77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.


x = 2 + √t, y = 2 - 4t; slope = -8

Verified step by step guidance

1. Identify the given parametric equations: \(x = 2 + \sqrt{t}\) and \(y = 2 - 4t\).
2. Recall that the slope of the tangent line to a parametric curve is given by \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
3. Compute the derivatives with respect to \(t\): \(\frac{dx}{dt} = \frac{d}{dt}(2 + \sqrt{t})\) and \(\frac{dy}{dt} = \frac{d}{dt}(2 - 4t)\).
4. Set the slope equal to the given value: \(\frac{dy}{dx} = -8\), which means \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -8\).
5. Solve the resulting equation for \(t\), then substitute back into the original parametric equations to find the corresponding points \((x, y)\).

Key Concepts

Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted as t. Here, x and y are given in terms of t, allowing us to analyze the curve's behavior by studying these functions.

Derivative of Parametric Curves

The slope of the tangent line to a parametric curve is found by computing dy/dx = (dy/dt) / (dx/dt). This requires differentiating both x(t) and y(t) with respect to t and then dividing the results to find the instantaneous rate of change.

Finding Points with a Given Slope

To find points where the curve has a specific slope, set the derivative dy/dx equal to the given slope and solve for the parameter t. Substituting t back into x(t) and y(t) gives the coordinates of the points with that slope.