| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Model comparison/critique |
| Difficulty | Standard +0.3 This is a standard C4 differential equations question covering separable variables and exponential growth. Part (i) is routine integration, (ii) is straightforward substitution, (iii) requires basic model evaluation, and (iv) involves separating variables with a slightly more complex integrand but still follows standard techniques. The question is slightly easier than average because it's highly structured with clear steps and uses familiar exponential growth context. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{P} \, dP = \int k \, dt\) | M1 | |
| \(\ln | P | = kt + c\) |
| \(t = 0\), \(P = 300 \Rightarrow \ln 300 = c\) | M1 | |
| \(\ln | P | = kt + \ln 300\) |
| \(\ln\left | \frac{P}{300}\right | = kt\), \(\frac{P}{300} = e^{kt}\), \(P = 300e^{kt}\) |
| Answer | Marks |
|---|---|
| \(t = 1\), \(P = 360 \Rightarrow 360 = 300e^k\) | M1 |
| \(k = \ln\frac{6}{5} = 0.182 \text{ (3sf)}\) | A1 |
| Answer | Marks |
|---|---|
| \(P = 300e^{0.1821t}\) | |
| when \(t = 2\), \(P = 432\); when \(t = 3\), \(P = 518\) | B1 |
| model does not seem suitable as data diverges from predictions | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{P} \, dP = \int (0.4 - 0.25\cos 0.5t) \, dt\) | M1 | |
| \(\ln | P | = 0.4t - 0.5\sin 0.5t + c\) |
| \(t = 0\), \(P = 300 \Rightarrow \ln 300 = c\) | M1 | |
| \(\ln\left | \frac{P}{300}\right | = 0.4t - 0.5\sin 0.5t\) |
| Answer | Marks | Guidance |
|---|---|---|
| second model: \(t = 1, 2, 3 \Rightarrow P = 352, 438, 605\) | B1 | |
| the second model seems more suitable as it fits the data better | B1 | (14) |
## (i)
$\int \frac{1}{P} \, dP = \int k \, dt$ | M1 |
$\ln|P| = kt + c$ | A1 |
$t = 0$, $P = 300 \Rightarrow \ln 300 = c$ | M1 |
$\ln|P| = kt + \ln 300$ | |
$\ln\left|\frac{P}{300}\right| = kt$, $\frac{P}{300} = e^{kt}$, $P = 300e^{kt}$ | M1 A1 |
## (ii)
$t = 1$, $P = 360 \Rightarrow 360 = 300e^k$ | M1 |
$k = \ln\frac{6}{5} = 0.182 \text{ (3sf)}$ | A1 |
## (iii)
$P = 300e^{0.1821t}$ | |
when $t = 2$, $P = 432$; when $t = 3$, $P = 518$ | B1 |
model does not seem suitable as data diverges from predictions | B1 |
## (iv)
$\int \frac{1}{P} \, dP = \int (0.4 - 0.25\cos 0.5t) \, dt$ | M1 |
$\ln|P| = 0.4t - 0.5\sin 0.5t + c$ | M1 |
$t = 0$, $P = 300 \Rightarrow \ln 300 = c$ | M1 |
$\ln\left|\frac{P}{300}\right| = 0.4t - 0.5\sin 0.5t$ | $[P = 300e^{0.4t - 0.5\sin 0.5t}]$ | A1 |
## (v)
second model: $t = 1, 2, 3 \Rightarrow P = 352, 438, 605$ | B1 |
the second model seems more suitable as it fits the data better | B1 | **(14)** |
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**Total (72)**An entomologist is studying the population of insects in a colony.
Initially there are 300 insects in the colony and in a model, the entomologist assumes that the population, $P$, at time $t$ weeks satisfies the differential equation
$$\frac{dP}{dt} = kP,$$
where $k$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $P$ in terms of $k$ and $t$. [5]
\end{enumerate}
Given that after one week there are 360 insects in the colony,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item find the value of $k$ to 3 significant figures. [2]
\end{enumerate}
Given also that after two and three weeks there are 440 and 600 insects respectively,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item comment on suitability of the modelling assumption. [2]
\end{enumerate}
An alternative model assumes that
$$\frac{dP}{dt} = P(0.4 - 0.25 \cos 0.5t).$$
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Using the initial data, $P = 300$ when $t = 0$, solve this differential equation. [3]
\item Compare the suitability of the two models. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q9 [14]}}