| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2015 |
| 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 standard related rates problem requiring students to: (1) use the 45° angle to establish r=h, (2) substitute into the cone volume formula to get V in terms of h only, (3) differentiate implicitly with respect to t, and (4) substitute given values. While it involves multiple steps, each step follows a well-established procedure taught in C3, making it slightly easier than average for this level. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = r\) so \(V = \pi h^3/3\) | B1 | o.e. e.g \(\pi h^2 \tan 45°/3\) |
| \(\frac{dV}{dt} = 5\) | B1 | soi (can be implied from \(V = 5t\)) |
| \(\frac{dV}{dh} = \pi h^2\) | B1 ft | must be \(dV/dh\) soi, ft where \(\pi h^3/3\) |
| \(\frac{dV}{dt} = \left(\frac{dV}{dh}\right) \cdot \frac{dh}{dt}\) | M1 | any correct chain rule in \(V, h\) and \(t\) (soi) |
| \(\Rightarrow 5 = 100\pi \frac{dh}{dt}\) | A1 | 0.016 or better; accept 1/(20π) o.e., but mark final answer |
| Total: [5] | e.g. from a correct chain rule; but must have substituted for \(r\); e.g. \(dh/dt = dh/dV \times dV/dt\); 0.01591549… penalise incorrect rounding |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = 5t\) so \(\pi h^3/3 = 5t\) | B1 | |
| \(\Rightarrow \pi h^2 dh/dt = 5\) | M1 | or 5 dt/dh = π h^2 o.e. |
| \(\Rightarrow dh/dt = 5/\pi h^2 = 5/100\pi = 0.016 \, \text{cm s}^{-1}\) | A1 | 0.016 or better; accept 1/(20π) o.e., but mark final answer |
| Total: [5] | penalise incorrect rounding |
$h = r$ so $V = \pi h^3/3$ | B1 | o.e. e.g $\pi h^2 \tan 45°/3$
$\frac{dV}{dt} = 5$ | B1 | soi (can be implied from $V = 5t$)
$\frac{dV}{dh} = \pi h^2$ | B1 ft | must be $dV/dh$ soi, ft where $\pi h^3/3$
$\frac{dV}{dt} = \left(\frac{dV}{dh}\right) \cdot \frac{dh}{dt}$ | M1 | any correct chain rule in $V, h$ and $t$ (soi)
$\Rightarrow 5 = 100\pi \frac{dh}{dt}$ | A1 | 0.016 or better; accept 1/(20π) o.e., but mark final answer
**Total: [5]** | | e.g. from a correct chain rule; but must have substituted for $r$; e.g. $dh/dt = dh/dV \times dV/dt$; 0.01591549… penalise incorrect rounding
**or**
$V = 5t$ so $\pi h^3/3 = 5t$ | B1 |
$\Rightarrow \pi h^2 dh/dt = 5$ | M1 | or 5 dt/dh = π h^2 o.e.
$\Rightarrow dh/dt = 5/\pi h^2 = 5/100\pi = 0.016 \, \text{cm s}^{-1}$ | A1 | 0.016 or better; accept 1/(20π) o.e., but mark final answer
**Total: [5]** | | penalise incorrect rounding
---4 Fig. 4 shows a cone with its axis vertical. The angle between the axis and the slant edge is $45 ^ { \circ }$. Water is poured into the cone at a constant rate of $5 \mathrm {~cm} ^ { 3 }$ per second. At time $t$ seconds, the height of the water surface above the vertex O of the cone is $h \mathrm {~cm}$, and the volume of water in the cone is $V \mathrm {~cm} ^ { 3 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{955bebfb-04a3-4cd9-a33e-a8ba4b73e2ba-2_296_405_1804_831}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Find $V$ in terms of $h$.\\
Hence find the rate at which the height of water is increasing when the height is 10 cm .\\[0pt]
[You are given that the volume $V$ of a cone of height $h$ and radius $r$ is $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$ ].
\hfill \mbox{\textit{OCR MEI C3 2015 Q4 [5]}}