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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

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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.

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Answer

To sketch the circle given by the polar equation r=ar = a, we find that this represents a circle of radius aa centered at the origin. The cardioid given by r=a(1cosθ)r = a(1 - \cos \theta) looks like a heart shape and touches the origin. On the same diagram, draw both figures with the cardioid starting at (0, 0) and its maximum at (2a,0)(2a, 0). Ensure the orientations are correct.

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(1cosθ)a = a(1 - \cos \theta) This simplifies to cosθ=0\cos \theta = 0, yielding points of intersection at:

  • θ=π2\theta = \frac{\pi}{2}, leading to coordinates (a,π2)(a, \frac{\pi}{2});
  • θ=3π2\theta = \frac{3\pi}{2}, leading to coordinates (a,3π2)(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 a^{2}$.

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Answer

To find the area between the two curves, we determine the area inside the cardioid and subtract the area inside the circle.

The area of the cardioid from θ=0\theta = 0 to θ=π\theta = \pi is: Acardioid=1202π(a(1cosθ))2dθ=1202π(a2(12cosθ+cos2θ))dθA_{cardioid} = \frac{1}{2} \int_0^{2\pi} (a(1 - \cos \theta))^2 d\theta = \frac{1}{2} \int_0^{2\pi} (a^{2}(1 - 2\cos \theta + \cos^2 \theta)) d\theta Using the half angle identity, this can be computed: =12a2(π+2)= \frac{1}{2} \cdot a^{2} \left(\pi + 2\right)

The area of the circle is simply: Acircle=1202πa2dθ=12×2πa2=πa2A_{circle} = \frac{1}{2} \int_0^{2\pi} a^{2} d\theta = \frac{1}{2} \times 2\pi a^{2} = \pi a^{2}

Thus, the final area is: Area=AcardioidAcircle=12a2(π+2)πa2=12πa2Area = A_{cardioid} - A_{circle} = \frac{1}{2}a^{2}(\pi + 2) - \pi a^{2} = \frac{1}{2}\pi a^{2}

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