| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple advanced techniques: hyperbolic function manipulation, arc length and surface area of revolution integrals, parametric differentiation, normals, and evolutes with curvature. While each individual part follows standard FP3 methods, the combination of topics, the algebraic manipulation required (especially simplifying the surface area integral), and the evolute calculation make this significantly harder than average A-level questions. However, it's still a structured multi-part question with clear signposting rather than requiring deep novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.03h Parametric equations: in modelling contexts4.08d Volumes of revolution: about x and y axes8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{1}{2}e^{\frac{3x}{2}} - \frac{1}{4}e^{-\frac{1}{2}x}\) | B1 | |
| \(\left(\frac{dy}{dx}\right)^2 = \frac{1}{4}e^{3x} - \frac{1}{2} + \frac{1}{4}e^{-x}\) | M1 | |
| \(1 + \left(\frac{dy}{dx}\right)^2 = \frac{1}{4}e^{3x} + \frac{1}{2} + \frac{1}{4}e^{-x} = \left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{x}{2}}\right)^2\) | A1 (ag) | Correct completion (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Length is \(\int_0^{\ln a} \left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{1}{2}x}\right) dx\) | M1 | For \(\int \left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{1}{2}x}\right) dx\) |
| \(= \left[e^{\frac{3x}{2}} - e^{-\frac{1}{2}x}\right]_0^{\ln a}\) | A1 | For \(e^{\frac{3x}{2}} - e^{-\frac{1}{2}x}\) |
| \(= \left(\sqrt{a} - \frac{1}{\sqrt{a}}\right) - (1 - 1) = \frac{a - 1}{\sqrt{a}}\) | A1 (ag) | Correctly shown (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Curved surface area is \(\int 2\pi y \, ds\) | M1 | For \(\int y \, ds\) |
| \(= \int_0^{\ln a} 2\pi \left(e^{\frac{3x}{2}} + e^{-\frac{1}{2}x}\right)\left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{1}{2}x}\right) dx\) | A1 | Correct integral form including limits |
| \(= \pi \int_0^{\ln a} (e^{3x} + 2 + e^{-x}) dx\) | M1 | Obtaining integrable expression |
| \(= \pi \left[e^x + 2x - e^{-x}\right]_0^{\ln a}\) | A1 | For \(e^x + 2x - e^{-x}\) |
| \(= \pi \left(a + 2\ln a - \frac{1}{a}\right)\) | A1 (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{\cos \theta}{-2\sin \theta}\) | B1 | |
| Gradient of normal is \(\frac{2\sin \theta}{\cos \theta} (= 2\tan \theta)\) | M1 | |
| Normal is \(y - \sin \theta = 2\tan \theta(x - 2\cos \theta)\) | ||
| \(y = 2x \tan \theta - 3\sin \theta\) | A1 (ag) | Correctly shown (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiating partially w.r.t. \(\theta\): \(0 = 2x\sec^2 \theta - 3\cos \theta\) | M1 | |
| \(x = \frac{3}{2}\cos^3 \theta\) | A1 | |
| \(y = 3\cos^3 \theta \tan \theta - 3\sin \theta = 3\sin \theta(\cos^2 \theta - 1) = -3\sin^3 \theta\) | M1 A1 | |
| \((2x)^3 + y^3 = (3\cos^3 \theta)^3 + (-3\sin^3 \theta)^3 = 3^3(\cos^2 \theta + \sin^2 \theta) = 3^3\) | M1 A1 (ag) | Using \(1 - \cos^2 \theta = \sin^2 \theta\) (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \((2, 0)\) has \(\theta = 0\) | M1 | Using param eqn with \(\theta = 0\) (or other method for \(\rho\) or cc) |
| Centre of curvature is \(\left(\frac{3}{2}, 0\right)\) | M1 A1 | |
| \(\rho = \frac{1}{2}\) | A1 (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \((0, 1)\) has \(\theta = \frac{1}{2}\pi\) | M1 | Using param eqn with \(\theta = \frac{1}{2}\pi\) (or other method for \(\rho\) or cc) |
| Centre of curvature is \((0, -3)\) | M1 A1 | |
| \(\rho = 4\) | A1 (4 marks) |
### 3(a)(i)
$\frac{dy}{dx} = \frac{1}{2}e^{\frac{3x}{2}} - \frac{1}{4}e^{-\frac{1}{2}x}$ | B1 |
$\left(\frac{dy}{dx}\right)^2 = \frac{1}{4}e^{3x} - \frac{1}{2} + \frac{1}{4}e^{-x}$ | M1 |
$1 + \left(\frac{dy}{dx}\right)^2 = \frac{1}{4}e^{3x} + \frac{1}{2} + \frac{1}{4}e^{-x} = \left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{x}{2}}\right)^2$ | A1 (ag) | Correct completion (3 marks)
### 3(a)(ii)
Length is $\int_0^{\ln a} \left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{1}{2}x}\right) dx$ | M1 | For $\int \left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{1}{2}x}\right) dx$
$= \left[e^{\frac{3x}{2}} - e^{-\frac{1}{2}x}\right]_0^{\ln a}$ | A1 | For $e^{\frac{3x}{2}} - e^{-\frac{1}{2}x}$
$= \left(\sqrt{a} - \frac{1}{\sqrt{a}}\right) - (1 - 1) = \frac{a - 1}{\sqrt{a}}$ | A1 (ag) | Correctly shown (3 marks)
### 3(a)(iii)
Curved surface area is $\int 2\pi y \, ds$ | M1 | For $\int y \, ds$
$= \int_0^{\ln a} 2\pi \left(e^{\frac{3x}{2}} + e^{-\frac{1}{2}x}\right)\left(\frac{1}{2}e^{\frac{3x}{2}} + \frac{1}{2}e^{-\frac{1}{2}x}\right) dx$ | A1 | Correct integral form including limits
$= \pi \int_0^{\ln a} (e^{3x} + 2 + e^{-x}) dx$ | M1 | Obtaining integrable expression
$= \pi \left[e^x + 2x - e^{-x}\right]_0^{\ln a}$ | A1 | For $e^x + 2x - e^{-x}$
$= \pi \left(a + 2\ln a - \frac{1}{a}\right)$ | A1 (5 marks)
### 3(b)(i)
$\frac{dy}{dx} = \frac{\cos \theta}{-2\sin \theta}$ | B1 |
Gradient of normal is $\frac{2\sin \theta}{\cos \theta} (= 2\tan \theta)$ | M1 |
Normal is $y - \sin \theta = 2\tan \theta(x - 2\cos \theta)$ |
$y = 2x \tan \theta - 3\sin \theta$ | A1 (ag) | Correctly shown (3 marks)
### 3(b)(ii)
Differentiating partially w.r.t. $\theta$: $0 = 2x\sec^2 \theta - 3\cos \theta$ | M1 |
$x = \frac{3}{2}\cos^3 \theta$ | A1 |
$y = 3\cos^3 \theta \tan \theta - 3\sin \theta = 3\sin \theta(\cos^2 \theta - 1) = -3\sin^3 \theta$ | M1 A1 |
$(2x)^3 + y^3 = (3\cos^3 \theta)^3 + (-3\sin^3 \theta)^3 = 3^3(\cos^2 \theta + \sin^2 \theta) = 3^3$ | M1 A1 (ag) | Using $1 - \cos^2 \theta = \sin^2 \theta$ (6 marks)
### 3(b)(iii) A
$(2, 0)$ has $\theta = 0$ | M1 | Using param eqn with $\theta = 0$ (or other method for $\rho$ or cc)
Centre of curvature is $\left(\frac{3}{2}, 0\right)$ | M1 A1 |
$\rho = \frac{1}{2}$ | A1 (4 marks)
### 3(b)(iii) B
$(0, 1)$ has $\theta = \frac{1}{2}\pi$ | M1 | Using param eqn with $\theta = \frac{1}{2}\pi$ (or other method for $\rho$ or cc)
Centre of curvature is $(0, -3)$ | M1 A1 |
$\rho = 4$ | A1 (4 marks)3
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \mathrm { e } ^ { \frac { 1 } { 2 } x } + \mathrm { e } ^ { - \frac { 1 } { 2 } x }$, show that $1 + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = \left( \frac { 1 } { 2 } \mathrm { e } ^ { \frac { 1 } { 2 } x } + \frac { 1 } { 2 } \mathrm { e } ^ { - \frac { 1 } { 2 } x } \right) ^ { 2 }$.
The arc of the curve $y = \mathrm { e } ^ { \frac { 1 } { 2 } x } + \mathrm { e } ^ { - \frac { 1 } { 2 } x }$ for $0 \leqslant x \leqslant \ln a$ (where $a > 1$ ) is denoted by $C$.
\item Show that the length of $C$ is $\frac { a - 1 } { \sqrt { a } }$.
\item Find the area of the surface formed when $C$ is rotated through $2 \pi$ radians about the $x$-axis.
\end{enumerate}\item An ellipse has parametric equations $x = 2 \cos \theta , y = \sin \theta$ for $0 \leqslant \theta < 2 \pi$.
\begin{enumerate}[label=(\roman*)]
\item Show that the normal to the ellipse at the point with parameter $\theta$ has equation
$$y = 2 x \tan \theta - 3 \sin \theta$$
\item Find parametric equations for the evolute of the ellipse, and show that the evolute has cartesian equation
$$( 2 x ) ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 3 ^ { \frac { 2 } { 3 } }$$
\item Using the evolute found in part (ii), or otherwise, find the radius of curvature of the ellipse\\
(A) at the point $( 2,0 )$,\\
(B) at the point $( 0,1 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2011 Q3 [24]}}