| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Arc length with substitution |
| Difficulty | Challenging +1.8 This FP3 question requires applying the surface area of revolution formula to parametric equations, then executing a hyperbolic substitution to evaluate the resulting integral. While the setup involves multiple calculus techniques (parametric differentiation, surface area formula, substitution method), each step follows a standard procedure. The hyperbolic substitution is given, reducing problem-solving demand. This is challenging for A-level but represents expected FP3 material rather than requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.08h Integration by substitution8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{d\theta} = -3\sin\theta\), \(\frac{dy}{d\theta} = 5\cos\theta\) | B1 | |
| \(S = 2\pi\int 5\sin\theta\sqrt{(-3\sin\theta)^2 + (5\cos\theta)^2}\,d\theta\) | M1 | |
| \(S = 2\pi\int 5\sin\theta\sqrt{9 - 9\cos^2\theta + 25\cos^2\theta}\,d\theta\) | M1 | |
| Let \(c = \cos\theta\), \(\frac{dc}{d\theta} = -\sin\theta\), limits \(0\) and \(\frac{\pi}{2}\) become \(1\) and \(0\) | M1 | |
| \(S = k\pi\int_{0}^{\alpha}\sqrt{16c^2 + 9}\,dc\), where \(k = 10\), and \(\alpha\) is \(1\) | A1, A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(c = \frac{3}{4}\sinh u\). Then \(\frac{dc}{du} = \frac{3}{4}\cosh u\) | M1 | |
| \(S = k\pi\int_{?}^{?}\sqrt{9\sinh^2 u + 9}\cdot\frac{3}{4}\cosh u\,du\) | A1 | |
| \(= k\pi\int_{?}^{?}\frac{9}{4}\cosh^2 u\,du = k\pi\int_{?}^{?}\frac{9}{8}(\cosh 2u + 1)\,du\) | M1 | |
| \(= k\pi\left[\frac{9}{16}\sinh 2u + \frac{9}{8}u\right]_{0}^{d}\) | A1 | |
| \(= \frac{45\pi}{4}\left[\frac{20}{9} + \ln 3\right]\) i.e. \(\underline{117}\) | B1 | (5) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{d\theta} = -3\sin\theta$, $\frac{dy}{d\theta} = 5\cos\theta$ | B1 | |
| $S = 2\pi\int 5\sin\theta\sqrt{(-3\sin\theta)^2 + (5\cos\theta)^2}\,d\theta$ | M1 | |
| $S = 2\pi\int 5\sin\theta\sqrt{9 - 9\cos^2\theta + 25\cos^2\theta}\,d\theta$ | M1 | |
| Let $c = \cos\theta$, $\frac{dc}{d\theta} = -\sin\theta$, limits $0$ and $\frac{\pi}{2}$ become $1$ and $0$ | M1 | |
| $S = k\pi\int_{0}^{\alpha}\sqrt{16c^2 + 9}\,dc$, where $k = 10$, and $\alpha$ is $1$ | A1, A1 | (6) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $c = \frac{3}{4}\sinh u$. Then $\frac{dc}{du} = \frac{3}{4}\cosh u$ | M1 | |
| $S = k\pi\int_{?}^{?}\sqrt{9\sinh^2 u + 9}\cdot\frac{3}{4}\cosh u\,du$ | A1 | |
| $= k\pi\int_{?}^{?}\frac{9}{4}\cosh^2 u\,du = k\pi\int_{?}^{?}\frac{9}{8}(\cosh 2u + 1)\,du$ | M1 | |
| $= k\pi\left[\frac{9}{16}\sinh 2u + \frac{9}{8}u\right]_{0}^{d}$ | A1 | |
| $= \frac{45\pi}{4}\left[\frac{20}{9} + \ln 3\right]$ i.e. $\underline{117}$ | B1 | (5) |
**Total: [11]**\begin{enumerate}
\item A curve, which is part of an ellipse, has parametric equations
\end{enumerate}
$$x = 3 \cos \theta , \quad y = 5 \sin \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 } .$$
The curve is rotated through $2 \pi$ radians about the $x$-axis.\\
(a) Show that the area of the surface generated is given by the integral
$$k \pi \int _ { 0 } ^ { \alpha } \sqrt { } \left( 16 c ^ { 2 } + 9 \right) \mathrm { d } c , \quad \text { where } c = \cos \theta$$
and where $k$ and $\alpha$ are constants to be found.\\
(b) Using the substitution $c = \frac { 3 } { 4 } \sinh u$, or otherwise, evaluate the integral, showing all of your working and giving the final answer to 3 significant figures.\\
\hfill \mbox{\textit{Edexcel FP3 2009 Q8 [11]}}