| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| 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 similar triangles to eliminate r, then differentiating V with respect to time. Part (a) is routine algebraic manipulation, and part (b) applies the chain rule dV/dt = (dV/dh)(dh/dt). While it requires multiple steps, this is a textbook C4 question with no novel insight needed—slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Similar shapes \(\Rightarrow\) either \(\dfrac{\frac{1}{3}\pi(16)^2 24}{V} = \left(\dfrac{24}{h}\right)^3\) or \(\dfrac{V}{\frac{1}{3}\pi(16)^2 24} = \left(\dfrac{h}{24}\right)^3\) | M1 | Uses similar shapes to find either one of these two expressions |
| \(V = 2048\pi \times \left(\dfrac{h}{24}\right)^3 = \dfrac{4\pi h^3}{27}\) AG | A1 | Substitutes their equation to give the correct formula for the volume of water \(V\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From question, \(\dfrac{dV}{dt} = 8 \Rightarrow V = 8t\ (+c)\) | B1 | \(\dfrac{dV}{dt} = 8\) or \(V = 8t\) |
| \(h = \left(\dfrac{27V}{4\pi}\right)^{\frac{1}{3}} \Rightarrow h = \left(\dfrac{27(8t)}{4\pi}\right)^{\frac{1}{3}} = \left(\dfrac{54t}{\pi}\right)^{\frac{1}{3}} = 3\left(\dfrac{2t}{\pi}\right)^{\frac{1}{3}}\) | B1 | \(\left(\dfrac{27(8t)}{4\pi}\right)^{\frac{1}{3}}\) or \(\left(\dfrac{54t}{\pi}\right)^{\frac{1}{3}}\) or \(3\left(\dfrac{2t}{\pi}\right)^{\frac{1}{3}}\) |
| \(\dfrac{dh}{dt} = 3\left(\dfrac{2}{\pi}\right)^{\frac{1}{3}} \cdot \dfrac{1}{3} t^{-\frac{2}{3}}\) | M1; A1 oe | \(\dfrac{dh}{dt} = \pm kt^{-\frac{2}{3}}\); \(\dfrac{dh}{dt} = 3\left(\dfrac{2}{\pi}\right)^{\frac{1}{3}} \dfrac{1}{3} t^{-\frac{2}{3}}\) |
| When \(h=12\), \(t = \left(\dfrac{12}{3}\right)^3 \times \dfrac{\pi}{2} = 32\pi\) | ||
| So when \(h=12\), \(\dfrac{dh}{dt} = \left(\dfrac{2}{\pi}\right)^{\frac{1}{3}}\left(\dfrac{1}{32\pi}\right)^{\frac{2}{3}} = \left(\dfrac{2}{1024\pi^3}\right)^{\frac{1}{3}} = \dfrac{1}{8\pi}\) | A1 oe | \(\dfrac{1}{8\pi}\) |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Similar shapes $\Rightarrow$ either $\dfrac{\frac{1}{3}\pi(16)^2 24}{V} = \left(\dfrac{24}{h}\right)^3$ or $\dfrac{V}{\frac{1}{3}\pi(16)^2 24} = \left(\dfrac{h}{24}\right)^3$ | M1 | Uses similar shapes to find either one of these two expressions |
| $V = 2048\pi \times \left(\dfrac{h}{24}\right)^3 = \dfrac{4\pi h^3}{27}$ **AG** | A1 | Substitutes their equation to give the correct formula for the volume of water $V$ |
*Note: Candidates simply writing $V = \frac{4}{9} \times \frac{1}{3}\pi h^3$ or $V = \frac{1}{3}\pi\left(\frac{16}{24}\right)^2 h^3$ would be awarded M0A0.*
**Total: [2]**
---
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| From question, $\dfrac{dV}{dt} = 8 \Rightarrow V = 8t\ (+c)$ | B1 | $\dfrac{dV}{dt} = 8$ or $V = 8t$ |
| $h = \left(\dfrac{27V}{4\pi}\right)^{\frac{1}{3}} \Rightarrow h = \left(\dfrac{27(8t)}{4\pi}\right)^{\frac{1}{3}} = \left(\dfrac{54t}{\pi}\right)^{\frac{1}{3}} = 3\left(\dfrac{2t}{\pi}\right)^{\frac{1}{3}}$ | B1 | $\left(\dfrac{27(8t)}{4\pi}\right)^{\frac{1}{3}}$ or $\left(\dfrac{54t}{\pi}\right)^{\frac{1}{3}}$ or $3\left(\dfrac{2t}{\pi}\right)^{\frac{1}{3}}$ |
| $\dfrac{dh}{dt} = 3\left(\dfrac{2}{\pi}\right)^{\frac{1}{3}} \cdot \dfrac{1}{3} t^{-\frac{2}{3}}$ | M1; A1 oe | $\dfrac{dh}{dt} = \pm kt^{-\frac{2}{3}}$; $\dfrac{dh}{dt} = 3\left(\dfrac{2}{\pi}\right)^{\frac{1}{3}} \dfrac{1}{3} t^{-\frac{2}{3}}$ |
| When $h=12$, $t = \left(\dfrac{12}{3}\right)^3 \times \dfrac{\pi}{2} = 32\pi$ | | |
| So when $h=12$, $\dfrac{dh}{dt} = \left(\dfrac{2}{\pi}\right)^{\frac{1}{3}}\left(\dfrac{1}{32\pi}\right)^{\frac{2}{3}} = \left(\dfrac{2}{1024\pi^3}\right)^{\frac{1}{3}} = \dfrac{1}{8\pi}$ | A1 oe | $\dfrac{1}{8\pi}$ |
**Total: [5]**
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-09_696_686_196_626}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm , as shown in Figure 2. Water is flowing into the container. When the height of water is $h \mathrm {~cm}$, the surface of the water has radius $r \mathrm {~cm}$ and the volume of water is $V \mathrm {~cm} ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = \frac { 4 \pi h ^ { 3 } } { 27 }$.\\[0pt]
[The volume $V$ of a right circular cone with vertical height $h$ and base radius $r$ is given by the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$.]
Water flows into the container at a rate of $8 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\item Find, in terms of $\pi$, the rate of change of $h$ when $h = 12$.
\section*{Question 5 continued}
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$\\
$\_\_\_\_$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q5 [7]}}