| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Convert Cartesian to polar equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths polar coordinates question requiring standard conversions (x²-y²=a to polar using x=rcosθ, y=rsinθ), a routine sketch identifying minimum r value, and direct application of the polar area formula ½∫r²dθ. All steps are mechanical applications of known techniques with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| 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 |
|---|---|---|
| Answer | Mark | Guidance |
| \(r^2(\cos^2\theta - \sin^2\theta) = a\) | B1 | Uses \(x = r\cos\theta\) and/or \(y = r\sin\theta\) |
| \(r^2\cos2\theta = a\) | M1 | Applies relevant double angle formulae |
| \(r^2 = a\sec2\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| [sketch: initial line drawn, correct domain and position, \(r\) strictly increasing] | B1* | Initial line drawn. Correct domain and position, \(r\) strictly increasing |
| [sketch: sloping to right, concave on opposite side to pole, correct gradient when \(\theta=0\) and \(\theta \to \pi/4\)] | dB1 | Also sloping to right, concave on opposite side to pole, correct gradient when \(\theta=0\) and \(\theta \to \pi/4\) |
| \(\sqrt{a}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}a\int_0^{\frac{1}{6}\pi}\sec2\theta\, d\theta\) | M1 | Uses \(\frac{1}{2}\int r^2\, d\theta\) with correct limits |
| \(\frac{1}{4}a\Big[\ln\tan(\theta+\frac{1}{4}\pi)\Big]_0^{\frac{1}{12}\pi}\) or \(\frac{1}{4}a\Big[\ln(\tan2\theta+\sec2\theta)\Big]_0^{\frac{1}{12}\pi}\) | M1A1 | Integrates |
| \(\frac{1}{4}a\ln\sqrt{3} = \frac{1}{8}a\ln 3\) | A1 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $r^2(\cos^2\theta - \sin^2\theta) = a$ | B1 | Uses $x = r\cos\theta$ and/or $y = r\sin\theta$ |
| $r^2\cos2\theta = a$ | M1 | Applies relevant double angle formulae |
| $r^2 = a\sec2\theta$ | A1 | AG |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| [sketch: initial line drawn, correct domain and position, $r$ strictly increasing] | B1* | Initial line drawn. Correct domain and position, $r$ strictly increasing |
| [sketch: sloping to right, concave on opposite side to pole, correct gradient when $\theta=0$ and $\theta \to \pi/4$] | dB1 | Also sloping to right, concave on opposite side to pole, correct gradient when $\theta=0$ and $\theta \to \pi/4$ |
| $\sqrt{a}$ | B1 | |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}a\int_0^{\frac{1}{6}\pi}\sec2\theta\, d\theta$ | M1 | Uses $\frac{1}{2}\int r^2\, d\theta$ with correct limits |
| $\frac{1}{4}a\Big[\ln\tan(\theta+\frac{1}{4}\pi)\Big]_0^{\frac{1}{12}\pi}$ or $\frac{1}{4}a\Big[\ln(\tan2\theta+\sec2\theta)\Big]_0^{\frac{1}{12}\pi}$ | M1A1 | Integrates |
| $\frac{1}{4}a\ln\sqrt{3} = \frac{1}{8}a\ln 3$ | A1 | |5
\begin{enumerate}[label=(\alph*)]
\item Show that the curve with Cartesian equation
$$x ^ { 2 } - y ^ { 2 } = a$$
where $a$ is a positive constant, has polar equation $r ^ { 2 } = a \sec 2 \theta$.\\
The curve $C$ has polar equation $r ^ { 2 } = \operatorname { asec } 2 \theta$, where $a$ is a positive constant, for $0 \leqslant \theta < \frac { 1 } { 4 } \pi$.
\item Sketch $C$ and state the minimum distance of $C$ from the pole.
\item Find, in terms of $a$, the exact value of the area of the region enclosed by $C$, the initial line, and the half-line $\theta = \frac { 1 } { 12 } \pi$. [You may use any result from the list of formulae (MF19) without proof.] [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q5 [10]}}