| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.2 This is a structured multi-part polar coordinates question requiring conversion from Cartesian to polar form, symmetry analysis, and curve sketching. While it involves Further Maths content (polar coordinates), the steps are methodical and well-scaffolded: substituting x=r cos θ, y=r sin θ, factoring out r, and using standard symmetry arguments. The curve (folium of Descartes) is a standard FP2 example. More routine than typical FP2 proof questions but requires more technique than basic C3/C4 work. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = r\cos\theta\), \(y = r\sin\theta\) | M1 | For substituting for \(x\) and \(y\) |
| \(\Rightarrow r = \frac{a\cos\theta\sin\theta}{\cos^3\theta + \sin^3\theta}\) | A1 | For correct equation oe (Must be \(r = \ldots\)) |
| for \(0 \leq \theta \leq \frac{1}{2}\pi\) | A1 | For correct limits for \(\theta\) (Condone \(<\)) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f\!\left(\frac{1}{2}\pi - \theta\right) = \frac{a\cos\!\left(\frac{1}{2}\pi-\theta\right)\sin\!\left(\frac{1}{2}\pi-\theta\right)}{\cos^3\!\left(\frac{1}{2}\pi-\theta\right)+\sin^3\!\left(\frac{1}{2}\pi-\theta\right)} = \frac{a\sin\theta\cos\theta}{\sin^3\theta+\cos^3\theta}\) | M1 | For replacing \(\theta\) by \(\left(\frac{1}{2}\pi - \theta\right)\) in their \(f(\theta)\) — N.B. answer given |
| A1 | For correct simplified form (Must be convincing) | |
| \(f(\theta) = f\!\left(\frac{1}{2}\pi - \theta\right) \Rightarrow \alpha = \frac{1}{4}\pi\) | A1 | For correct reason for \(\alpha = \frac{1}{4}\pi\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = \frac{a \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^3 + \left(\frac{1}{\sqrt{2}}\right)^3} = \frac{1}{2}\sqrt{2}\, a\) | B1 | For correct value of \(r\) oe |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Closed curve in 1st quadrant only, symmetrical about \(\theta = \frac{1}{4}\pi\) | B1 | |
| Diagram showing \(\theta = 0\), \(\frac{1}{2}\pi\) tangential at \(O\) | B1 | |
| Total: 2 |
# Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = r\cos\theta$, $y = r\sin\theta$ | M1 | For substituting for $x$ and $y$ |
| $\Rightarrow r = \frac{a\cos\theta\sin\theta}{\cos^3\theta + \sin^3\theta}$ | A1 | For correct equation **oe** (Must be $r = \ldots$) |
| for $0 \leq \theta \leq \frac{1}{2}\pi$ | A1 | For correct limits for $\theta$ (Condone $<$) |
| **Total: 3** | | |
---
# Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f\!\left(\frac{1}{2}\pi - \theta\right) = \frac{a\cos\!\left(\frac{1}{2}\pi-\theta\right)\sin\!\left(\frac{1}{2}\pi-\theta\right)}{\cos^3\!\left(\frac{1}{2}\pi-\theta\right)+\sin^3\!\left(\frac{1}{2}\pi-\theta\right)} = \frac{a\sin\theta\cos\theta}{\sin^3\theta+\cos^3\theta}$ | M1 | For replacing $\theta$ by $\left(\frac{1}{2}\pi - \theta\right)$ in their $f(\theta)$ — **N.B. answer given** |
| | A1 | For correct simplified form (Must be convincing) |
| $f(\theta) = f\!\left(\frac{1}{2}\pi - \theta\right) \Rightarrow \alpha = \frac{1}{4}\pi$ | A1 | For correct reason for $\alpha = \frac{1}{4}\pi$ |
| **Total: 3** | | |
---
# Question 4(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{a \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^3 + \left(\frac{1}{\sqrt{2}}\right)^3} = \frac{1}{2}\sqrt{2}\, a$ | B1 | For correct value of $r$ **oe** |
| **Total: 1** | | |
---
# Question 4(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Closed curve in 1st quadrant only, symmetrical about $\theta = \frac{1}{4}\pi$ | B1 | |
| Diagram showing $\theta = 0$, $\frac{1}{2}\pi$ tangential at $O$ | B1 | |
| **Total: 2** | | |
---4 A curve $C$ has the cartesian equation $x ^ { 3 } + y ^ { 3 } = a x y$, where $x \geqslant 0 , y \geqslant 0$ and $a > 0$.\\
(i) Express the polar equation of $C$ in the form $r = \mathrm { f } ( \theta )$ and state the limits between which $\theta$ lies.
The line $\theta = \alpha$ is a line of symmetry of $C$.\\
(ii) Find and simplify an expression for $\mathrm { f } \left( \frac { 1 } { 2 } \pi - \theta \right)$ and hence explain why $\alpha = \frac { 1 } { 4 } \pi$.\\
(iii) Find the value of $r$ when $\theta = \frac { 1 } { 4 } \pi$.\\
(iv) Sketch the curve $C$.
\hfill \mbox{\textit{OCR FP2 2011 Q4 [9]}}