For a0, t(0,2), let x=a^^-1t and y=a^^-1t.

Question

For $a>0$, $t\in\left(0,\dfrac\pi2\right)$, let $x=a^{\sin^{-1}t}$ and $y=a^{\cos^{-1}t}$. Then $1+\left(\dfrac{dy}{dx}\right)^2$ equals:

Accepted Answer

Taking logs and differentiating: $\dfrac{2}{x}=\dfrac{\ln a}{\sqrt{1-t^2}}\dfrac{dt}{dx}$, and similarly for $y$; combining gives $\dfrac{dy}{dx}=-\dfrac yx$.

So $1+\left(\dfrac{dy}{dx}\right)^2=1+\dfrac{y^2}{x^2}=\dfrac{x^2+y^2}{x^2}$.

Solutions · 1