| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: parametric curve |
| Difficulty | Challenging +1.2 This is a standard Further Maths surface of revolution question requiring parametric differentiation and integration of sin³t using a given identity. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the techniques are routine: apply the surface area formula, simplify using sin²t + cos²t = 1, then integrate using the provided triple angle formula. No novel insight required, just careful execution of standard methods. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = 1 - \cos 2t\), \(\frac{dy}{dt} = 2\sin t \cos t\) | B1 | B1 for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) |
| \(\frac{dx}{dt} = 1 - \cos 2t = 2\sin^2 t\) | M1 | Uses an appropriate identity |
| \(\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{4\sin^4 t + 4\sin^2 t \cos^2 t} = 2\sin t\) | A1 | |
| \(S = 2\pi \int_0^{\pi}(\sin^2 t)(2\sin t)\, dt \Rightarrow a = 4\) | M1 A1 | Uses the correct formula for \(S\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S = 4\pi \int_0^{\pi} \sin^3 t\, dt = \pi \int_0^{\pi} 3\sin t - \sin 3t\, dt\) | M1 | Uses the result \(4\sin^3 t = 3\sin t - \sin 3t\) |
| \(= \pi\left[-3\cos t + \frac{1}{3}\cos 3t\right]_0^{\pi}\) | A1 | |
| \(= \pi\left(3 - \frac{1}{3} - \left(-3 + \frac{1}{3}\right)\right) = \frac{16\pi}{3}\) | A1 | Answer must be exact |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = 1 - \cos 2t$, $\frac{dy}{dt} = 2\sin t \cos t$ | B1 | B1 for $\frac{dx}{dt}$ and $\frac{dy}{dt}$ |
| $\frac{dx}{dt} = 1 - \cos 2t = 2\sin^2 t$ | M1 | Uses an appropriate identity |
| $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{4\sin^4 t + 4\sin^2 t \cos^2 t} = 2\sin t$ | A1 | |
| $S = 2\pi \int_0^{\pi}(\sin^2 t)(2\sin t)\, dt \Rightarrow a = 4$ | M1 A1 | Uses the correct formula for $S$ |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S = 4\pi \int_0^{\pi} \sin^3 t\, dt = \pi \int_0^{\pi} 3\sin t - \sin 3t\, dt$ | M1 | Uses the result $4\sin^3 t = 3\sin t - \sin 3t$ |
| $= \pi\left[-3\cos t + \frac{1}{3}\cos 3t\right]_0^{\pi}$ | A1 | |
| $= \pi\left(3 - \frac{1}{3} - \left(-3 + \frac{1}{3}\right)\right) = \frac{16\pi}{3}$ | A1 | Answer must be exact |
---4 A curve is defined parametrically by
$$x = t - \frac { 1 } { 2 } \sin 2 t \quad \text { and } \quad y = \sin ^ { 2 } t$$
The arc of the curve joining the point where $t = 0$ to the point where $t = \pi$ is rotated through one complete revolution about the $x$-axis. The area of the surface generated is denoted by $S$.\\
(i) Show that
$$S = a \pi \int _ { 0 } ^ { \pi } \sin ^ { 3 } t \mathrm {~d} t$$
where the constant $a$ is to be found.\\
(ii) Using the result $\sin 3 t = 3 \sin t - 4 \sin ^ { 3 } t$, find the exact value of $S$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q4}}