| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.2 This is a standard Further Pure 2 surface area of revolution question with hyperbolic functions. Part (a) requires routine differentiation and use of hyperbolic identities (cosh²t - sinh²t = 1). Part (b)(i) applies the standard formula S = 2π∫y√((dx/dt)² + (dy/dt)²)dt, and part (b)(ii) involves straightforward integration by substitution (u = cosh t). While it requires knowledge of multiple techniques, each step follows a predictable pattern with no novel insight needed—typical of FP2 exam questions. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = \sinh 2t\) | B1 | |
| \(\frac{dy}{dt} = 2\cosh t\) | B1 | |
| \(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \sinh^2 2t + 4\cosh^2 t\) | M1 | |
| Use of \(\sinh 2t = 2\sinh t\cosh t\) | m1 | Or other correct formula for double angle |
| \(= 4\cosh^2 t(\sinh^2 t + 1)\) | A1 | For taking out factor |
| \(= 4\cosh^4 t\) | A1F | 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = 2\pi\int_0^1 2\sinh t \cdot 2\cosh^2 t \, dt\) | M1 | Using the value obtained in (a) |
| \(= 8\pi\int_0^1 \sinh t\cosh^2 t \, dt\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = 8\pi\left[\frac{\cosh^3 t}{3}\right]_0^1\) | M1 | |
| \(= \frac{8\pi}{3}[\cosh^3 1 - 1]\) | A1 | 2 marks |
### Part (a)
$\frac{dx}{dt} = \sinh 2t$ | B1 |
$\frac{dy}{dt} = 2\cosh t$ | B1 |
$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \sinh^2 2t + 4\cosh^2 t$ | M1 |
Use of $\sinh 2t = 2\sinh t\cosh t$ | m1 | Or other correct formula for double angle
$= 4\cosh^2 t(\sinh^2 t + 1)$ | A1 | For taking out factor
$= 4\cosh^4 t$ | A1F | 6 marks | ft errors of sign in $\frac{dx}{dt}$ or $\frac{dy}{dt}$
### Part (b)(i)
$S = 2\pi\int_0^1 2\sinh t \cdot 2\cosh^2 t \, dt$ | M1 | Using the value obtained in (a)
$= 8\pi\int_0^1 \sinh t\cosh^2 t \, dt$ | A1 | 2 marks | AG
### Part (b)(ii)
$S = 8\pi\left[\frac{\cosh^3 t}{3}\right]_0^1$ | M1 |
$= \frac{8\pi}{3}[\cosh^3 1 - 1]$ | A1 | 2 marks | OE eg $\frac{\pi}{3}\left(\left(e + \frac{1}{e}\right)^3 - 8\right)$4 A curve $C$ is given parametrically by the equations
$$x = \frac { 1 } { 2 } \cosh 2 t , \quad y = 2 \sinh t$$
(a) Express
$$\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} t } \right) ^ { 2 }$$
in terms of $\cosh t$.\\
(b) The arc of $C$ from $t = 0$ to $t = 1$ is rotated through $2 \pi$ radians about the $x$-axis.\\
(i) Show that $S$, the area of the curved surface generated, is given by
$$S = 8 \pi \int _ { 0 } ^ { 1 } \sinh t \cosh ^ { 2 } t \mathrm {~d} t$$
(ii) Find the exact value of $S$.
\hfill \mbox{\textit{AQA FP2 2010 Q4 [10]}}