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.