67–72. Derivatives Consider the following parametric curves.
b. Make a sketch of the curve showing the tangent line at the point corresponding to 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 coordinates of the point on the curve at \(t = 2\) by substituting \(t = 2\) into both equations: calculate \(x(2)\) and \(y(2)\).
3. Compute the derivatives of \(x\) and \(y\) with respect to \(t\): find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
4. Determine the slope of the tangent line at \(t = 2\) using the formula for parametric curves: \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) evaluated at \(t = 2\).
5. Write the equation of the tangent line in point-slope form using the point \((x(2), y(2))\) and the slope found in the previous step: \(y - y(2) = m (x - x(2))\).

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, usually 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

To find the slope of the tangent line to a parametric curve, compute dx/dt and dy/dt, then find dy/dx = (dy/dt) / (dx/dt). This derivative gives the instantaneous rate of change of y with respect to x at a given parameter value.

Equation of the Tangent Line

Once the slope of the tangent line at a point is known, use the point-slope form y - y₀ = m(x - x₀) to write the tangent line's equation. Here, (x₀, y₀) is the point on the curve at the given parameter t, and m is the slope.