| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric volume of revolution |
| Difficulty | Standard +0.8 This is a multi-part parametric question requiring: (a) routine parameter finding, (b) normal line calculation with parametric differentiation, and (c) volume of revolution using parametric integration with trigonometric substitution. Part (c) requires careful setup of the integral V = π∫y²dx with parametric conversion and non-trivial trigonometric integration. The combination of techniques and algebraic manipulation elevates this above average difficulty, though it follows standard C4 methods. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\tan\theta = \sqrt{3}\) or \(\sin\theta = \frac{\sqrt{3}}{2}\) | M1 | |
| \(\theta = \frac{\pi}{3}\) | A1 | awrt 1.05, (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{dx}{d\theta} = \sec^2\theta\), \(\frac{dy}{d\theta} = \cos\theta\) | ||
| \(\frac{dy}{dx} = \frac{\cos\theta}{\sec^2\theta} \left(= \cos^3\theta\right)\) | M1 A1 | |
| At \(P\): \(m = \cos^3\!\left(\frac{\pi}{3}\right) = \frac{1}{8}\) | A1 | Can be implied |
| Using \(mm' = -1\), \(m' = -8\) | M1 | |
| For normal: \(y - \frac{1}{2}\sqrt{3} = -8\!\left(x - \sqrt{3}\right)\) | dM1 | |
| At \(Q\), \(y=0\): \(-\frac{1}{2}\sqrt{3} = -8\!\left(x - \sqrt{3}\right)\) | ||
| \(x = \frac{17}{16}\sqrt{3}\) \(\left(k = \frac{17}{16}\right)\), \(1.0625\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\int y^2\,dx = \int y^2 \frac{dx}{d\theta}\,d\theta = \int \sin^2\theta\sec^2\theta\,d\theta\) | M1 A1 | |
| \(= \int \tan^2\theta\,d\theta\) | A1 | |
| \(= \int(\sec^2\theta - 1)\,d\theta\) | dM1 | |
| \(= \tan\theta - \theta \quad (+C)\) | A1 | |
| \(V = \pi\int_0^{\frac{\pi}{3}} y^2\,dx = \left[\tan\theta - \theta\right]_0^{\frac{\pi}{3}} = \pi\!\left[\left(\sqrt{3} - \frac{\pi}{3}\right)-(0-0)\right]\) | dM1 | |
| \(= \sqrt{3}\pi - \frac{1}{3}\pi^2 \quad \left(p=1,\, q=-\frac{1}{3}\right)\) | A1 | (7), (15 marks) |
## Question 10:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\tan\theta = \sqrt{3}$ or $\sin\theta = \frac{\sqrt{3}}{2}$ | M1 | |
| $\theta = \frac{\pi}{3}$ | A1 | awrt 1.05, (2) |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = \sec^2\theta$, $\frac{dy}{d\theta} = \cos\theta$ | | |
| $\frac{dy}{dx} = \frac{\cos\theta}{\sec^2\theta} \left(= \cos^3\theta\right)$ | M1 A1 | |
| At $P$: $m = \cos^3\!\left(\frac{\pi}{3}\right) = \frac{1}{8}$ | A1 | Can be implied |
| Using $mm' = -1$, $m' = -8$ | M1 | |
| For normal: $y - \frac{1}{2}\sqrt{3} = -8\!\left(x - \sqrt{3}\right)$ | dM1 | |
| At $Q$, $y=0$: $-\frac{1}{2}\sqrt{3} = -8\!\left(x - \sqrt{3}\right)$ | | |
| $x = \frac{17}{16}\sqrt{3}$ $\left(k = \frac{17}{16}\right)$, $1.0625$ | A1 | (6) |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\int y^2\,dx = \int y^2 \frac{dx}{d\theta}\,d\theta = \int \sin^2\theta\sec^2\theta\,d\theta$ | M1 A1 | |
| $= \int \tan^2\theta\,d\theta$ | A1 | |
| $= \int(\sec^2\theta - 1)\,d\theta$ | dM1 | |
| $= \tan\theta - \theta \quad (+C)$ | A1 | |
| $V = \pi\int_0^{\frac{\pi}{3}} y^2\,dx = \left[\tan\theta - \theta\right]_0^{\frac{\pi}{3}} = \pi\!\left[\left(\sqrt{3} - \frac{\pi}{3}\right)-(0-0)\right]$ | dM1 | |
| $= \sqrt{3}\pi - \frac{1}{3}\pi^2 \quad \left(p=1,\, q=-\frac{1}{3}\right)$ | A1 | (7), **(15 marks)** |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e375f6ad-4a76-42a0-b7bf-ae47e5cbdaeb-34_599_923_322_571}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows part of the curve $C$ with parametric equations
$$x = \tan \theta , \quad y = \sin \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
The point $P$ lies on $C$ and has coordinates $\left( \sqrt { 3 } , \frac { 1 } { 2 } \sqrt { 3 } \right)$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\theta$ at the point $P$.
The line $l$ is a normal to $C$ at $P$. The normal cuts the $x$-axis at the point $Q$.
\item Show that $Q$ has coordinates $( k \sqrt { 3 } , 0 )$, giving the value of the constant $k$.
The finite shaded region $S$ shown in Figure 3 is bounded by the curve $C$, the line $x = \sqrt { 3 }$ and the $x$-axis. This shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
\item Find the volume of the solid of revolution, giving your answer in the form $p \pi \sqrt { 3 } + q \pi ^ { 2 }$, where $p$ and $q$ are constants.
\includegraphics[max width=\textwidth, alt={}, center]{e375f6ad-4a76-42a0-b7bf-ae47e5cbdaeb-39_61_29_2608_1886}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 Q10 [15]}}