Question

If $x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$ then find $\frac{dy}{dx}$

Answer 1

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

$x={\sin }^{3}t{\left(\cos 2t\right)}^{-1/2};y={\cos }^{3}t{\left(\cos 2t\right)}^{-1/2}$

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

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

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

$=\frac{\frac{{\cos }^{2}t}{\sqrt{\cos 2t}}[-3\sin t+\frac{\cos t\sin 2t}{\cos 2t}]}{\frac{{\sin }^{2}t}{\sqrt{\cos 2t}}[3\cos t+\frac{\sin t\sin 2t}{\cos 2t}]}=\frac{1}{{\tan }^{2}t}\frac{[-3\tan t+\tan 2t]\cos t}{[3\cos t+\tan 2t]\sin t}$

$=\frac{1}{{\tan }^{2}t}\frac{[-3\tan t+\tan 2t]}{[3\cot t+\tan 2t]}\times \frac{1}{\tan t}=\frac{1}{{\tan }^{3}t}\frac{[-3\tan t+\tan 2t]}{[3\cot t+\tan 2t]}$

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

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