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 = 2 + 4t, y = 4 − 8t; t = 2

Verified step by step guidance

1. Identify the parametric equations given: \(x = 2 + 4t\) and \(y = 4 - 8t\).
2. Find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\): compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
3. Calculate \(\frac{dx}{dt}\) by differentiating \(x = 2 + 4t\) with respect to \(t\), and calculate \(\frac{dy}{dt}\) by differentiating \(y = 4 - 8t\) with respect to \(t\).
4. Use the formula for the derivative of \(y\) with respect to \(x\) in parametric form: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
5. Evaluate \(\frac{dy}{dx}\) at the given value \(t = 2\) by substituting \(t = 2\) into the expressions for \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), then compute the ratio.

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, usually denoted as t. Instead of y as a function of x, both x and y depend on t, allowing the description of more complex curves and motions.

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 gives the slope of the tangent line to the curve at a given parameter value.

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 calculation of the slope of the curve at that specific point. This step is essential for understanding the behavior of the curve at particular parameter values.