| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Intrinsic equation and curvature |
| Difficulty | Challenging +1.8 This FP3 question requires multiple advanced techniques: parametric arc length integration with trigonometric substitution, intrinsic equations relating s and ψ, radius of curvature calculations, and surface of revolution. While each component follows standard FP3 methods, the multi-part nature, algebraic manipulation of trigonometric identities, and integration of the surface area formula make this substantially harder than typical A-level questions but still within standard Further Maths scope. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dx}{d\theta} = 3a\cos^2\theta\sin\theta,\quad \dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta\) | B1 | |
| \(\left(\dfrac{dx}{d\theta}\right)^2 + \left(\dfrac{dy}{d\theta}\right)^2 = (3a\sin\theta\cos\theta)^2(\cos^2\theta + \sin^2\theta)\) | M1 | |
| \(= (3a\sin\theta\cos\theta)^2\) | A1 | |
| \(s = \int 3a\sin\theta\cos\theta\, d\theta\) | M1A1 | FT. A1 requires workable integral form |
| \(= \tfrac{3}{2}a\sin^2\theta\ (+c)\) | A1 | |
| \(s = 0\) when \(\theta = 0 \Rightarrow c = 0\) | Or \(\int_0^\theta \ldots\) used. Required for final E1 | |
| \(\tan\psi = \dfrac{dy}{dx} = \dfrac{3a\sin^2\theta\cos\theta}{3a\cos^2\theta\sin\theta}\) | M1 | |
| \(= \tan\theta\) | A1 | |
| Hence \(\psi = \theta\) and \(s = \tfrac{3}{2}a\sin^2\psi\) | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\rho = \dfrac{ds}{d\psi} = 3a\sin\psi\cos\psi\) | M1A1 | |
| When \(\theta = \dfrac{\pi}{6}\): \(\psi = \dfrac{\pi}{6}\), \(\rho = 3a\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)\) | M1 | |
| Radius of curvature is \(\dfrac{3\sqrt{3}}{4}a\) | A1 | Condone use of \(\psi = \frac{\pi}{6}\) even if \(\psi = \theta\) not established in (i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(\theta = \dfrac{\pi}{6}\): \(\dot{x} = \dfrac{9a}{8}\), \(\dot{y} = \dfrac{3\sqrt{3}a}{8}\) | M1 | Obtaining second derivatives |
| \(\ddot{x} = \dfrac{3\sqrt{3}a}{8}\), \(\ddot{y} = \dfrac{15a}{8}\) | A1 | May be implied by later work |
| \(\rho = \dfrac{(\dot{x}^2+\dot{y}^2)^{3/2}}{\dot{x}\ddot{y} - \dot{y}\ddot{x}} = \dfrac{\left(\frac{81a^2}{64}+\frac{27a^2}{64}\right)^{3/2}}{\left(\frac{9a}{8}\right)\left(\frac{15a}{8}\right) - \left(\frac{3\sqrt{3}a}{8}\right)\left(\frac{3\sqrt{3}a}{8}\right)}\) | M1 | Applying formula for \(\rho\) or \(\kappa\) |
| Radius of curvature is \(\dfrac{3\sqrt{3}}{4}a\) | A1 | |
| Answer | Marks | Guidance |
| Normal vector is \(\hat{\mathbf{n}} = \begin{pmatrix}-\sin\psi\\\cos\psi\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{pmatrix}\) | M1 | Obtaining gradient or normal vector |
| A1 | Correct normal vector. Not necessarily unit vector | |
| \(\mathbf{c} = \begin{pmatrix}a\!\left(1-\frac{3\sqrt{3}}{8}\right)\\\frac{1}{8}a\end{pmatrix} + \dfrac{3\sqrt{3}}{4}a\begin{pmatrix}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{pmatrix}\) | M1 | Must use unit normal here |
| Centre of curvature is \(\left(a\!\left(1-\dfrac{3\sqrt{3}}{4}\right),\ \dfrac{5a}{4}\right)\) | A1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Surface area is \(\int 2\pi x \, ds\) | M1 | |
| \(= \int_0^{\frac{\pi}{3}} 2\pi a(1-\cos^3\theta)(3a\sin\theta\cos\theta) \, d\theta\) | A1 | FT; Correct limits required |
| \(= 6\pi a^2 \int_0^{\frac{\pi}{3}} (\sin\theta\cos\theta - \sin\theta\cos^4\theta) \, d\theta\) | ||
| \(= 6\pi a^2 \left[\frac{1}{2}\sin^2\theta + \frac{1}{5}\cos^5\theta\right]_0^{\frac{\pi}{3}}\) | M1 A1A1 | For \(\frac{1}{2}\sin^2\theta\) and \(\frac{1}{5}\cos^5\theta\); at least one trigonometric term or equivalent |
| \(= \frac{87\pi a^2}{80}\) | A1 | |
| [6] |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dx}{d\theta} = 3a\cos^2\theta\sin\theta,\quad \dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$ | B1 | |
| $\left(\dfrac{dx}{d\theta}\right)^2 + \left(\dfrac{dy}{d\theta}\right)^2 = (3a\sin\theta\cos\theta)^2(\cos^2\theta + \sin^2\theta)$ | M1 | |
| $= (3a\sin\theta\cos\theta)^2$ | A1 | |
| $s = \int 3a\sin\theta\cos\theta\, d\theta$ | M1A1 | FT. A1 requires workable integral form |
| $= \tfrac{3}{2}a\sin^2\theta\ (+c)$ | A1 | |
| $s = 0$ when $\theta = 0 \Rightarrow c = 0$ | | Or $\int_0^\theta \ldots$ used. Required for final E1 |
| $\tan\psi = \dfrac{dy}{dx} = \dfrac{3a\sin^2\theta\cos\theta}{3a\cos^2\theta\sin\theta}$ | M1 | |
| $= \tan\theta$ | A1 | |
| Hence $\psi = \theta$ and $s = \tfrac{3}{2}a\sin^2\psi$ | E1 | Correctly shown |
**[9 marks]**
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\rho = \dfrac{ds}{d\psi} = 3a\sin\psi\cos\psi$ | M1A1 | |
| When $\theta = \dfrac{\pi}{6}$: $\psi = \dfrac{\pi}{6}$, $\rho = 3a\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)$ | M1 | |
| Radius of curvature is $\dfrac{3\sqrt{3}}{4}a$ | A1 | Condone use of $\psi = \frac{\pi}{6}$ even if $\psi = \theta$ not established in (i) |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $\theta = \dfrac{\pi}{6}$: $\dot{x} = \dfrac{9a}{8}$, $\dot{y} = \dfrac{3\sqrt{3}a}{8}$ | M1 | Obtaining second derivatives |
| $\ddot{x} = \dfrac{3\sqrt{3}a}{8}$, $\ddot{y} = \dfrac{15a}{8}$ | A1 | May be implied by later work |
| $\rho = \dfrac{(\dot{x}^2+\dot{y}^2)^{3/2}}{\dot{x}\ddot{y} - \dot{y}\ddot{x}} = \dfrac{\left(\frac{81a^2}{64}+\frac{27a^2}{64}\right)^{3/2}}{\left(\frac{9a}{8}\right)\left(\frac{15a}{8}\right) - \left(\frac{3\sqrt{3}a}{8}\right)\left(\frac{3\sqrt{3}a}{8}\right)}$ | M1 | Applying formula for $\rho$ or $\kappa$ |
| Radius of curvature is $\dfrac{3\sqrt{3}}{4}a$ | A1 | |
| Answer | Marks | Guidance |
|--------|-------|----------|
| Normal vector is $\hat{\mathbf{n}} = \begin{pmatrix}-\sin\psi\\\cos\psi\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{pmatrix}$ | M1 | Obtaining gradient or normal vector |
| | A1 | Correct normal vector. Not necessarily unit vector |
| $\mathbf{c} = \begin{pmatrix}a\!\left(1-\frac{3\sqrt{3}}{8}\right)\\\frac{1}{8}a\end{pmatrix} + \dfrac{3\sqrt{3}}{4}a\begin{pmatrix}-\frac{1}{2}\\\frac{\sqrt{3}}{2}\end{pmatrix}$ | M1 | Must use unit normal here |
| Centre of curvature is $\left(a\!\left(1-\dfrac{3\sqrt{3}}{4}\right),\ \dfrac{5a}{4}\right)$ | A1A1 | |
**[9 marks]**
# Question 3:
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Surface area is $\int 2\pi x \, ds$ | M1 | |
| $= \int_0^{\frac{\pi}{3}} 2\pi a(1-\cos^3\theta)(3a\sin\theta\cos\theta) \, d\theta$ | A1 | FT; Correct limits required |
| $= 6\pi a^2 \int_0^{\frac{\pi}{3}} (\sin\theta\cos\theta - \sin\theta\cos^4\theta) \, d\theta$ | | |
| $= 6\pi a^2 \left[\frac{1}{2}\sin^2\theta + \frac{1}{5}\cos^5\theta\right]_0^{\frac{\pi}{3}}$ | M1 A1A1 | For $\frac{1}{2}\sin^2\theta$ and $\frac{1}{5}\cos^5\theta$; at least one trigonometric term or equivalent |
| $= \frac{87\pi a^2}{80}$ | A1 | |
| | **[6]** | |
---3 A curve has parametric equations
$$x = a \left( 1 - \cos ^ { 3 } \theta \right) , \quad y = a \sin ^ { 3 } \theta , \quad \text { for } 0 \leqslant \theta \leqslant \frac { \pi } { 3 }$$
where $a$ is a positive constant.\\
The arc length from the origin to a general point on the curve is denoted by $s$, and $\psi$ is the acute angle defined by $\tan \psi = \frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(i) Express $s$ and $\psi$ in terms of $\theta$, and hence show that the intrinsic equation of the curve is
$$s = \frac { 3 } { 2 } a \sin ^ { 2 } \psi$$
(ii) For the point on the curve given by $\theta = \frac { \pi } { 6 }$, find the radius of curvature and the coordinates of the centre of curvature.\\
(iii) Find the area of the curved surface generated when the curve is rotated through $2 \pi$ radians about the $y$-axis.
\hfill \mbox{\textit{OCR MEI FP3 2012 Q3 [24]}}