| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| 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 simple translation to a differential equation (dx/dt = k√x), and part (b) involves routine separation, integration of y^(-1/2), and substituting initial conditions. The arithmetic is clean and the method is textbook-standard, making it slightly easier than average for A-level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = k\sqrt{x}\) | M1 | \(\frac{dx}{dt} = \ldots\) |
| A1 [2] | \(k\sqrt{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dt} = \frac{10000}{\sqrt{y}}\) | ||
| \(\Rightarrow \int\sqrt{y}\,dy = \int 10000\,dt\) | M1 | separating variables |
| \(\Rightarrow \frac{2}{3}y^{\frac{3}{2}} = 10000t + c\) | A1 | condone omission of \(c\) |
| When \(t=0\), \(y=900 \Rightarrow 18000 = c\) | B1 | evaluating constant for their integral |
| \(\Rightarrow y = \left[\frac{3}{2}(10000t+18000)\right]^{\frac{2}{3}}\) | A1 | any correct expression for \(y=\) |
| \(= (1500(10t+18))^{\frac{2}{3}}\) | ||
| When \(t=10\), \(y = 3152\) | M1 | for method: substituting \(t=10\) in their expression |
| A1 [6] | cao |
## Question 6:
### Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = k\sqrt{x}$ | M1 | $\frac{dx}{dt} = \ldots$ |
| | A1 [2] | $k\sqrt{x}$ |
### Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dt} = \frac{10000}{\sqrt{y}}$ | | |
| $\Rightarrow \int\sqrt{y}\,dy = \int 10000\,dt$ | M1 | separating variables |
| $\Rightarrow \frac{2}{3}y^{\frac{3}{2}} = 10000t + c$ | A1 | condone omission of $c$ |
| When $t=0$, $y=900 \Rightarrow 18000 = c$ | B1 | evaluating constant for their integral |
| $\Rightarrow y = \left[\frac{3}{2}(10000t+18000)\right]^{\frac{2}{3}}$ | A1 | any correct expression for $y=$ |
| $= (1500(10t+18))^{\frac{2}{3}}$ | | |
| When $t=10$, $y = 3152$ | M1 | for method: substituting $t=10$ in their expression |
| | A1 [6] | cao |6
\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 Q6 [8]}}