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A curve C has polar equation $r^2 = 8 \, ext{cosec} \, 2\theta$ for $0 < \theta < \frac{\pi}{2}$ - CIE - A-Level Further Maths - Question 4 - 2016 - Paper 1

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A curve C has polar equation $r^2 = 8 \, ext{cosec} \, 2\theta$ for $0 < \theta < \frac{\pi}{2}$. Find a cartesian equation of C. Sketch C. Determine the exac... show full transcript

Worked Solution & Example Answer:A curve C has polar equation $r^2 = 8 \, ext{cosec} \, 2\theta$ for $0 < \theta < \frac{\pi}{2}$ - CIE - A-Level Further Maths - Question 4 - 2016 - Paper 1

Step 1

Find a cartesian equation of C.

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Answer

To find the Cartesian equation, we start with the polar equation:

$r^2 = 8 \, \csc(2\theta)$

Substituting $x = r \cos \theta$ and $y = r \sin \theta$ :

Rewrite $\csc(2\theta)$ using the double angle formula: $\csc(2\theta) = \frac{1}{\sin(2\theta)} = \frac{2}{2 \sin \theta \cos \theta}$
Thus, we have: $r^2 = 8 \cdot \frac{2}{2 \sin \theta \cos \theta} = 4 \cdot \csc(2\theta)$
This leads to: $\cos \theta \sin \theta = 4 \Rightarrow xy = 4$

Therefore, the Cartesian equation of curve C is $xy = 4$ .

Step 2

Sketch C.

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The curve $xy = 4$ is a hyperbola. We will focus on the first quadrant where both $x$ and $y$ are positive. The curve is asymptotic to both axes, showing that as $x$ approaches 0, $y$ tends towards infinity, and similarly for $y$ approaching 0, $x$ tends towards infinity.

Step 3

Determine the exact area of the sector bounded by the arc of C between $\theta = \frac{\pi}{6}$ and $\theta = \frac{\pi}{3}$.

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Answer

To calculate the area, we need the integral of $\frac{1}{2} r^2 d\theta$ :

$\text{Area} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} r^2 \, d\theta$

Here, substitute $r^2 = 8 \, \csc(2\theta)$ :

$\text{Area} = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 8 \, \csc(2\theta) \, d\theta$

Calculate the integral:

$= 4 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc(2\theta) \, d\theta$

Using the identity for the integral: $\int \csc(u) \, du = \ln | \csc(u) - \cot(u) | + C$

Substituting back: $= 4 \left[ \frac{1}{2} \ln | 3 \sqrt{3} | - \frac{1}{2} \ln | \sqrt{3} | \right] = 2 \ln | \sqrt{3} |$

Therefore the exact area is: $= 2 \ln(3 \sqrt{3}) - 2 \ln(\sqrt{3})$ = $2 \ln(3)$ .

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A curve C has polar equation $r^2 = 8 \, ext{cosec} \, 2\theta$ for $0 < \theta < \frac{\pi}{2}$ - CIE - A-Level Further Maths - Question 4 - 2016 - Paper 1

Worked Solution & Example Answer:A curve C has polar equation $r^2 = 8 \, ext{cosec} \, 2\theta$ for $0 < \theta < \frac{\pi}{2}$ - CIE - A-Level Further Maths - Question 4 - 2016 - Paper 1

Find a cartesian equation of C.

Sketch C.

Determine the exact area of the sector bounded by the arc of C between $\theta = \frac{\pi}{6}$ and $\theta = \frac{\pi}{3}$.

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