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Property 6

If x=a cos 2 ...

Question

$If x = a (\cos 2 t + 2 t \sin 2 t) and y = a (\sin 2 t - 2 t \cos 2 t), then find \frac{d^{2} y}{d x^{2}} .$

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Solution

$We have, x = a (\cos 2 t + 2 t \sin 2 t) and y = a (\sin 2 t - 2 t \cos 2 t) On differentiating with respect to t, we get \frac{d x}{d t} = \frac{d}{d t} [a (\cos 2 t + 2 t \sin 2 t)] = a (- 2 \sin 2 t + 2 \sin 2 t + 4 t \cos 2 t) = a (4 t \cos 2 t) and \frac{d y}{d t} = \frac{d}{d t} [a (\sin 2 t - 2 t \cos 2 t)] = a (2 \cos 2 t - 2 \cos 2 t + 4 t \sin 2 t) = a (4 t \sin 2 t) Now, (\frac{d y}{d x}) = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{a (4 t \sin 2 t)}{a (4 t \cos 2 t)} = \tan 2 t Therefore, \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\tan 2 t) = \frac{d}{d t} (\tan 2 t) \times \frac{d t}{d x} = 2 \sec^{2} 2 t \times \frac{1}{a (4 t \cos 2 t)} = \frac{1}{2 a t \cos^{3} 2 t} = \frac{1}{2 a t} \sec^{3} 2 t Hence, \frac{d^{2} y}{d x^{2}} = \frac{1}{2 a t} \sec^{3} 2 t .$

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