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Mathematics Question on Continuity and differentiability

If x=3cos⁡t−2cos⁡3t,y=3sin⁡t−2sin⁡3t,x= 3 \cos t - 2 \cos^{3} t , y = 3\sin t - 2 \sin^{3} t , then d2ydx2t=π6 \frac{d^{2}y}{dx^{2}} t = \frac{\pi}{6} is

A

1623 \frac{16}{2 \sqrt{3}}

B

−163- \frac{16}{3}

C

163 \frac{16}{3}

D

−1623 \frac{ - 16}{2 \sqrt{3}}

Answer

−163- \frac{16}{3}

Explanation

Solution

x=3cos⁡t−2cos⁡3t,y=3sin⁡t−2sin⁡3tx= 3 \cos t - 2 \cos^{3} t , y = 3\sin t - 2 \sin^{3} t
dxdt=−3sin⁡t+6cos⁡2tsin⁡t\frac{dx}{dt} = -3 \sin t + 6 \cos^{2} t \sin t
=−3sin⁡t(1−2cos⁡2t)=3sin⁡tcos⁡2t=- 3 \sin t \left(1-2 \cos ^{2} t\right) = 3 \sin t \cos 2t
dydt=3cos⁡t−6sin⁡2tcos⁡t\frac{dy}{dt} = 3\cos t - 6 \sin^{2} t \cos t
=3cos⁡t(1−2sin⁡2t)=3cos⁡tcos⁡2t= 3\cos t\left(1- 2 \sin^{2} t \right) = 3\cos t \cos 2t
Now, dydx=dy/dtdx/dt=3cos⁡tcos⁡2t3sin⁡tcos⁡2t=cot⁡t\frac{dy}{dx} = \frac{dy /dt}{dx/dt} = \frac{3\cos t \cos 2t }{3 \sin t \cos 2t } = \cot t
d2ydx2=ddx[cot⁡t]=−cosec2tdtdx\frac{d^{2}y}{dx^{2}} = \frac{d}{dx} \left[\cot t\right] = - cosec^{2 } t \frac{dt}{dx}
=−cosec2t.13sin⁡tcos⁡2t=−13cosec3tsec⁡2t= - cosec^{2} t . \frac{1}{3 \sin t \cos2t} = - \frac{1}{3} cosec^{3} t \sec2t
Now, d2ydx2t=π6=−13(cosecπ6)3sec⁡(π3)=−13(8)(2)\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)
=−163=- \frac{16}{3}