| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear signposting through multiple parts. Part (a) involves routine algebraic manipulation of hyperbolic definitions (standard FP2 content). Part (b) requires parametric differentiation, applying the arc length formula, and integration using substitution—all standard techniques for FP2 students. While it requires multiple steps and careful algebra, each component follows predictable patterns without requiring novel insight or problem-solving beyond applying learned methods. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(2\left(\frac{e^\theta - e^{-\theta}}{2}\right)\left(\frac{e^\theta + e^{-\theta}}{2}\right) = \frac{e^{2\theta} - e^{-2\theta}}{2} = \sinh 2\theta\) | M1A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\left(\frac{e^\theta - e^{-\theta}}{2}\right)^2 + \left(\frac{e^\theta + e^{-\theta}}{2}\right)^2\) | M1 | |
| \(= \frac{e^{2\theta} - 2 + e^{-2\theta} + e^{2\theta} + 2 + e^{-2\theta}}{4}\) | A1 | |
| \(= \cosh 2\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\dot{x} = 3\cosh^2\theta\sinh\theta\) | M1A1 | Allow M1 for reasonable attempt at differentiation, but M0 for putting in terms of \(e^{k\theta}\) or \(\sinh 3\theta\) unless real progress made towards \(\dot{x}^2 + \dot{y}^2\) |
| \(\dot{y} = 3\sinh^2\theta\cosh\theta\) | A1 | Allow this M1 if not squared out, must be clear sum in question is \(\dot{x}^2 + \dot{y}^2\) |
| \(\dot{x}^2 + \dot{y}^2 = 9\cosh^4\theta\sinh^2\theta + 9\sinh^4\theta\cosh^2\theta\) | M1 | AG |
| \(= 9\sinh^2\theta\cosh^2\theta(\cosh^2\theta + \sinh^2\theta)\) | A1 | Accept \(\int_0^1\sqrt{\frac{9}{4}\sinh^2 2\theta \cosh 2\theta}\, d\theta\) but limits must appear somewhere |
| \(= \frac{9}{4}\sinh^2 2\theta\cosh 2\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(S = \int_0^1 \frac{3}{2}\sinh 2\theta\sqrt{\cosh 2\theta}\, d\theta\) | M1 | |
| \(u = \cosh 2\theta\), \(du = 2\sinh 2\theta\, d\theta\) | M1A1 | |
| \(I = \int \frac{3}{4}u^{\frac{1}{2}}\, du = \frac{3}{4} \times \frac{2}{3}u^{\frac{3}{2}}\) | A1F | |
| \(S = \left\{\frac{1}{2}(\cosh 2\theta)^{\frac{3}{2}}\right\}_0^1\) | A1F | |
| \(= \frac{1}{2}\left\{(\cosh 2)^{\frac{3}{2}} - 1\right\}\) | A1 | AG |
# Question 7:
## Part 7(a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $2\left(\frac{e^\theta - e^{-\theta}}{2}\right)\left(\frac{e^\theta + e^{-\theta}}{2}\right) = \frac{e^{2\theta} - e^{-2\theta}}{2} = \sinh 2\theta$ | M1A1 | AG |
## Part 7(a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\left(\frac{e^\theta - e^{-\theta}}{2}\right)^2 + \left(\frac{e^\theta + e^{-\theta}}{2}\right)^2$ | M1 | |
| $= \frac{e^{2\theta} - 2 + e^{-2\theta} + e^{2\theta} + 2 + e^{-2\theta}}{4}$ | A1 | |
| $= \cosh 2\theta$ | A1 | AG |
## Part 7(b)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\dot{x} = 3\cosh^2\theta\sinh\theta$ | M1A1 | Allow M1 for reasonable attempt at differentiation, but M0 for putting in terms of $e^{k\theta}$ or $\sinh 3\theta$ unless real progress made towards $\dot{x}^2 + \dot{y}^2$ |
| $\dot{y} = 3\sinh^2\theta\cosh\theta$ | A1 | Allow this M1 if not squared out, must be clear sum in question is $\dot{x}^2 + \dot{y}^2$ |
| $\dot{x}^2 + \dot{y}^2 = 9\cosh^4\theta\sinh^2\theta + 9\sinh^4\theta\cosh^2\theta$ | M1 | AG |
| $= 9\sinh^2\theta\cosh^2\theta(\cosh^2\theta + \sinh^2\theta)$ | A1 | Accept $\int_0^1\sqrt{\frac{9}{4}\sinh^2 2\theta \cosh 2\theta}\, d\theta$ but limits must appear somewhere |
| $= \frac{9}{4}\sinh^2 2\theta\cosh 2\theta$ | A1 | AG |
## Part 7(b)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $S = \int_0^1 \frac{3}{2}\sinh 2\theta\sqrt{\cosh 2\theta}\, d\theta$ | M1 | |
| $u = \cosh 2\theta$, $du = 2\sinh 2\theta\, d\theta$ | M1A1 | |
| $I = \int \frac{3}{4}u^{\frac{1}{2}}\, du = \frac{3}{4} \times \frac{2}{3}u^{\frac{3}{2}}$ | A1F | |
| $S = \left\{\frac{1}{2}(\cosh 2\theta)^{\frac{3}{2}}\right\}_0^1$ | A1F | |
| $= \frac{1}{2}\left\{(\cosh 2)^{\frac{3}{2}} - 1\right\}$ | A1 | AG |7
\begin{enumerate}[label=(\alph*)]
\item Use the definitions
$$\sinh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } \right) \quad \text { and } \quad \cosh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } \right)$$
to show that:
\begin{enumerate}[label=(\roman*)]
\item $2 \sinh \theta \cosh \theta = \sinh 2 \theta$;
\item $\cosh ^ { 2 } \theta + \sinh ^ { 2 } \theta = \cosh 2 \theta$.
\end{enumerate}\item A curve is given parametrically by
$$x = \cosh ^ { 3 } \theta , \quad y = \sinh ^ { 3 } \theta$$
\begin{enumerate}[label=(\roman*)]
\item Show that
$$\left( \frac { \mathrm { d } x } { \mathrm {~d} \theta } \right) ^ { 2 } + \left( \frac { \mathrm { d } y } { \mathrm {~d} \theta } \right) ^ { 2 } = \frac { 9 } { 4 } \sinh ^ { 2 } 2 \theta \cosh 2 \theta$$
\item Show that the length of the arc of the curve from the point where $\theta = 0$ to the point where $\theta = 1$ is
$$\frac { 1 } { 2 } \left[ ( \cosh 2 ) ^ { \frac { 3 } { 2 } } - 1 \right]$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q7 [17]}}