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

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(D) Given $x = 2 \cos t - \cos 2t$ and $y = 2 \sin t - \sin 2t$. First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = -2 \sin t + 2 \sin 2

Vedclass · Accepted Answer

$-3$

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(D) Given $x = 2 \cos t - \cos 2t$ and $y = 2 \sin t - \sin 2t$. First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = -2 \sin t + 2 \sin 2t$ $\frac{dy}{dt} = 2 \cos t - 2 \cos 2t$ Now,find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2 \cos t - 2 \cos 2t}{-2 \sin t + 2 \sin 2t} = \frac{\cos t - \cos 2t}{\sin 2t - \sin t}$. Using trigonometric identities $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$ and $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$: $\frac{dy}{dx} = \frac{2 \sin \frac{3t}{2} \sin \frac{t}{2}}{2 \cos \frac{3t}{2} \sin \frac{t}{2}} = 	an \frac{3t}{2}$. Now,differentiate with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dt} \left( 	an \frac{3t}{2} ight) \cdot \frac{dt}{dx} = \sec^2 \frac{3t}{2} \cdot \frac{3}{2} \cdot \frac{1}{-2 \sin t + 2 \sin 2t}$. At $t = \pi/2$: $\frac{d^2y}{dx^2} = \frac{3}{2} \sec^2 \left( \frac{3\pi}{4} ight) \cdot \frac{1}{-2 \sin(\pi/2) + 2 \sin(\pi)} = \frac{3}{2} (-\sqrt{2})^2 \cdot \frac{1}{-2(1) + 2(0)} = \frac{3}{2} (2) \cdot \frac{1}{-2} = -3/2$.

If $x=2 \cos t-\cos 2 t$ and $y=2 \sin t-\sin 2 t$,then the value of $\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}$ is (in $/2$)

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