67–72. Derivatives Consider the following parametric curves.
a. Determine dy/dx in terms of t and evaluate it at the given value of t.


x = t + 1/t, y = t − 1/t; t = 1

Verified step by step guidance

1. Identify the parametric equations given: \(x = t + \frac{1}{t}\) and \(y = t - \frac{1}{t}\).
2. Find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\). Compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) by differentiating each expression:
3. \[\frac{dx}{dt} = 1 - \frac{1}{t^2}\]
4. \[\frac{dy}{dt} = 1 + \frac{1}{t^2}\]
5. Use the chain rule for parametric curves to find \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\):
6. \[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}}\]
7. Evaluate \(\frac{dy}{dx}\) at the given value \(t = 1\) by substituting \(t = 1\) into the expression.

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, usually denoted t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves. Understanding how to work with these equations is essential for analyzing the curve's behavior.

Derivative of Parametric Curves (dy/dx)

For parametric curves, the derivative dy/dx is found by dividing the derivative of y with respect to t by the derivative of x with respect to t, i.e., (dy/dt) / (dx/dt). This method allows us to find the slope of the tangent line to the curve at any parameter value t.

Evaluating Derivatives at a Specific Parameter Value

After finding the general expression for dy/dx in terms of t, substituting the given value of t allows us to find the slope of the curve at that specific point. This step is crucial for understanding the curve's instantaneous rate of change or tangent line slope at that parameter.