| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 This is a standard Further Maths arc length question with hyperbolic functions. Part (a) requires differentiation of a composite hyperbolic function using chain rule and standard identities (routine for FP3). Part (b) applies the arc length formula with the given derivative, requiring integration of √(1 + cosech²2x) = coth 2x, which simplifies nicely. While it involves multiple steps and hyperbolic identities, it follows a predictable template for FP3 arc length problems without requiring novel insight. |
| Spec | 1.08g Integration as limit of sum: Riemann sums4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = \dfrac{1}{2}\times\dfrac{1}{\coth x}\times -\text{cosech}^2 x\) | M1A1 | M1: Correct use of chain rule; allow \(\dfrac{k}{\coth x}\times f(x)\) where \(f(x)\) is a hyperbolic function; A1: Correct differentiation |
| \(= \dfrac{-1}{2\sinh x \cosh x} = \dfrac{-1}{\sinh 2x} = -\text{cosech}\,2x\;*\) | A1* | Completes to printed answer with at least one line of working and no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{2y}=\coth x \Rightarrow 2e^{2y}\dfrac{dy}{dx}=-\text{cosech}^2 x\) | M1A1 | M1: Makes \(e^y\) the subject and attempts to differentiate w.r.t. \(x\); A1: Correct differentiation |
| \(\dfrac{dy}{dx}=\dfrac{-\text{cosech}^2 x}{2\coth x}=\dfrac{-1}{\sinh 2x}=-\text{cosech}\,2x\;*\) | A1* | Completes to printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S=\displaystyle\int_{(\ln 2)}^{(\ln 3)}\!\!\left(1+\text{cosech}^2 2x\right)^{\frac{1}{2}}dx\) | M1 | Substitutes \(\text{cosech}\,2x\) into a correct formula (limits not needed) |
| \(S=\displaystyle\int_{(\ln 2)}^{(\ln 3)}\coth 2x\,dx\) | M1 | Use of \(1+\text{cosech}^2 2x = \coth^2 2x\) |
| \(S=\left[\dfrac{1}{2}\ln(\sinh 2x)\right]_{\ln 2}^{\ln 3}\) | A1 | Correct integration |
| \(S=\dfrac{1}{2}\ln(\sinh(2\ln 3))-\dfrac{1}{2}\ln(\sinh(2\ln 2))\) | dM1 | Uses limits \(\ln 2\) and \(\ln 3\) and subtracts either way round; dependent on first M |
| \(S=\dfrac{1}{2}\ln\!\left(\dfrac{9-\frac{1}{9}}{2}\right)\!\left(\dfrac{2}{4-\frac{1}{4}}\right)\) | ddM1 | Uses exponential form of \(\sinh x\) and combines ln's to give expression in terms of ln only; dependent on first and 3rd M |
| \(S=\dfrac{1}{2}\ln\dfrac{64}{27}\) or \(\dfrac{3}{2}\ln\dfrac{4}{3}\) | A1 |
# Question 3: $y=\dfrac{1}{2}\ln(\coth x)$
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = \dfrac{1}{2}\times\dfrac{1}{\coth x}\times -\text{cosech}^2 x$ | M1A1 | M1: Correct use of chain rule; allow $\dfrac{k}{\coth x}\times f(x)$ where $f(x)$ is a hyperbolic function; A1: Correct differentiation |
| $= \dfrac{-1}{2\sinh x \cosh x} = \dfrac{-1}{\sinh 2x} = -\text{cosech}\,2x\;*$ | A1* | Completes to printed answer with at least one line of working and **no** errors |
**Part (a) Way 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{2y}=\coth x \Rightarrow 2e^{2y}\dfrac{dy}{dx}=-\text{cosech}^2 x$ | M1A1 | M1: Makes $e^y$ the subject and attempts to differentiate w.r.t. $x$; A1: Correct differentiation |
| $\dfrac{dy}{dx}=\dfrac{-\text{cosech}^2 x}{2\coth x}=\dfrac{-1}{\sinh 2x}=-\text{cosech}\,2x\;*$ | A1* | Completes to printed answer with no errors |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S=\displaystyle\int_{(\ln 2)}^{(\ln 3)}\!\!\left(1+\text{cosech}^2 2x\right)^{\frac{1}{2}}dx$ | M1 | Substitutes $\text{cosech}\,2x$ into a correct formula (limits not needed) |
| $S=\displaystyle\int_{(\ln 2)}^{(\ln 3)}\coth 2x\,dx$ | M1 | Use of $1+\text{cosech}^2 2x = \coth^2 2x$ |
| $S=\left[\dfrac{1}{2}\ln(\sinh 2x)\right]_{\ln 2}^{\ln 3}$ | A1 | Correct integration |
| $S=\dfrac{1}{2}\ln(\sinh(2\ln 3))-\dfrac{1}{2}\ln(\sinh(2\ln 2))$ | dM1 | Uses limits $\ln 2$ and $\ln 3$ and subtracts either way round; dependent on first M |
| $S=\dfrac{1}{2}\ln\!\left(\dfrac{9-\frac{1}{9}}{2}\right)\!\left(\dfrac{2}{4-\frac{1}{4}}\right)$ | ddM1 | Uses exponential form of $\sinh x$ **and** combines ln's to give expression in terms of ln only; **dependent on first and 3rd M** |
| $S=\dfrac{1}{2}\ln\dfrac{64}{27}$ or $\dfrac{3}{2}\ln\dfrac{4}{3}$ | A1 | |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { 1 } { 2 } \ln ( \operatorname { coth } x ) , \quad x > 0$$
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosech } 2 x$$
The points $A$ and $B$ lie on $C$.
The $x$ coordinates of $A$ and $B$ are $\ln 2$ and $\ln 3$ respectively.\\
(b) Find the length of the arc $A B$, giving your answer in the form $p \ln q$, where $p$ and $q$ are rational numbers.\\
(6)\\
\hfill \mbox{\textit{Edexcel FP3 2014 Q3 [9]}}