Question

(a) If \(\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y)\), prove that \(\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}\).

Hint:

Apply the chain rule carefully for derivatives of composite functions like \(\sqrt{1 - x^2}\) and \(\sqrt{1 - y^2}\). Rearrange terms systematically to isolate \(\frac{dy}{dx}\).

Answer 1

The provided equation is:\[\sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y).\]Differentiating both sides with respect to \(x\):\[\frac{d}{dx}\left(\sqrt{1 - x^2}\right) + \frac{d}{dx}\left(\sqrt{1 - y^2}\right) = \frac{d}{dx}[a(x - y)].\]Applying the chain rule:\[\frac{-x}{\sqrt{1 - x^2}} + \frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = a(1 - \frac{dy}{dx}).\]Rearranging terms:\[\frac{-y}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} + a\frac{dy}{dx} = a - \frac{x}{\sqrt{1 - x^2}}.\]Factoring out \(\frac{dy}{dx}\) on the left side:\[\frac{dy}{dx}\left(a - \frac{y}{\sqrt{1 - y^2}}\right) = a - \frac{x}{\sqrt{1 - x^2}}.\]Solving for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{a - \frac{x}{\sqrt{1 - x^2}}}{a - \frac{y}{\sqrt{1 - y^2}}}.\]Under the condition \(a = 1\), the expression simplifies to:\[\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}.\]Thus, it is demonstrated that:\[\frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}}.\]