| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Standard +0.3 This is a straightforward application of standard polar coordinate formulas for area (½∫r²dθ) and arc length (∫√(r² + (dr/dθ)²)dθ). The exponential function r = 2e^θ differentiates and integrates cleanly, requiring only routine calculus techniques with no conceptual challenges or problem-solving insight beyond formula recall. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Area} = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 4e^{2\theta}\,d\theta = \left[e^{2\theta}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\) | M1 | Uses area formula |
| \(= e^{\pi} - e^{\frac{\pi}{3}}\) \((= 20.3)\) | A1 | Obtains result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Arc length} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\sqrt{4e^{2\theta} + 4e^{2\theta}}\,d\theta = 2\sqrt{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} e^{\theta}\,d\theta\) | M1A1 | Uses arc length formula |
| \(= 2\sqrt{2}\left[e^{\theta}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = 2\sqrt{2}\left[e^{\frac{\pi}{2}} - e^{\frac{\pi}{6}}\right]\) \((= 8.83)\) | A1 | Obtains result |
## Question 1:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 4e^{2\theta}\,d\theta = \left[e^{2\theta}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$ | M1 | Uses area formula |
| $= e^{\pi} - e^{\frac{\pi}{3}}$ $(= 20.3)$ | A1 | Obtains result |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Arc length} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\sqrt{4e^{2\theta} + 4e^{2\theta}}\,d\theta = 2\sqrt{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} e^{\theta}\,d\theta$ | M1A1 | Uses arc length formula |
| $= 2\sqrt{2}\left[e^{\theta}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = 2\sqrt{2}\left[e^{\frac{\pi}{2}} - e^{\frac{\pi}{6}}\right]$ $(= 8.83)$ | A1 | Obtains result |
---1 The curve $C$ has polar equation $r = 2 \mathrm { e } ^ { \theta }$, for $\frac { 1 } { 6 } \pi \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$. Find\\
(i) the area of the region bounded by the half-lines $\theta = \frac { 1 } { 6 } \pi , \theta = \frac { 1 } { 2 } \pi$ and $C$,\\
(ii) the length of $C$.
\hfill \mbox{\textit{CAIE FP1 2013 Q1 [5]}}