Question

If $x=a(\cos t+log\tan \frac{t}{2})$ and $y=a\sin t$, then find $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{\pi }{3}$.

Answer 1

We are given, $x=a(\cos t+\log \tan \frac{t}{2})$

and $y=a\sin t$

Let differentiate $x$ & $y$ wrt $t$, we get,

$\frac{dy}{dt}=a\cos t$

& $\frac{dx}{dt}=a\left[\right(-\sin t)+\frac{1}{\tan \frac{t}{2}}.{\sec }^2\frac{t}{2}.\frac{1}{2}]$

( by chain rule)

$\frac{dx}{dt}=a[-\sin t+\frac{1\cos \frac{t}{2}}{2\sin \frac{t}{2}}\times \frac{1}{{\cos }^2\frac{t}{2}}]$

$=a[-\sin t+\frac{1}{2\sin \frac{t}{2}\cos \frac{t}{2}}]$

$=a[-\sin t+\frac{1}{\sin t}]$

$\frac{dx}{dt}=\frac{a(1-{\sin }^{2}t}{\sin t}=\frac{a{\cos }^{2}t}{\sin t}$

now, $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{a\cos t}{a{\cos }^{2}t}\times \sin t=\tan t$

again differentiate $\frac{dy}{dx}$ wrt $x$.

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^{2}y}{d{x}^2}={\sec }^{2}t.\frac{dt}{dx}$

$\frac{d^{2}y}{d{x}^2}={\sec }^{2}t\times \frac{\sin t}{a{\cos }^{2}t}$

${\left[\frac{d^{2}y}{d{x}^2}\right]}_{t=\frac{\pi }{3}}=\frac{{\sec }^2\left(\pi /3\right).\sin \left(\pi /3\right)}{a.{\cos }^2\left(\pi /3\right)}=\frac{(2{)}^2.(\sqrt{3}/2)}{a.(1/2{)}^2}$

${\left[\frac{d^{2}y}{d{x}^2}\right]}_{t=\frac{\pi }{3}}=\frac{8\sqrt{3}}{a}$