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

x = 4 cos t, y = 4 sin t; slope = 1/2

검증된 단계별 안내

Recall that the slope of the tangent line to a parametric curve given by \(x = f(t)\) and \(y = g(t)\) 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 \(x = 4 \cos t\) and \(\frac{dy}{dt}\) for \(y = 4 \sin t\).

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 \(\frac{1}{2}\) and solve the resulting equation for \(t\).

Use the values of \(t\) found to calculate the corresponding points on the curve by substituting back into \(x = 4 \cos t\) and \(y = 4 \sin t\).

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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 shape and behavior, which is essential for finding slopes of tangent lines at specific points.

Derivative of Parametric Equations (dy/dx)

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, which gives the instantaneous rate of change of y with respect to x.

Solving for Parameter Values Given a Slope

After finding the expression for dy/dx in terms of t, we set it equal to the given slope and solve for t. These parameter values correspond to points on the curve where the tangent line has the specified slope, enabling us to find the exact coordinates.

77–80. Slopes of tangent lines Find all points at which the following curves have the given slope.x = 4 cos t, y = 4 sin t; slope = 1/2