A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant. Draw sketches of the circle and the cardioid on the same diagram. Write down the polar coordinates of the points of intersection of the circle and the cardioid. Show that the area of the region that is both inside the circle and inside the cardioid is $rac{1}{2}(\pi - 2) a^2$.

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A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant - CIE - A-Level Further Maths - Question 8 - 2014 - Paper 1

Question 8

$A-circle-has-polar-equation-$r-=-a$,-for-$0-\leq-\theta-<-2\pi$,-and-a-cardioid-has-polar-equation-$r-=-a(1---\cos-\theta)$,-for-$0-\leq-\theta-<-2\pi$,-where-$a$-is-a-positive-constant-CIE-A-Level Further Maths-Question 8-2014-Paper 1.png$

A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is... show full transcript

Worked Solution & Example Answer:A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant - CIE - A-Level Further Maths - Question 8 - 2014 - Paper 1

Step 1

Draw sketches of the circle and the cardioid on the same diagram.

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Answer

To sketch the circle and the cardioid, plot the circle with radius $a$ centered at the pole (origin). This will appear as a perfect circle. For the cardioid, plot the function $r = a(1 - \cos \theta)$ . This cardioid opens to the right and has a cusp at the pole. Mark any critical points to ensure clear intersection when sketches are combined.

Step 2

Write down the polar coordinates of the points of intersection of the circle and the cardioid.

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Answer

To find the points of intersection, set the equations equal: $a = a(1 - \cos \theta)$ This simplifies to $\cos \theta = 0$ , giving points of intersection at $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ . Thus, the polar coordinates of the points of intersection are $(a, \frac{\pi}{2})$ and $(a, \frac{3\pi}{2})$ .

Step 3

Show that the area of the region that is both inside the circle and inside the cardioid is $rac{1}{2}(\pi - 2) a^2$.

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Answer

The area of the region can be calculated using the integral: $\text{Area} = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left( r_{\text{cardioid}}^2 - r_{\text{circle}}^2 \right) d\theta$ Substituting the functions. $r_{\text{cardioid}} = a(1 - \cos \theta)$ $r_{\text{circle}} = a$ The limits are from $\theta = 0$ to $\theta = \pi$ since that covers one complete loop of the cardioid. Evaluating: $\text{Area} = \frac{1}{2} \int_0^{\pi} \left[ \left( a(1 - \cos \theta) \right)^2 - a^2 \right] d\theta$ This expands to: $= \frac{1}{2} \int_0^{\pi} \left[ a^2(1 - 2\cos \theta + \cos^2 \theta) - a^2 \right] d\theta$ $= \frac{1}{2} a^2 \int_0^{\pi} ( -2\cos \theta + \cos^2 \theta) d\theta$ Using the half-angle formula, this leads to: After calculation, the area results in: $= \frac{1}{2}(\pi - 2)a^2$ .

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A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant - CIE - A-Level Further Maths - Question 8 - 2014 - Paper 1

Worked Solution & Example Answer:A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant - CIE - A-Level Further Maths - Question 8 - 2014 - Paper 1

Draw sketches of the circle and the cardioid on the same diagram.

Write down the polar coordinates of the points of intersection of the circle and the cardioid.

Show that the area of the region that is both inside the circle and inside the cardioid is $rac{1}{2}(\pi - 2) a^2$.

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Other A-Level Further Maths topics to explore

A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant - CIE - A-Level Further Maths - Question 8 - 2014 - Paper 1

Worked Solution & Example Answer:A circle has polar equation $r = a$, for $0 \leq \theta < 2\pi$, and a cardioid has polar equation $r = a(1 - \cos \theta)$, for $0 \leq \theta < 2\pi$, where $a$ is a positive constant - CIE - A-Level Further Maths - Question 8 - 2014 - Paper 1

Draw sketches of the circle and the cardioid on the same diagram.

Write down the polar coordinates of the points of intersection of the circle and the cardioid.

Show that the area of the region that is both inside the circle and inside the cardioid is $ rac{1}{2}(\pi - 2) a^2$.

Join the A-Level students using SimpleStudy...

Other A-Level Further Maths topics to explore

Show that the area of the region that is both inside the circle and inside the cardioid is $rac{1}{2}(\pi - 2) a^2$.