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


x = 2 cos t, y = 8 sin t; slope = -1

Verified step by step guidance

1. Recall that the slope of the tangent line to a parametric curve given by \(x = x(t)\) and \(y = y(t)\) is found by computing \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
2. Differentiate both parametric equations with respect to \(t\): compute \(\frac{dx}{dt}\) for \(x = 2 \cos t\) and \(\frac{dy}{dt}\) for \(y = 8 \sin t\).
3. Write the expression for the slope \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) using the derivatives found in the previous step.
4. Set the slope equal to the given value \(-1\) and solve the resulting equation for \(t\).
5. Substitute the values of \(t\) found back into the parametric equations \(x = 2 \cos t\) and \(y = 8 \sin t\) to find the corresponding points on the curve.

Key Concepts

Parametric Equations and Curves

Parametric equations express the coordinates of points on a curve as functions of a parameter, often denoted t. Understanding how x and y depend on t allows us to analyze the curve's behavior and find points corresponding to specific conditions, such as a given slope.

Derivative of Parametric Curves (dy/dx)

For parametric curves, the slope of the tangent line 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 ratio gives the instantaneous rate of change of y with respect to x at any parameter value t.

Finding Points with a Given Slope

To find points where the curve has a specific slope, set the expression for dy/dx equal to the given slope and solve for the parameter t. Then, substitute t back into the parametric equations to find the corresponding (x, y) points on the curve.