If x = sin t cos 2t and y = cos t sin 2t,then | vedclass

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(C) Given $x = \sin t \cos 2t$ $(i)$ and $y = \cos t \sin 2t$ $(ii)$.Differentiating $(i)$ with respect to $t$ using the product rule:$\frac{dx}{dt} =

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$1/2$

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(C) Given $x = \sin t \cos 2t$ $(i)$ and $y = \cos t \sin 2t$ $(ii)$.Differentiating $(i)$ with respect to $t$ using the product rule:$\frac{dx}{dt} = \cos t \cos 2t - 2 \sin t \sin 2t$ $(iii)$Differentiating $(ii)$ with respect to $t$ using the product rule:$\frac{dy}{dt} = - \sin t \sin 2t + 2 \cos t \cos 2t$ $(iv)$Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \cos t \cos 2t - \sin t \sin 2t}{\cos t \cos 2t - 2 \sin t \sin 2t}$.At $t = \frac{\pi}{4}$,we have $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cos 2(\frac{\pi}{4}) = \cos \frac{\pi}{2} = 0$,and $\sin 2(\frac{\pi}{4}) = \sin \frac{\pi}{2} = 1$.Substituting these values:$\frac{dy}{dx} = \frac{2(\frac{1}{\sqrt{2}})(0) - (\frac{1}{\sqrt{2}})(1)}{(\frac{1}{\sqrt{2}})(0) - 2(\frac{1}{\sqrt{2}})(1)} = \frac{-\frac{1}{\sqrt{2}}}{-\frac{2}{\sqrt{2}}} = \frac{1}{2}$.

If $x = \sin t \cos 2t$ and $y = \cos t \sin 2t$,then at $t = \frac{\pi}{4}$,the value of $\frac{dy}{dx}$ is equal to

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