If $x$ and $y$ are connected parametrically by the equations given in $x = \large\frac{\sin^3t}{\sqrt {\cos 2t}}, $$y = \large\frac{\cos^3t}{\sqrt {\cos 2t}} without eliminating the parameter, Find \large\frac{ dy}{ dx}. 1 Answer Toolbox: • \big(\Large\frac{u}{v}\big)'=\Large\frac{u'v-uv'}{v^2} • By chain rule we have \large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
Step 1:
$x=\large\frac{\sin^3t}{\sqrt{\cos 2t}}$
$\large\frac{dx}{dt}=\frac{3\sin^2t.\cos t\sqrt{\cos 2t}-\sin^3t.\Large\frac{1}{2\sqrt{\cos 2t}}\normalsize\times (\large-2\sin 2t)}{\cos 2t}$
$\quad\;\;=\large\frac{3\sin^2t\cos t+\sin^3t\sin 2t}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\sin^2t(3\cos t\cos 2t+\sin t\sin 2t)}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\sin^2t(2\cos t\cos 2t+\cos 2t\cos t+\sin t\sin 2t)}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\sin^2t(2\cos t\cos 2t+\cos t)}{(\cos2 t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\sin^2t\cos t(2\cos 2t+1)}{(\cos 2t)^{\Large\frac{3}{2}}}$
Step 2:
$y=\large\frac{\cos^3t}{\sqrt{\cos 2t}}$
$\large\frac{dy}{dt}=\frac{-3\cos^2t(\sin t)\sqrt{\cos 2t}-\cos ^3t\Large\frac{1}{2\sqrt{\cos 2t}}(\large-2\sin 2t)}{\cos 2t}$
$\quad\;\;=\large\frac{-3\cos^2t(\sin t)\sqrt{\cos 2t}+\Large\frac{\cos^3t\sin 2t}{\sqrt{\cos 2t}}}{\cos 2t}$
$\quad\;\;=\large\frac{-3\cos^2t\sin t\cos 2t+\cos^3t\sin 2t}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\cos^2t(-3\sin t\cos 2t+\cos t\sin 2t)}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\cos^2t(-2\sin t\cos 2t+\sin t\cos t-\cos 2t\sin t)}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\cos^2t(-2\sin t\cos 2t+\sin (2t-t))}{(\cos 2t)^{\Large\frac{3}{2}}}$
$\quad\;\;=\large\frac{\cos^2t\sin t(1-2\cos 2t)}{(\cos 2t)^{\Large\frac{3}{2}}}$
Step 3:
$\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\quad\;\;=\large\frac{\cos^2t\sin t(1-2\cos 2t)}{(\cos 2t)^{\Large\frac{3}{2}}}\large\frac{(\cos 2t)^{\Large\frac{3}{2}}}{\sin^2t\cos t(2\cos 2t+1)}$
$\quad\;\;=\large\frac{\cos^2t\sin t(1-2\cos 2t)}{\sin^2t\cos t(1+2\cos 2t)}$
$\quad\;\;=\large\frac{\cot t(1-2\cos 2t)}{1+2\cos 2t}$