| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| 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 with clear scaffolding through three parts. Part (i) is trivial recall, part (ii) requires basic trigonometry to relate h and r then substitution into the volume formula, and part (iii) applies the chain rule dV/dt = (dV/dr)(dr/dt). While it involves multiple steps, each is routine and the question structure guides students through the solution method, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| \(\frac{dV}{dt} = 2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan 30 = \frac{1}{\sqrt{3}} = \frac{r}{h} \Rightarrow h = \sqrt{3}\, r\) | M1 | Correct relationship between \(r\) and \(h\) in any form; from exact working only |
| \(\Rightarrow V = \frac{1}{3}\pi r^2 \cdot \sqrt{3}r = \frac{\sqrt{3}}{3}\pi r^3\) | E1 | o.e. e.g. \((3\sqrt{3}/3)\pi r^2\) |
| \(\frac{dV}{dr} = \sqrt{3}\pi r^2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| When \(r = 2\), \(dV/dr = 4\sqrt{3}\pi\) | ||
| \(\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}\) | M1 | or \(\frac{dr}{dt} = \frac{dr}{dV} \cdot \frac{dV}{dt}\), substituting 2 for \(dV/dt\) and \(r=2\) into their \(dV/dr\) |
| \(\Rightarrow 2 = 4\sqrt{3}\pi\, \frac{dr}{dt}\) | M1 | |
| \(\Rightarrow dr/dt = \frac{1}{2\sqrt{3}\pi}\) or \(0.092\) cm s\(^{-1}\) | A1cao | |
| [7] |
## Question 7:
### Part (i)
| $\frac{dV}{dt} = 2$ | B1 | |
|---|---|---|
### Part (ii)
| $\tan 30 = \frac{1}{\sqrt{3}} = \frac{r}{h} \Rightarrow h = \sqrt{3}\, r$ | M1 | Correct relationship between $r$ and $h$ in any form; from exact working only |
|---|---|---|
| $\Rightarrow V = \frac{1}{3}\pi r^2 \cdot \sqrt{3}r = \frac{\sqrt{3}}{3}\pi r^3$ | E1 | o.e. e.g. $(3\sqrt{3}/3)\pi r^2$ |
| $\frac{dV}{dr} = \sqrt{3}\pi r^2$ | B1 | |
### Part (iii)
| When $r = 2$, $dV/dr = 4\sqrt{3}\pi$ | | |
|---|---|---|
| $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$ | M1 | or $\frac{dr}{dt} = \frac{dr}{dV} \cdot \frac{dV}{dt}$, substituting 2 for $dV/dt$ and $r=2$ into their $dV/dr$ |
| $\Rightarrow 2 = 4\sqrt{3}\pi\, \frac{dr}{dt}$ | M1 | |
| $\Rightarrow dr/dt = \frac{1}{2\sqrt{3}\pi}$ or $0.092$ cm s$^{-1}$ | A1cao | |
| **[7]** | | |
---7 Fig. 4 shows a cone. The angle between the axis and the slant edge is $30 ^ { \circ }$. Water is poured into the cone at a constant rate of $2 \mathrm {~cm} ^ { 3 }$ per second. At time $t$ seconds, the radius of the water surface is $r \mathrm {~cm}$ 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-e12b074a7ba9-4_363_391_1447_887}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Write down the value of $\frac { \mathrm { d } V } { \mathrm {~d} t }$.\\
(ii) Show that $V = \frac { \sqrt { 3 } } { 3 } \pi r ^ { 3 }$, and find $\frac { \mathrm { d } V } { \mathrm {~d} r }$.\\[0pt]
[You may assume that the volume of a cone of height $h$ and radius $r$ is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
(iii) Use the results of parts (i) and (ii) to find the value of $\frac { \mathrm { d } r } { \mathrm {~d} t }$ when $r = 2$.
\hfill \mbox{\textit{OCR MEI C3 Q7 [7]}}