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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblema 12.3.11
Capitolo 12, Problema 12.3.11

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)

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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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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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05:32
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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Percorso guidato
05:13
Slopes of Tangent Lines

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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Percorso guidato
06:49
Differentiation of Parametric Curves
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31–38. Equations of parabolas Find an equation of the following parabolas. Unless otherwise specified, assume the vertex is at the origin.

A parabola that opens to the right with directrix x = -4

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15–22. Sets in polar coordinates Sketch the following sets of points.


r = 3

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85–87. Grazing goat problems Consider the following sequence of problems related to grazing goats tied to a rope. (See the Guided Project Grazing goat problems.)


A circular corral of unit radius is enclosed by a fence. A goat inside the corral is tied to the fence with a rope of length 0≤a≤2 (see figure). What is the area of the region (inside the corral) that the goat can graze? Check your answer with the special cases a=0 and a=2.


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81–88. Arc length Find the arc length of the following curves on the given interval.


x = sin t, y = t - cos t; 0 ≤ t ≤ π/2

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15–30. Working with parametric equations Consider the following parametric equations.

a. Eliminate the parameter to obtain an equation in x and y.

b. Describe the curve and indicate the positive orientation.


x = 8 + 2t, y = 1; −∞ < t < ∞

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Plot the points with polar coordinates (2, π/6) and (−3, −π/2). Give two alternative sets of coordinate pairs for both points.

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