Question

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

Answer 1

LHS $=\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}$

Now,

$\frac{dy}{dt}=\frac{d}{dt}\left(\sqrt{a^{{\cos }^{-1}t}}\right)$

$=\frac{1}{2\sqrt{a^{{\cos }^{-1}t}}}.a^{{\cos }^{-1}t}loga.\frac{d}{dt}\left({\cos }^{-1}t\right)$

$=\frac{1}{2\sqrt{a^{{\cos }^{-1}t}}}.a^{{\cos }^{-1}t}loga.\frac{-1}{\sqrt{1-t^2}}$

$\frac{dx}{dt}=\frac{d}{dt}\left(\sqrt{a^{{\sin }^{-1}t}}\right)$

$=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}.a^{{\sin }^{-1}t}loga.\frac{d}{dt}\left({\sin }^{-1}t\right)$

$=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}.a^{{\sin }^{-1}t}loga.\frac{1}{\sqrt{1-t^2}}$

$\frac{dy}{dx}=\frac{{-a}^{{\cos }^{-1}t}loga}{2\sqrt{a^{{\cos }^{-1}t}}.\sqrt{1-t^2}}.\frac{2\sqrt{a^{{\sin }^{-1}t}}.\sqrt{1-t^2}}{a^{{\sin }^{-1}t}loga}$

$=\frac{-\sqrt{a^{{\cos }^{-1}t}}}{\sqrt{a^{{\sin }^{-1}t}}}=\frac{-y}{x}=$ RHS