Advertisements
Advertisements
प्रश्न
\[\text { If x } = a\left( \cos2t + 2t \sin2t \right)\text { and y } = a\left( \sin2t - 2t \cos2t \right), \text { then find } \frac{d^2 y}{d x^2} \] ?
Advertisements
उत्तर
\[\text { We have}, \]
\[x = a\left( \cos2t + 2t \sin2t \right) \text { and y }= a\left( \sin2t - 2t \cos2t \right)\]
\[\text { On differentiating with respect to t, we get }\]
\[\frac{d x}{d t} = \frac{d}{d t}\left[ a\left( \cos2t + 2t \sin2t \right) \right] = a\left( - 2\sin2t + 2\sin2t + 4t \cos2t \right)\]
\[ = a\left( 4t \cos2t \right)\]
\[\text { and }\]
\[\frac{d y}{d t} = \frac{d}{d t}\left[ a\left( \sin2t - 2t \cos2t \right) \right] = a\left( 2\cos2t - 2\cos2t + 4t \sin2t \right)\]
\[ = a\left( 4t \sin2t \right)\]
\[\text { Now,} \left( \frac{d y}{d x} \right) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\left( 4t \sin2t \right)}{a\left( 4t \cos2t \right)} = \tan2t\]
\[\text { Therefore,} \]
\[\frac{d^2 y}{d x^2} = \frac{d}{d x}\left( \frac{d y}{d x} \right) = \frac{d}{d x}\left( \tan2t \right)\]
\[ = \frac{d}{d t}\left( \tan2t \right) \times \frac{dt}{dx} = 2 \sec^2 2t \times \frac{1}{a\left( 4t \cos2t \right)}\]
\[ = \frac{1}{2at \cos^3 2t} = \frac{1}{2at} \sec^3 2t\]
\[\text { Hence,} \frac{d^2 y}{d x^2} = \frac{1}{2at} \sec^3 2t .\]
