| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| 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, then substituting given values. It's slightly easier than average because it's a direct application of the chain rule with clear setup and only one substitution step, though it does require understanding the relationship between rates of change. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V = \pi h^2 \Rightarrow dV/dh = 2\pi h \Rightarrow\) | M1A1 | if derivative \(2\pi h\) seen without \(dV/dh = \ldots\) allow M1A0 |
| \(dV/dt = dV/dh \times dh/dt\) | M1 | soi; o.e. – any correct statement of the chain rule using \(V\), \(h\) and \(t\) – condone use of a letter other than \(t\) for time here |
| \(dV/dt = 10\) | B1 | soi; if a letter other than \(t\) used (and not defined) B0 |
| \(dh/dt = \frac{10}{(2\pi \times 5)} = \frac{1}{\pi}\) | A1 | or 0.32 or better, mark final answer |
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $V = \pi h^2 \Rightarrow dV/dh = 2\pi h \Rightarrow$ | M1A1 | if derivative $2\pi h$ seen without $dV/dh = \ldots$ allow M1A0 |
| $dV/dt = dV/dh \times dh/dt$ | M1 | soi; o.e. – any correct statement of the chain rule using $V$, $h$ and $t$ – condone use of a letter other than $t$ for time here |
| $dV/dt = 10$ | B1 | soi; if a letter other than $t$ used (and not defined) B0 |
| $dh/dt = \frac{10}{(2\pi \times 5)} = \frac{1}{\pi}$ | A1 | or 0.32 or better, mark final answer |
---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]{28ce1bcc-e9d5-4ae6-98c0-67b5b8c50bc6-3_422_385_1628_815}
\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 2013 Q4 [5]}}