Ch.12 - Parametric and Polar Curves

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch.12 - Parametric and Polar Curves

Problem 12.2.111

Chapter 12, Problem 12.2.111

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

Verified step by step guidance

Recall that the given lemniscate is expressed in polar coordinates as \(r^{2} = a^{2} \cos 2\theta\), where \(r\) is the radius and \(\theta\) is the angle.

Use the double-angle identity for cosine: \(\cos 2\theta = \cos^{2} \theta - \sin^{2} \theta\).

Express \(r\), \(\cos \theta\), and \(\sin \theta\) in terms of Cartesian coordinates: \(r = \sqrt{x^{2} + y^{2}}\), \(\cos \theta = \frac{x}{r}\), and \(\sin \theta = \frac{y}{r}\).

Substitute these into the equation: \(r^{2} = a^{2} \left( \frac{x^{2}}{r^{2}} - \frac{y^{2}}{r^{2}} \right)\), which simplifies to \(r^{2} = a^{2} \frac{x^{2} - y^{2}}{r^{2}}\).

Multiply both sides by \(r^{2}\) to eliminate the denominator, resulting in \(r^{4} = a^{2} (x^{2} - y^{2})\), and then replace \(r^{2}\) by \(x^{2} + y^{2}\) to get the Cartesian form: \((x^{2} + y^{2})^{2} = a^{2} (x^{2} - y^{2})\).

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Key Concepts

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

Polar to Cartesian Coordinate Conversion

Polar coordinates (r, θ) relate to Cartesian coordinates (x, y) through the formulas x = r cos θ and y = r sin θ. Understanding this conversion is essential to rewrite equations given in polar form into Cartesian form by expressing r and θ in terms of x and y.

Double-Angle Trigonometric Identities

The double-angle identity for cosine, cos 2θ = cos² θ - sin² θ, helps express trigonometric functions of 2θ in terms of sin θ and cos θ. This identity is crucial for manipulating the given polar equation to relate it to x and y coordinates.

Equation Manipulation and Substitution

After converting r and θ to x and y, algebraic manipulation and substitution are needed to eliminate θ and express the equation purely in terms of x and y. This process involves squaring, factoring, and rearranging terms to achieve the Cartesian form of the lemniscate.

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Related Practice

Textbook Question

Find the slope of the parametric curve x=−2t ³ +1, y=3t ², for −∞<t<∞, at the point corresponding to t=2.

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Textbook Question

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 = cos t, y = 1 + sin t; 0 ≤ t ≤ 2π

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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

29–32. Intersection points Use algebraic methods to find as many intersection points of the following curves as possible. Use graphical methods to identify the remaining intersection points.

r = 1 and r = 2 sin 2θ

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Textbook Question

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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