Question

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

Answer 1

$x=\sqrt{a^{si{n}^{-1}t}},y=\sqrt{a^{co{s}^{-1}t}}$

$\Rightarrow x^2=a^{si{n}^{-1}t},y^2=a^{co{s}^{-1}t}$

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

Now, $\frac{dy}{dt}\times 2y=a^{co{s}^{-1}t}\times \ln a\times \frac{-1}{\sqrt{1-t^2}}=y^2\times \ln a\times \frac{-1}{\sqrt{1-t^2}}...\left(1\right)$

$\frac{dx}{dt}\times 2x=a^{si{n}^{-1}t}\times \ln a\times \frac{1}{\sqrt{1-t^2}}=x^2\times \ln a\times \frac{1}{\sqrt{1-t^2}}...\left(2\right)$

Dividing equations (1) and (2), we have

$\frac{dy}{dx}\times \frac{y}{x}=\frac{-y^2}{x^2}$

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