Question

if $x=\sqrt{a^{{\sin }^{-1}t}}$ and $y=\sqrt{a^{{\cos }^{-1}t}}$ prove that $\frac{dy}{dx}$ + $\frac{y}{x}=0$

Answer 1

Given,

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

$\frac{dx}{dt}=\frac{\sqrt{a^{{\sin }^{-1}t}}\cdot \log a}{2\sqrt{1-t^2}}$

$\frac{dy}{dt}=\frac{-a^{{\cos }^{-1}t}\cdot \log a}{2\sqrt{a^{{\cos }^{-1}t}}\sqrt{1-t^2}}$

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

$=\frac{\frac{-a^{{\cos }^{-1}t}\cdot \log a}{2\sqrt{a^{{\cos }^{-1}t}}\sqrt{1-t^2}}}{\frac{\sqrt{a^{{\sin }^{-1}t}}\cdot \log a}{2\sqrt{1-t^2}}}$

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

$\frac{dy}{dx}=-\frac{y}{x}$

$\frac{dy}{dx}+\frac{y}{x}=0$

Hence proved.