Question

If $x= a\left(\cos \theta +\theta \sin \theta\right)$ and $ y = a \left(\sin \theta- \theta \cos \theta \right) $ where $0 < \theta < \frac{\pi}{2}$ , then $ \frac{d^{2}y}{dx^{2}}$ at $ \theta = \frac{\pi}{4}$ is equal to

• A. $\frac{8\sqrt{2}}{a\pi}$
• B. $\frac{4\sqrt{2}}{a\pi}$
• C. $\frac{4}{a\pi \sqrt{2}}$
• D. $none\, of\, these$

Answer 1

The Correct Option is A

We have, $x-a\left(\cos \theta +\theta \sin \theta\right) $
$y = a\left( \sin \theta - \cos \theta \right), 0 < \theta< \frac{\pi}{2}$
Differentiating w.r.t. 0, we get
$ \frac{dx}{d\theta} = a\left( - \sin \theta + \cos \theta +\theta \sin \theta \right) = a \theta \cos \theta $
$\frac{dy}{d\theta } = a\left(\cos \theta - \cos \theta +\theta \sin \theta \right) = a \theta \sin \theta $
$\frac{dy}{dx}= \frac{dy d\theta}{dx d\theta} = \frac{a \theta \sin \theta}{a \theta \cos \theta} = \tan \theta$
Now, $ \frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left( \tan \theta\right) $
$=\sec^{2} \theta . \frac{d\theta}{dx} = \sec ^{2} \theta \frac{1}{a\theta \cos\theta } = \frac{\sec^{3} \theta}{a\theta}$
$ \frac{d^{2} y}{dx^{2} } _{\theta =\pi 4} = \frac{\left(\sec \frac{\pi}{4}\right)^{3}}{a\left(\frac{\pi}{4}\right)} = \frac{4}{a\pi} \left(\sqrt{2}\right)^{3} = \frac{8\sqrt{2}}{a\pi}$