If x = 2cos t - cos 2t and y = 2sin t - sin 2 | vedclass

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(A) Given $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$.Differentiating with respect to $t$:$\frac{dx}{dt} = -2\sin t + 2\sin 2t$$\frac{dy}{dt}

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$\sqrt{2} + 1$

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(A) Given $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$.Differentiating with respect to $t$:$\frac{dx}{dt} = -2\sin t + 2\sin 2t$$\frac{dy}{dt} = 2\cos t - 2\cos 2t$Using the chain rule,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos t - 2\cos 2t}{-2\sin t + 2\sin 2t} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$.At $t = \frac{\pi}{4}$:$\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$,$\cos \frac{\pi}{2} = 0$,$\sin \frac{\pi}{2} = 1$,$\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.Substituting these values:$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{2}} - 0}{1 - \frac{1}{\sqrt{2}}} = \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{1}{\sqrt{2}-1}$.Rationalizing the denominator:$\frac{1}{\sqrt{2}-1} 	imes \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2} + 1$.

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.

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