Question

If $x=\sqrt{a^{si{n}^{-1}t}}$ and $y=\sqrt{a^{co{s}^{-1}t}}$, show that $\frac{dy}{dx}=\frac{-y}{x}$.

Answer 1

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

$logx=\frac{1}{2}log{a}^{{\sin }^{-1}t}$ $logy=\frac{1}{2}log{a}^{{\cos }^{-1}t}$

$=\frac{{\sin }^{-1}t}{2}lo{g}_{10}a$ $=\frac{{\cos }^{-1}t}{2}loga$

$\frac{1}{x}\frac{dx}{dt}=\frac{1}{2\sqrt{1-t^2}}$ $\frac{1}{y}\frac{dy}{dt}=\frac{-1}{2\sqrt{1-t^2}}$

$\Rightarrow \frac{dx}{dt}=\frac{x}{2\sqrt{1-t^2}}$ $\frac{dy}{dx}=\frac{-y}{2\sqrt{1-t^2}}$

$\therefore \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-y/2\sqrt{1-t^2}}{x/2\sqrt{1-t^2}}$

$\therefore \frac{dy}{dx}=\frac{-y}{x}$ proved.