11–20. Slopes of tangent lines Find the slope of the line tangent to the following polar curves at the given points.

r = 1 - sin θ; (1/2, π/6)

검증된 단계별 안내

Recall that for a polar curve given by \(r = f(\theta)\), the slope of the tangent line \(\frac{dy}{dx}\) can be found using the formulas for \(x\) and \(y\) in terms of \(r\) and \(\theta\): \(x = r \cos \theta\) and \(y = r \sin \theta\).

Express \(x\) and \(y\) as functions of \(\theta\): \(x(\theta) = r(\theta) \cos \theta\) and \(y(\theta) = r(\theta) \sin \theta\). For the given curve, \(r = 1 - \sin \theta\), so substitute this into \(x(\theta)\) and \(y(\theta)\).

Find the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) by applying the product rule: \(\frac{d}{d\theta}[r(\theta) \cos \theta]\) and \(\frac{d}{d\theta}[r(\theta) \sin \theta]\). Remember to differentiate both \(r(\theta)\) and the trigonometric functions.

Calculate the slope of the tangent line using the formula \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\). This gives the slope in terms of \(\theta\).

Evaluate \(\frac{dy}{dx}\) at the given point \(\theta = \frac{\pi}{6}\) to find the slope of the tangent line at that point.

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주요 개념

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Polar Coordinates and Polar Curves

Polar coordinates represent points using a radius and an angle (r, θ) instead of Cartesian coordinates (x, y). Polar curves are equations expressed as r = f(θ), describing how the radius changes with the angle. Understanding this system is essential for interpreting the given curve and point.

Slope of Tangent Line in Polar Coordinates

The slope of the tangent line to a polar curve at a point is found by converting the curve to parametric form (x = r cos θ, y = r sin θ) and then computing dy/dx = (dy/dθ) / (dx/dθ). This method links polar derivatives to Cartesian slopes.

Differentiation of Parametric Functions

To find dy/dx for parametric equations, differentiate x(θ) and y(θ) with respect to θ separately, then divide dy/dθ by dx/dθ. This technique is crucial for determining the slope of the tangent line to curves defined parametrically, such as polar curves.

11–20. Slopes of tangent lines Find the slope of the line tangent to the following polar curves at the given points.r = 1 - sin θ; (1/2, π/6)