| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Topic | Polar coordinates |
| Type | Show polar curve has Cartesian form |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring conversion from polar to Cartesian form (involving algebraic manipulation with r² = x² + y² and y = r sin θ), sketching a parabola, and evaluating a non-trivial integral. While the techniques are standard for Further Maths, the algebraic manipulation and the integral evaluation require careful work and are above average difficulty. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
**(i)** $r = 1 + r\sin\theta \Rightarrow \sqrt{x^2+y^2} = 1+y$ **M1 M1**
Squaring and cancelling: $x^2 + y^2 = y^2 + 2y + 1 \Rightarrow y = \frac{1}{2}(x^2-1)$ **A1**
**[3]**
**(ii)** Parabola. All correct: Crossing-points at $(\pm 1, 0)$ and $(0, -\frac{1}{2})$ **M1 A1**
**[2]**
**(iii)** $\displaystyle\int_\pi^{2\pi} \frac{1}{(1-\sin\theta)^2}\,\mathrm{d}\theta = 2\times\int_\pi^{2\pi}\frac{1}{2}r^2\,\mathrm{d}\theta$ Recognition that this is related to area **M1**
$= -2\displaystyle\int_{-1}^{1}\frac{1}{2}(x^2-1)\,\mathrm{d}x$ Matching up with parabola-related region **M1**
$= -\left[\dfrac{x^3}{3} - x\right]_{-1}^{1} = \dfrac{4}{3}$ Ignore $-$ve answer **A1**
**[3]**6 The curve $P$ has polar equation $r = \frac { 1 } { 1 - \sin \theta }$ for $0 \leqslant \theta < 2 \pi , \theta \neq \frac { 1 } { 2 } \pi$.\\
(i) Determine, in the form $y = \mathrm { f } ( x )$, the cartesian equation of $P$.\\
(ii) Sketch $P$.\\
(iii) Evaluate $\int _ { \pi } ^ { 2 \pi } \frac { 1 } { ( 1 - \sin \theta ) ^ { 2 } } \mathrm {~d} \theta$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2013 Q6 [8]}}