| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Moderate -0.3 This is a straightforward separable variables question with standard techniques: part (a) requires forming a simple differential equation from a verbal description, and part (b) involves routine separation, integration of power functions, and substitution of initial conditions. The algebra is uncomplicated and the methods are textbook-standard for C4, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| When \(t = 0\), \(e^{-kt} = 1\), so \(R = L + (U-L) = U\) | B1 | |
| This is consistent with Fig. 2 where the record starts at \(U\) (the initial upper value) | B1 | |
| \(R\) decreasing from \(U\) is consistent with the graph | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| As \(t \to \infty\), \(e^{-kt} \to 0\) | B1 | |
| So \(R \to L\), consistent with the asymptote at \(L\) in Fig. 2 | B1 |
# Question 4(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| When $t = 0$, $e^{-kt} = 1$, so $R = L + (U-L) = U$ | B1 | |
| This is consistent with Fig. 2 where the record starts at $U$ (the initial upper value) | B1 | |
| $R$ decreasing from $U$ is consistent with the graph | B1 | |
# Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| As $t \to \infty$, $e^{-kt} \to 0$ | B1 | |
| So $R \to L$, consistent with the asymptote at $L$ in Fig. 2 | B1 | |
---4
\begin{enumerate}[label=(\alph*)]
\item The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating $x$, the number of bacteria, to the time $t$.
\item In another colony, the number of bacteria, $y$, after time $t$ minutes is modelled by the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 10000 } { \sqrt { y } } .$$
Find $y$ in terms of $t$, given that $y = 900$ when $t = 0$. Hence find the number of bacteria after 10 minutes.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 2006 Q4 [13]}}