Question

\[\text { If x } = a\left( \cos2t + 2t \sin2t \right)\text {  and y } = a\left( \sin2t - 2t \cos2t \right), \text { then find } \frac{d^2 y}{d x^2} \] ?

Answer 1

\[\text { We have}, \]

\[x = a\left( \cos2t + 2t \sin2t \right) \text { and y }= a\left( \sin2t - 2t \cos2t \right)\]

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

\[\frac{d x}{d t} = \frac{d}{d t}\left[ a\left( \cos2t + 2t \sin2t \right) \right] = a\left( - 2\sin2t + 2\sin2t + 4t \cos2t \right)\]

\[ = a\left( 4t \cos2t \right)\]

\[\text { and }\]

\[\frac{d y}{d t} = \frac{d}{d t}\left[ a\left( \sin2t - 2t \cos2t \right) \right] = a\left( 2\cos2t - 2\cos2t + 4t \sin2t \right)\]

\[ = a\left( 4t \sin2t \right)\]

\[\text { Now,} \left( \frac{d y}{d x} \right) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\left( 4t \sin2t \right)}{a\left( 4t \cos2t \right)} = \tan2t\]

\[\text { Therefore,} \]

\[\frac{d^2 y}{d x^2} = \frac{d}{d x}\left( \frac{d y}{d x} \right) = \frac{d}{d x}\left( \tan2t \right)\]

\[ = \frac{d}{d t}\left( \tan2t \right) \times \frac{dt}{dx} = 2 \sec^2 2t \times \frac{1}{a\left( 4t \cos2t \right)}\]

\[ = \frac{1}{2at \cos^3 2t} = \frac{1}{2at} \sec^3 2t\]

\[\text { Hence,} \frac{d^2 y}{d x^2} = \frac{1}{2at} \sec^3 2t .\]