| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar curves question requiring multiple techniques (sketching, area integration, tangent finding, coordinate conversion) but all are routine applications of FP2 methods. The r = a sin(nθ) form is a textbook rose curve, and converting to Cartesian using standard substitutions is mechanical though algebraically involved. More challenging than typical A-level due to being Further Maths content, but straightforward for FP2 students. |
| Spec | 1.07m Tangents and normals: gradient and equations4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Enclosed loop sketch | B1 | Enclosed loop |
| Axes tangential | B1 | With axes tangential; ignore anything in other quadrants |
| \(\theta = \frac{\pi}{4}\) is a line of symmetry drawn and named | B1 | |
| P is at \(r=5\), \(\theta=\frac{\pi}{4}\) | B1 | For both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Area} = \frac{1}{2}\int_0^{\pi/2} r^2\,d\theta = \frac{1}{2}\int_0^{\pi/2} 25\sin^2 2\theta\,d\theta\) | M1 | Correct formula with \(r\) substituted |
| \(= \frac{25}{4}\int_0^{\pi/2}(1-\cos 4\theta)\,d\theta = \frac{25}{4}\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\pi/2}\) | M1 | Correct method of integration including limits |
| \(= \frac{25}{4}\left(\left(\frac{\pi}{2}-0\right)-(0)\right) = \frac{25\pi}{8}\) | A1 | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Equation is of the form \(x+y=c\) | B1 | |
| P is \(\left(\frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}}\right)\) oe | B1 | |
| \(\Rightarrow x+y = 5\sqrt{2}\) | B1 | Ft. \(x+y=c\) where \(c\) comes from their P |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = 5\sin 2\theta = 10\sin\theta\cos\theta\) | M1 | Square and convert \(r^2\) |
| \(\Rightarrow r^2 = 100\sin^2\theta\cos^2\theta = 100\left(\frac{y}{r}\right)^2\left(\frac{x}{r}\right)^2\) | M1 | Substitute for \(r\) and \(\theta\) |
| \(\Rightarrow (x^2+y^2)^3 = 100x^2y^2\) | A1 | NB Answer given |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Enclosed loop sketch | B1 | Enclosed loop |
| Axes tangential | B1 | With axes tangential; ignore anything in other quadrants |
| $\theta = \frac{\pi}{4}$ is a line of symmetry drawn and named | B1 | |
| P is at $r=5$, $\theta=\frac{\pi}{4}$ | B1 | For both |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Area} = \frac{1}{2}\int_0^{\pi/2} r^2\,d\theta = \frac{1}{2}\int_0^{\pi/2} 25\sin^2 2\theta\,d\theta$ | M1 | Correct formula with $r$ substituted |
| $= \frac{25}{4}\int_0^{\pi/2}(1-\cos 4\theta)\,d\theta = \frac{25}{4}\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\pi/2}$ | M1 | Correct method of integration including limits |
| $= \frac{25}{4}\left(\left(\frac{\pi}{2}-0\right)-(0)\right) = \frac{25\pi}{8}$ | A1 | www |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Equation is of the form $x+y=c$ | B1 | |
| P is $\left(\frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}}\right)$ oe | B1 | |
| $\Rightarrow x+y = 5\sqrt{2}$ | B1 | Ft. $x+y=c$ where $c$ comes from their P |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = 5\sin 2\theta = 10\sin\theta\cos\theta$ | M1 | Square and convert $r^2$ |
| $\Rightarrow r^2 = 100\sin^2\theta\cos^2\theta = 100\left(\frac{y}{r}\right)^2\left(\frac{x}{r}\right)^2$ | M1 | Substitute for $r$ and $\theta$ |
| $\Rightarrow (x^2+y^2)^3 = 100x^2y^2$ | A1 | **NB Answer given** |
---7 A curve has polar equation $r = 5 \sin 2 \theta$ for $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.\\
(i) Sketch the curve, indicating the line of symmetry and stating the polar coordinates of the point $P$ on the curve which is furthest away from the pole.\\
(ii) Calculate the area enclosed by the curve.\\
(iii) Find the cartesian equation of the tangent to the curve at $P$.\\
(iv) Show that a cartesian equation of the curve is $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 } = ( 10 x y ) ^ { 2 }$.
\hfill \mbox{\textit{OCR FP2 2013 Q7 [13]}}