| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Challenging +1.8 This is a comprehensive Further Maths polar coordinates question requiring sketching, Cartesian conversion, area calculation, and optimization. While it involves multiple techniques (polar area formula, calculus for maximum distance), each part follows standard procedures taught in FM Pure. The curve equation r²=sin2θcosθ requires careful analysis but no exceptional insight. The multi-part nature and FM content places it above average difficulty, but it's a straightforward application of polar coordinate methods rather than requiring novel problem-solving approaches. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_143_2014} | \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_714_2014} | \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_438_29_1283_2014} | \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_1852_2014} | \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_2423_2014} |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| One loop correct with initial line | B1 | One loop correct with initial line |
| Second loop correct with correct form at the pole | B1 | Second loop correct with correct form at the pole |
| \(\theta = \frac{1}{2}\pi\) | B1 | May be seen on their diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r^5 = 2(r\sin\theta)(r\cos\theta)^2\) | M1 | Use of \(\sin 2\theta = 2\sin\theta\cos\theta\), and \(x = r\cos\theta\) or \(y = r\sin\theta\) |
| \(r^5 = 2yx^2\) | A1 | |
| \(\left(x^2+y^2\right)^{\frac{5}{2}} = 2x^2 y\) | A1 | AEF |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\int_0^{\pi} \sin(2\theta)\cos\theta\ d\theta\) | M1 | Applies \(\frac{1}{2}\int r^2\ d\theta\) |
| \(\frac{1}{2}\int \sin(2\theta)\cos\theta\ d\theta\) | M1 | Attempt to integrate in a valid way. May apply \(\sin(2\theta) = 2\sin\theta\cos\theta\) |
| \(= \int \sin\theta\cos^2\theta\ d\theta = -\frac{1}{3}\cos^3\theta + [c]\) | A1 | Correct answer (there are alternative forms) |
| \(= \left[-\frac{1}{3}\cos^3\theta\right]_0^{\pi} = \frac{2}{3}\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dr}{d\theta} = \frac{1}{2}(\sin 2\theta \cos\theta)^{-\frac{1}{2}}(2\cos 2\theta\cos\theta - \sin 2\theta\sin\theta)\) | *M1 A1 | Differentiates with respect to \(\theta\) |
| \(2\cos 2\theta\cos\theta - \sin 2\theta\sin\theta = 6\cos^3\theta - 4\cos\theta\) | dM1 | Correct use of relevant identities to express in terms of a single trig function |
| \(6\cos^3\theta - 4\cos\theta = 0 \Rightarrow \cos^2\theta = \frac{2}{3}\) | dM1 A1 | Sets derivative equal to 0 and solves to find \(\sin^2\theta = \frac{1}{3}\), \(\cos^2\theta = \frac{2}{3}\), one of \(\tan^2\theta = \frac{1}{2}\) or \(\theta = 0.6155\) |
| \(r = \sqrt{\dfrac{4}{3\sqrt{3}}} = \sqrt{\dfrac{4\sqrt{3}}{9}} = 0.877\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2r\frac{dr}{d\theta} = 2\left(-2\sin^2\theta\cos\theta + \cos^3\theta\right)\) | *M1 A1 | Differentiates [RHS] with respect to \(\theta\) |
| \(2\left(-2(1-\cos^2\theta)\cos\theta + \cos^3\theta\right) = 6\cos^3\theta - 4\cos\theta\) | dM1 | Correct use of relevant identities in terms of a single trig function |
| \(6\cos^3\theta - 4\cos\theta = 0 \Rightarrow \cos^2\theta = \frac{2}{3}\) | dM1 A1 | Sets derivative equal to 0 and solves to find \(\sin^2\theta = \frac{1}{3}\), \(\cos^2\theta = \frac{2}{3}\), one of \(\tan^2\theta = \frac{1}{2}\) or \(\theta = 0.6155\) |
| \(r = \sqrt{\dfrac{4}{3\sqrt{3}}} = \sqrt{\dfrac{4\sqrt{3}}{9}} = 0.877\) | A1 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| One loop correct with initial line | B1 | One loop correct with initial line |
| Second loop correct with correct form at the pole | B1 | Second loop correct with correct form at the pole |
| $\theta = \frac{1}{2}\pi$ | B1 | May be seen on their diagram |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r^5 = 2(r\sin\theta)(r\cos\theta)^2$ | M1 | Use of $\sin 2\theta = 2\sin\theta\cos\theta$, and $x = r\cos\theta$ or $y = r\sin\theta$ |
| $r^5 = 2yx^2$ | A1 | |
| $\left(x^2+y^2\right)^{\frac{5}{2}} = 2x^2 y$ | A1 | AEF |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\int_0^{\pi} \sin(2\theta)\cos\theta\ d\theta$ | M1 | Applies $\frac{1}{2}\int r^2\ d\theta$ |
| $\frac{1}{2}\int \sin(2\theta)\cos\theta\ d\theta$ | M1 | Attempt to integrate in a valid way. May apply $\sin(2\theta) = 2\sin\theta\cos\theta$ |
| $= \int \sin\theta\cos^2\theta\ d\theta = -\frac{1}{3}\cos^3\theta + [c]$ | A1 | Correct answer (there are alternative forms) |
| $= \left[-\frac{1}{3}\cos^3\theta\right]_0^{\pi} = \frac{2}{3}$ | A1 | CAO |
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dr}{d\theta} = \frac{1}{2}(\sin 2\theta \cos\theta)^{-\frac{1}{2}}(2\cos 2\theta\cos\theta - \sin 2\theta\sin\theta)$ | *M1 A1 | Differentiates with respect to $\theta$ |
| $2\cos 2\theta\cos\theta - \sin 2\theta\sin\theta = 6\cos^3\theta - 4\cos\theta$ | dM1 | Correct use of relevant identities to express in terms of a single trig function |
| $6\cos^3\theta - 4\cos\theta = 0 \Rightarrow \cos^2\theta = \frac{2}{3}$ | dM1 A1 | Sets derivative equal to 0 and solves to find $\sin^2\theta = \frac{1}{3}$, $\cos^2\theta = \frac{2}{3}$, one of $\tan^2\theta = \frac{1}{2}$ or $\theta = 0.6155$ |
| $r = \sqrt{\dfrac{4}{3\sqrt{3}}} = \sqrt{\dfrac{4\sqrt{3}}{9}} = 0.877$ | A1 | |
**Alternative method for 7(d):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2r\frac{dr}{d\theta} = 2\left(-2\sin^2\theta\cos\theta + \cos^3\theta\right)$ | *M1 A1 | Differentiates [RHS] with respect to $\theta$ |
| $2\left(-2(1-\cos^2\theta)\cos\theta + \cos^3\theta\right) = 6\cos^3\theta - 4\cos\theta$ | dM1 | Correct use of relevant identities in terms of a single trig function |
| $6\cos^3\theta - 4\cos\theta = 0 \Rightarrow \cos^2\theta = \frac{2}{3}$ | dM1 A1 | Sets derivative equal to 0 and solves to find $\sin^2\theta = \frac{1}{3}$, $\cos^2\theta = \frac{2}{3}$, one of $\tan^2\theta = \frac{1}{2}$ or $\theta = 0.6155$ |
| $r = \sqrt{\dfrac{4}{3\sqrt{3}}} = \sqrt{\dfrac{4\sqrt{3}}{9}} = 0.877$ | A1 | |7 The curve $C$ has polar equation $r ^ { 2 } = \sin 2 \theta \cos \theta$, for $0 \leqslant \theta \leqslant \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and state the equation of the line of symmetry.
\item Find a Cartesian equation for $C$.\\
\begin{center}
\end{center}
\item Find the total area enclosed by $C$.
\item Find the greatest distance of a point on $C$ from the pole.\\
\includegraphics[max width=\textwidth, alt={}, center]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-16_2718_36_141_2011}
If you use the following page to complete the answer to any question, the question number must be clearly shown.\\
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\begin{tabular}{|l|l|l|l|l|}
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\includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_143_2014}
& \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_714_2014}
& \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_438_29_1283_2014}
& \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_1852_2014}
& \includegraphics[max width=\textwidth, alt={}]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-18_436_29_2423_2014}
\\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q7 [13]}}