Question

Find \[\frac{dy}{dx}\] , when  \[x = \cos^{- 1} \frac{1}{\sqrt{1 + t^2}} \text{ and y } = \sin^{- 1} \frac{t}{\sqrt{1 + t^2}}, t \in R\] ?

Answer 1

\[\text{ We have, x } = \cos^{- 1} \left( \frac{1}{\sqrt{1 + t^2}} \right)\]

\[\Rightarrow \frac{dx}{dt} = \frac{- 1}{\sqrt{1 - \left( \frac{1}{\sqrt{1 + t^2}} \right)^2}}\frac{d}{dt}\left( \frac{1}{\sqrt{1 + t^2}} \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{- 1}{\sqrt{1 - \frac{1}{\left( 1 + t^2 \right)}}}\left\{ \frac{- 1}{2 \left( 1 + t^2 \right)^\frac{3}{2}} \right\}\frac{d}{dt}\left( 1 + t^2 \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{\left( 1 + t^2 \right)^\frac{1}{2}}{\sqrt{1 + t^2 - 1}} \times \frac{1}{2 \left( 1 + t^2 \right)^\frac{3}{2}}\left( 2t \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{t}{\sqrt{t^2} \times \left( 1 + t^2 \right)}\]
\[ \Rightarrow \frac{dx}{dt} = \frac{1}{1 + t^2} . . . \left( i \right)\]
\[\text{ Now, y }= \sin^{- 1} \left( \frac{1}{\sqrt{1 + t^2}} \right)\]

\[\Rightarrow \frac{dy}{dt} = \frac{1}{\sqrt{1 - \left( \frac{1}{\sqrt{1 + t^2}} \right)^2}}\frac{d}{dt}\left( \frac{1}{\sqrt{1 + t^2}} \right)\]
\[ \Rightarrow \frac{dy}{dt} = \frac{1}{\sqrt{1 - \frac{1}{\left( 1 + t^2 \right)}}}\left\{ \frac{- 1}{2 \left( 1 + t^2 \right)^\frac{3}{2}} \right\}\frac{d}{dt}\left( 1 + t^2 \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{\left( 1 + t^2 \right)^\frac{1}{2}}{\sqrt{1 + t^2 - 1}} \times \frac{- 1}{2 \left( 1 + t^2 \right)^\frac{3}{2}}\left( 2t \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{- 1}{2\sqrt{t^2} \times \left( 1 + t^2 \right)}\left( 2t \right)\]
\[ \Rightarrow \frac{dx}{dt} = \frac{- 1}{1 + t^2} . . . \left( ii \right)\]
\[\text{ Dividing equation } \left( ii \right) \text{ by } \left( i \right), \]
\[\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{\left( 1 + t^2 \right)} \times \frac{\left( 1 + t^2 \right)}{- 1}\]
\[ \Rightarrow \frac{dy}{dx} = - 1\]