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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
Capitolo 12, Problema 12.2.82

80–83. Equations of circles Use the results of Exercises 78–79 to describe and graph the following circles.


r² - 8r cos(θ - π/2) = 9

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Recognize that the given equation is in polar form: \(r^{2} - 8r \cos(\theta - \frac{\pi}{2}) = 9\). Our goal is to rewrite this in a more recognizable form, such as the standard form of a circle in Cartesian coordinates.
Recall the polar to Cartesian coordinate conversions: \(x = r \cos \theta\) and \(y = r \sin \theta\). Also, note that \(\cos(\theta - \alpha) = \cos \theta \cos \alpha + \sin \theta \sin \alpha\).
Apply the cosine difference identity to expand \(\cos(\theta - \frac{\pi}{2})\): \(\cos\left(\theta - \frac{\pi}{2}\right) = \cos \theta \cos \frac{\pi}{2} + \sin \theta \sin \frac{\pi}{2} = 0 + \sin \theta = \sin \theta\).
Substitute this back into the original equation to get: \(r^{2} - 8r \sin \theta = 9\). Then, rewrite \(r^{2}\) as \(x^{2} + y^{2}\) and \(r \sin \theta\) as \(y\), giving: \(x^{2} + y^{2} - 8y = 9\).
Complete the square for the \(y\) terms to write the equation in standard circle form: \(x^{2} + (y^{2} - 8y + 16) = 9 + 16\) which simplifies to: \(x^{2} + (y - 4)^{2} = 25\). This represents a circle centered at \((0, 4)\) with radius \(5\).

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