FP3 June 2014 (R) Q7
7. The curve \(C\) has equation \[y = \mathrm{e}^{-x}, \quad x \in \mathbb{R}\]
The part of the curve \(C\) between \(x = 0\) and \(x = \ln 3\) is rotated through \(2\pi\) radians about the \(x\)-axis.
(a) Show that the area \(S\) of the curved surface generated is given by \[S = 2\pi\int_0^{\ln 3} \mathrm{e}^{-x}\sqrt{1 + \mathrm{e}^{-2x}}\,\mathrm{d}x\] (3)
(b) Use the substitution \(\mathrm{e}^{-x} = \sinh u\) to show that \[S = 2\pi\int_{\mathrm{arsinh}\,\alpha}^{\mathrm{arsinh}\,\beta} \cosh^2 u\,\mathrm{d}u\] where \(\alpha\) and \(\beta\) are constants to be determined. (5)
(c) Show that \[2\int \cosh^2 u\,\mathrm{d}u = \frac{1}{2}\sinh 2u + u + k\] where \(k\) is an arbitrary constant. (2)
(d) Hence find the value of \(S\), giving your answer to 3 decimal places. (2)
| Scheme | Marks |
|---|---|
| \(y = \mathrm{e}^{-x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\mathrm{e}^{-x}\) Correct derivative | B1 |
| \(S = 2\pi\displaystyle\int y\sqrt{1 + (y^{\prime})^2}\,\mathrm{d}x = 2\pi\int \mathrm{e}^{-x}\sqrt{1 + \mathrm{e}^{-2x}}\,\mathrm{d}x\) M1: Use of correct formula A1: Correct proof with no errors | M1A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{e}^{-x} = \sinh u \Rightarrow -\mathrm{e}^{-x} = \cosh u\dfrac{\mathrm{d}u}{\mathrm{d}x}\) Correct differentiation | B1 |
| \(S = 2\pi\displaystyle\int \mathrm{e}^{-x}\sqrt{1 + \mathrm{e}^{-2x}}\,\mathrm{d}x = 2\pi\int \sinh u\sqrt{1 + \sinh^2 u} \cdot \frac{\cosh u}{-\sinh u}\,\mathrm{d}u\) A complete substitution | M1 |
| \((2\pi)\displaystyle\int \sinh u\sqrt{1 + \sinh^2 u} \cdot \frac{\cosh u}{-\sinh u}\,\mathrm{d}u = -2\pi\int \cosh^2 u\,\mathrm{d}u\) | A1 |
| \(x = 0 \Rightarrow u = \mathrm{arsinh}(1)(= \ln(1 + \sqrt{2}))\) \(x = \ln 3 \Rightarrow u = \mathrm{arsinh}\left(\tfrac{1}{3}\right)\left(= \ln\left(\tfrac{1}{3} + \sqrt{1 + \tfrac{1}{9}}\right)\right)\) Both limits correct | B1 |
| \(S = 2\pi\displaystyle\int_{\mathrm{arsinh}\left(\frac{1}{3}\right)}^{\mathrm{arsinh}(1)} \cosh^2 u\,\mathrm{d}u\) Correct completion with no errors | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(2\displaystyle\int \cosh^2 u\,\mathrm{d}u = \int (\cosh 2u + 1)\,\mathrm{d}u\) Uses \(2\cosh^2 u = \pm\cosh 2u \pm 1\) | M1 |
| \(= \tfrac{1}{2}\sinh 2u + u\,(+k)\ ^*\) cso | A1* |
| (2) |
| Scheme | Marks |
|---|---|
| \(S = \pi\left(\tfrac{1}{2}\sinh 2(\mathrm{arsinh}\,\beta) + \mathrm{arsinh}\,\beta - \tfrac{1}{2}\sinh 2(\mathrm{arsinh}\,\alpha) - \mathrm{arsinh}\,\alpha\right)\) Attempt to use their limits (subtracting either way round) (allow the omission of \(\pi\) and allow \(2\pi\) instead of \(\pi\)) There must be some evidence of the use of their limits e.g. an answer of 5.08 with no working loses this mark | M1 |
| \(= 5.079\) Cao (Allow recovery from \(-5.079\)) | A1 |
| (2) | |
| (12 marks) |
Notes
NB \(S = \pi\left(\sqrt{2} + \ln(1 + \sqrt{2}) - \tfrac{1}{3}\tfrac{\sqrt{10}}{3} - \ln\left(\tfrac{1}{3} + \tfrac{\sqrt{10}}{3}\right)\right) = 5.079241597\)