| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring polar curve sketching, area integration using the standard formula ½∫r²dθ, and conversion to Cartesian form using triple angle formula and polar-Cartesian relationships. While the area calculation is routine for FP2, the Cartesian conversion requires careful algebraic manipulation across multiple steps. Moderately challenging for Further Maths but not requiring exceptional insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Enclosed loop in first quadrant with origin as pole | B1 | |
| Looking symmetric with line of symmetry around \(\theta = \frac{\pi}{6}\) | B1 | N.B. This means that \(\theta = \frac{\pi}{2}\) is not a tangent at the pole. |
Total: 2
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area} = \frac{1}{2}\int_0^{\pi/3} r^2\, d\theta = 2\int_0^{\pi/3} \sin^2 3\theta\, d\theta\) | M1 | Correct formula plus limits. Must include \(\frac{1}{2}\) |
| \(= \int_0^{\pi/3}(1 - \cos 6\theta)\, d\theta = \left[\theta - \frac{1}{6}\sin 6\theta\right]_0^{\pi/3}\) | M1 | For obtaining fn in form to integrate using double angle formulae |
| \(= \frac{\pi}{3}\) | A1 | Integral Ft lack of \(\frac{1}{2}\) |
| A1 | Answer www |
Total: 4
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\) | M1 | Obtaining \(\sin 3\theta\) as a function of \(\theta\) |
| \(y = r\sin\theta \Rightarrow \sin\theta = \dfrac{y}{r}\) and \(r^2 = x^2 + y^2\) | A1 | A correct expression |
| \(r = 2\sin 3\theta = 6\sin\theta - 8\sin^3\theta = \dfrac{6y}{r} - \dfrac{8y^3}{r^3}\) | M1 | Eliminate \(\theta\) |
| \(\Rightarrow r^4 = 6yr^2 - 8y^3\) | M1 | Eliminate \(r\) |
| \(\Rightarrow \left(x^2 + y^2\right)^2 = 6\left(x^2 + y^2\right)y - 8y^3 = 6x^2y - 2y^3\) | A1 | ag |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Eliminating \(x\) and \(y\) | M1 | |
| Get \(r\) | M1 | |
| \(r = 6\sin\theta - 8\sin^3\theta\) | A1 | |
| Obtain triple angle formula | M1 | |
| Ans | A1 |
## Question 9:
### Part (i):
Enclosed loop in first quadrant with origin as pole | **B1** |
Looking symmetric with line of symmetry around $\theta = \frac{\pi}{6}$ | **B1** | N.B. This means that $\theta = \frac{\pi}{2}$ is not a tangent at the pole.
Take one off full marks for more loops
Total: **2**
### Part (ii):
$\text{Area} = \frac{1}{2}\int_0^{\pi/3} r^2\, d\theta = 2\int_0^{\pi/3} \sin^2 3\theta\, d\theta$ | **M1** | Correct formula plus limits. Must include $\frac{1}{2}$
$= \int_0^{\pi/3}(1 - \cos 6\theta)\, d\theta = \left[\theta - \frac{1}{6}\sin 6\theta\right]_0^{\pi/3}$ | **M1** | For obtaining fn in form to integrate using double angle formulae
$= \frac{\pi}{3}$ | **A1** | Integral Ft lack of $\frac{1}{2}$
| **A1** | Answer www
Total: **4**
Alternative: Starting from given equation: Eliminating $x$ and $y$ **M1**, Get $r$ **M1**
## Question (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ | **M1** | Obtaining $\sin 3\theta$ as a function of $\theta$ |
| $y = r\sin\theta \Rightarrow \sin\theta = \dfrac{y}{r}$ and $r^2 = x^2 + y^2$ | **A1** | A correct expression |
| $r = 2\sin 3\theta = 6\sin\theta - 8\sin^3\theta = \dfrac{6y}{r} - \dfrac{8y^3}{r^3}$ | **M1** | Eliminate $\theta$ |
| $\Rightarrow r^4 = 6yr^2 - 8y^3$ | **M1** | Eliminate $r$ |
| $\Rightarrow \left(x^2 + y^2\right)^2 = 6\left(x^2 + y^2\right)y - 8y^3 = 6x^2y - 2y^3$ | **A1** | ag |
| **Total** | **5** | |
**Alternative: Starting from given equation:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Eliminating $x$ and $y$ | **M1** | |
| Get $r$ | **M1** | |
| $r = 6\sin\theta - 8\sin^3\theta$ | **A1** | |
| Obtain triple angle formula | **M1** | |
| Ans | **A1** | |9 The equation of a curve in polar coordinates is $r = 2 \sin 3 \theta$ for $0 \leqslant \theta \leqslant \frac { 1 } { 3 } \pi$.\\
(i) Sketch the curve.\\
(ii) Find the area of the region enclosed by this curve.\\
(iii) By expressing $\sin 3 \theta$ in terms of $\sin \theta$, show that a cartesian equation for the curve is
$$\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 6 x ^ { 2 } y - 2 y ^ { 3 } .$$
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR FP2 2015 Q9 [11]}}