Question

Find $dy/dx$ of : $x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$

Answer 1

Given,

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

Here,

$\frac{dy}{dx}=\frac{dy}{dx}\times \frac{dt}{dt}\Rightarrow \frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Now,

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

And

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

So,

$\frac{dy}{dx}=\frac{\frac{{\cos }^{2}t(-3\sin t\cdot \cos 2t+\cos t\cdot \sin 2t)}{{\left(\cos 2t\right)}^{\frac{3}{2}}}}{\frac{{\sin }^{2}t(3\cos t\cdot \cos 2t+\sin t\cdot \sin 2t)}{{\left(\cos 2t\right)}^{\frac{3}{2}}}}=\frac{{\cos }^{2}t(-3\sin t\cdot \cos 2t+\cos t\cdot \sin 2t)}{{\sin }^{2}t(3\cos t\cdot \cos 2t+\sin t\cdot \sin 2t)}={\cot }^{2}t\left(\frac{-3\sin t\cdot \cos 2t+\cos t\cdot \sin 2t}{3\cos t\cdot \cos 2t+\sin t\cdot \sin 2t}\right)={\cot }^{2}t\left(\frac{\cos 2t(-3\sin t+\cos t\cdot \frac{\sin 2t}{\cos 2t})}{\cos 2t(3\cos t+\sin t\cdot \frac{\sin 2t}{\cos 2t})}\right)={\cot }^{2}t\left(\frac{-3\sin t+\cos t\cdot \tan 2t}{3\cos t+\sin t\cdot \tan 2t}\right)={\cot }^{2}t\left(\frac{\cos t(-3\frac{\sin t}{\cos t}+\tan 2t)}{\cos t(3+\frac{\sin t}{\cos t}\cdot \tan 2t)}\right)={\cot }^{2}t\left(\frac{\tan 2t-3\tan t}{3+\tan t\cdot \tan 2t}\right)={\cot }^{2}t\left(\frac{\frac{2\tan t}{1-{\tan }^{2}t}-3\tan t}{3+\tan t\left(\frac{2\tan t}{1-{\tan }^{2}t}\right)}\right)[\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }]={\cot }^{2}t\left(\frac{2\tan t-3\tan t+3{\tan }^{3}t}{3-3{\tan }^{2}t+2{\tan }^{2}t}\right)=-{\cot }^{2}t\left(\frac{\tan t-3\tan t}{3-{\tan }^{2}t}\right)=-\left(\frac{{\cot }^{2}t\cdot \tan t-3{\cot }^{2}t\cdot {\tan }^{3}t}{3-{\tan }^{2}t}\right)=-\left(\frac{\frac{1}{{\tan }^{2}t}\cdot \tan t-3\frac{1}{{\tan }^{2}t}\cdot {\tan }^{3}t}{3-{\tan }^{2}t}\right)=-\left(\frac{\frac{1-3\tan t\cdot \left(\tan t\right)}{\tan t}}{3-{\tan }^{2}t}\right)=-\left(\frac{1-3{\tan }^{2}t}{3\tan t-{\tan }^{3}t}\right)=\frac{-1}{\left(\frac{3\tan t-{\tan }^{3}t}{1-3{\tan }^{2}t}\right)}=\frac{-1}{\tan 3t}[\because \tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }]=-\cot 3t\therefore \frac{dy}{dx}=-\cot 3t.$