FP3 June 2015 Q2
2. A curve has equation \[y = \cosh x, \qquad 1 \leqslant x \leqslant \ln 5\] Find the exact length of this curve. Give your answer in terms of e. (5)
| Scheme | Marks |
|---|---|
| \(y = \cosh x \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \sinh x\) | B1 |
| \(\displaystyle\int\sqrt{\left(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right)}\,\mathrm{d}x = \int\sqrt{1 + \sinh^2 x}\,\mathrm{d}x\) | M1 |
| \(\displaystyle = \int\cosh x\ \mathrm{d}x\) or \(\displaystyle\int\frac{e^x + e^{-x}}{2}\,\mathrm{d}x\) | A1 |
| \(= \left[\sinh x\right]_1^{\ln 5} = \sinh(\ln 5) - \sinh(1)\) | dM1 |
| \(= \dfrac{12}{5} - \dfrac{1}{2}\left(e - \dfrac{1}{e}\right)\) | A1cso |
| (5) | |
| (5 marks) |
Notes
B1: Correct derivative
M1: Uses the correct formula with their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: Correct integral (Condone omission of \(\mathrm{d}x\))
dM1: \(\int\cosh x\,\mathrm{d}x = \sinh x\) and correct use of the correct limits. Dependent on the first method mark.
A1cso: Or equivalent (must be in terms of e with no ln’s) Score when a correct answer is first seen and isw.
Alternative for first 2 marks
\(y = \dfrac{e^x + e^{-x}}{2} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{e^x - e^{-x}}{2}\) = B1
\(\displaystyle\int\sqrt{\left(1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2\right)}\,\mathrm{d}x = \int\sqrt{1 + \left(\frac{e^x - e^{-x}}{2}\right)^2}\,\mathrm{d}x\) = M1
Then apply the scheme
Some equivalent final answers
\(\dfrac{12}{5} - \dfrac{e}{2} + \dfrac{e^{-1}}{2},\quad 2.4 - \dfrac{e - e^{-1}}{2},\quad \dfrac{12}{5} - \dfrac{e^2 - 1}{2e},\quad \dfrac{24e - 5e^2 + 5}{10e}\)
Special Case: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\sinh x\) leads to a correct answer. This scores a maximum of 3/5 i.e. B0M1A1(recovery)dM1A0