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Ch.12 - Parametric and Polar Curves
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh.12 - Parametric and Polar CurvesProblem 12.2.82
Chapter 12, Problem 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

Verified step by step guidance
1
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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polar Coordinates and Equations

Polar coordinates represent points using a radius and an angle (r, θ) instead of Cartesian coordinates (x, y). Understanding how to interpret and manipulate equations in polar form is essential for analyzing curves like circles defined by r and θ.
Recommended video:
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Intro to Polar Coordinates

Conversion Between Polar and Cartesian Forms

Converting polar equations to Cartesian form (using x = r cos θ and y = r sin θ) helps visualize and graph curves. This conversion is crucial for recognizing standard geometric shapes, such as circles, from polar equations.
Recommended video:
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Intro to Polar Coordinates

Standard Equation of a Circle in Polar Coordinates

A circle in polar coordinates can often be expressed as r = 2a cos(θ - α) + b or similar forms. Recognizing these forms and relating them to the circle's center and radius aids in graphing and understanding the geometric properties of the circle.
Recommended video:
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Intro to Polar Coordinates
Related Practice
Textbook Question

Air drop A plane traveling horizontally at 80 m/s over flat ground at an elevation of 3000 m releases an emergency packet. The trajectory of the packet is given by

x = 80t, y = −4.9t² + 3000, t ≥ 0

where the origin is the point on the ground directly beneath the plane at the moment of the release (see figure). Graph the trajectory of the packet and find the coordinates of the point where the packet lands.

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Textbook Question

37–48. Polar-to-Cartesian coordinates Convert the following equations to Cartesian coordinates. Describe the resulting curve.


r = sin θ sec² θ

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Textbook Question

45–60. Areas of regions Find the area of the following regions.


The region outside the circle r = 1/2 and inside the circle r = cos θ

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Textbook Question

Cartesian lemniscate Find the equation in Cartesian coordinates of the lemniscate r² = a² cos 2θ, where a is a real number.

Textbook Question

What is the polar equation of the vertical line x = 5?

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Textbook Question

31–36. Eliminating the parameter Eliminate the parameter to express the following parametric equations as a single equation in x and y.


x=tan t, y=sec ² t−1 

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