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

Image

 

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

Image

 

\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.     

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