Question

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

Answer 1

\[\text { We have }, \]

\[x = a\left( \cos t + \log \tan\frac{t}{2} \right)\text { and y } = a \sin t\]

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

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

\[ = a\left( - \sin t + \frac{1}{2\sin\frac{t}{2}\cos\frac{t}{2}} \right) = a\left( - \sin t + \frac{1}{\sin t} \right)\]

\[ = a\left( \frac{- \sin^2 t + 1}{\sin t} \right) = a\left( \frac{\cos^2 t}{\sin t} \right)\]

\[\text { and }\]

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

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

\[\text { Therefore,} \]

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

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

\[ = \left( \frac{\sin t}{a \cos^4 t} \right)\]

\[ \left( \frac{d^2 y}{d x^2} \right)_{t = \frac{\pi}{3}} = \left( \frac{\sin\left( \frac{\pi}{3} \right)}{a \cos^4 \left( \frac{\pi}{3} \right)} \right) = \frac{\frac{\sqrt{3}}{2}}{a\left( \frac{1}{16} \right)} = \frac{8\sqrt{3}}{a}\]

\[\text { Hence, at t } = \frac{\pi}{3}, \frac{d^2 y}{d x^2} = \frac{8\sqrt{3}}{a} .\]