Textbook Question
Finding Lengths of Polar Curves
Find the lengths of the curves in Exercises 21–28.
The curve r = cos³(θ/3), 0 ≤ θ ≤ π/4
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16. Parametric Equations & Polar Coordinates
Calculus in Polar Coordinates
4:01 minutesProblem 12.3.11
Textbook Question
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)
Verified step by step guidance1
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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Video duration:
4mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
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Intro to Polar Coordinates
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.
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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.
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06:49Differentiation of Parametric Curves
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