# Solve this

Question:

Find $\frac{d y}{d x}$, when

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

Solution:

$\operatorname{as} x=\frac{\sin ^{3} t}{\sqrt{\cos 2 t}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$=-\frac{\cos 3 t}{\sin 3 t}\left[\cos 3 t=4 \cos ^{3} t-3 \cos t\right.$

$\left.\sin 3 t=3 \sin t-4 \sin ^{3} t\right]$

$=\cot 3 t$