Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x=a\left(\cos t+\log \tan \frac{t}{2}\right)\mathrm{and}y=a\sin t \\ \mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get} \\ \frac{dx}{dt}=\frac{d}{dt}\left[a\left(\cos t+\log \tan \frac{t}{2}\right)\right]=a\left(-\sin t+\frac{1}{\tan {\displaystyle \frac{t}{2}}}\times {\sec }^2\frac{t}{2}\times \frac{1}{2}\right) \\ =a\left(-\sin t+\frac{1}{2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}\right)=a\left(-\sin t+\frac{1}{\sin t}\right) \\ =a\left(\frac{-{\sin }^{2}t+1}{\sin t}\right)=a\left(\frac{{\cos }^{2}t}{\sin t}\right) \\ \mathrm{and} \\ \frac{dy}{dt}=\frac{d}{dt}\left(a\sin t\right)=a\cos t \\ \mathrm{Now},\left(\frac{dy}{dx}\right)=\frac{{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}}}{{\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}}}=\frac{a\cos t}{{\displaystyle a\frac{{{\displaystyle \cos }}^2{\displaystyle t}}{\sin t}}}=\tan t \\ \mathrm{Therefore}, \\ \frac{d^{2}y}{d{x}^2}=\frac{\mathrm{d}}{dx}\left(\frac{dy}{dx}\right)=\frac{\mathrm{d}}{dx}\left(\tan \left(t\right)\right) \\ =\frac{\mathrm{d}}{dt}\left(\tan \left(t\right)\right)\times \frac{\mathrm{d}t}{\mathrm{d}x}={\sec }^{2}t\times \frac{\sin t}{a{\cos }^{2}t} \\ =\left(\frac{\sin t}{a{\cos }^{4}t}\right) \\ {\left(\frac{d^{2}y}{d{x}^2}\right)}_{t=\frac{\mathrm{\pi }}{3}}=\left(\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}{a{\cos }^4\left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)}\right)=\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{a\left({\displaystyle \frac{1}{16}}\right)}=\frac{8\sqrt{3}}{a} \\ \mathrm{Hence},\mathrm{at}t=\frac{\mathrm{\pi }}{3},\frac{d^{2}y}{d{x}^2}=\frac{8\sqrt{3}}{a}. \end{gathered}$$