| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Related rates problems |
| Difficulty | Moderate -0.8 This is a straightforward related rates problem requiring basic differentiation of a polynomial (expanding and applying power rule) and simple substitution into the chain rule formula dh/dt = (dV/dt)/(dV/dh). Both parts are routine C3 techniques with no conceptual challenges. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \pi h^2 - \frac{1}{3}\pi h^3 \Rightarrow \frac{dV}{dh} = 2\pi h - \pi h^2\) | M1 | expanding brackets (correctly) or product rule |
| A1 | o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dV}{dt} = 0.02\) | B1 | soi |
| \(\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}\) | M1 | o.e. |
| \(\Rightarrow \frac{dh}{dt} = \frac{0.02}{dV/dh} = \frac{0.02}{2\pi h - \pi h^2}\) | M1dep | substituting \(h = 0.4\) into their \(\frac{dV}{dh}\) and \(\frac{dV}{dt} = 0.02\) |
| When \(h = 0.4\): \(\frac{dh}{dt} = \frac{0.02}{0.8\pi - 0.16\pi} = 0.0099\) m/min | A1cao | 0.01 or better, or \(\frac{1}{32\pi}\) |
| [6] |
## Question 8:
### Part (i)
| $V = \pi h^2 - \frac{1}{3}\pi h^3 \Rightarrow \frac{dV}{dh} = 2\pi h - \pi h^2$ | M1 | expanding brackets (correctly) or product rule |
|---|---|---|
| | A1 | o.e. |
### Part (ii)
| $\frac{dV}{dt} = 0.02$ | B1 | soi |
|---|---|---|
| $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$ | M1 | o.e. |
| $\Rightarrow \frac{dh}{dt} = \frac{0.02}{dV/dh} = \frac{0.02}{2\pi h - \pi h^2}$ | M1dep | substituting $h = 0.4$ into their $\frac{dV}{dh}$ and $\frac{dV}{dt} = 0.02$ |
| When $h = 0.4$: $\frac{dh}{dt} = \frac{0.02}{0.8\pi - 0.16\pi} = 0.0099$ m/min | A1cao | 0.01 or better, or $\frac{1}{32\pi}$ |
| **[6]** | | |8 Fig. 4 is a diagram of a garden pond.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{431d496a-a606-4b92-9f5c-e12b074a7ba9-5_295_742_410_693}
\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 Q8 [6]}}