| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Cone: related rates of dimensions |
| Difficulty | Standard +0.3 This is a standard C4 related rates problem involving a cone. Part (a) requires basic geometry (using the 60° angle to find the radius-height relationship) and the cone volume formula—straightforward with clear guidance. Part (b) applies implicit differentiation of V with respect to time, a core C4 technique practiced extensively. The calculations are routine once dV/dt = 120 is given, making this slightly easier than average for C4 material. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks |
|---|---|
| let radius \(= r, \therefore \tan 30° = \frac{r}{\sqrt{3}} = \frac{r}{h}, \quad r = \frac{h}{\sqrt{3}}\) | M1 |
| \(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h \times \frac{1}{3} = \frac{1}{9}\pi h^3\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\frac{dV}{dt} = 120, \quad \frac{dV}{dh} = \frac{1}{3}\pi h^2\) | B1 |
| \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}, \quad 120 = \frac{1}{3}\pi h^2 \frac{dh}{dt}, \quad \frac{dh}{dt} = \frac{360}{\pi h^2}\) | M1 A1 |
| when \(h=6, \quad \frac{dh}{dt} = 3.18 \text{ cm s}^{-1}\) (2dp) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = 8 \times 120 = 960 = \frac{1}{9}\pi h^3 \therefore h = \sqrt[3]{\frac{8 \times 960}{\pi}} = 14.011\) | M1 | |
| \(\therefore \frac{dh}{dt} = 0.58 \text{ cm s}^{-1}\) (2dp) | A1 | (10 marks) |
## (a)
let radius $= r, \therefore \tan 30° = \frac{r}{\sqrt{3}} = \frac{r}{h}, \quad r = \frac{h}{\sqrt{3}}$ | M1 |
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h \times \frac{1}{3} = \frac{1}{9}\pi h^3$ | M1 A1 |
## (b)
### (i)
$\frac{dV}{dt} = 120, \quad \frac{dV}{dh} = \frac{1}{3}\pi h^2$ | B1 |
$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}, \quad 120 = \frac{1}{3}\pi h^2 \frac{dh}{dt}, \quad \frac{dh}{dt} = \frac{360}{\pi h^2}$ | M1 A1 |
when $h=6, \quad \frac{dh}{dt} = 3.18 \text{ cm s}^{-1}$ (2dp) | M1 A1 |
### (ii)
$V = 8 \times 120 = 960 = \frac{1}{9}\pi h^3 \therefore h = \sqrt[3]{\frac{8 \times 960}{\pi}} = 14.011$ | M1 |
$\therefore \frac{dh}{dt} = 0.58 \text{ cm s}^{-1}$ (2dp) | A1 | **(10 marks)**
---\includegraphics{figure_2}
Figure 2 shows a vertical cross-section of a vase.
The inside of the vase is in the shape of a right-circular cone with the angle between the sides in the cross-section being $60°$. When the depth of water in the vase is $h$ cm, the volume of water in the vase is $V$ cm$^3$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \frac{1}{9}\pi h^3$. [3]
\end{enumerate}
The vase is initially empty and water is poured in at a constant rate of 120 cm$^3$ s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, to 2 decimal places, the rate at which $h$ is increasing
\begin{enumerate}[label=(\roman*)]
\item when $h = 6$,
\item after water has been poured in for 8 seconds. [7]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [10]}}