | = kt\) |
| \(\frac{N}{N_0} = e^{kt}\), \(N = N_0e^{kt}\) | A1 |
| Answer | Marks |
|---|---|
| \(2N_0 = N_0e^{0k}\) | M1 |
| \(k = \frac{1}{h}\ln 2 = 0.116\) (3sf) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(10N_0 = N_0e^{0.1155t}\) | ||
| \(t = \frac{1}{0.1155}\ln 10 = 19.932\) hours \(= 19\) hours \(56\) mins | M1 A1 | (12) |
| Answer | Marks |
|---|---|
| Total | (72) |
## (i)
$\frac{dN}{dt} = kN$ | B1 |
## (ii)
$\int \frac{1}{N} \, dN = \int k \, dt$ | M1 |
$\ln |N| = kt + c$ | M1 A1 |
$t = 0, N = N_0 \Rightarrow \ln |N_0| = c$ | M1 |
$\ln |N| = kt + \ln |N_0|$ | M1 |
$\ln \left|\frac{N}{N_0}\right| = kt$ | M1 |
$\frac{N}{N_0} = e^{kt}$, $N = N_0e^{kt}$ | A1 |
## (iii)
$2N_0 = N_0e^{0k}$ | M1 |
$k = \frac{1}{h}\ln 2 = 0.116$ (3sf) | M1 A1 |
## (iv)
$10N_0 = N_0e^{0.1155t}$ | |
$t = \frac{1}{0.1155}\ln 10 = 19.932$ hours $= 19$ hours $56$ mins | M1 A1 | (12) |
---
**Total** | (72) |The rate of increase in the number of bacteria in a culture, $N$, at time $t$ hours is proportional to $N$.
\begin{enumerate}[label=(\roman*)]
\item Write down a differential equation connecting $N$ and $t$. [1]
\end{enumerate}
Given that initially there are $N_0$ bacteria present in a culture,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $N = N_0 e^{kt}$, where $k$ is a positive constant. [6]
\end{enumerate}
Given also that the number of bacteria present doubles every six hours,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item find the value of $k$, [3]
\item Find how long it takes for the number of bacteria to increase by a factor of ten, giving your answer to the nearest minute. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q8 [12]}}