| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Related rates problems |
| Difficulty | Moderate -0.3 This is a straightforward related rates problem requiring basic differentiation of a polynomial (expanding then differentiating, or using product rule) and simple substitution with the chain rule. Part (i) is routine calculus, part (ii) applies dV/dt = (dV/dh)(dh/dt) with given values. Slightly easier than average due to clear structure and standard technique. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi h^2 - \frac{1}{3}\pi h^3\) | M1 A1 | expanding brackets (correctly) or product rule oe |
| \(\Rightarrow \frac{dV}{dh} = 2\pi h - \pi h^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dt} = -0.02\) | B1 | soi |
| \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\) | M1 | \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\) oe |
| \(\Rightarrow \frac{dh}{dt} = \frac{dV/dt}{dV/dh} = \frac{0.02}{2\pi h - \pi h^2}\) | M1dep | substituting \(h = 0.4\) into their \(\frac{dV}{dh}\) and \(\frac{dV}{dt} = -0.02\) |
| When \(h = 0.4, \Rightarrow \frac{dh}{dt} = \frac{0.02}{0.8\pi - 0.16\pi} = 0.0099 m/\text{min}\) | A1cao | \(0.01\) or better or \(\frac{1}{32\pi}\) |
| [6] |
**(i)**
$V = \pi h^2 - \frac{1}{3}\pi h^3$ | M1 A1 | expanding brackets (correctly) or product rule oe
$\Rightarrow \frac{dV}{dh} = 2\pi h - \pi h^2$ |
**(ii)**
$\frac{dV}{dt} = -0.02$ | B1 | soi
$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$ | M1 | $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$ oe
$\Rightarrow \frac{dh}{dt} = \frac{dV/dt}{dV/dh} = \frac{0.02}{2\pi h - \pi h^2}$ | M1dep | substituting $h = 0.4$ into their $\frac{dV}{dh}$ and $\frac{dV}{dt} = -0.02$
When $h = 0.4, \Rightarrow \frac{dh}{dt} = \frac{0.02}{0.8\pi - 0.16\pi} = 0.0099 m/\text{min}$ | A1cao | $0.01$ or better or $\frac{1}{32\pi}$
| [6] |4 Fig. 4 is a diagram of a garden pond.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{72de4365-c3f0-4106-8daf-3047afeab723-3_296_746_351_657}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
The volume $V \mathrm {~m} ^ { 3 }$ of water in the pond when the depth is $h$ metres is given by
$$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 3 - h )$$
(i) Find $\frac { \mathrm { d } V } { \mathrm {~d} h }$.
Water is poured into the pond at the rate of $0.02 \mathrm {~m} ^ { 3 }$ per minute.\\
(ii) Find the value of $\frac { \mathrm { d } h } { \mathrm {~d} t }$ when $h = 0.4$.
\hfill \mbox{\textit{OCR MEI C3 2006 Q4 [6]}}