Question

If $x=\sqrt{a^{{\sin }^{-1}t}},y=\sqrt{a^{{\cos }^{-1}t}}$, show that $\frac{dy}{dx}=-\frac{y}{x}$

Answer 1

Given, $x=\sqrt{a^{{\sin }^{-1}t}},y=\sqrt{a^{{\cos }^{-1}t}}$

$\therefore \frac{dx}{dt}=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}.a^{{\sin }^{-1}t}\log a.\frac{1}{\sqrt{1-t^2}}$ ..(i)

and $\frac{dy}{dt}=\frac{1}{2\sqrt{a^{{\cos }^{i}t}}}.a^{{\cos }^{-1}t}\log a.\frac{-1}{\sqrt{1-t^2}}$ ...(ii)

When we divide (ii) from (i), we will get
$\frac{dy}{dx}=\frac{\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}.a^{{\sin }^{-1}t}\log a.\frac{1}{\sqrt{1-t^2}}}{\frac{1}{2\sqrt{a^{{\cos }^{i}t}}}.a^{{\cos }^{-1}t}\log a.\frac{-1}{\sqrt{1-t^2}}}$
$\frac{dy}{dx}=-\frac{y}{x}$