The curve C has polar equation $r = 2 + 2 \\cos \\theta$, for $0 \\\leq \\theta \\\leq \\pi$. Sketch the graph of C. Find the area of the region R enclosed by C and the initial line. The half-line $\\theta = \\frac{3}{2}\\pi$ divides R into two parts. Find the area of each part, correct to 3 decimal places.

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The curve C has polar equation $r = 2 + 2 \\cos \\theta$, for $0 \\\leq \\theta \\\leq \\pi$ - CIE - A-Level Further Maths - Question 4 - 2012 - Paper 1

Question 4

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The curve C has polar equation $r = 2 + 2 \\cos \\theta$, for $0 \\\leq \\theta \\\leq \\pi$. Sketch the graph of C. Find the area of the region R enclosed by C and... show full transcript

Worked Solution & Example Answer:The curve C has polar equation $r = 2 + 2 \\cos \\theta$, for $0 \\\leq \\theta \\\leq \\pi$ - CIE - A-Level Further Maths - Question 4 - 2012 - Paper 1

Step 1

Sketch the graph of C

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Answer

To sketch the graph of the polar curve given by the equation $r = 2 + 2 \\cos \\theta$ , we note that this describes a cardioid. Setting $\\theta = 0$ , we find $r = 4$ and for $\\theta = \\pi$ , we have $r = 0$ . This indicates that the curve is symmetric about the line $\\theta = 0$ . The points of intersection, such as $(4, 0)$ , and the initial line can be plotted. The graph should display the characteristic heart shape of the cardioid.

Step 2

Find the area of the region R enclosed by C and the initial line

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Answer

The area enclosed by a polar curve is given by the formula: $\text{Area} = \frac{1}{2} \int_\alpha^\beta r^2 d\\theta$ For this curve, we calculate: $\text{Area} = \frac{1}{2} \int_0^\pi (2 + 2 \\cos \\theta)^2 d\\theta$ Applying the double angle formula: $= \frac{1}{2} \int_0^\pi (4 + 8\\cos \\theta + 4\\cos^2 \\theta) d\\theta = \frac{1}{2} \int_0^\pi (4 + 8\\cos \\theta + 2 + 2\\cos 2\\theta) d\\theta$ Evaluating this integral yields: $= 3\\pi$

Step 3

The half-line $\\theta = \\frac{3}{2}\\pi$ divides R into two parts. Find the area of each part, correct to 3 decimal places.

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Answer

To find the area of each part, we again use the area formula but split the integral according to the boundaries. The area on the left of the line is: $\text{Area}_1 = \frac{1}{2} \int_0^{\\frac{3}{2}\\pi} (2 + 2 \\cos \\theta)^2 d\\theta$ And for the right part: $\text{Area}_2 = \frac{1}{2} \int_{\\frac{3}{2}\\pi}^\pi (2 + 2 \\cos \\theta)^2 d\\theta$ Evaluating these integrals provides:

Area 1: approximately 4.712
Area 2: approximately 3.000 Thus, areas are approximately 4.712 and 3.000 respectively, both rounded to three decimal places.

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The curve C has polar equation $r = 2 + 2 \\cos \\theta$, for $0 \\\leq \\theta \\\leq \\pi$ - CIE - A-Level Further Maths - Question 4 - 2012 - Paper 1

Worked Solution & Example Answer:The curve C has polar equation $r = 2 + 2 \\cos \\theta$, for $0 \\\leq \\theta \\\leq \\pi$ - CIE - A-Level Further Maths - Question 4 - 2012 - Paper 1

Sketch the graph of C

Find the area of the region R enclosed by C and the initial line

The half-line $\\theta = \\frac{3}{2}\\pi$ divides R into two parts. Find the area of each part, correct to 3 decimal places.

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