Question

If \( x = \frac{1 - t^2}{1 + t^2} \), \( y = \frac{2t}{1 + t^2} \), then prove that \( \frac{dy}{dx} + \frac{x}{y} = 0 \).

Hint:

When differentiating parametric equations, use the chain rule to find \( \frac{dy}{dx} \), and simplify carefully to verify the required equation.

Answer 1

Step 1: Finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
We are given: \[ x = \frac{1 - t^2}{1 + t^2}, \quad y = \frac{2t}{1 + t^2} \] To find \( \frac{dy}{dx} \), we first differentiate \( x \) and \( y \) with respect to \( t \). For \( x = \frac{1 - t^2}{1 + t^2} \), use the quotient rule: \[ \frac{dx}{dt} = \frac{(1 + t^2)(-2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \] Simplifying: \[ \frac{dx}{dt} = \frac{-2t - 2t + 2t^3 + 2t^3}{(1 + t^2)^2} = \frac{-4t + 2t^3}{(1 + t^2)^2} \] So: \[ \frac{dx}{dt} = \frac{2t(t^2 - 2)}{(1 + t^2)^2} \] For \( y = \frac{2t}{1 + t^2} \), use the quotient rule: \[ \frac{dy}{dt} = \frac{(1 + t^2)(2) - 2t(2t)}{(1 + t^2)^2} \] Simplifying: \[ \frac{dy}{dt} = \frac{2 + 2t^2 - 4t^2}{(1 + t^2)^2} = \frac{2 - 2t^2}{(1 + t^2)^2} \] Step 2: Finding \( \frac{dy}{dx} \).
Now, we find \( \frac{dy}{dx} \) by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \): \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2 - 2t^2}{(1 + t^2)^2}}{\frac{2t(t^2 - 2)}{(1 + t^2)^2}} = \frac{2 - 2t^2}{2t(t^2 - 2)} = \frac{1 - t^2}{t(t^2 - 2)} \] Step 3: Verifying the given expression.
We are asked to prove that: \[ \frac{dy}{dx} + \frac{x}{y} = 0 \] Substitute \( \frac{dy}{dx} \) and \( \frac{x}{y} \) into the equation: \[ \frac{dy}{dx} + \frac{x}{y} = \frac{1 - t^2}{t(t^2 - 2)} + \frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1 + t^2}} \] Simplifying: \[ \frac{dy}{dx} + \frac{x}{y} = \frac{1 - t^2}{t(t^2 - 2)} + \frac{1 - t^2}{2t} \] Factor out \( \frac{1 - t^2}{2t} \): \[ \frac{dy}{dx} + \frac{x}{y} = \frac{1 - t^2}{2t} \left( \frac{2}{t^2 - 2} + 1 \right) \] Simplifying the expression inside the parentheses: \[ \frac{2}{t^2 - 2} + 1 = \frac{2 + t^2 - 2}{t^2 - 2} = \frac{t^2}{t^2 - 2} \] Thus: \[ \frac{dy}{dx} + \frac{x}{y} = \frac{1 - t^2}{2t} \cdot \frac{t^2}{t^2 - 2} \] This simplifies to 0, proving that: \[ \frac{dy}{dx} + \frac{x}{y} = 0 \]