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If $x$ and $y$ are connected parametrically by the given equations, without eliminating the parameter. Find $\frac{d y}{d x}$
$x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}} \& y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}$
$\sqrt{\cos 2 t}\left(3 \sin ^2 t \cdot \cos t\right)$
$x=\frac{\sin ^3 t}{\sqrt{\cos 2 t}} \& y=\frac{\cos ^3 t}{\sqrt{\cos 2 t}}$
$\sqrt{\cos 2 t}\left(3 \sin ^2 t \cdot \cos t\right)$
Solution:
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Verified Answer
$\begin{aligned} \frac{d x}{d t} &=\frac{-\sin ^3 t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \times(-2 \sin 2 t)}{\cos 2 t} \\=& \frac{\sin ^2 t \cos t\left[3-4 \sin ^2 t\right]}{(\cos 2 t)^{3 / 2}} \end{aligned}$
$\begin{aligned} \frac{d y}{d t} &=\frac{\sqrt{\cos 2 t} \cdot\left(3 \cos ^2 t\right)(-\sin t)-\cos ^3 t \cdot \frac{1(-2 \sin 2 t)}{2 \sqrt{\cos 2 t}}}{\cos 2 t} \\=& \frac{-\cos ^2 t \sin t\left(4 \cos ^2 t-3\right)}{(\cos 2 t)^{3 / 2}} \end{aligned}$
$\frac{d y}{d x}=-\cot t$
$\begin{aligned} \frac{d y}{d t} &=\frac{\sqrt{\cos 2 t} \cdot\left(3 \cos ^2 t\right)(-\sin t)-\cos ^3 t \cdot \frac{1(-2 \sin 2 t)}{2 \sqrt{\cos 2 t}}}{\cos 2 t} \\=& \frac{-\cos ^2 t \sin t\left(4 \cos ^2 t-3\right)}{(\cos 2 t)^{3 / 2}} \end{aligned}$
$\frac{d y}{d x}=-\cot t$
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