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Volume 3

Calculus Volume 3
SENIOR CONTRIBUTING AUTHORS
EDWIN "JED" HERMAN, UNIVERSITY OF WISCONSIN-STEVENS POINT GILBERT STRANG, MASSACHUSETTS INSTITUTE OF TECHNOLOGY

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Table of Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Chapter 1: Parametric Equations and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . 7
1.1 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.2 Calculus of Parametric Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 1.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 1.4 Area and Arc Length in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 1.5 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 Chapter 2: Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 2.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 2.2 Vectors in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 2.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 2.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 2.5 Equations of Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 2.6 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 2.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 Chapter 3: Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 3.1 Vector-Valued Functions and Space Curves . . . . . . . . . . . . . . . . . . . . . . . . . . 260 3.2 Calculus of Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 3.3 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283 3.4 Motion in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Chapter 4: Differentiation of Functions of Several Variables . . . . . . . . . . . . . . . . . . . . 333 4.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 4.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 4.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 4.4 Tangent Planes and Linear Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . 389 4.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 4.6 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422 4.7 Maxima/Minima Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438 4.8 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 458 Chapter 5: Multiple Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477 5.1 Double Integrals over Rectangular Regions . . . . . . . . . . . . . . . . . . . . . . . . . . 478 5.2 Double Integrals over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501 5.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526 5.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546 5.5 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . 566 5.6 Calculating Centers of Mass and Moments of Inertia . . . . . . . . . . . . . . . . . . . . . 592 5.7 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 610 Chapter 6: Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 641 6.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 642 6.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663 6.3 Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689 6.4 Green’s Theorem . . . . 

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Preface

1

PREFACE
Welcome to Calculus Volume 3, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
About OpenStax
OpenStax is a nonprofit based at Rice University, and it’s our mission to improve student access to education. Our first openly licensed college textbook was published in 2012, and our library has since scaled to over 25 books for college and AP® courses used by hundreds of thousands of students. OpenStax Tutor, our low-cost personalized learning tool, is being used in college courses throughout the country. Through our partnerships with philanthropic foundations and our alliance with other educational resource organizations, OpenStax is breaking down the most common barriers to learning and empowering students and instructors to succeed.
About OpenStax's resources
Customization
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Instructors also have the option of creating a customized version of their OpenStax book. The custom version can be made available to students in low-cost print or digital form through their campus bookstore. Visit your book page on OpenStax.org for more information.
Errata
All OpenStax textbooks undergo a rigorous review process. However, like any professional-grade textbook, errors sometimes occur. Since our books are web based, we can make updates periodically when deemed pedagogically necessary. If you have a correction to suggest, submit it through the link on your book page on OpenStax.org. Subject matter experts review all errata suggestions. OpenStax is committed to remaining transparent about all updates, so you will also find a list of past errata changes on your book page on OpenStax.org.
Format
You can access this textbook for free in web view or PDF through OpenStax.org, and for a low cost in print.
About Calculus Volume 3
Calculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to enhance student learning. The book guides students through the core concepts of calculus and helps them understand how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency. Volume 3 covers parametric equations and polar coordinates, vectors, functions of several variables, multiple integration, and second-order differential equations.
Coverage and scope
Our Calculus Volume 3 textbook adheres to the scope and sequence of most general calculus courses nationwide. We have worked to make calculus interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of the three volumes of Calculus have been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from mathematics educators dedicated to the project.
Volume 1

2

Preface

Chapter 1: Functions and Graphs
Chapter 2: Limits
Chapter 3: Derivatives
Chapter 4: Applications of Derivatives
Chapter 5: Integration
Chapter 6: Applications of Integration
Volume 2 Chapter 1: Integration
Chapter 2: Applications of Integration
Chapter 3: Techniques of Integration
Chapter 4: Introduction to Differential Equations
Chapter 5: Sequences and Series
Chapter 6: Power Series
Chapter 7: Parametric Equations and Polar Coordinates
Volume 3 Chapter 1: Parametric Equations and Polar Coordinates
Chapter 2: Vectors in Space
Chapter 3: Vector-Valued Functions
Chapter 4: Differentiation of Functions of Several Variables
Chapter 5: Multiple Integration
Chapter 6: Vector Calculus
Chapter 7: Second-Order Differential Equations
Pedagogical foundation
Throughout Calculus Volume 3 you will find examples and exercises that present classical ideas and techniques as well as modern applications and methods. Derivations and explanations are based on years of classroom experience on the part of long-time calculus professors, striving for a balance of clarity and rigor that has proven successful with their students. Motivational applications cover important topics in probability, biology, ecology, business, and economics, as well as areas of physics, chemistry, engineering, and computer science. Student Projects in each chapter give students opportunities to explore interesting sidelights in pure and applied mathematics, from navigating a banked turn to adapting a moon landing vehicle for a new mission to Mars. Chapter Opening Applications pose problems that are solved later in the chapter, using the ideas covered in that chapter. Problems include the average distance of Halley's Comment from the Sun, and the vector field of a hurricane. Definitions, Rules, and Theorems are highlighted throughout the text, including over 60 Proofs of theorems.
Assessments that reinforce key concepts
In-chapter Examples walk students through problems by posing a question, stepping out a solution, and then asking students to practice the skill with a “Checkpoint” question. The book also includes assessments at the end of each chapter so students can apply what they’ve learned through practice problems. Many exercises are marked with a [T] to indicate they are suitable for solution by technology, including calculators or Computer Algebra Systems (CAS). Answers for selected exercises are available in the Answer Key at the back of the book. The book also includes assessments at the end of each chapter so students can apply what they’ve learned through practice problems.
Early or late transcendentals
The three volumes of Calculus are designed to accommodate both Early and Late Transcendental approaches to calculus. Exponential and logarithmic functions are introduced informally in Chapter 1 of Volume 1 and presented in more rigorous terms in Chapter 6 in Volume 1 and Chapter 2 in Volume 2. Differentiation and integration of these functions is covered in Chapters 3–5 in Volume 1 and Chapter 1 in Volume 2 for instructors who want to include them with other types of functions. These discussions, however, are in separate sections that can be skipped for instructors who prefer to wait until the integral definitions are given before teaching the calculus derivations of exponentials and logarithms.

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Preface

3

Comprehensive art program
Our art program is designed to enhance students’ understanding of concepts through clear and effective illustrations, diagrams, and photographs.

Additional resources
Student and instructor resources
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To reach the Community Hubs, visit www.oercommons.org/hubs/OpenStax.
Partner resources
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4

Preface

About the authors
Senior contributing authors
Gilbert Strang, Massachusetts Institute of Technology Dr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus online textbook is one of eleven that he has published and is the basis from which our final product has been derived and updated for today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University.
Edwin “Jed” Herman, University of Wisconsin-Stevens Point Dr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in 1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University of Wisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student research mentor, is experienced in course development/design, and is also an avid board game designer and player.
Contributing authors
Catherine Abbott, Keuka College Nicoleta Virginia Bila, Fayetteville State University Sheri J. Boyd, Rollins College Joyati Debnath, Winona State University Valeree Falduto, Palm Beach State College Joseph Lakey, New Mexico State University Julie Levandosky, Framingham State University David McCune, William Jewell College Michelle Merriweather, Bronxville High School Kirsten R. Messer, Colorado State University - Pueblo Alfred K. Mulzet, Florida State College at Jacksonville William Radulovich (retired), Florida State College at Jacksonville Erica M. Rutter, Arizona State University David Smith, University of the Virgin Islands Elaine A. Terry, Saint Joseph’s University David Torain, Hampton University
Reviewers
Marwan A. Abu-Sawwa, Florida State College at Jacksonville Kenneth J. Bernard, Virginia State University John Beyers, University of Maryland Charles Buehrle, Franklin & Marshall College Matthew Cathey, Wofford College Michael Cohen, Hofstra University William DeSalazar, Broward County School System Murray Eisenberg, University of Massachusetts Amherst Kristyanna Erickson, Cecil College Tiernan Fogarty, Oregon Institute of Technology David French, Tidewater Community College Marilyn Gloyer, Virginia Commonwealth University Shawna Haider, Salt Lake Community College Lance Hemlow, Raritan Valley Community College Jerry Jared, The Blue Ridge School Peter Jipsen, Chapman University David Johnson, Lehigh University M.R. Khadivi, Jackson State University Robert J. Krueger, Concordia University Tor A. Kwembe, Jackson State University Jean-Marie Magnier, Springfield Technical Community College Cheryl Chute Miller, SUNY Potsdam Bagisa Mukherjee, Penn State University, Worthington Scranton Campus Kasso Okoudjou, University of Maryland College Park Peter Olszewski, Penn State Erie, The Behrend College Steven Purtee, Valencia College Alice Ramos, Bethel College

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Preface

5

Doug Shaw, University of Northern Iowa Hussain Elalaoui-Talibi, Tuskegee University Jeffrey Taub, Maine Maritime Academy William Thistleton, SUNY Polytechnic Institute A. David Trubatch, Montclair State University Carmen Wright, Jackson State University Zhenbu Zhang, Jackson State University

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Chapter 1 | Parametric Equations and Polar Coordinates

7

1 | PARAMETRIC EQUATIONS AND POLAR COORDINATES

Figure 1.1 The chambered nautilus is a marine animal that lives in the tropical Pacific Ocean. Scientists think they have existed mostly unchanged for about 500 million years.(credit: modification of work by Jitze Couperus, Flickr)

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Chapter 1 | Parametric Equations and Polar Coordinates

Chapter Outline
1.1 Parametric Equations
1.2 Calculus of Parametric Curves
1.3 Polar Coordinates
1.4 Area and Arc Length in Polar Coordinates
1.5 Conic Sections
Introduction
The chambered nautilus is a fascinating creature. This animal feeds on hermit crabs, fish, and other crustaceans. It has a hard outer shell with many chambers connected in a spiral fashion, and it can retract into its shell to avoid predators. When part of the shell is cut away, a perfect spiral is revealed, with chambers inside that are somewhat similar to growth rings in a tree.
The mathematical function that describes a spiral can be expressed using rectangular (or Cartesian) coordinates. However, if we change our coordinate system to something that works a bit better with circular patterns, the function becomes much simpler to describe. The polar coordinate system is well suited for describing curves of this type. How can we use this coordinate system to describe spirals and other radial figures? (See Example 1.14.)
In this chapter we also study parametric equations, which give us a convenient way to describe curves, or to study the position of a particle or object in two dimensions as a function of time. We will use parametric equations and polar coordinates for describing many topics later in this text.
1.1 | Parametric Equations
Learning Objectives
1.1.1 Plot a curve described by parametric equations. 1.1.2 Convert the parametric equations of a curve into the form y = f (x). 1.1.3 Recognize the parametric equations of basic curves, such as a line and a circle. 1.1.4 Recognize the parametric equations of a cycloid.
In this section we examine parametric equations and their graphs. In the two-dimensional coordinate system, parametric equations are useful for describing curves that are not necessarily functions. The parameter is an independent variable that both x and y depend on, and as the parameter increases, the values of x and y trace out a path along a plane curve. For example, if the parameter is t (a common choice), then t might represent time. Then x and y are defined as functions of time, and ⎛⎝x(t), y(t)⎞⎠ can describe the position in the plane of a given object as it moves along a curved path.
Parametric Equations and Their Graphs
Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra 1 day of orbiting time is built into the calendar. We call January 1 “day 1”
4 of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.
The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.

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Chapter 1 | Parametric Equations and Polar Coordinates

9

Figure 1.2 Earth’s orbit around the Sun in one year.
Figure 1.2 depicts Earth’s orbit around the Sun during one year. The point labeled F2 is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (Figure 1.3). Then each x value on the graph is a value of position as a function of time, and each y value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.

Figure 1.3 Coordinate axes superimposed on the orbit of Earth.
We can determine the functions for x(t) and y(t), thereby parameterizing the orbit of Earth around the Sun. The variable t is called an independent parameter and, in this context, represents time relative to the beginning of each year. A curve in the (x, y) plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.
Definition If x and y are continuous functions of t on an interval I, then the equations
x = x(t) and y = y(t) are called parametric equations and t is called the parameter. The set of points (x, y) obtained as t varies over the

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Chapter 1 | Parametric Equations and Polar Coordinates

interval I is called the graph of the parametric equations. The graph of parametric equations is called a parametric curve or plane curve, and is denoted by C.

Notice in this definition that x and y are used in two ways. The first is as functions of the independent variable t. As t varies over the interval I, the functions x(t) and y(t) generate a set of ordered pairs (x, y). This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, x and y are variables. It is important to distinguish the variables x and y from the functions x(t) and y(t).
Example 1.1
Graphing a Parametrically Defined Curve

Sketch the curves described by the following parametric equations: a. x(t) = t − 1, y(t) = 2t + 4, −3 ≤ t ≤ 2 b. x(t) = t2 − 3, y(t) = 2t + 1, −2 ≤ t ≤ 3 c. x(t) = 4 cos t, y(t) = 4 sin t, 0 ≤ t ≤ 2π

Solution
a. To create a graph of this curve, first set up a table of values. Since the independent variable in both x(t) and y(t) is t, let t appear in the first column. Then x(t) and y(t) will appear in the second and third columns of the table.

t

x(t)

y(t)

−3

−4

−2

−2

−3

0

−1

−2

2

0

−1

4

1

0

6

2

1

8

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in Figure 1.4. The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as t varies from −3 to 2.

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Chapter 1 | Parametric Equations and Polar Coordinates

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Figure 1.4 Graph of the plane curve described by the parametric equations in part a.

b. To create a graph of this curve, again set up a table of values.

t

x(t)

y(t)

−2

1

−3

−1

−2

−1

0

−3

1

1

−2

3

2

1

5

3

6

7

The second and third columns in this table give a set of points to be plotted (Figure 1.5). The first point on the graph (corresponding to t = −2) has coordinates (1, −3), and the last point (corresponding
to t = 3) has coordinates (6, 7). As t progresses from −2 to 3, the point on the curve travels along a
parabola. The direction the point moves is again called the orientation and is indicated on the graph.

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Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.5 Graph of the plane curve described by the parametric equations in part b.

c. In this case, use multiples of π/6 for t and create another table of values:

t

x(t)

y(t)

t

x(t)

0

4

0

7π 6

−2 3 ≈ −3.5

y(t) 2

π 6

2 3 ≈ 3.5

2

4π

−2

−2 3 ≈ −3.5

3

π 3

2

2 3 ≈ 3.5

3π 2

0

−4

π

0

4

2

5π

2

−2 3 ≈ −3.5

3

2π 3

−2

2 3 ≈ 3.5

11π

2 3 ≈ 3.5

6

2

5π

−2 3 ≈ −3.5

2

2π

4

0

6

π

−4

0

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13

The graph of this plane curve appears in the following graph.

Figure 1.6 Graph of the plane curve described by the parametric equations in part c.
This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates (4, 0).

1.1 Sketch the curve described by the parametric equations x(t) = 3t + 2, y(t) = t2 − 1, −3 ≤ t ≤ 2.

Eliminating the Parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables x and y. Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in Example 1.1b. are

x(t) = t2 − 3, y(t) = 2t + 1, −2 ≤ t ≤ 3.

Solving the second equation for t gives

This can be substituted into the first equation:

t

=

y

− 2

1.

x

=

⎛y ⎝

− 2

1⎞2 ⎠

−

3

=

y2

−

2y 4

+

1

−

3

=

y2

−

2y 4

−

11.

This equation describes x as a function of y. These steps give an example of eliminating the parameter. The graph of this function is a parabola opening to the right. Recall that the plane curve started at (1, −3) and ended at (6, 7). These
terminations were due to the restriction on the parameter t.

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Chapter 1 | Parametric Equations and Polar Coordinates

Example 1.2
Eliminating the Parameter

Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.
a. x(t) = 2t + 4, y(t) = 2t + 1, −2 ≤ t ≤ 6
b. x(t) = 4 cos t, y(t) = 3 sin t, 0 ≤ t ≤ 2π

Solution
a. To eliminate the parameter, we can solve either of the equations for t. For example, solving the first equation for t gives

x = 2t + 4

x2 = 2t + 4

x2 − 4 = 2t

t

=

x2

− 2

4

.

Note that when we square both sides it is important to observe that

x ≥ 0.

Substituting

t=

x2 − 4 2

this

into y(t) yields

y(t) = 2t + 1

y

=

2⎛⎝x

2

− 2

4

⎞ ⎠

+

1

y = x2 − 4 + 1

y = x2 − 3.

This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter t. When t = −2, x = 2(−2) + 4 = 0, and when t = 6, x = 2(6) + 4 = 4. The graph of this plane curve follows.

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15

Figure 1.7 Graph of the plane curve described by the parametric equations in part a.
b. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are
x(t) = 4 cos t and y(t) = 3 sin t.

Solving either equation for t directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the t) gives us

cos

t

=

x 4

and

sin

t

=

3y .

Now use the Pythagorean identity cos2 t + sin2 t = 1 and replace the expressions for sin t and cos t with the equivalent expressions in terms of x and y. This gives

⎛ x ⎞2 ⎝4⎠

+

⎛y ⎝3

⎞2 ⎠

=

1

x2 16

+

y2 9

=

1.

This is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.

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Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.8 Graph of the plane curve described by the parametric equations in part b.
As t progresses from 0 to 2π, a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.
1.2 Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.
x(t) = 2 + 3t , y(t) = t − 1, 2 ≤ t ≤ 6
So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.
Example 1.3
Parameterizing a Curve
Find two different pairs of parametric equations to represent the graph of y = 2x2 − 3. Solution First, it is always possible to parameterize a curve by defining x(t) = t, then replacing x with t in the equation for y(t). This gives the parameterization
x(t) = t, y(t) = 2t2 − 3.
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Since there is no restriction on the domain in the original graph, there is no restriction on the values of t. We have complete freedom in the choice for the second parameterization. For example, we can choose x(t) = 3t − 2. The only thing we need to check is that there are no restrictions imposed on x; that is, the range of x(t) is all real numbers. This is the case for x(t) = 3t − 2. Now since y = 2x2 − 3, we can substitute x(t) = 3t − 2 for x. This gives
y(t) = 2(3t − 2)2 − 2 = 2⎛⎝9t2 − 12t + 4⎞⎠ − 2 = 18t2 − 24t + 8 − 2 = 18t2 − 24t + 6.
Therefore, a second parameterization of the curve can be written as x(t) = 3t − 2 and y(t) = 18t2 − 24t + 6.
1.3 Find two different sets of parametric equations to represent the graph of y = x2 + 2x.
Cycloids and Other Parametric Curves
Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (Figure 1.9). A cycloid generated by a circle (or bicycle wheel) of radius a is given by the parametric equations
x(t) = a(t − sin t), y(t) = a(1 − cos t). To see why this is true, consider the path that the center of the wheel takes. The center moves along the x-axis at a constant height equal to the radius of the wheel. If the radius is a, then the coordinates of the center can be given by the equations
x(t) = at, y(t) = a for any value of t. Next, consider the ant, which rotates around the center along a circular path. If the bicycle is moving from left to right then the wheels are rotating in a clockwise direction. A possible parameterization of the circular motion of the ant (relative to the center of the wheel) is given by
x(t) = −a sin t, y(t) = −a cos t. (The negative sign is needed to reverse the orientation of the curve. If the negative sign were not there, we would have to imagine the wheel rotating counterclockwise.) Adding these equations together gives the equations for the cycloid.
x(t) = a(t − sin t), y(t) = a(1 − cos t).

Figure 1.9 A wheel traveling along a road without slipping; the point on the edge of the wheel traces out a cycloid.
Now suppose that the bicycle wheel doesn’t travel along a straight road but instead moves along the inside of a larger wheel, as in Figure 1.10. In this graph, the green circle is traveling around the blue circle in a counterclockwise direction. A point

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Chapter 1 | Parametric Equations and Polar Coordinates

on the edge of the green circle traces out the red graph, which is called a hypocycloid.

Figure 1.10 Graph of the hypocycloid described by the parametric equations shown.

The general parametric equations for a hypocycloid are

x(t)

=

(a

−

b)

cos

t

+

b

cos⎛⎝a

− b

b

⎞ ⎠

t

y(t)

=

(a

−

b)

sin

t

−

b

sin⎛⎝a

− b

b

⎞ ⎠

t.

These equations are a bit more complicated, but the derivation is somewhat similar to the equations for the cycloid. In this

case we assume the radius of the larger circle is a and the radius of the smaller circle is b. Then the center of the wheel

travels along a circle of radius a − b. This fact explains the first term in each equation above. The period of the second

trigonometric function in both

x(t)

and

y(t)

is equal to

2πb a−b

.

The ratio a is related to the number of cusps on the graph (cusps are the corners or pointed ends of the graph), as illustrated b
in Figure 1.11. This ratio can lead to some very interesting graphs, depending on whether or not the ratio is rational. Figure 1.10 corresponds to a = 4 and b = 1. The result is a hypocycloid with four cusps. Figure 1.11 shows some other possibilities. The last two hypocycloids have irrational values for ab. In these cases the hypocycloids have an infinite
number of cusps, so they never return to their starting point. These are examples of what are known as space-filling curves.

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19

Figure 1.11 Graph of various hypocycloids corresponding to different values of a/b.

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Chapter 1 | Parametric Equations and Polar Coordinates

The Witch of Agnesi
Many plane curves in mathematics are named after the people who first investigated them, like the folium of Descartes or the spiral of Archimedes. However, perhaps the strangest name for a curve is the witch of Agnesi. Why a witch?
Maria Gaetana Agnesi (1718–1799) was one of the few recognized women mathematicians of eighteenth-century Italy. She wrote a popular book on analytic geometry, published in 1748, which included an interesting curve that had been studied by Fermat in 1630. The mathematician Guido Grandi showed in 1703 how to construct this curve, which he later called the “versoria,” a Latin term for a rope used in sailing. Agnesi used the Italian term for this rope, “versiera,” but in Latin, this same word means a “female goblin.” When Agnesi’s book was translated into English in 1801, the translator used the term “witch” for the curve, instead of rope. The name “witch of Agnesi” has stuck ever since.
The witch of Agnesi is a curve defined as follows: Start with a circle of radius a so that the points (0, 0) and (0, 2a) are points on the circle (Figure 1.12). Let O denote the origin. Choose any other point A on the circle, and draw the secant line OA. Let B denote the point at which the line OA intersects the horizontal line through (0, 2a). The vertical line through B intersects the horizontal line through A at the point P. As the point A varies, the path that the point P travels is the witch of Agnesi curve for the given circle.
Witch of Agnesi curves have applications in physics, including modeling water waves and distributions of spectral lines. In probability theory, the curve describes the probability density function of the Cauchy distribution. In this project you will parameterize these curves.

Figure 1.12 As the point A moves around the circle, the point P traces out the witch of Agnesi curve for the given circle. 1. On the figure, label the following points, lengths, and angle: a. C is the point on the x-axis with the same x-coordinate as A. b. x is the x-coordinate of P, and y is the y-coordinate of P. c. E is the point (0, a). d. F is the point on the line segment OA such that the line segment EF is perpendicular to the line segment
OA. e. b is the distance from O to F. f. c is the distance from F to A. g. d is the distance from O to B. h. θ is the measure of angle ∠COA.
The goal of this project is to parameterize the witch using θ as a parameter. To do this, write equations for x and y in terms of only θ.
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21

2.

Show

that

d

=

2a sin θ

.

3. Note that x = d cos θ. Show that x = 2a cot θ. When you do this, you will have parameterized the x-coordinate of the curve with respect to θ. If you can get a similar equation for y, you will have parameterized the curve.

4. In terms of θ, what is the angle ∠EOA ?

5. Show that b + c = 2a cos⎛⎝π2 − θ⎞⎠.

6. Show that y = 2a cos⎛⎝π2 − θ⎞⎠ sin θ.

7. Show that y = 2a sin2 θ. You have now parameterized the y-coordinate of the curve with respect to θ.

8. Conclude that a parameterization of the given witch curve is

x = 2a cot θ, y = 2a sin2 θ, − ∞ < θ < ∞.

9.

Use your parameterization to show that the given witch curve is the graph of the function

f (x)

=

x2

8a 3 + 4a2

.

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Chapter 1 | Parametric Equations and Polar Coordinates

Travels with My Ant: The Curtate and Prolate Cycloids

Earlier in this section, we looked at the parametric equations for a cycloid, which is the path a point on the edge of a wheel traces as the wheel rolls along a straight path. In this project we look at two different variations of the cycloid, called the curtate and prolate cycloids.
First, let’s revisit the derivation of the parametric equations for a cycloid. Recall that we considered a tenacious ant trying to get home by hanging onto the edge of a bicycle tire. We have assumed the ant climbed onto the tire at the very edge, where the tire touches the ground. As the wheel rolls, the ant moves with the edge of the tire (Figure 1.13).
As we have discussed, we have a lot of flexibility when parameterizing a curve. In this case we let our parameter t represent the angle the tire has rotated through. Looking at Figure 1.13, we see that after the tire has rotated through an angle of t, the position of the center of the wheel, C = (xC, yC), is given by

xC = at and yC = a. Furthermore, letting A = (x A, y A) denote the position of the ant, we note that

Then

xC − x A = a sin t and yC − y A = a cos t.
x A = xC − a sin t = at − a sin t = a(t − sin t) y A = yC − a cos t = a − a cos t = a(1 − cos t).

Figure 1.13 (a) The ant clings to the edge of the bicycle tire as the tire rolls along the ground. (b) Using geometry to determine the position of the ant after the tire has rotated through an angle of t.
Note that these are the same parametric representations we had before, but we have now assigned a physical meaning to the parametric variable t. After a while the ant is getting dizzy from going round and round on the edge of the tire. So he climbs up one of the spokes toward the center of the wheel. By climbing toward the center of the wheel, the ant has changed his path of motion. The new path has less up-and-down motion and is called a curtate cycloid (Figure 1.14). As shown in the figure, we let b denote the distance along the spoke from the center of the wheel to the ant. As before, we let t represent the angle the tire has rotated through. Additionally, we let C = (xC, yC) represent the position of the center of the wheel and A = (x A, y A) represent the position of the ant.
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Figure 1.14 (a) The ant climbs up one of the spokes toward the center of the wheel. (b) The ant’s path of motion after he climbs closer to the center of the wheel. This is called a curtate cycloid. (c) The new setup, now that the ant has moved closer to the center of the wheel.
1. What is the position of the center of the wheel after the tire has rotated through an angle of t?
2. Use geometry to find expressions for xC − x A and for yC − y A.
3. On the basis of your answers to parts 1 and 2, what are the parametric equations representing the curtate cycloid? Once the ant’s head clears, he realizes that the bicyclist has made a turn, and is now traveling away from his home. So he drops off the bicycle tire and looks around. Fortunately, there is a set of train tracks nearby, headed back in the right direction. So the ant heads over to the train tracks to wait. After a while, a train goes by, heading in the right direction, and he manages to jump up and just catch the edge of the train wheel (without getting squished!). The ant is still worried about getting dizzy, but the train wheel is slippery and has no spokes to climb, so he decides to just hang on to the edge of the wheel and hope for the best. Now, train wheels have a flange to keep the wheel running on the tracks. So, in this case, since the ant is hanging on to the very edge of the flange, the distance from the center of the wheel to the ant is actually greater than the radius of the wheel (Figure 1.15). The setup here is essentially the same as when the ant climbed up the spoke on the bicycle wheel. We let b denote the distance from the center of the wheel to the ant, and we let t represent the angle the tire has rotated through. Additionally, we let C = (xC, yC) represent the position of the center of the wheel and
A = (x A, y A) represent the position of the ant (Figure 1.15).
When the distance from the center of the wheel to the ant is greater than the radius of the wheel, his path of motion is called a prolate cycloid. A graph of a prolate cycloid is shown in the figure.

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Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.15 (a) The ant is hanging onto the flange of the train wheel. (b) The new setup, now that the ant has jumped onto the train wheel. (c) The ant travels along a prolate cycloid.
4. Using the same approach you used in parts 1– 3, find the parametric equations for the path of motion of the ant.
5. What do you notice about your answer to part 3 and your answer to part 4? Notice that the ant is actually traveling backward at times (the “loops” in the graph), even though the train continues to move forward. He is probably going to be really dizzy by the time he gets home!

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25

1.1 EXERCISES
For the following exercises, sketch the curves below by eliminating the parameter t. Give the orientation of the curve. 1. x = t2 + 2t, y = t + 1 2. x = cos(t), y = sin(t), (0, 2π] 3. x = 2t + 4, y = t − 1 4. x = 3 − t, y = 2t − 3, 1.5 ≤ t ≤ 3 For the following exercises, eliminate the parameter and sketch the graphs. 5. x = 2t2, y = t4 + 1 For the following exercises, use technology (CAS or calculator) to sketch the parametric equations. 6. [T] x = t2 + t, y = t2 − 1
7. [T] x = e−t, y = e2t − 1 8. [T] x = 3 cos t, y = 4 sin t 9. [T] x = sec t, y = cos t For the following exercises, sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph. 10. x = et, y = e2t + 1 11. x = 6 sin(2θ), y = 4 cos(2θ) 12. x = cos θ, y = 2 sin(2θ) 13. x = 3 − 2 cos θ, y = −5 + 3 sin θ 14. x = 4 + 2 cos θ, y = −1 + sin θ 15. x = sec t, y = tan t
16. x = ln(2t), y = t2
17. x = et, y = e2t
18. x = e−2t, y = e3t
19. x = t3, y = 3 ln t

20. x = 4 sec θ, y = 3 tan θ

For the following exercises, convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form.

21.

x = t2 − 1,

y

=

t 2

22.

x=

t

1 +

1,

y

=

1

t +

t,

t

>

−1

23. x = 4 cos θ, y = 3 sin θ, t ∈ (0, 2π]

24. x = cosh t, y = sinh t

25. x = 2t − 3, y = 6t − 7

26. x = t2, y = t3

27. x = 1 + cos t, y = 3 − sin t

28. x = t, y = 2t + 4

29.

x = sec t,

y

=

tan t,

π

≤

t

<

3π 2

30. x = 2 cosh t, y = 4 sinh t

31. x = cos(2t), y = sin t

32. x = 4t + 3, y = 16t2 − 9

33. x = t2, y = 2 ln t, t ≥ 1

34. x = t3, y = 3 ln t, t ≥ 1
35. x = tn, y = n ln t, t ≥ 1, where n is a natural number
x = ln(5t) 36. y = ln(t2) where 1 ≤ t ≤ e

37.

x = 2 sin(8t) y = 2 cos(8t)

38.

x = tan t y = sec2 t − 1

For the following exercises, the pairs of parametric equations represent lines, parabolas, circles, ellipses, or hyperbolas. Name the type of basic curve that each pair of

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Chapter 1 | Parametric Equations and Polar Coordinates

equations represents.

39.

x = 3t + 4 y = 5t − 2

40.

x − 4 = 5t y+2=t

x = 2t + 1 41. y = t2 − 3

42.

x = 3 cos t y = 3 sin t

43.

x = 2 cos(3t) y = 2 sin(3t)

44.

x = cosh t y = sinh t

45.

x = 3 cos t y = 4 sin t

46.

x = 2 cos(3t) y = 5 sin(3t)

47.

x = 3 cosh(4t) y = 4 sinh(4t)

48.

x = 2 cosh t y = 2 sinh t

49.

Show that

x y

= =

h + r cos θ k + r sin θ

represents the

equation of

a circle.

50. Use the equations in the preceding problem to find a set of parametric equations for a circle whose radius is 5 and whose center is (−2, 3).

For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation.

51.

[T]

x y

= =

θ 1

+ −

sin θ cos θ

52.

[T]

x y

= =

2t − 2 sin t 2 − 2 cos t

53.

[T]

x y

= =

t − 0.5 sin t 1 − 1.5 cos t

54. An airplane traveling horizontally at 100 m/s over flat ground at an elevation of 4000 meters must drop an emergency package on a target on the ground. The trajectory of the package is given by
x = 100t, y = −4.9t2 + 4000, t ≥ 0 where the origin is
the point on the ground directly beneath the plane at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target?

55. The trajectory of a bullet is given by

x

=

v0

(cos

α)

ty

=

v0

(sin

α)

t

−

1 2

gt

2

where

v0 = 500 m/s,

g = 9.8 = 9.8 m/s2,

and

α = 30 degrees. When will the bullet hit the ground? How

far from the gun will the bullet hit the ground?

56. [T] Use technology to sketch the curve represented by x = sin(4t), y = sin(3t), 0 ≤ t ≤ 2π.

57.

[T] Use technology to

x = 2 tan(t), y = 3 sec(t), −π < t < π.

sketch

58. Sketch the curve known as an epitrochoid, which gives the path of a point on a circle of radius b as it rolls on the outside of a circle of radius a. The equations are

x

=

(a

+

b)cos

t

−

c

·

cos⎡⎣(a

+ b

b)t

⎤ ⎦

y

=

(a

+

b)sin

t

−

c

·

sin⎡⎣(a

+ b

b)t

⎤⎦.

Let a = 1, b = 2, c = 1.

59. [T] Use technology to sketch the spiral curve given by x = t cos(t), y = t sin(t) from −2π ≤ t ≤ 2π.

60. [T] Use technology to graph the curve given by the

parametric

equations

x = 2 cot(t), y = 1 − cos(2t), −π/2 ≤ t ≤ π/2. This

curve is known as the witch of Agnesi.

61. [T] Sketch the curve given by parametric equations

x = cosh(t) y = sinh(t),

where −2 ≤ t ≤ 2.

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27

1.2 | Calculus of Parametric Curves
Learning Objectives
1.2.1 Determine derivatives and equations of tangents for parametric curves. 1.2.2 Find the area under a parametric curve. 1.2.3 Use the equation for arc length of a parametric curve. 1.2.4 Apply the formula for surface area to a volume generated by a parametric curve.
Now that we have introduced the concept of a parameterized curve, our next step is to learn how to work with this concept in the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slope of a tangent line to the curve? How about the arc length of the curve? Or the area under the curve? Another scenario: Suppose we would like to represent the location of a baseball after the ball leaves a pitcher’s hand. If the position of the baseball is represented by the plane curve ⎛⎝x(t), y(t)⎞⎠, then we should be able to use calculus to find the speed of the ball at any given time. Furthermore, we should be able to calculate just how far that ball has traveled as a function of time.
Derivatives of Parametric Equations
We start by asking how to calculate the slope of a line tangent to a parametric curve at a point. Consider the plane curve defined by the parametric equations
x(t) = 2t + 3, y(t) = 3t − 4, −2 ≤ t ≤ 3.
The graph of this curve appears in Figure 1.16. It is a line segment starting at (−1, −10) and ending at (9, 5).

Figure 1.16 Graph of the line segment described by the given parametric equations.

28

Chapter 1 | Parametric Equations and Polar Coordinates

We can eliminate the parameter by first solving the equation x(t) = 2t + 3 for t:

Substituting this into y(t), we obtain

x(t) = 2t + 3

x − 3 = 2t

t

=

x

− 2

3

.

y(t) = 3t − 4

y

=

3⎛⎝x

− 2

3

⎞ ⎠

−

4

y

=

3x 2

−

9 2

−

4

y

=

3x 2

−

127.

The slope of this line is given by

dy dx

=

32.

Next we calculate

x′ (t)

and

y′ (t).

This gives

x′ (t)

=

2

and

y′ (t)

=

3.

Notice

that

dy dx

=

dy/dt dx/dt

=

3 2

.

This is no coincidence, as outlined in the following theorem.

Theorem 1.1: Derivative of Parametric Equations

Consider the plane curve defined by the parametric equations x = x(t) and y = y(t). Suppose that x′ (t) and y′ (t)

exist, and assume that

x′ (t) ≠ 0.

Then the derivative

dy dx

is given by

dy dx

=

dy/dt dx/dt

=

y′ x′

((tt)).

(1.1)

Proof
This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yielding a differentiable function y = F(x). Then y(t) = F(x(t)). Differentiating both sides of this equation using the Chain Rule yields

y′ (t) = F′ (x(t))x′ (t),

so

F′ ⎛⎝x(t)⎞⎠

=

y′ x′

((tt)).

But F′ ⎛⎝x(t)⎞⎠ = ddyx, which proves the theorem.
□
Equation 1.1 can be used to calculate derivatives of plane curves, as well as critical points. Recall that a critical point of a differentiable function y = f (x) is any point x = x0 such that either f ′ (x0) = 0 or f ′ (x0) does not exist. Equation 1.1 gives a formula for the slope of a tangent line to a curve defined parametrically regardless of whether the curve can be described by a function y = f (x) or not.

Example 1.4

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Chapter 1 | Parametric Equations and Polar Coordinates

29

Finding the Derivative of a Parametric Curve

Calculate the derivative dy for each of the following parametrically defined plane curves, and locate any critical dx
points on their respective graphs. a. x(t) = t2 − 3, y(t) = 2t − 1, −3 ≤ t ≤ 4
b. x(t) = 2t + 1, y(t) = t3 − 3t + 4, −2 ≤ t ≤ 5 c. x(t) = 5 cos t, y(t) = 5 sin t, 0 ≤ t ≤ 2π

Solution a. To apply Equation 1.1, first calculate x′ (t) and y′(t):
x′ (t) = 2t y′ (t) = 2.

Next substitute these into the equation:

dy dx

=

dy/dt dx/dt

dy dx

=

2 2t

dy dx

=

1t .

This derivative is undefined when t = 0. Calculating x(0) and y(0) gives x(0) = (0)2 − 3 = −3 and y(0) = 2(0) − 1 = −1, which corresponds to the point (−3, −1) on the graph. The graph of this curve is a parabola opening to the right, and the point (−3, −1) is its vertex as shown.

Figure 1.17 Graph of the parabola described by parametric equations in part a.
b. To apply Equation 1.1, first calculate x′ (t) and y′(t):
x′ (t) = 2 y′ (t) = 3t2 − 3.

30

Chapter 1 | Parametric Equations and Polar Coordinates

Next substitute these into the equation:

dy dx

=

dy/dt dx/dt

dy dx

=

3t2 − 2

3.

This derivative is zero when t = ±1. When t = −1 we have x(−1) = 2(−1) + 1 = −1 and y(−1) = (−1)3 − 3(−1) + 4 = −1 + 3 + 4 = 6,

which corresponds to the point (−1, 6) on the graph. When t = 1 we have x(1) = 2(1) + 1 = 3 and y(1) = (1)3 − 3(1) + 4 = 1 − 3 + 4 = 2,

which corresponds to the point (3, 2) on the graph. The point (3, 2) is a relative minimum and the point (−1, 6) is a relative maximum, as seen in the following graph.

Figure 1.18 Graph of the curve described by parametric equations in part b.
c. To apply Equation 1.1, first calculate x′ (t) and y′(t):
x′ (t) = −5 sin t y′ (t) = 5 cos t.

Next substitute these into the equation:

dy dx

=

dy/dt dx/dt

dy dx

=

5 cos t −5 sin t

dy dx

=

−cot

t.

This derivative is zero when cos t = 0 and is undefined when sin t = 0. This gives

t

=

0,

π 2

,

π,

3π 2

,

and

2π

as critical

points for

t.

Substituting

each

of

these into

x(t)

and

y(t),

we obtain

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Chapter 1 | Parametric Equations and Polar Coordinates

31

t

x(t)

y(t)

0

5

0

π

0

5

2

π

−5

0

3π 0

−5

2

2π

5

0

These points correspond to the sides, top, and bottom of the circle that is represented by the parametric equations (Figure 1.19). On the left and right edges of the circle, the derivative is undefined, and on the top and bottom, the derivative equals zero.

Figure 1.19 Graph of the curve described by parametric equations in part c.
1.4 Calculate the derivative dy/dx for the plane curve defined by the equations x(t) = t2 − 4t, y(t) = 2t3 − 6t, −2 ≤ t ≤ 3
and locate any critical points on its graph.

32

Chapter 1 | Parametric Equations and Polar Coordinates

Example 1.5
Finding a Tangent Line

Find the equation of the tangent line to the curve defined by the equations x(t) = t2 − 3, y(t) = 2t − 1, −3 ≤ t ≤ 4 when t = 2.

Solution First find the slope of the tangent line using Equation 1.1, which means calculating x′ (t) and y′(t):

x′ (t) = 2t y′ (t) = 2.

Next substitute these into the equation:

dy dx

=

dy/dt dx/dt

dy dx

=

2 2t

dy dx

=

1t .

When t = 2,

dy dx

=

12,

so this is the slope of the tangent line. Calculating x(2) and y(2) gives

x(2) = (2)2 − 3 = 1 and y(2) = 2(2) − 1 = 3,

which corresponds to the point (1, 3) on the graph (Figure 1.20). Now use the point-slope form of the equation of a line to find the equation of the tangent line:

y − y0 = m(x − x0)

y − 3 = 12(x − 1)

y−3

=

1 2

x

−

1 2

y

=

1 2

x

+

52.

Figure 1.20 Tangent line to the parabola described by the given parametric equations when t = 2.
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Chapter 1 | Parametric Equations and Polar Coordinates

33

1.5 Find the equation of the tangent line to the curve defined by the equations x(t) = t2 − 4t, y(t) = 2t3 − 6t, −2 ≤ t ≤ 3 when t = 5.

Second-Order Derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function y = f (x) is defined to be the derivative of the first derivative; that is,

d2 y dx2

=

d dx

⎡dy ⎣dx

⎤⎦.

Since

dy dx

=

dy/dt dx/dt

,

we can replace the y on both sides of this equation with ddyx. This gives us

d2 y dx2

=

d ⎛dy⎞ dx⎝dx⎠

=

(d

/dt)⎛⎝dy/d dx/dt

x⎞ ⎠

.

If we know dy/dx as a function of t, then this formula is straightforward to apply.

(1.2)

Example 1.6

Finding a Second Derivative

Calculate the second derivative d2 y/dx2 for the plane curve defined by the parametric equations x(t) = t2 − 3, y(t) = 2t − 1, −3 ≤ t ≤ 4.

Solution

From Example 1.4 we know that

dy dx

=

2 2t

=

1t .

Using Equation 1.2, we obtain

d2 y dx2

=

(d/dt)⎛⎝d

y/d

x⎞ ⎠

dx/dt

=

(d/dt)(1/t) 2t

=

−t −2 2t

=

− 21t3.

1.6 Calculate the second derivative d2 y/dx2 for the plane curve defined by the equations x(t) = t2 − 4t, y(t) = 2t3 − 6t, −2 ≤ t ≤ 3
and locate any critical points on its graph.
Integrals Involving Parametric Equations
Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations x(t) = t − sin t, y(t) = 1 − cos t. Suppose we want to find the area of the shaded region in the following graph.

34

Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.21 Graph of a cycloid with the arch over [0, 2π] highlighted.
To derive a formula for the area under the curve defined by the functions x = x(t), y = y(t), a ≤ t ≤ b,
we assume that x(t) is differentiable and start with an equal partition of the interval a ≤ t ≤ b. Suppose t0 = a < t1 < t2 < ⋯ < tn = b and consider the following graph.

Figure 1.22 Approximating the area under a parametrically defined curve.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is

y⎛⎝x⎛⎝ –t

⎞⎞
i⎠⎠

for some

value

–t i in the ith subinterval, and the width can be calculated as x(ti) − x(ti − 1). Thus the area of the

ith rectangle is given by

Ai

=

y⎛⎝x⎛⎝

–t

⎞⎞
i⎠⎠

⎛⎝x(t i)

−

x(ti

−

1)⎞⎠.

Then a Riemann sum for the area is

∑ An =

n

y⎛⎝x⎛⎝

–t

⎞⎞
i⎠⎠

⎛⎝x(t i)

−

x(ti

−

1)⎞⎠.

i=1

Multiplying and dividing each area by ti − ti − 1 gives

∑ ∑ An

=

i

n =

1

y⎛⎝x⎛⎝

–t

⎞⎞
i⎠⎠

⎛x(ti) ⎝ ti

− −

x(ti − ti − 1

1)⎞⎠(ti

−

ti

−

1)

=

i

n =

1

y⎛⎝x⎛⎝

–t

⎞⎞
i⎠⎠

⎛x(ti) ⎝

− x(ti Δt

−

1)⎞⎠Δt.

Taking the limit as n approaches infinity gives

b
A = nl→im∞An = ∫a y(t)x′ (t) dt.

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Chapter 1 | Parametric Equations and Polar Coordinates
This leads to the following theorem.
Theorem 1.2: Area under a Parametric Curve Consider the non-self-intersecting plane curve defined by the parametric equations
x = x(t), y = y(t), a ≤ t ≤ b and assume that x(t) is differentiable. The area under this curve is given by
b
A = ∫ y(t)x′ (t) dt. a

Example 1.7
Finding the Area under a Parametric Curve

Find the area under the curve of the cycloid defined by the equations x(t) = t − sin t, y(t) = 1 − cos t, 0 ≤ t ≤ 2π.

Solution Using Equation 1.3, we have

b
A = ∫a y(t)x′ (t) dt

2π
= ∫ (1 − cos t)(1 − cos t) dt 0

2π
= ∫ (1 − 2 cos t + cos2 t)dt 0

=

∫ 02π⎛⎝1

−

2

cos

t

+

1

+

cos 2

2t ⎞
⎠

dt

=

∫ 2π⎛3
0 ⎝2

−

2

cos

t

+

cos 2

2t

⎞ ⎠

dt

| =

3t 2

−

2

sin

t

+

sin 2t 4

2π 0

= 3π.

35
(1.3)

1.7 Find the area under the curve of the hypocycloid defined by the equations
x(t) = 3 cos t + cos 3t, y(t) = 3 sin t − sin 3t, 0 ≤ t ≤ π.
Arc Length of a Parametric Curve
In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

36

Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.23 Approximation of a curve by line segments.

Given a plane curve defined by the functions x = x(t), y = y(t), a ≤ t ≤ b, we start by partitioning the interval [a, b]
into n equal subintervals: t0 = a < t1 < t2 < ⋯ < tn = b. The width of each subinterval is given by Δt = (b − a)/n. We can calculate the length of each line segment:

d1 = ⎛⎝x(t1) − x(t0)⎞⎠2 + ⎛⎝y(t1) − y(t0)⎞⎠2 d2 = ⎛⎝x(t2) − x(t1)⎞⎠2 + ⎛⎝y(t2) − y(t1)⎞⎠2 etc.

Then add these up. We let s denote the exact arc length and sn denote the approximation by n line segments:

n

n

∑ ∑ s ≈ sk =

⎛⎝x(tk) − x(tk − 1)⎞⎠2 + ⎛⎝y(tk) − y(tk − 1)⎞⎠2.

k=1

k=1

(1.4)

If we assume that x(t) and y(t) are differentiable functions of t, then the Mean Value Theorem (Introduction to the Applications of Derivatives (http://cnx.org/content/m53602/latest/) ) applies, so in each subinterval [tk − 1, tk] there exist ^t k and ˜t k such that

x(tk) − x(tk − 1) = x′ ⎛⎝^t k⎞⎠(tk − tk − 1) = x′ ⎛⎝^t k⎞⎠Δt

y(tk) − y(tk − 1) = y′ ⎛⎝˜t k⎞⎠(tk − tk − 1) = y′ ⎛⎝˜t k⎞⎠Δt.

Therefore Equation 1.4 becomes

n
s ≈ ∑ sk k=1

∑n
=

⎛⎝x′ ⎛⎝^t k⎞⎠Δt⎞⎠2 + ⎛⎝y′ ⎛⎝˜t k⎞⎠Δt⎞⎠2

k=1

∑n
=

⎛⎝x′

⎛⎝^t

⎞⎞2 k⎠⎠

(Δt)

2

+

⎛⎝y′

⎛⎝˜t

⎞⎞2 k⎠⎠

(Δt)

2

k=1

∑ ⎛ n
=⎜ ⎝k = 1

⎛⎝x′

⎛⎝^t

⎞⎞2 k⎠⎠

+

⎛⎝y′

⎛⎝˜t

k⎞⎠⎞⎠2⎞⎠⎟Δt.

This is a Riemann sum that approximates the arc length over a partition of the interval [a, b]. If we further assume that
the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

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Chapter 1 | Parametric Equations and Polar Coordinates

37

n
∑ s = n l→im∞k = 1 sk

∑ ⎛ n
= n l→im∞⎝⎜k = 1

⎛⎝x′

⎛⎝^t

⎞⎞2 k⎠⎠

+

⎛⎝y′

⎛⎝˜t

k⎞⎠⎞⎠2⎞⎠⎟Δt

b

∫ =

(x′ (t))2 + ⎛⎝y′ (t)⎞⎠2dt.

a

When taking the limit, the values of ^t k and ˜t k are both contained within the same ever-shrinking interval of width Δt, so they must converge to the same value.
We can summarize this method in the following theorem.

Theorem 1.3: Arc Length of a Parametric Curve

Consider the plane curve defined by the parametric equations

x = x(t), y = y(t), t1 ≤ t ≤ t2 and assume that x(t) and y(t) are differentiable functions of t. Then the arc length of this curve is given by

t2
s=∫ t1

⎛dx⎞2 ⎝dt ⎠

+

⎛dy ⎝dt

⎞⎠2dt.

(1.5)

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function y = F(x). Then y(t) = F(x(t)) and the Chain Rule gives y′ (t) = F′ (x(t))x′ (t).
Substituting this into Equation 1.5 gives

s

= ∫ t2 t1

⎛dx ⎝ dt

⎞2 ⎠

+

⎛⎝ddyt ⎞⎠2dt

t2
=∫ t1

⎛dx ⎝ dt

⎞2 ⎠

+

⎛⎝F′

(x)ddxt

⎞⎠2dt

∫ =

t2 t1

⎛dx ⎝ dt

⎞2 ⎠

⎛⎝1

+

(F′

(x))

2⎞⎠dt

t2
= ∫ x′ (t) t1

1

+

⎛dy ⎝dx

⎞⎠2dt.

Here we have assumed that x′ (t) > 0, which is a reasonable assumption. The Chain Rule gives dx = x′ (t) dt, and letting a = x(t1) and b = x(t2) we obtain the formula

b
s = ∫a

1

+

⎛dy ⎝dx

⎞⎠2dx,

which is the formula for arc length obtained in the Introduction to the Applications of Integration (http://cnx.org/ content/m53638/latest/) .

Example 1.8

Finding the Arc Length of a Parametric Curve

Find the arc length of the semicircle defined by the equations

38

Chapter 1 | Parametric Equations and Polar Coordinates

x(t) = 3 cos t, y(t) = 3 sin t, 0 ≤ t ≤ π.

Solution The values t = 0 to t = π trace out the red curve in Figure 1.23. To determine its length, use Equation 1.5:

s

t2
=∫ t1

⎛dx ⎝ dt

⎞2 ⎠

+

⎛dy ⎝dt

⎞⎠2dt

π

= ∫ (−3 sin t)2 + (3 cos t)2dt

0

π

= ∫ 9 sin2 t + 9 cos2 t dt

0

π

∫ = 0 9⎛⎝sin2 t + cos2 t⎞⎠dt

π

=

∫ 0

3dt

=

3t| 0π

=

3π.

Note that the formula for the arc length of a semicircle is πr and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.

Figure 1.24 The arc length of the semicircle is equal to its radius times π.
1.8 Find the arc length of the curve defined by the equations x(t) = 3t2, y(t) = 2t3, 1 ≤ t ≤ 3.
We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can be written as
x(t) = 140t, y(t) = −16t2 + 2t where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions x(t) and y(t) using v as an independent variable, so as to eliminate any confusion with the parameter t:
x(v) = 140v, y(v) = −16v2 + 2v.
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Chapter 1 | Parametric Equations and Polar Coordinates

39

Then we write the arc length formula as follows:

s(t)

t
=∫ 0

⎛dx⎞2 ⎝dv⎠

+

⎛dy ⎝dv

⎞⎠2dv

t
= ∫ 1402 + (−32v + 2)2dv. 0

The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression we can use a formula from Appendix A,

| | ∫

a2

+

u 2du

=

u 2

a2 + u2 + a22ln u +

a2 + u2 + C.

We set a = 140 and u = −32v + 2. This gives du = −32dv,

so

dv =

−

1 32

du.

Therefore

∫ 1402 + (−32v + 2)2dv = − 312∫ a2 + u2du

| | =

−

1 32

⎡(−32v + 2) ⎢2

140 2

⎢ ⎣+

140 2 2

ln

(−32v

+

+ (−32v + 2)2 2) + 1402 + (−32v

+

2) 2

⎤ ⎥ ⎥ ⎦

+

C

and

| | s(t)

=

−

1 32

⎡(−32t ⎣2

+

2)

140 2

+

(−32t

+

2) 2

+

140 2

2

ln

(−32t

+

2)

+

140

2

+

(−32t

+

2)

2

⎤ ⎦

| | +312⎡⎣

1402 + 22 +

140 2

2

ln

2

+

140 2

+

2

2

⎤ ⎦

| | =

⎛t ⎝2

−

312 ⎞⎠

1024t2 − 128t + 19604 −

12425ln (−32t + 2) +

1024t2 − 128t + 19604

+

19604 32

+

1225 4

ln⎛⎝2

+

19604⎞⎠.

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

d dx

∫

x a

f

(u)

du

=

f (x).

Therefore

s′ (t)

=

d dt

⎡⎣s(t)⎤⎦

=

d dt

⎡⎣∫0t

1402 + (−32v + 2)2dv⎤⎦

= 1402 + (−32t + 2)2

= 1024t2 − 128t + 19604 = 2 256t2 − 32t + 4901.

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

40

Chapter 1 | Parametric Equations and Polar Coordinates

s⎛⎝13 ⎞⎠

=

⎛1/3 ⎝2

−

312 ⎞⎠

1024⎛⎝13 ⎞⎠2

−

128⎛⎝13

⎞ ⎠

+

19604

| | −

1225 4

ln

⎛⎝−32⎛⎝13

⎞ ⎠

+

2⎞⎠

+

1024⎛⎝13⎞⎠2 − 128⎛⎝13⎞⎠ + 19604

+

19604 32

+

1225 4

ln⎛⎝2

+

19604⎞⎠

≈ 46.69 feet.

This value is just over three quarters of the way to home plate. The speed of the ball is

s′ ⎛⎝13⎞⎠ = 2

256⎛⎝13

⎞2 ⎠

−

16⎛⎝13 ⎞⎠

+

4901

≈

140.34

ft/s.

This speed translates to approximately 95 mph—a major-league fastball.

Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area (http://cnx.org/content/m53644/latest/) , we derived a formula for finding the surface area of a volume generated by a function y = f (x) from x = a to x = b, revolved around the x-axis:

b

∫ S = 2π

f (x)

1

+

⎛ ⎝

f

′

(x)⎞⎠2dx.

a

We now consider a volume of revolution generated by revolving a parametrically defined curve x = x(t), y = y(t), a ≤ t ≤ b around the x-axis as shown in the following figure.

Figure 1.25 A surface of revolution generated by a parametrically defined curve. The analogous formula for a parametrically defined curve is
b
∫ S = 2π y(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt a
provided that y(t) is not negative on [a, b].
Example 1.9
Finding Surface Area
Find the surface area of a sphere of radius r centered at the origin.
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(1.6)

Chapter 1 | Parametric Equations and Polar Coordinates

41

Solution We start with the curve defined by the equations
x(t) = r cos t, y(t) = r sin t, 0 ≤ t ≤ π. This generates an upper semicircle of radius r centered at the origin as shown in the following graph.

Figure 1.26 A semicircle generated by parametric equations.

When this curve is revolved around the x-axis, it generates a sphere of radius r. To calculate the surface area of the sphere, we use Equation 1.6:

b
∫ S = 2π y(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt a

π
= 2π∫ r sin t (−r sin t)2 + (r cos t)2dt 0

π
= 2π∫ r sin t r2 sin2 t + r2 cos2 t dt 0

π

= 2π∫ r sin t 0

r2 ⎛⎝sin2 t + cos2 t⎞⎠dt

π
= 2π∫ r2 sin t dt 0

= 2πr2(−cos t|0π)

= 2πr2 (−cos π + cos 0)

= 4πr2.

This is, in fact, the formula for the surface area of a sphere.

1.9 Find the surface area generated when the plane curve defined by the equations x(t) = t3, y(t) = t2, 0 ≤ t ≤ 1
is revolved around the x-axis.

42

Chapter 1 | Parametric Equations and Polar Coordinates

1.2 EXERCISES

For the following exercises, each set of parametric equations represents a line. Without eliminating the parameter, find the slope of each line.

62. x = 3 + t, y = 1 − t

63. x = 8 + 2t, y = 1

64. x = 4 − 3t, y = −2 + 6t

65. x = −5t + 7, y = 3t − 1

For the following exercises, determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter.

66. x = 3 sin t, y = 3 cos t,

t

=

π 4

67.

x = cos t,

y

=

8

sin

t,

t

=

π 2

68. x = 2t, y = t3, t = −1

69.

x = t + 1t ,

y

=

t

−

1 t

,

t=1

70. x = t, y = 2t, t = 4

For the following exercises, find all points on the curve that have the given slope.

71. x = 4 cos t, y = 4 sin t, slope = 0.5

72. x = 2 cos t, y = 8 sin t, slope = −1

73.

x = t + 1t ,

y

=

t

−

1 t

,

slope = 1

74. x = 2 + t, y = 2 − 4t, slope = 0

For the following exercises, write the equation of the tangent line in Cartesian coordinates for the given parameter t.

75. x = e t, y = 1 − ln t2, t = 1

76.

x = t ln t,

y

=

sin2

t,

t

=

π 4

77. x = et, y = (t − 1)2, at(1, 1)

78. For x = sin(2t), y = 2 sin t where 0 ≤ t < 2π. Find all values of t at which a horizontal tangent line exists.

79. For x = sin(2t), y = 2 sin t where 0 ≤ t < 2π. Find all values of t at which a vertical tangent line exists.

80. Find all points on the curve x = 4 cos(t), y = 4 sin(t) that have the slope of 12.

81.

Find

dy dx

for x = sin(t), y = cos(t).

82. Find the equation of the tangent line to

x

=

sin(t),

y

=

cos(t)

at

t

=

π 4

.

83. For the curve x = 4t, y = 3t − 2, find the slope and concavity of the curve at t = 3.

84. For the parametric curve whose equation is

x = 4 cos θ, y = 4 sin θ, find the slope and concavity of

the

curve

at

θ

=

π 4

.

85. Find the slope and concavity for the curve whose

equation is

x

=

2 + sec θ,

y

=

1 + 2 tan θ

at

θ

=

π 6

.

86. Find all points on the curve x = t + 4, y = t3 − 3t at which there are vertical and horizontal tangents.

87. Find all points on the curve x = sec θ, y = tan θ at which horizontal and vertical tangents exist.

For the following exercises, find d2 y/dx2.

88. x = t4 − 1, y = t − t2

89. x = sin(πt), y = cos(πt)

90. x = e−t, y = te2t
For the following exercises, find points on the curve at which tangent line is horizontal or vertical.
91. x = t(t2 − 3), y = 3(t2 − 3)

92.

x

=

1

3t +

t

3,

y

=

3t 2 1 + t3

For the following exercises, find dy/dx at the value of the parameter.

93. x = cos t, y = sin t,

t

=

3π 4

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Chapter 1 | Parametric Equations and Polar Coordinates

43

94. x = t, y = 2t + 4, t = 9

95. x = 4 cos(2πs), y = 3 sin(2πs),

s=

−

1 4

For the following exercises, find d2 y/dx2 at the given point without eliminating the parameter.

96.

x

=

1 2

t

2,

y = 13t3,

t=2

97. x = t, y = 2t + 4, t = 1

98. Find t intervals on which the curve x = 3t2, y = t3 − t is concave up as well as concave down.
99. Determine the concavity of the curve x = 2t + ln t, y = 2t − ln t.

100. Sketch and find the area under one arch of the cycloid x = r(θ − sin θ), y = r(1 − cos θ).

101. Find the area bounded by the curve

x

=

cos

t,

y

=

et,

0

≤

t

≤

π 2

and the lines

y=1

and

x = 0.

102. Find the area enclosed by the ellipse x = a cos θ, y = b sin θ, 0 ≤ θ < 2π.

103. Find the area of the region bounded by

x = 2 sin2 θ, y = 2 sin2 θ tan θ,

for

0

≤

θ

≤

π 2

.

For the following exercises, find the area of the regions bounded by the parametric curves and the indicated values of the parameter.

104. x = 2 cot θ, y = 2 sin2 θ, 0 ≤ θ ≤ π

105.

[T]

x = 2a cos t − a cos(2t), y = 2a sin t − a sin(2t), 0 ≤ t < 2π

106. [T] x = a sin(2t), y = b sin(t), 0 ≤ t < 2π (the “hourglass”)

107.

[T]

x = 2a cos t − a sin(2t), y = b sin t, 0 ≤ t < 2π (the

“teardrop”)

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.

108. x = 4t + 3, y = 3t − 2, 0 ≤ t ≤ 2

109.

x

=

1 3

t

3,

y

=

1 2

t

2,

0≤t≤1

110.

x = cos(2t),

y = sin(2t),

0

≤

t

≤

π 2

111. x = 1 + t2, y = (1 + t)3, 0 ≤ t ≤ 1

112.

x = et cos t,

y = et sin t,

0

≤

t

≤

π 2

(express

answer as a decimal rounded to three places)

113. x = a cos3 θ, y = a sin3 θ on the interval [0, 2π) (the hypocycloid)
114. Find the length of one arch of the cycloid x = 4(t − sin t), y = 4(1 − cos t).

115. Find the distance traveled by a particle with position (x, y) as t varies in the given time interval:
x = sin2 t, y = cos2 t, 0 ≤ t ≤ 3π.

116. Find the length of one arch of the cycloid x = θ − sin θ, y = 1 − cos θ.

117. Show that the total x = 4 sin θ, y = 3 cos θ
π/2
L = 16∫ 1 − e2 sin2 θ dθ, 0
c = a2 − b2.

length of the ellipse is
where e = ac and

118. Find the length of the curve x = et − t, y = 4et/2, −8 ≤ t ≤ 3.

For the following exercises, find the area of the surface obtained by rotating the given curve about the x-axis.

119. x = t3, y = t2, 0 ≤ t ≤ 1

120.

x = a cos3 θ,

y = a sin3 θ,

0

≤

θ

≤

π 2

121. [T] Use a CAS to find the area of the surface

generated by rotating

x

=

t

+

t 3,

y

=

t

−

1 t2

,

1

≤

t

≤

2

about the x-axis. (Answer to three decimal places.)

122. Find the surface area obtained by rotating x = 3t2, y = 2t3, 0 ≤ t ≤ 5 about the y-axis.

123. Find the area of the surface generated by revolving x = t2, y = 2t, 0 ≤ t ≤ 4 about the x-axis.

124. Find the surface area generated by revolving x = t2, y = 2t2, 0 ≤ t ≤ 1 about the y-axis.

44

Chapter 1 | Parametric Equations and Polar Coordinates

1.3 | Polar Coordinates
Learning Objectives
1.3.1 Locate points in a plane by using polar coordinates. 1.3.2 Convert points between rectangular and polar coordinates. 1.3.3 Sketch polar curves from given equations. 1.3.4 Convert equations between rectangular and polar coordinates. 1.3.5 Identify symmetry in polar curves and equations.
The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a one-to-one mapping from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.
Defining Polar Coordinates
To find the coordinates of a point in the polar coordinate system, consider Figure 1.27. The point P has Cartesian coordinates (x, y). The line segment connecting the origin to the point P measures the distance from the origin to P and has length r. The angle between the positive x -axis and the line segment has measure θ. This observation suggests a natural correspondence between the coordinate pair (x, y) and the values r and θ. This correspondence is the basis of the polar coordinate system. Note that every point in the Cartesian plane has two values (hence the term ordered pair) associated with it. In the polar coordinate system, each point also two values associated with it: r and θ.

Figure 1.27 An arbitrary point in the Cartesian plane.

Using right-triangle trigonometry, the following equations are true for the point P:

cos θ

=

x r

so x

=

r cos θ

sin

θ

=

y r

so

y

=

r

sin

θ.

Furthermore,

r2

=

x2

+

y2

and

tan

θ

=

y x

.

Each point (x, y) in the Cartesian coordinate system can therefore be represented as an ordered pair (r, θ) in the polar
coordinate system. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate. Every point in the plane can be represented in this form.

Note that the equation tan θ = y/x has an infinite number of solutions for any ordered pair (x, y). However, if we restrict the solutions to values between 0 and 2π then we can assign a unique solution to the quadrant in which the original point (x, y) is located. Then the corresponding value of r is positive, so r2 = x2 + y2.

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Chapter 1 | Parametric Equations and Polar Coordinates

45

Theorem 1.4: Converting Points between Coordinate Systems

Given a point P in the plane with Cartesian coordinates (x, y) and polar coordinates (r, θ), the following conversion formulas hold true:

x = r cos θ and y = r sin θ, r2 = x2 + y2 and tan θ = yx.

(1.7) (1.8)

These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.

Example 1.10
Converting between Rectangular and Polar Coordinates

Convert each of the following points into polar coordinates. a. (1, 1) b. (−3, 4) c. (0, 3) d. (5 3, −5)
Convert each of the following points into rectangular coordinates. e. (3, π/3) f. (2, 3π/2) g. (6, −5π/6)

Solution a. Use x = 1 and y = 1 in Equation 1.8:

r2 = x2 + y2

tan θ

=

y x

= 12 + 12 and

=

1 1

=

1

r= 2

θ

=

π 4

.

Therefore this point can be represented as

⎛ ⎝

2,

π⎞ 4⎠

in polar coordinates.

b. Use x = −3 and y = 4 in Equation 1.8:

tan θ

=

y x

r2 = x2 + y2 = (−3)2 + (4)2 and
r=5

=

−

4 3

θ = −arctan⎛⎝43⎞⎠

≈ 2.21.

Therefore this point can be represented as (5, 2.21) in polar coordinates.

46

Chapter 1 | Parametric Equations and Polar Coordinates

c. Use x = 0 and y = 3 in Equation 1.8:

r2 = x2 + y2 = (3)2 + (0)2 = 9+0

and

tan θ

=

y x

= 30.

r=3

Direct application of the second equation leads to division by zero. Graphing the point (0, 3) on the

rectangular coordinate system reveals that the point is located on the positive y-axis. The angle between

the positive x-axis and the positive y-axis is

π2 .

Therefore this point can be represented as

⎛⎝3,

π⎞ 2⎠

in polar

coordinates.

d. Use x = 5 3 and y = −5 in Equation 1.8:

r2 = x2 + y2

tan θ

=

y x

= ⎛⎝5 3⎞⎠2 + (−5)2 and = 75 + 25 r = 10

=

−5 53

=

−

3 3

θ

=

−

π 6

.

Therefore this point can be represented as

⎛⎝10,

−

π⎞ 6⎠

in polar coordinates.

e.

Use

r

=3

and

θ=

π 3

in Equation 1.7:

x = r cos θ

y = r sin θ

=

3

cos⎛⎝π3

⎞ ⎠

and

=

3

sin⎛⎝π3

⎞ ⎠

=

3⎛⎝12 ⎞⎠

=

3 2

=

3⎛⎝

3 2

⎞ ⎠

=

33 2

.

Therefore this point can be represented as ⎛⎝32, 323⎞⎠ in rectangular coordinates.

f.

Use

r

=2

and

θ=

3π 2

in Equation 1.7:

x = r cos θ

y = r sin θ

=

2

cos⎛⎝32π

⎞ ⎠

and

=

2

sin⎛⎝32π

⎞ ⎠

= 2(0) = 0

= 2(−1) = −2.

Therefore this point can be represented as (0, −2) in rectangular coordinates.

g.

Use

r

=6

and

θ=

−

5π 6

in Equation 1.7:

x = r cos θ

y = r sin θ

=

6

cos⎛⎝−

5π 6

⎞ ⎠

= 6⎛⎝− 23⎞⎠

and

=

6

sin⎛⎝−

5π 6

⎞ ⎠

=

6⎛⎝−

1 2

⎞ ⎠

= −3 3

= −3.

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Chapter 1 | Parametric Equations and Polar Coordinates

47

Therefore this point can be represented as ⎛⎝−3 3, −3⎞⎠ in rectangular coordinates.

1.10

Convert

(−8,

−8)

into

polar

coordinates

and

⎛⎝4,

2π 3

⎞ ⎠

into

rectangular

coordinates.

The polar representation of a point is not unique. For example, the polar coordinates

⎛⎝2,

π⎞ 3⎠

and

⎛⎝2,

7π 3

⎞ ⎠

both represent the

point ⎛⎝1, 3⎞⎠ in the rectangular system. Also, the value of r can be negative. Therefore, the point with polar coordinates

⎛⎝−2,

4π 3

⎞ ⎠

also represents the point

⎛⎝1,

3⎞⎠ in the rectangular system, as we can see by using Equation 1.8:

x = r cos θ

y = r sin θ

=

−2

cos⎛⎝43π

⎞ ⎠

and

=

−2

sin⎛⎝43π

⎞ ⎠

= −2⎛⎝− 12⎞⎠ = 1

=

−2⎛⎝−

3 2

⎞ ⎠

=

3.

Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane has only one representation in the rectangular coordinate system.

Note that the polar representation of a point in the plane also has a visual interpretation. In particular, r is the directed distance that the point lies from the origin, and θ measures the angle that the line segment from the origin to the point makes with the positive x -axis. Positive angles are measured in a counterclockwise direction and negative angles are measured in a clockwise direction. The polar coordinate system appears in the following figure.

Figure 1.28 The polar coordinate system.
The line segment starting from the center of the graph going to the right (called the positive x-axis in the Cartesian system) is the polar axis. The center point is the pole, or origin, of the coordinate system, and corresponds to r = 0. The innermost circle shown in Figure 1.28 contains all points a distance of 1 unit from the pole, and is represented by the equation r = 1.

48

Chapter 1 | Parametric Equations and Polar Coordinates

Then r = 2 is the set of points 2 units from the pole, and so on. The line segments emanating from the pole correspond to fixed angles. To plot a point in the polar coordinate system, start with the angle. If the angle is positive, then measure the angle from the polar axis in a counterclockwise direction. If it is negative, then measure it clockwise. If the value of r is positive, move that distance along the terminal ray of the angle. If it is negative, move along the ray that is opposite the terminal ray of the given angle.
Example 1.11
Plotting Points in the Polar Plane

Plot each of the following points on the polar plane.

a.

⎛⎝2,

π⎞ 4⎠

b.

⎛⎝−3,

2π 3

⎞ ⎠

c.

⎛⎝4,

5π ⎞ 4⎠

Solution The three points are plotted in the following figure.

Figure 1.29 Three points plotted in the polar coordinate system.

1.11

Plot

⎛⎝4,

5π 3

⎞ ⎠

and

⎛⎝−3,

−

7π 2

⎞ ⎠

on the polar plane.

Polar Curves
Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function y = f (x) and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function r = f (θ).

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Chapter 1 | Parametric Equations and Polar Coordinates

49

The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable (θ in this case) and calculate the corresponding values of the dependent variable r. This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.
Problem-Solving Strategy: Plotting a Curve in Polar Coordinates 1. Create a table with two columns. The first column is for θ, and the second column is for r. 2. Create a list of values for θ. 3. Calculate the corresponding r values for each θ. 4. Plot each ordered pair (r, θ) on the coordinate axes.
5. Connect the points and look for a pattern.

Watch this video (http://www.openstaxcollege.org/l/20_polarcurves) for more information on sketching polar curves.

Example 1.12
Graphing a Function in Polar Coordinates
Graph the curve defined by the function r = 4 sin θ. Identify the curve and rewrite the equation in rectangular coordinates.
Solution Because the function is a multiple of a sine function, it is periodic with period 2π, so use values for θ between 0 and 2π. The result of steps 1–3 appear in the following table. Figure 1.30 shows the graph based on this table.

50

Chapter 1 | Parametric Equations and Polar Coordinates

θ

r = 4 sin θ

0

0

π 6

2

π 4

2 2 ≈ 2.8

π 3

2 3 ≈ 3.4

π 2

4

2π 3

2 3 ≈ 3.4

3π

2 2 ≈ 2.8

4

5π

2

6

θ

r = 4 sin θ

π

0

7π

−2

6

5π 4

−2 2 ≈ −2.8

4π

−2 3 ≈ −3.4

3

3π

4

2

5π 3

−2 3 ≈ −3.4

7π 4

−2 2 ≈ −2.8

11π

−2

6

2π

0

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Chapter 1 | Parametric Equations and Polar Coordinates

51

Figure 1.30 The graph of the function r = 4 sin θ is a circle.

This is the graph of a circle. The equation r = 4 sin θ can be converted into rectangular coordinates by first multiplying both sides by r. This gives the equation r2 = 4r sin θ. Next use the facts that r2 = x2 + y2 and
y = r sin θ. This gives x2 + y2 = 4y. To put this equation into standard form, subtract 4y from both sides of the equation and complete the square:

x2 + y2 − 4y = 0

x2 + ⎛⎝y2 − 4y⎞⎠ = 0

x2 + ⎛⎝y2 − 4y + 4⎞⎠ = 0 + 4

x2

+

y⎛
⎝

−

2⎞ ⎠

2

=

4.

This is the equation of a circle with radius 2 and center (0, 2) in the rectangular coordinate system.

1.12 Create a graph of the curve defined by the function r = 4 + 4 cos θ.

The graph in Example 1.12 was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in Equation 1.8. Example 1.14 gives some more examples of functions for transforming from polar to rectangular coordinates.
Example 1.13
Transforming Polar Equations to Rectangular Coordinates

Rewrite each of the following equations in rectangular coordinates and identify the graph.

a.

θ

=

π 3

52

Chapter 1 | Parametric Equations and Polar Coordinates

b. r = 3 c. r = 6 cos θ − 8 sin θ

Solution

a. Take the tangent of both sides. This gives tan θ = tan(π/3) = 3. Since tan θ = y/x we can replace the
left-hand side of this equation by y/x. This gives y/x = 3, which can be rewritten as y = x 3. This
is the equation of a straight line passing through the origin with slope 3. In general, any polar equation of the form θ = K represents a straight line through the pole with slope equal to tan K.

b. First, square both sides of the equation. This gives r2 = 9. Next replace r2 with x2 + y2. This gives

the equation x2 + y2 = 9, which is the equation of a circle centered at the origin with radius 3. In

general, any polar equation of the form r = k where k is a positive constant represents a circle of radius

k centered at the origin. (Note: when squaring both sides of an equation it is possible to introduce new

points unintentionally. This should always be taken into consideration. However, in this case we do not

introduce new points. For example,

⎛⎝−3,

π⎞ 3⎠

is the same point as

⎛⎝3,

4π 3

⎞⎠.)

c. Multiply both sides of the equation by r. This leads to r2 = 6r cos θ − 8r sin θ. Next use the formulas

r2 = x2 + y2, x = r cos θ, y = r sin θ.

This gives

r2 = 6(r cos θ) − 8(r sin θ) x2 + y2 = 6x − 8y.

To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.

x2 + y2 = 6x − 8y

x2 − 6x + y2 + 8y = 0

⎛⎝x2 − 6x⎞⎠ + ⎛⎝y2 + 8y⎞⎠ = 0

⎛⎝x2 − 6x + 9⎞⎠ + ⎛⎝y2 + 8y + 16⎞⎠ = 9 + 16

(x

−

3)

2

+

y⎛
⎝

+

4⎞ ⎠

2

=

25.

This is the equation of a circle with center at (3, −4) and radius 5. Notice that the circle passes through the origin since the center is 5 units away.

1.13 Rewrite the equation r = sec θ tan θ in rectangular coordinates and identify its graph.
We have now seen several examples of drawing graphs of curves defined by polar equations. A summary of some common curves is given in the tables below. In each equation, a and b are arbitrary constants.

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Chapter 1 | Parametric Equations and Polar Coordinates

53

Figure 1.31

54

Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.32 A cardioid is a special case of a limaçon (pronounced “lee-mah-son”), in which a = b or a = −b. The rose is a very interesting curve. Notice that the graph of r = 3 sin 2θ has four petals. However, the graph of r = 3 sin 3θ has three petals as shown.
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Chapter 1 | Parametric Equations and Polar Coordinates

55

Figure 1.33 Graph of r = 3 sin 3θ.
If the coefficient of θ is even, the graph has twice as many petals as the coefficient. If the coefficient of θ is odd, then the number of petals equals the coefficient. You are encouraged to explore why this happens. Even more interesting graphs emerge when the coefficient of θ is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends where it started (Figure 1.34(a)). However, if the coefficient is irrational, then the curve never closes (Figure 1.34(b)). Although it may appear that the curve is closed, a closer examination reveals that the petals just above the positive x axis are slightly thicker. This is because the petal does not quite match up with the starting point.

Figure 1.34 Polar rose graphs of functions with (a) rational coefficient and (b) irrational coefficient. Note that the rose in part (b) would actually fill the entire circle if plotted in full.
Since the curve defined by the graph of r = 3 sin(πθ) never closes, the curve depicted in Figure 1.34(b) is only a partial depiction. In fact, this is an example of a space-filling curve. A space-filling curve is one that in fact occupies a twodimensional subset of the real plane. In this case the curve occupies the circle of radius 3 centered at the origin.
Example 1.14

56

Chapter 1 | Parametric Equations and Polar Coordinates

Chapter Opener: Describing a Spiral
Recall the chambered nautilus introduced in the chapter opener. This creature displays a spiral when half the outer shell is cut away. It is possible to describe a spiral using rectangular coordinates. Figure 1.35 shows a spiral in rectangular coordinates. How can we describe this curve mathematically?

Figure 1.35 How can we describe a spiral graph mathematically?

Solution
As the point P travels around the spiral in a counterclockwise direction, its distance d from the origin increases. Assume that the distance d is a constant multiple k of the angle θ that the line segment OP makes with the positive x-axis. Therefore d(P, O) = kθ, where O is the origin. Now use the distance formula and some trigonometry:

d(P, O) = kθ

(x

−

0) 2

+

y⎛
⎝

−

0⎞⎠2

=

k

arctan⎛⎝yx

⎞ ⎠

x2 + y2

=

k

arctan⎛⎝yx

⎞ ⎠

arctan⎛⎝yx⎞⎠ =

x2 + y2 k

y

=

⎛ x tan⎜
⎝

x

2

+ k

y

2

⎞ ⎟. ⎠

Although this equation describes the spiral, it is not possible to solve it directly for either x or y. However, if we use polar coordinates, the equation becomes much simpler. In particular, d(P, O) = r, and θ is the second
coordinate. Therefore the equation for the spiral becomes r = kθ. Note that when θ = 0 we also have r = 0,
so the spiral emanates from the origin. We can remove this restriction by adding a constant to the equation. Then the equation for the spiral becomes r = a + kθ for arbitrary constants a and k. This is referred to as an Archimedean spiral, after the Greek mathematician Archimedes.

Another type of spiral is the logarithmic spiral, described by the function r = a · bθ. A graph of the function

r

=

1.2⎛⎝1.25

θ⎞ ⎠

is

given

in

Figure

1.36.

This

spiral

describes

the

shell

shape

of

the

chambered

nautilus.

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Chapter 1 | Parametric Equations and Polar Coordinates

57

Figure 1.36 A logarithmic spiral is similar to the shape of the chambered nautilus shell. (credit: modification of work by Jitze Couperus, Flickr)
Suppose a curve is described in the polar coordinate system via the function r = f (θ). Since we have conversion formulas from polar to rectangular coordinates given by
x = r cos θ y = r sin θ,
it is possible to rewrite these formulas using the function x = f (θ) cos θ y = f (θ) sin θ.
This step gives a parameterization of the curve in rectangular coordinates using θ as the parameter. For example, the spiral formula r = a + bθ from Figure 1.31 becomes
x = (a + bθ) cos θ y = (a + bθ) sin θ.
Letting θ range from −∞ to ∞ generates the entire spiral.
Symmetry in Polar Coordinates
When studying symmetry of functions in rectangular coordinates (i.e., in the form y = f (x)), we talk about symmetry with respect to the y-axis and symmetry with respect to the origin. In particular, if f (−x) = f (x) for all x in the domain of f , then f is an even function and its graph is symmetric with respect to the y-axis. If f (−x) = − f (x) for all x in the domain of f , then f is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.
Theorem 1.5: Symmetry in Polar Curves and Equations Consider a curve generated by the function r = f (θ) in polar coordinates.

58

Chapter 1 | Parametric Equations and Polar Coordinates

i. The curve is symmetric about the polar axis if for every point (r, θ) on the graph, the point (r, −θ) is also on the graph. Similarly, the equation r = f (θ) is unchanged by replacing θ with −θ.

ii. The curve is symmetric about the pole if for every point (r, θ) on the graph, the point (r, π + θ) is also on the graph. Similarly, the equation r = f (θ) is unchanged when replacing r with −r, or θ with π + θ.

iii.

The curve is symmetric about the vertical line

θ

=

π 2

if for every point

(r, θ)

on the graph, the point

(r, π − θ) is also on the graph. Similarly, the equation r = f (θ) is unchanged when θ is replaced by π − θ.

The following table shows examples of each type of symmetry.

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Chapter 1 | Parametric Equations and Polar Coordinates

59

Example 1.15
Using Symmetry to Graph a Polar Equation

Find the symmetry of the rose defined by the equation r = 3 sin(2θ) and create a graph.

Solution Suppose the point (r, θ) is on the graph of r = 3 sin(2θ).
i. To test for symmetry about the polar axis, first try replacing θ with −θ. This gives r = 3 sin(2(−θ)) = −3 sin(2θ). Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing r with −r and θ with π − θ yields
−r = 3 sin(2(π − θ)) −r = 3 sin(2π − 2θ) −r = 3 sin(−2θ) −r = −3 sin 2θ.

Multiplying both sides of this equation by −1 gives r = 3 sin 2θ, which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.
ii. To test for symmetry with respect to the pole, first replace r with −r, which yields −r = 3 sin(2θ). Multiplying both sides by −1 gives r = −3 sin(2θ), which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing θ with θ + π gives
r = 3 sin(2(θ + π)) = 3 sin(2θ + 2π) = 3(sin 2θ cos 2π + cos 2θ sin 2π) = 3 sin 2θ.

Since this agrees with the original equation, the graph is symmetric about the pole.

iii.

To test for

symmetry with respect to the vertical line

θ

=

π 2

,

first replace both r with −r and θ with

−θ.

−r = 3 sin(2(−θ)) −r = 3 sin(−2θ) −r = −3 sin 2θ.

Multiplying both sides of this equation by −1 gives r = 3 sin 2θ, which is the original equation.

Therefore

the

graph

is

symmetric

about

the

vertical

line

θ

=

π 2

.

This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of θ between 0 and π/2 and then reflect the resulting graph.

60

Chapter 1 | Parametric Equations and Polar Coordinates

θ

r

0

0

π 6

33 2

≈

2.6

π 4

3

π 3

33 2

≈

2.6

π 2

0

This gives one petal of the rose, as shown in the following graph.

Figure 1.37 The graph of the equation between θ = 0 and θ = π/2. Reflecting this image into the other three quadrants gives the entire graph as shown.
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Chapter 1 | Parametric Equations and Polar Coordinates

61

Figure 1.38 The entire graph of the equation is called a fourpetaled rose.
1.14 Determine the symmetry of the graph determined by the equation r = 2 cos(3θ) and create a graph.

62

Chapter 1 | Parametric Equations and Polar Coordinates

1.3 EXERCISES

In the following exercises, plot the point whose polar coordinates are given by first constructing the angle θ and
then marking off the distance r along the ray.

125.

⎛⎝3,

π⎞ 6⎠

126.

⎛⎝−2,

5π 3

⎞ ⎠

127.

⎛⎝0,

7π ⎞ 6⎠

128.

⎛⎝−4,

3π 4

⎞ ⎠

129.

⎛⎝1,

π⎞ 4⎠

130.

⎛⎝2,

5π 6

⎞ ⎠

131.

⎛⎝1,

π⎞ 2⎠

For the following exercises, consider the polar graph below. Give two sets of polar coordinates for each point.

132. Coordinates of point A. 133. Coordinates of point B. 134. Coordinates of point C. 135. Coordinates of point D. For the following exercises, the rectangular coordinates of a point are given. Find two sets of polar coordinates for the

point in (0, 2π]. Round to three decimal places.

136. (2, 2)

137. (3, −4) (3, −4)

138. (8, 15)

139. (−6, 8)

140. (4, 3)

141. ⎛⎝3, − 3⎞⎠

For the following exercises, find rectangular coordinates for the given point in polar coordinates.

142.

⎛⎝2,

5π ⎞ 4⎠

143.

⎛⎝−2,

π⎞ 6⎠

144.

⎛⎝5,

π⎞ 3⎠

145.

⎛⎝1,

7π ⎞ 6⎠

146.

⎛⎝−3,

3π 4

⎞ ⎠

147.

⎛⎝0,

π⎞ 2⎠

148. (−4.5, 6.5)

For the following exercises, determine whether the graphs of the polar equation are symmetric with respect to the x
-axis, the y -axis, or the origin.

149. r = 3 sin(2θ)

150. r2 = 9 cos θ

151.

r

=

cos⎛⎝5θ

⎞ ⎠

152. r = 2 sec θ

153. r = 1 + cos θ

For the following exercises, describe the graph of each polar equation. Confirm each description by converting into a rectangular equation.

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Chapter 1 | Parametric Equations and Polar Coordinates

63

154. r = 3

155.

θ

=

π 4

156. r = sec θ

157. r = csc θ

For the following exercises, convert the rectangular equation to polar form and sketch its graph.

158. x2 + y2 = 16

159. x2 − y2 = 16

160. x = 8
For the following exercises, convert the rectangular equation to polar form and sketch its graph.
161. 3x − y = 2

162. y2 = 4x

For the following exercises, convert the polar equation to rectangular form and sketch its graph. 163. r = 4 sin θ 164. r = 6 cos θ 165. r = θ 166. r = cot θ csc θ For the following exercises, sketch a graph of the polar equation and identify any symmetry. 167. r = 1 + sin θ 168. r = 3 − 2 cos θ 169. r = 2 − 2 sin θ 170. r = 5 − 4 sin θ 171. r = 3 cos(2θ)
172. r = 3 sin(2θ)
173. r = 2 cos(3θ)

174.

r

=

3

cos⎛⎝2θ

⎞ ⎠

175. r2 = 4 cos(2θ)

176. r2 = 4 sin θ

177. r = 2θ

178. [T] The graph of r = 2 cos(2θ)sec(θ). is called a
strophoid. Use a graphing utility to sketch the graph, and, from the graph, determine the asymptote.

179. [T] Use a graphing utility and sketch the graph of

r

=

2

sin

θ

6 −

3

cos

θ.

180.

[T]

Use

a

graphing

utility

to

graph

r

=

1

−

1 cos

θ.

181.

[T] Use technology to graph

r = esin(θ) − 2 cos(4θ).

182.

[T] Use technology

to plot

r

=

sin⎛⎝37θ

⎞ ⎠

(use the

interval 0 ≤ θ ≤ 14π).

183. Without using technology, sketch the polar curve

θ

=

2π 3

.

184. [T] Use a graphing utility to plot r = θ sin θ for −π ≤ θ ≤ π.

185. [T] Use technology to plot r = e−0.1θ for −10 ≤ θ ≤ 10.

186. [T] There is a curve known as the “Black Hole.” Use technology to plot r = e−0.01θ for −100 ≤ θ ≤ 100.

187. [T] Use the results of the preceding two problems to explore the graphs of r = e−0.001θ and r = e−0.0001θ for |θ| > 100.

64

Chapter 1 | Parametric Equations and Polar Coordinates

1.4 | Area and Arc Length in Polar Coordinates

Learning Objectives
1.4.1 Apply the formula for area of a region in polar coordinates. 1.4.2 Determine the arc length of a polar curve.

In the rectangular coordinate system, the definite integral provides a way to calculate the area under a curve. In particular, if we have a function y = f (x) defined from x = a to x = b where f (x) > 0 on this interval, the area between the curve

b
and the x-axis is given by A = ∫a f (x) dx. This fact, along with the formula for evaluating this integral, is summarized in

b

the Fundamental Theorem of Calculus. Similarly, the arc length of this curve is given by L = ∫

1

+

⎛ ⎝

f

′

(x)⎞⎠2d

x.

In this

a

section, we study analogous formulas for area and arc length in the polar coordinate system.

Areas of Regions Bounded by Polar Curves

We have studied the formulas for area under a curve defined in rectangular coordinates and parametrically defined curves. Now we turn our attention to deriving a formula for the area of a region bounded by a polar curve. Recall that the proof of the Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by using rectangles. For polar curves we use the Riemann sum again, but the rectangles are replaced by sectors of a circle.

Consider a curve defined by the function r = f (θ), where α ≤ θ ≤ β. Our first step is to partition the interval [α, β] into n equal-width subintervals. The width of each subinterval is given by the formula Δθ = (β − α)/n, and the ith partition point θi is given by the formula θi = α + iΔθ. Each partition point θ = θi defines a line with slope tanθi passing through the pole as shown in the following graph.

Figure 1.39 A partition of a typical curve in polar coordinates.
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Chapter 1 | Parametric Equations and Polar Coordinates

65

The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using a geometric formula. The area of each sector is then used to approximate the area between successive line segments. We then sum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the total area. The formula for the area of a sector of a circle is illustrated in the following figure.

Figure 1.40 The area of a sector of a circle is given by A = 12θr2.

Recall that the area of a circle is A = πr2. When measuring angles in radians, 360 degrees is equal to 2π radians.

Therefore a fraction of a circle can be measured by the central angle

θ.

The fraction of the circle is given by

θ 2π

,

so the

area of the sector is this fraction multiplied by the total area:

A

=

⎛θ ⎝2π

⎞ ⎠

πr

2

=

1 2

θr

2.

Since the radius of a typical sector in Figure 1.39 is given by ri = f ⎛⎝θi⎞⎠, the area of the ith sector is given by

Ai

=

1 2

(Δθ)⎛⎝

f

⎛⎝θ

⎞⎞
i⎠⎠

2.

Therefore a Riemann sum that approximates the area is given by

n

n

∑ ∑ An

=

i

=

1

Ai

≈

i

=

1

1 2

(Δθ)⎛⎝

f

⎛⎝θ

i⎞⎠⎞⎠2.

We take the limit as n → ∞ to get the exact area:

β

∫ A

=

n l→im∞An

=

1 2

⎛ ⎝

f

(θ)⎞⎠

2

dθ.

α

This gives the following theorem.

Theorem 1.6: Area of a Region Bounded by a Polar Curve

Suppose f is continuous and nonnegative on the interval α ≤ θ ≤ β with 0 < β − α ≤ 2π. The area of the region bounded by the graph of r = f (θ) between the radial lines θ = α and θ = β is

β

β

∫ ∫ A

=

1 2

α

⎡ ⎣

f

(θ)⎤⎦

2

dθ

=

1 2

r2 dθ.
α

(1.9)

66

Chapter 1 | Parametric Equations and Polar Coordinates

Example 1.16
Finding an Area of a Polar Region
Find the area of one petal of the rose defined by the equation r = 3 sin(2θ). Solution The graph of r = 3 sin(2θ) follows.

Figure 1.41 The graph of r = 3 sin(2θ).

When θ = 0 we have r = 3 sin(2(0)) = 0. The next value for which r = 0 is θ = π/2. This can be seen by solving the equation 3 sin(2θ) = 0 for θ. Therefore the values θ = 0 to θ = π/2 trace out the first petal of the rose. To find the area inside this petal, use Equation 1.9 with f (θ) = 3 sin(2θ), α = 0, and β = π/2:

β

∫ A

=

1 2

⎡ ⎣

f

(θ)⎤⎦

2

dθ

α

∫ =

1 2

π/2

3⎡
⎣

sin(2θ)⎤⎦

2

dθ

0

=

1 2

∫

π/2
9
0

sin2 (2θ)

dθ.

To evaluate this integral, use the formula sin2 α = (1 − cos(2α))/2 with α = 2θ:

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Chapter 1 | Parametric Equations and Polar Coordinates

67

A

=

1 2

π/2
∫9 0

sin2 (2θ)

dθ

=

9 2

π/2
∫ 0

(1

−

c2os(4θ))dθ

=

9 4

⎛ π/2
⎜∫ 1
⎝0

−

cos(4θ)

⎞ dθ⎟
⎠

=

94⎛⎝θ

−

sin(4θ)⎞π/2 4 ⎠0

=

9 4

⎛π ⎝2

−

sin 2π 4

⎞ ⎠

−

9 4

⎛⎝0

−

sin 4(0)⎞ 4⎠

=

9π 8

.

1.15 Find the area inside the cardioid defined by the equation r = 1 − cos θ.

Example 1.16 involved finding the area inside one curve. We can also use Area of a Region Bounded by a Polar Curve to find the area between two polar curves. However, we often need to find the points of intersection of the curves and determine which function defines the outer curve or the inner curve between these two points.
Example 1.17
Finding the Area between Two Polar Curves
Find the area outside the cardioid r = 2 + 2 sin θ and inside the circle r = 6 sin θ.
Solution First draw a graph containing both curves as shown.

Figure 1.42 The region between the curves r = 2 + 2 sin θ and r = 6 sin θ.

68

Chapter 1 | Parametric Equations and Polar Coordinates

To determine the limits of integration, first find the points of intersection by setting the two functions equal to each other and solving for θ:

6 sin θ = 2 + 2 sin θ 4 sin θ = 2
sin θ = 12.

This gives the solutions

θ

=

π 6

and

θ=

5π 6

,

which are the limits of integration. The circle r = 3 sin θ is the

red graph, which is the outer function, and the cardioid r = 2 + 2 sin θ is the blue graph, which is the inner

function. To calculate the area between the curves, start with the area inside the circle between

θ

=

π 6

and

θ

=

5π 6

,

then subtract the area inside the cardioid between

θ=

π 6

and

θ=

5π 6

:

A = circle − cardioid

=

∫1
2

5π/6
[6
π/6

sin

θ]2 dθ

−

∫1
2

5π/6
[2
π/6

+

2

sin

θ]2 dθ

=

∫1
2

5π/6
36
π/6

sin2 θ

dθ

−

∫1
2

5π/6
4
π/6

+

8

sin

θ

+

4

sin2 θ

dθ

=

18∫ 5π/61 π/6

−

c2os(2θ)dθ

−

5π/6
2∫ 1 π/6

+

2

sin

θ

+

1

−

cos(2θ) 2

dθ

=

9⎡⎣θ

−

sin(2θ) ⎤5π/6 2 ⎦π/6

−

2⎡⎣32θ

−

2

cos

θ

−

sin(2θ) ⎤5π/6 4 ⎦π/6

=

9⎛⎝56π

−

sin

2(5π/6)⎞ 2⎠

−

9⎛⎝π6

−

sin

2(π/6)⎞ 2⎠

−⎛⎝3⎛⎝56π

⎞ ⎠

−

4

cos

5π 6

−

sin

2(5π/6)⎞ 2⎠

+

⎛⎝3⎛⎝π6

⎞ ⎠

−

4

cos

π 6

−

sin

2(π/6)⎞ 2⎠

= 4π.

1.16 Find the area inside the circle r = 4 cos θ and outside the circle r = 2.

In Example 1.17 we found the area inside the circle and outside the cardioid by first finding their intersection points.

Notice that solving the equation directly for

θ

yielded two solutions:

θ=

π 6

and

θ=

5π 6

.

However, in the graph there are

three intersection points. The third intersection point is the origin. The reason why this point did not show up as a solution

is because the origin is on both graphs but for different values of θ. For example, for the cardioid we get

2 + 2 sin θ = 0 sin θ = −1,

so the values for

θ

that solve this equation are

θ

=

3π 2

+ 2nπ,

where n is any integer. For the circle we get

6 sin θ = 0.
The solutions to this equation are of the form θ = nπ for any integer value of n. These two solution sets have no points in common. Regardless of this fact, the curves intersect at the origin. This case must always be taken into consideration.

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Chapter 1 | Parametric Equations and Polar Coordinates

69

Arc Length in Polar Curves
Here we derive a formula for the arc length of a curve defined in polar coordinates. In rectangular coordinates, the arc length of a parameterized curve ⎛⎝x(t), y(t)⎞⎠ for a ≤ t ≤ b is given by

b
L=∫ a

⎛dx⎞2 ⎝dt ⎠

+

⎛dy ⎝dt

⎞⎠2dt.

In polar coordinates we define the curve by the equation r = f (θ), where α ≤ θ ≤ β. In order to adapt the arc length formula for a polar curve, we use the equations

x = r cos θ = f (θ) cos θ and y = r sin θ = f (θ) sin θ,

and we replace the parameter t by θ. Then

dx dθ

=

f ′ (θ) cos θ −

f (θ) sin θ

dy dθ

=

f ′ (θ) sin θ +

f (θ) cos θ.

We replace dt by dθ, and the lower and upper limits of integration are α and β, respectively. Then the arc length formula becomes

L

b
=∫ a

⎛dx ⎝ dt

⎞2 ⎠

+

⎛dy ⎝dt

⎞⎠2dt

β
= ∫α

⎛dx ⎝dθ

⎞2 ⎠

+

⎛dy ⎝dθ

⎞⎠2dθ

β

∫ =

⎛ ⎝

f

′

(θ)

cos

θ

−

f (θ)

sin

θ⎞ ⎠

2

+

⎛ ⎝

f

′

(θ)

sin

θ

+

f (θ)

cos

θ⎞ ⎠

2dθ

α

β

∫ =

α

⎛ ⎝

f

′

(θ)⎞⎠2

⎛⎝cos 2

θ

+

sin2

θ⎞⎠

+

⎛ ⎝

f

(θ)⎞⎠2

⎛⎝cos 2

θ

+

sin2

θ⎞⎠dθ

β

∫ =

⎛ ⎝

f

′

(θ)⎞⎠2

+

⎛ ⎝

f

(θ)⎞⎠2dθ

α

β
=∫ α

r

2

+

⎛dr ⎝dθ

⎞⎠2dθ.

This gives us the following theorem.

Theorem 1.7: Arc Length of a Curve Defined by a Polar Function

Let f be a function whose derivative is continuous on an interval α ≤ θ ≤ β. The length of the graph of r = f (θ) from θ = α to θ = β is

∫ ∫ L =

β

⎡ ⎣

f

(θ)⎤⎦2

+

⎡ ⎣

f

′

(θ)⎤⎦2dθ

=

α

β α

r

2

+

⎛dr ⎝dθ

⎞⎠2dθ.

(1.10)

Example 1.18
Finding the Arc Length of a Polar Curve Find the arc length of the cardioid r = 2 + 2 cos θ.

70

Chapter 1 | Parametric Equations and Polar Coordinates

Solution
When θ = 0, r = 2 + 2 cos 0 = 4. Furthermore, as θ goes from 0 to 2π, the cardioid is traced out exactly once. Therefore these are the limits of integration. Using f (θ) = 2 + 2 cos θ, α = 0, and β = 2π, Equation 1.10 becomes

β

∫ L =

⎡ ⎣

f

(θ)⎤⎦2

+

⎡ ⎣

f

′

(θ)⎤⎦2

dθ

α

2π
= ∫ [2 + 2cosθ]2 + [−2sinθ]2 dθ 0

2π
= ∫ 4 + 8cosθ + 4cos2 θ + 4sin2 θ dθ 0

2π

=∫ 0

4 + 8 cos θ + 4⎛⎝cos2 θ + sin2 θ⎞⎠ dθ

2π
= ∫ 8 + 8cosθ dθ 0

2π
= 2∫ 2 + 2cosθ dθ. 0

Next, using the identity cos(2α) = 2 cos2 α − 1, add 1 to both sides and multiply by 2. This gives 2 + 2 cos(2α) = 4 cos2 α. Substituting α = θ/2 gives 2 + 2 cos θ = 4 cos2(θ/2), so the integral becomes

2π
L = 2∫ 2 + 2 cos θdθ 0

2π
= 2∫ 0

4

cos2

⎛θ ⎝2

⎞⎠dθ

| | =

2π
2∫ 2 0

cos⎛⎝2θ ⎞⎠

dθ.

The absolute value is necessary because the cosine is negative for some values in its domain. To resolve this issue,

change the limits from 0 to π and double the answer. This strategy works because cosine is positive between 0

and

π 2

.

Thus,

| | L

=

2π
4∫ 0

cos⎛⎝2θ

⎞ ⎠

dθ

=

8∫0πcos⎛⎝2θ

⎞ ⎠

dθ

=

8⎛⎝2

sin⎛⎝2θ

⎞⎞π ⎠⎠0

= 16.

1.17 Find the total arc length of r = 3 sin θ.

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Chapter 1 | Parametric Equations and Polar Coordinates

71

1.4 EXERCISES
For the following exercises, determine a definite integral that represents the area. 188. Region enclosed by r = 4 189. Region enclosed by r = 3 sin θ 190. Region in the first quadrant within the cardioid r = 1 + sin θ 191. Region enclosed by one petal of r = 8 sin(2θ)
192. Region enclosed by one petal of r = cos(3θ)
193. Region below the polar axis and enclosed by r = 1 − sin θ 194. Region in the first quadrant enclosed by r = 2 − cos θ 195. Region enclosed by the inner loop of r = 2 − 3 sin θ 196. Region enclosed by the inner loop of r = 3 − 4 cos θ 197. Region enclosed by r = 1 − 2 cos θ and outside the inner loop 198. Region common to r = 3 sin θ and r = 2 − sin θ 199. Region common to r = 2 and r = 4 cos θ 200. Region common to r = 3 cos θ and r = 3 sin θ For the following exercises, find the area of the described region. 201. Enclosed by r = 6 sin θ 202. Above the polar axis enclosed by r = 2 + sin θ 203. Below the polar axis and enclosed by r = 2 − cos θ
204. Enclosed by one petal of r = 4 cos(3θ)
205. Enclosed by one petal of r = 3 cos(2θ)
206. Enclosed by r = 1 + sin θ 207. Enclosed by the inner loop of r = 3 + 6 cos θ 208. Enclosed by r = 2 + 4 cos θ and outside the inner loop

209. Common interior of r = 4 sin(2θ) and r = 2

210.

Common

interior

of

r = 3 − 2 sin θ and r = −3 + 2 sin θ

211. Common interior of r = 6 sin θ and r = 3

212. Inside r = 1 + cos θ and outside r = cos θ

213.

Common

interior

of

r = 2 + 2 cos θ and r = 2 sin θ

For the following exercises, find a definite integral that represents the arc length.

214.

r

=

4

cos

θ

on

the

interval

0

≤

θ

≤

π 2

215. r = 1 + sin θ on the interval 0 ≤ θ ≤ 2π

216.

r

=

2

sec

θ

on

the

interval

0

≤

θ

≤

π 3

217. r = eθ on the interval 0 ≤ θ ≤ 1

For the following exercises, find the length of the curve over the given interval.

218.

r

=

6

on

the

interval 0

≤

θ

≤

π 2

219. r = e3θ on the interval 0 ≤ θ ≤ 2

220.

r

=

6

cos

θ

on

the

interval

0

≤

θ

≤

π 2

221. r = 8 + 8 cos θ on the interval 0 ≤ θ ≤ π

222. r = 1 − sin θ on the interval 0 ≤ θ ≤ 2π

For the following exercises, use the integration capabilities of a calculator to approximate the length of the curve.

223.

[T]

r

=

3θ on

the interval 0 ≤

θ

≤

π 2

224.

[T]

r

=

2 θ

on the interval π

≤

θ

≤ 2π

225.

[T]

r

=

sin

2

⎛θ ⎝2

⎞ ⎠

on

the

interval 0

≤

θ

≤

π

226. [T] r = 2θ2 on the interval 0 ≤ θ ≤ π

227. [T] r = sin(3 cos θ) on the interval 0 ≤ θ ≤ π

For the following exercises, use the familiar formula from

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Chapter 1 | Parametric Equations and Polar Coordinates

geometry to find the area of the region described and then confirm by using the definite integral.

228. r = 3 sin θ on the interval 0 ≤ θ ≤ π

229. r = sin θ + cos θ on the interval 0 ≤ θ ≤ π

230. r = 6 sin θ + 8 cos θ on the interval 0 ≤ θ ≤ π

For the following exercises, use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.

231. r = 3 sin θ on the interval 0 ≤ θ ≤ π

232. r = sin θ + cos θ on the interval 0 ≤ θ ≤ π

233. r = 6 sin θ + 8 cos θ on the interval 0 ≤ θ ≤ π

234. Verify that if y = r sin θ = f (θ)sin θ then

dy dθ

=

f ′(θ)sin θ +

f (θ)cos θ.

For the following exercises, find the slope of a tangent line to a polar curve r = f (θ). Let x = r cos θ = f (θ)cos θ
and y = r sin θ = f (θ)sin θ, so the polar equation
r = f (θ) is now written in parametric form.

235.

Use the definition of the derivative

dy dx

=

dy/dθ dx/dθ

and

the product rule to derive the derivative of a polar equation.

236.

r = 1 − sin θ;

⎛⎝12 ,

π⎞ 6⎠

237.

r = 4 cos θ;

⎛⎝2,

π⎞ 3⎠

238.

r = 8 sin θ;

⎛⎝4,

5π ⎞ 6⎠

239.

r = 4 + sin θ;

⎛⎝3,

3π ⎞ 2⎠

240. r = 6 + 3 cos θ; (3, π)

241. r = 4 cos(2θ); tips of the leaves

242. r = 2 sin(3θ); tips of the leaves

243.

r = 2θ;

⎛π ⎝2

,

π⎞ 4⎠

244. Find the points on the interval −π ≤ θ ≤ π at which
the cardioid r = 1 − cos θ has a vertical or horizontal tangent line.

245. For the cardioid r = 1 + sin θ, find the slope of the tangent line when θ = π3.

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of θ.

246.

r

=

3

cos

θ,

θ

=

π 3

247.

r = θ,

θ

=

π 2

248. r = ln θ, θ = e

249.

[T] Use technology:

r

=

2 + 4 cos θ

at

θ

=

π 6

For the following exercises, find the points at which the following polar curves have a horizontal or vertical tangent line.

250. r = 4 cos θ

251. r2 = 4 cos(2θ)

252. r = 2 sin(2θ)

253. The cardioid r = 1 + sin θ

254. Show that the curve r = sin θ tan θ (called a cissoid of Diocles) has the line x = 1 as a vertical asymptote.

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73

1.5 | Conic Sections
Learning Objectives
1.5.1 Identify the equation of a parabola in standard form with given focus and directrix. 1.5.2 Identify the equation of an ellipse in standard form with given foci. 1.5.3 Identify the equation of a hyperbola in standard form with given foci. 1.5.4 Recognize a parabola, ellipse, or hyperbola from its eccentricity value. 1.5.5 Write the polar equation of a conic section with eccentricity e . 1.5.6 Identify when a general equation of degree two is a parabola, ellipse, or hyperbola.
Conic sections have been studied since the time of the ancient Greeks, and were considered to be an important mathematical concept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since then, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even architecture. In this section we discuss the three basic conic sections, some of their properties, and their equations.
Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what most people mean by “cone,” having the shape of a party hat. A right circular cone can be generated by revolving a line passing through the origin around the y-axis as shown.

Figure 1.43 A cone generated by revolving the line y = 3x around the y -axis.
Conic sections are generated by the intersection of a plane with a cone (Figure 1.44). If the plane is parallel to the axis of revolution (the y-axis), then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than 90°), then the conic section is an ellipse.

74

Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.44 The four conic sections. Each conic is determined by the angle the plane makes with the axis of the cone.
Parabolas
A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects only one of the nappes. A parabola can also be defined in terms of distances.
Definition A parabola is the set of all points whose distance from a fixed point, called the focus, is equal to the distance from a fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of the parabola.
A graph of a typical parabola appears in Figure 1.45. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates (x1, y1) and point Q with coordinates (x2, y2), the distance between them is given by the formula
d(P, Q) = (x2 − x1)2 + (y2 − y1)2. Then from the definition of a parabola and Figure 1.45, we get
d(F, P) = d(P, Q) (0 − x)2 + (p − y)2 = (x − x)2 + (−p − y)2. Squaring both sides and simplifying yields
x2 + (p − y)2 = 02 + (−p − y)2 x2 + p2 − 2py + y2 = p2 + 2py + y2
x2 − 2py = 2py x2 = 4py.
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75

Figure 1.45 A typical parabola in which the distance from the focus to the vertex is represented by the variable p.

Now suppose we want to relocate the vertex. We use the variables (h, k) to denote the coordinates of the vertex. Then if

the focus is directly above the vertex, it has coordinates

⎛⎝h,

k

+

p⎞ ⎠

and the directrix has the equation

y = k − p.

Going

through the same derivation yields the formula (x − h)2 = 4p⎛⎝y − k⎞⎠. Solving this equation for y leads to the following

theorem.

Theorem 1.8: Equations for Parabolas

Given a parabola opening upward with vertex located at (h, k) and focus located at ⎛⎝h, k + p⎞⎠, where p is a constant, the equation for the parabola is given by

y = 41p(x − h)2 + k.

(1.11)

This is the standard form of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

76

Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.46 Four parabolas, opening in various directions, along with their equations in standard form.
In addition, the equation of a parabola can be written in the general form, though in this form the values of h, k, and p are not immediately recognizable. The general form of a parabola is written as
ax2 + bx + cy + d = 0 or ay2 + bx + cy + d = 0. The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.
Example 1.19
Converting the Equation of a Parabola from General into Standard Form Put the equation x2 − 4x − 8y + 12 = 0 into standard form and graph the resulting parabola.
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Chapter 1 | Parametric Equations and Polar Coordinates

77

Solution
Since y is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y, which will put the equation into standard form. To do that, first add 8y to both sides of the equation:

8y = x2 − 4x + 12.

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the righthand side using parentheses:

8y = ⎛⎝x2 − 4x⎞⎠ + 12.

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses

a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives

⎛⎝−24

⎞2 ⎠

=

4.

Add 4

inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

8y = ⎛⎝x2 − 4x + 4⎞⎠ + 12 − 4.

Now combine like terms and factor the quantity inside the parentheses:

8y = (x − 2)2 + 8.

Finally, divide by 8:

y

=

1 8

(x

−

2) 2

+

1.

This equation is now in standard form. Comparing this to Equation 1.11 gives h = 2, k = 1, and p = 2. The parabola opens up, with vertex at (2, 1), focus at (2, 3), and directrix y = −1. The graph of this parabola appears as follows.

Figure 1.47 The parabola in Example 1.19. 1.18 Put the equation 2y2 − x + 12y + 16 = 0 into standard form and graph the resulting parabola.

78

Chapter 1 | Parametric Equations and Polar Coordinates

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

Consider a parabolic dish designed to collect signals from a satellite in space. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from a wide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (that is, the light bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam straight ahead. This allows a small light bulb to illuminate a wide angle of space in front of the flashlight or car.
Ellipses
An ellipse can also be defined in terms of distances. In the case of an ellipse, there are two foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more detail later in this section.
Definition
An ellipse is the set of all points for which the sum of their distances from two fixed points (the foci) is constant.

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Chapter 1 | Parametric Equations and Polar Coordinates

79

Figure 1.48 A typical ellipse in which the sum of the distances from any point on the ellipse to the foci is constant.
A graph of a typical ellipse is shown in Figure 1.48. In this figure the foci are labeled as F and F′. Both are the same fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of F are (c, 0) and the coordinates of F′ are (−c, 0). The points P and P′ are located at the ends of the major axis of the ellipse, and have coordinates (a, 0) and (−a, 0), respectively. The major axis is always the longest distance across the ellipse, and can be horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore, P and P′ are called the vertices of the ellipse. The points Q and Q′ are located at the ends of the minor axis of the ellipse, and have coordinates (0, b) and (0, −b), respectively. The minor axis is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis. According to the definition of the ellipse, we can choose any point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are (a, 0), the sum of the distances is
d(P, F) + d(P, F′) = (a − c) + (a + c) = 2a.
Therefore the sum of the distances from an arbitrary point A with coordinates (x, y) is also equal to 2a. Using the distance formula, we get
d(A, F) + d(A, F′) = 2a (x − c)2 + y2 + (x + c)2 + y2 = 2a.
Subtract the second radical from both sides and square both sides:
(x − c)2 + y2 = 2a − (x + c)2 + y2 (x − c)2 + y2 = 4a2 − 4a (x + c)2 + y2 + (x + c)2 + y2 x2 − 2cx + c2 + y2 = 4a2 − 4a (x + c)2 + y2 + x2 + 2cx + c2 + y2
−2cx = 4a2 − 4a (x + c)2 + y2 + 2cx.
Now isolate the radical on the right-hand side and square again:

80

Chapter 1 | Parametric Equations and Polar Coordinates

−2cx = 4a2 − 4a (x + c)2 + y2 + 2cx 4a (x + c)2 + y2 = 4a2 + 4cx

(x + c)2 + y2

=

a

+

cx a

(x + c)2 + y2

=

a2

+

2cx

+

c2 x2 a2

x2 + 2cx + c2 + y2

=

a2

+

2cx

+

c2 x2 a2

x2 + c2 + y2

=

a2

+

c

2
a

x
2

2

.

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

x2

−

c2 x2 a2

+

y2

=

a2 − c2

⎛⎝a 2

− c2⎞⎠x2 a2

+

y2

=

a2 − c2.

Divide both sides by a2 − c2. This gives the equation

x2 a2

+

y2 a2 − c2

=

1.

If we refer back to Figure 1.48, then the length of each of the two green line segments is equal to a. This is true because the sum of the distances from the point Q to the foci F and F′ is equal to 2a, and the lengths of these two line segments are equal. This line segment forms a right triangle with hypotenuse length a and leg lengths b and c. From the Pythagorean
theorem, a2 + b2 = c2 and b2 = a2 − c2. Therefore the equation of the ellipse becomes

x2 a2

+

y2 b2

=

1.

Finally, if the center of the ellipse is moved from the origin to a point (h, k), we have the following standard form of an ellipse.

Theorem 1.9: Equation of an Ellipse in Standard Form

Consider the ellipse with center (h, k), a horizontal major axis with length 2a, and a vertical minor axis with length 2b. Then the equation of this ellipse in standard form is

(x

− h)2 a2

+

y⎛
⎝

−

k⎞ ⎠

2

b2

=

1

(1.12)

and the foci are located at (h ± c, k), where c2 = a2 − b2. The equations of the directrices are x = h ± ac2.

If the major axis is vertical, then the equation of the ellipse becomes

(x

− h)2 b2

+

y⎛
⎝

−

k⎞ ⎠

2

a2

=

1

(1.13)

and the foci are located at (h, k ± c), where c2 = a2 − b2. The equations of the directrices in this case are

y

=

k

±

a2 c

.

If the major axis is horizontal, then the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is

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Chapter 1 | Parametric Equations and Polar Coordinates

81

called vertical. The equation of an ellipse is in general form if it is in the form Ax2 + By2 + Cx + Dy + E = 0, where A and B are either both positive or both negative. To convert the equation from general to standard form, use the method of completing the square.
Example 1.20
Finding the Standard Form of an Ellipse

Put the equation 9x2 + 4y2 − 36x + 24y + 36 = 0 into standard form and graph the resulting ellipse.

Solution First subtract 36 from both sides of the equation:
9x2 + 4y2 − 36x + 24y = −36.
Next group the x terms together and the y terms together, and factor out the common factor:
⎛⎝9x2 − 36x⎞⎠ + ⎛⎝4y2 + 24y⎞⎠ = −36 9⎛⎝x2 − 4x⎞⎠ + 4⎛⎝y2 + 6y⎞⎠ = −36.

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square.

In the first set of parentheses,

take half the

coefficient of x and square it. This

gives

⎛⎝−24

⎞2 ⎠

=

4.

In the second

set of parentheses, take half the coefficient of y

and square it. This

gives

⎛⎝62

⎞2 ⎠

=

9.

Add these

inside each pair

of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side.

Similarly, we are adding 36 to the second set as well. Therefore the equation becomes

9⎛⎝x2 − 4x + 4⎞⎠ + 4⎛⎝y2 + 6y + 9⎞⎠ = −36 + 36 + 36 9⎛⎝x2 − 4x + 4⎞⎠ + 4⎛⎝y2 + 6y + 9⎞⎠ = 36.

Now factor both sets of parentheses and divide by 36:

9(x − 2)2 + 4⎛⎝y + 3⎞⎠2 = 36

9(x − 2)2 36

+

4⎛⎝y

+

3⎞ ⎠

2

36

=

1

(x

− 2)2 4

+

y⎛
⎝

+

3⎞ ⎠

2

9

=

1.

The equation is now in standard form. Comparing this to Equation 1.14 gives h = 2, k = −3, a = 3, and b = 2. This is a vertical ellipse with center at (2, −3), major axis 6, and minor axis 4. The graph of this ellipse appears as follows.

82

Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.49 The ellipse in Example 1.20.
1.19 Put the equation 9x2 + 16y2 + 18x − 64y − 71 = 0 into standard form and graph the resulting ellipse.
According to Kepler’s first law of planetary motion, the orbit of a planet around the Sun is an ellipse with the Sun at one of the foci as shown in Figure 1.50(a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northern hemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axis in the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting the planets and satellites orbiting Earth. Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focus after mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure 1.50(b). This hall served as the meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semielliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they can hear someone standing close by. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Central Station in New York City.
Figure 1.50 (a) Earth’s orbit around the Sun is an ellipse with the Sun at one focus. (b) Statuary Hall in the U.S. Capitol is a whispering gallery with an elliptical cross section.
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83

Hyperbolas
A hyperbola can also be defined in terms of distances. In the case of a hyperbola, there are two foci and two directrices. Hyperbolas also have two asymptotes.
Definition A hyperbola is the set of all points where the difference between their distances from two fixed points (the foci) is constant.
A graph of a typical hyperbola appears as follows.

Figure 1.51 A typical hyperbola in which the difference of the distances from any point on the ellipse to the foci is constant. The transverse axis is also called the major axis, and the conjugate axis is also called the minor axis.

The derivation of the equation of a hyperbola in standard form is virtually identical to that of an ellipse. One slight hitch lies

in the definition: The difference between two numbers is always positive. Let P be a point on the hyperbola with coordinates

| | (x, y). Then the definition of the hyperbola gives

d⎛⎝P,

F

⎞
1⎠

−

d⎛⎝P,

F

⎞
2⎠

= constant.

To simplify the derivation, assume

that P is on the right branch of the hyperbola, so the absolute value bars drop. If it is on the left branch, then the subtraction

is reversed. The vertex of the right branch has coordinates (a, 0), so

d⎛⎝P,

F

⎞
1⎠

−

d⎛⎝P,

F

⎞
2⎠

=

(c

+

a)

−

(c

−

a)

=

2a.

This equation is therefore true for any point on the hyperbola. Returning to the coordinates (x, y) for P:

d⎛⎝P,

F

⎞
1⎠

−

d⎛⎝P,

F

⎞
2⎠

=

2a

(x + c)2 + y2 − (x − c)2 + y2 = 2a.

Add the second radical from both sides and square both sides:

(x − c)2 + y2 = 2a + (x + c)2 + y2 (x − c)2 + y2 = 4a2 + 4a (x + c)2 + y2 + (x + c)2 + y2 x2 − 2cx + c2 + y2 = 4a2 + 4a (x + c)2 + y2 + x2 + 2cx + c2 + y2
−2cx = 4a2 + 4a (x + c)2 + y2 + 2cx.

Now isolate the radical on the right-hand side and square again:

84

Chapter 1 | Parametric Equations and Polar Coordinates

−2cx = 4a2 + 4a (x + c)2 + y2 + 2cx 4a (x + c)2 + y2 = −4a2 − 4cx

(x + c)2 + y2

=

−a

−

cx a

(x + c)2 + y2

=

a2

+

2cx

+

c2 x2 a2

x2 + 2cx + c2 + y2

=

a2

+

2cx

+

c2 x2 a2

x2 + c2 + y2

=

a2

+

c

2
a

x
2

2

.

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

x2

−

c2 x2 a2

+

y2

=

a2 − c2

⎛⎝a 2

− c2⎞⎠x2 a2

+

y2

=

a2 − c2.

Finally, divide both sides by a2 − c2. This gives the equation

x2 a2

+

y2 a2 − c2

=

1.

We now define b so that b2 = c2 − a2. This is possible because c > a. Therefore the equation of the ellipse becomes

x2 a2

−

y2 b2

=

1.

Finally, if the center of the hyperbola is moved from the origin to the point (h, k), we have the following standard form of a hyperbola.

Theorem 1.10: Equation of a Hyperbola in Standard Form

Consider the hyperbola with center (h, k), a horizontal major axis, and a vertical minor axis. Then the equation of this ellipse is

(x

− h)2 a2

−

y⎛
⎝

−

k⎞ ⎠

2

b2

=

1

(1.14)

and the foci are located at (h ± c, k), where c2 = a2 + b2. The equations of the asymptotes are given by y = k ± ba(x − h). The equations of the directrices are

x=k±

a2 a2 +

b2

=

h

±

ac2 .

If the major axis is vertical, then the equation of the hyperbola becomes

y⎛
⎝

− k⎞⎠2 a2

−

(x

− h)2 b2

=

1

(1.15)

and the foci are located at (h, k ± c), where c2 = a2 + b2. The equations of the asymptotes are given by y = k ± ab(x − h). The equations of the directrices are

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Chapter 1 | Parametric Equations and Polar Coordinates

85

y=k±

a2 a2 +

b2

=

k

±

a2 c

.

If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general form if it is in the form Ax2 + By2 + Cx + Dy + E = 0, where A and B have opposite signs. In order to convert the equation from general to standard form, use the method of completing the square.
Example 1.21
Finding the Standard Form of a Hyperbola

Put the equation 9x2 − 16y2 + 36x + 32y − 124 = 0 into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Solution First add 124 to both sides of the equation:
9x2 − 16y2 + 36x + 32y = 124.
Next group the x terms together and the y terms together, then factor out the common factors:
⎛⎝9x2 + 36x⎞⎠ − ⎛⎝16y2 − 32y⎞⎠ = 124 9⎛⎝x2 + 4x⎞⎠ − 16⎛⎝y2 − 2y⎞⎠ = 124.

We need to determine the constant that, when added inside each set of parentheses, results in a perfect square. In
the first set of parentheses, take half the coefficient of x and square it. This gives ⎛⎝42⎞⎠2 = 4. In the second set
of parentheses, take half the coefficient of y and square it. This gives ⎛⎝−22⎞⎠2 = 1. Add these inside each pair of parentheses. Since the first set of parentheses has a 9 in front, we are actually adding 36 to the left-hand side. Similarly, we are subtracting 16 from the second set of parentheses. Therefore the equation becomes

9⎛⎝x2 + 4x + 4⎞⎠ − 16⎛⎝y2 − 2y + 1⎞⎠ = 124 + 36 − 16 9⎛⎝x2 + 4x + 4⎞⎠ − 16⎛⎝y2 − 2y + 1⎞⎠ = 144.

Next factor both sets of parentheses and divide by 144:

9(x + 2)2 − 16⎛⎝y − 1⎞⎠2 = 144

9(x + 2)2 144

−

16⎛⎝y

−

1⎞ ⎠

2

144

=

1

(x

+ 2)2 16

−

y⎛
⎝

−

1⎞ ⎠

2

9

=

1.

The equation is now in standard form. Comparing this to Equation 1.15 gives h = −2, k = 1, a = 4, and b = 3. This is a horizontal hyperbola with center at (−2, 1) and asymptotes given by the equations

y = 1 ± 34(x + 2). The graph of this hyperbola appears in the following figure.

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Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.52 Graph of the hyperbola in Example 1.21.
1.20 Put the equation 4y2 − 9x2 + 16y + 18x − 29 = 0 into standard form and graph the resulting hyperbola. What are the equations of the asymptotes? Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.
Figure 1.53 A hyperbolic mirror used to collect light from distant stars. This property of the hyperbola has important applications. It is used in radio direction finding (since the difference in signals from two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passes through the solar system is hyperbolic.
Eccentricity and Directrix
An alternative way to describe a conic section involves the directrices, the foci, and a new property called eccentricity. We
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87

will see that the value of the eccentricity of a conic section can uniquely define that conic.
Definition The eccentricity e of a conic section is defined to be the distance from any point on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic section, and can define the conic section as well:
1. If e = 1, the conic is a parabola. 2. If e < 1, it is an ellipse. 3. If e > 1, it is a hyperbola. The eccentricity of a circle is zero. The directrix of a conic section is the line that, together with the point known as the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associated directrices. Parabolas have one focus and one directrix.
The three conic sections with their directrices appear in the following figure.

Figure 1.54 The three conic sections with their foci and directrices.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the

directrices of a horizontal ellipse are

x

=

±

a2 c

.

The right vertex of the ellipse is located at

(a, 0)

and the right focus is

(c, 0). Therefore the distance from the vertex to the focus is a − c and the distance from the vertex to the right directrix

is

a2 c

−

c.

This gives the eccentricity as

e

=

a−c

a2 c

−

a

=

c(a − c) a2 − ac

=

c(a − c) a(a − c)

=

c a

.

Since c < a, this step proves that the eccentricity of an ellipse is less than 1. The directrices of a horizontal hyperbola are

also located at

x

=

±

a2 c

,

and a similar calculation shows that the eccentricity of a hyperbola is also e = ac. However in

this case we have c > a, so the eccentricity of a hyperbola is greater than 1.

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Chapter 1 | Parametric Equations and Polar Coordinates

Example 1.22
Determining Eccentricity of a Conic Section

Determine the eccentricity of the ellipse described by the equation

(x

− 3)2 16

+

y⎛
⎝

+

2⎞ ⎠

2

25

=

1.

Solution

From the equation we see that a = 5 and b = 4. The value of c can be calculated using the equation

a2 = b2 + c2 for an ellipse. Substituting the values of a and b and solving for c gives c = 3. Therefore the

eccentricity of the ellipse is

e

=

ac

=

3 5

=

0.6.

1.21 Determine the eccentricity of the hyperbola described by the equation

y⎛
⎝

−

3⎞ ⎠

2

49

−

(x

+ 2)2 25

=

1.

Polar Equations of Conic Sections

Sometimes it is useful to write or identify the equation of a conic section in polar form. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined as the distance from a focus to the nearest directrix. The following table gives the focal parameters for the different types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the case of a parabola, a represents the distance from the vertex to the focus.

Conic

e

p

Ellipse

0<e<1

a2

−c

c2

=

a⎛⎝1

− c

e 2⎞⎠

Parabola

e=1

2a

Hyperbola e > 1

c2

−c

a2

=

a⎛⎝e2 − e

1⎞⎠

Table 1.1 Eccentricities and Focal Parameters of the Conic Sections

Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conic section in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then using the definition of the various conic sections in terms of distances, it is possible to prove the following theorem.

Theorem 1.11: Polar Equation of Conic Sections

The polar equation of a conic section with focal parameter p is given by

r

=

1

±

ep e cos

θ

or

r

=

1

ep ± e sin

θ.

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89

In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar form, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is at the origin only if the conic is a circle (i.e., e = 0).
Example 1.23
Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described by the equation

r

=

1

+

3 2 cos

θ.

Solution
The constant term in the denominator is 1, so the eccentricity of the conic is 2. This is a hyperbola. The focal parameter p can be calculated by using the equation ep = 3. Since e = 2, this gives p = 32. The cosine function appears in the denominator, so the hyperbola is horizontal. Pick a few values for θ and create a table of values. Then we can graph the hyperbola (Figure 1.55).

θ

r

θ

r

0

1

π

−3

π 4

3 1+

2 ≈ 1.2426

5π 4

3 1−

2 ≈ −7.2426

π 2

3

3π

3

2

3π 4

3 1−

2 ≈ −7.2426

7π 4

3 1+

2 ≈ 1.2426

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Chapter 1 | Parametric Equations and Polar Coordinates

Figure 1.55 Graph of the hyperbola described in Example 1.23.

1.22 Identify and create a graph of the conic section described by the equation

r

=

1

−

4 0.8

sin

θ.

General Equations of Degree Two
A general equation of degree two can be written in the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
The graph of an equation of this form is a conic section. If B ≠ 0 then the coordinate axes are rotated. To identify the conic section, we use the discriminant of the conic section 4AC − B2. One of the following cases must be true:
1. 4AC − B2 > 0. If so, the graph is an ellipse. 2. 4AC − B2 = 0. If so, the graph is a parabola. 3. 4AC − B2 < 0. If so, the graph is a hyperbola. The simplest example of a second-degree equation involving a cross term is xy = 1. This equation can be solved for y to obtain y = 1x. The graph of this function is called a rectangular hyperbola as shown.

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91

Figure 1.56 Graph of the equation xy = 1; The red lines indicate the rotated axes.

The asymptotes of this hyperbola are the x and y coordinate axes. To determine the angle θ of rotation of the conic section,

we

use the formula

cot 2θ =

A

− B

C

.

In this case

A=C=0

and

B = 1,

so cot 2θ = (0 − 0)/1 = 0 and θ = 45°.

The method for graphing a conic section with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled A′, B′, C′, D′, E′, and F′, and are given by the formulas

A′ = A cos2 θ + B cos θ sin θ + C sin2 θ B′ = 0 C′ = A sin2 θ − B sin θ cos θ + C cos2 θ D′ = D cos θ + E sin θ E′ = −D sin θ + E cos θ F′ = F.
The procedure for graphing a rotated conic is the following:

1. Identify the conic section using the discriminant 4AC − B2.

2.

Determine

θ

using the formula

cot 2θ

=

A

− B

C

.

3. Calculate A′, B′, C′, D′, E′, and F′.

4. Rewrite the original equation using A′, B′, C′, D′, E′, and F′.

5. Draw a graph using the rotated equation.

Example 1.24

Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation 13x2 − 6 3xy + 7y2 − 256 = 0.
Solution

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Chapter 1 | Parametric Equations and Polar Coordinates

In this equation, A = 13, B = −6 3, C = 7, D = 0, E = 0, and F = −256. The discriminant of this

equation is 4AC − B2 = 4(13)(7) − ⎛⎝−6 3⎞⎠2 = 364 − 108 = 256. Therefore this conic is an ellipse. To

calculate the angle of rotation of the axes, use

cot 2θ

=

A

− B

C

.

This gives

cot 2θ

=

A−C B

=

13 − 7 −6 3

= − 33.

Therefore 2θ = 120o and θ = 60o, which is the angle of the rotation of the axes.

To determine the rotated coefficients, use the formulas given above:

A′ = A cos2 θ + B cos θ sin θ + C sin2 θ = 13cos2 60 + ⎛⎝−6 3⎞⎠ cos 60 sin 60 + 7sin2 60

=

13⎛⎝12

⎞2 ⎠

−

6

3⎛⎝12 ⎞⎠⎛⎝

3 2

⎞ ⎠

+

7⎛⎝

3 2

⎞2 ⎠

= 4,

B′ = 0,

C′ = A sin2 θ − B sin θ cos θ + C cos2 θ = 13sin2 60 + ⎛⎝−6 3⎞⎠ sin 60 cos 60 = 7cos2 60

=

⎛ ⎝

3 2

⎞2 ⎠

+

6

3⎛⎝

3 2

⎞⎠⎛⎝12 ⎞⎠

+

7⎛⎝12 ⎞⎠2

= 16,

D′ = D cos θ + E sin θ = (0) cos 60 + (0) sin 60

= 0,

E′ = −D sin θ + E cos θ = −(0) sin 60 + (0) cos 60

= 0,

F′ = F = −256.

The equation of the conic in the rotated coordinate system becomes

4(x′)2 + 16⎛⎝y′⎞⎠2 = 256

(x′)2 64

+

y′⎛ ⎞
⎝⎠

2

16

=

1.

A graph of this conic section appears as follows.

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