| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric volume of revolution |
| Difficulty | Challenging +1.2 This is a standard C4 parametric volume of revolution question requiring multiple techniques: finding a parameter value, computing dy/dx via the chain rule, finding a normal equation, and setting up the volume integral V = π∫y²dx with parametric substitution. While it involves several steps and careful algebraic manipulation (especially the trigonometric integration), each component is a well-practiced technique from the C4 syllabus with no novel insight required. The multi-part structure and integration complexity place it slightly above average difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07s Parametric and implicit differentiation4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\tan\theta = \sqrt{3}\) or \(\sin\theta = \frac{\sqrt{3}}{2}\) | M1 | |
| \(\theta = \frac{\pi}{3}\) | A1 | awrt 1.05 |
| (b) \(\frac{dx}{d\theta} = \sec^2\theta, \frac{dy}{d\theta} = \cos\theta\) | M1 A1 | |
| \(\frac{dy}{dx} = \frac{\cos\theta}{\sec^2\theta} (= \cos^3\theta)\) | M1 A1 | |
| At \(P\): \(m = \cos^3\left(\frac{\pi}{3}\right) = \frac{1}{8}\) Can be implied | A1 | |
| Using \(mm' = -1\): \(m' = -8\) | M1 | |
| For normal: \(y - \frac{1}{2}\sqrt{3} = -8(x - \sqrt{3})\) | M1 | |
| At \(Q\): \(y = 0\): \(-\frac{1}{2}\sqrt{3} = -8(x - \sqrt{3})\) | M1 | |
| leading to \(x = \frac{17}{16}\sqrt{3}\) (\(k = \frac{17}{16}\)) | A1 | 1.0625 |
| (c) \(\int y^2 dx = \int y^2\frac{dx}{d\theta} d\theta = \int\sin^2\theta\sec^2\theta d\theta\) | M1 A1 | |
| \(= \int\tan^2\theta d\theta\) | A1 | |
| \(= \int(\sec^2\theta - 1) d\theta\) | M1 | |
| \(= \tan\theta - \theta (+C)\) | A1 | |
| \(V = \pi\int_0^{\frac{\pi}{3}} y^2 dx = [\tan\theta - \theta]_0^{\frac{\pi}{3}} = \pi[(\sqrt{3} - \frac{\pi}{3}) - (0 - 0)]\) | M1 | |
| \(= \sqrt{3}\pi - \frac{1}{3}\pi^2\) (\(p = 1, q = -\frac{1}{3}\)) | A1 | (7) [15] |
**(a)** $\tan\theta = \sqrt{3}$ or $\sin\theta = \frac{\sqrt{3}}{2}$ | M1 |
$\theta = \frac{\pi}{3}$ | A1 | awrt 1.05 | (2)
**(b)** $\frac{dx}{d\theta} = \sec^2\theta, \frac{dy}{d\theta} = \cos\theta$ | M1 A1 |
$\frac{dy}{dx} = \frac{\cos\theta}{\sec^2\theta} (= \cos^3\theta)$ | M1 A1 |
At $P$: $m = \cos^3\left(\frac{\pi}{3}\right) = \frac{1}{8}$ Can be implied | A1 |
Using $mm' = -1$: $m' = -8$ | M1 |
For normal: $y - \frac{1}{2}\sqrt{3} = -8(x - \sqrt{3})$ | M1 |
At $Q$: $y = 0$: $-\frac{1}{2}\sqrt{3} = -8(x - \sqrt{3})$ | M1 |
leading to $x = \frac{17}{16}\sqrt{3}$ ($k = \frac{17}{16}$) | A1 | 1.0625 | (6)
**(c)** $\int y^2 dx = \int y^2\frac{dx}{d\theta} d\theta = \int\sin^2\theta\sec^2\theta d\theta$ | M1 A1 |
$= \int\tan^2\theta d\theta$ | A1 |
$= \int(\sec^2\theta - 1) d\theta$ | M1 |
$= \tan\theta - \theta (+C)$ | A1 |
$V = \pi\int_0^{\frac{\pi}{3}} y^2 dx = [\tan\theta - \theta]_0^{\frac{\pi}{3}} = \pi[(\sqrt{3} - \frac{\pi}{3}) - (0 - 0)]$ | M1 |
$= \sqrt{3}\pi - \frac{1}{3}\pi^2$ ($p = 1, q = -\frac{1}{3}$) | A1 | (7) [15] |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-10_643_999_276_475}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows part of the curve $C$ with parametric equations
$$x = \tan \theta , \quad y = \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
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.
Question 7 continued\\
8. (a) Find $\int ( 4 y + 3 ) ^ { - \frac { 1 } { 2 } } \mathrm {~d} y$\\
(b) Given that $y = 1.5$ at $x = - 2$, solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { } ( 4 y + 3 ) } { x ^ { 2 } }$$
giving your answer in the form $y = \mathrm { f } ( x )$.\\
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\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2011 Q7 [15]}}