| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 16 |
| 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 exponential growth models and separable variables. Part (a) is routine separation of variables with initial conditions, (b) is straightforward substitution and logarithms, (c) requires simple comparison of predicted vs actual values, and (d) involves separating variables with a slightly more complex integrand (sin term). Part (e) is qualitative comparison. While multi-part with 16 marks total, each component uses standard techniques without requiring novel insight or particularly challenging integration—slightly easier than the typical average A-level question. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| 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 | |
| \(\ln | P | = kt + \ln 300\) |
| \(\ln\left | \frac{P}{300}\right | = kt\), \(P = 300e^{kt}\) |
| Answer | Marks |
|---|---|
| \(t = 1, P = 360 \Rightarrow 360 = 300e^k\) | M1 |
| \(k = \ln\frac{6}{5} = 0.182\) (3sf) | A1 |
| Answer | Marks |
|---|---|
| \(P = 300e^{0.1823t}\) 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 | |
| \(\ln | P | = 0.4t - 0.5\sin 0.5t + c\) |
| \(t = 0, P = 300 \Rightarrow \ln 300 = c\) | ||
| \(\ln\left | \frac{P}{300}\right | = 0.4t - 0.5\sin 0.5t\) |
| Answer | Marks | Guidance |
|---|---|---|
| second model: \(t = 1, 2, 3 \Rightarrow P = 352, 438, 605\) | M1 A1 | |
| the second model seems more suitable as it fits the data better | B1 | (16) |
**(a)**
$\int \frac{1}{P} \, dP = \int k \, dt$ | M1 |
$\ln|P| = kt + c$ | A1 |
$t = 0, P = 300 \Rightarrow \ln 300 = c$ | M1 |
$\ln|P| = kt + \ln 300$ | |
$\ln\left|\frac{P}{300}\right| = kt$, $P = 300e^{kt}$ | M1 A1 |
**(b)**
$t = 1, P = 360 \Rightarrow 360 = 300e^k$ | M1 |
$k = \ln\frac{6}{5} = 0.182$ (3sf) | A1 |
**(c)**
$P = 300e^{0.1823t}$ when $t = 2, P = 432$; when $t = 3, P = 518$ | B1 |
model does not seem suitable as data diverges from predictions | B1 |
**(d)**
$\int \frac{1}{P} \, dP = \int (0.4 - 0.25\cos 0.5t) \, dt$ | M1 |
$\ln|P| = 0.4t - 0.5\sin 0.5t + c$ | A1 |
$t = 0, P = 300 \Rightarrow \ln 300 = c$ | |
$\ln\left|\frac{P}{300}\right| = 0.4t - 0.5\sin 0.5t$ | M1 A1 | $[P = 300e^{0.4t - 0.5\sin 0.5t}]$ |
**(e)**
second model: $t = 1, 2, 3 \Rightarrow P = 352, 438, 605$ | M1 A1 |
the second model seems more suitable as it fits the data better | B1 | (16)
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**Total: (75)**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=(\alph*)]
\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=(\alph*)]
\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=(\alph*)]
\setcounter{enumi}{2}
\item comment on suitability of the model. [2]
\end{enumerate}
An alternative model assumes that
$$\frac{dP}{dt} = P(0.4 - 0.25\cos 0.5t).$$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Using the initial data, $P = 300$ when $t = 0$, solve this differential equation. [4]
\item Compare the suitability of the two models. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q8 [16]}}