Question

If \( x = \frac{4t}{1+t^2} \), \( y = 3\left(\frac{1-t^2}{1+t^2}\right) \), then show that \( \frac{dy}{dx} = -\frac{9x}{4y} \).

Hint:

For parametric differentiation, the key formula is \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). After finding the derivative in terms of the parameter \(t\), you may need to substitute the original expressions for \(x\) and \(y\) to show equivalence.

Answer 1

This is a parametric differentiation problem. We will find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) first.
Step 1: Find \( \frac{dx}{dt} \).
Using the quotient rule: \[ \frac{dx}{dt} = \frac{(4)(1+t^2) - (4t)(2t)}{(1+t^2)^2} = \frac{4+4t^2-8t^2}{(1+t^2)^2} = \frac{4(1-t^2)}{(1+t^2)^2} \] Step 2: Find \( \frac{dy}{dt} \).
\( y = \frac{3-3t^2}{1+t^2} \). Using the quotient rule: \[ \frac{dy}{dt} = \frac{(-6t)(1+t^2) - (3-3t^2)(2t)}{(1+t^2)^2} = \frac{-6t-6t^3-6t+6t^3}{(1+t^2)^2} = \frac{-12t}{(1+t^2)^2} \] Step 3: Find \( \frac{dy}{dx} \). \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-12t / (1+t^2)^2}{4(1-t^2) / (1+t^2)^2} = \frac{-12t}{4(1-t^2)} = \frac{-3t}{1-t^2} \] Step 4: Evaluate the right-hand side. \[ -\frac{9x}{4y} = -\frac{9(\frac{4t}{1+t^2})}{4(3\frac{1-t^2}{1+t^2})} = -\frac{36t/(1+t^2)}{12(1-t^2)/(1+t^2)} = -\frac{36t}{12(1-t^2)} = \frac{-3t}{1-t^2} \] Since \( \frac{dy}{dx} \) and \( -\frac{9x}{4y} \) both simplify to the same expression in terms of \(t\), we have shown that \( \frac{dy}{dx} = -\frac{9x}{4y} \).