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Question
If x = 2 cos t − cos 2t, y = 2 sin t − sin 2t, find \[\frac{d^2 y}{d x^2}\text{ at } t = \frac{\pi}{2}\] ?
Solution
Here,
\[x = 2 \cos t - \cos2t \text { and }y = 2 \sin t - \sin2t\]
\[\text{ Differentiating w . r . t . t, we get} \]
\[\frac{d x}{d t} = - 2 \sin t + 2 \sin2t \text { and }\frac{d y}{d t} = 2 \cos t - 2 \cos2t\]
\[ \therefore \frac{d y}{d x} = \frac{2 \cos t - 2 \cos2t}{- 2 \sin t + 2 \sin2t} = \frac{\cos t - \cos2t}{- \sin t + \sin2t}\]
\[\text { Differentiating w . r . t . x, we get}\]
\[\frac{d^2 y}{d x^2} = \frac{\left( - \sin t + 2 \sin2t \right)\left( - \sin t + \sin2t \right) - \left( \cos t - \cos2t \right)\left( - \cos t + 2 \cos2t \right)}{\left( - \sin t + \sin2t \right)^2}\frac{dt}{dx}\]
\[ = \frac{\left( - \sin t + 2 \sin2t \right)\left( - \sin t + \sin2t \right) - \left( \cos t - \cos2t \right)\left( - \cos t + 2 \cos2t \right)}{\left( - \sin t + \sin2t \right)^2 \left( - 2 \sin t + 2 \sin2t \right)}\]
\[\text { At } t = \frac{\pi}{2}: \]
\[\frac{d^2 y}{d x^2} = \frac{\left( - 1 + 0 \right)\left( - 1 + 0 \right) - \left( 0 + 1 \right)\left( - 0 - 2 \right)}{\left( - 1 + 0 \right)^2 \left( - 2 + 0 \right)} = \frac{1 + 2}{- 2} = \frac{- 3}{2}\]
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