Question

If $x$ and $y$ are connected parametrically by the given equation, then without eliminating the parameter, find $\frac{dy}{dx}$.
$x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$

Answer 1

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

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

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

$=\frac{\sqrt{\cos 2t}.3{\sin }^{2}t\frac{d}{dt}\left(\sin t\right)-{\sin }^{3}t\times \frac{1}{2\sqrt{\cos 2t}}\frac{d}{dt}\left(\cos 2t\right)}{\cos 2t}$

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

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

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

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

$=\frac{\sqrt{\cos 2t}.3{\cos }^{2}t.\frac{d}{dt}\left(\cos t\right)-{\cos }^{3}t.\frac{1}{2\sqrt{\cos 2t}}.\frac{d}{dt}\left(\cos 2t\right)}{\cos 2t}$

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

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

$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{-3\cos 2t.{\cos }^{2}t.\sin t+{\cos }^{3}t\sin 2t}{3\cos 2t.{\sin }^{2}t.\cos t+{\sin }^{3}t\sin 2t}$

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

$=\frac{\sin t\cos t[-3\cos 2t.\cos t+2{\cos }^{3}t]}{\sin t\cos t[3\cos 2t.\sin t+2{\sin }^{3}t]}$

$=\frac{[-3(2{\cos }^{2}t-1)\cos t+2{\cos }^{3}t]}{\left[3\right(1-2{\sin }^{2}t)\sin t+2{\sin }^{3}t]}$

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

$=\frac{-\cos 3t}{\sin 3t}=-\cot 3t$