| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| 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 requiring chain rule application (dV/dt = dV/dr × dr/dt) with straightforward geometry using the 30° angle to relate height and radius. The steps are routine for C3: identify the rate, derive volume formula using similar triangles, differentiate, then substitute. Slightly easier than average due to clear scaffolding and standard technique. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07t Construct differential equations: in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(\frac{dV}{dt} = 2\) | B1 | |
| (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 |
| \(V = \frac{1}{3}\pi r^2 \cdot \sqrt{3}r = \frac{\sqrt{3}}{3}\pi r^3\) * | E1 | e.g. \(\frac{3\sqrt{3}}{3}\pi r^2\) |
| \(\frac{dV}{dr} = \sqrt{3}\pi r^2\) | B1 | |
| (iii) When \(r=2\), \(\frac{dV}{dr} = 4\sqrt{3}\pi\) | M1 | or \(\frac{dr}{dt} = \frac{dr}{dV}\cdot\frac{dV}{dt}\) |
| \(\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt} \Rightarrow 2 = 4\sqrt{3}\pi\,\frac{dr}{dt}\) | M1 | Substituting 2 for \(\frac{dV}{dt}\) and \(r=2\) into their \(\frac{dV}{dr}\) |
| \(\frac{dr}{dt} = \frac{1}{2\sqrt{3}\pi}\) or \(0.092\) cm s\(^{-1}\) | A1cao |
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $\frac{dV}{dt} = 2$ | B1 | |
| **(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 |
| $V = \frac{1}{3}\pi r^2 \cdot \sqrt{3}r = \frac{\sqrt{3}}{3}\pi r^3$ * | E1 | e.g. $\frac{3\sqrt{3}}{3}\pi r^2$ |
| $\frac{dV}{dr} = \sqrt{3}\pi r^2$ | B1 | |
| **(iii)** When $r=2$, $\frac{dV}{dr} = 4\sqrt{3}\pi$ | M1 | or $\frac{dr}{dt} = \frac{dr}{dV}\cdot\frac{dV}{dt}$ |
| $\frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt} \Rightarrow 2 = 4\sqrt{3}\pi\,\frac{dr}{dt}$ | M1 | Substituting 2 for $\frac{dV}{dt}$ and $r=2$ into their $\frac{dV}{dr}$ |
| $\frac{dr}{dt} = \frac{1}{2\sqrt{3}\pi}$ or $0.092$ cm s$^{-1}$ | A1cao | |
---4 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]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-3_369_401_431_831}
\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 2006 Q4 [7]}}