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Ch. 11 - Parametric Equations and Polar Coordinates
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 11, Problem 11.3.48

Polar to Cartesian Equations


Replace the polar equations in Exercises 27–52 with equivalent Cartesian equations. Then describe or identify the graph.


r = 3 cos θ

Verified step by step guidance
1
Recall the relationships between polar and Cartesian coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\), and also \(r^2 = x^2 + y^2\).
Start with the given polar equation: \(r = 3 \cos \theta\).
Multiply both sides of the equation by \(r\) to eliminate the \(\cos \theta\) term: \(r \cdot r = 3r \cos \theta\), which gives \(r^2 = 3r \cos \theta\).
Substitute the Cartesian equivalents: replace \(r^2\) with \(x^2 + y^2\) and \(r \cos \theta\) with \(x\), resulting in the equation \(x^2 + y^2 = 3x\).
Rewrite the equation in standard Cartesian form by moving all terms to one side: \(x^2 - 3x + y^2 = 0\). Then, complete the square for the \(x\) terms to identify the graph type.

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Key Concepts

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

Polar and Cartesian Coordinate Systems

Polar coordinates represent points using a radius and an angle (r, θ), while Cartesian coordinates use (x, y) positions on a plane. Understanding how these systems relate is essential for converting equations between them.
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Intro to Polar Coordinates

Conversion Formulas Between Polar and Cartesian Coordinates

The key formulas are x = r cos θ and y = r sin θ, with r = √(x² + y²) and θ = arctan(y/x). These allow rewriting polar equations in terms of x and y, enabling analysis in the Cartesian plane.
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Intro to Polar Coordinates

Graphing and Identifying Curves from Equations

After converting to Cartesian form, recognizing the type of curve (circle, line, etc.) involves analyzing the equation's structure. For example, r = 3 cos θ corresponds to a circle in Cartesian coordinates.
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Determining Concavity from the Graph of f