Question

If $x=a(\theta +sin\theta )andy=a(1-cos\theta )$ then find the value of $\frac{d^{2}y}{{dx}^2}at\theta =\frac{\pi }{3}$

Answer 1

$x=a(\theta +\sin \theta )$

$\frac{dx}{d\theta }=a(1+\cos \theta )$

$y=a(1-\cos \theta )$

$\frac{dy}{d\theta }=a\sin \theta$

$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$

$=\frac{a\sin \theta }{a(1+\cos \theta )}$

$=\frac{\sin \theta }{(1+\cos \theta )}$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}$

$=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}$

$=\tan \frac{\theta }{2}$

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\tan \frac{\theta }{2}\right)$

$={\sec }^2\frac{\theta }{2}\frac{d}{dx}\left(\frac{\theta }{2}\right)$

$=\frac{1}{2}{\sec }^2\frac{\theta }{2}\frac{d\theta }{dx}$

$=\frac{1}{2}{\sec }^2\frac{\theta }{2}\frac{1}{a(1+\cos \theta )}$ since $\frac{dx}{d\theta }=a(1+\cos \theta )$

$=\frac{1}{2a}{\sec }^2\frac{\theta }{2}2{\sec }^2\frac{\theta }{2}$

$=\frac{1}{a}{\sec }^4\frac{\theta }{2}$

$\frac{d^{2}y}{d{x}^2}$ at $\theta =\frac{\pi }{3}=\frac{{\sec }^4\frac{\frac{\pi }{3}}{2}}{a}$

$=\frac{1}{a}{\sec }^4\frac{\pi }{6}$

$=\frac{1}{a}{\left(\frac{2}{\sqrt{3}}\right)}^4$

$=\frac{1}{a}\times \frac{16}{9}$

$=\frac{16}{9a}$