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Question
\[\text { If x } = a\left( \cos t + t \sin t \right) \text { and y} = a\left( \sin t - t \cos t \right),\text { then find the value of } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
Solution
\[\text { We have, }\]
\[x = a\left( \cos t + t \sin t \right) \text { and y } = a\left( \sin t - t \cos t \right)\]
\[\text { On differentiating with respect to t, we get }\]
\[\frac{d x}{d t} = \frac{d}{d t}\left[ a\left( \cos t + t \sin t \right) \right] = - a\sin t + a \ sin t + at \ cos t \]
\[ = at\cos t\]
\[\text { and }\]
\[\frac{d y}{d t} = \frac{d}{d t}\left[ a\left( \sin t - t \cos t \right) \right] = a\cos t - a\cos t + at \ sin t\]
\[ = at \sin t\]
\[\text { Now,} \left( \frac{d y}{d x} \right) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at\cos t} = \tan t\]
\[\frac{d^2 y}{d x^2} = \frac{d}{d x}\left( \frac{d y}{d x} \right) = \frac{d}{d x}\left( \tan t \right)\]
\[ = \frac{d}{d t}\left( \tan t \right) \times \frac{dt}{dx} = \sec^2 t \times \frac{1}{at\cos t}\]
\[ = \frac{1}{at \cos^3 t}\]
\[ \left( \frac{d^2 y}{d x^2} \right)_{t = \frac{\pi}{4}} = \frac{1}{a\left( \frac{\pi}{4} \right) \cos^3 \left( \frac{\pi}{4} \right)} = \frac{8\sqrt{2}}{a\pi}\]
\[\text { Hence, at t } = \frac{\pi}{4}, \frac{d^2 y}{d x^2} = \frac{8\sqrt{2}}{a\pi} .\]
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