Question

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

Answer 1

\[\text { We have }, \]

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

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

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

\[\text { and }\]

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

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

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

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

\[\text { Therefore}, \]

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

\[ = \frac{d}{d t}\left( \cot\left( t \right) \right) \times \frac{dt}{dx} = - {cosec}^2 t \times \frac{1}{a \cos t}\]

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

\[\text { Hence }, \frac{d^2 y}{d x^2} = - \left( \frac{1}{a \sin^2 t \cos t} \right) .\]