Question

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

Answer 1

Given : $x=\sqrt{a^{{\sin }^{-1}t}}$ and $y=\sqrt{a^{{\cos }^{-1}t}}$

$⟹\frac{dx}{dt}=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}d\left(a^{{\sin }^{-1}t}\right)$

$=\frac{1}{2\sqrt{a^{{\sin }^{-1}t}}}\ln \left(a\right)a^{{\sin }^{-1}t}\times \frac{1}{\sqrt{1-t^2}}$

$⟹\frac{dx}{dt}=\frac{\ln \left(a\right)a^{{\sin }^{-1}t}}{2\sqrt{a^{{\sin }^{-1}t}}\sqrt{1-t^2}}$ ......... $\left(i\right)$

$\frac{dy}{dt}=\frac{1}{2\sqrt{a^{{\cos }^{-1}t}}}d\left(a^{{\cos }^{-1}t}\right)$

$=\frac{1}{2\sqrt{a^{{\cos }^{-1}t}}}\ln \left(a\right)a^{{\cos }^{-1}t}\times \frac{-1}{\sqrt{1-t^2}}$

$⟹\frac{dy}{dt}=\frac{-\ln \left(a\right)a^{{\cos }^{-1}t}}{2\sqrt{a^{{\cos }^{-1}t}}\sqrt{1-t^2}}$ ......... $\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-\ln \left(a\right)a^{{\cos }^{-1}t}}{2\sqrt{a^{{\cos }^{-1}t}}\sqrt{1-t^2}}\times \frac{2\sqrt{a^{{\sin }^{-1}t}}\sqrt{1-t^2}}{\ln \left(a\right)a^{{\sin }^{-1}t}}$

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

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

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