Question

If $x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos t2t}}$, find $\frac{dy}{dx}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}}\mathrm{and}y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}} \\ \Rightarrow \frac{dx}{dt}=\frac{d}{dt}\left[\frac{{\sin }^{3}t}{\sqrt{\cos 2t}}\right] \\ \Rightarrow \frac{dx}{dt}=\frac{\sqrt{\cos 2t}{\displaystyle \frac{d}{dt}}\left({\sin }^{3}t\right)-{\sin }^{3}t{\displaystyle \frac{d}{dt}}\sqrt{\cos 2t}}{\cos 2t}\left[\mathrm{Using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dx}{dt}=\frac{\sqrt{\cos 2t}\left(3{\sin }^{2}t\right){\displaystyle \frac{d}{dt}}\left(\sin t\right)-{\sin }^{3}t\times {\displaystyle \frac{1}{2\sqrt{\cos 2t}}}{\displaystyle \frac{d}{dt}}\left(\cos 2t\right)}{\cos 2t} \\ \Rightarrow \frac{dx}{dt}=\frac{3\sqrt{\cos 2t}\left({\sin }^{2}t\cos t\right)-{\displaystyle \frac{{\sin }^{3}t}{2\sqrt{\cos 2t}}}\left(-2\sin 2t\right)}{\cos 2t} \\ \Rightarrow \frac{dx}{dt}=\frac{3\cos 2t{\sin }^{2}t\cos t+{\sin }^{3}t\sin 2t}{\cos 2t\sqrt{\cos 2t}} \\ \mathrm{Now},\frac{dy}{dt}=\frac{d}{dt}\left[\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}\right] \\ \Rightarrow \frac{dy}{dt}=\frac{\sqrt{\cos 2t}{\displaystyle \frac{d}{dt}}\left({\cos }^{3}t\right)-{\cos }^{3}t{\displaystyle \frac{d}{dt}}\sqrt{\cos 2t}}{\cos 2t}\left[\mathrm{Using}\mathrm{quotient}\mathrm{rule}\right] \\ \Rightarrow \frac{dy}{dt}=\frac{\sqrt{\cos 2t}\left(3{\cos }^{2}t\right){\displaystyle \frac{d}{dt}}\left(\cos t\right)-{\cos }^{3}t\times {\displaystyle \frac{1}{2\sqrt{\cos 2t}}}{\displaystyle \frac{d}{dt}}\left(\cos 2t\right)}{\cos 2t} \\ \Rightarrow \frac{dy}{dt}=\frac{3\sqrt{\cos 2t}{\cos }^{2}t\left(-\sin t\right)-{\displaystyle \frac{{\cos }^{3}t}{2\sqrt{\cos 2t}}}\left(-2\sin 2t\right)}{\cos 2t} \\ \Rightarrow \frac{dy}{dt}=\frac{-3\cos 2t{\cos }^{2}t\sin t+{\cos }^{3}t\sin 2t}{\cos 2t\sqrt{\cos 2t}} \\ \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}=\frac{-3\cos 2t{\cos }^{2}t\sin t+{\cos }^{3}t\sin 2t}{\cos 2t\sqrt{\cos 2t}}\times \frac{\cos 2t\sqrt{\cos 2t}}{3\cos 2t{\sin }^{2}t\cos t+{\sin }^{3}t\sin 2t} \\ \Rightarrow \frac{dy}{dx}=\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} \\ \Rightarrow \frac{dy}{dx}=\frac{\sin t\cos t\left[-3\cos 2t\cos t+2{\cos }^{3}t\right]}{\sin t\cos t\left[3\cos 2t\sin t+2{\sin }^{3}t\right]} \\ \Rightarrow \frac{dy}{dx}=\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]}\left[\begin{array}{c}\cos 2t=2{\cos }^{2}t-1\\ \cos 2t=1-2{\sin }^{2}t\end{array}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{-4{\cos }^{3}t+3\cos t}{3\sin t-4{\sin }^{3}t} \\ \Rightarrow \frac{dy}{dx}=\frac{-\cos 3t}{\sin 3t}\left[\begin{array}{c}\cos 3t=4{\cos }^{3}t-3\cos t\\ \sin 3t=3\sin t-4{\sin }^{3}t\end{array}\right] \\ \therefore \frac{dy}{dx}=-\cot 3t \end{gathered}$$