FP3 June 2013 Q3
3. The curve with parametric equations \[x = \cosh 2\theta, \quad y = 4\sinh\theta, \quad 0 \leqslant \theta \leqslant 1\] is rotated through \(2\pi\) radians about the \(x\)-axis.
Show that the area of the surface generated is \(\lambda(\cosh^3\alpha - 1)\), where \(\alpha = 1\) and \(\lambda\) is a constant to be found. (7)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\right) = 2\sinh 2\theta\) and \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\right) = 4\cosh\theta\) Or equivalent correct derivatives | B1 |
| \(A = (2\pi)\displaystyle\int 4\sinh\theta\sqrt{\text{"}2\sinh 2\theta\text{"}^2 + \text{"}4\cosh\theta\text{"}^2}\,\mathrm{d}\theta\) or \(A = (2\pi)\displaystyle\int 4\sinh\theta\sqrt{\left(1 + \left(\frac{\text{"}4\cosh\theta\text{"}}{\text{"}2\sinh 2\theta\text{"}}\right)^2\right)} \cdot 2\sinh 2\theta\,\mathrm{d}\theta\) Use of correct formula including replacing \(\mathrm{d}x\) with "\(2\sinh 2\theta\)"\(\mathrm{d}\theta\) if chain rule used. Allow the omission of the \(2\pi\) here. | M1 |
| \(A = 32\pi\displaystyle\int \sinh\theta\cosh^2\theta\,\mathrm{d}\theta\) \(A = 32\pi\displaystyle\int (\sinh\theta + \sinh^3\theta)\,\mathrm{d}\theta\) Completely correct expression for \(A\) with the square root removed This mark may be recovered later if the \(2\pi\) is introduced later | B1 |
| \(A = \dfrac{32\pi}{3}\left[\cosh^3\theta\right]_0^1\) M1: Valid attempt to integrate a correct expression or a multiple of a correct expression – dependent on the first M1 A1: Correct expression | dM1A1 |
| \(= \dfrac{32\pi}{3}\left[\cosh^3 1 - 1\right]\) M1: Uses the limits 0 and 1 correctly. Dependent on both previous M’s A1: Cao and cso (no errors seen) | ddM1A1 |
| (7) | |
| (7 marks) |
Notes
Example Alternative Integration for last 4 marks
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \sinh\theta\cosh^2\theta\,\mathrm{d}\theta = \int \sinh\theta(1 + \sinh^2\theta)\,\mathrm{d}\theta = \int (\sinh\theta + \sinh^3\theta)\,\mathrm{d}\theta\) \(\displaystyle\int \left(\sinh\theta + \frac{1}{4}\sinh 3\theta - \frac{3}{4}\sinh\theta\right)\mathrm{d}\theta = \frac{1}{4}\int (\sinh\theta + \sinh 3\theta)\,\mathrm{d}\theta\) \(= \dfrac{1}{4}\cosh\theta + \dfrac{1}{12}\cosh 3\theta\) dM1: \(\displaystyle\int \sinh\theta\cosh^2\theta\,\mathrm{d}\theta = p\cosh\theta + q\cosh 3\theta\) A1: \(32\pi\left[\dfrac{1}{4}\cosh\theta + \dfrac{1}{12}\cosh 3\theta\right]\) | dM1A1 |
| \(A = 8\pi\left[\cosh\theta + \dfrac{1}{3}\cosh 3\theta\right]_0^1\) \(= 8\pi\left(\cosh 1 + \dfrac{1}{3}\cosh 3 - \cosh 0 - \dfrac{1}{3}\cosh 0\right)\) \(\ldots\ldots\) \(\dfrac{32\pi}{3}\left[\cosh^3 1 - 1\right]\) M1: Uses the limits 0 and 1 correctly. Dependent on both previous M’s A1: Cao | ddM1A1 |
Alternative Cartesian Approach
| Scheme | Marks |
|---|---|
| \(x = 1 + \dfrac{y^2}{8}\) Any correct Cartesian equation | B1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{y}{4}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{2}}{(x - 1)^{\frac{1}{2}}}\) Correct Derivative | B1 |
| \(A = \displaystyle\int 2\pi \cdot y\sqrt{\left(1 + \left(\frac{y}{4}\right)^2\right)}\,\mathrm{d}y\) or \(A = \displaystyle\int 2\pi \cdot \sqrt{8}(x - 1)^{\frac{1}{2}}\sqrt{\left(1 + \left(\frac{2}{x - 1}\right)\right)}\,\mathrm{d}x\) Use of a correct formula | M1 |
| \(A = 2\pi \times \dfrac{2}{3} \times 8\left(1 + \dfrac{y^2}{16}\right)^{\frac{3}{2}}\) or \(A = \dfrac{4\pi\sqrt{8}}{3}(x + 1)^{\frac{3}{2}}\) M1: Convincing attempt to integrate a relevant expression – dependent on the first M1 but allow the omission of \(2\pi\) A1: Completely correct expression for A | dM1 A1 |
| \(A = 2\pi \times \dfrac{2}{3} \times 8(1 + \sinh^2 1)^{\frac{3}{2}} - 2\pi \times \dfrac{2}{3} \times 8\) or \(2\pi \times \dfrac{2}{3} \times \sqrt{8}(1 + \cosh 2)^{\frac{3}{2}} - \dfrac{32\pi}{3}\) Correct use of limits (\(0 \to 4\sinh 1\) for \(y\) or \(1 \to \cosh 2\) for \(x\)) | ddM1 |
| Use \(1 + \sinh^2 1 = \cosh^2 1\) to give \(\dfrac{32\pi}{3}\left[\cosh^3 1 - 1\right]\) or Use \(\cosh 2 = 2\cosh^2 1 - 1\) to give \(\dfrac{32\pi}{3}\left[\cosh^3 1 - 1\right]\) | A1 |