| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with explicitly given non-geometric algebraic relationships |
| Difficulty | Moderate -0.3 This is a straightforward related rates problem with clear scaffolding. Part (i) is trivial substitution, part (ii) is basic differentiation of k/V, and part (iii) applies the chain rule dP/dt = (dP/dV)(dV/dt) with given values. The question guides students through each step and requires only standard C3 techniques with no problem-solving insight needed. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 = k/100 \Rightarrow k = 500\) | E1 | NB answer given |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dP}{dV} = -500V^{-2} = -\frac{500}{V^2}\) | M1 | \((-1)V^{-2}\) o.e. allow \(-k/V^2\) |
| A1 | ||
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dP}{dt} = \frac{dP}{dV} \cdot \frac{dV}{dt}\) | M1 | chain rule (any correct version) |
| When \(V = 100\), \(dP/dV = -500/10000 = -0.05\) | B1ft | (soi) |
| \(dV/dt = 10\) | B1 | (soi) |
| \(\Rightarrow dP/dt = -0.05 \times 10 = -0.5\) | A1 | \(-0.5\) ca |
| So P is decreasing at 0.5 Atm/s | ||
| [4] |
## Question 6:
### Part (i)
| $5 = k/100 \Rightarrow k = 500$ | E1 | NB answer given |
|---|---|---|
| **[1]** | | |
### Part (ii)
| $\frac{dP}{dV} = -500V^{-2} = -\frac{500}{V^2}$ | M1 | $(-1)V^{-2}$ o.e. allow $-k/V^2$ |
|---|---|---|
| | A1 | |
| **[2]** | | |
### Part (iii)
| $\frac{dP}{dt} = \frac{dP}{dV} \cdot \frac{dV}{dt}$ | M1 | chain rule (any correct version) |
|---|---|---|
| When $V = 100$, $dP/dV = -500/10000 = -0.05$ | B1ft | (soi) |
| $dV/dt = 10$ | B1 | (soi) |
| $\Rightarrow dP/dt = -0.05 \times 10 = -0.5$ | A1 | $-0.5$ ca |
| So P is decreasing at 0.5 Atm/s | | |
| **[4]** | | |
---6 When the gas in a balloon is kept at a constant temperature, the pressure $P$ in atmospheres and the volume $V \mathrm {~m} ^ { 3 }$ are related by the equation
$$P = \frac { k } { V } ,$$
where $k$ is a constant. [This is known as Boyle's Law.]\\
When the volume is $100 \mathrm {~m} ^ { 3 }$, the pressure is 5 atmospheres, and the volume is increasing at a rate of $10 \mathrm {~m} ^ { 3 }$ per second.\\
(i) Show that $k = 500$.\\
(ii) Find $\frac { \mathrm { d } P } { \mathrm {~d} V }$ in terms of $V$.\\
(iii) Find the rate at which the pressure is decreasing when $V = 100$.
\hfill \mbox{\textit{OCR MEI C3 Q6 [7]}}