Question

If $x=a(cos\theta +logtan\frac{\theta }{2}$ and $y=asin\theta$, then find the values of $\frac{d^{2}y}{d{x}^2}at\theta =\frac{\pi }{4}$

Answer 1

$x=a(\cos \theta +\log \tan \frac{\theta }{2})$

$\Rightarrow \frac{dx}{d\theta }=a(-\sin \theta +\frac{1}{\tan \frac{\theta }{2}}{\sec }^2\frac{\theta }{2}\times \frac{1}{2})$

$=a(-\sin \theta +\frac{1}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}})$

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

$\frac{d^{2}x}{d{\theta }^2}=a(-\cos \theta +\frac{-1\left(\cos \theta \right)}{{\sin }^2\theta })$

$=-a\left(\cos \theta (1+\frac{1}{{\sin }^2\theta })\right)$

$y=a\sin \theta$ ; $\Rightarrow \frac{dy}{d\theta }=a\cos \theta$ ;

$\frac{d^{2}y}{d{\theta }^2}=-a\sin \theta$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d^{2}y}{d{\theta }^2}\times \frac{d{\theta }^2}{d{x}^2}=\frac{-a\sin \theta }{-a\cos \theta (1+\frac{1}{{\sin }^2\theta })}$

$=\frac{\tan \theta {\sin }^2\theta }{1+{\sin }^2\theta }$

$\Rightarrow \frac{d^{2}y}{d{x}^2}|\theta =\frac{\pi }{4}=\frac{{\sin }^2\frac{\pi }{4}}{1+{\sin }^2\frac{\pi }{4}}$

$=\frac{\frac{1}{2}}{1+\frac{1}{2}}$

$=\frac{1}{3}$ .