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Calculus Volume 3










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Table of ContentsPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 1: Parametric Equations and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . 51.1 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Calculus of Parametric Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.3 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421.4 Area and Arc Length in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.5 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71Chapter 2: Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 992.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1002.2 Vectors in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1442.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1632.5 Equations of Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1842.6 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2092.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226Chapter 3: Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2573.1 Vector-Valued Functions and Space Curves . . . . . . . . . . . . . . . . . . . . . . . . . . 2583.2 Calculus of Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2683.3 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2813.4 Motion in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303Chapter 4: Differentiation of Functions of Several Variables . . . . . . . . . . . . . . . . . . . . 3314.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3324.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3504.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3674.4 Tangent Planes and Linear Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . 3874.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4044.6 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4204.7 Maxima/Minima Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4364.8 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456Chapter 5: Multiple Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4755.1 Double Integrals over Rectangular Regions . . . . . . . . . . . . . . . . . . . . . . . . . . 4765.2 Double Integrals over General Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4995.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5245.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5445.5 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . 5645.6 Calculating Centers of Mass and Moments of Inertia . . . . . . . . . . . . . . . . . . . . . 5905.7 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 608Chapter 6: Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6396.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6406.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6616.3 Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6876.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7096.5 Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7356.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7516.7 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7876.8 The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 805Chapter 7: Second-Order Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 8297.1 Second-Order Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8307.2 Nonhomogeneous Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8477.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8617.4 Series Solutions of Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 882Appendix A: Table of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895Appendix B: Table of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 901Appendix C: Review of Pre-Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 903Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011




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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 20 books for college and AP courses used by hundreds of thousands of students. Our adaptive learning technology, designed to improve learning outcomes through personalized educational paths, is being piloted 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 ResourcesCustomization
Calculus Volume 3 is licensed under a Creative Commons Attribution 4.0 International (CC BY) license, which means that you can distribute, remix, and build upon the content, as long as you provide attribution to OpenStax and its content contributors.
Because our books are openly licensed, you are free to use the entire book or pick and choose the sections that are most relevant to the needs of your course. Feel free to remix the content by assigning your students certain chapters and sections in your syllabus, in the order that you prefer. You can even provide a direct link in your syllabus to the sections in the web view of your book.
Faculty also have the option of creating a customized version of their OpenStax book through the aerSelect platform. 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 a link to your book on aerSelect.
Errata
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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 1Chapter 1: Functions and Graphs


Preface 1




Chapter 2: Limits
Chapter 3: Derivatives
Chapter 4: Applications of Derivatives
Chapter 5: Integration
Chapter 6: Applications of Integration


Volume 2Chapter 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 3Chapter 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 asmodern applications and methods. Derivations and explanations are based on years of classroom experience on the partof 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 areasof physics, chemistry, engineering, and computer science. Student Projects in each chapter give students opportunities toexplore interesting sidelights in pure and applied mathematics, from navigating a banked turn to adapting a moon landingvehicle for a new mission to Mars. Chapter Opening Applications pose problems that are solved later in the chapter, usingthe ideas covered in that chapter. Problems include the average distance of Halley's Comment from the Sun, and the vectorfield of a hurricane. Definitions, Rules, and Theorems are highlighted throughout the text, including over 60 Proofs oftheorems.
Assessments That Reinforce Key Concepts
In-chapter Exampleswalk students through problems by posing a question, stepping out a solution, and then asking studentsto practice the skill with a “Checkpoint” question. The book also includes assessments at the end of each chapter sostudents can apply what they’ve learned through practice problems. Many exercises are marked with a [T] to indicate theyare suitable for solution by technology, including calculators or Computer Algebra Systems (CAS). Answers for selectedexercises are available in the Answer Key at the back of the book. The book also includes assessments at the end of eachchapter 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 rigorousterms in Chapter 6 in Volume 1 and Chapter 2 in Volume 2. Differentiation and integration of these functions is covered inChapters 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 integraldefinitions are given before teaching the calculus derivations of exponentials and logarithms.
Comprehensive Art Program
Our art program is designed to enhance students’ understanding of concepts through clear and effective illustrations,


2 Preface


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diagrams, and photographs.


Additional ResourcesStudent and Instructor Resources
We’ve compiled additional resources for both students and instructors, including Getting Started Guides, an instructorsolution manual, and PowerPoint slides. Instructor resources require a verified instructor account, which can be requestedon your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.
Partner Resources
OpenStax Partners are our allies in the mission to make high-quality learning materials affordable and accessible to studentsand instructors everywhere. Their tools integrate seamlessly with our OpenStax titles at a low cost. To access the partnerresources for your text, visit your book page on openstax.org.
About the AuthorsSenior Contributing Authors
Gilbert Strang, Massachusetts Institute of TechnologyDr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus onlinetextbook is one of eleven that he has published and is the basis from which our final product has been derived and updatedfor today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University.
Edwin “Jed” Herman, University of Wisconsin-Stevens PointDr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University ofWisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student researchmentor, is experienced in course development/design, and is also an avid board game designer and player.


Preface 3




Contributing Authors
Catherine Abbott, Keuka CollegeNicoleta Virginia Bila, Fayetteville State UniversitySheri J. Boyd, Rollins CollegeJoyati Debnath, Winona State UniversityValeree Falduto, Palm Beach State CollegeJoseph Lakey, New Mexico State UniversityJulie Levandosky, Framingham State UniversityDavid McCune, William Jewell CollegeMichelle Merriweather, Bronxville High SchoolKirsten R. Messer, Colorado State University - PuebloAlfred K. Mulzet, Florida State College at JacksonvilleWilliam Radulovich (retired), Florida State College at JacksonvilleErica M. Rutter, Arizona State UniversityDavid Smith, University of the Virgin IslandsElaine A. Terry, Saint Joseph’s UniversityDavid Torain, Hampton University
Reviewers
Marwan A. Abu-Sawwa, Florida State College at JacksonvilleKenneth J. Bernard, Virginia State UniversityJohn Beyers, University of MarylandCharles Buehrle, Franklin & Marshall CollegeMatthew Cathey, Wofford CollegeMichael Cohen, Hofstra UniversityWilliam DeSalazar, Broward County School SystemMurray Eisenberg, University of Massachusetts AmherstKristyanna Erickson, Cecil CollegeTiernan Fogarty, Oregon Institute of TechnologyDavid French, Tidewater Community CollegeMarilyn Gloyer, Virginia Commonwealth UniversityShawna Haider, Salt Lake Community CollegeLance Hemlow, Raritan Valley Community CollegeJerry Jared, The Blue Ridge SchoolPeter Jipsen, Chapman UniversityDavid Johnson, Lehigh UniversityM.R. Khadivi, Jackson State UniversityRobert J. Krueger, Concordia UniversityTor A. Kwembe, Jackson State UniversityJean-Marie Magnier, Springfield Technical Community CollegeCheryl Chute Miller, SUNY PotsdamBagisa Mukherjee, Penn State University, Worthington Scranton CampusKasso Okoudjou, University of Maryland College ParkPeter Olszewski, Penn State Erie, The Behrend CollegeSteven Purtee, Valencia CollegeAlice Ramos, Bethel CollegeDoug Shaw, University of Northern IowaHussain Elalaoui-Talibi, Tuskegee UniversityJeffrey Taub, Maine Maritime AcademyWilliam Thistleton, SUNY Polytechnic InstituteA. David Trubatch, Montclair State UniversityCarmen Wright, Jackson State UniversityZhenbu Zhang, Jackson State University


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1 | PARAMETRICEQUATIONS AND POLARCOORDINATES


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


Chapter 1 | Parametric Equations and Polar Coordinates 5




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 ahard outer shell with many chambers connected in a spiral fashion, and it can retract into its shell to avoid predators. Whenpart of the shell is cut away, a perfect spiral is revealed, with chambers inside that are somewhat similar to growth rings ina 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 muchsimpler to describe. The polar coordinate system is well suited for describing curves of this type. How can we use thiscoordinate 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 theposition of a particle or object in two dimensions as a function of time. We will use parametric equations and polarcoordinates 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, parametricequations are useful for describing curves that are not necessarily functions. The parameter is an independent variable thatboth x and y depend on, and as the parameter increases, the values of x and y trace out a path along a plane curve. Forexample, if the parameter is t (a common choice), then tmight 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 use365 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


4
day of orbiting time is built into the calendar. We call January 1 “day 1”


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 revolvesaround the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and anew year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at onefocus of the ellipse. We study this idea in more detail in Conic Sections.


6 Chapter 1 | Parametric Equations and Polar Coordinates


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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 toeach 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 eachy value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’sposition as a function of time.


Figure 1.3 Coordinate axes superimposed on the orbit ofEarth.


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


Chapter 1 | Parametric Equations and Polar Coordinates 7




interval I is called the graph of the parametric equations. The graph of parametric equations is called a parametriccurve 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 variesover 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 importantto 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 pointsappears in Figure 1.4. The arrows on the graph indicate the orientation of the graph, that is, the directionthat a point moves on the graph as t varies from −3 to 2.


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Figure 1.4 Graph of the plane curve described by theparametric 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 pointon 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.


Chapter 1 | Parametric Equations and Polar Coordinates 9




Figure 1.5 Graph of the plane curve described by theparametric 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) y(t)
0 4 0 7π


6
−2 3 ≈ −3.5 2


π
6 2 3 ≈ 3.5


2 4π
3


−2 −2 3 ≈ −3.5


π
3


2 2 3 ≈ 3.5 3π
2


0 −4


π
2


0 4 5π
3


2 −2 3 ≈ −3.5



3


−2 2 3 ≈ 3.5 11π
6


2 3 ≈ 3.5 2



6


−2 3 ≈ −3.5 2 2π 4 0


π −4 0


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1.1


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


Figure 1.6 Graph of the plane curve described by theparametric equations in part c.


This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. Thestarting point and ending points of the curve both have coordinates (4, 0).


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 singleequation relating the variables x and y. Then we can apply any previous knowledge of equations of curves in the plane toidentify 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
t =


y − 1
2


.


This can be substituted into the first equation:
x =


y − 1
2



2


− 3 =
y2 − 2y + 1


4
− 3 =


y2 − 2y − 11
4


.


This equation describes x as a function of y. These steps give an example of eliminating the parameter. The graph of thisfunction 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.


Chapter 1 | Parametric Equations and Polar Coordinates 11




Example 1.2
Eliminating the Parameter
Eliminate the parameter for each of the plane curves described by the following parametric equations and describethe 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 firstequation for t gives


x = 2t + 4


x2 = 2t + 4


x2 − 4 = 2t


t = x
2 − 4
2


.


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


x2 − 4


2

⎠+ 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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Figure 1.7 Graph of the plane curve described by theparametric equations in part a.
b. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations forthis 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 =


y
3
.


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
4



2
+ ⎛⎝


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 semiminoraxis 3 as shown in the following graph.


Chapter 1 | Parametric Equations and Polar Coordinates 13




1.2


Figure 1.8 Graph of the plane curve described by theparametric 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 aperfect example of using parameterized curves to model a real-world phenomenon.


Eliminate the parameter for the plane curve defined by the following parametric equations and describethe 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 describea plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations forthat curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The processis 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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1.3


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⎛⎝9t
2 − 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.


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 predictablepattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side ofthe tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (Figure 1.9). A cycloidgenerated 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 constantheight 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 ofthe 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 toimagine 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 onthe 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


Chapter 1 | Parametric Equations and Polar Coordinates 15




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 parametricequations 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 thiscase we assume the radius of the larger circle is a and the radius of the smaller circle is b. Then the center of the wheeltravels 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
b
is related to the number of cusps on the graph (cusps are the corners or pointed ends of the graph), as illustrated


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 a


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


Chapter 1 | Parametric Equations and Polar Coordinates 17




The Witch of Agnesi
Many plane curves in mathematics are named after the people who first investigated them, like the folium of Descartesor 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 beenstudied by Fermat in 1630. The mathematician Guido Grandi showed in 1703 how to construct this curve, which helater 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, thetranslator 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 thesecant 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 Ptravels 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 spectrallines. In probability theory, the curve describes the probability density function of the Cauchy distribution. In thisproject you will parameterize these curves.


Figure 1.12 As the point A moves around the circle, the point P traces out the witch ofAgnesi 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 segmentOA.
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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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) = 8a3
x2 + 4a2


.


Chapter 1 | Parametric Equations and Polar Coordinates 19




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 awheel 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 anttrying to get home by hanging onto the edge of a bicycle tire. We have assumed the ant climbed onto the tire at the veryedge, 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 trepresent the angle the tire has rotated through. Looking at Figure 1.13, we see that after the tire has rotated throughan 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 = (xA, yA) denote the position of the ant, we note that
xC − xA = a sin t and yC − yA = a cos t.


Then
xA = xC − a sin t = at − a sin t = a(t − sin t)


yA = 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 alongthe ground. (b) Using geometry to determine the position of the ant after the tire hasrotated through an angle of t.


Note that these are the same parametric representations we had before, but we have now assigned a physical meaningto 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 thespokes toward the center of the wheel. By climbing toward the center of the wheel, the ant has changed his path ofmotion. The new path has less up-and-down motion and is called a curtate cycloid (Figure 1.14). As shown in thefigure, we let b denote the distance along the spoke from the center of the wheel to the ant. As before, we let t representthe angle the tire has rotated through. Additionally, we let C = (xC, yC) represent the position of the center of the
wheel and A = (xA, yA) 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 acurtate cycloid. (c) The new setup, now that the ant has moved closer to the center of thewheel.
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 − xA and for yC − yA.
3. On the basis of your answers to parts 1 and 2, what are the parametric equations representing the curtatecycloid?Once the ant’s head clears, he realizes that the bicyclist has made a turn, and is now traveling away from hishome. So he drops off the bicycle tire and looks around. Fortunately, there is a set of train tracks nearby, headedback 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 (withoutgetting squished!).The ant is still worried about getting dizzy, but the train wheel is slippery and has no spokes to climb, so hedecides to just hang on to the edge of the wheel and hope for the best. Now, train wheels have a flange to keepthe wheel running on the tracks. So, in this case, since the ant is hanging on to the very edge of the flange, thedistance 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 letb denote the distance from the center of the wheel to the ant, and we let t represent the angle the tire hasrotated through. Additionally, we let C = (xC, yC) represent the position of the center of the wheel and


A = (xA, yA) 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 ofmotion is called a prolate cycloid. A graph of a prolate cycloid is shown in the figure.


Chapter 1 | Parametric Equations and Polar Coordinates 21




Figure 1.15 (a) The ant is hanging onto the flange of the train wheel. (b) The newsetup, now that the ant has jumped onto the train wheel. (c) The ant travels along aprolate cycloid.
4. Using the same approach you used in parts 1– 3, find the parametric equations for the path of motion of theant.
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 traincontinues to move forward. He is probably going to be really dizzy by the time he gets home!


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1.1 EXERCISES
For the following exercises, sketch the curves below byeliminating the parameter t. Give the orientation of thecurve.
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 andsketch the graphs.
5. x = 2t2, y = t4 + 1
For the following exercises, use technology (CAS orcalculator) 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 parametricequations by eliminating the parameter. Indicate anyasymptotes 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 parametricequations of a curve into rectangular form. No sketch isnecessary. State the domain of the rectangular form.
21. x = t2 − 1, y = t2
22. x = 1


t + 1
, y = t


1 + 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
36. x = ln(5t)y = ln(t2) where 1 ≤ t ≤ e


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


38. x = tan ty = sec2 t − 1
For the following exercises, the pairs of parametricequations represent lines, parabolas, circles, ellipses, orhyperbolas. Name the type of basic curve that each pair of


Chapter 1 | Parametric Equations and Polar Coordinates 23




equations represents.
39. x = 3t + 4y = 5t − 2


40. x − 4 = 5ty + 2 = t


41. x = 2t + 1y = t2 − 3


42. x = 3 cos ty = 3 sin t


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


44. x = cosh ty = sinh t


45. x = 3 cos ty = 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 ty = 2 sinh t


49. Show that x = h + r cos θy = k + r sin θ represents the equation of
a circle.
50. Use the equations in the preceding problem to find aset of parametric equations for a circle whose radius is 5and whose center is (−2, 3).
For the following exercises, use a graphing utility to graphthe curve represented by the parametric equations andidentify the curve from its equation.
51. [T] x = θ + sin θy = 1 − cos θ


52. [T] x = 2t − 2 sin ty = 2 − 2 cos t


53. [T] x = t − 0.5 sin ty = 1 − 1.5 cos t


54. An airplane traveling horizontally at 100 m/s overflat ground at an elevation of 4000 meters must drop anemergency package on a target on the ground. Thetrajectory 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 themoment of release. How many horizontal meters before thetarget should the package be released in order to hit thetarget?
55. The trajectory of a bullet is given by
x = v0 (cos α) ty = v0 (sin α) t −


1
2
gt2 where


v0 = 500 m/s, g = 9.8 = 9.8 m/s
2, 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 sketch
x = 2 tan(t), y = 3 sec(t), −π < t < π.


58. Sketch the curve known as an epitrochoid, which givesthe path of a point on a circle of radius b as it rolls onthe outside of a circle of radius a. The equations are
x = (a + b)cos t − c · cos




(a + b)t


b



y = (a + b)sin t − c · sin


(a + b)t


b

⎦.


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 theparametric 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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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 conceptin the context of calculus. For example, if we know a parameterization of a given curve, is it possible to calculate the slopeof 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. Ifthe 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 afunction 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 curvedefined 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 givenparametric equations.


Chapter 1 | Parametric Equations and Polar Coordinates 25




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


x − 3 = 2t


t = x − 3
2


.


Substituting this into y(t), we obtain
y(t) = 3t − 4


y = 3⎛⎝
x − 3
2

⎠− 4


y = 3x
2


− 9
2
− 4


y = 3x
2


− 17
2
.


The slope of this line is given by dy
dx


= 3
2
. 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


(1.1)dy
dx


=
dy/dt
dx/dt


=
y′ (t)
x′ (t)


.


Proof
This theorem can be proven using the Chain Rule. In particular, assume that the parameter t can be eliminated, yieldinga 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′ (t)
x′ (t)


.


But F′ ⎛⎝x(t)⎞⎠ = dy
dx


, 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 ofa 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 bedescribed by a function y = f (x) or not.
Example 1.4


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Finding the Derivative of a Parametric Curve
Calculate the derivative dy


dx
for each of the following parametrically defined plane curves, and locate any critical


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 parametricequations in part a.
b. To apply Equation 1.1, first calculate x′ (t) and y′(t):


x′ (t) = 2
y′ (t) = 3t2 − 3.


Chapter 1 | Parametric Equations and Polar Coordinates 27




Next substitute these into the equation:
dy
dx


=
dy/dt
dx/dt


dy
dx


= 3t
2 − 3
2


.


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 parametricequations 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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1.4


t x(t) y(t)
0 5 0
π
2


0 5


π −5 0



2


0 −5


2π 5 0


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


Figure 1.19 Graph of the curve described by parametricequations in part c.


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.


Chapter 1 | Parametric Equations and Polar Coordinates 29




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


= 1
2
, 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 = 1
2
(x − 1)


y − 3 = 1
2
x − 1


2


y = 1
2
x + 5


2
.


Figure 1.20 Tangent line to the parabola described by thegiven parametric equations when t = 2.


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1.5


1.6


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 afunction 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 dy
dx


. This gives us
(1.2)d2 y


dx2
= d


dx


dy
dx

⎠ =


(d/dt)⎛⎝dy/dx⎞⎠
dx/dt


.


If we know dy/dx as a function of t, then this formula is straightforward to apply.
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)⎛⎝dy/dx⎞⎠
dx/dt


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


= −t
−2


2t
= − 1


2t3
.


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 thearea 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.


Chapter 1 | Parametric Equations and Polar Coordinates 31




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 parametricallydefined 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(ti) − x(ti − 1)



⎠.


Then a Riemann sum for the area is
An = ∑


i = 1


n


y⎛⎝x

⎝ t

i





⎝x(ti) − x(ti − 1)



⎠.


Multiplying and dividing each area by ti − ti − 1 gives
An = ∑


i = 1


n


y⎛⎝x

⎝ t

i






x(ti) − x(ti − 1)


ti − ti − 1

⎠(ti − ti − 1) = ∑


i = 1


n


y⎛⎝x

⎝ t

i






x(ti) − x(ti − 1)


Δt

⎠Δt.


Taking the limit as n approaches infinity gives
A = lim


n → ∞
An = ∫


a


b
y(t)x′ (t) dt.


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1.7


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
(1.3)


A = ∫
a


b
y(t)x′ (t) dt.


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


A = ∫
a


b
y(t)x′ (t) dt


= ∫
0



(1 − cos t)(1 − cos t) dt


= ∫
0



(1 − 2 cos t + cos2 t)dt


= ∫
0




⎝1 − 2 cos t +


1 + cos 2t
2



⎠dt


= ∫
0





3
2
− 2 cos t + cos 2t


2

⎠dt


= 3t
2


− 2 sin t + sin 2t
4 |0




= 3π.


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. Inthe case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A topoint B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we startwith an approximation by line segments as shown in the following graph.


Chapter 1 | Parametric Equations and Polar Coordinates 33




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:
(1.4)


s ≈ ∑
k = 1


n


sk = ∑
k = 1


n

⎝x(tk) − x(tk − 1)




2 + ⎛⎝y(tk) − y(tk − 1)




2.


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 t̃k such that


x(tk) − x(tk − 1) = x′

⎝t
^
k

⎠(tk − tk − 1) = x′



⎝t
^
k

⎠Δt


y(tk) − y(tk − 1) = y′

⎝t̃k

⎠(tk − tk − 1) = y′



⎝t̃k

⎠Δt.


Therefore Equation 1.4 becomes
s ≈ ∑


k = 1


n


sk


= ∑
k = 1


n

⎝x′

⎝t
^
k

⎠Δt



2
+ ⎛⎝y′



⎝t̃k

⎠Δt



2


= ∑
k = 1


n

⎝x′

⎝t
^
k





2
(Δt)2 + ⎛⎝y′



⎝t̃k





2
(Δt)2


=



⎜∑
k = 1


n

⎝x′

⎝t
^
k





2
+ ⎛⎝y′



⎝t̃k





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 approximationapproaches the exact arc length. This gives


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s = lim
n → ∞



k = 1


n


sk


= lim
n → ∞





⎜∑
k = 1


n

⎝x′

⎝t
^
k





2
+ ⎛⎝y′



⎝t̃k





2⎞



⎟Δt


= ∫
a


b
(x′ (t))2 + ⎛⎝y′ (t)⎞⎠2dt.


When taking the limit, the values of t^ k and t̃k 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
(1.5)


s = ∫
t1


t2 ⎛

dx
dt



2
+


dy
dt



2


dt.


At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter canbe 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 = ∫
t1


t2 ⎛

dx
dt



2
+


dy
dt



2


dt


= ∫
t1


t2 ⎛

dx
dt



2
+ ⎛⎝F′ (x)


dx
dt



2
dt


= ∫
t1


t2 ⎛

dx
dt



2

⎝1 + (F′ (x))


2⎞
⎠dt


= ∫
t1


t2
x′ (t) 1 +




dy
dx



2


dt.


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


s = ∫
a


b
1 +


dy
dx



2


dx,


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


Chapter 1 | Parametric Equations and Polar Coordinates 35




1.8


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 = ∫
t1


t2 ⎛

dx
dt



2
+


dy
dt



2


dt


= ∫
0


π
(−3 sin t)2 + (3 cos t)2dt


= ∫
0


π
9 sin2 t + 9 cos2 t dt


= ∫
0


π
9⎛⎝sin


2 t + cos2 t⎞⎠dt


= ∫
0


π
3dt = 3t|0


π = 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 itsradius times π.


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 theeffect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at theorigin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can bewritten 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 representedby 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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Then we write the arc length formula as follows:
s(t) = ∫


0


t ⎛

dx
dv



2
+


dy
dv



2


dv


= ∫
0


t
1402 + (−32v + 2)2dv.


The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. Tointegrate this expression we can use a formula from Appendix A,
∫ a2 + u2du = u


2
a2 + u2 + a


2


2
ln|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 = − 1


32
∫ a2 + u2du


= − 1
32







(−32v + 2)


2
1402 + (−32v + 2)2


+140
2


2
ln|(−32v + 2) + 1402 + (−32v + 2)2|







+ C


and
s(t) = − 1


32


(−32t + 2)


2
1402 + (−32t + 2)2 + 140


2


2
ln|(−32t + 2) + 1402 + (−32t + 2)2|⎤⎦


+ 1
32

⎣ 140


2 + 22 + 140
2


2
ln|2 + 1402 + 22|⎤⎦


= ⎛⎝
t
2
− 1


32

⎠ 1024t


2 − 128t + 19604 − 1225
4


ln|(−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 ofthis function with respect to t.While this may seem like a daunting task, it is possible to obtain the answer directly from theFundamental Theorem of Calculus:
d
dx

a


x
f (u) du = f (x).


Therefore
s′ (t) = d


dt

⎣s(t)⎤⎦


= d
dt

⎣∫0


t
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


Chapter 1 | Parametric Equations and Polar Coordinates 37




s⎛⎝
1
3

⎠ =


1/3
2


− 1
32

⎠ 1024




1
3



2
− 128⎛⎝


1
3

⎠+ 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′ ⎛⎝


1
3

⎠ = 2 256




1
3



2
− 16⎛⎝


1
3

⎠+ 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 bya function y = f (x) from x = a to x = b, revolved around the x-axis:


S = 2π∫
a


b
f (x) 1 + ⎛⎝ f ′ (x)⎞⎠2dx.


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 aparametrically defined curve.


The analogous formula for a parametrically defined curve is
(1.6)


S = 2π∫
a


b
y(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt


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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1.9


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 ofthe sphere, we use Equation 1.6:
S = 2π∫


a


b
y(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt


= 2π∫
0


π
r sin t (−r sin t)2 + (r cos t)2dt


= 2π∫
0


π
r sin t r2 sin2 t + r2 cos2 t dt


= 2π∫
0


π
r sin t r2 ⎛⎝sin


2 t + cos2 t⎞⎠dt


= 2π∫
0


π
r2 sin t dt


= 2πr2(−cos t|0
π)


= 2πr2 (−cos π + cos 0)


= 4πr2.


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


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.


Chapter 1 | Parametric Equations and Polar Coordinates 39




1.2 EXERCISES
For the following exercises, each set of parametricequations represents a line. Without eliminating theparameter, 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 thetangent line, then find the equation of the tangent line at thegiven value of the parameter.
66. x = 3 sin t, y = 3 cos t, t = π4
67. x = cos t, y = 8 sin t, t = π


2


68. x = 2t, y = t3, t = −1
69. x = t + 1t , y = t − 1t , t = 1
70. x = t, y = 2t, t = 4
For the following exercises, find all points on the curve thathave 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 − 1t , slope = 1
74. x = 2 + t, y = 2 − 4t, slope = 0
For the following exercises, write the equation of thetangent line in Cartesian coordinates for the givenparameter t.
75. x = e t, y = 1 − ln t2, t = 1
76. x = t ln t, y = sin2 t, t = π4
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 1


2
.


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 whoseequation 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 atwhich tangent line is horizontal or vertical.
91. x = t(t2 − 3), y = 3(t2 − 3)
92. x = 3t


1 + t3
, 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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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 = 12t2, 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 regionsbounded by the parametric curves and the indicated valuesof 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 curveon the indicated interval of the parameter.
108. x = 4t + 3, y = 3t − 2, 0 ≤ t ≤ 2


109. x = 13t3, y = 12t2, 0 ≤ t ≤ 1
110. x = cos(2t), y = sin(2t), 0 ≤ t ≤ π2
111. x = 1 + t2, y = (1 + t)3, 0 ≤ t ≤ 1
112. x = et cos t, y = et sin t, 0 ≤ t ≤ π2 (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 length of the ellipse
x = 4 sin θ, y = 3 cos θ is
L = 16∫


0


π/2
1 − e2 sin2 θ dθ, where e = ca and


c = a2 − b2.


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 surfaceobtained 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 ≤ π2
121. [T] Use a CAS to find the area of the surface
generated by rotating x = t + t3, 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.


Chapter 1 | Parametric Equations and Polar Coordinates 41




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 orderedpairs to points. This is called a one-to-one mapping from points in the plane to ordered pairs. The polar coordinate systemprovides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polarcoordinates 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 = xr 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 angularcoordinate. 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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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:


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


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


= 12 + 12


r = 2


and


tan θ =
y
x


= 1
1
= 1


θ = π
4
.


Therefore this point can be represented as ⎛⎝ 2, π4⎞⎠ in polar coordinates.
b. Use x = −3 and y = 4 in Equation 1.8:


r2 = x2 + y2


= (−3)2 + (4)2


r = 5


and


tan θ =
y
x


= −4
3


θ = −arctan⎛⎝
4
3



≈ 2.21.


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


Chapter 1 | Parametric Equations and Polar Coordinates 43




c. Use x = 0 and y = 3 in Equation 1.8:
r2 = x2 + y2


= (3)2 + (0)2


= 9 + 0
r = 3


and
tan θ =


y
x


= 3
0
.


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 π


2
. Therefore this point can be represented as ⎛⎝3, π2⎞⎠ in polar


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


r2 = x2 + y2


= ⎛⎝5 3


2 + (−5)2


= 75 + 25
r = 10


and


tan θ =
y
x


= −5
5 3


= − 3
3


θ = −π
6
.


Therefore this point can be represented as ⎛⎝10, − π6⎞⎠ in polar coordinates.
e. Use r = 3 and θ = π


3
in Equation 1.7:


x = r cos θ


= 3 cos⎛⎝
π
3



= 3⎛⎝
1
2

⎠ =


3
2


and


y = r sin θ


= 3 sin⎛⎝
π
3



= 3⎛⎝
3
2

⎠ =


3 3
2


.


Therefore this point can be represented as ⎛⎝32, 3 32 ⎞⎠ in rectangular coordinates.
f. Use r = 2 and θ = 3π


2
in Equation 1.7:
x = r cos θ


= 2 cos⎛⎝

2



= 2(0) = 0


and


y = r sin θ


= 2 sin⎛⎝

2



= 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


= 6 cos⎛⎝−

6



= 6⎛⎝−
3
2



= −3 3


and


y = r sin θ


= 6 sin⎛⎝−

6



= 6⎛⎝−
1
2



= −3.


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1.10


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


Convert (−8, −8) into polar coordinates and ⎛⎝4, 2π3 ⎞⎠ into rectangular coordinates.


The polar representation of a point is not unique. For example, the polar coordinates ⎛⎝2, π3⎞⎠ and ⎛⎝2, 7π3 ⎞⎠ 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,



3

⎠ also represents the point ⎛⎝1, 3⎞⎠ in the rectangular system, as we can see by using Equation 1.8:


x = r cos θ


= −2 cos⎛⎝

3



= −2⎛⎝−
1
2

⎠ = 1


and


y = r sin θ


= −2 sin⎛⎝

3



= −2⎛⎝−
3
2

⎠ = 3.


Every point in the plane has an infinite number of representations in polar coordinates. However, each point in the plane hasonly 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.


Chapter 1 | Parametric Equations and Polar Coordinates 45




1.11


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 measurethe 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 theterminal 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, π4⎞⎠
b. ⎛⎝−3, 2π3 ⎞⎠
c. ⎛⎝4, 5π4 ⎞⎠


Solution
The three points are plotted in the following figure.


Figure 1.29 Three points plotted in the polar coordinatesystem.


Plot ⎛⎝4, 5π3 ⎞⎠ and ⎛⎝−3, − 7π2 ⎞⎠ 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 rectangularcoordinate 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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The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangularcoordinates. 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, forexample, 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 sketchingpolar 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.


Chapter 1 | Parametric Equations and Polar Coordinates 47




θ r = 4 sinθ θ r = 4 sinθ


0 0 π 0
π
6


2 7π
6


−2


π
4 2 2 ≈ 2.8



4


−2 2 ≈ −2.8


π
3 2 3 ≈ 3.4



3


−2 3 ≈ −3.4


π
2


4 3π
2


4



3


2 3 ≈ 3.4 5π
3


−2 3 ≈ −3.4



4


2 2 ≈ 2.8 7π
4


−2 2 ≈ −2.8



6


2 11π
6


−2


2π 0


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1.12


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 + ⎛⎝y
2 − 4y⎞⎠ = 0


x2 + ⎛⎝y
2 − 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.


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 coordinatesusing the coordinate transformation formulas in Equation 1.8. Example 1.14 gives some more examples of functionsfor 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


Chapter 1 | Parametric Equations and Polar Coordinates 49




1.13


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 tanK.


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 newpoints unintentionally. This should always be taken into consideration. However, in this case we do not
introduce new points. For example, ⎛⎝−3, π3⎞⎠ is the same point as ⎛⎝3, 4π3 ⎞⎠.)


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 equationto the left-hand side, then complete the square.
x2 + y2 = 6x − 8y


x2 − 6x + y2 + 8y = 0

⎝x


2 − 6x⎞⎠+

⎝y


2 + 8y⎞⎠ = 0



⎝x


2 − 6x + 9⎞⎠+

⎝y


2 + 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.


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 commoncurves is given in the tables below. In each equation, a and b are arbitrary constants.


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Figure 1.31


Chapter 1 | Parametric Equations and Polar Coordinates 51




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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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 interestinggraphs 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 abovethe 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 thatthe 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 two-dimensional subset of the real plane. In this case the curve occupies the circle of radius 3 centered at the origin.
Example 1.14


Chapter 1 | Parametric Equations and Polar Coordinates 53




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


Figure 1.35 How can we describe a spiral graphmathematically?
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⎛⎝
y
x



x2 + y2 = k arctan⎛⎝
y
x



arctan⎛⎝
y
x

⎠ =


x2 + y2


k


y = x tan



⎜ x


2 + y2


k





⎟.


Although this equation describes the spiral, it is not possible to solve it directly for either x or y. However, ifwe 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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Figure 1.36 A logarithmic spiral is similar to the shape of the chambered nautilus shell. (credit: modification ofwork 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 revealother 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.


Chapter 1 | Parametric Equations and Polar Coordinates 55




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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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 originalequation 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.


Chapter 1 | Parametric Equations and Polar Coordinates 57




θ r


0 0


π
6 3 32


≈ 2.6


π
4


3


π
3 3 32


≈ 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.


58 Chapter 1 | Parametric Equations and Polar Coordinates


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1.14


Figure 1.38 The entire graph of the equation is called a four-petaled rose.


Determine the symmetry of the graph determined by the equation r = 2 cos(3θ) and create a graph.


Chapter 1 | Parametric Equations and Polar Coordinates 59




1.3 EXERCISES
In the following exercises, plot the point whose polarcoordinates are given by first constructing the angle θ and
then marking off the distance r along the ray.
125. ⎛⎝3, π6⎞⎠
126. ⎛⎝−2, 5π3 ⎞⎠
127. ⎛⎝0, 7π6 ⎞⎠
128. ⎛⎝−4, 3π4 ⎞⎠
129. ⎛⎝1, π4⎞⎠
130. ⎛⎝2, 5π6 ⎞⎠
131. ⎛⎝1, π2⎞⎠
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 ofa 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 coordinatesfor the given point in polar coordinates.
142. ⎛⎝2, 5π4 ⎞⎠
143. ⎛⎝−2, π6⎞⎠
144. ⎛⎝5, π3⎞⎠
145. ⎛⎝1, 7π6 ⎞⎠
146. ⎛⎝−3, 3π4 ⎞⎠
147. ⎛⎝0, π2⎞⎠
148. (−4.5, 6.5)
For the following exercises, determine whether the graphsof 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⎛⎝θ5⎞⎠
152. r = 2 sec θ
153. r = 1 + cos θ
For the following exercises, describe the graph of eachpolar equation. Confirm each description by convertinginto a rectangular equation.


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154. r = 3
155. θ = π


4


156. r = sec θ
157. r = csc θ
For the following exercises, convert the rectangularequation to polar form and sketch its graph.
158. x2 + y2 = 16
159. x2 − y2 = 16
160. x = 8
For the following exercises, convert the rectangularequation to polar form and sketch its graph.
161. 3x − y = 2
162. y2 = 4x
For the following exercises, convert the polar equation torectangular 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 polarequation 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⎛⎝θ2⎞⎠
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 = 6


2 sin θ − 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⎛⎝3θ7 ⎞⎠ (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.


Chapter 1 | Parametric Equations and Polar Coordinates 61




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
and the x-axis is given by A = ∫


a


b
f (x) dx. This fact, along with the formula for evaluating this integral, is summarized in


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


b
1 + ⎛⎝ f ′ (x)⎞⎠2dx. In this


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 ofthe Fundamental Theorem of Calculus used the concept of a Riemann sum to approximate the area under a curve by usingrectangles. 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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The line segments are connected by arcs of constant radius. This defines sectors whose areas can be calculated by using ageometric formula. The area of each sector is then used to approximate the area between successive line segments. We thensum the areas of the sectors to approximate the total area. This approach gives a Riemann sum approximation for the totalarea. 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 = 1


2
θr2.


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



, so the


area of the sector is this fraction multiplied by the total area:
A = ⎛⎝


θ


⎠πr


2 = 1
2
θr2.


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 ⎛⎝θi






2.


Therefore a Riemann sum that approximates the area is given by
An = ∑


i = 1


n


Ai ≈ ∑
i = 1


n
1
2
(Δθ)⎛⎝ f ⎛⎝θi






2.


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


n → ∞
An = 12



α


β

⎝ f (θ)⎞⎠2dθ.


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


(1.9)
A = 1


2

α


β

⎣ f (θ)⎤⎦2dθ = 12



α


β
r2dθ.


Chapter 1 | Parametric Equations and Polar Coordinates 63




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:


A = 1
2

α


β

⎣ f (θ)⎤⎦2dθ


= 1
2


0


π/2

⎣3 sin(2θ)⎤⎦2dθ


= 1
2


0


π/2
9 sin2 (2θ) dθ.


To evaluate this integral, use the formula sin2α = (1 − cos(2α))/2 with α = 2θ:


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1.15


A = 1
2


0


π/2
9 sin2 (2θ) dθ


= 9
2


0


π/2(1 − cos(4θ))
2




= 9
4





⎜∫


0


π/2
1 − cos(4θ) dθ







= 9
4

⎝θ −


sin(4θ)
4 |0


π/2


= 9
4


π
2
− sin 2π


4

⎠−


9
4

⎝0 −


sin 4(0)
4



= 9π
8
.


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 PolarCurve to find the area between two polar curves. However, we often need to find the points of intersection of the curvesand 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.


Chapter 1 | Parametric Equations and Polar Coordinates 65




1.16


To determine the limits of integration, first find the points of intersection by setting the two functions equal toeach other and solving for θ:
6 sin θ = 2 + 2 sin θ
4 sin θ = 2


sin θ = 1
2
.


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

π/6


5π/6
[6 sin θ]2dθ − 1


2

π/6


5π/6
[2 + 2 sin θ]2dθ


= 1
2

π/6


5π/6
36 sin2 θ dθ − 1


2

π/6


5π/6
4 + 8 sin θ + 4 sin2 θ dθ


= 18∫
π/6


5π/61 − cos(2θ)
2


dθ − 2∫
π/6


5π/6
1 + 2 sin θ + 1 − cos(2θ)


2


= 9

⎣θ −


sin(2θ)
2

⎦π/6
5π/6


− 2



2


− 2 cos θ − sin(2θ)
4

⎦π/6
5π/6


= 9⎛⎝

6


− sin 2(5π/6)
2



⎠− 9


π
6
− sin 2(π/6)


2



−⎛⎝3



6

⎠− 4 cos



6


− sin 2(5π/6)
2



⎠+

⎝3


π
6

⎠− 4 cos


π
6
− sin 2(π/6)


2



= 4π.


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 solutionis 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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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


L = ∫
a


b ⎛

dx
dt



2
+


dy
dt



2


dt.


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


= f ′ (θ) cos θ − f (θ) sin θ


dy


= 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 = ∫
a


b ⎛

dx
dt



2
+


dy
dt



2


dt


= ∫
α


β ⎛

dx




2
+


dy




2




= ∫
α


β

⎝ f ′ (θ) cos θ − f (θ) sin θ⎞⎠2 + ⎛⎝ f ′ (θ) sin θ + f (θ) cos θ⎞⎠2dθ


= ∫
α


β

⎝ f ′ (θ)⎞⎠2 ⎛⎝cos


2 θ + sin2 θ⎞⎠+

⎝ f (θ)⎞⎠2 ⎛⎝cos


2 θ + sin2 θ⎞⎠dθ


= ∫
α


β

⎝ f ′ (θ)⎞⎠2 + ⎛⎝ f (θ)⎞⎠2dθ


= ∫
α


β
r2 + ⎛⎝


dr




2
dθ.


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


(1.10)
L = ∫


α


β

⎣ f (θ)⎤⎦2 + ⎡⎣ f ′ (θ)⎤⎦2dθ = ∫


α


β
r2 + ⎛⎝


dr




2
dθ.


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


Chapter 1 | Parametric Equations and Polar Coordinates 67




1.17


Solution
When θ = 0, r = 2 + 2cos0 = 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 + 2cosθ, α = 0, and β = 2π, Equation
1.10 becomes


L = ∫
α


β

⎣ f (θ)⎤⎦2 + ⎡⎣ f ′ (θ)⎤⎦2 dθ


= ∫
0



[2 + 2cosθ]2 + [−2sinθ]2 dθ


= ∫
0



4 + 8cosθ + 4cos2 θ + 4sin2 θ dθ


= ∫
0



4 + 8cosθ + 4⎛⎝cos


2 θ + sin2 θ⎞⎠ dθ


= ∫
0



8 + 8cosθ dθ


= 2∫
0



2 + 2cosθ dθ.


Next, using the identity cos(2α) = 2cos2α − 1, add 1 to both sides and multiply by 2. This gives
2 + 2cos(2α) = 4cos2α. Substituting α = θ/2 gives 2 + 2cosθ = 4cos2(θ/2), so the integral becomes


L = 2∫
0



2 + 2 cos θdθ


= 2∫
0



4 cos2 ⎛⎝


θ
2

⎠dθ


= 2∫
0




|cos⎛⎝θ2⎞⎠|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 = 4∫
0




|cos⎛⎝θ2⎞⎠|dθ
= 8∫


0


π
cos⎛⎝


θ
2

⎠dθ


= 8⎛⎝2 sin


θ
2

⎠|0
π


= 16.


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


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1.4 EXERCISES
For the following exercises, determine a definite integralthat 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 describedregion.
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 thatrepresents 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 curveover 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 capabilitiesof 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 = sin2 ⎛⎝θ2⎞⎠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


Chapter 1 | Parametric Equations and Polar Coordinates 69




geometry to find the area of the region described and thenconfirm 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 fromgeometry to find the length of the curve and then confirmusing 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


= f ′(θ)sin θ + f (θ)cos θ.


For the following exercises, find the slope of a tangent lineto 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, π6⎞⎠
237. r = 4 cos θ; ⎛⎝2, π3⎞⎠
238. r = 8 sin θ; ⎛⎝4, 5π6 ⎞⎠
239. r = 4 + sin θ; ⎛⎝3, 3π2 ⎞⎠
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; ⎛⎝π2, π4⎞⎠
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 = π


3
.


For the following exercises, find the slope of the tangentline to the given polar curve at the point given by the valueof 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 thefollowing polar curves have a horizontal or vertical tangentline.
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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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 mathematicalconcept. As early as 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinatedby these curves. Appollonius wrote an entire eight-volume treatise on conic sections in which he was, for example, able toderive a specific method for identifying a conic section through the use of geometry. Since then, important applications ofconic 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 theirproperties, and their equations.
Conic sections get their name because they can be generated by intersecting a plane with a cone. A cone has two identicallyshaped parts called nappes. One nappe is what most people mean by “cone,” having the shape of a party hat. A right circularcone 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 ofrevolution (the y-axis), then the conic section is a hyperbola. If the plane is parallel to the generating line, the conic sectionis a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects onenappe at an angle to the axis (other than 90°), then the conic section is an ellipse.


Chapter 1 | Parametric Equations and Polar Coordinates 71




Figure 1.44 The four conic sections. Each conic is determined by the angle the plane makes with the axis ofthe cone.
Parabolas
A parabola is generated when a plane intersects a cone parallel to the generating line. In this case, the plane intersects onlyone 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 froma fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of theparabola.


A graph of a typical parabola appears in Figure 1.45. Using this diagram in conjunction with the distance formula, we canderive 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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Figure 1.45 A typical parabola in which the distance from thefocus 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


(1.11)y = 1
4p


(x − h)2 + k.


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 casescan also be written in standard form as shown in the following graphs.


Chapter 1 | Parametric Equations and Polar Coordinates 73




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 arenot 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 openseither 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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1.18


Solution
Since y is not squared in this equation, we know that the parabola opens either upward or downward. Thereforewe 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 right-hand side using parentheses:
8y = ⎛⎝x


2 − 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 ⎛⎝−42 ⎞⎠


2
= 4. Add 4


inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:
8y = ⎛⎝x


2 − 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.


Put the equation 2y2 − x + 12y + 16 = 0 into standard form and graph the resulting parabola.


Chapter 1 | Parametric Equations and Polar Coordinates 75




The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. Theparabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam ofelectromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to theaxis 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, anda receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of theparabola to the receiver, which collects and decodes the digital signals. This allows a small receiver to gather signals from awide angle of sky. Flashlights and headlights in a car work on the same principle, but in reverse: the source of the light (thatis, 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 twodirectrices (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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Figure 1.48 A typical ellipse in which the sum of the distances from anypoint 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 thispoint 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:


Chapter 1 | Parametric Equations and Polar Coordinates 77




−2cx = 4a2 − 4a (x + c)2 + y2 + 2cx


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


(x + c)2 + y2 = a + cxa


(x + c)2 + y2 = a2 + 2cx + c
2 x2


a2


x2 + 2cx + c2 + y2 = a2 + 2cx + c
2 x2


a2


x2 + c2 + y2 = a2 + c
2 x2


a2
.


Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
x2 − c


2 x2


a2
+ y2 = a2 − c2



⎝a


2 − c2⎞⎠x
2


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 becausethe 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


(1.12)(x − h)2
a2


+

⎝y − k⎞⎠2


b2
= 1


and the foci are located at (h ± c, k), where c2 = a2 − b2. The equations of the directrices are x = h ± a2c .
If the major axis is vertical, then the equation of the ellipse becomes


(1.13)(x − h)2
b2


+

⎝y − k⎞⎠2


a2
= 1


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


2


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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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 ofcompleting 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:

⎝9x


2 − 36x⎞⎠+

⎝4y


2 + 24y⎞⎠ = −36


9⎛⎝x
2 − 4x⎞⎠+ 4



⎝y


2 + 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 ⎛⎝−42 ⎞⎠


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⎛⎝x


2 − 4x + 4⎞⎠+ 4

⎝y


2 + 6y + 9⎞⎠ = −36 + 36 + 36


9⎛⎝x
2 − 4x + 4⎞⎠+ 4



⎝y


2 + 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.


Chapter 1 | Parametric Equations and Polar Coordinates 79




1.19


Figure 1.49 The ellipse in Example 1.20.


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 oneof the foci as shown in Figure 1.50(a). Because Earth’s orbit is an ellipse, the distance from the Sun varies throughout theyear. A commonly held misconception is that Earth is closer to the Sun in the summer. In fact, in summer for the northernhemisphere, Earth is farther from the Sun than during winter. The difference in season is caused by the tilt of Earth’s axisin the orbital plane. Comets that orbit the Sun, such as Halley’s Comet, also have elliptical orbits, as do moons orbiting theplanets and satellites orbiting Earth.
Ellipses also have interesting reflective properties: A light ray emanating from one focus passes through the other focusafter mirror reflection in the ellipse. The same thing occurs with a sound wave as well. The National Statuary Hall in theU.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure 1.50(b). This hall served asthe meeting place for the U.S. House of Representatives for almost fifty years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand onthese spots and speak to each other, they can hear each other much more clearly than they can hear someone standing closeby. Legend has it that John Quincy Adams had his desk located on one of the foci and was able to eavesdrop on everyoneelse in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because theoriginal ceiling produced so many echoes that the entire room had to be hung with carpets to dampen the noise. The ceilingwas rebuilt in 1902 and only then did the now-famous whispering effect emerge. Another famous whispering gallery—thesite 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 awhispering gallery with an elliptical cross section.


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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) isconstant.


A graph of a typical hyperbola appears as follows.


Figure 1.51 A typical hyperbola in which the difference of the distances from anypoint on the ellipse to the foci is constant. The transverse axis is also called the majoraxis, 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 liesin 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, F1⎞⎠− d⎛⎝P, F2⎞⎠| = 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 subtractionis reversed. The vertex of the right branch has coordinates (a, 0), so


d⎛⎝P, F1

⎠− d⎛⎝P, F2



⎠ = (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, F1



⎠− d⎛⎝P, F2



⎠ = 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:


Chapter 1 | Parametric Equations and Polar Coordinates 81




−2cx = 4a2 + 4a (x + c)2 + y2 + 2cx


4a (x + c)2 + y2 = −4a2 − 4cx


(x + c)2 + y2 = −a − cxa


(x + c)2 + y2 = a2 + 2cx + c
2 x2


a2


x2 + 2cx + c2 + y2 = a2 + 2cx + c
2 x2


a2


x2 + c2 + y2 = a2 + c
2 x2


a2
.


Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:
x2 − c


2 x2


a2
+ y2 = a2 − c2



⎝a


2 − c2⎞⎠x
2


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


(1.14)(x − h)2
a2




⎝y − k⎞⎠2


b2
= 1


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 ± a
2


a2 + b2
= h ± a


2


c .


If the major axis is vertical, then the equation of the hyperbola becomes
(1.15)⎛


⎝y − k⎞⎠2


a2
− (x − h)


2


b2
= 1


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


b
(x − h). The equations of the directrices are


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y = k ± a
2


a2 + b2
= k ± a


2


c .


If the major axis (transverse axis) is horizontal, then the hyperbola is called horizontal, and if the major axis is verticalthen 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:

⎝9x


2 + 36x⎞⎠−

⎝16y


2 − 32y⎞⎠ = 124


9⎛⎝x
2 + 4x⎞⎠− 16



⎝y


2 − 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⎛⎝x


2 + 4x + 4⎞⎠− 16

⎝y


2 − 2y + 1⎞⎠ = 124 + 36 − 16


9⎛⎝x
2 + 4x + 4⎞⎠− 16



⎝y


2 − 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 ± 3


4
(x + 2). The graph of this hyperbola appears in the following figure.


Chapter 1 | Parametric Equations and Polar Coordinates 83




1.20


Figure 1.52 Graph of the hyperbola in Example 1.21.


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 ahyperbolic 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 signalsfrom two towers is constant along hyperbolas), and in the construction of mirrors inside telescopes (to reflect light comingfrom the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solarsystem, if the speed is great enough to escape the Sun’s gravitational pull, then the path that the comet takes as it passesthrough 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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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 conicsection, 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 knownas the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have two foci and two associateddirectrices. 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 distancefrom 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 = ±a2c . 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 a2c − c. This gives the eccentricity as


e = a − c
a2
c − a


= c(a − c)
a2 − ac


= c(a − c)
a(a − c)


= ca.


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 = ±a2c , and a similar calculation shows that the eccentricity of a hyperbola is also e = ca. However in
this case we have c > a, so the eccentricity of a hyperbola is greater than 1.


Chapter 1 | Parametric Equations and Polar Coordinates 85




1.21


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 = ca = 35 = 0.6.


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 ofthe 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 ofa parabola, a represents the distance from the vertex to the focus.


Conic e p
Ellipse 0 < e < 1


a2 − c2
c =


a⎛⎝1 − e
2⎞


c


Parabola e = 1 2a


Hyperbola e > 1
c2 − a2


c =
a⎛⎝e


2 − 1⎞⎠
e


Table 1.7 Eccentricities and Focal Parameters of theConic Sections
Using the definitions of the focal parameter and eccentricity of the conic section, we can derive an equation for any conicsection in polar coordinates. In particular, we assume that one of the foci of a given conic section lies at the pole. Then usingthe 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 =
ep


1 ± e cos θ
or r =


ep
1 ± e sin θ


.


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In the equation on the left, the major axis of the conic section is horizontal, and in the equation on the right, the major axisis 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 ofthe plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. Thisvalue identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conicis vertical. If both appear then the axes are rotated. The center of the conic is not necessarily at the origin. The center is atthe 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 = 3
1 + 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 = 3


2
. 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
2


3



4


3
1 − 2


≈ −7.2426 7π
4


3
1 + 2


≈ 1.2426


Chapter 1 | Parametric Equations and Polar Coordinates 87




1.22


Figure 1.55 Graph of the hyperbola described in Example1.23.


Identify and create a graph of the conic section described by the equation
r = 4


1 − 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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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 − C


B
. 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 rotatedcoordinate 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 − C


B
.


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


Chapter 1 | Parametric Equations and Polar Coordinates 89




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 − C


B
. This gives


cot 2θ = A − C
B


= 13 − 7
−6 3


= − 3
3
.


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


= 13cos260 + ⎛⎝−6 3

⎠ cos 60 sin 60 + 7sin


260


= 13⎛⎝
1
2



2
− 6 3⎛⎝


1
2





3
2

⎠+ 7



3
2



2


= 4,
B′ = 0,


C′ = A sin2 θ − B sin θ cos θ + C cos2 θ


= 13sin260 + ⎛⎝−6 3

⎠ sin 60 cos 60 = 7cos


260


= ⎛⎝
3
2



2
+ 6 3⎛⎝


3
2




1
2

⎠+ 7


1
2



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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1.23


Figure 1.57 Graph of the ellipse described by the equation
13x2 − 6 3xy + 7y2 − 256 = 0. The axes are rotated 60°.
The red dashed lines indicate the rotated axes.


Identify the conic and calculate the angle of rotation of axes for the curve described by the equation
3x2 + 5xy − 2y2 − 125 = 0.


Chapter 1 | Parametric Equations and Polar Coordinates 91




1.5 EXERCISES
For the following exercises, determine the equation of theparabola using the information given.
255. Focus (4, 0) and directrix x = −4
256. Focus (0, −3) and directrix y = 3
257. Focus (0, 0.5) and directrix y = −0.5
258. Focus (2, 3) and directrix x = −2
259. Focus (0, 2) and directrix y = 4
260. Focus (−1, 4) and directrix x = 5
261. Focus (−3, 5) and directrix y = 1
262. Focus ⎛⎝52, −4⎞⎠ and directrix x = 72
For the following exercises, determine the equation of theellipse using the information given.
263. Endpoints of major axis at (4, 0), (−4, 0) and foci
located at (2, 0), (−2, 0)
264. Endpoints of major axis at (0, 5), (0, −5) and foci
located at (0, 3), (0, −3)
265. Endpoints of major axis at (0, 2), (0, −2) and foci
located at (3, 0), (−3, 0)
266. Endpoints of major axis at (−3, 3), (7, 3) and foci
located at (−2, 3), (6, 3)
267. Endpoints of major axis at (−3, 5), (−3, −3) and
foci located at (−3, 3), (−3, −1)
268. Endpoints of major axis at (0, 0), (0, 4) and foci
located at (5, 2), (−5, 2)
269. Foci located at (2, 0), (−2, 0) and eccentricity of
1
2


270. Foci located at (0, −3), (0, 3) and eccentricity of
3
4


For the following exercises, determine the equation of thehyperbola using the information given.


271. Vertices located at (5, 0), (−5, 0) and foci located
at (6, 0), (−6, 0)
272. Vertices located at (0, 2), (0, −2) and foci located
at (0, 3), (0, −3)
273. Endpoints of the conjugate axis located at
(0, 3), (0, −3) and foci located (4, 0), (−4, 0)
274. Vertices located at (0, 1), (6, 1) and focus located
at (8, 1)
275. Vertices located at (−2, 0), (−2, −4) and focus
located at (−2, −8)
276. Endpoints of the conjugate axis located at
(3, 2), (3, 4) and focus located at (3, 7)
277. Foci located at (6, −0), (6, 0) and eccentricity of 3
278. (0, 10), (0, −10) and eccentricity of 2.5
For the following exercises, consider the following polarequations of conics. Determine the eccentricity and identifythe conic.
279. r = −1


1 + cos θ


280. r = 8
2 − sin θ


281. r = 5
2 + sin θ


282. r = 5
−1 + 2 sin θ


283. r = 3
2 − 6 sin θ


284. r = 3
−4 + 3 sin θ


For the following exercises, find a polar equation of theconic with focus at the origin and eccentricity and directrixas given.
285. Directrix: x = 4; e = 1


5


286. Directrix: x = −4; e = 5
287. Directrix: y = 2; e = 2


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288. Directrix: y = −2; e = 1
2


For the following exercises, sketch the graph of each conic.
289. r = 1


1 + sin θ


290. r = 1
1 − cos θ


291. r = 4
1 + cos θ


292. r = 10
5 + 4 sin θ


293. r = 15
3 − 2 cos θ


294. r = 32
3 + 5 sin θ


295. r(2 + sin θ) = 4
296. r = 3


2 + 6 sin θ


297. r = 3
−4 + 2 sin θ


298. x29 + y
2


4
= 1


299. x2
4


+
y2


16
= 1


300. 4x2 + 9y2 = 36
301. 25x2 − 4y2 = 100


302. x2
16



y2


9
= 1


303. x2 = 12y
304. y2 = 20x
305. 12x = 5y2
For the following equations, determine which of the conicsections is described.
306. xy = 4
307. x2 + 4xy − 2y2 − 6 = 0


308. x2 + 2 3xy + 3y2 − 6 = 0
309. x2 − xy + y2 − 2 = 0
310. 34x2 − 24xy + 41y2 − 25 = 0
311. 52x2 − 72xy + 73y2 + 40x + 30y − 75 = 0
312. The mirror in an automobile headlight has a paraboliccross section, with the lightbulb at the focus. On aschematic, the equation of the parabola is given as
x2 = 4y. At what coordinates should you place the
lightbulb?
313. A satellite dish is shaped like a paraboloid ofrevolution. The receiver is to be located at the focus. If thedish is 12 feet across at its opening and 4 feet deep at itscenter, where should the receiver be placed?
314. Consider the satellite dish of the preceding problem.If the dish is 8 feet across at the opening and 2 feet deep,where should we place the receiver?
315. A searchlight is shaped like a paraboloid ofrevolution. A light source is located 1 foot from the basealong the axis of symmetry. If the opening of thesearchlight is 3 feet across, find the depth.
316. Whispering galleries are rooms designed withelliptical ceilings. A person standing at one focus canwhisper and be heard by a person standing at the otherfocus because all the sound waves that reach the ceiling arereflected to the other person. If a whispering gallery has alength of 120 feet and the foci are located 30 feet from thecenter, find the height of the ceiling at the center.
317. A person is standing 8 feet from the nearest wall ina whispering gallery. If that person is at one focus and theother focus is 80 feet away, what is the length and the heightat the center of the gallery?
For the following exercises, determine the polar equationform of the orbit given the length of the major axis andeccentricity for the orbits of the comets or planets. Distanceis given in astronomical units (AU).
318. Halley’s Comet: length of major axis = 35.88,eccentricity = 0.967
319. Hale-Bopp Comet: length of major axis = 525.91,eccentricity = 0.995
320. Mars: length of major axis = 3.049, eccentricity =0.0934
321. Jupiter: length of major axis = 10.408, eccentricity =0.0484


Chapter 1 | Parametric Equations and Polar Coordinates 93




angular coordinate
cardioid
conic section
cusp
cycloid
directrix
discriminant


eccentricity
focal parameter
focus
general form
limaçon
major axis


minor axis
nappe
orientation
parameter
parameterization of a curve
parametric curve
parametric equations
polar axis
polar coordinate system


polar equation
pole


CHAPTER 1 REVIEW
KEY TERMS


θ the angle formed by a line segment connecting the origin to a point in the polar coordinate
system with the positive radial (x) axis, measured counterclockwise


a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius;the equation of a cardioid is r = a(1 + sin θ) or r = a(1 + cos θ)
a conic section is any curve formed by the intersection of a plane with a cone of two nappes


a pointed end or part where two curves meet
the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage
a directrix (plural: directrices) is a line used to construct and define a conic section; a parabola has one directrix;ellipses and hyperbolas have two


the value 4AC − B2, which is used to identify a conic when the equation contains a term involving xy,
is called a discriminant


the eccentricity is defined as the distance from any point on the conic section to its focus divided by theperpendicular distance from that point to the nearest directrix
the focal parameter is the distance from a focus of a conic section to the nearest directrix


a focus (plural: foci) is a point used to construct and define a conic section; a parabola has one focus; an ellipse anda hyperbola have two
an equation of a conic section written as a general second-degree equation


the graph of the equation r = a + b sin θ or r = a + b cos θ. If a = b then the graph is a cardioid
the major axis of a conic section passes through the vertex in the case of a parabola or through the twovertices in the case of an ellipse or hyperbola; it is also an axis of symmetry of the conic; also called the transverseaxis
the minor axis is perpendicular to the major axis and intersects the major axis at the center of the conic, or atthe vertex in the case of the parabola; also called the conjugate axis


a nappe is one half of a double cone
the direction that a point moves on a graph as the parameter increases
an independent variable that both x and y depend on in a parametric curve; usually represented by the variablet


rewriting the equation of a curve defined by a function y = f (x) as parametric
equations


the graph of the parametric equations x(t) and y(t) over an interval a ≤ t ≤ b combined with the
equations


the equations x = x(t) and y = y(t) that define a parametric curve
the horizontal axis in the polar coordinate system corresponding to r ≥ 0


a system for locating points in the plane. The coordinates are r, the radial coordinate, and
θ, the angular coordinate


an equation or function relating the radial coordinate to the angular coordinate in the polar coordinatesystem
the central point of the polar coordinate system, equivalent to the origin of a Cartesian system


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radial coordinate


rose
space-filling curve
standard form
vertex


r the coordinate in the polar coordinate system that measures the distance from a point in the plane to
the pole
graph of the polar equation r = a cos 2θ or r = a sin 2θ for a positive constant a


a curve that completely occupies a two-dimensional subset of the real plane
an equation of a conic section showing its properties, such as location of the vertex or lengths of majorand minor axes


a vertex is an extreme point on a conic section; a parabola has one vertex at its turning point. An ellipse has twovertices, one at each end of the major axis; a hyperbola has two vertices, one at the turning point of each branch
KEY EQUATIONS


• Derivative of parametric equations
dy
dx


=
dy/dt
dx/dt


=
y′ (t)
x′ (t)


• Second-order derivative of parametric equations
d2 y


dx2
= d


dx


dy
dx

⎠ =


(d/dt)⎛⎝dy/dx⎞⎠
dx/dt


• Area under a parametric curve
A = ∫


a


b
y(t)x′ (t) dt


• Arc length of a parametric curve
s = ∫


t1


t2 ⎛

dx
dt



2
+


dy
dt



2


dt


• Surface area generated by a parametric curve
S = 2π∫


a


b
y(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt


• Area of a region bounded by a polar curve
A = 1


2

α


β

⎣ f (θ)⎤⎦2dθ = 12



α


β
r2dθ


• Arc length of a polar curve
L = ∫


α


β

⎣ f (θ)⎤⎦2 + ⎡⎣ f ′ (θ)⎤⎦2dθ = ∫


α


β
r2 + ⎛⎝


dr




2


KEY CONCEPTS
1.1 Parametric Equations


• Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some othermeaningful quantity.
• It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describingthat curve.
• There is always more than one way to parameterize a curve.
• Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe usingrectangular coordinates.


Chapter 1 | Parametric Equations and Polar Coordinates 95




1.2 Calculus of Parametric Curves
• The derivative of the parametrically defined curve x = x(t) and y = y(t) can be calculated using the formula


dy
dx


=
y′(t)
x′(t)


. Using the derivative, we can find the equation of a tangent line to a parametric curve.


• The area between a parametric curve and the x-axis can be determined by using the formula A = ∫
t1


t2
y(t)x′ (t) dt.


• The arc length of a parametric curve can be calculated by using the formula s = ∫
t1


t2 ⎛

dx
dt



2
+


dy
dt



2


dt.


• The surface area of a volume of revolution revolved around the x-axis is given by
S = 2π∫


a


b
y(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt. If the curve is revolved around the y-axis, then the formula is


S = 2π∫
a


b
x(t) ⎛⎝x′ (t)⎞⎠2 + ⎛⎝y′ (t)⎞⎠2dt.


1.3 Polar Coordinates
• The polar coordinate system provides an alternative way to locate points in the plane.
• Convert points between rectangular and polar coordinates using the formulas


x = r cos θ and y = r sin θ


and
r = x2 + y2 and tan θ =


y
x.


• To sketch a polar curve from a given polar function, make a table of values and take advantage of periodicproperties.
• Use the conversion formulas to convert equations between rectangular and polar coordinates.
• Identify symmetry in polar curves, which can occur through the pole, the horizontal axis, or the vertical axis.


1.4 Area and Arc Length in Polar Coordinates
• The area of a region in polar coordinates defined by the equation r = f (θ) with α ≤ θ ≤ β is given by the integral


A = 1
2

α


β

⎣ f (θ)⎤⎦


2


dθ.


• To find the area between two curves in the polar coordinate system, first find the points of intersection, then subtractthe corresponding areas.
• The arc length of a polar curve defined by the equation r = f (θ) with α ≤ θ ≤ β is given by the integral


L = ∫
α


β

⎣ f (θ)⎤⎦2 + ⎡⎣ f ′ (θ)⎤⎦2dθ = ∫


α


β
r2 + ⎛⎝


dr




2
dθ.


1.5 Conic Sections
• The equation of a vertical parabola in standard form with given focus and directrix is y = 1


4p
(x − h)2 + k where p


is the distance from the vertex to the focus and (h, k) are the coordinates of the vertex.


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• The equation of a horizontal ellipse in standard form is (x − h)2
a2


+

⎝y − k⎞⎠2


b2
= 1 where the center has coordinates


(h, k), the major axis has length 2a, the minor axis has length 2b, and the coordinates of the foci are (h ± c, k),
where c2 = a2 − b2.


• The equation of a horizontal hyperbola in standard form is (x − h)2
a2




⎝y − k⎞⎠2


b2
= 1 where the center has


coordinates (h, k), the vertices are located at (h ± a, k), and the coordinates of the foci are (h ± c, k), where
c2 = a2 + b2.


• The eccentricity of an ellipse is less than 1, the eccentricity of a parabola is equal to 1, and the eccentricity of ahyperbola is greater than 1. The eccentricity of a circle is 0.
• The polar equation of a conic section with eccentricity e is r = ep


1 ± e cos θ
or r = ep


1 ± e sin θ
, where p


represents the focal parameter.
• To identify a conic generated by the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, first calculate the
discriminant D = 4AC − B2. If D > 0 then the conic is an ellipse, if D = 0 then the conic is a parabola, and if
D < 0 then the conic is a hyperbola.


CHAPTER 1 REVIEW EXERCISES
True or False? Justify your answer with a proof or acounterexample.
322. The rectangular coordinates of the point ⎛⎝4, 5π6 ⎞⎠ are

⎝2 3, −2



⎠.


323. The equations x = cosh(3t), y = 2 sinh(3t)
represent a hyperbola.
324. The arc length of the spiral given by r = θ


2
for


0 ≤ θ ≤ 3π is 9
4
π3.


325. Given x = f (t) and y = g(t), if dx
dy


=
dy
dx


, then
f (t) = g(t) + C, where C is a constant.
For the following exercises, sketch the parametric curveand eliminate the parameter to find the Cartesian equationof the curve.
326. x = 1 + t, y = t2 − 1, −1 ≤ t ≤ 1


327. x = et, y = 1 − e3t, 0 ≤ t ≤ 1
328. x = sin θ, y = 1 − csc θ, 0 ≤ θ ≤ 2π
329. x = 4 cos ϕ, y = 1 − sin ϕ, 0 ≤ ϕ ≤ 2π


For the following exercises, sketch the polar curve anddetermine what type of symmetry exists, if any.
330. r = 4 sin⎛⎝θ3⎞⎠
331. r = 5 cos(5θ)
For the following exercises, find the polar equation for thecurve given as a Cartesian equation.
332. x + y = 5


333. y2 = 4 + x2


For the following exercises, find the equation of the tangentline to the given curve. Graph both the function and itstangent line.
334. x = ln(t), y = t2 − 1, t = 1


335. r = 3 + cos(2θ), θ = 3π
4


336. Find dy
dx


, dx
dy


, and d2 x
dy2


of y = ⎛⎝2 + e−t⎞⎠,
x = 1 − sin(t)


For the following exercises, find the area of the region.
337. x = t2, y = ln(t), 0 ≤ t ≤ e


Chapter 1 | Parametric Equations and Polar Coordinates 97




338. r = 1 − sin θ in the first quadrant
For the following exercises, find the arc length of the curveover the given interval.
339. x = 3t + 4, y = 9t − 2, 0 ≤ t ≤ 3
340. r = 6 cos θ, 0 ≤ θ ≤ 2π. Check your answer by
geometry.
For the following exercises, find the Cartesian equationdescribing the given shapes.
341. A parabola with focus (2, −5) and directrix x = 6
342. An ellipse with a major axis length of 10 and foci at
(−7, 2) and (1, 2)
343. A hyperbola with vertices at (3, −2) and (−5, −2)
and foci at (−2, −6) and (−2, 4)
For the following exercises, determine the eccentricity andidentify the conic. Sketch the conic.
344. r = 6


1 + 3 cos(θ)


345. r = 4
3 − 2 cos θ


346. r = 7
5 − 5 cos θ


347. Determine the Cartesian equation describing the orbitof Pluto, the most eccentric orbit around the Sun. Thelength of the major axis is 39.26 AU and minor axis is38.07 AU. What is the eccentricity?
348. The C/1980 E1 comet was observed in 1980. Givenan eccentricity of 1.057 and a perihelion (point of closestapproach to the Sun) of 3.364 AU, find the Cartesianequations describing the comet’s trajectory. Are weguaranteed to see this comet again? (Hint: Consider the Sunat point (0, 0).)


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2 | VECTORS IN SPACE


Figure 2.1 The Karl G. Jansky Very Large Array, located in Socorro, New Mexico, consists of a large number of radiotelescopes that can collect radio waves and collate them as if they were gathering waves over a huge area with no gaps incoverage. (credit: modification of work by CGP Grey, Wikimedia Commons)


Chapter Outline
2.1 Vectors in the Plane
2.2 Vectors in Three Dimensions
2.3 The Dot Product
2.4 The Cross Product
2.5 Equations of Lines and Planes in Space
2.6 Quadric Surfaces
2.7 Cylindrical and Spherical Coordinates


Introduction
Modern astronomical observatories often consist of a large number of parabolic reflectors, connected by computers, usedto analyze radio waves. Each dish focuses the incoming parallel beams of radio waves to a precise focal point, wherethey can be synchronized by computer. If the surface of one of the parabolic reflectors is described by the equation
x2
100


+
y2


100
= z


4
, where is the focal point of the reflector? (See Example 2.58.)


We are now about to begin a new part of the calculus course, when we study functions of two or three independent variables


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in multidimensional space. Many of the computations are similar to those in the study of single-variable functions, but thereare also a lot of differences. In this first chapter, we examine coordinate systems for working in three-dimensional space,along with vectors, which are a key mathematical tool for dealing with quantities in more than one dimension. Let’s starthere with the basic ideas and work our way up to the more general and powerful tools of mathematics in later chapters.
2.1 | Vectors in the Plane


Learning Objectives
2.1.1 Describe a plane vector, using correct notation.
2.1.2 Perform basic vector operations (scalar multiplication, addition, subtraction).
2.1.3 Express a vector in component form.
2.1.4 Explain the formula for the magnitude of a vector.
2.1.5 Express a vector in terms of unit vectors.
2.1.6 Give two examples of vector quantities.


When describing the movement of an airplane in flight, it is important to communicate two pieces of information: thedirection in which the plane is traveling and the plane’s speed. When measuring a force, such as the thrust of the plane’sengines, it is important to describe not only the strength of that force, but also the direction in which it is applied. Somequantities, such as or force, are defined in terms of both size (also called magnitude) and direction. A quantity that hasmagnitude and direction is called a vector. In this text, we denote vectors by boldface letters, such as v.
Definition
A vector is a quantity that has both magnitude and direction.


Vector Representation
A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initialpoint and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction ofthe vector. The length of the line segment represents its magnitude. We use the notation ‖ v ‖ to denote the magnitude of
the vector v. A vector with an initial point and terminal point that are the same is called the zero vector, denoted 0. The
zero vector is the only vector without a direction, and by convention can be considered to have any direction convenient tothe problem at hand.
Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even ifthey have different initial points. Thus, if v and w are equivalent, we write


v = w.


Definition
Vectors are said to be equivalent vectors if they have the same magnitude and direction.


The arrows in Figure 2.2(b) are equivalent. Each arrow has the same length and direction. A closely related concept is theidea of parallel vectors. Two vectors are said to be parallel if they have the same or opposite directions. We explore this ideain more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point islocated.


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2.1


Figure 2.2 (a) A vector is represented by a directed line segment from its initialpoint to its terminal point. (b) Vectors v1 through v5 are equivalent.


The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternativenotations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to
simulate boldface type: v→ . When a vector has initial point P and terminal point Q, the notation PQ→ is useful because
it indicates the direction and location of the vector.
Example 2.1
Sketching Vectors
Sketch a vector in the plane from initial point P(1, 1) to terminal point Q(8, 5).
Solution
See Figure 2.3. Because the vector goes from point P to point Q, we name it PQ→ .


Figure 2.3 The vector with initial point (1, 1) and terminal
point (8, 5) is named PQ→ .


Sketch the vector ST→ where S is point (3, −1) and T is point (−2, 3).


Combining Vectors
Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces


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acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates aforce in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into accountif we want to know where the boat will go.
A second example that involves vectors is a quarterback throwing a football. The quarterback does not throw the ballparallel to the ground; instead, he aims up into the air. The velocity of his throw can be represented by a vector. If we knowhow hard he throws the ball (magnitude—in this case, speed), and the angle (direction), we can tell how far the ball willtravel down the field.
A real number is often called a scalar in mathematics and physics. Unlike vectors, scalars are generally considered to havea magnitude only, but no direction. Multiplying a vector by a scalar changes the vector’s magnitude. This is called scalarmultiplication. Note that changing the magnitude of a vector does not indicate a change in its direction. For example, windblowing from north to south might increase or decrease in speed while maintaining its direction from north to south.
Definition
The product kv of a vector v and a scalar k is a vector with a magnitude that is |k| times the magnitude of v, and
with a direction that is the same as the direction of v if k > 0, and opposite the direction of v if k < 0. This is
called scalar multiplication. If k = 0 or v = 0, then kv = 0.


As you might expect, if k = −1, we denote the product kv as
kv = (−1)v = −v.


Note that −v has the same magnitude as v, but has the opposite direction (Figure 2.4).


Figure 2.4 (a) The original vector v has length n units. (b)The length of 2v equals 2n units. (c) The length of v/2 is
n/2 units. (d) The vectors v and −v have the same length but
opposite directions.


Another operation we can perform on vectors is to add them together in vector addition, but because each vector may haveits own direction, the process is different from adding two numbers. The most common graphical method for adding twovectors is to place the initial point of the second vector at the terminal point of the first, as in Figure 2.5(a). To see why thismakes sense, suppose, for example, that both vectors represent displacement. If an object moves first from the initial pointto the terminal point of vector v, then from the initial point to the terminal point of vector w, the overall displacement
is the same as if the object had made just one movement from the initial point to the terminal point of the vector v + w.
For obvious reasons, this approach is called the triangle method. Notice that if we had switched the order, so that w was
our first vector and v was our second vector, we would have ended up in the same place. (Again, see Figure 2.5(a).) Thus,
v + w = w + v.


A second method for adding vectors is called the parallelogram method. With this method, we place the two vectors sothey have the same initial point, and then we draw a parallelogram with the vectors as two adjacent sides, as in Figure2.5(b). The length of the diagonal of the parallelogram is the sum. Comparing Figure 2.5(b) and Figure 2.5(a), we cansee that we get the same answer using either method. The vector v + w is called the vector sum.
Definition
The sum of two vectors v and w can be constructed graphically by placing the initial point of w at the terminal point


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of v. Then, the vector sum, v + w, is the vector with an initial point that coincides with the initial point of v and
has a terminal point that coincides with the terminal point of w. This operation is known as vector addition.


Figure 2.5 (a) When adding vectors by the triangle method,the initial point of w is the terminal point of v. (b) When
adding vectors by the parallelogram method, the vectors v and
w have the same initial point.


It is also appropriate here to discuss vector subtraction. We define v − w as v + (−w) = v + (−1)w. The vector v − w
is called the vector difference. Graphically, the vector v − w is depicted by drawing a vector from the terminal point of
w to the terminal point of v (Figure 2.6).


Figure 2.6 (a) The vector difference v − w is depicted by
drawing a vector from the terminal point of w to the terminal
point of v. (b) The vector v − w is equivalent to the vector
v + (−w).


In Figure 2.5(a), the initial point of v + w is the initial point of v. The terminal point of v + w is the terminal point of
w. These three vectors form the sides of a triangle. It follows that the length of any one side is less than the sum of the
lengths of the remaining sides. So we have


‖ v + w ‖ ≤ ‖ v ‖ + ‖ w ‖ .


This is known more generally as the triangle inequality. There is one case, however, when the resultant vector u + v has
the same magnitude as the sum of the magnitudes of u and v. This happens only when u and v have the same direction.


Example 2.2
Combining Vectors
Given the vectors v and w shown in Figure 2.7, sketch the vectors


a. 3w
b. v + w


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c. 2v − w


Figure 2.7 Vectors v and w lie in the same plane.
Solution


a. The vector 3w has the same direction as w; it is three times as long as w.


Vector 3w has the same direction as w and is three times as long.
b. Use either addition method to find v + w.


Figure 2.8 To find v + w, align the vectors at their initial
points or place the initial point of one vector at the terminalpoint of the other. (a) The vector v + w is the diagonal of the
parallelogram with sides v and w (b) The vector v + w is the
third side of a triangle formed with w placed at the terminal
point of v.


c. To find 2v − w, we can first rewrite the expression as 2v + (−w). Then we can draw the vector −w,
then add it to the vector 2v.


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2.2


Figure 2.9 To find 2v − w, simply add 2v + (−w).


Using vectors v and w from Example 2.2, sketch the vector 2w − v.


Vector Components
Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminalpoints of vectors are given in Cartesian coordinates, computations become straightforward.
Example 2.3
Comparing Vectors
Are v and w equivalent vectors?


a. v has initial point (3, 2) and terminal point (7, 2)
w has initial point (1, −4) and terminal point (1, 0)


b. v has initial point (0, 0) and terminal point (1, 1)
w has initial point (−2, 2) and terminal point (−1, 3)


Solution
a. The vectors are each 4 units long, but they are oriented in different directions. So v and w are not


equivalent (Figure 2.10).


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2.3


Figure 2.10 These vectors are not equivalent.
b. Based on Figure 2.11, and using a bit of geometry, it is clear these vectors have the same length and thesame direction, so v and w are equivalent.


Figure 2.11 These vectors are equivalent.


Which of the following vectors are equivalent?


We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can


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be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides withthe origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of anyvector in standard position is known to be (0, 0), we can describe the vector by looking at the coordinates of its terminal
point. Thus, if vector v has its initial point at the origin and its terminal point at (x, y), we write the vector in component
form as


v = 〈 x, y 〉 .


When a vector is written in component form like this, the scalars x and y are called the components of v.


Definition
The vector with initial point (0, 0) and terminal point (x, y) can be written in component form as


v = 〈 x, y 〉 .


The scalars x and y are called the components of v.


Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have alsolearned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets.However, when writing the component form of a vector, it is important to distinguish between 〈 x, y 〉 and (x, y). The
first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane.The initial point of 〈 x, y 〉 is (0, 0); the terminal point of 〈 x, y 〉 is (x, y).
When we have a vector not already in standard position, we can determine its component form in one of two ways. We canuse a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point.To find it algebraically, we subtract the x-coordinate of the initial point from the x-coordinate of the terminal point to getthe x component, and we subtract the y-coordinate of the initial point from the y-coordinate of the terminal point to get they component.
Rule: Component Form of a Vector
Let v be a vector with initial point (xi, yi) and terminal point (xt, yt). Then we can express v in component form as
v = 〈 xt − xi, yt − yi 〉 .


Example 2.4
Expressing Vectors in Component Form
Express vector v with initial point (−3, 4) and terminal point (1, 2) in component form.
Solution


a. Geometric
1. Sketch the vector in the coordinate plane (Figure 2.12).
2. The terminal point is 4 units to the right and 2 units down from the initial point.
3. Find the point that is 4 units to the right and 2 units down from the origin.
4. In standard position, this vector has initial point (0, 0) and terminal point (4, −2):


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2.4


v = 〈 4, −2 〉 .


Figure 2.12 These vectors are equivalent.
b. AlgebraicIn the first solution, we used a sketch of the vector to see that the terminal point lies 4 units to the right.We can accomplish this algebraically by finding the difference of the x-coordinates:


xt − xi = 1 − (−3) = 4.


Similarly, the difference of the y-coordinates shows the vertical length of the vector.
yt − yi = 2 − 4 = −2.


So, in component form,
v = 〈 xt − xi, yt − yi 〉


= 〈 1 − (−3), 2 − 4 〉


= 〈 4, −2 〉 .


Vector w has initial point (−4, −5) and terminal point (−1, 2). Express w in component form.


To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitudeof vector v = 〈 x, y 〉 is denoted ‖ v ‖ , or |v|, and can be computed using the formula
‖ v ‖ = x2 + y2.


Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initialpoint at the origin and terminal point (x, y). Thus, it suffices to calculate the magnitude of the vector in standard position.
Using the distance formula to calculate the distance between initial point (0, 0) and terminal point (x, y), we have


‖ v ‖ = (x − 0)2 + ⎛⎝y − 0⎞⎠2


= x2 + y2.


Based on this formula, it is clear that for any vector v, ‖ v ‖ ≥ 0, and ‖ v ‖ = 0 if and only if v = 0.


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The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.


Figure 2.13 If you use the components of a vector to define aright triangle, the magnitude of the vector is the length of thetriangle’s hypotenuse.


We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows usto perform these same operations algebraically.
Definition
Let v = 〈 x1, y1 〉 and w = 〈 x2, y2 〉 be vectors, and let k be a scalar.
Scalar multiplication: kv = 〈 kx1, ky1 〉
Vector addition: v + w = 〈 x1, y1 〉 + 〈 x2, y2 〉 = 〈 x1 + x2, y1 + y2 〉


Example 2.5
Performing Operations in Component Form
Let v be the vector with initial point (2, 5) and terminal point (8, 13), and let w = 〈 −2, 4 〉 .


a. Express v in component form and find ‖ v ‖ . Then, using algebra, find
b. v + w,
c. 3v, and
d. v − 2w.


Solution
a. To place the initial point of v at the origin, we must translate the vector 2 units to the left and


5 units down (Figure 2.15). Using the algebraic method, we can express v as
v = 〈 8 − 2, 13 − 5 〉 = 〈 6, 8 〉 :


‖ v ‖ = 62 + 82 = 36 + 64 = 100 = 10.


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2.5


Figure 2.14 In component form, v = 〈 6, 8 〉 .
b. To find v + w, add the x-components and the y-components separately:


v + w = 〈 6, 8 〉 + 〈 −2, 4 〉 = 〈 4, 12 〉 .


c. To find 3v, multiply v by the scalar k = 3:
3v = 3 · 〈 6, 8 〉 = 〈 3 · 6, 3 · 8 〉 = 〈 18, 24 〉 .


d. To find v − 2w, find −2w and add it to v:
v − 2w = 〈 6, 8 〉 − 2 · 〈 −2, 4 〉 = 〈 6, 8 〉 + 〈 4, −8 〉 = 〈 10, 0 〉 .


Let a = 〈 7, 1 〉 and let b be the vector with initial point (3, 2) and terminal point (−1, −1).
a. Find ‖ a ‖ .
b. Express b in component form.
c. Find 3a − 4b.


Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We willprove two of these properties. The others can be proved in a similar manner.
Theorem 2.1: Properties of Vector Operations
Let u, v, andw be vectors in a plane. Let r and s be scalars.


i. u + v = v + u Commutative property
ii. (u + v) + w = u + (v + w) Associative property
iii. u + 0 = u Additive identity property
iv. u + (−u) = 0 Additive inverse property
v. r(su) = (rs)u Associativity of scalar multiplication
vi. (r + s)u = ru + su Distributive property
vii. r(u + v) = ru + rv Distributive property
viii. 1u = u, 0u = 0 Identity and zero properties


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2.6


Proof of Commutative Property
Let u = 〈 x1, y1 〉 and v = 〈 x2, y2 〉 . Apply the commutative property for real numbers:


u + v = 〈 x1 + x2, y1 + y2 〉 = 〈 x2 + x1, y2 + y1 〉 = v + u.



Proof of Distributive Property
Apply the distributive property for real numbers:


r(u + v) = r · 〈 x1 + x2, y1 + y2 〉


= 〈 r(x1 + x2), r(y1 + y2) 〉


= 〈 rx1 + rx2, ry1 + ry2 〉


= 〈 rx1, ry1 〉 + 〈 rx2, ry2 〉
= ru + rv.



Prove the additive inverse property.


We have found the components of a vector given its initial and terminal points. In some cases, we may only have themagnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical componentsusing trigonometry (Figure 2.15).


Figure 2.15 The components of a vector form the legs of aright triangle, with the vector as the hypotenuse.


Consider the angle θ formed by the vector v and the positive x-axis. We can see from the triangle that the components
of vector v are 〈 ‖ v ‖ cos θ, ‖ v ‖ sin θ 〉 . Therefore, given an angle and the magnitude of a vector, we can use the
cosine and sine of the angle to find the components of the vector.
Example 2.6
Finding the Component Form of a Vector Using Trigonometry
Find the component form of a vector with magnitude 4 that forms an angle of −45° with the x-axis.
Solution
Let x and y represent the components of the vector (Figure 2.16). Then x = 4 cos(−45°) = 2 2 and
y = 4 sin(−45°) = −2 2. The component form of the vector is 〈 2 2, −2 2 〉 .


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2.7


Figure 2.16 Use trigonometric ratios, x = ‖ v ‖ cos θ and
y = ‖ v ‖ sin θ, to identify the components of the vector.


Find the component form of vector v with magnitude 10 that forms an angle of 120° with the positive
x-axis.


Unit Vectors
A unit vector is a vector with magnitude 1. For any nonzero vector v, we can use scalar multiplication to find a unit
vector u that has the same direction as v. To do this, we multiply the vector by the reciprocal of its magnitude:


u = 1
‖ v ‖


v.


Recall that when we defined scalar multiplication, we noted that ‖ kv ‖ = |k| · ‖ v ‖ . For u = 1‖ v ‖ v, it follows that
‖ u ‖ = 1


‖ v ‖

⎝ ‖ v ‖ ⎞⎠ = 1. We say that u is the unit vector in the direction of v (Figure 2.17). The process of using


scalar multiplication to find a unit vector with a given direction is called normalization.


Figure 2.17 The vector v and associated unit vector
u = 1


‖ v ‖
v. In this case, ‖ v ‖ > 1.


Example 2.7
Finding a Unit Vector
Let v = 〈 1, 2 〉 .


a. Find a unit vector with the same direction as v.
b. Find a vector w with the same direction as v such that ‖ w ‖ = 7.


Solution


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2.8


a. First, find the magnitude of v, then divide the components of v by the magnitude:
‖ v ‖ = 12 + 22 = 1 + 4 = 5


u = 1
‖ v ‖


v = 1
5
〈 1, 2 〉 = 〈 1


5
, 2


5
〉 .


b. The vector u is in the same direction as v and ‖ u ‖ = 1. Use scalar multiplication to increase the
length of u without changing direction:


w = 7u = 7 〈 1
5
, 2


5
〉 = 〈 7


5
, 14


5
〉 .


Let v = 〈 9, 2 〉 . Find a vector with magnitude 5 in the opposite direction as v.


We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient towrite a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let’s look at standard unit vectors.The standard unit vectors are the vectors i = 〈 1, 0 〉 and j = 〈 0, 1 〉 (Figure 2.18).


Figure 2.18 The standard unit vectors i and j.


By applying the properties of vectors, it is possible to express any vector in terms of i and j in what we call a linear
combination:


v = 〈 x, y 〉 = 〈 x, 0 〉 + 〈 0, y 〉 = x 〈 1, 0 〉 + y 〈 0, 1 〉 = xi + yj.


Thus, v is the sum of a horizontal vector with magnitude x, and a vertical vector with magnitude y, as in the following
figure.


Figure 2.19 The vector v is the sum of xi and yj.


Example 2.8


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2.9


Using Standard Unit Vectors
a. Express the vector w = 〈 3, −4 〉 in terms of standard unit vectors.
b. Vector u is a unit vector that forms an angle of 60° with the positive x-axis. Use standard unit vectors


to describe u.
Solution


a. Resolve vector w into a vector with a zero y-component and a vector with a zero x-component:
w = 〈 3, −4 〉 = 3i − 4j.


b. Because u is a unit vector, the terminal point lies on the unit circle when the vector is placed in standard
position (Figure 2.20).


u = 〈 cos 60°, sin 60° 〉


= 〈 1
2
, 3


2


= 1
2
i + 3


2
j.


Figure 2.20 The terminal point of u lies on the unit circle
(cos θ, sin θ).


Let a = 〈 16, −11 〉 and let b be a unit vector that forms an angle of 225° with the positive x-axis.
Express a and b in terms of the standard unit vectors.


Applications of Vectors
Because vectors have both direction and magnitude, they are valuable tools for solving problems involving suchapplications as motion and force. Recall the boat example and the quarterback example we described earlier. Here we lookat two other examples in detail.
Example 2.9


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Finding Resultant Force
Jane’s car is stuck in the mud. Lisa and Jed come along in a truck to help pull her out. They attach one end of a towstrap to the front of the car and the other end to the truck’s trailer hitch, and the truck starts to pull. Meanwhile,Jane and Jed get behind the car and push. The truck generates a horizontal force of 300 lb on the car. Jane and Jed
are pushing at a slight upward angle and generate a force of 150 lb on the car. These forces can be represented by
vectors, as shown in Figure 2.21. The angle between these vectors is 15°. Find the resultant force (the vector
sum) and give its magnitude to the nearest tenth of a pound and its direction angle from the positive x-axis.


Figure 2.21 Two forces acting on a car in different directions.
Solution
To find the effect of combining the two forces, add their representative vectors. First, express each vector incomponent form or in terms of the standard unit vectors. For this purpose, it is easiest if we align one of thevectors with the positive x-axis. The horizontal vector, then, has initial point (0, 0) and terminal point (300, 0).
It can be expressed as 〈 300, 0 〉 or 300i.
The second vector has magnitude 150 and makes an angle of 15° with the first, so we can express it as
〈 150 cos(15°), 150 sin(15°) 〉 , or 150 cos(15°)i + 150 sin(15°)j. Then, the sum of the vectors, or resultant
vector, is r = 〈 300, 0 〉 + 〈 150 cos(15°), 150 sin(15°) 〉 , and we have


‖ r ‖ = ⎛⎝300 + 150 cos(15°)⎞⎠2 + ⎛⎝150 sin(15°)⎞⎠2


≈ 446.6.


The angle θ made by r and the positive x-axis has tan θ = 150 sin 15°
(300 + 150 cos 15°)


≈ 0.09, so
θ ≈ tan−1 (0.09) ≈ 5°, which means the resultant force r has an angle of 5° above the horizontal axis.


Example 2.10
Finding Resultant Velocity
An airplane flies due west at an airspeed of 425 mph. The wind is blowing from the northeast at 40 mph. What
is the ground speed of the airplane? What is the bearing of the airplane?
Solution
Let’s start by sketching the situation described (Figure 2.22).


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2.10


Figure 2.22 Initially, the plane travels due west. The wind isfrom the northeast, so it is blowing to the southwest. The anglebetween the plane’s course and the wind is 45°. (Figure not
drawn to scale.)


Set up a sketch so that the initial points of the vectors lie at the origin. Then, the plane’s velocity vector is
p = −425i. The vector describing the wind makes an angle of 225° with the positive x-axis:


w = 〈 40 cos(225°), 40 sin(225°) 〉 = 〈 −40
2
, − 40


2
〉 = − 40


2
i − 40


2
j.


When the airspeed and the wind act together on the plane, we can add their vectors to find the resultant force:
p + w = −425i + ⎛⎝−


40
2
i − 40


2
j⎞⎠ =

⎝−425 −


40
2

⎠i −


40
2
j.


The magnitude of the resultant vector shows the effect of the wind on the ground speed of the airplane:
‖ p + w ‖ = ⎛⎝−425 −


40
2



2


+

⎝−


40
2



2


≈ 454.17 mph


As a result of the wind, the plane is traveling at approximately 454 mph relative to the ground.
To determine the bearing of the airplane, we want to find the direction of the vector p + w:


tan θ =
− 40


2

⎝−425 −


40
2





≈ 0.06


θ ≈ 3.57°.


The overall direction of the plane is 3.57° south of west.


An airplane flies due north at an airspeed of 550 mph. The wind is blowing from the northwest at 50
mph. What is the ground speed of the airplane?


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2.1 EXERCISES
For the following exercises, consider points P(−1, 3),
Q(1, 5), and R(−3, 7). Determine the requested vectors
and express each of them a. in component form and b. byusing the standard unit vectors.
1. PQ→


2. PR→


3. QP→


4. RP→


5. PQ→ + PR→


6. PQ→ − PR→


7. 2PQ→ − 2PR→


8. 2PQ→ + 1
2
PR


9. The unit vector in the direction of PQ→


10. The unit vector in the direction of PR→
11. A vector v has initial point (−1, −3) and terminal
point (2, 1). Find the unit vector in the direction of v.
Express the answer in component form.
12. A vector v has initial point (−2, 5) and terminal
point (3, −1). Find the unit vector in the direction of v.
Express the answer in component form.
13. The vector v has initial point P(1, 0) and terminal
point Q that is on the y-axis and above the initial point.
Find the coordinates of terminal point Q such that the
magnitude of the vector v is 5.
14. The vector v has initial point P(1, 1) and terminal
point Q that is on the x-axis and left of the initial point.
Find the coordinates of terminal point Q such that the
magnitude of the vector v is 10.
For the following exercises, use the given vectors a and
b.


a. Determine the vector sum a + b and express it in


both the component form and by using the standardunit vectors.
b. Find the vector difference a − b and express it in


both the component form and by using the standardunit vectors.
c. Verify that the vectors a, b, and a + b, and,


respectively, a, b, and a − b satisfy the
triangle inequality.


d. Determine the vectors 2a, −b, and 2a − b.
Express the vectors in both the component formand by using standard unit vectors.


15. a = 2i + j, b = i + 3j
16. a = 2i, b = −2i + 2j
17. Let a be a standard-position vector with terminal
point (−2, −4). Let b be a vector with initial point
(1, 2) and terminal point (−1, 4). Find the magnitude of
vector −3a + b − 4i + j.
18. Let a be a standard-position vector with terminal
point at (2, 5). Let b be a vector with initial point
(−1, 3) and terminal point (1, 0). Find the magnitude of
vector a − 3b + 14i − 14j.
19. Let u and v be two nonzero vectors that are
nonequivalent. Consider the vectors a = 4u + 5v and
b = u + 2v defined in terms of u and v. Find the scalar
λ such that vectors a + λb and u − v are equivalent.
20. Let u and v be two nonzero vectors that are
nonequivalent. Consider the vectors a = 2u − 4v and
b = 3u − 7v defined in terms of u and v. Find the
scalars α and β such that vectors αa + βb and u − v
are equivalent.
21. Consider the vector a(t) = 〈 cos t, sin t 〉 with
components that depend on a real number t. As the number
t varies, the components of a(t) change as well,
depending on the functions that define them.a. Write the vectors a(0) and a(π) in component


form.b. Show that the magnitude ‖ a(t) ‖ of vector a(t)
remains constant for any real number t.


c. As t varies, show that the terminal point of vector
a(t) describes a circle centered at the origin of
radius 1.


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22. Consider vector a(x) = 〈 x, 1 − x2 〉 with
components that depend on a real number x ∈ [−1, 1]. As
the number x varies, the components of a(x) change as
well, depending on the functions that define them.a. Write the vectors a(0) and a(1) in component


form.b. Show that the magnitude ‖ a(x) ‖ of vector
a(x) remains constant for any real number x


c. As x varies, show that the terminal point of vector
a(x) describes a circle centered at the origin of
radius 1.


23. Show that vectors a(t) = 〈 cos t, sin t 〉 and
a(x) = 〈 x, 1 − x2 〉 are equivalent for x = r and
t = 2kπ, where k is an integer.
24. Show that vectors a(t) = 〈 cos t, sin t 〉 and
a(x) = 〈 x, 1 − x2 〉 are opposite for x = r and
t = π + 2kπ, where k is an integer.
For the following exercises, find vector v with the given
magnitude and in the same direction as vector u.
25. ‖ v ‖ = 7, u = 〈 3, 4 〉
26. ‖ v ‖ = 3, u = 〈 −2, 5 〉
27. ‖ v ‖ = 7, u = 〈 3, −5 〉
28. ‖ v ‖ = 10, u = 〈 2, −1 〉
For the following exercises, find the component form ofvector u, given its magnitude and the angle the vector
makes with the positive x-axis. Give exact answers whenpossible.
29. ‖ u ‖ = 2, θ = 30°
30. ‖ u ‖ = 6, θ = 60°
31. ‖ u ‖ = 5, θ = π


2


32. ‖ u ‖ = 8, θ = π
33. ‖ u ‖ = 10, θ = 5π


6


34. ‖ u ‖ = 50, θ = 3π
4


For the following exercises, vector u is given. Find the


angle θ ∈ [0, 2π) that vector u makes with the positive
direction of the x-axis, in a counter-clockwise direction.
35. u = 5 2i − 5 2j
36. u = − 3i − j
37. Let a = 〈 a1, a2 〉 , b = 〈 b1, b2 〉 , and
c = 〈 c1, c2 〉 be three nonzero vectors. If
a1b2 − a2b1 ≠ 0, then show there are two scalars, α
and β, such that c = αa + βb.
38. Consider vectors a = 〈 2, −4 〉 , b = 〈 −1, 2 〉 ,
and 0 Determine the scalars α and β such that
c = αa + βb.


39. Let P⎛⎝x0, f (x0)⎞⎠ be a fixed point on the graph of the
differential function f with a domain that is the set of real
numbers.a. Determine the real number z0 such that point


Q⎛⎝x0 + 1, z0

⎠ is situated on the line tangent to the


graph of f at point P.
b. Determine the unit vector u with initial point P


and terminal point Q.
40. Consider the function f (x) = x4, where x ∈ ℝ .


a. Determine the real number z0 such that point
Q⎛⎝2, z0



⎠ s situated on the line tangent to the graph


of f at point P(1, 1).
b. Determine the unit vector u with initial point P


and terminal point Q.
41. Consider f and g two functions defined on the
same set of real numbers D. Let a = 〈 x, f (x) 〉 and
b = 〈 x, g(x) 〉 be two vectors that describe the graphs
of the functions, where x ∈ D. Show that if the graphs of
the functions f and g do not intersect, then the vectors a
and b are not equivalent.
42. Find x ∈ ℝ such that vectors a = 〈 x, sin x 〉 and
b = 〈 x, cos x 〉 are equivalent.
43. Calculate the coordinates of point D such that
ABCD is a parallelogram, with A(1, 1), B(2, 4), and
C(7, 4).


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44. Consider the points A(2, 1), B(10, 6), C(13, 4),
and D(16, −2). Determine the component form of vector
AD


.


45. The speed of an object is the magnitude of its relatedvelocity vector. A football thrown by a quarterback hasan initial speed of 70 mph and an angle of elevation of
30°. Determine the velocity vector in mph and express it
in component form. (Round to two decimal places.)


46. A baseball player throws a baseball at an angle of
30° with the horizontal. If the initial speed of the ball is
100 mph, find the horizontal and vertical components of
the initial velocity vector of the baseball. (Round to twodecimal places.)
47. A bullet is fired with an initial velocity of 1500 ft/sec
at an angle of 60° with the horizontal. Find the horizontal
and vertical components of the velocity vector of the bullet.(Round to two decimal places.)


48. [T] A 65-kg sprinter exerts a force of 798 N at a 19°
angle with respect to the ground on the starting block at theinstant a race begins. Find the horizontal component of theforce. (Round to two decimal places.)
49. [T] Two forces, a horizontal force of 45 lb and
another of 52 lb, act on the same object. The angle
between these forces is 25°. Find the magnitude and
direction angle from the positive x-axis of the resultantforce that acts on the object. (Round to two decimal places.)


50. [T] Two forces, a vertical force of 26 lb and another
of 45 lb, act on the same object. The angle between these
forces is 55°. Find the magnitude and direction angle from
the positive x-axis of the resultant force that acts on theobject. (Round to two decimal places.)
51. [T] Three forces act on object. Two of the forces havethe magnitudes 58 N and 27 N, and make angles 53°
and 152°, respectively, with the positive x-axis. Find the
magnitude and the direction angle from the positive x-axisof the third force such that the resultant force acting on theobject is zero. (Round to two decimal places.)
52. Three forces with magnitudes 80 lb, 120 lb, and
60 lb act on an object at angles of 45°, 60° and 30°,
respectively, with the positive x-axis. Find the magnitudeand direction angle from the positive x-axis of the resultantforce. (Round to two decimal places.)


53. [T] An airplane is flying in the direction of 43°
east of north (also abbreviated as N43E) at a speed of
550 mph. A wind with speed 25 mph comes from the
southwest at a bearing of N15E. What are the ground
speed and new direction of the airplane?


Chapter 2 | Vectors in Space 119




54. [T] A boat is traveling in the water at 30 mph in a
direction of N20E (that is, 20° east of north). A strong
current is moving at 15 mph in a direction of N45E.
What are the new speed and direction of the boat?


55. [T] A 50-lb weight is hung by a cable so that thetwo portions of the cable make angles of 40° and 53°,
respectively, with the horizontal. Find the magnitudes ofthe forces of tension T1 and T2 in the cables if the
resultant force acting on the object is zero. (Round to twodecimal places.)


56. [T] A 62-lb weight hangs from a rope that makes theangles of 29° and 61°, respectively, with the horizontal.
Find the magnitudes of the forces of tension T1 and T2 in
the cables if the resultant force acting on the object is zero.(Round to two decimal places.)
57. [T] A 1500-lb boat is parked on a ramp that makesan angle of 30° with the horizontal. The boat’s weight
vector points downward and is a sum of two vectors: ahorizontal vector v1 that is parallel to the ramp and a
vertical vector v2 that is perpendicular to the inclined
surface. The magnitudes of vectors v1 and v2 are the
horizontal and vertical component, respectively, of theboat’s weight vector. Find the magnitudes of v1 and v2.
(Round to the nearest integer.)


58. [T] An 85-lb box is at rest on a 26° incline.
Determine the magnitude of the force parallel to the inclinenecessary to keep the box from sliding. (Round to thenearest integer.)
59. A guy-wire supports a pole that is 75 ft high. One end
of the wire is attached to the top of the pole and the otherend is anchored to the ground 50 ft from the base of the
pole. Determine the horizontal and vertical components ofthe force of tension in the wire if its magnitude is 50 lb.
(Round to the nearest integer.)


60. A telephone pole guy-wire has an angle of elevationof 35° with respect to the ground. The force of tension
in the guy-wire is 120 lb. Find the horizontal and vertical
components of the force of tension. (Round to the nearestinteger.)


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2.2 | Vectors in Three Dimensions
Learning Objectives


2.2.1 Describe three-dimensional space mathematically.
2.2.2 Locate points in space using coordinates.
2.2.3 Write the distance formula in three dimensions.
2.2.4 Write the equations for simple planes and spheres.
2.2.5 Perform vector operations in ℝ3 .


Vectors are useful tools for solving two-dimensional problems. Life, however, happens in three dimensions. To expand theuse of vectors to more realistic applications, it is necessary to create a framework for describing three-dimensional space.For example, although a two-dimensional map is a useful tool for navigating from one place to another, in some cases thetopography of the land is important. Does your planned route go through the mountains? Do you have to cross a river?To appreciate fully the impact of these geographic features, you must use three dimensions. This section presents a naturalextension of the two-dimensional Cartesian coordinate plane into three dimensions.
Three-Dimensional Coordinate Systems
As we have learned, the two-dimensional rectangular coordinate system contains two perpendicular axes: the horizontalx-axis and the vertical y-axis. We can add a third dimension, the z-axis, which is perpendicular to both the x-axis andthe y-axis. We call this system the three-dimensional rectangular coordinate system. It represents the three dimensions weencounter in real life.
Definition
The three-dimensional rectangular coordinate system consists of three perpendicular axes: the x-axis, the y-axis,and the z-axis. Because each axis is a number line representing all real numbers in ℝ , the three-dimensional system
is often denoted by ℝ3 .


In Figure 2.23(a), the positive z-axis is shown above the plane containing the x- and y-axes. The positive x-axis appears tothe left and the positive y-axis is to the right. A natural question to ask is: How was arrangement determined? The systemdisplayed follows the right-hand rule. If we take our right hand and align the fingers with the positive x-axis, then curlthe fingers so they point in the direction of the positive y-axis, our thumb points in the direction of the positive z-axis. Inthis text, we always work with coordinate systems set up in accordance with the right-hand rule. Some systems do follow aleft-hand rule, but the right-hand rule is considered the standard representation.


Chapter 2 | Vectors in Space 121




Figure 2.23 (a) We can extend the two-dimensional rectangular coordinate system byadding a third axis, the z-axis, that is perpendicular to both the x-axis and the y-axis. (b) Theright-hand rule is used to determine the placement of the coordinate axes in the standardCartesian plane.


In two dimensions, we describe a point in the plane with the coordinates (x, y). Each coordinate describes how the point
aligns with the corresponding axis. In three dimensions, a new coordinate, z, is appended to indicate alignment with the
z-axis: (x, y, z). A point in space is identified by all three coordinates (Figure 2.24). To plot the point (x, y, z), go x
units along the x-axis, then y units in the direction of the y-axis, then z units in the direction of the z-axis.


Figure 2.24 To plot the point (x, y, z) go x units along the
x-axis, then y units in the direction of the y-axis, then z units
in the direction of the z-axis.


Example 2.11
Locating Points in Space
Sketch the point (1, −2, 3) in three-dimensional space.
Solution
To sketch a point, start by sketching three sides of a rectangular prism along the coordinate axes: one unit in thepositive x direction, 2 units in the negative y direction, and 3 units in the positive z direction. Complete the
prism to plot the point (Figure 2.25).


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2.11


Figure 2.25 Sketching the point (1, −2, 3).


Sketch the point (−2, 3, −1) in three-dimensional space.


In two-dimensional space, the coordinate plane is defined by a pair of perpendicular axes. These axes allow us toname any location within the plane. In three dimensions, we define coordinate planes by the coordinate axes, just asin two dimensions. There are three axes now, so there are three intersecting pairs of axes. Each pair of axes forms acoordinate plane: the xy-plane, the xz-plane, and the yz-plane (Figure 2.26). We define the xy-plane formally as thefollowing set: ⎧⎩⎨⎛⎝x, y, 0⎞⎠ : x, y ∈ ℝ ⎫⎭⎬. Similarly, the xz-plane and the yz-plane are defined as {(x, 0, z) : x, z ∈ ℝ } and





⎝0, y, z⎞⎠ : y, z ∈ ℝ





⎬, respectively.


To visualize this, imagine you’re building a house and are standing in a room with only two of the four walls finished.(Assume the two finished walls are adjacent to each other.) If you stand with your back to the corner where the two finishedwalls meet, facing out into the room, the floor is the xy-plane, the wall to your right is the xz-plane, and the wall to your leftis the yz-plane.


Figure 2.26 The plane containing the x- and y-axes is calledthe xy-plane. The plane containing the x- and z-axes is called thexz-plane, and the y- and z-axes define the yz-plane.


In two dimensions, the coordinate axes partition the plane into four quadrants. Similarly, the coordinate planes divide space
between them into eight regions about the origin, called octants. The octants fill ℝ3 in the same way that quadrants fill


Chapter 2 | Vectors in Space 123




ℝ2 , as shown in Figure 2.27.


Figure 2.27 Points that lie in octants have three nonzerocoordinates.


Most work in three-dimensional space is a comfortable extension of the corresponding concepts in two dimensions. Inthis section, we use our knowledge of circles to describe spheres, then we expand our understanding of vectors to threedimensions. To accomplish these goals, we begin by adapting the distance formula to three-dimensional space.
If two points lie in the same coordinate plane, then it is straightforward to calculate the distance between them. We that thedistance d between two points (x1, y1) and (x2, y2) in the xy-coordinate plane is given by the formula


d = (x2 − x1)
2 + (y2 − y1)


2.


The formula for the distance between two points in space is a natural extension of this formula.
Theorem 2.2: The Distance between Two Points in Space
The distance d between points (x1, y1, z1) and (x2, y2, z2) is given by the formula


(2.1)d = (x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.


The proof of this theorem is left as an exercise. (Hint: First find the distance d1 between the points (x1, y1, z1) and
(x2, y2, z1) as shown in Figure 2.28.)


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2.12


Figure 2.28 The distance between P1 and P2 is the length
of the diagonal of the rectangular prism having P1 and P2 as
opposite corners.


Example 2.12
Distance in Space
Find the distance between points P1 = (3, −1, 5) and P2 = (2, 1, −1).


Figure 2.29 Find the distance between the two points.
Solution
Substitute values directly into the distance formula:


d⎛⎝P1, P2

⎠ = (x2 − x1)


2 + (y2 − y1)
2 + (z2 − z1)


2


= (2 − 3)2 + (1 − (−1))2 + (−1 − 5)2


= 12 + 22 + (−6)2


= 41.


Find the distance between points P1 = (1, −5, 4) and P2 = (4, −1, −1).


Chapter 2 | Vectors in Space 125




Before moving on to the next section, let’s get a feel for how ℝ3 differs from ℝ2 . For example, in ℝ2 , lines that
are not parallel must always intersect. This is not the case in ℝ3 . For example, consider the line shown in Figure 2.30.
These two lines are not parallel, nor do they intersect.


Figure 2.30 These two lines are not parallel, but still do notintersect.


You can also have circles that are interconnected but have no points in common, as in Figure 2.31.


Figure 2.31 These circles are interconnected, but have nopoints in common.


We have a lot more flexibility working in three dimensions than we do if we stuck with only two dimensions.


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Writing Equations in ℝ3
Now that we can represent points in space and find the distance between them, we can learn how to write equations of
geometric objects such as lines, planes, and curved surfaces in ℝ3 . First, we start with a simple equation. Compare the
graphs of the equation x = 0 in ℝ , ℝ2 , and ℝ3 (Figure 2.32). From these graphs, we can see the same equation
can describe a point, a line, or a plane.


Figure 2.32 (a) In ℝ , the equation x = 0 describes a single point. (b) In ℝ2 , the equation x = 0 describes a line, the
y-axis. (c) In ℝ3 , the equation x = 0 describes a plane, the yz-plane.


In space, the equation x = 0 describes all points ⎛⎝0, y, z⎞⎠. This equation defines the yz-plane. Similarly, the xy-plane
contains all points of the form ⎛⎝x, y, 0⎞⎠. The equation z = 0 defines the xy-plane and the equation y = 0 describes the
xz-plane (Figure 2.33).


Figure 2.33 (a) In space, the equation z = 0 describes the xy-plane. (b) All points in the
xz-plane satisfy the equation y = 0.


Understanding the equations of the coordinate planes allows us to write an equation for any plane that is parallel to one ofthe coordinate planes. When a plane is parallel to the xy-plane, for example, the z-coordinate of each point in the plane hasthe same constant value. Only the x- and y-coordinates of points in that plane vary from point to point.


Chapter 2 | Vectors in Space 127




2.13


Rule: Equations of Planes Parallel to Coordinate Planes
1. The plane in space that is parallel to the xy-plane and contains point (a, b, c) can be represented by the


equation z = c.
2. The plane in space that is parallel to the xz-plane and contains point (a, b, c) can be represented by the


equation y = b.
3. The plane in space that is parallel to the yz-plane and contains point (a, b, c) can be represented by the


equation x = a.


Example 2.13
Writing Equations of Planes Parallel to Coordinate Planes


a. Write an equation of the plane passing through point (3, 11, 7) that is parallel to the yz-plane.
b. Find an equation of the plane passing through points (6, −2, 9), (0, −2, 4), and (1, −2, −3).


Solution
a. When a plane is parallel to the yz-plane, only the y- and z-coordinates may vary. The x-coordinate has thesame constant value for all points in this plane, so this plane can be represented by the equation x = 3.
b. Each of the points (6, −2, 9), (0, −2, 4), and (1, −2, −3) has the same y-coordinate. This plane


can be represented by the equation y = −2.


Write an equation of the plane passing through point (1, −6, −4) that is parallel to the xy-plane.


As we have seen, in ℝ2 the equation x = 5 describes the vertical line passing through point (5, 0). This line is parallel
to the y-axis. In a natural extension, the equation x = 5 in ℝ3 describes the plane passing through point (5, 0, 0),
which is parallel to the yz-plane. Another natural extension of a familiar equation is found in the equation of a sphere.
Definition
A sphere is the set of all points in space equidistant from a fixed point, the center of the sphere (Figure 2.34), justas the set of all points in a plane that are equidistant from the center represents a circle. In a sphere, as in a circle, thedistance from the center to a point on the sphere is called the radius.


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Figure 2.34 Each point (x, y, z) on the surface of a sphere is
r units away from the center (a, b, c).


The equation of a circle is derived using the distance formula in two dimensions. In the same way, the equation of a sphereis based on the three-dimensional formula for distance.
Rule: Equation of a Sphere
The sphere with center (a, b, c) and radius r can be represented by the equation


(2.2)(x − a)2 + ⎛⎝y − b⎞⎠2 + (z − c)2 = r2.
This equation is known as the standard equation of a sphere.


Chapter 2 | Vectors in Space 129




2.14


Example 2.14
Finding an Equation of a Sphere
Find the standard equation of the sphere with center (10, 7, 4) and point (−1, 3, −2), as shown in Figure
2.35.


Figure 2.35 The sphere centered at (10, 7, 4) containing
point (−1, 3, −2).


Solution
Use the distance formula to find the radius r of the sphere:


r = (−1 − 10)2 + (3 − 7)2 + (−2 − 4)2


= (−11)2 + (−4)2 + (−6)2


= 173.


The standard equation of the sphere is
(x − 10)2 + ⎛⎝y − 7⎞⎠2 + (z − 4)2 = 173.


Find the standard equation of the sphere with center (−2, 4, −5) containing point (4, 4, −1).


Example 2.15
Finding the Equation of a Sphere
Let P = (−5, 2, 3) and Q = (3, 4, −1), and suppose line segment PQ forms the diameter of a sphere


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2.15


(Figure 2.36). Find the equation of the sphere.


Figure 2.36 Line segment PQ.


Solution
Since PQ is a diameter of the sphere, we know the center of the sphere is the midpoint of PQ. Then,


C = ⎛⎝
−5 + 3


2
, 2 + 4


2
, 3 + (−1)


2



= (−1, 3, 1).


Furthermore, we know the radius of the sphere is half the length of the diameter. This gives
r = 1


2
(−5 − 3)2 + (2 − 4)2 + (3 − (−1))2


= 1
2


64 + 4 + 16


= 21.


Then, the equation of the sphere is (x + 1)2 + ⎛⎝y − 3⎞⎠2 + (z − 1)2 = 21.


Find the equation of the sphere with diameter PQ, where P = (2, −1, −3) and Q = (−2, 5, −1).


Example 2.16
Graphing Other Equations in Three Dimensions
Describe the set of points that satisfies (x − 4)(z − 2) = 0, and graph the set.


Chapter 2 | Vectors in Space 131




2.16


Solution
We must have either x − 4 = 0 or z − 2 = 0, so the set of points forms the two planes x = 4 and z = 2
(Figure 2.37).


Figure 2.37 The set of points satisfying (x − 4)(z − 2) = 0 forms the two planes
x = 4 and z = 2.


Describe the set of points that satisfies ⎛⎝y + 2⎞⎠(z − 3) = 0, and graph the set.


Example 2.17
Graphing Other Equations in Three Dimensions
Describe the set of points in three-dimensional space that satisfies (x − 2)2 + ⎛⎝y − 1⎞⎠2 = 4, and graph the set.
Solution
The x- and y-coordinates form a circle in the xy-plane of radius 2, centered at (2, 1). Since there is no
restriction on the z-coordinate, the three-dimensional result is a circular cylinder of radius 2 centered on the line
with x = 2 and y = 1. The cylinder extends indefinitely in the z-direction (Figure 2.38).


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2.17


Figure 2.38 The set of points satisfying
(x − 2)2 + ⎛⎝y − 1⎞⎠2 = 4. This is a cylinder of radius 2 centered
on the line with x = 2 and y = 1.


Describe the set of points in three dimensional space that satisfies x2 + (z − 2)2 = 16, and graph the
surface.


Working with Vectors in ℝ3
Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they arerepresented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.
Three-dimensional vectors can also be represented in component form. The notation v = 〈 x, y, z 〉 is a natural extension
of the two-dimensional case, representing a vector with the initial point at the origin, (0, 0, 0), and terminal point
(x, y, z). The zero vector is 0 = 〈 0, 0, 0 〉 . So, for example, the three dimensional vector v = 〈 2, 4, 1 〉 is
represented by a directed line segment from point (0, 0, 0) to point (2, 4, 1) (Figure 2.39).


Chapter 2 | Vectors in Space 133




Figure 2.39 Vector v = 〈 2, 4, 1 〉 is represented by a
directed line segment from point (0, 0, 0) to point (2, 4, 1).


Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If v = 〈 x1, y1, z1 〉 and
w = 〈 x2, y2, z2 〉 are vectors, and k is a scalar, then


v + w = 〈 x1 + x2, y1 + y2, z1 + z2 〉 and kv = 〈 kx1, ky1, kz1 〉 .


If k = −1, then kv = (−1)v is written as −v, and vector subtraction is defined by v − w = v + (−w) = v + (−1)w.
The standard unit vectors extend easily into three dimensions as well— i = 〈 1, 0, 0 〉 , j = 〈 0, 1, 0 〉 , and
k = 〈 0, 0, 1 〉 —and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can
represent a vector in ℝ3 in the following ways:


v = 〈 x, y, z 〉 = xi + yj + zk.


Example 2.18
Vector Representations
Let PQ→ be the vector with initial point P = (3, 12, 6) and terminal point Q = (−4, −3, 2) as shown in
Figure 2.40. Express PQ→ in both component form and using standard unit vectors.


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2.18


Figure 2.40 The vector with initial point P = (3, 12, 6) and
terminal point Q = (−4, −3, 2).


Solution
In component form,


PQ


= 〈 x2 − x1, y2 − y1, z2 − z1 〉


= 〈 −4 − 3, −3 − 12, 2 − 6 〉 = 〈 −7, −15, −4 〉 .


In standard unit form,
PQ


= −7i − 15j − 4k.


Let S = (3, 8, 2) and T = (2, −1, 3). Express ST→ in component form and in standard unit form.


As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretationof vector addition, for example, is the same in both two- and three-dimensional space (Figure 2.41).


Figure 2.41 To add vectors in three dimensions, we followthe same procedures we learned for two dimensions.


We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication,can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our


Chapter 2 | Vectors in Space 135




reference.
Rule: Properties of Vectors in Space
Let v = 〈 x1, y1, z1 〉 and w = 〈 x2, y2, z2 〉 be vectors, and let k be a scalar.
Scalar multiplication: kv = 〈 kx1, ky1, kz1 〉
Vector addition: v + w = 〈 x1, y1, z1 〉 + 〈 x2, y2, z2 〉 = 〈 x1 + x2, y1 + y2, z1 + z2 〉
Vector subtraction: v − w = 〈 x1, y1, z1 〉 − 〈 x2, y2, z2 〉 = 〈 x1 − x2, y1 − y2, z1 − z2 〉
Vector magnitude: ‖ v ‖ = x1 2 + y1 2 + z1 2
Unit vector in the direction of v: 1


‖ v ‖
v = 1


‖ v ‖
〈 x1, y1, z1 〉 = 〈


x1
‖ v ‖


,
y1


‖ v ‖
,


z1
‖ v ‖


〉 , if v ≠ 0


We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties.These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectorssatisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties inthree dimensions are straightforward extensions of the proofs in two dimensions.
Example 2.19
Vector Operations in Three Dimensions
Let v = 〈 −2, 9, 5 〉 and w = 〈 1, −1, 0 〉 (Figure 2.42). Find the following vectors.


a. 3v − 2w
b. 5 ‖ w ‖
c. ‖ 5w ‖
d. A unit vector in the direction of v


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2.19


Figure 2.42 The vectors v = 〈 −2, 9, 5 〉 and
w = 〈 1, −1, 0 〉 .


Solution
a. First, use scalar multiplication of each vector, then subtract:


3v − 2w = 3 〈 −2, 9, 5 〉 − 2 〈 1, −1, 0 〉


= 〈 −6, 27, 15 〉 − 〈 2, −2, 0 〉


= 〈 −6 − 2, 27 − (−2), 15 − 0 〉


= 〈 −8, 29, 15 〉 .


b. Write the equation for the magnitude of the vector, then use scalar multiplication:
5 ‖ w ‖ = 5 12 + (−1)2 + 02 = 5 2.


c. First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the sameas for part b.:
‖ 5w ‖ = ‖ 〈 5, −5, 0 〉 ‖ = 52 + (−5)2 + 02 = 50 = 5 2.


d. Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure isthe same in three dimensions:
v


‖ v ‖
= 1


‖ v ‖
〈 −2, 9, 5 〉


= 1
(−2)2 + 92 + 52


〈 −2, 9, 5 〉


= 1
110


〈 −2, 9, 5 〉


= 〈 −2
110


, 9
110


, 5
110


〉 .


Let v = 〈 −1, −1, 1 〉 and w = 〈 2, 0, 1 〉 . Find a unit vector in the direction of 5v + 3w.


Chapter 2 | Vectors in Space 137




Example 2.20
Throwing a Forward Pass
A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd downthe field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward thereceiver at an upward angle of 30° (see the following figure). Write the initial velocity vector of the ball, v, in
component form.


Solution
The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We thenscale the vector appropriately so that it has the right magnitude. Consider the vector w extending from the
quarterback’s arm to a point directly above the receiver’s head at an angle of 30° (see the following figure). This
vector would have the same direction as v, but it may not have the right magnitude.


The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance fromthe quarterback to the receiver is
Dist from QB to receiver = 152 + 202 = 225 + 400 = 625 = 25 yd.


We have 25
‖ w ‖


= cos 30°. Then the magnitude of w is given by
‖ w ‖ = 25


cos 30°
= 25 · 2


3
= 50


3
yd


and the vertical distance from the receiver to the terminal point of w is
Vert dist from receiver to terminal point of w = ‖ w ‖ sin 30° = 50


3
· 1
2
= 25


3
yd.


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2.20


Then w = 〈 20, 15, 25
3
〉 , and has the same direction as v.


Recall, though, that we calculated the magnitude of w to be ‖ w ‖ = 50
3
, and v has magnitude 60 mph.


So, we need to multiply vector w by an appropriate constant, k. We want to find a value of k so that
‖ kw ‖ = 60 mph. We have


‖ kw ‖ = k ‖ w ‖ = k50
3
mph,


so we want
k50


3
= 60


k = 60 3
50


k = 6 3
5


.


Then
v = kw = k 〈 20, 15, 25


3
〉 = 6 3


5
〈 20, 15, 25


3
〉 = 〈 24 3, 18 3, 30 〉 .


Let’s double-check that ‖ v ‖ = 60. We have
‖ v ‖ = ⎛⎝24 3




2 + ⎛⎝18 3




2 + (30)2 = 1728 + 972 + 900 = 3600 = 60 mph.


So, we have found the correct components for v.


Assume the quarterback and the receiver are in the same place as in the previous example. This time,however, the quarterback throws the ball at velocity of 40 mph and an angle of 45°. Write the initial velocity
vector of the ball, v, in component form.


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2.2 EXERCISES
61. Consider a rectangular box with one of the verticesat the origin, as shown in the following figure. If point
A(2, 3, 5) is the opposite vertex to the origin, then find


a. the coordinates of the other six vertices of the boxandb. the length of the diagonal of the box determined bythe vertices O and A.


62. Find the coordinates of point P and determine its
distance to the origin.


For the following exercises, describe and graph the set ofpoints that satisfies the given equation.
63. ⎛⎝y − 5⎞⎠(z − 6) = 0
64. (z − 2)(z − 5) = 0
65. ⎛⎝y − 1⎞⎠2 + (z − 1)2 = 1
66. (x − 2)2 + (z − 5)2 = 4
67. Write the equation of the plane passing through point
(1, 1, 1) that is parallel to the xy-plane.
68. Write the equation of the plane passing through point
(1, −3, 2) that is parallel to the xz-plane.


69. Find an equation of the plane passing through points
(1, −3, −2), (0, 3, −2), and (1, 0, −2).
70. Find an equation of the plane passing through points
(1, 9, 2), (1, 3, 6), and (1, −7, 8).
For the following exercises, find the equation of the spherein standard form that satisfies the given conditions.
71. Center C(−1, 7, 4) and radius 4
72. Center C(−4, 7, 2) and radius 6
73. Diameter PQ, where P(−1, 5, 7) and
Q(−5, 2, 9)


74. Diameter PQ, where P(−16, −3, 9) and
Q(−2, 3, 5)


For the following exercises, find the center and radius ofthe sphere with an equation in general form that is given.
75. P(1, 2, 3) x2 + y2 + z2 − 4z + 3 = 0
76. x2 + y2 + z2 − 6x + 8y − 10z + 25 = 0
For the following exercises, express vector PQ→ with the
initial point at P and the terminal point at Q


a. in component form and
b. by using standard unit vectors.


77. P(3, 0, 2) and Q(−1, −1, 4)
78. P(0, 10, 5) and Q(1, 1, −3)
79. P(−2, 5, −8) and M(1, −7, 4), where M is the
midpoint of the line segment PQ
80. Q(0, 7, −6) and M(−1, 3, 2), where M is the
midpoint of the line segment PQ
81. Find terminal point Q of vector
PQ


= 〈 7, −1, 3 〉 with the initial point at
P(−2, 3, 5).


82. Find initial point P of vector PQ→ = 〈 −9, 1, 2 〉
with the terminal point at Q(10, 0, −1).
For the following exercises, use the given vectors a and


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b to find and express the vectors a + b, 4a, and
−5a + 3b in component form.
83. a = 〈 −1, −2, 4 〉 , b = 〈 −5, 6, −7 〉
84. a = 〈 3, −2, 4 〉 , b = 〈 −5, 6, −9 〉
85. a = −k, b = −i
86. a = i + j + k, b = 2i − 3j + 2k
For the following exercises, vectors u and v are given. Findthe magnitudes of vectors u − v and −2u.
87. u = 2i + 3j + 4k, v = −i + 5j − k
88. u = i + j, v = j − k
89. u = 〈 2 cos t, −2 sin t, 3 〉 , v = 〈 0, 0, 3 〉 ,
where t is a real number.
90. u = 〈 0, 1, sinh t 〉 , v = 〈 1, 1, 0 〉 , where t
is a real number.
For the following exercises, find the unit vector in thedirection of the given vector a and express it using
standard unit vectors.
91. a = 3i − 4j
92. a = 〈 4, −3, 6 〉
93. a = PQ→ , where P(−2, 3, 1) and Q(0, −4, 4)
94. a = OP→ , where P(−1, −1, 1)
95. a = u − v + w, where u = i − j − k,
v = 2i − j + k, and w = −i + j + 3k
96. a = 2u + v − w, where u = i − k, v = 2j, and
w = i − j


97. Determine whether AB→ and PQ→ are equivalent
vectors, where A(1, 1, 1), B(3, 3, 3), P(1, 4, 5), and
Q(3, 6, 7).


98. Determine whether the vectors AB→ and PQ→ are
equivalent, where A(1, 4, 1), B(−2, 2, 0),
P(2, 5, 7), and Q(−3, 2, 1).
For the following exercises, find vector u with a
magnitude that is given and satisfies the given conditions.


99. v = 〈 7, −1, 3 〉 , ‖ u ‖ = 10, u and v have
the same direction
100. v = 〈 2, 4, 1 〉 , ‖ u ‖ = 15, u and v have
the same direction
101. v = 〈 2 sin t, 2 cos t, 1 〉 , ‖ u ‖ = 2, u and
v have opposite directions for any t, where t is a real
number
102. v = 〈 3 sinh t, 0, 3 〉 , ‖ u ‖ = 5, u and v
have opposite directions for any t, where t is a real
number
103. Determine a vector of magnitude 5 in the direction
of vector AB→ , where A(2, 1, 5) and B(3, 4, −7).
104. Find a vector of magnitude 2 that points in the
opposite direction than vector AB→ , where A(−1, −1, 1)
and B(0, 1, 1). Express the answer in component form.
105. Consider the points A(2, α, 0), B⎛⎝0, 1, β⎞⎠, and
C⎛⎝1, 1, β⎞⎠, where α and β are negative real numbers.
Find α and β such that
‖ OA



− OB



+ OC



‖ = ‖ OB



‖ = 4.


106. Consider points A(α, 0, 0), B⎛⎝0, β, 0⎞⎠, and
C⎛⎝α, β, β⎞⎠, where α and β are positive real numbers.
Find α and β such that
‖ OA



+ OB



‖ = 2 and ‖ OC



‖ = 3.


107. Let P(x, y, z) be a point situated at an equal
distance from points A(1, −1, 0) and B(−1, 2, 1). Show
that point P lies on the plane of equation
−2x + 3y + z = 2.


108. Let P(x, y, z) be a point situated at an equal
distance from the origin and point A(4, 1, 2). Show that
the coordinates of point P satisfy the equation
8x + 2y + 4z = 21.


109. The points A, B, and C are collinear (in this order)
if the relation ‖ AB→ ‖ + ‖ BC→ ‖ = ‖ AC→ ‖ is
satisfied. Show that A(5, 3, −1), B(−5, −3, 1), and
C(−15, −9, 3) are collinear points.
110. Show that points A(1, 0, 1), B(0, 1, 1), and
C(1, 1, 1) are not collinear.


Chapter 2 | Vectors in Space 141




111. [T] A force F of 50 N acts on a particle in the
direction of the vector OP→ , where P(3, 4, 0).


a. Express the force as a vector in component form.b. Find the angle between force F and the positive
direction of the x-axis. Express the answer indegrees rounded to the nearest integer.


112. [T] A force F of 40 N acts on a box in the direction
of the vector OP→ , where P(1, 0, 2).


a. Express the force as a vector by using standard unitvectors.b. Find the angle between force F and the positive
direction of the x-axis.


113. If F is a force that moves an object from point
P1 (x1, y1, z1) to another point P2 (x2, y2, z2), then the
displacement vector is defined as
D = (x2 − x1)i + (y2 − y1)j + (z2 − z1)k. A metal
container is lifted 10 m vertically by a constant force F.
Express the displacement vector D by using standard unit
vectors.
114. A box is pulled 4 yd horizontally in the x-direction
by a constant force F. Find the displacement vector in
component form.
115. The sum of the forces acting on an object is calledthe resultant or net force. An object is said to be in staticequilibrium if the resultant force of the forces that act on itis zero. Let F1 = 〈 10, 6, 3 〉 , F2 = 〈 0, 4, 9 〉 , and
F3 = 〈 10, −3, −9 〉 be three forces acting on a box.
Find the force F4 acting on the box such that the box is in
static equilibrium. Express the answer in component form.
116. [T] Let Fk = 〈 1, k, k2 〉 , k = 1,..., n be n
forces acting on a particle, with n ≥ 2.


a. Find the net force F = ∑
k = 1


n


Fk. Express the
answer using standard unit vectors.b. Use a computer algebra system (CAS) to find nsuch that ‖ F ‖ < 100.


117. The force of gravity F acting on an object is given
by F = mg, where m is the mass of the object (expressed
in kilograms) and g is acceleration resulting from gravity,
with ‖ g ‖ = 9.8 N/kg. A 2-kg disco ball hangs by a
chain from the ceiling of a room.a. Find the force of gravity F acting on the disco ball


and find its magnitude.b. Find the force of tension T in the chain and its
magnitude.Express the answers using standard unit vectors.


Figure 2.43 (credit: modification of work by Kenneth Lu,Flickr)
118. A 5-kg pendant chandelier is designed such that thealabaster bowl is held by four chains of equal length, asshown in the following figure.a. Find the magnitude of the force of gravity acting onthe chandelier.b. Find the magnitudes of the forces of tension foreach of the four chains (assume chains areessentially vertical).


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119. [T] A 30-kg block of cement is suspended by threecables of equal length that are anchored at points
P(−2, 0, 0), Q⎛⎝1, 3, 0



⎠, and R⎛⎝1, − 3, 0⎞⎠. The load


is located at S⎛⎝0, 0, −2 3⎞⎠, as shown in the following
figure. Let F1, F2, and F3 be the forces of tension
resulting from the load in cables RS, QS, and PS,
respectively.a. Find the gravitational force F acting on the block


of cement that counterbalances the sum
F1 + F2 + F3 of the forces of tension in the
cables.b. Find forces F1, F2, and F3. Express the
answer in component form.


120. Two soccer players are practicing for an upcominggame. One of them runs 10 m from point A to point B. Shethen turns left at 90° and runs 10 m until she reaches point
C. Then she kicks the ball with a speed of 10 m/sec at anupward angle of 45° to her teammate, who is located at
point A. Write the velocity of the ball in component form.


121. Let r(t) = 〈 x(t), y(t), z(t) 〉 be the position vector
of a particle at the time t ∈ [0, T], where x, y, and
z are smooth functions on [0, T]. The instantaneous
velocity of the particle at time t is defined by vector
v(t) = 〈 x′(t), y′(t), z′(t) 〉 , with components that are
the derivatives with respect to t, of the functions x, y,
and z, respectively. The magnitude ‖ v(t) ‖ of the
instantaneous velocity vector is called the speed of theparticle at time t. Vector a(t) = 〈 x″(t), y″(t), z″(t) 〉 ,
with components that are the second derivatives withrespect to t, of the functions x, y, and z, respectively,
gives the acceleration of the particle at time t. Consider
r(t) = 〈 cos t, sin t, 2t 〉 the position vector of a particle
at time t ∈ [0, 30], where the components of r are
expressed in centimeters and time is expressed in seconds.a. Find the instantaneous velocity, speed, andacceleration of the particle after the first second.Round your answer to two decimal places.b. Use a CAS to visualize the path of theparticle—that is, the set of all points of coordinates


(cos t, sin t, 2t), where t ∈ [0, 30].
122. [T] Let r(t) = 〈 t, 2t2, 4t2 〉 be the position vector
of a particle at time t (in seconds), where t ∈ [0, 10]
(here the components of r are expressed in centimeters).


a. Find the instantaneous velocity, speed, andacceleration of the particle after the first twoseconds. Round your answer to two decimal places.b. Use a CAS to visualize the path of the particle
defined by the points ⎛⎝t, 2t2, 4t2⎞⎠, where
t ∈ [0, 60].


Chapter 2 | Vectors in Space 143




2.3 | The Dot Product
Learning Objectives


2.3.1 Calculate the dot product of two given vectors.
2.3.2 Determine whether two given vectors are perpendicular.
2.3.3 Find the direction cosines of a given vector.
2.3.4 Explain what is meant by the vector projection of one vector onto another vector, anddescribe how to compute it.
2.3.5 Calculate the work done by a given force.


If we apply a force to an object so that the object moves, we say that work is done by the force. In Introduction toApplications of Integration (http://cnx.org/content/m53638/latest/) on integration applications, we looked at aconstant force and we assumed the force was applied in the direction of motion of the object. Under those conditions, workcan be expressed as the product of the force acting on an object and the distance the object moves. In this chapter, however,we have seen that both force and the motion of an object can be represented by vectors.
In this section, we develop an operation called the dot product, which allows us to calculate work in the case when the forcevector and the motion vector have different directions. The dot product essentially tells us how much of the force vector isapplied in the direction of the motion vector. The dot product can also help us measure the angle formed by a pair of vectorsand the position of a vector relative to the coordinate axes. It even provides a simple test to determine whether two vectorsmeet at a right angle.
The Dot Product and Its Properties
We have already learned how to add and subtract vectors. In this chapter, we investigate two types of vector multiplication.The first type of vector multiplication is called the dot product, based on the notation we use for it, and it is defined asfollows:
Definition
The dot product of vectors u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 is given by the sum of the products of the
components


(2.3)u · v = u1 v1 + u2 v2 + u3 v3.
Note that if u and v are two-dimensional vectors, we calculate the dot product in a similar fashion. Thus, if u = 〈 u1, u2 〉
and v = 〈 v1, v2 〉 , then


u · v = u1 v1 + u2 v2.


When two vectors are combined under addition or subtraction, the result is a vector. When two vectors are combined usingthe dot product, the result is a scalar. For this reason, the dot product is often called the scalar product. It may also be calledthe inner product.
Example 2.21
Calculating Dot Products


a. Find the dot product of u = 〈 3, 5, 2 〉 and v = 〈 −1, 3, 0 〉 .
b. Find the scalar product of p = 10i − 4j + 7k and q = −2i + j + 6k.


Solution


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2.21


a. Substitute the vector components into the formula for the dot product:
u · v = u1 v1 + u2 v2 + u3 v3


= 3(−1) + 5(3) + 2(0) = −3 + 15 + 0 = 12.


b. The calculation is the same if the vectors are written using standard unit vectors. We still have threecomponents for each vector to substitute into the formula for the dot product:
p ·q = u1 v1 + u2 v2 + u3 v3


= 10(−2) + (−4)(1) + (7)(6) = −20 − 4 + 42 = 18.


Find u · v, where u = 〈 2, 9, −1 〉 and v = 〈 −3, 1, −4 〉 .


Like vector addition and subtraction, the dot product has several algebraic properties. We prove three of these propertiesand leave the rest as exercises.
Theorem 2.3: Properties of the Dot Product
Let u, v, and w be vectors, and let c be a scalar.


i. u · v = v ·u Commutative property
ii. u · (v + w) = u · v + u ·w Distributive property
iii. c(u · v) = (cu) · v = u · (cv) Associative property


iv. v · v = ‖ v ‖ 2 Property of magnitude


Proof
Let u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 . Then


u · v = 〈 u1, u2, u3 〉 · 〈 v1, v2, v3 〉
= u1 v1 + u2 v2 + u3 v3
= v1u1 + v2u2 + v3u3
= 〈 v1, v2, v3 〉 · 〈 u1, u2, u3 〉
= v ·u.


The associative property looks like the associative property for real-number multiplication, but pay close attention to thedifference between scalar and vector objects:
c(u · v) = c(u1 v1 + u2 v2 + u3 v3)


= c(u1 v1) + c(u2 v2) + c(u3 v3)


= (cu1)v1 + (cu2)v2 + (cu3)v3
= 〈 cu1, cu2, cu3 〉 · 〈 v1, v2, v3 〉


= c 〈 u1, u2, u3 〉 · 〈 v1, v2, v3 〉


= (cu) · v.


The proof that c(u · v) = u · (cv) is similar.
The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:


Chapter 2 | Vectors in Space 145




2.22


v · v = 〈 v1, v2, v3 〉 · 〈 v1, v2, v3 〉


= (v1)
2 + (v2)


2 + (v3)
2


= ⎡⎣ (v1)
2 + (v2)


2 + (v3)
2⎤


2


= ‖ v ‖ 2.



Note that by property iv. we have 0 · v = 0. Also by property iv. if v · v = 0, then v = 0.
Example 2.22
Using Properties of the Dot Product
Let a = 〈 1, 2, −3 〉 , b = 〈 0, 2, 4 〉 , and c = 〈 5, −1, 3 〉 . Find each of the following products.


a. (a ·b)c
b. a · (2c)
c. ‖ b ‖ 2


Solution
a. Note that this expression asks for the scalar multiple of c by a ·b:


(a ·b)c = ⎛⎝ 〈 1, 2, −3 〉 · 〈 0, 2, 4 〉 ⎞⎠ 〈 5, −1, 3 〉


= (1(0) + 2(2) + (−3)(4)) 〈 5, −1, 3 〉


= −8 〈 5, −1, 3 〉


= 〈 −40, 8, −24 〉 .


b. This expression is a dot product of vector a and scalar multiple 2c:
a · (2c) = 2(a · c)


= 2⎛⎝ 〈 1, 2, −3 〉 · 〈 5, −1, 3 〉 ⎞⎠
= 2⎛⎝1(5) + 2(−1) + (−3)(3)⎞⎠
= 2(−6) = −12.


c. Simplifying this expression is a straightforward application of the dot product:
‖ b ‖ 2 = b ·b = 〈 0, 2, 4 〉 · 〈 0, 2, 4 〉 = 02 + 22 + 42 = 0 + 4 + 16 = 20.


Find the following products for p = 〈 7, 0, 2 〉 , q = 〈 −2, 2, −2 〉 , and r = 〈 0, 2, −3 〉 .
a. (r ·p)q
b. ‖ p ‖ 2


Using the Dot Product to Find the Angle between Two Vectors
When two nonzero vectors are placed in standard position, whether in two dimensions or three dimensions, they form anangle between them (Figure 2.44). The dot product provides a way to find the measure of this angle. This property is aresult of the fact that we can express the dot product in terms of the cosine of the angle formed by two vectors.


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Figure 2.44 Let θ be the angle between two nonzero vectors uand v such that 0 ≤ θ ≤ π.


Theorem 2.4: Evaluating a Dot Product
The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle betweenthem:


(2.4)u · v = ‖ u ‖ ‖ v ‖ cos θ.


Proof
Place vectors u and v in standard position and consider the vector v − u (Figure 2.45). These three vectors form a triangle
with side lengths ‖ u ‖ , ‖ v ‖ , and ‖ v − u ‖ .


Figure 2.45 The lengths of the sides of the triangle are givenby the magnitudes of the vectors that form the triangle.


Recall from trigonometry that the law of cosines describes the relationship among the side lengths of the triangle and theangle θ. Applying the law of cosines here gives
‖ v − u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ‖ u ‖ ‖ v ‖ cos θ.


The dot product provides a way to rewrite the left side of this equation:
‖ v − u ‖ 2 = (v − u) · (v − u)


= (v − u) · v − (v − u) ·u
= v · v − u · v − v ·u + u ·u
= v · v − u · v − u · v + u ·u


= ‖ v ‖ 2 − 2u · v + ‖ u ‖ 2.


Substituting into the law of cosines yields
‖ v − u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ‖ u ‖ ‖ v ‖ cos θ


‖ v ‖ 2 − 2u · v + ‖ u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ‖ u ‖ ‖ v ‖ cos θ
−2u · v = −2 ‖ u ‖ ‖ v ‖ cos θ


u · v = ‖ u ‖ ‖ v ‖ cos θ.



We can use this form of the dot product to find the measure of the angle between two nonzero vectors. The followingequation rearranges Equation 2.3 to solve for the cosine of the angle:


(2.5)cos θ = u · v
‖ u ‖ ‖ v ‖


.


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2.23


Using this equation, we can find the cosine of the angle between two nonzero vectors. Since we are considering the smallestangle between the vectors, we assume 0° ≤ θ ≤ 180° (or 0 ≤ θ ≤ π if we are working in radians). The inverse cosine is
unique over this range, so we are then able to determine the measure of the angle θ.
Example 2.23
Finding the Angle between Two Vectors
Find the measure of the angle between each pair of vectors.


a. i + j + k and 2i – j – 3k
b. 〈 2, 5, 6 〉 and 〈 −2, −4, 4 〉


Solution
a. To find the cosine of the angle formed by the two vectors, substitute the components of the vectors intoEquation 2.5:


cos θ =
(i + j + k) · ⎛⎝2i − j − 3k⎞⎠


‖ i + j + k ‖ · ‖ 2i − j − 3k ‖


= 1(2) + (1)(−1) + (1)(−3)


12 + 12 + 12 22 + (−1)2 + (−3)2


= −2
3 14


= −2
42


.


Therefore, θ = arccos −2
42


rad.
b. Start by finding the value of the cosine of the angle between the vectors:


cos θ =
〈 2, 5, 6 〉 · 〈 −2, −4, 4 〉


‖ 〈 2, 5, 6 〉 ‖ · ‖ 〈 −2, −4, 4 〉 ‖


= 2(−2) + (5)(−4) + (6)(4)


22 + 52 + 62 (−2)2 + (−4)2 + 42


= 0
65 36


= 0.


Now, cos θ = 0 and 0 ≤ θ ≤ π, so θ = π/2.


Find the measure of the angle, in radians, formed by vectors a = 〈 1, 2, 0 〉 and b = 〈 2, 4, 1 〉 .
Round to the nearest hundredth.


The angle between two vectors can be acute (0 < cos θ < 1), obtuse (−1 < cos θ < 0), or straight (cos θ = −1). If
cos θ = 1, then both vectors have the same direction. If cos θ = 0, then the vectors, when placed in standard position,
form a right angle (Figure 2.46). We can formalize this result into a theorem regarding orthogonal (perpendicular) vectors.


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Figure 2.46 (a) An acute angle has 0 < cos θ < 1. (b) An obtuse angle has
−1 < cos θ < 0. (c) A straight line has cos θ = −1. (d) If the vectors have the same
direction, cos θ = 1. (e) If the vectors are orthogonal (perpendicular), cos θ = 0.


Theorem 2.5: Orthogonal Vectors
The nonzero vectors u and v are orthogonal vectors if and only if u · v = 0.


Proof
Let u and v be nonzero vectors, and let θ denote the angle between them. First, assume u · v = 0. Then


‖ u ‖ ‖ v ‖ cos θ = 0.


However, ‖ u ‖ ≠ 0 and ‖ v ‖ ≠ 0, so we must have cos θ = 0. Hence, θ = 90°, and the vectors are orthogonal.
Now assume u and v are orthogonal. Then θ = 90° and we have


u · v = ‖ u ‖ ‖ v ‖ cos θ = ‖ u ‖ ‖ v ‖ cos 90° = ‖ u ‖ ‖ v ‖ (0) = 0.



The terms orthogonal, perpendicular, and normal each indicate that mathematical objects are intersecting at right angles.The use of each term is determined mainly by its context. We say that vectors are orthogonal and lines are perpendicular.The term normal is used most often when measuring the angle made with a plane or other surface.


Example 2.24
Identifying Orthogonal Vectors
Determine whether p = 〈 1, 0, 5 〉 and q = 〈 10, 3, −2 〉 are orthogonal vectors.


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2.24


Solution
Using the definition, we need only check the dot product of the vectors:


p ·q = 1(10) + (0)(3) + (5)(−2) = 10 + 0 − 10 = 0.


Because p ·q = 0, the vectors are orthogonal (Figure 2.47).


Figure 2.47 Vectors p and q form a right angle when theirinitial points are aligned.


For which value of x is p = 〈 2, 8, −1 〉 orthogonal to q = 〈 x, −1, 2 〉 ?


Example 2.25
Measuring the Angle Formed by Two Vectors
Let v = 〈 2, 3, 3 〉 . Find the measures of the angles formed by the following vectors.


a. v and i
b. v and j
c. v and k


Solution
a. Let α be the angle formed by v and i:


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2.25


cos α = v · i
‖ v ‖ · ‖ i ‖


=
〈 2, 3, 3 〉 · 〈 1, 0, 0 〉


22 + 32 + 32 1


= 2
22


.


α = arccos 2
22


≈ 1.130 rad.


b. Let β represent the angle formed by v and j:
cos β =


v · j
‖ v ‖ · ‖ j ‖


=
〈 2, 3, 3 〉 · 〈 0, 1, 0 〉


22 + 32 + 32 1


= 3
22


.


β = arccos 3
22


≈ 0.877 rad.


c. Let γ represent the angle formed by v and k:
cos γ = v ·k


‖ v ‖ · ‖ k ‖


=
〈 2, 3, 3 〉 · 〈 0, 0, 1 〉


22 + 32 + 32 1


= 3
22


.


γ = arccos 3
22


≈ 0.877 rad.


Let v = 〈 3, −5, 1 〉 . Find the measure of the angles formed by each pair of vectors.
a. v and i
b. v and j
c. v and k


The angle a vector makes with each of the coordinate axes, called a direction angle, is very important in practicalcomputations, especially in a field such as engineering. For example, in astronautical engineering, the angle at which arocket is launched must be determined very precisely. A very small error in the angle can lead to the rocket going hundredsof miles off course. Direction angles are often calculated by using the dot product and the cosines of the angles, called thedirection cosines. Therefore, we define both these angles and their cosines.
Definition
The angles formed by a nonzero vector and the coordinate axes are called the direction angles for the vector (Figure2.48). The cosines for these angles are called the direction cosines.


Chapter 2 | Vectors in Space 151




Figure 2.48 Angle α is formed by vector v and unit vector i.Angle β is formed by vector v and unit vector j. Angle γ isformed by vector v and unit vector k.


In Example 2.25, the direction cosines of v = 〈 2, 3, 3 〉 are cos α = 2
22


, cos β = 3
22


, and cos γ = 3
22


. The
direction angles of v are α = 1.130 rad, β = 0.877 rad, and γ = 0.877 rad.
So far, we have focused mainly on vectors related to force, movement, and position in three-dimensional physical space.However, vectors are often used in more abstract ways. For example, suppose a fruit vendor sells apples, bananas,and oranges. On a given day, he sells 30 apples, 12 bananas, and 18 oranges. He might use a quantity vector,
q = 〈 30, 12, 18 〉 , to represent the quantity of fruit he sold that day. Similarly, he might want to use a price vector,
p = 〈 0.50, 0.25, 1 〉 , to indicate that he sells his apples for 50¢ each, bananas for 25¢ each, and oranges for $1 apiece.
In this example, although we could still graph these vectors, we do not interpret them as literal representations of positionin the physical world. We are simply using vectors to keep track of particular pieces of information about apples, bananas,and oranges.
This idea might seem a little strange, but if we simply regard vectors as a way to order and store data, we find they can bequite a powerful tool. Going back to the fruit vendor, let’s think about the dot product, q ·p. We compute it by multiplying
the number of apples sold (30) by the price per apple (50¢), the number of bananas sold by the price per banana, and thenumber of oranges sold by the price per orange. We then add all these values together. So, in this example, the dot producttells us how much money the fruit vendor had in sales on that particular day.
When we use vectors in this more general way, there is no reason to limit the number of components to three. What ifthe fruit vendor decides to start selling grapefruit? In that case, he would want to use four-dimensional quantity and pricevectors to represent the number of apples, bananas, oranges, and grapefruit sold, and their unit prices. As you might expect,to calculate the dot product of four-dimensional vectors, we simply add the products of the components as before, but thesum has four terms instead of three.
Example 2.26
Using Vectors in an Economic Context
AAA Party Supply Store sells invitations, party favors, decorations, and food service items such as paper platesand napkins. When AAA buys its inventory, it pays 25¢ per package for invitations and party favors. Decorationscost AAA 50¢ each, and food service items cost 20¢ per package. AAA sells invitations for $2.50 per package andparty favors for $1.50 per package. Decorations sell for $4.50 each and food service items for $1.25 per package.
During the month of May, AAA Party Supply Store sells 1258 invitations, 342 party favors, 2426 decorations,and 1354 food service items. Use vectors and dot products to calculate how much money AAA made in salesduring the month of May. How much did the store make in profit?


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2.26


Solution
The cost, price, and quantity vectors are


c = 〈 0.25, 0.25, 0.50, 0.20 〉


p = 〈 2.50, 1.50, 4.50, 1.25 〉


q = 〈 1258, 342, 2426, 1354 〉 .


AAA sales for the month of May can be calculated using the dot product p ·q. We have
p ·q = 〈 2.50, 1.50, 4.50, 1.25 〉 · 〈 1258, 342, 2426, 1354 〉


= 3145 + 513 + 10917 + 1692.5
= 16267.5.


So, AAA took in $16,267.50 during the month of May.
To calculate the profit, we must first calculate how much AAA paid for the items sold. We use the dot product
c ·q to get


c ·q = 〈 0.25, 0.25, 0.50, 0.20 〉 · 〈 1258, 342, 2426, 1354 〉
= 314.5 + 85.5 + 1213 + 270.8
= 1883.8.


So, AAA paid $1,883.30 for the items they sold. Their profit, then, is given by
p ·q − c ·q = 16267.5 − 1883.8


= 14383.7.


Therefore, AAA Party Supply Store made $14,383.70 in May.


On June 1, AAA Party Supply Store decided to increase the price they charge for party favors to $2 perpackage. They also changed suppliers for their invitations, and are now able to purchase invitations for only10¢ per package. All their other costs and prices remain the same. If AAA sells 1408 invitations, 147 partyfavors, 2112 decorations, and 1894 food service items in the month of June, use vectors and dot products tocalculate their total sales and profit for June.


Projections
As we have seen, addition combines two vectors to create a resultant vector. But what if we are given a vector and we needto find its component parts? We use vector projections to perform the opposite process; they can break down a vector intoits components. The magnitude of a vector projection is a scalar projection. For example, if a child is pulling the handle ofa wagon at a 55° angle, we can use projections to determine how much of the force on the handle is actually moving thewagon forward (Figure 2.49). We return to this example and learn how to solve it after we see how to calculate projections.


Figure 2.49 When a child pulls a wagon, only the horizontalcomponent of the force propels the wagon forward.


Chapter 2 | Vectors in Space 153




Definition
The vector projection of v onto u is the vector labeled projuv in Figure 2.50. It has the same initial point as u and vand the same direction as u, and represents the component of v that acts in the direction of u. If θ represents the angle
between u and v, then, by properties of triangles, we know the length of projuv is ‖ projuv ‖ = ‖ v ‖ cos θ.
When expressing cos θ in terms of the dot product, this becomes


‖ projuv ‖ = ‖ v ‖ cos θ


= ‖ v ‖



u · v
‖ u ‖ ‖ v ‖





= u · v
‖ u ‖


.


We now multiply by a unit vector in the direction of u to get projuv:
(2.6)


projuv = u · v‖ u ‖



1
‖ u ‖


u

⎠ =


u · v
‖ u ‖ 2


u.


The length of this vector is also known as the scalar projection of v onto u and is denoted by
(2.7)‖ projuv ‖ = compuv = u · v‖ u ‖ .


Figure 2.50 The projection of v onto u shows the componentof vector v in the direction of u.


Example 2.27
Finding Projections
Find the projection of v onto u.


a. v = 〈 3, 5, 1 〉 and u = 〈 −1, 4, 3 〉
b. v = 3i − 2j and u = i + 6j


Solution
a. Substitute the components of v and u into the formula for the projection:


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projuv = u · v
‖ u ‖ 2


u


=
〈 −1, 4, 3 〉 · 〈 3, 5, 1 〉


‖ 〈 −1, 4, 3 〉 ‖ 2
〈 −1, 4, 3 〉


= −3 + 20 + 3
(−1)2 + 42 + 32


〈 −1, 4, 3 〉


= 20
26


〈 −1, 4, 3 〉


= 〈 −10
13


, 40
13


, 30
13


〉 .


b. To find the two-dimensional projection, simply adapt the formula to the two-dimensional case:
projuv = u · v


‖ u ‖ 2
u


=

⎝i + 6j⎞⎠ · ⎛⎝3i − 2j⎞⎠


‖ i + 6j ‖ 2

⎝i + 6j⎞⎠


= 1(3) + 6(−2)
12 + 62



⎝i + 6j⎞⎠


= − 9
37

⎝i + 6j⎞⎠


= − 9
37


i − 54
37


j.


Sometimes it is useful to decompose vectors—that is, to break a vector apart into a sum. This process is called the resolutionof a vector into components. Projections allow us to identify two orthogonal vectors having a desired sum. For example, let
v = 〈 6, −4 〉 and let u = 〈 3, 1 〉 . We want to decompose the vector v into orthogonal components such that one of
the component vectors has the same direction as u.
We first find the component that has the same direction as u by projecting v onto u. Let p = projuv. Then, we have


p = u · v
‖ u ‖ 2


u


= 18 − 4
9 + 1


u


= 7
5
u = 7


5
〈 3, 1 〉 = 〈 21


5
, 7
5
〉 .


Now consider the vector q = v − p. We have
q = v − p


= 〈 6, −4 〉 − 〈 21
5
, 7
5


= 〈 9
5
, − 27


5
〉 .


Clearly, by the way we defined q, we have v = q + p, and
q ·p = 〈 9


5
, − 27


5
〉 · 〈 21


5
, 7
5


= 9(21)
25


+ −27(7)
25


= 189
25


− 189
25


= 0.


Therefore, q and p are orthogonal.


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2.27


Example 2.28
Resolving Vectors into Components
Express v = 〈 8, −3, −3 〉 as a sum of orthogonal vectors such that one of the vectors has the same direction
as u = 〈 2, 3, 2 〉 .
Solution
Let p represent the projection of v onto u:


p = projuv


= u · v
‖ u ‖ 2


u


=
〈 2, 3, 2 〉 · 〈 8, −3, −3 〉


‖ 〈 2, 3, 2 〉 ‖ 2
〈 2, 3, 2 〉


= 16 − 9 − 6
22 + 32 + 22


〈 2, 3, 2 〉


= 1
17


〈 2, 3, 2 〉


= 〈 2
17


, 3
17


, 2
17


〉 .


Then,
q = v − p = 〈 8, −3, −3 〉 − 〈 2


17
, 3
17


, 2
17


〉 = 〈 134
17


, − 54
17


, − 53
17


〉 .


To check our work, we can use the dot product to verify that p and q are orthogonal vectors:
p ·q = 〈 2


17
, 3
17


, 2
17


〉 · 〈 134
17


, − 54
17


, − 53
17


〉 = 268
17


− 162
17


− 106
17


= 0.


Then,
v = p + q = 〈 2


17
, 3
17


, 2
17


〉 + 〈 134
17


, − 54
17


, − 53
17


〉 .


Express v = 5i − j as a sum of orthogonal vectors such that one of the vectors has the same direction as
u = 4i + 2j.


Example 2.29
Scalar Projection of Velocity
A container ship leaves port traveling 15° north of east. Its engine generates a speed of 20 knots along that
path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of 2 knots.Considering both the engine and the current, how fast is the ship moving in the direction 15° north of east?
Round the answer to two decimal places.


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2.28


Solution
Let v be the velocity vector generated by the engine, and let w be the velocity vector of the current. We alreadyknow ‖ v ‖ = 20 along the desired route. We just need to add in the scalar projection of w onto v. We get


compvw = v ·w‖ v ‖


= ‖ v ‖ ‖ w ‖ cos(30°)
‖ v ‖


= ‖ w ‖ cos(30°)


= 2 3
2


= 3 ≈ 1.73 knots.


The ship is moving at 21.73 knots in the direction 15° north of east.


Repeat the previous example, but assume the ocean current is moving southeast instead of northeast, asshown in the following figure.


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Work
Now that we understand dot products, we can see how to apply them to real-life situations. The most common applicationof the dot product of two vectors is in the calculation of work.
From physics, we know that work is done when an object is moved by a force. When the force is constant and applied in thesame direction the object moves, then we define the work done as the product of the force and the distance the object travels:
W = Fd. We saw several examples of this type in earlier chapters. Now imagine the direction of the force is different from
the direction of motion, as with the example of a child pulling a wagon. To find the work done, we need to multiply thecomponent of the force that acts in the direction of the motion by the magnitude of the displacement. The dot product allowsus to do just that. If we represent an applied force by a vector F and the displacement of an object by a vector s, then thework done by the force is the dot product of F and s.
Definition
When a constant force is applied to an object so the object moves in a straight line from point P to point Q, the workW done by the force F, acting at an angle θ from the line of motion, is given by


(2.8)W = F · PQ→ = ‖ F ‖ ‖ PQ→ ‖ cos θ.
Let’s revisit the problem of the child’s wagon introduced earlier. Suppose a child is pulling a wagon with a force havinga magnitude of 8 lb on the handle at an angle of 55°. If the child pulls the wagon 50 ft, find the work done by the force(Figure 2.51).


Figure 2.51 The horizontal component of the force is theprojection of F onto the positive x-axis.


We have
W = ‖ F ‖ ‖ PQ



‖ cos θ = 8(50)⎛⎝cos(55°)⎞⎠ ≈ 229 ft · lb.


In U.S. standard units, we measure the magnitude of force ‖ F ‖ in pounds. The magnitude of the displacement vector
‖ PQ



‖ tells us how far the object moved, and it is measured in feet. The customary unit of measure for work, then, is


the foot-pound. One foot-pound is the amount of work required to move an object weighing 1 lb a distance of 1 ft straightup. In the metric system, the unit of measure for force is the newton (N), and the unit of measure of magnitude for work isa newton-meter (N·m), or a joule (J).
Example 2.30
Calculating Work
A conveyor belt generates a force F = 5i − 3j + k that moves a suitcase from point (1, 1, 1) to point (9, 4, 7)
along a straight line. Find the work done by the conveyor belt. The distance is measured in meters and the forceis measured in newtons.
Solution


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2.29


The displacement vector PQ→ has initial point (1, 1, 1) and terminal point (9, 4, 7):
PQ


= 〈 9 − 1, 4 − 1, 7 − 1 〉 = 〈 8, 3, 6 〉 = 8i + 3j + 6k.


Work is the dot product of force and displacement:
W = F · PQ




= ⎛⎝5i − 3j + k⎞⎠ · ⎛⎝8i + 3j + 6k⎞⎠
= 5(8) + (−3)(3) + 1(6)
= 37N ·m
= 37 J.


A constant force of 30 lb is applied at an angle of 60° to pull a handcart 10 ft across the ground (Figure2.52). What is the work done by this force?


Figure 2.52


Chapter 2 | Vectors in Space 159




2.3 EXERCISES
For the following exercises, the vectors u and v are given.Calculate the dot product u · v.
123. u = 〈 3, 0 〉 , v = 〈 2, 2 〉
124. u = 〈 3, −4 〉 , v = 〈 4, 3 〉
125. u = 〈 2, 2, −1 〉 , v = 〈 −1, 2, 2 〉
126. u = 〈 4, 5, −6 〉 , v = 〈 0, −2, −3 〉
For the following exercises, the vectors a, b, and c aregiven. Determine the vectors (a ·b)c and (a · c)b.
Express the vectors in component form.
127. a = 〈 2, 0, −3 〉 , b = 〈 −4, −7, 1 〉 ,
c = 〈 1, 1, −1 〉


128. a = 〈 0, 1, 2 〉 , b = 〈 −1, 0, 1 〉 ,
c = 〈 1, 0, −1 〉


129. a = i + j, b = i − k, c = i − 2k
130. a = i − j + k, b = j + 3k, c = −i + 2j − 4k
For the following exercises, the two-dimensional vectors aand b are given.


a. Find the measure of the angle θ between a and
b. Express the answer in radians rounded to twodecimal places, if it is not possible to express itexactly.


b. Is θ an acute angle?
131. [T] a = 〈 3, −1 〉 , b = 〈 −4, 0 〉
132. [T] a = 〈 2, 1 〉 , b = 〈 −1, 3 〉
133. u = 3i, v = 4i + 4j
134. u = 5i, v = −6i + 6j
For the following exercises, find the measure of the anglebetween the three-dimensional vectors a and b. Express theanswer in radians rounded to two decimal places, if it is notpossible to express it exactly.
135. a = 〈 3, −1, 2 〉 , b = 〈 1, −1, −2 〉
136. a = 〈 0, −1, −3 〉 , b = 〈 2, 3, −1 〉
137. a = i + j, b = j − k


138. a = i − 2j + k, b = i + j − 2k
139. [T] a = 3i − j − 2k, b = v + w, where
v = −2i − 3j + 2k and w = i + 2k
140. [T] a = 3i − j + 2k, b = v − w, where
v = 2i + j + 4k and w = 6i + j + 2k
For the following exercises determine whether the givenvectors are orthogonal.
141. a = 〈 x, y 〉 , b = 〈 −y, x 〉 , where x and y are
nonzero real numbers
142. a = 〈 x, x 〉 , b = 〈 −y, y 〉 , where x and y are
nonzero real numbers
143. a = 3i − j − 2k, b = −2i − 3j + k
144. a = i − j, b = 7i + 2j − k
145. Find all two-dimensional vectors a orthogonal tovector b = 〈 3, 4 〉 . Express the answer in component
form.
146. Find all two-dimensional vectors a orthogonal tovector b = 〈 5, −6 〉 . Express the answer by using
standard unit vectors.
147. Determine all three-dimensional vectors uorthogonal to vector v = 〈 1, 1, 0 〉 . Express the answer
by using standard unit vectors.
148. Determine all three-dimensional vectors uorthogonal to vector v = i − j − k. Express the answer in
component form.
149. Determine the real number α such that vectors
a = 2i + 3j and b = 9i + αj are orthogonal.
150. Determine the real number α such that vectors
a = −3i + 2j and b = 2i + αj are orthogonal.
151. [T] Consider the points P(4, 5) and Q(5, −7).


a. Determine vectors OP→ and OQ→ . Express the
answer by using standard unit vectors.b. Determine the measure of angle O in triangle OPQ.Express the answer in degrees rounded to twodecimal places.


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152. [T] Consider points A(1, 1), B(2, −7), and
C(6, 3).


a. Determine vectors BA→ and BC→ . Express the
answer in component form.b. Determine the measure of angle B in triangle ABC.Express the answer in degrees rounded to twodecimal places.


153. Determine the measure of angle A in triangle ABC,where A(1, 1, 8), B(4, −3, −4), and C(−3, 1, 5).
Express your answer in degrees rounded to two decimalplaces.
154. Consider points P(3, 7, −2) and Q(1, 1, −3).
Determine the angle between vectors OP→ and OQ→ .
Express the answer in degrees rounded to two decimalplaces.
For the following exercises, determine which (if any) pairsof the following vectors are orthogonal.
155. u = 〈 3, 7, −2 〉 , v = 〈 5, −3, −3 〉 ,
w = 〈 0, 1, −1 〉


156. u = i − k, v = 5j − 5k, w = 10j
157. Use vectors to show that a parallelogram with equaldiagonals is a square.
158. Use vectors to show that the diagonals of a rhombusare perpendicular.
159. Show that u · (v + w) = u · v + u ·w is true for any
vectors u, v, and w.
160. Verify the identity u · (v + w) = u · v + u ·w for
vectors u = 〈 1, 0, 4 〉 , v = 〈 −2, 3, 5 〉 , and
w = 〈 4, −2, 6 〉 .


For the following problems, the vector u is given.
a. Find the direction cosines for the vector u.
b. Find the direction angles for the vector u expressedin degrees. (Round the answer to the nearestinteger.)


161. u = 〈 2, 2, 1 〉
162. u = i − 2j + 2k
163. u = 〈 −1, 5, 2 〉
164. u = 〈 2, 3, 4 〉


165. Consider u = 〈 a, b, c 〉 a nonzero three-
dimensional vector. Let cos α, cos β, and cos γ be
the directions of the cosines of u. Show that
cos2α + cos2 β + cos2 γ = 1.


166. Determine the direction cosines of vector
u = i + 2j + 2k and show they satisfy
cos2α + cos2 β + cos2 γ = 1.


For the following exercises, the vectors u and v are given.
a. Find the vector projection w = projuv of vector v


onto vector u. Express your answer in componentform.
b. Find the scalar projection compuv of vector v


onto vector u.
167. u = 5i + 2j, v = 2i + 3j
168. u = 〈 −4, 7 〉 , v = 〈 3, 5 〉
169. u = 3i + 2k, v = 2j + 4k
170. u = 〈 4, 4, 0 〉 , v = 〈 0, 4, 1 〉
171. Consider the vectors u = 4i − 3j and v = 3i + 2j.


a. Find the component form of vector w = projuv
that represents the projection of v onto u.b. Write the decomposition v = w + q of vector v
into the orthogonal components w and q, wherew is the projection of v onto u and q is a vectororthogonal to the direction of u.


172. Consider vectors u = 2i + 4j and v = 4j + 2k.
a. Find the component form of vector w = projuv


that represents the projection of v onto u.b. Write the decomposition v = w + q of vector v
into the orthogonal components w and q, wherew is the projection of v onto u and q is a vectororthogonal to the direction of u.


Chapter 2 | Vectors in Space 161




173. A methane molecule has a carbon atom situated atthe origin and four hydrogen atoms located at points
P(1, 1, −1), Q(1, −1, 1), R(−1, 1, 1), and S(−1, −1, −1)


(see figure).a. Find the distance between the hydrogen atomslocated at P and R.
b. Find the angle between vectors OS→ and OR→ that


connect the carbon atom with the hydrogen atomslocated at S and R, which is also called the bondangle. Express the answer in degrees rounded totwo decimal places.


174. [T] Find the vectors that join the center of a clock tothe hours 1:00, 2:00, and 3:00. Assume the clock is circularwith a radius of 1 unit.
175. Find the work done by force F = 〈 5, 6, −2 〉
(measured in Newtons) that moves a particle from point
P(3, −1, 0) to point Q(2, 3, 1) along a straight line (the
distance is measured in meters).
176. [T] A sled is pulled by exerting a force of 100 N on arope that makes an angle of 25° with the horizontal. Find
the work done in pulling the sled 40 m. (Round the answerto one decimal place.)
177. [T] A father is pulling his son on a sled at an angleof 20° with the horizontal with a force of 25 lb (see the
following image). He pulls the sled in a straight path of 50ft. How much work was done by the man pulling the sled?(Round the answer to the nearest integer.)


178. [T] A car is towed using a force of 1600 N. Therope used to pull the car makes an angle of 25° with thehorizontal. Find the work done in towing the car 2 km.Express the answer in joules (1J = 1N ·m) rounded to the
nearest integer.


179. [T] A boat sails north aided by a wind blowing ina direction of N30°E with a magnitude of 500 lb. How
much work is performed by the wind as the boat moves 100ft? (Round the answer to two decimal places.)
180. Vector p = 〈 150, 225, 375 〉 represents the price
of certain models of bicycles sold by a bicycle shop. Vector
n = 〈 10, 7, 9 〉 represents the number of bicycles sold
of each model, respectively. Compute the dot product p ·n
and state its meaning.
181. [T] Two forces F1 and F2 are represented by
vectors with initial points that are at the origin. The firstforce has a magnitude of 20 lb and the terminal point ofthe vector is point P(1, 1, 0). The second force has a
magnitude of 40 lb and the terminal point of its vector ispoint Q(0, 1, 1). Let F be the resultant force of forces F1
and F2.


a. Find the magnitude of F. (Round the answer to onedecimal place.)b. Find the direction angles of F. (Express the answerin degrees rounded to one decimal place.)
182. [T] Consider r(t) = 〈 cos t, sin t, 2t 〉 the position
vector of a particle at time t ∈ [0, 30], where the
components of r are expressed in centimeters and time in
seconds. Let OP→ be the position vector of the particle after
1 sec.


a. Show that all vectors PQ→ , where Q(x, y, z) is
an arbitrary point, orthogonal to the instantaneousvelocity vector v(1) of the particle after 1 sec,
can be expressed as
PQ


= 〈 x − cos 1, y − sin 1, z − 2 〉 , where
x sin 1 − y cos 1 − 2z + 4 = 0. The set of point Q
describes a plane called the normal plane to thepath of the particle at point P.b. Use a CAS to visualize the instantaneous velocityvector and the normal plane at point P along withthe path of the particle.


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2.4 | The Cross Product
Learning Objectives


2.4.1 Calculate the cross product of two given vectors.
2.4.2 Use determinants to calculate a cross product.
2.4.3 Find a vector orthogonal to two given vectors.
2.4.4 Determine areas and volumes by using the cross product.
2.4.5 Calculate the torque of a given force and position vector.


Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies a force at the end of the wrench. This createsrotation, or torque, which tightens the bolt. We can use vectors to represent the force applied by the mechanic, and thedistance (radius) from the bolt to the end of the wrench. Then, we can represent torque by a vector oriented along the axisof rotation. Note that the torque vector is orthogonal to both the force vector and the radius vector.
In this section, we develop an operation called the cross product, which allows us to find a vector orthogonal to two givenvectors. Calculating torque is an important application of cross products, and we examine torque in more detail later in thesection.
The Cross Product and Its Properties
The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product oftwo vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let
u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 be nonzero vectors. We want to find a vector w = 〈 w1, w2, w3 〉
orthogonal to both u and v—that is, we want to find w such that u ·w = 0 and v ·w = 0. Therefore, w1, w2, and
w3 must satisfy


u1w1 + u2w2 + u3w3 = 0


v1w1 + v2w2 + v3w3 = 0.


If we multiply the top equation by v3 and the bottom equation by u3 and subtract, we can eliminate the variable w3,
which gives


(u1 v3 − v1u3)w1 + (u2 v3 − v2u3)w2 = 0.


If we select
w1 = u2 v3 − u3 v2
w2 = −(u1 v3 − u3 v1),


we get a possible solution vector. Substituting these values back into the original equations gives
w3 = u1 v2 − u2 v1.


That is, vector
w = 〈 u2 v3 − u3 v2, −(u1 v3 − u3 v1), u1 v2 − u2 v1 〉


is orthogonal to both u and v, which leads us to define the following operation, called the cross product.


Definition
Let u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 . Then, the cross product u × v is vector


(2.9)u × v = (u2 v3 − u3 v2)i − (u1 v3 − u3 v1)j + (u1 v2 − u2 v1)k
= 〈 u2 v3 − u3 v2, −(u1 v3 − u3 v1), u1 v2 − u2 v1 〉 .


From the way we have developed u × v, it should be clear that the cross product is orthogonal to both u and v. However,


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2.30


it never hurts to check. To show that u × v is orthogonal to u, we calculate the dot product of u and u × v.
u · (u × v) = 〈 u1, u2, u3 〉 · 〈 u2 v3 − u3 v2, −u1 v3 + u3 v1, u1 v2 − u2 v1 〉


= u1 (u2 v3 − u3 v2) + u2 (−u1 v3 + u3 v1) + u3 (u1 v2 − u2 v1)
= u1u2 v3 − u1u3 v2 − u1u2 v3 + u2u3 v1 + u1u3 v2 − u2u3 v1
= (u1u2 v3 − u1u2 v3) + (−u1u3 v2 + u1u3 v2) + (u2u3 v1 − u2u3 v1)


= 0


In a similar manner, we can show that the cross product is also orthogonal to v.
Example 2.31
Finding a Cross Product
Let p = 〈 −1, 2, 5 〉 and q = 〈 4, 0, −3 〉 (Figure 2.53). Find p × q.


Figure 2.53 Finding a cross product to two given vectors.
Solution
Substitute the components of the vectors into Equation 2.9:


p × q = 〈 −1, 2, 5 〉 × 〈 4, 0, −3 〉


= 〈 p2q3 − p3q2, p1q3 − p3q1, p1q2 − p2q1 〉


= 〈 2(−3) − 5(0), −(−1)(−3) + 5(4), (−1)(0) − 2(4) 〉


= 〈 −6, 17, −8 〉 .


Find p × q for p = 〈 5, 1, 2 〉 and q = 〈 −2, 0, 1 〉 . Express the answer using standard unit
vectors.


Although it may not be obvious from Equation 2.9, the direction of u × v is given by the right-hand rule. If we hold the
right hand out with the fingers pointing in the direction of u, then curl the fingers toward vector v, the thumb points in


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the direction of the cross product, as shown.


Figure 2.54 The direction of u × v is determined by the right-hand rule.


Notice what this means for the direction of v × u. If we apply the right-hand rule to v × u, we start with our fingers
pointed in the direction of v, then curl our fingers toward the vector u. In this case, the thumb points in the opposite
direction of u × v. (Try it!)
Example 2.32
Anticommutativity of the Cross Product
Let u = 〈 0, 2, 1 〉 and v = 〈 3, −1, 0 〉 . Calculate u × v and v × u and graph them.


Figure 2.55 Are the cross products u × v and v × u in the same
direction?


Chapter 2 | Vectors in Space 165




2.31


Solution
We have


u × v = 〈 (0 + 1), −(0 − 3), (0 − 6) 〉 = 〈 1, 3, −6 〉


v × u = 〈 (−1 − 0), −(3 − 0), (6 − 0) 〉 = 〈 −1, −3, 6 〉 .


We see that, in this case, u × v = −(v × u) (Figure 2.56). We prove this in general later in this section.


Figure 2.56 The cross products u × v and v × u are both
orthogonal to u and v, but in opposite directions.


Suppose vectors u and v lie in the xy-plane (the z-component of each vector is zero). Now suppose the
x- and y-components of u and the y-component of v are all positive, whereas the x-component of v is
negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does u × v
point?


The cross products of the standard unit vectors i, j, and k can be useful for simplifying some calculations, so let’s
consider these cross products. A straightforward application of the definition shows that


i × i = j × j = k × k = 0.


(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar 0.) It’s up
to you to verify the calculations on your own.
Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product


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2.32


of i and j is parallel to k. Similarly, the vector product of i and k is parallel to j, and the vector product of j and k
is parallel to i. We can use the right-hand rule to determine the direction of each product. Then we have


i × j = k j × i = −k
j × k = i k × j = −i
k × i = j i × k = −j.


These formulas come in handy later.
Example 2.33
Cross Product of Standard Unit Vectors
Find i × ⎛⎝j × k⎞⎠.
Solution
We know that j × k = i. Therefore, i × ⎛⎝j × k⎞⎠ = i × i = 0.


Find ⎛⎝i × j⎞⎠× (k × i).


As we have seen, the dot product is often called the scalar product because it results in a scalar. The cross product results ina vector, so it is sometimes called the vector product. These operations are both versions of vector multiplication, but theyhave very different properties and applications. Let’s explore some properties of the cross product. We prove only a few ofthem. Proofs of the other properties are left as exercises.
Theorem 2.6: Properties of the Cross Product
Let u, v, and w be vectors in space, and let c be a scalar.


i. u × v = −(v × u) Anticommutative property
ii. u × (v + w) = u × v + u × w Distributive property
iii. c(u × v) = (cu) × v = u × (cv) Multiplication by a constant
iv. u × 0 = 0 × u = 0 Cross product of the zero vector
v. v × v = 0 Cross product of a vector with itself
vi. u · (v × w) = (u × v) ·w Scalar triple product


Proof
For property i., we want to show u × v = −(v × u). We have


u × v = 〈 u1, u2, u3 〉 × 〈 v1, v2, v3 〉


= 〈 u2 v3 − u3 v2, −u1 v3 + u3 v1, u1 v2 − u2 v1 〉


= − 〈 u3 v2 − u2 v3, −u3 v1 + u1 v3, u2 v1 − u1 v2 〉


= − 〈 v1, v2, v3 〉 × 〈 u1, u2, u3 〉


= −(v × u).


Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-handrule.
For property iv., this follows directly from the definition of the cross product. We have


Chapter 2 | Vectors in Space 167




2.33


u × 0 = 〈 u2 (0) − u3 (0), −

⎝u2 (0) − u3 (0)



⎠, u1 (0) − u2 (0) 〉


= 〈 0, 0, 0 〉 = 0.


Then, by property i., 0 × u = 0 as well. Remember that the dot product of a vector and the zero vector is the scalar 0,
whereas the cross product of a vector with the zero vector is the vector 0.
Property vi. looks like the associative property, but note the change in operations:


u · (v × w) = u · 〈 v2w3 − v3w2, −v1w3 + v3w1, v1w2 − v2w1 〉


= u1 (v2w3 − v3w2) + u2 (−v1w3 + v3w1) + u3 (v1w2 − v2w1)
= u1 v2w3 − u1 v3w2 − u2 v1w3 + u2 v3w1 + u3 v1w2 − u3 v2w1
= (u2 v3 − u3 v2)w1 + (u3 v1 − u1 v3)w2 + (u1 v2 − u2 v1)w3
= 〈 u2 v3 − u3 v2, u3 v1 − u1 v3, u1 v2 − u2 v1 〉 · 〈 w1, w2, w3 〉


= (u × v) ·w.



Example 2.34
Using the Properties of the Cross Product
Use the cross product properties to calculate ⎛⎝2i × 3j⎞⎠× j.
Solution



⎝2i × 3j⎞⎠× j = 2⎛⎝i × 3j⎞⎠× j


= 2(3)⎛⎝i × j⎞⎠× j
= (6k) × j
= 6⎛⎝k × j⎞⎠
= 6(−i) = −6i.


Use the properties of the cross product to calculate (i × k) × ⎛⎝k × j⎞⎠.


So far in this section, we have been concerned with the direction of the vector u × v, but we have not discussed its
magnitude. It turns out there is a simple expression for the magnitude of u × v involving the magnitudes of u and v,
and the sine of the angle between them.
Theorem 2.7: Magnitude of the Cross Product
Let u and v be vectors, and let θ be the angle between them. Then, ‖ u × v ‖ = ‖ u ‖ · ‖ v ‖ · sin θ.


Proof
Let u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 be vectors, and let θ denote the angle between them. Then


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2.34


‖ u × v ‖ 2 = (u2 v3 − u3 v2)
2 + (u3 v1 − u1 v3)


2 + (u1 v2 − u2 v1)
2


= u2
2 v3


2 − 2u2u3 v2 v3 + u3
2 v2


2 + u3
2 v1


2 − 2u1u3 v1 v3 + u1
2 v3


2 + u1
2 v2


2 − 2u1u2 v1 v2 + u2
2 v1


2


= u1
2 v1


2 + u1
2 v2


2 + u1
2 v3


2 + u2
2 v1


2 + u2
2 v2


2 + u2
2 v3


2 + u3
2 v1


2 + u3
2 v2


2 + u3
2 v3


2


−⎛⎝u1
2 v1


2 + u2
2 v2


2 + u3
2 v3


2 + 2u1u2 v1 v2 + 2u1u3 v1 v3 + 2u2u3 v2 v3



= ⎛⎝u1
2 + u2


2 + u3
2⎞


⎝v1


2 + v2
2 + v3


2⎞
⎠− (u1 v1 + u2 v2 + u3 v3)


2


= ‖ u ‖ 2 ‖ v ‖ 2 − (u · v)2


= ‖ u ‖ 2 ‖ v ‖ 2 − ‖ u ‖ 2 ‖ v ‖ 2 cos2 θ


= ‖ u ‖ 2 ‖ v ‖ 2 ⎛⎝1 − cos
2 θ⎞⎠


= ‖ u ‖ 2 ‖ v ‖ 2 ⎛⎝sin
2 θ⎞⎠.


Taking square roots and noting that sin2 θ = sin θ for 0 ≤ θ ≤ 180°, we have the desired result:
‖ u × v ‖ = ‖ u ‖ ‖ v ‖ sin θ.



This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, thatthe cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it isimpossible to generate a vector simultaneously orthogonal to two nonparallel vectors.
Example 2.35
Calculating the Cross Product
Use Properties of the Cross Product to find the magnitude of the cross product of u = 〈 0, 4, 0 〉 and
v = 〈 0, 0, −3 〉 .


Solution
We have


‖ u × v ‖ = ‖ u ‖ · ‖ v ‖ · sin θ


= 02 + 42 + 02 · 02 + 02 + (−3)2 · sin π
2


= 4(3)(1) = 12.


Use Properties of the Cross Product to find the magnitude of u × v, where u = 〈 −8, 0, 0 〉
and v = 〈 0, 2, 0 〉 .


Determinants and the Cross Product
Using Equation 2.9 to find the cross product of two vectors is straightforward, and it presents the cross product inthe useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have analternative. We can calculate the cross product of two vectors using determinant notation.
A 2 × 2 determinant is defined by


|a1 a2b1 b2| = a1b2 − b1a2.
For example,


Chapter 2 | Vectors in Space 169




2.35


|3 −25 1| = 3(1) − 5(−2) = 3 + 10 = 13.
A 3 × 3 determinant is defined in terms of 2 × 2 determinants as follows:


(2.10)|
a1 a2 a3
b1 b2 b3
c1 c2 c3| = a1 |b2 b3c2 c3| − a2 |b1 b3c1 c3| + a3 |b1 b2c1 c2|.


Equation 2.10 is referred to as the expansion of the determinant along the first row. Notice that the multipliers of eachof the 2 × 2 determinants on the right side of this expression are the entries in the first row of the 3 × 3 determinant.
Furthermore, each of the 2 × 2 determinants contains the entries from the 3 × 3 determinant that would remain if you
crossed out the row and column containing the multiplier. Thus, for the first term on the right, a1 is the multiplier, and the
2 × 2 determinant contains the entries that remain if you cross out the first row and first column of the 3 × 3 determinant.
Similarly, for the second term, the multiplier is a2, and the 2 × 2 determinant contains the entries that remain if you cross
out the first row and second column of the 3 × 3 determinant. Notice, however, that the coefficient of the second term is
negative. The third term can be calculated in similar fashion.
Example 2.36
Using Expansion Along the First Row to Compute a 3 × 3 Determinant


Evaluate the determinant | 2 5 −1−1 1 3−2 3 4|.
Solution
We have


| 2 5 −1−1 1 3−2 3 4| = 2|1 33 4| − 5|−1 3−2 4| − 1|−1 1−2 3|
= 2(4 − 9) − 5(−4 + 6) − 1(−3 + 2)
= 2(−5) − 5(2) − 1(−1) = −10 − 10 + 1
= −19.


Evaluate the determinant |1 −2 −13 2 −31 5 4|.
Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation providesa useful mnemonic device for the cross product formula.
Rule: Cross Product Calculated by a Determinant
Let u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 be vectors. Then the cross product u × v is given by


u × v = |i j ku1 u2 u3v1 v2 v3| = |u2 u3v2 v3|i − |u1 u3v1 v3|j + |u1 u2v1 v2|k.


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2.36


Example 2.37
Using Determinant Notation to find p × q
Let p = 〈 −1, 2, 5 〉 and q = 〈 4, 0, −3 〉 . Find p × q.
Solution
We set up our determinant by putting the standard unit vectors across the first row, the components of u in the
second row, and the components of v in the third row. Then, we have


p × q = | i j k−1 2 54 0 −3| = |2 50 −3|i − |−1 54 −3|j + |−1 24 0|k
= (−6 − 0)i − (3 − 20)j + (0 − 8)k
= −6i + 17j − 8k.


Notice that this answer confirms the calculation of the cross product in Example 2.31.


Use determinant notation to find a × b, where a = 〈 8, 2, 3 〉 and b = 〈 −1, 0, 4 〉 .


Using the Cross Product
The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors,computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometricshape made of parallelograms known as a parallelepiped. The following examples illustrate these calculations.
Example 2.38
Finding a Unit Vector Orthogonal to Two Given Vectors
Let a = 〈 5, 2, −1 〉 and b = 〈 0, −1, 4 〉 . Find a unit vector orthogonal to both a and b.
Solution
The cross product a × b is orthogonal to both vectors a and b. We can calculate it with a determinant:


a × b = | i j k5 2 −10 −1 4| = | 2 −1−1 4|i − |5 −10 4|j + |5 20 −1|k
= (8 − 1)i − (20 − 0)j + (−5 − 0)k
= 7i − 20j − 5k.


Normalize this vector to find a unit vector in the same direction:
‖ a × b ‖ = (7)2 + (−20)2 + (−5)2 = 474.


Thus, 〈 7
474


, −20
474


, −5
474


〉 is a unit vector orthogonal to a and b.


Chapter 2 | Vectors in Space 171




2.37 Find a unit vector orthogonal to both a and b, where a = 〈 4, 0, 3 〉 and b = 〈 1, 1, 4 〉 .


To use the cross product for calculating areas, we state and prove the following theorem.
Theorem 2.8: Area of a Parallelogram
If we locate vectors u and v such that they form adjacent sides of a parallelogram, then the area of the parallelogram
is given by ‖ u × v ‖ (Figure 2.57).


Figure 2.57 The parallelogram with adjacent sides u and v
has base ‖ u ‖ and height ‖ v ‖ sin θ.


Proof
We show that the magnitude of the cross product is equal to the base times height of the parallelogram.


Area of a parallelogram = base × height


= ‖ u ‖ ⎛⎝ ‖ v ‖ sin θ⎞⎠
= ‖ u × v ‖



Example 2.39
Finding the Area of a Triangle
Let P = (1, 0, 0), Q = (0, 1, 0), and R = (0, 0, 1) be the vertices of a triangle (Figure 2.58). Find its area.


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2.38


Figure 2.58 Finding the area of a triangle by using the crossproduct.
Solution
We have PQ→ = 〈 0 − 1, 1 − 0, 0 − 0 〉 = 〈 −1, 1, 0 〉 and
PR


= 〈 0 − 1, 0 − 0, 1 − 0 〉 = 〈 −1, 0, 1 〉 . The area of the parallelogram with adjacent sides PQ→ and
PR
→ is given by ‖ PQ→ × PR→ ‖ :


PQ


× PR


= | i j k−1 1 0−1 0 1| = (1 − 0)i − (−1 − 0)j + ⎛⎝0 − (−1)⎞⎠k = i + j + k
‖ PQ



× PR



‖ = ‖ 〈 1, 1, 1 〉 ‖ = 12 + 12 + 12 = 3.


The area of ΔPQR is half the area of the parallelogram, or 3/2.


Find the area of the parallelogram PQRS with vertices P(1, 1, 0), Q(7, 1, 0), R(9, 4, 2), and
S(3, 4, 2).


The Triple Scalar Product
Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. Thedot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is ascalar.
Definition
The triple scalar product of vectors u, v, and w is u · (v × w).


Chapter 2 | Vectors in Space 173




2.39


Theorem 2.9: Calculating a Triple Scalar Product
The triple scalar product of vectors u = u1 i + u2 j + u3k, v = v1 i + v2 j + v3k, and w = w1 i + w2 j + w3k
is the determinant of the 3 × 3 matrix formed by the components of the vectors:


u · (v × w) = |
u1 u2 u3
v1 v2 v3
w1 w2 w3|.


Proof
The calculation is straightforward.


u · (v × w) = 〈 u1, u2, u3 〉 · 〈 v2w3 − v3w2, −v1w3 + v3w1, v1w2 − v2w1 〉


= u1 (v2w3 − v3w2) + u2 (−v1w3 + v3w1) + u3 (v1w2 − v2w1)


= u1 (v2w3 − v3w2) − u2 (v1w3 − v3w1) + u3 (v1w2 − v2w1)


= |
u1 u2 u3
v1 v2 v3
w1 w2 w3|



Example 2.40
Calculating the Triple Scalar Product
Let u = 〈 1, 3, 5 〉 , v = 〈 2, −1, 0 〉 andw = 〈 −3, 0, −1 〉 . Calculate the triple scalar product
u · (v × w).


Solution
Apply Calculating a Triple Scalar Product directly:


u · (v × w) = | 1 3 52 −1 0−3 0 −1|
= 1|−1 00 −1| − 3| 2 0−3 −1| + 5| 2 −1−3 0|
= (1 − 0) − 3(−2 − 0) + 5(0 − 3)
= 1 + 6 − 15 = −8.


Calculate the triple scalar product a · (b × c), where a = 〈 2, −4, 1 〉 , b = 〈 0, 3, −1 〉 , and
c = 〈 5, −3, 3 〉 .


When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list themin a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However,each time two rows switch places, the determinant changes sign:


|
a1 a2 a3
b1 b2 b3
c1 c2 c3| = d |b1 b2 b3a1 a2 a3c1 c2 c3| = −d |b1 b2 b3c1 c2 c3a1 a2 a3| = d |


c1 c2 c3
b1 b2 b3
a1 a2 a3| = −d.


Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:


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| 1 2 1−2 0 34 1 −1| = |0 31 −1| − 2|−2 34 −1| + |−2 04 1|
= (0 − 3) − 2(2 − 12) + (−2 − 0) = −3 + 20 − 2 = 15.


Switching the top two rows we have


|−2 0 31 2 14 1 −1| = −2|2 11 −1| + 3|1 24 1| = −2(−2 − 1) + 3(1 − 8) = 6 − 21 = −15.
Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let
u = u1 i + u2 j + u3k, v = v1 i + v2 j + v3k, and w = w1 i + w2 j + w3k. Applying Calculating a Triple Scalar
Product, we have


u · (v × w) = |
u1 u2 u3
v1 v2 v3
w1 w2 w3| and u · (w × v) = |


u1 u2 u3
w1 w2 w3
v1 v2 v3 |.


We can obtain the determinant for calculating u · (w × v) by switching the bottom two rows of u · (v × w). Therefore,
u · (v × w) = −u · (w × v).


Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead tothe following identities:
u · (v × w) = −u · (w × v)


u · (v × w) = v · (w × u) = w · (u × v).


Let u and v be two vectors in standard position. If u and v are not scalar multiples of each other, then these vectors form
adjacent sides of a parallelogram. We saw in Area of a Parallelogram that the area of this parallelogram is ‖ u × v ‖ .
Now suppose we add a third vector w that does not lie in the same plane as u and v but still shares the same initial
point. Then these vectors form three edges of a parallelepiped, a three-dimensional prism with six faces that are eachparallelograms, as shown in Figure 2.59. The volume of this prism is the product of the figure’s height and the area of itsbase. The triple scalar product of u, v, and w provides a simple method for calculating the volume of the parallelepiped
defined by these vectors.
Theorem 2.10: Volume of a Parallelepiped
The volume of a parallelepiped with adjacent edges given by the vectors u, v, andw is the absolute value of the
triple scalar product:


V = |u · (v × w)|.


See Figure 2.59.


Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses theabsolute value of a scalar quantity.


Chapter 2 | Vectors in Space 175




Figure 2.59 The height of the parallelepiped is given by
‖ projv×wu ‖ .


Proof
The area of the base of the parallelepiped is given by ‖ v × w ‖ . The height of the figure is given by ‖ projv×wu ‖ .
The volume of the parallelepiped is the product of the height and the area of the base, so we have


V = ‖ projv × wu ‖ ‖ v × w ‖


= | u · (v × w)‖ v × w ‖ | ‖ v × w ‖
= |u · (v × w)|.



Example 2.41
Calculating the Volume of a Parallelepiped
Let u = 〈 −1, −2, 1 〉 , v = 〈 4, 3, 2 〉 , andw = 〈 0, −5, −2 〉 . Find the volume of the parallelepiped
with adjacent edges u, v, andw (Figure 2.60).


Figure 2.60
Solution


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2.40


2.41


We have
u · (v × w) = |−1 −2 14 3 20 −5 −2| = (−1)| 3 2−5 −2| + 2|4 20 −2| + |4 30 −5|


= (−1)(−6 + 10) + 2(−8 − 0) + (−20 − 0)
= −4 − 16 − 20
= −40.


Thus, the volume of the parallelepiped is |−40| = 40 units3.


Find the volume of the parallelepiped formed by the vectors a = 3i + 4j − k, b = 2i − j − k, and
c = 3j + k.


Applications of the Cross Product
The cross product appears in many practical applications in mathematics, physics, and engineering. Let’s examine someof these applications here, including the idea of torque, with which we began this section. Other applications show up inlater chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (Introduction toVector Calculus).
Example 2.42
Using the Triple Scalar Product
Use the triple scalar product to show that vectors u = 〈 2, 0, 5 〉 , v = 〈 2, 2, 4 〉 , andw = 〈 1, −1, 3 〉
are coplanar—that is, show that these vectors lie in the same plane.
Solution
Start by calculating the triple scalar product to find the volume of the parallelepiped defined by u, v, andw:


u · (v × w) = |2 0 52 2 41 −1 3|
= ⎡⎣2(2)(3) + (0)(4)(1) + 5(2)(−1)⎤⎦− ⎡⎣5(2)(1) + (2)(4)(−1) + (0)(2)(3)⎤⎦
= 2 − 2
= 0.


The volume of the parallelepiped is 0 units3, so one of the dimensions must be zero. Therefore, the three vectors
all lie in the same plane.


Are the vectors a = i + j − k, b = i − j + k, and c = i + j + k coplanar?


Chapter 2 | Vectors in Space 177




Example 2.43
Finding an Orthogonal Vector
Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the planecontaining points P = (9, −3, −2), Q = (1, 3, 0), and R = (−2, 5, 0).
Solution
The plane must contain vectors PQ→ and QR→ :


PQ


= 〈 1 − 9, 3 − (−3), 0 − (−2) 〉 = 〈 −8, 6, 2 〉


QR


= 〈 −2 − 1, 5 − 3, 0 − 0 〉 = 〈 −3, 2, 0 〉 .


The cross product PQ→ × QR→ produces a vector orthogonal to both PQ→ and QR→ . Therefore, the cross product
is orthogonal to the plane that contains these two vectors:


PQ


× QR


= | i j k−8 6 2−3 2 0|
= 0i − 6j − 16k − ⎛⎝−18k + 4i + 0j⎞⎠
= −4i − 6j + 2k.


We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the crossproduct to real-world situations.
Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotationaleffect, called torque.
Definition
Torque, τ (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let r
be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where theforce is applied, and let vector F represent the force. Then torque is equal to the cross product of r and F:


τ = r × F.


See Figure 2.61.


Figure 2.61 Torque measures how a force causes an object torotate.


Think about using a wrench to tighten a bolt. The torque τ applied to the bolt depends on how hard we push the wrench
(force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrenchat a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is


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2.42


dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in thecontext of rotation, whereas work typically involves motion along a line.
Example 2.44
Evaluating Torque
A bolt is tightened by applying a force of 6 N to a 0.15-m wrench (Figure 2.62). The angle between the wrench
and the force vector is 40°. Find the magnitude of the torque about the center of the bolt. Round the answer to
two decimal places.


Figure 2.62 Torque describes the twisting action of thewrench.
Solution
Substitute the given information into the equation defining torque:


‖ τ ‖ = ‖ r × F ‖ = ‖ r ‖ ‖ F ‖ sin θ = (0.15 m)(6 N)sin 40° ≈ 0.58 N ·m.


Calculate the force required to produce 15 N ·m torque at an angle of 30 º from a 150-cm rod.


Chapter 2 | Vectors in Space 179




2.4 EXERCISES
For the following exercises, the vectors u and v are
given.


a. Find the cross product u × v of the vectors u and
v. Express the answer in component form.


b. Sketch the vectors u, v, and u × v.
183. u = 〈 2, 0, 0 〉 , v = 〈 2, 2, 0 〉
184. u = 〈 3, 2, −1 〉 , v = 〈 1, 1, 0 〉
185. u = 2i + 3j, v = j + 2k
186. u = 2j + 3k, v = 3i + k
187. Simplify ⎛⎝i × i − 2i × j − 4i × k + 3j × k⎞⎠× i.
188. Simplify j × ⎛⎝k × j + 2j × i − 3j × j + 5i × k⎞⎠.
In the following exercises, vectors u and v are given.
Find unit vector w in the direction of the cross product
vector u × v. Express your answer using standard unit
vectors.
189. u = 〈 3, −1, 2 〉 , v = 〈 −2, 0, 1 〉
190. u = 〈 2, 6, 1 〉 , v = 〈 3, 0, 1 〉
191. u = AB→ , v = AC→ , where A(1, 0, 1),
B(1, −1, 3), and C(0, 0, 5)
192. u = OP→ , v = PQ→ , where P(−1, 1, 0) and
Q(0, 2, 1)


193. Determine the real number α such that u × v and
i are orthogonal, where u = 3i + j − 5k and
v = 4i − 2j + αk.


194. Show that u × v and 2i − 14j + 2k cannot be
orthogonal for any α real number, where u = i + 7j − k
and v = αi + 5j + k.
195. Show that u × v is orthogonal to u + v and
u − v, where u and v are nonzero vectors.
196. Show that v × u is orthogonal to
(u · v)(u + v) + u, where u and v are nonzero vectors.


197. Calculate the determinant |i j k1 −1 72 0 3|.
198. Calculate the determinant |i j k0 3 −41 6 −1|.
For the following exercises, the vectors u and v are
given. Use determinant notation to find vector w
orthogonal to vectors u and v.
199. u = 〈 −1, 0, et 〉 , v = 〈 1, e−t, 0 〉 , where t
is a real number
200. u = 〈 1, 0, x 〉 , v = 〈 2x , 1, 0 〉 , where x is a
nonzero real number


201. Find vector (a − 2b) × c, where a = |i j k2 −1 50 1 8|,
b = |i j k0 1 12 −1 −2|, and c = i + j + k.
202. Find vector c × (a + 3b), where a = |i j k5 0 90 1 0|,
b = |i j k0 −1 17 1 −1|, and c = i − k.
203. [T] Use the cross product u × v to find the acute
angle between vectors u and v, where u = i + 2j and
v = i + k. Express the answer in degrees rounded to the
nearest integer.
204. [T] Use the cross product u × v to find the obtuse
angle between vectors u and v, where u = −i + 3j + k
and v = i − 2j. Express the answer in degrees rounded to
the nearest integer.
205. Use the sine and cosine of the angle between twononzero vectors u and v to prove Lagrange’s identity:
‖ u × v ‖ 2 = ‖ u ‖ 2 ‖ v ‖ 2 − (u · v)2.


206. Verify Lagrange’s identity
‖ u × v ‖ 2 = ‖ u ‖ 2 ‖ v ‖ 2 − (u · v)2 for vectors
u = −i + j − 2k and v = 2i − j.


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207. Nonzero vectors u and v are called collinear if
there exists a nonzero scalar α such that v = αu. Show
that u and v are collinear if and only if u × v = 0.
208. Nonzero vectors u and v are called collinear if
there exists a nonzero scalar α such that v = αu. Show
that vectors AB→ and AC→ are collinear, where
A(4, 1, 0), B(6, 5, −2), and C(5, 3, −1).
209. Find the area of the parallelogram with adjacent sides
u = 〈 3, 2, 0 〉 and v = 〈 0, 2, 1 〉 .
210. Find the area of the parallelogram with adjacent sides
u = i + j and v = i + k.
211. Consider points A(3, −1, 2), B(2, 1, 5), and
C(1, −2, −2).


a. Find the area of parallelogram ABCD with
adjacent sides AB→ and AC→ .


b. Find the area of triangle ABC.
c. Find the distance from point A to line BC.


212. Consider points A(2, −3, 4), B(0, 1, 2), and
C(−1, 2, 0).


a. Find the area of parallelogram ABCD with
adjacent sides AB→ and AC→ .


b. Find the area of triangle ABC.
c. Find the distance from point B to line AC.


In the following exercises, vectors u, v, andw are given.
a. Find the triple scalar product u · (v × w).
b. Find the volume of the parallelepiped with theadjacent edges u, v, andw.


213. u = i + j, v = j + k, and w = i + k
214. u = 〈 −3, 5, −1 〉 , v = 〈 0, 2, −2 〉 , and
w = 〈 3, 1, 1 〉


215. Calculate the triple scalar products v · (u × w) and
w · (u × v), where u = 〈 1, 1, 1 〉 ,
v = 〈 7, 6, 9 〉 , and w = 〈 4, 2, 7 〉 .
216. Calculate the triple scalar products w · (v × u) and
u · (w × v), where u = 〈 4, 2, −1 〉 ,
v = 〈 2, 5, −3 〉 , and w = 〈 9, 5, −10 〉 .


217. Find vectors a, b, and c with a triple scalar product
given by the determinant |1 2 30 2 58 9 2|. Determine their triple
scalar product.
218. The triple scalar product of vectors a, b, and c
is given by the determinant |0 −2 10 1 41 −3 7|. Find vector
a − b + c.


219. Consider the parallelepiped with edges OA, OB,
and OC, where A(2, 1, 0), B(1, 2, 0), and
C(0, 1, α).


a. Find the real number α > 0 such that the volume
of the parallelepiped is 3 units3.


b. For α = 1, find the height h from vertex C of
the parallelepiped. Sketch the parallelepiped.


220. Consider points A(α, 0, 0), B⎛⎝0, β, 0⎞⎠, and
C⎛⎝0, 0, γ⎞⎠, with α, β, and γ positive real numbers.


a. Determine the volume of the parallelepiped with
adjacent sides OA→ , OB→ , and OC→ .


b. Find the volume of the tetrahedron with vertices
O, A, B, andC. (Hint: The volume of the
tetrahedron is 1/6 of the volume of the
parallelepiped.)c. Find the distance from the origin to the planedetermined by A, B, andC. Sketch the
parallelepiped and tetrahedron.


221. Let u, v, andw be three-dimensional vectors and
c be a real number. Prove the following properties of the
cross product.a. u × u = 0


b. u × (v + w) = (u × v) + (u × w)
c. c(u × v) = (cu) × v = u × (cv)
d. u · (u × v) = 0


222. Show that vectors u = 〈 1, 0, −8 〉 ,
v = 〈 0, 1, 6 〉 , and w = 〈 −1, 9, 3 〉 satisfy the
following properties of the cross product.a. u × u = 0


b. u × (v + w) = (u × v) + (u × w)
c. c(u × v) = (cu) × v = u × (cv)
d. u · (u × v) = 0


Chapter 2 | Vectors in Space 181




223. Nonzero vectors u, v, andw are said to be linearly
dependent if one of the vectors is a linear combination ofthe other two. For instance, there exist two nonzero realnumbers α and β such that w = αu + βv. Otherwise,
the vectors are called linearly independent. Show that
u, v, andw are coplanar if and only if they are linear
dependent.
224. Consider vectors u = 〈 1, 4, −7 〉 ,
v = 〈 2, −1, 4 〉 , w = 〈 0, −9, 18 〉 , and
p = 〈 0, −9, 17 〉 .


a. Show that u, v, andw are coplanar by using their
triple scalar productb. Show that u, v, andw are coplanar, using the
definition that there exist two nonzero real numbers
α and β such that w = αu + βv.


c. Show that u, v, and p are linearly
independent—that is, none of the vectors is a linearcombination of the other two.


225. Consider points A(0, 0, 2), B(1, 0, 2),
C(1, 1, 2), and D(0, 1, 2). Are vectors AB→ , AC→ ,
and AD→ linearly dependent (that is, one of the vectors is a
linear combination of the other two)?
226. Show that vectors i + j, i − j, and i + j + k
are linearly independent—that is, there exist two nonzeroreal numbers α and β such that
i + j + k = α⎛⎝i + j⎞⎠+ β⎛⎝i − j⎞⎠.


227. Let u = 〈 u1, u2 〉 and v = 〈 v1, v2 〉 be two-
dimensional vectors. The cross product of vectors u and
v is not defined. However, if the vectors are regarded
as the three-dimensional vectors ũ = 〈 u1, u2, 0 〉 and
ṽ = 〈 v1, v2, 0 〉 , respectively, then, in this case, we
can define the cross product of ũ and ṽ. In particular,
in determinant notation, the cross product of ũ and ṽ is
given by


ũ × ṽ = |i j ku1 u2 0v1 v2 0|.
Use this result to compute
(icos θ + jsin θ) × (isinθ − jcosθ), where θ is a real
number.
228. Consider points P(2, 1), Q(4, 2), and R(1, 2).


a. Find the area of triangle P, Q, and R.
b. Determine the distance from point R to the line


passing through P andQ.


229. Determine a vector of magnitude 10 perpendicular
to the plane passing through the x-axis and point
P(1, 2, 4).


230. Determine a unit vector perpendicular to the planepassing through the z-axis and point A(3, 1, −2).
231. Consider u and v two three-dimensional vectors.
If the magnitude of the cross product vector u × v is k
times larger than the magnitude of vector u, show that the
magnitude of v is greater than or equal to k, where k is
a natural number.
232. [T] Assume that the magnitudes of two nonzerovectors u and v are known. The function
f (θ) = ‖ u ‖ ‖ v ‖ sin θ defines the magnitude of the
cross product vector u × v, where θ ∈ [0, π] is the
angle between u and v.


a. Graph the function f .
b. Find the absolute minimum and maximum offunction f . Interpret the results.
c. If ‖ u ‖ = 5 and ‖ v ‖ = 2, find the angle


between u and v if the magnitude of their cross
product vector is equal to 9.


233. Find all vectors w = 〈 w1, w2, w3 〉 that satisfy
the equation 〈 1, 1, 1 〉 × w = 〈 −1, −1, 2 〉 .
234. Solve the equation
w × 〈 1, 0, −1 〉 = 〈 3, 0, 3 〉 , where
w = 〈 w1, w2, w3 〉 is a nonzero vector with a
magnitude of 3.
235. [T] A mechanic uses a 12-in. wrench to turn a bolt.The wrench makes a 30° angle with the horizontal. If the
mechanic applies a vertical force of 10 lb on the wrench
handle, what is the magnitude of the torque at point P (see
the following figure)? Express the answer in foot-poundsrounded to two decimal places.


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236. [T] A boy applies the brakes on a bicycle by applyinga downward force of 20 lb on the pedal when the 6-in.
crank makes a 40° angle with the horizontal (see the
following figure). Find the torque at point P. Express your
answer in foot-pounds rounded to two decimal places.


237. [T] Find the magnitude of the force that needs tobe applied to the end of a 20-cm wrench located on thepositive direction of the y-axis if the force is applied in thedirection 〈 0, 1, −2 〉 and it produces a 100 N·m torque
to the bolt located at the origin.
238. [T] What is the magnitude of the force required to beapplied to the end of a 1-ft wrench at an angle of 35° to
produce a torque of 20 N·m?
239. [T] The force vector F acting on a proton with an
electric charge of 1.6 × 10−19C (in coulombs) moving
in a magnetic field B where the velocity vector v is
given by F = 1.6 × 10−19 (v × B) (here, v is expressed
in meters per second, B is in tesla [T], and F is in newtons
[N]). Find the force that acts on a proton that moves in
the xy-plane at velocity v = 105 i + 105 j (in meters per
second) in a magnetic field given by B = 0.3j.
240. [T] The force vector F acting on a proton with an
electric charge of 1.6 × 10−19C moving in a magnetic
field B where the velocity vector v is given by
F = 1.6 × 10−19 (v × B) (here, v is expressed in meters
per second, B in T, and F in N). If the magnitude
of force F acting on a proton is 5.9 × 10−17 N and the
proton is moving at the speed of 300 m/sec in magneticfield B of magnitude 2.4 T, find the angle between velocity
vector v of the proton and magnetic field B. Express the
answer in degrees rounded to the nearest integer.


241. [T] Consider r(t) = 〈 cos t, sin t, 2t 〉 the position
vector of a particle at time t ∈ [0, 30], where the
components of r are expressed in centimeters and time in
seconds. Let OP→ be the position vector of the particle after
1 sec.


a. Determine unit vector B(t) (called the binormal
unit vector) that has the direction of cross productvector v(t) × a(t), where v(t) and a(t) are the
instantaneous velocity vector and, respectively, theacceleration vector of the particle after t seconds.


b. Use a CAS to visualize vectors v(1), a(1), and
B(1) as vectors starting at point P along with the
path of the particle.


242. A solar panel is mounted on the roof of a house.The panel may be regarded as positioned at the pointsof coordinates (in meters) A(8, 0, 0), B(8, 18, 0),
C(0, 18, 8), and D(0, 0, 8) (see the following figure).


a. Find vector n = AB→ × AD→ perpendicular to the
surface of the solar panels. Express the answerusing standard unit vectors.


b. Assume unit vector s = 1
3
i + 1


3
j + 1


3
k points


toward the Sun at a particular time of the day andthe flow of solar energy is F = 900s (in watts per
square meter [W/m2 ]). Find the predicted amount
of electrical power the panel can produce, whichis given by the dot product of vectors F and n
(expressed in watts).c. Determine the angle of elevation of the Sun abovethe solar panel. Express the answer in degreesrounded to the nearest whole number. (Hint: Theangle between vectors n and s and the angle of
elevation are complementary.)


Chapter 2 | Vectors in Space 183




2.5 | Equations of Lines and Planes in Space
Learning Objectives


2.5.1 Write the vector and parametric equations, and the general form, of a line through a givenpoint in a given direction, and a line through two given points.
2.5.2 Find the distance from a point to a given line.
2.5.3 Write the vector and scalar equations of a plane through a given point with a given normal.
2.5.4 Find the distance from a point to a given plane.
2.5.5 Find the angle between two planes.


By now, we are familiar with writing equations that describe a line in two dimensions. To write an equation for a line,we must know two points on the line, or we must know the direction of the line and at least one point through which theline passes. In two dimensions, we use the concept of slope to describe the orientation, or direction, of a line. In threedimensions, we describe the direction of a line using a vector parallel to the line. In this section, we examine how to useequations to describe lines and planes in space.
Equations for a Line in Space
Let’s first explore what it means for two vectors to be parallel. Recall that parallel vectors must have the same or oppositedirections. If two nonzero vectors, u and v, are parallel, we claim there must be a scalar, k, such that u = kv. If u
and v have the same direction, simply choose k = ‖ u ‖


‖ v ‖
. If u and v have opposite directions, choose k = − ‖ u ‖


‖ v ‖
.


Note that the converse holds as well. If u = kv for some scalar k, then either u and v have the same direction (k > 0)
or opposite directions (k < 0), so u and v are parallel. Therefore, two nonzero vectors u and v are parallel if and only
if u = kv for some scalar k. By convention, the zero vector 0 is considered to be parallel to all vectors.
As in two dimensions, we can describe a line in space using a point on the line and the direction of the line, or a parallelvector, which we call the direction vector (Figure 2.63). Let L be a line in space passing through point P(x0, y0, z0).
Let v = 〈 a, b, c 〉 be a vector parallel to L. Then, for any point on line Q(x, y, z), we know that PQ→ is parallel to
v. Thus, as we just discussed, there is a scalar, t, such that PQ→ = tv, which gives


(2.11)PQ→ = tv
〈 x − x0, y − y0, z − z0 〉 = t 〈 a, b, c 〉


〈 x − x0, y − y0, z − z0 〉 = 〈 ta, tb, tc 〉 .


Figure 2.63 Vector v is the direction vector for PQ→ .


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Using vector operations, we can rewrite Equation 2.11 as
〈 x − x0, y − y0, z − z0 〉 = 〈 ta, tb, tc 〉


〈 x, y, z 〉 − 〈 x0, y0, z0 〉 = t 〈 a, b, c 〉


〈 x, y, z 〉 = 〈 x0, y0, z0 〉 + t 〈 a, b, c 〉 .


Setting r = 〈 x, y, z 〉 and r0 = 〈 x0, y0, z0 〉 , we now have the vector equation of a line:
(2.12)r = r0 + tv.


Equating components, Equation 2.11 shows that the following equations are simultaneously true: x − x0 = ta,
y − y0 = tb, and z − z0 = tc. If we solve each of these equations for the component variables x, y, and z, we get a set
of equations in which each variable is defined in terms of the parameter t and that, together, describe the line. This set ofthree equations forms a set of parametric equations of a line:


x = x0 + ta y = y0 + tb z = z0 + tc.


If we solve each of the equations for t assuming a, b, and c are nonzero, we get a different description of the same line:
x − x0
a = t


y − y0
b


= t
z − z0
c = t.


Because each expression equals t, they all have the same value. We can set them equal to each other to create symmetricequations of a line:
x − x0
a =


y − y0
b


=
z − z0
c .


We summarize the results in the following theorem.
Theorem 2.11: Parametric and Symmetric Equations of a Line
A line L parallel to vector v = 〈 a, b, c 〉 and passing through point P(x0, y0, z0) can be described by the
following parametric equations:


(2.13)x = x0 + ta, y = y0 + tb, and z = z0 + tc.
If the constants a, b, and c are all nonzero, then L can be described by the symmetric equation of the line:


(2.14)x − x0
a =


y − y0
b


=
z − z0
c .


The parametric equations of a line are not unique. Using a different parallel vector or a different point on the line leads toa different, equivalent representation. Each set of parametric equations leads to a related set of symmetric equations, so itfollows that a symmetric equation of a line is not unique either.
Example 2.45
Equations of a Line in Space
Find parametric and symmetric equations of the line passing through points (1, 4, −2) and (−3, 5, 0).
Solution
First, identify a vector parallel to the line:


Chapter 2 | Vectors in Space 185




2.43


v = 〈 −3 − 1, 5 − 4, 0 − (−2) 〉 = 〈 −4, 1, 2 〉 .


Use either of the given points on the line to complete the parametric equations:
x = 1 − 4t, y = 4 + t, and z = −2 + 2t.


Solve each equation for t to create the symmetric equation of the line:
x − 1
−4


= y − 4 = z + 2
2


.


Find parametric and symmetric equations of the line passing through points (1, −3, 2) and (5, −2, 8).


Sometimes we don’t want the equation of a whole line, just a line segment. In this case, we limit the values of ourparameter t. For example, let P(x0, y0, z0) and Q(x1, y1, z1) be points on a line, and let p = 〈 x0, y0, z0 〉 and
q = 〈 x1, y1, z1 〉 be the associated position vectors. In addition, let r = 〈 x, y, z 〉 . We want to find a vector
equation for the line segment between P and Q. Using P as our known point on the line, and
PQ


= 〈 x1 − x0, y1 − y0, z1 − z0 〉 as the direction vector equation, Equation 2.12 gives
r = p + t⎛⎝PQ


→ ⎞
⎠.


Using properties of vectors, then
r = p + t⎛⎝PQ


→ ⎞


= 〈 x0, y0, z0 〉 + t 〈 x1 − x0, y1 − y0, z1 − z0 〉


= 〈 x0, y0, z0 〉 + t

⎝ 〈 x1, y1, z1 〉 − 〈 x0, y0, z0 〉





= 〈 x0, y0, z0 〉 + t 〈 x1, y1, z1 〉 − t 〈 x0, y0, z0 〉


= (1 − t) 〈 x0, y0, z0 〉 + t 〈 x1, y1, z1 〉


= (1 − t)p + tq.


Thus, the vector equation of the line passing through P and Q is
r = (1 − t)p + tq.


Remember that we didn’t want the equation of the whole line, just the line segment between P and Q. Notice that when
t = 0, we have r = p, and when t = 1, we have r = q. Therefore, the vector equation of the line segment between
P and Q is


(2.15)r = (1 − t)p + tq, 0 ≤ t ≤ 1.
Going back to Equation 2.12, we can also find parametric equations for this line segment. We have


r = p + t⎛⎝PQ
→ ⎞


〈 x, y, z 〉 = 〈 x0, y0, z0 〉 + t 〈 x1 − x0, y1 − y0, z1 − z0 〉


= 〈 x0 + t(x1 − x0), y0 + t(y1 − y0), z0 + t(z1 − z0) 〉 .


Then, the parametric equations are
(2.16)x = x0 + t(x1 − x0), y = y0 + t(y1 − y0), z = z0 + t(z1 − z0), 0 ≤ t ≤ 1.


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2.44


Example 2.46
Parametric Equations of a Line Segment
Find parametric equations of the line segment between the points P(2, 1, 4) and Q(3, −1, 3).
Solution
By Equation 2.16, we have


x = x0 + t(x1 − x0), y = y0 + t(y1 − y0), z = z0 + t(z1 − z0), 0 ≤ t ≤ 1.


Working with each component separately, we get
x = x0 + t(x1 − x0)


= 2 + t(3 − 2)
= 2 + t,


y = y0 + t(y1 − y0)


= 1 + t(−1 − 1)
= 1 − 2t,


and
z = z0 + t(z1 − z0)


= 4 + t(3 − 4)
= 4 − t.


Therefore, the parametric equations for the line segment are
x = 2 + t, y = 1 − 2t, z = 4 − t, 0 ≤ t ≤ 1.


Find parametric equations of the line segment between points P(−1, 3, 6) and Q(−8, 2, 4).


Distance between a Point and a Line
We already know how to calculate the distance between two points in space. We now expand this definition to describe thedistance between a point and a line in space. Several real-world contexts exist when it is important to be able to calculatethese distances. When building a home, for example, builders must consider “setback” requirements, when structures orfixtures have to be a certain distance from the property line. Air travel offers another example. Airlines are concerned aboutthe distances between populated areas and proposed flight paths.
Let L be a line in the plane and let M be any point not on the line. Then, we define distance d from M to L as the
length of line segment MP— , where P is a point on L such that MP— is perpendicular to L (Figure 2.64).


Chapter 2 | Vectors in Space 187




Figure 2.64 The distance from point M to line L is the
length of MP— .


When we’re looking for the distance between a line and a point in space, Figure 2.64 still applies. We still define thedistance as the length of the perpendicular line segment connecting the point to the line. In space, however, there is no clearway to know which point on the line creates such a perpendicular line segment, so we select an arbitrary point on the lineand use properties of vectors to calculate the distance. Therefore, let P be an arbitrary point on line L and let v be a
direction vector for L (Figure 2.65).


Figure 2.65 Vectors PM→ and v form two sides of a
parallelogram with base ‖ v ‖ and height d, which is the
distance between a line and a point in space.


By Area of a Parallelogram, vectors PM→ and v form two sides of a parallelogram with area ‖ PM→ × v ‖ . Using a
formula from geometry, the area of this parallelogram can also be calculated as the product of its base and height:


‖ PM


× v ‖ = ‖ v ‖ d.


We can use this formula to find a general formula for the distance between a line in space and any point not on the line.
Theorem 2.12: Distance from a Point to a Line
Let L be a line in space passing through point P with direction vector v. If M is any point not on L, then the
distance from M to L is


d = ‖ PM


× v ‖
‖ v ‖


.


Example 2.47
Calculating the Distance from a Point to a Line


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2.45


Find the distance between t point M = (1, 1, 3) and line x − 3
4


=
y + 1
2


= z − 3.


Solution
From the symmetric equations of the line, we know that vector v = 〈 4, 2, 1 〉 is a direction vector for the line.
Setting the symmetric equations of the line equal to zero, we see that point P(3, −1, 3) lies on the line. Then,


PM


= 〈 1 − 3, 1 − (−1), 3 − 3 〉 = 〈 −2, 2, 0 〉 .


To calculate the distance, we need to find PM→ × v:
PM


× v = | i j k−2 2 04 2 1|
= (2 − 0)i − (−2 − 0)j + (−4 − 8)k
= 2i + 2j − 12k.


Therefore, the distance between the point and the line is (Figure 2.66)
d = ‖ PM



× v ‖


‖ v ‖


= 2
2 + 22 + 122


42 + 22 + 12


= 2 38
21


.


Figure 2.66 Point (1, 1, 3) is approximately 2.7 units from
the line with symmetric equations x − 3


4
=


y + 1
2


= z − 3.


Find the distance between point (0, 3, 6) and the line with parametric equations
x = 1 − t, y = 1 + 2t, z = 5 + 3t.


Chapter 2 | Vectors in Space 189




Relationships between Lines
Given two lines in the two-dimensional plane, the lines are equal, they are parallel but not equal, or they intersect in a singlepoint. In three dimensions, a fourth case is possible. If two lines in space are not parallel, but do not intersect, then the linesare said to be skew lines (Figure 2.67).


Figure 2.67 In three dimensions, it is possible that two linesdo not cross, even when they have different directions.


To classify lines as parallel but not equal, equal, intersecting, or skew, we need to know two things: whether the directionvectors are parallel and whether the lines share a point (Figure 2.68).


Figure 2.68 Determine the relationship between two lines based on whethertheir direction vectors are parallel and whether they share a point.


Example 2.48
Classifying Lines in Space
For each pair of lines, determine whether the lines are equal, parallel but not equal, skew, or intersecting.


a. L1 : x = 2s − 1, y = s − 1, z = s − 4
L2 : x = t − 3, y = 3t + 8, z = 5 − 2t


b. L1 : x = −y = z
L2 :


x − 3
2


= y = z − 2


c. L1 : x = 6s − 1, y = −2s, z = 3s + 1
L2 :


x − 4
6


=
y + 3
−2


= z − 1
3


Solution
a. Line L1 has direction vector v1 = 〈 2, 1, 1 〉 ; line L2 has direction vector v2 = 〈 1, 3, −2 〉 .


Because the direction vectors are not parallel vectors, the lines are either intersecting or skew. To


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determine whether the lines intersect, we see if there is a point, (x, y, z), that lies on both lines. To find
this point, we use the parametric equations to create a system of equalities:


2s − 1 = t − 3; s − 1 = 3t + 8; s − 4 = 5 − 2t.


By the first equation, t = 2s + 2. Substituting into the second equation yields
s − 1 = 3(2s + 2) + 8


s − 1 = 6s + 6 + 8


5s = −15
s = −3.


Substitution into the third equation, however, yields a contradiction:
s − 4 = 5 − 2(2s + 2)


s − 4 = 5 − 4s − 4


5s = 5
s = 1.


There is no single point that satisfies the parametric equations for L1 and L2 simultaneously. These lines
do not intersect, so they are skew (see the following figure).


b. Line L1 has direction vector v1 = 〈 1, −1, 1 〉 and passes through the origin, (0, 0, 0). Line L2 has
a different direction vector, v2 = 〈 2, 1, 1 〉 , so these lines are not parallel or equal. Let r represent
the parameter for line L1 and let s represent the parameter for L2 :


x = r
y = −r
z = r


x = 2s + 3
y = s
z = s + 2.


Solve the system of equations to find r = 1 and s = − 1. If we need to find the point of intersection, we
can substitute these parameters into the original equations to get (1, −1, 1) (see the following figure).


Chapter 2 | Vectors in Space 191




2.46


c. Lines L1 and L2 have equivalent direction vectors: v = 〈 6, −2, 3 〉 . These two lines are parallel
(see the following figure).


Describe the relationship between the lines with the following parametric equations:
x = 1 − 4t, y = 3 + t, z = 8 − 6t


x = 2 + 3s, y = 2s, z = −1 − 3s.


Equations for a Plane
We know that a line is determined by two points. In other words, for any two distinct points, there is exactly one line thatpasses through those points, whether in two dimensions or three. Similarly, given any three points that do not all lie on thesame line, there is a unique plane that passes through these points. Just as a line is determined by two points, a plane isdetermined by three.
This may be the simplest way to characterize a plane, but we can use other descriptions as well. For example, given twodistinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any


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point that does not lie on the line. These characterizations arise naturally from the idea that a plane is determined by threepoints. Perhaps the most surprising characterization of a plane is actually the most useful.
Imagine a pair of orthogonal vectors that share an initial point. Visualize grabbing one of the vectors and twisting it.As you twist, the other vector spins around and sweeps out a plane. Here, we describe that concept mathematically. Let
n = 〈 a, b, c 〉 be a vector and P = (x0, y0, z0) be a point. Then the set of all points Q = (x, y, z) such that PQ→ is
orthogonal to n forms a plane (Figure 2.69). We say that n is a normal vector, or perpendicular to the plane. Remember,
the dot product of orthogonal vectors is zero. This fact generates the vector equation of a plane: n · PQ→ = 0. Rewriting
this equation provides additional ways to describe the plane:


n · PQ


= 0


〈 a, b, c 〉 · 〈 x − x0, y − y0, z − z0 〉 = 0


a(x − x0) + b(y − y0) + c(z − z0) = 0.


Figure 2.69 Given a point P and vector n, the set of all
points Q with PQ→ orthogonal to n forms a plane.


Definition
Given a point P and vector n, the set of all points Q satisfying the equation n · PQ→ = 0 forms a plane. The
equation


(2.17)n · PQ→ = 0
is known as the vector equation of a plane.
The scalar equation of a plane containing point P = (x0, y0, z0) with normal vector n = 〈 a, b, c 〉 is


(2.18)a(x − x0) + b(y − y0) + c(z − z0) = 0.
This equation can be expressed as ax + by + cz + d = 0, where d = −ax0 − by0 − cz0. This form of the equation
is sometimes called the general form of the equation of a plane.


As described earlier in this section, any three points that do not all lie on the same line determine a plane. Given three suchpoints, we can find an equation for the plane containing these points.
Example 2.49
Writing an Equation of a Plane Given Three Points in the Plane
Write an equation for the plane containing points P = (1, 1, −2), Q = (0, 2, 1), and R = (−1, −1, 0) in
both standard and general forms.


Chapter 2 | Vectors in Space 193




Solution
To write an equation for a plane, we must find a normal vector for the plane. We start by identifying two vectorsin the plane:


PQ


= 〈 0 − 1, 2 − 1, 1 − (−2) 〉 = 〈 −1, 1, 3 〉


QR


= 〈 −1 − 0, −1 − 2, 0 − 1 〉 = 〈 −1, −3, −1 〉 .


The cross product PQ→ × QR→ is orthogonal to both PQ→ and QR→ , so it is normal to the plane that contains
these two vectors:


n = PQ


× QR


= | i j k−1 1 3−1 −3 −1|
= (−1 + 9)i − (1 + 3)j + (3 + 1)k
= 8i − 4j + 4k.


Thus, n = 〈 8, −4, 4 〉 , and we can choose any of the three given points to write an equation of the plane:
8(x − 1) − 4(y − 1) + 4(z + 2) = 0


8x − 4y + 4z + 4 = 0.


The scalar equations of a plane vary depending on the normal vector and point chosen.
Example 2.50
Writing an Equation for a Plane Given a Point and a Line
Find an equation of the plane that passes through point (1, 4, 3) and contains the line given by
x =


y − 1
2


= z + 1.


Solution
Symmetric equations describe the line that passes through point (0, 1, −1) parallel to vector v1 = 〈 1, 2, 1 〉
(see the following figure). Use this point and the given point, (1, 4, 3), to identify a second vector parallel to
the plane:


v2 = 〈 1 − 0, 4 − 1, 3 − (−1) 〉 = 〈 1, 3, 4 〉 .


Use the cross product of these vectors to identify a normal vector for the plane:
n = v1 × v2


= | i j k1 2 11 3 4|
= (8 − 3)i − (4 − 1)j + (3 − 2)k
= 5i − 3j + k.


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2.47


The scalar equations for the plane are 5x − 3⎛⎝y − 1⎞⎠+ (z + 1) = 0 and 5x − 3y + z + 4 = 0.


Find an equation of the plane containing the lines L1 and L2 :
L1 : x = −y = z


L2 :
x − 3
2


= y = z − 2.


Now that we can write an equation for a plane, we can use the equation to find the distance d between a point P and the
plane. It is defined as the shortest possible distance from P to a point on the plane.


Figure 2.70 We want to find the shortest distance from pointP to the plane. Let point R be the point in the plane such that,
for any other point in the plane Q, ‖ RP→ ‖ < ‖ QP→ ‖ .


Just as we find the two-dimensional distance between a point and a line by calculating the length of a line segmentperpendicular to the line, we find the three-dimensional distance between a point and a plane by calculating the length of a
line segment perpendicular to the plane. Let R bet the point in the plane such that RP→ is orthogonal to the plane, and let


Chapter 2 | Vectors in Space 195




Q be an arbitrary point in the plane. Then the projection of vector QP→ onto the normal vector describes vector RP→ , as
shown in Figure 2.70.
Theorem 2.13: The Distance between a Plane and a Point
Suppose a plane with normal vector n passes through point Q. The distance d from the plane to a point P not in
the plane is given by


(2.19)
d = ‖ projn QP



‖ = |compn QP→ | = |


QP


·n|
‖ n ‖


.


Example 2.51
Distance between a Point and a Plane
Find the distance between point P = (3, 1, 2) and the plane given by x − 2y + z = 5 (see the following figure).


Solution
The coefficients of the plane’s equation provide a normal vector for the plane: n = 〈 1, −2, 1 〉 . To find vector
QP


, we need a point in the plane. Any point will work, so set y = z = 0 to see that point Q = (5, 0, 0) lies
in the plane. Find the component form of the vector from Q to P:


QP


= 〈 3 − 5, 1 − 0, 2 − 0 〉 = 〈 −2, 1, 2 〉 .


Apply the distance formula from Equation 2.19:


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2.48


d = |QP


·n|
‖ n ‖


= | 〈 −2, 1, 2 〉 · 〈 1, −2, 1 〉 |
12 + (−2)2 + 12


= |−2 − 2 + 2|
6


= 2
6
.


Find the distance between point P = (5, −1, 0) and the plane given by 4x + 2y − z = 3.


Parallel and Intersecting Planes
We have discussed the various possible relationships between two lines in two dimensions and three dimensions. When wedescribe the relationship between two planes in space, we have only two possibilities: the two distinct planes are parallel orthey intersect. When two planes are parallel, their normal vectors are parallel. When two planes intersect, the intersection isa line (Figure 2.71).


Figure 2.71 The intersection of two nonparallel planes isalways a line.


We can use the equations of the two planes to find parametric equations for the line of intersection.
Example 2.52
Finding the Line of Intersection for Two Planes
Find parametric and symmetric equations for the line formed by the intersection of the planes given by
x + y + z = 0 and 2x − y + z = 0 (see the following figure).


Chapter 2 | Vectors in Space 197




2.49


Solution
Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies eachequation, so we know the line of intersection passes through the origin. Add the plane equations so we caneliminate the one of the variables, in this case, y:


x + y + z = 0
2x − y + z = 0____________________
3x + 2z = 0.


This gives us x = − 2
3
z. We substitute this value into the first equation to express y in terms of z:


x + y + z = 0


−2
3
z + y + z = 0


y + 1
3
z = 0


y = −1
3
z.


We now have the first two variables, x and y, in terms of the third variable, z. Now we define z in terms
of t. To eliminate the need for fractions, we choose to define the parameter t as t = − 1


3
z. Then, z = −3t.


Substituting the parametric representation of z back into the other two equations, we see that the parametric
equations for the line of intersection are x = 2t, y = t, z = −3t. The symmetric equations for the line are
x
2
= y = z


−3
.


Find parametric equations for the line formed by the intersection of planes x + y − z = 3 and
3x − y + 3z = 5.


In addition to finding the equation of the line of intersection between two planes, we may need to find the angle formed bythe intersection of two planes. For example, builders constructing a house need to know the angle where different sections


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of the roof meet to know whether the roof will look good and drain properly. We can use normal vectors to calculate theangle between the two planes. We can do this because the angle between the normal vectors is the same as the angle betweenthe planes. Figure 2.72 shows why this is true.


Figure 2.72 The angle between two planes has the samemeasure as the angle between the normal vectors for the planes.


We can find the measure of the angle θ between two intersecting planes by first finding the cosine of the angle, using thefollowing equation:
cos θ = |n1 ·n2|


‖ n1 ‖ ‖ n2 ‖
.


We can then use the angle to determine whether two planes are parallel or orthogonal or if they intersect at some other angle.
Example 2.53
Finding the Angle between Two Planes
Determine whether each pair of planes is parallel, orthogonal, or neither. If the planes are intersecting, but notorthogonal, find the measure of the angle between them. Give the answer in radians and round to two decimalplaces.


a. x + 2y − z = 8 and 2x + 4y − 2z = 10
b. 2x − 3y + 2z = 3 and 6x + 2y − 3z = 1
c. x + y + z = 4 and x − 3y + 5z = 1


Solution
a. The normal vectors for these planes are n1 = 〈 1, 2, −1 〉 and n2 = 〈 2, 4, −2 〉 . These two vectors


are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.
b. The normal vectors for these planes are n1 = 〈 2, −3, 2 〉 and n2 = 〈 6, 2, −3 〉 . Taking the dot


product of these vectors, we have
n1 ·n2 = 〈 2, −3, 2 〉 · 〈 6, 2, −3 〉 = 2(6) − 3(2) + 2(−3) = 0.


The normal vectors are orthogonal, so the corresponding planes are orthogonal as well.
c. The normal vectors for these planes are n1 = 〈 1, 1, 1 〉 and n2 = 〈 1, −3, 5 〉 :


Chapter 2 | Vectors in Space 199




2.50


cos θ = |n1 ·n2|
‖ n1 ‖ ‖ n2 ‖


= | 〈 1, 1, 1 〉 · 〈 1, −3, 5 〉 |
12 + 12 + 12 12 + (−3)2 + 52


= 3
105


.


The angle between the two planes is 1.27 rad, or approximately 73°.


Find the measure of the angle between planes x + y − z = 3 and 3x − y + 3z = 5. Give the answer in
radians and round to two decimal places.


When we find that two planes are parallel, we may need to find the distance between them. To find this distance, we simplyselect a point in one of the planes. The distance from this point to the other plane is the distance between the planes.
Previously, we introduced the formula for calculating this distance in Equation 2.19:


d = QP


·n
‖ n ‖


,


where Q is a point on the plane, P is a point not on the plane, and n is the normal vector that passes through point Q.
Consider the distance from point (x0, y0, z0) to plane ax + by + cz + k = 0. Let (x1, y1, z1) be any point in the plane.
Substituting into the formula yields


d = |a(x0 − x1) + b(y0 − y1) + c(z0 − z1)|
a2 + b2 + c2


= |ax0 + by0 + cz0 + k|
a2 + b2 + c2


.


We state this result formally in the following theorem.
Theorem 2.14: Distance from a Point to a Plane
Let P(x0, y0, z0) be a point. The distance from P to plane ax + by + cz + k = 0 is given by


d = |ax0 + by0 + cz0 + k|
a2 + b2 + c2


.


Example 2.54
Finding the Distance between Parallel Planes
Find the distance between the two parallel planes given by 2x + y − z = 2 and 2x + y − z = 8.
Solution


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2.51


Point (1, 0, 0) lies in the first plane. The desired distance, then, is
d = |ax0 + by0 + cz0 + k|


a2 + b2 + c2


= |2(1) + 1(0) + (−1)(0) + (−8)|
22 + 12 + (−1)2


= 6
6
= 6.


Find the distance between parallel planes 5x − 2y + z = 6 and 5x − 2y + z = −3.


Chapter 2 | Vectors in Space 201




Distance between Two Skew Lines


Figure 2.73 Industrial pipe installations often feature pipes running indifferent directions. How can we find the distance between two skew pipes?


Finding the distance from a point to a line or from a line to a plane seems like a pretty abstract procedure. But, if thelines represent pipes in a chemical plant or tubes in an oil refinery or roads at an intersection of highways, confirmingthat the distance between them meets specifications can be both important and awkward to measure. One way is tomodel the two pipes as lines, using the techniques in this chapter, and then calculate the distance between them. Thecalculation involves forming vectors along the directions of the lines and using both the cross product and the dotproduct.
The symmetric forms of two lines, L1 and L2, are


L1 :
x − x1
a1


=
y − y1
b1


=
z − z1
c1


L2 :
x − x2
a2


=
y − y2
b2


=
z − z2
c2


.


You are to develop a formula for the distance d between these two lines, in terms of the values
a1, b1, c1; a2, b2, c2; x1, y1, z1; and x2, y2, z2. The distance between two lines is usually taken to mean the
minimum distance, so this is the length of a line segment or the length of a vector that is perpendicular to both linesand intersects both lines.


1. First, write down two vectors, v1 and v2, that lie along L1 and L2, respectively.
2. Find the cross product of these two vectors and call it N. This vector is perpendicular to v1 and v2, and


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hence is perpendicular to both lines.
3. From vector N, form a unit vector n in the same direction.
4. Use symmetric equations to find a convenient vector v12 that lies between any two points, one on each line.


Again, this can be done directly from the symmetric equations.
5. The dot product of two vectors is the magnitude of the projection of one vector onto the other—that is,


A ·B = ‖ A ‖ ‖ B ‖ cos θ, where θ is the angle between the vectors. Using the dot product, find the
projection of vector v12 found in step 4 onto unit vector n found in step 3. This projection is perpendicular
to both lines, and hence its length must be the perpendicular distance d between them. Note that the value of
d may be negative, depending on your choice of vector v12 or the order of the cross product, so use absolute
value signs around the numerator.


6. Check that your formula gives the correct distance of |−25|/ 198 ≈ 1.78 between the following two lines:
L1 :


x − 5
2


=
y − 3
4


= z − 1
3


L2 :
x − 6
3


=
y − 1
5


= z
7
.


7. Is your general expression valid when the lines are parallel? If not, why not? (Hint:What do you know aboutthe value of the cross product of two parallel vectors? Where would that result show up in your expression for
d ?)


8. Demonstrate that your expression for the distance is zero when the lines intersect. Recall that two lines intersectif they are not parallel and they are in the same plane. Hence, consider the direction of n and v12. What is
the result of their dot product?


9. Consider the following application. Engineers at a refinery have determined they need to install support strutsbetween many of the gas pipes to reduce damaging vibrations. To minimize cost, they plan to install these strutsat the closest points between adjacent skewed pipes. Because they have detailed schematics of the structure,they are able to determine the correct lengths of the struts needed, and hence manufacture and distribute themto the installation crews without spending valuable time making measurements.The rectangular frame structure has the dimensions 4.0 × 15.0 × 10.0 m (height, width, and depth). One
sector has a pipe entering the lower corner of the standard frame unit and exiting at the diametrically opposedcorner (the one farthest away at the top); call this L1. A second pipe enters and exits at the two different
opposite lower corners; call this L2 (Figure 2.74).


Figure 2.74 Two pipes cross through a standard frame unit.


Chapter 2 | Vectors in Space 203




Write down the vectors along the lines representing those pipes, find the cross product between them fromwhich to create the unit vector n, define a vector that spans two points on each line, and finally determine the
minimum distance between the lines. (Take the origin to be at the lower corner of the first pipe.) Similarly, youmay also develop the symmetric equations for each line and substitute directly into your formula.


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2.5 EXERCISES
In the following exercises, points P and Q are given. Let
L be the line passing through points P and Q.


a. Find the vector equation of line L.
b. Find parametric equations of line L.
c. Find symmetric equations of line L.
d. Find parametric equations of the line segmentdetermined by P and Q.


243. P(−3, 5, 9), Q(4, −7, 2)
244. P(4, 0, 5), Q(2, 3, 1)
245. P(−1, 0, 5), Q(4, 0, 3)
246. P(7, −2, 6), Q(−3, 0, 6)
For the following exercises, point P and vector v are
given. Let L be the line passing through point P with
direction v.


a. Find parametric equations of line L.
b. Find symmetric equations of line L.
c. Find the intersection of the line with the xy-plane.


247. P(1, −2, 3), v = 〈 1, 2, 3 〉
248. P(3, 1, 5), v = 〈 1, 1, 1 〉
249. P(3, 1, 5), v = QR→ , where Q(2, 2, 3) and
R(3, 2, 3)


250. P(2, 3, 0), v = QR→ , where Q(0, 4, 5) and
R(0, 4, 6)


For the following exercises, line L is given.
a. Find point P that belongs to the line and direction


vector v of the line. Express v in component
form.


b. Find the distance from the origin to line L.
251. x = 1 + t, y = 3 + t, z = 5 + 4t, t ∈ ℝ
252. −x = y + 1, z = 2
253. Find the distance between point A(−3, 1, 1) and the
line of symmetric equations x = −y = −z.


254. Find the distance between point A(4, 2, 5) and the
line of parametric equations x = −1 − t, y = −t, z = 2,
t ∈ ℝ .


For the following exercises, lines L1 and L2 are given.
a. Verify whether lines L1 and L2 are parallel.
b. If the lines L1 and L2 are parallel, then find the


distance between them.
255. L1 : x = 1 + t, y = t, z = 2 + t, t ∈ ℝ ,
L2 : x − 3 = y − 1 = z − 3


256. L1 : x = 2, y = 1, z = t,
L2 : x = 1, y = 1, z = 2 − 3t, t ∈ ℝ


257. Show that the line passing through points P(3, 1, 0)
and Q(1, 4, −3) is perpendicular to the line with equation
x = 3t, y = 3 + 8t, z = −7 + 6t, t ∈ ℝ .


258. Are the lines of equations
x = −2 + 2t, y = −6, z = 2 + 6t and
x = −1 + t, y = 1 + t, z = t, t ∈ ℝ , perpendicular to
each other?
259. Find the point of intersection of the lines of equations
x = −2y = 3z and x = −5 − t, y = −1 + t, z = t − 11,
t ∈ ℝ .


260. Find the intersection point of the x-axis with theline of parametric equations
x = 10 + t, y = 2 − 2t, z = −3 + 3t, t ∈ ℝ .


For the following exercises, lines L1 and L2 are given.
Determine whether the lines are equal, parallel but notequal, skew, or intersecting.
261. L1 : x = y − 1 = −z and L2 : x − 2 = −y = z2
262. L1 : x = 2t, y = 0, z = 3, t ∈ ℝ and
L2 : x = 0, y = 8 + s, z = 7 + s, s ∈ ℝ


263. L1 : x = −1 + 2t, y = 1 + 3t, z = 7t, t ∈ ℝ
and L2 : x − 1 = 23 ⎛⎝y − 4⎞⎠ = 27z − 2
264. L1 : 3x = y + 1 = 2z and
L2 : x = 6 + 2t, y = 17 + 6t, z = 9 + 3t, t ∈ ℝ


Chapter 2 | Vectors in Space 205




265. Consider line L of symmetric equations
x − 2 = −y = z


2
and point A(1, 1, 1).


a. Find parametric equations for a line parallel to L
that passes through point A.


b. Find symmetric equations of a line skew to L and
that passes through point A.


c. Find symmetric equations of a line that intersects
L and passes through point A.


266. Consider line L of parametric equations
x = t, y = 2t, z = 3, t ∈ ℝ .


a. Find parametric equations for a line parallel to L
that passes through the origin.b. Find parametric equations of a line skew to L that
passes through the origin.c. Find symmetric equations of a line that intersects
L and passes through the origin.


For the following exercises, point P and vector n are
given.


a. Find the scalar equation of the plane that passesthrough P and has normal vector n.
b. Find the general form of the equation of the planethat passes through P and has normal vector n.


267. P(0, 0, 0), n = 3i − 2j + 4k
268. P(3, 2, 2), n = 2i + 3j − k
269. P(1, 2, 3), n = 〈 1, 2, 3 〉
270. P(0, 0, 0), n = 〈 −3, 2, −1 〉
For the following exercises, the equation of a plane isgiven.


a. Find normal vector n to the plane. Express n
using standard unit vectors.


b. Find the intersections of the plane with the axes ofcoordinates.
c. Sketch the plane.


271. [T] 4x + 5y + 10z − 20 = 0
272. 3x + 4y − 12 = 0
273. 3x − 2y + 4z = 0
274. x + z = 0


275. Given point P(1, 2, 3) and vector n = i + j, find
point Q on the x-axis such that PQ→ and n are
orthogonal.
276. Show there is no plane perpendicular to n = i + j
that passes through points P(1, 2, 3) and Q(2, 3, 4).
277. Find parametric equations of the line passing throughpoint P(−2, 1, 3) that is perpendicular to the plane of
equation 2x − 3y + z = 7.
278. Find symmetric equations of the line passing throughpoint P(2, 5, 4) that is perpendicular to the plane of
equation 2x + 3y − 5z = 0.
279. Show that line x − 1


2
=


y + 1
3


= z − 2
4


is parallel to
plane x − 2y + z = 6.
280. Find the real number α such that the line of
parametric equations x = t, y = 2 − t, z = 3 + t,
t ∈ ℝ is parallel to the plane of equation
αx + 5y + z − 10 = 0.


For the following exercises, points P, Q, and R are given.
a. Find the general equation of the plane passingthrough P, Q, and R.
b. Write the vector equation n · PS→ = 0 of the plane


at a., where S(x, y, z) is an arbitrary point of the
plane.


c. Find parametric equations of the line passingthrough the origin that is perpendicular to the planepassing through P, Q, and R.
281. P(1, 1, 1), Q(2, 4, 3), and R(−1, −2, −1)
282. P(−2, 1, 4), Q(3, 1, 3), and R(−2, 1, 0)
283. Consider the planes of equations x + y + z = 1 and
x + z = 0.


a. Show that the planes intersect.b. Find symmetric equations of the line passingthrough point P(1, 4, 6) that is parallel to the line
of intersection of the planes.


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284. Consider the planes of equations −y + z − 2 = 0
and x − y = 0.


a. Show that the planes intersect.b. Find parametric equations of the line passingthrough point P(−8, 0, 2) that is parallel to the
line of intersection of the planes.


285. Find the scalar equation of the plane that passesthrough point P(−1, 2, 1) and is perpendicular to the line
of intersection of planes x + y − z − 2 = 0 and
2x − y + 3z − 1 = 0.


286. Find the general equation of the plane that passesthrough the origin and is perpendicular to the line ofintersection of planes −x + y + 2 = 0 and z − 3 = 0.
287. Determine whether the line of parametric equations
x = 1 + 2t, y = −2t, z = 2 + t, t ∈ ℝ intersects the
plane with equation 3x + 4y + 6z − 7 = 0. If it does
intersect, find the point of intersection.
288. Determine whether the line of parametric equations
x = 5, y = 4 − t, z = 2t, t ∈ ℝ intersects the plane
with equation 2x − y + z = 5. If it does intersect, find the
point of intersection.
289. Find the distance from point P(1, 5, −4) to the
plane of equation 3x − y + 2z − 6 = 0.
290. Find the distance from point P(1, −2, 3) to the
plane of equation (x − 3) + 2⎛⎝y + 1⎞⎠− 4z = 0.
For the following exercises, the equations of two planes aregiven.


a. Determine whether the planes are parallel,orthogonal, or neither.
b. If the planes are neither parallel nor orthogonal,then find the measure of the angle between theplanes. Express the answer in degrees rounded tothe nearest integer.


291. [T] x + y + z = 0, 2x − y + z − 7 = 0
292. 5x − 3y + z = 4, x + 4y + 7z = 1
293. x − 5y − z = 1, 5x − 25y − 5z = −3
294. [T] x − 3y + 6z = 4, 5x + y − z = 4


295. Show that the lines of equations
x = t, y = 1 + t, z = 2 + t, t ∈ ℝ , and
x
2
=


y − 1
3


= z − 3 are skew, and find the distance
between them.
296. Show that the lines of equations
x = −1 + t, y = −2 + t, z = 3t, t ∈ ℝ , and
x = 5 + s, y = −8 + 2s, z = 7s, s ∈ ℝ are skew, and
find the distance between them.
297. Consider point C(−3, 2, 4) and the plane of
equation 2x + 4y − 3z = 8.


a. Find the radius of the sphere with center C tangent
to the given plane.b. Find point P of tangency.


298. Consider the plane of equation x − y − z − 8 = 0.
a. Find the equation of the sphere with center C at


the origin that is tangent to the given plane.b. Find parametric equations of the line passingthrough the origin and the point of tangency.
299. Two children are playing with a ball. The girl throwsthe ball to the boy. The ball travels in the air, curves 3
ft to the right, and falls 5 ft away from the girl (see the
following figure). If the plane that contains the trajectory ofthe ball is perpendicular to the ground, find its equation.


300. [T] John allocates d dollars to consume monthly
three goods of prices a, b, and c. In this context, the
budget equation is defined as ax + by + cz = d, where
x ≥ 0, y ≥ 0, and z ≥ 0 represent the number of items
bought from each of the goods. The budget set is given by



⎨(x, y, z)|ax + by + cz ≤ d, x ≥ 0, y ≥ 0, z ≥ 0⎫⎭⎬, and
the budget plane is the part of the plane of equation
ax + by + cz = d for which x ≥ 0, y ≥ 0, and z ≥ 0.
Consider a = $8, b = $5, c = $10, and d = $500.


a. Use a CAS to graph the budget set and budgetplane.b. For z = 25, find the new budget equation and
graph the budget set in the same system ofcoordinates.


Chapter 2 | Vectors in Space 207




301. [T] Consider r(t) = 〈 sin t, cos t, 2t 〉 the position
vector of a particle at time t ∈ [0, 3], where the
components of r are expressed in centimeters and time is
measured in seconds. Let OP→ be the position vector of the
particle after 1 sec.


a. Determine the velocity vector v(1) of the particle
after 1 sec.


b. Find the scalar equation of the plane that isperpendicular to v(1) and passes through point P.
This plane is called the normal plane to the path ofthe particle at point P.


c. Use a CAS to visualize the path of the particlealong with the velocity vector and normal plane atpoint P.
302. [T] A solar panel is mounted on the roof of a house.The panel may be regarded as positioned at the pointsof coordinates (in meters) A(8, 0, 0), B(8, 18, 0),
C(0, 18, 8), and D(0, 0, 8) (see the following figure).


a. Find the general form of the equation of the planethat contains the solar panel by using points
A, B, andC, and show that its normal vector is
equivalent to AB→ × AD→ .


b. Find parametric equations of line L1 that passes
through the center of the solar panel and has
direction vector s = 1


3
i + 1


3
j + 1


3
k, which


points toward the position of the Sun at a particulartime of day.c. Find symmetric equations of line L2 that passes
through the center of the solar panel and isperpendicular to it.d. Determine the angle of elevation of the Sun abovethe solar panel by using the angle between lines L1
and L2.


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2.6 | Quadric Surfaces
Learning Objectives


2.6.1 Identify a cylinder as a type of three-dimensional surface.
2.6.2 Recognize the main features of ellipsoids, paraboloids, and hyperboloids.
2.6.3 Use traces to draw the intersections of quadric surfaces with the coordinate planes.


We have been exploring vectors and vector operations in three-dimensional space, and we have developed equations todescribe lines, planes, and spheres. In this section, we use our knowledge of planes and spheres, which are examples ofthree-dimensional figures called surfaces, to explore a variety of other surfaces that can be graphed in a three-dimensionalcoordinate system.
Identifying Cylinders
The first surface we’ll examine is the cylinder. Although most people immediately think of a hollow pipe or a soda strawwhen they hear the word cylinder, here we use the broad mathematical meaning of the term. As we have seen, cylindricalsurfaces don’t have to be circular. A rectangular heating duct is a cylinder, as is a rolled-up yoga mat, the cross-section ofwhich is a spiral shape.
In the two-dimensional coordinate plane, the equation x2 + y2 = 9 describes a circle centered at the origin with radius 3.
In three-dimensional space, this same equation represents a surface. Imagine copies of a circle stacked on top of each othercentered on the z-axis (Figure 2.75), forming a hollow tube. We can then construct a cylinder from the set of lines parallel
to the z-axis passing through circle x2 + y2 = 9 in the xy-plane, as shown in the figure. In this way, any curve in one of the
coordinate planes can be extended to become a surface.


Figure 2.75 In three-dimensional space, the graph of equation
x2 + y2 = 9 is a cylinder with radius 3 centered on the
z-axis. It continues indefinitely in the positive and negativedirections.


Definition
A set of lines parallel to a given line passing through a given curve is known as a cylindrical surface, or cylinder. Theparallel lines are called rulings.


From this definition, we can see that we still have a cylinder in three-dimensional space, even if the curve is not a circle.Any curve can form a cylinder, and the rulings that compose the cylinder may be parallel to any given line (Figure 2.76).


Chapter 2 | Vectors in Space 209




Figure 2.76 In three-dimensional space, the graph of equation
z = x3 is a cylinder, or a cylindrical surface with rulings
parallel to the y-axis.


Example 2.55
Graphing Cylindrical Surfaces
Sketch the graphs of the following cylindrical surfaces.


a. x2 + z2 = 25
b. z = 2x2 − y
c. y = sin x


Solution
a. The variable y can take on any value without limit. Therefore, the lines ruling this surface are parallel


to the y-axis. The intersection of this surface with the xz-plane forms a circle centered at the origin withradius 5 (see the following figure).


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Figure 2.77 The graph of equation x2 + z2 = 25 is a
cylinder with radius 5 centered on the y-axis.


b. In this case, the equation contains all three variables — x, y, and z — so none of the variables can
vary arbitrarily. The easiest way to visualize this surface is to use a computer graphing utility (see thefollowing figure).


Figure 2.78
c. In this equation, the variable z can take on any value without limit. Therefore, the lines composingthis surface are parallel to the z-axis. The intersection of this surface with the yz-plane outlines curve


y = sin x (see the following figure).


Chapter 2 | Vectors in Space 211




2.52


Figure 2.79 The graph of equation y = sin x is formed by a set of
lines parallel to the z-axis passing through curve y = sin x in the
xy-plane.


Sketch or use a graphing tool to view the graph of the cylindrical surface defined by equation z = y2.


When sketching surfaces, we have seen that it is useful to sketch the intersection of the surface with a plane parallel to oneof the coordinate planes. These curves are called traces. We can see them in the plot of the cylinder in Figure 2.80.
Definition
The traces of a surface are the cross-sections created when the surface intersects a plane parallel to one of thecoordinate planes.


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Figure 2.80 (a) This is one view of the graph of equation z = sin x. (b) To find the trace of the graph in the
xz-plane, set y = 0. The trace is simply a two-dimensional sine wave.


Traces are useful in sketching cylindrical surfaces. For a cylinder in three dimensions, though, only one set of traces isuseful. Notice, in Figure 2.80, that the trace of the graph of z = sin x in the xz-plane is useful in constructing the graph.
The trace in the xy-plane, though, is just a series of parallel lines, and the trace in the yz-plane is simply one line.
Cylindrical surfaces are formed by a set of parallel lines. Not all surfaces in three dimensions are constructed so simply,however. We now explore more complex surfaces, and traces are an important tool in this investigation.
Quadric Surfaces
We have learned about surfaces in three dimensions described by first-order equations; these are planes. Some othercommon types of surfaces can be described by second-order equations. We can view these surfaces as three-dimensionalextensions of the conic sections we discussed earlier: the ellipse, the parabola, and the hyperbola. We call these graphsquadric surfaces.
Definition
Quadric surfaces are the graphs of equations that can be expressed in the form


Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0.


When a quadric surface intersects a coordinate plane, the trace is a conic section.
An ellipsoid is a surface described by an equation of the form x2


a2
+


y2


b2
+ z


2


c2
= 1. Set x = 0 to see the trace of the


ellipsoid in the yz-plane. To see the traces in the y- and xz-planes, set z = 0 and y = 0, respectively. Notice that, if
a = b, the trace in the xy-plane is a circle. Similarly, if a = c, the trace in the xz-plane is a circle and, if b = c, then
the trace in the yz-plane is a circle. A sphere, then, is an ellipsoid with a = b = c.


Example 2.56


Chapter 2 | Vectors in Space 213




Sketching an Ellipsoid
Sketch the ellipsoid x2


22
+


y2


32
+ z


2


52
= 1.


Solution
Start by sketching the traces. To find the trace in the xy-plane, set z = 0: x2


22
+


y2


32
= 1 (see Figure 2.81). To


find the other traces, first set y = 0 and then set x = 0.


Figure 2.81 (a) This graph represents the trace of equation x2
22


+
y2


32
+ z


2


52
= 1 in the xy-plane, when we


set z = 0. (b) When we set y = 0, we get the trace of the ellipsoid in the xz-plane, which is an ellipse. (c)
When we set x = 0, we get the trace of the ellipsoid in the yz-plane, which is also an ellipse.


Now that we know what traces of this solid look like, we can sketch the surface in three dimensions (Figure2.82).


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Figure 2.82 (a) The traces provide a framework for thesurface. (b) The center of this ellipsoid is the origin.


The trace of an ellipsoid is an ellipse in each of the coordinate planes. However, this does not have to be the case for allquadric surfaces. Many quadric surfaces have traces that are different kinds of conic sections, and this is usually indicated
by the name of the surface. For example, if a surface can be described by an equation of the form x2


a2
+


y2


b2
= zc, then we


call that surface an elliptic paraboloid. The trace in the xy-plane is an ellipse, but the traces in the xz-plane and yz-planeare parabolas (Figure 2.83). Other elliptic paraboloids can have other orientations simply by interchanging the variables
to give us a different variable in the linear term of the equation x2


a2
+ z


2


c2
=


y
b
or y2


b2
+ z


2


c2
= xa.


Figure 2.83 This quadric surface is called an ellipticparaboloid.


Chapter 2 | Vectors in Space 215




Example 2.57
Identifying Traces of Quadric Surfaces
Describe the traces of the elliptic paraboloid x2 + y2


22
= z


5
.


Solution
To find the trace in the xy-plane, set z = 0: x2 + y2


22
= 0. The trace in the plane z = 0 is simply one point, the


origin. Since a single point does not tell us what the shape is, we can move up the z-axis to an arbitrary plane tofind the shape of other traces of the figure.
The trace in plane z = 5 is the graph of equation x2 + y2


22
= 1, which is an ellipse. In the xz-plane, the equation


becomes z = 5x2. The trace is a parabola in this plane and in any plane with the equation y = b.
In planes parallel to the yz-plane, the traces are also parabolas, as we can see in the following figure.


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2.53


Figure 2.84 (a) The paraboloid x2 + y2
22


= z
5
. (b) The trace in plane z = 5. (c) The


trace in the xz-plane. (d) The trace in the yz-plane.


A hyperboloid of one sheet is any surface that can be described with an equation of the form
x2


a2
+


y2


b2
− z


2


c2
= 1. Describe the traces of the hyperboloid of one sheet given by equation x2


32
+


y2


22
− z


2


52
= 1.


Hyperboloids of one sheet have some fascinating properties. For example, they can be constructed using straight lines, suchas in the sculpture in Figure 2.85(a). In fact, cooling towers for nuclear power plants are often constructed in the shapeof a hyperboloid. The builders are able to use straight steel beams in the construction, which makes the towers very strongwhile using relatively little material (Figure 2.85(b)).


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Figure 2.85 (a) A sculpture in the shape of a hyperboloid can be constructed of straight lines.(b) Cooling towers for nuclear power plants are often built in the shape of a hyperboloid.


Example 2.58
Chapter Opener: Finding the Focus of a Parabolic Reflector
Energy hitting the surface of a parabolic reflector is concentrated at the focal point of the reflector (Figure 2.86).
If the surface of a parabolic reflector is described by equation x2


100
+


y2


100
= z


4
, where is the focal point of the


reflector?


Figure 2.86 Energy reflects off of the parabolic reflector and is collected at the focal point. (credit: modification ofCGP Grey, Wikimedia Commons)
Solution
Since z is the first-power variable, the axis of the reflector corresponds to the z-axis. The coefficients of x2 and
y2 are equal, so the cross-section of the paraboloid perpendicular to the z-axis is a circle. We can consider a
trace in the xz-plane or the yz-plane; the result is the same. Setting y = 0, the trace is a parabola opening up
along the z-axis, with standard equation x2 = 4pz, where p is the focal length of the parabola. In this case,
this equation becomes x2 = 100 · z


4
= 4pz or 25 = 4p. So p is 6.25 m, which tells us that the focus of the


paraboloid is 6.25 m up the axis from the vertex. Because the vertex of this surface is the origin, the focal point
is (0, 0, 6.25).


Seventeen standard quadric surfaces can be derived from the general equation


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Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0.


The following figures summarizes the most important ones.


Figure 2.87 Characteristics of Common Quadratic Surfaces: Ellipsoid, Hyperboloid of One Sheet, Hyperboloid of TwoSheets.


Chapter 2 | Vectors in Space 219




Figure 2.88 Characteristics of Common Quadratic Surfaces: Elliptic Cone, Elliptic Paraboloid, Hyperbolic Paraboloid.


Example 2.59
Identifying Equations of Quadric Surfaces
Identify the surfaces represented by the given equations.


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2.54


a. 16x2 + 9y2 + 16z2 = 144
b. 9x2 − 18x + 4y2 + 16y − 36z + 25 = 0


Solution
a. The x, y, and z terms are all squared, and are all positive, so this is probably an ellipsoid. However,


let’s put the equation into the standard form for an ellipsoid just to be sure. We have
16x2 + 9y2 + 16z2 = 144.


Dividing through by 144 gives
x2
9


+
y2


16
+ z


2


9
= 1.


So, this is, in fact, an ellipsoid, centered at the origin.
b. We first notice that the z term is raised only to the first power, so this is either an elliptic paraboloid or a


hyperbolic paraboloid. We also note there are x terms and y terms that are not squared, so this quadric
surface is not centered at the origin. We need to complete the square to put this equation in one of thestandard forms. We have


9x2 − 18x + 4y2 + 16y − 36z + 25 = 0


9x2 − 18x + 4y2 + 16y + 25 = 36z


9⎛⎝x
2 − 2x⎞⎠+ 4



⎝y


2 + 4y⎞⎠+ 25 = 36z


9⎛⎝x
2 − 2x + 1 − 1⎞⎠+ 4



⎝y


2 + 4y + 4 − 4⎞⎠+ 25 = 36z


9(x − 1)2 − 9 + 4⎛⎝y + 2⎞⎠2 − 16 + 25 = 36z


9(x − 1)2 + 4⎛⎝y + 2⎞⎠2 = 36z


(x − 1)2


4
+

⎝y − 2⎞⎠2


9
= z.


This is an elliptic paraboloid centered at (1, 2, 0).


Identify the surface represented by equation 9x2 + y2 − z2 + 2z − 10 = 0.


Chapter 2 | Vectors in Space 221




2.6 EXERCISES
For the following exercises, sketch and describe thecylindrical surface of the given equation.
303. [T] x2 + z2 = 1
304. [T] x2 + y2 = 9
305. [T] z = cos⎛⎝π2 + x⎞⎠
306. [T] z = ex
307. [T] z = 9 − y2
308. [T] z = ln(x)
For the following exercises, the graph of a quadric surfaceis given.


a. Specify the name of the quadric surface.
b. Determine the axis of symmetry of the quadricsurface.


309.


310.


311.


312.


For the following exercises, match the given quadricsurface with its corresponding equation in standard form.
a. x2


4
+
y2


9
− z


2


12
= 1


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b. x2
4



y2


9
− z


2


12
= 1


c. x2
4


+
y2


9
+ z


2


12
= 1


d. z2 = 4x2 + 3y2
e. z = 4x2 − y2
f. 4x2 + y2 − z2 = 0


313. Hyperboloid of two sheets
314. Ellipsoid
315. Elliptic paraboloid
316. Hyperbolic paraboloid
317. Hyperboloid of one sheet
318. Elliptic cone
For the following exercises, rewrite the given equation ofthe quadric surface in standard form. Identify the surface.
319. −x2 + 36y2 + 36z2 = 9
320. −4x2 + 25y2 + z2 = 100
321. −3x2 + 5y2 − z2 = 10
322. 3x2 − y2 − 6z2 = 18
323. 5y = x2 − z2
324. 8x2 − 5y2 − 10z = 0
325. x2 + 5y2 + 3z2 − 15 = 0
326. 63x2 + 7y2 + 9z2 − 63 = 0
327. x2 + 5y2 − 8z2 = 0
328. 5x2 − 4y2 + 20z2 = 0
329. 6x = 3y2 + 2z2
330. 49y = x2 + 7z2
For the following exercises, find the trace of the givenquadric surface in the specified plane of coordinates andsketch it.


331. [T] x2 + z2 + 4y = 0, z = 0
332. [T] x2 + z2 + 4y = 0, x = 0
333. [T] −4x2 + 25y2 + z2 = 100, x = 0
334. [T] −4x2 + 25y2 + z2 = 100, y = 0


335. [T] x2 + y2
4


+ z
2


100
= 1, x = 0


336. [T] x2 − y − z2 = 1, y = 0
337. Use the graph of the given quadric surface to answerthe questions.


a. Specify the name of the quadric surface.b. Which of the equations—
16x2 + 9y2 + 36z2 = 3600, 9x2 + 36y2 + 16z2 = 3600,


or 36x2 + 9y2 + 16z2 = 3600—corresponds to
the graph?c. Use b. to write the equation of the quadric surfacein standard form.


338. Use the graph of the given quadric surface to answerthe questions.


a. Specify the name of the quadric surface.b. Which of the equations—
36z = 9x2 + y2, 9x2 + 4y2 = 36z, or − 36z = −81x2 + 4y2


—corresponds to the graph above?c. Use b. to write the equation of the quadric surfacein standard form.


Chapter 2 | Vectors in Space 223




For the following exercises, the equation of a quadricsurface is given.
a. Use the method of completing the square to writethe equation in standard form.
b. Identify the surface.


339. x2 + 2z2 + 6x − 8z + 1 = 0
340. 4x2 − y2 + z2 − 8x + 2y + 2z + 3 = 0
341. x2 + 4y2 − 4z2 − 6x − 16y − 16z + 5 = 0
342. x2 + z2 − 4y + 4 = 0


343. x2 + y2
4


− z
2


3
+ 6x + 9 = 0


344. x2 − y2 + z2 − 12z + 2x + 37 = 0
345. Write the standard form of the equation of theellipsoid centered at the origin that passes through points
A(2, 0, 0), B(0, 0, 1), and C⎛⎝12, 11, 12⎞⎠.
346. Write the standard form of the equation of theellipsoid centered at point P(1, 1, 0) that passes through
points A(6, 1, 0), B(4, 2, 0) and C(1, 2, 1).
347. Determine the intersection points of elliptic cone
x2 − y2 − z2 = 0 with the line of symmetric equations
x − 1
2


=
y + 1
3


= z.


348. Determine the intersection points of parabolic
hyperboloid z = 3x2 − 2y2 with the line of parametric
equations x = 3t, y = 2t, z = 19t, where t ∈ ℝ .
349. Find the equation of the quadric surface with points
P(x, y, z) that are equidistant from point Q(0, −1, 0)
and plane of equation y = 1. Identify the surface.
350. Find the equation of the quadric surface with points
P(x, y, z) that are equidistant from point Q(0, 2, 0) and
plane of equation y = −2. Identify the surface.
351. If the surface of a parabolic reflector is described
by equation 400z = x2 + y2, find the focal point of the
reflector.
352. Consider the parabolic reflector described by
equation z = 20x2 + 20y2. Find its focal point.


353. Show that quadric surface
x2 + y2 + z2 + 2xy + 2xz + 2yz + x + y + z = 0 reduces
to two parallel planes.
354. Show that quadric surface
x2 + y2 + z2 − 2xy − 2xz + 2yz − 1 = 0 reduces to two
parallel planes passing.
355. [T] The intersection between cylinder
(x − 1)2 + y2 = 1 and sphere x2 + y2 + z2 = 4 is called
a Viviani curve.


a. Solve the system consisting of the equations ofthe surfaces to find the equation of the intersectioncurve. (Hint: Find x and y in terms of z.)
b. Use a computer algebra system (CAS) to visualize


the intersection curve on sphere x2 + y2 + z2 = 4.
356. Hyperboloid of one sheet 25x2 + 25y2 − z2 = 25
and elliptic cone −25x2 + 75y2 + z2 = 0 are represented
in the following figure along with their intersection curves.Identify the intersection curves and find their equations(Hint: Find y from the system consisting of the equations ofthe surfaces.)


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357. [T] Use a CAS to create the intersection between
cylinder 9x2 + 4y2 = 18 and ellipsoid
36x2 + 16y2 + 9z2 = 144, and find the equations of the
intersection curves.
358. [T] A spheroid is an ellipsoid with two equalsemiaxes. For instance, the equation of a spheroid withthe z-axis as its axis of symmetry is given by
x2


a2
+


y2


a2
+ z


2


c2
= 1, where a and c are positive real


numbers. The spheroid is called oblate if c < a, and
prolate for c > a.


a. The eye cornea is approximated as a prolatespheroid with an axis that is the eye, where
a = 8.7 mm and c = 9.6 mm. Write the equation
of the spheroid that models the cornea and sketchthe surface.b. Give two examples of objects with prolate spheroidshapes.


359. [T] In cartography, Earth is approximated by anoblate spheroid rather than a sphere. The radii at the equatorand poles are approximately 3963 mi and 3950 mi,
respectively.a. Write the equation in standard form of the ellipsoidthat represents the shape of Earth. Assume thecenter of Earth is at the origin and that the traceformed by plane z = 0 corresponds to the equator.


b. Sketch the graph.c. Find the equation of the intersection curve of thesurface with plane z = 1000 that is parallel to the
xy-plane. The intersection curve is called a parallel.d. Find the equation of the intersection curve of thesurface with plane x + y = 0 that passes through
the z-axis. The intersection curve is called ameridian.


360. [T] A set of buzzing stunt magnets (or “rattlesnakeeggs”) includes two sparkling, polished, superstrongspheroid-shaped magnets well-known for children’sentertainment. Each magnet is 1.625 in. long and 0.5 in.
wide at the middle. While tossing them into the air, theycreate a buzzing sound as they attract each other.a. Write the equation of the prolate spheroid centeredat the origin that describes the shape of one of themagnets.b. Write the equations of the prolate spheroids thatmodel the shape of the buzzing stunt magnets. Usea CAS to create the graphs.


361. [T] A heart-shaped surface is given by equation

⎝x


2 + 9
4
y2 + z2 − 1⎞⎠


3
− x2 z3 − 9


80
y2 z3 = 0.


a. Use a CAS to graph the surface that models thisshape.b. Determine and sketch the trace of the heart-shapedsurface on the xz-plane.
362. [T] The ring torus symmetric about the z-axis is aspecial type of surface in topology and its equation is given
by ⎛⎝x2 + y2 + z2 + R2 − r2⎞⎠2 = 4R2 ⎛⎝x2 + y2⎞⎠, where
R > r > 0. The numbers R and r are called are the
major and minor radii, respectively, of the surface. Thefollowing figure shows a ring torus for which
R = 2 and r = 1.


a. Write the equation of the ring torus with
R = 2 and r = 1, and use a CAS to graph the
surface. Compare the graph with the figure given.b. Determine the equation and sketch the trace of thering torus from a. on the xy-plane.c. Give two examples of objects with ring torusshapes.


Chapter 2 | Vectors in Space 225




2.7 | Cylindrical and Spherical Coordinates
Learning Objectives


2.7.1 Convert from cylindrical to rectangular coordinates.
2.7.2 Convert from rectangular to cylindrical coordinates.
2.7.3 Convert from spherical to rectangular coordinates.
2.7.4 Convert from rectangular to spherical coordinates.


The Cartesian coordinate system provides a straightforward way to describe the location of points in space. Some surfaces,however, can be difficult to model with equations based on the Cartesian system. This is a familiar problem; recall thatin two dimensions, polar coordinates often provide a useful alternative system for describing the location of a point in theplane, particularly in cases involving circles. In this section, we look at two different ways of describing the location ofpoints in space, both of them based on extensions of polar coordinates. As the name suggests, cylindrical coordinates areuseful for dealing with problems involving cylinders, such as calculating the volume of a round water tank or the amount ofoil flowing through a pipe. Similarly, spherical coordinates are useful for dealing with problems involving spheres, such asfinding the volume of domed structures.
Cylindrical Coordinates
When we expanded the traditional Cartesian coordinate system from two dimensions to three, we simply added a newaxis to model the third dimension. Starting with polar coordinates, we can follow this same process to create a new three-dimensional coordinate system, called the cylindrical coordinate system. In this way, cylindrical coordinates provide anatural extension of polar coordinates to three dimensions.
Definition
In the cylindrical coordinate system, a point in space (Figure 2.89) is represented by the ordered triple (r, θ, z),
where


• (r, θ) are the polar coordinates of the point’s projection in the xy-plane
• z is the usual z-coordinate in the Cartesian coordinate system


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Figure 2.89 The right triangle lies in the xy-plane. The lengthof the hypotenuse is r and θ is the measure of the angle
formed by the positive x-axis and the hypotenuse. Thez-coordinate describes the location of the point above or belowthe xy-plane.


In the xy-plane, the right triangle shown in Figure 2.89 provides the key to transformation between cylindrical andCartesian, or rectangular, coordinates.
Theorem 2.15: Conversion between Cylindrical and Cartesian Coordinates
The rectangular coordinates (x, y, z) and the cylindrical coordinates (r, θ, z) of a point are related as follows:


x = r cos θ These equations are used to convert from


y = r sin θ cylindrical coordinates to rectangular


z = z coordinates.
and


r2 = x2 + y2 These equations are used to convert from


tan θ =
y
x rectangular coordinates to cylindrical


z = z coordinates.


As when we discussed conversion from rectangular coordinates to polar coordinates in two dimensions, it should be noted
that the equation tan θ = yx has an infinite number of solutions. However, if we restrict θ to values between 0 and 2π,
then we can find a unique solution based on the quadrant of the xy-plane in which original point (x, y, z) is located. Note
that if x = 0, then the value of θ is either π


2
, 3π


2
, or 0, depending on the value of y.


Notice that these equations are derived from properties of right triangles. To make this easy to see, consider point P in
the xy-plane with rectangular coordinates ⎛⎝x, y, 0⎞⎠ and with cylindrical coordinates (r, θ, 0), as shown in the following
figure.


Chapter 2 | Vectors in Space 227




Figure 2.90 The Pythagorean theorem provides equation
r2 = x2 + y2. Right-triangle relationships tell us that
x = r cos θ, y = r sin θ, and tan θ = y/x.


Let’s consider the differences between rectangular and cylindrical coordinates by looking at the surfaces generated wheneach of the coordinates is held constant. If c is a constant, then in rectangular coordinates, surfaces of the form x = c,
y = c, or z = c are all planes. Planes of these forms are parallel to the yz-plane, the xz-plane, and the xy-plane,
respectively. When we convert to cylindrical coordinates, the z-coordinate does not change. Therefore, in cylindricalcoordinates, surfaces of the form z = c are planes parallel to the xy-plane. Now, let’s think about surfaces of the form
r = c. The points on these surfaces are at a fixed distance from the z-axis. In other words, these surfaces are vertical circular
cylinders. Last, what about θ = c? The points on a surface of the form θ = c are at a fixed angle from the x-axis, which
gives us a half-plane that starts at the z-axis (Figure 2.91 and Figure 2.92).


Figure 2.91 In rectangular coordinates, (a) surfaces of the form x = c are planes parallel
to the yz-plane, (b) surfaces of the form y = c are planes parallel to the xz-plane, and (c)
surfaces of the form z = c are planes parallel to the xy-plane.


Figure 2.92 In cylindrical coordinates, (a) surfaces of the form r = c are vertical cylinders
of radius r, (b) surfaces of the form θ = c are half-planes at angle θ from the x-axis, and
(c) surfaces of the form z = c are planes parallel to the xy-plane.


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2.55


Example 2.60
Converting from Cylindrical to Rectangular Coordinates
Plot the point with cylindrical coordinates ⎛⎝4, 2π3 , −2⎞⎠ and express its location in rectangular coordinates.


Solution
Conversion from cylindrical to rectangular coordinates requires a simple application of the equations listed inConversion between Cylindrical and Cartesian Coordinates:


x = r cos θ = 4 cos 2π
3


= −2


y = r sin θ = 4 sin 2π
3


= 2 3


z = −2.


The point with cylindrical coordinates ⎛⎝4, 2π3 , −2⎞⎠ has rectangular coordinates ⎛⎝−2, 2 3, −2⎞⎠ (see the
following figure).


Figure 2.93 The projection of the point in the xy-plane is 4units from the origin. The line from the origin to the point’s
projection forms an angle of 2π


3
with the positive x-axis. The


point lies 2 units below the xy-plane.


Point R has cylindrical coordinates ⎛⎝5, π6, 4⎞⎠ . Plot R and describe its location in space using
rectangular, or Cartesian, coordinates.


If this process seems familiar, it is with good reason. This is exactly the same process that we followed in Introductionto Parametric Equations and Polar Coordinates to convert from polar coordinates to two-dimensional rectangularcoordinates.
Example 2.61


Chapter 2 | Vectors in Space 229




2.56


Converting from Rectangular to Cylindrical Coordinates
Convert the rectangular coordinates (1, −3, 5) to cylindrical coordinates.
Solution
Use the second set of equations from Conversion between Cylindrical and Cartesian Coordinates totranslate from rectangular to cylindrical coordinates:


r2 = x2 + y2


r = ± 12 + (−3)2 = ± 10.


We choose the positive square root, so r = 10. Now, we apply the formula to find θ. In this case, y is negative
and x is positive, which means we must select the value of θ between 3π


2
and 2π:


tan θ =
y
x =


−3
1


θ = arctan(−3) ≈ 5.03 rad.


In this case, the z-coordinates are the same in both rectangular and cylindrical coordinates:
z = 5.


The point with rectangular coordinates (1, −3, 5) has cylindrical coordinates approximately equal to

⎝ 10, 5.03, 5



⎠.


Convert point (−8, 8, −7) from Cartesian coordinates to cylindrical coordinates.


The use of cylindrical coordinates is common in fields such as physics. Physicists studying electrical charges and thecapacitors used to store these charges have discovered that these systems sometimes have a cylindrical symmetry. Thesesystems have complicated modeling equations in the Cartesian coordinate system, which make them difficult to describeand analyze. The equations can often be expressed in more simple terms using cylindrical coordinates. For example, the
cylinder described by equation x2 + y2 = 25 in the Cartesian system can be represented by cylindrical equation r = 5.
Example 2.62
Identifying Surfaces in the Cylindrical Coordinate System
Describe the surfaces with the given cylindrical equations.


a. θ = π
4


b. r2 + z2 = 9
c. z = r


Solution


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a. When the angle θ is held constant while r and z are allowed to vary, the result is a half-plane (see the
following figure).


Figure 2.94 In polar coordinates, the equation θ = π/4
describes the ray extending diagonally through the firstquadrant. In three dimensions, this same equation describes ahalf-plane.


b. Substitute r2 = x2 + y2 into equation r2 + z2 = 9 to express the rectangular form of the equation:
x2 + y2 + z2 = 9. This equation describes a sphere centered at the origin with radius 3 (see the
following figure).


Chapter 2 | Vectors in Space 231




Figure 2.95 The sphere centered at the origin with radius 3
can be described by the cylindrical equation r2 + z2 = 9.


c. To describe the surface defined by equation z = r, is it useful to examine traces parallel to the xy-plane.
For example, the trace in plane z = 1 is circle r = 1, the trace in plane z = 3 is circle r = 3, and so
on. Each trace is a circle. As the value of z increases, the radius of the circle also increases. The resulting
surface is a cone (see the following figure).


Figure 2.96 The traces in planes parallel to the xy-plane arecircles. The radius of the circles increases as z increases.


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2.57 Describe the surface with cylindrical equation r = 6.


Spherical Coordinates
In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which eachcoordinate represents a distance. In the cylindrical coordinate system, location of a point in space is described using twodistances (r and z) and an angle measure (θ). In the spherical coordinate system, we again use an ordered triple to describe
the location of a point in space. In this case, the triple describes one distance and two angles. Spherical coordinates makeit simple to describe a sphere, just as cylindrical coordinates make it easy to describe a cylinder. Grid lines for sphericalcoordinates are based on angle measures, like those for polar coordinates.
Definition
In the spherical coordinate system, a point P in space (Figure 2.97) is represented by the ordered triple (ρ, θ, φ)
where


• ρ (the Greek letter rho) is the distance between P and the origin ⎛⎝ρ ≠ 0⎞⎠;
• θ is the same angle used to describe the location in cylindrical coordinates;
• φ (the Greek letter phi) is the angle formed by the positive z-axis and line segment OP— , where O is the
origin and 0 ≤ φ ≤ π.


Figure 2.97 The relationship among spherical, rectangular,and cylindrical coordinates.


By convention, the origin is represented as (0, 0, 0) in spherical coordinates.


Theorem 2.16: Converting among Spherical, Cylindrical, and Rectangular Coordinates
Rectangular coordinates (x, y, z) and spherical coordinates ⎛⎝ρ, θ, φ⎞⎠ of a point are related as follows:


Chapter 2 | Vectors in Space 233




x = ρ sin φ cos θ These equations are used to convert from


y = ρ sin φ sin θ spherical coordinates to rectangular


z = ρ cos φ coordinates.
and


ρ2 = x2 + y2 + z2 These equations are used to convert from


tan θ =
y
x rectangular coordinates to spherical


φ = arccos



⎜ z
x2 + y2 + z2





⎟. coordinates.


If a point has cylindrical coordinates (r, θ, z), then these equations define the relationship between cylindrical and
spherical coordinates.


r = ρ sin φ These equations are used to convert from


θ = θ spherical coordinates to rectangular


z = ρ cos φ coordinates.
and


ρ = r2 + z2 These equations are used to convert from


θ = θ cylindrical coordinates to spherical


φ = arccos



⎜ z


r2 + z2





⎟ coordinates.


The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they arestraightforward applications of trigonometry. Looking at Figure 2.98, it is easy to see that r = ρ sin φ. Then, looking at
the triangle in the xy-plane with r as its hypotenuse, we have x = r cos θ = ρ sin φ cos θ. The derivation of the formula
for y is similar. Figure 2.96 also shows that ρ2 = r2 + z2 = x2 + y2 + z2 and z = ρ cos φ. Solving this last equation
for φ and then substituting ρ = r2 + z2 (from the first equation) yields φ = arccos⎛



⎜ z


r2 + z2





⎟. Also, note that, as


before, we must be careful when using the formula tan θ = yx to choose the correct value of θ.


Figure 2.98 The equations that convert from one system toanother are derived from right-triangle relationships.


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As we did with cylindrical coordinates, let’s consider the surfaces that are generated when each of the coordinates is heldconstant. Let c be a constant, and consider surfaces of the form ρ = c. Points on these surfaces are at a fixed distance
from the origin and form a sphere. The coordinate θ in the spherical coordinate system is the same as in the cylindrical
coordinate system, so surfaces of the form θ = c are half-planes, as before. Last, consider surfaces of the form φ = 0.
The points on these surfaces are at a fixed angle from the z-axis and form a half-cone (Figure 2.99).


Figure 2.99 In spherical coordinates, surfaces of the form ρ = c are spheres of
radius ρ (a), surfaces of the form θ = c are half-planes at an angle θ from the
x-axis (b), and surfaces of the form ϕ = c are half-cones at an angle ϕ from the
z-axis (c).


Example 2.63
Converting from Spherical Coordinates
Plot the point with spherical coordinates ⎛⎝8, π3, π6⎞⎠ and express its location in both rectangular and cylindrical
coordinates.
Solution
Use the equations in Converting among Spherical, Cylindrical, and Rectangular Coordinates totranslate between spherical and cylindrical coordinates (Figure 2.100):


x = ρ sin φ cos θ = 8 sin⎛⎝
π
6

⎠cos


π
3

⎠ = 8


1
2


1
2
= 2


y = ρ sin φ sin θ = 8 sin⎛⎝
π
6

⎠sin


π
3

⎠ = 8


1
2



3
2


= 2 3


z = ρ cos φ = 8 cos⎛⎝
π
6

⎠ = 8



3
2

⎠ = 4 3.


Chapter 2 | Vectors in Space 235




2.58


Figure 2.100 The projection of the point in the xy-plane is 4 units from the origin.
The line from the origin to the point’s projection forms an angle of π/3 with the positive
x-axis. The point lies 4 3 units above the xy-plane.


The point with spherical coordinates ⎛⎝8, π3, π6⎞⎠ has rectangular coordinates ⎛⎝2, 2 3, 4 3⎞⎠.
Finding the values in cylindrical coordinates is equally straightforward:


r = ρ sin φ = 8 sin π
6
= 4


θ = θ
z = ρ cos φ = 8 cos π


6
= 4 3.


Thus, cylindrical coordinates for the point are ⎛⎝4, π3, 4 3⎞⎠.


Plot the point with spherical coordinates ⎛⎝2, − 5π6 , π6⎞⎠ and describe its location in both rectangular and
cylindrical coordinates.


Example 2.64
Converting from Rectangular Coordinates
Convert the rectangular coordinates ⎛⎝−1, 1, 6⎞⎠ to both spherical and cylindrical coordinates.


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Solution
Start by converting from rectangular to spherical coordinates:


ρ2 = x2 + y2 + z2 = (−1)2 + 12 + ⎛⎝ 6


2 = 8


ρ = 2 2


tan θ = 1
−1


θ = arctan(−1) = 3π
4
.


Because (x, y) = (−1, 1), then the correct choice for θ is 3π
4
.


There are actually two ways to identify φ. We can use the equation φ = arccos⎛

⎜ z
x2 + y2 + z2





⎟. Amore simple


approach, however, is to use equation z = ρ cos φ. We know that z = 6 and ρ = 2 2, so
6 = 2 2 cos φ, so cos φ = 6


2 2
= 3


2


and therefore φ = π
6
. The spherical coordinates of the point are ⎛⎝2 2, 3π4 , π6⎞⎠.


To find the cylindrical coordinates for the point, we need only find r:
r = ρ sin φ = 2 2 sin⎛⎝


π
6

⎠ = 2.


The cylindrical coordinates for the point are ⎛⎝ 2, 3π4 , 6⎞⎠.


Example 2.65
Identifying Surfaces in the Spherical Coordinate System
Describe the surfaces with the given spherical equations.


a. θ = π
3


b. φ = 5π
6


c. ρ = 6
d. ρ = sin θ sin φ


Solution
a. The variable θ represents the measure of the same angle in both the cylindrical and spherical coordinate


systems. Points with coordinates ⎛⎝ρ, π3, φ⎞⎠ lie on the plane that forms angle θ = π3 with the positive
x-axis. Because ρ > 0, the surface described by equation θ = π


3
is the half-plane shown in Figure


2.101.


Chapter 2 | Vectors in Space 237




Figure 2.101 The surface described by equation θ = π
3
is a


half-plane.


b. Equation φ = 5π
6


describes all points in the spherical coordinate system that lie on a line from the origin
forming an angle measuring 5π


6
rad with the positive z-axis. These points form a half-cone (Figure


2.102). Because there is only one value for φ that is measured from the positive z-axis, we do not get
the full cone (with two pieces).


Figure 2.102 The equation φ = 5π
6


describes a cone.


To find the equation in rectangular coordinates, use equation φ = arccos⎛

⎜ z
x2 + y2 + z2





⎟.


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6


= arccos



⎜ z
x2 + y2 + z2







cos 5π
6


= z


x2 + y2 + z2


− 3
2


= z


x2 + y2 + z2


3
4


= z
2


x2 + y2 + z2


3x2
4


+
3y2


4
+ 3z


2


4
= z2


3x2
4


+
3y2


4
− z


2


4
= 0.


This is the equation of a cone centered on the z-axis.
c. Equation ρ = 6 describes the set of all points 6 units away from the origin—a sphere with radius 6


(Figure 2.103).


Figure 2.103 Equation ρ = 6 describes a sphere with radius
6.


d. To identify this surface, convert the equation from spherical to rectangular coordinates, using equations
y = ρ sin φ sin θ and ρ2 = x2 + y2 + z2 :


Chapter 2 | Vectors in Space 239




2.59


ρ = sin θ sin φ


ρ2 = ρ sin θ sin φ Multiply both sides of the equation by ρ.


x2 + y2 + z2 = y Substitute rectangular variables using the equations above.


x2 + y2 − y + z2 = 0 Subtract y from both sides of the equation.


x2 + y2 − y + 1
4
+ z2 = 1


4
Complete the square.


x2 + ⎛⎝y −
1
2



2
+ z2 = 1


4
. Rewrite the middle terms as a perfect square.


The equation describes a sphere centered at point ⎛⎝0, 12, 0⎞⎠ with radius 12.


Describe the surfaces defined by the following equations.
a. ρ = 13
b. θ = 2π


3


c. φ = π
4


Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as thevolume of the space inside a domed stadium or wind speeds in a planet’s atmosphere. A sphere that has Cartesian equation
x2 + y2 + z2 = c2 has the simple equation ρ = c in spherical coordinates.
In geography, latitude and longitude are used to describe locations on Earth’s surface, as shown in Figure 2.104. Althoughthe shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth.Let’s assume Earth has the shape of a sphere with radius 4000 mi. We express angle measures in degrees rather than
radians because latitude and longitude are measured in degrees.


Figure 2.104 In the latitude–longitude system, anglesdescribe the location of a point on Earth relative to the equatorand the prime meridian.


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2.60


Let the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing thepositive z-axis. The prime meridian represents the trace of the surface as it intersects the xz-plane. The equator is the traceof the sphere intersecting the xy-plane.
Example 2.66
Converting Latitude and Longitude to Spherical Coordinates
The latitude of Columbus, Ohio, is 40° N and the longitude is 83° W, which means that Columbus is 40°
north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earththrough the equator directly south of Columbus. The measure of the angle formed by the rays is 40°. In the same
way, measuring from the prime meridian, Columbus lies 83° to the west. Express the location of Columbus in
spherical coordinates.
Solution
The radius of Earth is 4000 mi, so ρ = 4000. The intersection of the prime meridian and the equator lies on
the positive x-axis. Movement to the west is then described with negative angle measures, which shows that
θ = −83°, Because Columbus lies 40° north of the equator, it lies 50° south of the North Pole, so φ = 50°.
In spherical coordinates, Columbus lies at point (4000, −83°, 50°).


Sydney, Australia is at 34°S and 151°E. Express Sydney’s location in spherical coordinates.


Cylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand.A thoughtful choice of coordinate system can make a problem much easier to solve, whereas a poor choice can lead tounnecessarily complex calculations. In the following example, we examine several different problems and discuss how toselect the best coordinate system for each one.
Example 2.67
Choosing the Best Coordinate System
In each of the following situations, we determine which coordinate system is most appropriate and describe howwe would orient the coordinate axes. There could be more than one right answer for how the axes should beoriented, but we select an orientation that makes sense in the context of the problem. Note: There is not enoughinformation to set up or solve these problems; we simply select the coordinate system (Figure 2.105).


a. Find the center of gravity of a bowling ball.
b. Determine the velocity of a submarine subjected to an ocean current.
c. Calculate the pressure in a conical water tank.
d. Find the volume of oil flowing through a pipeline.
e. Determine the amount of leather required to make a football.


Chapter 2 | Vectors in Space 241




Figure 2.105 (credit: (a) modification of work by scl hua, Wikimedia, (b) modification ofwork by DVIDSHUB, Flickr, (c) modification of work by Michael Malak, Wikimedia, (d)modification of work by Sean Mack, Wikimedia, (e) modification of work by Elvert Barnes,Flickr)


Solution
a. Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The originshould be located at the physical center of the ball. There is no obvious choice for how the x-, y- andz-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choiceis to align the z-axis with the axis of symmetry of the weight block.
b. A submarine generally moves in a straight line. There is no rotational or spherical symmetry that appliesin this situation, so rectangular coordinates are a good choice. The z-axis should probably point upward.The x- and y-axes could be aligned to point east and north, respectively. The origin should be someconvenient physical location, such as the starting position of the submarine or the location of a particularport.
c. A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation


z = kr, where k is a constant. In spherical coordinates, we have seen that surfaces of the form φ = c
are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented
by equations of the form z2 = x2


a2
+


y2


b2
. In this case, we could choose any of the three. However, the


equation for the surface is more complicated in rectangular coordinates than in the other two systems, sowe might want to avoid that choice. In addition, we are talking about a water tank, and the depth of thewater might come into play at some point in our calculations, so it might be nice to have a componentthat represents height and depth directly. Based on this reasoning, cylindrical coordinates might be thebest choice. Choose the z-axis to align with the axis of the cone. The orientation of the other two axes isarbitrary. The origin should be the bottom point of the cone.
d. A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, wewould likely choose to orient our z-axis with the center axis of the pipeline. The x-axis could be chosento point straight downward or to some other logical direction. The origin should be chosen based on theproblem statement. Note that this puts the z-axis in a horizontal orientation, which is a little different from


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2.61


what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense forthe problem.
e. A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. Thez-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one ofthe ends. The position of the x-axis is arbitrary.


Which coordinate system is most appropriate for creating a star map, as viewed from Earth (see thefollowing figure)?


How should we orient the coordinate axes?


Chapter 2 | Vectors in Space 243




2.7 EXERCISES
Use the following figure as an aid in identifying therelationship between the rectangular, cylindrical, andspherical coordinate systems.


For the following exercises, the cylindrical coordinates
(r, θ, z) of a point are given. Find the rectangular
coordinates (x, y, z) of the point.
363. ⎛⎝4, π6, 3⎞⎠
364. ⎛⎝3, π3, 5⎞⎠
365. ⎛⎝4, 7π6 , 3⎞⎠
366. (2, π, −4)
For the following exercises, the rectangular coordinates
(x, y, z) of a point are given. Find the cylindrical
coordinates (r, θ, z) of the point.
367. ⎛⎝1, 3, 2⎞⎠
368. (1, 1, 5)
369. (3, −3, 7)
370. ⎛⎝−2 2, 2 2, 4⎞⎠
For the following exercises, the equation of a surface incylindrical coordinates is given.
Find the equation of the surface in rectangular coordinates.Identify and graph the surface.
371. [T] r = 4


372. [T] z = r2 cos2 θ
373. [T] r2 cos(2θ) + z2 + 1 = 0
374. [T] r = 3 sin θ
375. [T] r = 2 cos θ
376. [T] r2 + z2 = 5
377. [T] r = 2 sec θ
378. [T] r = 3 csc θ
For the following exercises, the equation of a surface inrectangular coordinates is given. Find the equation of thesurface in cylindrical coordinates.
379. z = 3
380. x = 6
381. x2 + y2 + z2 = 9
382. y = 2x2
383. x2 + y2 − 16x = 0
384. x2 + y2 − 3 x2 + y2 + 2 = 0
For the following exercises, the spherical coordinates

⎝ρ, θ, φ⎞⎠ of a point are given. Find the rectangular
coordinates (x, y, z) of the point.
385. (3, 0, π)
386. ⎛⎝1, π6, π6⎞⎠
387. ⎛⎝12, − π4, π4⎞⎠
388. ⎛⎝3, π4, π6⎞⎠
For the following exercises, the rectangular coordinates
(x, y, z) of a point are given. Find the spherical
coordinates ⎛⎝ρ, θ, φ⎞⎠ of the point. Express the measure of
the angles in degrees rounded to the nearest integer.
389. (4, 0, 0)


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390. (−1, 2, 1)
391. (0, 3, 0)
392. ⎛⎝−2, 2 3, 4⎞⎠
For the following exercises, the equation of a surface inspherical coordinates is given. Find the equation of thesurface in rectangular coordinates. Identify and graph thesurface.
393. [T] ρ = 3
394. [T] φ = π


3


395. [T] ρ = 2 cos φ
396. [T] ρ = 4 csc φ
397. [T] φ = π


2


398. [T] ρ = 6 csc φ sec θ
For the following exercises, the equation of a surface inrectangular coordinates is given. Find the equation of thesurface in spherical coordinates. Identify the surface.
399. x2 + y2 − 3z2 = 0, z ≠ 0
400. x2 + y2 + z2 − 4z = 0
401. z = 6
402. x2 + y2 = 9
For the following exercises, the cylindrical coordinates ofa point are given. Find its associated spherical coordinates,with the measure of the angle φ in radians rounded to four
decimal places.
403. [T] ⎛⎝1, π4, 3⎞⎠
404. [T] (5, π, 12)
405. ⎛⎝3, π2, 3⎞⎠
406. ⎛⎝3, − π6, 3⎞⎠
For the following exercises, the spherical coordinates of apoint are given. Find its associated cylindrical coordinates.


407. ⎛⎝2, − π4, π2⎞⎠
408. ⎛⎝4, π4, π6⎞⎠
409. ⎛⎝8, π3, π2⎞⎠
410. ⎛⎝9, − π6, π3⎞⎠
For the following exercises, find the most suitable systemof coordinates to describe the solids.
411. The solid situated in the first octant with a vertex atthe origin and enclosed by a cube of edge length a, where
a > 0


412. A spherical shell determined by the region betweentwo concentric spheres centered at the origin, of radii of a
and b, respectively, where b > a > 0
413. A solid inside sphere x2 + y2 + z2 = 9 and outside
cylinder ⎛⎝x − 32⎞⎠


2
+ y2 = 9


4


414. A cylindrical shell of height 10 determined by the
region between two cylinders with the same center, parallelrulings, and radii of 2 and 5, respectively
415. [T] Use a CAS to graph in cylindrical coordinates the
region between elliptic paraboloid z = x2 + y2 and cone
x2 + y2 − z2 = 0.


416. [T] Use a CAS to graph in spherical coordinatesthe “ice cream-cone region” situated above the xy-plane
between sphere x2 + y2 + z2 = 4 and elliptical cone
x2 + y2 − z2 = 0.


Chapter 2 | Vectors in Space 245




417. Washington, DC, is located at 39° N and 77° W
(see the following figure). Assume the radius of Earth is
4000 mi. Express the location of Washington, DC, in
spherical coordinates.


418. San Francisco is located at 37.78°N and
122.42°W. Assume the radius of Earth is 4000 mi.
Express the location of San Francisco in sphericalcoordinates.
419. Find the latitude and longitude of Rio de Janeiro if itsspherical coordinates are (4000, −43.17°, 102.91°).
420. Find the latitude and longitude of Berlin if itsspherical coordinates are (4000, 13.38°, 37.48°).
421. [T] Consider the torus of equation

⎝x


2 + y2 + z2 + R2 − r2⎞⎠
2
= 4R2 ⎛⎝x


2 + y2⎞⎠, where
R ≥ r > 0.


a. Write the equation of the torus in sphericalcoordinates.b. If R = r, the surface is called a horn torus. Show
that the equation of a horn torus in sphericalcoordinates is ρ = 2R sin φ.


c. Use a CAS to graph the horn torus with R = r = 2
in spherical coordinates.


422. [T] The “bumpy sphere” with an equation inspherical coordinates is ρ = a + b cos(mθ)sin(nφ), with
θ ∈ [0, 2π] and φ ∈ [0, π], where a and b are
positive numbers and m and n are positive integers, may
be used in applied mathematics to model tumor growth.a. Show that the “bumpy sphere” is contained insidea sphere of equation ρ = a + b. Find the values of


θ and φ at which the two surfaces intersect.
b. Use a CAS to graph the surface for a = 14,


b = 2, m = 4, and n = 6 along with sphere
ρ = a + b.


c. Find the equation of the intersection curve of thesurface at b. with the cone φ = π
12


. Graph the
intersection curve in the plane of intersection.


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component
coordinate plane
cross product


cylinder
cylindrical coordinate system


determinant
direction angles
direction cosines
direction vector
dot product or scalar product
ellipsoid


elliptic cone


elliptic paraboloid


equivalent vectors
general form of the equation of a plane


hyperboloid of one sheet


hyperboloid of two sheets


initial point
magnitude
normal vector
normalization
octants


CHAPTER 2 REVIEW
KEY TERMS


a scalar that describes either the vertical or horizontal direction of a vector
a plane containing two of the three coordinate axes in the three-dimensional coordinate system,named by the axes it contains: the xy-plane, xz-plane, or the yz-plane


u × v = (u2 v3 − u3 v2)i − (u1 v3 − u3 v1)j + (u1 v2 − u2 v1)k, where u = 〈 u1, u2, u3 〉 and
v = 〈 v1, v2, v3 〉


a set of lines parallel to a given line passing through a given curve
a way to describe a location in space with an ordered triple (r, θ, z), where (r, θ)


represents the polar coordinates of the point’s projection in the xy-plane, and z represents the point’s projection onto
the z-axis


a real number associated with a square matrix
the angles formed by a nonzero vector and the coordinate axes
the cosines of the angles formed by a nonzero vector and the coordinate axes


a vector parallel to a line that is used to describe the direction, or orientation, of the line in space
u · v = u1 v1 + u2 v2 + u3 v3 where u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉


a three-dimensional surface described by an equation of the form x2
a2


+
y2


b2
+ z


2


c2
= 1; all traces of this surface


are ellipses
a three-dimensional surface described by an equation of the form x2


a2
+


y2


b2
− z


2


c2
= 0; traces of this


surface include ellipses and intersecting lines
a three-dimensional surface described by an equation of the form z = x2


a2
+


y2


b2
; traces of this


surface include ellipses and parabolas
vectors that have the same magnitude and the same direction


an equation in the form ax + by + cz + d = 0, where n = 〈 a, b, c 〉
is a normal vector of the plane, P = (x0, y0, z0) is a point on the plane, and d = −ax0 − by0 − cz0


a three-dimensional surface described by an equation of the form x2
a2


+
y2


b2
− z


2


c2
= 1;


traces of this surface include ellipses and hyperbolas
a three-dimensional surface described by an equation of the form z2


c2
− x


2


a2


y2


b2
= 1;


traces of this surface include ellipses and hyperbolas
the starting point of a vector
the length of a vector


a vector perpendicular to a plane
using scalar multiplication to find a unit vector with a given direction


the eight regions of space created by the coordinate planes


Chapter 2 | Vectors in Space 247




orthogonal vectors
parallelepiped
parallelogram method


parametric equations of a line


quadric surfaces
right-hand rule


rulings
scalar
scalar equation of a plane


scalar multiplication
scalar projection
skew lines
sphere
spherical coordinate system


standard equation of a sphere


standard unit vectors
standard-position vector
symmetric equations of a line


terminal point
three-dimensional rectangular coordinate system


torque
trace
triangle inequality
triangle method


vectors that form a right angle when placed in standard position
a three-dimensional prism with six faces that are parallelograms


a method for finding the sum of two vectors; position the vectors so they share the same initialpoint; the vectors then form two adjacent sides of a parallelogram; the sum of the vectors is the diagonal of thatparallelogram
the set of equations x = x0 + ta, y = y0 + tb, and z = z0 + tc describing the


line with direction vector v = 〈 a, b, c 〉 passing through point (x0, y0, z0)
surfaces in three dimensions having the property that the traces of the surface are conic sections(ellipses, hyperbolas, and parabolas)


a common way to define the orientation of the three-dimensional coordinate system; when the righthand is curved around the z-axis in such a way that the fingers curl from the positive x-axis to the positive y-axis, thethumb points in the direction of the positive z-axis
parallel lines that make up a cylindrical surface
a real number


the equation a(x − x0) + b(y − y0) + c(z − z0) = 0 used to describe a plane containing
point P = (x0, y0, z0) with normal vector n = 〈 a, b, c 〉 or its alternate form ax + by + cz + d = 0, where
d = −ax0 − by0 − cz0


a vector operation that defines the product of a scalar and a vector
the magnitude of the vector projection of a vector


two lines that are not parallel but do not intersect
the set of all points equidistant from a given point known as the center


a way to describe a location in space with an ordered triple ⎛⎝ρ, θ, φ⎞⎠, where ρ is the
distance between P and the origin ⎛⎝ρ ≠ 0⎞⎠, θ is the same angle used to describe the location in cylindrical
coordinates, and φ is the angle formed by the positive z-axis and line segment OP— , where O is the origin and
0 ≤ φ ≤ π


(x − a)2 + ⎛⎝y − b⎞⎠2 + (z − c)2 = r2 describes a sphere with center (a, b, c) and
radius r


unit vectors along the coordinate axes: i = 〈 1, 0 〉 , j = 〈 0, 1 〉
a vector with initial point (0, 0)


the equations x − x0a = y − y0b = z − z0c describing the line with direction vector
v = 〈 a, b, c 〉 passing through point (x0, y0, z0)


the endpoint of a vector
a coordinate system defined by three lines that intersect atright angles; every point in space is described by an ordered triple (x, y, z) that plots its location relative to the


defining axes
the effect of a force that causes an object to rotate


the intersection of a three-dimensional surface with a coordinate plane
the length of any side of a triangle is less than the sum of the lengths of the other two sides


a method for finding the sum of two vectors; position the vectors so the terminal point of one vector isthe initial point of the other; these vectors then form two sides of a triangle; the sum of the vectors is the vector that


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triple scalar product
unit vector
vector
vector addition
vector difference
vector equation of a line


vector equation of a plane


vector product
vector projection
vector sum


work done by a force


zero vector


forms the third side; the initial point of the sum is the initial point of the first vector; the terminal point of the sum isthe terminal point of the second vector
the dot product of a vector with the cross product of two other vectors: u · (v × w)


a vector with margnitude 1
a mathematical object that has both magnitude and direction


a vector operation that defines the sum of two vectors
the vector difference v − w is defined as v + (−w) = v + (−1)w


the equation r = r0 + tv used to describe a line with direction vector v = 〈 a, b, c 〉
passing through point P = (x0, y0, z0), where r0 = 〈 x0, y0, z0 〉 , is the position vector of point P


the equation n · PQ→ = 0, where P is a given point in the plane, Q is any point in the
plane, and n is a normal vector of the plane


the cross product of two vectors
the component of a vector that follows a given direction


the sum of two vectors, v and w, can be constructed graphically by placing the initial point of w at the
terminal point of v; then the vector sum v + w is the vector with an initial point that coincides with the initial point
of v, and with a terminal point that coincides with the terminal point of w


work is generally thought of as the amount of energy it takes to move an object; if we representan applied force by a vector F and the displacement of an object by a vector s, then the work done by the force is thedot product of F and s.
the vector with both initial point and terminal point (0, 0)


KEY EQUATIONS
• Distance between two points in space:


d = (x2 − x1)
2 + (y2 − y1)


2 + (z2 − z1)
2


• Sphere with center (a, b, c) and radius r:
(x − a)2 + ⎛⎝y − b⎞⎠2 + (z − c)2 = r2


• Dot product of u and v
u · v = u1 v1 + u2 v2 + u3 v3


= ‖ u ‖ ‖ v ‖ cos θ


• Cosine of the angle formed by u and v
cos θ = u · v


‖ u ‖ ‖ v ‖


• Vector projection of v onto u
projuv = u · v


‖ u ‖ 2
u


• Scalar projection of v onto u
compuv = u · v‖ u ‖


• Work done by a force F to move an object through displacement vector PQ→
W = F · PQ



= ‖ F ‖ ‖ PQ



‖ cos θ


• The cross product of two vectors in terms of the unit vectors


Chapter 2 | Vectors in Space 249




u × v = (u2 v3 − u3 v2)i − (u1 v3 − u3 v1)j + (u1 v2 − u2 v1)k


• Vector Equation of a Line
r = r0 + tv


• Parametric Equations of a Line
x = x0 + ta, y = y0 + tb, and z = z0 + tc


• Symmetric Equations of a Line
x − x0
a =


y − y0
b


=
z − z0
c


• Vector Equation of a Plane
n · PQ



= 0


• Scalar Equation of a Plane
a(x − x0) + b(y − y0) + c(z − z0) = 0


• Distance between a Plane and a Point
d = ‖ projn QP



‖ = |compn QP→ | = |


QP


·n|
‖ n ‖


KEY CONCEPTS
2.1 Vectors in the Plane


• Vectors are used to represent quantities that have both magnitude and direction.
• We can add vectors by using the parallelogram method or the triangle method to find the sum. We can multiply avector by a scalar to change its length or give it the opposite direction.
• Subtraction of vectors is defined in terms of adding the negative of the vector.
• A vector is written in component form as v = 〈 x, y 〉 .
• The magnitude of a vector is a scalar: ‖ v ‖ = x2 + y2.
• A unit vector u has magnitude 1 and can be found by dividing a vector by its magnitude: u = 1


‖ v ‖
v. The


standard unit vectors are i = 〈 1, 0 〉 and j = 〈 0, 1 〉 . A vector v = 〈 x, y 〉 can be expressed in terms of the
standard unit vectors as v = xi + yj.


• Vectors are often used in physics and engineering to represent forces and velocities, among other quantities.
2.2 Vectors in Three Dimensions


• The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a singlepoint, the origin. Ordered triples (x, y, z) are used to describe the location of a point in space.
• The distance d between points (x1, y1, z1) and (x2, y2, z2) is given by the formula


d = (x2 − x1)
2 + (y2 − y1)


2 + (z2 − z1)
2.


• In three dimensions, the equations x = a, y = b, and z = c describe planes that are parallel to the coordinate
planes.


• The standard equation of a sphere with center (a, b, c) and radius r is
(x − a)2 + ⎛⎝y − b⎞⎠2 + (z − c)2 = r2.


• In three dimensions, as in two, vectors are commonly expressed in component form, v = 〈 x, y, z 〉 , or in terms


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of the standard unit vectors, xi + yj + zk.
• Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let


v = 〈 x1, y1, z1 〉 and w = 〈 x2, y2, z2 〉 be vectors, and let k be a scalar.
◦ Scalar multiplication: kv = 〈 kx1, ky1, kz1 〉
◦ Vector addition: v + w = 〈 x1, y1, z1 〉 + 〈 x2, y2, z2 〉 = 〈 x1 + x2, y1 + y2, z1 + z2 〉
◦ Vector subtraction: v − w = 〈 x1, y1, z1 〉 − 〈 x2, y2, z2 〉 = 〈 x1 − x2, y1 − y2, z1 − z2 〉
◦ Vector magnitude: ‖ v ‖ = x1 2 + y1 2 + z1 2
◦ Unit vector in the direction of v: v


‖ v ‖
= 1


‖ v ‖
〈 x1, y1, z1 〉 = 〈


x1
‖ v ‖


,
y1


‖ v ‖
,


z1
‖ v ‖


〉 ,


v ≠ 0


2.3 The Dot Product
• The dot product, or scalar product, of two vectors u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 is


u · v = u1 v1 + u2 v2 + u3 v3.


• The dot product satisfies the following properties:
◦ u · v = v ·u
◦ u · (v + w) = u · v + u ·w
◦ c(u · v) = (cu) · v = u · (cv)
◦ v · v = ‖ v ‖ 2


• The dot product of two vectors can be expressed, alternatively, as u · v = ‖ u ‖ ‖ v ‖ cos θ. This form of the dot
product is useful for finding the measure of the angle formed by two vectors.


• Vectors u and v are orthogonal if u · v = 0.
• The angles formed by a nonzero vector and the coordinate axes are called the direction angles for the vector. Thecosines of these angles are known as the direction cosines.
• The vector projection of v onto u is the vector projuv = u · v


‖ u ‖ 2
u. The magnitude of this vector is known as the


scalar projection of v onto u, given by compuv = u · v‖ u ‖ .
• Work is done when a force is applied to an object, causing displacement. When the force is represented by thevector F and the displacement is represented by the vector s, then the work done W is given by the formula


W = F · s = ‖ F ‖ ‖ s ‖ cos θ.


2.4 The Cross Product
• The cross product u × v of two vectors u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 is a vector orthogonal to
both u and v. Its length is given by ‖ u × v ‖ = ‖ u ‖ · ‖ v ‖ · sin θ, where θ is the angle between u and
v. Its direction is given by the right-hand rule.


• The algebraic formula for calculating the cross product of two vectors,
u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 , is
u × v = (u2 v3 − u3 v2)i − (u1 v3 − u3 v1)j + (u1 v2 − u2 v1)k.


Chapter 2 | Vectors in Space 251




• The cross product satisfies the following properties for vectors u, v, andw, and scalar c:
◦ u × v = −(v × u)
◦ u × (v + w) = u × v + u × w
◦ c(u × v) = (cu) × v = u × (cv)
◦ u × 0 = 0 × u = 0
◦ v × v = 0
◦ u · (v × w) = (u × v) ·w


• The cross product of vectors u = 〈 u1, u2, u3 〉 and v = 〈 v1, v2, v3 〉 is the determinant |i j ku1 u2 u3v1 v2 v3|.
• If vectors u and v form adjacent sides of a parallelogram, then the area of the parallelogram is given by


‖ u × v ‖ .


• The triple scalar product of vectors u, v, and w is u · (v × w).
• The volume of a parallelepiped with adjacent edges given by vectors u, v, andw is V = |u · (v × w)|.
• If the triple scalar product of vectors u, v, andw is zero, then the vectors are coplanar. The converse is also true:
If the vectors are coplanar, then their triple scalar product is zero.


• The cross product can be used to identify a vector orthogonal to two given vectors or to a plane.
• Torque τ measures the tendency of a force to produce rotation about an axis of rotation. If force F is acting at a
distance r from the axis, then torque is equal to the cross product of r and F: τ = r × F.


2.5 Equations of Lines and Planes in Space
• In three dimensions, the direction of a line is described by a direction vector. The vector equation of a linewith direction vector v = 〈 a, b, c 〉 passing through point P = (x0, y0, z0) is r = r0 + tv, where


r0 = 〈 x0, y0, z0 〉 is the position vector of point P. This equation can be rewritten to form the parametric
equations of the line: x = x0 + ta, y = y0 + tb, and z = z0 + tc. The line can also be described with the
symmetric equations x − x0a = y − y0b = z − z0c .


• Let L be a line in space passing through point P with direction vector v. If Q is any point not on L, then the
distance from Q to L is d = ‖ PQ→ × v ‖


‖ v ‖
.


• In three dimensions, two lines may be parallel but not equal, equal, intersecting, or skew.
• Given a point P and vector n, the set of all points Q satisfying equation n · PQ→ = 0 forms a plane. Equation


n · PQ


= 0 is known as the vector equation of a plane.
• The scalar equation of a plane containing point P = (x0, y0, z0) with normal vector n = 〈 a, b, c 〉 is


a(x − x0) + b(y − y0) + c(z − z0) = 0. This equation can be expressed as ax + by + cz + d = 0, where
d = −ax0 − by0 − cz0. This form of the equation is sometimes called the general form of the equation of a plane.


• Suppose a plane with normal vector n passes through point Q. The distance D from the plane to point P not in
the plane is given by


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D = ‖ projn QP


‖ = |compn QP→ | = |
QP


·n|
‖ n ‖


.


• The normal vectors of parallel planes are parallel. When two planes intersect, they form a line.
• The measure of the angle θ between two intersecting planes can be found using the equation:


cos θ = |n1 ·n2|
‖ n1 ‖ ‖ n2 ‖


, where n1 and n2 are normal vectors to the planes.
• The distance D from point (x0, y0, z0) to plane ax + by + cz + d = 0 is given by


D = |a(x0 − x1) + b(y0 − y1) + c(z0 − z1)|
a2 + b2 + c2


= |ax0 + by0 + cz0 + d|
a2 + b2 + c2


.


2.6 Quadric Surfaces
• A set of lines parallel to a given line passing through a given curve is called a cylinder, or a cylindrical surface. Theparallel lines are called rulings.
• The intersection of a three-dimensional surface and a plane is called a trace. To find the trace in the xy-, yz-, orxz-planes, set z = 0, x = 0, or y = 0, respectively.
• Quadric surfaces are three-dimensional surfaces with traces composed of conic sections. Every quadric surface can
be expressed with an equation of the form Ax2 + By2 + Cz2 + Dxy + Exz + Fyz + Gx + Hy + Jz + K = 0.


• To sketch the graph of a quadric surface, start by sketching the traces to understand the framework of the surface.
• Important quadric surfaces are summarized in Figure 2.87 and Figure 2.88.


2.7 Cylindrical and Spherical Coordinates
• In the cylindrical coordinate system, a point in space is represented by the ordered triple (r, θ, z), where (r, θ)
represents the polar coordinates of the point’s projection in the xy-plane and z represents the point’s projection onto
the z-axis.


• To convert a point from cylindrical coordinates to Cartesian coordinates, use equations x = r cos θ, y = r sin θ,
and z = z.


• To convert a point from Cartesian coordinates to cylindrical coordinates, use equations r2 = x2 + y2, tan θ = yx,
and z = z.


• In the spherical coordinate system, a point P in space is represented by the ordered triple ⎛⎝ρ, θ, φ⎞⎠, where ρ is
the distance between P and the origin ⎛⎝ρ ≠ 0⎞⎠, θ is the same angle used to describe the location in cylindrical
coordinates, and φ is the angle formed by the positive z-axis and line segment OP— , where O is the origin and
0 ≤ φ ≤ π.


• To convert a point from spherical coordinates to Cartesian coordinates, use equations x = ρ sin φ cos θ,
y = ρ sin φ sin θ, and z = ρ cos φ.


• To convert a point from Cartesian coordinates to spherical coordinates, use equations ρ2 = x2 + y2 + z2,
tan θ =


y
x, and φ = arccos





⎜ z
x2 + y2 + z2





⎟.


• To convert a point from spherical coordinates to cylindrical coordinates, use equations r = ρ sin φ, θ = θ, and
z = ρ cos φ.


Chapter 2 | Vectors in Space 253




• To convert a point from cylindrical coordinates to spherical coordinates, use equations ρ = r2 + z2, θ = θ,
and φ = arccos⎛



⎜ z


r2 + z2





⎟.


CHAPTER 2 REVIEW EXERCISES
For the following exercises, determine whether thestatement is true or false. Justify the answer with a proof ora counterexample.
423. For vectors a and b and any given scalar c,
c(a ·b) = (ca) ·b.


424. For vectors a and b and any given scalar c,
c(a × b) = (ca) × b.


425. The symmetric equation for the line of intersectionbetween two planes x + y + z = 2 and x + 2y − 4z = 5
is given by − x − 1


6
=


y − 1
5


= z.


426. If a ·b = 0, then a is perpendicular to b.
For the following exercises, use the given vectors to findthe quantities.
427. a = 9i − 2j, b = −3i + j


a. 3a + b
b. |a|
c. a × |b × |a
d. b × |a


428. a = 2i + j − 9k, b = −i + 2k, c = 4i − 2j + k
a. 2a − b
b. |b × c|
c. b × |b × c|
d. c × |b × a|
e. projab


429. Find the values of a such that vectors 〈 2, 4, a 〉
and 〈 0, −1, a 〉 are orthogonal.
For the following exercises, find the unit vectors.
430. Find the unit vector that has the same direction asvector v that begins at (0, −3) and ends at (4, 10).
431. Find the unit vector that has the same direction asvector v that begins at (1, 4, 10) and ends at (3, 0, 4).


For the following exercises, find the area or volume of thegiven shapes.
432. The parallelogram spanned by vectors
a = 〈 1, 13 〉 and b = 〈 3, 21 〉


433. The parallelepiped formed by
a = 〈 1, 4, 1 〉 and b = 〈 3, 6, 2 〉 , and
c = 〈 −2, 1, −5 〉


For the following exercises, find the vector and parametricequations of the line with the given properties.
434. The line that passes through point (2, −3, 7) that is
parallel to vector 〈 1, 3, −2 〉
435. The line that passes through points (1, 3, 5) and
(−2, 6, −3)


For the following exercises, find the equation of the planewith the given properties.
436. The plane that passes through point (4, 7, −1) and
has normal vector n = 〈 3, 4, 2 〉
437. The plane that passes through points
(0, 1, 5), (2, −1, 6), and (3, 2, 5).


For the following exercises, find the traces for the surfacesin planes x = k, y = k, and z = k. Then, describe and
draw the surfaces.
438. 9x2 + 4y2 − 16y + 36z2 = 20


439. x2 = y2 + z2


For the following exercises, write the given equation incylindrical coordinates and spherical coordinates.
440. x2 + y2 + z2 = 144


441. z = x2 + y2 − 1
For the following exercises, convert the given equationsfrom cylindrical or spherical coordinates to rectangular


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coordinates. Identify the given surface.
442. ρ2 ⎛⎝sin2 (φ) − cos2 (φ)⎞⎠ = 1


443. r2 − 2r cos(θ) + z2 = 1
For the following exercises, consider a small boat crossinga river.
444. If the boat velocity is 5 km/h due north in still water
and the water has a current of 2 km/h due west (see the
following figure), what is the velocity of the boat relativeto shore? What is the angle θ that the boat is actually
traveling?


445. When the boat reaches the shore, two ropes arethrown to people to help pull the boat ashore. One rope is atan angle of 25° and the other is at 35°. If the boat must be
pulled straight and at a force of 500N, find the magnitude
of force for each rope (see the following figure).


446. An airplane is flying in the direction of 52° east ofnorth with a speed of 450 mph. A strong wind has a bearing33° east of north with a speed of 50 mph. What is theresultant ground speed and bearing of the airplane?
447. Calculate the work done by moving a particle fromposition (1, 2, 0) to (8, 4, 5) along a straight line with a
force F = 2i + 3j − k.
The following problems consider your unsuccessfulattempt to take the tire off your car using a wrench to loosenthe bolts. Assume the wrench is 0.3 m long and you are
able to apply a 200-N force.
448. Because your tire is flat, you are only able to applyyour force at a 60° angle. What is the torque at the center
of the bolt? Assume this force is not enough to loosen thebolt.
449. Someone lends you a tire jack and you are now ableto apply a 200-N force at an 80° angle. Is your resulting
torque going to be more or less? What is the new resultingtorque at the center of the bolt? Assume this force is notenough to loosen the bolt.


Chapter 2 | Vectors in Space 255




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3 | VECTOR-VALUEDFUNCTIONS


Figure 3.1 Halley’s Comet appeared in view of Earth in 1986 and will appear again in 2061.


Chapter Outline
3.1 Vector-Valued Functions and Space Curves
3.2 Calculus of Vector-Valued Functions
3.3 Arc Length and Curvature
3.4 Motion in Space


Introduction
In 1705, using Sir Isaac Newton’s new laws of motion, the astronomer Edmond Halley made a prediction. He stated thatcomets that had appeared in 1531, 1607, and 1682 were actually the same comet and that it would reappear in 1758. Halleywas proved to be correct, although he did not live to see it. However, the comet was later named in his honor.
Halley’s Comet follows an elliptical path through the solar system, with the Sun appearing at one focus of the ellipse. Thismotion is predicted by Johannes Kepler’s first law of planetary motion, which we mentioned briefly in the Introductionto Parametric Equations and Polar Coordinates. In Example 3.15, we show how to use Kepler’s third law ofplanetary motion along with the calculus of vector-valued functions to find the average distance of Halley’s Comet from theSun.
Vector-valued functions provide a useful method for studying various curves both in the plane and in three-dimensionalspace. We can apply this concept to calculate the velocity, acceleration, arc length, and curvature of an object’s trajectory.In this chapter, we examine these methods and show how they are used.


Chapter 3 | Vector-Valued Functions 257




3.1 | Vector-Valued Functions and Space Curves
Learning Objectives


3.1.1 Write the general equation of a vector-valued function in component form and unit-vectorform.
3.1.2 Recognize parametric equations for a space curve.
3.1.3 Describe the shape of a helix and write its equation.
3.1.4 Define the limit of a vector-valued function.


Our study of vector-valued functions combines ideas from our earlier examination of single-variable calculus with ourdescription of vectors in three dimensions from the preceding chapter. In this section we extend concepts from earlierchapters and also examine new ideas concerning curves in three-dimensional space. These definitions and theorems supportthe presentation of material in the rest of this chapter and also in the remaining chapters of the text.
Definition of a Vector-Valued Function
Our first step in studying the calculus of vector-valued functions is to define what exactly a vector-valued function is. Wecan then look at graphs of vector-valued functions and see how they define curves in both two and three dimensions.
Definition
A vector-valued function is a function of the form


(3.1)r(t) = f (t) i + g(t) j or r(t) = f (t) i + g(t) j + h(t)k,
where the component functions f, g, and h, are real-valued functions of the parameter t. Vector-valued functions arealso written in the form


(3.2)r(t) = 〈 f (t), g(t) 〉 or r(t) = 〈 f (t), g(t), h(t) 〉 .
In both cases, the first form of the function defines a two-dimensional vector-valued function; the second formdescribes a three-dimensional vector-valued function.


The parameter t can lie between two real numbers: a ≤ t ≤ b. Another possibility is that the value of t might take on all
real numbers. Last, the component functions themselves may have domain restrictions that enforce restrictions on the valueof t.We often use t as a parameter because t can represent time.
Example 3.1
Evaluating Vector-Valued Functions and Determining Domains
For each of the following vector-valued functions, evaluate r(0), r⎛⎝π2⎞⎠, and r⎛⎝2π3 ⎞⎠. Do any of these functions
have domain restrictions?


a. r(t) = 4cos t i + 3sin t j
b. r(t) = 3tan t i + 4sec t j + 5tk


Solution
a. To calculate each of the function values, substitute the appropriate value of t into the function:


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3.1


r(0) = 4cos(0) i + 3sin(0) j


= 4 i + 0 j = 4 i


r⎛⎝
π
2

⎠ = 4cos




π
2

⎠i + 3sin




π
2

⎠ j


= 0 i + 3 j = 3 j


r⎛⎝

3

⎠ = 4cos





3

⎠i + 3sin





3

⎠ j


= 4⎛⎝−
1
2

⎠i + 3



3
2

⎠j = −2 i +


3 3
2


j.


To determine whether this function has any domain restrictions, consider the component functionsseparately. The first component function is f (t) = 4cos t and the second component function is
g(t) = 3sin t. Neither of these functions has a domain restriction, so the domain of
r(t) = 4cos t i + 3sin t j is all real numbers.


b. To calculate each of the function values, substitute the appropriate value of t into the function:
r(0) = 3tan(0) i + 4sec(0) j + 5(0)k


= 0 i + 4 j + 0k = 4 j


r⎛⎝
π
2

⎠ = 3tan




π
2

⎠i + 4sec




π
2

⎠ j + 5




π
2

⎠k, which does not exist


r⎛⎝

3

⎠ = 3tan





3

⎠i + 4sec





3

⎠ j + 5





3

⎠k


= 3⎛⎝− 3

⎠i + 4(−2) j + 10π3


k


= −3 3 i − 8 j + 10π
3


k.


To determine whether this function has any domain restrictions, consider the component functionsseparately. The first component function is f (t) = 3tan t, the second component function is
g(t) = 4sec t, and the third component function is h(t) = 5t. The first two functions are not defined
for odd multiples of π/2, so the function is not defined for odd multiples of π/2. Therefore,
dom(r(t)) =





⎨t |t ≠ (2n + 1)π2 ⎫⎭⎬, where n is any integer.


For the vector-valued function r(t) = ⎛⎝t2 − 3t⎞⎠i + (4t + 1) j, evaluate r(0), r(1), and r(−4). Does this
function have any domain restrictions?


Example 3.1 illustrates an important concept. The domain of a vector-valued function consists of real numbers. Thedomain can be all real numbers or a subset of the real numbers. The range of a vector-valued function consists of vectors.Each real number in the domain of a vector-valued function is mapped to either a two- or a three-dimensional vector.
Graphing Vector-Valued Functions
Recall that a plane vector consists of two quantities: direction and magnitude. Given any point in the plane (the initial point),if we move in a specific direction for a specific distance, we arrive at a second point. This represents the terminal point ofthe vector. We calculate the components of the vector by subtracting the coordinates of the initial point from the coordinates


Chapter 3 | Vector-Valued Functions 259




of the terminal point.
A vector is considered to be in standard position if the initial point is located at the origin. When graphing a vector-valuedfunction, we typically graph the vectors in the domain of the function in standard position, because doing so guarantees theuniqueness of the graph. This convention applies to the graphs of three-dimensional vector-valued functions as well. Thegraph of a vector-valued function of the form r(t) = f (t) i + g(t) j consists of the set of all (t, r(t)), and the path it traces
is called a plane curve. The graph of a vector-valued function of the form r(t) = f (t) i + g(t) j + h(t)k consists of the set
of all (t, r(t)), and the path it traces is called a space curve. Any representation of a plane curve or space curve using a
vector-valued function is called a vector parameterization of the curve.
Example 3.2
Graphing a Vector-Valued Function
Create a graph of each of the following vector-valued functions:


a. The plane curve represented by r(t) = 4cos t i + 3sin t j, 0 ≤ t ≤ 2π
b. The plane curve represented by r(t) = 4cos t3 i + 3sin t3 j, 0 ≤ t ≤ 2π
c. The space curve represented by r(t) = cos t i + sin t j + tk, 0 ≤ t ≤ 4π


Solution
a. As with any graph, we start with a table of values. We then graph each of the vectors in the second columnof the table in standard position and connect the terminal points of each vector to form a curve (Figure3.2). This curve turns out to be an ellipse centered at the origin.


t r(t) t r(t)


0 4 i π −4 i


π
4 2 2 i +


3 2
2


j 5π4 −2 2 i −
3 2
2


j


π
2 3 j



2


−3 j



4 −2 2 i +


3 2
2


j 7π4 2 2 i −
3 2
2


j


2π 4 i


Table 3.1Table of Values for r(t) = 4cos t i + 3sin t j, 0 ≤ t ≤ 2π


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Figure 3.2 The graph of the first vector-valued function is anellipse.
b. The table of values for r(t) = 4cos t i + 3sin t j, 0 ≤ t ≤ 2π is as follows:


t r(t) t r(t)


0 4 i π −4 i


π
4 2 2 i +


3 2
2


j 5π4 −2 2 i −
3 2
2


j


π
2 3 j



2


−3 j



4 −2 2 i +


3 2
2


j 7π4 2 2 i −
3 2
2


j


2π 4 i


Table 3.2Table of Values for r(t) = 4cos t i + 3sin t j, 0 ≤ t ≤ 2π


The graph of this curve is also an ellipse centered at the origin.


Chapter 3 | Vector-Valued Functions 261




Figure 3.3 The graph of the second vector-valued function isalso an ellipse.
c. We go through the same procedure for a three-dimensional vector function.


t r(t) t r(t)


0 4 i π −4 j + πk


π
4


2 2 i + 2 2 j + π
4
k 5π


4
−2 2 i − 2 2 j + 5π


4
k


π
2


4 j + π
2
k 3π


2
−4 j + 3π


2
k



4


−2 2 i + 2 2 j + 3π
4
k 7π


4
2 2 i − 2 2 j + 7π


4
k


2π 4 i + 2πk


Table 3.3Table of Values for r(t) = cos t i + sin t j + tk, 0 ≤ t ≤ 4π


The values then repeat themselves, except for the fact that the coefficient of k is always increasing(Figure 3.4). This curve is called a helix. Notice that if the k component is eliminated, then the functionbecomes r(t) = cos t i + sin t j, which is a unit circle centered at the origin.


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3.2


Figure 3.4 The graph of the third vector-valued function is ahelix.


You may notice that the graphs in parts a. and b. are identical. This happens because the function describing curveb is a so-called reparameterization of the function describing curve a. In fact, any curve has an infinite number ofreparameterizations; for example, we can replace t with 2t in any of the three previous curves without changing the shape
of the curve. The interval over which t is defined may change, but that is all. We return to this idea later in this chapter whenwe study arc-length parameterization.
As mentioned, the name of the shape of the curve of the graph in Example 3.2c. is a helix (Figure 3.4). The curveresembles a spring, with a circular cross-section looking down along the z-axis. It is possible for a helix to be elliptical incross-section as well. For example, the vector-valued function r(t) = 4cos t i + 3sin t j + tk describes an elliptical helix.
The projection of this helix into the x, y-plane is an ellipse. Last, the arrows in the graph of this helix indicate the
orientation of the curve as t progresses from 0 to 4π.


Create a graph of the vector-valued function r(t) = ⎛⎝t2 − 1⎞⎠i + (2t − 3) j, 0 ≤ t ≤ 3.


At this point, you may notice a similarity between vector-valued functions and parameterized curves. Indeed, given a vector-valued function r(t) = f (t) i + g(t) j, we can define x = f (t) and y = g(t). If a restriction exists on the values of t (for
example, t is restricted to the interval ⎡⎣a, b⎤⎦ for some constants a < b), then this restriction is enforced on the parameter.
The graph of the parameterized function would then agree with the graph of the vector-valued function, except that thevector-valued graph would represent vectors rather than points. Since we can parameterize a curve defined by a function
y = f (x), it is also possible to represent an arbitrary plane curve by a vector-valued function.
Limits and Continuity of a Vector-Valued Function
We now take a look at the limit of a vector-valued function. This is important to understand to study the calculus of vector-valued functions.


Chapter 3 | Vector-Valued Functions 263




Definition
A vector-valued function r approaches the limit L as t approaches a, written


lim
t → a


r(t) = L,


provided
lim
t → a|r(t) − L| = 0.


This is a rigorous definition of the limit of a vector-valued function. In practice, we use the following theorem:
Theorem 3.1: Limit of a Vector-Valued Function
Let f, g, and h be functions of t. Then the limit of the vector-valued function r(t) = f (t) i + g(t) j as t approaches a is
given by


(3.3)lim
t → a


r(t) = ⎡⎣ limt → a f (t)

⎦i +

⎣ limt → ag(t)



⎦ j,


provided the limits lim
t → a


f (t) and lim
t → a


g(t) exist. Similarly, the limit of the vector-valued function
r(t) = f (t) i + g(t) j + h(t)k as t approaches a is given by


(3.4)lim
t → a


r(t) = ⎡⎣ limt → a f (t)

⎦i +

⎣ limt → ag(t)



⎦ j +

⎣ limt → ah(t)



⎦k,


provided the limits lim
t → a


f (t), lim
t → a


g(t)and lim
t → a


h(t) exist.


In the following example, we show how to calculate the limit of a vector-valued function.
Example 3.3
Evaluating the Limit of a Vector-Valued Function
For each of the following vector-valued functions, calculate lim


t → 3
r(t) for


a. r(t) = ⎛⎝t2 − 3t + 4⎞⎠i + (4t + 3) j
b. r(t) = 2t − 4


t + 1
i + t


t2 + 1
j + (4t − 3)k


Solution
a. Use Equation 3.3 and substitute the value t = 3 into the two component expressions:


lim
t → 3


r(t) = lim
t → 3





⎝t
2 − 3t + 4⎞⎠i + (4t + 3) j





= ⎡⎣ limt → 3

⎝t
2 − 3t + 4⎞⎠



⎦i +

⎣ limt → 3


(4t + 3)⎤⎦ j


= 4 i + 15 j.


b. Use Equation 3.4 and substitute the value t = 3 into the three component expressions:


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3.3


lim
t → 3


r(t) = lim
t → 3




2t − 4
t + 1


i + t
t2 + 1


j + (4t − 3)k



= ⎡⎣ limt → 3


2t − 4
t + 1



⎦i +

⎣ limt → 3



t
t2 + 1





⎦j +

⎣ limt → 3


(4t − 3)⎤⎦k


= 1
2
i + 3


10
j + 9k.


Calculate lim
t → −2


r(t) for the function r(t) = t2 − 3t − 1 i + (4t + 3) j + sin (t + 1)π
2


k.


Now that we know how to calculate the limit of a vector-valued function, we can define continuity at a point for such afunction.
Definition
Let f, g, and h be functions of t. Then, the vector-valued function r(t) = f (t) i + g(t) j is continuous at point t = a if
the following three conditions hold:


1. r(a) exists
2. lim


t → a
r(t) exists


3. lim
t → a


r(t) = r(a)


Similarly, the vector-valued function r(t) = f (t) i + g(t) j + h(t)k is continuous at point t = a if the following three
conditions hold:


1. r(a) exists
2. lim


t → a
r(t) exists


3. lim
t → a


r(t) = r(a)


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3.1 EXERCISES
1. Give the component functions x = f (t) and y = g(t)
for the vector-valued function r(t) = 3sec t i + 2tan t j.
2. Given r(t) = 3sec t i + 2tan t j, find the following
values (if possible).


a. r⎛⎝π4⎞⎠
b. r(π)
c. r⎛⎝π2⎞⎠


3. Sketch the curve of the vector-valued function
r(t) = 3sec t i + 2tan t j and give the orientation of the
curve. Sketch asymptotes as a guide to the graph.
4. Evaluate lim


t → 0
〈 et i + sin tt j + e


−t k 〉 .


5. Given the vector-valued function
r(t) = 〈 cos t, sin t 〉 , find the following values:


a. lim
t → π


4


r(t)


b. r⎛⎝π3⎞⎠
c. Is r(t) continuous at t = π


3
?


d. Graph r(t).
6. Given the vector-valued function
r(t) = 〈 t, t2 + 1 〉 , find the following values:


a. lim
t → −3


r(t)


b. r(−3)
c. Is r(t) continuous at x = −3?
d. r(t + 2) − r(t)


7. Let r(t) = et i + sin t j + ln tk. Find the following
values:


a. r⎛⎝π4⎞⎠
b. lim


t → π/4
r(t)


c. Is r(t) continuous at t = t = π
4
?


Find the limit of the following vector-valued functions atthe indicated value of t.
8. lim


t → 4
〈 t − 3, t − 2


t − 4
, tan⎛⎝


π
t

⎠ 〉


9. lim
t → π/2


r(t) for r(t) = et i + sin t j + ln tk


10. lim
t → ∞


〈 e−2t, 2t + 3
3t − 1


, arctan(2t) 〉


11. lim
t → e2


〈 t ln(t), ln t
t2


, ln ⎛⎝t
2⎞
⎠ 〉


12. lim
t → π/6


〈 cos2 t, sin2 t, 1 〉


13. lim
t → ∞


r(t) for r(t) = 2e−t i + e−t j + ln(t − 1)k
14. Describe the curve defined by the vector-valuedfunction r(t) = (1 + t) i + (2 + 5t) j + (−1 + 6t)k.
Find the domain of the vector-valued functions.
15. Domain: r(t) = 〈 t2, tan t, ln t 〉
16. Domain: r(t) = 〈 t2, t − 3, 3


2t + 1


17. Domain: r(t) = 〈 csc(t), 1
t − 3


, ln(t − 2) 〉


Let r(t) = 〈 cos t, t, sin t 〉 and use it to answer the
following questions.
18. For what values of t is r(t) continuous?
19. Sketch the graph of r(t).
20. Find the domain of
r(t) = 2e−t i + e−t j + ln(t − 1)k.


21. For what values of t is
r(t) = 2e−t i + e−t j + ln(t − 1)k continuous?
Eliminate the parameter t, write the equation in Cartesiancoordinates, then sketch the graphs of the vector-valued
functions. (Hint: Let x = 2t and y = t2. Solve the first
equation for x in terms of t and substitute this result into thesecond equation.)
22. r(t) = 2t i + t2 j
23. r(t) = t3 i + 2t j
24. r(t) = 2(sinh t) i + 2(cosh t) j, t > 0
25. r(t) = 3(cos t) i + 3(sin t) j
26. r(t) = 〈 3sin t, 3cos t 〉


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Use a graphing utility to sketch each of the followingvector-valued functions:
27. [T] r(t) = 2cos t2 i + (2 − t) j
28. [T] r(t) = 〈 ecos(3t), e−sin(t) 〉
29. [T] r(t) = 〈 2 − sin(2t), 3 + 2cos t 〉
30. 4x2 + 9y2 = 36; clockwise and counterclockwise
31. r(t) = 〈 t, t2 〉 ; from left to right
32. The line through P and Q where P is (1, 4, −2) and
Q is (3, 9, 6)
Consider the curve described by the vector-valued function
r(t) = ⎛⎝50e


−t cos t⎞⎠i +

⎝50e


−t sin t⎞⎠ j + (5 − 5e
−t)k.


33. What is the initial point of the path corresponding to
r(0)?


34. What is lim
t → ∞


r(t)?


35. [T] Use technology to sketch the curve.
36. Eliminate the parameter t to show that z = 5 − r


10


where r = x2 + y2.
37. [T] Let r(t) = cos t i + sin t j + 0.3sin(2t)k. Use
technology to graph the curve (called the roller-coastercurve) over the interval [0, 2π). Choose at least two views
to determine the peaks and valleys.
38. [T] Use the result of the preceding problem toconstruct an equation of a roller coaster with a steep dropfrom the peak and steep incline from the “valley.” Then, usetechnology to graph the equation.
39. Use the results of the preceding two problems toconstruct an equation of a path of a roller coaster with morethan two turning points (peaks and valleys).
40. a. Graph the curve


r(t) = (4 + cos(18t))cos(t) i + ⎛⎝4 + cos(18t)sin(t)⎞⎠ j + 0.3sin(18t)k


using two viewing angles of your choice to see theoverall shape of the curve.b. Does the curve resemble a “slinky”?c. What changes to the equation should be made toincrease the number of coils of the slinky?


Chapter 3 | Vector-Valued Functions 267




3.2 | Calculus of Vector-Valued Functions
Learning Objectives


3.2.1 Write an expression for the derivative of a vector-valued function.
3.2.2 Find the tangent vector at a point for a given position vector.
3.2.3 Find the unit tangent vector at a point for a given position vector and explain itssignificance.
3.2.4 Calculate the definite integral of a vector-valued function.


To study the calculus of vector-valued functions, we follow a similar path to the one we took in studying real-valuedfunctions. First, we define the derivative, then we examine applications of the derivative, then we move on to definingintegrals. However, we will find some interesting new ideas along the way as a result of the vector nature of these functionsand the properties of space curves.
Derivatives of Vector-Valued Functions
Now that we have seen what a vector-valued function is and how to take its limit, the next step is to learn how to differentiatea vector-valued function. The definition of the derivative of a vector-valued function is nearly identical to the definition ofa real-valued function of one variable. However, because the range of a vector-valued function consists of vectors, the sameis true for the range of the derivative of a vector-valued function.
Definition
The derivative of a vector-valued function r(t) is


(3.5)r′(t) = lim
Δt → 0


r(t + Δt) − r(t)
Δt


,


provided the limit exists. If r′(t) exists, then r is differentiable at t. If r′(t) exists for all t in an open interval (a, b),
then r is differentiable over the interval (a, b). For the function to be differentiable over the closed interval ⎡⎣a, b⎤⎦,
the following two limits must exist as well:


r′(a) = lim
Δt → 0+


r(a + Δt) − r(a)
Δt


and r′(b) = lim
Δt → 0−


r(b + Δt) − r(b)
Δt


.


Many of the rules for calculating derivatives of real-valued functions can be applied to calculating the derivatives of vector-valued functions as well. Recall that the derivative of a real-valued function can be interpreted as the slope of a tangent lineor the instantaneous rate of change of the function. The derivative of a vector-valued function can be understood to be aninstantaneous rate of change as well; for example, when the function represents the position of an object at a given point intime, the derivative represents its velocity at that same point in time.
We now demonstrate taking the derivative of a vector-valued function.
Example 3.4
Finding the Derivative of a Vector-Valued Function
Use the definition to calculate the derivative of the function


r(t) = (3t + 4) i + ⎛⎝t
2 − 4t + 3⎞⎠ j.


Solution
Let’s use Equation 3.5:


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3.4


r′(t) = lim
Δt → 0


r(t + Δt) − r(t)
Δt


= lim
Δt → 0



⎣(3(t + Δt) + 4) i +



⎝(t + Δt)


2 − 4(t + Δt) + 3⎞⎠ j

⎦−

⎣(3t + 4) i +



⎝t
2 − 4t + 3⎞⎠ j





Δt


= lim
Δt → 0


(3t + 3Δt + 4) i − (3t + 4) i + ⎛⎝t
2 + 2tΔt + (Δt)2 − 4t − 4Δt + 3⎞⎠ j −



⎝t
2 − 4t + 3⎞⎠ j


Δt


= lim
Δt → 0


(3Δt) i + ⎛⎝2tΔt + (Δt)
2 − 4Δt⎞⎠ j


Δt
= lim


Δt → 0

⎝3 i + (2t + Δt − 4) j⎞⎠


= 3 i + (2t − 4) j.


Use the definition to calculate the derivative of the function r(t) = ⎛⎝2t2 + 3⎞⎠i + (5t − 6) j.


Notice that in the calculations in Example 3.4, we could also obtain the answer by first calculating the derivative of eachcomponent function, then putting these derivatives back into the vector-valued function. This is always true for calculatingthe derivative of a vector-valued function, whether it is in two or three dimensions. We state this in the following theorem.The proof of this theorem follows directly from the definitions of the limit of a vector-valued function and the derivative ofa vector-valued function.
Theorem 3.2: Differentiation of Vector-Valued Functions
Let f, g, and h be differentiable functions of t.


i. If r(t) = f (t) i + g(t) j, then r′(t) = f ′ (t) i + g′ (t) j.
ii. If r(t) = f (t) i + g(t) j + h(t)k, then r′(t) = f ′ (t) i + g′ (t) j + h′ (t)k.


Example 3.5
Calculating the Derivative of Vector-Valued Functions
Use Differentiation of Vector-Valued Functions to calculate the derivative of each of the followingfunctions.


a. r(t) = (6t + 8) i + ⎛⎝4t2 + 2t − 3⎞⎠ j
b. r(t) = 3cos t i + 4sin t j
c. r(t) = et sin t i + et cos t j − e2t k


Solution
We use Differentiation of Vector-Valued Functions and what we know about differentiating functions ofone variable.


Chapter 3 | Vector-Valued Functions 269




3.5


a. The first component of r(t) = (6t + 8) i + ⎛⎝4t2 + 2t − 3⎞⎠ j is f (t) = 6t + 8. The second component
is g(t) = 4t2 + 2t − 3. We have f ′ (t) = 6 and g′ (t) = 8t + 2, so the theorem gives
r′(t) = 6 i + (8t + 2) j.


b. The first component is f (t) = 3cos t and the second component is g(t) = 4sin t. We have
f ′ (t) = −3sin t and g′ (t) = 4cos t, so we obtain r′(t) = −3sin t i + 4cos t j.


c. The first component of r(t) = et sin t i + et cos t j − e2t k is f (t) = et sin t, the second component
is g(t) = et cos t, and the third component is h(t) = − e2t. We have f ′ (t) = et (sin t + cos t),
g′ (t) = et (cos t − sin t), and h′ (t) = −2e2t, so the theorem gives
r′(t) = et (sin t + cos t) i + et (cos t − sin t) j − 2e2t k.


Calculate the derivative of the function
r(t) = (t ln t) i + ⎛⎝5et



⎠ j + (cos t − sin t)k.


We can extend to vector-valued functions the properties of the derivative that we presented in the Introduction toDerivatives (http://cnx.org/content/m53494/latest/) . In particular, the constant multiple rule, the sum and differencerules, the product rule, and the chain rule all extend to vector-valued functions. However, in the case of the product rule,there are actually three extensions: (1) for a real-valued function multiplied by a vector-valued function, (2) for the dotproduct of two vector-valued functions, and (3) for the cross product of two vector-valued functions.
Theorem 3.3: Properties of the Derivative of Vector-Valued Functions
Let r and u be differentiable vector-valued functions of t, let f be a differentiable real-valued function of t, and let c bea scalar.


i. d
dt

⎣cr(t)⎤⎦ = cr′(t) Scalar multiple


ii. d
dt

⎣r(t) ± u(t)⎤⎦ = r′(t) ± u′(t) Sum and diffe ence


iii. d
dt

⎣ f (t)u(t)⎤⎦ = f ′ (t)u(t) + f (t)u′(t) Scalar product


iv. d
dt

⎣r(t) · u(t)⎤⎦ = r′(t) · u(t) + r(t) · u′(t) Dot product


v. d
dt

⎣r(t) × u(t)⎤⎦ = r′(t) × u(t) + r(t) × u′(t) Cross product


vi. d
dt

⎣r⎛⎝ f (t)⎞⎠⎤⎦ = r′⎛⎝ f (t)⎞⎠ · f ′ (t) Chain rule


vii. If r(t) · r(t) = c, then r(t) · r′(t) = 0.


Proof
The proofs of the first two properties follow directly from the definition of the derivative of a vector-valued function.The third property can be derived from the first two properties, along with the product rule from the Introduction toDerivatives (http://cnx.org/content/m53494/latest/) . Let u(t) = g(t) i + h(t) j. Then


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d
dt

⎣ f (t)u(t)⎤⎦ = ddt



⎣ f (t)⎛⎝g(t) i + h(t) j⎞⎠⎤⎦


= d
dt

⎣ f (t)g(t) i + f (t)h(t) j⎤⎦


= d
dt

⎣ f (t)g(t)⎤⎦i + ddt



⎣ f (t)h(t)⎤⎦ j


= ⎛⎝ f ′ (t)g(t) + f (t)g′ (t)⎞⎠i + ⎛⎝ f ′ (t)h(t) + f (t)h′ (t)⎞⎠ j
= f ′ (t)u(t) + f (t)u′(t).


To prove property iv. let r(t) = f1 (t) i + g1 (t) j and u(t) = f2 (t) i + g2 (t) j. Then
d
dt

⎣r(t) · u(t)⎤⎦ = d


dt

⎣ f1 (t) f2 (t) + g1 (t)g2 (t)





= f1 ′ (t) f2 (t) + f1 (t) f2 ′ (t) + g1 ′ (t)g2 (t) + g1 (t)g2 ′ (t)


= f1 ′ (t) f2 (t) + g1 ′ (t)g2 (t) + f1 (t) f2 ′ (t) + g1 (t)g2 ′ (t)


= ⎛⎝ f1 ′ i + g1 ′ j

⎠ · ⎛⎝ f2 i + g2 j



⎠+ ⎛⎝ f1 i + g1 j



⎠ · ⎛⎝ f2 ′ i + g2 ′ j





= r′(t) · u(t) + r(t) · u′(t).


The proof of property v. is similar to that of property iv. Property vi. can be proved using the chain rule. Last, property vii.follows from property iv:
d
dt

⎣r(t) · r(t)⎤⎦ = d


dt
[c]


r′(t) · r(t) + r(t) · r′(t) = 0
2r(t) · r′(t) = 0
r(t) · r′(t) = 0.



Now for some examples using these properties.
Example 3.6
Using the Properties of Derivatives of Vector-Valued Functions
Given the vector-valued functions


r(t) = (6t + 8) i + ⎛⎝4t
2 + 2t − 3⎞⎠ j + 5tk


and
u(t) = ⎛⎝t


2 − 3⎞⎠i + (2t + 4) j +

⎝t
3 − 3t⎞⎠k,


calculate each of the following derivatives using the properties of the derivative of vector-valued functions.
a. d


dt

⎣r(t) · u(t)⎤⎦


b. d
dt

⎣u(t) × u′(t)⎤⎦


Solution
a. We have r′(t) = 6 i + (8t + 2) j + 5k and u′(t) = 2t i + 2 j + ⎛⎝3t2 − 3⎞⎠k. Therefore, according to


property iv.:


Chapter 3 | Vector-Valued Functions 271




3.6


d
dt

⎣r(t) · u(t)⎤⎦ = r′(t) · u(t) + r(t) · u′(t)


= ⎛⎝6 i + (8t + 2) j + 5k⎞⎠ · ⎛⎝

⎝t
2 − 3⎞⎠i + (2t + 4) j +



⎝t
3 − 3t⎞⎠k





+⎛⎝(6t + 8) i +

⎝4t


2 + 2t − 3⎞⎠ j + 5tk

⎠ ·

⎝2t i + 2 j +



⎝3t


2 − 3⎞⎠k



= 6⎛⎝t
2 − 3⎞⎠+ (8t + 2)(2t + 4) + 5



⎝t
3 − 3t⎞⎠


+2t(6t + 8) + 2⎛⎝4t
2 + 2t − 3⎞⎠+ 5t



⎝3t


2 − 3⎞⎠


= 20t3 + 42t2 + 26t − 16.
b. First, we need to adapt property v. for this problem:


d
dt

⎣u(t) × u′(t)⎤⎦ = u′(t) × u′(t) + u(t) × u″(t).


Recall that the cross product of any vector with itself is zero. Furthermore, u″(t) represents the second
derivative of u(t):


u″(t) = d
dt

⎣u′(t)⎤⎦ = ddt



⎣2t i + 2 j +



⎝3t


2 − 3⎞⎠k

⎦ = 2 i + 6tk.


Therefore,
d
dt

⎣u(t) × u′(t)⎤⎦ = 0 + ⎛⎝



⎝t
2 − 3⎞⎠i + (2t + 4) j +



⎝t
3 − 3t⎞⎠k



⎠× (2 i + 6tk)


= | i j kt2 − 3 2t + 4 t3 − 3t2 0 6t |
= 6t(2t + 4) i − ⎛⎝6t



⎝t
2 − 3⎞⎠− 2



⎝t
3 − 3t⎞⎠



⎠ j − 2(2t + 4)k


= ⎛⎝12t
2 + 24t⎞⎠i +



⎝12t − 4t


3⎞
⎠ j − (4t + 8)k.


Given the vector-valued functions r(t) = cos t i + sin t j − e2t k and u(t) = t i + sin t j + cos tk,
calculate d


dt

⎣r(t) · r′(t)⎤⎦ and ddt ⎡⎣u(t) × r(t)⎤⎦.


Tangent Vectors and Unit Tangent Vectors
Recall from the Introduction to Derivatives (http://cnx.org/content/m53494/latest/) that the derivative at a pointcan be interpreted as the slope of the tangent line to the graph at that point. In the case of a vector-valued function,the derivative provides a tangent vector to the curve represented by the function. Consider the vector-valued function
r(t) = cos t i + sin t j. The derivative of this function is r′(t) = − sin t i + cos t j. If we substitute the value t = π/6 into
both functions we get


r⎛⎝
π
6

⎠ =


3
2
i + 1


2
j and r′⎛⎝


π
6

⎠ = −


1
2
i + 3


2
j.


The graph of this function appears in Figure 3.5, along with the vectors r⎛⎝π6⎞⎠ and r′⎛⎝π6⎞⎠.


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Figure 3.5 The tangent line at a point is calculated from thederivative of the vector-valued function r(t).


Notice that the vector r′⎛⎝π6⎞⎠ is tangent to the circle at the point corresponding to t = π/6. This is an example of a tangent
vector to the plane curve defined by r(t) = cos t i + sin t j.


Definition
Let C be a curve defined by a vector-valued function r, and assume that r′(t) exists when t = t0. A tangent vector v
at t = t0 is any vector such that, when the tail of the vector is placed at point r(t0) on the graph, vector v is tangent
to curve C. Vector r′(t0) is an example of a tangent vector at point t = t0. Furthermore, assume that r′(t) ≠ 0. The
principal unit tangent vector at t is defined to be


(3.6)T(t) = r′(t)
‖r′(t)‖


,


provided ‖r′(t)‖ ≠ 0.


The unit tangent vector is exactly what it sounds like: a unit vector that is tangent to the curve. To calculate a unit tangentvector, first find the derivative r′(t). Second, calculate the magnitude of the derivative. The third step is to divide the
derivative by its magnitude.
Example 3.7
Finding a Unit Tangent Vector
Find the unit tangent vector for each of the following vector-valued functions:


a. r(t) = cos t i + sin t j
b. u(t) = ⎛⎝3t2 + 2t⎞⎠i + ⎛⎝2 − 4t3⎞⎠ j + (6t + 5)k


Solution
a.


Chapter 3 | Vector-Valued Functions 273




3.7


First step: r′(t) = −sin t i + cos t j


Second step: ‖r′(t)‖ = (−sin t)2 + (cos t)2 = 1


Third step: T(t) = r′(t)
‖r′(t)‖


=
−sin t i + cos t j


1
= −sin t i + cos t j


b.
First step: u′(t) = (6t + 2) i − 12t2 j + 6k


Second step: ‖u′(t)‖ = (6t + 2)2 + ⎛⎝−12t
2⎞


2
+ 62


= 144t4 + 36t2 + 24t + 40


= 2 36t4 + 9t2 + 6t + 10


Third step: T(t) = u′(t)
‖u′(t)‖


=
(6t + 2) i − 12t2 j + 6k


2 36t4 + 9t2 + 6t + 10


= 3t + 1


36t4 + 9t2 + 6t + 10
i − 6t


2


36t4 + 9t2 + 6t + 10
j + 3


36t4 + 9t2 + 6t + 10
k


Find the unit tangent vector for the vector-valued function
r(t) = ⎛⎝t


2 − 3⎞⎠i + (2t + 1) j + (t − 2)k.


Integrals of Vector-Valued Functions
We introduced antiderivatives of real-valued functions in Antiderivatives (http://cnx.org/content/m53621/latest/)and definite integrals of real-valued functions in The Definite Integral (http://cnx.org/content/m53631/latest/) .Each of these concepts can be extended to vector-valued functions. Also, just as we can calculate the derivative of a vector-valued function by differentiating the component functions separately, we can calculate the antiderivative in the samemanner. Furthermore, the Fundamental Theorem of Calculus applies to vector-valued functions as well.
The antiderivative of a vector-valued function appears in applications. For example, if a vector-valued function representsthe velocity of an object at time t, then its antiderivative represents position. Or, if the function represents the accelerationof the object at a given time, then the antiderivative represents its velocity.
Definition
Let f, g, and h be integrable real-valued functions over the closed interval ⎡⎣a, b⎤⎦.


1. The indefinite integral of a vector-valued function r(t) = f (t) i + g(t) j is
(3.7)∫ ⎡⎣ f (t) i + g(t) j⎤⎦dt = ⎡⎣∫ f (t)dt⎤⎦i + ⎡⎣∫ g(t)dt⎤⎦ j.


The definite integral of a vector-valued function is
(3.8)



a


b

⎣ f (t) i + g(t) j⎤⎦dt =





⎢∫


a


b
f (t)dt



⎥i +



⎢∫


a


b
g(t)dt





⎥ j.


2. The indefinite integral of a vector-valued function r(t) = f (t) i + g(t) j + h(t)k is
(3.9)∫ ⎡⎣ f (t) i + g(t) j + h(t)k⎤⎦dt = ⎡⎣∫ f (t)dt⎤⎦i + ⎡⎣∫ g(t)dt⎤⎦ j + ⎡⎣∫ h(t)dt⎤⎦k.


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The definite integral of the vector-valued function is
(3.10)



a


b

⎣ f (t) i + g(t) j + h(t)k⎤⎦dt =





⎢∫


a


b
f (t)dt



⎥i +



⎢∫


a


b
g(t)dt





⎥ j +



⎢∫


a


b
h(t)dt





⎥k.


Since the indefinite integral of a vector-valued function involves indefinite integrals of the component functions, each ofthese component integrals contains an integration constant. They can all be different. For example, in the two-dimensionalcase, we can have
∫ f (t)dt = F(t) + C1 and ∫ g(t)dt = G(t) + C2,


where F and G are antiderivatives of f and g, respectively. Then
∫ ⎡⎣ f (t) i + g(t) j⎤⎦dt = ⎡⎣∫ f (t)dt



⎦i +

⎣∫ g(t)dt



⎦ j


= ⎛⎝F(t) + C1

⎠i + ⎛⎝G(t) + C2



⎠ j


= F(t) i + G(t) j + C1 i + C2 j


= F(t) i + G(t) j + C,


where C = C1 i + C2 j. Therefore, the integration constant becomes a constant vector.
Example 3.8
Integrating Vector-Valued Functions
Calculate each of the following integrals:


a. ∫ ⎡⎣⎛⎝3t2 + 2t⎞⎠i + (3t − 6) j + ⎛⎝6t3 + 5t2 − 4⎞⎠k⎤⎦dt
b. ∫ ⎡⎣ 〈 t, t2, t3 〉 × 〈 t3, t2, t 〉 ⎤⎦dt
c. ∫


0


π/3

⎣sin2t i + tan t j + e


−2t k⎤⎦dt


Solution
a. We use the first part of the definition of the integral of a space curve:


∫ ⎡⎣⎛⎝3t2 + 2t⎞⎠i + (3t − 6) j + ⎛⎝6t3 + 5t2 − 4⎞⎠k⎤⎦dt
= ⎡⎣∫ 3t2 + 2t dt



⎦i +

⎣∫ 3t − 6dt



⎦ j +

⎣∫ 6t3 + 5t2 − 4dt



⎦k


= ⎛⎝t
3 + t2⎞⎠i +




3
2
t2 − 6t⎞⎠ j +




3
2
t4 + 5


3
t3 − 4t⎞⎠k + C.


b. First calculate 〈 t, t2, t3 〉 × 〈 t3, t2, t 〉 :


〈 t, t2, t3 〉 × 〈 t3, t2, t 〉 = | i j kt t2 t3t3 t2 t |
= ⎛⎝t


2 (t) − t3 ⎛⎝t
2⎞


⎠i −

⎝t
2 − t3 ⎛⎝t


3⎞


⎠ j +



⎝t

⎝t
2⎞
⎠− t


2 ⎛
⎝t
3⎞


⎠k


= ⎛⎝t
3 − t5⎞⎠i +



⎝t
6 − t2⎞⎠ j +



⎝t
3 − t5⎞⎠k.


Chapter 3 | Vector-Valued Functions 275




3.8


Next, substitute this back into the integral and integrate:
∫ ⎡⎣ 〈 t, t2, t3 〉 × 〈 t3, t2, t 〉 ⎤⎦dt = ∫ ⎛⎝t3 − t5⎞⎠i + ⎛⎝t6 − t2⎞⎠ j + ⎛⎝t3 − t5⎞⎠kdt


=


t4
4


− t
6


6

⎠i +


t7
7


− t
3


3

⎠j +


t4
4


− t
6


6

⎠k + C.


c. Use the second part of the definition of the integral of a space curve:


0


π/3

⎣sin2t i + tan t j + e


−2t k⎤⎦dt


=



⎢∫


0


π/3
sin2tdt





⎥i +



⎢∫


0


π/3
tan tdt





⎥ j +



⎢∫


0


π/3
e−2tdt





⎥k


= ⎛⎝−
1
2
cos2t⎞⎠|0


π/3
i − ⎛⎝ln(cos t)⎞⎠|0


π/3 j − ⎛⎝
1
2
e−2t⎞⎠|0


π/3
k


= ⎛⎝−
1
2
cos 2π


3
+ 1


2
cos0⎞⎠i −



⎝ln

⎝cos


π
3

⎠− ln(cos0)



⎠ j −


1
2
e−2π/3 − 1


2
e−2


(0)⎞
⎠k


= ⎛⎝
1
4
+ 1


2

⎠i − (−ln2) j −




1
2
e−2π/3 − 1


2

⎠k


= 3
4
i + (ln2) j + ⎛⎝


1
2
− 1


2
e−2π/3⎞⎠k.


Calculate the following integral:


1


3

⎣(2t + 4) i +



⎝3t


2 − 4t⎞⎠ j

⎦dt.


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3.2 EXERCISES
Compute the derivatives of the vector-valued functions.
41. r(t) = t3 i + 3t2 j + t3


6
k


42. r(t) = sin(t) i + cos(t) j + et k
43. r(t) = e−t i + sin(3t) j + 10 tk. A sketch of the
graph is shown here. Notice the varying periodic nature ofthe graph.


44. r(t) = et i + 2et j + k
45. r(t) = i + j + k
46. r(t) = tet i + t ln(t) j + sin(3t)k
47. r(t) = 1


t + 1
i + arctan(t) j + ln t3 k


48. r(t) = tan(2t) i + sec(2t) j + sin2(t)k
49. r(t) = 3 i + 4sin(3t) j + tcos(t)k
50. r(t) = t2 i + te−2t j − 5e−4t k
For the following problems, find a tangent vector at theindicated value of t.


51. r(t) = t i + sin(2t) j + cos(3t)k; t = π
3


52. r(t) = 3t3 i + 2t2 j + 1t k; t = 1
53. r(t) = 3et i + 2e−3t j + 4e2t k; t = ln(2)
54. r(t) = cos(2t) i + 2sin t j + t2 k; t = π


2


Find the unit tangent vector for the following parameterizedcurves.
55. r(t) = 6 i + cos(3t) j + 3sin(4t)k, 0 ≤ t < 2π
56. r(t) = cos t i + sin t j + sin tk, 0 ≤ t < 2π. Two
views of this curve are presented here:


57. r(t) = 3cos(4t) i + 3sin(4t) j + 5tk, 1 ≤ t ≤ 2
58. r(t) = t i + 3t j + t2 k
Let r(t) = t i + t2 j − t4 k and
s(t) = sin(t) i + et j + cos(t)k. Here is the graph of the
function:


Chapter 3 | Vector-Valued Functions 277




Find the following.
59. d


dt

⎣r

⎝t
2⎞




60. d
dt

⎣t
2 · s(t)⎤⎦


61. d
dt

⎣r(t) · s(t)⎤⎦


62. Compute the first, second, and third derivatives of
r(t) = 3t i + 6ln(t) j + 5e−3t k.


63. Find r′(t) · r″(t) for r(t) = −3t5 i + 5t j + 2t2 k.
64. The acceleration function, initial velocity, and initialposition of a particle are
a(t) = −5cos t i − 5sin t j, v(0) = 9 i + 2 j, and r(0) = 5 i.


Find v(t) and r(t).
65. The position vector of a particle is
r(t) = 5sec(2t) i − 4tan(t) j + 7t2 k.


a. Graph the position function and display a view ofthe graph that illustrates the asymptotic behavior ofthe function.b. Find the velocity as t approaches but is not equal to
π/4 (if it exists).


66. Find the velocity and the speed of a particle with the
position function r(t) = ⎛⎝2t − 12t + 1⎞⎠i + ln(1 − 4t2) j. The
speed of a particle is the magnitude of the velocity and is
represented by ‖r'(t)‖.
A particle moves on a circular path of radius b according tothe function r(t) = bcos(ωt) i + bsin(ωt) j, where ω is


the angular velocity, dθ/dt.


67. Find the velocity function and show that v(t) is
always orthogonal to r(t).
68. Show that the speed of the particle is proportional tothe angular velocity.
69. Evaluate d


dt

⎣u(t) × u′(t)⎤⎦ given


u(t) = t2 i − 2t j + k.


70. Find the antiderivative of
r '(t) = cos(2t) i − 2sin t j + 1


1 + t2
k that satisfies the


initial condition r(0) = 3 i − 2 j + k.


71. Evaluate ∫
0


3
‖t i + t2 j‖dt.


72. An object starts from rest at point P(1, 2, 0) and
moves with an acceleration of a(t) = j + 2k, where
‖a(t)‖ is measured in feet per second per second. Find the
location of the object after t = 2 sec.
73. Show that if the speed of a particle traveling along acurve represented by a vector-valued function is constant,then the velocity function is always perpendicular to theacceleration function.
74. Given r(t) = t i + 3t j + t2 k and
u(t) = 4t i + t2 j + t3 k, find d


dt
(r(t) × u(t)).


75. Given r(t) = 〈 t + cos t, t − sin t 〉 , find the
velocity and the speed at any time.
76. Find the velocity vector for the function
r(t) = 〈 et, e−t, 0 〉 .


77. Find the equation of the tangent line to the curve
r(t) = 〈 et, e−t, 0 〉 at t = 0.


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78. Describe and sketch the curve represented by the
vector-valued function r(t) = 〈 6t, 6t − t2 〉 .
79. Locate the highest point on the curve
r(t) = 〈 6t, 6t − t2 〉 and give the value of the function
at this point.
The position vector for a particle is r(t) = t i + t2 j + t3 k.
The graph is shown here:


80. Find the velocity vector at any time.
81. Find the speed of the particle at time t = 2 sec.
82. Find the acceleration at time t = 2 sec.
A particle travels along the path of a helix with the equation
r(t) = cos(t) i + sin(t) j + tk. See the graph presented
here:


Find the following:
83. Velocity of the particle at any time
84. Speed of the particle at any time
85. Acceleration of the particle at any time
86. Find the unit tangent vector for the helix.
A particle travels along the path of an ellipse with theequation r(t) = cos t i + 2sin t j + 0k. Find the following:
87. Velocity of the particle
88. Speed of the particle at t = π


4


89. Acceleration of the particle at t = π
4


Given the vector-valued function
r(t) = 〈 tan t, sec t, 0 〉 (graph is shown here), find the
following:


Chapter 3 | Vector-Valued Functions 279




90. Velocity
91. Speed
92. Acceleration
93. Find the minimum speed of a particle traveling alongthe curve r(t) = 〈 t + cos t, t − sin t 〉 t ∈ [0, 2π).
Given r(t) = t i + 2sin t j + 2cos tk and
u(t) = 1t i + 2sin t j + 2cos tk, find the following:
94. r(t) × u(t)
95. d


dt
(r(t) × u(t))


96. Now, use the product rule for the derivative of thecross product of two vectors and show this result is thesame as the answer for the preceding problem.
Find the unit tangent vector T(t) for the following vector-valued functions.


97. r(t) = 〈 t, 1t 〉 . The graph is shown here:


98. r(t) = 〈 tcos t, t sin t 〉
99. r(t) = 〈 t + 1, 2t + 1, 2t + 2 〉
Evaluate the following integrals:
100. ∫ ⎛⎝et i + sin t j + 12t − 1k⎞⎠dt


101. ∫
0


1
r(t)dt, where r(t) = 〈 t3 , 1


t + 1
, e−t 〉


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3.3 | Arc Length and Curvature
Learning Objectives


3.3.1 Determine the length of a particle’s path in space by using the arc-length function.
3.3.2 Explain the meaning of the curvature of a curve in space and state its formula.
3.3.3 Describe the meaning of the normal and binormal vectors of a curve in space.


In this section, we study formulas related to curves in both two and three dimensions, and see how they are related to variousproperties of the same curve. For example, suppose a vector-valued function describes the motion of a particle in space.We would like to determine how far the particle has traveled over a given time interval, which can be described by the arclength of the path it follows. Or, suppose that the vector-valued function describes a road we are building and we want todetermine how sharply the road curves at a given point. This is described by the curvature of the function at that point. Weexplore each of these concepts in this section.
Arc Length for Vector Functions
We have seen how a vector-valued function describes a curve in either two or three dimensions. Recall AlternativeFormulas for Curvature, which states that the formula for the arc length of a curve defined by the parametric functions
x = x(t), y = y(t), t1 ≤ t ≤ t2 is given by


s = ∫
t1


t2
(x′ (t))2 + ⎛⎝y′ (t)⎞⎠2dt.


In a similar fashion, if we define a smooth curve using a vector-valued function r(t) = f (t) i + g(t) j, where a ≤ t ≤ b,
the arc length is given by the formula


s = ∫
a


b

⎝ f ′ (t)⎞⎠2 + ⎛⎝g′ (t)⎞⎠2dt.


In three dimensions, if the vector-valued function is described by r(t) = f (t) i + g(t) j + h(t)k over the same interval
a ≤ t ≤ b, the arc length is given by


s = ∫
a


b

⎝ f ′ (t)⎞⎠2 + ⎛⎝g′ (t)⎞⎠2 + ⎛⎝h′ (t)⎞⎠2dt.


Theorem 3.4: Arc-Length Formulas
i. Plane curve: Given a smooth curve C defined by the function r(t) = f (t) i + g(t) j, where t lies within the


interval ⎡⎣a, b⎤⎦, the arc length of C over the interval is
(3.11)


s = ∫
a


b

⎣ f ′ (t)⎤⎦2 + ⎡⎣g′ (t)⎤⎦2dt = ∫


a


b
‖ r′(t) ‖ dt.


ii. Space curve: Given a smooth curve C defined by the function r(t) = f (t) i + g(t) j + h(t) k, where t lies
within the interval ⎡⎣a, b⎤⎦, the arc length of C over the interval is


(3.12)
s = ∫


a


b

⎣ f ′ (t)⎤⎦2 + ⎡⎣g′ (t)⎤⎦2 + ⎡⎣h′ (t)⎤⎦2dt = ∫


a


b
‖ r′(t) ‖ dt.


The two formulas are very similar; they differ only in the fact that a space curve has three component functions instead oftwo. Note that the formulas are defined for smooth curves: curves where the vector-valued function r(t) is differentiable
with a non-zero derivative. The smoothness condition guarantees that the curve has no cusps (or corners) that could makethe formula problematic.


Chapter 3 | Vector-Valued Functions 281




3.9


Example 3.9
Finding the Arc Length
Calculate the arc length for each of the following vector-valued functions:


a. r(t) = (3t − 2) i + (4t + 5) j, 1 ≤ t ≤ 5
b. r(t) = 〈 tcos t, t sin t, 2t 〉 , 0 ≤ t ≤ 2π


Solution
a. Using Equation 3.11, r′(t) = 3i + 4j, so


s = ∫
a


b
‖ r′(t) ‖ dt


= ∫
a


5
32 + 42dt


= ∫
1


5
5 dt = 5t|1


5 = 20.


b. Using Equation 3.12, r′(t) = 〈 cos t − t sin t, sin t + tcos t, 2 〉 , so
s = ∫


a


b
‖ r′(t) ‖ dt


= ∫
0



(cos t − t sin t)2 + (sin t + tcos t)2 + 22dt


= ∫
0




⎝cos


2 t − 2t sin tcos t + t2 sin2 t⎞⎠+

⎝sin


2 t + 2t sin tcos t + t2 cos2 t⎞⎠+ 4 dt


= ∫
0



cos2 t + sin2 t + t2 ⎛⎝cos


2 t + sin2 t⎞⎠+ 4 dt


= ∫
0



t2 + 5 dt.


Here we can use a table integration formula
∫ u2 + a2 du = u


2
u2 + a2 + a


2


2
ln |u + u2 + a2| + C,


so we obtain


0



t2 + 5 dt = 1


2

⎝t t


2 + 5 + 5ln|t + t2 + 5|⎞⎠0


= 1
2

⎝2π 4π


2 + 5 + 5ln⎛⎝2π + 4π
2 + 5⎞⎠



⎠−


5
2
ln 5


≈ 25.343.


Calculate the arc length of the parameterized curve
r(t) = 〈 2t2 + 1, 2t2 − 1, t3 〉 , 0 ≤ t ≤ 3.


We now return to the helix introduced earlier in this chapter. A vector-valued function that describes a helix can be writtenin the form


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r(t) = Rcos⎛⎝
2πNt
h

⎠i + Rsin




2πNt
h

⎠ j + t k, 0 ≤ t ≤ h,


where R represents the radius of the helix, h represents the height (distance between two consecutive turns), and the helixcompletes N turns. Let’s derive a formula for the arc length of this helix using Equation 3.12. First of all,
r′(t) = − 2πNR


h
sin⎛⎝


2πNt
h

⎠i +


2πNR
h


cos⎛⎝
2πNt
h

⎠ j + k.


Therefore,
s = ∫


a


b
‖ r′(t) ‖ dt


= ∫
0


h

⎝−


2πNR
h


sin⎛⎝
2πNt
h





2
+ ⎛⎝


2πNR
h


cos⎛⎝
2πNt
h





2
+ 12dt


= ∫
0


h
4π2N 2R2


h2

⎝sin


2 ⎛

2πNt
h

⎠+ cos


2 ⎛

2πNt
h



⎠+ 1dt


= ∫
0


h
4π2N 2R2


h2
+ 1dt


=



⎢ t 4π


2N 2R2


h2
+ 1



⎥ 0


h


= h 4π
2N 2R2 + h2


h2


= 4π2N 2R2 + h2.


This gives a formula for the length of a wire needed to form a helix with N turns that has radius R and height h.
Arc-Length Parameterization
We now have a formula for the arc length of a curve defined by a vector-valued function. Let’s take this one step furtherand examine what an arc-length function is.
If a vector-valued function represents the position of a particle in space as a function of time, then the arc-length functionmeasures how far that particle travels as a function of time. The formula for the arc-length function follows directly fromthe formula for arc length:


(3.13)
s(t) = ∫


a


t

⎝ f ′ (u)⎞⎠2 + ⎛⎝g′ (u)⎞⎠2 + ⎛⎝h′ (u)⎞⎠2du.


If the curve is in two dimensions, then only two terms appear under the square root inside the integral. The reason for usingthe independent variable u is to distinguish between time and the variable of integration. Since s(t) measures distance
traveled as a function of time, s′ (t) measures the speed of the particle at any given time. Since we have a formula for s(t)
in Equation 3.13, we can differentiate both sides of the equation:


s′ (t) = d
dt

⎣∫a


t

⎝ f ′ (u)⎞⎠2 + ⎛⎝g′ (u)⎞⎠2 + ⎛⎝h′ (u)⎞⎠2du





= d
dt

⎣∫a


t
‖ r′(u) ‖ du





= ‖ r′(t) ‖ .


If we assume that r(t) defines a smooth curve, then the arc length is always increasing, so s′ (t) > 0 for t > a. Last, if
r(t) is a curve on which ‖ r′(t) ‖ = 1 for all t, then


s(t) = ∫
a


t
‖ r′(u) ‖ du = ∫


a


t
1 du = t − a,


which means that t represents the arc length as long as a = 0.


Chapter 3 | Vector-Valued Functions 283




Theorem 3.5: Arc-Length Function
Let r(t) describe a smooth curve for t ≥ a. Then the arc-length function is given by


(3.14)
s(t) = ∫


a


t
‖ r′(u) ‖ du.


Furthermore, ds
dt


= ‖ r′(t) ‖ > 0. If ‖ r′(t) ‖ = 1 for all t ≥ a, then the parameter t represents the arc length
from the starting point at t = a.


A useful application of this theorem is to find an alternative parameterization of a given curve, called an arc-lengthparameterization. Recall that any vector-valued function can be reparameterized via a change of variables. For example,if we have a function r(t) = 〈 3cos t, 3sin t 〉 , 0 ≤ t ≤ 2π that parameterizes a circle of radius 3, we can change the
parameter from t to 4t, obtaining a new parameterization r(t) = 〈 3cos4t, 3sin4t 〉 . The new parameterization still
defines a circle of radius 3, but now we need only use the values 0 ≤ t ≤ π/2 to traverse the circle once.
Suppose that we find the arc-length function s(t) and are able to solve this function for t as a function of s. We can then
reparameterize the original function r(t) by substituting the expression for t back into r(t). The vector-valued function
is now written in terms of the parameter s. Since the variable s represents the arc length, we call this an arc-lengthparameterization of the original function r(t). One advantage of finding the arc-length parameterization is that the distance
traveled along the curve starting from s = 0 is now equal to the parameter s. The arc-length parameterization also appears
in the context of curvature (which we examine later in this section) and line integrals, which we study in the Introductionto Vector Calculus.


Example 3.10
Finding an Arc-Length Parameterization
Find the arc-length parameterization for each of the following curves:


a. r(t) = 4cos t i + 4sin t j, t ≥ 0
b. r(t) = 〈 t + 3, 2t − 4, 2t 〉 , t ≥ 3


Solution
a. First we find the arc-length function using Equation 3.14:


s(t) = ∫
a


t
‖ r′(u) ‖ du


= ∫
0


t
‖ 〈 −4sinu, 4cosu 〉 ‖ du


= ∫
0


t
(−4sinu)2 + (4cosu)2 du


= ∫
0


t
16sin2u + 16cos2u du


= ∫
0


t
4 du = 4t,


which gives the relationship between the arc length s and the parameter t as s = 4t; so, t = s/4. Next we


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3.10


replace the variable t in the original function r(t) = 4cos t i + 4sin t j with the expression s/4 to obtain
r(s) = 4cos⎛⎝


s
4

⎠i + 4sin




s
4

⎠ j.


This is the arc-length parameterization of r(t). Since the original restriction on t was given by t ≥ 0,
the restriction on s becomes s/4 ≥ 0, or s ≥ 0.


b. The arc-length function is given by Equation 3.14:
s(t) = ∫


a


t
‖ r′(u) ‖ du


= ∫
3


t
‖ 〈 1, 2, 2 〉 ‖ du


= ∫
3


t
12 + 22 + 22 du


= ∫
3


t
3 du


= 3t − 9.


Therefore, the relationship between the arc length s and the parameter t is s = 3t − 9, so t = s
3
+ 3.


Substituting this into the original function r(t) = 〈 t + 3, 2t − 4, 2t 〉 yields
r(s) = 〈 ⎛⎝


s
3
+ 3⎞⎠+ 3, 2




s
3
+ 3⎞⎠− 4, 2




s
3
+ 3⎞⎠ 〉 = 〈


s
3
+ 6, 2s


3
+ 2, 2s


3
+ 6 〉 .


This is an arc-length parameterization of r(t). The original restriction on the parameter t was t ≥ 3,
so the restriction on s is (s/3) + 3 ≥ 3, or s ≥ 0.


Find the arc-length function for the helix
r(t) = 〈 3cos t, 3sin t, 4t 〉 , t ≥ 0.


Then, use the relationship between the arc length and the parameter t to find an arc-length parameterization of
r(t).


Curvature
An important topic related to arc length is curvature. The concept of curvature provides a way to measure how sharply asmooth curve turns. A circle has constant curvature. The smaller the radius of the circle, the greater the curvature.
Think of driving down a road. Suppose the road lies on an arc of a large circle. In this case you would barely have to turnthe wheel to stay on the road. Now suppose the radius is smaller. In this case you would need to turn more sharply to stayon the road. In the case of a curve other than a circle, it is often useful first to inscribe a circle to the curve at a given pointso that it is tangent to the curve at that point and “hugs” the curve as closely as possible in a neighborhood of the point(Figure 3.6). The curvature of the graph at that point is then defined to be the same as the curvature of the inscribed circle.


Chapter 3 | Vector-Valued Functions 285




Figure 3.6 The graph represents the curvature of a function
y = f (x). The sharper the turn in the graph, the greater the
curvature, and the smaller the radius of the inscribed circle.


Definition
Let C be a smooth curve in the plane or in space given by r(s), where s is the arc-length parameter. The curvature
κ at s is


κ = ‖ d T
ds


‖ = ‖ T′(s) ‖ .


Visit this website (http://www.openstaxcollege.org/l/20_spacecurve) for more information about thecurvature of a space curve.


The formula in the definition of curvature is not very useful in terms of calculation. In particular, recall that T(t) represents
the unit tangent vector to a given vector-valued function r(t), and the formula for T(t) is T(t) = r′(t)


‖ r′(t) ‖
. To use the


formula for curvature, it is first necessary to express r(t) in terms of the arc-length parameter s, then find the unit tangent
vector T(s) for the function r(s), then take the derivative of T(s) with respect to s. This is a tedious process. Fortunately,
there are equivalent formulas for curvature.
Theorem 3.6: Alternative Formulas for Curvature
If C is a smooth curve given by r(t), then the curvature κ of C at t is given by


(3.15)
κ = ‖ T′(t) ‖


‖ r′(t) ‖
.


If C is a three-dimensional curve, then the curvature can be given by the formula
(3.16)


κ = ‖ r′(t) × r″(t) ‖
‖ r′(t) ‖ 3


.


If C is the graph of a function y = f (x) and both y′ and y″ exist, then the curvature κ at point (x, y) is given by
(3.17)


κ = |y″|

⎣1 +



⎝y′⎞⎠2⎤⎦


3/2
.


Proof
The first formula follows directly from the chain rule:


dT
dt


= dT
ds


ds
dt


,


where s is the arc length along the curve C. Dividing both sides by ds/dt, and taking the magnitude of both sides gives


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‖ dT
ds


‖ = ‖ T′(t)
ds
dt


‖ .


Since ds/dt = ‖ r′(t) ‖ , this gives the formula for the curvature κ of a curve C in terms of any parameterization of C:
κ = ‖ T′(t) ‖


‖ r′(t) ‖
.


In the case of a three-dimensional curve, we start with the formulas T(t) = ⎛⎝r′(t)⎞⎠/ ‖ r′(t) ‖ and ds/dt = ‖ r′(t) ‖ .
Therefore, r′(t) = (ds/dt)T(t). We can take the derivative of this function using the scalar product formula:


r″(t) = d
2 s
dt2


T(t) + ds
dt
T′(t).


Using these last two equations we get
r′(t) × r″(t) = ds


dt
T(t) ×


d2 s
dt2


T(t) + ds
dt
T′(t)



= ds
dt


d2 s
dt2


T(t) × T(t) + ⎛⎝
ds
dt



2
T(t) × T′(t).


Since T(t) × T(t) = 0, this reduces to
r′(t) × r″(t) = ⎛⎝


ds
dt



2
T(t) × T′(t).


Since T′ is parallel to N, and T is orthogonal to N, it follows that T and T′ are orthogonal. This means that
‖ T × T′ ‖ = ‖ T ‖ ‖ T′ ‖ sin(π/2) = ‖ T′ ‖ , so


r′(t) × r″(t) = ⎛⎝
ds
dt



2
‖ T′(t) ‖ .


Now we solve this equation for ‖ T′(t) ‖ and use the fact that ds/dt = ‖ r′(t) ‖ :
‖ T′(t) ‖ = ‖ r′(t) × r″(t) ‖


‖ r′(t) ‖ 2
.


Then, we divide both sides by ‖ r′(t) ‖ . This gives
κ = ‖ T′(t) ‖


‖ r′(t) ‖
= ‖ r′(t) × r″(t) ‖


‖ r′(t) ‖ 3
.


This proves Equation 3.16. To prove Equation 3.17, we start with the assumption that curve C is defined by the function
y = f (x). Then, we can define r(t) = x i + f (x) j + 0 k. Using the previous formula for curvature:


r′(t) = i + f ′ (x) j
r″(t) = f ″(x) j


r′(t) × r″(t) = | i j k1 f ′ (x) 00 f ″(x) 0| = f ″(x) k.
Therefore,


κ = ‖ r′(t) × r″(t) ‖
‖ r′(t) ‖ 3


= | f ″(x)|

⎝1 + ⎡⎣ f ′ (x)⎤⎦⎞⎠3/2


.




Chapter 3 | Vector-Valued Functions 287




Example 3.11
Finding Curvature
Find the curvature for each of the following curves at the given point:


a. r(t) = 4cos t i + 4sin t j + 3t k, t = 4π
3


b. f (x) = 4x − x2, x = 2
Solution


a. This function describes a helix.


The curvature of the helix at t = (4π)/3 can be found by using Equation 3.15. First, calculate T(t):
T(t) = r′(t)


‖ r′(t) ‖


=
〈 −4sin t, 4cos t, 3 〉


(−4sin t)2 + (4cos t)2 + 32


= 〈 −4
5
sin t, 4


5
cos t, 3


5
〉 .


Next, calculate T′(t):
T′(t) = 〈 −4


5
cos t, − 4


5
sin t, 0 〉 .


Last, apply Equation 3.15:


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κ = ‖ T′(t) ‖
‖ r′(t) ‖


=
‖ 〈 − 4


5
cos t, − 4


5
sin t, 0 〉 ‖


‖ 〈 −4sin t, 4cos t, 3 〉 ‖


=

⎝−


4
5
cos t⎞⎠


2
+ ⎛⎝−


4
5
sin t⎞⎠


2
+ 02


(−4sin t)2 + (4cos t)2 + 32


= 4/5
5


= 4
25


.


The curvature of this helix is constant at all points on the helix.
b. This function describes a semicircle.


To find the curvature of this graph, we must use Equation 3.16. First, we calculate y′ and y″:
y = 4x − x2 = ⎛⎝4x − x


2⎞


1/2


y′ = 1
2

⎝4x − x


2⎞


−1/2
(4 − 2x) = (2 − x)⎛⎝4x − x


2⎞


−1/2


y″ = − ⎛⎝4x − x
2⎞


−1/2
+ (2 − x)⎛⎝−


1
2



⎝4x − x


2⎞


−3/2
(4 − 2x)


= − 4x − x
2



⎝4x − x


2⎞


3/2
− (2 − x)


2



⎝4x − x


2⎞


3/2


=
x2 − 4x − ⎛⎝4 − 4x + x


2⎞



⎝4x − x


2⎞


3/2


= − 4

⎝4x − x


2⎞


3/2
.


Then, we apply Equation 3.17:


Chapter 3 | Vector-Valued Functions 289




3.11


κ = |y″|

⎣1 +



⎝y′⎞⎠2⎤⎦


3/2


= |− 4⎛⎝4x − x2⎞⎠3/2 |



⎢1 + ⎛⎝(2 − x)



⎝4x − x


2⎞


−1/2⎞


2⎤




3/2


= | 4⎛⎝4x − x2⎞⎠3/2 |

⎣1 +


(2 − x)2


4x − x2



3/2


= | 4⎛⎝4x − x2⎞⎠3/2 |


4x − x2 + x2 − 4x + 4


4x − x2



3/2
= | 4⎛⎝4x − x2⎞⎠3/2 | · ⎛⎝4x − x2⎞⎠


3/2


8


= 1
2
.


The curvature of this circle is equal to the reciprocal of its radius. There is a minor issue with the absolutevalue in Equation 3.16; however, a closer look at the calculation reveals that the denominator is positivefor any value of x.


Find the curvature of the curve defined by the function
y = 3x2 − 2x + 4


at the point x = 2.


The Normal and Binormal Vectors
We have seen that the derivative r′(t) of a vector-valued function is a tangent vector to the curve defined by r(t), and the
unit tangent vector T(t) can be calculated by dividing r′(t) by its magnitude. When studying motion in three dimensions,
two other vectors are useful in describing the motion of a particle along a path in space: the principal unit normal vectorand the binormal vector.
Definition
Let C be a three-dimensional smooth curve represented by r over an open interval I. If T′(t) ≠ 0, then the principal
unit normal vector at t is defined to be


(3.18)N(t) = T′(t)
‖ T′(t) ‖


.


The binormal vector at t is defined as
(3.19)B(t) = T(t) × N(t),


where T(t) is the unit tangent vector.


Note that, by definition, the binormal vector is orthogonal to both the unit tangent vector and the normal vector.Furthermore, B(t) is always a unit vector. This can be shown using the formula for the magnitude of a cross product


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‖ B(t) ‖ = ‖ T(t) × N(t) ‖ = ‖ T(t) ‖ ‖ N(t) ‖ sinθ,


where θ is the angle between T(t) and N(t). Since N(t) is the derivative of a unit vector, property (vii) of the derivative
of a vector-valued function tells us that T(t) and N(t) are orthogonal to each other, so θ = π/2. Furthermore, they are
both unit vectors, so their magnitude is 1. Therefore, ‖ T(t) ‖ ‖ N(t) ‖ sinθ = (1)(1)sin(π/2) = 1 and B(t) is a unit
vector.
The principal unit normal vector can be challenging to calculate because the unit tangent vector involves a quotient, andthis quotient often has a square root in the denominator. In the three-dimensional case, finding the cross product of the unittangent vector and the unit normal vector can be even more cumbersome. Fortunately, we have alternative formulas forfinding these two vectors, and they are presented in Motion in Space.
Example 3.12
Finding the Principal Unit Normal Vector and Binormal Vector
For each of the following vector-valued functions, find the principal unit normal vector. Then, if possible, findthe binormal vector.


a. r(t) = 4cos t i − 4sin t j
b. r(t) = (6t + 2) i + 5t2 j − 8t k


Solution
a. This function describes a circle.


To find the principal unit normal vector, we first must find the unit tangent vector T(t):


Chapter 3 | Vector-Valued Functions 291




T(t) = r′(t)
‖ r′(t) ‖


=
−4sin t i − 4cos t j


(−4sin t)2 + (−4cos t)2


=
−4sin t i − 4cos t j


16sin2 t + 16cos2 t


=
−4sin t i − 4cos t j


16⎛⎝sin
2 t + cos2 t⎞⎠


=
−4sin t i − 4cos t j


4
= − sin t i − cos t j.


Next, we use Equation 3.18:
N(t) = T′(t)


‖ T′(t) ‖


=
−cos t i + sin t j


(−cos t)2 + (sin t)2


=
−cos t i + sin t j


cos2 t + sin2 t
= − cos t i + sin t j.


Notice that the unit tangent vector and the principal unit normal vector are orthogonal to each other forall values of t:
T(t) ·N(t) = 〈 −sin t, − cos t 〉 · 〈 −cos t, sin t 〉


= sin tcos t − cos t sin t
= 0.


Furthermore, the principal unit normal vector points toward the center of the circle from every point onthe circle. Since r(t) defines a curve in two dimensions, we cannot calculate the binormal vector.


b. This function looks like this:


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To find the principal unit normal vector, we first find the unit tangent vector T(t):
T(t) = r′(t)


‖ r′(t) ‖


=
6 i + 10t j − 8 k


62 + (10t)2 + (−8)2


=
6 i + 10t j − 8 k


36 + 100t2 + 64


=
6 i + 10t j − 8k


100⎛⎝t
2 + 1⎞⎠


=
3 i − 5t j − 4k


5 t2 + 1


= 3
5

⎝t
2 + 1⎞⎠


−1/2
i − t⎛⎝t


2 + 1⎞⎠
−1/2


j − 4
5

⎝t
2 + 1⎞⎠


−1/2
k.


Next, we calculate T′(t) and ‖ T′(t) ‖ :


Chapter 3 | Vector-Valued Functions 293




T′(t) = 3
5

⎝−


1
2



⎝t
2 + 1⎞⎠


−3/2
(2t) i −





⎝t
2 + 1⎞⎠


−1/2
− t⎛⎝


1
2



⎝t
2 + 1⎞⎠


−3/2
(2t)

⎠j


−4
5

⎝−


1
2



⎝t
2 + 1⎞⎠


−3/2
(2t)k


= − 3t


5⎛⎝t
2 + 1⎞⎠


3/2
i − 1

⎝t
2 + 1⎞⎠


3/2
j + 4t


5⎛⎝t
2 + 1⎞⎠


3/2
k


‖ T′(t) ‖ =







− 3t


5⎛⎝t
2 + 1⎞⎠


3/2







2


+







− 1

⎝t
2 + 1⎞⎠


3/2







2


+






⎜ 4t


5⎛⎝t
2 + 1⎞⎠


3/2







2


= 9t
2


25⎛⎝t
2 + 1⎞⎠


3
+ 1

⎝t
2 + 1⎞⎠


3
+ 16t


2


25⎛⎝t
2 + 1⎞⎠


3


= 25t
2 + 25


25⎛⎝t
2 + 1⎞⎠


3


= 1

⎝t
2 + 1⎞⎠


2


= 1
t2 + 1


.


Therefore, according to Equation 3.18:
N(t) = T′(t)


‖ T′(t) ‖


=







− 3t


5⎛⎝t
2 + 1⎞⎠


3/2
i − 1

⎝t
2 + 1⎞⎠


3/2
j + 4t


5⎛⎝t
2 + 1⎞⎠


3/2
k






⎟⎛
⎝t
2 + 1⎞⎠


= − 3t


5⎛⎝t
2 + 1⎞⎠


1/2
i − 5


5⎛⎝t
2 + 1⎞⎠


1/2
j + 4t


5⎛⎝t
2 + 1⎞⎠


1/2
k


= −
3t i + 5j − 4t k


5 t2 + 1
.


Once again, the unit tangent vector and the principal unit normal vector are orthogonal to each other forall values of t:
T(t) ·N(t) =





⎜3 i − 5t j − 4k


5 t2 + 1





⎟ ·



⎜−


3t i + 5j − 4t k


5 t2 + 1







= 3(−3t) − 5t(−5) − 4(4t)


5 t2 + 1


= −9t + 25t − 16t


5 t2 + 1
= 0.


Last, since r(t) represents a three-dimensional curve, we can calculate the binormal vector using
Equation 3.17:


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3.12


B(t) = T(t) × N(t)


= |
i j k
3


5 t2 + 1
− 5t


5 t2 + 1
− 4
5 t2 + 1


− 3t


5 t2 + 1
− 5


5 t2 + 1
4t


5 t2 + 1 |
=







⎜− 5t


5 t2 + 1









⎜ 4t
5 t2 + 1





⎟−



⎜− 4


5 t2 + 1









⎜− 5


5 t2 + 1









⎟i










⎜ 3


5 t2 + 1









⎜ 4t
5 t2 + 1





⎟−



⎜− 4


5 t2 + 1









⎜− 3t


5 t2 + 1









⎟ j


+







⎜ 3


5 t2 + 1









⎜− 5


5 t2 + 1





⎟−



⎜− 5t


5 t2 + 1









⎜− 3t


5 t2 + 1









⎟k


=



⎜−20t


2 − 20
25⎛⎝t


2 + 1⎞⎠





⎟i +



⎜−15 − 15t


2


25⎛⎝t
2 + 1⎞⎠





⎟k


= −20



⎜ t


2 + 1
25⎛⎝t


2 + 1⎞⎠





⎟i − 15





⎜ t


2 + 1
25⎛⎝t


2 + 1⎞⎠





⎟k


= − 4
5
i − 3


5
k.


Find the unit normal vector for the vector-valued function r(t) = ⎛⎝t2 − 3t⎞⎠ i + (4t + 1) j and evaluate it
at t = 2.


For any smooth curve in three dimensions that is defined by a vector-valued function, we now have formulas for the unittangent vector T, the unit normal vector N, and the binormal vector B. The unit normal vector and the binormal vector forma plane that is perpendicular to the curve at any point on the curve, called the normal plane. In addition, these three vectorsform a frame of reference in three-dimensional space called the Frenet frame of reference (also called the TNB frame)(Figure 3.7). Lat, the plane determined by the vectors T and N forms the osculating plane of C at any point P on the curve.


Figure 3.7 This figure depicts a Frenet frame of reference. At every point P on a three-dimensional curve, the unit tangent, unit normal, and binormal vectors form a three-dimensional frame of reference.


Suppose we form a circle in the osculating plane of C at point P on the curve. Assume that the circle has the same curvature


Chapter 3 | Vector-Valued Functions 295




as the curve does at point P and let the circle have radius r. Then, the curvature of the circle is given by 1/r. We call r the
radius of curvature of the curve, and it is equal to the reciprocal of the curvature. If this circle lies on the concave sideof the curve and is tangent to the curve at point P, then this circle is called the osculating circle of C at P, as shown in thefollowing figure.


Figure 3.8 In this osculating circle, the circle is tangent to curve C at point P and shares thesame curvature.
For more information on osculating circles, see this demonstration (http://www.openstaxcollege.org/l/20_OsculCircle1) on curvature and torsion, this article (http://www.openstaxcollege.org/l/20_OsculCircle3) on osculating circles, and this discussion (http://www.openstaxcollege.org/l/20_OsculCircle2) of Serret formulas.


To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.
Example 3.13
Finding the Equation of an Osculating Circle
Find the equation of the osculating circle of the helix defined by the function y = x3 − 3x + 1 at t = 1.
Solution
Figure 3.9 shows the graph of y = x3 − 3x + 1.


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Figure 3.9 We want to find the osculating circle of this graphat the point where t = 1.


First, let’s calculate the curvature at x = 1:
κ = | f ″(x)|

⎝1 +



⎣ f ′ (x)⎤⎦2⎞⎠


3/2
= |6x|

⎝1 +



⎣3x


2 − 3⎤⎦
2⎞


3/2
.


This gives κ = 6. Therefore, the radius of the osculating circle is given by R = 1κ = 16. Next, we then calculate
the coordinates of the center of the circle. When x = 1, the slope of the tangent line is zero. Therefore, the center
of the osculating circle is directly above the point on the graph with coordinates (1, −1). The center is located
at ⎛⎝1, − 56⎞⎠. The formula for a circle with radius r and center (h, k) is given by (x − h)2 + ⎛⎝y − k⎞⎠2 = r2.
Therefore, the equation of the osculating circle is (x − 1)2 + ⎛⎝y + 56⎞⎠


2
= 1


36
. The graph and its osculating circle


appears in the following graph.


Figure 3.10 The osculating circle has radius R = 1/6.


Chapter 3 | Vector-Valued Functions 297




3.13 Find the equation of the osculating circle of the curve defined by the vector-valued function
y = 2x2 − 4x + 5 at x = 1.


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3.3 EXERCISES
Find the arc length of the curve on the given interval.
102. r(t) = t2 i + 14tj, 0 ≤ t ≤ 7. This portion of the
graph is shown here:


103. r(t) = t2 i + (2t2 + 1)j, 1 ≤ t ≤ 3
104. r(t) = 〈 2sin t, 5t, 2cos t 〉 , 0 ≤ t ≤ π. This
portion of the graph is shown here:


105. r(t) = 〈 t2 + 1, 4t3 + 3 〉 , − 1 ≤ t ≤ 0


106. r(t) = 〈 e−t cos t, e−t sin t 〉 over the interval

⎣0,


π
2

⎦. Here is the portion of the graph on the indicated


interval:


107. Find the length of one turn of the helix given by
r(t) = 1


2
cos ti + 1


2
sin tj + 3


4
t k.


108. Find the arc length of the vector-valued function
r(t) = − ti + 4tj + 3tk over [0, 1].
109. A particle travels in a circle with the equation ofmotion r(t) = 3cos ti + 3sin tj + 0k. Find the distance
traveled around the circle by the particle.
110. Set up an integral to find the circumference of theellipse with the equation r(t) = cos ti + 2sin tj + 0k.
111. Find the length of the curve r(t) = 〈 2t, et, e−t 〉
over the interval 0 ≤ t ≤ 1. The graph is shown here:


112. Find the length of the curve
r(t) = 〈 2sin t, 5t, 2cos t 〉 for t ∈ [−10, 10].
113. The position function for a particle is
r(t) = acos(ωt)i + bsin(ωt)j. Find the unit tangent
vector and the unit normal vector at t = 0.


Chapter 3 | Vector-Valued Functions 299




114. Given r(t) = acos(ωt)i + bsin(ωt)j, find the
binormal vector B(0).
115. Given r(t) = 〈 2et, et cos t, et sin t 〉 , determine
the tangent vector T(t).
116. Given r(t) = 〈 2et, et cos t, et sin t 〉 , determine
the unit tangent vector T(t) evaluated at t = 0.
117. Given r(t) = 〈 2et, et cos t, et sin t 〉 , find the
unit normal vector N(t) evaluated at t = 0, N(0).
118. Given r(t) = 〈 2et, et cos t, et sin t 〉 , find the
unit normal vector evaluated at t = 0.
119. Given r(t) = ti + t2 j + tk, find the unit tangent
vector T(t). The graph is shown here:


120. Find the unit tangent vector T(t) and unit normal
vector N(t) at t = 0 for the plane curve
r(t) = 〈 t3 − 4t, 5t2 − 2 〉 . The graph is shown here:


121. Find the unit tangent vector T(t) for
r(t) = 3ti + 5t2 j + 2tk


122. Find the principal normal vector to the curve
r(t) = 〈 6cos t, 6sin t 〉 at the point determined by
t = π/3.


123. Find T(t) for the curve
r(t) = ⎛⎝t


3 − 4t⎞⎠i +

⎝5t


2 − 2⎞⎠ j.


124. Find N(t) for the curve
r(t) = ⎛⎝t


3 − 4t⎞⎠i +

⎝5t


2 − 2⎞⎠ j.


125. Find the unit normal vector N(t) for
r(t) = 〈 2sin t, 5t, 2cos t 〉 .


126. Find the unit tangent vector T(t) for
r(t) = 〈 2sin t, 5t, 2cos t 〉 .


127. Find the arc-length function s(t) for the line segment
given by r(t) = 〈 3 − 3t, 4t 〉 . Write r as a parameter of
s.
128. Parameterize the helix r(t) = cos ti + sin tj + tk
using the arc-length parameter s, from t = 0.
129. Parameterize the curve using the arc-lengthparameter s, at the point at which t = 0 for
r(t) = et sin ti + et cos tj.


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130. Find the curvature of the curve
r(t) = 5cos ti + 4sin tj at t = π/3. (Note: The graph is
an ellipse.)


131. Find the x-coordinate at which the curvature of thecurve y = 1/x is a maximum value.
132. Find the curvature of the curve
r(t) = 5cos ti + 5sin tj. Does the curvature depend upon
the parameter t?
133. Find the curvature κ for the curve y = x − 1


4
x2 at


the point x = 2.
134. Find the curvature κ for the curve y = 1


3
x3 at the


point x = 1.
135. Find the curvature κ of the curve
r(t) = ti + 6t2 j + 4tk. The graph is shown here:


136. Find the curvature of r(t) = 〈 2sin t, 5t, 2cos t 〉 .


137. Find the curvature of r(t) = 2ti + et j + e−tk at
point P(0, 1, 1).
138. At what point does the curve y = ex have maximum
curvature?
139. What happens to the curvature as x → ∞ for the
curve y = ex?
140. Find the point of maximum curvature on the curve
y = lnx.


141. Find the equations of the normal plane and theosculating plane of the curve
r(t) = 〈 2sin(3t), t, 2cos(3t) 〉 at point (0, π, −2).
142. Find equations of the osculating circles of the ellipse
4y2 + 9x2 = 36 at the points (2, 0) and (0, 3).
143. Find the equation for the osculating plane at point
t = π/4 on the curve r(t) = cos(2t)i + sin(2t)j + t.
144. Find the radius of curvature of 6y = x3 at the point

⎝2,


4
3

⎠.


145. Find the curvature at each point (x, y) on the
hyperbola r(t) = 〈 acosh(t), bsinh(t) 〉 .
146. Calculate the curvature of the circular helix
r(t) = r sin(t)i + rcos(t)j + tk.


147. Find the radius of curvature of y = ln(x + 1) at
point (2, ln3).
148. Find the radius of curvature of the hyperbola xy = 1
at point (1, 1).
A particle moves along the plane curve C described by
r(t) = ti + t2 j. Solve the following problems.
149. Find the length of the curve over the interval [0, 2].


150. Find the curvature of the plane curve at t = 0, 1, 2.


151. Describe the curvature as t increases from t = 0 to
t = 2.


The surface of a large cup is formed by revolving the graph
of the function y = 0.25x1.6 from x = 0 to x = 5 about
the y-axis (measured in centimeters).


Chapter 3 | Vector-Valued Functions 301




152. [T] Use technology to graph the surface.
153. Find the curvature κ of the generating curve as a
function of x.
154. [T] Use technology to graph the curvature function.


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3.4 | Motion in Space
Learning Objectives


3.4.1 Describe the velocity and acceleration vectors of a particle moving in space.
3.4.2 Explain the tangential and normal components of acceleration.
3.4.3 State Kepler’s laws of planetary motion.


We have now seen how to describe curves in the plane and in space, and how to determine their properties, such as arclength and curvature. All of this leads to the main goal of this chapter, which is the description of motion along plane curvesand space curves. We now have all the tools we need; in this section, we put these ideas together and look at how to usethem.
Motion Vectors in the Plane and in Space
Our starting point is using vector-valued functions to represent the position of an object as a function of time. All of thefollowing material can be applied either to curves in the plane or to space curves. For example, when we look at the orbit ofthe planets, the curves defining these orbits all lie in a plane because they are elliptical. However, a particle traveling alonga helix moves on a curve in three dimensions.
Definition
Let r(t) be a twice-differentiable vector-valued function of the parameter t that represents the position of an object as
a function of time. The velocity vector v(t) of the object is given by


(3.20)Velocity = v(t) = r′(t).
The acceleration vector a(t) is defined to be


(3.21)Acceleration = a(t) = v′(t) = r″(t).
The speed is defined to be


(3.22)Speed = v(t) = ‖ v(t) ‖ = ‖ r′(t) ‖ = ds
dt


.


Since r(t) can be in either two or three dimensions, these vector-valued functions can have either two or three components.
In two dimensions, we define r(t) = x(t) i + y(t) j and in three dimensions r(t) = x(t) i + y(t) j + z(t)k. Then the velocity,
acceleration, and speed can be written as shown in the following table.


Quantity Two Dimensions Three Dimensions
Position r(t) = x(t) i + y(t) j r(t) = x(t) i + y(t) j + z(t)k


Velocity v(t) = x′ (t) i + y′ (t) j v(t) = x′ (t) i + y′ (t) j + z′ (t)k


Acceleration a(t) = x″(t) i + y″(t) j a(t) = x″(t) i + y″(t) j + z″(t)k


Speed v(t) = (x′ (t))2 + ⎛⎝y′ (t)⎞⎠2 v(t) = (x′ (t))2 + ⎛⎝y′ (t)⎞⎠2 + (z′ (t))2


Table 3.4 Formulas for Position, Velocity, Acceleration, and Speed


Chapter 3 | Vector-Valued Functions 303




Example 3.14
Studying Motion Along a Parabola
A particle moves in a parabolic path defined by the vector-valued function r(t) = t2 i + 5 − t2j, where t
measures time in seconds.


a. Find the velocity, acceleration, and speed as functions of time.
b. Sketch the curve along with the velocity vector at time t = 1.


Solution
a. We use Equation 3.20, Equation 3.21, and Equation 3.22:


v(t) = r′(t) = 2t i − t
5 − t2


j


a(t) = v′(t) = 2i − 5⎛⎝5 − t
2⎞


−3
2
j


v(t) = ‖ r′(t) ‖


= (2t)2 +



⎜− t


5 − t2






2


= 4t2 + t
2


5 − t2


= 21t
2 − 4t4


5 − t2
.


b. The graph of r(t) = t2 i + 5 − t2j is a portion of a parabola (Figure 3.11). The velocity vector at
t = 1 is


v(1) = r′(1) = 2(1) i − 1
5 − (1)2


j = 2i − 1
2
j


and the acceleration vector at t = 1 is
a(1) = v′(1) = 2i − 5⎛⎝5 − (1)


2⎞


−3/2
j = 2i − 5


8
j.


Notice that the velocity vector is tangent to the path, as is always the case.


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3.14


Figure 3.11 This graph depicts the velocity vector at time
t = 1 for a particle moving in a parabolic path.


A particle moves in a path defined by the vector-valued function
r(t) = ⎛⎝t


2 − 3t⎞⎠i + (2t − 4) j + (t + 2)k, where t measures time in seconds and where distance is measured in
feet. Find the velocity, acceleration, and speed as functions of time.


To gain a better understanding of the velocity and acceleration vectors, imagine you are driving along a curvy road. If youdo not turn the steering wheel, you would continue in a straight line and run off the road. The speed at which you aretraveling when you run off the road, coupled with the direction, gives a vector representing your velocity, as illustrated inthe following figure.


Figure 3.12 At each point along a road traveled by a car, the velocity vector of the car istangent to the path traveled by the car.


However, the fact that you must turn the steering wheel to stay on the road indicates that your velocity is always changing(even if your speed is not) because your direction is constantly changing to keep you on the road. As you turn to the right,your acceleration vector also points to the right. As you turn to the left, your acceleration vector points to the left. Thisindicates that your velocity and acceleration vectors are constantly changing, regardless of whether your actual speed varies(Figure 3.13).


Chapter 3 | Vector-Valued Functions 305




Figure 3.13 The dashed line represents the trajectory of an object (acar, for example). The acceleration vector points toward the inside of theturn at all times.
Components of the Acceleration Vector
We can combine some of the concepts discussed in Arc Length and Curvature with the acceleration vector to gain adeeper understanding of how this vector relates to motion in the plane and in space. Recall that the unit tangent vector Tand the unit normal vector N form an osculating plane at any point P on the curve defined by a vector-valued function r(t).
The following theorem shows that the acceleration vector a(t) lies in the osculating plane and can be written as a linear
combination of the unit tangent and the unit normal vectors.
Theorem 3.7: The Plane of the Acceleration Vector
The acceleration vector a(t) of an object moving along a curve traced out by a twice-differentiable function r(t) lies
in the plane formed by the unit tangent vector T(t) and the principal unit normal vector N(t) to C. Furthermore,


a(t) = v′ (t)T(t) + ⎡⎣v(t)⎤⎦2 κN(t).


Here, v(t) is the speed of the object and κ is the curvature of C traced out by r(t).


Proof
Because v(t) = r′(t) and T(t) = r′(t)


‖ r′(t) ‖
, we have v(t) = ‖ r′(t) ‖ T(t) = v(t)T(t). Now we differentiate this


equation:
a(t) = v′(t) = d


dt

⎝v(t)T(t)⎞⎠ = v′ (t)T(t) + v(t)T′(t).


Since N(t) = T′(t)
‖ T′(t) ‖


, we know T′(t) = ‖ T′(t) ‖ N(t), so
a(t) = v′ (t)T(t) + v(t) ‖ T′(t) ‖ N(t).


A formula for curvature is κ = ‖ T′(t) ‖
‖ r′(t) ‖


, so ‖ T′(t) ‖ = κ ‖ r′(t) ‖ = κv(t). This gives
a(t) = v′ (t)T(t) + κ⎛⎝v(t)⎞⎠2N(t).



The coefficients of T(t) and N(t) are referred to as the tangential component of acceleration and the normal component


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of acceleration, respectively. We write aT to denote the tangential component and aN to denote the normal component.


Theorem 3.8: Tangential and Normal Components of Acceleration
Let r(t) be a vector-valued function that denotes the position of an object as a function of time. Then a(t) = r″(t) is
the acceleration vector. The tangential and normal components of acceleration aT and aN are given by the formulas


(3.23)aT = a ·T = v · a‖ v ‖
and


(3.24)aN = a ·N = ‖ v × a ‖‖ v ‖ = ‖ a ‖ 2 − aT2 .
These components are related by the formula


(3.25)a(t) = aTT(t) + aNN(t).
Here T(t) is the unit tangent vector to the curve defined by r(t), and N(t) is the unit normal vector to the curve
defined by r(t).


The normal component of acceleration is also called the centripetal component of acceleration or sometimes the radialcomponent of acceleration. To understand centripetal acceleration, suppose you are traveling in a car on a circular track ata constant speed. Then, as we saw earlier, the acceleration vector points toward the center of the track at all times. As arider in the car, you feel a pull toward the outside of the track because you are constantly turning. This sensation acts in theopposite direction of centripetal acceleration. The same holds true for noncircular paths. The reason is that your body tendsto travel in a straight line and resists the force resulting from acceleration that push it toward the side. Note that at point B inFigure 3.14 the acceleration vector is pointing backward. This is because the car is decelerating as it goes into the curve.


Figure 3.14 The tangential and normal components of acceleration can be used todescribe the acceleration vector.


The tangential and normal unit vectors at any given point on the curve provide a frame of reference at that point. Thetangential and normal components of acceleration are the projections of the acceleration vector onto T and N, respectively.
Example 3.15
Finding Components of Acceleration
A particle moves in a path defined by the vector-valued function r(t) = t2 i + (2t − 3) j + ⎛⎝3t2 − 3t⎞⎠k, where t
measures time in seconds and distance is measured in feet.


a. Find aT and aN as functions of t.


Chapter 3 | Vector-Valued Functions 307




b. Find aT and aN at time t = 2.
Solution


a. Let’s start with Equation 3.23:
v(t) = r′(t) = 2t i + 2j + (6t − 3)k


a(t) = v′(t) = 2i + 6k


aT =
v · a
‖ v ‖


=

⎝2t i + 2j + (6t − 3)k⎞⎠ · (2i + 6k)


‖ 2t i + 2j + (6t − 3)k ‖


= 4t + 6(6t − 3)


(2t)2 + 22 + (6t − 3)2


= 40t − 18


40t2 − 36t + 13
.


Then we apply Equation 3.24:
aN = ‖ a ‖


2 − aT


= ‖ 2i + 6k ‖ 2 −



⎜ 40t − 18


40t2 − 36t + 13






2


= 4 + 36 − (40t − 18)
2


40t2 − 36t + 13


=
40⎛⎝40t


2 − 36t + 13⎞⎠−

⎝1600t


2 − 1440t + 324⎞⎠


40t2 − 36t + 13


= 196
40t2 − 36t + 13


= 14
40t2 − 36t + 13


.


b. We must evaluate each of the answers from part a. at t = 2:
aT (2) =


40(2) − 18


40(2)2 − 36(2) + 13


= 80 − 18
160 − 72 + 13


= 62
101


aN (2) =
14


40(2)2 − 36(2) + 13


= 14
160 − 72 + 13


= 14
101


.


The units of acceleration are feet per second squared, as are the units of the normal and tangentialcomponents of acceleration.


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3.15 An object moves in a path defined by the vector-valued function r(t) = 4t i + t2 j, where t measures
time in seconds.


a. Find aT and aN as functions of t.
b. Find aT and aN at time t = −3.


Projectile Motion
Now let’s look at an application of vector functions. In particular, let’s consider the effect of gravity on the motion of anobject as it travels through the air, and how it determines the resulting trajectory of that object. In the following, we ignorethe effect of air resistance. This situation, with an object moving with an initial velocity but with no forces acting on it otherthan gravity, is known as projectile motion. It describes the motion of objects from golf balls to baseballs, and from arrowsto cannonballs.
First we need to choose a coordinate system. If we are standing at the origin of this coordinate system, then we choose thepositive y-axis to be up, the negative y-axis to be down, and the positive x-axis to be forward (i.e., away from the throwerof the object). The effect of gravity is in a downward direction, so Newton’s second law tells us that the force on the objectresulting from gravity is equal to the mass of the object times the acceleration resulting from to gravity, or Fg = mg,
where Fg represents the force from gravity and g represents the acceleration resulting from gravity at Earth’s surface.
The value of g in the English system of measurement is approximately 32 ft/sec2 and it is approximately 9.8 m/sec2 in themetric system. This is the only force acting on the object. Since gravity acts in a downward direction, we can write the forceresulting from gravity in the form Fg = −mg j, as shown in the following figure.


Figure 3.15 An object is falling under the influence ofgravity.
Visit thiswebsite (http://www.openstaxcollege.org/l/20_projectile) for a video showing projectile motion.


Newton’s second law also tells us that F = ma, where a represents the acceleration vector of the object. This force must
be equal to the force of gravity at all times, so we therefore know that


F = Fg
ma = −mg j
a = −g j.


Now we use the fact that the acceleration vector is the first derivative of the velocity vector. Therefore, we can rewrite thelast equation in the form


Chapter 3 | Vector-Valued Functions 309




v′(t) = −g j.


By taking the antiderivative of each side of this equation we obtain
v(t) = ∫ −g jdt


= −gt j + C1


for some constant vector C1. To determine the value of this vector, we can use the velocity of the object at a fixed time, say
at time t = 0. We call this velocity the initial velocity: v(0) = v0. Therefore, v(0) = −g(0) j + C1 = v0 and C1 = v0.
This gives the velocity vector as v(t) = −gt j + v0.
Next we use the fact that velocity v(t) is the derivative of position s(t). This gives the equation


s′(t) = −gt j + v0.


Taking the antiderivative of both sides of this equation leads to
s(t) = ∫ −gt j + v0dt


= − 1
2
gt2 j + v0 t + C2,


with another unknown constant vector C2. To determine the value of C2, we can use the position of the object
at a given time, say at time t = 0. We call this position the initial position: s(0) = s0. Therefore,
s(0) = −(1/2)g(0)2 j + v0 (0) + C2 = s0 and C2 = s0. This gives the position of the object at any time as


s(t) = − 1
2
gt2 j + v0 t + s0.


Let’s take a closer look at the initial velocity and initial position. In particular, suppose the object is thrown upward from theorigin at an angle θ to the horizontal, with initial speed v0. How can we modify the previous result to reflect this scenario?
First, we can assume it is thrown from the origin. If not, then we can move the origin to the point from where it is thrown.Therefore, s0 = 0, as shown in the following figure.


Figure 3.16 Projectile motion when the object is thrown upward at an angle θ.
The horizontal motion is at constant velocity and the vertical motion is at constantacceleration.


We can rewrite the initial velocity vector in the form v0 = v0 cosθ i + v0 sinθ j. Then the equation for the position function
s(t) becomes


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s(t) = − 1
2
gt2 j + v0 tcosθ i + v0 t sinθ j


= v0 tcosθ i + v0 t sinθ j −
1
2
gt2 j


= v0 tcosθ i +

⎝v0 t sinθ −


1
2
gt2⎞⎠ j.


The coefficient of i represents the horizontal component of s(t) and is the horizontal distance of the object from the origin
at time t. The maximum value of the horizontal distance (measured at the same initial and final altitude) is called the rangeR. The coefficient of j represents the vertical component of s(t) and is the altitude of the object at time t. The maximum
value of the vertical distance is the height H.
Example 3.16
Motion of a Cannonball
During an Independence Day celebration, a cannonball is fired from a cannon on a cliff toward the water. Thecannon is aimed at an angle of 30° above horizontal and the initial speed of the cannonball is 600 ft/sec. The
cliff is 100 ft above the water (Figure 3.17).


a. Find the maximum height of the cannonball.
b. How long will it take for the cannonball to splash into the sea?
c. How far out to sea will the cannonball hit the water?


Figure 3.17 The flight of a cannonball (ignoring air resistance) is projectile motion.
Solution
We use the equation


s(t) = v0 tcosθ i +

⎝v0 t sinθ −


1
2
gt2⎞⎠ j


with θ = 30°, g = 32 ft/sec2, and v0 = 600 ft/sec. Then the position equation becomes
s(t) = 600t(cos30) i + ⎛⎝600t sin30 −


1
2
(32)t2⎞⎠ j


= 300t 3i + ⎛⎝300t − 16t
2⎞
⎠ j.


a. The cannonball reaches its maximum height when the vertical component of its velocity is zero, because


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3.16


the cannonball is neither rising nor falling at that point. The velocity vector is
v(t) = s′(t)


= 300 3i + (300 − 32t) j.


Therefore, the vertical component of velocity is given by the expression 300 − 32t. Setting this
expression equal to zero and solving for t gives t = 9.375 sec. The height of the cannonball at this time
is given by the vertical component of the position vector, evaluated at t = 9.375.


s(9.375) = 300(9.375) 3i + ⎛⎝300(9.375) − 16(9.375)
2⎞
⎠ j


= 4871.39i + 1406.25j


Therefore, the maximum height of the cannonball is 1406.39 ft above the cannon, or 1506.39 ft above sealevel.
b. When the cannonball lands in the water, it is 100 ft below the cannon. Therefore, the vertical componentof the position vector is equal to −100. Setting the vertical component of s(t) equal to −100 and


solving, we obtain
300t − 16t2 = −100


16t2 − 300t − 100 = 0


4t2 − 75t − 25 = 0


t = 75 ± (−75)
2 − 4(4)(−25)


2(4)


= 75 ± 6025
8


= 75 ± 5 241
8


.


The positive value of t that solves this equation is approximately 19.08. Therefore, the cannonball hits thewater after approximately 19.08 sec.
c. To find the distance out to sea, we simply substitute the answer from part (b) into s(t):


s(19.08) = 300(19.08) 3i + ⎛⎝300(19.08) − 16(19.08)
2⎞
⎠ j


= 9914.26 i − 100.7424 j.


Therefore, the ball hits the water about 9914.26 ft away from the base of the cliff. Notice that the verticalcomponent of the position vector is very close to −100, which tells us that the ball just hit the water.
Note that 9914.26 feet is not the true range of the cannon since the cannonball lands in the ocean at alocation below the cannon. The range of the cannon would be determined by finding how far out thecannonball is when its height is 100 ft above the water (the same as the altitude of the cannon).


An archer fires an arrow at an angle of 40° above the horizontal with an initial speed of 98 m/sec. Theheight of the archer is 171.5 cm. Find the horizontal distance the arrow travels before it hits the ground.


One final question remains: In general, what is the maximum distance a projectile can travel, given its initial speed? Todetermine this distance, we assume the projectile is fired from ground level and we wish it to return to ground level. In otherwords, we want to determine an equation for the range. In this case, the equation of projectile motion is


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s(t) = v0 tcosθ i +

⎝v0 t sinθ −


1
2
gt2⎞⎠ j.


Setting the second component equal to zero and solving for t yields
v0 t sinθ −


1
2
gt2 = 0


t⎛⎝v0 sinθ −
1
2
gt⎞⎠ = 0.


Therefore, either t = 0 or t = 2v0 sinθg . We are interested in the second value of t, so we substitute this into s(t), which
gives


s


2v0 sinθ


g

⎠ = v0




2v0 sinθ


g

⎠cosθ i +





⎜v0


2v0 sinθ


g

⎠sinθ −


1
2
g


2v0 sinθ


g



2⎞



⎟j


=




2v0


2 sinθcosθ
g





⎟i


=
v0
2 sin2θ
g i.


Thus, the expression for the range of a projectile fired at an angle θ is
R =


v0
2 sin2θ
g i.


The only variable in this expression is θ. To maximize the distance traveled, take the derivative of the coefficient of i with
respect to θ and set it equal to zero:


d






v0
2 sin2θ
g



⎟ = 0


2v0
2 cos2θ
g = 0


θ = 45°.


This value of θ is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air
resistance, the best angle to fire a projectile (to maximize the range) is at a 45° angle. The distance it travels is given by


s


2v0 sin45


g

⎠ =


v0
2 sin90
g i =


v0
2


g j.


Therefore, the range for an angle of 45° is v02/g.
Kepler’s Laws
During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Brahe toformulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion. These laws also apply toother objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations ofthese laws apply to satellites in orbit around Earth.
Theorem 3.9: Kepler’s Laws of Planetary Motion


i. The path of any planet about the Sun is elliptical in shape, with the center of the Sun located at one focus ofthe ellipse (the law of ellipses).
ii. A line drawn from the center of the Sun to the center of a planet sweeps out equal areas in equal time intervals


Chapter 3 | Vector-Valued Functions 313




(the law of equal areas) (Figure 3.18).
iii. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of the lengths oftheir semimajor orbital axes (the law of harmonies).


Figure 3.18 Kepler’s first and second laws are pictured here.The Sun is located at a focus of the elliptical orbit of any planet.Furthermore, the shaded areas are all equal, assuming that theamount of time measured as the planet moves is the same foreach region.


Kepler’s third law is especially useful when using appropriate units. In particular, 1 astronomical unit is defined to be theaverage distance from Earth to the Sun, and is now recognized to be 149,597,870,700 m or, approximately 93,000,000 mi.We therefore write 1 A.U. = 93,000,000 mi. Since the time it takes for Earth to orbit the Sun is 1 year, we use Earth yearsfor units of time. Then, substituting 1 year for the period of Earth and 1 A.U. for the average distance to the Sun, Kepler’sthird law can be written as
T p
2 = D p


3


for any planet in the solar system, where TP is the period of that planet measured in Earth years and DP is the average
distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from aplanet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.
Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir IsaacNewton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems,such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun.
Proof
Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’splace the Sun at the origin of the coordinate system and let the vector-valued function r(t) represent the location of a planet
as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation.Newton’s second law of motion can be written as F = ma, where F represents the net force acting on the planet. His law
of universal gravitation can be written in the form F = − GmM


‖ r ‖ 2
· r
‖ r ‖


, which indicates that the force resulting from
the gravitational attraction of the Sun points back toward the Sun, and has magnitude GmM


‖ r ‖ 2
(Figure 3.19).


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Figure 3.19 The gravitational force between Earth and theSun is equal to the mass of the earth times its acceleration.


Setting these two forces equal to each other, and using the fact that a(t) = v′(t), we obtain
mv′(t) = − GmM


‖ r ‖ 2
· r
‖ r ‖


,


which can be rewritten as
dv
dt


= − GM
‖ r ‖ 3


r.


This equation shows that the vectors dv/dt and r are parallel to each other, so dv/dt × r = 0. Next, let’s differentiate
r × v with respect to time:


d
dt
(r × v) = dr


dt
× v + r × dv


dt
= v × v + 0 = 0.


This proves that r × v is a constant vector, which we call C. Since r and v are both perpendicular to C for all values of t,
they must lie in a plane perpendicular to C. Therefore, the motion of the planet lies in a plane.
Next we calculate the expression dv/dt × C:


(3.26)dv
dt


× C = − GM
‖ r ‖ 3


r × (r × v) = − GM
‖ r ‖ 3



⎣(r · v)r − (r · r)v⎤⎦.


The last equality in Equation 3.26 is from the triple cross product formula (Introduction to Vectors in Space). Weneed an expression for r · v. To calculate this, we differentiate r · r with respect to time:
(3.27)d


dt
(r · r) = dr


dt
· r + r · dr


dt
= 2r · dr


dt
= 2r · v.


Since r · r = ‖ r ‖ 2, we also have
(3.28)d


dt
(r · r) = d


dt
‖ r ‖ 2 = 2 ‖ r ‖ d


dt
‖ r ‖ .


Combining Equation 3.27 and Equation 3.28, we get
2r · v = 2 ‖ r ‖ d


dt
‖ r ‖


r · v = ‖ r ‖ d
dt


‖ r ‖ .


Substituting this into Equation 3.26 gives us


Chapter 3 | Vector-Valued Functions 315




(3.29)dv
dt


× C = − GM
‖ r ‖ 3



⎣(r · v)r − (r · r)v⎤⎦


= − GM
‖ r ‖ 3



⎣ ‖ r ‖




d
dt


‖ r ‖ ⎞⎠r − ‖ r ‖
2v⎤⎦


= −GM



⎢ 1
‖ r ‖ 2




d
dt


‖ r ‖ ⎞⎠r −
1


‖ r ‖
v





= GM



⎢ v
‖ r ‖


− r
‖ r ‖ 2




d
dt


‖ r ‖ ⎞⎠



⎥.


However,
d
dt


r
‖ r ‖


=
d
dt
(r) ‖ r ‖ − r d


dt
‖ r ‖


‖ r ‖ 2


=
dr
dt


‖ r ‖
− r


‖ r ‖ 2
d
dt


‖ r ‖


= v
‖ r ‖


− r
‖ r ‖ 2


d
dt


‖ r ‖ .


Therefore, Equation 3.29 becomes
dv
dt


× C = GM


d
dt


r
‖ r ‖

⎠.


Since C is a constant vector, we can integrate both sides and obtain
v × C = GM r


‖ r ‖
+ D,


where D is a constant vector. Our goal is to solve for ‖ r ‖ . Let’s start by calculating r · (v × C):
r · (v × C) = r ·



⎝GM


r
‖ r ‖


+ D

⎠ = GM


‖ r ‖ 2


‖ r ‖
+ r ·D = GM ‖ r ‖ + r ·D.


However, r · (v × C) = (r × v) ·C, so
(r × v) ·C = GM ‖ r ‖ + r ·D.


Since r × v = C, we have
‖ C ‖ 2 = GM ‖ r ‖ + r ·D.


Note that r ·D = ‖ r ‖ ‖ D ‖ cosθ, where θ is the angle between r and D. Therefore,
‖ C ‖ 2 = GM ‖ r ‖ + ‖ r ‖ ‖ D ‖ cosθ.


Solving for ‖ r ‖ ,
‖ r ‖ = ‖ C ‖


2


GM + ‖ D ‖ cosθ
= ‖ C ‖


2


GM



1
1 + ecosθ



⎠,


where e = ‖ D ‖ /GM. This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is
a hyperbola if e > 1, a parabola if e = 1, or an ellipse if e < 1. Since planets have closed orbits, the only possibility is
an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merelypassing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough tothe Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.


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(3.30)


3.17


Example 3.17
Using Kepler’s Third Law for Nonheliocentric Orbits
Kepler’s third law of planetary motion can be modified to the case of one object in orbit around an object otherthan the Sun, such as the Moon around the Earth. In this case, Kepler’s third law becomes


P2 = 4π
2a3


G(m + M)
,


where m is the mass of the Moon and M is the mass of Earth, a represents the length of the major axis of theelliptical orbit, and P represents the period.
Given that the mass of the Moon is 7.35 × 1022 kg, the mass of Earth is 5.97 × 1024 kg,
G = 6.67 × 10−11m /kg · sec2, and the period of the moon is 27.3 days, let’s find the length of the major axis
of the orbit of the Moon around Earth.
Solution
It is important to be consistent with units. Since the universal gravitational constant contains seconds in the units,we need to use seconds for the period of the Moon as well:


27.3 days × 24 hr
1 day


× 3600sec
1hour


= 2,358,720sec.


Substitute all the data into Equation 3.30 and solve for a:
(2,358,720sec)2 = 4π


2a3



⎝6.67 × 10


−11 m


kg · sec2



⎝7.35 x 10


22kg + 5.97 x 1024kg⎞⎠


5.563 × 1012 = 4π
2a3



⎝6.67 × 10


−11 m3⎞⎠

⎝6.04 x 10


24⎞



⎝5.563 × 10


12⎞


⎝6.67 × 10


−11 m3⎞⎠

⎝6.04 × 10


24⎞
⎠ = 4π


2a3


a3 = 2.241 × 10
27


4π2
m3


a = 3.84 × 108m
≈ 384,000 km.


Analysis
According to solarsystem.nasa.gov, the actual average distance from the Moon to Earth is 384,400 km. This iscalculated using reflectors left on the Moon by Apollo astronauts back in the 1960s.


Titan is the largest moon of Saturn. The mass of Titan is approximately 1.35 × 1023 kg. The mass of
Saturn is approximately 5.68 × 1026 kg. Titan takes approximately 16 days to orbit Saturn. Use this
information, along with the universal gravitation constant G = 6.67 × 10−11m /kg · sec2 to estimate the
distance from Titan to Saturn.


Example 3.18
Chapter Opener: Halley’s Comet


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We now return to the chapter opener, which discusses the motion of Halley’s comet around the Sun. Kepler’sfirst law states that Halley’s comet follows an elliptical path around the Sun, with the Sun as one focus of theellipse. The period of Halley’s comet is approximately 76.1 years, depending on how closely it passes by Jupiterand Saturn as it passes through the outer solar system. Let’s use T = 76.1 years. What is the average distance of
Halley’s comet from the Sun?
Solution
Using the equation T 2 = D3 with T = 76.1, we obtain D3 = 5791.21, so D ≈ 17.96 A.U. This comes out
to approximately 1.67 × 109 mi.
A natural question to ask is: What are the maximum (aphelion) and minimum (perihelion) distances from Halley’sComet to the Sun? The eccentricity of the orbit of Halley’s Comet is 0.967 (Source: http://nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html). Recall that the formula for the eccentricity of an ellipse is e = c/a, where
a is the length of the semimajor axis and c is the distance from the center to either focus. Therefore,
0.967 = c/17.96 and c ≈ 17.37 A.U. Subtracting this from a gives the perihelion distance
p = a − c = 17.96 − 17.37 = 0.59 A.U. According to the National Space Science Data Center (Source:
http://nssdc.gsfc.nasa.gov/planetary/factsheet/cometfact.html), the perihelion distance for Halley’s comet is 0.587A.U. To calculate the aphelion distance, we add


P = a + c = 17.96 + 17.37 = 35.33 A.U.


This is approximately 3.3 × 109 mi. The average distance from Pluto to the Sun is 39.5 A.U. (Source:
http://www.oarval.org/furthest.htm), so it would appear that Halley’s Comet stays just within the orbit of Pluto.


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Navigating a Banked Turn
How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could dependon several factors:


• The weight of the car;
• The friction between the tires and the road;
• The radius of the circle;
• The “steepness” of the turn.


In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Beforeconsidering this track in particular, we use vector functions to develop the mathematics and physics necessary foranswering questions such as this.
A car of mass m moves with constant angular speed ω around a circular curve of radius R (Figure 3.20). The curve
is banked at an angle θ. If the height of the car off the ground is h, then the position of the car at time t is given by the
function r(t) = 〈 Rcos(ωt), Rsin(ωt), h 〉 .


Figure 3.20 Views of a race car moving around a track.
1. Find the velocity function v(t) of the car. Show that v is tangent to the circular curve. This means that, without


a force to keep the car on the curve, the car will shoot off of it.
2. Show that the speed of the car is ωR. Use this to show that (2π4)/|v| = (2π)/ω.
3. Find the acceleration a. Show that this vector points toward the center of the circle and that |a| = Rω2.
4. The force required to produce this circular motion is called the centripetal force, and it is denoted Fcent. Thisforce points toward the center of the circle (not toward the ground). Show that |Fcent| = ⎛⎝m|v|2⎞⎠/R.


As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this forceis perpendicular to the ground), and the friction force (Figure 3.21). Because describing the frictional forcegenerated by the tires and the road is complex, we use a standard approximation for the frictional force.Assume that f = µN for some positive constant µ. The constant µ is called the coefficient of friction.


Chapter 3 | Vector-Valued Functions 319




Figure 3.21 The car has three forces acting on it: gravity(denoted by mg), the friction force f, and the force exerted by theroad N.


Let vmax denote the maximum speed the car can attain through the curve without skidding. In other words,
vmax is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the
magnitude of the centripetal force is


|Fcent| =
mvmax


2


R
.


The next three questions deal with developing a formula that relates the speed vmax to the banking angle θ.
5. Show that Ncosθ = mg + f sinθ. Conclude that N = (mg)/⎛⎝cosθ − µ sinθ⎞⎠.
6. The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force pointstoward the center of the circular curve. Show that


Fcent = Nsinθ + f cosθ.


Conclude that
Fcent =


sinθ + µcosθ
cosθ − µ sinθ


mg.


7. Show that vmax2 = ⎛⎝⎛⎝sinθ + µcosθ⎞⎠/⎛⎝cosθ − µ sinθ⎞⎠⎞⎠gR. Conclude that the maximum speed does not actually
depend on the mass of the car.Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a positionto answer the questions like the one posed at the beginning of the project.The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximateshape shown in Figure 3.22. Each end of the track is approximately semicircular, so when cars make turnsthey are traveling along an approximately circular curve. If a car takes the inside track and speeds along thebottom of turn 1, the car travels along a semicircle of radius approximately 211 ft with a banking angle of24°. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along asemicircle with a banking angle of 28°. (The track has variable angle banking.)


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Figure 3.22 At the Bristol Motor Speedway, Bristol, Tennessee (a), the turns have an inner radius of about 211 ft anda width of 40 ft (b). (credit: part (a) photo by Raniel Diaz, Flickr)


The coefficient of friction for a normal tire in dry conditions is approximately 0.7. Therefore, we assume the coefficientfor a NASCAR tire in dry conditions is approximately 0.98.
Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, andthen convert the answers to miles per hour as a final step.


8. In dry conditions, how fast can the car travel through the bottom of the turn without skidding?
9. In dry conditions, how fast can the car travel through the top of the turn without skidding?


10. In wet conditions, the coefficient of friction can become as low as 0.1. If this is the case, how fast can the cartravel through the bottom of the turn without skidding?
11. Suppose the measured speed of a car going along the outside edge of the turn is 105 mph. Estimate thecoefficient of friction for the car’s tires.


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3.4 EXERCISES
155. Given r(t) = (3t2 − 2)i + (2t − sin(t))j, find the
velocity of a particle moving along this curve.


156. Given r(t) = (3t2 − 2)i + (2t − sin(t))j, find the
acceleration vector of a particle moving along the curve inthe preceding exercise.
Given the following position functions, find the velocity,acceleration, and speed in terms of the parameter t.
157. r(t) = 〈 3cos t, 3sin t, t2 〉
158. r(t) = e−t i + t2 j + tan tk
159. r(t) = 2cos t j + 3sin tk. The graph is shown here:


Find the velocity, acceleration, and speed of a particle withthe given position function.
160. r(t) = 〈 t2 − 1, t 〉
161. r(t) = 〈 et, e−t 〉


162. r(t) = 〈 sin t, t, cos t 〉 . The graph is shown here:


163. The position function of an object is given by
r(t) = 〈 t2, 5t, t2 − 16t 〉 . At what time is the speed a
minimum?
164. Let r(t) = rcosh(ωt)i + r sinh(ωt)j. Find the
velocity and acceleration vectors and show that theacceleration is proportional to r(t).
Consider the motion of a point on the circumference of arolling circle. As the circle rolls, it generates the cycloid
r(t) = ⎛⎝ωt − sin(ωt)⎞⎠i + (1 − cos(ωt)) j, where ω is the
angular velocity of the circle and b is the radius of thecircle:


165. Find the equations for the velocity, acceleration, andspeed of the particle at any time.
A person on a hang glider is spiraling upward as a resultof the rapidly rising air on a path having position vector
r(t) = (3cos t)i + (3sin t)j + t2k. The path is similar to
that of a helix, although it is not a helix. The graph is shownhere:


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Find the following quantities:
166. The velocity and acceleration vectors
167. The glider’s speed at any time
168. The times, if any, at which the glider’s acceleration isorthogonal to its velocity
Given that r(t) = 〈 e−5t sin t, e−5t cos t, 4e−5t 〉 is the
position vector of a moving particle, find the followingquantities:
169. The velocity of the particle
170. The speed of the particle
171. The acceleration of the particle
172. Find the maximum speed of a point on thecircumference of an automobile tire of radius 1 ft when theautomobile is traveling at 55 mph.
A projectile is shot in the air from ground level with aninitial velocity of 500 m/sec at an angle of 60° with thehorizontal. The graph is shown here:


173. At what time does the projectile reach maximumheight?
174. What is the approximate maximum height of theprojectile?


175. At what time is the maximum range of the projectileattained?
176. What is the maximum range?
177. What is the total flight time of the projectile?
A projectile is fired at a height of 1.5 m above the groundwith an initial velocity of 100 m/sec and at an angle of 30°above the horizontal. Use this information to answer thefollowing questions:
178. Determine the maximum height of the projectile.
179. Determine the range of the projectile.
180. A golf ball is hit in a horizontal direction off the topedge of a building that is 100 ft tall. How fast must the ballbe launched to land 450 ft away?
181. A projectile is fired from ground level at an angle of8° with the horizontal. The projectile is to have a range of50 m. Find the minimum velocity necessary to achieve thisrange.
182. Prove that an object moving in a straight line at aconstant speed has an acceleration of zero.
183. The acceleration of an object is given by
a(t) = t j + tk. The velocity at t = 1 sec is v(1) = 5j
and the position of the object at t = 1 sec is
r(1) = 0i + 0j + 0k. Find the object’s position at any
time.
184. Find r(t) given that a(t) = −32j,
v(0) = 600 3i + 600j, and r(0) = 0.
185. Find the tangential and normal components ofacceleration for r(t) = acos(ωt)i + bsin(ωt)j at t = 0.
186. Given r(t) = t2 i + 2t j and t = 1, find the
tangential and normal components of acceleration.
For each of the following problems, find the tangential andnormal components of acceleration.


Chapter 3 | Vector-Valued Functions 323




187. r(t) = 〈 et cos t, et sin t, et 〉 . The graph is shown
here:


188. r(t) = 〈 cos(2t), sin(2t), 1 〉
189. r(t) = 〈 2t, t2, t3


3


190. r(t) = 〈 2
3
(1 + t)3/2, 2


3
(1 − t)3/2, 2t 〉


191. r(t) = 〈 6t, 3t2, 2t3 〉
192. r(t) = t2 i + t2 j + t3k
193. r(t) = 3cos(2πt) i + 3sin(2πt) j
194. Find the position vector-valued function r(t), given
that a(t) = i + et j, v(0) = 2j, and r(0) = 2i.
195. The force on a particle is given by
f(t) = (cos t) i + (sin t) j. The particle is located at point
(c, 0) at t = 0. The initial velocity of the particle is given
by v(0) = v0 j. Find the path of the particle of mass m.
(Recall, F = m · a.)
196. An automobile that weighs 2700 lb makes a turn ona flat road while traveling at 56 ft/sec. If the radius of theturn is 70 ft, what is the required frictional force to keep thecar from skidding?


197. Using Kepler’s laws, it can be shown that
v0 =


2GM
r0


is the minimum speed needed when θ = 0
so that an object will escape from the pull of a central forceresulting from massM. Use this result to find the minimumspeed when θ = 0 for a space capsule to escape from the
gravitational pull of Earth if the probe is at an altitude of300 km above Earth’s surface.
198. Find the time in years it takes the dwarf planet Plutoto make one orbit about the Sun given that a = 39.5 A.U.
Suppose that the position function for an object in threedimensions is given by the equation
r(t) = tcos(t)i + t sin(t)j + 3tk.


199. Show that the particle moves on a circular cone.
200. Find the angle between the velocity and accelerationvectors when t = 1.5.
201. Find the tangential and normal components ofacceleration when t = 1.5.


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acceleration vector
arc-length function
arc-length parameterization
binormal vector
component functions


curvature
definite integral of a vector-valued function


derivative of a vector-valued function


Frenet frame of reference
helix
indefinite integral of a vector-valued function
Kepler’s laws of planetary motion
limit of a vector-valued function
normal component of acceleration
normal plane
osculating circle
osculating plane
plane curve


principal unit normal vector
principal unit tangent vector
projectile motion
radius of curvature
reparameterization
smooth


CHAPTER 3 REVIEW
KEY TERMS


the second derivative of the position vector
a function s(t) that describes the arc length of curve C as a function of t


a reparameterization of a vector-valued function in which the parameter is equal to thearc length
a unit vector orthogonal to the unit tangent vector and the unit normal vector


the component functions of the vector-valued function r(t) = f (t) i + g(t) j are f (t) and
g(t), and the component functions of the vector-valued function r(t) = f (t) i + g(t) j + h(t)k are f (t), g(t) and
h(t)


the derivative of the unit tangent vector with respect to the arc-length parameter
the vector obtained by calculating the definite integral of each of thecomponent functions of a given vector-valued function, then using the results as the components of the resultingfunction


the derivative of a vector-valued function r(t) is
r′(t) = lim


Δt → 0


r(t + Δt) − r(t)
Δt


, provided the limit exists
(TNB frame) a frame of reference in three-dimensional space formed by the unit tangentvector, the unit normal vector, and the binormal vector


a three-dimensional curve in the shape of a spiral
a vector-valued function with a derivative that is equal to a givenvector-valued function


three laws governing the motion of planets, asteroids, and comets in orbit aroundthe Sun
a vector-valued function r(t) has a limit L as t approaches a if lim


t → a|r(t) − L| = 0


the coefficient of the unit normal vector N when the acceleration vector iswritten as a linear combination of T and N
a plane that is perpendicular to a curve at any point on the curve


a circle that is tangent to a curve C at a point P and that shares the same curvature
the plane determined by the unit tangent and the unit normal vector


the set of ordered pairs ⎛⎝ f (t), g(t)⎞⎠ together with their defining parametric equations x = f (t) and
y = g(t)


a vector orthogonal to the unit tangent vector, given by the formula T′(t)
‖ T′(t) ‖


a unit vector tangent to a curve C
motion of an object with an initial velocity but no force acting on it other than gravity
the reciprocal of the curvature
an alternative parameterization of a given vector-valued function


curves where the vector-valued function r(t) is differentiable with a non-zero derivative


Chapter 3 | Vector-Valued Functions 325




space curve


tangent vector
tangential component of acceleration
vector parameterization
vector-valued function
velocity vector


the set of ordered triples ⎛⎝ f (t), g(t), h(t)⎞⎠ together with their defining parametric equations x = f (t),
y = g(t) and z = h(t)


to r(t) at t = t0 any vector v such that, when the tail of the vector is placed at point r(t0) on the graph,
vector v is tangent to curve C


the coefficient of the unit tangent vector T when the acceleration vector iswritten as a linear combination of T and N
any representation of a plane or space curve using a vector-valued function
a function of the form r(t) = f (t) i + g(t) j or r(t) = f (t) i + g(t) j + h(t)k, where the


component functions f, g, and h are real-valued functions of the parameter t
the derivative of the position vector


KEY EQUATIONS
• Vector-valued function


r(t) = f (t) i + g(t) j or r(t) = f (t) i + g(t) j + h(t)k, or r(t) = 〈 f (t), g(t) 〉 or r(t) = 〈 f (t), g(t), h(t) 〉


• Limit of a vector-valued function
lim
t → a


r(t) = ⎡⎣ limt → a f (t)

⎦i +

⎣ limt → ag(t)



⎦ j or limt → ar(t) =



⎣ limt → a f (t)



⎦i +

⎣ limt → ag(t)



⎦ j +

⎣ limt → ah(t)



⎦k


• Derivative of a vector-valued function
r′(t) = lim


Δt → 0


r(t + Δt) − r(t)
Δt


• Principal unit tangent vector
T(t) = r′(t)


‖r′(t)‖


• Indefinite integral of a vector-valued function
∫ ⎡⎣ f (t) i + g(t) j + h(t)k⎤⎦dt = ⎡⎣∫ f (t)dt



⎦i +

⎣∫ g(t)dt



⎦ j +

⎣∫ h(t)dt



⎦k


• Definite integral of a vector-valued function

a


b

⎣ f (t) i + g(t) j + h(t)k⎤⎦dt =





⎢∫


a


b
f (t)dt



⎥i +



⎢∫


a


b
g(t)dt





⎥ j +



⎢∫


a


b
h(t)dt





⎥k


• Arc length of space curve
s = ∫


a


b

⎣ f ′ (t)⎤⎦2 + ⎡⎣g′ (t)⎤⎦2 + ⎡⎣h′ (t)⎤⎦2 dt = ∫


a


b
‖ r′(t) ‖ dt


• Arc-length function
s(t) = ∫


a


t

⎝ f ′ (u)⎞⎠2 + ⎛⎝g′ (u)⎞⎠2 + ⎛⎝h′ (u)⎞⎠2du or s(t) = ∫


a


t
‖ r′(u) ‖ du


• Curvature
κ = ‖ T′(t) ‖


‖ r′(t) ‖
or κ = ‖ r′(t) × r″(t) ‖


‖ r′(t) ‖ 3
or κ = |y″|



⎣1 +



⎝y′⎞⎠2⎤⎦


3/2


• Principal unit normal vector
N(t) = T′(t)


‖ T′(t) ‖


• Binormal vector
B(t) = T(t) × N(t)


• Velocity


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v(t) = r′(t)


• Acceleration
a(t) = v′(t) = r″(t)


• Speed
v(t) = ‖ v(t) ‖ = ‖ r′(t) ‖ = ds


dt


• Tangential component of acceleration
aT = a ·T =


v · a
‖ v ‖


• Normal component of acceleration
aN = a ·N =


‖ v × a ‖
‖ v ‖


= ‖ a ‖ 2 − aT


KEY CONCEPTS
3.1 Vector-Valued Functions and Space Curves


• A vector-valued function is a function of the form r(t) = f (t) i + g(t) j or r(t) = f (t) i + g(t) j + h(t)k, where
the component functions f, g, and h are real-valued functions of the parameter t.


• The graph of a vector-valued function of the form r(t) = f (t) i + g(t) j is called a plane curve. The graph of a
vector-valued function of the form r(t) = f (t) i + g(t) j + h(t)k is called a space curve.


• It is possible to represent an arbitrary plane curve by a vector-valued function.
• To calculate the limit of a vector-valued function, calculate the limits of the component functions separately.


3.2 Calculus of Vector-Valued Functions
• To calculate the derivative of a vector-valued function, calculate the derivatives of the component functions, thenput them back into a new vector-valued function.
• Many of the properties of differentiation from the Introduction to Derivatives (http://cnx.org/content/m53494/latest/) also apply to vector-valued functions.
• The derivative of a vector-valued function r(t) is also a tangent vector to the curve. The unit tangent vector T(t)
is calculated by dividing the derivative of a vector-valued function by its magnitude.


• The antiderivative of a vector-valued function is found by finding the antiderivatives of the component functions,then putting them back together in a vector-valued function.
• The definite integral of a vector-valued function is found by finding the definite integrals of the componentfunctions, then putting them back together in a vector-valued function.


3.3 Arc Length and Curvature
• The arc-length function for a vector-valued function is calculated using the integral formula


s(t) = ∫
a


t
‖ r′(u) ‖ du. This formula is valid in both two and three dimensions.


• The curvature of a curve at a point in either two or three dimensions is defined to be the curvature of the inscribedcircle at that point. The arc-length parameterization is used in the definition of curvature.
• There are several different formulas for curvature. The curvature of a circle is equal to the reciprocal of its radius.
• The principal unit normal vector at t is defined to be


N(t) = T′(t)
‖ T′(t) ‖


.


Chapter 3 | Vector-Valued Functions 327




• The binormal vector at t is defined as B(t) = T(t) × N(t), where T(t) is the unit tangent vector.
• The Frenet frame of reference is formed by the unit tangent vector, the principal unit normal vector, and the binormalvector.
• The osculating circle is tangent to a curve at a point and has the same curvature as the tangent curve at that point.


3.4 Motion in Space
• If r(t) represents the position of an object at time t, then r′(t) represents the velocity and r″(t) represents the
acceleration of the object at time t. The magnitude of the velocity vector is speed.


• The acceleration vector always points toward the concave side of the curve defined by r(t). The tangential and
normal components of acceleration aT and aN are the projections of the acceleration vector onto the unit tangent
and unit normal vectors to the curve.


• Kepler’s three laws of planetary motion describe the motion of objects in orbit around the Sun. His third law can bemodified to describe motion of objects in orbit around other celestial objects as well.
• Newton was able to use his law of universal gravitation in conjunction with his second law of motion and calculusto prove Kepler’s three laws.


CHAPTER 3 REVIEW EXERCISES
True or False? Justify your answer with a proof or acounterexample.
202. A parametric equation that passes through points
P and Q can be given by r(t) = 〈 t2, 3t + 1, t − 2 〉 ,
where P(1, 4, −1) and Q(16, 11, 2).


203. d
dt

⎣u(t) × u(t)⎤⎦ = 2u′(t) × u(t)


204. The curvature of a circle of radius r is constant
everywhere. Furthermore, the curvature is equal to 1/r.
205. The speed of a particle with a position function r(t)
is ⎛⎝r′(t)⎞⎠/⎛⎝|r′(t)|⎞⎠.
Find the domains of the vector-valued functions.
206. r(t) = 〈 sin(t), ln(t), t 〉


207. r(t) = 〈 et, 1
4 − t


, sec(t) 〉


Sketch the curves for the following vector equations. Use acalculator if needed.
208. [T] r(t) = 〈 t2, t3 〉
209. [T] r(t) = 〈 sin(20t)e−t, cos(20t)e−t, e−t 〉
Find a vector function that describes the following curves.


210. Intersection of the cylinder x2 + y2 = 4 with the
plane x + z = 6


211. Intersection of the cone z = x2 + y2 and plane
z = y − 4


Find the derivatives of u(t), u′(t), u′(t) × u(t),
u(t) × u′(t), and u(t) ·u′(t). Find the unit tangent vector.
212. u(t) = 〈 et, e−t 〉


213. u(t) = 〈 t2, 2t + 6, 4t5 − 12 〉
Evaluate the following integrals.
214. ∫ ⎛⎝tan(t)sec(t)i − te3t j⎞⎠dt


215. ∫
1


4


u(t)dt, with u(t) = 〈 ln(t)t , 1t , sin⎛⎝tπ4 ⎞⎠ 〉


Find the length for the following curves.
216. r(t) = 〈 3(t), 4cos(t), 4sin(t) 〉 for 1 ≤ t ≤ 4


217. r(t) = 2i + t j + 3t2k for 0 ≤ t ≤ 1
Reparameterize the following functions with respect totheir arc length measured from t = 0 in direction of


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increasing t.
218. r(t) = 2t i + (4t − 5)j + (1 − 3t)k
219. r(t) = cos(2t)i + 8t j − sin(2t)k
Find the curvature for the following vector functions.
220. r(t) = (2sin t)i − 4t j + (2cos t)k
221. r(t) = 2et i + 2e−t j + 2tk
222. Find the unit tangent vector, the unit normal vector,and the binormal vector for
r(t) = 2cos t i + 3t j + 2sin tk.


223. Find the tangential and normal accelerationcomponents with the position vector
r(t) = 〈 cos t, sin t, et 〉 .


224. A Ferris wheel car is moving at a constant speed v
and has a constant radius r. Find the tangential and normal
acceleration of the Ferris wheel car.
225. The position of a particle is given by
r(t) = 〈 t2, ln(t), sin(πt) 〉 , where t is measured in
seconds and r is measured in meters. Find the velocity,
acceleration, and speed functions. What are the position,velocity, speed, and acceleration of the particle at 1 sec?
The following problems consider launching a cannonballout of a cannon. The cannonball is shot out of the cannonwith an angle θ and initial velocity v0. The only force
acting on the cannonball is gravity, so we begin with aconstant acceleration a(t) = −g j.
226. Find the velocity vector function v(t).
227. Find the position vector r(t) and the parametric
representation for the position.
228. At what angle do you need to fire the cannonballfor the horizontal distance to be greatest? What is the totaldistance it would travel?


Chapter 3 | Vector-Valued Functions 329




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4 | DIFFERENTIATION OFFUNCTIONS OF SEVERALVARIABLES


Figure 4.1 Americans use (and lose) millions of golf balls a year, which keeps golf ball manufacturers in business. In thischapter, we study a profit model and learn methods for calculating optimal production levels for a typical golf ball manufacturingcompany. (credit: modification of work by oatsy40, Flickr)


Chapter 4 | Differentiation of Functions of Several Variables 331




Chapter Outline
4.1 Functions of Several Variables
4.2 Limits and Continuity
4.3 Partial Derivatives
4.4 Tangent Planes and Linear Approximations
4.5 The Chain Rule
4.6 Directional Derivatives and the Gradient
4.7 Maxima/Minima Problems
4.8 Lagrange Multipliers


Introduction
In Introduction to Applications of Derivatives (http://cnx.org/content/m53602/latest/) , we studied how todetermine the maximum and minimum of a function of one variable over a closed interval. This function might representthe temperature over a given time interval, the position of a car as a function of time, or the altitude of a jet plane as it travelsfrom New York to San Francisco. In each of these examples, the function has one independent variable.
Suppose, however, that we have a quantity that depends on more than one variable. For example, temperature can dependon location and the time of day, or a company’s profit model might depend on the number of units sold and the amountof money spent on advertising. In this chapter, we look at a company that produces golf balls. We develop a profit modeland, under various restrictions, we find that the optimal level of production and advertising dollars spent determines themaximum possible profit. Depending on the nature of the restrictions, both the method of solution and the solution itselfchanges (see Example 4.41).
When dealing with a function of more than one independent variable, several questions naturally arise. For example, howdo we calculate limits of functions of more than one variable? The definition of derivative we used before involved a limit.Does the new definition of derivative involve limits as well? Do the rules of differentiation apply in this context? Can wefind relative extrema of functions using derivatives? All these questions are answered in this chapter.
4.1 | Functions of Several Variables


Learning Objectives
4.1.1 Recognize a function of two variables and identify its domain and range.
4.1.2 Sketch a graph of a function of two variables.
4.1.3 Sketch several traces or level curves of a function of two variables.
4.1.4 Recognize a function of three or more variables and identify its level surfaces.


Our first step is to explain what a function of more than one variable is, starting with functions of two independent variables.This step includes identifying the domain and range of such functions and learning how to graph them. We also examineways to relate the graphs of functions in three dimensions to graphs of more familiar planar functions.
Functions of Two Variables
The definition of a function of two variables is very similar to the definition for a function of one variable. The maindifference is that, instead of mapping values of one variable to values of another variable, we map ordered pairs of variablesto another variable.
Definition
A function of two variables z = (x, y) maps each ordered pair (x, y) in a subset D of the real plane ℝ2 to a unique
real number z. The set D is called the domain of the function. The range of f is the set of all real numbers z that


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has at least one ordered pair (x, y) ∈ D such that f (x, y) = z as shown in the following figure.


Figure 4.2 The domain of a function of two variables consistsof ordered pairs (x, y).


Determining the domain of a function of two variables involves taking into account any domain restrictions that may exist.Let’s take a look.
Example 4.1
Domains and Ranges for Functions of Two Variables
Find the domain and range of each of the following functions:


a. f (x, y) = 3x + 5y + 2
b. g(x, y) = 9 − x2 − y2


Solution
a. This is an example of a linear function in two variables. There are no values or combinations of x and


y that cause f (x, y) to be undefined, so the domain of f is ℝ2. To determine the range, first pick
a value for z. We need to find a solution to the equation f (x, y) = z, or 3x − 5y + 2 = z. One such
solution can be obtained by first setting y = 0, which yields the equation 3x + 2 = z. The solution
to this equation is x = z − 2


3
, which gives the ordered pair ⎛⎝z − 23 , 0⎞⎠ as a solution to the equation


f (x, y) = z for any value of z. Therefore, the range of the function is all real numbers, or ℝ.
b. For the function g(x, y) to have a real value, the quantity under the square root must be nonnegative:


9 − x2 − y2 ≥ 0.


This inequality can be written in the form
x2 + y2 ≤ 9.


Therefore, the domain of g(x, y) is ⎧

⎨(x, y) ∈ ℝ2|x2 + y2 ≤ 9⎫⎭⎬. The graph of this set of points can be


described as a disk of radius 3 centered at the origin. The domain includes the boundary circle as shown
in the following graph.


Chapter 4 | Differentiation of Functions of Several Variables 333




4.1


Figure 4.3 The domain of the function
g(x, y) = 9 − x2 − y2 is a closed disk of radius 3.


To determine the range of g(x, y) = 9 − x2 − y2 we start with a point (x0, y0) on the boundary of the
domain, which is defined by the relation x2 + y2 = 9. It follows that x02 + y02 = 9 and


g(x0, y0) = 9 − x0
2 − y0


2 = 9 − ⎛⎝x0
2 + y0


2⎞
⎠ = 9 − 9 = 0.


If x02 + y02 = 0 (in other words, x0 = y0 = 0), then
g(x0, y0) = 9 − x0


2 − y0
2 = 9 − ⎛⎝x0


2 + y0
2⎞
⎠ = 9 − 0 = 3.


This is the maximum value of the function. Given any value c between 0 and 3, we can find an entire
set of points inside the domain of g such that g(x, y) = c:


9 − x2 − y2 = c


9 − x2 − y2 = c2


x2 + y2 = 9 − c2.


Since 9 − c2 > 0, this describes a circle of radius 9 − c2 centered at the origin. Any point on this
circle satisfies the equation g(x, y) = c. Therefore, the range of this function can be written in interval
notation as [0, 3].


Find the domain and range of the function f (x, y) = 36 − 9x2 − 9y2.


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Graphing Functions of Two Variables
Suppose we wish to graph the function z = (x, y). This function has two independent variables ⎛⎝x and y⎞⎠ and one
dependent variable (z). When graphing a function y = f (x) of one variable, we use the Cartesian plane. We are able to
graph any ordered pair (x, y) in the plane, and every point in the plane has an ordered pair (x, y) associated with it. With a
function of two variables, each ordered pair (x, y) in the domain of the function is mapped to a real number z. Therefore,
the graph of the function f consists of ordered triples (x, y, z). The graph of a function z = (x, y) of two variables is
called a surface.
To understand more completely the concept of plotting a set of ordered triples to obtain a surface in three-dimensionalspace, imagine the (x, y) coordinate system laying flat. Then, every point in the domain of the function f has a unique
z-value associated with it. If z is positive, then the graphed point is located above the xy-plane, if z is negative, then the
graphed point is located below the xy-plane. The set of all the graphed points becomes the two-dimensional surface that is
the graph of the function f .
Example 4.2
Graphing Functions of Two Variables
Create a graph of each of the following functions:


a. g(x, y) = 9 − x2 − y2
b. f (x, y) = x2 + y2


Solution
a. In Example 4.1, we determined that the domain of g(x, y) = 9 − x2 − y2 is





⎨(x, y) ∈ ℝ2|x2 + y2 ≤ 9⎫⎭⎬ and the range is ⎧⎩⎨z ∈ ℝ2|0 ≤ z ≤ 3⎫⎭⎬. When x2 + y2 = 9 we have
g(x, y) = 0. Therefore any point on the circle of radius 3 centered at the origin in the x, y-plane maps
to z = 0 in ℝ3. If x2 + y2 = 8, then g(x, y) = 1, so any point on the circle of radius 2 2 centered
at the origin in the x, y-plane maps to z = 1 in ℝ3. As x2 + y2 gets closer to zero, the value of z
approaches 3. When x2 + y2 = 0, then g(x, y) = 3. This is the origin in the x, y-plane. If x2 + y2 is
equal to any other value between 0 and 9, then g(x, y) equals some other constant between 0 and 3.
The surface described by this function is a hemisphere centered at the origin with radius 3 as shown in
the following graph.


Chapter 4 | Differentiation of Functions of Several Variables 335




Figure 4.4 Graph of the hemisphere represented by the given function oftwo variables.
b. This function also contains the expression x2 + y2. Setting this expression equal to various values


starting at zero, we obtain circles of increasing radius. The minimum value of f (x, y) = x2 + y2 is
zero (attained when x = y = 0.). When x = 0, the function becomes z = y2, and when y = 0,
then the function becomes z = x2. These are cross-sections of the graph, and are parabolas. Recall from
Introduction to Vectors in Space that the name of the graph of f (x, y) = x2 + y2 is a paraboloid.
The graph of f appears in the following graph.


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Figure 4.5 A paraboloid is the graph of the given function of twovariables.


Example 4.3
Nuts and Bolts
A profit function for a hardware manufacturer is given by


f (x, y) = 16 − (x − 3)2 − ⎛⎝y − 2⎞⎠2,


where x is the number of nuts sold per month (measured in thousands) and y represents the number of bolts sold
per month (measured in thousands). Profit is measured in thousands of dollars. Sketch a graph of this function.
Solution
This function is a polynomial function in two variables. The domain of f consists of (x, y) coordinate pairs that
yield a nonnegative profit:


16 − (x − 3)2 − ⎛⎝y − 2⎞⎠2 ≥ 0


(x − 3)2 + ⎛⎝y − 2⎞⎠2 ≤ 16.


This is a disk of radius 4 centered at (3, 2). A further restriction is that both x and y must be nonnegative.
When x = 3 and y = 2, f (x, y) = 16. Note that it is possible for either value to be a noninteger; for example,
it is possible to sell 2.5 thousand nuts in a month. The domain, therefore, contains thousands of points, so we


Chapter 4 | Differentiation of Functions of Several Variables 337




can consider all points within the disk. For any z < 16, we can solve the equation f (x, y) = 16:
16 − (x − 3)2 − ⎛⎝y − 2⎞⎠2 = z


(x − 3)2 + ⎛⎝y − 2⎞⎠2 = 16 − z.


Since z < 16, we know that 16 − z > 0, so the previous equation describes a circle with radius 16 − z
centered at the point (3, 2). Therefore. the range of f (x, y) is {z ∈ ℝ|z ≤ 16}. The graph of f (x, y) is also a
paraboloid, and this paraboloid points downward as shown.


Figure 4.6 The graph of the given function of two variables isalso a paraboloid.


Level Curves
If hikers walk along rugged trails, they might use a topographical map that shows how steeply the trails change. Atopographical map contains curved lines called contour lines. Each contour line corresponds to the points on the map thathave equal elevation (Figure 4.7). A level curve of a function of two variables f (x, y) is completely analogous to a
contour line on a topographical map.


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Figure 4.7 (a) A topographical map of Devil’s Tower, Wyoming. Lines that are close together indicate very steep terrain. (b) Aperspective photo of Devil’s Tower shows just how steep its sides are. Notice the top of the tower has the same shape as thecenter of the topographical map.


Definition
Given a function f (x, y) and a number c in the range of f , a level curve of a function of two variables for the
value c is defined to be the set of points satisfying the equation f (x, y) = c.


Returning to the function g(x, y) = 9 − x2 − y2, we can determine the level curves of this function. The range of g is
the closed interval [0, 3]. First, we choose any number in this closed interval—say, c = 2. The level curve corresponding
to c = 2 is described by the equation


9 − x2 − y2 = 2.


To simplify, square both sides of this equation:
9 − x2 − y2 = 4.


Now, multiply both sides of the equation by −1 and add 9 to each side:
x2 + y2 = 5.


This equation describes a circle centered at the origin with radius 5. Using values of c between 0 and 3 yields other
circles also centered at the origin. If c = 3, then the circle has radius 0, so it consists solely of the origin. Figure 4.8
is a graph of the level curves of this function corresponding to c = 0, 1, 2, and 3. Note that in the previous derivation it
may be possible that we introduced extra solutions by squaring both sides. This is not the case here because the range of thesquare root function is nonnegative.


Chapter 4 | Differentiation of Functions of Several Variables 339




Figure 4.8 Level curves of the function
g(x, y) = 9 − x2 − y2, using c = 0, 1, 2, and 3 (c = 3
corresponds to the origin).


A graph of the various level curves of a function is called a contour map.
Example 4.4
Making a Contour Map
Given the function f (x, y) = 8 + 8x − 4y − 4x2 − y2, find the level curve corresponding to c = 0. Then
create a contour map for this function. What are the domain and range of f ?
Solution
To find the level curve for c = 0, we set f (x, y) = 0 and solve. This gives


0 = 8 + 8x − 4y − 4x2 − y2.


We then square both sides and multiply both sides of the equation by −1:
4x2 + y2 − 8x + 4y − 8 = 0.


Now, we rearrange the terms, putting the x terms together and the y terms together, and add 8 to each side:
4x2 − 8x + y2 + 4y = 8.


Next, we group the pairs of terms containing the same variable in parentheses, and factor 4 from the first pair:
4⎛⎝x


2 − 2x⎞⎠+

⎝y


2 + 4y⎞⎠ = 8.


Then we complete the square in each pair of parentheses and add the correct value to the right-hand side:
4⎛⎝x


2 − 2x + 1⎞⎠+

⎝y


2 + 4y + 4⎞⎠ = 8 + 4(1) + 4.


Next, we factor the left-hand side and simplify the right-hand side:
4(x − 1)2 + ⎛⎝y + 2⎞⎠2 = 16.


Last, we divide both sides by 16:


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(4.1)(x − 1)2
4


+

⎝y + 2⎞⎠2


16
= 1.


This equation describes an ellipse centered at (1, −2). The graph of this ellipse appears in the following graph.


Figure 4.9 Level curve of the function
f (x, y) = 8 + 8x − 4y − 4x2 − y2 corresponding to
c = 0.


We can repeat the same derivation for values of c less than 4. Then, Equation 4.1 becomes
4(x − 1)2


16 − c2
+

⎝y + 2⎞⎠2


16 − c2
= 1


for an arbitrary value of c. Figure 4.10 shows a contour map for f (x, y) using the values c = 0, 1, 2, and 3.
When c = 4, the level curve is the point (−1, 2).


Chapter 4 | Differentiation of Functions of Several Variables 341




4.2


Figure 4.10 Contour map for the function
f (x, y) = 8 + 8x − 4y − 4x2 − y2 using the values
c = 0, 1, 2, 3, and 4.


Find and graph the level curve of the function g(x, y) = x2 + y2 − 6x + 2y corresponding to c = 15.


Another useful tool for understanding the graph of a function of two variables is called a vertical trace. Level curves arealways graphed in the xy-plane, but as their name implies, vertical traces are graphed in the xz - or yz-planes.


Definition
Consider a function z = f (x, y) with domain D ⊆ ℝ2. A vertical trace of the function can be either the set of points
that solves the equation f (a, y) = z for a given constant x = a or f (x, b) = z for a given constant y = b.


Example 4.5
Finding Vertical Traces
Find vertical traces for the function f (x, y) = sin x cos y corresponding to x = − π


4
, 0, and π


4
, and


y = − π
4
, 0, and π


4
.


Solution
First set x = − π


4
in the equation z = sin x cos y:


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z = sin⎛⎝−
π
4

⎠cos y = −


2 cos y
2


≈ −0.7071 cos y.


This describes a cosine graph in the plane x = − π
4
. The other values of z appear in the following table.


c Vertical Trace for x= c
−π
4 z = −


2 cos y
2


0 z = 0


π
4 z =


2 cos y
2


Table 4.1Vertical Traces Parallel to the xz-Plane
for the Function f (x, y) = sin x cos y


In a similar fashion, we can substitute the y-values in the equation f (x, y) to obtain the traces in the yz-plane,
as listed in the following table.


d Vertical Trace for y = d
−π
4 z = − 2 sin x2


0 z = sin x


π
4 z = 2 sin x2


Table 4.2Vertical Traces Parallel to the yz-Plane
for the Function f (x, y) = sin x cos y


The three traces in the xz-plane are cosine functions; the three traces in the yz-plane are sine functions.
These curves appear in the intersections of the surface with the planes x = − π


4
, x = 0, x = π


4
and


y = − π
4
, y = 0, y = π


4
as shown in the following figure.


Chapter 4 | Differentiation of Functions of Several Variables 343




4.3


Figure 4.11 Vertical traces of the function f (x, y) are cosine curves in the xz-planes (a) and sine curves in the
yz-planes (b).


Determine the equation of the vertical trace of the function g(x, y) = −x2 − y2 + 2x + 4y − 1
corresponding to y = 3, and describe its graph.


Functions of two variables can produce some striking-looking surfaces. The following figure shows two examples.


Figure 4.12 Examples of surfaces representing functions of two variables: (a) a combination of a power function and a sinefunction and (b) a combination of trigonometric, exponential, and logarithmic functions.
Functions of More Than Two Variables
So far, we have examined only functions of two variables. However, it is useful to take a brief look at functions of morethan two variables. Two such examples are


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f (x, y, z) = x2 − 2xy + y2 + 3yz − z2 + 4x − 2y + 3x − 6 (a polynomial in three variables)


and
g(x, y, t) = ⎛⎝x


2 − 4xy + y2⎞⎠sin t −

⎝3x + 5y⎞⎠cos t.


In the first function, (x, y, z) represents a point in space, and the function f maps each point in space to a fourth quantity,
such as temperature or wind speed. In the second function, (x, y) can represent a point in the plane, and t can represent
time. The function might map a point in the plane to a third quantity (for example, pressure) at a given time t. The method
for finding the domain of a function of more than two variables is analogous to the method for functions of one or twovariables.
Example 4.6
Domains for Functions of Three Variables
Find the domain of each of the following functions:


a. f (x, y, z) = 3x − 4y + 2z
9 − x2 − y2 − z2


b. g(x, y, t) = 2t − 4
x2 − y2


Solution
a. For the function f (x, y, z) = 3x − 4y + 2z


9 − x2 − y2 − z2
to be defined (and be a real value), two conditions must


hold:
1. The denominator cannot be zero.
2. The radicand cannot be negative.


Combining these conditions leads to the inequality
9 − x2 − y2 − z2 > 0.


Moving the variables to the other side and reversing the inequality gives the domain as
domain⎛⎝ f ⎞⎠ =





⎨(x, y, z) ∈ ℝ3|x2 + y2 + z2 < 9⎫⎭⎬,


which describes a ball of radius 3 centered at the origin. (Note: The surface of the ball is not included in
this domain.)


b. For the function g(x, y, t) = 2t − 4
x2 − y2


to be defined (and be a real value), two conditions must hold:
1. The radicand cannot be negative.
2. The denominator cannot be zero.


Since the radicand cannot be negative, this implies 2t − 4 ≥ 0, and therefore that t ≥ 2. Since the
denominator cannot be zero, x2 − y2 ≠ 0, or x2 ≠ y2, Which can be rewritten as y = ±x, which
are the equations of two lines passing through the origin. Therefore, the domain of g is


domain(g) = ⎧⎩⎨(x, y, t)|y ≠ ±x, t ≥ 2⎫⎭⎬.


Chapter 4 | Differentiation of Functions of Several Variables 345




4.4 Find the domain of the function h(x, y, t) = (3t − 6) y − 4x2 + 4.


Functions of two variables have level curves, which are shown as curves in the xy-plane. However, when the function has
three variables, the curves become surfaces, so we can define level surfaces for functions of three variables.
Definition
Given a function f (x, y, z) and a number c in the range of f , a level surface of a function of three variables is
defined to be the set of points satisfying the equation f (x, y, z) = c.


Example 4.7
Finding a Level Surface
Find the level surface for the function f (x, y, z) = 4x2 + 9y2 − z2 corresponding to c = 1.
Solution
The level surface is defined by the equation 4x2 + 9y2 − z2 = 1. This equation describes a hyperboloid of one
sheet as shown in the following figure.


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4.5


Figure 4.13 A hyperboloid of one sheet with some of its level surfaces.


Find the equation of the level surface of the function
g(x, y, z) = x2 + y2 + z2 − 2x + 4y − 6z


corresponding to c = 2, and describe the surface, if possible.


Chapter 4 | Differentiation of Functions of Several Variables 347




4.1 EXERCISES
For the following exercises, evaluate each function at theindicated values.
1. W(x, y) = 4x2 + y2. Find W(2, −1), W(−3, 6).
2. W(x, y) = 4x2 + y2. Find W(2 + h, 3 + h).
3. The volume of a right circular cylinder is calculated by
a function of two variables, V(x, y) = πx2 y, where x is
the radius of the right circular cylinder and y represents the
height of the cylinder. Evaluate V(2, 5) and explain what
this means.
4. An oxygen tank is constructed of a right cylinder ofheight y and radius x with two hemispheres of radius x
mounted on the top and bottom of the cylinder. Expressthe volume of the cylinder as a function of two variables,
x and y, find V(10, 2), and explain what this means.
For the following exercises, find the domain of thefunction.
5. V(x, y) = 4x2 + y2


6. f (x, y) = x2 + y2 − 4
7. f (x, y) = 4 ln(y2 − x)
8. g(x, y) = 16 − 4x2 − y2
9. z(x, y) = y2 − x2


10. f (x, y) = y + 2
x2


Find the range of the functions.
11. g(x, y) = 16 − 4x2 − y2
12. V(x, y) = 4x2 + y2
13. z = y2 − x2
For the following exercises, find the level curves of eachfunction at the indicated value of c to visualize the given
function.
14. z(x, y) = y2 − x2, c = 1


15. z(x, y) = y2 − x2, c = 4
16. g(x, y) = x2 + y2; c = 4, c = 9
17. g(x, y) = 4 − x − y; c = 0, 4
18. f (x, y) = xy; c = 1; c = −1
19. h(x, y) = 2x − y; c = 0, −2, 2
20. f (x, y) = x2 − y; c = 1, 2
21. g(x, y) = xx + y; c = −1, 0, 2


22. g(x, y) = x3 − y; c = −1, 0, 2
23. g(x, y) = exy; c = 1


2
, 3


24. f (x, y) = x2; c = 4, 9
25. f (x, y) = xy − x; c = −2, 0, 2
26. h(x, y) = ln(x2 + y2); c = −1, 0, 1
27. g(x, y) = ln⎛⎝ yx2



⎠; c = −2, 0, 2


28. z = f (x, y) = x2 + y2, c = 3
29. f (x, y) = y + 2


x2
, c = any constant


For the following exercises, find the vertical traces of thefunctions at the indicated values of x and y, and plot the
traces.
30. z = 4 − x − y; x = 2
31. f (x, y) = 3x + y3, x = 1
32. z = cos x2 + y2 x = 1
Find the domain of the following functions.
33. z = 100 − 4x2 − 25y2
34. z = ln⎛⎝x − y2⎞⎠


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35. f (x, y, z) = 1
36 − 4x2 − 9y2 − z2


36. f (x, y, z) = 49 − x2 − y2 − z2


37. f (x, y, z) = 16 − x2 − y2 − z23


38. f (x, y) = cos x2 + y2
For the following exercises, plot a graph of the function.
39. z = f (x, y) = x2 + y2
40. z = x2 + y2
41. Use technology to graph z = x2 y.
Sketch the following by finding the level curves. Verify thegraph using technology.
42. f (x, y) = 4 − x2 − y2


43. f (x, y) = 2 − x2 + y2


44. z = 1 + e−x2 − y2


45. z = cos x2 + y2
46. z = y2 − x2
47. Describe the contour lines for several values of c for
z = x2 + y2 − 2x − 2y.


Find the level surface for the functions of three variablesand describe it.
48. w(x, y, z) = x − 2y + z, c = 4
49. w(x, y, z) = x2 + y2 + z2, c = 9
50. w(x, y, z) = x2 + y2 − z2, c = −4
51. w(x, y, z) = x2 + y2 − z2, c = 4
52. w(x, y, z) = 9x2 − 4y2 + 36z2, c = 0
For the following exercises, find an equation of the levelcurve of f that contains the point P.
53. f (x, y) = 1 − 4x2 − y2, P(0, 1)


54. g(x, y) = y2 arctan x, P(1, 2)
55. g(x, y) = exy(x2 + y2), P(1, 0)
56. The strength E of an electric field at point (x, y, z)
resulting from an infinitely long charged wire lying along
the y-axis is given by E(x, y, z) = k/ x2 + y2, where k
is a positive constant. For simplicity, let k = 1 and find the
equations of the level surfaces for E = 10 and E = 100.
57. A thin plate made of iron is located in the xy-plane.
The temperature T in degrees Celsius at a point P(x, y) is
inversely proportional to the square of its distance from theorigin. Express T as a function of x and y.
58. Refer to the preceding problem. Using the temperaturefunction found there, determine the proportionalityconstant if the temperature at point P(1, 2) is 50°C. Use
this constant to determine the temperature at point
Q(3, 4).


59. Refer to the preceding problem. Find the level curvesfor T = 40°C and T = 100°C, and describe what the
level curves represent.


Chapter 4 | Differentiation of Functions of Several Variables 349




4.2 | Limits and Continuity
Learning Objectives


4.2.1 Calculate the limit of a function of two variables.
4.2.2 Learn how a function of two variables can approach different values at a boundary point,depending on the path of approach.
4.2.3 State the conditions for continuity of a function of two variables.
4.2.4 Verify the continuity of a function of two variables at a point.
4.2.5 Calculate the limit of a function of three or more variables and verify the continuity of thefunction at a point.


We have now examined functions of more than one variable and seen how to graph them. In this section, we see how to takethe limit of a function of more than one variable, and what it means for a function of more than one variable to be continuousat a point in its domain. It turns out these concepts have aspects that just don’t occur with functions of one variable.
Limit of a Function of Two Variables
Recall from Section 2.2 the definition of a limit of a function of one variable:
Let f (x) be defined for all x ≠ a in an open interval containing a. Let L be a real number. Then


lim
x → a


f (x) = L


if for every ε > 0, there exists a δ > 0, such that if 0 < |x − a| < δ for all x in the domain of f , then
| f (x) − L| > ε.


Before we can adapt this definition to define a limit of a function of two variables, we first need to see how to extend theidea of an open interval in one variable to an open interval in two variables.
Definition
Consider a point (a, b) ∈ ℝ2. A δ disk centered at point (a, b) is defined to be an open disk of radius δ centered
at point (a, b)—that is,





⎨(x, y) ∈ ℝ2|(x − a)2 + (y − b)2 < δ2⎫⎭⎬


as shown in the following graph.


Figure 4.14 A δ disk centered around the point (2, 1).


The idea of a δ disk appears in the definition of the limit of a function of two variables. If δ is small, then all the points
(x, y) in the δ disk are close to (a, b). This is completely analogous to x being close to a in the definition of a limit of
a function of one variable. In one dimension, we express this restriction as


a − δ < x < a + δ.


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In more than one dimension, we use a δ disk.


Definition
Let f be a function of two variables, x and y. The limit of f (x, y) as (x, y) approaches (a, b) is L, written


lim
(x, y) → (a, b)


f (x, y) = L


if for each ε > 0 there exists a small enough δ > 0 such that for all points (x, y) in a δ disk around (a, b), except
possibly for (a, b) itself, the value of f (x, y) is no more than ε away from L (Figure 4.15). Using symbols, we
write the following: For any ε > 0, there exists a number δ > 0 such that


| f (x, y) − L| < εwhenever 0 < (x − a)2 + ⎛⎝y − b⎞⎠2 < δ.


Figure 4.15 The limit of a function involving two variables requires that f (x, y)
be within ε of L whenever (x, y) is within δ of (a, b). The smaller the value of
ε, the smaller the value of δ.


Proving that a limit exists using the definition of a limit of a function of two variables can be challenging. Instead, we usethe following theorem, which gives us shortcuts to finding limits. The formulas in this theorem are an extension of theformulas in the limit laws theorem in The Limit Laws (http://cnx.org/content/m53492/latest/) .
Theorem 4.1: Limit laws for functions of two variables
Let f (x, y) and g(x, y) be defined for all (x, y) ≠ (a, b) in a neighborhood around (a, b), and assume the
neighborhood is contained completely inside the domain of f . Assume that L and M are real numbers such
that lim


(x, y) → (a, b)
f (x, y) = L and lim


(x, y) → (a, b)
g(x, y) = M, and let c be a constant. Then each of the following


Chapter 4 | Differentiation of Functions of Several Variables 351




statements holds:
Constant Law:


(4.2)lim
(x, y) → (a, b)


c = c


Identity Laws:
(4.3)lim


(x, y) → (a, b)
x = a


(4.4)lim
(x, y) → (a, b)


y = b


Sum Law:
(4.5)lim


(x, y) → (a, b)

⎝ f (x, y) + g(x, y)⎞⎠ = L + M


Difference Law:
(4.6)lim


(x, y) → (a, b)

⎝ f (x, y) − g(x, y)⎞⎠ = L − M


Constant Multiple Law:
(4.7)lim


(x, y) → (a, b)

⎝c f (x, y)⎞⎠ = cL


Product Law:
(4.8)lim


(x, y) → (a, b)

⎝ f (x, y)g(x, y)⎞⎠ = LM


Quotient Law:
(4.9)lim


(x, y) → (a, b)


f (x, y)
g(x, y)


= L
M


for M ≠ 0


Power Law:
(4.10)lim


(x, y) → (a, b)

⎝ f (x, y)⎞⎠n = Ln


for any positive integer n.
Root Law:


(4.11)lim
(x, y) → (a, b)


f (x, y)
n


= L
n


for all L if n is odd and positive, and for L ≥ 0 if n is even and positive.


The proofs of these properties are similar to those for the limits of functions of one variable. We can apply these laws tofinding limits of various functions.
Example 4.8
Finding the Limit of a Function of Two Variables
Find each of the following limits:


a. lim
(x, y) → (2, −1)



⎝x


2 − 2xy + 3y2 − 4x + 3y − 6⎞⎠


b. lim
(x, y) → (2, −1)


2x + 3y
4x − 3y


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Solution
a. First use the sum and difference laws to separate the terms:


lim
(x, y) → (2, −1)



⎝x


2 − 2xy + 3y2 − 4x + 3y − 6⎞⎠


=

⎝ lim(x, y) → (2, −1)x


2⎞
⎠−

⎝ lim(x, y) → (2, −1)2xy



⎠+

⎝ lim(x, y) → (2, −1)3y


2⎞
⎠−

⎝ lim(x, y) → (2, −1)4x





+

⎝ lim(x, y) → (2, −1)3y



⎠−

⎝ lim(x, y) → (2, −1)6



⎠.


Next, use the constant multiple law on the second, third, fourth, and fifth limits:
=

⎝ lim(x, y) → (2, −1)x


2⎞
⎠− 2

⎝ lim(x, y) → (2, −1)xy



⎠+ 3

⎝ lim(x, y) → (2, −1)y


2⎞
⎠− 4

⎝ lim(x, y) → (2, −1)x





+3

⎝ lim(x, y) → (2, −1)y



⎠− lim(x, y) → (2, −1)6.


Now, use the power law on the first and third limits, and the product law on the second limit:
=

⎝ lim(x, y) → (2, −1)x





2


− 2

⎝ lim(x, y) → (2, −1)x





⎝ lim(x, y) → (2, −1)y



⎠+ 3

⎝ lim(x, y) → (2, −1)y





2


−4

⎝ lim(x, y) → (2, −1)x



⎠+ 3

⎝ lim(x, y) → (2, −1)y



⎠− lim(x, y) → (2, −1)6.


Last, use the identity laws on the first six limits and the constant law on the last limit:
lim


(x, y) → (2, −1)

⎝x


2 − 2xy + 3y2 − 4x + 3y − 6⎞⎠ = (2)
2 − 2(2)(−1) + 3(−1)2 − 4(2) + 3(−1) − 6


= −6.


b. Before applying the quotient law, we need to verify that the limit of the denominator is nonzero. Usingthe difference law, constant multiple law, and identity law,
lim


(x, y) → (2, −1)

⎝4x − 3y⎞⎠ = lim


(x, y) → (2, −1)
4x − lim


(x, y) → (2, −1)
3y


= 4

⎝ lim(x, y) → (2, −1)x



⎠− 3

⎝ lim(x, y) → (2, −1)y





= 4(2) − 3(−1) = 11.


Since the limit of the denominator is nonzero, the quotient law applies. We now calculate the limit of thenumerator using the difference law, constant multiple law, and identity law:
lim


(x, y) → (2, −1)

⎝2x + 3y⎞⎠ = lim


(x, y) → (2, −1)
2x + lim


(x, y) → (2, −1)
3y


= 2

⎝ lim(x, y) → (2, −1)x



⎠+ 3

⎝ lim(x, y) → (2, −1)y





= 2(2) + 3(−1)
= 1.


Therefore, according to the quotient law we have


Chapter 4 | Differentiation of Functions of Several Variables 353




4.6


lim
(x, y) → (2, −1)


2x + 3y
4x − 3y


=
lim


(x, y) → (2, −1)

⎝2x + 3y⎞⎠


lim
(x, y) → (2, −1)



⎝4x − 3y⎞⎠


= 1
11


.


Evaluate the following limit:
lim


(x, y) → (5, −2)


x2 − y


y2 + x − 1


3
.


Since we are taking the limit of a function of two variables, the point (a, b) is in ℝ2, and it is possible to approach this
point from an infinite number of directions. Sometimes when calculating a limit, the answer varies depending on the pathtaken toward (a, b). If this is the case, then the limit fails to exist. In other words, the limit must be unique, regardless of
path taken.
Example 4.9
Limits That Fail to Exist
Show that neither of the following limits exist:


a. lim
(x, y) → (0, 0)


2xy


3x2 + y2


b. lim
(x, y) → (0, 0)


4xy2


x2 + 3y4


Solution
a. The domain of the function f (x, y) = 2xy


3x2 + y2
consists of all points in the xy-plane except for the


point (0, 0) (Figure 4.16). To show that the limit does not exist as (x, y) approaches (0, 0), we note
that it is impossible to satisfy the definition of a limit of a function of two variables because of the factthat the function takes different values along different lines passing through point (0, 0). First, consider
the line y = 0 in the xy-plane. Substituting y = 0 into f (x, y) gives


f (x, 0) = 2x(0)
3x2 + 02


= 0


for any value of x. Therefore the value of f remains constant for any point on the x-axis, and as y
approaches zero, the function remains fixed at zero.Next, consider the line y = x. Substituting y = x into f (x, y) gives


f (x, x) = 2x(x)
3x2 + x2


= 2x
2


4x2
= 1


2
.


This is true for any point on the line y = x. If we let x approach zero while staying on this line, the


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value of the function remains fixed at 1
2
, regardless of how small x is.


Choose a value for ε that is less than 1/2—say, 1/4. Then, no matter how small a δ disk we
draw around (0, 0), the values of f (x, y) for points inside that δ disk will include both 0 and 1


2
.


Therefore, the definition of limit at a point is never satisfied and the limit fails to exist.


Figure 4.16 Graph of the function
f (x, y) = ⎛⎝2xy⎞⎠/⎛⎝3x


2 + y2⎞⎠. Along the line y = 0, the
function is equal to zero; along the line y = x, the function is
equal to 1


2
.


In a similar fashion to a., we can approach the origin along any straight line passing through the origin. Ifwe try the x-axis (i.e., y = 0), then the function remains fixed at zero. The same is true for the y-axis.
Suppose we approach the origin along a straight line of slope k. The equation of this line is y = kx.
Then the limit becomes


lim
(x, y) → (0, 0)


4xy2


x2 + 3y4
= lim


(x, y) → (0, 0)


4x(kx)2


x2 + 3(kx)4


= lim
(x, y) → (0, 0)


4k2 x3


x2 + 3k4 x4


= lim
(x, y) → (0, 0)


4k2 x
1 + 3k4 x2


=
lim


(x, y) → (0, 0)

⎝4k


2 x⎞⎠


lim
(x, y) → (0, 0)



⎝1 + 3k


4 x2⎞⎠


= 0


regardless of the value of k. It would seem that the limit is equal to zero. What if we chose a curve
passing through the origin instead? For example, we can consider the parabola given by the equation
x = y2. Substituting y2 in place of x in f (x, y) gives


Chapter 4 | Differentiation of Functions of Several Variables 355




4.7


lim
(x, y) → (0, 0)


4xy2


x2 + 3y4
= lim


(x, y) → (0, 0)


4⎛⎝y
2⎞
⎠y


2



⎝y


2⎞


2
+ 3y4


= lim
(x, y) → (0, 0)


4y4


y4 + 3y4


= lim
(x, y) → (0, 0)


1


= 1.


By the same logic in a., it is impossible to find a δ disk around the origin that satisfies the definition of
the limit for any value of ε < 1. Therefore, lim


(x, y) → (0, 0)


4xy2


x2 + 3y4
does not exist.


Show that
lim


(x, y) → (2, 1)


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


does not exist.


Interior Points and Boundary Points
To study continuity and differentiability of a function of two or more variables, we first need to learn some new terminology.
Definition
Let S be a subset of ℝ2 (Figure 4.17).


1. A point P0 is called an interior point of S if there is a δ disk centered around P0 contained completely in
S.


2. A point P0 is called a boundary point of S if every δ disk centered around P0 contains points both inside
and outside S.


356 Chapter 4 | Differentiation of Functions of Several Variables


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Figure 4.17 In the set S shown, (−1, 1) is an interior point
and (2, 3) is a boundary point.


Definition
Let S be a subset of ℝ2 (Figure 4.17).


1. S is called an open set if every point of S is an interior point.
2. S is called a closed set if it contains all its boundary points.


An example of an open set is a δ disk. If we include the boundary of the disk, then it becomes a closed set. A set that
contains some, but not all, of its boundary points is neither open nor closed. For example if we include half the boundary ofa δ disk but not the other half, then the set is neither open nor closed.


Definition
Let S be a subset of ℝ2 (Figure 4.17).


1. An open set S is a connected set if it cannot be represented as the union of two or more disjoint, nonempty
open subsets.


2. A set S is a region if it is open, connected, and nonempty.


The definition of a limit of a function of two variables requires the δ disk to be contained inside the domain of the function.
However, if we wish to find the limit of a function at a boundary point of the domain, the δ disk is not contained inside
the domain. By definition, some of the points of the δ disk are inside the domain and some are outside. Therefore, we need
only consider points that are inside both the δ disk and the domain of the function. This leads to the definition of the limit
of a function at a boundary point.
Definition
Let f be a function of two variables, x and y, and suppose (a, b) is on the boundary of the domain of f . Then,
the limit of f (x, y) as (x, y) approaches (a, b) is L, written


lim
(x, y) → (a, b)


f (x, y) = L,


Chapter 4 | Differentiation of Functions of Several Variables 357




if for any ε > 0, there exists a number δ > 0 such that for any point (x, y) inside the domain of f and within a
suitably small distance positive δ of (a, b), the value of f (x, y) is no more than ε away from L (Figure 4.15).
Using symbols, we can write: For any ε > 0, there exists a number δ > 0 such that


| f (x, y) − L| < εwhenever 0 < (x − a)2 + ⎛⎝y − b⎞⎠2 < δ.


Example 4.10
Limit of a Function at a Boundary Point
Prove lim


(x, y) → (4, 3)
25 − x2 − y2 = 0.


Solution
The domain of the function f (x, y) = 25 − x2 − y2 is ⎧



⎨(x, y) ∈ ℝ2 |x2 + y2 ≤ 25⎫⎭⎬, which is a circle of


radius 5 centered at the origin, along with its interior as shown in the following graph.


Figure 4.18 Domain of the function
f (x, y) = 25 − x2 − y2.


We can use the limit laws, which apply to limits at the boundary of domains as well as interior points:
lim


(x, y) → (4, 3)
25 − x2 − y2 = lim


(x, y) → (4, 3)

⎝25 − x


2 − y2⎞⎠


= lim
(x, y) → (4, 3)


25 − lim
(x, y) → (4, 3)


x2 − lim
(x, y) → (4, 3)


y2


= 25 − 42 − 32


= 0.


See the following graph.


358 Chapter 4 | Differentiation of Functions of Several Variables


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4.8


Figure 4.19 Graph of the function
f (x, y) = 25 − x2 − y2.


Evaluate the following limit:
lim


(x, y) → (5, −2)
29 − x2 − y2.


Continuity of Functions of Two Variables
In Continuity (http://cnx.org/content/m53489/latest/) , we defined the continuity of a function of one variable andsaw how it relied on the limit of a function of one variable. In particular, three conditions are necessary for f (x) to be
continuous at point x = a:


1. f (a) exists.
2. lim


x → a
f (x) exists.


3. lim
x → a


f (x) = f (a).


These three conditions are necessary for continuity of a function of two variables as well.
Definition
A function f (x, y) is continuous at a point (a, b) in its domain if the following conditions are satisfied:


1. f (a, b) exists.
2. lim


(x, y) → (a, b)
f (x, y) exists.


3. lim
(x, y) → (a, b)


f (x, y) = f (a, b).


Chapter 4 | Differentiation of Functions of Several Variables 359




4.9


Example 4.11
Demonstrating Continuity for a Function of Two Variables
Show that the function f (x, y) = 3x + 2y


x + y + 1
is continuous at point (5, −3).


Solution
There are three conditions to be satisfied, per the definition of continuity. In this example, a = 5 and b = −3.


1. f (a, b) exists. This is true because the domain of the function f consists of those ordered pairs
for which the denominator is nonzero (i.e., x + y + 1 ≠ 0). Point (5, −3) satisfies this condition.
Furthermore,


f (a, b) = f (5, −3) = 3(5) + 2(−3)
5 + (−3) + 1


= 15 − 6
2 + 1


= 3.


2. lim
(x, y) → (a, b)


f (x, y) exists. This is also true:
lim


(x, y) → (a, b)
f (x, y) = lim


(x, y) → (5, −3)


3x + 2y
x + y + 1


=
lim


(x, y) → (5, −3)

⎝3x + 2y⎞⎠


lim
(x, y) → (5, −3)



⎝x + y + 1⎞⎠


= 15 − 6
5 − 3 + 1


= 3.
3. lim


(x, y) → (a, b)
f (x, y) = f (a, b). This is true because we have just shown that both sides of this equation


equal three.


Show that the function f (x, y) = 26 − 2x2 − y2 is continuous at point (2, −3).


Continuity of a function of any number of variables can also be defined in terms of delta and epsilon. A function of twovariables is continuous at a point (x0, y0) in its domain if for every ε > 0 there exists a δ > 0 such that, whenever
(x − x0)


2 + (y − y0)
2 < δ it is true, | f (x, y) − f (a, b)| < ε. This definition can be combined with the formal definition


(that is, the epsilon–delta definition) of continuity of a function of one variable to prove the following theorems:
Theorem 4.2: The Sum of Continuous Functions Is Continuous
If f (x, y) is continuous at (x0, y0), and g(x, y) is continuous at (x0, y0), then f (x, y) + g(x, y) is continuous
at (x0, y0).


Theorem 4.3: The Product of Continuous Functions Is Continuous
If g(x) is continuous at x0 and h(y) is continuous at y0, then f (x, y) = g(x)h(y) is continuous at (x0, y0).


360 Chapter 4 | Differentiation of Functions of Several Variables


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4.10


Theorem 4.4: The Composition of Continuous Functions Is Continuous
Let g be a function of two variables from a domain D ⊆ ℝ2 to a range R ⊆ ℝ. Suppose g is continuous at some
point (x0, y0) ∈ D and define z0 = g(x0, y0). Let f be a function that maps ℝ to ℝ such that z0 is in the domain
of f . Last, assume f is continuous at z0. Then f ∘g is continuous at (x0, y0) as shown in the following figure.


Figure 4.20 The composition of two continuous functions is continuous.


Let’s now use the previous theorems to show continuity of functions in the following examples.
Example 4.12
More Examples of Continuity of a Function of Two Variables
Show that the functions f (x, y) = 4x3 y2 and g(x, y) = cos⎛⎝4x3 y2⎞⎠ are continuous everywhere.
Solution
The polynomials g(x) = 4x3 and h(y) = y2 are continuous at every real number, and therefore by the product
of continuous functions theorem, f (x, y) = 4x3 y2 is continuous at every point (x, y) in the xy-plane. Since
f (x, y) = 4x3 y2 is continuous at every point (x, y) in the xy-plane and g(x) = cos x is continuous at every
real number x, the continuity of the composition of functions tells us that g(x, y) = cos⎛⎝4x3 y2⎞⎠ is continuous
at every point (x, y) in the xy-plane.


Show that the functions f (x, y) = 2x2 y3 + 3 and g(x, y) = ⎛⎝2x2 y3 + 3⎞⎠4 are continuous everywhere.


Functions of Three or More Variables
The limit of a function of three or more variables occurs readily in applications. For example, suppose we have afunction f (x, y, z) that gives the temperature at a physical location (x, y, z) in three dimensions. Or perhaps a function
g(x, y, z, t) can indicate air pressure at a location (x, y, z) at time t. How can we take a limit at a point in ℝ3? What
does it mean to be continuous at a point in four dimensions?
The answers to these questions rely on extending the concept of a δ disk into more than two dimensions. Then, the ideas of
the limit of a function of three or more variables and the continuity of a function of three or more variables are very similarto the definitions given earlier for a function of two variables.


Chapter 4 | Differentiation of Functions of Several Variables 361




Definition
Let (x0, y0, z0) be a point in ℝ3. Then, a δ ball in three dimensions consists of all points in ℝ3 lying at a distance
of less than δ from (x0, y0, z0)—that is,





⎨(x, y, z) ∈ ℝ3| (x − x0)2 + (y − y0)2 + (z − z0)2 < δ⎫⎭⎬.


To define a δ ball in higher dimensions, add additional terms under the radical to correspond to each additional
dimension. For example, given a point P = (w0, x0, y0, z0) in ℝ4, a δ ball around P can be described by





⎨(w, x, y, z) ∈ ℝ4| (w − w0)2 + (x − x0)2 + (y − y0)2 + (z − z0)2 < δ⎫⎭⎬.


To show that a limit of a function of three variables exists at a point (x0, y0, z0), it suffices to show that for any point in a
δ ball centered at (x0, y0, z0), the value of the function at that point is arbitrarily close to a fixed value (the limit value).
All the limit laws for functions of two variables hold for functions of more than two variables as well.
Example 4.13
Finding the Limit of a Function of Three Variables
Find lim


(x, y, z) → (4, 1, −3)


x2 y − 3z
2x + 5y − z


.


Solution
Before we can apply the quotient law, we need to verify that the limit of the denominator is nonzero. Using thedifference law, the identity law, and the constant law,


lim
(x, y, z) → (4, 1, −3)



⎝2x + 5y − z⎞⎠ = 2



⎝ lim(x, y, z) → (4, 1, −3)x



⎠+ 5

⎝ lim(x, y, z) → (4, 1, −3)y



⎠−

⎝ lim(x, y, z) → (4, 1, −3)z





= 2(4) + 5(1) − (−3)
= 16.


Since this is nonzero, we next find the limit of the numerator. Using the product law, difference law, constantmultiple law, and identity law,
lim


(x, y, z) → (4, 1, −3)

⎝x


2 y − 3z⎞⎠ =

⎝ lim(x, y, z) → (4, 1, −3)x





2 ⎛
⎝ lim(x, y, z) → (4, 1, −3)y



⎠− 3 lim(x, y, z) → (4, 1, −3)z


= ⎛⎝4
2⎞
⎠(1) − 3(−3)


= 16 + 9
= 25.


Last, applying the quotient law:
lim


(x, y, z) → (4, 1, −3)


x2 y − 3z
2x + 5y − z


=
lim


(x, y, z) → (4, 1, −3)

⎝x


2 y − 3z⎞⎠


lim
(x, y, z) → (4, 1, −3)



⎝2x + 5y − z⎞⎠


= 25
16


.


362 Chapter 4 | Differentiation of Functions of Several Variables


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4.11 Find lim
(x, y, z) → (4, −1, 3)


13 − x2 − 2y2 + z2.


Chapter 4 | Differentiation of Functions of Several Variables 363




4.2 EXERCISES
For the following exercises, find the limit of the function.
60. lim


(x, y) → (1, 2)
x


61. lim
(x, y) → (1, 2)


5x2 y


x2 + y2


62. Show that the limit lim
(x, y) → (0, 0)


5x2 y


x2 + y2
exists and is


the same along the paths: y-axis and x-axis, and along
y = x.


For the following exercises, evaluate the limits at theindicated values of x and y. If the limit does not exist, state
this and explain why the limit does not exist.
63. lim


(x, y) → (0, 0)


4x2 + 10y2 + 4


4x2 − 10y2 + 6


64. lim
(x, y) → (11, 13)


1
xy


65. lim
(x, y) → (0, 1)


y2 sin x
x


66. lim
(x, y) → (0, 0)


sin



⎜ x


8 + y7


x − y + 10







67. lim
(x, y) → (π/4, 1)


y tan x
y + 1


68. lim
(x, y) → (0, π/4)


sec x + 2
3x − tan y


69. lim
(x, y) → (2, 5)




1
x −


5
y



70. lim
(x, y) → (4, 4)


x ln y


71. lim
(x, y) → (4, 4)


e
−x2 − y2


72. lim
(x, y) → (0, 0)


9 − x2 − y2


73. lim
(x, y) → (1, 2)



⎝x


2 y3 − x3 y2 + 3x + 2y⎞⎠


74. lim
(x, y) → (π, π)


x sin⎛⎝
x + y
4



75. lim
(x, y) → (0, 0)


xy + 1


x2 + y2 + 1


76. lim
(x, y) → (0, 0)


x2 + y2


x2 + y2 + 1 − 1


77. lim
(x, y) → (0, 0)


ln⎛⎝x
2 + y2⎞⎠


For the following exercises, complete the statement.
78. A point (x0, y0) in a plane region R is an interior
point of R if _________________.
79. A point (x0, y0) in a plane region R is called a
boundary point of R if ___________.
For the following exercises, use algebraic techniques toevaluate the limit.
80. lim


(x, y) → (2, 1)


x − y − 1
x − y − 1


81. lim
(x, y) → (0, 0)


x4 − 4y4


x2 + 2y2


82. lim
(x, y) → (0, 0)


x3 − y3
x − y


83. lim
(x, y) → (0, 0)


x2 − xy
x − y


For the following exercises, evaluate the limits of thefunctions of three variables.
84. lim


(x, y, z) → (1, 2, 3)


xz2 − y2 z
xyz − 1


85. lim
(x, y, z) → (0, 0, 0)


x2 − y2 − z2


x2 + y2 − z2


For the following exercises, evaluate the limit of thefunction by determining the value the function approachesalong the indicated paths. If the limit does not exist, explainwhy not.


364 Chapter 4 | Differentiation of Functions of Several Variables


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86. lim
(x, y) → (0, 0)


xy + y3


x2 + y2


a. Along the x-axis (y = 0)
b. Along the y-axis (x = 0)
c. Along the path y = 2x


87. Evaluate lim
(x, y) → (0, 0)


xy + y3


x2 + y2
using the results of


previous problem.
88. lim


(x, y) → (0, 0)


x2 y


x4 + y2


a. Along the x-axis (y = 0)
b. Along the y-axis (x = 0)
c. Along the path y = x2


89. Evaluate lim
(x, y) → (0, 0)


x2 y


x4 + y2
using the results of


previous problem.
Discuss the continuity of the following functions. Find thelargest region in the xy-plane in which the following
functions are continuous.
90. f (x, y) = sin(xy)
91. f (x, y) = ln(x + y)
92. f (x, y) = e3xy


93. f (x, y) = 1xy
For the following exercises, determine the region in whichthe function is continuous. Explain your answer.
94. f (x, y) = x2 y


x2 + y2


95. f (x, y) =









x2 y


x2 + y2
if (x, y) ≠ (0, 0)


0 if (x, y) = (0, 0)










(Hint:


Show that the function approaches different values alongtwo different paths.)
96. f (x, y) = sin(x2 + y2)


x2 + y2


97. Determine whether g(x, y) = x2 − y2
x2 + y2


is continuous
at (0, 0).
98. Create a plot using graphing software to determinewhere the limit does not exist. Determine the region of
the coordinate plane in which f (x, y) = 1


x2 − y
is


continuous.
99. Determine the region of the xy-plane in which the
composite function g(x, y) = arctan⎛



⎜ xy


2


x + y



⎟ is


continuous. Use technology to support your conclusion.
100. Determine the region of the xy-plane in which
f (x, y) = ln(x2 + y2 − 1) is continuous. Use technology
to support your conclusion. (Hint: Choose the range ofvalues for x and y carefully!)
101. At what points in space is
g(x, y, z) = x2 + y2 − 2z2 continuous?
102. At what points in space is g(x, y, z) = 1


x2 + z2 − 1


continuous?
103. Show that lim


(x, y) → (0, 0)
1


x2 + y2
does not exist at


(0, 0) by plotting the graph of the function.


104. [T] Evaluate lim
(x, y) → (0, 0)


−xy2


x2 + y4
by plotting the


function using a CAS. Determine analytically the limit
along the path x = y2.
105. [T]a. Use a CAS to draw a contour map of


z = 9 − x2 − y2.


b. What is the name of the geometric shape of thelevel curves?c. Give the general equation of the level curves.d. What is the maximum value of z?
e. What is the domain of the function?f. What is the range of the function?


106. True or False: If we evaluate lim
(x, y) → (0, 0)


f (x)


along several paths and each time the limit is 1, we can
conclude that lim


(x, y) → (0, 0)
f (x) = 1.


Chapter 4 | Differentiation of Functions of Several Variables 365




107. Use polar coordinates to find
lim


(x, y) → (0, 0)


sin x2 + y2


x2 + y2
. You can also find the limit using


L’Hôpital’s rule.
108. Use polar coordinates to find


lim
(x, y) → (0, 0)


cos⎛⎝x
2 + y2⎞⎠.


109. Discuss the continuity of f (g(x, y)) where
f (t) = 1/t and g(x, y) = 2x − 5y.
110. Given f (x, y) = x2 − 4y, find
lim
h → 0


f (x + h, y) − f (x, y)
h


.


111. Given f (x, y) = x2 − 4y, find
lim
h → 0


f (1 + h, y) − f (1, y)
h


.


366 Chapter 4 | Differentiation of Functions of Several Variables


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4.3 | Partial Derivatives
Learning Objectives


4.3.1 Calculate the partial derivatives of a function of two variables.
4.3.2 Calculate the partial derivatives of a function of more than two variables.
4.3.3 Determine the higher-order derivatives of a function of two variables.
4.3.4 Explain the meaning of a partial differential equation and give an example.


Now that we have examined limits and continuity of functions of two variables, we can proceed to study derivatives.Finding derivatives of functions of two variables is the key concept in this chapter, with as many applications inmathematics, science, and engineering as differentiation of single-variable functions. However, we have already seen thatlimits and continuity of multivariable functions have new issues and require new terminology and ideas to deal with them.This carries over into differentiation as well.
Derivatives of a Function of Two Variables
When studying derivatives of functions of one variable, we found that one interpretation of the derivative is an instantaneousrate of change of y as a function of x. Leibniz notation for the derivative is dy/dx, which implies that y is the dependent
variable and x is the independent variable. For a function z = f (x, y) of two variables, x and y are the independent
variables and z is the dependent variable. This raises two questions right away: How do we adapt Leibniz notation for
functions of two variables? Also, what is an interpretation of the derivative? The answer lies in partial derivatives.
Definition
Let f (x, y) be a function of two variables. Then the partial derivative of f with respect to x, written as ∂ f /∂ x,
or fx, is defined as


(4.12)∂ f
∂ x


= lim
h → 0


f ⎛⎝x + h, y⎞⎠− f (x, y)
h


.


The partial derivative of f with respect to y, written as ∂ f /∂ y, or fy, is defined as
(4.13)∂ f


∂ y
= lim


k → 0


f ⎛⎝x, y + k⎞⎠− f (x, y)
k


.


This definition shows two differences already. First, the notation changes, in the sense that we still use a version of Leibniznotation, but the d in the original notation is replaced with the symbol ∂ . (This rounded “d” is usually called “partial,” so
∂ f /∂ x is spoken as the “partial of f with respect to x.”) This is the first hint that we are dealing with partial derivatives.
Second, we now have two different derivatives we can take, since there are two different independent variables. Dependingon which variable we choose, we can come up with different partial derivatives altogether, and often do.
Example 4.14
Calculating Partial Derivatives from the Definition
Use the definition of the partial derivative as a limit to calculate ∂ f /∂ x and ∂ f /∂ y for the function


f (x, y) = x2 − 3xy + 2y2 − 4x + 5y − 12.


Solution


Chapter 4 | Differentiation of Functions of Several Variables 367




4.12


First, calculate f ⎛⎝x + h, y⎞⎠.
f ⎛⎝x + h, y⎞⎠ = (x + h)2 − 3(x + h)y + 2y2 − 4(x + h) + 5y − 12


= x2 + 2xh + h2 − 3xy − 3hy + 2y2 − 4x − 4h + 5y − 12.


Next, substitute this into Equation 4.12 and simplify:
∂ f
∂ x


= lim
h → 0


f ⎛⎝x + h, y⎞⎠− f (x, y)
h


= lim
h → 0



⎝x


2 + 2xh + h2 − 3xy − 3hy + 2y2 − 4x − 4h + 5y − 12⎞⎠−

⎝x


2 − 3xy + 2y2 − 4x + 5y − 12⎞⎠
h


= lim
h → 0


x2 + 2xh + h2 − 3xy − 3hy + 2y2 − 4x − 4h + 5y − 12 − x2 + 3xy − 2y2 + 4x − 5y + 12
h


= lim
h → 0


2xh + h2 − 3hy − 4h
h


= lim
h → 0


h⎛⎝2x + h − 3y − 4⎞⎠
h


= lim
h → 0



⎝2x + h − 3y − 4⎞⎠


= 2x − 3y − 4.


To calculate ∂ f
∂ y


, first calculate f ⎛⎝x, y + h⎞⎠:
f ⎛⎝x + h, y⎞⎠ = x2 − 3x⎛⎝y + h⎞⎠+ 2⎛⎝y + h⎞⎠2 − 4x + 5⎛⎝y + h⎞⎠− 12


= x2 − 3xy − 3xh + 2y2 + 4yh + 2h2 − 4x + 5y + 5h − 12.


Next, substitute this into Equation 4.13 and simplify:
∂ f
∂ y


= lim
h → 0


f (x, y + h) − f (x, y)
h


= lim
h → 0



⎝x


2 − 3xy − 3xh + 2y2 + 4yh + 2h2 − 4x + 5y + 5h − 12⎞⎠−

⎝x


2 − 3xy + 2y2 − 4x + 5y − 12⎞⎠
h


= lim
h → 0


x2 − 3xy − 3xh + 2y2 + 4yh + 2h2 − 4x + 5y + 5h − 12 − x2 + 3xy − 2y2 + 4x − 5y + 12
h


= lim
h → 0


−3xh + 4yh + 2h2 + 5h
h


= lim
h → 0


h⎛⎝−3x + 4y + 2h + 5⎞⎠
h


= lim
h → 0



⎝−3x + 4y + 2h + 5⎞⎠


= −3x + 4y + 5.


Use the definition of the partial derivative as a limit to calculate ∂ f /∂ x and ∂ f /∂ y for the function
f (x, y) = 4x2 + 2xy − y2 + 3x − 2y + 5.


The idea to keep in mind when calculating partial derivatives is to treat all independent variables, other than the variablewith respect to which we are differentiating, as constants. Then proceed to differentiate as with a function of a singlevariable. To see why this is true, first fix y and define g(x) = f (x, y) as a function of x. Then


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g′ (x) = lim
h → 0


g(x + h) − g(x)
h


= lim
h → 0


f ⎛⎝x + h, y⎞⎠− f (x, y)
h


=
∂ f
∂ x


.


The same is true for calculating the partial derivative of f with respect to y. This time, fix x and define h(y) = f (x, y)
as a function of y. Then


h′ (x) = lim
k → 0


h(x + k) − h(x)
k


= lim
k → 0


f ⎛⎝x, y + k⎞⎠− f (x, y)
k


=
∂ f
∂ y


.


All differentiation rules from Introduction to Derivatives (http://cnx.org/content/m53494/latest/) apply.
Example 4.15
Calculating Partial Derivatives
Calculate ∂ f /∂ x and ∂ f /∂ y for the following functions by holding the opposite variable constant then
differentiating:


a. f (x, y) = x2 − 3xy + 2y2 − 4x + 5y − 12
b. g(x, y) = sin⎛⎝x2 y − 2x + 4⎞⎠


Solution
a. To calculate ∂ f /∂ x, treat the variable y as a constant. Then differentiate f (x, y) with respect to x


using the sum, difference, and power rules:
∂ f
∂ x


= ∂
∂ x

⎣x


2 − 3xy + 2y2 − 4x + 5y − 12⎤⎦


= ∂
∂ x

⎣x


2⎤
⎦−



∂ x

⎣3xy⎤⎦+ ∂∂ x



⎣2y


2⎤
⎦−



∂ x


[4x] + ∂
∂ x

⎣5y⎤⎦− ∂∂ x


[12]


= 2x − 3y + 0 − 4 + 0 − 0
= 2x − 3y − 4.


The derivatives of the third, fifth, and sixth terms are all zero because they do not contain the variable x,
so they are treated as constant terms. The derivative of the second term is equal to the coefficient of x,
which is −3y. Calculating ∂ f /∂ y:


∂ f
∂ y


= ∂
∂ y

⎣x


2 − 3xy + 2y2 − 4x + 5y − 12⎤⎦


= ∂
∂ y

⎣x


2⎤
⎦−



∂ y

⎣3xy⎤⎦+ ∂∂ y



⎣2y


2⎤
⎦−



∂ y


[4x] + ∂
∂ y

⎣5y⎤⎦− ∂∂ y


[12]


= −3x + 4y − 0 + 5 − 0


= −3x + 4y + 5.


These are the same answers obtained in Example 4.14.
b. To calculate ∂g/∂ x, treat the variable y as a constant. Then differentiate g(x, y) with respect to x using


the chain rule and power rule:
∂g
∂ x


= ∂
∂ x

⎣sin

⎝x


2 y − 2x + 4⎞⎠



= cos⎛⎝x
2 y − 2x + 4⎞⎠



∂ x

⎣x


2 y − 2x + 4⎤⎦


= ⎛⎝2xy − 2⎞⎠cos⎛⎝x
2 y − 2x + 4⎞⎠.


Chapter 4 | Differentiation of Functions of Several Variables 369




4.13


To calculate ∂g/∂ y, treat the variable x as a constant. Then differentiate g(x, y) with respect to y
using the chain rule and power rule:


∂g
∂ y


= ∂
∂ y

⎣sin

⎝x


2 y − 2x + 4⎞⎠



= cos⎛⎝x
2 y − 2x + 4⎞⎠



∂ y

⎣x


2 y − 2x + 4⎤⎦


= x2 cos⎛⎝x
2 y − 2x + 4⎞⎠.


Calculate ∂ f /∂ x and ∂ f /∂ y for the function f (x, y) = tan⎛⎝x3 − 3x2 y2 + 2y4⎞⎠ by holding the
opposite variable constant, then differentiating.


How can we interpret these partial derivatives? Recall that the graph of a function of two variables is a surface in ℝ3. If
we remove the limit from the definition of the partial derivative with respect to x, the difference quotient remains:


f ⎛⎝x + h, y⎞⎠− f (x, y)
h


.


This resembles the difference quotient for the derivative of a function of one variable, except for the presence of the y
variable. Figure 4.21 illustrates a surface described by an arbitrary function z = f (x, y).


Figure 4.21 Secant line passing through the points

⎝x, y, f (x, y)⎞⎠ and ⎛⎝x + h, y, f ⎛⎝x + h, y⎞⎠⎞⎠.


In Figure 4.21, the value of h is positive. If we graph f (x, y) and f ⎛⎝x + h, y⎞⎠ for an arbitrary point (x, y), then the
slope of the secant line passing through these two points is given by


f ⎛⎝x + h, y⎞⎠− f (x, y)
h


.


This line is parallel to the x-axis. Therefore, the slope of the secant line represents an average rate of change of the function
f as we travel parallel to the x-axis. As h approaches zero, the slope of the secant line approaches the slope of the tangent
line.


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If we choose to change y instead of x by the same incremental value h, then the secant line is parallel to the y-axis and
so is the tangent line. Therefore, ∂ f /∂ x represents the slope of the tangent line passing through the point ⎛⎝x, y, f (x, y)⎞⎠
parallel to the x-axis and ∂ f /∂ y represents the slope of the tangent line passing through the point ⎛⎝x, y, f (x, y)⎞⎠ parallel
to the y-axis. If we wish to find the slope of a tangent line passing through the same point in any other direction, then we
need what are called directional derivatives, which we discuss in Directional Derivatives and the Gradient.
We now return to the idea of contour maps, which we introduced in Functions of Several Variables. We can use acontour map to estimate partial derivatives of a function g(x, y).
Example 4.16
Partial Derivatives from a Contour Map
Use a contour map to estimate ∂g/∂ x at the point ⎛⎝ 5, 0⎞⎠ for the function g(x, y) = 9 − x2 − y2.
Solution
The following graph represents a contour map for the function g(x, y) = 9 − x2 − y2.


Figure 4.22 Contour map for the function
g(x, y) = 9 − x2 − y2, using c = 0, 1, 2, and 3
(c = 3 corresponds to the origin).


The inner circle on the contour map corresponds to c = 2 and the next circle out corresponds to c = 1. The first
circle is given by the equation 2 = 9 − x2 − y2; the second circle is given by the equation 1 = 9 − x2 − y2.
The first equation simplifies to x2 + y2 = 5 and the second equation simplifies to x2 + y2 = 8. The
x-intercept of the first circle is ⎛⎝ 5, 0⎞⎠ and the x-intercept of the second circle is ⎛⎝2 2, 0⎞⎠. We can estimate
the value of ∂g/∂ x evaluated at the point ⎛⎝ 5, 0⎞⎠ using the slope formula:


∂g
∂ x |(x, y) = ⎛⎝ 5, 0⎞⎠ ≈ g



⎝ 5, 0



⎠− g



⎝2 2, 0





5 − 2 2
= 2 − 1


5 − 2 2
= 1


5 − 2 2
≈ −1.688.


To calculate the exact value of ∂g/∂ x evaluated at the point ⎛⎝ 5, 0⎞⎠, we start by finding ∂g/∂ x using the


Chapter 4 | Differentiation of Functions of Several Variables 371




4.14


chain rule. First, we rewrite the function as g(x, y) = 9 − x2 − y2 = ⎛⎝9 − x2 − y2⎞⎠1/2 and then differentiate
with respect to x while holding y constant:


∂g
∂ x


= 1
2

⎝9 − x


2 − y2⎞⎠
−1/2


(−2x) = − x
9 − x2 − y2


.


Next, we evaluate this expression using x = 5 and y = 0:
∂g
∂ x |(x, y) = ⎛⎝ 5, 0⎞⎠ = − 59 − ⎛⎝ 5⎞⎠2 − (0)2 = − 54 = − 52 ≈ −1.118.


The estimate for the partial derivative corresponds to the slope of the secant line passing through the points

⎝ 5, 0, g



⎝ 5, 0





⎠ and ⎛⎝2 2, 0, g⎛⎝2 2, 0⎞⎠⎞⎠. It represents an approximation to the slope of the tangent line to the


surface through the point ⎛⎝ 5, 0, g⎛⎝ 5, 0⎞⎠⎞⎠, which is parallel to the x-axis.


Use a contour map to estimate ∂ f /∂ y at point ⎛⎝0, 2⎞⎠ for the function
f (x, y) = x2 − y2.


Compare this with the exact answer.


Functions of More Than Two Variables
Suppose we have a function of three variables, such as w = f (x, y, z). We can calculate partial derivatives of w with
respect to any of the independent variables, simply as extensions of the definitions for partial derivatives of functions of twovariables.
Definition
Let f (x, y, z) be a function of three variables. Then, the partial derivative of f with respect to x, written as ∂ f /∂ x,
or fx, is defined to be


(4.14)∂ f
∂ x


= lim
h → 0


f ⎛⎝x + h, y, z⎞⎠− f (x, y, z)
h


.


The partial derivative of f with respect to y, written as ∂ f /∂ y, or fy, is defined to be
(4.15)∂ f


∂ y
= lim


k → 0


f ⎛⎝x, y + k, z⎞⎠− f (x, y, z)
k


.


The partial derivative of f with respect to z, written as ∂ f /∂z, or fz, is defined to be
(4.16)∂ f


∂z
= lim


m → 0


f (x, y, z + m) − f (x, y, z)
m .


We can calculate a partial derivative of a function of three variables using the same idea we used for a function of twovariables. For example, if we have a function f of x, y, and z, and we wish to calculate ∂ f /∂ x, then we treat the other
two independent variables as if they are constants, then differentiate with respect to x.


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Example 4.17
Calculating Partial Derivatives for a Function of Three Variables
Use the limit definition of partial derivatives to calculate ∂ f /∂ x for the function


f (x, y, z) = x2 − 3xy + 2y2 − 4xz + 5yz2 − 12x + 4y − 3z.


Then, find ∂ f /∂ y and ∂ f /∂z by setting the other two variables constant and differentiating accordingly.
Solution
We first calculate ∂ f /∂ x using Equation 4.14, then we calculate the other two partial derivatives by holding
the remaining variables constant. To use the equation to find ∂ f /∂ x, we first need to calculate f ⎛⎝x + h, y, z⎞⎠:


f ⎛⎝x + h, y, z⎞⎠ = (x + h)2 − 3(x + h)y + 2y2 − 4(x + h)z + 5yz2 − 12(x + h) + 4y − 3z


= x2 + 2xh + h2 − 3xy − 3xh + 2y2 − 4xz − 4hz + 5yz2 − 12x − 12h + 4y − 3z


and recall that f (x, y, z) = x2 − 3xy + 2y2 − 4zx + 5yz2 − 12x + 4y − 3z. Next, we substitute these two
expressions into the equation:


∂ f
∂ x


= lim
h → 0





⎢ x


2 + 2xh + h2 − 3xy − 3hy + 2y2 − 4xz − 4hz + 5yz2 − 12x − 12h + 4y − 3z
h



x2 − 3xy + 2y2 − 4xz + 5yz2 − 12x + 4y − 3z


h







= lim
h → 0





⎢ 2xh + h


2 − 3hy − 4hz − 12h
h







= lim
h → 0




h⎛⎝2x + h − 3y − 4z − 12⎞⎠


h



= lim
h → 0



⎝2x + h − 3y − 4z − 12⎞⎠


= 2x − 3y − 4z − 12.


Then we find ∂ f /∂ y by holding x and z constant. Therefore, any term that does not include the variable y
is constant, and its derivative is zero. We can apply the sum, difference, and power rules for functions of onevariable:



∂ y

⎣x


2 − 3xy + 2y2 − 4xz + 5yz2 − 12x + 4y − 3z⎤⎦


= ∂
∂ y

⎣x


2⎤
⎦−



∂ y

⎣3xy⎤⎦+ ∂∂ y



⎣2y


2⎤
⎦−



∂ y


[4xz] + ∂
∂ y

⎣5yz


2⎤
⎦−



∂ y


[12x] + ∂
∂ y

⎣4y⎤⎦− ∂∂ y


[3z]


= 0 − 3x + 4y − 0 + 5z2 − 0 + 4 − 0


= −3x + 4y + 5z2 + 4.


To calculate ∂ f /∂z, we hold x and y constant and apply the sum, difference, and power rules for functions of
one variable:



∂z

⎣x


2 − 3xy + 2y2 − 4xz + 5yz2 − 12x + 4y − 3z⎤⎦
= ∂


∂z

⎣x


2⎤
⎦−



∂z

⎣3xy⎤⎦+ ∂∂z



⎣2y


2⎤
⎦−



∂z


[4xz] + ∂
∂z

⎣5yz


2⎤
⎦−



∂z


[12x] + ∂
∂z

⎣4y⎤⎦− ∂∂z


[3z]


= 0 − 0 + 0 − 4x + 10yz − 0 + 0 − 3
= −4x + 10yz − 3.


Chapter 4 | Differentiation of Functions of Several Variables 373




4.15 Use the limit definition of partial derivatives to calculate ∂ f /∂ x for the function
f (x, y, z) = 2x2 − 4x2 y + 2y2 + 5xz2 − 6x + 3z − 8.


Then find ∂ f /∂ y and ∂ f /∂z by setting the other two variables constant and differentiating accordingly.


Example 4.18
Calculating Partial Derivatives for a Function of Three Variables
Calculate the three partial derivatives of the following functions.


a. f (x, y, z) = x2 y − 4xz + y2
x − 3yz


b. g(x, y, z) = sin⎛⎝x2 y − z⎞⎠+ cos⎛⎝x2 − yz⎞⎠
Solution
In each case, treat all variables as constants except the one whose partial derivative you are calculating.


a.


∂ f
∂ x


= ∂
∂ x





⎢ x


2 y − 4xz + y2


x − 3yz







=

∂ x

⎝x


2 y − 4xz + y2⎞⎠

⎝x − 3yz⎞⎠− ⎛⎝x


2 y − 4xz + y2⎞⎠

∂ x

⎝x − 3yz⎞⎠



⎝x − 3yz⎞⎠2


=

⎝2xy − 4z⎞⎠⎛⎝x − 3yz⎞⎠− ⎛⎝x


2 y − 4xz + y2⎞⎠(1)

⎝x − 3yz⎞⎠2


=
2x2 y − 6xy2 z − 4xz + 12yz2 − x2 y + 4xz − y2



⎝x − 3yz⎞⎠2


=
x2 y − 6xy2 z − 4xz + 12yz2 + 4xz − y2



⎝x − 3yz⎞⎠2


∂ f
∂ y


= ∂
∂ y





⎢ x


2 y − 4xz + y2


x − 3yz







=

∂ y

⎝x


2 y − 4xz + y2⎞⎠

⎝x − 3yz⎞⎠− ⎛⎝x


2 y − 4xz + y2⎞⎠

∂ y

⎝x − 3yz⎞⎠



⎝x − 3yz⎞⎠2


=

⎝x


2 + 2y⎞⎠

⎝x − 3yz⎞⎠− ⎛⎝x


2 y − 4xz + y2⎞⎠(−3z)

⎝x − 3yz⎞⎠2


=
x3 − 3x2 yz + 2xy − 6y2 z + 3x2 yz − 12xz2 + 3y2 z



⎝x − 3yz⎞⎠2


=
x3 + 2xy − 3y2 z − 12xz2



⎝x − 3yz⎞⎠2


374 Chapter 4 | Differentiation of Functions of Several Variables


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4.16


∂ f
∂z


= ∂
∂z





⎢ x


2 y − 4xz + y2


x − 3yz







=

∂z

⎝x


2 y − 4xz + y2⎞⎠

⎝x − 3yz⎞⎠− ⎛⎝x


2 y − 4xz + y2⎞⎠

∂z

⎝x − 3yz⎞⎠



⎝x − 3yz⎞⎠2


=
(−4x)⎛⎝x − 3yz⎞⎠− ⎛⎝x


2 y − 4xz + y2⎞⎠

⎝−3y⎞⎠



⎝x − 3yz⎞⎠2


=
−4x2 + 12xyz + 3x2 y2 − 12xyz + 3y3



⎝x − 3yz⎞⎠2


=
−4x2 + 3x2 y2 + 3y3



⎝x − 3yz⎞⎠2


b.


∂ f
∂ x


= ∂
∂ x

⎣sin

⎝x


2 y − z⎞⎠+ cos

⎝x


2 − yz⎞⎠



= ⎛⎝cos

⎝x


2 y − z⎞⎠



∂ x

⎝x


2 y − z⎞⎠−

⎝sin

⎝x


2 − yz⎞⎠



∂ x

⎝x


2 − yz⎞⎠


= 2xy cos⎛⎝x
2 y − z⎞⎠− 2x sin



⎝x


2 − yz⎞⎠
∂ f
∂ y


= ∂
∂ y

⎣sin

⎝x


2 y − z⎞⎠+ cos

⎝x


2 − yz⎞⎠



= ⎛⎝cos

⎝x


2 y − z⎞⎠



∂ y

⎝x


2 y − z⎞⎠−

⎝sin

⎝x


2 − yz⎞⎠



∂ y

⎝x


2 − yz⎞⎠


= x2 cos⎛⎝x
2 y − z⎞⎠+ z sin



⎝x


2 − yz⎞⎠
∂ f
∂z


= ∂
∂z

⎣sin

⎝x


2 y − z⎞⎠+ cos

⎝x


2 − yz⎞⎠



= ⎛⎝cos

⎝x


2 y − z⎞⎠



∂z

⎝x


2 y − z⎞⎠−

⎝sin

⎝x


2 − yz⎞⎠



∂z

⎝x


2 − yz⎞⎠


= −cos⎛⎝x
2 y − z⎞⎠+ y sin



⎝x


2 − yz⎞⎠


Calculate ∂ f /∂ x, ∂ f /∂ y, and ∂ f /∂z for the function f (x, y, z) = sec⎛⎝x2 y⎞⎠− tan⎛⎝x3 yz2⎞⎠.


Higher-Order Partial Derivatives
Consider the function


f (x, y) = 2x3 − 4xy2 + 5y3 − 6xy + 5x − 4y + 12.


Its partial derivatives are
∂ f
∂ x


= 6x2 − 4y2 − 6y + 5 and
∂ f
∂ y


= −8xy + 15y2 − 6x − 4.


Each of these partial derivatives is a function of two variables, so we can calculate partial derivatives of these functions.Just as with derivatives of single-variable functions, we can call these second-order derivatives, third-order derivatives, andso on. In general, they are referred to as higher-order partial derivatives. There are four second-order partial derivativesfor any function (provided they all exist):
∂2 f


∂ x2
= ∂


∂ x


∂ f
∂ x

⎦,


∂2 f
∂ x∂ y


= ∂
∂ y


∂ f
∂ x

⎦,


∂2 f
∂ y∂ x


= ∂
∂ x


∂ f
∂ y

⎦,


∂2 f


∂ y2
= ∂


∂ y


∂ f
∂ y

⎦.


Chapter 4 | Differentiation of Functions of Several Variables 375




An alternative notation for each is fxx, fxy, fyx, and fyy, respectively. Higher-order partial derivatives calculated with
respect to different variables, such as fxy and fyx, are commonly called mixed partial derivatives.


Example 4.19
Calculating Second Partial Derivatives
Calculate all four second partial derivatives for the function


f (x, y) = xe
−3y


+ sin⎛⎝2x − 5y⎞⎠.


Solution
To calculate ∂2 f /dx2 and ∂2 f /∂ x∂ y, we first calculate ∂ f /∂ x:


∂ f
∂ x


= e
−3y


+ 2 cos⎛⎝2x − 5y⎞⎠.


To calculate ∂2 f /dx2, differentiate ∂ f /∂ x with respect to x:
∂2 f


∂ x2
= ∂


∂ x


∂ f
∂ x



= ∂
∂ x

⎣e


−3y
+ 2 cos⎛⎝2x − 5y⎞⎠





= −4 sin⎛⎝2x − 5y⎞⎠.


To calculate ∂2 f /∂ x∂ y, differentiate ∂ f /∂ x with respect to y:
∂2 f
∂ x∂ y


= ∂
∂ y


∂ f
∂ x



= ∂
∂ y

⎣e


−3y
+ 2 cos⎛⎝2x − 5y⎞⎠





= −3e
−3y


+ 10 sin⎛⎝2x − 5y⎞⎠.


To calculate ∂2 f /∂ x∂ y and ∂2 f /dy2, first calculate ∂ f /∂ y:
∂ f
∂ y


= −3xe
−3y


− 5 cos⎛⎝2x − 5y⎞⎠.


To calculate ∂2 f /∂ y∂ x, differentiate ∂ f /∂ y with respect to x:
∂2 f
∂ y∂ x


= ∂
∂ x


∂ f
∂ y



= ∂
∂ x

⎣−3xe


−3y
− 5 cos⎛⎝2x − 5y⎞⎠





= −3e
−3y


+ 10 sin⎛⎝2x − 5y⎞⎠.


To calculate ∂2 f /∂ y2, differentiate ∂ f /∂ y with respect to y:


376 Chapter 4 | Differentiation of Functions of Several Variables


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4.17


∂2 f


∂ y2
= ∂


∂ y


∂ f
∂ y



= ∂
∂ y

⎣−3xe


−3y
− 5 cos⎛⎝2x − 5y⎞⎠





= 9xe
−3y


− 25 sin⎛⎝2x − 5y⎞⎠.


Calculate all four second partial derivatives for the function
f (x, y) = sin⎛⎝3x − 2y⎞⎠+ cos⎛⎝x + 4y⎞⎠.


At this point we should notice that, in both Example 4.19 and the checkpoint, it was true that ∂2 f /∂ x∂ y = ∂2 f /∂ y∂ x.
Under certain conditions, this is always true. In fact, it is a direct consequence of the following theorem.
Theorem 4.5: Equality of Mixed Partial Derivatives (Clairaut’s Theorem)
Suppose that f (x, y) is defined on an open disk D that contains the point (a, b). If the functions fxy and fyx are
continuous on D, then fxy = fyx.


Clairaut’s theorem guarantees that as long as mixed second-order derivatives are continuous, the order in which we chooseto differentiate the functions (i.e., which variable goes first, then second, and so on) does not matter. It can be extended tohigher-order derivatives as well. The proof of Clairaut’s theorem can be found in most advanced calculus books.
Two other second-order partial derivatives can be calculated for any function f (x, y). The partial derivative fxx is equal
to the partial derivative of fx with respect to x, and fyy is equal to the partial derivative of fy with respect to y.
Partial Differential Equations
In Introduction to Differential Equations (http://cnx.org/content/m53696/latest/) , we studied differentialequations in which the unknown function had one independent variable. A partial differential equation is an equation thatinvolves an unknown function of more than one independent variable and one or more of its partial derivatives. Examplesof partial differential equations are


(4.17)ut = c2 ⎛⎝uxx + uyy⎞⎠
( heat equation in two dimensions)


(4.18)utt = c2 ⎛⎝uxx + uyy⎞⎠
( wave equation in two dimensions)


(4.19)uxx + uyy = 0
( Laplace’s equation in two dimensions)
In the first two equations, the unknown function u has three independent variables— t, x, and y—and c is an arbitrary
constant. The independent variables x and y are considered to be spatial variables, and the variable t represents time. In
Laplace’s equation, the unknown function u has two independent variables x and y.


Chapter 4 | Differentiation of Functions of Several Variables 377




(4.20)


4.18


Example 4.20
A Solution to the Wave Equation
Verify that


u(x, y, t) = 5 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt)


is a solution to the wave equation
utt = 4



⎝uxx + uyy



⎠.


Solution
First, we calculate utt, uxx, and uyy :


utt =

∂ t


∂u
∂ t



= ∂
∂ t

⎣5 sin(3πx)sin⎛⎝4πy⎞⎠⎛⎝−10π sin(10πt)⎞⎠⎤⎦


= ∂
∂ t

⎣−50π sin(3πx)sin⎛⎝4πy⎞⎠sin(10πt)⎤⎦


= −500π2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt)


uxx = ∂∂ x


∂u
∂ x



= ∂
∂ x

⎣15π cos(3πx)sin⎛⎝4πy⎞⎠cos(10πt)⎤⎦


= −45π2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt)


uyy = ∂∂ y


∂u
∂ y



= ∂
∂ y

⎣5 sin(3πx)⎛⎝4π cos⎛⎝4πy⎞⎠⎞⎠cos(10πt)⎤⎦


= ∂
∂ y

⎣20π sin(3πx)cos⎛⎝4πy⎞⎠cos(10πt)⎤⎦


= −80π2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt).


Next, we substitute each of these into the right-hand side of Equation 4.20 and simplify:
4⎛⎝uxx + uyy



⎠ = 4



⎝−45π


2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt) + − 80π2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt)⎞⎠


= 4⎛⎝−125π
2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt)⎞⎠


= −500π2 sin(3πx)sin⎛⎝4πy⎞⎠cos(10πt)
= utt.


This verifies the solution.


Verify that u(x, y, t) = 2 sin⎛⎝x3⎞⎠sin⎛⎝y4⎞⎠e−25t/16 is a solution to the heat equation
(4.21)ut = 9⎛⎝uxx + uyy⎞⎠.


Since the solution to the two-dimensional heat equation is a function of three variables, it is not easy to create a visualrepresentation of the solution. We can graph the solution for fixed values of t, which amounts to snapshots of the heatdistributions at fixed times. These snapshots show how the heat is distributed over a two-dimensional surface as timeprogresses. The graph of the preceding solution at time t = 0 appears in the following figure. As time progresses, the
extremes level out, approaching zero as t approaches infinity.


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Figure 4.23


If we consider the heat equation in one dimension, then it is possible to graph the solution over time. The heat equation inone dimension becomes
ut = c


2uxx,


where c2 represents the thermal diffusivity of the material in question. A solution of this differential equation can be written
in the form


(4.22)
um (x, t) = e


−π2m2 c2 t sin(mπx)


where m is any positive integer. A graph of this solution using m = 1 appears in Figure 4.24, where the initial
temperature distribution over a wire of length 1 is given by u(x, 0) = sin πx. Notice that as time progresses, the wire
cools off. This is seen because, from left to right, the highest temperature (which occurs in the middle of the wire) decreasesand changes color from red to blue.


Chapter 4 | Differentiation of Functions of Several Variables 379




Figure 4.24 Graph of a solution of the heat equation in one dimension over time.


380 Chapter 4 | Differentiation of Functions of Several Variables


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Lord Kelvin and the Age of Earth


Figure 4.25 (a) William Thomson (Lord Kelvin), 1824-1907, was a British physicist andelectrical engineer; (b) Kelvin used the heat diffusion equation to estimate the age of Earth(credit: modification of work by NASA).


During the late 1800s, the scientists of the new field of geology were coming to the conclusion that Earth must be“millions and millions” of years old. At about the same time, Charles Darwin had published his treatise on evolution.Darwin’s view was that evolution needed many millions of years to take place, and he made a bold claim that theWeald chalk fields, where important fossils were found, were the result of 300 million years of erosion.
At that time, eminent physicist William Thomson (Lord Kelvin) used an important partial differential equation, knownas the heat diffusion equation, to estimate the age of Earth by determining how long it would take Earth to cool frommolten rock to what we had at that time. His conclusion was a range of 20 to 400 million years, but most likely
about 50 million years. For many decades, the proclamations of this irrefutable icon of science did not sit well with
geologists or with Darwin.


Read Kelvin’s paper (http://www.openstaxcollege.org/l/20_KelEarthAge) on estimating the age ofthe Earth.


Kelvin made reasonable assumptions based on what was known in his time, but he also made several assumptionsthat turned out to be wrong. One incorrect assumption was that Earth is solid and that the cooling was thereforevia conduction only, hence justifying the use of the diffusion equation. But the most serious error was a forgivableone—omission of the fact that Earth contains radioactive elements that continually supply heat beneath Earth’s mantle.The discovery of radioactivity came near the end of Kelvin’s life and he acknowledged that his calculation would haveto be modified.
Kelvin used the simple one-dimensional model applied only to Earth’s outer shell, and derived the age from graphsand the roughly known temperature gradient near Earth’s surface. Let’s take a look at a more appropriate version ofthe diffusion equation in radial coordinates, which has the form


(4.23)∂T
∂ t


= K


∂2T
∂2r


+ 2r
∂T
∂r

⎦.


Chapter 4 | Differentiation of Functions of Several Variables 381




Here, T(r, t) is temperature as a function of r (measured from the center of Earth) and time t. K is the heat
conductivity—for molten rock, in this case. The standard method of solving such a partial differential equation is byseparation of variables, where we express the solution as the product of functions containing each variable separately.In this case, we would write the temperature as


T(r, t) = R(r) f (t).


1. Substitute this form into Equation 4.13 and, noting that f (t) is constant with respect to distance (r) and
R(r) is constant with respect to time (t), show that


1
f
∂ f
∂ t


= K
R


∂2R
∂r2


+ 2r
∂R
∂r

⎦.


2. This equation represents the separation of variables we want. The left-hand side is only a function of t and
the right-hand side is only a function of r, and they must be equal for all values of r and t. Therefore, they
both must be equal to a constant. Let’s call that constant −λ2. (The convenience of this choice is seen on
substitution.) So, we have


1
f
∂ f
∂ t


= −λ2 and K
R


∂2R
∂r2


+ 2r
∂R
∂r

⎦ = −λ


2.


Now, we can verify through direct substitution for each equation that the solutions are f (t) = Ae−λ2 t and
R(r) = B⎛⎝


sin αr
r

⎠+ C




cos αr


r

⎠, where α = λ/ K. Note that f (t) = Ae+λn2 t is also a valid solution, so we


could have chosen +λ2 for our constant. Can you see why it would not be valid for this case as time increases?
3. Let’s now apply boundary conditions.


a. The temperature must be finite at the center of Earth, r = 0. Which of the two constants, B or C,
must therefore be zero to keep R finite at r = 0? (Recall that sin(αr)/r → α = as r → 0, but
cos(αr)/r behaves very differently.)


b. Kelvin argued that when magma reaches Earth’s surface, it cools very rapidly. A person can often touchthe surface within weeks of the flow. Therefore, the surface reached a moderate temperature very earlyand remained nearly constant at a surface temperature Ts. For simplicity, let’s set T = 0 at r = RE
and find α such that this is the temperature there for all time t. (Kelvin took the value to be
300 K ≈ 80°F. We can add this 300 K constant to our solution later.) For this to be true, the sine
argument must be zero at r = RE. Note that α has an infinite series of values that satisfies this
condition. Each value of α represents a valid solution (each with its own value for A). The total or
general solution is the sum of all these solutions.


c. At t = 0, we assume that all of Earth was at an initial hot temperature T0 (Kelvin took this to be
about 7000 K.) The application of this boundary condition involves the more advanced application of
Fourier coefficients. As noted in part b. each value of αn represents a valid solution, and the general
solution is a sum of all these solutions. This results in a series solution:


T(r, t) =


T0RE
π

⎠∑n


(−1)n − 1
n e


−λn2 t sin(αn r)
r , where αn = nπ/RE.


Note how the values of αn come from the boundary condition applied in part b. The term −1n − 1n is the constant
An for each term in the series, determined from applying the Fourier method. Letting β = πRE, examine the first
few terms of this solution shown here and note how λ2 in the exponential causes the higher terms to decrease quickly
as time progresses:


382 Chapter 4 | Differentiation of Functions of Several Variables


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T(r, t) =
T0RE
πr







⎜e


−Kβ2 t ⎛
⎝sin βr⎞⎠− 12


e
−4Kβ2 t ⎛


⎝sin 2βr⎞⎠+ 13
e
−9Kβ2 t ⎛


⎝sin 3βr⎞⎠


−1
4
e
−16Kβ2 t ⎛


⎝sin 4βr⎞⎠+ 15
e
−25Kβ2 t ⎛


⎝sin 5βr⎞⎠...








.


Near time t = 0, many terms of the solution are needed for accuracy. Inserting values for the conductivity K and
β = π/RE for time approaching merely thousands of years, only the first few terms make a significant contribution.
Kelvin only needed to look at the solution near Earth’s surface (Figure 4.26) and, after a long time, determine whattime best yielded the estimated temperature gradient known during his era (1°F increase per 50 ft). He simply chose
a range of times with a gradient close to this value. In Figure 4.26, the solutions are plotted and scaled, with the
300 − K surface temperature added. Note that the center of Earth would be relatively cool. At the time, it was thought
Earth must be solid.


Figure 4.26 Temperature versus radial distance from the center of Earth. (a) Kelvin’s results, plotted to scale. (b) Aclose-up of the results at a depth of 4.0 mi below Earth’s surface.


Epilog
On May 20, 1904, physicist Ernest Rutherford spoke at the Royal Institution to announce a revised calculation that
included the contribution of radioactivity as a source of Earth’s heat. In Rutherford’s own words:
“I came into the room, which was half-dark, and presently spotted Lord Kelvin in the audience, and realised that I wasin for trouble at the last part of my speech dealing with the age of the Earth, where my views conflicted with his. Tomy relief, Kelvin fell fast asleep, but as I came to the important point, I saw the old bird sit up, open an eye and cock abaleful glance at me.
Then a sudden inspiration came, and I said Lord Kelvin had limited the age of the Earth, provided no new source [ofheat] was discovered. That prophetic utterance referred to what we are now considering tonight, radium! Behold! Theold boy beamed upon me.”
Rutherford calculated an age for Earth of about 500 million years. Today’s accepted value of Earth’s age is about 4.6
billion years.


Chapter 4 | Differentiation of Functions of Several Variables 383




4.3 EXERCISES
For the following exercises, calculate the partial derivativeusing the limit definitions only.
112. ∂z


∂ x
for z = x2 − 3xy + y2


113. ∂z
∂ y


for z = x2 − 3xy + y2


For the following exercises, calculate the sign of the partialderivative using the graph of the surface.


114. fx(1, 1)
115. fx(−1, 1)
116. fy(1, 1)
117. fx(0, 0)
For the following exercises, calculate the partialderivatives.
118. ∂z


∂ x
for z = sin(3x)cos(3y)


119. ∂z
∂ y


for z = sin(3x)cos(3y)


120. ∂z
∂ x


and ∂z
∂ y


for z = x8 e3y


121. ∂z
∂ x


and ∂z
∂ y


for z = ln⎛⎝x6 + y4⎞⎠
122. Find fy(x, y) for f (x, y) = exy cos(x)sin(y).
123. Let z = exy. Find ∂z


∂ x
and ∂z


∂ y
.


124. Let z = ln⎛⎝xy⎞⎠. Find ∂z∂ x and ∂z∂ y.


125. Let z = tan(2x − y). Find ∂z
∂ x


and ∂z
∂ y


.


126. Let z = sinh⎛⎝2x + 3y⎞⎠. Find ∂z∂ x and ∂z∂ y.


127. Let f (x, y) = arctan⎛⎝yx⎞⎠. Evaluate fx(2, −2) and
fy(2, −2).


128. Let f (x, y) = xyx − y. Find fx(2, −2) and
fy(2, −2).


Evaluate the partial derivatives at point P(0, 1).
129. Find ∂z


∂ x
at (0, 1) for z = e−x cos(y).


130. Given f (x, y, z) = x3 yz2, find ∂2 f
∂ x∂ y


and
fz(1, 1, 1).


131. Given f (x, y, z) = 2 sin(x + y), find
fx

⎝0,


π
2
, −4⎞⎠, fy



⎝0,


π
2
, −4⎞⎠, and fz ⎛⎝0, π2, −4⎞⎠.


132. The area of a parallelogram with adjacent sidelengths that are a and b, and in which the angle between
these two sides is θ, is given by the function
A(a, b, θ) = ba sin(θ). Find the rate of change of the area
of the parallelogram with respect to the following:a. Side ab. Side bc. Angle θ
133. Express the volume of a right circular cylinder as afunction of two variables:a. its radius r and its height h.


b. Show that the rate of change of the volume of thecylinder with respect to its radius is the product ofits circumference multiplied by its height.c. Show that the rate of change of the volume of thecylinder with respect to its height is equal to thearea of the circular base.
134. Calculate ∂w


∂z
for w = z sin(xy2 + 2z).


Find the indicated higher-order partial derivatives.
135. fxy for z = ln(x − y)


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136. fyx for z = ln(x − y)


137. Let z = x2 + 3xy + 2y2. Find ∂2z
∂ x2


and ∂2z
∂ y2


.


138. Given z = ex tan y, find ∂2z
∂ x∂ y


and ∂2z
∂ y∂ x


.


139. Given f (x, y, z) = xyz, find fxyy, fyxy, and
fyyx.


140. Given f (x, y, z) = e−2x sin⎛⎝z2 y⎞⎠, show that
fxyy = fyxy.


141. Show that z = 1
2

⎝e
y − e


−y⎞
⎠sin x is a solution of the


differential equation ∂2z
∂ x2


+ ∂
2z


∂ y2
= 0.


142. Find fxx(x, y) for f (x, y) = 4x2y + y
2


2x
.


143. Let f (x, y, z) = x2 y3 z − 3xy2 z3 + 5x2 z − y3 z.
Find fxyz.
144. Let F(x, y, z) = x3 yz2 − 2x2 yz + 3xz − 2y3 z.
Find Fxyz.
145. Given f (x, y) = x2 + x − 3xy + y3 − 5, find all
points at which fx = fy = 0 simultaneously.
146. Given f (x, y) = 2x2 + 2xy + y2 + 2x − 3, find
all points at which ∂ f


∂ x
= 0 and ∂ f


∂ y
= 0 simultaneously.


147. Given f (x, y) = y3 − 3yx2 − 3y2 − 3x2 + 1, find
all points on f at which fx = fy = 0 simultaneously.
148. Given f (x, y) = 15x3 − 3xy + 15y3, find all
points at which fx(x, y) = fy(x, y) = 0 simultaneously.
149. Show that z = ex sin y satisfies the equation
∂2z
∂ x2


+ ∂
2z


∂ y2
= 0.


150. Show that f (x, y) = ln⎛⎝x2 + y2⎞⎠ solves Laplace’s
equation ∂2z


∂ x2
+ ∂


2z
∂ y2


= 0.


151. Show that z = e−t cos⎛⎝xc⎞⎠ satisfies the heat equation
∂z
∂ t


= −e−t cos⎛⎝xc

⎠.


152. Find lim
Δx → 0


f (x + Δx) − f (x, y)
Δx


for
f (x, y) = −7x − 2xy + 7y.


153. Find lim
Δy → 0


f (x, y + Δy) − f (x, y)
Δy


for
f (x, y) = −7x − 2xy + 7y.


154. Find lim
Δx → 0


Δ f
Δx


= lim
Δx → 0


f (x + Δx, y) − f (x, y)
Δx


for f (x, y) = x2 y2 + xy + y.
155. Find lim


Δx → 0


Δ f
Δx


= lim
Δx → 0


f (x + Δx, y) − f (x, y)
Δx


for f (x, y) = sin(xy).
156. The function P(T , V) = nRT


V
gives the pressure at


a point in a gas as a function of temperature T and volume
V . The letters n and R are constants. Find ∂P


∂V
and ∂P


∂T
,


and explain what these quantities represent.
157. The equation for heat flow in the xy-plane is
∂ f
∂ t


=
∂2 f


∂ x2
+


∂2 f


∂ y2
. Show that


f (x, y, t) = e−2t sin x sin y is a solution.
158. The basic wave equation is ftt = fxx. Verify that
f (x, t) = sin(x + t) and f (x, t) = sin(x − t) are
solutions.
159. The law of cosines can be thought of as a functionof three variables. Let x, y, and θ be two sides of any
triangle where the angle θ is the included angle between
the two sides. Then, F(x, y, θ) = x2 + y2 − 2xy cos θ
gives the square of the third side of the triangle. Find ∂F


∂θ


and ∂F
∂ x


when x = 2, y = 3, and θ = π
6
.


Chapter 4 | Differentiation of Functions of Several Variables 385




160. Suppose the sides of a rectangle are changing withrespect to time. The first side is changing at a rate of 2
in./sec whereas the second side is changing at the rate of 4
in/sec. How fast is the diagonal of the rectangle changingwhen the first side measures 16 in. and the second side
measures 20 in.? (Round answer to three decimal places.)
161. A Cobb-Douglas production function is
f (x, y) = 200x0.7 y0.3, where x and y represent the
amount of labor and capital available. Let x = 500 and
y = 1000. Find δ f


δx
and δ f


δy
at these values, which


represent the marginal productivity of labor and capital,respectively.
162. The apparent temperature index is a measure of howthe temperature feels, and it is based on two variables:
h, which is relative humidity, and t, which is the air
temperature. A = 0.885t − 22.4h + 1.20th − 0.544. Find
∂A
∂ t


and ∂A
∂h


when t = 20°F and h = 0.90.


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4.4 | Tangent Planes and Linear Approximations
Learning Objectives


4.4.1 Determine the equation of a plane tangent to a given surface at a point.
4.4.2 Use the tangent plane to approximate a function of two variables at a point.
4.4.3 Explain when a function of two variables is differentiable.
4.4.4 Use the total differential to approximate the change in a function of two variables.


In this section, we consider the problem of finding the tangent plane to a surface, which is analogous to finding the equationof a tangent line to a curve when the curve is defined by the graph of a function of one variable, y = f (x). The slope of the
tangent line at the point x = a is given by m = f ′(a); what is the slope of a tangent plane? We learned about the equation
of a plane in Equations of Lines and Planes in Space; in this section, we see how it can be applied to the problem athand.
Tangent Planes
Intuitively, it seems clear that, in a plane, only one line can be tangent to a curve at a point. However, in three-dimensionalspace, many lines can be tangent to a given point. If these lines lie in the same plane, they determine the tangent plane at thatpoint. A more intuitive way to think of a tangent plane is to assume the surface is smooth at that point (no corners). Then,a tangent line to the surface at that point in any direction does not have any abrupt changes in slope because the directionchanges smoothly. Therefore, in a small-enough neighborhood around the point, a tangent plane touches the surface at thatpoint only.
Definition
Let P0 = (x0, y0, z0) be a point on a surface S, and let C be any curve passing through P0 and lying entirely in
S. If the tangent lines to all such curves C at P0 lie in the same plane, then this plane is called the tangent plane to
S at P0 (Figure 4.27).


Chapter 4 | Differentiation of Functions of Several Variables 387




Figure 4.27 The tangent plane to a surface S at a point P0 contains all the tangent lines
to curves in S that pass through P0.


For a tangent plane to a surface to exist at a point on that surface, it is sufficient for the function that defines the surface tobe differentiable at that point. We define the term tangent plane here and then explore the idea intuitively.
Definition
Let S be a surface defined by a differentiable function z = f (x, y), and let P0 = (x0, y0) be a point in the domain
of f . Then, the equation of the tangent plane to S at P0 is given by


(4.24)z = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0).
To see why this formula is correct, let’s first find two tangent lines to the surface S. The equation of the tangent
line to the curve that is represented by the intersection of S with the vertical trace given by x = x0 is
z = f (x0, y0) + fy (x0, y0)(y − y0). Similarly, the equation of the tangent line to the curve that is represented by the
intersection of S with the vertical trace given by y = y0 is z = f (x0, y0) + fx (x0, y0)(x − x0). A parallel vector to the
first tangent line is a = j + fy (x0, y0)k; a parallel vector to the second tangent line is b = i + fx (x0, y0)k. We can take
the cross product of these two vectors:


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a × b = ⎛⎝j + fy (x0, y0)k

⎠×

⎝i + fx (x0, y0)k





= |i j k0 1 fy (x0, y0)1 0 fx (x0, y0)|
= fx (x0, y0)i + fy (x0, y0)j − k.


This vector is perpendicular to both lines and is therefore perpendicular to the tangent plane. We can use this vector as anormal vector to the tangent plane, along with the point P0 = ⎛⎝x0, y0, f (x0, y0)⎞⎠ in the equation for a plane:
n · ⎛⎝(x − x0)i + (y − y0)j +



⎝z − f (x0, y0)



⎠k⎞⎠ = 0



⎝ fx (x0, y0)i + fy (x0, y0)j-k



⎠ ·

⎝(x − x0)i + (y − y0)j +



⎝z − f (x0, y0)



⎠k⎞⎠ = 0


fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0) −

⎝z − f (x0, y0)



⎠ = 0.


Solving this equation for z gives Equation 4.24.
Example 4.21
Finding a Tangent Plane
Find the equation of the tangent plane to the surface defined by the function
f (x, y) = 2x2 − 3xy + 8y2 + 2x − 4y + 4 at point (2, −1).
Solution
First, we must calculate fx (x, y) and fy (x, y), then use Equation 4.24 with x0 = 2 and y0 = −1:


fx (x, y) = 4x − 3y + 2


fy (x, y) = −3x + 16y − 4


f (2, −1) = 2(2)2 − 3(2)(−1) + 8(−1)2 + 2(2) − 4(−1) + 4 = 34.


fx (2, −1) = 4(2) − 3(−1) + 2 = 13


fy (2, −1) = −3(2) + 16(−1) − 4 = −26.


Then Equation 4.24 becomes
z = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0)


z = 34 + 13(x − 2) − 26⎛⎝y − (−1)⎞⎠
z = 34 + 13x − 26 − 26y − 26


z = 13x − 26y − 18.


(See the following figure).


Chapter 4 | Differentiation of Functions of Several Variables 389




4.19


Figure 4.28 Calculating the equation of a tangent plane to a given surface at a given point.


Find the equation of the tangent plane to the surface defined by the function
f (x, y) = x3 − x2 y + y2 − 2x + 3y − 2 at point (−1, 3).


Example 4.22
Finding Another Tangent Plane
Find the equation of the tangent plane to the surface defined by the function f (x, y) = sin(2x)cos⎛⎝3y⎞⎠ at the point
(π/3, π/4).


Solution
First, calculate fx (x, y) and fy (x, y), then use Equation 4.24 with x0 = π/3 and y0 = π/4:


fx (x, y) = 2 cos(2x)cos

⎝3y⎞⎠


fy (x, y) = −3 sin(2x)sin

⎝3y⎞⎠


f ⎛⎝
π
3
, π
4

⎠ = sin



⎝2


π
3



⎠cos

⎝3


π
4



⎠ =



3
2



⎝−


2
2

⎠ = −


6
4


fx


π
3
, π
4

⎠ = 2 cos



⎝2


π
3



⎠cos

⎝3


π
4



⎠ = 2

⎝−


1
2



⎝−


2
2

⎠ =


2
2


fy


π
3
, π
4

⎠ = −3 sin



⎝2


π
3



⎠sin

⎝3


π
4



⎠ = −3





3
2





2
2

⎠ = −


3 6
4


.


Then Equation 4.24 becomes


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z = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0)


z = − 6
4


+ 2
2

⎝x −


π
3

⎠−


3 6
4

⎝y −


π
4



z = 2
2
x − 3 6


4
y − 6


4
− π 2


6
+ 3π 6


16
.


A tangent plane to a surface does not always exist at every point on the surface. Consider the function
f (x, y) =







xy


x2 + y2
(x, y) ≠ (0, 0)


0 (x, y) = (0, 0).


The graph of this function follows.


Figure 4.29 Graph of a function that does not have a tangent plane at theorigin.


If either x = 0 or y = 0, then f (x, y) = 0, so the value of the function does not change on either the x- or y-axis.
Therefore, fx (x, 0) = fy ⎛⎝0, y⎞⎠ = 0, so as either x or y approach zero, these partial derivatives stay equal to zero.
Substituting them into Equation 4.24 gives z = 0 as the equation of the tangent line. However, if we approach the origin
from a different direction, we get a different story. For example, suppose we approach the origin along the line y = x. If
we put y = x into the original function, it becomes


f (x, x) = x(x)


x2 + (x)2
= x


2


2x2
= |x|


2
.


When x > 0, the slope of this curve is equal to 2/2; when x < 0, the slope of this curve is equal to −⎛⎝ 2/2⎞⎠. This
presents a problem. In the definition of tangent plane, we presumed that all tangent lines through point P (in this case, the
origin) lay in the same plane. This is clearly not the case here. When we study differentiable functions, we will see that thisfunction is not differentiable at the origin.
Linear Approximations
Recall from Linear Approximations and Differentials (http://cnx.org/content/m53605/latest/) that the formula


Chapter 4 | Differentiation of Functions of Several Variables 391




for the linear approximation of a function f (x) at the point x = a is given by
y ≈ f (a) + f ′(a)(x − a).


The diagram for the linear approximation of a function of one variable appears in the following graph.


Figure 4.30 Linear approximation of a function in onevariable.


The tangent line can be used as an approximation to the function f (x) for values of x reasonably close to x = a. When
working with a function of two variables, the tangent line is replaced by a tangent plane, but the approximation idea is muchthe same.
Definition
Given a function z = f (x, y) with continuous partial derivatives that exist at the point (x0, y0), the linear
approximation of f at the point (x0, y0) is given by the equation


(4.25)L(x, y) = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0).
Notice that this equation also represents the tangent plane to the surface defined by z = f (x, y) at the point (x0, y0). The
idea behind using a linear approximation is that, if there is a point (x0, y0) at which the precise value of f (x, y) is known,
then for values of (x, y) reasonably close to (x0, y0), the linear approximation (i.e., tangent plane) yields a value that
is also reasonably close to the exact value of f (x, y) (Figure 4.31). Furthermore the plane that is used to find the linear
approximation is also the tangent plane to the surface at the point (x0, y0).


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Figure 4.31 Using a tangent plane for linear approximation at a point.


Example 4.23
Using a Tangent Plane Approximation
Given the function f (x, y) = 41 − 4x2 − y2, approximate f (2.1, 2.9) using point (2, 3) for (x0, y0).
What is the approximate value of f (2.1, 2.9) to four decimal places?
Solution
To apply Equation 4.25, we first must calculate f (x0, y0), fx (x0, y0), and fy (x0, y0) using x0 = 2 and
y0 = 3:


f (x0, y0) = f (2, 3) = 41 − 4(2)
2 − (3)2 = 41 − 16 − 9 = 16 = 4


fx (x, y) = − 4x
41 − 4x2 − y2


so fx (x0, y0) = −
4(2)


41 − 4(2)2 − (3)2
= −2


fy (x, y) = −
y


41 − 4x2 − y2
so fy (x0, y0) = −


3


41 − 4(2)2 − (3)2
= − 3


4
.


Now we substitute these values into Equation 4.25:


Chapter 4 | Differentiation of Functions of Several Variables 393




4.20


L(x, y) = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0)


= 4 − 2(x − 2) − 3
4

⎝y − 3⎞⎠


= 41
4


− 2x − 3
4
y.


Last, we substitute x = 2.1 and y = 2.9 into L(x, y):
L(2.1, 2.9) = 41


4
− 2(2.1) − 3


4
(2.9) = 10.25 − 4.2 − 2.175 = 3.875.


The approximate value of f (2.1, 2.9) to four decimal places is
f (2.1, 2.9) = 41 − 4(2.1)2 − (2.9)2 = 14.95 ≈ 3.8665,


which corresponds to a 0.2% error in approximation.


Given the function f (x, y) = e5 − 2x + 3y, approximate f (4.1, 0.9) using point (4, 1) for (x0, y0).
What is the approximate value of f (4.1, 0.9) to four decimal places?


Differentiability
When working with a function y = f (x) of one variable, the function is said to be differentiable at a point x = a if f ′ (a)
exists. Furthermore, if a function of one variable is differentiable at a point, the graph is “smooth” at that point (i.e., nocorners exist) and a tangent line is well-defined at that point.
The idea behind differentiability of a function of two variables is connected to the idea of smoothness at that point. In thiscase, a surface is considered to be smooth at point P if a tangent plane to the surface exists at that point. If a function is
differentiable at a point, then a tangent plane to the surface exists at that point. Recall the formula for a tangent plane at apoint (x0, y0) is given by


z = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0),


For a tangent plane to exist at the point (x0, y0), the partial derivatives must therefore exist at that point. However, this is
not a sufficient condition for smoothness, as was illustrated in Figure 4.29. In that case, the partial derivatives existed atthe origin, but the function also had a corner on the graph at the origin.
Definition
A function f (x, y) is differentiable at a point P(x0, y0) if, for all points (x, y) in a δ disk around P, we can
write


(4.26)f (x, y) = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0) + E(x, y),
where the error term E satisfies


lim
(x, y) → (x0, y0)


E(x, y)


(x − x0)
2 + (y − y0)


2
= 0.


The last term in Equation 4.26 is referred to as the error term and it represents how closely the tangent plane comes to thesurface in a small neighborhood (δ disk) of point P. For the function f to be differentiable at P, the function must be
smooth—that is, the graph of f must be close to the tangent plane for points near P.


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4.21


Example 4.24
Demonstrating Differentiability
Show that the function f (x, y) = 2x2 − 4y is differentiable at point (2, −3).
Solution
First, we calculate f (x0, y0), fx (x0, y0), and fy (x0, y0) using x0 = 2 and y0 = −3, then we use Equation
4.26:


f (2, −3) = 2(2)2 − 4(−3) = 8 + 12 = 20


fx (2, −3) = 4(2) = 8
fy (2, −3) = −4.


Therefore m1 = 8 and m2 = −4, and Equation 4.26 becomes
f (x, y) = f (2, −3) + fx (2, −3)(x − 2) + fy (2, −3)



⎝y + 3⎞⎠+ E(x, y)


2x2 − 4y = 20 + 8(x − 2) − 4⎛⎝y + 3⎞⎠+ E(x, y)


2x2 − 4y = 20 + 8x − 16 − 4y − 12 + E(x, y)


2x2 − 4y = 8x − 4y − 8 + E(x, y)


E(x, y) = 2x2 − 8x + 8.


Next, we calculate lim
(x, y) → (x0, y0)


E(x, y)


(x − x0)
2 + (y − y0)


2
:


lim
(x, y) → (x0, y0)


E(x, y)


(x − x0)
2 + (y − y0)


2
= lim


(x, y) → (2, −3)
2x2 − 8x + 8


(x − 2)2 + ⎛⎝y + 3⎞⎠2


= lim
(x, y) → (2, −3)


2⎛⎝x
2 − 4x + 4⎞⎠


(x − 2)2 + ⎛⎝y + 3⎞⎠2


= lim
(x, y) → (2, −3)


2(x − 2)2


(x − 2)2 + ⎛⎝y + 3⎞⎠2


≤ lim
(x, y) → (2, −3)


2⎛⎝(x − 2)
2 + ⎛⎝y + 3⎞⎠2⎞⎠


(x − 2)2 + ⎛⎝y + 3⎞⎠2


= lim
(x, y) → (2, −3)


2 (x − 2)2 + ⎛⎝y + 3⎞⎠2


= 0.


Since E(x, y) ≥ 0 for any value of x or y, the original limit must be equal to zero. Therefore,
f (x, y) = 2x2 − 4y is differentiable at point (2, −3).


Show that the function f (x, y) = 3x − 4y2 is differentiable at point (−1, 2).


Chapter 4 | Differentiation of Functions of Several Variables 395




The function f (x, y) = ⎧



xy


x2 + y2
(x, y) ≠ (0, 0)


0 (x, y) = (0, 0)
is not differentiable at the origin. We can see this by calculating


the partial derivatives. This function appeared earlier in the section, where we showed that fx (0, 0) = fy (0, 0) = 0.
Substituting this information into Equation 4.26 using x0 = 0 and y0 = 0, we get


f (x, y) = f (0, 0) + fx (0, 0)(x − 0) + fy (0, 0)

⎝y − 0⎞⎠+ E(x, y)


E(x, y) =
xy


x2 + y2
.


Calculating lim
(x, y) → (x0, y0)


E(x, y)


(x − x0)
2 + (y − y0)


2
gives


lim
(x, y) → (x0, y0)


E(x, y)


(x − x0)
2 + (y − y0)


2
= lim


(x, y) → (0, 0)


xy


x2 + y2


x2 + y2


= lim
(x, y) → (0, 0)


xy


x2 + y2
.


Depending on the path taken toward the origin, this limit takes different values. Therefore, the limit does not exist and thefunction f is not differentiable at the origin as shown in the following figure.


Figure 4.32 This function f (x, y) is not differentiable at the origin.


Differentiability and continuity for functions of two or more variables are connected, the same as for functions of onevariable. In fact, with some adjustments of notation, the basic theorem is the same.
Theorem 4.6: Differentiability Implies Continuity
Let z = f (x, y) be a function of two variables with (x0, y0) in the domain of f . If f (x, y) is differentiable at
(x0, y0), then f (x, y) is continuous at (x0, y0).


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Differentiability Implies Continuity shows that if a function is differentiable at a point, then it is continuous there.However, if a function is continuous at a point, then it is not necessarily differentiable at that point. For example,
f (x, y) =







xy


x2 + y2
(x, y) ≠ (0, 0)


0 (x, y) = (0, 0)


is continuous at the origin, but it is not differentiable at the origin. This observation is also similar to the situation in single-variable calculus.
Continuity of First Partials Implies Differentiability further explores the connection between continuity anddifferentiability at a point. This theorem says that if the function and its partial derivatives are continuous at a point, thefunction is differentiable.
Theorem 4.7: Continuity of First Partials Implies Differentiability
Let z = f (x, y) be a function of two variables with (x0, y0) in the domain of f . If f (x, y), fx (x, y), and
fy (x, y) all exist in a neighborhood of (x0, y0) and are continuous at (x0, y0), then f (x, y) is differentiable there.


Recall that earlier we showed that the function
f (x, y) =







xy


x2 + y2
(x, y) ≠ (0, 0)


0 (x, y) = (0, 0)


was not differentiable at the origin. Let’s calculate the partial derivatives fx and fy :
∂ f
∂ x


=
y3



⎝x


2 + y2⎞⎠
3/2


and
∂ f
∂ y


= x
3



⎝x


2 + y2⎞⎠
3/2


.


The contrapositive of the preceding theorem states that if a function is not differentiable, then at least one of the hypothesesmust be false. Let’s explore the condition that fx (0, 0) must be continuous. For this to be true, it must be true that
lim


(x, y) → (0, 0)
fx (0, 0) = fx (0, 0):


lim
(x, y) → (0, 0)


fx (x, y) = lim
(x, y) → (0, 0)


y3



⎝x


2 + y2⎞⎠
3/2


.


Let x = ky. Then
lim


(x, y) → (0, 0)


y3



⎝x


2 + y2⎞⎠
3/2


= lim
y → 0


y3





⎝ky⎞⎠2 + y2⎞⎠


3/2


= lim
y → 0


y3



⎝k


2 y2 + y2⎞⎠
3/2


= lim
y → 0


y3


|y|3 ⎛⎝k
2 + 1⎞⎠


3/2


= 1

⎝k


2 + 1⎞⎠
3/2


lim
y → 0


|y|
y .


If y > 0, then this expression equals 1/⎛⎝k2 + 1⎞⎠3/2; if y < 0, then it equals −⎛⎝1/⎛⎝k2 + 1⎞⎠
3/2⎞
⎠. In either case, the value


Chapter 4 | Differentiation of Functions of Several Variables 397




depends on k, so the limit fails to exist.
Differentials
In Linear Approximations and Differentials (http://cnx.org/content/m53605/latest/) we first studied the conceptof differentials. The differential of y, written dy, is defined as f ′ (x)dx. The differential is used to approximate
Δy = f (x + Δx) − f (x), where Δx = dx. Extending this idea to the linear approximation of a function of two variables
at the point (x0, y0) yields the formula for the total differential for a function of two variables.


Definition
Let z = f (x, y) be a function of two variables with (x0, y0) in the domain of f , and let Δx and Δy be chosen so
that ⎛⎝x0 + Δx, y0 + Δy⎞⎠ is also in the domain of f . If f is differentiable at the point (x0, y0), then the differentials
dx and dy are defined as


dx = Δx and dy = Δy.


The differential dz, also called the total differential of z = f (x, y) at (x0, y0), is defined as
(4.27)dz = fx (x0, y0)dx + fy (x0, y0)dy.


Notice that the symbol ∂ is not used to denote the total differential; rather, d appears in front of z. Now, let’s define
Δz = f ⎛⎝x + Δx, y + Δy⎞⎠− f (x, y). We use dz to approximate Δz, so


Δz ≈ dz = fx (x0, y0)dx + fy (x0, y0)dy.


Therefore, the differential is used to approximate the change in the function z = f (x0, y0) at the point (x0, y0) for
given values of Δx and Δy. Since Δz = f ⎛⎝x + Δx, y + Δy⎞⎠− f (x, y), this can be used further to approximate
f ⎛⎝x + Δx, y + Δy⎞⎠:


f ⎛⎝x + Δx, y + Δy⎞⎠ = f (x, y) + Δz


≈ f (x, y) + fx (x0, y0)Δx + fy (x0, y0)Δy.


See the following figure.


Figure 4.33 The linear approximation is calculated via theformula
f ⎛⎝x + Δx, y + Δy⎞⎠ ≈ f (x, y) + fx (x0, y0)Δx + fy (x0, y0)Δy.


One such application of this idea is to determine error propagation. For example, if we are manufacturing a gadget and areoff by a certain amount in measuring a given quantity, the differential can be used to estimate the error in the total volume


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4.22


of the gadget.
Example 4.25
Approximation by Differentials
Find the differential dz of the function f (x, y) = 3x2 − 2xy + y2 and use it to approximate Δz at point
(2, −3). Use Δx = 0.1 and Δy = −0.05. What is the exact value of Δz?
Solution
First, we must calculate f (x0, y0), fx (x0, y0), and fy (x0, y0) using x0 = 2 and y0 = −3:


f (x0, y0) = f (2, −3) = 3(2)
2 − 2(2)(−3) + (−3)2 = 12 + 12 + 9 = 33


fx (x, y) = 6x − 2y
fy (x, y) = −2x + 2y


fx (x0, y0) = fx (2, −3) = 6(2) − 2(−3) = 12 + 6 = 18


fy (x0, y0) = fy (2, −3) = −2(2) + 2(−3) = −4 − 6 = −10.


Then, we substitute these quantities into Equation 4.27:
dz = fx (x0, y0)dx + fy (x0, y0)dy


dz = 18(0.1) − 10(−0.05) = 1.8 + 0.5 = 2.3.


This is the approximation to Δz = f ⎛⎝x0 + Δx, y0 + Δy⎞⎠− f (x0, y0). The exact value of Δz is given by
Δz = f ⎛⎝x0 + Δx, y0 + Δy



⎠− f (x0, y0)


= f (2 + 0.1, −3 − 0.05) − f (2, −3)


= f (2.1, −3.05) − f (2, −3)


= 2.3425.


Find the differential dz of the function f (x, y) = 4y2 + x2 y − 2xy and use it to approximate Δz at
point (1, −1). Use Δx = 0.03 and Δy = −0.02. What is the exact value of Δz?


Differentiability of a Function of Three Variables
All of the preceding results for differentiability of functions of two variables can be generalized to functions of threevariables. First, the definition:
Definition
A function f (x, y, z) is differentiable at a point P(x0, y0, z0) if for all points (x, y, z) in a δ disk around P we
can write


(4.28)f (x, y) = f (x0, y0, z0) + fx (x0, y0, z0)(x − x0) + fy (x0, y0, z0)(y − y0)
+ fz (x0, y0, z0)(z − z0) + E(x, y, z),


where the error term E satisfies


Chapter 4 | Differentiation of Functions of Several Variables 399




lim
(x, y, z) → (x0, y0, z0)


E(x, y, z)


(x − x0)
2 + (y − y0)


2 + (z − z0)
2
= 0.


If a function of three variables is differentiable at a point (x0, y0, z0), then it is continuous there. Furthermore, continuity
of first partial derivatives at that point guarantees differentiability.


400 Chapter 4 | Differentiation of Functions of Several Variables


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4.4 EXERCISES
For the following exercises, find a unit normal vector to thesurface at the indicated point.
163. f (x, y) = x3, (2, −1, 8)
164. ln⎛⎝ xy − z⎞⎠ = 0 when x = y = 1
For the following exercises, as a useful review fortechniques used in this section, find a normal vector and atangent vector at point P.
165. x2 + xy + y2 = 3, P(−1, −1)


166. ⎛⎝x2 + y2⎞⎠2 = 9⎛⎝x2 − y2⎞⎠, P( 2, 1)
167. xy2 − 2x2 + y + 5x = 6, P(4, 2)
168. 2x3 − x2 y2 = 3x − y − 7, P(1, −2)


169. zex2 − y2 − 3 = 0, P(2, 2, 3)
For the following exercises, find the equation for thetangent plane to the surface at the indicated point. (Hint:Solve for z in terms of x and y.)
170. −8x − 3y − 7z = −19, P(1, −1, 2)
171. z = −9x2 − 3y2, P(2, 1, −39)
172. x2 + 10xyz + y2 + 8z2 = 0, P(−1, −1, −1)
173. z = ln(10x2 + 2y2 + 1), P(0, 0, 0)


174. z = e7x2 + 4y2, P(0, 0, 1)
175. xy + yz + zx = 11, P(1, 2, 3)
176. x2 + 4y2 = z2, P(3, 2, 5)
177. x3 + y3 = 3xyz, P⎛⎝1, 2, 32⎞⎠
178. z = axy, P⎛⎝1, 1a, 1⎞⎠
179. z = sin x + sin y + sin(x + y), P(0, 0, 0)
180. h(x, y) = ln x2 + y2, P(3, 4)


181. z = x2 − 2xy + y2, P(1, 2, 1)
For the following exercises, find parametric equations forthe normal line to the surface at the indicated point. (Recallthat to find the equation of a line in space, you need apoint on the line, P0 (x0, y0, z0), and a vector
n = 〈 a, b, c 〉 that is parallel to the line. Then the
equation of the line is
x − x0 = at, y − y0 = bt, z − z0 = ct.)


182. −3x + 9y + 4z = −4, P(1, −1, 2)
183. z = 5x2 − 2y2, P(2, 1, 18)
184. x2 − 8xyz + y2 + 6z2 = 0, P(1, 1, 1)
185. z = ln⎛⎝3x2 + 7y2 + 1⎞⎠, P(0, 0, 0)


186. z = e4x2 + 6y2, P(0, 0, 1)
187. z = x2 − 2xy + y2 at point P(1, 2, 1)
For the following exercises, use the figure shown here.


188. The length of line segment AC is equal to what
mathematical expression?
189. The length of line segment BC is equal to what
mathematical expression?
190. Using the figure, explain what the length of linesegment AB represents.


Chapter 4 | Differentiation of Functions of Several Variables 401




For the following exercises, complete each task.
191. Show that f (x, y) = exy x is differentiable at point
(1, 0).


192. Find the total differential of the function
w = ey cos(x) + z2.


193. Show that f (x, y) = x2 + 3y is differentiable at
every point. In other words, show that
Δz = f (x + Δx, y + Δy) − f (x, y) = fxΔx + fyΔy + ε1Δx + ε2Δy,


where both ε1 and ε2 approach zero as ⎛⎝Δx, Δy⎞⎠
approaches (0, 0).
194. Find the total differential of the function z = xyy + x
where x changes from 10 to 10.5 and y changes from
15 to 13.


195. Let z = f (x, y) = xey. Compute Δz from P(1, 2)
to Q(1.05, 2.1) and then find the approximate change
in z from point P to point Q. Recall
Δz = f (x + Δx, y + Δy) − f (x, y), and dz and Δz are
approximately equal.
196. The volume of a right circular cylinder is given by
V(r, h) = πr2h. Find the differential dV . Interpret the
formula geometrically.
197. See the preceding problem. Use differentials toestimate the amount of aluminum in an enclosed aluminumcan with diameter 8.0 cm and height 12 cm if the
aluminum is 0.04 cm thick.
198. Use the differential dz to approximate the change in
z = 4 − x2 − y2 as (x, y) moves from point (1, 1) to
point (1.01, 0.97). Compare this approximation with the
actual change in the function.
199. Let z = f (x, y) = x2 + 3xy − y2. Find the exact
change in the function and the approximate change in thefunction as x changes from 2.00 to 2.05 and y changes
from 3.00 to 2.96.


200. The centripetal acceleration of a particle moving
in a circle is given by a(r, v) = v2r , where v is the
velocity and r is the radius of the circle. Approximate
the maximum percent error in measuring the accelerationresulting from errors of 3% in v and 2% in r. (Recall
that the percentage error is the ratio of the amount of errorover the original amount. So, in this case, the percentage
error in a is given by daa .)
201. The radius r and height h of a right circular
cylinder are measured with possible errors of 4% and 5%,
respectively. Approximate the maximum possiblepercentage error in measuring the volume (Recall that thepercentage error is the ratio of the amount of error over theoriginal amount. So, in this case, the percentage error in V
is given by dV


V
.)


202. The base radius and height of a right circular coneare measured as 10 in. and 25 in., respectively, with
a possible error in measurement of as much as 0.1 in.
each. Use differentials to estimate the maximum error in thecalculated volume of the cone.
203. The electrical resistance R produced by wiring
resistors R1 and R2 in parallel can be calculated from the
formula 1


R
= 1


R1
+ 1
R2


. If R1 and R2 are measured to be
7Ω and 6Ω, respectively, and if these measurements are
accurate to within 0.05Ω, estimate the maximum possible
error in computing R. (The symbol Ω represents an ohm,
the unit of electrical resistance.)
204. The area of an ellipse with axes of length 2a and
2b is given by the formula A = πab. Approximate the
percent change in the area when a increases by 2% and
b increases by 1.5%.
205. The period T of a simple pendulum with small
oscillations is calculated from the formula T = 2π Lg ,
where L is the length of the pendulum and g is the
acceleration resulting from gravity. Suppose that L and g
have errors of, at most, 0.5% and 0.1%, respectively.
Use differentials to approximate the maximum percentageerror in the calculated value of T .


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206. Electrical power P is given by P = V 2
R


, where
V is the voltage and R is the resistance. Approximate
the maximum percentage error in calculating power if 120
V is applied to a 2000 − Ω resistor and the possible
percent errors in measuring V and R are 3% and 4%,
respectively.
For the following exercises, find the linear approximationof each function at the indicated point.
207. f (x, y) = x y, P(1, 4)
208. f (x, y) = ex cos y; P(0, 0)
209. f (x, y) = arctan(x + 2y), P(1, 0)
210. f (x, y) = 20 − x2 − 7y2, P(2, 1)
211. f (x, y, z) = x2 + y2 + z2, P(3, 2, 6)
212. [T] Find the equation of the tangent plane to the
surface f (x, y) = x2 + y2 at point (1, 2, 5), and graph
the surface and the tangent plane at the point.
213. [T] Find the equation for the tangent plane to thesurface at the indicated point, and graph the surface and the
tangent plane: z = ln(10x2 + 2y2 + 1), P(0, 0, 0).
214. [T] Find the equation of the tangent plane to the
surface z = f (x, y) = sin(x + y2) at point ⎛⎝π4, 0, 22 ⎞⎠,
and graph the surface and the tangent plane.


Chapter 4 | Differentiation of Functions of Several Variables 403




4.5 | The Chain Rule
Learning Objectives


4.5.1 State the chain rules for one or two independent variables.
4.5.2 Use tree diagrams as an aid to understanding the chain rule for several independent andintermediate variables.
4.5.3 Perform implicit differentiation of a function of two or more variables.


In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us tofind the derivative of the composition of two functions. The same thing is true for multivariable calculus, but this time wehave to deal with more than one form of the chain rule. In this section, we study extensions of the chain rule and learn howto take derivatives of compositions of functions of more than one variable.
Chain Rules for One or Two Independent Variables
Recall that the chain rule for the derivative of a composite of two functions can be written in the form


d
dx

⎝ f ⎛⎝g(x)⎞⎠⎞⎠ = f ′⎛⎝g(x)⎞⎠g′(x).


In this equation, both f (x) and g(x) are functions of one variable. Now suppose that f is a function of two variables and
g is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we calculate
the derivative in these cases? The following theorem gives us the answer for the case of one independent variable.
Theorem 4.8: Chain Rule for One Independent Variable
Suppose that x = g(t) and y = h(t) are differentiable functions of t and z = f (x, y) is a differentiable function of
x and y. Then z = f ⎛⎝x(t), y(t)⎞⎠ is a differentiable function of t and


(4.29)dz
dt


= ∂z
∂ x


· dx
dt


+ ∂z
∂ y


·
dy
dt


,


where the ordinary derivatives are evaluated at t and the partial derivatives are evaluated at (x, y).


Proof
The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that f isdifferentiable at the point P(x0, y0), where x0 = g(t0) and y0 = h(t0) for a fixed value of t0. We wish to prove that
z = f ⎛⎝x(t), y(t)⎞⎠ is differentiable at t = t0 and that Equation 4.29 holds at that point as well.
Since f is differentiable at P, we know that


(4.30)z(t) = f (x, y) = f (x0, y0) + fx (x0, y0)(x − x0) + fy (x0, y0)(y − y0) + E(x, y),
where lim


(x, y) → (x0, y0)


E(x, y)


(x − x0)
2 + (y − y0)


2
= 0. We then subtract z0 = f (x0, y0) from both sides of this equation:


z(t) − z(t0) = f

⎝x(t), y(t)⎞⎠− f ⎛⎝x(t0), y(t0)





= fx (x0, y0)

⎝x(t) − x(t0)



⎠+ fy (x0, y0)



⎝y(t) − y(t0)



⎠+ E⎛⎝x(t), y(t)⎞⎠.


Next, we divide both sides by t − t0 :
z(t) − z(t0)


t − t0
= fx (x0, y0)




x(t) − x(t0)


t − t0

⎠+ fy (x0, y0)




y(t) − y(t0)


t − t0

⎠+


E⎛⎝x(t), y(t)⎞⎠
t − t0


.


404 Chapter 4 | Differentiation of Functions of Several Variables


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Then we take the limit as t approaches t0 :
lim
t → t0


z(t) − z(t0)
t − t0


= fx (x0, y0) limt → t0




x(t) − x(t0)


t − t0

⎠+ fy (x0, y0) limt → t0




y(t) − y(t0)


t − t0



+ lim
t → t0


E⎛⎝x(t), y(t)⎞⎠
t − t0


.


The left-hand side of this equation is equal to dz/dt, which leads to
dz
dt


= fx (x0, y0)
dx
dt


+ fy (x0, y0)
dy
dt


+ lim
t → t0


E⎛⎝x(t), y(t)⎞⎠
t − t0


.


The last term can be rewritten as
lim
t → t0


E⎛⎝x(t), y(t)⎞⎠
t − t0


= lim
t → t0





⎜ E(x, y)


(x − x0)
2 + (y − y0)


2


(x − x0)
2 + (y − y0)


2


t − t0







= lim
t → t0





⎜ E(x, y)


(x − x0)
2 + (y − y0)


2





⎟ lim
t → t0





⎜ (x − x0)


2 + (y − y0)
2


t − t0





⎟.


As t approaches t0, ⎛⎝x(t), y(t)⎞⎠ approaches ⎛⎝x(t0), y(t0)⎞⎠, so we can rewrite the last product as


lim
(x, y) → (x0, y0)





⎜ E(x, y)


(x − x0)
2 + (y − y0)


2





⎟ lim
(x, y) → (x0, y0)





⎜ (x − x0)


2 + (y − y0)
2


t − t0





⎟.


Since the first limit is equal to zero, we need only show that the second limit is finite:
lim


(x, y) → (x0, y0)





⎜ (x − x0)


2 + (y − y0)
2


t − t0





⎟ = lim


(x, y) → (x0, y0)





⎜ (x − x0)


2 + (y − y0)
2


(t − t0)
2







= lim
(x, y) → (x0, y0)





⎜ ⎛⎝


x − x0
t − t0



2


+


y − y0
t − t0



2⎞





=

⎝ lim(x, y) → (x0, y0)




x − x0
t − t0





2


+

⎝ lim(x, y) → (x0, y0)




y − y0
t − t0





2


.


Since x(t) and y(t) are both differentiable functions of t, both limits inside the last radical exist. Therefore, this value
is finite. This proves the chain rule at t = t0; the rest of the theorem follows from the assumption that all functions are
differentiable over their entire domains.

Closer examination of Equation 4.29 reveals an interesting pattern. The first term in the equation is ∂ f


∂ x
· dx
dt


and the
second term is ∂ f


∂ y
·
dy
dt


. Recall that when multiplying fractions, cancelation can be used. If we treat these derivatives
as fractions, then each product “simplifies” to something resembling ∂ f /dt. The variables x and y that disappear in
this simplification are often called intermediate variables: they are independent variables for the function f , but are
dependent variables for the variable t. Two terms appear on the right-hand side of the formula, and f is a function of two
variables. This pattern works with functions of more than two variables as well, as we see later in this section.
Example 4.26


Chapter 4 | Differentiation of Functions of Several Variables 405




Using the Chain Rule
Calculate dz/dt for each of the following functions:


a. z = f (x, y) = 4x2 + 3y2, x = x(t) = sin t, y = y(t) = cos t
b. z = f (x, y) = x2 − y2, x = x(t) = e2t, y = y(t) = e−t


Solution
a. To use the chain rule, we need four quantities— ∂z/∂ x, ∂z/∂ y, dx/dt, and dy/dt:


∂z
∂ x


= 8x ∂z
∂ y


= 6y


dx
dt


= cos t
dy
dt


= −sin t


Now, we substitute each of these into Equation 4.29:
dz
dt


= ∂z
∂ x


· dx
dt


+ ∂z
∂ y


·
dy
dt


= (8x)(cos t) + ⎛⎝6y⎞⎠(−sin t)


= 8x cos t − 6y sin t.


This answer has three variables in it. To reduce it to one variable, use the fact that
x(t) = sin t and y(t) = cos t. We obtain


dz
dt


= 8x cos t − 6y sin t


= 8(sin t)cos t − 6(cos t)sin t
= 2 sin t cos t.


This derivative can also be calculated by first substituting x(t) and y(t) into f (x, y), then
differentiating with respect to t:


z = f (x, y)


= f ⎛⎝x(t), y(t)⎞⎠


= 4(x(t))2 + 3⎛⎝y(t)⎞⎠2


= 4sin2 t + 3cos2 t.


Then
dz
dt


= 2(4 sin t)(cos t) + 2(3 cos t)(−sin t)


= 8 sin t cos t − 6 sin t cos t
= 2 sin t cos t,


which is the same solution. However, it may not always be this easy to differentiate in this form.
b. To use the chain rule, we again need four quantities— ∂z/∂ x, ∂z/dy, dx/dt, and dy/dt:


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∂z
∂ x


= x
x2 − y2


∂z
∂ y


=
−y


x2 − y2


dx
dt


= 2e2t dx
dt


= −e−t.


We substitute each of these into Equation 4.29:
dz
dt


= ∂z
∂ x


· dx
dt


+ ∂z
∂ y


·
dy
dt


=



⎜ x
x2 − y2





⎟⎛⎝2e


2t⎞
⎠+



⎜ −y


x2 − y2





⎟⎛⎝−e−t⎞⎠


=
2xe2t − ye−t


x2 − y2
.


To reduce this to one variable, we use the fact that x(t) = e2t and y(t) = e−t. Therefore,
dz
dt


=
2xe2t + ye−t


x2 − y2


=
2⎛⎝e


2t⎞
⎠e


2t + ⎛⎝e
−t⎞
⎠e


−t


e4t − e−2t


= 2e
4t + e−2t


e4t − e−2t
.


To eliminate negative exponents, we multiply the top by e2t and the bottom by e4t:
dz
dt


= 2e
4t + e−2t


e4t − e−2t
· e


2t


e4t


= 2e
6t + 1


e8t − e2t


= 2e
6t + 1


e2t ⎛⎝e
6t − 1⎞⎠


= 2e
6t + 1


et e6t − 1
.


Again, this derivative can also be calculated by first substituting x(t) and y(t) into f (x, y), then
differentiating with respect to t:


z = f (x, y)


= f ⎛⎝x(t), y(t)⎞⎠


= (x(t))2 − ⎛⎝y(t)⎞⎠2


= e4t − e−2t


= ⎛⎝e
4t − e−2t⎞⎠


1/2
.


Then


Chapter 4 | Differentiation of Functions of Several Variables 407




4.23


dz
dt


= 1
2

⎝e


4t − e−2t⎞⎠
−1/2 ⎛
⎝4e


4t + 2e−2t⎞⎠


= 2e
4t + e−2t


e4t − e−2t
.


This is the same solution.


Calculate dz/dt given the following functions. Express the final answer in terms of t.
z = f (x, y) = x2 − 3xy + 2y2, x = x(t) = 3 sin 2t, y = y(t) = 4 cos 2t


It is often useful to create a visual representation of Equation 4.29 for the chain rule. This is called a tree diagram for thechain rule for functions of one variable and it provides a way to remember the formula (Figure 4.34). This diagram can beexpanded for functions of more than one variable, as we shall see very shortly.


Figure 4.34 Tree diagram for the case
dz
dt


= ∂z
∂ x


· dx
dt


+ ∂z
∂ y


·
dy
dt


.


In this diagram, the leftmost corner corresponds to z = f (x, y). Since f has two independent variables, there are two lines
coming from this corner. The upper branch corresponds to the variable x and the lower branch corresponds to the variable
y. Since each of these variables is then dependent on one variable t, one branch then comes from x and one branch
comes from y. Last, each of the branches on the far right has a label that represents the path traveled to reach that branch.
The top branch is reached by following the x branch, then the t branch; therefore, it is labeled (∂z/∂ x) × (dx/dt). The
bottom branch is similar: first the y branch, then the t branch. This branch is labeled ⎛⎝∂z/∂ y⎞⎠× ⎛⎝dy/dt⎞⎠. To get the formula
for dz/dt, add all the terms that appear on the rightmost side of the diagram. This gives us Equation 4.29.
In Chain Rule for Two Independent Variables, z = f (x, y) is a function of x and y, and both x = g(u, v) and
y = h(u, v) are functions of the independent variables u and v.


Theorem 4.9: Chain Rule for Two Independent Variables
Suppose x = g(u, v) and y = h(u, v) are differentiable functions of u and v, and z = f (x, y) is a differentiable
function of x and y. Then, z = f ⎛⎝g(u, v), h(u, v)⎞⎠ is a differentiable function of u and v, and


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(4.31)∂z
∂u


= ∂z
∂ x


∂ x
∂u


+ ∂z
∂ y


∂ x
∂u


and
(4.32)∂z


∂v
= ∂z


∂ x
∂ x
∂v


+ ∂z
∂ y


∂ y
∂v


.


We can draw a tree diagram for each of these formulas as well as follows.


Figure 4.35 Tree diagram for ∂z
∂u


= ∂z
∂ x


· ∂ x
∂u


+ ∂z
∂ y


·
∂ y
∂u


and
∂z
∂v


= ∂z
∂ x


· ∂ x
∂v


+ ∂z
∂ y


·
∂ y
∂v


.


To derive the formula for ∂z/∂u, start from the left side of the diagram, then follow only the branches that end with u
and add the terms that appear at the end of those branches. For the formula for ∂z/∂v, follow only the branches that end
with v and add the terms that appear at the end of those branches.
There is an important difference between these two chain rule theorems. InChain Rule for One Independent Variable,the left-hand side of the formula for the derivative is not a partial derivative, but in Chain Rule for Two IndependentVariables it is. The reason is that, in Chain Rule for One Independent Variable, z is ultimately a function of t
alone, whereas in Chain Rule for Two Independent Variables, z is a function of both u and v.
Example 4.27
Using the Chain Rule for Two Variables
Calculate ∂z/∂u and ∂z/∂v using the following functions:


z = f (x, y) = 3x2 − 2xy + y2, x = x(u, v) = 3u + 2v, y = y(u, v) = 4u − v.


Solution
To implement the chain rule for two variables, we need six partial derivatives—
∂z/∂ x, ∂z/∂ y, ∂ x/∂u, ∂ x/∂v, ∂ y/∂u, and ∂ y/∂v:


Chapter 4 | Differentiation of Functions of Several Variables 409




4.24


∂z
∂ x


= 6x − 2y ∂z
∂ y


= −2x + 2y


∂ x
∂u


= 3 ∂ x
∂v


= 2


∂ y
∂u


= 4
∂ y
∂v


= −1.


To find ∂z/∂u, we use Equation 4.31:
∂z
∂u


= ∂z
∂ x


· ∂ x
∂u


+ ∂z
∂ y


·
∂ y
∂u


= 3⎛⎝6x − 2y⎞⎠+ 4⎛⎝−2x + 2y⎞⎠
= 10x + 2y.


Next, we substitute x(u, v) = 3u + 2v and y(u, v) = 4u − v:
∂z
∂u


= 10x + 2y


= 10(3u + 2v) + 2(4u − v)
= 38u + 18v.


To find ∂z/∂v, we use Equation 4.32:
∂z
∂v


= ∂z
∂ x


∂ x
∂v


+ ∂z
∂ y


∂ y
∂v


= 2⎛⎝6x − 2y⎞⎠+ (−1)⎛⎝−2x + 2y⎞⎠
= 14x − 6y.


Then we substitute x(u, v) = 3u + 2v and y(u, v) = 4u − v:
∂z
∂v


= 14x − 6y


= 14(3u + 2v) − 6(4u − v)
= 18u + 34v.


Calculate ∂z/∂u and ∂z/∂v given the following functions:
z = f (x, y) =


2x − y
x + 3y


, x(u, v) = e2u cos 3v, y(u, v) = e2u sin 3v.


The Generalized Chain Rule
Now that we’ve see how to extend the original chain rule to functions of two variables, it is natural to ask: Can we extendthe rule to more than two variables? The answer is yes, as the generalized chain rule states.
Theorem 4.10: Generalized Chain Rule
Let w = f (x1, x2 ,…, xm) be a differentiable function of m independent variables, and for each i ∈ {1,…,m}, let
xi = xi(t1, t2 ,…, tn) be a differentiable function of n independent variables. Then


(4.33)∂w
∂ t j


= ∂w
∂ x1


∂ x1
∂ t j


+ ∂w
∂ x2


∂ x2
∂ t j


+⋯+ ∂w
∂ xm


∂ xm
∂ t j


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for any j ∈ ⎧⎩⎨1, 2,…, n⎫⎭⎬.


In the next example we calculate the derivative of a function of three independent variables in which each of the threevariables is dependent on two other variables.
Example 4.28
Using the Generalized Chain Rule
Calculate ∂w/∂u and ∂w/∂v using the following functions:


w = f (x, y, z) = 3x2 − 2xy + 4z2


x = x(u, v) = eu sin v


y = y(u, v) = eu cos v


z = z(u, v) = eu.


Solution
The formulas for ∂w/∂u and ∂w/∂v are


∂w
∂u


= ∂w
∂ x


· ∂ x
∂u


+ ∂w
∂ y


·
∂ y
∂u


+ ∂w
∂z


· ∂z
∂u


∂w
∂v


= ∂w
∂ x


· ∂ x
∂v


+ ∂w
∂ y


·
∂ y
∂v


+ ∂w
∂z


· ∂z
∂v


.


Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need tocalculate each of them:
∂w
∂ x


= 6x − 2y ∂w
∂ y


= −2x ∂w
∂z


= 8z


∂ x
∂u


= eu sin v
∂ y
∂u


= eu cos v ∂z
∂u


= eu


∂ x
∂v


= eu cos v
∂ y
∂v


= −eu sin v ∂z
∂v


= 0.


Now, we substitute each of them into the first formula to calculate ∂w/∂u:
∂w
∂u


= ∂w
∂ x


· ∂ x
∂u


+ ∂w
∂ y


·
∂ y
∂u


+ ∂w
∂z


· ∂z
∂u


= ⎛⎝6x − 2y⎞⎠eu sin v − 2xeu cos v + 8zeu,


then substitute x(u, v) = eu sin v, y(u, v) = eu cos v, and z(u, v) = eu into this equation:
∂w
∂u


= ⎛⎝6x − 2y⎞⎠eu sin v − 2xeu cos v + 8zeu


= (6eu sin v − 2eu cos v)eu sin v − 2(eu sin v)eu cos v + 8e2u


= 6e2u sin2 v − 4e2u sin v cos v + 8e2u


= 2e2u ⎛⎝3 sin
2 v − 2 sin v cos v + 4⎞⎠.


Next, we calculate ∂w/∂v:


Chapter 4 | Differentiation of Functions of Several Variables 411




4.25


∂w
∂v


= ∂w
∂ x


· ∂ x
∂v


+ ∂w
∂ y


·
∂ y
∂v


+ ∂w
∂z


· ∂z
∂v


= ⎛⎝6x − 2y⎞⎠eu cos v − 2x(−eu sin v) + 8z(0),


then we substitute x(u, v) = eu sin v, y(u, v) = eu cos v, and z(u, v) = eu into this equation:
∂w
∂v


= ⎛⎝6x − 2y⎞⎠eu cos v − 2x(−eu sin v)


= (6eu sin v − 2eu cos v)eu cos v + 2(eu sin v)(eu sin v)


= 2e2u sin2 v + 6e2u sin v cos v − 2e2u cos2 v


= 2e2u ⎛⎝sin
2 v + sin v cos v − cos2 v⎞⎠.


Calculate ∂w/∂u and ∂w/∂v given the following functions:


w = f (x, y, z) =
x + 2y − 4z
2x − y + 3z


x = x(u, v) = e2u cos 3v


y = y(u, v) = e2u sin 3v


z = z(u, v) = e2u.


Example 4.29
Drawing a Tree Diagram
Create a tree diagram for the case when


w = f (x, y, z), x = x(t, u, v), y = y(t, u, v), z = z(t, u, v)


and write out the formulas for the three partial derivatives of w.
Solution
Starting from the left, the function f has three independent variables: x, y, and z. Therefore, three branches
must be emanating from the first node. Each of these three branches also has three branches, for each of thevariables t, u, and v.


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4.26


Figure 4.36 Tree diagram for a function of three variables, each of which isa function of three independent variables.


The three formulas are
∂w
∂ t


= ∂w
∂ x


∂ x
∂ t


+ ∂w
∂ y


∂ y
∂ t


+ ∂w
∂z


∂z
∂ t


∂w
∂u


= ∂w
∂ x


∂ x
∂u


+ ∂w
∂ y


∂ y
∂u


+ ∂w
∂z


∂z
∂u


∂w
∂v


= ∂w
∂ x


∂ x
∂v


+ ∂w
∂ y


∂ y
∂v


+ ∂w
∂z


∂z
∂v


.


Create a tree diagram for the case when
w = f (x, y), x = x(t, u, v), y = y(t, u, v)


and write out the formulas for the three partial derivatives of w.


Implicit Differentiation
Recall from Implicit Differentiation (http://cnx.org/content/m53585/latest/) that implicit differentiation provides amethod for finding dy/dx when y is defined implicitly as a function of x. The method involves differentiating both sides
of the equation defining the function with respect to x, then solving for dy/dx. Partial derivatives provide an alternative
to this method.
Consider the ellipse defined by the equation x2 + 3y2 + 4y − 4 = 0 as follows.


Chapter 4 | Differentiation of Functions of Several Variables 413




Figure 4.37 Graph of the ellipse defined by
x2 + 3y2 + 4y − 4 = 0.


This equation implicitly defines y as a function of x. As such, we can find the derivative dy/dx using the method of
implicit differentiation:


d
dx

⎝x


2 + 3y2 + 4y − 4⎞⎠ =
d
dx


(0)


2x + 6y
dy
dx


+ 4
dy
dx


= 0



⎝6y + 4⎞⎠


dy
dx


= −2x


dy
dx


= − x
3y + 2


.


We can also define a function z = f (x, y) by using the left-hand side of the equation defining the ellipse. Then
f (x, y) = x2 + 3y2 + 4y − 4. The ellipse x2 + 3y2 + 4y − 4 = 0 can then be described by the equation f (x, y) = 0.
Using this function and the following theorem gives us an alternative approach to calculating dy/dx.


Theorem 4.11: Implicit Differentiation of a Function of Two or More Variables
Suppose the function z = f (x, y) defines y implicitly as a function y = g(x) of x via the equation f (x, y) = 0.
Then


(4.34)dy
dx


= −
∂ f /∂ x
∂ f /∂ y


provided fy (x, y) ≠ 0.
If the equation f (x, y, z) = 0 defines z implicitly as a differentiable function of x and y, then


(4.35)dz
dx


= −
∂ f /∂ x
∂ f /∂z


and dz
dy


= −
∂ f /∂ y
∂ f /∂z


as long as fz (x, y, z) ≠ 0.


Equation 4.34 is a direct consequence of Equation 4.31. In particular, if we assume that y is defined implicitly as a
function of x via the equation f (x, y) = 0, we can apply the chain rule to find dy/dx:


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d
dx


f (x, y) = d
dx


(0)


∂ f
∂ x


· dx
dx


+
∂ f
∂ y


·
dy
dx


= 0


∂ f
∂ x


+
∂ f
∂ y


·
dy
dx


= 0.


Solving this equation for dy/dx gives Equation 4.34. Equation 4.35 can be derived in a similar fashion.
Let’s now return to the problem that we started before the previous theorem. Using Implicit Differentiation of a
Function of Two or More Variables and the function f (x, y) = x2 + 3y2 + 4y − 4, we obtain


∂ f
∂ x


= 2x


∂ f
∂ y


= 6y + 4.


Then Equation 4.34 gives
dy
dx


= −
∂ f /∂ x
∂ f /∂ y


= − 2x
6y + 4


= − x
3y + 2


,


which is the same result obtained by the earlier use of implicit differentiation.
Example 4.30
Implicit Differentiation by Partial Derivatives


a. Calculate dy/dx if y is defined implicitly as a function of x via the equation
3x2 − 2xy + y2 + 4x − 6y − 11 = 0. What is the equation of the tangent line to the graph of this curve
at point (2, 1)?


b. Calculate ∂z/∂ x and ∂z/∂ y, given x2 ey − yzex = 0.
Solution


a. Set f (x, y) = 3x2 − 2xy + y2 + 4x − 6y − 11 = 0, then calculate fx and fy : fx = 6x − 2y + 4fy = −2x + 2y − 6.
The derivative is given by


dy
dx


= −
∂ f /∂ x
∂ f /∂ y


= −
6x − 2y + 4
−2x + 2y − 6


=
3x − y + 2
x − y + 3


.


The slope of the tangent line at point (2, 1) is given by
dy
dx |(x, y) = (2, 1) = 3(2) − 1 + 22 − 1 + 3 = 74.


To find the equation of the tangent line, we use the point-slope form (Figure 4.38):
y − y0 = m(x − x0)


y − 1 = 7
4
(x − 2)


y = 7
4
x − 7


2
+ 1


y = 7
4
x − 5


2
.


Chapter 4 | Differentiation of Functions of Several Variables 415




4.27


Figure 4.38 Graph of the rotated ellipse defined by
3x2 − 2xy + y2 + 4x − 6y − 11 = 0.


b. We have f (x, y, z) = x2 ey − yzex. Therefore,
∂ f
∂ x


= 2xey − yzex


∂ f
∂ y


= x2 ey − zex


∂ f
∂z


= −yex.


Using Equation 4.35,
∂z
∂ x


= −
∂ f /∂ x
∂ f /∂ y


= −
2xey − yzex


−yex


=
2xey − yzex


yex


and


∂z
∂ y


= −
∂ f /∂ y
∂ f /∂z


= − x
2 ey − zex


−yex


= x
2 ey − zex


yex
.


Find dy/dx if y is defined implicitly as a function of x by the equation
x2 + xy − y2 + 7x − 3y − 26 = 0. What is the equation of the tangent line to the graph of this curve at point
(3, −2)?


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4.5 EXERCISES
For the following exercises, use the information providedto solve the problem.
215. Let w(x, y, z) = xy cos z, where x = t, y = t2,
and z = arcsin t. Find dw


dt
.


216. Let w(t, v) = etv where t = r + s and v = rs.
Find ∂w


∂r
and ∂w


∂s
.


217. If w = 5x2 + 2y2, x = −3s + t, and y = s − 4t,
find ∂w


∂s
and ∂w


∂ t
.


218. If w = xy2, x = 5 cos(2t), and y = 5 sin(2t),
find ∂w


∂ t
.


219. If f (x, y) = xy, x = r cos θ, and y = r sin θ,
find ∂ f


∂r
and express the answer in terms of r and θ.


220. Suppose f (x, y) = x + y, u = ex sin y, x = t2,
and y = πt, where x = r cos θ and y = r sin θ. Find
∂ f
∂θ


.


For the following exercises, find d f
dt


using the chain rule
and direct substitution.
221. f (x, y) = x2 + y2, x = t, y = t2


222. f (x, y) = x2 + y2, y = t2, x = t
223. f (x, y) = xy, x = 1 − t, y = 1 + t
224. f (x, y) = xy, x = et, y = 2et


225. f (x, y) = ln(x + y), x = et, y = et
226. f (x, y) = x4, x = t, y = t
227. Let w(x, y, z) = x2 + y2 + z2,
x = cos t, y = sin t, and z = et. Express w as a
function of t and find dw


dt
directly. Then, find dw


dt
using


the chain rule.


228. Let z = x2 y, where x = t2 and y = t3. Find dz
dt


.


229. Let u = ex sin y, where x = t2 and y = πt. Find
du
dt


when x = ln 2 and y = π
4
.


For the following exercises, find dy
dx


using partial
derivatives.
230. sin(6x) + tan⎛⎝8y⎞⎠+ 5 = 0
231. x3 + y2 x − 3 = 0
232. sin(x + y) + cos(x − y) = 4
233. x2 − 2xy + y4 = 4
234. xey + yex − 2x2 y = 0
235. x2/3 + y2/3 = a2/3
236. x cos(xy) + y cos x = 2
237. exy + yey = 1
238. x2 y3 + cos y = 0
239. Find dz


dt
using the chain rule where


z = 3x2 y3, x = t4, and y = t2.
240. Let z = 3 cos x − sin(xy), x = 1t , and y = 3t.
Find dz


dt
.


241. Let z = e1 − xy, x = t1/3, and y = t3. Find dz
dt


.


242. Find dz
dt


by the chain rule where
z = cosh2(xy), x = 1


2
t, and y = et.


243. Let z = xy, x = 2 cos u, and y = 3 sin v. Find ∂z∂u
and ∂z


∂v
.


Chapter 4 | Differentiation of Functions of Several Variables 417




244. Let z = ex2 y, where x = uv and y = 1v . Find
∂z
∂u


and ∂z
∂v


.


245. If z = xyex/y, x = r cos θ, and y = r sin θ, find
∂z
∂r


and ∂z
∂θ


when r = 2 and θ = π
6
.


246. Find ∂w
∂s


if
w = 4x + y2 + z3, x = ers


2
, y = ln⎛⎝


r + s
t

⎠, and


z = rst2.


247. If w = sin(xyz), x = 1 − 3t, y = e1 − t, and
z = 4t, find ∂w


∂ t
.


For the following exercises, use this information: Afunction f (x, y) is said to be homogeneous of degree n
if f (tx, ty) = tn f (x, y). For all homogeneous functions
of degree n, the following equation is true:
x
∂ f
∂ x


+ y
∂ f
∂ y


= n f (x, y). Show that the given function is
homogeneous and verify that x∂ f


∂ x
+ y


∂ f
∂ y


= n f (x, y).


248. f (x, y) = 3x2 + y2


249. f (x, y) = x2 + y2
250. f (x, y) = x2 y − 2y3
251. The volume of a right circular cylinder is given by
V(x, y) = πx2 y, where x is the radius of the cylinder
and y is the cylinder height. Suppose x and y are
functions of t given by x = 1


2
t and y = 1


3
t so that


x and y are both increasing with time. How fast is the
volume increasing when x = 2 and y = 5?
252. The pressure P of a gas is related to the volume and
temperature by the formula PV = kT , where temperature
is expressed in kelvins. Express the pressure of the gas as
a function of both V and T . Find dP


dt
when k = 1,


dV
dt


= 2 cm3/min, dT
dt


= 1
2


K/min, V = 20 cm3, and
T = 20°F.


253. The radius of a right circular cone is increasing at 3
cm/min whereas the height of the cone is decreasing at 2
cm/min. Find the rate of change of the volume of the conewhen the radius is 13 cm and the height is 18 cm.
254. The volume of a frustum of a cone is given by the
formula V = 1


3
πz⎛⎝x


2 + y2 + xy⎞⎠, where x is the radius
of the smaller circle, y is the radius of the larger circle,
and z is the height of the frustum (see figure). Find the
rate of change of the volume of this frustum when
x = 10 in., y = 12 in., and z = 18 in.


255. A closed box is in the shape of a rectangular solidwith dimensions x, y, and z. (Dimensions are in inches.)
Suppose each dimension is changing at the rate of 0.5
in./min. Find the rate of change of the total surface area ofthe box when x = 2 in., y = 3 in., and z = 1 in.
256. The total resistance in a circuit that has threeindividual resistances represented by x, y, and z is given
by the formula R(x, y, z) = xyzyz + xz + xy. Suppose at a
given time the x resistance is 100Ω, the y resistance
is 200Ω, and the z resistance is 300Ω. Also, suppose
the x resistance is changing at a rate of 2Ω/min, the y
resistance is changing at the rate of 1Ω/min, and the z
resistance has no change. Find the rate of change of thetotal resistance in this circuit at this time.
257. The temperature T at a point (x, y) is T(x, y) and
is measured using the Celsius scale. A fly crawls so that
its position after t seconds is given by x = 1 + t and
y = 2 + 1


3
t, where x and y are measured in centimeters.


The temperature function satisfies Tx (2, 3) = 4 and
Ty (2, 3) = 3. How fast is the temperature increasing on
the fly’s path after 3 sec?


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258. The x and y components of a fluid moving in two
dimensions are given by the following functions:
u(x, y) = 2y and v(x, y) = −2x; x ≥ 0; y ≥ 0. The
speed of the fluid at the point (x, y) is
s(x, y) = u(x, y)2 + v(x, y)2. Find ∂s


∂ x
and ∂s


∂ y
using


the chain rule.
259. Let u = u(x, y, z), where
x = x(w, t), y = y(w, t), z = z(w, t), w = w(r, s), and t = t(r, s).


Use a tree diagram and the chain rule to find an expression
for ∂u


∂r
.


Chapter 4 | Differentiation of Functions of Several Variables 419




4.6 | Directional Derivatives and the Gradient
Learning Objectives


4.6.1 Determine the directional derivative in a given direction for a function of two variables.
4.6.2 Determine the gradient vector of a given real-valued function.
4.6.3 Explain the significance of the gradient vector with regard to direction of change along asurface.
4.6.4 Use the gradient to find the tangent to a level curve of a given function.
4.6.5 Calculate directional derivatives and gradients in three dimensions.


In Partial Derivatives we introduced the partial derivative. A function z = f (x, y) has two partial derivatives: ∂z/∂ x
and ∂z/∂ y. These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates
of change (that is, as slopes of a tangent line). For example, ∂z/∂ x represents the slope of a tangent line passing through
a given point on the surface defined by z = f (x, y), assuming the tangent line is parallel to the x-axis. Similarly, ∂z/∂ y
represents the slope of the tangent line parallel to the y-axis. Now we consider the possibility of a tangent line parallel to
neither axis.
Directional Derivatives
We start with the graph of a surface defined by the equation z = f (x, y). Given a point (a, b) in the domain of
f , we choose a direction to travel from that point. We measure the direction using an angle θ, which is measured
counterclockwise in the x, y-plane, starting at zero from the positive x-axis (Figure 4.39). The distance we travel is h and
the direction we travel is given by the unit vector u = (cos θ)i + (sin θ)j. Therefore, the z-coordinate of the second point
on the graph is given by z = f (a + h cos θ, b + h sin θ).


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Figure 4.39 Finding the directional derivative at a point on the graph of z = f (x, y). The
slope of the black arrow on the graph indicates the value of the directional derivative at thatpoint.


We can calculate the slope of the secant line by dividing the difference in z-values by the length of the line segment
connecting the two points in the domain. The length of the line segment is h. Therefore, the slope of the secant line is


msec =
f (a + h cos θ, b + h sin θ) − f (a, b)


h
.


To find the slope of the tangent line in the same direction, we take the limit as h approaches zero.


Definition
Suppose z = f (x, y) is a function of two variables with a domain of D. Let (a, b) ∈ D and define
u = cos θi + sin θj. Then the directional derivative of f in the direction of u is given by


(4.36)Du f (a, b) = lim
h → 0


f (a + h cos θ, b + h sin θ) − f (a, b)
h


,


provided the limit exists.


Equation 4.36 provides a formal definition of the directional derivative that can be used in many cases to calculate adirectional derivative.
Example 4.31
Finding a Directional Derivative from the Definition


Chapter 4 | Differentiation of Functions of Several Variables 421




Let θ = arccos(3/5). Find the directional derivative Du f (x, y) of f (x, y) = x2 − xy + 3y2 in the direction of
u = (cos θ)i + (sin θ)j. What is Du f (−1, 2)?
Solution
First of all, since cos θ = 3/5 and θ is acute, this implies


sin θ = 1 − ⎛⎝
3
5



2
= 16


25
= 4


5
.


Using f (x, y) = x2 − xy + 3y2, we first calculate f ⎛⎝x + h cos θ, y + h sin θ⎞⎠:
f ⎛⎝x + h cos θ, y + h sin θ⎞⎠ = (x + h cos θ)2 − (x + h cos θ)⎛⎝y + h sin θ⎞⎠+ 3⎛⎝y + h sin θ⎞⎠2


= x2 + 2xh cos θ + h2 cos2 θ − xy − xh sin θ − yh cos θ


−h2 sin θ cos θ + 3y2 + 6yh sin θ + 3h2 sin2 θ


= x2 + 2xh⎛⎝
3
5

⎠+


9h2
25


− xy − 4xh
5



3yh
5


− 12h
2


25
+ 3y2


+6yh⎛⎝
4
5

⎠+ 3h


2 ⎛

16
25



= x2 − xy + 3y2 + 2xh
5


+ 9h
2


5
+


21yh
5


.


We substitute this expression into Equation 4.36:
Du f (a, b) = lim


h → 0


f (a + h cos θ, b + h sin θ) − f (a, b)
h


= lim
h → 0



⎝x


2 − xy + 3y2 + 2xh
5


+ 9h
2


5
+ 21yh


5

⎠−

⎝x


2 − xy + 3y2⎞⎠


h


= lim
h → 0


2xh
5


+ 9h
2


5
+ 21yh


5
h


= lim
h → 0


2x
5


+ 9h
5


+
21y
5


=
2x + 21y


5
.


To calculate Du f (−1, 2), we substitute x = −1 and y = 2 into this answer:
Du f (−1, 2) =


2(−1) + 21(2)
5


= −2 + 42
5


= 8.


(See the following figure.)


422 Chapter 4 | Differentiation of Functions of Several Variables


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Figure 4.40 Finding the directional derivative in a given direction u at a given point on a surface.
The plane is tangent to the surface at the given point (−1, 2, 15).


Another approach to calculating a directional derivative involves partial derivatives, as outlined in the following theorem.
Theorem 4.12: Directional Derivative of a Function of Two Variables
Let z = f (x, y) be a function of two variables x and y, and assume that fx and fy exist. Then the directional
derivative of f in the direction of u = cos θi + sin θj is given by


(4.37)Du f (x, y) = fx (x, y)cos θ + fy (x, y)sin θ.


Proof
Equation 4.36 states that the directional derivative of f in the direction of u = cos θi + sin θj is given by


Du f (a, b) = lim
t → 0


f (a + t cos θ, b + t sin θ) − f (a, b)
t .


Let x = a + t cos θ and y = b + t sin θ, and define g(t) = f (x, y). Since fx and fy both exist, we can use the chain
rule for functions of two variables to calculate g′ (t):


g′ (t) =
∂ f
∂ x


dx
dt


+
∂ f
∂ y


dy
dt


= fx (x, y)cos θ + fy (x, y)sin θ.


If t = 0, then x = x0 and y = y0, so
g′ (0) = fx (x0, y0)cos θ + fy (x0, y0)sin θ.


Chapter 4 | Differentiation of Functions of Several Variables 423




4.28


By the definition of g′ (t), it is also true that
g′ (0) = lim


t → 0


g(t) − g(0)
t


= lim
t → 0


f ⎛⎝x0 + t cos θ, y0 + t sin θ

⎠− f (x0, y0)


t .


Therefore, Du f (x0, y0) = fx (x, y)cos θ + fy (x, y)sin θ.

Example 4.32
Finding a Directional Derivative: Alternative Method
Let θ = arccos(3/5). Find the directional derivative Du f (x, y) of f (x, y) = x2 − xy + 3y2 in the direction of
u = (cos θ)i + (sin θ)j. What is Du f (−1, 2)?
Solution
First, we must calculate the partial derivatives of f :


fx = 2x − y
fy = −x + 6y,


Then we use Equation 4.37 with θ = arccos(3/5):
Du f (x, y) = fx (x, y)cos θ + fy (x, y)sin θ


= ⎛⎝2x − y⎞⎠35
+ ⎛⎝−x + 6y⎞⎠45


= 6x
5



3y
5


− 4x
5


+
24y
5


=
2x + 21y


5
.


To calculate Du f (−1, 2), let x = −1 and y = 2:
Du f (−1, 2) =


2(−1) + 21(2)
5


= −2 + 42
5


= 8.


This is the same answer obtained in Example 4.31.


Find the directional derivative Du f (x, y) of f (x, y) = 3x2 y − 4xy3 + 3y2 − 4x in the direction of
u = ⎛⎝cos


π
3

⎠i +

⎝sin


π
3

⎠j using Equation 4.37. What is Du f (3, 4)?


If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the normof the vector. For example, if we wished to find the directional derivative of the function in Example 4.32 in the directionof the vector 〈 −5, 12 〉 , we would first divide by its magnitude to get u. This gives us u = 〈 −(5/13), 12/13 〉 .
Then


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Du f (x, y) = ∇ f (x, y) ·u


= − 5
13

⎝2x − y⎞⎠+ 1213



⎝−x + 6y⎞⎠


= − 22
13


x + 17
13


y.


Gradient
The right-hand side of Equation 4.37 is equal to fx (x, y)cos θ + fy (x, y)sin θ, which can be written as the dot
product of two vectors. Define the first vector as ∇ f (x, y) = fx (x, y)i + fy (x, y)j and the second vector as
u = (cos θ)i + (sin θ)j. Then the right-hand side of the equation can be written as the dot product of these two vectors:


(4.38)Du f (x, y) = ∇ f (x, y) ·u.
The first vector in Equation 4.38 has a special name: the gradient of the function f . The symbol ∇ is called nabla and
the vector ∇ f is read “del f .”


Definition
Let z = f (x, y) be a function of x and y such that fx and fy exist. The vector ∇ f (x, y) is called the gradient of
f and is defined as


(4.39)∇ f (x, y) = fx (x, y)i + fy (x, y)j.
The vector ∇ f (x, y) is also written as “grad f .”


Example 4.33
Finding Gradients
Find the gradient ∇ f (x, y) of each of the following functions:


a. f (x, y) = x2 − xy + 3y2
b. f (x, y) = sin 3x cos 3y


Solution
For both parts a. and b., we first calculate the partial derivatives fx and fy, then use Equation 4.39.


a.
fx (x, y) = 2x − y and fy (x, y) = −x + 6y, so


∇ f (x, y) = fx (x, y)i + fy (x, y)j


= ⎛⎝2x − y⎞⎠i + ⎛⎝−x + 6y⎞⎠j.


b.
fx (x, y) = 3 cos 3x cos 3y and fy (x, y) = −3 sin 3x sin 3y, so


∇ f (x, y) = fx (x, y)i + fy (x, y)j


= ⎛⎝3 cos 3x cos 3y⎞⎠i − ⎛⎝3 sin 3x sin 3y⎞⎠j.


Chapter 4 | Differentiation of Functions of Several Variables 425




4.29 Find the gradient ∇ f (x, y) of f (x, y) = ⎛⎝x2 − 3y2⎞⎠/⎛⎝2x + y⎞⎠.


The gradient has some important properties. We have already seen one formula that uses the gradient: the formula forthe directional derivative. Recall from The Dot Product that if the angle between two vectors a and b is φ, then
a · b = ‖ a ‖ ‖ b ‖ cos φ. Therefore, if the angle between ∇ f (x0, y0) and u = (cos θ)i + (sin θ)j is φ, we have


Du f (x0, y0) = ∇ f (x0, y0) ·u = ‖ ∇ f (x0, y0) ‖ ‖ u ‖ cos φ = ‖ ∇ f (x0, y0) ‖ cos φ.


The ‖ u ‖ disappears because u is a unit vector. Therefore, the directional derivative is equal to the magnitude of the
gradient evaluated at (x0, y0) multiplied by cos φ. Recall that cos φ ranges from −1 to 1. If φ = 0, then cos φ = 1
and ∇ f (x0, y0) and u both point in the same direction. If φ = π, then cos φ = −1 and ∇ f (x0, y0) and u point in
opposite directions. In the first case, the value of Du f (x0, y0) is maximized; in the second case, the value of Du f (x0, y0)
is minimized. If ∇ f (x0, y0) = 0, then Du f (x0, y0) = ∇ f (x0, y0) ·u = 0 for any vector u. These three cases are
outlined in the following theorem.
Theorem 4.13: Properties of the Gradient
Suppose the function z = f (x, y) is differentiable at (x0, y0) (Figure 4.41).


i. If ∇ f (x0, y0) = 0, then Du f (x0, y0) = 0 for any unit vector u.
ii. If ∇ f (x0, y0) ≠ 0, then Du f (x0, y0) is maximized when u points in the same direction as ∇ f (x0, y0).


The maximum value of Du f (x0, y0) is ‖ ∇ f (x0, y0) ‖ .
iii. If ∇ f (x0, y0) ≠ 0, then Du f (x0, y0) is minimized when u points in the opposite direction from


∇ f (x0, y0). The minimum value of Du f (x0, y0) is − ‖ ∇ f (x0, y0) ‖ .


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Figure 4.41 The gradient indicates the maximum and minimum values of thedirectional derivative at a point.


Example 4.34
Finding a Maximum Directional Derivative
Find the direction for which the directional derivative of f (x, y) = 3x2 − 4xy + 2y2 at (−2, 3) is a maximum.
What is the maximum value?
Solution
The maximum value of the directional derivative occurs when ∇ f and the unit vector point in the same direction.
Therefore, we start by calculating ∇ f (x, y):


fx (x, y) = 6x − 4y and fy (x, y) = −4x + 4y, so


∇ f (x, y) = fx (x, y)i + fy (x, y)j =

⎝6x − 4y⎞⎠i + ⎛⎝−4x + 4y⎞⎠j.


Next, we evaluate the gradient at (−2, 3):
∇ f (−2, 3) = ⎛⎝6(−2) − 4(3)⎞⎠i + ⎛⎝−4(−2) + 4(3)⎞⎠j = −24i + 20j.


We need to find a unit vector that points in the same direction as ∇ f (−2, 3), so the next step is to divide
∇ f (−2, 3) by its magnitude, which is (−24)2 + (20)2 = 976 = 4 61. Therefore,


∇ f (−2, 3)
‖ ∇ f (−2, 3) ‖


= −24
4 61


i + 20
4 61


j = −6 61
61


i + 5 61
61


j.


This is the unit vector that points in the same direction as ∇ f (−2, 3). To find the angle corresponding to this


Chapter 4 | Differentiation of Functions of Several Variables 427




4.30


unit vector, we solve the equations
cos θ = −6 61


61
and sin θ = 5 61


61


for θ. Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore,
θ = π − arcsin⎛⎝



⎝5 61



⎠/61



⎠ ≈ 2.45 rad.


The maximum value of the directional derivative at (−2, 3) is ‖ ∇ f (−2, 3) ‖ = 4 61 (see the following
figure).


Figure 4.42 The maximum value of the directional derivative at
(−2, 3) is in the direction of the gradient.


Find the direction for which the directional derivative of g(x, y) = 4x − xy + 2y2 at (−2, 3) is a
maximum. What is the maximum value?


Figure 4.43 shows a portion of the graph of the function f (x, y) = 3 + sin x sin y. Given a point (a, b) in the domain
of f , the maximum value of the gradient at that point is given by ‖ ∇ f (a, b) ‖ . This would equal the rate of greatest
ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatestdescent.


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Figure 4.43 A typical surface in ℝ3. Given a point on the surface, the directional derivative
can be calculated using the gradient.


When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together(see Figure 4.44). This is analogous to the contour map of a function, assuming the level curves are obtained for equallyspaced values throughout the range of that function.


Figure 4.44 Contour map for the function
f (x, y) = x2 − y2 using level values between −5 and 5.


Gradients and Level Curves
Recall that if a curve is defined parametrically by the function pair ⎛⎝x(t), y(t)⎞⎠, then the vector x′ (t)i + y′ (t)j is tangent
to the curve for every value of t in the domain. Now let’s assume z = f (x, y) is a differentiable function of x and y, and
(x0, y0) is in its domain. Let’s suppose further that x0 = x(t0) and y0 = y(t0) for some value of t, and consider the
level curve f (x, y) = k. Define g(t) = f ⎛⎝x(t), y(t)⎞⎠ and calculate g′ (t) on the level curve. By the chain Rule,


g′ (t) = fx

⎝x(t), y(t)⎞⎠x′ (t) + fy



⎝x(t), y(t)⎞⎠y′ (t).


Chapter 4 | Differentiation of Functions of Several Variables 429




But g′ (t) = 0 because g(t) = k for all t. Therefore, on the one hand,
fx

⎝x(t), y(t)⎞⎠x′ (t) + fy



⎝x(t), y(t)⎞⎠y′ (t) = 0;


on the other hand,
fx

⎝x(t), y(t)⎞⎠x′ (t) + fy



⎝x(t), y(t)⎞⎠y′ (t) = ∇ f (x, y) · 〈 x′ (t), y′ (t) 〉 .


Therefore,
∇ f (x, y) · 〈 x′ (t), y′ (t) 〉 = 0.


Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector istangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the followingtheorem.
Theorem 4.14: Gradient Is Normal to the Level Curve
Suppose the function z = f (x, y) has continuous first-order partial derivatives in an open disk centered at a point
(x0, y0). If ∇ f (x0, y0) ≠ 0, then ∇ f (x0, y0) is normal to the level curve of f at (x0, y0).


We can use this theorem to find tangent and normal vectors to level curves of a function.
Example 4.35
Finding Tangents to Level Curves
For the function f (x, y) = 2x2 − 3xy + 8y2 + 2x − 4y + 4, find a tangent vector to the level curve at point
(−2, 1). Graph the level curve corresponding to f (x, y) = 18 and draw in ∇ f (−2, 1) and a tangent vector.
Solution
First, we must calculate ∇ f (x, y):


fx (x, y) = 4x − 3y + 2 and fy = −3x + 16y − 4 so ∇ f (x, y) =

⎝4x − 3y + 2⎞⎠i + ⎛⎝−3x + 16y − 4⎞⎠j.


Next, we evaluate ∇ f (x, y) at (−2, 1):
∇ f (−2, 1) = ⎛⎝4(−2) − 3(1) + 2⎞⎠i + ⎛⎝−3(−2) + 16(1) − 4⎞⎠j = −9i + 18j.


This vector is orthogonal to the curve at point (−2, 1). We can obtain a tangent vector by reversing the
components and multiplying either one by −1. Thus, for example, −18i − 9j is a tangent vector (see the
following graph).


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4.31


Figure 4.45 Tangent and normal vectors to
2x2 − 3xy + 8y2 + 2x − 4y + 4 = 18 at point (−2, 1).


For the function f (x, y) = x2 − 2xy + 5y2 + 3x − 2y + 4, find the tangent to the level curve at point
(1, 1). Draw the graph of the level curve corresponding to f (x, y) = 8 and draw ∇ f (1, 1) and a tangent
vector.


Three-Dimensional Gradients and Directional Derivatives
The definition of a gradient can be extended to functions of more than two variables.
Definition
Let w = f (x, y, z) be a function of three variables such that fx, fy, and fz exist. The vector ∇ f (x, y, z) is called
the gradient of f and is defined as


(4.40)∇ f (x, y, z) = fx (x, y, z)i + fy (x, y, z)j + fz (x, y, z)k.
∇ f (x, y, z) can also be written as grad f (x, y, z).


Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in twovariables. First, we calculate the partial derivatives fx, fy, and fz, and then we use Equation 4.40.
Example 4.36
Finding Gradients in Three Dimensions
Find the gradient ∇ f (x, y, z) of each of the following functions:


a. f (x, y) = 5x2 − 2xy + y2 − 4yz + z2 + 3xz
b. f (x, y, z) = e−2z sin 2x cos 2y


Solution


Chapter 4 | Differentiation of Functions of Several Variables 431




4.32


For both parts a. and b., we first calculate the partial derivatives fx, fy, and fz, then use Equation 4.40.
a.


fz (x, y, z) = 10x − 2y + 3z, fy (x, y, z) = −2x + 2y − 4z and fz (x, y, z) = 3x − 4y + 2z, so


∇ f (x, y, z) = fx (x, y, z)i + fy (x, y, z)j + fz (x, y, z)k


= ⎛⎝10x − 2y + 3z⎞⎠i + ⎛⎝−2x + 2y − 4z⎞⎠j + ⎛⎝−4x + 3y + 2z⎞⎠k.


b.
fx (x, y, z) = −2e


−2z cos 2x cos 2y, fy (x, y, z) = −2e
−2z sin 2x sin 2y and


fz (x, y, z) = −2e
−2z sin 2x cos 2y, so


∇ f (x, y, z) = fx (x, y, z)i + fy (x, y, z)j + fz (x, y, z)k


= ⎛⎝2e
−2z cos 2x cos 2y⎞⎠i +



⎝−2e


−2z⎞
⎠j +

⎝−2e


−2z⎞


= 2e−2z ⎛⎝cos 2x cos 2y i − sin 2x sin 2y j − sin 2x cos 2y k⎞⎠.


Find the gradient ∇ f (x, y, z) of f (x, y, z) = x2 − 3y2 + z2
2x + y − 4z


.


The directional derivative can also be generalized to functions of three variables. To determine a direction in threedimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector arecalled directional cosines. Given a three-dimensional unit vector u in standard form (i.e., the initial point is at the origin),
this vector forms three different angles with the positive x − , y − , and z-axes. Let’s call these angles α, β, and γ.
Then the directional cosines are given by cos α, cos β, and cos γ. These are the components of the unit vector u; since
u is a unit vector, it is true that cos2α + cos2 β + cos2 γ = 1.


Definition
Suppose w = f (x, y, z) is a function of three variables with a domain of D. Let (x0, y0, z0) ∈ D and let
u = cos αi + cos βj + cos γk be a unit vector. Then, the directional derivative of f in the direction of u is given by


(4.41)
Du f (x0, y0, z0) = limt → 0


f ⎛⎝x0 + t cos α, y0 + t cos β, z0 + t cos γ

⎠− f (x0, y0, z0)


t ,


provided the limit exists.


We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that isanalogous to Equation 4.38.
Theorem 4.15: Directional Derivative of a Function of Three Variables
Let f (x, y, z) be a differentiable function of three variables and let u = cos αi + cos βj + cos γk be a unit vector.
Then, the directional derivative of f in the direction of u is given by


(4.42)Du f (x, y, z) = ∇ f (x, y, z) ·u
= fx (x, y, z)cos α + fy (x, y, z)cos β + fz (x, y, z)cos γ.


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4.33


The three angles α, β, and γ determine the unit vector u. In practice, we can use an arbitrary (nonunit) vector, then divide
by its magnitude to obtain a unit vector in the desired direction.
Example 4.37
Finding a Directional Derivative in Three Dimensions
Calculate Du f (1, −2, 3) in the direction of v = −i + 2j + 2k for the function


f (x, y, z) = 5x2 − 2xy + y2 − 4yz + z2 + 3xz.


Solution
First, we find the magnitude of v:


‖ v ‖ = (−1)2 + (2)2 = 3.


Therefore, v
‖ v ‖


=
−i + 2j + 2k


3
= − 1


3
i + 2


3
j + 2


3
k is a unit vector in the direction of v, so


cos α = − 1
3
, cos β = 2


3
, and cos γ = 2


3
. Next, we calculate the partial derivatives of f :


fx (x, y, z) = 10x − 2y + 3z
fy (x, y, z) = −2x + 2y − 4z


fz (x, y, z) = −4y + 2z + 3x,


then substitute them into Equation 4.42:
Du f (x, y, z) = fx (x, y, z)cos α + fy (x, y, z)cos β + fz (x, y, z)cos γ


= ⎛⎝10x − 2y + 3z⎞⎠

⎝−


1
3

⎠+



⎝−2x + 2y − 4z⎞⎠




2
3

⎠+



⎝−4y + 2z + 3x⎞⎠




2
3



= − 10x
3


+
2y
3


− 3z
3


− 4x
3


+
4y
3


− 8z
3



8y
3


+ 4z
3


+ 6x
3


= − 8x
3



2y
3


− 7z
3
.


Last, to find Du f (1, −2, 3), we substitute x = 1, y = −2, and z = 3:
Du f (1, −2, 3) = −


8(1)
3


− 2(−2)
3


− 7(3)
3


= − 8
3
+ 4


3
− 21


3


= − 25
3
.


Calculate Du f (x, y, z) and Du f (0, −2, 5) in the direction of v = −3i + 12j − 4k for the function
f (x, y, z) = 3x2 + xy − 2y2 + 4yz − z2 + 2xz.


Chapter 4 | Differentiation of Functions of Several Variables 433




4.6 EXERCISES
For the following exercises, find the directional derivativeusing the limit definition only.
260. f (x, y) = 5 − 2x2 − 1


2
y2 at point P(3, 4) in the


direction of u = ⎛⎝cos π4⎞⎠i + ⎛⎝sin π4⎞⎠j


261. f (x, y) = y2 cos(2x) at point P⎛⎝π3, 2⎞⎠ in the
direction of u = ⎛⎝cos π4⎞⎠i + ⎛⎝sin π4⎞⎠j
262. Find the directional derivative of
f (x, y) = y2 sin(2x) at point P⎛⎝π4, 2⎞⎠ in the direction of
u = 5i + 12j.


For the following exercises, find the directional derivativeof the function at point P in the direction of v.
263. f (x, y) = xy, P(0, −2), v = 1


2
i + 3


2
j


264. h(x, y) = ex sin y, P⎛⎝1, π2⎞⎠, v = −i
265. h(x, y, z) = xyz, P(2, 1, 1), v = 2i + j − k
266. f (x, y) = xy, P(1, 1), u = 〈 2


2
, 2


2


267. f (x, y) = x2 − y2, u = 〈 3
2
, 1
2
〉 , P(1, 0)


268. f (x, y) = 3x + 4y + 7, u = 〈 3
5
, 4
5
〉 , P⎛⎝0,


π
2



269. f (x, y) = ex cos y, u = 〈 0, 1 〉 , P = ⎛⎝0, π2⎞⎠
270. f (x, y) = y10, u = 〈 0, −1 〉 , P = (1, −1)
271. f (x, y) = ln(x2 + y2), u = 〈 3


5
, 4
5
〉 , P(1, 2)


272. f (x, y) = x2 y, P(−5, 5), v = 3i − 4j
273. f (x, y) = y2 + xz, P(1, 2, 2), v = 〈 2, −1, 2 〉
For the following exercises, find the directional derivativeof the function in the direction of the unit vector
u = cos θi + sin θj.


274. f (x, y) = x2 + 2y2, θ = π
6


275. f (x, y) = y
x + 2y


, θ = − π
4


276. f (x, y) = cos⎛⎝3x + y⎞⎠, θ = π4
277. w(x, y) = yex, θ = π


3


278. f (x, y) = x arctan(y), θ = π2
279. f (x, y) = ln(x + 2y), θ = π3
For the following exercises, find the gradient.
280. Find the gradient of f (x, y) = 14 − x2 − y2


3
. Then,


find the gradient at point P(1, 2).
281. Find the gradient of f (x, y, z) = xy + yz + xz at
point P(1, 2, 3).
282. Find the gradient of f (x, y, z) at P and in the
direction of u:
f (x, y, z) = ln(x2 + 2y2 + 3z2), P(2, 1, 4), u = −313


i − 4
13


j − 12
13


k.


283.
f (x, y, z) = 4x5 y2 z3, P(2, −1, 1), u = 13


i + 2
3
j − 2


3
k


For the following exercises, find the directional derivativeof the function at point P in the direction of Q.
284. f (x, y) = x2 + 3y2, P(1, 1), Q(4, 5)
285. f (x, y, z) = yx + z, P(2, 1, −1), Q(−1, 2, 0)
For the following exercises, find the derivative of thefunction at P in the direction of u.
286. f (x, y) = −7x + 2y, P(2, −4), u = 4i − 3j
287. f (x, y) = ln(5x + 4y), P(3, 9), u = 6i + 8j
288. [T] Use technology to sketch the level curve of
f (x, y) = 4x − 2y + 3 that passes through P(1, 2) and
draw the gradient vector at P.


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289. [T] Use technology to sketch the level curve of
f (x, y) = x2 + 4y2 that passes through P(−2, 0) and
draw the gradient vector at P.
For the following exercises, find the gradient vector at theindicated point.
290. f (x, y) = xy2 − yx2, P(−1, 1)
291. f (x, y) = xey − ln(x), P(−3, 0)
292. f (x, y, z) = xy − ln(z), P(2, −2, 2)
293. f (x, y, z) = x y2 + z2, P(−2, −1, −1)
For the following exercises, find the derivative of thefunction.
294. f (x, y) = x2 + xy + y2 at point (−5, −4) in the
direction the function increases most rapidly
295. f (x, y) = exy at point (6, 7) in the direction the
function increases most rapidly
296. f (x, y) = arctan⎛⎝yx⎞⎠ at point (−9, 9) in the
direction the function increases most rapidly
297. f (x, y, z) = ln(xy + yz + zx) at point
(−9, −18, −27) in the direction the function increases
most rapidly
298. f (x, y, z) = xy + yz + zx at point (5, −5, 5) in the
direction the function increases most rapidly
For the following exercises, find the maximum rate ofchange of f at the given point and the direction in which
it occurs.
299. f (x, y) = xe−y, (1, 0)
300. f (x, y) = x2 + 2y, (4, 10)
301. f (x, y) = cos(3x + 2y), ⎛⎝π6, − π8⎞⎠
For the following exercises, find equations of


a. the tangent plane and
b. the normal line to the given surface at the givenpoint.


302. The level curve f (x, y, z) = 12 for
f (x, y, z) = 4x2 − 2y2 + z2 at point (2, 2, 2).


303. f (x, y, z) = xy + yz + xz = 3 at point (1, 1, 1)
304. f (x, y, z) = xyz = 6 at point (1, 2, 3)
305. f (x, y, z) = xey cos z − z = 1 at point (1, 0, 0)
For the following exercises, solve the problem.
306. The temperature T in a metal sphere is inversely
proportional to the distance from the center of the sphere(the origin: (0, 0, 0)). The temperature at point (1, 2, 2)
is 120°C.


a. Find the rate of change of the temperature at point
(1, 2, 2) in the direction toward point (2, 1, 3).


b. Show that, at any point in the sphere, the directionof greatest increase in temperature is given by avector that points toward the origin.
307. The electrical potential (voltage) in a certain regionof space is given by the function
V(x, y, z) = 5x2 − 3xy + xyz.


a. Find the rate of change of the voltage at point
(3, 4, 5) in the direction of the vector
〈 1, 1, −1 〉 .


b. In which direction does the voltage change mostrapidly at point (3, 4, 5)?
c. What is the maximum rate of change of the voltageat point (3, 4, 5)?


308. If the electric potential at a point (x, y) in the
xy-plane is V(x, y) = e−2x cos(2y), then the electric
intensity vector at (x, y) is E = −∇V(x, y).


a. Find the electric intensity vector at ⎛⎝π4, 0⎞⎠.
b. Show that, at each point in the plane, the electricpotential decreases most rapidly in the direction ofthe vector E.


309. In two dimensions, the motion of an ideal fluid isgoverned by a velocity potential φ. The velocity
components of the fluid u in the x-direction and v in
the y-direction, are given by 〈 u, v 〉 = ∇φ. Find the
velocity components associated with the velocity potential
φ(x, y) = sin πx sin 2πy.


Chapter 4 | Differentiation of Functions of Several Variables 435




4.7 | Maxima/Minima Problems
Learning Objectives


4.7.1 Use partial derivatives to locate critical points for a function of two variables.
4.7.2 Apply a second derivative test to identify a critical point as a local maximum, local minimum,or saddle point for a function of two variables.
4.7.3 Examine critical points and boundary points to find absolute maximum and minimum valuesfor a function of two variables.


One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/orminimum values. This application is also important for functions of two or more variables, but as we have seen in earliersections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations.The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessingthe results.
Critical Points
For functions of a single variable, we defined critical points as the values of the function when the derivative equals zeroor does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we arenow working with partial derivatives.
Definition
Let z = f (x, y) be a function of two variables that is differentiable on an open set containing the point (x0, y0).
The point (x0, y0) is called a critical point of a function of two variables f if one of the two following conditions
holds:


1. fx (x0, y0) = fy (x0, y0) = 0
2. Either fx (x0, y0) or fy (x0, y0) does not exist.


Example 4.38
Finding Critical Points
Find the critical points of each of the following functions:


a. f (x, y) = 4y2 − 9x2 + 24y + 36x + 36
b. g(x, y) = x2 + 2xy − 4y2 + 4x − 6y + 4


Solution
a. First, we calculate fx (x, y) and fy (x, y):


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fx (x, y) = 12
(−18x + 36)⎛⎝4y


2 − 9x2 + 24y + 36x + 36⎞⎠
−1/2


= −9x + 18


4y2 − 9x2 + 24y + 36x + 36


fy (x, y) = 12

⎝8y + 24⎞⎠⎛⎝4y


2 − 9x2 + 24y + 36x + 36⎞⎠
−1/2


=
4y + 12


4y2 − 9x2 + 24y + 36x + 36
.


Next, we set each of these expressions equal to zero:
−9x + 18


4y2 − 9x2 + 24y + 36x + 36
= 0


4y + 12


4y2 − 9x2 + 24y + 36x + 36
= 0.


Then, multiply each equation by its common denominator:
−9x + 18 = 0


4y + 12 = 0.


Therefore, x = 2 and y = −3, so (2, −3) is a critical point of f .
We must also check for the possibility that the denominator of each partial derivative can equal zero, thuscausing the partial derivative not to exist. Since the denominator is the same in each partial derivative, weneed only do this once:


4y2 − 9x2 + 24y + 36x + 36 = 0.


This equation represents a hyperbola. We should also note that the domain of f consists of points
satisfying the inequality


4y2 − 9x2 + 24y + 36x + 36 ≥ 0.


Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of thedomain. To put the hyperbola in standard form, we use the method of completing the square:
4y2 − 9x2 + 24y + 36x + 36 = 0


4y2 − 9x2 + 24y + 36x = −36


4y2 + 24y − 9x2 + 36x = −36


4⎛⎝y
2 + 6y⎞⎠− 9



⎝x


2 − 4x⎞⎠ = −36


4⎛⎝y
2 + 6y + 9⎞⎠− 9



⎝x


2 − 4x + 4⎞⎠ = −36 + 36 − 36


4⎛⎝y + 3⎞⎠2 − 9(x − 2)2 = −36.


Dividing both sides by −36 puts the equation in standard form:


Chapter 4 | Differentiation of Functions of Several Variables 437




4.34


4⎛⎝y + 3⎞⎠2


−36
− 9(x − 2)


2


−36
= 1


(x − 2)2


4


⎝y + 3⎞⎠2


9
= 1.


Notice that point (2, −3) is the center of the hyperbola.
b. First, we calculate gx (x, y) and gy (x, y):


gx (x, y) = 2x + 2y + 4


gy (x, y) = 2x − 8y − 6.


Next, we set each of these expressions equal to zero, which gives a system of equations in x and y:
2x + 2y + 4 = 0
2x − 8y − 6 = 0.


Subtracting the second equation from the first gives 10y + 10 = 0, so y = −1. Substituting this into
the first equation gives 2x + 2(−1) + 4 = 0, so x = −1. Therefore (−1, −1) is a critical point of g
(Figure 4.46). There are no points in ℝ2 that make either partial derivative not exist.


Figure 4.46 The function g(x, y) has a critical point at (−1, −1, 6).


Find the critical point of the function f (x, y) = x3 + 2xy − 2x − 4y.


The main purpose for determining critical points is to locate relative maxima and minima, as in single-variable calculus.When working with a function of one variable, the definition of a local extremum involves finding an interval around thecritical point such that the function value is either greater than or less than all the other function values in that interval.When working with a function of two or more variables, we work with an open disk around the point.


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Definition
Let z = f (x, y) be a function of two variables that is defined and continuous on an open set containing the point
(x0, y0). Then f has a local maximum at (x0, y0) if


f (x0, y0) ≥ f (x, y)


for all points (x, y) within some disk centered at (x0, y0). The number f (x0, y0) is called a local maximum value.
If the preceding inequality holds for every point (x, y) in the domain of f , then f has a global maximum (also
called an absolute maximum) at (x0, y0).
The function f has a local minimum at (x0, y0) if


f (x0, y0) ≤ f (x, y)


for all points (x, y) within some disk centered at (x0, y0). The number f (x0, y0) is called a local minimum value. If
the preceding inequality holds for every point (x, y) in the domain of f , then f has a global minimum (also called
an absolute minimum) at (x0, y0).
If f (x0, y0) is either a local maximum or local minimum value, then it is called a local extremum (see the following
figure).


Figure 4.47 The graph of z = 16 − x2 − y2 has a
maximum value when (x, y) = (0, 0). It attains its minimum
value at the boundary of its domain, which is the circle
x2 + y2 = 16.


In Maxima and Minima (http://cnx.org/content/m53611/latest/) , we showed that extrema of functions of onevariable occur at critical points. The same is true for functions of more than one variable, as stated in the following theorem.
Theorem 4.16: Fermat’s Theorem for Functions of Two Variables
Let z = f (x, y) be a function of two variables that is defined and continuous on an open set containing the point


Chapter 4 | Differentiation of Functions of Several Variables 439




(x0, y0). Suppose fx and fy each exists at (x0, y0). If f has a local extremum at (x0, y0), then (x0, y0) is a
critical point of f .


Second Derivative Test
Consider the function f (x) = x3. This function has a critical point at x = 0, since f ′(0) = 3(0)2 = 0. However, f
does not have an extreme value at x = 0. Therefore, the existence of a critical value at x = x0 does not guarantee a local
extremum at x = x0. The same is true for a function of two or more variables. One way this can happen is at a saddle
point. An example of a saddle point appears in the following figure.


Figure 4.48 Graph of the function z = x2 − y2. This graph
has a saddle point at the origin.


In this graph, the origin is a saddle point. This is because the first partial derivatives of f (x, y) = x2 − y2 are both equal to
zero at this point, but it is neither a maximum nor a minimum for the function. Furthermore the vertical trace corresponding
to y = 0 is z = x2 (a parabola opening upward), but the vertical trace corresponding to x = 0 is z = −y2 (a parabola
opening downward). Therefore, it is both a global maximum for one trace and a global minimum for another.
Definition
Given the function z = f (x, y), the point ⎛⎝x0, y0, f (x0, y0)⎞⎠ is a saddle point if both f0 (x0, y0) = 0 and
fy (x0, y0) = 0, but f does not have a local extremum at (x0, y0).


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The second derivative test for a function of one variable provides a method for determining whether an extremum occurs ata critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact thatthere are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three.The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant D that
replaces f ″(x0) in the second derivative test for a function of one variable.


Theorem 4.17: Second Derivative Test
Let z = f (x, y) be a function of two variables for which the first- and second-order partial derivatives are continuous
on some disk containing the point (x0, y0). Suppose fx (x0, y0) = 0 and fy (x0, y0) = 0. Define the quantity


(4.43)D = fxx (x0, y0) fyy (x0, y0) − ⎛⎝ fxy (x0, y0)⎞⎠2.
i. If D > 0 and fxx (x0, y0) > 0, then f has a local minimum at (x0, y0).
ii. If D > 0 and fxx (x0, y0) < 0, then f has a local maximum at (x0, y0).
iii. If D < 0, , then f has a saddle point at (x0, y0).
iv. If D = 0, then the test is inconclusive.


See Figure 4.49.


Figure 4.49 The second derivative test can often determine whether a function of two variables has a local minima (a), alocal maxima (b), or a saddle point (c).


To apply the second derivative test, it is necessary that we first find the critical points of the function. There are severalsteps involved in the entire procedure, which are outlined in a problem-solving strategy.
Problem-Solving Strategy: Using the Second Derivative Test for Functions of Two Variables
Let z = f (x, y) be a function of two variables for which the first- and second-order partial derivatives are continuous
on some disk containing the point (x0, y0). To apply the second derivative test to find local extrema, use the following
steps:


1. Determine the critical points (x0, y0) of the function f where fx (x0, y0) = fy (x0, y0) = 0. Discard any
points where at least one of the partial derivatives does not exist.


2. Calculate the discriminant D = fxx (x0, y0) fyy (x0, y0) − ⎛⎝ fxy (x0, y0)⎞⎠2 for each critical point of f .


Chapter 4 | Differentiation of Functions of Several Variables 441




3. Apply Second Derivative Test to determine whether each critical point is a local maximum, localminimum, or saddle point, or whether the theorem is inconclusive.


Example 4.39
Using the Second Derivative Test
Find the critical points for each of the following functions, and use the second derivative test to find the localextrema:


a. f (x, y) = 4x2 + 9y2 + 8x − 36y + 24
b. g(x, y) = 1


3
x3 + y2 + 2xy − 6x − 3y + 4


Solution
a. Step 1 of the problem-solving strategy involves finding the critical points of f . To do this, we first


calculate fx (x, y) and fy (x, y), then set each of them equal to zero:
fx (x, y) = 8x + 8


fy (x, y) = 18y − 36.


Setting them equal to zero yields the system of equations
8x + 8 = 0


18y − 36 = 0.


The solution to this system is x = −1 and y = 2. Therefore (−1, 2) is a critical point of f .
Step 2 of the problem-solving strategy involves calculating D. To do this, we first calculate the second
partial derivatives of f :


fxx (x, y) = 8
fxy (x, y) = 0


fyy (x, y) = 18.


Therefore, D = fxx (−1, 2) fyy (−1, 2) − ⎛⎝ fxy (−1, 2)⎞⎠2 = (8)(18) − (0)2 = 144.
Step 3 states to check Fermat’s Theorem for Functions of Two Variables. Since D > 0 and
fxx (−1, 2) > 0, this corresponds to case 1. Therefore, f has a local minimum at (−1, 2) as shown in
the following figure.


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Figure 4.50 The function f (x, y) has a local minimum at (−1, 2, −16).
b. For step 1, we first calculate gx (x, y) and gy (x, y), then set each of them equal to zero:


gx (x, y) = x
2 + 2y − 6


gy (x, y) = 2y + 2x − 3.


Setting them equal to zero yields the system of equations
x2 + 2y − 6 = 0
2y + 2x − 3 = 0.


To solve this system, first solve the second equation for y. This gives y = 3 − 2x
2


. Substituting this into
the first equation gives


x2 + 3 − 2x − 6 = 0


x2 − 2x − 3 = 0
(x − 3)(x + 1) = 0.


Therefore, x = −1 or x = 3. Substituting these values into the equation y = 3 − 2x
2


yields the critical
points ⎛⎝−1, 52⎞⎠ and ⎛⎝3, − 32⎞⎠.
Step 2 involves calculating the second partial derivatives of g:


gxx (x, y) = 2x


gxy (x, y) = 2


gyy (x, y) = 2.


Then, we find a general formula for D:
D = gxx (x0, y0)gyy (x0, y0) −



⎝gxy (x0, y0)




2


= ⎛⎝2x0

⎠(2) − 22


= 4x0 − 4.


Chapter 4 | Differentiation of Functions of Several Variables 443




4.35


Next, we substitute each critical point into this formula:
D⎛⎝−1,


5
2

⎠ =



⎝2(−1)⎞⎠(2) − (2)2 = −4 − 4 = −8


D⎛⎝3, −
3
2

⎠ =



⎝2(3)⎞⎠(2) − (2)2 = 12 − 4 = 8.


In step 3, we note that, applying Fermat’s Theorem for Functions of Two Variables to point

⎝−1,


5
2

⎠ leads to case 3, which means that ⎛⎝−1, 52⎞⎠ is a saddle point. Applying the theorem to point



⎝3, −


3
2

⎠ leads to case 1, which means that ⎛⎝3, − 32⎞⎠ corresponds to a local minimum as shown in the


following figure.


Figure 4.51 The function g(x, y) has a local minimum and a saddle point.


Use the second derivative to find the local extrema of the function
f (x, y) = x3 + 2xy − 6x − 4y2.


Absolute Maxima and Minima
When finding global extrema of functions of one variable on a closed interval, we start by checking the critical values overthat interval and then evaluate the function at the endpoints of the interval. When working with a function of two variables,the closed interval is replaced by a closed, bounded set. A set is bounded if all the points in that set can be contained withina ball (or disk) of finite radius. First, we need to find the critical points inside the set and calculate the corresponding criticalvalues. Then, it is necessary to find the maximum and minimum value of the function on the boundary of the set. Whenwe have all these values, the largest function value corresponds to the global maximum and the smallest function valuecorresponds to the absolute minimum. First, however, we need to be assured that such values exist. The following theoremdoes this.


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Theorem 4.18: Extreme Value Theorem
A continuous function f (x, y) on a closed and bounded set D in the plane attains an absolute maximum value at
some point of D and an absolute minimum value at some point of D.


Now that we know any continuous function f defined on a closed, bounded set attains its extreme values, we need to know
how to find them.
Theorem 4.19: Finding Extreme Values of a Function of Two Variables
Assume z = f (x, y) is a differentiable function of two variables defined on a closed, bounded set D. Then f will
attain the absolute maximum value and the absolute minimum value, which are, respectively, the largest and smallestvalues found among the following:


i. The values of f at the critical points of f in D.
ii. The values of f on the boundary of D.


The proof of this theorem is a direct consequence of the extreme value theorem and Fermat’s theorem. In particular, ifeither extremum is not located on the boundary of D, then it is located at an interior point of D. But an interior point
(x0, y0) of D that’s an absolute extremum is also a local extremum; hence, (x0, y0) is a critical point of f by Fermat’s
theorem. Therefore the only possible values for the global extrema of f on D are the extreme values of f on the interior
or boundary of D.


Problem-Solving Strategy: Finding Absolute Maximum and Minimum Values
Let z = f (x, y) be a continuous function of two variables defined on a closed, bounded set D, and assume f is
differentiable on D. To find the absolute maximum and minimum values of f on D, do the following:


1. Determine the critical points of f in D.
2. Calculate f at each of these critical points.
3. Determine the maximum and minimum values of f on the boundary of its domain.
4. The maximum and minimum values of f will occur at one of the values obtained in steps 2 and 3.


Finding the maximum and minimum values of f on the boundary of D can be challenging. If the boundary is a rectangle
or set of straight lines, then it is possible to parameterize the line segments and determine the maxima on each of thesesegments, as seen in Example 4.40. The same approach can be used for other shapes such as circles and ellipses.
If the boundary of the set D is a more complicated curve defined by a function g(x, y) = c for some constant c, and
the first-order partial derivatives of g exist, then the method of Lagrange multipliers can prove useful for determining the
extrema of f on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers.


Example 4.40
Finding Absolute Extrema


Chapter 4 | Differentiation of Functions of Several Variables 445




Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema ofeach of the following functions:
a. f (x, y) = x2 − 2xy + 4y2 − 4x − 2y + 24 on the domain defined by 0 ≤ x ≤ 4 and 0 ≤ y ≤ 2
b. g(x, y) = x2 + y2 + 4x − 6y on the domain defined by x2 + y2 ≤ 16


Solution
a. Using the problem-solving strategy, step 1 involves finding the critical points of f on its domain.


Therefore, we first calculate fx (x, y) and fy (x, y), then set them each equal to zero:
fx (x, y) = 2x − 2y − 4
fy (x, y) = −2x + 8y − 2.


Setting them equal to zero yields the system of equations
2x − 2y − 4 = 0


−2x + 8y − 2 = 0.


The solution to this system is x = 3 and y = 1. Therefore (3, 1) is a critical point of f . Calculating
f (3, 1) gives f (3, 1) = 17.
The next step involves finding the extrema of f on the boundary of its domain. The boundary of its
domain consists of four line segments as shown in the following graph:


Figure 4.52 Graph of the domain of the function
f (x, y) = x2 − 2xy + 4y2 − 4x − 2y + 24.


L1 is the line segment connecting (0, 0) and (4, 0), and it can be parameterized by the equations
x(t) = t, y(t) = 0 for 0 ≤ t ≤ 4. Define g(t) = f ⎛⎝x(t), y(t)⎞⎠. This gives g(t) = t2 − 4t + 24.
Differentiating g leads to g′ (t) = 2t − 4. Therefore, g has a critical value at t = 2, which corresponds
to the point (2, 0). Calculating f (2, 0) gives the z-value 20.
L2 is the line segment connecting (4, 0) and (4, 2), and it can be parameterized by the equations
x(t) = 4, y(t) = t for 0 ≤ t ≤ 2. Again, define g(t) = f ⎛⎝x(t), y(t)⎞⎠. This gives g(t) = 4t2 − 10t + 24.
Then, g′ (t) = 8t − 10. g has a critical value at t = 5


4
, which corresponds to the point ⎛⎝0, 54⎞⎠.


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Calculating f ⎛⎝0, 54⎞⎠ gives the z-value 27.75.
L3 is the line segment connecting (0, 2) and (4, 2), and it can be parameterized by the equations
x(t) = t, y(t) = 2 for 0 ≤ t ≤ 4. Again, define g(t) = f ⎛⎝x(t), y(t)⎞⎠. This gives g(t) = t2 − 8t + 36.
The critical value corresponds to the point (4, 2). So, calculating f (4, 2) gives the z-value 20.
L4 is the line segment connecting (0, 0) and (0, 2), and it can be parameterized by the equations
x(t) = 0, y(t) = t for 0 ≤ t ≤ 2. This time, g(t) = 4t2 − 2t + 24 and the critical value t = 1


4


correspond to the point ⎛⎝0, 14⎞⎠. Calculating f ⎛⎝0, 14⎞⎠ gives the z-value 23.75.
We also need to find the values of f (x, y) at the corners of its domain. These corners are located at
(0, 0), (4, 0), (4, 2) and (0, 2):


f (0, 0) = (0)2 − 2(0)(0) + 4(0)2 − 4(0) − 2(0) + 24 = 24


f (4, 0) = (4)2 − 2(4)(0) + 4(0)2 − 4(4) − 2(0) + 24 = 24


f (4, 2) = (4)2 − 2(4)(2) + 4(2)2 − 4(4) − 2(2) + 24 = 20


f (0, 2) = (0)2 − 2(0)(2) + 4(2)2 − 4(0) − 2(2) + 24 = 36.


The absolute maximum value is 36, which occurs at (0, 2), and the global minimum value is 20,
which occurs at both (4, 2) and (2, 0) as shown in the following figure.


Figure 4.53 The function f (x, y) has two global minima and one global maximum over its
domain.


Chapter 4 | Differentiation of Functions of Several Variables 447




b. Using the problem-solving strategy, step 1 involves finding the critical points of g on its domain.
Therefore, we first calculate gx (x, y) and gy (x, y), then set them each equal to zero:


gx (x, y) = 2x + 4


gy (x, y) = 2y − 6.


Setting them equal to zero yields the system of equations
2x + 4 = 0
2y − 6 = 0.


The solution to this system is x = −2 and y = 3. Therefore, (−2, 3) is a critical point of g.
Calculating g(−2, 3), we get


g(−2, 3) = (−2)2 + 32 + 4(−2) − 6(3) = 4 + 9 − 8 − 18 = −13.


The next step involves finding the extrema of g on the boundary of its domain. The boundary of itsdomain consists of a circle of radius 4 centered at the origin as shown in the following graph.


Figure 4.54 Graph of the domain of the function
g(x, y) = x2 + y2 + 4x − 6y.


The boundary of the domain of g can be parameterized using the functions x(t) = 4 cos t, y(t) = 4 sin t
for 0 ≤ t ≤ 2π. Define h(t) = g⎛⎝x(t), y(t)⎞⎠:


h(t) = g⎛⎝x(t), y(t)⎞⎠


= (4 cos t)2 + (4 sin t)2 + 4(4 cos t) − 6(4 sin t)


= 16cos2 t + 16sin2 t + 16 cos t − 24 sin t
= 16 + 16 cos t − 24 sin t.


448 Chapter 4 | Differentiation of Functions of Several Variables


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Setting h′ (t) = 0 leads to
−16 sin t − 24 cos t = 0


−16 sin t = 24 cos t


−16 sin t
−16 cos t


= 24 cos t
−16 cos t


tan t = −4
3
.


−16 sin t − 24 cos t = 0


−16 sin t = 24 cos t


−16 sin t
−16 cos t


= 24 cos t
−16 cos t


tan t = −3
2
.


This equation has two solutions over the interval 0 ≤ t ≤ 2π. One is t = π − arctan⎛⎝32⎞⎠ and the other is
t = 2π − arctan⎛⎝


3
2

⎠. For the first angle,


sin t = sin⎛⎝π − arctan


3
2



⎠ = sin



⎝arctan




3
2



⎠ =


3 13
13


cos t = cos⎛⎝π − arctan


3
2



⎠ = −cos



⎝arctan




3
2



⎠ = −


2 13
13


.


Therefore, x(t) = 4 cos t = − 8 13
13


and y(t) = 4 sin t = 12 13
13


, so ⎛⎝−8 1313 , 12 1313 ⎞⎠ is a critical
point on the boundary and


g

⎝−


8 13
13


, 12 13
13

⎠ =

⎝−


8 13
13



2
+ ⎛⎝


12 13
13



2
+ 4⎛⎝−


8 13
13

⎠− 6


12 13
13



= 144
13


+ 64
13


− 32 13
13


− 72 13
13


= 208 − 104 13
13


≈ −12.844.


For the second angle,
sin t = sin⎛⎝2π − arctan




3
2



⎠ = −sin



⎝arctan




3
2



⎠ = −


3 13
13


cos t = cos⎛⎝2π − arctan


3
2



⎠ = cos



⎝arctan




3
2



⎠ =


2 13
13


.


Therefore, x(t) = 4 cos t = 8 13
13


and y(t) = 4 sin t = − 12 13
13


, so ⎛⎝8 1313 , − 12 1313 ⎞⎠ is a critical
point on the boundary and


Chapter 4 | Differentiation of Functions of Several Variables 449




4.36


g


8 13
13


, − 12 13
13

⎠ =


8 13
13



2
+ ⎛⎝−


12 13
13



2
+ 4⎛⎝


8 13
13

⎠− 6

⎝−


12 13
13



= 144
13


+ 64
13


+ 32 13
13


+ 72 13
13


= 208 + 104 13
13


≈ 44.844.


The absolute minimum of g is −13, which is attained at the point (−2, 3), which is an interior point
of D. The absolute maximum of g is approximately equal to 44.844, which is attained at the boundary
point ⎛⎝8 1313 , − 12 1313 ⎞⎠. These are the absolute extrema of g on D as shown in the following figure.


Figure 4.55 The function f (x, y) has a local minimum and a local maximum.


Use the problem-solving strategy for finding absolute extrema of a function to find the absolute extremaof the function
f (x, y) = 4x2 − 2xy + 6y2 − 8x + 2y + 3


on the domain defined by 0 ≤ x ≤ 2 and −1 ≤ y ≤ 3.


Example 4.41
Chapter Opener: Profitable Golf Balls


450 Chapter 4 | Differentiation of Functions of Several Variables


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Figure 4.56 (credit: modification of work by oatsy40, Flickr)


Pro- T company has developed a profit model that depends on the number x of golf balls sold per month
(measured in thousands), and the number of hours per month of advertising y, according to the function


z = f (x, y) = 48x + 96y − x2 − 2xy − 9y2,


where z is measured in thousands of dollars. The maximum number of golf balls that can be produced and sold
is 50,000, and the maximum number of hours of advertising that can be purchased is 25. Find the values of x
and y that maximize profit, and find the maximum profit.
Solution
Using the problem-solving strategy, step 1 involves finding the critical points of f on its domain. Therefore, we
first calculate fx (x, y) and fy (x, y), then set them each equal to zero:


fx (x, y) = 48 − 2x − 2y


fy (x, y) = 96 − 2x − 18y.


Setting them equal to zero yields the system of equations
48 − 2x − 2y = 0


96 − 2x − 18y = 0.


The solution to this system is x = 21 and y = 3. Therefore (21, 3) is a critical point of f . Calculating
f (21, 3) gives f (21, 3) = 48(21) + 96(3) − 212 − 2(21)(3) − 9(3)2 = 648.
The domain of this function is 0 ≤ x ≤ 50 and 0 ≤ y ≤ 25 as shown in the following graph.


Chapter 4 | Differentiation of Functions of Several Variables 451




Figure 4.57 Graph of the domain of the function
f (x, y) = 48x + 96y − x2 − 2xy − 9y2.


L1 is the line segment connecting (0, 0) and (50, 0), and it can be parameterized by the equations
x(t) = t, y(t) = 0 for 0 ≤ t ≤ 50. We then define g(t) = f ⎛⎝x(t), y(t)⎞⎠:


g(t) = f ⎛⎝x(t), y(t)⎞⎠


= f (t, 0)


= 48t + 96(0) − y2 − 2(t)(0) − 9(0)2


= 48t − t2.


Setting g′ (t) = 0 yields the critical point t = 24, which corresponds to the point (24, 0) in the domain of f .
Calculating f (24, 0) gives 576.
L2 is the line segment connecting and (50, 25), and it can be parameterized by the equations
x(t) = 50, y(t) = t for 0 ≤ t ≤ 25. Once again, we define g(t) = f ⎛⎝x(t), y(t)⎞⎠:


g(t) = f ⎛⎝x(t), y(t)⎞⎠


= f (50, t)


= 48(50) + 96t − 502 − 2(50)t − 9t2


= −9t2 − 4t − 100.


This function has a critical point at t = − 2
9
, which corresponds to the point ⎛⎝50, − 29⎞⎠. This point is not in


the domain of f .
L3 is the line segment connecting (0, 25) and (50, 25), and it can be parameterized by the equations
x(t) = t, y(t) = 25 for 0 ≤ t ≤ 50. We define g(t) = f ⎛⎝x(t), y(t)⎞⎠:


g(t) = f ⎛⎝x(t), y(t)⎞⎠


= f (t, 25)


= 48t + 96(25) − t2 − 2t(25) − 9⎛⎝25
2⎞


= −t2 − 2t − 3225.


This function has a critical point at t = −1, which corresponds to the point (−1, 25), which is not in the


452 Chapter 4 | Differentiation of Functions of Several Variables


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domain.
L4 is the line segment connecting (0, 0) to (0, 25), and it can be parameterized by the equations
x(t) = 0, y(t) = t for 0 ≤ t ≤ 25. We define g(t) = f ⎛⎝x(t), y(t)⎞⎠:


g(t) = f ⎛⎝x(t), y(t)⎞⎠


= f (0, t)


= 48(0) + 96t − (0)2 − 2(0)t − 9t2


= 96t − t2.


This function has a critical point at t = 16
3
, which corresponds to the point ⎛⎝0, 163 ⎞⎠, which is on th