| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric surface area of revolution |
| Difficulty | Challenging +1.2 This is a standard Further Maths surface area of revolution question requiring the formula S = 2π∫y√((dx/dt)² + (dy/dt)²)dt. While it involves exponential functions and algebraic manipulation to simplify the integrand, the technique is routine for F3 students and the 'show that' format provides a clear target, reducing problem-solving demands. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dx}{dt} = 2e^{\frac{1}{2}t}\), \(\frac{dy}{dt} = e^t - 1\) | B1 | Correct derivatives |
| \(S = (2\pi)\int y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt = (2\pi)\int(e^t-t)\sqrt{\left(4e^{\frac{1}{2}t}\right)^2+(e^t-t)^2}\,dt\) \(= (2\pi)\int(e^t-t)\sqrt{4e^t+e^{2t}-2e^t+1}\,dt\) | M1 | Applies the surface area formula with or without the \(2\pi\) |
| \(= (2\pi)\int(e^t-t)(e^t+1)\,dt\) | A1 | Correct simplified integral. Brackets must be present unless implied by subsequent work, but award by implication if \((2\pi)\int(e^{2t}+e^t-te^t-t)\,dt\) is seen |
| \(= (2\pi)\int(e^{2t}+e^t-te^t-t)\,dt = (2\pi)\left[\frac{1}{2}e^{2t}+e^t-te^t+e^t-\frac{1}{2}t^2\right]\) | B1 | For \(\int te^t\,dt = te^t - e^t\ (+c)\) |
| Fully correct integration | A1 | Integration may be shown as 2 separate parts, score B1A1 if both parts correct |
| \(= 2\pi\left[\frac{1}{2}e^{2t}+2e^t-te^t-\frac{1}{2}t^2\right]_0^4 = 2\pi\left\{\left(\frac{1}{2}e^8+2e^4-4e^4-8\right)-\left(\frac{1}{2}+2\right)\right\}\) | dM1 | Applies limits 0 and 4. Must include \(2\pi\) now. If 2 integrals used, limits must be applied to both and results added. Depends on first M mark (and some valid integration) |
| \(\pi(e^8 - 4e^4 - 21)\) | A1 | cao |
# Question 5:
$$x = 4e^{\frac{1}{2}t}, \quad y = e^t - t, \quad 0 \leqslant t \leqslant 4$$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 2e^{\frac{1}{2}t}$, $\frac{dy}{dt} = e^t - 1$ | B1 | Correct derivatives |
| $S = (2\pi)\int y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt = (2\pi)\int(e^t-t)\sqrt{\left(4e^{\frac{1}{2}t}\right)^2+(e^t-t)^2}\,dt$ $= (2\pi)\int(e^t-t)\sqrt{4e^t+e^{2t}-2e^t+1}\,dt$ | M1 | Applies the surface area formula with or without the $2\pi$ |
| $= (2\pi)\int(e^t-t)(e^t+1)\,dt$ | A1 | Correct simplified integral. Brackets must be present unless implied by subsequent work, but award by implication if $(2\pi)\int(e^{2t}+e^t-te^t-t)\,dt$ is seen |
| $= (2\pi)\int(e^{2t}+e^t-te^t-t)\,dt = (2\pi)\left[\frac{1}{2}e^{2t}+e^t-te^t+e^t-\frac{1}{2}t^2\right]$ | B1 | For $\int te^t\,dt = te^t - e^t\ (+c)$ |
| Fully correct integration | A1 | Integration may be shown as 2 separate parts, score B1A1 if both parts correct |
| $= 2\pi\left[\frac{1}{2}e^{2t}+2e^t-te^t-\frac{1}{2}t^2\right]_0^4 = 2\pi\left\{\left(\frac{1}{2}e^8+2e^4-4e^4-8\right)-\left(\frac{1}{2}+2\right)\right\}$ | dM1 | Applies limits 0 and 4. Must include $2\pi$ now. If 2 integrals used, limits must be applied to both and results added. Depends on first M mark (and some valid integration) |
| $\pi(e^8 - 4e^4 - 21)$ | A1 | cao |
---\begin{enumerate}
\item A curve has parametric equations
\end{enumerate}
$$x = 4 \mathrm { e } ^ { \frac { 1 } { 2 } t } \quad y = \mathrm { e } ^ { t } - t \quad 0 \leqslant t \leqslant 4$$
The curve is rotated through $2 \pi$ radians about the $x$-axis.\\
Show that the area of the curved surface generated is
$$\pi \left( \mathrm { e } ^ { 8 } + A \mathrm { e } ^ { 4 } + B \right)$$
where $A$ and $B$ are constants to be determined.
\hfill \mbox{\textit{Edexcel F3 2022 Q5 [7]}}