| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (polynomial/exponential x-side) |
| Difficulty | Standard +0.3 This is a standard C4 separable differential equation question with straightforward integration and substitution. Part (a) requires routine separation of variables and applying a boundary condition, while part (b) involves simple evaluation and comparison. The context is slightly unusual but the mathematical techniques are entirely standard for this module. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int x\,dx = \int k(s-t)\,dt\) | M1 | |
| \(\frac{1}{3}x^3 = k(st - \frac{1}{2}t^2) + c\) | M1 A1 | |
| \(t = 0, x = 0 \Rightarrow c = 0\) | B1 | |
| \(t = 2, x = 96 \Rightarrow 4608 = 8k\), \(k = 576\) | M1 A1 | |
| \(t = 1 \Rightarrow \frac{1}{3}x^3 = 576 \times \frac{3}{2}\), \(x = \sqrt[3]{5184} = 72\) | M1 A1 | |
| \(3\text{ hours }5\text{ mins} \Rightarrow t = 3.0833, x = \sqrt{12284} = 110.83\) | M1 A1 | |
| \(\therefore \frac{dx}{dt} = \frac{576(s - 3.0833)}{110.83} = 9.96\), \(\frac{dx}{dt} < 10\) so she should have left | M1 A1 | (12 marks) |
$\int x\,dx = \int k(s-t)\,dt$ | M1 |
$\frac{1}{3}x^3 = k(st - \frac{1}{2}t^2) + c$ | M1 A1 |
$t = 0, x = 0 \Rightarrow c = 0$ | B1 |
$t = 2, x = 96 \Rightarrow 4608 = 8k$, $k = 576$ | M1 A1 |
$t = 1 \Rightarrow \frac{1}{3}x^3 = 576 \times \frac{3}{2}$, $x = \sqrt[3]{5184} = 72$ | M1 A1 |
$3\text{ hours }5\text{ mins} \Rightarrow t = 3.0833, x = \sqrt{12284} = 110.83$ | M1 A1 |
$\therefore \frac{dx}{dt} = \frac{576(s - 3.0833)}{110.83} = 9.96$, $\frac{dx}{dt} < 10$ so she should have left | M1 A1 | (12 marks)
---A mathematician is selling goods at a car boot sale. She believes that the rate at which she makes sales depends on the length of time since the start of the sale, $t$ hours, and the total value of sales she has made up to that time, £$x$.
She uses the model
$$\frac{dx}{dt} = \frac{k(5-t)}{x},$$
where $k$ is a constant.
Given that after two hours she has made sales of £96 in total,
\begin{enumerate}[label=(\alph*)]
\item solve the differential equation and show that she made £72 in the first hour of the sale. [8]
\end{enumerate}
The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than £10 per hour.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Verify that at 3 hours and 5 minutes after the start of the sale, she should have already left. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q4 [12]}}