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(A) Given equations are $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$.First,find $\frac{dx}{dt}$ using the quotient

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$-\cot 3t$

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(A) Given equations are $x = \frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$.First,find $\frac{dx}{dt}$ using the quotient rule:$\frac{dx}{dt} = \frac{\sqrt{\cos 2t} \cdot \frac{d}{dt}(\sin^3 t) - \sin^3 t \cdot \frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t}$$= \frac{\sqrt{\cos 2t} \cdot 3\sin^2 t \cos t - \sin^3 t \cdot \frac{1}{2\sqrt{\cos 2t}} \cdot (-2\sin 2t)}{\cos 2t}$$= \frac{3\cos 2t \sin^2 t \cos t + \sin^3 t \sin 2t}{(\cos 2t)^{3/2}}$Next,find $\frac{dy}{dt}$ using the quotient rule:$\frac{dy}{dt} = \frac{\sqrt{\cos 2t} \cdot \frac{d}{dt}(\cos^3 t) - \cos^3 t \cdot \frac{d}{dt}(\sqrt{\cos 2t})}{\cos 2t}$$= \frac{\sqrt{\cos 2t} \cdot 3\cos^2 t (-\sin t) - \cos^3 t \cdot \frac{1}{2\sqrt{\cos 2t}} \cdot (-2\sin 2t)}{\cos 2t}$$= \frac{-3\cos 2t \cos^2 t \sin t + \cos^3 t \sin 2t}{(\cos 2t)^{3/2}}$Now,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-3\cos 2t \cos^2 t \sin t + \cos^3 t \sin 2t}{3\cos 2t \sin^2 t \cos t + \sin^3 t \sin 2t}$Substitute $\sin 2t = 2\sin t \cos t$:$= \frac{\sin t \cos t [-3\cos 2t \cos t + 2\cos^3 t]}{\sin t \cos t [3\cos 2t \sin t + 2\sin^3 t]}$$= \frac{-3(2\cos^2 t - 1)\cos t + 2\cos^3 t}{3(1 - 2\sin^2 t)\sin t + 2\sin^3 t}$$= \frac{-6\cos^3 t + 3\cos t + 2\cos^3 t}{3\sin t - 6\sin^3 t + 2\sin^3 t} = \frac{-4\cos^3 t + 3\cos t}{3\sin t - 4\sin^3 t}$Using identities $\cos 3t = 4\cos^3 t - 3\cos t$ and $\sin 3t = 3\sin t - 4\sin^3 t$:$= \frac{-(\cos 3t)}{\sin 3t} = -\cot 3t$.

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{d y}{d x}$:
$x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$

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If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{d y}{d x}$:$x = \frac{\sin^3 t}{\sqrt{\cos 2t}}, y = \frac{\cos^3 t}{\sqrt{\cos 2t}}$