Question

$\mathrm{If}x=\cos t+\log \tan \frac{t}{2},y=\sin t,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{d^{2}y}{d{t}^2}\mathrm{and}\frac{d^{2}y}{d{x}^2}\mathrm{at}t=\frac{\mathrm{\pi }}{4}.$

Answer 1

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