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1992 Paper 3 Q10
D: 1700.0 B: 1484.8
polar coordinates ellipse arc length area parametric curve sketching perimeter comparison trigonometric

Sketch the curve \(C\) whose polar equation is \[ r=4a\cos2\theta\qquad\mbox{ for }-\tfrac{1}{4}\pi<\theta<\tfrac{1}{4}\pi. \] The ellipse \(E\) has parametric equations \[ x=2a\cos\phi,\qquad y=a\sin\phi. \] Show, without evaluating the integrals, that the perimeters of \(C\) and \(E\) are equal. Show also that the areas of the regions enclosed by \(C\) and \(E\) are equal.

Solution:

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\begin{align*} && \text{Perimeter}(C) &= \int_{-\pi/4}^{\pi/4} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} \sqrt{16a^2 \cos^2 2 \theta + 64a^2 \sin^2 2 \theta } \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} 4a\sqrt{1 + 3 \sin^2 2 \theta } \d \theta \\ \\ \\ && \text{Perimeter}(D) &= \int_0^{2 \pi} \sqrt{\left ( \frac{\d x}{\d \phi}\right)^2+\left ( \frac{\d y}{\d \phi}\right)^2} \d \phi \\ &&&= \int_0^{2 \pi} \sqrt{ 4a^2 \sin^2 \phi+a^2 \cos^2 \phi} \d \phi \\ &&&= a^2\int_0^{2 \pi} \sqrt{ 1+3 \sin^2 \phi} \d \phi \\ \end{align*} But clearly these two integrals are equal. \begin{align*} && \text{A}(C) &= \frac12 \int_{-\pi/4}^{\pi/4} r^2 \d \theta \\ &&&= \frac12 \int_{-\pi/4}^{\pi/4} 16a^2 \cos^2 2 \theta \d \theta \\ &&&= 8a^2\int_{-\pi/4}^{\pi/4} \cos^2 2 \theta \d \theta \\ &&&= 8a^2 \frac{\pi}{4} = 2\pi a^2 \\ && \text{A}(D) &= 2\pi a^2 \end{align*}

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