| 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 standard related rates problem requiring students to: (1) use geometry (45° angle means r=h) to express V in terms of h alone, (2) differentiate implicitly with respect to t, and (3) substitute given values. While it involves multiple steps, each step follows a well-established procedure taught in C3, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| B1 | \(h = r\) so \(V = \frac{h^3}{3}\) |
| B1 | \(\frac{dV}{dt} = 5\) o.e. e.g. \(\frac{h^3 \tan 45°}{3}\) |
| B1 | \(\frac{dV}{dh} = h^2\) |
| M1 | \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\) (any correct chain rule in \(V\), \(h\) and \(t\) soi) or e.g. \(\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}\) or \(5 \frac{dt}{dh} = h^2\) o.e. |
| M1 | must be \(\frac{dV}{dh}\) soi, ft their \(\frac{h^3}{3}\) but must have substituted for \(r\) |
| A1 | \(5 = 100 \frac{dh}{dt}\) and \(\frac{dh}{dt} = 5/100 = 0.016\) cm s\(^{-1}\) or 0.016 or better; accept \(\frac{1}{20}\) o.e., but mark final answer and penalise incorrect rounding |
| Answer | Marks |
|---|---|
| B1 | \(V = 5t\) so \(h^3/3 = 5t\) |
| M1 | \(h^2 \frac{dh}{dt} = 5\) |
| A1 | \(\frac{dh}{dt} = \frac{5}{h^2} = \frac{5}{100} = 0.016\) cm s\(^{-1}\) or 0.016 or better; accept \(\frac{1}{20}\) o.e., but mark final answer and penalise incorrect rounding |
# Question 1
B1 | $h = r$ so $V = \frac{h^3}{3}$
B1 | $\frac{dV}{dt} = 5$ o.e. e.g. $\frac{h^3 \tan 45°}{3}$
B1 | $\frac{dV}{dh} = h^2$
M1 | $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$ (any correct chain rule in $V$, $h$ and $t$ soi) or e.g. $\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$ or $5 \frac{dt}{dh} = h^2$ o.e.
M1 | must be $\frac{dV}{dh}$ soi, ft their $\frac{h^3}{3}$ but must have substituted for $r$
A1 | $5 = 100 \frac{dh}{dt}$ and $\frac{dh}{dt} = 5/100 = 0.016$ cm s$^{-1}$ or 0.016 or better; accept $\frac{1}{20}$ o.e., but mark final answer and penalise incorrect rounding
*Alternative:*
B1 | $V = 5t$ so $h^3/3 = 5t$
M1 | $h^2 \frac{dh}{dt} = 5$
A1 | $\frac{dh}{dt} = \frac{5}{h^2} = \frac{5}{100} = 0.016$ cm s$^{-1}$ or 0.016 or better; accept $\frac{1}{20}$ o.e., but mark final answer and penalise incorrect rounding1 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]{431d496a-a606-4b92-9f5c-e12b074a7ba9-1_295_403_542_871}
\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 Q1 [5]}}