CAIE FP1 2014 June — Question 4 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeArea enclosed by polar curve
DifficultyStandard +0.8 This question requires converting between Cartesian and polar coordinates (a standard technique), sketching a lemniscate curve (requiring understanding of when r² is positive/negative), and applying the polar area formula with a trigonometric integral. While each component is accessible, the combination of coordinate conversion, curve analysis, and integration with appropriate limits makes this moderately challenging for Further Maths students.
Spec4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve
4 The curve \(C\) has cartesian equation \(\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2 a ^ { 2 } x y\), where \(a\) is a positive constant. Show that the polar equation of \(C\) is \(r ^ { 2 } = a ^ { 2 } \sin 2 \theta\). Sketch \(C\) for \(- \pi < \theta \leqslant \pi\). Find the area enclosed by one loop of \(C\).
Question 4:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Use of \(r^2=x^2+y^2\)B1
Use of \(x=r\cos\theta\) and \(y=r\sin\theta\) (both)B1
Obtains \(r^2=a^2\sin 2\theta\)B1 [3] AG
Sketch with two loops, approximately symmetrical about \(\theta=\frac{1}{4}\pi\) and \(\theta=-\frac{3}{4}\pi\)B1B1 [2]
\(\frac{1}{2}\int_0^{\frac{1}{2}\pi}a^2\sin 2\theta\,d\theta=\left[-\frac{a^2}{4}\cos 2\theta\right]_0^{\frac{1}{2}\pi}\)M1 LNR
\(=\frac{1}{2}a^2\)A1 (2) 
## Question 4:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $r^2=x^2+y^2$ | B1 | |
| Use of $x=r\cos\theta$ and $y=r\sin\theta$ (both) | B1 | |
| Obtains $r^2=a^2\sin 2\theta$ | B1 | [3] AG |
| Sketch with two loops, approximately symmetrical about $\theta=\frac{1}{4}\pi$ and $\theta=-\frac{3}{4}\pi$ | B1B1 | [2] |
| $\frac{1}{2}\int_0^{\frac{1}{2}\pi}a^2\sin 2\theta\,d\theta=\left[-\frac{a^2}{4}\cos 2\theta\right]_0^{\frac{1}{2}\pi}$ | M1 | LNR |
| $=\frac{1}{2}a^2$ | A1 | (2) |

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4 The curve $C$ has cartesian equation $\left( x ^ { 2 } + y ^ { 2 } \right) ^ { 2 } = 2 a ^ { 2 } x y$, where $a$ is a positive constant. Show that the polar equation of $C$ is $r ^ { 2 } = a ^ { 2 } \sin 2 \theta$.

Sketch $C$ for $- \pi < \theta \leqslant \pi$.

Find the area enclosed by one loop of $C$.

\hfill \mbox{\textit{CAIE FP1 2014 Q4 [7]}}
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