| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Arc length of polar curve |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring arc length calculation in polar coordinates, then multiplication by the width to find surface area. Students must recall the polar arc length formula, perform non-trivial integration involving √(r² + (dr/dθ)²), and interpret the 3D geometry correctly. The integration itself requires careful algebraic manipulation and is more demanding than standard FP2 questions. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete overall strategy: finding area of two sides and curved surface via arc length | M1 | Must include both sides and attempt curved surface area using arc length formula |
| Area of each side is \(\int \frac{1}{2}r^2\,d\theta = 450\int_0^1 (1-\theta^2)^2\,d\theta\) | B1 | Uses polar area formula for at least one side; may use \(2\times\int\frac{1}{2}r^2\,d\theta\) |
| \(= 450\int_0^1 1 - 2\theta^2 + \theta^4\,d\theta = 450\left[\theta - \frac{2}{3}\theta^3 + \frac{1}{5}\theta^5\right]_0^1\) | M1 | Expands \(r^2\) and integrates, powers raised by 1 |
| \(= 450\left(1 - \frac{2}{3} + \frac{1}{5}\right) = 240 \text{ cm}^2\) | A1 | Applies limits; 240 cm² for one side, 480 cm² for both |
| \(r^2 + \left(\frac{dr}{d\theta}\right)^2 = 900(1-2\theta^2+\theta^4) + (30\times -2\theta)^2\) | M1 | Attempts \(r^2 + \left(\frac{dr}{d\theta}\right)^2\) with correct differentiation |
| \(= 900(1+\theta^2)^2\) | A1 | Correct factorised expression |
| \(\int_0^1\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta = 30\int_0^1 1+\theta^2\,d\theta = 30\left[\theta + \frac{1}{3}\theta^3\right]_0^1\) | M1 | Applies arc length formula; valid attempt to take square root |
| \(= 30\left(1 + \frac{1}{3} - 0\right) = 40 \text{ cm}\) | A1 | Correct arc length |
| Surface area \(= 2\times 240 + 235\times 40 = \ldots\) | M1 | Uses surface area = arc length \(\times\) width, adds areas of sides |
| \(= 9880 \text{ cm}^2\) | A1 | cao |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete overall strategy: finding area of two sides and curved surface via arc length | **M1** | Must include both sides and attempt curved surface area using arc length formula |
| Area of each side is $\int \frac{1}{2}r^2\,d\theta = 450\int_0^1 (1-\theta^2)^2\,d\theta$ | **B1** | Uses polar area formula for at least one side; may use $2\times\int\frac{1}{2}r^2\,d\theta$ |
| $= 450\int_0^1 1 - 2\theta^2 + \theta^4\,d\theta = 450\left[\theta - \frac{2}{3}\theta^3 + \frac{1}{5}\theta^5\right]_0^1$ | **M1** | Expands $r^2$ and integrates, powers raised by 1 |
| $= 450\left(1 - \frac{2}{3} + \frac{1}{5}\right) = 240 \text{ cm}^2$ | **A1** | Applies limits; 240 cm² for one side, 480 cm² for both |
| $r^2 + \left(\frac{dr}{d\theta}\right)^2 = 900(1-2\theta^2+\theta^4) + (30\times -2\theta)^2$ | **M1** | Attempts $r^2 + \left(\frac{dr}{d\theta}\right)^2$ with correct differentiation |
| $= 900(1+\theta^2)^2$ | **A1** | Correct factorised expression |
| $\int_0^1\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta = 30\int_0^1 1+\theta^2\,d\theta = 30\left[\theta + \frac{1}{3}\theta^3\right]_0^1$ | **M1** | Applies arc length formula; valid attempt to take square root |
| $= 30\left(1 + \frac{1}{3} - 0\right) = 40 \text{ cm}$ | **A1** | Correct arc length |
| Surface area $= 2\times 240 + 235\times 40 = \ldots$ | **M1** | Uses surface area = arc length $\times$ width, adds areas of sides |
| $= 9880 \text{ cm}^2$ | **A1** | cao |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{868aedc8-6afb-4419-ae29-2ecad3461999-12_213_684_257_221}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{868aedc8-6afb-4419-ae29-2ecad3461999-12_193_736_258_1053}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 1 shows a sketch of a design for a road speed bump of width 2.35 metres. The speed bump has a uniform cross-section with vertical ends and its length is 30 cm . A side profile of the speed bump is shown in Figure 2.
The curve $C$ shown in Figure 2 is modelled by the polar equation
$$r = 30 \left( 1 - \theta ^ { 2 } \right) \quad 0 \leqslant \theta \leqslant 1$$
The units for $r$ are centimetres and the initial line lies along the road surface, which is assumed to be horizontal.
Once the speed bump has been fixed to the road, the visible surfaces of the speed bump are to be painted.
Determine, in $\mathrm { cm } ^ { 2 }$, the area that is to be painted, according to the model.
\hfill \mbox{\textit{Edexcel FP2 2020 Q4 [10]}}