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

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Identify the parametric equations given: \(x = 2 + 4t\) and \(y = 4 - 8t\).

Find the derivatives of \(x\) and \(y\) with respect to the parameter \(t\): compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

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 \(t\).

Use the formula for the derivative of \(y\) with respect to \(x\) in parametric form: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

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.

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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.

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