| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with explicitly given non-geometric algebraic relationships |
| Difficulty | Moderate -0.5 This is a straightforward related rates problem requiring only the chain rule (dp/dt = dp/dx × dx/dt) with simple differentiation of p = 100/x. Given all necessary values, it's a direct substitution exercise with minimal steps, making it slightly easier than average for C3. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p = 100/x = 100x^{-1}\) | ||
| \(dp/dx = -100x^{-2} = -100/x^2\) | M1 | attempt to differentiate |
| A1 | \(-100x^{-2}\) o.e. | |
| \(dp/dt = dp/dx \times dx/dt\); \(dx/dt = 10\) | M1 | o.e. soi; condone poor notation if chain rule correct |
| B1 | soi; or \(x = 50 + 10t\): B1 | |
| When \(x=50\), \(dp/dx = -100/50^2\) | M1dep | substituting \(x=50\) into their \(dp/dx\), dep 2nd M1 |
| \(\Rightarrow dp/dt = 10 \times -0.04 = -0.4\) | A1cao [6] | o.e. e.g. decreasing at 0.4 |
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p = 100/x = 100x^{-1}$ | | |
| $dp/dx = -100x^{-2} = -100/x^2$ | M1 | attempt to differentiate |
| | A1 | $-100x^{-2}$ o.e. |
| $dp/dt = dp/dx \times dx/dt$; $dx/dt = 10$ | M1 | o.e. soi; condone poor notation if chain rule correct |
| | B1 | soi; or $x = 50 + 10t$: B1 |
| When $x=50$, $dp/dx = -100/50^2$ | M1dep | substituting $x=50$ into their $dp/dx$, dep 2nd M1 |
| $\Rightarrow dp/dt = 10 \times -0.04 = -0.4$ | A1cao [6] | o.e. e.g. decreasing at 0.4 |
---4 A piston can slide inside a tube which is closed at one end and encloses a quantity of gas (see Fig. 4).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{30d0d728-d6d6-4a54-baf9-a6df8646bf64-2_154_1003_1080_571}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
The pressure of the gas in atmospheric units is given by $p = \frac { 100 } { x }$, where $x \mathrm {~cm}$ is the distance of the piston from the closed end. At a certain moment, $x = 50$, and the piston is being pulled away from the closed end at 10 cm per minute. At what rate is the pressure changing at that time?
\hfill \mbox{\textit{OCR MEI C3 2010 Q4 [6]}}