Question

If \(x = t + \frac{1}{t}\) and \(y = t - \frac{1}{t}\), find \(\frac{dy}{dx}\) at \(t = 2\).

Hint:

Always simplify your derivatives of \(x\) and \(y\) before dividing them to avoid complex fractions. Note that \(\frac{dy}{dx}\) is the slope of the curve at the point defined by the parameter \(t\).

Answer 1

Concept:
• This is a problem involving parametric differentiation.
• If \(x\) and \(y\) are given as functions of a parameter \(t\), then \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
• Differentiation of \(t^n\) is \(n t^{n-1}\).

Step 1: Differentiate \(x\) with respect to \(t\)
Given \(x = t + \frac{1}{t} = t + t^{-1}\). Differentiating both sides with respect to \(t\): \[ \frac{dx}{dt} = \frac{d}{dt}(t) + \frac{d}{dt}(t^{-1}) \] \[ \frac{dx}{dt} = 1 + (-1)t^{-2} = 1 - \frac{1}{t^2} \] Simplifying: \[ \frac{dx}{dt} = \frac{t^2 - 1}{t^2} \]

Step 2: Differentiate \(y\) with respect to \(t\)
Given \(y = t - \frac{1}{t} = t - t^{-1}\). Differentiating both sides with respect to \(t\): \[ \frac{dy}{dt} = \frac{d}{dt}(t) - \frac{d}{dt}(t^{-1}) \] \[ \frac{dy}{dt} = 1 - (-1)t^{-2} = 1 + \frac{1}{t^2} \] Simplifying: \[ \frac{dy}{dt} = \frac{t^2 + 1}{t^2} \]

Step 3: Find the expression for \(\frac{dy}{dx}\)
Using the parametric formula: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{t^2 + 1}{t^2}}{\frac{t^2 - 1}{t^2}} \] \[ \frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1} \]

Step 4: Substitute \(t = 2\) to find the numerical value
\[ \left( \frac{dy}{dx} \right)_{t=2} = \frac{2^2 + 1}{2^2 - 1} \] \[ \left( \frac{dy}{dx} \right)_{t=2} = \frac{4 + 1}{4 - 1} = \frac{5}{3} \]