| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 This is a standard arc length calculation with hyperbolic functions requiring knowledge of the formula L = ∫√(1+(dy/dx)²)dx, differentiation of cosh, and the hyperbolic identity cosh²-sinh²=1. While it involves Further Maths content (hyperbolic functions), the question follows a routine template with straightforward algebraic manipulation and integration, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \sinh 3x\) | B1 | cao |
| \(s = \int \sqrt{1 + \sinh^2 3x}\, dx\) | M1 | Use of arc length formula, need both \(\sqrt{\phantom{x}}\) and \(\left(\frac{dy}{dx}\right)^2\) |
| \(\therefore s = \int \cosh 3x\, dx\) | A1 | \(\int \cosh 3x\, dx\) cao |
| \(= \left[\frac{1}{3}\sinh 3x\right]_0^{\ln a}\) | M1 | Attempt to integrate, getting a hyperbolic function |
| \(= \frac{1}{3}\sinh 3\ln a = \frac{1}{6}[e^{3\ln a} - e^{-3\ln a}]\) | DM1 | Depends on previous M. Correct use of \(\ln a\) and \(0\) as limits. Must see some exponentials |
| \(= \frac{1}{6}\left(a^3 - \frac{1}{a^3}\right)\) (so \(k = 1/6\)) | A1 | cao |
## Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \sinh 3x$ | B1 | cao |
| $s = \int \sqrt{1 + \sinh^2 3x}\, dx$ | M1 | Use of arc length formula, need both $\sqrt{\phantom{x}}$ and $\left(\frac{dy}{dx}\right)^2$ |
| $\therefore s = \int \cosh 3x\, dx$ | A1 | $\int \cosh 3x\, dx$ cao |
| $= \left[\frac{1}{3}\sinh 3x\right]_0^{\ln a}$ | M1 | Attempt to integrate, getting a hyperbolic function |
| $= \frac{1}{3}\sinh 3\ln a = \frac{1}{6}[e^{3\ln a} - e^{-3\ln a}]$ | DM1 | Depends on previous M. Correct use of $\ln a$ and $0$ as limits. Must see some exponentials |
| $= \frac{1}{6}\left(a^3 - \frac{1}{a^3}\right)$ (so $k = 1/6$) | A1 | cao |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb604886-6671-441a-b03d-427b5176df6e-03_606_1271_212_335}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$, shown in Figure 1, has equation
$$y = \frac { 1 } { 3 } \cosh 3 x , \quad 0 \leqslant x \leqslant \ln a$$
where $a$ is a constant and $a > 1$
Using calculus, show that the length of curve $C$ is
$$k \left( a ^ { 3 } - \frac { 1 } { a ^ { 3 } } \right)$$
and state the value of the constant $k$.\\
\hfill \mbox{\textit{Edexcel FP3 2012 Q2 [6]}}