| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Exponential growth/decay - approach to limit (dN/dt = k(N - N₀)) |
| Difficulty | Moderate -0.3 This is a standard C4 differential equations question with straightforward setup and solution. Part (a) requires simple algebraic manipulation of rates, part (b) is routine separation of variables with given initial condition, and part (c) involves finding a limit. The question follows a predictable template with no novel insights required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Rate in \(= 30\), rate out \(= \frac{2}{15}V\) | B1 | Identifying both rates |
| \(\frac{dV}{dt} = 30 - \frac{2}{15}V\) | M1 | Forming DE |
| Multiply both sides by \(-15\): \(-15\frac{dV}{dt} = 2V - 450\) | A1 | Correct rearrangement shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{dV}{2V-450} = \int \frac{dt}{-15}\) | M1 | Separating variables |
| \(\frac{1}{2}\ln | 2V-450 | = -\frac{t}{15} + c\) |
| \(t=0, V=1000\): \(\frac{1}{2}\ln(1550) = c\) | M1 | Applying initial condition |
| \(\ln | 2V-450 | = -\frac{2t}{15} + \ln(1550)\) |
| \(2V - 450 = 1550e^{-2t/15}\) | A1 | Correct exponential form |
| \(V = 225 + 775e^{-2t/15}\) | A1 A1 | Correct final solution |
| Answer | Marks | Guidance |
|---|---|---|
| As \(t \to \infty\), \(e^{-2t/15} \to 0\), so \(V \to 225\) | B1 | Correct limiting value |
# Question 6:
## Part (a)
| Rate in $= 30$, rate out $= \frac{2}{15}V$ | B1 | Identifying both rates |
| $\frac{dV}{dt} = 30 - \frac{2}{15}V$ | M1 | Forming DE |
| Multiply both sides by $-15$: $-15\frac{dV}{dt} = 2V - 450$ | A1 | Correct rearrangement shown |
## Part (b)
| $\int \frac{dV}{2V-450} = \int \frac{dt}{-15}$ | M1 | Separating variables |
| $\frac{1}{2}\ln|2V-450| = -\frac{t}{15} + c$ | M1 A1 | Integrating both sides correctly |
| $t=0, V=1000$: $\frac{1}{2}\ln(1550) = c$ | M1 | Applying initial condition |
| $\ln|2V-450| = -\frac{2t}{15} + \ln(1550)$ | | |
| $2V - 450 = 1550e^{-2t/15}$ | A1 | Correct exponential form |
| $V = 225 + 775e^{-2t/15}$ | A1 A1 | Correct final solution |
## Part (c)
| As $t \to \infty$, $e^{-2t/15} \to 0$, so $V \to 225$ | B1 | Correct limiting value |
---6. Liquid is poured into a container at a constant rate of $30 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$. At time $t$ seconds liquid is leaking from the container at a rate of $\frac { 2 } { 15 } V \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$, where $V \mathrm {~cm} ^ { 3 }$ is the volume of liquid in the container at that time.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$- 15 \frac { \mathrm {~d} V } { \mathrm {~d} t } = 2 V - 450$$
Given that $V = 1000$ when $t = 0$,
\item find the solution of the differential equation, in the form $V = \mathrm { f } ( t )$.
\item Find the limiting value of $V$ as $t \rightarrow \infty$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [11]}}