Question

\[\text { If x } = \cos t + \log \tan\frac{t}{2}, y = \sin t, \text { then find the value of } \frac{d^2 y}{d t^2} \text { and } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?

Answer 1

\[\text { We have}, \]

\[x = \cos t + \log \tan\frac{t}{2} \text { and y } = \sin t \]

\[\text { On differentiating with respect to t, we get }\]

\[\frac{d x}{d t} = \frac{d}{d t}\left( \cos t + \log \tan\frac{t}{2} \right) = - \sin t + \frac{1}{\tan\frac{t}{2}} \times \sec^2 \frac{t}{2} \times \frac{1}{2}\]

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

\[ = \frac{- \sin^2 t + 1}{\sin t} = \frac{- \sin^2 t + 1}{\sin t}\]

\[ = \frac{\cos^2 t}{\sin t}\]

\[\text { and }\]

\[\frac{d y}{d t} = \frac{d}{d t}\left( \sin t \right) = \cos t\]

\[\text { Now}, \frac{d^2 y}{d t^2} = \frac{d}{d t}\left( \frac{d y}{d t} \right) = \frac{d}{d t}\left( \cos t \right) = - \sin t\]

\[ \left( \frac{d^2 y}{d t^2} \right)_{t = \frac{\pi}{4}} = - \sin\left( \frac{\pi}{4} \right) = - \frac{1}{\sqrt{2}} . . . (1)\]

\[\text { Also }, \left( \frac{d y}{d x} \right) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{\frac{\cos^2 t}{\sin t}} = \frac{\sin t}{\cos t} = \tan t\]

\[\text { Now,} \frac{d^2 y}{d x^2} = \frac{d}{d x}\left( \frac{d y}{d x} \right) = \frac{d}{d x}\left( \tan t \right)\]

\[ = \frac{d}{d t}\left( \tan t \right) \times \frac{dt}{dx} = \sec^2 t \times \frac{\sin t}{\cos^2 t}\]

\[ = \frac{\sin t}{\cos^4 t}\]

\[ \left( \frac{d^2 y}{d x^2} \right)_{t = \frac{\pi}{4}} = \frac{\sin\left( \frac{\pi}{4} \right)}{\cos^4 \left( \frac{\pi}{4} \right)} = 2\sqrt{2} . . . (2)\]

\[\text { Hence, at } t = \frac{\pi}{4}, \frac{d^2 y}{d t^2} = - \frac{1}{\sqrt{2}} \text { and } \frac{d^2 y}{d x^2} = 2\sqrt{2} .\]