| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve differential equations with hyperbolics |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring multiple sophisticated techniques: arc length formula, separable differential equations, hyperbolic function integration, and connecting multiple relationships. However, it's highly structured with clear signposting through parts (a)(i)-(iii) and (b), making it more accessible than unguided multi-step problems. The hyperbolic integration is standard FP2 content, but the arc length setup and conceptual understanding of the relationships elevate it above routine exercises. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges4.10a General/particular solutions: of differential equations8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(\frac{ds}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \left(\frac{s}{2}\right)^2}\) | M1A1 | Allow M1 for \(s = \int\sqrt{1 + \left(\frac{s}{2}\right)} dx\); then A1 for \(\frac{dy}{dx}\) |
| \(= \frac{1}{2}\sqrt{4 + s^2}\) | A1 | 3 marks |
| (ii) \(\int\frac{ds}{\sqrt{4 + s^2}} = \int\frac{1}{2}dx\) | M1 | For separation of variables; allow without integral sign |
| \(\sinh^{-1}\frac{s}{2} = \frac{1}{2}x + C\) | A1 | Allow if \(C\) is missing |
| \(C = 0\) | A1 | |
| \(s = 2\sinh\frac{1}{2}x\) | A1 | 4 marks |
| (iii) \(\frac{dy}{dx} = \sinh\frac{1}{2}x\) | M1 | |
| \(y = 2\cosh\frac{1}{2}x + C\) | A1 | Allow if \(C\) is missing |
| \(C = 0\) | A1 | 3 marks |
| (b) \(y^2 = 4\left(1 + \sinh^2\frac{x}{2}\right)\) | M1 | Use of \(\cosh^2 = 1 + \sinh^2\) |
| \(= 4 + s^2\) | A1 | 2 marks |
**(a)(i)** $\frac{ds}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \left(\frac{s}{2}\right)^2}$ | M1A1 | Allow M1 for $s = \int\sqrt{1 + \left(\frac{s}{2}\right)} dx$; then A1 for $\frac{dy}{dx}$
$= \frac{1}{2}\sqrt{4 + s^2}$ | A1 | 3 marks | AG
**(ii)** $\int\frac{ds}{\sqrt{4 + s^2}} = \int\frac{1}{2}dx$ | M1 | For separation of variables; allow without integral sign
$\sinh^{-1}\frac{s}{2} = \frac{1}{2}x + C$ | A1 | Allow if $C$ is missing
$C = 0$ | A1 |
$s = 2\sinh\frac{1}{2}x$ | A1 | 4 marks | AG if $C$ not mentioned allow $\frac{3}{4}$. SC incomplete proof of (a)(ii), differentiating: $s = 2\sinh\frac{x}{2}$ to arrive at $\frac{ds}{dx} = \frac{1}{2}\sqrt{4 + s^2}$; allow M1A1 only $\left(\frac{2}{4}\right)$
**(iii)** $\frac{dy}{dx} = \sinh\frac{1}{2}x$ | M1 |
$y = 2\cosh\frac{1}{2}x + C$ | A1 | Allow if $C$ is missing
$C = 0$ | A1 | 3 marks | Must be shown to be zero and CAO
**(b)** $y^2 = 4\left(1 + \sinh^2\frac{x}{2}\right)$ | M1 | Use of $\cosh^2 = 1 + \sinh^2$
$= 4 + s^2$ | A1 | 2 marks | AG
**TOTAL: 12 marks**
**TOTAL FOR PAPER: 75 marks**7 The diagram shows a curve which starts from the point $A$ with coordinates ( 0,2 ). The curve is such that, at every point $P$ on the curve,
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 } s$$
where $s$ is the length of the $\operatorname { arc } A P$.\\
\includegraphics[max width=\textwidth, alt={}, center]{587aac5c-fbc2-41d2-b1b3-16f3f7851d9d-4_399_764_1324_605}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that
$$\frac { \mathrm { d } s } { \mathrm {~d} x } = \frac { 1 } { 2 } \sqrt { 4 + s ^ { 2 } }$$
(3 marks)
\item Hence show that
$$s = 2 \sinh \frac { x } { 2 }$$
\item Hence find the cartesian equation of the curve.
\end{enumerate}\item Show that
$$y ^ { 2 } = 4 + s ^ { 2 }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q7 [12]}}