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Question
\[\text { If x } = a \sin t - b \cos t, y = a \cos t + b \sin t, \text { prove that } \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} \] ?
Solution
\[\text { We have }, \]
\[x = a \sin t - b \cos t, y = a \cos t + b \sin t\]
\[\text { On differentiating with respect to t, we get }\]
\[\frac{d x}{d t} = \frac{d}{d t}\left( a \sin t - b \cos t \right) = a \cos t + b \sin t\]
\[\text { and }\]
\[\frac{d y}{d t} = \frac{d}{d t}\left( a \cos t + b \sin t \right) = - a \sin t + b \cos t\]
\[\text { Now,} \left( \frac{d y}{d x} \right) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- a \sin t + b \cos t}{a \cos t + b \sin t}\]
\[\text { Therefore, } \]
\[\frac{d^2 y}{d x^2} = \frac{d}{d x}\left( \frac{d y}{d x} \right) = \frac{d}{d x}\left( \frac{- a \sin t + b \cos t}{a \cos t + b \sin t} \right)\]
\[ = \frac{d}{d t}\left( \frac{- a \sin t + b \cos t}{a \cos t + b \sin t} \right) \times \frac{dt}{dx}\]
\[ = \frac{\left( a \cos t + b \sin t \right)\frac{d}{dt}\left( - a \sin t + b \cos t \right) - \left( - a \sin t + b \cos t \right)\frac{d}{dt}\left( a \cos t + b \sin t \right)}{\left( a \cos t + b \sin t \right)^2} \times \frac{1}{a \cos t + b \sin t}\]
\[ = \frac{\left( a \cos t + b \sin t \right)\left( - a \cos t - b \sin t \right) - \left( - a \sin t + b \cos t \right)\left( - a \sin t + b \cos t \right)}{\left( a \cos t + b \sin t \right)^3}\]
\[ = \frac{- \left( a \cos t + b \sin t \right)^2 - \left( - a \sin t + b \cos t \right)^2}{\left( a \cos t + b \sin t \right)^3}\]
\[ = \frac{- \left( a \cos t + b \sin t \right)^2 - \left( a \sin t - b \cos t \right)^2}{\left( a \cos t + b \sin t \right)^3}\]
\[ = \frac{- y^2 - x^2}{y^3}\]
\[\text{Hence,} \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} .\]
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