Standard +0.8 This is a standard arc length problem requiring the formula L = ∫√(1 + (dy/dx)²) dx, knowledge that d/dx(cosh x) = sinh x, and the identity cosh²x - sinh²x = 1 which simplifies the integrand to cosh x. The integration and evaluation at limits ln 5 and 1 is straightforward. While it requires multiple techniques and careful algebra, it's a textbook application of arc length to hyperbolic functions without novel insight—moderately above average for Further Maths.
Must be in terms of \(e\) with no \(\ln\)s. Score when correct answer first seen
*Special Case:* \(\frac{dy}{dx} = -\sinh x\) leads to correct answer — maximum 3/5 i.e. B0M1A1(recovery)dM1A0
## Question 2:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $y = \cosh x \Rightarrow \frac{dy}{dx} = \sinh x$ | B1 | Correct derivative |
| $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int\sqrt{1+\sinh^2 x}\,dx$ | M1 | Uses correct arc length formula with their $\frac{dy}{dx}$ |
| $= \int \cosh x\,dx$ or $\int\frac{e^x+e^{-x}}{2}\,dx$ | A1 | Correct integral (condone omission of $dx$) |
| $= [\sinh x]_1^{\ln 5} = \sinh(\ln 5) - \sinh(1)$ | dM1 | $\int \cosh x\,dx = \sinh x$ and correct use of limits. Dependent on first M mark |
| $= \frac{12}{5} - \frac{1}{2}\left(e - \frac{1}{e}\right)$ | A1cso | Must be in terms of $e$ with no $\ln$s. Score when correct answer first seen |
*Special Case:* $\frac{dy}{dx} = -\sinh x$ leads to correct answer — maximum **3/5** i.e. B0M1A1(recovery)dM1A0
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2. A curve has equation
$$y = \cosh x , \quad 1 \leqslant x \leqslant \ln 5$$
Find the exact length of this curve. Give your answer in terms of e .\\
\hfill \mbox{\textit{Edexcel FP3 2015 Q2 [5]}}