| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (tan/sec/cot/cosec identities) |
| Difficulty | Standard +0.8 This is a multi-part parametric equations question requiring differentiation using the chain rule, conversion to Cartesian form using trigonometric identities, and volume of revolution. While each individual step uses standard techniques, the combination of skills and the need to work with sec/tan identities and set up the volume integral correctly makes this moderately challenging, above average difficulty. |
| 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 |
# Question 1
## (i)
$x = \sec\theta$, $y = 2\tan\theta$
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\sec^2\theta}{\sec\theta\tan\theta}$
$= \frac{2\sec\theta}{\tan\theta} = \frac{2\cos\theta}{\sin\theta} = 2\csc\theta$
**M1** for their $\frac{dy/d\theta}{dx/d\theta}$ in terms of $\theta$
**A1** cao (oe) allow for unsimplified form even if subsequently cancelled incorrectly (i.e. can isw)
**A1** cao www (NB AG) — must be at least one intermediate step between $\frac{2\sec^2\theta}{\sec\theta\tan\theta}$ or $2\csc\theta$
[3]
## (ii)
$x^2 = \sec^2\theta = 1 + \tan^2\theta = 1 + \frac{1}{4}y^2$
$y^2 = 4(x^2 - 1) = 4x^2 - 4$
**M1** for $\sec^2\theta = 1 + \tan^2\theta$ (oe) used
**A1** www NB AG
**OR**
$4\tan^2\theta = 4 \cdot \frac{\sec^2\theta - 1}{\tan^2\theta}\tan^2\theta$ is true
$1 + \tan^2\theta = \sec^2\theta$ which is true
**B1** Correct substitution of $x$ and $y$ into the given answer
**B1 dep** Dependent on previous mark — must simplify/remove the factor of 4 from each term and state that the correctly derived trig identity is true
[2]
## (iii)
$V = \int_1^2 \pi y^2 \, dx = \int_1^2 \pi(4x^2 - 4) \, dx$
$= \pi\left[\frac{4x^3}{3} - 4x\right]_1^2$
$= \pi\left(\frac{32}{3} - 8 - \frac{4}{3} + 4\right) = \frac{16\pi}{3}$
**M1** for $\int_1^2 k(4x^2 - 4) \, dx$ with $k = \pi$ or $k = 1/\pi$, allow correct limits later (condone lack of $dx$)
**B1** for $(4/3)x^3 - 4x$ or $(2/3)x^3 - 2x$
**A1** exact — mark final answer
[3]1 A curve has parametric equations $x = \sec \theta , y = 2 \tan \theta$.\\
(i) Given that the derivative of $\sec \theta$ is $\sec \theta \tan \theta$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \operatorname { cosec } \theta$.\\
(ii) Verify that the cartesian equation of the curve is $y ^ { 2 } = 4 x ^ { 2 } - 4$.
Fig. 5 shows the region enclosed by the curve and the line $x = 2$. This region is rotated through $180 ^ { \circ }$ about the $x$-axis.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1e788cb0-36b0-42a9-9e0c-077022d410ae-1_556_867_580_588}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(iii) Find the volume of revolution produced, giving your answer in exact form.
\hfill \mbox{\textit{OCR MEI C4 Q1 [8]}}