| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 Part (a) is straightforward application of the arc length formula requiring only differentiation of a power function. Part (b) involves a guided hyperbolic substitution with standard integration techniques. While the algebraic manipulation and simplification to the exact form requires care, the substitution is provided and the methods are standard for FP3. This is moderately above average difficulty due to the multi-step nature and need for careful algebraic work, but not exceptionally challenging for Further Maths students. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08h Integration by substitution4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = x^{-\frac{1}{2}}\) | B1 | Correct derivative (may be unsimplified) |
| \(s = \int\sqrt{1+(x^{-\frac{1}{2}})^2}\,dx = \int_1^8\sqrt{1+\frac{1}{x}}\,dx\) | B1 | Correct formula quoted or implied; must be some working before printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \sinh^2 u \Rightarrow \frac{dx}{du} = 2\sinh u\cosh u\) | B1 | Correct derivative |
| \(\left(1+\frac{1}{x}\right) = 1 + \text{cosech}^2 u = \coth^2 u\) or \(\frac{\cosh^2 u}{\sinh^2 u}\) | B1 | May be implied by later work |
| \(s = \int\coth u \cdot 2\sinh u\cosh u\,du = \int 2\cosh^2 u\,du\) | M1 A1 | M1: Complete substitution; A1: \(\int 2\cosh^2 u\,du\) |
| \(= u + \frac{1}{2}\sinh 2u\) or \(\frac{1}{4}e^{2u} + u - \frac{1}{4}e^{-2u}\) | dM1 A1 | M1: Uses \(\cosh 2u = \pm 2\cosh^2 u \pm 1\) or changes to exponentials; A1: Correct integration |
| \(x=8 \Rightarrow u = \text{arsinh}\sqrt{8} = \ln(3+2\sqrt{2})\), \(x=1 \Rightarrow u = \text{arsinh}1 = \ln(1+\sqrt{2})\) | ||
| \(\left[u + \frac{1}{2}\sinh 2u\right]_{\text{arsinh}1}^{\text{arsinh}\sqrt{8}}\) | ddM1 A1 | M1: Limits \(\text{arsinh}\sqrt{8}\) and \(\text{arsinh}1\) (or \(\ln\) forms) used correctly; A1: Completely correct expression; depends on both previous M's |
| \(\ln(1+\sqrt{2}) + 5\sqrt{2}\) | A1 |
## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = x^{-\frac{1}{2}}$ | B1 | Correct derivative (may be unsimplified) |
| $s = \int\sqrt{1+(x^{-\frac{1}{2}})^2}\,dx = \int_1^8\sqrt{1+\frac{1}{x}}\,dx$ | B1 | Correct formula quoted or implied; must be some working before printed answer |
**(2 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \sinh^2 u \Rightarrow \frac{dx}{du} = 2\sinh u\cosh u$ | B1 | Correct derivative |
| $\left(1+\frac{1}{x}\right) = 1 + \text{cosech}^2 u = \coth^2 u$ or $\frac{\cosh^2 u}{\sinh^2 u}$ | B1 | May be implied by later work |
| $s = \int\coth u \cdot 2\sinh u\cosh u\,du = \int 2\cosh^2 u\,du$ | M1 A1 | M1: Complete substitution; A1: $\int 2\cosh^2 u\,du$ |
| $= u + \frac{1}{2}\sinh 2u$ or $\frac{1}{4}e^{2u} + u - \frac{1}{4}e^{-2u}$ | dM1 A1 | M1: Uses $\cosh 2u = \pm 2\cosh^2 u \pm 1$ or changes to exponentials; A1: Correct integration |
| $x=8 \Rightarrow u = \text{arsinh}\sqrt{8} = \ln(3+2\sqrt{2})$, $x=1 \Rightarrow u = \text{arsinh}1 = \ln(1+\sqrt{2})$ | | |
| $\left[u + \frac{1}{2}\sinh 2u\right]_{\text{arsinh}1}^{\text{arsinh}\sqrt{8}}$ | ddM1 A1 | M1: Limits $\text{arsinh}\sqrt{8}$ and $\text{arsinh}1$ (or $\ln$ forms) used correctly; A1: Completely correct expression; depends on both previous M's |
| $\ln(1+\sqrt{2}) + 5\sqrt{2}$ | A1 | |
**(9 marks) — Total 11**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{094b3c91-1460-44a2-b9d6-4de90d3adfa0-15_590_855_210_548}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The curve $C$, shown in Figure 2, has equation
$$y = 2 x ^ { \frac { 1 } { 2 } } , \quad 1 \leqslant x \leqslant 8$$
\begin{enumerate}[label=(\alph*)]
\item Show that the length $s$ of curve $C$ is given by the equation
$$s = \int _ { 1 } ^ { 8 } \sqrt { } \left( 1 + \frac { 1 } { x } \right) \mathrm { d } x$$
\item Using the substitution $x = \sinh ^ { 2 } u$, or otherwise, find an exact value for $s$.
Give your answer in the form $a \sqrt { } 2 + \ln ( b + c \sqrt { } 2 )$ where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2013 Q8 [11]}}