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

검증된 단계별 안내

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 \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

Differentiate both parametric equations with respect to \(t\): compute \(\frac{dx}{dt}\) for \(x = 2 \cos t\) and \(\frac{dy}{dt}\) for \(y = 8 \sin t\).

Write the expression for the slope \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) using the derivatives found in the previous step.

Set the slope equal to the given value \(-1\) and solve the resulting equation for \(t\).

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.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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