| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Intrinsic equation and curvature |
| Difficulty | Challenging +1.8 This is a substantial FP3 question requiring multiple advanced techniques: parametric arc length integration with trigonometric simplification, intrinsic equations, curvature formulas, and surface of revolution. While each part follows established methods, the multi-step nature, need for careful trigonometric manipulation (especially the half-angle identities), and integration of several Further Maths concepts makes this significantly harder than typical A-level questions but not exceptionally difficult for FP3 standard. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08e Mean value of function: using integral8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^2 = [a(1+\cos\theta)]^2+(a\sin\theta)^2\) | M1, A1 | Forming \(\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^2\) |
| \(= a^2(2+2\cos\theta) = 4a^2\cos^2\frac{1}{2}\theta\) | M1 | Using half-angle formula |
| \(s = \int 2a\cos\frac{1}{2}\theta\,\mathrm{d}\theta = 4a\sin\frac{1}{2}\theta + C\) | M1, A1 | Integrating to obtain \(k\sin\frac{1}{2}\theta\); correctly obtained (\(+C\) not needed) |
| \(s=0\) when \(\theta=0 \Rightarrow C=0\) | A1 ag | 6 Dependent on all previous marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{a\sin\theta}{a(1+\cos\theta)}\) | M1 | |
| \(= \frac{2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta}{2a\cos^2\frac{1}{2}\theta} = \tan\frac{1}{2}\theta\) | M1, A1 | Using half-angle formulae |
| \(\psi = \frac{1}{2}\theta\), and so \(s = 4a\sin\psi\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\rho = \frac{\mathrm{d}s}{\mathrm{d}\psi} = 4a\cos\psi = 4a\cos\frac{1}{2}\theta\) | M1, A1 ft, A1 ag | Differentiating intrinsic equation; 3 marks |
| OR: \(\rho = \frac{(4a^2\cos^2\frac{1}{2}\theta)^{3/2}}{a(1+\cos\theta)(a\cos\theta)-(-a\sin\theta)(a\sin\theta)}\) | M1A1 ft | Correct expression for \(\rho\) or \(\kappa\) |
| \(= \frac{8a^3\cos^3\frac{1}{2}\theta}{a^2(1+\cos\theta)} = \frac{8a^3\cos^3\frac{1}{2}\theta}{2a^2\cos^2\frac{1}{2}\theta} = 4a\cos\frac{1}{2}\theta\) | A1 ag |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(\theta=\frac{2}{3}\pi\): \(\psi=\frac{1}{3}\pi\), \(x=a(\frac{2}{3}\pi+\frac{1}{2}\sqrt{3})\), \(y=\frac{3}{2}a\), \(\rho=2a\) | B1 | |
| \(\hat{\mathbf{n}} = \begin{pmatrix}-\sin\psi\\\cos\psi\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}\sqrt{3}\\\frac{1}{2}\end{pmatrix}\) | M1, A1 | Obtaining a normal vector; correct unit normal (possibly in terms of \(\theta\)) |
| \(\mathbf{c} = \begin{pmatrix}a(\frac{2}{3}\pi+\frac{1}{2}\sqrt{3})\\\frac{3}{2}a\end{pmatrix} + 2a\begin{pmatrix}-\frac{1}{2}\sqrt{3}\\\frac{1}{2}\end{pmatrix}\) | M1 | |
| Centre of curvature is \(\left(a(\frac{2}{3}\pi-\frac{1}{2}\sqrt{3}),\ \frac{5}{2}a\right)\) | A1A1 | Accept \((1.23a,\ 2.5a)\); 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Curved surface area is \(\int 2\pi y \, ds\) | M1 | |
| \(= \int_0^{\pi} 2\pi a(1-\cos\theta) \cdot 2a\cos\frac{1}{2}\theta \, d\theta\) | A1 ft | Correct integral expression in any form (including limits; may be implied by later working) |
| \(= \int_0^{\pi} 8\pi a^2 \sin^2\frac{1}{2}\theta \cos\frac{1}{2}\theta \, d\theta\) | M1 | Obtaining an integrable form |
| \(= \left[\frac{16}{3}\pi a^2 \sin^3\frac{1}{2}\theta\right]_0^{\pi}\) | M1 | Obtaining \(k\sin^3\frac{1}{2}\theta\) or equivalent |
| \(= \frac{16}{3}\pi a^2\) | A1 | |
| Total: 5 |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^2 = [a(1+\cos\theta)]^2+(a\sin\theta)^2$ | M1, A1 | Forming $\left(\frac{\mathrm{d}x}{\mathrm{d}\theta}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}\theta}\right)^2$ |
| $= a^2(2+2\cos\theta) = 4a^2\cos^2\frac{1}{2}\theta$ | M1 | Using half-angle formula |
| $s = \int 2a\cos\frac{1}{2}\theta\,\mathrm{d}\theta = 4a\sin\frac{1}{2}\theta + C$ | M1, A1 | Integrating to obtain $k\sin\frac{1}{2}\theta$; correctly obtained ($+C$ not needed) |
| $s=0$ when $\theta=0 \Rightarrow C=0$ | A1 ag | **6** Dependent on all previous marks |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{a\sin\theta}{a(1+\cos\theta)}$ | M1 | |
| $= \frac{2\sin\frac{1}{2}\theta\cos\frac{1}{2}\theta}{2a\cos^2\frac{1}{2}\theta} = \tan\frac{1}{2}\theta$ | M1, A1 | Using half-angle formulae |
| $\psi = \frac{1}{2}\theta$, and so $s = 4a\sin\psi$ | A1 | **4 marks** |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\rho = \frac{\mathrm{d}s}{\mathrm{d}\psi} = 4a\cos\psi = 4a\cos\frac{1}{2}\theta$ | M1, A1 ft, A1 ag | Differentiating intrinsic equation; **3 marks** |
| OR: $\rho = \frac{(4a^2\cos^2\frac{1}{2}\theta)^{3/2}}{a(1+\cos\theta)(a\cos\theta)-(-a\sin\theta)(a\sin\theta)}$ | M1A1 ft | Correct expression for $\rho$ or $\kappa$ |
| $= \frac{8a^3\cos^3\frac{1}{2}\theta}{a^2(1+\cos\theta)} = \frac{8a^3\cos^3\frac{1}{2}\theta}{2a^2\cos^2\frac{1}{2}\theta} = 4a\cos\frac{1}{2}\theta$ | A1 ag | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $\theta=\frac{2}{3}\pi$: $\psi=\frac{1}{3}\pi$, $x=a(\frac{2}{3}\pi+\frac{1}{2}\sqrt{3})$, $y=\frac{3}{2}a$, $\rho=2a$ | B1 | |
| $\hat{\mathbf{n}} = \begin{pmatrix}-\sin\psi\\\cos\psi\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}\sqrt{3}\\\frac{1}{2}\end{pmatrix}$ | M1, A1 | Obtaining a normal vector; correct unit normal (possibly in terms of $\theta$) |
| $\mathbf{c} = \begin{pmatrix}a(\frac{2}{3}\pi+\frac{1}{2}\sqrt{3})\\\frac{3}{2}a\end{pmatrix} + 2a\begin{pmatrix}-\frac{1}{2}\sqrt{3}\\\frac{1}{2}\end{pmatrix}$ | M1 | |
| Centre of curvature is $\left(a(\frac{2}{3}\pi-\frac{1}{2}\sqrt{3}),\ \frac{5}{2}a\right)$ | A1A1 | Accept $(1.23a,\ 2.5a)$; **6 marks** |
# Question (v) - Curved Surface Area:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Curved surface area is $\int 2\pi y \, ds$ | M1 | |
| $= \int_0^{\pi} 2\pi a(1-\cos\theta) \cdot 2a\cos\frac{1}{2}\theta \, d\theta$ | A1 ft | Correct integral expression in any form (including limits; may be implied by later working) |
| $= \int_0^{\pi} 8\pi a^2 \sin^2\frac{1}{2}\theta \cos\frac{1}{2}\theta \, d\theta$ | M1 | Obtaining an integrable form |
| $= \left[\frac{16}{3}\pi a^2 \sin^3\frac{1}{2}\theta\right]_0^{\pi}$ | M1 | Obtaining $k\sin^3\frac{1}{2}\theta$ or equivalent |
| $= \frac{16}{3}\pi a^2$ | A1 | |
| **Total: 5** | | |
---3 A curve has parametric equations $x = a ( \theta + \sin \theta ) , y = a ( 1 - \cos \theta )$, for $0 \leqslant \theta \leqslant \pi$, where $a$ is a positive constant.\\
(i) Show that the arc length $s$ from the origin to a general point on the curve is given by $s = 4 a \sin \frac { 1 } { 2 } \theta$.\\
(ii) Find the intrinsic equation of the curve giving $s$ in terms of $a$ and $\psi$, where $\tan \psi = \frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(iii) Hence, or otherwise, show that the radius of curvature at a point on the curve is $4 a \cos \frac { 1 } { 2 } \theta$.\\
(iv) Find the coordinates of the centre of curvature corresponding to the point on the curve where $\theta = \frac { 2 } { 3 } \pi$.\\
(v) Find the area of the surface generated when the curve is rotated through $2 \pi$ radians about the $x$-axis.
\hfill \mbox{\textit{OCR MEI FP3 2009 Q3 [24]}}