| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring arc length integration with hyperbolic functions, finding conditions from boundary values, and numerical solving. While the arc length formula is standard (√(1+(dy/dx)²)), applying it to cosh requires knowing hyperbolic identities (sinh²+1=cosh²) and integrating sinh. Part (a) is structured proof, parts (b-c) require solving transcendental equations numerically. The multi-part nature, Further Maths content, and need for both algebraic manipulation and numerical methods place it well above average difficulty, though the steps are relatively guided. |
| Spec | 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -\sinh 2x\) | B1 | Finds correct derivative |
| \(S = \int\sqrt{1 + \sinh^2 2x}\, dx\) | M1 | Uses derivative in arc length formula |
| \(\therefore s = \int \cosh 2x\, dx\) | A1 | Uses identity to simplify integrand |
| \(= \left[\frac{1}{2}\sinh 2x\right]_{-\ln a}^{\ln a}\) or \(\left[\sinh 2x\right]_0^{\ln a}\) | M1 | Integrates with appropriate limits |
| \(= \sinh 2\ln a = \frac{1}{2}[e^{2\ln a} - e^{-2\ln a}] = \frac{1}{2}\left(a^2 - \frac{1}{a^2}\right)\), so \(k = \frac{1}{2}\) | A1 | Uses definition of sinh to complete proof, identifies \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}\left(a^2 - \frac{1}{a^2}\right) = 2\), so \(a^4 - 4a^2 - 1 = 0\) | M1 | Uses formula and arc length to create quartic |
| \(a^2 = 2 + \sqrt{5}\) (and \(a = 2.06\) approx.) | M1 | Solves to obtain quadratic, finds values of \(a\) |
| When \(x = \ln a\), \(y = 0\) so \(A = \frac{1}{2}\cosh(2\ln a)\) | M1 | Attempts value of \(A\) to find \(h\) |
| Height \(= A - 0.5 =\) approx. \(0.62\)m | A1 | Correct height to 2sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Width of base \(= 2\ln a = 1.4\)m | B1 | Correct width to 2sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Parabola of form \(y = 0.62 - 1.19x^2\), or other symmetric curve e.g. \(0.62\cos(2.2x)\) | M1A1 | Even function with maximum on \(y\)-axis; passes through \((0, 0.62)\), \((0.7, 0)\), \((-0.7, 0)\) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\sinh 2x$ | B1 | Finds correct derivative |
| $S = \int\sqrt{1 + \sinh^2 2x}\, dx$ | M1 | Uses derivative in arc length formula |
| $\therefore s = \int \cosh 2x\, dx$ | A1 | Uses identity to simplify integrand |
| $= \left[\frac{1}{2}\sinh 2x\right]_{-\ln a}^{\ln a}$ or $\left[\sinh 2x\right]_0^{\ln a}$ | M1 | Integrates with appropriate limits |
| $= \sinh 2\ln a = \frac{1}{2}[e^{2\ln a} - e^{-2\ln a}] = \frac{1}{2}\left(a^2 - \frac{1}{a^2}\right)$, so $k = \frac{1}{2}$ | A1 | Uses definition of sinh to complete proof, identifies $k$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\left(a^2 - \frac{1}{a^2}\right) = 2$, so $a^4 - 4a^2 - 1 = 0$ | M1 | Uses formula and arc length to create quartic |
| $a^2 = 2 + \sqrt{5}$ (and $a = 2.06$ approx.) | M1 | Solves to obtain quadratic, finds values of $a$ |
| When $x = \ln a$, $y = 0$ so $A = \frac{1}{2}\cosh(2\ln a)$ | M1 | Attempts value of $A$ to find $h$ |
| Height $= A - 0.5 =$ approx. $0.62$m | A1 | Correct height to 2sf |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Width of base $= 2\ln a = 1.4$m | B1 | Correct width to 2sf |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Parabola of form $y = 0.62 - 1.19x^2$, or other symmetric curve e.g. $0.62\cos(2.2x)$ | M1A1 | Even function with maximum on $y$-axis; passes through $(0, 0.62)$, $(0.7, 0)$, $(-0.7, 0)$ |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1c262813-4160-4eda-9a36-e4ba38182c8a-14_480_588_210_740}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
An engineering student makes a miniature arch as part of the design for a piece of coursework.
The cross-section of this arch is modelled by the curve with equation
$$y = A - \frac { 1 } { 2 } \cosh 2 x , \quad - \ln a \leqslant x \leqslant \ln a$$
where $a > 1$ and $A$ is a positive constant. The curve begins and ends on the $x$-axis, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the length of this curve is $k \left( a ^ { 2 } - \frac { 1 } { a ^ { 2 } } \right)$, stating the value of the constant $k$.
The length of the curved cross-section of the miniature arch is required to be 2 m long.
\item Find the height of the arch, according to this model, giving your answer to 2 significant figures.
\item Find also the width of the base of the arch giving your answer to 2 significant figures.
\item Give the equation of another curve that could be used as a suitable model for the cross-section of an arch, with approximately the same height and width as you found using the first model.\\
(You do not need to consider the arc length of your curve)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 Q5 [12]}}