Let x(t) = 2 sqrt{2} cos t sqrt{sin 2t} and y | vedclass

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

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$\frac{-2}{3}$

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(D) Given $x = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$.First,find $\frac{dx}{dt}$ and $\frac{dy}{dt}$:$\frac{dx}{dt} = 2 \sqrt{2} [-\sin t \sqrt{\sin 2t} + \cos t \cdot \frac{1}{2\sqrt{\sin 2t}} \cdot 2 \cos 2t] = \frac{2 \sqrt{2} [-\sin t \sin 2t + \cos t \cos 2t]}{\sqrt{\sin 2t}} = \frac{2 \sqrt{2} \cos 3t}{\sqrt{\sin 2t}}$.Similarly,$\frac{dy}{dt} = 2 \sqrt{2} [\cos t \sqrt{\sin 2t} + \sin t \cdot \frac{1}{2\sqrt{\sin 2t}} \cdot 2 \cos 2t] = \frac{2 \sqrt{2} [\cos t \sin 2t + \sin t \cos 2t]}{\sqrt{\sin 2t}} = \frac{2 \sqrt{2} \sin 3t}{\sqrt{\sin 2t}}$.Thus,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\sin 3t}{\cos 3t} = 	an 3t$.At $t = \frac{\pi}{4}$,$\frac{dy}{dx} = 	an(\frac{3\pi}{4}) = -1$.Now,find $\frac{d^2y}{dx^2} = \frac{d}{dx}(	an 3t) = \frac{d}{dt}(	an 3t) \cdot \frac{dt}{dx} = 3 \sec^2 3t \cdot \frac{\sqrt{\sin 2t}}{2 \sqrt{2} \cos 3t} = \frac{3 \sec^3 3t \sqrt{\sin 2t}}{2 \sqrt{2}}$.At $t = \frac{\pi}{4}$,$\sec 3t = \sec(\frac{3\pi}{4}) = -\sqrt{2}$,so $\sec^3 3t = -2 \sqrt{2}$.$\frac{d^2y}{dx^2} = \frac{3(-2 \sqrt{2}) \sqrt{\sin(\pi/2)}}{2 \sqrt{2}} = -3$.Finally,$\frac{1 + (dy/dx)^2}{d^2y/dx^2} = \frac{1 + (-1)^2}{-3} = \frac{2}{-3} = -\frac{2}{3}$.

Let $x(t) = 2 \sqrt{2} \cos t \sqrt{\sin 2t}$ and $y(t) = 2 \sqrt{2} \sin t \sqrt{\sin 2t}$,$t \in (0, \frac{\pi}{2})$. Then $\frac{1 + (\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}$ at $t = \frac{\pi}{4}$ is equal to:

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