| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 11 |
| 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 straightforward separable variables question with a real-world context. The separation and integration are routine (leading to x²/2 = k(5t - t²/2) + C), and finding constants from given conditions is standard C4 technique. Part (ii) is simple substitution and arithmetic. Slightly easier than average due to the clean algebraic manipulation required. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context |
| Answer | Marks |
|---|---|
| \(\int x\, dx = \int k(5-t)\, dt\) | M1 |
| \(\frac{1}{2}x^2 = k(5t - \frac{1}{2}t^2) + c\) | M1 A1 |
| \(t=0,\ x=0 \Rightarrow c=0\) | B1 |
| \(t=2,\ x=96 \Rightarrow 4608 = 8k,\quad k=576\) | M1 |
| \(t=1 \Rightarrow \frac{1}{2}x^2 = 576\times\frac{9}{2},\quad x=\sqrt{5184}=72\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 hours 5 mins \(\Rightarrow t=3.0833,\ x=\sqrt{12284}=110.83\) | M1 A1 | |
| \(\therefore \frac{dx}{dt} = \frac{576(5-3.0833)}{110.83} = 9.96\), \(\quad\frac{dx}{dt}<10\) so she should have left | M1 A1 | (11) |
# Question 7:
## Part (i)
| $\int x\, dx = \int k(5-t)\, dt$ | M1 | |
|---|---|---|
| $\frac{1}{2}x^2 = k(5t - \frac{1}{2}t^2) + c$ | M1 A1 | |
| $t=0,\ x=0 \Rightarrow c=0$ | B1 | |
| $t=2,\ x=96 \Rightarrow 4608 = 8k,\quad k=576$ | M1 | |
| $t=1 \Rightarrow \frac{1}{2}x^2 = 576\times\frac{9}{2},\quad x=\sqrt{5184}=72$ | M1 A1 | |
## Part (ii)
| 3 hours 5 mins $\Rightarrow t=3.0833,\ x=\sqrt{12284}=110.83$ | M1 A1 | |
|---|---|---|
| $\therefore \frac{dx}{dt} = \frac{576(5-3.0833)}{110.83} = 9.96$, $\quad\frac{dx}{dt}<10$ so she should have left | M1 A1 | **(11)** |
---7. 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, $\pounds x$.
She uses the model
$$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k ( 5 - t ) } { x }$$
where $k$ is a constant.\\
Given that after two hours she has made sales of $\pounds 96$ in total,\\
(i) solve the differential equation and show that she made $\pounds 72$ in the first hour of the sale.
The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than $\pounds 10$ per hour.\\
(ii) Verify that at 3 hours and 5 minutes after the start of the sale, she should have already left.\\
\hfill \mbox{\textit{OCR C4 Q7 [11]}}