If x and y are connected parametrically by the equation,
Question:

If $x$ and $y$ are connected parametrically by the equation, without eliminating the parameter, find $\frac{d y}{d x}$.

$x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}, y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

Solution:

The given equations are $x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}$ and $y=\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}$

Then, $\frac{d x}{d t}=\frac{d}{d t}\left[\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}\right]$

$=\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}\left(\sin ^{3} t\right)-\sin ^{3} t \cdot \frac{d}{d t} \sqrt{\cos 2 t}}{\cos 2 t}$

$=\frac{\sqrt{\cos 2 t} \cdot 3 \sin ^{2} t \cdot \frac{d}{d t}(\sin t)-\sin ^{3} t \times \frac{1}{2 \sqrt{\cos 2 t}} \cdot \frac{d}{d t}(\cos 2 t)}{\cos 2 t}$

$=\frac{3 \sqrt{\cos 2 t} \cdot \sin ^{2} t \cos t-\frac{\sin ^{3} t}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}$

$=\frac{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}{\cos 2 t \sqrt{\cos 2 t}}$

$\frac{d y}{d t}=\frac{d}{d t}\left[\frac{\cos ^{3} t}{\sqrt{\cos 2 t}}\right]$

$=\frac{\sqrt{\cos 2 t} \cdot \frac{d}{d t}\left(\cos ^{3} t\right)-\cos ^{3} t \cdot \frac{d}{d t}(\sqrt{\cos 2 t})}{\cos 2 t}$

$=\frac{\sqrt{\cos 2 t} \cdot 3 \cos ^{2} t \cdot \frac{d}{d t}(\cos t)-\cos ^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot \frac{d}{d t}(\cos 2 t)}{\cos 2 t}$

$=\frac{3 \sqrt{\cos 2 t} \cdot \cos ^{2} t(-\sin t)-\cos ^{3} t \cdot \frac{1}{2 \sqrt{\cos 2 t}} \cdot(-2 \sin 2 t)}{\cos 2 t}$

$=\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{\cos 2 t \cdot \sqrt{\cos 2 t}}$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t \sin 2 t}$

$=\frac{-3 \cos 2 t \cdot \cos ^{2} t \cdot \sin t+\cos ^{3} t(2 \sin t \cos t)}{3 \cos 2 t \sin ^{2} t \cos t+\sin ^{3} t(2 \sin t \cos t)}$

$=\frac{\sin t \cos t\left[-3 \cos 2 t \cdot \cos t+2 \cos ^{3} t\right]}{\sin t \cos t\left[3 \cos 2 t \sin t+2 \sin ^{3} t\right]}$’

$=\frac{\left[-3\left(2 \cos ^{2} t-1\right) \cos t+2 \cos ^{3} t\right]}{\left[3\left(1-2 \sin ^{2} t\right) \sin t+2 \sin ^{3} t\right]} \quad\left[\begin{array}{l}\cos 2 t=\left(2 \cos ^{2} t-1\right), \\ \cos 2 t=\left(1-2 \sin ^{2} t\right)\end{array}\right]$

$=\frac{-4 \cos ^{3} t+3 \cos t}{3 \sin t-4 \sin ^{3} t}$

$=\frac{-\cos 3 t}{\sin 3 t} \quad\left[\begin{array}{l}\cos 3 t=4 \cos ^{3} t-3 \cos t \\ \sin 3 t=3 \sin t-4 \sin ^{3} t\end{array}\right]$

$=-\cot 3 t$