Question

If \[ x = \sqrt{1 + t^2}, \quad y = \sqrt{1 - t^2}, \] then find \(\frac{dy}{dx}\).

Hint:

For parametric derivatives, find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) first, then compute \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).

Answer 1

Differentiate \(x\) and \(y\) with respect to \(t\): \[ \frac{dx}{dt} = \frac{1}{2\sqrt{1 + t^2}} \times 2t = \frac{t}{\sqrt{1 + t^2}}, \] \[ \frac{dy}{dt} = \frac{1}{2\sqrt{1 - t^2}} \times (-2t) = \frac{-t}{\sqrt{1 - t^2}}. \] Using the chain rule for parametric functions: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-t / \sqrt{1 - t^2}}{t / \sqrt{1 + t^2}} = - \frac{\sqrt{1 + t^2}}{\sqrt{1 - t^2}}. \]
Final answer: \[ \boxed{ \frac{dy}{dx} = - \frac{\sqrt{1 + t^2}}{\sqrt{1 - t^2}}. } \]