| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| 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 FP2 surface area of revolution question requiring the formula S = 2π∫x√(1+(dx/dy)²)dy, conversion from parametric form, algebraic manipulation to reach the given expression, then routine integration and arithmetic. While it involves multiple steps and parametric equations (making it harder than typical A-level), it follows a well-practiced template for FP2 students with no novel insight required. |
| Spec | 8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Surface of revolution \(= 2\pi\int_0^1 x\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt\) giving \(2\pi\int(10+15t-5t^3)\sqrt{(15-15t^2)^2+(30t)^2}\,dt\) | M1, A1 | |
| \(= \{2\pi\}\int\sqrt{225-450t^2+225t^4+900t^2}\,dt = \{2\pi\}\int\sqrt{225+450t^2+225t^4}\,dt\) | M1 | |
| \(= \{2\pi\}\int\sqrt{(15+15t^2)^2}\,dt = \{2\pi\}\int(15+15t^2)\,dt\) | M1 | |
| \(= 2\pi\int(10+15t-5t^3)(15+15t^2)\,dt\) leading to \(150\pi\int_0^1(2+3t+2t^2+2t^3-t^5)\,dt\) | A1* | No errors seen; evidence of 75 taken as factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(150\pi\left[2t+\frac{3}{2}t^2+\frac{2}{3}t^3+\frac{2}{4}t^4-\frac{1}{6}t^6\right]_0^1 = 150\pi\left[\left(2+\frac{3}{2}+\frac{2}{3}+\frac{2}{4}-\frac{1}{6}\right)-(0)\right]\) | M1 | Integrates polynomial, uses limits \(t=0\) and \(t=1\) |
| Surface of revolution \(= 675\pi \approx 2120\) | A1 | |
| Adds surface of revolution to area of circular base: \(675\pi + \pi \times 10^2 = \ldots\) | M1 | Must add circular base |
| Total inner surface area \(= 775\pi\) or \(\approx 2430\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Number of plant pots \(= \dfrac{120\,000}{2 \times \text{their total surface area}}\) or \(\dfrac{12}{2\times\frac{\text{their total surface area}}{100^2}}\) | M1 | |
| 24 (complete plant pots) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. The outside surface area will be more than the inside; there will be a rim not considered; there is thickness to the pot; the surface area may not perfectly fit the curve; pot may not be smooth | B1 | Any correct limitation of the model |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Surface of revolution $= 2\pi\int_0^1 x\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$ giving $2\pi\int(10+15t-5t^3)\sqrt{(15-15t^2)^2+(30t)^2}\,dt$ | M1, A1 | |
| $= \{2\pi\}\int\sqrt{225-450t^2+225t^4+900t^2}\,dt = \{2\pi\}\int\sqrt{225+450t^2+225t^4}\,dt$ | M1 | |
| $= \{2\pi\}\int\sqrt{(15+15t^2)^2}\,dt = \{2\pi\}\int(15+15t^2)\,dt$ | M1 | |
| $= 2\pi\int(10+15t-5t^3)(15+15t^2)\,dt$ leading to $150\pi\int_0^1(2+3t+2t^2+2t^3-t^5)\,dt$ | A1* | No errors seen; evidence of 75 taken as factor |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $150\pi\left[2t+\frac{3}{2}t^2+\frac{2}{3}t^3+\frac{2}{4}t^4-\frac{1}{6}t^6\right]_0^1 = 150\pi\left[\left(2+\frac{3}{2}+\frac{2}{3}+\frac{2}{4}-\frac{1}{6}\right)-(0)\right]$ | M1 | Integrates polynomial, uses limits $t=0$ and $t=1$ |
| Surface of revolution $= 675\pi \approx 2120$ | A1 | |
| Adds surface of revolution to area of circular base: $675\pi + \pi \times 10^2 = \ldots$ | M1 | Must add circular base |
| Total inner surface area $= 775\pi$ or $\approx 2430$ | A1 | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Number of plant pots $= \dfrac{120\,000}{2 \times \text{their total surface area}}$ or $\dfrac{12}{2\times\frac{\text{their total surface area}}{100^2}}$ | M1 | |
| 24 (complete plant pots) | A1 | |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. The outside surface area will be more than the inside; there will be a rim not considered; there is thickness to the pot; the surface area may not perfectly fit the curve; pot may not be smooth | B1 | Any correct limitation of the model |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9516df6d-0e85-45d8-afb0-281c80450159-28_387_474_340_324}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9516df6d-0e85-45d8-afb0-281c80450159-28_448_716_315_1023}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 2 shows a picture of a plant pot.\\
The plant pot has
\begin{itemize}
\item a flat circular base of radius 10 cm
\item a height of 15 cm
\end{itemize}
Figure 3 shows a sketch of the curve $C$ with parametric equations
$$x = 10 + 15 t - 5 t ^ { 3 } \quad y = 15 t ^ { 2 } \quad 0 \leqslant t \leqslant 1$$
The curved inner surface of the plant pot is modelled by the surface of revolution formed by rotating curve $C$ through $2 \pi$ radians about the $y$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model, the area of the curved inner surface of the plant pot is given by
$$150 \pi \int _ { 0 } ^ { 1 } \left( 2 + 3 t + 2 t ^ { 2 } + 2 t ^ { 3 } - t ^ { 5 } \right) \mathrm { d } t$$
\item Determine, according to the model, the total area of the inner surface of the plant pot.
Each plant pot will be painted with one coat of paint, both inside and outside. The paint in one tin will cover an area of $12 \mathrm {~m} ^ { 2 }$
\item Use the answer to part (b) to estimate how many plant pots can be painted using one tin of paint.
\item Give a reason why the model might not give an accurate answer to part (c).
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2022 Q10 [12]}}