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 $\frac{d^{2}y}{d{x}^2}$ 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. $noneofthese$

Answer 1

Answer: (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
$\frac{dx}{d\theta }=a\left(-\sin \theta +\cos \theta +\theta \sin \theta \right)=a\theta \cos \theta$
$\frac{dy}{d\theta }=a\left(\cos \theta -\cos \theta +\theta \sin \theta \right)=a\theta \sin \theta$
$\frac{dy}{dx}=\frac{dyd\theta }{dxd\theta }=\frac{a\theta \sin \theta }{a\theta \cos \theta }=\tan \theta$
Now, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\tan \theta \right)$
$={\sec }^2\theta .\frac{d\theta }{dx}={\sec }^2\theta \frac{1}{a\theta \cos \theta }=\frac{{\sec }^3\theta }{a\theta }$
${\frac{d^{2}y}{d{x}^2}}_{\theta =\pi 4}=\frac{{\left(\sec \frac{\pi }{4}\right)}^3}{a\left(\frac{\pi }{4}\right)}=\frac{4}{a\pi }{\left(\sqrt{2}\right)}^3=\frac{8\sqrt{2}}{a\pi }$