Question

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

Answer 1

The given equations are $x=\sqrt{a^{sin^{-1}t}},y=\sqrt{a^{cos^{-1}t}}$
$x=\sqrt{a^{sin^{-1}t}}\,and\,y=\sqrt{a^{cos^{-1}t}}$
$⇒x=(a^{sin^{-1}t})^{\frac{1}{2}} \,and\, y=(a^{cos^{-1}t})^{\frac{1}{2}}$
$⇒x=a^{\frac{1}{2}sin^{-1}t}\, and\, y=a^{\frac{1}{2}cos^{-1}t}$
consider $x=a^{\frac{1}{2}sin^{-1}t}$
Taking logarithm on both sides,we obtain
$logx=\frac{1}{2}sin^{-1}t\,loga$
$∴\frac{1}{x}.\frac{dx}{dt}=\frac{1}{2}log\,a.\frac{d}{dt}(sin^{-1}t)$
$⇒\frac{dx}{dt}=\frac{x}{2}loga.\frac{1}{\sqrt{1-t^2}}$
$⇒\frac{dx}{dt}=\frac{xloga}{2\sqrt{1-t^2}}$
Then,consider $y=a^{\frac{1}{2}cos^{-1}t}$
Taking logarithm on both sides,we obtain
$logy=\frac{1}{2}cos^{-1}t\,loga$
$∴\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}loga.\frac{d}{dt}(cos^{-1}t)$
$⇒\frac{dy}{dt}=\frac{yloga}{2}.(\frac{-1}{\sqrt{1-t^2}})$
$⇒\frac{dy}{dt}=\frac{-yloga}{2\sqrt{1-t^2}}$
$∴\frac{dy}{dx}=\frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}=\frac{(\frac{-yloga}{2\sqrt{1-t^2)}}}{(\frac{xloga}{2\sqrt{1-t^2}}})$
$=\frac{-y}{x}$
Hence,proved.