| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find stationary points of parametric curve |
| Difficulty | Standard +0.8 This is a substantial multi-part parametric equations question requiring dy/dx via chain rule, finding stationary points, algebraic manipulation to eliminate the parameter, and volume of revolution integration. While each individual technique is standard C4 material, the question demands sustained accuracy across multiple steps, particularly the cartesian conversion and the final volume integral, placing it moderately 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{d\theta} = 2\cos 2\theta - 2\sin\theta\), \(\frac{dx}{d\theta} = 2\cos\theta\) | B1, B1 | |
| \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\) | M1 | Substituting for theirs |
| \(\frac{dy}{dx} = \frac{2\cos 2\theta - 2\sin\theta}{2\cos\theta} = \frac{\cos 2\theta - \sin\theta}{\cos\theta}\) | A1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(\theta = \frac{\pi}{6}\): \(\frac{dy}{dx} = \frac{\cos(\pi/3) - \sin(\pi/6)}{\cos(\pi/6)} = \frac{1/2 - 1/2}{\sqrt{3}/2} = 0\) | E1 | |
| Coords of B: \(x = 2 + 2\sin(\pi/6) = 3\); \(y = 2\cos(\pi/6) + \sin(\pi/3) = 3\sqrt{3}/2\) | M1, A1, A1 | For either; exact |
| \(BC = 2 \times \frac{3\sqrt{3}}{2} = 3\sqrt{3}\) | B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (A) \(y = 2\cos\theta + \sin 2\theta = 2\cos\theta + 2\sin\theta\cos\theta = 2\cos\theta(1+\sin\theta) = x\cos\theta\) | M1 | \(\sin 2\theta = 2\sin\theta\cos\theta\) |
| Result \(y = x\cos\theta\) * | E1 | |
| (B) \(\sin\theta = \frac{1}{2}(x-2)\); \(\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{4}(x-2)^2 = 1 - \frac{1}{4}x^2 + x - 1 = x - \frac{1}{4}x^2\) | B1, M1, E1 | |
| Result \(\cos^2\theta = x - \frac{1}{4}x^2\) * | ||
| (C) Cartesian equation: \(y^2 = x^2\cos^2\theta = x^2(x - \frac{1}{4}x^2) = x^3 - \frac{1}{4}x^4\) * | M1, E1 | Squaring and substituting for \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = \int_0^4 \pi y^2\, dx\) | M1 | Need limits |
| \(= \pi\int_0^4\left(x^3 - \frac{1}{4}x^4\right)dx = \pi\left[\frac{1}{4}x^4 - \frac{1}{20}x^5\right]_0^4\) | B1 | \(\left[\frac{1}{4}x^4 - \frac{1}{20}x^5\right]\) |
| \(= \pi(64 - 51.2) = 12.8\pi \approx 40.2\ (\text{m}^3)\) | A1 | \(12.8\pi\) or 40 or better |
## Question 6:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{d\theta} = 2\cos 2\theta - 2\sin\theta$, $\frac{dx}{d\theta} = 2\cos\theta$ | B1, B1 | |
| $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ | M1 | Substituting for theirs |
| $\frac{dy}{dx} = \frac{2\cos 2\theta - 2\sin\theta}{2\cos\theta} = \frac{\cos 2\theta - \sin\theta}{\cos\theta}$ | A1 | oe |
**[4 marks]**
---
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $\theta = \frac{\pi}{6}$: $\frac{dy}{dx} = \frac{\cos(\pi/3) - \sin(\pi/6)}{\cos(\pi/6)} = \frac{1/2 - 1/2}{\sqrt{3}/2} = 0$ | E1 | |
| Coords of B: $x = 2 + 2\sin(\pi/6) = 3$; $y = 2\cos(\pi/6) + \sin(\pi/3) = 3\sqrt{3}/2$ | M1, A1, A1 | For either; exact |
| $BC = 2 \times \frac{3\sqrt{3}}{2} = 3\sqrt{3}$ | B1ft | |
**[5 marks]**
---
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| **(A)** $y = 2\cos\theta + \sin 2\theta = 2\cos\theta + 2\sin\theta\cos\theta = 2\cos\theta(1+\sin\theta) = x\cos\theta$ | M1 | $\sin 2\theta = 2\sin\theta\cos\theta$ |
| Result $y = x\cos\theta$ * | E1 | |
| **(B)** $\sin\theta = \frac{1}{2}(x-2)$; $\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{4}(x-2)^2 = 1 - \frac{1}{4}x^2 + x - 1 = x - \frac{1}{4}x^2$ | B1, M1, E1 | |
| Result $\cos^2\theta = x - \frac{1}{4}x^2$ * | | |
| **(C)** Cartesian equation: $y^2 = x^2\cos^2\theta = x^2(x - \frac{1}{4}x^2) = x^3 - \frac{1}{4}x^4$ * | M1, E1 | Squaring and substituting for $x$ |
**[7 marks]**
---
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^4 \pi y^2\, dx$ | M1 | Need limits |
| $= \pi\int_0^4\left(x^3 - \frac{1}{4}x^4\right)dx = \pi\left[\frac{1}{4}x^4 - \frac{1}{20}x^5\right]_0^4$ | B1 | $\left[\frac{1}{4}x^4 - \frac{1}{20}x^5\right]$ |
| $= \pi(64 - 51.2) = 12.8\pi \approx 40.2\ (\text{m}^3)$ | A1 | $12.8\pi$ or 40 or better |
**[3 marks]**6 Fig. 8 illustrates a hot air balloon on its side. The balloon is modelled by the volume of revolution about the $x$-axis of the curve with parametric equations
$$x = 2 + 2 \sin \theta , \quad y = 2 \cos \theta + \sin 2 \theta , \quad ( 0 \leqslant \theta \leqslant 2 \pi ) .$$
The curve crosses the $x$-axis at the point $\mathrm { A } ( 4,0 )$. B and C are maximum and minimum points on the curve. Units on the axes are metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{252453c9-9afa-435c-b64b-5ea37ec69eed-6_812_801_517_706}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $\theta$.
\item Verify that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $\theta = \frac { 1 } { 6 } \pi$, and find the exact coordinates of B .
Hence find the maximum width BC of the balloon.
\item (A) Show that $y = x \cos \theta$.\\
(B) Find $\sin \theta$ in terms of $x$ and show that $\cos ^ { 2 } \theta = x - \frac { 1 } { 4 } x ^ { 2 }$.\\
(C) Hence show that the cartesian equation of the curve is $y ^ { 2 } = x ^ { 3 } - \frac { 1 } { 4 } x ^ { 4 }$.
\item Find the volume of the balloon.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C4 Q6 [19]}}