# A Polar Equation Problem

The following is a useful problem at advanced Further Pure Maths level, involving a curve described by a polar equation. I give my suggested solution below.

The Problem

Suggested solution to part (a)

The tangent to $C$ at $P$ will be parallel to the initial line $\theta = 0$, i.e., horizontal, when $\frac{dy}{dx} = 0$ where

$y = r\sin\phi = a\sin\phi \sin2\phi$

$x = r\cos\phi = a\cos\phi \sin2\phi$

$\frac{dy}{dx} = \frac{dy/d\phi }{dx/d\phi }$

Since $\frac{dy}{dx} = 0$ we must have $\frac{dy}{d\phi } = 0$, i.e.,

$\frac{dy}{d\phi } = a\cos\phi \sin2\phi + 2a\sin\phi \cos2\phi = 0$

$\Longrightarrow$

$\frac{\tan\phi }{\tan2\phi } = -\frac{1}{2}$

But

$\tan2\phi = \tan(\phi + \phi) = \frac{2\tan\phi }{1 - \tan^2\phi }$

Using this in the previous expression gives

$\frac{\tan\phi }{2\tan\phi }(1 - \tan^2\phi) = \frac{1}{2}(1 - \tan^2\phi) = - \frac{1}{2}$

$\Longrightarrow$

$\tan\phi = \sqrt{2}$

(positive since we are working in the first quadrant). So

$\tan\phi = \frac{2}{\sqrt{2}}$

and therefore

$\cos\phi = \frac{\sqrt{2}}{\sqrt{6}} = \frac{1}{\sqrt{3}}$

as required.

Suggested solution to part (b)

We have $\sin\phi = \sqrt{\frac{2}{3}}$ and we know that at $P$ we have

$R = a\sin2\phi$

But

$\sin2\phi = \sin(\phi + \phi) = 2\sin\phi\cos\phi$

Therefore

$R = a2\sin\phi\cos\phi = 2a\frac{\sqrt{2}}{3}$

Suggested solution to part (c)

At the initial line we have $\theta = 0$. At $P$ we have $\theta = \arccos\big(\frac{1}{\sqrt{3}}\big)$. Therefore the limits of integration are $\theta = 0$ and $\theta = \arccos\big(\frac{1}{\sqrt{3}}\big)$, and the required area is

$\int_{\theta = 0}^{\arccos(\frac{1}{\sqrt{3}})}\frac{1}{2}r^2d\theta = \int_{\theta = 0}^{\arccos(\frac{1}{\sqrt{3}})}\frac{1}{2}a^2\sin^22\theta d\theta$

Using the double-angle formula

$\cos2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta$

we have

$\sin^2\theta = \frac{1 - \cos2\theta}{2}$

$\Longrightarrow$

$\sin^22\theta = \frac{1 - \cos4\theta}{2}$

The above integral becomes

$\frac{1}{4}a^2 \int_{\theta = 0}^{\arccos(\frac{1}{\sqrt{3}})}d\theta - \frac{1}{4}a^2 \int_{\theta = 0}^{\arccos(\frac{1}{\sqrt{3}})}\cos4\theta d\theta$

$= \frac{1}{4}a^2\big(\arccos\big(\frac{1}{\sqrt{3}}\big)\big) - \frac{1}{4}a^2 \big[\frac{1}{4} \sin4 \theta \big]_{\theta = 0}^{\arccos(\frac{1}{\sqrt{3}})}$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big)\big) - \frac{1}{16}a^2\sin\big(4\arccos\big(\frac{1}{\sqrt{3}}\big)\big)$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big)\big) - \frac{1}{16}a^2\sin\big(2\arccos\big(\frac{1}{\sqrt{3}}\big) + 2\arccos\big(\frac{1}{\sqrt{3}}\big)\big)$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big)\big) - \frac{1}{16}a^22\sin\big(2\arccos\big(\frac{1}{\sqrt{3}}\big)\big)\cos\big(2\arccos\big(\frac{1}{\sqrt{3}}\big)\big)$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big)\big)$

$- \frac{1}{16}a^24\sin\big(\arccos\big(\frac{1}{\sqrt{3}}\big)\big)\cos\big(\arccos\big(\frac{1}{\sqrt{3}}\big)\big)\times\big(\cos^2\big(\arccos\big(\frac{1}{\sqrt{3}}\big)\big) - \sin^2\big(\arccos\big(\frac{1}{\sqrt{3}}\big)\big)\big)$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big)\big) + \frac{1}{4}a^2\sqrt{\frac{2}{3}}\frac{1}{\sqrt{3}}\times\big(\frac{1}{3} - \frac{2}{3}\big)$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big)\big) + \frac{\sqrt{2}}{36}a^2$

$= \frac{1}{36}a^2\big(9\arccos\big(\frac{1}{\sqrt{3}}\big) + \sqrt{2}\big)$

as claimed.