Question:

If $x= 3 \cos t - 2 \cos^{3} t , y = 3\sin t - 2 \sin^{3} t ,$ then $ \frac{d^{2}y}{dx^{2}} t = \frac{\pi}{6}$ is

Updated On: Jun 21, 2022
  • $ \frac{16}{2 \sqrt{3}}$
  • $- \frac{16}{3}$
  • $ \frac{16}{3}$
  • $ \frac{ - 16}{2 \sqrt{3}}$
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The Correct Option is B

Solution and Explanation

$x= 3 \cos t - 2 \cos^{3} t , y = 3\sin t - 2 \sin^{3} t $
$ \frac{dx}{dt} = -3 \sin t + 6 \cos^{2} t \sin t$
$ =- 3 \sin t \left(1-2 \cos ^{2} t\right) = 3 \sin t \cos 2t$
$ \frac{dy}{dt} = 3\cos t - 6 \sin^{2} t \cos t $
$= 3\cos t\left(1- 2 \sin^{2} t \right) = 3\cos t \cos 2t $
Now, $\frac{dy}{dx} = \frac{dy /dt}{dx/dt} = \frac{3\cos t \cos 2t }{3 \sin t \cos 2t } = \cot t$
$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left[\cot t\right] = - cosec^{2 } t \frac{dt}{dx} $
$= - cosec^{2} t . \frac{1}{3 \sin t \cos2t} = - \frac{1}{3} cosec^{3} t \sec2t $
Now, $\frac{d^{2}y}{dx^{2}} _{t=\frac{\pi}{6}} =- \frac{1}{3} \left(cosec \frac{\pi}{6}\right)^{3} \sec\left(\frac{\pi}{3}\right) = -\frac{1}{3} \left(8\right)\left(2\right)$
$ =- \frac{16}{3}$
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