| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with cones, hemispheres, and bowls (variable depth) |
| Difficulty | Standard +0.3 This is a straightforward related rates problem requiring differentiation of V = πh² with respect to time, substitution of given values (dV/dt = 10, h = 5), and solving for dh/dt. It's slightly easier than average as it involves only one differentiation step and direct substitution with no geometric complexity or multi-step reasoning. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi h^2 \Rightarrow \frac{dV}{dh} = 2\pi h\) | M1A1 | if derivative \(2\pi h\) seen without \(dV/dh = \ldots\) allow M1A0 |
| \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\) | M1 | soi; any correct statement of chain rule using \(V\), \(h\) and \(t\) |
| \(\frac{dV}{dt} = 10\) | B1 | soi; if letter other than \(t\) used (and not defined) B0 |
| \(\frac{dh}{dt} = \frac{10}{2\pi \times 5} = \frac{1}{\pi}\) | A1 | or 0.32 or better, mark final answer |
| [5] |
## Question 4:
| $V = \pi h^2 \Rightarrow \frac{dV}{dh} = 2\pi h$ | M1A1 | if derivative $2\pi h$ seen without $dV/dh = \ldots$ allow M1A0 |
|---|---|---|
| $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$ | M1 | soi; any correct statement of chain rule using $V$, $h$ and $t$ |
| $\frac{dV}{dt} = 10$ | B1 | soi; if letter other than $t$ used (and not defined) B0 |
| $\frac{dh}{dt} = \frac{10}{2\pi \times 5} = \frac{1}{\pi}$ | A1 | or 0.32 or better, mark final answer |
| **[5]** | | |
---4 Water flows into a bowl at a constant rate of $10 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ (see Fig. 4).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-2_414_379_485_838}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
When the depth of water in the bowl is $h \mathrm {~cm}$, the volume of water is $V \mathrm {~cm} ^ { 3 }$, where $V = \pi h ^ { 2 }$. Find the rate at which the depth is increasing at the instant in time when the depth is 5 cm .
\hfill \mbox{\textit{OCR MEI C3 Q4 [5]}}