| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.2 This is a Further Maths FP2 question requiring conversion to polar form (straightforward using r²=x²+y², x=r cos θ), sketching a cardioid (standard polar curve), and finding area ratios using polar integration. While multi-part and requiring several techniques, each step follows standard FP2 procedures without requiring novel insight or particularly complex manipulation. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(r^2=x^2+y^2\), \(x=r\cos\theta\) | M1, A1 | |
| \(\Rightarrow r^2-r\cos\theta=r \Rightarrow r=1+\cos\theta\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch of cardioid] | B1 | Cardioid (general shape). e.g. cusp clearly at pole, vertical tangent at \(r=2\) |
| B1 | Correct shape at pole, \(r=2\) and symmetric |
| Answer | Marks | Guidance |
|---|---|---|
| Line cuts curve at \((0,1)\) and \((2,0)\) | B1 | |
| Total area \(= 2\times\frac{1}{2}\times\int_0^{\pi}(1+\cos\theta)^2\,d\theta\) | ||
| \(=\int_0^{\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta = \int_0^{\pi}\left(1+2\cos\theta+\frac{1+\cos 2\theta}{2}\right)d\theta\) | M1 | Formula for area used. Sight of expansion and attempt to integrate |
| \(=\left[\frac{3}{2}\theta+2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\pi}=\frac{3}{2}\pi\) | A1 |
| Answer | Marks |
|---|---|
| \(=\frac{1}{2}\left[\frac{3}{2}\theta+2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{2}}=\frac{3}{8}\pi+1\) | A1 |
| Area under line in 1st quadrant \(= 1\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow\) ratio \(=\left(\frac{3}{2}\pi-\frac{3}{8}\pi\right):\frac{3}{8}\pi = 3:1\) | A1 | Or ratio \(1:3\) |
# Question 8(i):
Substitute $r^2=x^2+y^2$, $x=r\cos\theta$ | M1, A1 |
$\Rightarrow r^2-r\cos\theta=r \Rightarrow r=1+\cos\theta$ | A1 | **cao** |
**[3]**
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# Question 8(ii):
[Sketch of cardioid] | B1 | Cardioid (general shape). e.g. cusp clearly at pole, vertical tangent at $r=2$ |
| B1 | Correct shape at pole, $r=2$ and symmetric |
**[2]**
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# Question 8(iii):
Line cuts curve at $(0,1)$ and $(2,0)$ | B1 |
Total area $= 2\times\frac{1}{2}\times\int_0^{\pi}(1+\cos\theta)^2\,d\theta$ | |
$=\int_0^{\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta = \int_0^{\pi}\left(1+2\cos\theta+\frac{1+\cos 2\theta}{2}\right)d\theta$ | M1 | Formula for area used. Sight of expansion and attempt to integrate |
$=\left[\frac{3}{2}\theta+2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\pi}=\frac{3}{2}\pi$ | A1 |
Area in 1st quadrant $=\frac{1}{2}\times\int_0^{\frac{1}{2}\pi}(1+\cos\theta)^2\,d\theta$
$=\frac{1}{2}\left[\frac{3}{2}\theta+2\sin\theta+\frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{2}}=\frac{3}{8}\pi+1$ | A1 |
Area under line in 1st quadrant $= 1$ | M1 |
$\Rightarrow$ Area enclosed by line and curve $=\frac{3}{8}\pi+1-1=\frac{3}{8}\pi$
$\Rightarrow$ ratio $=\left(\frac{3}{2}\pi-\frac{3}{8}\pi\right):\frac{3}{8}\pi = 3:1$ | A1 | Or ratio $1:3$ |
**[6]**8 The equation of a curve is $x ^ { 2 } + y ^ { 2 } - x = \sqrt { x ^ { 2 } + y ^ { 2 } }$.\\
(i) Find the polar equation of this curve in the form $r = \mathrm { f } ( \theta )$.\\
(ii) Sketch the curve.\\
(iii) The line $x + 2 y = 2$ divides the region enclosed by the curve into two parts. Find the ratio of the two areas.
\hfill \mbox{\textit{OCR FP2 2013 Q8 [11]}}