Question

If $x=\frac{{\sin }^{3}t}{\sqrt{\left(\cos 2t\right)}},y=\frac{{\cos }^{3}t}{\sqrt{\left(\cos 2t\right)}},$ find $\frac{dy}{dx}$ at $t=\frac{\pi }{6}.$

Answer 1

$x=\frac{{\sin }^{3}t}{\sqrt{\left(\cos 2t\right)}},y=\frac{{\cos }^{3}t}{\sqrt{\left(\cos 2t\right)}}$
$\Rightarrow \log x=3\log \sin t-\frac{1}{2}\log \cos 2t,\log y=3\log \cos t-\frac{1}{2}\log \cos 2t$
Differentiating w.r.t $t$
$\frac{1}{x}\frac{dx}{dt}=3\cot t+\tan 2t,\frac{1}{y}\frac{dy}{dt}=-3\tan t+\tan 2t$
$\therefore \frac{dy}{dx}=\frac{y(-3\tan t+\tan 2t)}{x(3\cot t+\tan 2t)}$
Now at $t=\frac{\pi }{6},-3\tan t+\tan 2t=-\frac{3}{\sqrt{3}}+\sqrt{3}=0$

$\therefore \frac{dy}{dx}=0$