Student Study Guide
for
Calculus, 9th Edition
with Late Transcendentals
by
Brian Camp
Preface
Welcome to the Student Study Guide to accompany Calculus with Late Transcendentals by Anton, Bivens, and Davis. Rather than producing a study guide that is
another exhaustive textbook of the subject, this book is primarily intended to be a
supplement to the main textbook. If this book is read by itself, many useful ideas
will be found but there are many others which this study guide does not attempt
to cover. Instead, this book will compliment the text by offering several key features that are aimed at students who are learning calculus so that they may become
proficient in the subject matter.
A Different Perspective
This book is not exhaustive, nor is it intended to be. An effort has been made to
keep each section short so that they are accessible without spending a great deal
of time reading them. While the sections are short, they strive to focus on what
is important so that a student can keep their attention on the necessary details in
the section. The book is also written to give a different perspective than the text.
Often seeing or hearing a concept in two or more ways can be a great help in
learning the material.
Summaries, Objectives, and Checklists
At the beginning of each chapter, a brief summary of the chapter is given. This is
followed by a list of objectives for the chapter. Each objective is also listed with
an appropriate section where material related to the objective may be found.
Each section in a chapter begins with a purpose for that section. At the end of each
section is a checklist of important ideas (in no particular order) that are discussed
in the section. Some of these ideas are explained in this book while some ideas
are left for the main text.
iii
iv
Focuses on Key Concepts
The goal of each written summary in this book is to focus upon key concepts and
to cut to the chase. Some of the more important ideas are placed in highlighted
boxes to set them apart. In a similar fashion, common pitfalls and things to be
cautious of are highlighted as well.
Room for Notes
Space is left on the pages of this book on purpose so that there is room to write or
take notes. Some key ideas are highlighted in the margin as well to make it easier
to find some topics within a section.
Quizzes for Self-Testing
At the end of each chapter, there is material that a student may use to self-test
themselves. Quizzes are provided for material in each section and a test at the end
of the chapter is also included.
Learning the subject of Calculus may be a challenging endeavor but in the end,
the rewards are well worth the effort that is put in. The goal of the author has
been to try to simplify the process of learning calculus so that some of its more
confusing aspects do not become a hindrance. Calculus is a very powerful subject
especially because it has applications in so many areas of life. At the same time,
this almost universal applicability can cause calculus to be more intimidating than
it should be. Hopefully this book can help the reader keep the focus of calculus
on the basics as they learn the material.
B. Camp
Ruston, Louisiana
July 2009
Contents
Chapter 0: Before Calculus
1
0.1
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
0.2
New Functions from Old . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
0.3
Families of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
0.4
Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
Chapter 0 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Chapter 1: Limits and Continuity
13
1.1
Limits (An Intuitive Approach) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
1.2
Computing Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
1.3
Limits at Infinity; End Behavior of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.4
Limits (Discussed More Rigorously) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
1.5
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
1.6
Continuity of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
Chapter 1 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
Chapter 2: The Derivative
29
2.1
Tangent Lines and Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.2
The Derivative Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
2.3
Introduction to Techniques of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
2.4
The Product and Quotient Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
2.5
Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
2.6
The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
2.7
Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
2.8
Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
2.9
Local Linear Approximations; Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
Chapter 2 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
v
vi
Chapter 3: The Derivative in Graphing and Applications
53
3.1
Analysis of Functions I: Increase, Decrease, and Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
3.2
Analysis of Functions II: Relative Extrema; Graphing Polynomials . . . . . . . . . . . . . . . . . . . . . . . .
56
3.3
Analysis of Functions III: Rational Functions, Cusps, and Vertical Tangents . . . . . . . . . . . . . . . . . . .
57
3.4
Absolute Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58
3.5
Applied Maximum and Minimum Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
3.6
Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
3.7
Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
3.8
Rolle’s Theorem; Mean-Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
Chapter 3 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
Chapter 4: Integration
77
4.1
An Overview of the Area Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
78
4.2
The Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
79
4.3
Integration By Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
4.4
The Definition of Area as a Limit; Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
82
4.5
The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
4.6
The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
4.7
Rectilinear Motion Revisited Using Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
4.8
Average Value of a Function and Its Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
4.9
Evaluating Definite Integrals by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
Chapter 4 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
Chapter 5: Applications of the Definite Integral in Geometry, Science and Engineering
97
5.1
Area Between Two Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
5.2
Volumes by Slicing; Disks and Washers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
99
5.3
Volumes by Cylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.4
Length of a Plane Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.5
Area of a Surface of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
5.6
Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.7
Moments, Centers of Gravity, and Centroids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.8
Fluid Pressure and Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
Chapter 5 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
vii
Chapter 6: Exponential, Logarithmic, and Inverse Trigonometric Functions
119
6.1
Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.2
Derivatives and Integrals Involving Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
6.3
Derivatives of Inverse Functions; Derivatives and Integrals Involving Exponential Functions . . . . . . . . . . 122
6.4
Graphs and Applications Involving Logarithmic and Exponential Functions . . . . . . . . . . . . . . . . . . . 123
6.5
L’Hôpital’s Rule; Indeterminate Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
6.6
Logarithmic Functions from the Integral Point of View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
6.7
Derivatives and Integrals Involving Inverse
Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6.8
Hyperbolic Functions and Hanging Cables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
Chapter 6 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
Chapter 7: Principles of Integral Evaluation
141
7.1
An Overview of Integration Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
7.2
Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
7.3
Integrating Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
7.4
Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
7.5
Integrating Rational Functions by Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
7.6
Using Computer Algebra Systems and Tables of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
7.7
Numerical Integration; Simpson’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
7.8
Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151
Chapter 7 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
Chapter 8: Mathematical Modeling with Differential Equations
165
8.1
Modeling With Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
8.2
Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
8.3
Slope Fields; Euler’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
8.4
First-Order Differential Equations and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
Chapter 8 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
viii
Chapter 9: Infinite Series
179
9.1
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
9.2
Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
9.3
Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
9.4
Convergence Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
9.5
The Comparison, Ratio, and Root Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
9.6
Alternating Series; Conditional Convergence
9.7
Maclaurin and Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
9.8
Maclaurin and Taylor Series; Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
9.9
Convergence of Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
9.10 Differentiating and Integrating Power Series; Modeling with Taylor Series . . . . . . . . . . . . . . . . . . . . 192
Chapter 9 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
Chapter 10: Parametric And Polar Curves; Conic Sections
207
10.1 Parametric Equations; Tangent Lines And Arc Length For Parametric Curves . . . . . . . . . . . . . . . . . . 208
10.2 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
10.3 Tangent Lines, Arc Length, And Area For Polar Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
10.4 Conic Sections in Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
10.5 Rotation of Axes; Second-Degree Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
10.6 Conic Sections in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
Chapter 10 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
Chapter 11: Three Dimensional Space; Vectors
227
11.1 Rectangular Coordinates in 3-Space; Spheres; Cylindrical Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 228
11.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229
11.3 Dot Products; Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
11.4 Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
11.5 Parametric Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
11.6 Planes in 3-Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
11.7 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
11.8 Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242
Chapter 11 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
ix
Chapter 12: Vector-Valued Functions
255
12.1 Introduction to Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
12.2 Calculus of Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
12.3 Change of Parameter; Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260
12.4 Unit Tangent, Normal, and Binormal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
12.5 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
12.6 Motion Along a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264
12.7 Kepler’s Laws of Planetary Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
Chapter 12 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
Chapter 13: Partial Derivatives
279
13.1 Functions of Two or More Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
13.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
13.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
13.4 Differentiability, Differentials, and Local Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
13.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
13.6 Directional Derivatives and Gradients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290
13.7 Tangent Planes and Normal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
13.8 Maxima and Minima of Functions of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
13.9 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
Chapter 13 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
Chapter 14: Multiple Integrals
307
14.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308
14.2 Double Integrals over Nonrectangular Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310
14.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311
14.4 Parametric Surfaces; Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
14.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
14.6 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
14.7 Change of Variables in Multiple Integrals;
Jacobians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317
14.8 Centroid, Center of Gravity, Theorem of
Pappus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
Chapter 14 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321
x
Chapter 15: Topics in Vector Calculus
333
15.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334
15.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
15.3 Independence of Path; Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338
15.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
15.5 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341
15.6 Applications of Surface Integrals; Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
15.7 The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
15.8 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344
Chapter 15 Sample Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346
Chapter 0: Functions
Summary: This chapter sets out to describe mathematical functions. This idea
is central to the rest of the book as virtually all concepts will be framed in the
context of a function. The important concepts that are covered here are the domain and range of functions, creating new functions by combining functions, and
inverse functions. Along the way, families of functions such as polynomials, rational functions, and trigonometric functions are introduced.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Identify a function and its domain and range (§0.1).
2. Combine functions arithmetically or through composition to create new
functions (§0.2).
3. Modify functions through translation, stretching either the inputs or the outputs or using an appropriate reflection (§0.2).
4. Know the properties of power functions, polynomials, rational functions,
and trigonometric functions (§0.3).
5. Find the inverse of a function if it is possible to do so (§0.4).
0.1 Functions
PURPOSE: To describe what functions are.
In this section, the main idea of a function is developed. Here the main discussion
is about the domain and range of a function. In particular, functions are described
as having the relationship that for each input to a function there is a unique output value. A clear distinction is made between the inputs and outputs in a function
relationship. Inputs are referred to as the independent variable and outputs are the
dependent variable.
input → unique output
Typically, the domain of a function is found by realizing what inputs (or x-values)
domain and range
1
2
√
will give valid outputs (or y-values). For example,
√ since the square root x does
not give a real number value for x < 0 then y = x would only have a domain of
x ≥ 0. The range of the function is directly dependent upon the domain (hence
y is the dependent variable). Whatever values are obtained using the valid inputs
(the domain) results in the appropriate range.
Things to watch out for when inspecting domains are:
1. Division by zero,
2. Functions that only accept certain inputs (i.e.,
√
x or ln x),
3. Any restrictions given explicitly.
vertical line test
The vertical line test is used to determine if a relationship between x- and yvalues is actually a function. The convention is that the inputs are usually given
along the horizontal axis (x-axis) and that the outputs are given along the vertical
axis (y-axis). So the vertical line test will indicate when any given input will give
you more than one output which would√result in a nonfunction.
An example is
√
x2 + y2 = 1. When x = 1/2 then y = 3/2 or y = − 3/2. So this fails the
vertical line test and is not a function.
Checklist of Key Ideas:
function
input x and output y
independent variable
dependent variable
graph
zero/root of a graph (also x-intercept)
vertical line test
absolute value or magnitude
piecewise functions
domain and range
3
0.2 New Functions from Old
PURPOSE: To use combinations of “simple” functions to create
and graph more complex functions.
The primary focus in this section is upon the combination of functions to create
new functions. The arithmetic combination of functions (such as f (x) + g(x) or
f (x)g(x)) is discussed as well as the composition of functions (such as f ◦ g(x) =
f (g(x))). Translations and stretching of functions are considered where a constant value c is used in some way to modify either the input, x, of a function or
the output, f (x), of a function. Pay special attention to how the constant value
of c is used. In cases where c is used to modify the input of the function such
as f (cx) or f (x + c) you will see the function change in its horizontal behavior.
This is because the inputs are given along the horizontal axis. Likewise, when c
is applied to the output value such as c f (x) or f (x) + c then this will affect the
vertical behavior of the function.
The behavior of some functions is also described in terms of symmetry. The
main types of symmetry to be aware of are even functions which are symmetric
across the y-axis and odd functions which are symmetric across the origin. Even
functions have the property that f (x) = f (−x) for all x-values while odd functions
have the property that f (−x) = − f (x) for all x−values.
The function y = x2 is an example of simple even function and y = x3 is an example
of a simple odd function. These are also good functions to experiment with when
learning about translations and stretching. For example, using either y = f (x) = x2
or y = f (x) = x3 try plotting the following functions:
1. f (x + 2) or f (x − 2)
2. f (2x) or f (x/2)
3. f (x) + 2 or f (x) − 2
4. 2 f (x) or f (x)/2
Composition is also easy to learn if you use some simple functions and remember
the difference between inputs and outputs. The idea of composition is to use the
output of one function as the input for another. Try it with f (x) = x − 2 and
g(x) = 1/(x + 1). What are f ◦ g(x) and g ◦ f (x)? How do these compositions
change the domains and ranges of the functions?
Checklist of Key Ideas:
composition
“inside” and “outside” functions
vertical and horizontal shifts or translations
vertical and horizontal stretching or compressing
symmetry
even and odd functions
composition: f (g(x))
translation: f (x) +C and f (x +C)
stretching: c f (x) and f (cx)
symmetry, even, odd
4
0.3 Families of Functions
PURPOSE: To describe some common groups of functions that
are similar to one another.
Many functions have similar definitions and so they will have similar behaviors.
A group of functions with similar definitions may be called a family of functions
(i.e., they are similar to each other and so they are in the same family). Types of
functions that are described here are power functions (with both positive and negative exponents), polynomials, rational functions and some basic trigonometric
functions.
power functions: y = xn
Power functions y = xn where n is an integer, will either be even functions (if n is
even) or odd functions (if n is odd). Even power functions will all go through the
points (−1, 1) and (1, 1) (showing symmetry across the y-axis) while odd power
functions will all go through the points (−1, −1) and (1, 1).
If n > 0 then power functions will approach +∞ as x grows large positively. The
behavior as x grows large negatively can be determined by their symmetry.
IDEA: As x gets large, polynomials and rational functions also behave like
power functions.
rational function =
continuity, see §1.5
period, amplitude
numerator
denominator
Power functions, polynomials and rational functions are very closely related. This
is because polynomials are made up of power functions with positive integer powers and rational functions are a ratio of polynomial functions. The dominant term
of a polynomial (the one with the highest power of x) will control the behavior
of the polynomial as x becomes very large negatively or positively. Likewise,
the behavior of a rational function is controlled by its dominant terms in both the
numerator and the denominator.
While polynomials are unbroken curves (or continuous as described in Section
1.5), rational functions can have either holes or vertical asymptotes at x-values
which cause division by zero.
The trigonometric functions y = A sin (Bx −C) and y = A cos (Bx −C) are periodic functions meaning that they repeat themselves after a certain period of input
values. This period is equal to 2π /|B|. They also oscillate about the x-axis with
an amplitude of |A|. These graphs are also shifted horizontally to the left or right
by C/B units depending upon whether C/B < 0 or C/B > 0 respectively.
Checklist of Key Ideas:
family of functions
parameters
power functions
proportionality constant
5
polynomials
coefficients
degree
linear, quadratic, cubic, quartic and quintic
rational functions
vertical and horizontal asymptotes
algebraic functions
amplitude, period and frequency
0.4 Inverse Functions
PURPOSE: To describe what an inverse function is and how it
relates to a function.
Functions can often have a related function, f −1 , called an inverse function. Inverse here is meant to be a function that inverts its outputs back to the original
inputs. For example, f ( f −1 (x)) = f −1 ( f (x)) = x. This is called the cancellation
equation. In other words, if y = f (x) goes through the point (a, b) then y = f −1 (x)
will go through the point (b, a). Another way of describing this property is to say
that f and f −1 are reflections of each other across the line y = x. This kind of
reflection does not always result in a function. If the resulting curve is a function
then it is called the inverse function and denoted by f −1 .
This section discusses when to expect that a function can have an inverse function
and how to find its inverse. One particular strategy that is employed with trigonometric functions is to first restrict the domain of the function so that the resulting
reflection across the line y = x will be a function.
Checklist of Key Ideas:
reflection across y = x
If y = f (x) goes through (a, b)
then y = f −1 (x) goes through (b, a).
Finding y = f −1 (x)
1. solve for x in y = f (x)
→ x = g(y)
2. swap x and y → y = g(x)
3. then g(x) = f −1 (x)
inverse function
cancellation equations
domain and range of inverse functions
finding the inverse of a function
one-to-one or invertible
horizontal line test
increasing and decreasing functions
restricting the domain
6
Chapter 0 Sample Tests
Section 0.1
1. Answer true or false. If f (x) = 4x2 − 1, then f (0) = 0.
1
2. Answer true or false. If f (x) = , then f (0) = 0.
x
1
. The natural domain of the function is
3. f (x) = 2
x −1
(a) all real numbers
(b) all real numbers except 1
(c) all real numbers except −1 and 1
(d) all real numbers except −1, 0 and 1
4. Use a graphing utility to determine the natural domain of
1
.
h(x) =
|x| − 1
(a) all real numbers
(b) all real numbers except 1
(c) all real numbers except −1 and 1
(d) all real numbers except −1, 0 and 1
5. Use a graphing
utility to determine the natural domain of
p
g(x) = 9 − x2 .
(a) {x : −3 ≤ x ≤ 3}
(b) 0.707
(c) 1.414
(d) 3
10. A box is made from a piece of sheet metal by cutting a square
whose sides measure x from each corner. If the sheet of metal
initially measured 20 cm by 10 cm, the volume of the box is
given by
(a) V (x) = (20 − 2x)(10)x cm3
(b) V (x) = (20 − 2x)(10 − x)x cm3
(c) V (x) = (20 − 2x)(10 − 2x)x cm3
(d) V (x) = (20 − 2x)(10 − 2x)(2x) cm3
11. Answer true or false. At a given location on a certain day
the temperature
T in ◦ F changed according to the equation
tπ + 4 where t represents time in hours startT (t) = t sin
12
ing at midnight. The temperature at 6 P.M. was approximately 58◦ F.
12. The speed of a truck in miles/hour for the first 10 seconds afx2
ter leaving a red light is given by f (x) = . Find the speed
2
of the truck 6 seconds after leaving the red light.
(a) 18 miles/hour
(b) {x : −9 ≤ x ≤ 3}
(b) 36 miles/hour
(d) {x : x ≥ −3}
(d) 6 miles/hour
(c) {x : x ≥ −9}
6. Answer true or false. f (x) = x
− |x + 1| can be represented in
2x + 1,
if x ≤ −1
.
the piecewise form by f (x) =
−1,
if x > −1
7. Find the x-coordinate of any hole(s) in the graph of
(x2 − 9)(x + 4)
f (x) =
.
(x + 3)(x + 4)
(c) 3 miles/hour
13. Find f (2) if
4
f (x) =
, ifx < 2 and 3x if x ≥ 2 .
x
(a) 2
(b) 4
(a) 3
(c) 6
(b) −3 and −4
(d) it cannot be determined
(c) 3 and 4
(d) −12
x2 − 4
and g(x) = x − 2 are
x+2
identical except f (x) has a hole at x = −2.
8. Answer true or false. f (x) =
9. Use a graphing device to plot the function, then find the
indicated value from observing the graph. A flying object
attains a height h(t) over time according
π to the equation
π
h(t) = 3 sint − cost. If t = , then h
is approximately
4
4
(a) 0
14. Determine all x−values where there are holes in the graph of
x2 − 4
f (x) =
.
(x + 2)2 (x − 2)
(a) −2, 2
(b) −2
(c) 2
(d) none
15. Assume the hourly temperature in ◦ F starting at midnight is
given by T (x) = 36 − (x − 15)2 + x. Find the temperature at
3 P.M.
7
(a) 36◦ F
(b)
5. Answer true or false. f (x) = x2 and g(x) = x − 2. Then
( f − g)(x) = x2 − x − 2.
51◦ F
6. Answer true or false. f (x) = x2 − 9 and g(x) = x3 + 8. f /g
has the same domain as g/ f .
√
7. f (x) = x2 + 1 and g(x) = x2 − 2. f ◦ g(x) =
√
(a) x4 − 1
√
(b) x2 − 1
√
(c) x4 − 4x2 + 5
√
(d) x2 + 2x − 1
(c) 39◦ F
(d) 27◦ F
Section 0.2
y
y
4
4
f (x) =
√
x
-3
1.
9
-2
x
-3
9
x
-2
The graph on the left is the graph of f (x) =
on the right is the graph of
√
x. The graph
(a) y = f (x + 4)
(b) y = f (x − 4)
(c) y = f (x) − 4
(d) y = f (x) + 4
2. The graph of y = 1 + (x + 2)2 is obtained from the graph of
y = x2 by
(a) translating horizontally 2 units to the right, then translating vertically 1 unit up
(b) translating horizontally 2 units to the left, then translating vertically 1 unit up
(c) translating horizontally 2 units to the right, then translating vertically 1 unit down
(d) translating horizontally 2 units to the left, then translating vertically 1 unit down
√
√
3. The √
graph of y = x and y = −x are related. The graph of
y = −x is obtained by
√
(a) reflecting the graph of y = x about the x-axis
√
(b) reflecting the graph of y = x about the y-axis
√
(c) reflecting the graph of y = x about the origin
(d) The equations are not both defined.
4. The graphs of y = x3 and y = 4 − 2(x − 1)3 are related. Of reflection, stretching, vertical translation, and horizontal translation, which should be done first?
8. h(x) = |x + 3|3 can be written as the composition
h(x) = f ◦ g(x) if
(a) f (x) = x + 3 and g(x) = |x3 |
(b) f (x) = (x + 3)3 and g(x) = |x3 |
√
(c) f (x) = 3 x + 3 and g(x) = |x3 |
(d) f (x) = x3 and g(x) = |x + 3|
√
9. f (x) = 3 x. Find f (3x).
√
(a) 3 3 x
√
(b) 3 3x
√
3
x
(c)
3
r
x
(d) 3
3
10. f (x) = x2 + 1. Find f ( f (x)).
(a) x4 + 2
(b) 2x2 + 2
(c) x4 + 2x2 + 2
(d) x4 + 2x2
11. f (x) = |x + 2| is
(a) an even function only
(b) an odd function only
(c) both an even and an odd function
(d) neither an even nor an odd function
12. f (x) = 0 is
(a) reflection
(a) an even function only
(b) stretching
(b) an odd function only
(c) vertical translation
(c) both an even and an odd function
(d) horizontal translation
(d) neither an even nor an odd function
8
y
(d) They cross the x-axis at the point (5, 0).
5
2. What points do all graphs of the form y = xn , n is odd, have
in common?
(a) (0, 0) only
(b) (0, 0) and (1, 1)
(c) (−1, −1), (0, 0) and (1, 1)
(d) none
-5
5
y
x
3
-5
13.
5
x
-5
The function graphed above is
(a) an even function only
(b) an odd function only
(c) both an even and an odd function
(d) neither an even nor an odd function
14. Answer true or false. f (x) = |x| + cos x is an even function.
15. f (x) = 3|x| + 2 cos x is symmetric about
(a) the x-axis
(b) the y-axis
(c) the origin
(d) nothing
16. f (x) = 5x3 − 2x is symmetric about
(a) the x-axis
(b) the y-axis
(c) the origin
(d) nothing
3.
-3
The equation whose graph is given is
√
(a) y = x
√
(b) y = 3 x
1
(c) y = 2
x
1
(d) y = 3
x
4. Answer true or false. The graph of y = −3(x − 4)3 can be
obtained by making transformations to the graph y = x3 .
5. Answer true or false. The graph of y = x2 + 6x + 9 can be
obtained by transforming the graph of y = x2 to the left three
units.
6. Answer
There is no difference in the graphs of
p true or false.
√
y = |x| and y = | x|.
x+3
.
7. Determine the vertical asymptote(s) of y = 2
x + 2x − 8
(a) x = −4, x = 2
(b) x = 4
(c) x = −2, x − 4
(d) x = 8
Section 0.3
1. What do all members of the family of lines of the form
y = 5x + b have in common?
(a) Their slope is 5.
(b) Their slope is −5.
(c) They go through the origin.
8. Find the vertical asymptote(s) of y =
(a) x = 0
(b) x = −1, x = 1
(c) x = 3
1
(d) x =
3
x6
.
3x6 − 3
9
9. For which of the given angles, if any, is all of the trigonometric functions negative?
π
3
2π
(b)
3
4π
(c)
3
(d) No such angle exists
(a)
10. Use the trigonometric function of a calculating utility set to
π
.
the radian mode to evaluate sin
7
(a) 0.0078
(b) 0.4339
(c) 0.1424
(b) y2 has the greatest phase shift.
(c) y1 and y2 have the same phase shift.
(d) Neither y1 nor y2 have a phase shift.
π
14. Answer true or false. The period of y = cos 5x −
is 10π .
3
k
15. Answer true or false. A force acting on an object, F = 2 ,
x
that is inversely proportional to the square of the distance
from the object to the source of the force is found to be 25 N
when x = 1 m. The force will be 100 N if x becomes 2 m.
√
1. Answer true or false. The functions f (x) = 3 x + 3 and
g(x) = x3 − 3 are inverses of each other.
√
2. Answer true or false. The functions f (x) = 6 x and g(x) = x6
are inverses of each other.
3. Answer true or false. The trigonometric function y = sec x is
a one-to-one function.
x
(c) x9 + 3
1
(d) √
9
x+3
7. For the function defined by
−x2 ,
x<0
f (x) =
x2 ,
x≥0
which of the following is the inverse function f −1 (x) (if it
exists)?
√
− −x,
x<0
√
(a) f −1 (x) =
x,
x≥0
−1/x,
x<0
(b) f −1 (x) =
1/x,
x≥0
p
(c) f −1 (x) = |x|
(d) The inverse function does not exist.
8. Answer true or false. If f has a domain of 0 ≤ x ≤ 10, then
f −1 has a domain of 0 ≤ x ≤ 10.
9. The graphs of f and f −1 are reflections of each other about
the
(a) x-axis
Section 0.4
(a)
√
9
x + 3.
(b) (x + 3)9
(a) y1 has the greatest phase shift.
√
7
1
3x − 4
x+4
(b)
3
x
(c) + 4
3
1
(d)
−4
3x
(a)
(a) x9 − 3
π
11. Answer true or false. The amplitude of 5 cos 3x −
is 3.
3
π
π
12. Answer true or false. The phase shift of 2 sin x −
is .
3
3
π
and
13. Use a graphing utility to graph y1 = sin x −
3
π y2 = sin 2 x −
.
3
4. Find
5. Find f −1 (x) if f (x) = 3x − 4.
6. Find f −1 (x) if f (x) =
(d) 0.1433
f −1 (x)
1
x7
√
(c) − 7 x
1
(d) − 7
x
(b)
if
f (x) = x7 .
(b) y-axis
(c) line y = x
(d) origin
10. Answer true or false. A 200 foot fence is used as a perimeter
about a rectangular plot. The formula for the length of the
fence is the inverse of the formula for the width of the fence.
11. Find the domain of f −1 (x) if f (x) = (x + 3)2 for x ≥ −3.
(a) x ≥ −3
(b) x ≥ 3
10
(c) x ≥ 0
(b) x = 1
(c) x = 1, 4
(d) x ≤ 0
√
12. Find the domain of f −1 (x) if f (x) = − x − 5.
(a) x ≤ 0
(b) x ≥ 0
(c) x ≤ 5
(d) x = −4
7. The cumulative number of electrons passing through an
experiment over time, given in seconds, is given by
n(t) = t 2 + 4t + 6. How many electrons pass through the experiment in the first 5 seconds?
(a) 13
(d) x ≥ 5
13. Let f (x) = x2 + 6x + 9. Find the smallest value of k such that
f (x) is a one-to-one function on the interval [k, ∞).
(b) 39
(c) 15
(d) 51
(a) 0
8. What is the entire domain of f (x) =
(b) −3
(c) 3
(a) all real numbers
(d) −9
(b) 0 ≤ x ≤ 7
14. Answer true or false. The function f (x) = −x is its own inverse.
Chapter 0 Test
1. Answer true or false. For the equation y = x2 + 10x + 21, the
values of x that cause y to be zero are 3 and 7.
2. Answer true or false. The graph of y = x2 − 4x + 9 has a
minimum value.
3. A company has a profit/loss given by P(x) = 0.1x2 − 2x −
10, 000, where x is the time in years, good for the first 20
years. After how many years will the graph of the profit/loss
equation first begin to rise?
(c) −7 ≤ x ≤ 7
(d) 0 ≤ x ≤ 49
9. Answer true or false. The graph of f (x) = |x + 8| touches the
x-axis.
x2 − 8
10. Answer true or false. The graph of y =
would prox−9
duce a false line segment on a graphing utility on the domain
of −10 ≤ x ≤ 10.
11. The graph of y = (x −2)3 is obtained from the graph of y = x3
by
(a) translating vertically 2 units upward
(b) translating vertically 2 units downward
(c) translating horizontally 2 units to the left
(d) translating horizontally 2 units to the right
(b) 10 years
(a) x3 + 2
(c) 15 years
(b) (x3 + 2)2
(d) 0 years
(c) x3 + x2 + 2
1
.
|x − 2|
(a) all real numbers
(b) all real numbers except 2
(c) all real numbers except −2
(d) all real numbers except −2 and 2
2x
, then f (1) = 1.
x2 + 1
x−4
.
6. Find the hole(s) in the graph of f (x) = 2
x − 5x + 4
5. Answer true or false. If f (x) =
(a) x = 4
49 − x2 ?
12. If f (x) = x3 + 2 and g(x) = x2 , then g ◦ f (x) =
(a) 5 years
4. Find the natural domain of g(x) =
√
(d) x6 + 2
13. Answer true or false. f (x) = |x| + sin x is an odd function.
14. A particle initially at (1, 3) moves along a line of slope m = 4
to a new position at (x, y). Find y if x = 4.
(a) 12
(b) 15
(c) 16
(d) 19
15. The slope-intercept form of a line having a slope of 5 and a
y-intercept of 3 is
(a) x = 5y + 3
11
(a) y = x2
(b) x = 5y − 3
(c) y = 5x + 3
(d) y = 5x − 3
(b) y = (−x)2
16. A spring is initially 3 m long. When 2 kg are suspended from
the spring it stretches 4 cm. How long will the spring be if
10 kg are suspended from it?
(a)
(b)
(c)
(d)
3.20 m
3.02 m
3.40 m
3.04 m
(c) y = −x2
(d) y = x−2
18. Answer true or false. The only asymptote of y =
is y = −1.
19. Answer true or false. The functions f (x) =
g(x) = x3 + 3 are inverses of each other.
17. The equation whose graph is given is
y
20. If f (x) =
5
5
x
√
3
x−2
r
3 1 − 2x
(b)
x
√
3
(c) 1 + 2x
r
3 1 + 2x
(d)
x
(a)
-5
1
then find f −1 (x).
x3 + 2
21. Find the domain of f −1 (x) if f (x) =
(a) x ≥ 0
(b) x ≤ 0
(c) x ≥ 6
-5
(d) x ≥ −6
√
x − 6.
x+5
x2 + 6x + 5
√
3
x − 3 and
12
Chapter 0: Answers to Sample Tests
Section 0.1
1. false
9. c
2. false
10. c
3. c
11. false
4. c
12. a
5. a
13. c
6. true
14. c
7. b
15. b
8. true
2. b
10. c
3. b
11. d
4. d
12. c
5. false
13. b
6. false
14. true
7. c
15. b
8. d
16. c
2. c
10. b
3. b
11. false
4. true
12. true
5. true
13. c
6. false
14. false
7. a
15. false
8. b
2. false
10. true
3. false
11. c
4. a
12. a
5. b
13. b
6. a
14. true
7. a
15. false
8. false
2. true
10. true
18. false
3. b
11. d
19. true
4. b
12. b
20. b
5. true
13. false
21. a
6. a
14. b
7. d
15. c
8. c
16. a
Section 0.2
1. b
9. b
Section 0.3
1. a
9. d
Section 0.4
1. true
9. c
Chapter 0 Test
1. false
9. true
17. c
Chapter 1: Limits and Continuity
Summary: This chapter explores the behavior of functions as they approach
certain x-values. First, some graphical and numerical methods are used to try to
ascertain the behavior of a functions output values near a particular input value.
Later, the chapter introduces some algebraic techniques for finding exact answers
to this question. This idea is made more precise using ε and δ to help represent the
tolerances in the outputs and inputs respectively while identifying how a function
behaves.
Generally, if a function approaches a consistent output value as the input values
become closer to one value (such as x = a), then a limit of the function values
is said to exist (i.e., lim f (x) = L). Naturally, such limits can be discussed while
x→a
input values approach x = a from the positive side, the negative side, or both. This
directional idea is described accordingly as one-sided or two-sided limits.
Limits become an even more useful tool when they are used to describe the notion of a function being continuous. A function that is continuous can be said
to have the property of lim f (x) = f (a) which implies a limiting value, a funcx→a
tion definition, and that the two values are equal. This formalizes the idea from
the last chapter of polynomials having unbroken curves. Now polynomials can be
described as being continuous everywhere. As it turns out, the trigonometric functions sin x and cos x are continuous everywhere, which greatly aids in studying the
behavior of all six trigonometric ratios.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Describe the behavior of a function using limits (§1.1, 1.4).
2. Evaluate the limiting behavior of a function graphically, numerically or algebraically (§1.1, 1.2).
3. Discuss the relationship between vertical asymptotes and infinite limits (§1.1).
4. Discuss the relationship between horizontal asymptotes and the end behavior of a function (§1.3).
5. Describe the connection between limits and continuity (§1.5).
13
14
6. Use the continuity of a function to evaluate the limit of a function at a point
(§1.2, 1.5).
7. Determine on what intervals a function is continuous (§1.5).
8. Use the Squeezing Theorem to evaluate limits (§1.6).
9. Use the Intermediate-Value Theorem to approximate roots of a function
(§1.5).
1.1 Limits (An Intuitive Approach)
PURPOSE: To introduce the notion of a limit.
This section introduces the idea of a limit by discussing how to find the slope of a
tangent line to a curve. This is shown using the slopes of secant lines to a function
(lines through two points on the function) as the points become closer together.
The emphasis here is on what value a function approaches as the inputs become
closer and closer to a prescribed value (say x = a). The actual function value at
the prescribed value, x = a, is irrelevant and may not even be defined.
CAUTION: A function, f (x), does not have to be defined at x = a for lim f (x)
x→a
to exist.
informal definition of a limit
The informal definition of a limit then is to determine what value the output of a
function can be made “arbitrarily close” to a value, L, provided that the inputs are
“sufficiently close” to the prescribed input (x = a). This is written in the following
mathematical expression.
lim f (x) = L
x→a
one-sided limits
two-sided limits
The input values can be controlled to only approach x = a from the positive side
or the negative side giving rise the terminology of one-sided limits. Two-sided
limits occur when the inputs are allowed to approach from both sides.
IDEA: Both one-sided limits (as x → a− and x → a+ ) must exist for the twosided limit to exist.
Limits do not have to exist. If a particular value is not approached by the function
or if different values are approached from either side of the point x = a then the
limit is said to not exist. An example is the function
2x,
x<0
f (x) =
x+3 x ≥ 0
In this case, lim f (x) = 0 and lim f (x) = 3. Since different values are apx→0−
x→0+
proached at x = 0 from the left and right, the limit at x = 0 does not exist (i.e.,
the two-sided limit does not exist).
15
#
If a limit does not exist but instead appears to approach a value of lim f (x) =
x→a
0
then this is an indication that the function has a vertical asymptote at this point. If
this same behavior is seen at either x → a+ or x → a− then this is also an indication
of a vertical asymptote at x = a. An example of this is the function f (x) = 1/x at
x = 0.
vertical asymptote
CAUTION: Saying that lim f (x) = ∞ does not mean that f (a) = ∞.
x→a−
It should also be pointed out that if lim f (x) = ∞ then this does not mean that
x→a−
f (a) = ∞. Instead ∞ here is an indication of the limit not existing. Instead, the ∞
says that the function does not approach a limit but that it does have the consistent
behavior of getting arbitrarily large as x approaches a from the negative side (or
the left).
Checklist of Key Ideas:
tangent line problem
secant and tangent lines
slope of a secant line
concept of a limit
limit of function value
informal definition of limit
sampling pitfalls
one-sided vs. two-sided limits
limits that do not exist
infinite limits and vertical asymptotes
1.2 Computing Limits
PURPOSE: To compute the values of limits of algebraically.
This section goes beyond the graphical and numerical determination or approximation of limits and starts to develop some basic limit rules and laws which can
be used to obtain the values of more complicated limits. The most important laws
are summarized in Theorem 1.2.2 which essentially says that limits can be easily
added, subtracted, multiplied, divided (if the numerator doesn’t approach a limit
of zero) or raised to certain powers. “Easily” here means that some basic arithmetic is all that is required to evaluate the limits. The biggest difficulty arises
when division occurs as follows.
lim
x→a
f (x)
g(x)
limit does not exist
16
Here if lim g(x) = 0 then these “easy” rules cannot be directly applied. If this is
x→a
the situation, then there are few things that may happen especially if f and g are
polynomials.
IDEA: Finding limits of polynomials and rational functions will depend upon
whether division by zero occurs at x = a:
1. if no division by zero? → evaluate at x = a
2. division by zero:
(a) if #/0 → no limit; vertical asymptote
(b) if 0/0 → try canceling a factor
vertical asymptote
hole or removable discontinuity
piecewise functions
Essentially, rational functions can be evaluated at x = a just like polynomials to
find the limit as x → a. If division by zero occurs, however, then you may either
obtain #/0 or 0/0. In the first case, the limit does not exist and there is a vertical
asymptote at that point. In the second case, the numerator and denominator share
a factor which may be canceled. Then the new rational function should be tested
in the limit again. This process should end in either obtaining #/0 (a vertical
asymptote) or a number (which would be the limit). Obtaining a number means
that there is a hole (or removable discontinuity) in the graph at that point. Sometimes, through algebraic manipulation (such as rationalizing a radical), zeros in
the denominator can also be removed with the same results as canceling a factor.
To determine the limits of piecewise functions at a breakpoint, the one-sided
limits of the function need to be evaluated on either side of the breakpoint. If they
have the same value then that is the value of the two-sided limit. Otherwise, no
limiting value exists for the two-sided limit.
Checklist of Key Ideas:
algebraic techniques for finding limits
sums, differences, products, quotients and n-th roots of limits
limits of polynomials and rational functions
method of test points
canceling all common factors of x − a in rational functions
indeterminate form of type 0/0
one-sided and two-sided limits and piecewise functions
17
1.3 Limits at Infinity; End Behavior of a Function
PURPOSE: To find limits of functions as x → ±∞.
Up until now, limits have been restricted towards looking at x = a, which is some
finite value. Here, limits are considered where x → ∞ or x → −∞. This is called
the end behavior of the function.
end behavior of a function
Limits at infinity obey the same limit laws as limits as x → a (see previous section).
Limits at infinity can fail to exist if the function is oscillating continuously (i.e.,
trigonometric functions) or if the function increases (or decreases) without bound
(for example, y = x2 ). Any limit at infinity that does exist means that there is a
horizontal asymptote of the function.
horizontal asymptote
IDEA: If lim f (x) = L for some finite value L then the line y = L is a horizontal
x→∞
asymptote of the function y = f (x).
The limits at infinity for rational functions are controlled by their dominant terms.
Depending upon the ratio of the highest powers of x on the top and bottom of the
expression then the limit will either be ±∞ or L.
Checklist of Key Ideas:
end behavior
limits at infinity
horizontal asymptotes
limit laws at infinity
how limits at infinity can fail to exist
infinite limits at infinity
end behavior of polynomials and rational functions
dominant terms
end behavior of trigonometric functions
1.4 Limits (Discussed More Rigorously)
PURPOSE: To give a more rigorous (or thorough) definition of
limit.
The goal of this section is to explore limits more rigorously. With this in mind, the
ideas of “arbitrarily close” and “sufficiently close” from Section 1.1 are replaced
by the phrases | f (x) − L| < ε and 0 < |x − a| < δ respectively. The aim here is to
dominant terms
18
first describe how close a function should be to value L (within ε ) and then try to
show that if x is close enough to a (within δ ) that this will hold.
IDEA: The goal is to find δ so that f (x) is within ε of a some value L (ε is
some known value).
1. Pick ε : how close to L should f (x) be?
2. Find δ so that this works if |x − a| < δ .
Most problems in this section have the following format: it must be shown that
given any positive number ε , a positive number δ can be found such that
| f (x) − L| < ε if 0 < |x − a| < δ .
Then the process to be followed is to start with | f (x) − L| and try to transform this
into an inequality involving x of the form |x − a| < δ . When starting with
−ε < f (x) − L < ε
it very often occurs that an inequality involving x turns out to be
−δ1 < x − a < δ2
Then to satisfy the problem δ is chosen as the smaller of δ1 or δ2 .
Similar ideas hold for working with infinite limits.
Checklist of Key Ideas:
formal limit definition
“epsilon”, ε , and “delta”, δ
definition of limits as x → ±∞
definition of infinite limits
1.5 Continuity
PURPOSE: To establish what is meant by a “continuous function.”
Continuity is what has allowed the easy computation of limits for polynomial and
rational functions (when there is not division by zero). That is to say, the limit of
a polynomial is equal to its function value.
IDEA: Find limits of polynomials by evaluating the polynomial.
19
This idea of continuity can be summarized by the following.
lim f (x) = f (a)
x→a
This statement implies that the function is defined at x = a, the limit exists at x = a,
and that the value of the limit and the function value are equal. In other words,
the limiting value that the function approaches is the value that it is defined to
have. Visually this amounts to having an unbroken graph with no holes or vertical
asymptotes.
IDEA: The limits of continuous functions may be found by evaluating the function if the point is within the domain of the function.
Continuous functions have limits that are easy to evaluate within their domains.
Rational functions for example are continuous whenever their denominator is not
zero and then their limit is equivalent to their function values. If the denominator
was zero, this would be a hole or a vertical asymptote which would not be in their
domain.
The Intermediate-Value Theorem capitalizes on this idea. If a function is continuous on an interval then the function must attain all function values between
the function values at the endpoints. In particular, if the function crosses from
negative to positive, or from positive to negative, then it must have at least one
root in this interval. By smaller and smaller intervals where this occurs, better and
better approximations of roots can be found in this way.
Checklist of Key Ideas:
definition of continuity
three conditions of continuity
continuous on an interval (a, b)
continuous everywhere
continuity from the left or right at a point
continuous on a closed interval
continuity of f ± g, f g and f /g
continuity of polynomials and rational functions
continuity of compositions
the Intermediate-Value Theorem
approximating roots
Intermediate-Value Theorem
20
1.6 Continuity of Trigonometric Functions
PURPOSE: To discuss the continuity of trigonometric functions.
trigonometric functions are continuous on their domains
Squeeze Theorem
Simply stated, trigonometric functions are continuous in their domains. In particular, both sin (x) and cos (x) are continuous along the whole real line. The
other four trigonometric functions are each defined as ratios involving sin (x) and
cos (x). Consequently, they will be continuous throughout their domains or wherever there is no division by zero. Since the denominator will be sin (x) or cos (x)
in each case, then division by zero is determined by where these functions are
equal to zero. For example, when cos (x) = 0 there will be a point of discontinuity for both tan (x) and sec (x) since both of these functions have sin (x) in the
denominator.
The Squeezing Theorem offers a powerful tool for finding other limits such as
sin x
lim
. This is accomplished by finding two functions that have the same limit
x→0 x
at a point and which also bound the function in question from above and below.
Then the function is squeezed in between the two functions and must have the
same limit. This theorem is not always easy to apply because the two bounding
functions must be obtained before the theorem can be applied.
Checklist of Key Ideas:
continuity of sin x and cos x
continuity of general trigonometric functions
The Squeezing Theorem
21
Chapter 1 Sample Tests
x2 − 4
x2 − 4
by evaluating f (x) =
at
x−2
x→2 x − 2
x = 3, 2.5, 2.1, 2.01, 2.001, 1, 1.5, 1.9, 1.99, and 1.999.
Section 1.1
3. Approximate lim−
y
5
(a) 2
(b) −2
(c) 0
(d) 4
4. Answer true or false. If lim+ f (x) = 4 and lim− f (x) = 4,
x→0
x→0
then lim f (x) = 4.
-5
5
x→0
x
x
x
by evaluating f (x) =
at apx−2
x−2
propriate values of x.
5. Approximate lim−
x→2
(a) 1
(b) 0
(c) ∞
(d) −∞
-5
1.
The function f (x) is graphed above. lim− f (x) =
sin x
then approximate lim− f (x) by evaluating
5x
x→0
f (x) at appropriate values of x.
6. If f (x) =
x→0
(a) 1
(b) 2
3
(c)
2
(d) undefined
(a) 1
(b) 5
1
5
(d) ∞
(c)
2. Answer true or false.
y
5
x
then approximate lim f (x) by evaluating f (x)
sin x
x→0
at appropriate values of x.
7. If f (x) =
(a) 1
(b) −1
(c) 0
-5
5
x
(d) ∞
√
x+4−2
then approximate lim− f (x) by evalux
x→0
ating f (x) at appropriate values of x.
8. If f (x) =
1
4
(b) 0
(a)
-5
For the function graphed lim f (x) is undefined.
x→2
(c) ∞
(d) −∞
22
Section 1.2
(a) 6
(b) 4
1. Given that lim f (x) = 3 and lim g(x) = 5, then find
x→a
2
(c) 3
x→a
lim [ f (x) − g(x)] if it exists.
x→a
(a) −4
(b) 4
(d) It does not exist.
2
x + 4,
9. Let g(x) =
x3 ,
(c) 22
(a) 5
(d) It does not exist.
(b) 1
x≤1
. Then lim g(x) =.
x>1
x→1
(c) 3
2. lim 4 =
x→3
(d) It does not exist.
(a) 4
10. Answer true or false. lim
(b) 3
√
x→0
1
x2 + 25 − 5
= .
x
10
(c) 7
(d) 12
Section 1.3
3. Answer true or false. lim 4x = 8.
x→2
4. lim
x→−3
x2 − 9
x+3
1. lim
10x
x→∞ x2 − 5x + 3
=.
=
(a) 0
(a) −∞
(b) 2
(c) 6
(d) It does not exist.
(c) 5
(b) −6
(d) 1
2.
4
=.
x→5 x − 5
5. lim
2x2 − x
=
x2
(a) 2
lim
x→−∞
(a) ∞
(b) ∞
(b) −∞
(c) −∞
(c) 0
(d) It does not exist.
6. lim
4x
x→1 x2 − 6x + 5
=
(a) ∞
(b) −∞
(c) 0
(d) It does not exist.
x−4
7. lim √
=
x→1 x − 4
(d) It does not exist.
s
8
5
4 32x − 6x + 2
=
3. lim
8
3
x→−∞
2x − 3x + 1
(a) ∞
(b) −∞
(c) 2
(d) It does not exist.
4. lim (x4 − 500x3 ) =
x→∞
(a) ∞
(a) ∞
(b) −∞
(b) −∞
(d) It does not exist.
(c) 1
(d) It does not exist.
x + 4,
x≤2
8. Let f (x) =
. Then lim f (x) =.
x2 ,
x>2
x→2
(c) −500
√
x2 + 9 − 3
does not exist.
x
6. Use a graphing utility to approximate the y-coordinates of any
horizontal asymptotes of y = 4x−9
x+3 .
5. Answer true or false. lim
x→∞
23
(a) y = 4
(a) 0.033
(b) y = 1
(b) 0.33
(c) None exist
(c) 3.0
(d) y = −3
(d) 0.3
7. Use a graphing utility to approximate the y-coordinates of any
horizontal asymptotes of y = cosx x .
(a) y = 0
x2 − 25
= 10 and ε = 0.001 then find a least number
x→5 x − 5
δ such that | f (x) − L| < ε when 0 < |x − a| < δ .
4. If lim
(b) only y = 1
(c) both y = −1 and y = 1
(a) 0.001
(d) only y = −1
(b) 0.000001
8. Use a graphing utility to approximate the y-coordinates of any
2
+4
horizontal asymptotes of y = xx−2
.
(a) y = 0
(b) None exist
(c) y = 1
(b) N = 1, 000
9. Answertrue or false. A graphing utility can be used to show
2 x
has a horizontal asymptote.
f (x) = 1 +
x
10. Answer trueor false.
A graphing utility can be used to show
1 x
has a horizontal asymptote.
that f (x) = 6 +
x
2+x
11. Answer true or false. lim
is equivalent to lim+
x→∞ 1 − x
x→0
lim+ sin (2π x).
(c) 0.005
(d) 0.025
12
5. If lim 3 = 0 and ε = 0.1 then find the least positive integer
x→∞ x
N such that | f (x) − L| < ε when x > N.
(a) N = 100
(d) Both y = −1 and y = 1
12. Answer true or false.
3. Answer true or false. If f (x) = x3 , L = 27, a = 3 and
ε = 0.05 then a least number
δ such that | f (x) − L| < ε if
√
0 < |x − a| < δ is δ = 3 27.05 − 3.
2
x
1
x
+1
−1
.
sin (2π x)
is equivalent to
lim
x→∞ x − 2
x→0
x
has no hor13. Answer true or false. The function f (x) = 2
x −4
izontal asymptotes.
(c) N = 4
(d) N = 5
1
6. If lim 5 = 0 and ε = 0.1 then find the greatest negative
x→−∞ x
integer N such that | f (x) − L| < ε when x < N.
(a) N = −100, 000
(b) N = −10, 000
(c) N = −1
(d) N = −2
7. Answer true or false.
1
It is reasonable to prove that lim
x→∞ x2 + 1
= 0.
8. Answer true or false.
It is reasonable to prove that lim
x→−∞
1
= 0.
x+6
9. Answer true or false.
Section 1.4
1. If lim 4x = 20 and ε = 0.1 then find a least number δ such
x→5
that | f (x) − L| < ε when 0 < |x − a| < δ .
(a) 0.1
(b) 0.25
(c) 0.5
(d) 0.025
2. If lim 3x − 4 = 2 and ε = 0.1 then find a least number δ such
x→2
that | f (x) − L| < ε when 0 < |x − a| < δ .
It is reasonable to prove that lim
x→∞
10. Answer true or false.
1
= 0.
x−5
1
= ∞.
x→5 x − 5
11. To prove that lim (x + 2) = 7, a reasonable relationship beIt is reasonable to prove that lim
x→5
tween δ and ε would be
(a) δ = 5ε
(b) δ = ε
√
(c) δ = ε
1
(d) δ =
ε
24
12. Answer true or false. To use a δ -ε approach to show that
1
lim
= ∞, a reasonable first step would be to change the
x→0+ x3
limit to lim x3 = 0.
y
5
x→∞
13. Answer true or false.
It is possible to show that lim−
x→4
1
= −∞.
x−4
-8
8
x
14. To prove that lim f (x) = 6 where
x→3
f (x) =
2x,
x + 3,
x<3
x≥3
-5
2.
On the interval [−8, 8], where is f not continuous?
a reasonable relationship between δ and ε would be
(a) 3
(b) 0, 3
(a) δ = ε /2
(c) 0
(b) δ = ε
(d) nowhere
(c) δ = ε + 3
3. Answer true or false. f (x) = x6 − 4x4 + 2x2 − 8 has no point
of discontinuity.
(d) δ = ε /2 + 3
x
= 5.
x→∞ 5
15. Answer true or false. It is possible to show that lim
4. Answer true or false. f (x) = |x + 4| has a point of discontinuity at x = −4.
5. Find the x-coordinates for all points of discontinuity for the
x+6
function f (x) = 2
.
x + 8x + 12
(a) −6, −2
Section 1.5
(b) −2
(c) 2, 6
y
(d) 2
5
6. Find the x-coordinates for all points of discontinuity for the
|x + 5|
.
function f (x) = 2
x + 5x
(a) 0
-7
7
x
(b) −5
(c) −5, 0
(d) −5, 0, 5
-5
1.
On the interval [−7, 7], where is f not continuous?
(a) −2, 2
(b) 2
(c) −2
(d) nowhere
7. Find the x-coordinates
2 for all points of discontinuity for the
x − 5,
x≤1
function f (x) =
.
−4,
x>1
(a) 1
√ √
(b) − 5, 5
√ √
(c) −4, − 5, 5
(d) None exists.
x − 2,
x≤1
8. If f (x) =
, then find the value k, if possikx3 ,
x>1
ble, that will make f (x) continuous everywhere.
25
(a) 1
(a) 0
(b) −1
(b) 1
(d) None exists.
(d) ∞
(c) −1
(c) 2
9. Answer true or false. The function f (x) =
movable discontinuity at x = 1.
x3 − 1
has a rex−1
10. Answer true or false. The function
x2 ,
x≤2
f (x) =
x − 4,
x>2
is continuous everywhere.
11. Answer true or false. If f and g are continuous at x = c, then
f + g may be discontinuous at x = c.
12. Answer true or false. The Intermediate-Value Theorem can
be used to approximate the locations of all discontinuities for
x
f (x) = 3
. [Hint: It may be applied to the denomx − 5x + 13
inator only.]
13. Answer true or false. f (x) = x5 − 6x2 + 2 = 0 has at least one
solution on the interval [−1, 0].
14. Answer true or false. f (x) = x3 + 2x + 9 = 0 has at least one
solution on the interval [0, 1].
√
15. √
Use the fact that 8 is a solution of x2 −8 = 0 to approximate
8 with an error of at most 0.005.
(a) 1.25
4. Find the limit. lim−
x→0
(a) ∞
(b) 0
(c) 1
(d) −∞
5. Find the limit. lim
x→0
sin (3x)
=
sin (8x)
(a) ∞
(b) 0
3
(c)
8
(d) 1
x→0
4
=
1 − sin x
7. Find the limit. lim
1 + cos x
=
1 − cos x
6. Find the limit. lim
(a) 1
(b) −1
(c) 4
(d) 0
x→0
(b) 2.81
(a) 0
(c) 2.83
(b) ∞
(d) 2.84
sin x
=
x3
(c) −∞
(d) 2
Section 1.6
1. Answer true or false. f (x) = cos (x2 + 5) has no point of discontinuity.
1
2. A point of discontinuity of f (x) =
is at
|−1 + 2 cos x|
π
2
π
(b)
3
π
(c)
4
π
(d)
6
(a)
4
3. Find the limit. lim cos
=
x→∞
x
tan x
=
x→0 cos x
8. Find the limit. lim
(a) 0
(b) 1
(c) ∞
(d) −∞
1
=
9. Find the limit. lim− cos
x
x→0
(a) 1
(b) −1
(c) −∞
(d) does not exist
10. Find the limit. lim+
x→0
(a) 0
−1.3x + cos x
=
x
26
(b) 1
(d) undefined
(c) −1
x2 − 25
2
−25
by evaluating f (x) = xx+5
at
x+5
x = −4, −4.5, −4.9, −4.99, −4.999, −6, −5.5, −5.1,
−5.01, and −5.001.
2. Approximate lim
x→−5
(d) ∞
( cos x − 1
,
x≤0
11. Answer true or false. If f (x) =
, then
x
cos x + k,
x>0
f is continuous if k = 0.
1 − cos x
= 0 and that
12. Answer true or false. The fact that lim
x
x→0
(1 − cos x)2
lim x = 0 guarantees that lim
= 0 by the Squeeze
x
x→0
x→0
Theorem.
13. Answer true or false. The Squeezing Theorem can be used
sin (4x)
sin (4x)
to show lim
= 1 by utilizing lim
= 1 and
6x
4x
x→0
x→0
sin (6x)
= 1.
lim
6x
x→0
14. Answer true or false. The Intermediate-Value Theorem can
be used to show that the equation y3 = sin2 x has at least one
solution on the interval [−5π /6, 5π /6].
sin x
x
15. Find the limit. lim
+
=
2x
2 sin x
x→0
(a) 1
(b) 2
1
(c)
2
(d) 0
(a) 5
(b) −5
(c) 10
(d) −10
3. Use a graphing utility to approximate the y-coordinate of the
9x + 3
.
horizontal asymptote of y = f (x) =
4x + 2
9
(a)
4
4
(b)
9
3
(c)
2
2
(d)
3
4. Answer true or false. A graphing utility can be used to show
2 2x
has a horizontal asymptote.
that f (x) = 8 +
2x
5
5. Answer true or false. lim 2 is equivalent to lim+ 5x2 .
x→∞ x
x→0
6. Find the value of lim [2 f (x) + 4g(x)] if it is given that
x→a
lim f (x) = 5 and lim g(x) = −5.
x→a
(a) 0
Chapter 1 Test
(b) −5
(c) −10
y
(d) 30
5
7. lim 4 =
x→6
(a) 6
-8
8
x
(b) −6
(c) 4
(d) does not exist
x8 − 1
=
x→1 x − 1
(a) 0
8. lim
-5
1.
The function f is graphed. lim f (x) =
x→−1
(a) 2
(b) −2
(c) 0
(b) ∞
(c) 4
(d) 8
9. lim
x→6
1
=
x−6
x→a
27
1
12
(b) 0
(c) ∞
(d) does not exist
2
x ,
x≤1
10. Let f (x) =
. lim f (x) =
x,
x > 1 x→1
(a) 0
(a)
(a)
(b)
(c)
(d)
(b) −2
(c) 2
(d) nowhere
17. Find the x-coordinate of each point of discontinuity of
x+2
.
f (x) = 2
x − 3x − 10
1
−1
0
does not exist
(a) 5
11. If lim 4x = 20 and ε = 0.01 then find a least number δ such
x→5
(c) 2, 5
that | f (x) − L| < ε when 0 < |x − a| < δ .
(a)
(b)
(c)
(d)
(d) −5, 2
0.01
0.025
0.05
0.0025
x2 − 36
= −12 and ε = 0.001 then find a least number
x→6 x + 6
δ such that | f (x) − L| < ε when 0 < |x − a| < δ .
12. If lim
(a)
(b)
(c)
(d)
(b) −2, 5
0.001
0.000001
0.006
0.03
18. Find the value k, if possible, that will make the function
kx + 3,
x≤3
f (x) =
continuous everywhere.
x2 ,
x>3
(a) k = 3
(b) k = 0
(c) k = 2
(d) No such k exists.
1
13. Answer true or false. It is possible to prove that lim 7 = 0.
x→∞ x
14. To prove lim (3x + 1) = 7, a reasonable relationship between
x→2
δ and ε would be
ε
(a) δ =
3
(b) δ = 3ε
(c) δ = ε
(d) δ = ε − 1
19. Answer true or false. The function f (x) =
able discontinuity at x = 6.
20. Answer true or false. The equation f (x) = x5 + 6 = 0 has at
least one solution on the interval [−2, −1].
21. Find lim
x→0
sin (−3x)
.
sin (2x)
(a) 0
15. Answer true or false. It is possible to prove that lim (x − 3) =
x→∞
−3.
16. The graph of the function f (x) is shown below. On the interval [−5, 5], where is f not continuous?
(b) −
3
2
3
2
(d) undefined
(c)
y
22. Find lim
5
x→0
sin x
.
tan x
(a) 0
(b) −1
(c) 1
(d) undefined
-5
5
-3
x
1
has a removx−6
23. Answer true or false. lim
x→0
sin x
= 0.
1 − cos x
28
Chapter 1: Answers to Sample Tests
Section 1.1
1. a
2. false
3. d
4. true
5. d
6. c
7. a
8. a
2. a
10. false
3. true
4. b
5. d
6. d
7. c
8. d
2. a
10. false
3. c
11. true
4. a
12. false
5. false 6. a
13. false
7. a
8. b
2. a
10. false
3. false
11. b
4. a
12. false
5. c
13. true
6. c
14. a
7. true
15. false
8. true
2. c
10. false
3. true
11. false
4. false
12. true
5. a
13. true
6. c
14. false
7. d
15. c
8. b
2. b
10. d
3. b
11. false
4. a
12. true
5. c
13. false
6. c
14. false
7. b
15. a
8. a
2. d
10. a
18. c
3. a
11. d
19. false
4. false
12. a
20. true
5. true
13. true
21. b
6. c
14. a
22. c
7. c
15. false
23. false
8. d
16. a
Section 1.2
1. b
9. d
Section 1.3
1. a
9. true
Section 1.4
1. d
9. true
Section 1.5
1. a
9. true
Section 1.6
1. true
9. d
Chapter 1 Test
1. d
9. d
17. b
Chapter 2: The Derivative
Summary: Chapter 2 builds upon the ideas of limits and continuity discussed
in the previous chapter. By using limits, the instantaneous rate at which a function
changes with respect to its inputs can be investigated. This leads to the ability
to find tangent lines to a function at a given point. These two concepts are encompassed by the derivative which is a new function that can be used to find the
slope of a curve at a given point. A progression from continuous functions to
those which have tangent lines to those which are differentiable can be seen. The
concept of the derivative also provides some valuable tools for determining the
shape of a given function and how a function behaves. These tools are a direct
result of a derivative’s ability to describe the slope of a function at a point. Many
tools for finding the derivatives of functions are also discussed in this chapter so
that derivatives of more complicated functions can be found.
In the latter part of this chapter, it is discussed whether or not the slope of a general
curve can be found if the curve does not have an explicit function representation.
The process of implicit differentiation allows dy/dx to be found even when an
expression for y = f (x) is not explicitly known. Implicit differentiation can then
be applied to find a relationship between the rate of change of variables in many
situations. Finally, the ideas of tangent lines are more thoroughly described as
linear approximations. Linear approximations then lead to the idea of differentials
which allow for the estimation of a small change in a function given a small change
in its inputs.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Understand the definition of the derivative and its implications (§2.1, 2.2)
2. Be able to read and understand different notations for the derivative (§2.2)
3. Find derivatives of simple functions like polynomials (§2.3) or trigonometric functions (§2.5)
4. Use the product and/or quotient rules to find derivatives of functions involving products or quotients of other functions (§2.4)
5. Use the chain rule (either form) to find the derivatives of functions which
involve the compositions of other functions (§2.6)
29
30
6. Find the slope of a curve that is defined implicitly (§2.7).
7. Use derivatives to relate the rates of different quantities that depend upon
the same parameter such as time (§2.8).
8. Approximate functions using linearizations or differentials (§2.9).
2.1 Tangent Lines and Rates of Change
PURPOSE: To use limits and continuity to investigate the slope
of a function at a point.
slope of a secant line
instantaneous rate of change
This section attempts to bridge the gap between limits and continuity and the slope
of a graph. The process is one where the slope of a secant line can be seen as
the average rate of change of a function. This becomes the instantaneous rate of
change as the two points defining the secant line approach each other in the limit.
The slope of the tangent line at a point x0 is defined as a limit:
mtan = lim
h→0
f (x0 + h) − f (x0 )
h
This slope can then be used to construct the tangent line to a function f (x) at the
point (x0 , f (x0 )):
y − f (x0 ) = mtan (x − x0 )
This is nothing more than the point-slope form of a line. In this case, the slope
mtan is determined by a limiting process. The big idea in this section (and the next)
is the equivalence of the following three things:
1. The slope of a tangent line to y = f (x) at the point x = x0 .
2. The instantaneous rate of change of y = f (x) at the point x = x0 .
3. The derivative of the function y = f (x) at the point x = x0 .
rectilinear motion
velocity
In terms of rectilinear motion and velocity, this section shows the specific example where instantaneous velocity is the instantaneous rate of change of the position
function of a particle. Be careful to make the distinction between displacement
and distance (displacement implies a direction) and also the difference between
velocity and speed (velocity implies a direction).
Checklist of Key Ideas:
finding the equation of a tangent line to a function
speed or velocity of a particle
31
rectilinear motion
graphing position versus time
position function of a particle
displacement and distance
average velocity and average speed
instantaneous velocity and speed
rate of change of y with respect to x
average and instantaneous rate of change of y with respect to x
units of rate of change of y with respect to x
2.2 The Derivative Function
PURPOSE: To define the derivative, discuss its meaning and introduce some notation for the derivative.
In this section, the limit that defines the slope of the tangent line at a point is
referred to more generally as the derivative at a point. The derivative function is
defined as:
f ′ (x) = lim
h→0
f (x + h) − f (x)
h
The concepts of the previous section are extended in this section by the derivative.
The instantaneous rate of change of a function at any point x (where this limit
exists) is the derivative. Similarly, the derivative function can be used to forecast
the behavior of f (x) since it describes the slope of f (x) (or equivalently the slope
of the tangent line) at any given point.
The biggest ideas to come out of this section are the relationships between limits,
continuity, tangent lines, and derivatives.
IDEA: Continuity, differentiability, and tangent lines are related in the following
ways:
1. If a function f (x) is differentiable at a point, then it must have a tangent
line at that point.
2. If a function f (x) has a tangent line at a point, then it must be continuous
at that point.
3. If a function f (x) is continuous at a point, then its limit must exist at that
point.
Notice that the relationships described above do not go the other direction. Here
are some counterexamples:
definition of derivative
32
1. If a function has a limit at a point, it does not have to be continuous at a
point. Consider y = (x2 − 4)/(x + 2) which has a hole at x = −2 and so is
not continuous there.
2. If a function is continuous, it does not have to have a tangent line. Consider
y = |x| at x = 0 where it has a corner.
3. If a function has a tangent line, it does not have to be differentiable. Consider y = x1/3 which has a vertical tangent line at x = 0.
the derivative f ′ (x)
and
the graph of f (x)
Another key concept in this section is the relationship that exists between the
graphs of the function f (x) and its derivative f ′ (x). As you continue on in this
chapter, always try to be aware of how the graph of f ′ (x) relates to the slope of
the function f (x). Thinking of f ′ (x) as the slope of the function f (x) will establish
a natural connection between the behavior of f (x) and f ′ (x).
Checklist of Key Ideas:
derivative of f with respect to x
f ′ (x) and the slope of f (x)
graphing f and f ′ together
f ′ (t) and instantaneous velocity v(t)
when a function does not have a derivative at a point
differentiable at a point and on an interval
relationship between derivatives, continuity, and tangent lines
derivatives of piecewise functions and one-sided derivatives
different notations for derivatives
evaluating the derivative at a point
derivatives and increment notation
2.3 Introduction to Techniques of Differentiation
PURPOSE: To introduce basic rules for calculating derivatives.
Evaluating derivatives by using the definition (i.e. the limit of the difference quotient from the previous two sections) can be tedious and may require some algebraic manipulation. Several basic derivatives are introduced in this section.
First, the derivatives of constant functions and powers of x are developed. Next,
since the derivative is really a limit, it follows that derivatives can be added and
subtracted and also multiplied by constants with ease. Unfortunately, the same
33
straightforward results do not hold when functions are multiplied and divided (see
the next section).
The most important result to come out of this section is the ability to calculate the
derivative of a polynomial or the derivative of a sum of functions. Essentially, a
sum of functions can be differentiated by taking the derivative of each function
separately and then adding the derivatives back together.
Lastly, the idea of higher order derivatives is introduced. If we can take a derivative of a function, then why not take the derivative of the derivative? In this way,
so-called second order and higher derivatives are introduced. Order here simply
refers to how many times a derivative is taken beginning with the original function. Pay close attention also to the different notations that are introduced for the
derivative. There are (perhaps confusingly) several ways to say the same thing
with derivatives. To this end, it may be useful to write these things down on a note
card and have them close at hand as you work on problems in the text.
derivative of xn
derivative of c · f (x)
derivative of f (x) + g(x)
write new derivative rules on notecards
Checklist of Key Ideas:
derivatives of constants, the power rule
derivatives of constant multiples, sums, and differences
derivatives of polynomials
finding horizontal tangent lines
higher derivatives and order of a derivative
notation of higher order derivatives
2.4 The Product and Quotient Rules
PURPOSE: To develop rules for differentiating products or quotients of functions.
As indicated in the previous section, the derivatives of products of functions and
quotients of functions are not straightforward. This is primarily because the limit
definition involves a difference of functions. When these limits are multiplied
together, it is not generally true that these differences will behave nicely.
Some simple examples can show that we may not simply multiply derivatives
together. For example, consider f (x) = x and g(x) = x2 (see p.142-143 in the
text). Since y = f (x)g(x) = x3 we know that its derivative is y′ = 3x2 . Notice that
this is not equivalent to f ′ (x)g′ (x). These same two functions can be used to show
that if y = g(x)/ f (x) that the derivative is not simply g′ (x)/ f ′ (x). Can you see
why?
Write the product and quotient rules
on note cards for easy reference.
They are used quite frequently.
34
This section introduces two rules that are frequently used throughout calculus: the
product rule and the quotient rule for calculating derivatives of y = f (x)g(x) and
y = f (x)/g(x) respectively.
In words we have:
Product Rule:
Product Rule: If y = f (x)g(x) then
dy
= f (x)g′ (x) + f ′ (x)g(x)
dx
(first) × (derivative of second)
+ (second) × (derivative of first)
Quotient Rule:
(bottom) × (derivative of top)
− (top) × (derivative of bottom)
bottom2
Quotient Rule: If y = f (x)/g(x) and g(x) 6= 0 then
dy g(x) f ′ (x) − f (x)g′ (x)
=
dx
(g(x))2
Checklist of Key Ideas:
derivative of a product of functions, i.e., y = f (x)g(x)
derivative of a quotient of functions, i.e., y = f (x)/g(x)
2.5 Derivatives of Trigonometric Functions
PURPOSE: To give rules for calculating derivatives of the six basic trigonometric functions.
This section develops the derivatives of the six basic trigonometric functions starting with the derivatives of sin x and cos x. All trigonometric derivatives can be
traced back to these two.
d
[sin x] = cos x
dx
Write these six derivative rules on a
notecard.
Memorize them in pairs
• sin (x) and cos (x)
• tan (x) and sec (x)
• cot (x) and csc (x)
d
[cos x] = − sin x
dx
From these two derivatives, the other four trigonometric ratios can be obtained using the quotient rule. Knowing these six derivatives is important. There are many
patterns which can aide in the memorization of these derivatives. For example, the
graphs of y = sin x and y = cos x can help remind you what the derivatives of each
of these functions are. Since cos x begins at x = 0 by decreasing, this indicates
why the negative should be in front of its derivative. On the other hand, since sin x
starts by increasing with a positive slope, this indicates that positive cos x should
be its derivative.
The derivatives of tan x and sec x are easy to remember since they are related to
each other.
d
[sec x] = sec x tan x
dx
d
[tan x] = sec2 x
dx
35
The last two derivatives cot x and csc x are “co”-functions which follow the same
patterns as tan x and sec x with their derivatives except that they have negative
signs.
All of the “co”-functions have a negative sign in their derivatives.
d
[csc x] = − csc x cot x
dx
d
[cot x] = − csc2 x
dx
Knowing the graphs of these other four trigonometric ratios can also be helpful in
remembering these definitions. In particular, matching where a derivative crosses
the horizontal axis with places where a function has a horizontal asymptote is a
useful strategy in identifying any function with its derivative. For example, since
tan (0) = 0, this indicates that sec x will have a horizontal tangent at x = 0 because
the derivative of sec x is sec x tan x.
Checklist of Key Ideas:
derivatives of sin x and cos x
derivatives of the other trigonometric functions (using the quotient rule)
2.6 The Chain Rule
PURPOSE: To give a rule for calculating the derivative of a function that is a composition of one or more functions.
In the previous sections, the derivatives of basic algebraic combinations of functions such as sums, products, and quotients have been described. Additionally,
the derivatives of common functions such as polynomials, rational functions and
trigonometric functions have been established. In this section, the goal is to establish the derivative of a composition of functions.
The idea is very simple and yet easy to confuse. If a function can be written as a
composition of one function inside of another, then the derivative of the composition can be found by a product of the derivatives. A composition of functions
may be easiest to think of as an “inside function” and an “outside function.” The
derivative of the one function inside of the other is essentially the derivative of the
inside function times the derivative of the outside function.
In other words if y = f (u) and u = g(x) (or y = f (g(x))) then
dy
d f du
= f ′ (u)g′ (x) =
·
dx
du dx
This now allows the derivatives of simple looking functions such as sin (2x) and
(x + 13)100 to be taken in a quick and simple fashion. Both of the functions mentioned here could be differentiated using previous methods, but the process may



 
derivative
derivative

×
of
of
inside
outside
f ′ (u)
×
g′ (x)
36
have been tedious or may not have been straightforward. For example, the product rule can be used if it is seen that sin (2x) = 2 sin x cos x. On the other hand, the
derivative of (x + 13)100 can be found by first expanding the polynomial (“foil”ing
99 times!) and then taking the derivatives of each term. The chain rule makes this
process much simpler.
Checklist of Key Ideas:
derivatives of compositions
two versions of the chain rule
inside and outside functions
derivatives of the inside and outside functions
generalized derivative formulas
du
d
[ f (u)] = f ′ (u)
dx
dx
multiple applications of the chain rule
2.7 Implicit Differentiation
PURPOSE: To give rules for expressing the derivative of a function that is written implicitly.
implicit means that y = f (x) is not
known
This section approaches the problem of being able to find the slope of a curve even
if an explicit formula for the curve is unknown. In other words, most curves are
described by a formula of the form y = f (x). However, it may not always be practical or possible to find such an expression for y (i.e., consider y + sin (xy) = 3).
In such cases, the chain rule can be used to find the slope dy/dx.
1. take derivatives using the
chain rule
The basic idea in this section is to assume that y is a function of x, that is y = f (x),
and then to differentiate the relationship describing the curve with respect to x by
using the chain rule. In the end, the derivative dy/dx may be expressed in terms
of x and y.
2. solve for dy/dx
This brings up two questions.
Implicit Differentiation
1. Can we find dy/dx?
differentiate with the chain rule
differentiate
implicitly
=
use the
chain rule
2. How is dy/dx evaluated?
The first question can be answered using the chain rule. In fact, the phrase “differentiate implicity” is just another way of saying “use the chain rule.”
For example, if we have a curve described by x3 + xy + y3 = 1, then it is assumed
that y = f (x) is a function that corresponds to at least a portion of the graph of this
curve. Then the relationship can be rewritten as follows.
x3 + x f (x) + [ f (x)]3 = 1
37
Now to differentiate, we take the derivative of each term with respect to x on both
sides of the equation. This leads us to the following.
3x2 + f (x) + x f ′ (x) + 3[ f (x)]2 f ′ (x) = 0
d
Here we have used the product rule to find
[x f (x)] and the chain rule to find
dx
d
[ f (x)]3 as shown. But since y = f (x) and dy/dx = f ′ (x) we usually write
dx
this statement as
3x2 + y + x
dy
dy
+ 3y2
=0
dx
dx
Now we only have to solve a linear equation to find dy/dx. This is a key feature
of implicit differentiation.
IDEA: After differentiating an implicit equation, dy/dx is found by solving a
linear equation.
Once the equation of the curve has been differentiated, there will only remain a
linear equation for dy/dx to be solved. To solve for dy/dx we put all of the
dy/dx terms on one side, factor and divide. In the example above, this looks like
the following:
dy
dy
+ 3y2
= −3x2 − y
dx
dx
⇓
dy
x + 3y2 = −3x2 − y
dx
⇓
dy −3x2 − y
=
dx
x + 3y2
x
solve a linear equation for dy/dx
Put dy/dx terms on one side.
Factor out dy/dx.
Solve for dy/dx.
The second key concern is how to evaluate the derivative dy/dx. The answer is
that you now need both an x and a y value since dy/dx will often depend upon
both. However, arbitrary values of x and y cannot be used. Instead, the values of
x and y must satisfy the initial relationship that implicitly defined y in terms of
x. In the example above, this means that the ordered pair (x, y) must satisfy the
relationship x3 + xy + y3 = 3. One way to find a valid point is to pick an x value
(i.e., x = 0 or some other value) and then solve the resulting equation for y.
3
x + xy + y3 = 3 x=0
⇓
3
y =3
⇓
y = 3(1/3) ≈ 1.4422
Once dy/dx is known at a particular point (x0 , y0 ) then it is a simple matter to find
a tangent line using, for example, the point-slope form of a line.
evaluating dy/dx
both a value for y and x are needed
to find dy/dx
38
Because of the way that dy/dx is solved for, the derivative is often written as some
sort of fraction. This information may be used to find where the curve has either
horizontal or vertical tangents. To find horizontal tangents, the top of the fraction
is set equal to zero. To find vertical tangents, the bottom of the fraction is set equal
to zero. It is not quite as simple as this as in some cases, the top and bottom of the
fraction may be zero simultaneously. This may be indicative of a vertical tangent
and it may not.
In the above example, to find horizontal tangents to the curve x3 + xy + y3 = 3, we
set the top of the derivative equal to zero.
−3x2 − y = 0
Then the curve may have a horizontal tangent where y = −3x2 . This does not give
very specific information about x and y but it is enough. This relationship may be
substituted back into the original curve to find the coordinates of a point with a
horizontal tangent.
Checklist of Key Ideas:
y defined explicitly and implicitly
finding the slope of an implicitly defined curve
implicit differentiation
expressing dy/dx in terms of x and y
finding tangent lines to implicitly defined curves
2.8 Related Rates
PURPOSE: To use derivatives to relate the rates of change of
multiple functions.
In this section, derivatives are applied to problems that may involve more than one
function of an independent variable. Most often, the independent variable that is
considered is time, t. The main idea here is that a real situation may exist such that
two or more quantities can be related to each other using algebraic equations (such
as area, a, and radius, r of a circle). Then both of these quantities are assumed to
be functions of time, t. The most confusing aspect of this is that t does not show
up explicitly in the equations and definitions for the quantities are not known (i.e.,
a(t) =? and r(t) =? in the previous circle mentioned).
As it turns out, all that is needed is the information that these quantities are functions of t. Then the equation that relates them can be differentiated. The result is
an equation that relates their rates or relates their derivatives.
It may be helpful to write π r(t)2 as
π [r(t)]2
For example, if a = π r2 in a circle then we assume that area and radius are func-
39
tions of time, t.
a(t) = π r(t)2
Then we differentiate this equation.
a′ (t) = π 2r(t)r′ (t)
On the right hand side of the last equation, the derivative of r(t)2 was obtained
using the chain rule: let y = u2 and y = r(t) and then apply the chain rule.
dy dy du
=
·
= 2u · r′ (t) = 2r(t)r′ (t)
dt
du dt
Formulas are still not known for a(t), r(t) or their derivatives. But if numerical
information is known about them at a particular time, then frequently we can
discover the value of how one of the quantities is changing with respect to time.
Checklist of Key Ideas:
relating the rates of change of different variables
strategy for solving related rates problem
identifying equations to relate the variables
identifying variables as functions of time, t
2.9 Local Linear Approximations; Differentials
PURPOSE: To use a tangent line to approximate a function.
This section builds upon the notion at the beginning of this chapter that we can use
the derivative of a function to find the equation of a tangent line to the function.
The tangent line can then be used to approximate the values of the function nearby
the point of tangency. This is the idea of a local linear approximation. The tangent
line of a function is this local linear approximation. In some cases, it may be
simpler to obtain the tangent line of a function and then use this in place of actual
function values. For example, if the value of a function and its derivative are
known at a point but a definition of the actual function is not known, then a tangent
line approximation can be used.
IDEA: The derivative f ′ (x) is the slope of the tangent line.
Differentials take the approximation idea a step further. The notion of a differential for the function y = f (x) given by
dy = f ′ (x)dx
40
is just another way of using a local linear approximation. The derivative f ′ (x) is
also the slope of the tangent line. But near a point of tangency, the slope should
be equal to a small change in y divided by a small change in x. Then dy = f ′ (x)dx
says that a small change in y equals the slope of the tangent line times a small
change in x. This would be exact for a line but this is only an approximation for a
general function.
In this section ∆x and ∆y are used to denote actual (exact) changes in x-values and
function values. Usually, dx is assumed to be equal to a desired ∆x. But then dy
becomes an approximation to ∆y. Sometimes ∆y is called the error in y and so dy
∆y
would be an approximation to this error. On the other hand,
gives a percentage
y
dy
would be approximation of this
error or relative error in the value of y. Then
y
relative error.
Checklist of Key Ideas:
differentiable functions are locally linear
general equation of tangent line at x = x0
using tangent line as a linear approximation
differentials and differential form of derivatives
actual change, ∆y, and estimated change, dy
error, relative error and percentage error
general differential formulas
41
Chapter 2 Sample Tests
Section 2.1
1. Find the average rate of change of y with respect to x over the
1
interval [1, 5] if y = f (x) = 2 .
x
(a) 0.24
(b) −0.24
(c) −
3
x02
(d) −
2
x02
7. Find the slope of the tangent to the graph of f (x) = x3 − 2 at
a general point x0 .
(a) 2x0 − 2
(c) 0.48
(b) 3x02 − 2
(d) −0.48
2. Find the average rate of change of y with respect to x over the
interval [1, 4] if y = f (x) = x3 .
(c) 2x0
(d) 3x02
8. Answer true or false. The slope of the tangent line to the
graph of f (x) = x2 − 4 at x0 = 3 is 2.
(a) 21
(b) −21
(c) 31.5
(d) −31.5
3. Find the instantaneous rate of change of y = x4 with respect
to x at x0 = 3.
(a) 108
(b) 27
9. Answer true or false. Use a graphing utility to graph y = x3
on [0, 5]. If this graph represents a position versus time curve
for a particle, then the instantaneous velocity of the particle
is increasing over the graphed domain.
10. Use a graphing utility to graph y = x2 − 6x + 4 on [0, 10]. If
this graph represents a position versus time curve for a particle, then the instantaneous velocity of the particle is zero at
what time? Assume time is in seconds.
(c) 54
(a) 0 s
(d) 13.5
(b) 3 s
1
4. Find the instantaneous rate of change of y = with respect
x
to x at x0 = −2.
(a) 0.25
(b) 0.5
(c) −0.25
(d) −0.5
5. Find the instantaneous rate of change of
to x at a general point x0 .
(c)
6x02
4x02
2x02
(d)
x02
(a)
(b)
11. A rock is dropped from a height of 64 feet and falls toward
the earth in a straight line. What is the instantaneous velocity
downward when it hits the ground?
(a) 64 ft/s
(b) 32 ft/s
y = 2x3
with respect
3
2
6. Find the instantaneous rate of change of y = with respect
x
to x at a general point x0 .
3
(a) −
x0
3x
(b) − 0
2
(c) 5 s
(d) 10 s
(c) 2 ft/s
(d) 16 ft/s
12. Answer true or false. The magnitude of the instantaneous velocity is never less than the magnitude of the average velocity.
13. Answer true or false. If a rock is thrown straight upward from
the ground, when it returns to earth its average velocity will
be zero.
14. Answer true or false. If an object is thrown straight upward
with a positive instantaneous velocity, its instantaneous velocity when it returns to the ground will be negative.
15. An object moves in a straight line so that after t s its distance
in mm from its original position is given by s = t 3 + t. Its
instantaneous velocity at t = 5 s is
(a) 128 mm/s
42
3. If y =
(b) 28 mm/s
(c) 27 mm/s
(d) 76 mm/s
16. A particle moving along an x-axis with a constant velocity is
at the point x = 3 when t = 1 and x = 8 when t = 2. The
velocity of the particle if x is in meters and t is in seconds is
(a) 5 m/s
1
(b)
m/s
2
8
(c)
m/s
3
3
m/s
(d)
8
17. A family travels north along a highway at 60 mi/hr, then turns
back and travels south at 65 mi/hr until returning to the starting point. Their average velocity is
√
x then dy/dx =
√
x
(a)
2x
√
x
(b)
x
√
x
(c)
2
√
2 x
(d)
x
4. Answer true or false.
y
y
5
-5
5
5
x
-5
5
(a) 62.5 mi/hr
(b) 125 mi/hr
-5
(c) 5 mi/hr
-5
The derivative of the function graphed on the left is graphed
on the right.
(d) 0 mi/hr
18. Find the equation of the tangent line to y = f (x) = 4x2 at
x = 2.
(a) y = 8x
5. Answer true or false. Use a graphing utility to help in obtaining the graph of y = f (x) = |x|. The derivative f ′ (x) is not
defined at x = 0.
6. Find f ′ (t) if f (t) = 4t 3 − 2.
(b) y = 16x − 16
(a) 12t 3 − 2
(c) y = 8x − 16
(b) 3t 2
(d) y = 16x + 16
(c) 4t 2
(d) 12t 2
Section 2.2
1. Find the equation of the tangent line to y = f (x) =
x = 7.
x 11
(a) y = +
6
6
x
(b) y = − 4
6
x 11
(c) y = +
3
6
(d) y = 3x − 4
2. If y = x2 then dy/dx =
(a) 2
(b) 2x2
(c) 2x
x
(d)
2
(4 + h)3 − 64
represents the derivative of f (x) = x3 at
h
x = a. Find a.
7. lim
√
x + 2 at
h→0
(a) 4
(b) −4
(c) 64
(d) −64
√
3
√
27 + h − 3
8. lim
represents the derivative of f (x) = 3 x at
h
h→0
x = a. Find a.
(a) 27
(b) 3
(c) −3
(d) −27
9. Find an equation for the tangent line to the curve
y = x3 − x2 + x + 1 at (1, 2).
x
43
(a) 3e2
(a) y = −2
(b) y = 2
(b) 2e2
(c) y = x − 2
(c) 2e3
(d) 0
(d) y = 2x
π 10. Let f (x) = cos x. Estimate f ′
4
ity.
(a)
by using a graphing util-
1
4
√
2
(b) −
2
1
(c)
2
π
(d)
4
11. An air source constantly increases the air supply rate of a balloon. The volume V in cubic feet is given by V (t) = t 2 for
0 ≤ t ≤ 5, where t is time in seconds. How fast is the volume
of the balloon increasing at t = 3 s?
(a) 6
ft3 /s
(b) 9 ft3 /s
(c) 18
ft3 /s
(d) 3 ft3 /s
12. Answer true or false. Using a graphing utility it can be shown
√
that f (x) = 5 x is differentiable everywhere on [−10, 10].
13. Answer true or false.
graphing utility can be used to deter A
x3 ,
x≤1
mine that f (x) =
is differentiable at x = 1.
x2 ,
x>1
3. Find dy/dx if y = 3(x2 − 2x + 4).
(a) 6x − 6
(b) 3x2 − 6x + 4
(c) 9x3 − 12x2 + 4x
(d) 2x − 2
4. Answer true or false.
√
√
x
If f (x) = x + x2 , then f ′ (x) =
+ 2x.
2
2
2
2
5. Find d y/dx if y = 9x + 4x + 6.
(a) 19
(b) 18x + 4
(c) 18
(d) 9
6. Find y′′′ if y = x−6 + x3 .
(a) −120x−4 + 6
(b) −336x−9 + 6
(c) 120x−4 + 6
(d) 336x−9 + 6
7. Answer true or false. The equation y = y′′′ + 12y′ − 48x3 is
satisfied by the function y = x3 + 6x2 + 2.
8. Use a graphing utility to locate all horizontal tangent lines to
the curve y = x3 + 6x2 + 2.
(a) x = 0, 4
14. Answer true or false.
graphing utility can be used to deter A
x3 ,
x≤0
is differentiable at x = 0.
mine that f (x) =
x5 ,
x>0
(b) x = −4, 0
15. Answer true or false. If a function y = f (x) has a vertical
tangent at x = a then y = f (x) is differentiable at x = a.
(d) x = −4
16. Answer true or false. The function y = |x| is continuously
differentiable.
Section 2.3
1. Find dy/dx if y = 7x9 .
(a) 16x9
(b) 63x9
(c) 16x8
(d) 63x8
2. Find dy/dx if y = e3 .
(c) x = 0
9. Find the x-coordinate of the point on the graph of y = x4
where the tangent line is parallel to the secant line that cuts
the curve at x = 0 and at x = 1.
1
(a) √
3
4
(b) 4
(c) 1
1
(d)
2
10. The position of a moving particle is given by
s(t) = 3t 2 + 2t + 1 where t is time in seconds. The velocity in m/s is given by ds/dt. Find the velocity at t = 2 s.
(a) 6 m/s
(b) 8 m/s
44
Section 2.4
(c) 16 m/s
(d) 14 m/s
11. If f (x) =
x2 ,
x≥2
then what value of b makes
4x + b,
x<2
f (x) differentiable everywhere?
(a) b = −4
1
1
1. Answer true or false. If y =
then y′ (x) = .
4x
+
2
4
2x
dy 2. If y =
then
=
x−4
dx x=1
(a) −
(b) b = 0
(b)
(c) b = 4
(d) b can be any real number
12. If y′′ = 12 x3 and y = cx5 then the value of c is
(a) 20
4
. Then y′ (0) =
x+5
(a) 0
4
(b)
5
x3 + x + 1
13. If y =
then dy/dx =
x
(a) 3x2 + 1
4
25
4
(d) −
5
(c) −
4. Suppose that g(x) = x2 f (x). Find g′ (2) given that f (2) = 4
and f ′ (2) = 8.
(b) 2x
(c) 2x − x−2
(a) 48
2x3 − x
(d)
x2
14. If
8
9
8
9
3. Let y =
(b) 24
1
(c)
40
(d) any real number
y = (2x + 3)2
8
3
(c) −
(d)
8
3
(b) −16
(c) 16
then
y′
=
(a) 4x + 6
(b) 4x2 + 12x + 9
(c) 8x + 6
(d) 8x + 12
15. Let h(x) = 3 f (x) − 2g(x). If f ′ (3) = 3 and h(x) has a horizontal tangent line at x = 3, then g′ (3) =
(d) 32
5. Answer true or false. If f , g, and h are differentiable functions,
h 6= 0 anywhere on its domain, then
and
f g ′ h f ′ g′ − f gh′
.
=
h
h2
6. Let h(x) = f (x)g(x). If f ′ (x) = g(x) and g′ (x) = − f (x) then
which of the following expressions could represent h′′ (x)?
(a) 0
(b) f (x)g(x)
(a) 0
(c) 2 f (x)g(x)
(b) −3
(d) −4 f (x)g(x)
(c) 3
(d) 9/2
√
16. If y′ = 4x−1/2 and y = c x then c =
7. If y = (x − 2) f (x) has a tangent line with a slope of 3 at x = 3
then find the value of f ′ (3) when f (3) = 7.
(a) f ′ (3) = −4
(a) 2
(b) f ′ (3) = −7/2
(b) 4
(d) f ′ (3) can be any real number
(c) 8
(d) any real number
(c) f ′ (3) = 3/2
8. A line of the form y = x + b could be tangent to the graph of
1
y=
at which of the following points?
1−x
45
(a) (−2, 1/3)
(c) 2 sin x − cos x
(d) 3 sin x
(b) (0, 1)
4. Find d 2 y/dx2 if y = x sin x.
(c) (2, 1)
(d) No such points exist.
x
9. If y = 2
then y has a horizontal tangent line at
x +1
(a) x = 0
(b) x = −1 or x = 1
(c) only x = 1
(d) y has no horizontal tangents.
10. If y =
1
then y has a horizontal tangent line at
x2 + 1
(b) 0
(c) − cos x
(d) cos x
5. Answer true or false. If y = sec x then d 2 y/dx2 = sec x.
6. Find the equation of the line tangent to the graph of y = sin x
at the point where x = 0.
(a) y = −1
(b) y = −x
(a) x = 0
(b) only x = −1 or x = 1
(c) x = −1, 0, or 1
(d) y has no horizontal tangents.
11. If y = f (x)g(x) then
(a) −x sin x + 2 cos x
d2y
=
dx2
(a) f ′′ (x)g′′ (x)
(b) f ′ (x)g′′ (x) + f ′′ (x)g′ (x)
(c) f (x)g′′ (x) + f ′′ (x)g(x)
(d) f ′′ (x)g(x) + 2 f ′ (x)g′ (x) + f (x)g′′ (x)
12. Answer true or false. If y = f (x)g(x) and f ′ (2) = 0 then y
has a horizontal tangent line at x = 2.
(c) y = x
(d) y = 1
7. Find the x-coordinates of all points in the interval [−2π , 2π ]
at which the graph of f (x) = csc x has a horizontal tangent
line.
(a) −3π /2, −π /2, π /2, 3π /2
(b) −π , π
(c) −π , 0, π
(d) −3π /2, 0, 3π /2
8. Find d 93 (cos x)/dx93 .
(a) cos x
(b) − cos x
(c) sin x
Section 2.5
1. Find f ′ (x) if f (x) = x4 sin x.
(a) 4x3 cos x
(b) −4x3 cos x
(c) 4x3 sin x + x4 cos x
(d) 4x3 sin x − x4 cos x
2. Find f ′ (x) if f (x) = sin x tan x.
(a) cos x
(b) (sin x)(1 + sec2 x)
(c) (sin x)(1 − sec2 x)
(d) − cos x
3. Find f ′ (x) if f (x) = sin2 x + cos x.
(a) sin x
(b) 2 sin x cos x − sin x
(d) − sin x
9. Find all x-values on (0, 2π ) where f (x) is not differentiable
if f (x) = tan x cos x.
(a) π /2, 3π /2
(b) π
(c) π /2, π , 3π /2
(d) None
10. Answer true or false. If x is given in radians, the derivative
π
sec2 x.
formula for y = tan x in degrees is y′ =
180
11. A rock at an elevation angle of θ is falling in a straight line.
If at a given instant it has an angle of elevation of θ = π /4
and has a horizontal distance s from an observer, find the rate
at which the rock is falling with respect to θ .
π (a) sec
4
2 π
(b) s sec
4
46
√
(b) 3x2 x2 − 2
sec (π /4)
s
sπ (d) sec
4
(c)
12. Answer true or false. If f (x) = tan x cos x − cot x sin x, then
f ′ (x) = cos x − sin x.
1
then f ′ (x) = − csc2 x.
13. Answer true or false. If f (x) =
tan x
sin x
is differ14. Answer true or false. The function f (x) =
1 − cos x
entiable everywhere.
15. If y = x3 sin x then find d 2 y/dx2 .
3x4
(c) √
x2 − 2
p
x4
(d) √
+ 3x2 x2 − 2
2
x −2
6. y = sin (cos x). Find dy/dx.
(a) − cos (cos x) sin x
(b) cos (cos x) sin x
(c) sin (cos x) cos x
(d) − sin (cos x) cos x
7. y = x4 tan (6x). Find dy/dx.
(a) 6x sin x
(b)
6x sin x + 6x2 cos x + x3 sin x
(a) 24x3 sec2 (6x)
(c)
6x sin x + 6x2 cos x − x3 sin x
(b) 24x3 sec (6x) tan (6x)
(d)
6x sin x − x3 sin x
(c) 6x4 sec2 (6x) + 4x3 tan (6x)
Section 2.6
1. Let f (x) =
√
x2 − 4x + 3. f ′ (x) =
x−2
(a) √
x2 − 4x + 3
x−4
(b) √
2
x − 4x + 3
1
(c) √
2
2 x − 4x + 3
(d) 2x − 4
10. Answer true or false. If y = sin x3 − cos x2 then
y′′ = 6x cos x3 − 9x4 sin x3 + 2 sin x2 + 4x2 cos x2 .
(a) 20(x5 − 2)19
(b) 100x4 (x5 − 2)19
(c) 100x5 (x − 2)20
(d) 100x4
3. Let f (x) = sin (3x). f ′ (x) =
(a) cos (3x)
(b) 3 cos (3x)
(c) − cos (3x)
(d) −3 cos (3x)
(a)
x3
√
2 x2 − 2
p
+ 3x x2 − 2
2
2 sin x + sin x cos2 x
cos2 x
sin x + 2 sin x cos2 x − sin3 x
(b)
cos2 x
2 cos x
(c)
sin x
2 cos x
(d) −
sin x
(a)
9. Answer true or false. If y = cos (5x3 ) then
d 2 y/dx2 = − cos (5x3 ).
2. Let f (x) = (x5 − 2)20 . f ′ (x) =
4. Answer true or false. If f (x) =
cos x
.
f ′ (x) = p
sin2 x + 3
√
5. Let f (x) = x3 x2 − 2. f ′ (x) =
(d) 6x4 sec (6x) tan (6x) + 4x3 tan (6x)
1 + sin2 x
. Find dy/dx.
8. y =
cos x
p
sin2 x + 3 then
11. Find an equation for the tangent line to the graph of y = x tan x
at x = π /4.
π π
π
(a) y − =
+1 x−
4
2
4
π
(b) y − 1 = x −
4
π
(c) y − = x − 1
4
√ π
2
π
x−
(d) y − =
4
2
4
12. y = sin3 (π − 2θ ). Find dy/d θ .
(a) −6 sin2 (π − 2θ ) cos (π − 2θ )
(b) −6 sin2 (π − 2θ )
(c) −6 sin3 (π − 2θ )
(d) 3 sin2 (π − 2θ ) cos (π − 2θ )
47
13. Use a graphing utility to obtain the graph of f (x) = (x +
√
2)3 x. Determine the slope of the tangent line to the graph
at x = 1.
(a) 1
(b)
d 2 y 16x2 + 4y
=
dx2
6y3
(c)
(c) 27
(d) 40.5
14. Find the value of the constant A so that y = A cos 3t satisfies
the equation d 2 y/dt 2 + 3y = cos 3t.
(b)
d 2 y 16x2 + 4y
=
dx2
36y3
d 2 y 16x2 − 4y
=
dx2
36y3
2
d y −6y2 − 4x2
=
(d)
dx2
9y3
(b) 54
(a) −
(a)
1
12
7. Find the slope of the tangent line to x2 + y2 = 5 at the point
(1, 2).
(a)
1
6
1
2
(b) −
1
2
5
2
−5
(d)
2
1
6
9
(d) −
2
(c)
(c) −
√
15. Answer true or false. Suppose that f ′ (x) =
√ x + 2 and
3
′
2
g(x) = x . If F(x) = f (g(x)) then F (x) = 3x x3 + 2.
8. Find the slope of the tangent line to xy2 = 4 at the point (1, 2).
(a) 4
(b) −4
(c) 1
(d) −1
Section 2.7
9. Find dy/dx if xy4 = x3 .
√
5
4
dy
= √
1. Answer true or false. If y = 4x + 2 then
.
dx 5 4x + 2
1
dy
.
=
2. Answer true or false. If y3 = x then
dx 3y2
√
3. Find dy/dx if 3 y − cos x = 2.
(a) dy/dx = −3y2/3 sin x
(b) dy/dx = 3y2/3 sin x
(c) dy/dx = −6y2/3 sin x
3x2 − y4
4xy3
3x2
(c) − 4
4y
3x2 + y4
(d)
4xy3
(b)
1
sin (xy)
1
(b)
sin (xy)
1 + y sin (xy)
(c)
x sin (xy)
1 + y sin (xy)
(d) −
x sin (xy)
(a) −
4. Find dy/dx if x2 + y2 = 25.
25x
y
x
(b)
y
x
(c) −
y
25x
(d) −
y
(a)
6. Suppose that 2x2 + 3y2 = 9. Find d 2 y/dx2 .
3x
4y4
10. Find dy/dx if x = cos (xy).
(d) dy/dx = 6y2/3 sin x
5. Answer true or false. If y2 + 2xy = 5x then
(a)
11. Answer true or false. If cos x = sin y then dy/dx = tan x.
5
dy
=
.
dx 2y + 2x
dy
y sec2 (xy)
=−
.
dx
x
2
3
13. xy = x − 2x has a tangent line parallel to the x-axis at
12. Answer true or false. If tan (xy) = 4 then
(a) (−1, 1)
48
(b) (0, 0)
(c) (2, 2)
(d) (1, 2)
14. x2 + y2 = 16 has tangent lines parallel to the y-axis at
(a) (0, −16) and (0, 16)
(b) (0, −4) and (0, 4)
(c) (−4, 0) and (4, 0)
(d) (−16, 0) and (16, 0)
√
15. If y = 3 x, find the formula for ∆y.
√
√
(a) ∆y = 3 x + ∆x − 3 x
√
(b) ∆y = 3 x + ∆x
1
1
(c) ∆y = p
− √
3
3 3 (x + ∆x)2 3 x2
1
(d) ∆y = p
3 3 (x + ∆x)2
dy
x1/3 + x
then
=
16. If y = √
x
dx
1
1
(a) − x−7/6 + x−1/2
6
2
(b) −6x−7/6 + 2x−1/2
√ 1 −2/3
(c) 2 x
x
+1
3
√ 1 −2/3
+ 1 + 12 x−1/2 x1/3 + 1
x 3x
(d)
x
17. Find the equation of the tangent line to the function
y = (x2 + 4)1/3 at x = 2.
1
(x − 2) + 2
12
1
(b) y = (x − 2) + 8
12
1
(c) y = (x − 2) + 2
3
1
(d) y = (x − 2) + 8
3
(a) y =
Section 2.8
4
dV
in
1. The volume of a sphere is given by V = π r3 . Find
3
dt
dr
terms of .
dt
(a)
dV
dr
= 4π r2
dt
dt
dV
4
dr
= π r3
dt
3
dt
4 2 dr
dV
= πr
(c)
dt
3
dt
dV
2 dr
(d)
= 3r
dt
dt
2. A cylinder of length 2 m and radius 1 m is expanding such
that dl/dt = 0.01 m/sec and dr/dt = 0.02 m/sec. Find
dV /dt.
(b)
(a) 0.0013 m/s3
(b) 0.2827 m/s3
(c) 0.015 m/s3
(d) 0.03 m/s3
3. A 10-ft ladder rests against a wall. If it were to slip so that
when the bottom of the ladder is 6 feet from the wall it will
be moving at 0.02 ft/s, how fast would the ladder be moving
down the wall?
(a) 0.02 ft/s
(b) 0.0025 ft/s
(c) 0.015 ft/s
(d) 0.12 ft/s
4. A plane is approaching an observer with a horizontal speed of
200 ft/s and is currently 10, 000 ft from being directly overhead at an altitude of 20, 000 ft. Find the rate at which the
angle of elevation, θ , is changing with respect to time, d θ /dt.
(a) 0.020 rad/s
(b) 0.010 rad/s
(c) 0.08 rad/s
(d) 0.009 rad/s
5. Answer true or false. Suppose that z = y5 x3 .
Then dz/dt = (dy/dt)5 + (dx/dt)3 .
p
6. Suppose that z = x2 + y2 . Find dz/dt.
dx
dy
+ 2y
dt
dt
dx dy
+
(b)
dt
dt
s
2 2
dy
dx
(c)
+
dt
dt
(a) 2x
x dx + y dy
dt
(d) pdt
x2 + y2
7. The power in watts for a certain circuit is given by P = I 2 R.
How fast is the power changing if the resistance, R, of the
circuit is 1, 000 Ω, the current, I is 2 A, and the current is
decreasing with respect to time at a rate of 0.03 A/s.
(a) −0.09 w/s
(b) −60 w/s
49
(c) −120 w/s
(d) −1.8 w/s
8. Gravitational force is inversely proportional to the distance
5
between two objects squared. If F = 2 N at a distance
d
d = 3 m, how fast is the force diminishing if the objects are
moving away from each other at 2 m/s?
(a) 2 N/s
(b) −0.74 N/s
(c) −6.7 N/s
(d) −1.1 N/s
9. A point√P is moving along a curve whose equation is given
by y = x4 + 9. When P = (2, 5), y is increasing at a rate of
2 units/s. How fast is x changing?
(a) 2 units/s
5
units/s
(b)
8
(c) 64 units/s
5
(d)
units/s
16
10. Water is running out of an inverted conical tank at a rate of 3
ft3 /s. How fast is the height of the water in the tank changing
if the height is currently 5 ft and the radius is 5 ft?
(a) −0.038 ft/s
(b) 0.377 ft/s
(c) −1.131 ft/s
(d) 9.425 ft/s
11.
12.
13.
14.
dz
dy
dx
Answer true or false. If z = x ln y then
.
=
dt
dt
dt
dθ
dy
dx
Answer true or false. If sin θ = 3xy then
= 3x + 3y .
dt
dt
dt
dx
dy dV
2
=
− 6x .
Answer true or false. If V = 3x y then
dt
dt
dt
dV
dx
1
Answer true or false. If V = 10x3 then
=
.
dt
30x2 dt
Section 2.9
1. If
y = x4 ,
(a)
find the formula for ∆y.
∆y = (x + ∆x)4
(b) ∆y = 4x3 ∆x
(c) ∆y = 4(x − ∆x)3
(a) ∆y = cos (x + ∆x) − cos x
(b) ∆y = cos (x + ∆x)
(c) ∆y = − sin x∆x
(d) ∆y = ∆x + cos x
3. Answer true or false. The formula for dy is obtained from the
formula for ∆y by replacing ∆x with dx.
4. Let y = x5 . Find the formula for dy.
(a) dy = (x + dx)5
(b) dy = (x + dx)5 − x5
(c) dy = x5 + (dx)5
(d) dy = 5x4 dx
5. Let y = tan x. Find the formula for dy.
(a) dy = (sec2 x)dx
(b) dy = (sec x tan x)dx
(c) dy = tan (x + dx) − tan x
(d) dy = tan (x + dx)
6. Let y = x3 sin x. Find the formula for dy.
(a) dy = (3x2 sin x + x3 cos x)dx
(b) dy = 3(x + dx)3 sin (x + dx)
(c) dy = (3x2 cos x)dx
(d) dy = (x + dx)3 sin (x + dx) − x3 sin x
1
. Find dy at x = 1 if dx = 0.01.
x2
(a) 0.02
7. Let y =
(b) −0.02
(c) −0.001
(d) 0.001
8. Let y = x5 . Find dy at x = 1 if dx = −0.01.
(a) 0.00000005
(b) −0.00000005
(c) 0.05
(d) −0.05
√
9. Let y = x. Find ∆y at x = 3 if ∆x = 1.
(a) −0.268
(b) 0.268
(c) 0.289
(d) 0.250
10. Use dy to approximate
∆y = (x + ∆x)4 − x4
(a) 2.01
2. If y = cos x, find the formula for ∆y.
(b) 1.99
(d)
√
3.96 starting at x = 4.
50
(d) 11.5
(c) 4.01
(d) 3.99
11. Answer true or false. A circular spill is spreading so that
when its radius r is 2 m, dr = 0.05 m. The corresponding
change in the area covered by the spill, A, is 0.63 m2 (to the
nearest hundredth).
12. A small suspended droplet of radius 10 microns is evaporating. If dr = −0.001 micron, find the change in the volume,
dV , to the nearest thousandth of a cubic micron.
4. Find the equation of the tangent line to y = f (x) = 3x3 at
x = 3.
(a) y = 27x − 60
(b) y = 27x + 60
(c) y = 81x + 162
(d) y = 81x − 162
5. If y = x8 then dy/dx =
(a) −420.237 cubic microns
(a) 8x7
(b) −1.257 cubic microns
(b) 8x8
(c) −0.419 cubic microns
(c) 7x7
(d) −4.189 cubic microns
13. Answer true or false. A cube is expanding as temperature
increases. If the length of the cube is changing at a rate of
∆x = 2 mm when x is 1 m, the volume is experiencing a corresponding change of 0.006 mm3 (to the nearest thousandth).
14. Answer true or false. The radius of the base of a cylinder is
2 mm with a possible error of ±0.01 mm. The height of the
cylinder is exactly 4 m. Using differentials to estimate the
maximum error of the volume, it is found to be 502.65 mm3
(to the nearest hundredth).
(d) 7x8
6. Answer true or false.
y
5
y
2
-5
5
-3
-5
Chapter 2 Test
1. Find the average rate of change of y with respect to x over the
interval [1, 2] if y = f (x) = 3x2 .
(a) 9
(b) −9
(c) 11
(d) −11
2. Find the instantaneous rate of change of y = 2x4 with respect
to x at x0 = 3.
(a) 216
(b) 54
(c) 108
(d) 27
3. An object moves in a straight line so that after t s its distance
from its original position is given by s = t 3 − 2t. Its instantaneous velocity at t = 4 s is
x
3
-2
The derivative of the function graphed on the left is graphed
on the right.
(6 + 2h)2 − 36
represents the derivative of f (x) = (2x)2
h
h→0
at x =
7. lim
(a) 6
(b) 3
(c) 0
(d) −3
8. Let f (x) = tan x. Estimate f ′ (5π /4) by using a graphing utility.
(a) 1
(b) −1
(c) 0
(d) 2
9. Find dy/dx if y = π 5 .
(a) 5π 4
(a) 56
(b) π 5
(b) 46
(c) 0
(c) 14
(d) 4π 5
x
51
10. Answer true or false. If f (x) =
1
√ + 3x2 .
2 x3
5x
dy =
11. If y =
then
x−2
dx x=1
√
x3 + x3 , then f ′ (x) =
(a) 10
(b) −10
(c) 8
(d) −8
12. Let g(x) =
f ′ (4) = 6.
√
x f (x). Find g′ (4) given that f (4) = 4 and
18. If f (x) = sin (6x) then f ′ (x) =
(a) 6 cos (6x)
(b) −6 cos (6x)
(c) cos (6x)
(d) − cos (6x)
sin (3x),
x≥0
19. Let y =
. Find the value of a so
3x2 + ax + 3,
x<0
that y is differentiable everywhere.
(a) a = −6
(b) a = −3
(a) 24
(c) a = 3
(b) 1.5
(d) No such value of a exists.
(c) 13
20. If y = ( f (x))3 then
(d) 14
13. Find f ′ (x) if f (x) = x3 tan x.
(a) 3x2 sec2 x
(b) 3x2 sec x tan x
(c) 3x2 tan x + x3 sec2 x
(d) 3x2 tan x + x3 sec x tan x
14. Find d 2 y/dx2 if y = −2(sin x)(cos x).
(a) 8(cos x)(sin x)
(b) −8(cos x)(sin x)
(c) 2(cos x)(sin x)
(d) −2(cos x)(sin x)
15. Find the x-coordinates of all the points over the interval (0, π )
where the graph of f (x) = cot x has a horizontal tangent line.
dy
=
dx
(a) 3[ f (x)]2 f ′ (x)
(b) 3[ f (x)]2
(c) [ f ′ (x)]3
(d) 3[ f ′ (x)]2
√
21. If y = 9 x then find the formula for ∆y.
√
√
(a) ∆y = 9 x − ∆x + 9 x
√
√
(b) ∆y = 9 x + ∆x − 9 x
∆x
(c) ∆y = √
9
9 x8
√
√
(d) ∆y = 9 x + ∆x + 9 x
22. Answer true or false. If x2 + y2 = 4 then y′′ = −4y−3 .
23. If y = x2 tan x, find the formula for dy.
(a) π /4, π /2, 3π /4
(a) dy = (2x tan x)dx
(b) π /4, 3π /4
(b) dy = (x2 sec2 x)dx
(c) π /2
(c) dy = (2x tan x + x2 sec2 x)dx
(d) None exist.
(d) dy = (2x tan x − x2 sec2 x)dx
d 105
[sin x] = cos x.
dx105
√
17. Answer true or false. If f (x) = x3 − 2x2 then f ′ (x) =
3x2 − 4x
√
.
x3 − 2x2
16. Answer true or false.
1
5
and x = 2 then dy = − dx.
16
x5
25. Answer true or false. A spherical balloon is deflating. The
rate at which the volume is changing when r = 2 m is given
by dV = −16π dr m3 /s.
24. Answer true or false. If y =
52
Chapter 2: Answers to Sample Tests
Section 2.1
1. b
9. true
17. d
2. a
10. b
18. b
3. a
11. a
4. c
12. false
5. a
13. true
6. d
14. true
7. d
15. d
8. false
16. a
2. c
10. b
3. a
11. a
4. true
12. false
5. true
13. false
6. d
14. true
7. a
15. false
8. a
16. false
2. d
10. d
3. a
11. a
4. false
12. c
5. c
13. c
6. b
14. d
7. false
15. d
8. b
16. c
2. c
10. a
3. c
11. d
4. a
12. false
5. false
6. d
7. a
8. b
2. b
10. false
3. b
11. b
4. a
12. false
5. false
13. true
6. c
14. false
7. a
15. c
8. d
2. b
10. true
3. b
11. a
4. false
12. a
5. d
13. d
6. a
14. c
7. c
15. true
8. a
2. true
10. d
3. a
11. false
4. c
12. false
5. false
13. a
6. d
14. c
7. b
15. a
8. d
16. a
2. b
10. a
3. c
11. false
4. c
12. false
5. false
13. false
6. d
14. true
7. c
8. b
2. a
10. b
3. false
11. true
4. d
12. b
5. a
13. true
6. a
14. true
7. b
8. d
2. a
10. false
18. a
3. b
11. b
19. d
4. d
12. c
20. a
5. a
13. c
21. b
6. false
14. a
22. true
7. b
15. d
23. c
8. d
16. true
24. false
Section 2.2
1. a
9. d
Section 2.3
1. d
9. a
Section 2.4
1. false
9. b
Section 2.5
1. c
9. a
Section 2.6
1. a
9. false
Section 2.7
1. false
9. b
17. c
Section 2.8
1. a
9. b
Section 2.9
1. d
9. b
Chapter 2 Test
1. a
9. c
17. false
25. true
Chapter 3: The Derivative in Graphing and
Applications
Summary: The main purpose of this chapter is to use the derivative as a tool
to assist in the graphing of functions and for solving optimization problems. The
most prominent use of the derivative is to help determine the overall behavior of
a function. Types of behavior that can be described by using the derivative are
when the function is increasing or decreasing and where the function is concave
up or down. From these features, conclusions can be obtained about the extrema
and inflection points of a function.
This chapter, then, intends to show some of the many uses of derivatives. For
example, derivatives can be used to find the roots of functions. This is the primary motivation behind Newton’s Method. In motion problems where a position
function describes the location of an object or a particle, the derivatives of the
position function have special meanings such as the velocity and acceleration of
the particle. Derivatives also find their way into many application problems where
certain quantities either need to be maximized or minimized. Because derivatives
can be used to find the extreme points of a function, they can be instrumental in
optimizing quantities in many application problems.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Determine where a function is increasing or decreasing (§3.1, 3.2).
2. Determine the concavity of a function (§3.1).
3. Locate critical points (§3.2), relative extrema (§3.2), and points of inflection
of a function (§3.1).
4. Sketch the curve of a function based upon information from its derivatives
together with information about asymptotes and intercepts (§3.2, 3.3).
5. Find absolute extrema of a continuous function on a closed interval (§3.4).
6. Find the maximizing or minimizing value in various application problems
(§3.5).
53
54
7. Use derivatives to discuss the motion of a particle that has a position function for its location along a line (§3.6).
8. Apply Newton’s Method to find roots of functions (§3.7).
9. Draw conclusions about the value of a functions derivative at a point by
using the Mean-Value Theorem and Rolle’s Theorem (§3.8).
3.1 Analysis of Functions I: Increase, Decrease, and
Concavity
PURPOSE: To relate the derivative of a function to the ideas of
increasing, decreasing and concavity of functions.
The primary focus of this section is to introduce the ideas of the increase and
decrease of functions, the concavity of functions and how all of these relate to
the derivative of a function. As it turns out, there are some convenient ways to
remember these concepts that depend upon the tangent line of a function at any
given point. Since the derivative can be used to find the slope of a tangent line, it
becomes an important method for discussing the increase, decrease and concavity
of a function.
increasing and decreasing
Increasing and decreasing describe whether the values of the function are getting
larger or smaller as the inputs increase. The tangent line of a function can help
keep all of this in the right perspective. A curve will follow the direction of its
tangent line at any given point. So if a tangent line has a positive slope, then the
function should be increasing at that point. Since the derivative is what allows us
to determine the slope of the tangent line, it can help determine whether a function
is increasing or decreasing.
IDEA: Increasing functions have a positive derivative and decreasing functions
have a negative derivative.
To summarize, increasing functions have a derivative that is positive, while decreasing functions have a derivative that is negative. Of course, this is all assuming that the function being discussed is differentiable. If it is not differentiable
then these assertions cannot be made.
concavity
“if it’s a frown, it’s down”
Concavity is the other big topic. Simply put, concavity describes how the curve is
bending. If it is bending upward in a U shape then the concavity of the function is
positive. If it is bending downward then the concavity of the function is negative.
The phrase “if it’s a frown, it’s down” is meant to describe the concavity. If the
shape of a curve makes a frown (i.e., it is bending downward) then the concavity
of the curve is negative (or downward).
IDEA: Concavity describes whether the derivative is increasing or decreasing.
What concavity is really describing is when the derivative of a function is increasing or decreasing. If a function has a derivative that is increasing, then it will
55
be bending upward because its slopes will be gradually becoming more positive.
Unfortunately, the sign of the derivative of f does not provide curvature direction. The second derivative is usually related to concavity in the way that the first
derivative is related to increase and decrease of a function. If f ′′ > 0 then the
function f will have a concave upward shape while if f ′′ < 0 then the function f
will have a concave downward shape.
One interesting point that can be identified on a curve is called a point of inflection. This is a point on the curve where the bending of the curve changes
directions. In other words, if the concavity changes from positive to negative or
vice versa then the point where this happens is a point of inflection. Similar points
can be identified where a function changes from increasing to decreasing or vice
versa although these points are not called inflection points. They are described in
more detail in the next section.
Determining where a function is increasing or decreasing, and how its concavity
is behaving can allow a rough sketch of a function to be drawn. This is where the
analysis of the signs of the derivatives becomes important. By knowing where
the first derivative is positive or negative, the increase or decrease of a function
can be found. Similarly, by knowing where the second derivative is positive or
negative can determine the concavity of a function.
IDEA: Finding where derivatives equal zero or are undefined can help determine intervals of increase or decrease and concavity.
Sign analysis first involves finding locations where the derivatives are zero or undefined. These points are used to determine intervals on which to check whether
each derivative is positive or negative. Then conclusions can be made about the
behavior of the function.
One important thing to note about the zeros of the derivatives is that simply because f ′ (a) = 0 does not mean that f (x) will change from increasing to decreasing
at x = a. For example, consider y = x3 . Since y′ = 3x2 , then y′ = 0 when x = 0.
However, the function y = x3 is increasing for all x-values. Similarly, even though
f ′′ (a) = 0, it does not mean that there must be a point of inflection at x = a. For
example, consider y = x4 . Since y′′ = 12x2 then y′′ = 0 when x = 0. However,
the function y = x4 is concave upwards for all x-values meaning that there is not a
point of inflection at x = 0.
Checklist of Key Ideas:
increasing and decreasing functions
constant functions
critical numbers and relative extrema
concavity
inflection points
sign analysis
point of inflection
sign analysis
56
3.2 Analysis of Functions II: Relative Extrema; Graphing Polynomials
PURPOSE: To use the first and second derivative tests to find
relative extrema and to sketch polynomials.
critical points
stationary points
In this section, the analysis of derivatives are used to find relative or local extrema
of a function. Two tests are introduced: the first derivative test and the second
derivative test. First, it is noticed that if a relative extrema occurs on a continuous
function then it is required that the derivative of the function at the extrema must
either be equal to zero or not defined at that point. The list of all points where
the derivative is undefined or equal to zero are called critical points (stationary
points are simply critical points where the derivative is defined and equal to zero).
Then the basic process for graphing a function includes finding all of the critical
points of the function since these are the locations where the function may have
relative extrema.
Think of critical points as all of the possible places that we may have a relative
extrema. Once the critical points have been determined the first derivative and
second derivative tests may be applied to determine if there is a relative extrema
or not.
IDEA: The first derivative test finds relative extrema based upon changes in
the sign of the first derivative.
first derivative test
A relative extrema indicates that a function is defined at that point and that the
function must change from increasing to decreasing at that point (or vice versa).
This is essentially what the first derivative test does; it is a sign analysis of the
first derivative. If the derivative changes signs at the critical point then there is
a relative extrema located there. If the function changes from increasing to decreasing (from up to down) then the point is a relative maximum. Changing from
decreasing to increasing (the sign of the derivative would change from negative to
positive) would indicate a relative minimum.
IDEA: The second derivative test finds relative extrema based upon the sign
of f ′′ (x) at critical points.
second derivative test
The second derivative test goes one step further. If there is a critical point then
a relative maximum can only occur when the concavity is negative at that point.
This is essentially the second derivative test. If the second derivative is positive
at a critical point (concavity is positive) then there is a relative minimum. On the
other hand if the second derivative is negative at a critical point then there is a
relative maximum at that point.
sketching polynomials
Polynomials are relatively easy to analyze since their derivatives are easy to find.
Items of interest for polynomials are x-intercepts (roots), y-intercepts, intervals of
increase/decrease, intervals of concavity, critical points, relative extrema, points
of inflection, and end behavior. By putting all of these things together, a sketch of
the polynomial can be easily obtained.
57
To start the sketch, it may be easiest to first find the interesting points on the curve
such as intercepts and critical points. The multiplicity of roots can also be used
to find out information about the function. Generally, if a root has a multiplicity
of greater than one, then the function will have a critical point at that root. If
the multiplicity is even at that root, then it will be a relative extrema. Then sign
analysis can also give the general shape of the curve.
CAUTION: Several of the techniques listed here may give redundant information but there should not be any contradictions. If a contradiction is found then
there may be an error in some calculation.
Starting a sketch
1. Find interesting points
→ roots/intercepts
→ critical points
→ inflection points
2. determine the shape
→ increasing or decreasing?
→ concavity?
Checklist of Key Ideas:
relative maxima and minima
critical points
stationary points
First Derivative Test
Second Derivative Test
multiplicity of roots
graphing polynomials
properties of polynomials
3.3 Analysis of Functions III: Rational Functions,
Cusps, and Vertical Tangents
PURPOSE: To graph more challenging functions.
Sketching rational functions is very similar to sketching polynomial functions. All
of the same information can be used. The one thing to be cautious about is that
some rational functions may have vertical asymptotes. These are places of discontinuity which are interesting but they are not critical points (since the function is
not defined at these points). But when determining intervals of increase/decrease
and concavity, the points of discontinuity should be listed as possible endpoints of
intervals.
In addition to intercepts, critical points, points of inflection, and intervals of increase/decrease and concavity, rational functions will also potentially have vertical
and/or horizontal asymptotes. Vertical asymptotes may be determined by factors
in the denominator which do not cancel with the numerator (if they cancel completely then there is a hole at the zero of the factor). If a factor in the denominator
does not cancel then there will be a vertical asymptote at the zero of the factor.
vertical asymptotes
(see also §1.3)
58
horizontal asymptotes
(see also §1.3)
Horizontal asymptotes may only occur when the degree of the numerator is equal
to or less than the degree of the denominator. The position of any horizontal
asymptotes may then by found by taking limits as x → ±∞ of the leading terms
of the numerator and the denominator. These limits may sometimes be calculated
using L’Hôpital’s rule (see §6.8).
Checklist of Key Ideas:
graphing rational functions
symmetries
intercepts
periodicity
relative extrema
concavity
intervals of increase and decrease
inflection points
asymptotes
vertical tangents
cusps
3.4 Absolute Maxima and Minima
PURPOSE: To locate the absolute extrema of a function.
In this section, information about relative extrema and the derivatives of a function can be used to determine if a function may have an absolute maximum or
minimum. In general, there is no guarantee that a function will actually have an
absolute maximum or minimum on a given interval. However, if certain things are
known about both the function and the interval then there may be guarantees about
finding the maxima and minima of the function upon the interval in question.
IDEA: If a function is continuous on a closed interval then the absolute extrema
of the function can be found.
closed interval
For example, if a continuous function is on a closed interval, then there is guaranteed to be an absolute maximum and minimum of the function (the highest and
lowest points on the interval). These points have to occur either at the endpoints
or at any critical points in the interior of the interval. This guarantee immediately
disappears if the function is discontinuous or the interval is not finite or closed.
IDEA: If absolute extrema exist, they will occur at the endpoints of an interval
or at critical points.
59
On an infinite interval, absolute extrema can sometimes be found depending upon
the end behavior of a function. For example, if lim f (x) = ∞, the function will
infinite interval
x→±∞
have an absolute minimum if it is continuous. The absolute minimum will have to
occur at one of the functions relative minimum. A similar statement can be made
about an absolute maximum if the functions end behavior approaches −∞ all the
time.
Open intervals or half-open intervals can cause problems and there are no guarantees. Each situation may be different. If there are going to be any absolute
maximums or minimums on an open interval, however, they will have to occur either at the closed endpoints (if there are any) or the relative extrema of the function
(if there are any). A simple example can show the problems that can arise. The
function y = x on the interval (0, 1) has no absolute maximum or minimum. The
highest value approaches y = 1 but it never gets there. Likewise with the lowest
value and y = 0. To determine if there are absolute extrema, compare the limits at
the open endpoints with the function values at the critical points of the function.
open interval
Checklist of Key Ideas:
absolute extrema
extreme value theorem
closed interval
open interval
infinite interval
3.5 Applied Maximum and Minimum Problems
PURPOSE: To solve optimization problems by using the techniques of this chapter.
Optimization problems or applied maximum and minimum problems, follow from
the methods discussed earlier in this chapter. The approach to finding maximum
and minimum values of a function in an applied setting is the same as earlier in
this chapter.
In an applied problem, the first order of business is determining the variables
involved and the equations that need to be considered. There may often be more
than one equation. In these cases, one variable should be solved for in terms of the
other(s) and substituted into the appropriate equations. The goal is to reduce all of
the equations to a single equation with only one independent variable. Anything
else cannot be considered using the techniques presented here. Sometimes drawing pictures and using geometric information may be helpful in reducing the number of equations.
determine the variables involved
obtain a single equation
→ geometry?
→ draw a picture?
60
determine the interval
After a single equation is obtained with one variable, the interval over which the
optimization needs to occur has to be determined. This can usually be determined
by statements in the problem. For example, if one of the variables represents a
physical quantity like the height of a boy, then it cannot have any negative values.
That information may help to determine the interval of the function.
what is to be found?
Finally, it should be determined what is needed (i.e., a maximum or minimum).
Then taking derivatives begins so that the function can be analyzed using the techniques of this chapter. First find critical points, then apply the First and Second
derivative tests to find local extrema. Finally, determine which point of interest
satisfies the problem statement. The values of other variables may need to be determined at the end (especially if there were more than one equation at the start
with more than one variable).
Finding Maximums/Minimums
→ identify variables
→ get one equation
→ identify interval
→ find critical points
→ behavior at endpoints and critical
points
Here are the general steps: (1) list the variables, (2) draw a picture if necessary,
(3) list equations, (4) reduce to one equation with one variable, (5) determine
the appropriate interval, (6) analyze the function by taking derivatives, applying
derivative tests, etc., (7) find all relevant information.
Checklist of Key Ideas:
optimization over finite closed interval
optimization over infinite interval or open finite interval
steps for solving optimization problems
economics application
3.6 Rectilinear Motion
PURPOSE: To discuss motion along a straight line.
position and velocity
v(t) = s′ (t)
speed = |v(t)|
speed is always nonnegative
The convention with rectilinear motion (or motion along a line) is that movement
to the right is in the positive direction and movement to the left is in the negative
direction. The sign of the position function, s(t), determines on which side of the
origin a particle is located. The sign of the velocity function, v(t), determines in
which direction a particle is moving. A positive value for v(t) indicates movement
to the right, for example. If v(t) = 0, then the particle is momentarily stationary.
Speed is often confused with velocity. Speed only indicates how large the velocity
is (in an absolute value sense ignoring any positive or negative signs). A velocity
of +50 m/s and −50 m/s both have a speed of 50 meters per second. Because
speed ignores negative signs, it is possible that the speed may be increasing while
the velocity is decreasing.
IDEA: Speed tells how fast an object is moving but does not tell what direction
it is moving.
61
A particle with rectilinear motion is speeding up if its velocity and acceleration
both have the same sign. If the velocity and acceleration both have opposite signs,
then the particle is slowing down (approaching a velocity of zero). Another way
to interpret this is that speed only describes how fast an object is moving but not
what direction. If a particle is slowing down, then its velocity will be close to zero.
If the particle is speeding up, then its velocity is moving away from zero (either in
a positive or negative sense).
speeding up → speed increases
slowing down → speed decreases
Checklist of Key Ideas:
rectilinear motion
position
displacement
velocity
speed
acceleration
3.7 Newton’s Method
PURPOSE: To find the roots of a function using Newton’s
Method.
This section builds off of the idea that the derivative can be used to find a linear
approximation of a function near a point that was first discussed in Chapter 2.
This linear approximation is just a tangent line to the function. The idea in this
section is that if a tangent line is a good approximation of a function, then finding
the zero of a tangent line may help to approximate the zero of a function. This is
the idea that is called Newton’s Method.
linear approximation
(see §2.9)
IDEA: Newton’s Method uses a sequence of tangent lines to approximate the
root of a function.
If a tangent line to a function y = f (x) at the point x0 is given by y = y0 + m0 (x − x0 )
then it is a straightforward process to find the zero of this line. For example, if
y = 0 then x = x0 − y0 /m0 . But this is exactly the formula for the iteration of
Newton’s method when y0 = f (x0 ) and m0 = f ′ (x0 ).
xn+1 = xn − yn /mn = xn −
f (xn )
f ′ (xn )
Newton’s method is repeating this process of finding the zeros for tangent lines
over and over again. A tangent line is constructed for the function at the location
of each zero that was found in the previous step. Eventually, the zeros should
approach the value of a zero of the function.
Newton’s method iteration
x0 is original guess of root
xn is the n-th root approximation
62
try to pick x0 close to a root
horizontal tangents can cause failure
failure can occur if xn and xn+1 have
the same tangent line equation
There are some drawbacks to this method. Since the basic idea is to use a linear
approximation (i.e., the tangent line), there will be problems when trying to approximate the function farther away from the point of tangency. What does this
mean? Linear approximations only work well within a small area. The moral of
all of this is that a poorly chosen starting point can cause the method to fail.
Another way that the method can fail is if the tangent line at any given step is
horizontal. The formula shown above, then has us dividing by zero which will
not lead to any good result. Visually this is just reaffirming the fact that horizontal
lines are parallel to the x-axis and will not have roots since they will never intersect
the x-axis.
One last way that Newton’s method can fail is if the starting point leads to a new
point that returns us to the original starting point. This may happen for example if
two consecutive points xn and xn+1 both have the same equation for a tangent
line. In this type of situation, iteration will never stop. This problem can arise
in even simple problems. For example, y = x(x2 − 20) with a starting point of
x0 = 2. The iterations will bounce back and forth between 2 and −2. To correct
this problem, simply try a different starting point.
Checklist of Key Ideas:
Newton’s Method
root finding
linear approximation
tangent line approximation
pitfalls with Newton’s Method
3.8 Rolle’s Theorem; Mean-Value Theorem
PURPOSE: To introduce two theorems that relate to tangent
lines.
This section states two important theorems: Rolle’s Theorem and the Mean-Value
Theorem. The Mean-Value Theorem is actually a more general case of Rolle’s
Theorem. Essentially Rolle’s Theorem states a result about the existence of horizontal tangent lines for a function on a particular interval. In short, if the function
values of a differentiable function are the same at the ends of an interval, then
there will be a horizontal tangent line somewhere within the interval.
ROLLE’S THEOREM: If f (a) = f (b) then there is a horizontal tangent line
between x = a and x = b (if f is differentiable).
Rolle’s Theorem
Here is another version of Rolle’s Theorem: if the graph of a differentiable function intersects the x-axis at two places x = a and x = b, then somewhere between
63
x = a and x = b there will be at least one point, x = c, where there is a horizontal
tangent line. The value of c does not have to be unique. For example, the function
y = x4 − 8x2 − 9 = (x2 + 1)(x2 − 9) has zeros along the x-axis at x = −3 and x = 3.
However, when Rolle’s Theorem is applied to the zeros at x = −3 and x = 3, there
are three places where this function has horizontal tangents in between these two
points.
Rolle’s Theorem is a specific application of the Mean-Value Theorem. The MeanValue Theorem essentially says that on a particular interval there will be a tangent
line that is parallel to the secant line through the function values at the endpoints
of the interval. In the case of Rolle’s Theorem, this is a horizontal line.
Mean-Value Theorem
MEAN-VALUE THEOREM (MVT): There is a tangent line between x = a and
x = b that is parallel to the line between the points on f (x) at x = a and x = b
(if f is differentiable).
Here is a restatement of the Mean-Value Theorem (MVT): between any two points,
A (a, f (a)) and B (b, f (b)), on the graph of a differentiable function y = f (x), there
is at least one point where the tangent line to the graph is parallel to the secant line
that joins the points A and B.
Here is a another restatement of the MVT: over the interval [a, b], the instantaneous
velocity of a differentiable position function must equal the average velocity at
least once.
The Constant Difference Theorem says, “If f and g have the same derivative on
an open interval, then the graphs of f and g are vertical translations of one another
over that interval.” In other words, two functions that have the same derivative
must differ from each other by a constant value.
IDEA:A function must be differentiable for these theorems to be applied.
The main thing to watch out for with each of these theorems is that the function
must be differentiable over the entire interval that is being discussed. Discontinuous functions cause problems with these theorems from the start since not being
continuous immediately indicates that the function cannot be differentiable. For
example, no conclusions about the function y = 1/x can be drawn over any interval
that contains x = 0 since it is discontinuous at that point.
Checklist of Key Ideas:
Mean-Value Theorem
Rolle’s Theorem
velocity
constant difference theorem
Constant Difference Theorem
64
Chapter 3 Sample Tests
Section 3.1
1. Answer true or false. If f ′ (x) > 0 for all x on the interval I,
then f (x) is concave up on the interval I.
2. Answer true or false. A point of inflection always has an xcoordinate where f ′′ (x) = 0.
3. The largest interval over which f is increasing for
f (x) = (x − 5)4 is
(a) [5, ∞)
(b) [−5, ∞)
(c) (−∞, 5]
(d) (−∞, −5]
4. The largest interval over which f is decreasing for
f (x) = x3 − 12x + 7 is
(a) (−∞, −2]
(b) [2, ∞)
(c) [−2, 2]
(d) [−2, ∞)
5. The largest
interval over which f is increasing for
√
f (x) = 3 x − 2 is
(b) [π , 2π ]
(c) [π /2, 3π /2]
(d) [0, 2π ]
9. Answer true or false. y = cot x has a point of inflection on the
interval (0, π ).
10. Answer true or false. All functions of the form f (x) = axn
have an inflection point.
11. The function f (x) = x4 − 24x2 + 6 is concave down over
which of the following intervals?
(a) (−∞, 0) ∪ (0, ∞)
√
(b) (−∞, 0) ∪ ( 12, ∞)
(c) (−∞, −2]
(d) (−2, 2)
12. Answer true or false. If f ′′ (−2) = −3 and f ′′ (2) = 3, then
there must be a point of inflection on (−2, 2).
x2
has
x2 − 4
(a) points of inflection at x = −4 and x = 4.
13. The function f (x) =
(b) points of inflection at x = −2 and x = 2.
(c) a point of inflection at x = 0.
(d) no points of inflection.
(a) [2, ∞)
(b) (−∞, 2)
(c) (−∞, ∞)
(d) nowhere
6. The largest
√ open interval over which f is concave up for
f (x) = 3 x − 5 is
(a) (−∞, 5)
(b) (5, ∞)
(c) (−∞, ∞)
(d) nowhere
7. The function f (x) = x4/5 has a point of inflection with an
x-coordinate of
(a) 0
(b) 4/5
(c) −4/5
(d) None exist.
8. Use a graphing utility to determine where f (x) = sin x is decreasing on [0, 2π ].
(a) [0, π ]
Section 3.2
1. Determine the x-coordinate(s) for any stationary point(s) of
the function f (x) = 2x3 − 3x2 − 72x + 6.
(a) x = −4 and x = 3
(b) x = −3 and x = 4
(c) x = −6
(d) No stationary points.
2. Determine the x-coordinate(s)
of the critical point(s) of the
√
function f (x) = 5 x − 3.
(a) x = 0
(b) x = 3
(c) x = −3
(d) No critical points.
3. Answer true or false. f (x) = x2/7 has a critical point.
4. Answer true or false. A function has a relative extrema at
every critical point.
5. The function f (x) = x2 + 6x + 8 has a
(a) relative maximum at x = −3.
65
(b) relative minimum at x = −3.
(c) relative maximum at x = 3.
(d) relative minimum at x = 3.
6. The function f (x) = cos2 x on 0 < x < 2π has
(a) a relative maximum at x = π ; relative minima at
x = π /2 and x = 3π /2
(b) relative maxima at x = π /2 and x = 3π /2; a relative
minimum at x = π
(c) a relative maximum at x = π ; no relative minima
(d) no relative maxima; a relative minimum at x = π .
7. The function f (x) = x4 − 4x3 has
(a) a relative maximum at x = 0; no relative minima
(b) no relative maxima; a relative minimum at x = 3
(c) a relative maximum at x = 0; a relative minima at x = 3
(d) a relative maximum at x = 0; relative minima at x = −3
and x = 3
8. Answer true or false. f (x) = |tan x| has no relative extrema
on (−π /2, π /2).
9. f (x) = |x2 − 9| has
(a) no relative maxima; a relative minimum at x = 3
(b) a relative maximum at x = 3; no relative minima
(c) relative minima at x = −3 and x = 3; a relative maximum at x = 0
(d) relative minima at x = −9 and x = 9; a relative maximum at x = 0
10. f (x) = ln (x2 + 2) has
(a) a relative maximum only
(b) a relative minimum only
Section 3.3
1. Answer true or false. If f ′ (−2) = −1 and f ′ (2) = 1, then
there must be a relative minimum on (−2, 2).
2. The polynomial function y = x2 − 6x + 8 has
(a) one stationary point at x = 3.
(b) two stationary points, one at x = 0 and one at x = 3.
(c) one stationary point that is at x = −3.
(d) one stationary point that is at x = 0.
3. The rational function y =
3x + 6
has
x2 − 1
(a) a horizontal asymptote at y = 0.
(b) a horizontal asymptote at y = −2.
(c) horizontal asymptotes at x = −1 and x = 1.
(d) no horizontal asymptotes.
4. Determine the x-coordinates of all the stationary points of the
3x + 6
rational function y = 2
.
x −1
√
(a) x = −2 ± 3
(b) x = −2 only
(c) both x = −1 and x = 1
(d) x = −2, x = −1 and x = 1
5. Answer true or false. The rational function y = x3 −
no vertical asymptotes.
1
has
x2
6. On a [−10, 10] by [−10, 10] window on a graphing utility the
x3 + 8
rational function y = 3
can be determined to have
x −8
(a) one horizontal asymptote and no vertical asymptote.
(c) both a relative maximum and minimum
(b) no horizontal asymptotes and one vertical asymptote.
(d) no relative extrema
(c) one horizontal asymptote and one vertical asymptote.
11. On the interval (0, 2π ), the function f (x) = sin x cos (2x) has
(a) a relative maximum only
(b) a relative minimum only
(c) both a relative maximum and a minimum
(d) no relative extrema
12. Answer true or false. The function f (x) = ex ln x has a relative
minimum on (0, ∞).
13. Answer true or false. A graphing utility can be used to show
that f (x) = |x| has a relative minimum.
14. Answer true or false. A graphing utility can be used to show
that f (x) = x4 −3x2 +3 has two relative minima on [−10, 10].
(d) one horizontal asymptote and three vertical asymptotes.
7. Use a graphing utility to graph f (x) = x1/7 . How many points
of inflection does the function have?
(a) 0
(b) 1
(c) 2
(d) 3
8. Use a graphing utility to graph f (x) = x−1/7 . How many
points of inflection are there?
(a) 0
66
Section 3.4
(b) 1
(c) 2
1. In the interval [−2, 2], the function f (x) = 3x2 − x + 2 has an
absolute maximum value of
(d) 3
9. Determine which function is graphed below.
(a) 16
y
(b) 2
10
(c) 12
(d) 4
2. The function f (x) = |5 − 2x| has an absolute minimum value
of
5
(a) 0
(b) 3
-10
-5
5
10
x
(c) 1
(d) 5
3. Answer true or false. f (x) = x3 − x2 + 2 has an absolute maximum and an absolute minimum.
-5
4. Answer true or false. f (x) = x3 − 18x2 + 20x + 2 restricted
to the domain of [0, 20] has an absolute maximum at x = 2 of
−22, and an absolute minimum at x = 10 of −598.
√
5. The function f (x) = x − 2 has an absolute minimum value
of
-10
(a) f (x) = x1/2
(a) 0 at x = 2.
(b) f (x) = x−1/3
(b) 0 at x = 0.
(c) f (x) = x−1/2
(c) −2 at x = 0.
(d) f (x) = x1/3
10. Use a graphing utility to generate the graph of f (x) = x2 e3x ,
then determine the x-coordinates of all relative extrema on
(−10, 10) and identify them as relative maxima or minima.
(a) There is a relative maximum at x = 0 and a relative
minimum at x = −2/3.
(b) There is a relative minimum at x = 0 and a relative maximum at x = −2/3.
(c) There is a relative minimum at x = 0 and relative maxima at x = −1 and x = 1.
(d) There are no relative extrema.
12. Answer true or false. A fence is used to enclose a rectangular
plot of land. If there are 160 feet of fencing, it can be shown
that 40 ft by 40 ft square is the rectangle that can be enclosed
with the greatest area. (A square is considered a rectangle).
oblique asymptote.
6. The function f (x) =
one exists, at
√
x2 + 5 has an absolute maximum, if
(a) x = −5
(b) x = 5
(c) x = 0
(d) No maximum exists.
7. Find the location of the absolute maximum of y = tan x on
[0, π ], if it exists.
(a) x = 0
11. Answer true or false. Using a graphing utility, it can be shown
that f (x) = x2 tan x has a maximum on 0 < x < 2π .
13. Answer true or false. The function f (x) =
(d) 0 at x = −2
x2 + 3x + 4
has an
x−1
(b) x = π
(c) x = π /2
(d) No maximum exists.
8. f (x) = x2 − 3x + 2 on (−∞, ∞) has
(a) only an absolute maximum.
(b) only an absolute minimum.
(c) both an absolute maximum and minimum.
67
(d) neither an absolute maximum nor an absolute minimum.
1
9. f (x) = 2 on [1, 3] has
x
(a) an absolute maximum at x = 1 and an absolute minimum at x = 3.
(b) an absolute minimum at x = 1 and an absolute maximum at x = 3.
(a) a = 5, b = 15
(b) a = 1, b = 19
(c) a = 10, b = 10
(d) a = 0, b = 20
2. A right triangle has a perimeter of 16. What are the lengths
of each side if the area contained within the triangle is to be
maximized?
(c) no absolute extrema.
(a) 16/3, 16/3, 16/3
(d) an absolute minimum at x = 2 and absolute maxima at
x = 1 and x = 3.
(b) 5, 5, 6
10. Answer true or false. f (x) = sin x cos x on [0, π ] has an absolute maximum at x = π /2.
11. Use a graphing utility to assist in determining the location of
the absolute maximum of the function f (x) = −(x2 − 3)2 on
(−∞, ∞), if it exists.
√
√
(a) x = 3 and x = − 3
√
(b) x = 3 only
(c) x = 0
(d) No absolute maximum exists.
12. Answer true or false. If y = f (x) has an absolute minimum at
x = 2, then y = − f (x) also has an absolute minimum at x = 2.
13. Answer true or false. Every function has an absolute maximum and an absolute minimum if its domain is restricted to
where f is defined on the interval [−a, a], where a is finite.
14. Use a graphing utility to locate the value of x where
f (x) = x4 − 3x + 2 has an absolute minimum, if it exists.
(a) x = 1
p
(b) x = 3 3/4
(c) x = 0
(d) No absolute minimum exists.
15. Use a graphing utility to estimate the absolute maximum
value of f (x) = (x − 5)2 on [0, 6], if it exists.
(a) 25
√
√
√
(c) 16 − 8 2, 16 − 8 2, −16 + 16 2
(d) 12/3, 16/3, 20/3
3. A rectangular sheet of cardboard 2 m by 1 m is used to make
an open box by cutting squares of equal size from the four
corners and folding up the sides. What size squares should
be cut to obtain the largest possible volume?
√
3+ 3
m
(a)
6
√
3− 3
m
(b)
6
(c) 1/2 m
(d) 1/4 m
4. An object moves a distance s away from the origin according
to the equation s(t) = 4t 4 − 2t + 1, where 0 ≤ t ≤ 10. At what
time is the object farthest from the origin?
(a) t = 0
(b) t = 2
(c) t = 10
(d) t = 1/8
5. An electrical generator produces a current, I(t), starting at
t = 0 s and running until t = 6π s. If I(t) = sin (2t) A, then
find the maximum current that is produced.
(a) 1 A
(b) 0 A
(b) 0
(c) 2 A
(c) 1
(d)
(d) No maximum exists.
Section 3.5
1
2
A
6. A passing storm has a wind speed, v(t), in mph that changes
over time. If v(t) = −t 2 + 10t + 55 for 0 ≤ t ≤ 10 hours, then
find the highest wind speed that occurs during the storm.
(a) 55 mph
(b) 80 mph
1. Express the number 20 as the sum of two nonnegative numbers, a and b (with a ≤ b), whose product is as large as possible.
(c) 110 mph
(d) 30 mph
68
7. A company has a cost of operation function given by
C(t) = 0.01t 2 − 6t + 1, 000 for 0 ≤ t ≤ 500. Find the minimum cost of operation.
(a) The velocity is increasing.
(b) The velocity is decreasing.
(c) The velocity is constant.
(a) $1, 000
(d) Not enough information is given.
(b) $100
2. The graph below represents the position function of a particle
moving along a line.
(c) $500
(d) $0
8. Find the point on the curve x2 + y2 = 4 closest to the point
(0, 3).
y
5
(a) (0, 4)
(b) (0, 2)
(c) (2, 0)
(d) (4, 0)
9. Answer true or false. The point on the parabola y = x2 closest
to (0, 9) is (0, 0).
10. For a triangle with sides 3 m, 4 m and 5 m, the smallest circle
that contains the triangle has a diameter of
(a) 3 m
(b) 4 m
5
10
t
Which of the following statements is true about the acceleration of the particle at t = 5?
(a) Acceleration is positive.
(c) 5 m
(b) Acceleration is negative.
(d) 10 m
11. Answer true or false. The rectangle with the largest area that
can be inscribed inside a circle is a square.
12. Answer true or false. The rectangle with the largest area that
can be inscribed in a semi-circle is a square.
(c) Acceleration is zero.
(d) Not enough information is given.
3. The graph below represents the velocity function of an object.
y
13. Answer true or false. An object that is thrown upward and
will have a height of s(t) = 50 + 120t − 16t 2 for 0 ≤ t ≤ 2.
The object is highest at t = 2.
10
5
Section 3.6
1. The graph below represents the position function of a particle
moving along a line.
-10
y
-5
5
5
-5
-10
5
10
t
Determine what is happening to the velocity of the particle at
t = 5.
The acceleration of the object at t = 5 is
(a) positive.
(b) negative.
10
t
69
y
(c) zero.
y
5
5
(d) not able to be determined.
-5
5
t
-5
5
4. Answer true or false.
y
-5
-5
If the graph on the left is the position function, then the graph
on the right represents the corresponding velocity function.
7. Let s(t) = cost be a position function of a particle. At t = π /2
the particle’s velocity is
5
(a) positive
(b) negative
(c) zero
(d) unknown.
8. Let s(t) = t 3 − t be a position function of a particle. At t = 0
the particle’s acceleration is
t
5
(a) positive
(b) negative
(c) zero
(d) unknown.
If the graph above is the position function of a particle, then
the particle is moving to the right at t = 0.
9. Consider the position function s(t) = t 4 − 2t for t ≥ 0. The
velocity function, v(t), is given by
(a) t 3 − 2
5. Answer true or false.
(b) 4t 3 − 2
y
(c) 12t 2
5
(d) 12t 2 − 2
10. Consider the position function s(t) = t 4 − 2t for t ≥ 0. The
acceleration function, a(t), is given by
(a) t 3 − 2
-5
1
5
t
(b) 4t 3 − 2
(c) 12t 2
(d) 12t 2 − 2
11. Given the position function s(t) = t 4 − 4t 2 for t ≥ 0, find t
when the acceleration is zero.
(a) t = 12
-5
For the position function graphed above, the acceleration at
t = 1 is positive.
6. Answer true or false.
(b) t = −12
p
(c) t = 2/3
(d) t = −2/3
12. Given the position function s(t) = t 3 − 3t for t ≥ 0, find t
when the acceleration is zero.
t
70
(a) t = 1
(a) 1.7325
(b) t = 2
(b) 1.000
(c) t = 0
(c) 1.7319
(d) t = −1
√
13. Let s(t) = 3t 2 − 2 be a position function. Find the velocity,
v(t), when t = 1.
(a) v(1) = 3
(b) v(1) = 6
(c) v(1) = 1
(d) v(1) = 0
Section 3.7
√
1. Approximate 3 by applying Newton’s Method to the equation x2 − 3 = 0. Use x0 = 1 as a starting point and use x2 to
approximate the root.
(a) 2
(b) 1.75
(c) 1.73214
(d) 1.73205
√
2. Approximate 3 9 by applying Newton’s Method to the equation x3 − 9 = 0.
(a) 2.08008381347
(d) 1.7316
6. Use Newton’s Method to find the largest positive solution of
x4 + 6x3 − x2 − 6 = 0.
(a) 6.000
(b) 1.000
(c) 1.732
(d) 1.412
7. Use Newton’s Method to find the largest positive solution of
x4 + x3 − 4x − 4 = 0.
(a) 4.000
(b) 1.000
(c) 0.500
(d) 1.587
8. Use Newton’s Method to find the largest positive solution of
x5 − 2x3 − 14x2 + 28 = 0.
(a) 3.742
(b) 2.410
(c) 1.414
(d) 1.260
9. Use Newton’s Method to find the largest positive solution of
x5 + 2x3 − 2x2 − 4 = 0.
(b) 2.08008382305
(a) 1.260
(c) 2.08008397111
(b) 1.414
(d) 2.08008382176
(c) 1.587
3. Use Newton’s Method to approximate the solutions of
x3 + 2x2 − 5x − 10 = 0.
(a) −2.000, 2.236
(d) 2.000
10. Use Newton’s Method to find the largest positive solution of
x4 − 13x2 + 30 = 0.
(a) 3.162
(b) −5, 0, 5
(b) 2.340
(d) −3.1623, 3.1623
(d) 1.732
(c) −3.1623, 0, 3.1623
4. Use Newton’s Method to find the largest positive solution of
x3 − x2 + 2x − 4 = 0.
(a) 1.478
(b) 1.000
(c) 5.477
11. Use Newton’s Method to find the largest positive solution of
x5 + x4 + x3 − 5x2 − 5x − 5 = 0.
(a) 2.236
(b) 1.380
(c) 2.828
(c) 1.710
(d) 3.721
(d) 1.621
5. Use Newton’s Method to find the largest positive solution of
x3 − x2 + 3x − 3 = 0.
12. Use Newton’s Method to find the x-coordinate of the intersection of y = 2x3 − 2x2 and y = −x5 + 4.
71
(a) 3.742
(b) 2.410
(c) 1.414
10. Find the x-value for which f (x) = x2 + 3 on [1, 3] satisfies the
Mean-Value Theorem.
(a) x = 2
(d) 1.260
13. Use Newton’s Method to find the greatest x-coordinate of the
intersection of y = x4 − 7x2 and y = 6x2 − 30.
(a) 3.162
(b) x = 9/4
(c) x = 7/3
(d) x = 11/4
(b) 2.340
(c) 5.477
(d) 1.732
11. Find the x-value for which f (x) = x3 on [2, 3] satisfies the
Mean-Value Theorem.
(a) 2.517
Section 3.8
1
1. Answer true or false. f (x) = on [−1, 1] satisfies the hyx
potheses of Rolle’s Theorem.
2. Find the value c such that the conclusion of Rolle’s Theorem
are satisfied for f (x) = x2 − 4 on [−2, 2].
(a) 0
(b) −1
(c) 1
(d) 1/2
3. Answer true or false. The Mean-Value Theorem is used to
find the average of a function.
4. Answer true or false. The Mean-Value Theorem can be used
on f (x) = |x| on [−2, 1].
5. Answer true or false. The Mean-Value Theorem guarantees
there is at least one such c on [0, 1] such that f ′ (c) = 1 when
√
f (x) = x.
√
6. If f (x) = 3 x on [0, 1], find the value c that satisfies the MeanValue Theorem.
(a) 1
(b) 1/3
3/2
1
(c)
3
(b) 2.500
(c) 2.250
(d) 2.125
12. Answer true or false. A graphing utility can be used with
Rolle’s Theorem to show that the function f (x) = (x − 2)2
has a point where f ′ (x) = 0.
13. Answer true or false. According to Rolle’s Theorem, if a
function does not cross the x-axis, then its derivative cannot
be zero anywhere.
14. Find the value c that satisfies Rolle’s Theorem for f (x) = sin x
on [0, π ].
(a) π /4
(b) π /2
(c) 3π /4
(d) π /3
15. Find the value c that satisfies the Mean-Value Theorem for
f (x) = x3 + 3x on [0, 1].
(a)
(d) 1/9
7. Answer true or false. The hypotheses
of the Mean-Value Thep
orem are satisfied for f (x) = 3 |x| on [−1, 1].
(b)
8. Answer true or false. The hypotheses of the Mean-Value Theorem are satisfied for f (x) = sin x on [0, 4π ].
(c)
9. Answer true or false. The hypotheses of the Mean-Value The1
on [0, 4π ].
orem are satisfied for f (x) =
sin x
(d)
√
3
3
√
3
2
√
2
2
√
2
3
72
Chapter 3 Test
1. The largest interval on which f (x) = x2 − 4x + 7 is increasing
is
(a) [0, ∞)
(b) (−∞, 0]
(c) [2, ∞)
(d) (−∞, 2]
√
2. Answer true or false. The function f (x) = x − 2 is concave
down on its entire domain, except at x = 2.
3. The function f (x) = x3 − 8 is concave down if
(a) x < −2
(b) x = 0
(c) x = 16
(d) x = 1
11. Answer true or false. The function f (x) = x4/9 has a critical
point at x = 0.
12. The curve y = x2 − 8x + 9 has
(a) a relative maximum at x = 4
(b) a relative minimum at x = 4
(c) a relative maximum at x = −4
(d) a relative minimum at x = −4
13. The function f (x) = 16x2 has
(a) no relative maxima; a relative minimum at x = 4
(b) x > 2
(b) a relative maximum at x = 4; no relative minima
(c) x < 0
(c) a relative maximum at x = 0; relative minima at x = −4
and x = 4
(d) x > 0
4. Answer true or false. The function f (x) = x4 − 2x + 3 has a
point of inflection.
5. The function f (x) = |x2 − 4| is concave up on
(a) (−∞, −2) ∪ (2, ∞)
(b) (−∞, −4) ∪ (4, ∞)
(c) (−2, 2)
(d) (−4, 4)
6. Determine where f (x) = sin x is increasing on the interval
[0, 2π ].
(a) 0 ≤ x ≤ π
(b) π ≤ x ≤ 2π
(c) π /2 ≤ x ≤ 3π /2
(d) 0 ≤ x ≤ π /2 or 3π /2 ≤ x ≤ 2π
(d) no relative maxima; a relative minima at x = 0
4x + 20
14. The rational function f (x) = 2
has
x − 25
(a) a horizontal asymptote at y = 0.
(b) horizontal asymptotes at y = 5 and y = −5.
(c) a horizontal asymptote at y = 4.
(d) a horizontal asymptote at y = −4/5.
3
has a vertical
15. Answer true or false. The function f (x) =
x−5
asymptote.
16. A weekly profit function for a company that manufactures
umbrellas is P(x) = −0.01x2 + 3x − 2, 000, where x is the
number of umbrellas that are made and sold each week. How
many individual umbrellas must the company make and sell
each week to maximize their profit?
(a) 300 umbrellas
7. Answer true or false. The function f (x) = x4 − 2x2 + 10 has
at least one point of inflection.
(b) 150 umbrellas
8. The function f (x) = −x4 + 6x3 − 12x2 is concave down on
(d) 60 umbrellas
(a) (−∞, ∞)
(b) (−∞, 1) ∪ (2, ∞)
(c) 600 umbrellas
17. The function f (x) = 6x2 − 2 has an absolute minimum value
on [−3, 3] of
(c) (1, 2)
(a) 2
(d) nowhere.
(b) −2
9. Answer true or false. If f ′′ (−1) = 4 and f ′′ (1) = 4, and if f
is continuous on [−1, 1], then there is a point of inflection on
(−1, 1).
10. Determine the x-coordinate of each stationary point of
f (x) = 4x4 − 16.
(a) x = −1
(c) 52
(d) −52
18. The function f (x) = x3 + 3 has an absolute maximum value
on [−2, 2] of
(a) 0
(b) 6
73
(c) 11
y
5
(d) 8
19. The function f (x) = 3 sin (x + 2) has an absolute minimum
value of
(a) −2
(b) −3
(c) −1/3
2
5
10
x
(d) −2/3
20. The function f (x) = x−5 has an absolute maximum value on
[1, 3] of
(a) 1
(b) 1/243
Which of the following statements is true about the motion of
the particle?
(c) 243
(a) The particle is at the origin at t = 2.
(d) There is no absolute maximum on [1, 3].
(b) The particle is moving to the right for 0 ≤ t < 2.
21. Answer true or false. The function f (x) = x−7 has an absolute maximum value of 1 on [−1, 1].
22. Express the number 40 as the sum of two nonnegative numbers, a and b (with a < b), whose product is as large as possible.
(c) The particle is always moving to the left.
(d) The particle is slowing down when t > 2.
26. The graph below represents the velocity function of a particle
along a line.
y
5
(a) a = 5, b = 35
(b) a = 10, b = 30
(c) a = 20, b = 20
(d) a = 1, b = 39
23. The position of an object relative to the origin is given by
s(t) = t 4 − 2 for 0 ≤ t ≤ 10. At what time is the object farthest from the origin?
2
5
10
x
(a) t = 0
(b) t = 2
(c) t = 8
(d) t = 10
24. Find the point on the curve x2 + y2 = 16 closest to (0, 5).
(a) (0, 4)
Which of the following statements is true about the acceleration of the particle?
(a) The acceleration is positive for t ≥ 0.
(b) The acceleration is negative for t ≥ 0.
(b) (4, 0)
(c) The acceleration is positive on [0, 2) and negative on
(2, ∞)
(c) (−4, 0)
(d) Not enough information is given.
(d) (0, −4)
25. The graph below represents the velocity function of a particle
moving along a line.
27. Let s(t) = t 4 − 2 be the position function of a particle. The
particle’s acceleration for t > 0 is
(a) positive.
74
(b) negative.
(a) 3.31662479036
(c) zero.
(b) 3.31662478727
(d) Not enough information is given.
4 − t2
28. Let s(t) =
be the position function of a particle. The
particle’s acceleration for t > 0 is
(a) positive.
(b) negative.
(c) zero.
(d) Not enough information is given.
29. Let the position function for a particle be s(t) = 4t 2 − 8. Find
t when the acceleration of the particle is zero.
(a) t = 0
(b) t = 8
(c) t = 2
(d) Acceleration is never zero.
√
30. Approximate 11 using Newton’s Method.
(c) 3.31662479002
(d) 3.31662478841
31. Use Newton’s Method to approximate the greatest positive
solution of x3 + 4x2 − 5x − 20 = 0.
(a) 4.000
(b) 2.236
(c) 5.292
(d) 3.037
32. Answer true or false. The hypotheses of Rolle’s Theorem are
satisfied for f (x) = x−6 − 1 on [−1, 1].
33. Answer true or false. Given f (x) = x2 − 16 on [−4, 4], the
value c that satisfies Rolle’s Theorem is c = 0.
34. Answer true or false. Let f (x) = x5 on [−1, 1]. The value c
that satisfies Rolle’s Theorem is c = 0.
75
Chapter 3: Answers to Sample Tests
Section 3.1
1. false
9. true
2. false
10. false
3. a
11. d
4. c
12. false
5. c
13. d
6. a
7. d
8. c
2. b
10. b
3. true
11. c
4. false
12. false
5. b
13. true
6. a
14. false
7. b
8. false
2. a
10. b
3. a
11. false
4. a
12. true
5. false
13. true
6. c
7. b
8. a
2. a
10. false
3. false
11. a
4. false
12. false
5. a
13. false
6. d
14. b
7. d
15. a
8. b
2. c
10. c
3. b
11. true
4. c
12. false
5. a
13. false
6. b
7. b
8. b
2. b
10. c
3. b
11. c
4. false
12. c
5. true
13. a
6. true
7. b
8. c
2. b
10. a
3. a
11. c
4. a
12. d
5. b
13. a
6. b
7. d
8. b
2. a
10. a
3. false
11. a
4. false
12. false
5. true
13. false
6. c
14. b
7. false
15. a
8. true
2. true
9. false
15. true
22. c
29. d
3. c
10. b
16. b
23. d
30. a
4. false
11. true
17. b
24. a
31. b
5. a
12. b
18. c
25. b
32. false
6. d
13. d
19. b
26. b
33. true
7. false
Section 3.2
1. b
9. c
Section 3.3
1. false
9. c
Section 3.4
1. a
9. a
Section 3.5
1. c
9. false
Section 3.6
1. b
9. b
Section 3.7
1. b
9. a
Section 3.8
1. false
9. false
Chapter 3 Test
1. c
8. b
14. a
21. false
28. b
20. a
27. a
34. false
76
Chapter 4: Integration
Summary: The central concept introduced in this chapter is that of integration
and how to find the area under a curve. There are two varieties of integration
represented by indefinite integrals and definite integrals. One of the main features
of this chapter, the Fundamental Theorem of Calculus, shows how integrals are
tied very closely to the notion of a derivative of a function. In fact, the indefinite
integral is actually a way of representing the antiderivative of a function and so
integration is often seen as the opposite of differentiation. Towards the end of
the chapter, definite integrals are related to rectilinear motion and the concepts of
position, velocity and speed.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Approximate a definite integral using the rectangle method (§4.1).
2. Find general antiderivatives of functions by evaluating indefinite integrals
(§4.1).
3. Evaluating basic indefinite integrals (§4.2).
4. Evaluate constants of integration given an initial condition (§4.2).
5. Use u-substitution to evaluate indefinite integrals (§4.3).
6. Represent the area under a curve as a limit using sigma notation (§4.4).
7. Numerically approximate the area under a curve (§4.4).
8. Identify definite integrals as representing the net signed area between a
function and an interval (§4.4, 4.5).
9. Identify definite integrals as limits of Riemann sums (§4.5).
10. Evaluate definite integrals geometrically (§4.5, 4.6).
11. Understand the difference between net-signed area (§4.4, 4.5) and total area
(§4.6).
12. Evaluate definite integrals using the Fundamental Theorem of Calculus (§4.6).
77
78
13. Define functions using integrals and be able to take their derivatives by the
Fundamental Theorem of Calculus (§4.6).
14. Calculate displacement of a particle by a definite integral of the velocity
(§4.7).
15. Find the average value of a function (§4.8).
16. Evaluate definite integrals that require u-substitution (§4.9).
4.1 An Overview of the Area Problem
PURPOSE: To find the area under a curve over an interval.
The primary problem considered in this section is to find the area under a curve
over a particular interval. Specifically, only positive functions are considered
here (or at least intervals where the functions considered are positive). Two primary methods of calculating areas are introduced: using rectangles and using
antiderivatives.
partitioning an interval
The area underneath a curve on an interval may be approximated by first dividing
up the interval into pieces (this is called partitioning). Then an appropriate rectangle is drawn on each of the smaller intervals. The areas of all of the resulting
rectangles added together should approximate the area under the curve.
rectangle method of finding area
If the interval is divided into more and more pieces then the approximation of the
area can become more and more accurate. Drawing the rectangles is quite simple. The height of each rectangle should be determined by one particular function
value on the curve in any given interval. Typically function values used for the
height are chosen at a convenient location within each interval such as (i) the left
endpoint, (ii) the right endpoint, or (iii) the midpoint of the interval.
antiderivative method of finding
area
The antiderivative method of finding the area under the curve is a means of
finding an exact value for the area under the curve. Put simply, the area under a
curve y = f (x) on the interval [a, x] can be considered as a function, A(x). Then
A′ (x) = f (x) or if F(x) is an antiderivative of f (x) then A(x) = F(x) +C. A value
for C can be obtained by letting A(a) = 0 (since the area under one point is zero).
Checklist of Key Ideas:
the area problem
the rectangle method for finding area
the antiderivative method for finding area
79
4.2 The Indefinite Integral
PURPOSE: To find a general expression for the antiderivative of
a function.
After the last section, the importance of antiderivatives has been seen for finding
the area under a curve. In this section, the indefinite integral is
Z introduced as a notation for a general antiderivative of a function. For example,
f (x) dx represents
Z
f (x) dx = F(x) +C
any antiderivative of the function y = f (x). For this reason,
has a constant of integration included. In fact, any antiderivative of f (x) must
differ from any other antiderivative of f (x) by only a constant.
Indefinite integrals have properties similar to those of derivatives and limits. They
can be added and subtracted and also multiplied by constants. For example,
Z
(A · f (x) + B · g(x)) dx = A
Z
f (x) dx + B
Z
g(x) dx
Be very careful to notice however that integrands may not be multiplied together
and then split into separate integrals (these same rules also apply to derivatives).
For example, consider the following statement:
Z
Z
Z
g(x) dx
f (x) dx
( f (x) · g(x)) dx 6=
⇐= CAUTION
Integral curves are a way of showing all of the antiderivatives of a function on
a graph. Each curve that is shown on the graph is simply a vertical translation
of the other curves on the graph. Slope fields are another way of showing the
integral curves (or antiderivatives) of a function on a graph. In the case of a slope
field however, the curves themselves are not shown. Instead, the slope of any
particular antiderivative at a point is shown. If a slope field is placed over a graph
of the integral curves then the slope lines of the slope fields should show up as
tangent segments on each of the integral curves. Integral curves and slope fields
are discussed again in §8.2.
integral curves
slope fields
Differential equations are the opposite of finding the derivative of a function.
Instead, information about the derivative of a function, y′ (x), is given. Then the
goal is to find the function y(x) (in earlier chapters y(x) was given and y′ (x) was to
be found). Since y(x) is an antiderivative of y′ (x) then y(x) = F(x) +C. Usually a
value of y(x) is given at some location so that a specific value of C may be found.
This given value is called an initial condition and then the differential equation is
referred to as an Initial Value Problem or IVP.
differential equations
(see also §8.1ff)
IVP
80
IDEA: The following is a basic list of indefinite integrals. It is beneficial to
memorize these formulas as they show up in much of the remaining material
in Chapters 4, 5, 6, and 7.
Z
Z
xn dx =
Z
Z
Z
1 n+1
x +C (for n 6= −1)
n+1
sin x dx = − cos x +C
cos x dx = sin x +C
sec2 x dx = tan x +C
csc2 x dx = − cot x +C
Z
sec x tan x dx = sec x +C
Z
csc x cot x dx = − csc x +C
These rules can each be found by looking at the derivatives of the individual functions (in particular, derivatives that we know at this point without having to use
the chain rule, the product rule or the quotient rule). So by taking the derivative
of a given function, another rule could be added to this list.
For example, suppose that y = x sin (x). Then y′ (x) = sin (x) + x cos (x) (by the
product rule). Then we have just come up with another indefinite integral:
Z
know these basic integrals
(sin (x) + x cos (x)) dx = x sin (x) +C
Instead of listing every derivative that can be found using the chain, product or
quotient rules, new methods of integration will be introduced that will build upon
the above list of basic integrals. For this reason, knowing the above integral rules
is essential. It may be helpful to write each one on a notecard and have them
handy as you work through problems.
Checklist of Key Ideas:
antiderivatives
integral notation: indefinite integrals, integral sign, integrand, constant of
integration, variable of integration
properties of indefinite integrals
integral curves
differential equations: Initial Value Problems and initial conditions
slope field or direction field
4.3 Integration By Substitution
PURPOSE: To introduce the method of u-substitution for finding
antiderivatives of functions.
81
The method of u-substitution is a means by which many indefinite integrals can
be transformed into integrals which resemble our basic table or integrals from the
previous section. The goal is to pick an appropriate function u = g(x) and then
calculate du = g′ (x)dx. By substituting for u and du, a more complicated integral
can hopefully be rewritten as an integral from Section 4.2 (only in terms of u
instead of x).
know the integral rules from
Section 4.2
Using u should pose no problem, however. For example,
Z
xn dx =
1 n+1
x
+C
n+1
or using u
Z
un du =
1 n+1
u
+C
n+1
Both of these integral rules are the same except for the variable used in each.
IDEA: u-substitution will be used over and over again. A common theme over
the next few chapters of the book will be to take more complicated integrals
and rewrite them as the simple integrals in Section 4.2. This is why that table
of integrals is so crucial.
The most complicated part of u-substitution is to find an appropriate u and then
to correctly substitute for u and du in the integral. Here is a simple method of
finding an appropriate u: pick a u and try it. This may seem like a haphazard way
of approaching this method but it will verify itself for you.
pick a u and try it
IDEA: Here is the general process of u-substitution:
1. pick u = g(x)
2. calculate du = g′ (x) dx
After substituting for u, everything
remaining in the integral should be
equal to a constant times du
3. substitute for u
4. substitute for du
5. check the integral - if it does not resemble any of the basic integral rules
then u has been chosen incorrectly. Go back to step 1 and try a different u
if this is the case. (see comment below)
see Section 4.2 for the basic rules
6. Evaluate the integral in terms of u (or go back to step 1 and try picking a
different u).
7. rewrite the antiderivative in terms of x using u = g(x).
Note: Step 5 is not as simple as it seems. There may be times when some additional algebra or substitution is necessary to change the integral to the basic form.
The basic rule of thumb is that after substituting for u and du, if any x’s remain
then the choice of u may not be appropriate.
if u = g(x)
→ then try to find x = h(u)
this may allow everything to be written with u instead of x
82
IDEA: Try to pick u so that the resulting integral is one of the basic integral
rules from §4.2.
The process of picking a u is not completely arbitrary. The goal is to find an
integral in terms of u that resembles one of the basic integrals.
Z For thisZ reason, u
should be picked so that the resulting integral has forms like
Z
un du,
sin u du,
cos u du and so forth.
Checklist of Key Ideas:
method of u-substitution
picking an appropriate u
calculating du
rewriting the integral in terms of u
when a substitution will not work
4.4 The Definition of Area as a Limit; Sigma Notation
PURPOSE: To introduce sigma notation; to write the area under
a curve in sigma notation.
Sigma notation is a simple way of representing several terms that are being added
together. Instead of having to write out all of the terms in the sum, a pattern is
found which can be used to represent each term in the sum depending upon an
index of summation.
Sums are equivalent as long as they have the same value and as long as they
involve the same terms. It is important to note that sums can be written with
different indices and still be the same. For example,
4
5
5
9
i=0
i=1
k=1
j=5
∑ (−1)i 2i = ∑ (−1)i−1 2i−1 = ∑ (−1)k+1 2k−1 = ∑ (−1) j+1 2 j−5 = 1−2+4−8+16
Even though the upper limit and lower limit of summation change in each case, or
different letters are used for the index, each of these is an equivalent expression of
the sum given.
Some summation formulas would be hard to write out because they involve too
many terms. In some of these cases, formulas are introduced so that the sum
can be calculated quickly. For example, in Theorem 4.4.2 the following sum is
introduced:
n
∑i=
i=1
n(n + 1)
2
83
There are other formulas similar to this one that are introduced in the section that
the reader should also be acquainted with.
At the beginning of this chapter, the area under a curve was approximated using
rectangles underneath a curve. In this section, this same exact idea is brought
up again. This time, however, the notion is made more precise by using sigma
notation. As an interval is cut into a partition with smaller and smaller intervals,
the resulting rectangles will eventually approach the exact area underneath the
curve as a limit.
Lastly, the idea of area is extended in this section to include the area that is below the horizontal axis. This is the main difference between the area idea in this
section and the idea presented at the beginning of the chapter. Now if a function
is negative, the rectangles are still drawn between a point on the function and the
horizontal axis. In this case, these rectangles are said to have negative area. Then
rectangles above the x-axis are added, rectangles below the x-axis are subtracted
and the result is said to be the net signed area under the function.
rectangle method
(see also §4.1)
negative area
net signed area
Checklist of Key Ideas:
sigma notation: lower/upper limits of summation and index of summation
properties of sums
some common summation formulas
definition of area under a curve
subintervals; left/right endpoints and midpoints
net signed area
left/right endpoint approximations and midpoint approximation
4.5 The Definite Integral
PURPOSE: To use integrals to represent the area under a curve.
The focus of this section is to give an integral representation of the area under a
curve. In the last section, the area under the curve was defined as a limit of a sum
of rectangles. In this section, this limit is called the definite integral of a function.
In other words, the definite integral of a function on an interval represents the net
signed area under the curve on that interval. The sums of rectangles are now given
a special name and called Riemann sums.
IDEA: Definite integrals give a concise way of representing the net signed area
under a function on an interval.
definite integral
Riemann sums
84
Definite integrals have the same properties as indefinite integrals in terms of being
added together or subtracted when different integrands are involved or if they are
multiplied by a constant.
Z b
a
(k · f (x) ± l · g(x)) dx = k
Z b
a
f (x) dx ± l
Z b
g(x) dx
a
Additionally, since intervals can be combined, then integrals over different intervals can be combined:
Z b
a
f (x) dx =
Z c
f (x) dx +
a
Z b
f (x) dx
c
Typically this equation will be used for c ∈ (a, b). However, as long as f (x) is
continuous on the intervals (a, c) and (c, b), then c does not need to be in the
interval (a, b).
Checklist of Key Ideas:
Riemann sums; regular partitions
integrable functions
definite integral; lower/upper limits of integration, integrand
relationship between definite integrals and net signed area
properties of definite integrals
evaluating definite integrals using geometry
4.6 The Fundamental Theorem of Calculus
PURPOSE: To describe how definite and indefinite integrals are
related using the Fundamental Theorem of Calculus.
FTC means Fundamental Theorem
of Calculus
In this section, The Fundamental Theorem of Calculus (or FTC) is introduced
in two parts. Both parts describe how definite and indefinite integrals are related to
each other. The first part is for evaluating definite integrals using antiderivatives.
The second part is for defining functions using definite integrals and taking their
derivatives.
FTC, part I
f (x) dx = f (b) − f (a)
Part I of the Fundamental Theorem of Calculus essentially says that the definite integral of a function can be evaluated exactly by taking the difference between the antiderivative of the function evaluated at the right and left endpoints of
the interval.
total change = f (b) − f (a)
In application terms, the definite integral of a derivative, f ′ (x), is equal to the total
change in the function, f (x). The definite integral is a way of summing together
the individual changes in the function represented by f ′ (x) dx.
Z b
a
′
85
The Mean-Value Theorem of Integrals says that for a continuous function there
is an average function value within the interval (more will be said about this in
the next section). The area under this “average function value” over the interval is
equal to the definite integral over the same interval.
Mean-Value Theorem
for Integrals (See also §4.8)
Part II of the Fundamental Theorem of Calculus gives a quick way for constructing a continuous function that has y = f (x) as its derivative. This function is
defined as
FTC, part II
F(x) =
Z x
f (t) dt
a
Then F(x) is an antiderivative of f (x) or F ′ (x) = f (x). This use of the FTC provides a unique and powerful tool for creating functions that have known derivatives. For example, if the derivative of a function is known to be f (x) = x2 then
the function could be defined as
F(x) =
Z x
t 2 dt.
a
Checklist of Key Ideas:
the Fundamental Theorem of Calculus (Parts I and II)
evaluating definite integrals using antiderivatives
total area between a curve y = f (x) and an interval [a, b]
the relationship between indefinite and definite integrals
the Mean-Value Theorem for Integrals
functions defined as integrals and their derivatives
integrating a rate of change
4.7 Rectilinear Motion Revisited Using Integration
PURPOSE: To talk about motion along a straight line in terms of
integration.
The velocity of a particle is the derivative of the position function of the particle.
The velocity is also the instantaneous rate of change of the position of the function.
Now the position function can be written as an integral function of the velocity,
s(t) =
Z t
v(x) dx,
0
where x is just a dummy variable.
recall: v(t) = s′ (t)
(see also §3.6)
86
velocity and speed
speed = |v(t)|
Two functions which are closely related but different are velocity and speed.
Speed is equal to the absolute value of the velocity. The definite integrals of
speed and velocity are different also. The definite integral of velocity will represent the change in position of displacement over the time interval considered. The
definite integral of the speed function will represent the total distance traveled by
the particle over the time interval.
Checklist of Key Ideas:
displacement vs. distance traveled
integrating velocity and speed
uniformly accelerated motion
initial conditions
free-fall motion and acceleration due to gravity
4.8 Average Value of a Function and Its Applications
PURPOSE: To define the average value of a function in terms of
definite integrals.
continuous functions have
an average value
This section is developed from the Mean-Value Theorem for Integrals (see also
§4.6). This theorem said that a continuous function has an average value over
an interval. Another way of describing the average value of a function over an
interval is to talk about the net signed area beneath a function. Suppose that the
area beneath a horizontal line at y = k on an interval is equal to the net signed area
beneath a function on the same interval. Then the value y = k is the average value,
fave , of the function. In terms of equations this means
fave · (b − a) =
Z b
f (x) dx
a
or
fave =
1
b−a
Z b
f (x) dx
a
IDEA: The average value of f over an interval (a, b) times the length of the
interval is equal to the net signed area under the function on the interval.
87
Checklist of Key Ideas:
arithmetic mean/average
average value of a function on an interval; geometric interpretation
average velocity
4.9 Evaluating Definite Integrals by Substitution
PURPOSE: To evaluate definite integrals using u-substitution.
Up until this point in the chapter, definite integrals have only been considered
which have antiderivatives from the basic integrals that were introduced earlier.
When the antiderivatives are not from this basic list, u-substitution needs to be
used.
see §4.2 for a basic list
of integrals
Using u-substitution with definite integrals is straightforward although there are
two strategies that can be followed. Both strategies involve deciding how to handle
the limits of integration.
IDEA: Limits of integration can either be (a) evaluated in terms of x or (b)
evaluated in terms of u.
In the first strategy, the limits of integration are left unchanged. The u-substitution
is performed, an antiderivative is found in terms of u and then the antiderivative is
rewritten in terms of x. Finally the antiderivative is evaluated at the limits (which
are written in terms of x) and the Fundamental Theorem of Calculus is applied.
leave limits of integration
in terms of x
In the second strategy, the limits of integration are changed so that they are written
in terms of u rather than x. Then after an antiderivative is found in terms of u, the
Fundamental Theorem of Calculus can be applied directly without rewriting the
antiderivative in terms of x.
write limits of integration
in terms of u
Both strategies are equivalent and will arrive at the same answer.
Checklist of Key Ideas:
definite integrals and u-substitution
evaluate limits of integration in terms of x
evaluate limits of integration in terms of u
88
Chapter 4 Sample Tests
Section 4.1
1. Use the rectangle method to approximate the area between
the function f (x) = 2x and the interval [0, 1] using 4 rectangles evaluated at the left endpoints.
(a) 1
1 1 1 1
+ + +
2 3 4 5
7. If the rectangle method is used to approximate the area under
the curve y = 5x + 2 on [2, 5] with n rectangles, then which
of the following choices for the location of rectangle heights
will give an exact answer?
(d)
(a) at left endpoints
(b) 0.75
(b) at right endpoints
(c) 1.25
(c) at midpoints
(d) 0.875
2. Use the rectangle method to approximate the area between
the function f (x) = 5 + x and the interval [0, 2] using 4 rectangles evaluated at the right endpoints.
(d) No rectangle method will give an exact answer.
8. The graph below shows the rectangle method being used to
calculate the area under the function y = f (x) on [a, b].
y
(a) 11.0
(b) 11.5
f (x)
(c) 12.0
(d) 12.5
3. Use the rectangle method
to approximate the area between
√
the function f (x) = 1 + x and the interval [0, 1] using 4 rectangles evaluated at the left endpoints.
a
(b) 1.219
(a) at left endpoints
(c) 1.250
(b) at right endpoints
(d) 1.270
(c) at midpoints
(d) It cannot be determined.
4. If f (x) = x2 + 3 is an antiderivative of g(x) and g(x) > 0 then
find the area beneath g(x) on the interval [2, 4].
(b) 12
(c) 4
Section 4.2
1.
(d) It cannot be determined.
Z
6. Which of the following represents an approximation of the
1
area under the function y = on the interval [1, 5] using the
x
rectangle method with 4 rectangles and right endpoints?
(c)
2 2 2 2
+ + +
3 5 7 9
x6 dx =
x5
5
x7
(b)
7
x7
(c)
6
x6
(d)
7
(a)
5. Answer true or false. If the rectangle method is used with left
endpoints to approximate the area beneath y = x5/4 on the
interval [2, 10], it will underestimate the actual area.
1 1 1 1
(a) 1 + + + +
2 3 4 5
1 1
(b) 2
+
3 5
x
Which of the following best describes the heights of the rectangles being used?
(a) 1.166
(a) 19
b
2.
Z
+C
+C
+C
+C
x2/5 dx =
5
+C
2x3/5
5
(b) x7/5 +C
7
(a)
89
x3
+1
3
x3
(b) y(x) =
3
x3
−1
(c) y(x) =
3
x3 − 1
(d) y(x) =
3
5
(c) − x7/5 +C
7
5
(d) − 3/5 +C
2x
Z
√
3
3.
x dx =
(a) y(x) =
3
+C
2x2/3
3
(b) x4/3 +C
4
3
(c) − x4/3 +C
4
3
(d) − 2/3 +C
x
(a)
4.
Z
Section 4.3
1.
Z
x−2 dx =
3
+C
x3
1
(b) − +C
x
1
(c) 3 +C
x
3
(d) 3 +C
x
Z
2.
1
2
3.
(c) − cos x +C
cos x
dx =
sin2 x
1
(a) − 3 +C
sin x
1
(b) −
+C
sin x
1
(c)
+C
sin x
1
+C
(d)
sin3 x
7. Answer true or false.
8. Answer true or false.
x5/4 +C
9. Answer true or false.
sin3 x
+C
3
3
cos x
(b)
+C
3
3
(c) cos x sin2 x +C
(d) 3 sin x +C
Z
(x + 9)5 dx =
(a) 5(x + 9) +C
2
5
x
(b)
+ 9 +C
2
(x + 9)6
(c)
+C
6
4
(x + 9)
(d)
+C
4
4. Z
Answer true or false.
√
2
4
x x + 2 dx = (x + 2)5/2 − (x + 2)3/2 +C
5
3
Z
(d) − 21 sin (x2 ) +C
Z
sin2 x cos x dx =
(a)
sin2 x +C
2
(b) − cos x2 +C
6.
Z
sin x dx =
(a)
(x2 − 5)31
+C
31
(x2 − 5)31
(b)
+C
29
2
31
(c) 31(x − 5) +C
(d) 29(x2 − 5)29 +C
(a)
(a) −
5.
2x(x2 − 5)30 dx =
5. Answer true or false. For
Z
Z
Z
x sin (x2 ) dx a good choice for u
is x2 .
sin x cos x dx = sin x cos x +C
√
√
√ x + 3 x + 4 x dx = x3/2 + x4/3 +
[sin x + cos x] dx = sin x − cos x +C
dy
10. Find y(x) given that
= x2 and y(0) = 1.
dx
Z
6. Answer true or false.
by letting u = x3 .
7. Answer true or false.
by letting u = x3 − 4.
8. Answer true or false.
ting u = sin x.
x
p
3x3 − 2 dx can be easily solved
Z
x2 (x3 − 4)2/5 dx can be easily solved
Z
sin3 x dx can be easily solved by let-
90
Section 4.4
5
(d)
i=2
5
1.
∑ k2 =
7. Which of the following expressions represents the sum
2 + 8 + 18 + 32 in sigma notation?
k=1
(a) 15
4
(a)
(b) 55
4
(b)
(d) 6
2.
∑3
j
∑ i2
i=1
(c) 26
4
∑
∑ 2i2
i=1
4
=
(c)
j=2
∑ 2i + 6
i=1
4
(a) 90
(d)
(b) 18
∑ 2(i + 6)
i=1
8. Answer true or false. 4 + 9 + 16 + 25 + 36 can be expressed
(c) 120
6
(d) 117
in sigma notation as
i=2
4
3.
∑ sin (kπ ) =
9. Answer true or false. The sum 8 + 27 + 64 can be expressed
3
k=1
in sigma notation
(a) 4
∑ i3
i=1
(b) 0
20
10.
(c) 2
∑ 10(i − 1) =
i=5
(d) 3
(a) 1900
4
4. Answer true or false.
(b) 1710
∑ (i + 1) = 14.
(c) 1984
i=1
(d) 1840
5. Which of the following expressions represents the sum
1 + 2 + 3 + 4 in sigma notation?
n
11. Answer true or false.
3
(a)
∑ i2 .
∑i
i=0
12. Answer true or false.
4
(b)
∑i
(c)
∑i
13. Answer true or false.
i=1
4
(d)
∑ 4−i
Section 4.5
i=1
6. Which of the following expressions represents the sum
1 + 4 + 9 + 16 + 25 in sigma notation?
1.
1
(b) 6
∑i
i=1
(c) 24
5
(b)
(c)
(d) 4
∑ i+1
i=0
5
∑i
i=1
x dx =
(a) 12
5
(a)
Z 5
2.
2
Z 4
0
3 dx =
(a) 3
=
∑ xi
i=1
!3
n
i=1
n
∑ 2ai = 2 ∑ ai
i=1
2
n
n
∑ (ai + bi ) = ∑ ai + ∑ bi
i=1
n
i=1
4
∑
i=1
n
xi3
i=1
i=1
91
(b) 12
12. Answer true or false.
a
(c) 6
3.
−5
13.
|x − 10| dx =
(d) 10
14.
4 − x2 dx =
x3 dx =
(x − 1) dx =
(a) 0.5
(b) −0.5
(b) 8π
(c) 1
(c) 4π
(d) 2π
−2
Z 1
0
(a) 0
f (x) dx = 2,
−1
(d) 9
Z 2 p
Z 3
Z 1
(c) 13.5
(c) 0
5. If
i=1
(b) 3
(b) −100
−2
∑ 3xi ∆xi where
max ∆x→0
lim
(a) 0
(a) 100
4.
k
3x dx =
the interval [a, b] is split into a partition of k subintervals.
(d) 24
Z 5
Z b
Z 3
f (x) dx = 7 then find
0
Z 0
−2
4 f (x) dx.
(d) −1
Z 2 p
5 + 4x − x2 dx =
15.
−1
(a) 3π
(a) 5
(b) 9π
(b) −5
(c) 9π /4
(c) −20
(d) 9π /2
(d) Cannot be determined.
6.
Z π /4
−π /4
sin x dx =
Section 4.6
(a) 0
(b) 0.134
1. Answer true or false.
2
(c) 0.707
2. Answer true or false.
(d) 1.414
then
1
8.
Z 4
1
Z 3
1
f (x) dx = −1 and
Z 3
1
[ f (x) + 3g(x)] dx = 5
Z 2
−2
2
sin x dx = cos x]π0
x4 dx =
(c) 32
(d) 64
4. Find the area under the curve y = x2 − 2 on [2, 3].
(b) 162.75
(c) 3
(a) 5
(d) 4.5
(b) 21
Z 5 2
x
0
11. Answer true or false.
3.
4
(b) 64/5
(a) 71.25
10. Answer true or false.
g(x) dx = 2
x2
x dx =
2
(a) 0
(x + x3 ) dx =
9. Answer true or false.
Z π
0
7. Answer true or false. If
Z 3
Z 4
Z 0
1+x
|x + 2| dx is negative.
−2
Z −1
−2
dx is positive.
dx
is negative.
x2
(c) 13/3
(d) 20/3
5. Find the area under the curve y = −(x − 3)(x + 2) and above
the x-axis.
(a) 125/6
92
(b) 13/6
(d) 1
(c) 125/3
4. Find the displacement of a particle between t = 0 and t = 3 if
v(t) = t 2 − 2.
(d) 5
6. Z
Use the Fundamental Theorem of Calculus to evaluate
2
1
3x−3/7 dx.
(a) 2.551
(d) 6
(c) −0.850
7. Answer true or false.
8. Answer true or false.
9. Answer true or false.
5. Find the total area between the curve y = x2 −4 and the x-axis
over the interval [0, 4].
Z 5
−5
Z 1
−1
Z 2
1
Z 8
x5 dx = 0.
|x|dx =
x2 dx =
Z 0
(a) 16/3
−1
Z 4
−x dx +
Z 1
x dx.
0
x2 dx.
3
x2 dx = x∗ 8 − (−8) when x∗ = 0.
Z x
d
11. Answer true or false.
t 3 dt = x3 .
dx 0
Z x
d
sint dt = cos x.
12. Answer true or false.
dx 0
10. Answer true or false.
−8
Section 4.7
1. Find the displacement of a particle between t = 0 and t = 2π
if v(t) = cost.
(a) 0
(b) 1
(c) 2
(d) 2π
2. Find the displacement of a particle between t = 0 and t = π /2
if v(t) = sint.
(a) 1
(b) 0
(c) 2
(d) 2π
3. Find the displacement of a particle between t = 0 and t = 1 if
v(t) = t 5 .
(a) 1/5
(b) −1/5
(c) 1/6
(b) 9
(c) 3/2
(b) 0.850
(d) −0.771
(a) 3
(b) 32/3
(c) 16
(d) 8
6. Answer true or false. The area between the curve y = x3 − 2
and the x-axis on [0, 2] is given by
Z 2
0
(x3 − 2) dx.
7. Answer true or false. If the velocity of a particle is v(t) = t 5
for −2 ≤ t ≤ 2, then the displacement of the particle over this
time interval is given by
Z 0
−2
−t 5 dt +
Z 2
t 5 dt.
0
8. Answer true or false. The area between y =
on [2, 3] is given by A = −
Z 3
1
2
x
1
and the x-axis
x
dx.
9. Answer true or false. The area between y = x4 + sin x and the
x-axis
son [0, 7] is given by
A=
Z 7
0
x4 + sin x
2
dx.
10. Answer true or false. The area between y =
on [−2, −1] is given by A = −
Z −1
1
−2
x3
1
and the x-axis
x3
dx.
11. Answer true or false. The area between y = x4 − x3 and the
x-axis on [1, 2] is given by A =
Z 2
1
x4 dx +
Z 2
x4 dx.
1
12. If the velocity of a particle is given by v(t) = 4 for [0, 2] then
the displacement of the particle over the same time interval is
(a) 0
(b) 4
(c) 8
(d) 2
93
−2
3π
(c) 1
3π
(d)
2
Section 4.8
(b)
1. The average value of y = 3x + 6 on the interval [−1, 4] is
(a) 15/2
12. If a particle has a velocity given by v(t) = t 2 − 1 for 0 ≤ t ≤ 8
then the average velocity of the particle on [0, 3] is
(b) 153/10
(c) 105/2
(a) 2
(d) 21/2
(b) 6
2. Answer true or false. If the average value of y = f (x) is 0 on
the interval [a, b] then
Z b
(c) 8
f (x) dx = 0.
(d) 24
a
3. Answer true or false. If the average value of y = f (x) is
greater than or equal to 0 on the interval [a, b] then f (x) ≥ 0
for a ≤ x ≤ b.
13. If a particle has a velocity of v(t) = t 2 − 1 for 0 ≤ t ≤ 6 then
the average speed of the particle on [0, 6] is
(a) 11
4. Answer true or false. If the average value of y = f (x) is equal
to 1 on the interval [a, b] then the average value of f (x)2 on
[a, b] is also equal to 1.
5. Answer true or false. If the average value of y = f (x)2 is
equal to 0 on the interval [a, b] then f (x) = 0 for a ≤ x ≤ b.
6. Answer true or false. If the average value of y = f (x) on the
interval [a, b] is equal to 0 then f (x) = 0 for a ≤ x ≤ b.
7. If fave = −3 on [0, 2] and fave = 6 on [2, 5] then fave on [0, 5]
is equal to
(b) 101/9
(c) 202/3
(d) 208/15
14. Suppose that a function y = f (x) is continuously differentiable on the interval [0, 4]. If the absolute maximum of y is
15 and the absolute minimum of y is −4 then which of the
following statements is true on the interval [0, 4]?
(a) fave = 11/2
(b) −4 ≤ fave ≤ 15
(a) 12
(c) fave = 11/8
(b) 12/5
(d) −1 ≤ fave ≤ 15/4
(c) 3/2
(d) 3/5
8. If fave = 5 on [0, 8] and fave = −2 on [0, 3] then fave on [3, 8]
is equal to
(a) 46/5
Section 4.9
1.
0
(b) 2, 604.167
(c) 10
(c) 2, 882
(d) 7
(a) 8
(x + 3)5 dx =
(a) 2, 482.67
(b) 34/5
9. If the average value of y = f (x)2 is 16 on the interval [0, 4]
then fave on [0, 4] is equal to which of the following?
Z 2
(d) 544
2. Z
Answer true or false.Z
tan3 x sec2 x dx = u3 du if u = tan x.
(b) 4 or −4
(c) 2
3. Z
Answer true or false.
(d) Cannot be determined.
4.
1
0
Z 1
0
10. Answer true or false. If f (1) = 2 and f (x) is increasing on
[0, 4] then fave on [1, 3] is greater than 2.
11. The average value of y = sin x on the interval [π /2, 2π ] is
(a) 0
(x + 4)(x − 2)15 dx =
(5x + 2)3 dx =
(a) 1.75
(b) 119.25
(c) 477
(d) 596.25
Z −1
−2
u15 du if u = x − 2.
94
5. Answer true or false.
6. Answer true or false.
7.
Z π /2
0
Z 2
−2
Z 4
−4
sin2 x dx = 2
4
x dx = 2
Z 2
Z 0
−4
5
sin2 x dx
5.
0
i=2
4
(a) 50
x dx.
(b) 4
2 sin (4x) dx =
(c) 86
(d) 100
(a) 2.359
5
(b) 2
6. Answer true or false. 3 − 6 + 9 − 12 + 15 =
(c) 0
(d) 1
8.
7. Answer true or false.
Z 1 √
0
∑ (i + 1)2 =
x x + 4 dx =
Z
2x(x2 + 6)5 dx =
8. Answer true or false. For
is u = x + 2.
(a) 3.53
4
(b) 1.08
9. Answer true or false.
(c) 0
Z
∑ (−1)i+1 3i
i=1
(x2 + 6)6
+C
6
√
x x + 2 dx, a good choice for u
4
∑ 3i = 3 ∑ i
i=1
i=1
k
2
10. lim ∑
=
n→∞
k=1 3
n
(d) 7.06
(a) 2/3
Chapter 4 Test
(b) 1/2
1. If f (x) = x then approximate the area below f (x) on [0, 4]
using the rectangle method with 4 rectangles. Use rectangles
with heights determined by the functions value at the left end
of the interval.
(c) 2
(d) 3/2
11.
6
(a) 60
(b) 8
(b) 52
(c) 10
(c) 5
(d) 20
2. Use the antiderivative method to find the area under y =
on [0, 1].
(a) 2/5
x2
12. Answer true or false.
Z 6
5
13.
(b) 1/3
Z 3
−3
5
x
dx is positive.
2+x
3
(x − x + 3x) dx =
(a) 0
(c) 2/3
(b) 114.75
(d) 1
3. Answer true or false.
4.
|x + 5| dx =
(a) 6
(d) 4
Z
Z 10
3 cos x dx =
Z
(c) 155.25
x9 dx = 9x8 +C
(d) 229.5
14. Find the displacement of a particle over the interval [0, 2] if
the velocity of the particle is given by v(t) = t 1/3 .
(a) 3 sin x +C
(a) 2
(b) −3 sin x +C
(b) 1.26
(c)
3
2
(c) 1.89
(d)
− 23
cos2 x +C
cos2 x +C
(d) 2.51
95
Chapter 4: Answers to Sample Tests
Section 4.1
1. b
2. d
3. a
4. b
5. true
6. d
7. c
8. a
2. b
10. a
3. b
4. b
5. c
6. b
7. false
8. false
2. a
3. c
4. true
5. true
6. false
7. true
8. false
2. d
10. d
3. b
11. false
4. true
12. true
5. b
13. true
6. c
7. b
8. true
2. b
10. false
3. a
11. false
4. d
12. true
5. c
13. a
6. a
14. b
7. true
15. c
8. a
2. false
10. true
3. b
11. true
4. c
12. false
5. a
6. a
7. true
8. true
2. a
10. true
3. c
11. false
4. a
12. c
5. c
6. false
7. false
8. false
2. true
10. true
3. false
11. b
4. false
12. a
5. true
13. b
6. false
14. b
7. b
8. a
2. true
3. false
4. b
5. true
6. true
7. c
8. b
2. b
10. c
3. false
11. b
4. a
12. true
5. c
13. a
6. true
14. c
7. true
8. true
Section 4.2
1. b
9. true
Section 4.3
1. a
Section 4.4
1. b
9. false
Section 4.5
1. a
9. true
Section 4.6
1. true
9. false
Section 4.7
1. a
9. false
Section 4.8
1. d
9. d
Section 4.9
1. a
Chapter 4 Test
1. a
9. true
96
Chapter 5: Applications of the Definite
Integral in Geometry, Science and
Engineering
Summary: This chapter focuses upon using the methods of evaluating definite
integrals and applying them in various problems. The first problem considered
is that of finding the area between two curves. This extends the idea of finding
the area underneath a curve or the total area between a function and an interval.
The next application is to find the volumes of various objects or solids. Three
basic methods are introduced: the method of slicing, the method of washers (or
disks) and the method of shells. The latter two methods may only be applied in the
case of a volume of revolution when a curve is revolved around a particular axis.
After that, finding the length of a curve or the arc length is discussed. Next the
surface area of a solid of revolution is investigated and then the average value of a
function. Towards the end of the chapter, two physical applications are discussed:
work and force from fluid pressure.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Use definite integrals to find the area between two curves (§5.1).
2. Use the method of slicing to find the volume of a solid (§5.2).
3. Use the method of disks/washers to find the volume of a solid of revolution
(§5.2).
4. Use the method of cylindrical shells to find the volume of a solid of revolution (§5.3).
5. Find the arc length of a plane curve (§5.4).
6. Find the surface area of a solid of revolution (§5.5).
7. Calculate work done by constant and variable forces over a distance (§5.6).
8. Find the center of gravity and centroid of a two dimensional region or thin
lamina (§5.7).
97
98
9. Calculate the force due to fluid pressure on objects submerged in liquid
(§5.8).
5.1 Area Between Two Curves
PURPOSE: To use definite integrals to calculate the area between two curves.
area problem
(see §4.1, 4.4 − 4.6)
Now that area can be described using definite integrals (from the previous chapter), the next step is to use definite integrals to describe the area between two
curves. The process is rather straightforward. In the case of finding the area under
a single function, this can be thought of as finding the areas of many infinitesmal rectangles between the function and the axis. Now, these rectangles will be
between the two curves. Then the general area function can be stated as
Z b
a
( f1 (x) − f2 (x)) dx
where f1 (x) > f2 (x).
IDEA: Finding the area between two curves is like adding together the areas of
many small rectangles that lie between a top function and a bottom function.
However, if the two functions trade places then their position in the formula needs
to change (i.e., if f2 > f1 then use
Z b
a
( f2 (x) − f1 (x)) dx). The function on the bot-
tom is always being subtracted from the one on the top. This may require breaking
the integral into several smaller integrals with different limits of integration and
slightly different integrands.
IDEA: Limits of integration may depend upon where the two functions intersect. Limits are chosen so that the “bottom” function is always being subtracted from the “top.”
Limits of integration will either be given or they must be found. It always must
be determined if the functions intersect. The points of intersection will help to
determine the limits of integration necessary on the integrals as well as helping to
determine which function is the top and which is the bottom.
Another way of thinking of the area between two functions is to define a new
function: h(x) = f1 (x) − f2 (x). For example, h(x) might represent the height of
one of the small rectangles between the two functions. Then the area between the
two functions can be written as
Z b
a
|h(x)| dx
The integrand, |h(x)| dx represents the area of one of the infinitesmal rectangles
and the integral is then summing together all of the rectangle areas. The absolute
99
value ensures that the function on the bottom is being subtracted from the function
on the top so that each rectangle has a positive height.
It is also a simple matter to change these formulas so that they are written with
respect to y instead of x. The functions used as integrands need to be written as
functions of y instead of x and the limits of integration need to be changed to y
values instead of x values. Otherwise, the process is the same even though visually
it may appear to be different.
vertical rectangles between f1 (x)
and f2 (x) → dx width
horizontal rectangles between g1 (y)
and g2 (y) → dy width
Checklist of Key Ideas:
area between two curves
calculating area using definite integrals; finding the limits of integration
area formulas with respect to x and/or y
5.2 Volumes by Slicing; Disks and Washers
PURPOSE: To calculate the volumes of solids oriented along a
particular axis.
Finding the volume of different solids is very closely related to the process of
finding the area under a curve or the area between two curves. In each case, the
idea is to find a way to slice the solid into smaller shapes that can be described
by a definite integral. The slices that are used will all be perpendicular to some
common axis.
IDEA: Each object studied here will be divided up into slices so that each slice
is perpendicular to either the x or y-axis.
The most general is the method of volume by slicing. In this case, either the
cross-section of each slice is given or can be found geometrically. Then the area
of the cross-section, A(x), is the integrand.
V=
Z b
volume by slicing
A(x) dx
a
The volume of each slice can be thought of as A(x) dx and then the integral is
simply adding together all of the slices.
The method of disks and the method of washers are more specific applications
of the method of slicing. If the solid is the result of revolving a curve about an axis
then the cross-sections will all be circular in shape. Then A(x) = π [ f1 (x)]2 (in
the method of disks) or A(x) = π [ f1 (x)]2 − [ f2 (x)]2 (in the method of washers).
Here the function values are the radius of each of the circles (i.e., the distance to
the x-axis).
disk and washers
100
IDEA: The method of disks is a specific case of the method of washers where
the inner function is zero, i.e., f2 (x) = 0.
The reader may notice that the method of disks is just a specific case of the method
of washers where the f2 (x) = 0. Also, both the disk and washer methods are directly related to the area problem from the previous section. If h1 (x) = π [ f1 (x)]2
and h2 (x) = π [ f2 (x)]2 then the method of washers is just the total area between
the functions h1 and h2 .
revolving about y-axis
All of the above descriptions are given in terms of revolving a function about the
x-axis. If a function is revolving about the y-axis then the integrals need to be
written with respect to y and the functions need to be functions of y. Also the
limits need to be written as values of y.
Here is a general summary of the method of washers (and disks if f2 or g2 is zero):
Revolving around x-axis:
V=
Z x=b
x=a
π [ f1 (x)]2 − [ f2 (x)]2 dx
if f1 ≥ f2 ≥ 0
Revolving around the line y = k (parallel to the x-axis):
V=
Z x=b
x=a
V=
Z x=b
x=a
π [ f1 (x) − k]2 − [ f2 (x) − k]2 dx
if f1 ≥ f2 ≥ k
π [k − f2 (x)]2 − [k − f1 (x)]2 dx
if k ≥ f1 ≥ f2
Revolving around y-axis:
V=
Z y=d
y=c
π [g1 (y)]2 − [g2 (y)]2 dy
if g1 ≥ g2 ≥ 0
Revolving around the linex = k (parallel to y-axis):
V=
Z y=d
y=c
V=
Z y=d
y=c
π [g1 (y) − k]2 − [g2 (y) − k]2 dy
if g1 ≥ g2 ≥ k
π [k − g2 (y)]2 − [k − g1 (y)]2 dy
if k ≥ g1 ≥ g2
Checklist of Key Ideas:
volume by slicing; general volume formulas (with respect to x and/or y)
volume of a right cylinder; volume of a disk
axis perpendicular to disks
solids of revolution
method of disks; method of washers
101
5.3 Volumes by Cylindrical Shells
PURPOSE: To use integrals to compute the volume of a solid
using cylindrical shells.
The method of disks and shells are designed specifically for the situation where
a solid is obtained by a revolution of a curve about a particular axis (usually the
x- or y-axis or some line that is parallel to one of them). The method of shells
is based upon the volume of a right circular cylinder. The volume of one shell is
given by
method of shells
shell volume = 2π (x − a) f (x) dx
where the shell has a center along the line at x = a. Then the volume formula will
be given by
V=
Z x=b
x=a
2π (x − a) f (x) dx
Notice that in this example the solid is revolved about a line x = a which is parallel
to the y-axis.
IDEA: Washers and shells require integration with respect to the opposite variable. In some cases this may make one method easier than the other.
Using the method of washers in this case would have used integration with respect to y. Because of this difference, it may be easier to apply the method of
shells than the method of washers in some cases. Another difference between the
method of washers (or disks) and shells is that the method of washers requires that
the function be squared while the method of shells requires that the function be
multiplied by a distance (usually x, for example).
Checklist of Key Ideas:
volume of a cylindrical shell; difference between shells and disks
method of cylindrical shells
volume formula about the x- or y-axis
units of the integral
5.4 Length of a Plane Curve
PURPOSE: To use integrals to find the length of a piece of a
curve.
There is only one idea that is introduced in this section: how to find the length of
a piece of a curve. The formula that is used is straightforward and requires that
washers vs. shells
102
some integrals may require
numerical methods (see §4.4, 7.7) or
CAS (see §7.6)
the derivative of the function be obtained. Often the integral that results cannot be
evaluated without special means such as a numerical method or CAS. The fact
that the derivative is required indicates when this formula may be applied. The
function in question must have a derivative over the entire length of the curve or
else the formula may return faulty values.
IDEA: To calculate arc length requires that the appropriate derivatives of the
curve can be calculated.
Three arc length formulas
1. with respect to x, y = f (x)
2. with respect to y, x = g(y)
3. parametrically
§10.1)
(see
also
Three formulas are given for finding the length of a curve. One is with respect to
x and one is with respect to y. Arc length can also be calculated parametrically.
This will also be discussed later in Section 10.1.
Z bq
1 + (dy/dx)2 dx
with respect to x
a
Z dq
1 + (dx/dy)2 dy
with respect to x
(dx/dt)2 + (dy/dt)2 dt
parametrically
c
Z bq
a
In some cases, writing the length with a different integral can be the difference
between getting an exact answer and having to evaluate the definite integral numerically.
Checklist of Key Ideas:
arc length problem
smooth curves/functions
formula for arc length (with respect to x and/or y)
arc length formula for curves defined parametrically
units of the integral
5.5 Area of a Surface of Revolution
PURPOSE: To use integrals to compute the area of a surface of
revolution.
After discussing solids of revolution and the length of an arc, the area of a surface
of revolution can be discussed. There is only one formula that is introduced in this
section: finding the surface area of a solid of revolution.
frustum
slant height
The area formula is based upon the surface area of a frustum of slant height
l: area = π (r1 + r2 )l. The slant
p height of the frustum can be found using the
incremental arc length of l = 1 + [ f ′ (x)]2 dx.
103
IDEA: The slant height of the frustum is found using the arc length ideas from
the previous section. This means that the appropriate derivative needs to
exist.
The radius of the top and bottom of the frustum are approximately the same and
so the area becomes
q
area = π (2 f (x)) 1 + [ f ′ (x)]2 dx
Then summing these areas, the surface area can be written as
A=
Z x=b
x=a
q
2π f (x) 1 + [ f ′ (x)]2 dx
Often this formula cannot be evaluated directly or will need u-substitution.
numerical methods (see §4.4, 7.7)
CAS (see §7.6)
Checklist of Key Ideas:
surface area problem
surface of revolution
formula for finding area of a surface of revolution
5.6 Work
PURPOSE: To use integrals to compute the work done by a
force.
The units of work are energy and work is done by applying a force over a distance.
The units of energy (and work) can be found by multiplying together the units of
force and the units of distance.
units of work
IDEA: Three types of force:
1. constant force
2. variable force
3. many cases of constant force
In this section, three types of work are considered: with a constant force, with a
variable force, and adding together many cases where the force is constant. When
the force is constant, then the work done is simply the product of the force, F,
and the distance, d, over which the force is applied.
constant force
W = F ·d
When the force is variable, then the work done is the integral of the force, F(x):
variable force
104
work =
Z x=b
F(x) dx
x=a
Notice that the units of F(x) dx will be force times length which should equal
energy.
IDEA: Two special types of work:
1. work done by springs
2. work of pumping a liquid
work and springs
pumping liquids
lifting=applying a force
Two special types of work are considered in this section: work done on a spring
and work done pumping liquid. For springs, the force that is applied is a variable
force that is proportional to the amount that the spring is compressed or stretched,
F(x) = kx, where k is the spring constant.
For pumping liquids, the total work can be thought of as many separate instances
of work being done with a constant force that are then added together. This is
very similar to the volume by slicing method introduced earlier in this chapter
(see §5.2). A volume of liquid can be divided into thin layers of liquid that have a
cross-sectional area of A(x) and a thickness of dx so that each layer has a volume
of A(x) dx. Then multiplying the volume of each layer by its weight density will
give the weight of the layer. The process of pumping is then to lift each layer of
liquid a certain distance.
IDEA: Thinking of the liquid as many different layers is similar to finding the
volume of an object by slicing. In this problem, the slices or layers are also
being moved some distance. So two key ideas are
1. the size of each “layer” of liquid
2. how far the “layer” is being moved
For both pumping and spring problems, care should be taken when choosing what
the x-axis represents. In spring problems, x = 0 represents when the spring is at its
natural length. Then x = 1 would represent when the spring is stretched beyond,
or compressed from, its natural length by one unit.
IDEA: Picking an x-axis can usually be done in more than one way. The
important thing is to be consistent throughout the problem.
In pumping problems, one convention (as used by this book) is to have the positive
x direction to be in the downward direction. Where x = 0 is located is arbitrary
and is often left up to the reader. Usually x = 0 is chosen to be in a position that
makes other pieces of the integrand to be relatively simple expressions. However
the pumping is done, the lower limit of integration should be less than the upper
limit of integration and the distance that each layer of liquid is pumped should be
positive. The result should be a positive amount of work done. Also, the limits of
integration are chosen based upon where the liquid is located on the x-axis, not on
where the liquid is being pumped to.
105
IDEA: Be careful with the distance and the limits of integration. The overall
work in lifting liquid should be a positive amount of energy.
Lastly, work is equivalent to the change in kinetic energy of an object. This is
shown by the equation
work is change in
kinetic energy
1
1
W = mv2f − mv2i
2
2
where v f is the final velocity of an object and vi is the initial velocity of an object. When calculating work based upon change in velocity, no integrals have to
be evaluated. The work may be calculated by evaluating the above expression
directly.
Checklist of Key Ideas:
work-energy relationship; kinetic energy
work of a constant force over a distance
work of a variable force over a distance
springs and spring constants
pumping liquid
5.7 Moments, Centers of Gravity, and Centroids
PURPOSE: To find the center of gravity of an inhomogeneous
lamina or the centroid of a homogeneous lamina.
This section discusses how to find the center of gravity of a thin, two-dimensional
region called a lamina. Since this region is two-dimensional we will describe the
region using equations in the xy-plane. The lamina may have some non-constant
density function, δ (x), in which case the lamina is said to be inhomogeneous.
Otherwise the lamina is called homogeneous. The center of gravity is the point
in the region upon which the region can be balanced. In other words, the force
due to gravity on the region can be seen as a single force that acts on the region
through its center of gravity. When a lamina is homogeneous, the center of gravity
is called the centroid.
lamina
inhomogeneous
homogeneous
center of gravity
centroid
IDEA: The integrals for M , Mx and My are required to calculate the center of
gravity or centroid.
x̄ =
My
Mx
and ȳ =
M
M
Three integral formulas are necessary to calculate the center of gravity of a lamina:
the moment about the x-axis (Mx ), the moment about the y-axis (My ), and the
M, mass
Mx , moment about x-axis
My , moment about y-axis
106
mass (M) of the region. The integrals required to calculate these are very similar
to integrals used in other applications in this chapter.
IDEA: Setting up the integral for mass, M , of a laminar region R is very similar
to setting up the integral for finding the area of the region.
For example, finding the mass of a region is very similar to finding the area of a
region. Finding the moments about the x-axis and y-axis are very similar to the
integrals that arise when finding the volume of a solid of revolution.
IDEA: Setting up the integrals for Mx and My are very similar to the integrals
that are used to find volumes for solids of revolution about the x-axis and y-axis
respectively.
Suppose that a region R is bounded by the curves y = f (x) at the bottom, y = g(x)
at the top, and x = a and x = b at the left and right. When integrating with respect
to x, the area of this region is given by the following integral
Z b
a
(g(x) − f (x)) dx.
Calculating the mass simply inserts the density, δ , into this integral:
M=
Z b
a
δ · (g(x) − f (x)) dx.
Also, the moment about the x-axis,
Mx =
1
2
Z b
a
δ · (g(x)2 − f (x)2 ) dx,
looks remarkably similar to integration the washer method when it is used to find
the volume of the solid of revolution about the x-axis. Likewise, the moment about
the y-axis,
My =
Z b
a
x · δ · (g(x) − f (x)) dx
looks much like the volume when revolving about the y-axis if the shell method
is used. If integration with respect to y is used, then Mx will look like the shell
method being used around the x-axis and My will resemble the washer method
being used around the y-axis.
centroids depends only
upon the shape of lamina
The centroid is found in essentially the same way as the center of gravity. There
will be no real difference in the calculation other than the density of the region
will be constant. This will cause the density to divide out of the calculation of x̄
and ȳ. Since the centroid is then independent of the density, the location of the
centroid is considered to be a geometric property of the region. That is, only the
shape of the region has an impact upon the location of the centroid.
IDEA: The Theorem of Pappus allows the centroid of a region to be used as a
convenient tool for finding the volume of a solid of revolution.
volume = (circumference travelled by centroid) × (area of region)
Theorem of Pappus
solids of revolution
A useful application of the centroid is found in calculating the volume of a solid of
revolution. The Theorem of Pappus essentially says that the volume is equal to
107
the area of the region times the distance that the centroid travels when the region
is revolved. To find the distance that the centroid travels, imagine its path as a
circle. The distance it is revolved will be equal to the circumference of the circle.
The radius of this circle is the straight distance from the centroid to the line being
revolved around. Any line can be considered, not just vertical and horizontal lines.
Checklist of Key Ideas:
inhomogeneous lamina
homogeneous lamina
moments about the x-axis and y-axis
center of gravity
centroid
mass of a lamina
Theorem of Pappus
5.8 Fluid Pressure and Force
PURPOSE: To use integrals to calculate the total force that is
applied by a fluid over an area.
The idea presented in this section is force applied over an area and finding the
total force that is applied by a fluid on a submerged object or surface. As in the
previous section, the units that are being used can help to straighten out how to
write the integrals involved.
units
The basic formula being used is the relationship between force, F, pressure, P,
and area A.
F = PA
or
P=
F
A
Thus the units of pressure, P, are force per area. In the SI system this is typically
given as Pascals or Pa. In the BE system, this is usually given as pounds per square
inch or PSI.
The main problem considered here is how to calculate the force on a submerged
surface whether it is horizontal, vertical or inclined at some angle (see the exercises). The direction of the force on the surface is known to be perpendicular to
the surface in question.
If it is a horizontal surface then the force is just equal to the pressure times the
area. The pressure is shown to be the weight density, ρ , times the distance, h, at
which the surface is submerged. Then the force is given as follows.
F = ρ hA
horizontal surface
108
vertical surface
If the surface is vertical then the force is calculated using an integral. The x-axis
is taken to be positive in the downward direction. x = 0 does not necessarily need
to be at the surface of the liquid. The submerged surface is divided into horizontal
strips which have a thickness of dx and a width given by w(x) giving each horizontal strip an area of w(x) dx. Each strip has a depth of h(x) below the surface
of the liquid. Then each strip will experience a force of ρ h(x)w(x) dx. This is
integrated with limits of integration based upon where the submerged surface is
on the x-axis.
Z x=b
ρ h(x)w(x) dx
x=a
Here x = a is at the top of the surface and x = b is at the bottom of the submerged
surface. As a check, the units of ρ h(x)w(x) dx should be units of force.
Checklist of Key Ideas:
definition of pressure
fluid density; mass/weight density
fluid pressure on horizontal or vertical surface
formula for finding fluid force
units of the integral
109
Chapter 5 Sample Tests
Section 5.1
1. Find the area of the region enclosed by the curves y = x2 and
y = −x by integrating with respect to x.
(a) 1/6
(a) 130.5
(b) 120.75
(b) 1
(c) 140.5
(c) 1/4
(d) 125.25
(d) 1/16
√
(y − 3 y) dy
√
3. Find the area enclosed by the curves y = −x5 , y = − 3 x, x = 0
and x = 1/2.
2. Answer true or false.
8. Use a graphing utility to find the area enclosed by the curves
y = −x5 , y = x2 , x = 0 and x = 3.
R2
0
(8x − x3 ) dx =
R2
0
9. Use a graphing
√ utility to find the area enclosed by the curves
x = 2y4 , x = 4y.
(a) 2
(a) 0.295
(b) 0.315
(b) 1
(c) 0.273
(c) 0.76
(d) 0.279
(d) 0.93
4. Find the area enclosed by the curves y = sin (3x), y = 2x,
x = 0 and x = π .
(a) 4.27
11. Answer true or false. The curves x = y2 + 3 and x = 11y intersect at y = 3 and y = 6.
(b) 2.38
(c) 9.32
(d) 10.68
5. Find the area between the curves y = |x − 2|, y =
x
+ 2.
2
(a) 3.0240
(b) 12.000
(c) 3.0251
(d) 3.0262
6. Find the area between the curves x = 2|y|, x = −2y + 4 and
y = 0.
(a) 1
(b) 2
(c) 0.5
(d) 0.3
7. Use a graphing utility to find the area of the region enclosed
by the curves y = x3 − 2x2 + 5x + 2, y = 0, x = 0 and x = 2.
(a) 34/3
(b) 37/3
10. Answer true or false. The curves y = x2 + 5 and y = 6x intersect at x = 1 and x = 2.
12. Answer true or false. The curves y = cos (x) − 1 and y = x2
intersect at x = 0 and x = π .
13. Answer true or false. The curves y = 2 sin (π x/2) and y = 2x3
intersect at x = 0 and x = 1.
14. Find a vertical line x = k that divides the area enclosed by
√
y = − x, y = 0 and x = 4 into two equal areas.
(a) k = 4
(b) k = 42/3
(c) k = 43/2
(d) k = 2
15. Approximate the area of the region that lies below
y = 3 cos (x/3) and above y = 0.3x, where 0 ≤ x ≤ π .
(a) 6.314
(b) 2.442
(c) 38/3
(c) 0.558
(d) 40/3
(d) 1.118
110
Section 5.2
(a) 56.5
1. Use the method of disks to find the volume of the solid that results by revolving the region enclosed by the curves y = −x3 ,
x = −4, x = 0 and y = 0 about the x-axis (round to the nearest
whole number).
(a) 7, 353
(c) 15.3
(d) 18.0
8. Use the method of disks to find the volume of the solid that
results by revolving the region enclosed by the curves x = y2 ,
x = −y + 6 about the y-axis. The approximate volume is
(b) 3, 677
(c) 14, 706
(a) 20.83
(d) 46, 201
2. Use the method of disks to find the volume of the solid
that √
results by revolving the region enclosed by the curves
y = cos x, x = 0, x = π /2 and y = 0 about the x-axis.
(b) 65.45
(c) 523.60
(d) 1, 028.08
(a) π /4
(b) 2π
9. Answer true or false. The volume of the solid that results
when the region enclosed by the curves x = y6 and x = y8 is
(c) π /2
revolved about the y-axis is given by
(d) π
Z 1
0
3. Use the method of disks to find the volume of the solid
that results by revolving the region enclosed by the curves
y = x2 − 4, y = 0, and x = 0 about the y-axis (round to the
nearest whole number).
(a) 17
(y6 − y8 )2 dy.
10. Find the volume of the solid whose base is enclosed by the
circle (x − 2)2 + (y + 3)2 = 9 and whose cross sections taken
perpendicular to the base are semicircles. The approximate
volume is
(a) 113.10
(b) 54
(b) 355.31
(c) 64
(c) 56.5
(d) 201
4. Answer true or false. The volume of the solid that results
when the region enclosed by the curves y = x10 , y = 0, x = 0
and x = 2 is revolved about the x-axis is given by
(b) 48.2
Z 2
0
π x20 dx.
5. Answer true or false. The volume of the solid that results
√
when the region enclosed by the curves y = 3 x, y = 0,
x = 0 and x = 3 is revolved about the x-axis is given by
Z 3
2
√
π 3 x dx .
0
6. Use the method of disks to find the volume of the solid that results by revolving the region enclosed by the curves y = −x5 ,
x = 0, y = −1 about the y-axis. The approximate volume is
(a) 0.83
(b) 2.62
(c) 8.23
(d) 2.24
7. Use the method of disks to find the volume of the solid
that √
results by revolving the region enclosed by the curves
x = 4y + 16, x = 0, and y = −1 about the y-axis. The approximate volume is
(d) 117.65
11. Answer true or false A right-circular cylinder of radius 8 cm
contains a hollow sphere of radius 4 cm. If the cylinder is
filled to a height of h cm with water and the sphere floats so
that its highest point is 1 cm above the water level, there is
16π h − 8π /3 cm3 of water in the cylinder.
12. Use the method of disks to find the volume of the solid
that results by revolving the region enclosed by the curves
y = cos6 x, x = 2π , and x = 5π /2 about the x-axis. The approximate volume is
(a) 0.35
(b) 1.11
(c) 0.49
(d) 0.76
13. Answer true or false. The volume of the solid that results
when the region enclosed by the curves y = x2 and x = y
is revolved about x = 1 is V = 0.133 (rounded to 3 decimal
places).
111
Section 5.3
1. Use cylindrical shells to find the volume of the solid when
the region enclosed by y = x2 , x = −2, x = −1 and y = 0 is
revolved about the y-axis.
15π 2
4
15π
(b)
4
15π
(c)
8
15π
(d)
2
2. Use cylindrical shells to find
√ the volume of the solid when
the region enclosed by y = −x, x = 0, x = −1 and y = 0 is
revolved about the y-axis.
(a)
(a) 0.4π
(b) 0.8π
(c) 0.2π
(d) 0.2π 2
3. Use cylindrical shells to find the volume of the solid when
the region enclosed by y = −x−3 , x = 1, x = 2 and y = 0 is
revolved about the y-axis.
(a) π 2
(b) 0.5π
(c) π
(d) 0.05π 2
4. Use cylindrical shells to find the volume of the solid when the
region enclosed by y = − sin (−x2 ), y = 0, x = 0 and x = 1 is
revolved about the y-axis. The approximate volume is
(a) 0.560π
(b) 0.520π
(c) 0.460π
(d) 0.500π
5. Use cylindrical shells to find the volume of the solid when
the region enclosed by y = 4 − x, y = x, and y = 0 is revolved
about the y-axis.
(a) 80π
(b) 32π
(c) 16π
(d) 8π
6. Use cylindrical shells to find the volume of the solid when the
region enclosed by y = 2x2 − 6x and y = 0 is revolved about
the y-axis.
27π
2
(b) 27π
27π
(c)
8
27π
(d)
4
7. Use cylindrical shells to find the volume of the solid when
the region enclosed by x = −y2 , x = 0 and y = −2 is revolved
about the x-axis.
(a)
(a) 4
(b) 4π
(c) 8
(d) 8π
8. Use cylindrical shells to find
√ the volume of the solid when
the region enclosed by y = 3 8x, x = 1 and y = 0 is revolved
about the x-axis.
19π
(a)
20
12π
(b)
5
10π
(c)
3
8π
(d)
5
9. Use cylindrical shells to find the volume of the solid when the
region enclosed by xy = 7 and x + y = −6 is revolved about
the x-axis. The approximate volume is
(a) 1.32π
(b) 16.4π
(c) 7.54π
(d) 65.6π
10. Use cylindrical shells to find the volume of the solid when
the region enclosed by y = −x2 , x = 1, x = 2 and y = 0 is
revolved about the line x = 1.
17π
(a)
6
15π
(b)
2
5π
(c)
6
14π
(d)
3
11. Use cylindrical shells to find the volume of the solid when
the region enclosed by y = x2 , x = 0, x = −2 and y = 0 is
revolved about the line x = 1.
(a) 8π
112
40π
3
8π
(c)
3
16π
(d)
3
(b)
12. Answer true or false. The volume resulting
√ from revolving
the region enclosed by the semicircle y = 16 − x2 about the
32π
.
x-axis is
3
6. Use a CAS or a calculator with integration capabilities to
approximate the arc length of the curve x = sin (−3y) from
y = 0 to y = π .
(a)
(b)
(c)
(d)
2.042
6.987
2.051
2.916
7. Answer true or false. The arc length of y = x cos x from x = 0
to x = π can be approximated by a CAS or a calculator with
integration capabilities to be 4.698.
Section 5.4
1. Find the arc length of the curve y = −2x3/2 from x = 0 to
x = 3. The approximate arc length is
(a) 10.9
(b) 10.9π
(c) 6.8
(d) 6.8π
1
2. Find the arc length of the curve y = (x2 + 3)3/2 from x = 0
2
to x = 2. The approximate arc length is
(a) 7.17
(b) 14.34
(c) 28.68
(d) 51.96
3. Answer true or false.
The arc length of the curve
y = (x − 2)5/2 from x = 0 to x = 5 is given by
Z 5q
1 + (x − 2)5 dx.
0
1
4. The arc length of the curve x = (y2 + 4)3/2 from y = −1 to
6
y = 0 is
(a) 5/6
(b) 7/6
(c) 5/3
(d) 4
5. Use a CAS or a calculator with integration capabilities to approximate the arc length of the curve y = sin (−x) from x = 0
to x = π /2.
(a) 1.43
(b) 1.74
(c) 1.86
(d) 1.91
Section 5.5
1. Find the area of the surface generated by revolving y = −2x,
0 ≤ x ≤ 1 about the x-axis. The approximate surface area is
(a)
(b)
(c)
(d)
4.47
14.05
28.10
88.28
2. Find√ the area of the surface generated by revolving
y = 1 + x, −1 ≤ x ≤ 0 about the x-axis. The approximate
surface area is
(a)
(b)
(c)
(d)
4.47
28.07
5.33
7.02
3. Find√ the area of the surface generated by revolving
x = y + 1, 0 ≤ y ≤ 1 about the y-axis. The approximate surface area is
(a)
(b)
(c)
(d)
67.88
3.44
8.28
21.60
4. Answer true or√false. The area of the surface generated
revolving
x = 3y, 1 ≤ y ≤ 5 about the y-axis is given
Z 5
3
dy.
2π y 1 + √
1
4 3x
5. Answer true or false. the area of the surface generated
revolving x = sin y, 0 ≤ y ≤ π about the y-axis is given
Z
p
π
2π y 1 − cos2 x dx.
by
by
by
by
0
6. Use a CAS or a scientific calculator with numerical integration capabilities to approximate the area of the surface generated by revolving the curve xy = 1, 1 ≤ y ≤ 2 about the xaxis.
(a) 5.016
113
(b) 5.394
(a) 7, 225.7 J
(c) 7.678
(b) 7, 125.8 J
(d) 10.502
7. Answer true or false. A CAS or a calculator with numerical integration capabilities can be used to approximate the
area of the surface generated by revolving the curve y = cos x,
0 ≤ x ≤ π /2 about the x-axis to be 1.
8. Answer true or false. A CAS or a calculator with numerical integration capabilities can be used to approximate the
area of the surface generated by revolving the curve y = sin x,
0 ≤ x ≤ π /2 about the x-axis to be 1.
9. Answer true or false. A CAS or a calculator with numerical integration capabilities can be used to approximate the
area of the surface generated by revolving the curve x = cos y,
0 ≤ y ≤ π /2 about the y-axis to be 8.08.
Section 5.6
1. Find the work done when a constant force of 20 lb in the positive x direction moves an object from x = 3 to x = 4 ft.
(c) 7, 334.1 J
(d) 7, 310.2 J
6. Answer true or false. The amount of work needed to pump a
liquid of density 0.95 kg/m3 from a spherical tank of radius
4 m is
Z 8
0
0.95(8 − x)π x2 dx.
7. An object in deep space is initially considered to be stationary. If a force of 250 N acts on the object over a distance of
400 m, how much work is done on the object?
(a) 0 J
(b) 100, 000 J
(c) 50, 000 J
(d) 25, 000 J
3
N
x2
in the positive x-direction moves an object from x = 2 m to
x = 8 m.
8. Find the work done when a variable force of F(x) =
(a) 5.64 J
(a) 20 ft-lb
(b) 4.50 J
(b) 140 ft-lb
(c) 2.25 J
(c) 40 ft-lb
(d) 1.13 J
(d) 100 ft-lb
2. A spring whose natural length is 35 cm is stretched to a length
of 40 cm by a 2 N force. Find the work done in stretching the
spring.
1
N
x2
in the positive x-direction moves an object from x = −5 m to
x = −4 m.
9. Find the work done when a variable force of F(x) =
(a) 0.113 J
(a) 0.05 J
(b) 1.00 J
(b) 0.4 J
(c) 0.05 J
(c) 0.7 J
(d) 0.25 J
(d) 0.3 J
3. Assuming that 20 J of work stretches a spring from its natural
length of 60 cm to a length of 64 cm, find the spring constant
in N/cm.
10. Find the work done when a variable force of F(x) = 30x N
in the positive x-direction moves an object from x = −4 m to
x = 0 m.
(a) 240 J
(a) 4.03
(b) 320 J
(b) 8.06
(c) 80 J
(c) 125
(d) 160 J
(d) 250
4. Answer true or false. Assume a spring is stretched from
100 cm to 140 cm by a force of 500 N. The work needed
to do this is 200 J.
5. A cylindrical tank of radius 5 m and height 10 m is filled
with a liquid whose density is 1.84 kg/m3 . How much work
is needed to pump the liquid out of the tank?
11. If the Coulomb force is proportional to x−2 , the work it does
is proportional to
(a) x−1
(b) x−3
(c) x
(d) x−2
114
12. Answer true or false. It takes the same amount of work to
move an object from 100, 000 km above the earth to 200, 000
km above the earth as it does to move the abject from 200, 000
km above the earth to 300, 000 km above the earth.
13. Answer true or false. It takes twice as much work to elevate
an object to 120 m above the earth as it does to elevate the
same object 60 m above the earth.
14. Answer true or false. It takes twice as much work to stretch a
spring 100 cm as it does to stretch the same spring 50 cm.
15. A 1 kg object is moving at 10.0 m/s. If a force in the direction
of motion does 40.0 J of work on the object, then what is the
object’s final speed?
(a) 13.4 m/s
(b) 5.5 m/s
(c) 5.0 m/s
(d) 11 m/s
a5
(6 + 5a)
60
4. A triangular region in the first quadrant is bounded by the xaxis, y-axis, and the line between the points (0, a) and (b, 0)
where a, b > 0. Find the centroid of this region.
ab ab
(a)
,
2 2
b a
,
(b)
2 2
2
a b ab2
(c)
,
6
6
b a
,
(d)
3 3
(d)
5. Consider the region R in the first quadrant that is bounded by
y = 4 − x2 with a constant density δ . If the centroid of the
region is located at (x̄, ȳ), then which of the following statements is true?
(a) M = 38 δ
Section 5.7
1. Consider the lamina bounded by the curves y = x2 , x = a
(with a > 0) in the first quadrant. If δ = x + 1 then find an
expression for the mass of the lamina in terms of a.
1
(a) a3
3
a3 a4
(b)
+
3
4
a4
(1 + (4/5)a)
(c)
4
a5
(6 + 5a)
(d)
60
2. For the lamina in problem #1 find Mx in terms of a.
1 3
a
3
a3 a4
+
(b)
3
4
a4
(c)
(1 + (4/5)a)
4
a5
(d)
(6 + 5a)
60
3. For the lamina in problem #1 find My in terms of a.
(c) x̄ =
(d) Mx = 4δ
6. Consider the region in the first quadrant that is bounded by
the circle x2 + y2 = r2 with a density of δ = 1. Which of the
following statements is true?
(a) M = π r2
(b) My = π r/2
(c) Mx = My
(d) The centroid of the region cannot be found.
7. Consider the lamina bounded by the curves x = y2 − 4 and
y = x/3 with a constant density δ . Which of the following
expressions represents the mass of the region?
(a)
1 3
a
3
a3 a4
(b)
+
3
4
4
a
(1 + (4/5)a)
(c)
4
(a)
8
5
128
15
(b) ȳ =
(a)
(b)
(c)
Z 12
−3
Z 4
−1
Z 4
−1
√
δ · ( x + 4 − x/3) dx
δ · (3y − y2 + 4) dy
δ · y · (3y − y2 + 4) dy
(d) None of the above
8. For the region described in Problem 7, which of the the following expressions represents the moment about the x-axis,
Mx .
(a)
(b)
Z 4
δ · (3y − y2 + 4) dy
−1
Z 12
−4
√
δ · x · ( x + 4 − x/3) dx
115
(c)
Z 12 δ
−4
2
x+4−
x2
9
(d) None of the above
dx
9. Answer true or false. For the region described in Problem 7,
the x-coordinate of the centroid of the region is negative.
√
10. Find the centroid of the region bounded by the curves y = x,
y = 1, and x = 4.
(a) (2.94, 1.35)
(b) (1.35, 2.94)
(c) (2.25, 4.9)
(d) (4.9, 2.25)
11. The quadrilateral ABCD region has corners at the points
A(1, 5), B(8, 4), C(6, 0), D(2, 2). Find the centroid of the region.
(a) (9/2, 17/6)
(b) (9/2, 5/2)
(c) (44/9, 5/2)
(d) (17/4, 11/4)
Section 5.8
1. A flat rectangular plate is submerged horizontally in water to
a depth of 6.0 ft. If the top surface of the plate has an area of
50 ft2 , and the liquid in which it is submerged is water, then
find the force on the top of the plate. Neglect the effect of the
atmosphere above the liquid. (The density of water is 62.4
lb/ft3 .
(a) 300 lb
(b) 33.4 lb
(c) 2, 080 lb
(d) 18, 720 lb
2. Find the force (in N) on the top of a submerged object if its
surface is 10.0 m2 and the pressure acting on it is 3.2 × 105
Pa. Neglect the effect of the atmosphere above the liquid.
(a) 3.7 × 104 N
(b) 3.2 × 106 N
(c) 6.4 × 106 N
(d) 1.7 × 106 N
3. Find the force on a 100 ft wide by 5 ft deep wall of a swimming pool filled with water. Neglect the effect of the atmosphere above the liquid. (The density of water is 62.4 lb/ft3 .)
(a) 78, 000 lb
(b) 124, 800 lb
(c) 62, 400 lb
(d) 624 lb
4. Answer true or false. The force a liquid of density ρ exerts
on an equilateral triangle with edges h in length submerged
Z h
ρ 2
x dx.
point down is given by
0 3
5. A right triangle is submerged vertically with one side at the
surface in a liquid of density ρ . The triangle has a leg that is
20 m long located at the surface and a leg 10 m long straight
down. Find the force exerted on the triangular surface, in
terms of density. Neglect the effect of the atmosphere above
the liquid.
(a)
(b)
(c)
(d)
677ρ
500ρ
600ρ
200ρ
N
N
N
N
6. Answer true or false. A glass circular window on the side of
a submarine has the same force acting on the top half as on
the bottom half.
7. Find the force on a 30 ft2 horizontal surface 20 ft deep in
water. Neglect the effect of the atmosphere above the liquid.
(The density of water is 62.4 lb/ft3 .)
(a)
(b)
(c)
(d)
600 lb
37, 440 lb
1, 200 lb
30, 000 lb
8. Answer true or false. A flat sheet of material is submerged
vertically in water. The force acting on each side must be the
same.
9. Answer true or false. If a submerged horizontal object is elevated to half its original depth, the force exerted on the top
of the object will be half the force originally exerted on the
object. Assume there is a vacuum above the liquid surface.
10. Answer true or false. If a square, flat surface is suspended
vertically in water and its center is 20 m deep, the force on
the object will double if the object is relocated to a depth of
40 m. Neglect the effect of the atmosphere above the liquid.
11. Answer true or false. The force on r
a semicircular, vertical
Z 1/2
d2
2ρ x
wall with top d is given by
− x2 dx.
4
0
12. Answer true or false. The force exerted by water on a surface
of a square, vertical plate with edges of 3 m if it is suspended
with its top 2 m below the surface is 18 lb. (The density of
water is 62.4 lb/ft3 .)
13. Answer true or false. If a submerged rectangle is rotated 90◦
about an axis through its center and perpendicular to its surface, the force exerted on one side of it will be the same,
provided the entire rectangle remains submerged.
116
Chapter 5 Test
(b) 6, 120 J
1. Find the area of the region enclosed by y = x2 and y = x by
integrating with respect to x.
(c) 6, 000 J
(d) 240 J
9. Find the work done when a constant force F(x) = 15 N in the
positive x-direction moves an object from x = 4 m to 10 m.
(a) 1/6
(b) 1
(c) 1/4
(a) 45 J
(d) 1/16
(b) 90 J
2. Find the area of the region enclosed by
y = cos (x − π /2), y = −x, x = 0 and x = π /2. The approximate area is
(c) 180 J
(d) 150 J
4
N
x2
in the positive x-direction moves an object from x = 1 m to
x = 3 m.
10. Find the work done when a variable force of F(x) =
(a) 1.1169
(b) 2.2337
(c) 4.4674
(a) 0 J
(d) 1
3. Find the volume of the solid
p that results when the region enclosed by the curves y = − sin (−x), y = 0 and x = π /4 is
revolved about the x-axis. The approximate volume is
(a) 0.143
(c) 1.408
(d) 2.816
4. Answer true or false. Cylindrical shells can be used to find
√
the volume of the solid when the region enclosed by y = 3 x,
x = −3, x = 0 and y = 0 is revolved about the y-axis and the
volume of the solid is 5.563π .
5. Answer true or false. Cylindrical shells can be used to find
the volume of the solid when the region enclosed by x = y2 ,
x = 0 and y = −2 is revolved about the x-axis and the volume
of the solid is 4π .
6. Answer true or false. The arc length of y = cos (−x) from
x = 0 to x = π /2 is 1.
7. Answer true or false. The surface area of the curve
y = sin (xZ+ π ), −
qπ ≤ y ≤ 0 revolved about the x-axis is
π
0
2π x
1 + sin2 (x + π ) dx.
8. Assume a spring whose natural length is 2.0 m is stretched
0.8 m by a 150 N force. How much work is done in stretching the spring?
(a) 60 J
(c) 0.6 J
(d) 1.79 J
11. Answer true or false. A semicircular wall 20 ft across at the
top forms one end of a tank. The total force exerted on this
wall by a liquid that fills the tank is 24, 800 lb. Ignore the
force of air above the liquid. (The density of the liquid is
124.8 lb/ft3 .)
(b) 0.920
given by
(b) 2.67 J
12. A horizontal table top is submerged in 10 ft of water. If the
dimensions of the table are 6 ft by 1 ft, find the force on the
top of the table that exceeds the force that would be exerted
by the atmosphere if the table were at the surface of the water.
(The density of water is 62.4 lb/ft3 .)
(a) 3, 744 lb
(b) 1, 872 lb
(c) 4, 000 lb
(d) 60 lb
13. Find the centroid of the region bounded between the curves
y = |x| and x + 2y = 3.
(a) (−4/3, 2/3)
(b) (−2/3, 4/3)
(c) (−3/2, 4/3)
(d) (−3/2, 2)
117
Chapter 5: Answers to Sample Tests
Section 5.1
1. a
9. d
2. false
10. false
3. a
11. false
4. c
12. false
5. b
13. true
6. b
14. b
7. c
15. a
8. a
2. d
10. c
3. b
11. false
4. true
12. b
5. false
13. b
6. d
14. false
7. a
8. c
2. b
10. a
3. c
11. b
4. c
12. false
5. a
6. b
7. d
8. b
2. a
3. false
4. b
5. d
6. b
7. false
2. c
3. c
4. false
5. false
6. d
7. false
8. false
2. a
10. a
3. d
11. a
4. false
12. false
5. a
13. false
6. false
14. false
7. b
15. a
8. d
2. d
10. a
3. c
11. a
4. d
5. b
6. c
7. b
8. d
2. b
10. true
3. a
11. false
4. false
12. false
5. a
13. true
6. false
7. b
8. true
2. b
10. b
3. b
11. false
4. false
12. a
5. false
13. b
6. false
7. false
8. a
Section 5.2
1. a
9. false
Section 5.3
1. d
9. c
Section 5.4
1. a
Section 5.5
1. b
9. false
Section 5.6
1. a
9. c
Section 5.7
1. b
9. false
Section 5.8
1. d
9. true
Chapter 5 Test
1. a
9. b
118
Chapter 6: Exponential, Logarithmic, and
Inverse Trigonometric Functions
Summary: The main focus of this chapter is to introduce both logarithmic and
exponential functions. In the process, the derivative and integration rules for both
of these will be developed (making some use of implicit differentiation along the
way). Then these derivative and integration rules will be used in the applications
that were discussed in both Chapters 3 and 5. Later in the chapter, L’Hôpital’s
rule is introduced and discussed for evaluating limits that have an appropriate
indeterminate form.
The latter parts of the chapter look into describing the natural logarithmic function as a function that is defined in terms of the Fundamental Theorem of Calculus. Also, implicit differentiation is used again to discuss the derivatives and
integrals that are associated with the inverse trigonometric functions. The chapter
concludes by defining the hyperbolic trigonometric functions and discussing their
properties and uses.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Define logarithmic and exponential functions (§6.1).
2. Take derivatives of functions involving logarithms (§6.2).
3. Use logarithmic differentiation to find derivatives of complicated functions
(§6.2).
4. Evaluate integrals that involve logarithms (§6.2).
5. Relate the derivative of a function with the derivative of its inverse function
(§6.3).
6. Take derivatives of exponential functions (§6.3).
7. Evaluate integrals that involve exponential functions (§6.3).
8. Solve various application problems that involve either logarithmic or exponential functions (§6.4).
119
120
9. Use L’Hôpital’s rule to evaluate limits that have indeterminate forms such
0 ∞
as , , 0 · ∞, ∞ − ∞, 1∞ , 00 and ∞0 (§6.5).
0 ∞
10. Define the logarithmic function using the Fundamental Theorem of Calculus (§6.6).
11. Define the inverse trigonometric functions (§6.7).
12. Differentiate and integrate inverse trigonometric functions (§6.7).
13. Learn the definitions and properties of the hyperbolic functions (§6.8).
6.1 Exponential and Logarithmic Functions
PURPOSE: To introduce the functions ax and loga x.
exponential: y = ax
logarithmic: y = loga (x)
Note:
ax = b ⇔ loga (b) = x
Exponential and Logarithmic functions are introduced in this section. These are
two types of functions that arise naturally as inverses of each other. One key idea
to remember well is that a logarithm is simply a way of representing an exponent.
For example, if y = ax then x = loga y is a way of expressing the exponent x in the
first expression. Also because logarithms and exponential functions are inverses of
each other, the cancelation equations for inverse functions result in these functions
undoing each other. For example,
aloga (x) = x and loga (ax ) = x
There are some common properties of logarithms and exponentials that are easy to
remember when paired together. To remember the properties of logarithms, it may
be helpful to remember how exponents behave and then to think of logarithms as
representing exponents.
logb (x1 ) + logb (x2 ) = logb (x1 x2 )
logb (x1 ) − logb (x2 ) = logb (x1 /x2 )
logb b = 1
logb 1 = 0
⇔
⇔
⇔
⇔
by1 +y2 = by1 by2
by1 −y2 = by1 /by2
b1 = b
b0 = 1
IDEA: Any logarithmic function may be written in terms of the natural logarithm. Any exponential function may be written in terms of the natural exponential function.
The natural logarithm and the natural exponential functions are both important
since they can be used to express other logarithms and other exponential functions.
For example,
logb (x) =
ln x
ln b
ax = eln a
x
= ex ln a
These relationships will be useful in the next two sections where the derivatives
and integrals of logarithmic and exponential functions are discussed.
121
Checklist of Key Ideas:
integer, rational and irrational powers
exponential functions
representing exponents with logarithms
logarithm with base b
bx and logb x are inverse functions
the number e and the natural exponential and logarithmic functions
change of base formulas for logarithms
exponential and logarithmic growth
6.2 Derivatives and Integrals Involving Logarithmic
Functions
PURPOSE: To give differentiate and integrate logarithmic functions and to discuss the process of logarithmic differentiation.
This section develops the derivatives of logarithmic functions by using the limit
description of e (see margin), the properties of logarithms and the definition of the
derivative. The derivatives of logarithms give rise to a process called logarithmic
differentiation. The process can be summarized by taking a logarithm of a given
expression and then taking derivatives of the resulting expression.
IDEA: The log property of log (ab) = log a + log b can remove the need for the
product and/or quotient rules.
The idea behind logarithmic differentiation is to take advantage of the property of
logarithms to turn multiplication inside a logarithm into a sum of terms. This also
simplifies the process of taking derivatives of functions with powers and extends
the power rule to cases with arbitrary real exponents.
IDEA: Use the relationship logb x = ln x/ ln b to remember the derivative and
integration rules for logb x.
Checklist of Key Ideas:
derivatives of logarithmic functions
using properties of logarithmic functions before differentiation
logarithmic differentiation
extended power rule for general powers
integration rules for logarithmic functions
lim (1 + h)1/h = e
h→0
1. take logarithms of the expression
2. differentiate
3. solve for dy/dx
logarithm property
ln (ab) = ln a + ln b
122
6.3 Derivatives of Inverse Functions; Derivatives and
Integrals Involving Exponential Functions
PURPOSE: To relate the derivative of a function with the derivative of its inverse and to give rules for differentiating and integrating exponential functions.
differentiability of f −1
In this section, implicit differentiation and the properties of inverse functions are
both used to find the derivatives of exponential functions. The first observation is
that if a function is differentiable at a point (a, b) and its slope is not zero then the
inverse of the function should be differentiable at the point (b, a). In particular,
the slopes of the function and its inverse at these points should be the reciprocals
of each other.
IDEA: If f (a) = b then f goes through the point (a, b) and f −1 goes through
the point (b, a). The slope of f at (a, b) is the reciprocal of the slope of f −1 at
the point (b, a).
increasing and decreasing
derivatives of ax and ex
This result can be seen as a geometric description of inverses since f and f −1 are
reflections of each other across the line y = x. The concepts of increasing and
decreasing are briefly mentioned in order to discuss the invertibility of a given
function (see Chapter 4 for more on increasing and decreasing). This is important
because when a function is strictly decreasing or increasing then it will be one-toone (or passes the horizontal line test) and so will have an inverse function. This
is the case with both ex and ln x which are both strictly increasing functions.
Using implicit differentiation and the relationship between inverses, the derivatives of exponential functions may now be found. For example, the function y = ex
may also be written as x = ln y. Then dy/dx may be found using implicit differentiation together with the derivatives of logarithms from the previous section.
IDEA: Use the relationship ax = ex ln a to remember the derivative and integration rules for ax .
Checklist of Key Ideas:
differentiability of inverse functions
finding derivatives of inverse functions
increasing and decreasing functions and f ′ (x)
derivatives of exponential functions
derivatives of functions of the form y = f (x)g(x)
integration rules of exponential functions
123
6.4 Graphs and Applications Involving Logarithmic
and Exponential Functions
PURPOSE: To examine graphs of logarithmic and exponential
functions and to show some applications of logarithmic and exponential functions.
The main purpose of this section is to use the derivative and integral rules that
have now been developed for logarithmic and exponential functions in various
applications. There are many applications that have been discussed so far (see
Chapters 3 and 5). These applications may be separated into two broad groups:
those that involve differentiation and those that involve integration.
Applications involving differentiation are discussed in more detail in Chapter 3.
These applications are now applied to logarithmic and exponential functions. For
example, sketching exponential and logarithmic functions may be done in the
usual fashion by taking the first and second derivatives and looking for relevant
information such as intervals of increase and decrease, concavity, relative extrema
and points of inflection. On a closed interval, if the given function is continuous,
then an absolute maximum or minimum may be found by the usual methods as
well.
applications involving
differentiation → Chapter 3
Applications involving integration are discussed in Chapter 5. These applications include finding the area under a curve, finding the volume that results when
a given curve is revolved around an axis, finding the length of a given curve between two points, finding the surface area of a solid of revolution, finding the
average value of a function and others. All of these may now be discussed using
the integration and derivative rules that have been developed for both logarithmic
and exponential functions.
applications involving
integration → Chapter 5
Checklist of Key Ideas:
graphing techniques for functions
relative and absolute extrema
critical points and inflection points
intervals of increase and decrease
concavity of a function
area between two curves
calculating volume of a solid of revolution by disks
calculating volume of a solid of revolution by shells
arc length of a curve defined either explicitly or parametrically
surface area of solid of revolution
124
rectilinear motion
average value of a function on an interval
work of a variable force
6.5 L’Hôpital’s Rule; Indeterminate Forms
PURPOSE: To introduce a method for finding limiting values
when a limit evaluates as an indeterminate form.
Note:
Multiple applications of L’Hôpital’s
rule may be needed
L’Hôpital’s rule cannot be applied
directly to indetermine products,
quotients or powers
0∞ not indeterminate
The purpose of this section is to develop a method for finding limiting values when
a limit evaluates as an indeterminate form. The method in this section is called
L’Hôpital’s rule. The primary usefulness of L’Hôpital’s rule is when a limit has
the indeterminate form of 0/0 or ∞/∞. In either of these cases, the rule says that
f (x)
, will be equivalent to the limit
the limit of the ratio of the functions, i.e., lim
x→a g(x)
f ′ (x)
of the ratio of their derivatives, i.e., lim ′ .
x→a g (x)
The power of L’Hôpital’s rule is often abused. If a limit does not have one of the
indeterminate forms 0/0 or ∞/∞ then the rule cannot be used. The consequence
of this is that the rule cannot be directly applied to cases where a limit has a indeterminate form such as 0 · ∞, ∞ − ∞, 00 , ∞0 or 1∞ . In these situations, the limit
expressions need to be rewritten before applying the rule. This can often be done
by algebraic techniques (using a common denominator, rationalizing an expression with a radical or writing an expression as the reciprocal of its reciprocal, i.e.,
f (x) = 1/ 1f (x) ) or by using the properties of logarithms (ln ab = b ln a). It should
be noted that 0∞ is not an indeterminate form, nor is a0 where |a| < ∞ and a 6= 0.
Checklist of Key Ideas:
indeterminate forms of type 0/0 and ∞/∞
using L’Hôpital’s rule
growth of exponential functions versus polynomials and rational functions
indeterminate forms of types 0 · ∞ and ∞ − ∞ and L’Hôpital’s rule
indeterminate forms of types 00 , ∞0 and 1∞ and L’Hôpital’s rule
125
6.6 Logarithmic Functions from the Integral Point
of View
PURPOSE: To define logarithmic functions using the FTC, part
II.
There is only one main idea introduced in this section. Here, the natural logarithmic function is defined as an integral.
ln x =
Z x
1
1
t
recall: FTC, part II
(see §4.6)
dt
The rest of the section is spent showing the usual properties of the natural logarithmic function and the natural exponential function in terms of integrals. All of the
usual results can be found. For example, consider the following interesting result
that comes from the basic properties of logarithms. The logarithm relationship
ln (ab) = ln (a) + ln (b) written in terms of integrals means
Z ab
1
1
t
dt =
Z a
1
t
1
dt +
Z b
1
1
t
dt.
This is not something that is obvious at first glance.
Some other functions are introduced in this section that are defined in terms of integrals. An important function that is defined in this fashion is the error function.
2
erf(x) = √
π
Z x
0
2
e−t dt
One reason that this function is defined in terms of an integral is that it does not
have a simple expression for its antiderivative.
Checklist of Key Ideas:
integral definition of ln x
properties of ln x
defining the natural exponential function
changing base of logarithms
some limits involving e
functions defined by integrals; the error function
functions as limits of integration; the chain rule
error function
126
6.7 Derivatives and Integrals Involving Inverse
Trigonometric Functions
PURPOSE: To define the inverse trigonometric functions and to
give rules for their derivatives and integrals.
sin−1 (x)
sec−1 (x)
tan−1 (x)
In this section, the inverse trigonometric functions are introduced. The three most
important inverse trigonometric functions are sin−1 (x), tan−1 (x), and sec−1 (x).
It is important to realize that these are inverse functions of the respective trigonometric functions and not reciprocal functions. Because they are inverse functions,
the domain of sin x, tan x, and sec x have to be appropriately defined so that these
inverse functions are well defined.
y = sin−1 (x) means
“y is the angle whose sine is x”
The derivatives and integrals of the inverse trigonometric functions are defined using the inverse relationship and implicit differentiation together with some properties of right triangles and their angles. For example, y = sin−1 (x) which should
be read “y is the angle whose sine is x.” In other words, x = sin (y), meaning that
dy/dx may be found using implicit differentiation.
The three inverse
functions, sin−1 (x), sec−1 (x), and tan−1 (x), in√
√ trigonometric
2
2
volve the terms 1 − x , x − 1, and x2 + 1, respectively. It is important to know
which one of these goes with which function. When first becoming familiar with
these functions, it can be helpful to have this information written on a note card
or in a notebook as the material is studied.
Checklist of Key Ideas:
definitions of inverse trigonometric functions
domain restrictions of inverse trigonometric functions
derivative rules for inverse trigonometric functions
integration rules for inverse trigonometric functions
6.8 Hyperbolic Functions and Hanging Cables
PURPOSE: To define the various hyperbolic trigonometric functions and to discuss their properties.
sinh u =
eu − e−u
2
cosh u =
eu + e−u
2
tanh u =
sinh u
cosh u
Hyperbolic functions are introduced and defined in this section. These functions
are not really an application of integrals but rather are a new group of functions
which are defined as combinations of exponential functions. Everything that is
shown in this section can be found by remembering the definitions of each funcex − e−x
tion. For example, since sinh x =
then its derivatives and antiderivatives
2
can be found by simply knowing this information for ex and e−x . Once this type
127
of information is known for sinh x and cosh x, then similar results can be obtained
sinh x
for tanh x =
.
cosh x
Other important features can be remembered by thinking of sinh x and cosh x in
pieces:
sinh x
=
cosh x
=
ex e−x
−
2
2
ex e−x
+
2
2
Recall that ex → 0 as x → −∞ and e−x → 0 as x → ∞. Then sinh x and cosh x
become asymptotically close to ex /2 as x → ∞. On the other hand, sinh x becomes
asymptotically close to −e−x /2 as x → −∞ and cosh x becomes asymptotically
close to e−x /2 as x → −∞. Similar properties of tanh x and the other hyperbolic
functions are found by either defining them in terms of ex and e−x or in terms of
sinh x and cosh x.
IDEA: All the definitions can be written in terms of sinh x and cosh x.
For example,
tanh x =
sinh x
cosh x
sechx =
1
cosh x
coth x =
cosh x
sinh x
cschx =
1
sinh x
IDEA: Make notecards with the various definitions and properties of the hyperbolic functions to help remember them.
If the reader is not familiar with the material in this section, then it is suggested
that the reader make a few notecards with any new information and have them
handy as they attempt the exercises at the end of the section.
Checklist of Key Ideas:
ex as a sum of even and odd functions
hyperbolic functions
curvilinear asymptotes
catenary; hanging cables
hyperbolic identities
hyperbolic inverses and derivatives
logarithmic forms of inverse hyperbolic functions
128
Chapter 6 Sample Tests
Section 6.1
(b) log 22
log 24
(c)
5
(d) log 44
1. 2−5 =
(a)
1
10
(b) −
(c)
9. Solve log10 (x + 5) = 1 for x.
1
10
(a) 5
1
32
(d) −
1
32
2. Use a calculating utility to approximate
decimal places.
(b) −5
(c) 0
(d) no solution
√
6
29. Round to four
10. If log10 x5/2 − log10 x3/2 = 4 then find x.
(a) 4
(a) 1.7528
(b) 40
(b) 5.3852
(c) 1, 000
(c) 1.7530
(d) 10, 000
(d) 5.3854
11. Solve 4−2x = 6 for x to four decimal places.
(a) 0.6462
3. Use a calculating utility to approximate log 28.4. Round to
four decimal places.
(b) −0.6462
(c) 1.2925
(a) 3.3464
(d) −1.2925
(b) 3.3462
(c) 1.4533
(d) 1.4535
12. Solve for x if 3ex − xex = 0.
(a) 3
4. Find the exact value of log3 81.
(b) −3
1
(c)
3
1
(d) −
3
(a) 12
3
(b)
4
1
(c)
4
(d) 4
y
7
5. Use a calculating utility to approximate ln 39.1 to four decimal places.
(a) 1.5920
(b) 3.6661
(c) 1.5922
(d) 3.6663
6. Answer true or false.
a3 b
ln 2 = 3 ln a + ln b − 2 ln c.
c
√
7. Answer true or false. log (5x x − 2) = (log 5)(log x)(log1/2 (x − 2)).
8. Rewrite 5 log 2 − log 12 + log 24 as a single logarithm.
(a) log 64
-6
13.
6
-3
This is a graph of
x
129
(a) 4 − ln (2 + x)
(d)
(b) 4 + ln (2 + x)
(c) 4 − ln (x − 2)
(d) 4 + ln (x − 2)
14. Use a calculating utility and the change of base formula to
find log3 4.
2 ln x2
p
x 3 + ln2 x2
dy 3
= .
4. Answer true or false. If y = ln (x3 ) then
dx
x
1
5. Find dy/dx if y = ln
.
sin x
(a) cos x sin x
(a) 0.2007
(b) cot x
(b) 0.7925
(c) tan x
(c) 1.2619
(d) 0.4621
15. The equation Q = 6e−0.052t gives the mass Q in grams of a
certain radioactive substance remaining after t hours. How
much remains after 6 hours?
(a) 4.3919 g
(b) 4.3920 g
(c) 4.3921 g
(d) 2.3922 g
Section 6.2
1. If y = ln 6x then find dy/dx.
(a)
(b)
(c)
(d)
1
6x
6
x
1
x
6 ln 6x
x
2. If y = ln (cos x) then find dy/dx.
(a) tan x
(b) − tan x
1
(c)
cos x
1
(d) −
cos x
p
3. If y = 3 + ln2 x2 then dy/dx =
2
p
x 3 + ln2 x2
2
(b) p
3 + ln2 x2
1
(c) p
3 + ln2 x2
(a)
(d) − cot x
r
x+2
dy
then find
by logarithmic differentiation.
6. If y = 8
x+3
dx
1 x + 2 −7/8
(a)
8 x+3
1
(b)
8(x + 3)2
r
1
1
1
8 x+2
(c)
−
8 x+2 x+3
x+3
r
x+2
(d) 7
x+3
7. If y = ln (kx) then
dny
=
dxn
1
k n xn
(−1)n
(b) n n
k x
(−1)n
(c)
(k − 1)!
xn
1
(d)
x
(a)
8. Suppose that f (x) = (cos (x))π . What is dy/dx?
(a) π (cos (x))π −1
(b) (cos (x))π −1
(c) π (− sin (x))π −1
(d) −π sin (x) (cos (x))π −1
9. If y2 + ln (xy) = y3 then dy/dx =
x
(a)
1
2y + y − 3y2
(b)
(c)
−1
x(2y + 1y − 3y2 )
x+y
xy 2y − 3y2
(d) none of the above
10. If y = sec (ln x) then y′ =
130
1
sec (ln x) tan (ln x)
x
(b) sec (ln x) tan (ln x)
1
1
tan
(c) sec
x
x
1
(d) sec2 (ln x)
x
dy
1
then
=
11. If y =
ln x
dx
(a) x
1
(b) −
(ln x)2
1
(c) −
x (ln x)2
x
(d) −
(ln x)2
(a) ln x +C
(a)
x2 (x2 + 1)3 sin x
then which of the following exprescos (3x)
y′
sions is ?
y
12. If y =
(a)
(b)
(c)
(d)
13. If y =
(a)
(b)
(c)
(d)
14.
Z
6x
2
+
+ cot x + 3 tan (3x)
x x2 + 1
3
1
+
+ csc x − sec (3x)
2x x2 + 1
−3 sin (3x)
4x2 (x2 + 1) cos x
1
2
2x
+
+ 2
x x + 1 tan (−2x)
ln x
then dy/dx =
x
1
− 2
x
1
x2
ln x
1
− 2
x2
x
1
− 3
x
5
dx =
x
5
(a)
+C
2x2
5
(b) − 2 +C
2x
(c) −5 ln x +C
(d) 5 ln x +C
15.
Z
ln x
dx =
x
(b) (ln x)2 +C
(ln x)2
+C
2
(d) 2(ln x)2 +C
(c)
16.
Z e
1
1
x
dx =
(a) 1
(b) e
1
(c)
e
(d) 0
17. Answer true or false.
Z 2
1
dx
ln 7 − ln 4
=
.
3x + 1
3
Section 6.3
1. Answer true or false. If y = x8 e7x then dy/dx = 56x7 e7x .
2. If y = (ln x)e2x then dy/dx =
(a) 2(ln x)e2x +
e2x
x
(b) 2(ln x)e2x
(c) 2(ln x)e2x
(d)
e2x−1
x
e2x
x
3. Answer true or false. If x + exy = 2 then
4. Let f (x) = 5x . Find
dy −yexy − 1
=
.
dx
xexy
d
[ f (x)].
dx
(a) 5x ln 5
(b) 5x−1
(c) x ln 5x
(d) 5x ln x
5. Answer true or false. If f (x) = π sin x+cos x then
dy/dx = (sin x + cos x)π sin x+cos x−1 .
6. Answer true or false. If y = xcos x then
dy cos x
=
− sin x ln x xcos x .
dx
x
7. The equation y′ x = −yx is satisfied by
(a) y = ex
(b) y = cos x
(c) y = sin x
(d) y = e−x
131
Section 6.4
3h − 1
=
h→0 2h
8. Evaluate the limit. lim
1. Answer true or false. If the position of a particle along a line
is given by s(t) = lnt for t > 0, then the limiting velocity of
the particle as t → ∞ is zero.
(a) 1
(b) 0
(c) ∞
ln 3
(d)
2
9.
Z
2. If y = ln (x) − x then y has a horizontal tangent line at which
of the following x-values?
(a) x = −1
(b) x = 0
3ex dx =
(c) x = 1
(a) 3ex +C
(d) y has no horizontal tangents.
3e2x
(b)
+C
2
dz
dy
dx
.
=
dt
dt
dt
4. The largest open interval over which f is concave up for
4
f (x) = ex is
3. Answer true or false. If z = x ln y then
ex+1
+C
x+1
(d) None of these.
(c)
(a) (−∞, 0)
dy
= ex and y(0) = 2.
10. Find y(x) given that
dx
(a) y(x) = ex + 1
5. f (x) = ln (x2 + 2) has
y(x) = ex
(a) a relative maximum only
(d) y(x) = ex − 2
(b) a relative minimum only
(c)
Z
(c) both a relative maximum and minimum
2
2xex dx =
(d) no relative extrema
2
(a)
6. Use a graphing utility to generate the graph of f (x) = x2 e3x ,
then determine the x-coordinates of all relative extrema on
(−10, 10) and identify them as relative maxima or minima.
ex
+C
2x
2
(b) 2ex +C
(a) There is a relative maximum at x = 0.
2
(c) x2 + ex +C
(b) There is a relative minimum at x = 0.
2
(d) ex +C
12. Answer true or false. For
ex − 2.
Z
ex
dx a good choice for u is
ex − 2
13. Answer true or false. To evaluate
good choice for u is u = ex .
14. Answer true or false. To evaluate
choice for u is u = 1 + ex .
15.
(c) (−∞, ∞)
(d) nowhere
(b) y(x) = ex + 2
11.
(b) (0, ∞)
Z 1
0
e−3x dx =
(a) 0.216
(b) 0.148
Z 1
0
ex (1 + 5ex )12 dx, a
Z 1
0
(c) There is a relative minimum at x = 0 and relative maxima at x = −1 and x = 1.
(d) There are no relative extrema.
7. Find the displacement of a particle between t = 0 and t = 2 if
v(t) = et + t 2 .
(a) 7.437
ex
dx, a good
1 + ex
(b) 9.056
(c) 11.389
(d) 12.389
8. Use the method of disks to find the volume of the solid
that results by revolving the region enclosed by the curves
y = x + ex , y = 0, x = 0 and x = 1 about the x-axis.
(a) 5.53
(c) 0.317
(b) 11.05
(d) 0.519
(c) 17.37
132
(a) (5.17581, 0.14168)
(d) 34.73
9. Use cylindrical shells to find the volume of the solid when
2
the region enclosed by y = 2ex , x = 1, x = 2 and y = 0 is
revolved about the y-axis.
(b) (6.60199, 0.09556)
(c) (6.72727, 0.09091)
(d) (3.90865, 0.19543)
(a) 103.8π
(b) 12.970π
Section 6.5
(c) 6.485π
(d) 25.940π
10. Answer true or false. The arc length of the curve y = ex + e2x
Z 4q
1 + (e3x )2 dx.
from x = 0 to x = 4 is given by
0
11. Answer true or false. The arc length of the parametric curve
Z 3√
x = 3et and y = et for 0 ≤ t ≤ 3 is given by
4et dt.
0
12. Use a CAS or a calculator with integration capabilities to approximate the arc length of the curve y = −xex from x = 0 to
x = 2.
x→0
sin 7x
=
sin x
(a) 7
1
(b)
7
(c) −7
1
(d) −
7
2. lim
x2 − 9
x→∞ x2 − 3x
(a) 21.02
=
(a) 1
(b) 4.17
(b) ∞
(c) 15.04
(c) −∞
(d) 19.71
13. Answer true or false. The area of the surface generated by
revolving x = ey+2 , 0 ≤ y ≤ 1 about the y-axis is given by
Z 1 p
ey 1 + e2y+4 dy.
2π e2
0
14. Answer true or false. The area of the surface generated by
revolving the parametric curve x = t 2 and y = et for 0 ≤ t ≤ 1
Z 1 p
about the x-axis is given by 2π
et e2t + 4t 2 dt.
0
15. The average value of y =
1. lim
1
on the interval [1, 10] is
x
(a) ln 10
(b) ln 9
1
(c) ln 10
9
(d) 99/100
16. Find the volume of the solid that results when the region enclosed by the curves x = −ey , x = −1 and y = 1 is revolved
about the y-axis.
(a) 6.894
(d) 0
tan2 x
=
x→0 x
3. lim
(a) 1
(b) ∞
(c) −∞
(d) 0
4. lim+
x→0
ln (x + 1)
=
ex − 1
(a) 0
(b) 1
(c) ∞
(d) −∞
ex
=
x→∞ x4
(a) 1
5. lim
(b) 0
(b) 3.195
(c) ∞
(c) 10.205
(d) −∞
(d) 32.060
17. A laminar region is bounded by the curves x = 1, x = 10,
y = 0, and y = 1/x with δ = x2 + 1. Find the center of gravity
of the region.
6. lim e−x ln x =
x→∞
(a) 0
(b) 1
133
Section 6.6
(c) ∞
(d) −∞
1/x
7. lim (1 + 3x)
x→0
1. Simplify: e3 ln x =
=
(a) x3
(a) 3
(b) 3x
(b) ∞
(c) x/3
(c) ln 3
(d) e3
2. Simplify: ln (e−6x ) =
(d) e3
8. lim+
x→0
(a) x6
sin x
=
ln (x + 1)
(b) x−6
(c) −6x
(a) 10
(d) −6 − x
(b) 1
3. Simplify: ln (xe4x ) =
(c) ∞
(d) −∞
(a) 4
9. Answer true or false. lim
x→0
cos
cos
1
=
10. lim e−x −
x→∞
x
1
x
2
x
1
= .
2
(b) 4 + ln x
(c) 4x + ln x
(d) 4x2
4. Which of the following expressions is equivalent to ln (7/2)?
(a) ∞
(a)
(b) −∞
Z 7
1
0
(c) 1
(b)
t
1
(d) 0
(c)
11. lim+ (1 − ln x)x =
dt
Z 7
1
dt
Z 14
1
dt
1
x→0
dt
2t
Z 7/2
1
2t
ln 7
ln 2
5. Which of the of the following expressions is equivalent to
ln (7)?
(d)
(a) 0
(b) 1
(c) ∞
(d) −∞
(a)
sin 3x
12. Answer true or false. lim
= 1.
x→0 1 − cos x
2x3 − 2x2 + x − 3
= 2.
x3 + 2x2 − x + 1
p
14. Answer true or false. lim
x2 − 2x − x = 0.
x→∞
sin x 1
15. lim+
=
−
x
x
x→0
13. Answer true or false. lim
x→∞
(b)
0
(c)
dt
t
(d) ln 28 − ln 21
6. Which of the following expressions is equivalent to ln (6.5)?
Z 65
dx
(a)
1
10
(b)
Z 13
dx
(c)
(d) −∞
16. Answer true or false. lim+
x→0
√
x ln x = 0.
(d)
x
1
x
2
(c) ∞
dt
t
Z 2
7
1
(a) 0
(b) 1
t
2
Z 7
1
Z
5 13 dx
2
x
1
Z 65
dx
10
x
134
7. Let f (x) = e−3x . Which of the following is the exact value of
f (ln 2)?
1. If y = sin−1 (2x) then find dy/dx.
(a) 6−6
(b) 1/8
(c) −8
(d) −1/8
8. Answer true or false. If F(x) =
9. Answer true or false. If F(x) =
10. Answer true or false. lim
x→∞
Z x2
2
1
t
2
dt, then F ′ (x) = .
x
Z x3
2
6
dt then F ′ (x) = .
x
1
x
= 0.
1+
1
t
3x
11. Answer true or false. lim (1 + 5x)1/(5x) = e.
x→0
12. Answer true or false. lim (1 + 2x)1/(2x) = 0.
x→0
13. Which of the following expressions is equivalent to ln 9.8?
1
ln 100 − ln 2
10
1
ln 98
(b)
10
(c) ln 2 + ln 49 − ln 10
2 ln 49
(d)
ln 10
(a)
14. Which of the following expressions is equivalent to ln 42?
(a)
Z 7
dx
x
1
(b)
+
Z 50
dx
Z 40
dx
x
1
(d) 3
Z 6
dx
x
1
+
Z 14
dx
15. If F(x) =
Z 2
dx
1
x
Z e2x
1
is FALSE?
1
3. If y = esin
−1
x
then dy/dx =
−1
esin x
(a) √
1 − x2
(b) −(sin−2 x)esin
1
(c)
(cos x)esin x
(d) −
(a)
−1
(cos−1 x)esin
√
1 − x2
x
−1
x
t
dt then which of the following statements
(a) F(x) is always increasing.
(b) F(x) has no critical points.
2
e2x
(d) F(x) has no points of inflection.
(c) F ′ (x) =
1
x cos−1 x
√
1 − x2
√
−x + (cos−1 x) 1 − x2
√
(1 − x2 ) cos−1 x − x 1 − x2
(c)
x(1 − x2 ) cos−1 x
1
(d)
x
p
5. Let y = sin−1 x. Then dy/dx =
(b)
x
1
2
(a) √
1 − 2x2
2
(b) √
1 − x2
1
(c) √
1 − 4x2
2
(d) √
1 − 4x2
√
2. If y = tan−1 x then find dy/dx.
√
x
(a)
2x(1 + x)
x
(b)
1+x
x
(c)
2(1 + x2 )
r
1
(d)
1 + x2
4. Find dy/dx if y = ln (x cos−1 x).
x
8
(c)
Section 6.7
(a) y = √
4
1 − x2
1
p
√
2( sin−1 x)( 1 − x2 )
1
(c) y = √
4
2 1 − x2
(b) y =
16. Answer
true
or false.
Z x
Z x3
dt
dt
3
=
for all x > 0.
t
1 t
1
1
135
(d) y = −
6. If
1
p
√
4
−1
2( sin x)( 1 − x2 )
x2 − sin−1 y = ln x
(a)
1
− 2x
x
then find dy/dx.
q
1 − y2
13. Answer true or false. tan−1 (1) + tan−1 (2) = tan−1 (−3).
14.
Z
(a) cos−1 (x) +C
√
(b) 2 1 − x2 +C
q
1
1 − y2
(b) − + 2x
x
(c) ln |sin (x)| +C
1 −1 2
(d)
sin (x) +C
2
sin−2 y − 2x2
(c)
x sin−2 y
(d)
− sin−2 y + 2x2
x sin−2 y
7. Answer true or false. To have an inverse, a trigonometric
function must have its domain restricted to [0, 2π ].
sin−1 (x)
√
dx =
1 − x2
15. Find the volume of the solid that is generated by revolving
the curve y = (1 + x2 )−1/2 about the x-axis between x = 0
and x = 1.
(a)
π2
4
(b) π /2
(b)
π2
2
(c) π
(c) π ln 2
8. Find the exact value of sin−1 (1).
(a) 0
(d) 3π /2
(d)
9. Find the exact value of cos−1 (cos (3π /4)).
(a) 3π /4
16. Evaluate
(b) π /4
(c) −π /4
10. Use a calculating utility to approximate x if sin x = 0.15 and
π /2 < x < 3π /2.
(a) 0.1506
(c) 3.2932
17.
Z
(d) no solution
1
11. Answer true or false. cos−1 x =
for all x.
cos x
12. A ball is thrown at 5 m/s and travels 245 m horizontally before coming back to its original height. Given that the acceleration due to gravity is 9.8 m/s2 , and air resistance is negv2
ligible, the range formula is R =
sin 2θ , where θ is the
9.8
angle above the horizontal at which the ball is thrown. Find
all possible positive angles in radians above the horizontal at
which the ball can be thrown.
(a) 1.5708
(b) 1.5708 and 3.1416
(c) 0.7854 and 1.5708
(d) 0.7854
Z
sin (x)
dx.
1 + cos2 (x)
(a) ln |sin (x)| +C
(d) 5π /4
(b) 2.9910
π
ln 2
2
(b) ln |1 + cos2 (x)| +C
(c) − tan−1 cos (x) +C
(d) − sin−1 cos (x) +C
x
√
dx =
1 − x2
(a) sec−1 (1/x) +C
1 2 −1
x sin (x) +C
2
√
(c) − 1 − x2 +C
√
(d) x2 1 − x2 +C
(b)
18.
Z
dr
√
=
er 1 − e−2r
(a) r + 2 ln (1 − e−2r ) +C
(b) sec−1 (e−r ) +C
(c) − sin−1 (e−r ) +C
√
(d) −2e−r 1 − e−2r +C
136
Section 6.8
(a)
(b) 8 cosh8 x +C
1. Evaluate sinh (7).
(c) 7 cosh6 x +C
(a) Not defined.
(b) 551.1614
(d)
(c) 548.3161
(d) 549.4283
8.
2. Evaluate cosh−1 (2).
Z
(a)
(b) 1.3165
(b) 10 sinh10 x +C
(c) 1.3152
(c) 9 sinh8 x +C
(a) (5x + 1) cosh (5x + 1)
(d)
(d) −5 cosh (5x + 1)
4. Find dy/dx if y = sinh (3x2 ).
(a) 6x cosh (3x2 )
(b) −6x cosh (3x2 )
(c) 6 cosh (6x)
p
5. Find dy/dx if y = 2 sech(x + 5) − x3 .
(a)
(b)
−sech(x + 5) tanh (x + 5) − 3x2
p
sech(x + 5) − x3
(x + 5) cosh (x + 5) − 3x2
p
sinh (x + 5) − x3
− cosh (x + 5) + 3x2
(c) p
sinh (x + 5) − x3
(d)
sech(x + 5) tanh (x + 5) + 3x2
p
sech(x + 5) − x3
sinh (3x + 6) dx =
(a) 3 cosh (3x + 6) +C
1
(b) cosh (3x + 6) +C
3
(c) −3 cosh (3x + 6) +C
1
(d) − cosh (3x + 6) +C
3
7.
cosh7 x sinh x dx =
x
6
1
(a) √
36 + x2
1
(b) √
6 36 + x2
1
(c) √
36 − x2
1
(d) √
6 36 − x2
(c) −(5x + 1) cosh (5x + 1)
(d) −6 cosh (6x)
1
sinh8 x +C
8
9. Find dy/dx if y = sinh−1
(b) 5 cosh (5x + 1)
Z
1
sinh10 x +C
10
(a) 1.3170
3. Find dy/dx if y = sinh (5x + 1).
Z
1
cosh6 x +C
6
sinh9 x cosh x dx =
(d) 1.3174
6.
1
cosh8 x +C
8
.
10. Answer true or false. If y = − coth−1 (x + 3) when |x| > 0,
1
.
then dy/dx = − 2
x + 6x + 8
11.
Z
dx
√
=
1 + 16x2
1
sinh−1 (4x) +C
4
1
(b) coth−1 (4x) +C
4
1
(c) cosh−1 (4x) +C
4
1
(d) tanh−1 (4x) +C
4
(a)
12. Answer true or false.
Z
4dx
= 4 sinh−1 (ex ) +C
1 + e2x
13. Answer true or false.
Z
√
ex dx
1 + e2x
= sinh−1 (e2x ) +C
14. Answer true or false. lim (cosh x)2 = 0.
x→∞
15. Answer true or false. lim (coth x)2 = 1.
x→−∞
137
Chapter 6 Test
(d) ∞
1 x
8. Answer true or false. lim 8 +
= e8 .
x
x→0
1. If y = ln (5x2 ) then find dy/dx.
2
x
2
(b) 2
x
2
(c)
5x2
1
(d) 2
x
2xy − yexy
9. Answer true or false. If exy = yx2 then y′ = xy
.
xe − x2
1
1
−
10. Evaluate the limit: lim+
x sin x
x→0
(a)
(a) ∞
(b) 0
(c) 2
2. Answer true or false. If y = 3 ln xe3x then
dy 3e3x
=
+ 9 ln xe3x .
dx
x
d
3. If f (x) = 8x then find
[ f (x)].
dx
(d) Cannot be determined.
11. If y = ln (2x ) then
1
2
x
(b)
2
(c) ln (2)
ln (2)
(d)
2x
12. Answer true or false. If y = sin (arctan (ex )) then
dy
= ex (1 + e2x )−1/2 − e3x (1 + e2x )−3/2 .
dx
13. Evaluate the limit: lim (cos x)sin x
(a)
(a) 8x ln x
(b) x ln 8x
(c) x8x−1
(d) 8x ln 8
√
4. Find dy/dx if y = tan−1 4 x.
1
√
√
4
4( x3 + x 4 x)
√
x
√
(b)
1+ x
√
x
(c)
2(1 + x)
r
1
(d) 4
1+x
(a)
x→0
(a) 0
(b) 1
(c) ∞
(d) Cannot be determined.
√
sin 9x
=
x→0 sin 8x
(a) 1
(b) ∞
(c) −∞
9
(d)
8
sin 2x
=
7. lim
x→0 2x
14. Evaluate the limit: lim (ln (3x − 2) − ln x)
x→∞
cos−1 x + 1
5. Answer true or false. If y =
−1
dy
√
= √
.
dx 2( cos−1 x + 1)( 4 x2 − 1)
6. lim
then
(a) 0
(b) ln (2)
(c) ln (3)
(d) ∞
2x − 1
= e2
x
x→0
15. Answer true or false. lim
9
−1
16. Evaluate the limit: limx→1 xx−1
(a) 0
(b) 8
(c) 9
(d) ∞
(a) 0
(b) 1
1
(c)
2
dy
=
dx
17. If cos
(a)
y
x = ln
x
y
x
y
then
dy
=
dx
138
y−1
y sin (y/x)
y
(c) −
x
y
(d)
x
26. Answer true or false.
(b)
27. Answer true or false.
u = x2 .
18. Answer true or false. The function y = xe−x has no horizontal
tangent lines.
19. Evaluate the limit: lim+ x ln x
x→0
(b) −1
(c) −∞
(d) ∞
(b)
(c)
=
(c) 693
(d) 2, 178
+ ln x)
1
ab3
21. Answer true or false. log √ = log a + 3 log b − log c
c
2
22. If 52x = 8 then solve for x.
(a) 1.292
0
32. Use a CAS or a scientific calculator with numerical integration capabilities to approximate the area of the surface generated by revolving the curve y = ex+1 , −1 ≤ x ≤ −0.5 about
the x-axis.
(b) 9.27
(c) 0.204
(c) 1.48
(d) 0.102
(d) 6.78
23. Suppose that the number of bacteria present in a bacteria culture at time t is given by N = 10, 000e−t/10 . Find the smallest number of bacteria in the culture during the time interval
0 ≤ t ≤ 50.
(a) 67
33. Answer true or false. The area of the surface generated by
revolving the parametric curve x = t 2 and y = et for 0 ≤ t ≤ 1
Z 1 p
t 2 e2t + 4t 2 dt.
about the y-axis is given by 2π
0
34. Use a CAS to find the surface area of the solid that results
when the curve y = −ex , 0 ≤ x ≤ 0.5 is revolved about the
x-axis.
(b) 10, 000
(c) 3, 679
(a) 18.54
(d) 73, 891
(b) 9.27
dx
=
5x
(c) 1.48
(d) 6.78
(a) ln (5x) +C
ln x
(b)
+C
5
(c) 5 ln x +C
35. Use cylindrical shells to find the volume of the solid when the
3
region enclosed by y = 2 , x = 1, x = 2 and y = 0 is revolved
x
about the y-axis.
(d) ln x +C
25. Answer true or false. For
√
u is u = 2 + x.
31. Answer true or false. The arc length of the parametric curve
Z 2√
x = e3t and y = e3t for 0 ≤ t ≤ 2 is given by
3et dt.
(a) 18.54
(b) 0.646
24.
28. Answer true or false.
dx
dz ex dy
=
+ ex ln y .
If z = ex ln y then
dt
y dt
dt
29. Answer true or false. The function f (x) = ex ln x has a relative
minimum on (0, ∞).
(b) 2, 334
(d) Cannot be determined.
Z
2
ex dx can be easily solved by letting
(a) 574, 698
dy
dx
(a) (x + 1)xx
xx+1 (1 + 1x
ln (x)xx+1
Z
1
+ 3ex dx = ln |x| + 3ex +C
x
30. Use the method of disks to find the volume of the solid that results by revolving the region enclosed by the curves y = −e2x ,
x = 2, and y = −1 about the x-axis (round to the nearest whole
number).
(a) 0
20. If y = xx+1 then
Z
Z
√
dx
√ , a good choice for
x(2 + x)
(a) 2.08π
(b) 4.16π
(c) 2.08π 2
139
(a) 4 tanh4 x +C
(d) 4.16π 2
(b) 5 tanh6 x +C
36. Evaluate the limit: lim+ x3/2 ln x =
x→0
(c) 6 tanh6 x +C
1
(d) tanh7 x +C
7
(a) 0
(b) 1
(c) ∞
(d) The limit does not exist.
37. Which of the following expressions is equivalent to ln 12.8?
(a)
Z 8
2
dt
t
Z 8
2
dt
(b)
10 t
Z 128
1
(c)
dt
10 t
Z 12
Z 0.8
1
1
(d)
dt +
dt
t
t
1
1
1
38. Use a calculating utility to approximate x if sin x = 0.42 and
3π /2 < x < 5π /2.
41. Answer true or false.
43. Answer true or false.
p
2
F ′ (x) = 2x 1 + e−x .
x
G(x) =
by
45.
Z
(d) sech2 (5x4 )
40.
Z
tanh6 xsech2 x dx =
If F(x) =
Z π
0
ex
0
Z x2 p
2
1 + e−t dt then
(t + 1)2 − 1 dt between x = 0 and x = π is given
(x + 1) dx.
dx
=
1 + e2x
(a) ex ln |1 + e2x | +C
(b) ln |1 + e2x | +C
(c) tan−1 (ex ) +C
(d) 2x +C
(d) 6.723
(c) 5x4 tanh (x5 )
= 4 cosh−1 (ex )
44. AnswerZtrueqor false. The arc length of the curve defined by
(c) 6.719
(b) −5x4 sech2 (x5 )
4dx
e2x − 1
x→∞
(b) 6.717
(a) 5x4 sech2 (x5 )
√
42. Answer true or false. lim (coth x)2 = 1.
(a) 6.715
39. Find dy/dx if y = tanh (x5 ).
Z
46.
Z
dx
√
=
x 3x2 − 1
√
(a) sec−1 (x 3) +C
√
√
(b) 3 sin−1 (x 3) +C
(c) 2(3x2 − 1)1/2 ln |x| +C
1
(d) ln |x| + ln |3x2 − 1| +C
2
140
Chapter 6: Answers to Sample Tests
Section 6.1
1. c
9. a
2. a
10. d
3. c
11. b
4. d
12. a
5. b
13. b
6. true
14. c
7. false
15. a
8. a
2. b
10. a
3. d
11. c
4. true
12. a
5. d
13. d
6. c
14. d
7. c
15. c
8. d
16. a
2. a
10. a
3. true
11. d
4. a
12. true
5. false
13. false
6. true
14. true
7. d
15. c
8. d
2. c
10. false
3. false
11. false
4. c
12. c
5. b
13. true
6. b
14. true
7. b
15. c
8. c
16. a
2. a
10. d
3. d
11. b
4. b
12. false
5. c
13. true
6. a
14. false
7. d
15. d
8. b
16. true
2. c
10. false
3. c
11. true
4. b
12. false
5. a
13. c
6. d
14. a
7. b
15. c
8. false
16. true
2. a
10. b
18. c
3. a
11. false
4. c
12. d
5. b
13. false
6. b
14. d
7. false
15. a
8. b
16. c
2. a
10. false
3. b
11. a
4. a
12. false
5. a
13. false
6. b
14. false
7. a
15. false
8. a
2. false
10. b
18. false
26. true
34. d
42. true
3. d
11. c
19. a
27. false
35. b
43. true
4. a
12. false
20. b
28. true
36. a
44. true
5. false
13. b
21. true
29. false
37. c
45. c
6. d
14. c
22. b
30. b
38. b
46. a
7. b
15. false
23. a
31. false
39. a
8. false
16. c
24. b
32. d
40. d
Section 6.2
1. c
9. b
17. true
Section 6.3
1. false
9. a
Section 6.4
1. true
9. a
17. b
Section 6.5
1. a
9. false
Section 6.6
1. a
9. true
Section 6.7
1. d
9. a
17. c
Section 6.8
1. c
9. a
Chapter 6 Test
1. a
9. true
17. d
25. true
33. true
41. false
Chapter 7: Principles of Integral Evaluation
Summary: The primary focus of this chapter is to introduce and explain more
advanced integration techniques. With all of the techniques that are introduced,
however, the focus is usually to try and find a way to use the basic integral rules (or
basic list of antiderivatives) that were introduced in Chapter 4. Many of the ideas
in this chapter are methods to use algebraic manipulation or special substitutions
so that integrals may be simplified into an integral resembling a basic integral rule.
IDEA: At this point the reader should review the integration techniques from
Chapter 4. Particular emphasis should be placed upon the basic antiderivatives (§4.2) and u-substitution (§4.3).
One very powerful method that is introduced in this chapter is the method of integration by parts. This method can simply be thought of as the product rule for
antiderivatives. Next various trigonometric integrals are considered and then some
special trigonometric substitutions that can be used to simplify a variety of integrals that involve square roots and differences and sums of squares. In the case of
rational integrands, the method of partial fraction decomposition may sometimes
be used to simplify the integrand. In cases where antiderivatives cannot be expressed using simple functions, the idea of numerical integration is considered for
definite integrals. Finally at the end of the chapter, integrals are considered that
may involve either infinite limits of integration or integrands that have an infinite
discontinuity (such as a vertical asymptote).
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. be familiar with the basic integration techniques (see §4.2 and §7.1)
2. be aware of the different resources for evaluating integrals (§7.1)
3. be able to use integration by parts and recognize when it will be useful
(§7.2)
4. manipulate integrands using trigonometric identities and formulas to obtain
more simple integrals (§7.3)
5. be able to use trigonometric substitution and recognize when it can be used
(§7.4)
141
basic antiderivatives
→ §4.2
142
6. use partial fraction expansion to evaluate integrals (§7.5)
7. use CAS and tables of integrals where appropriate (§7.6)
8. use numerical integration techniques where appropriate (§7.7 and also §4.4)
9. identify definite integrals as improper and determine whether they converge
(to a value) or diverge (§7.8)
7.1 An Overview of Integration Methods
PURPOSE: To review general integration techniques.
REVIEW:
1. basic antiderivatives (§4.2)
2. u-substitution (§4.2 and §4.9)
3. F.T.C (§4.6)
There are no new ideas in this section but this section does serve as a good checkpoint for the reader. There are several concepts and methods that the reader should
be comfortable with before attempting to learn the material in this chapter. First,
this section should serve as a reminder to the reader of all of the integration rules
that have been introduced in the text so far. It is also a good idea at this point to review the integration techniques that have been introduced so far. This essentially
means to review the basic integral formulas that are related to polynomials, logarithms, exponential functions, trigonometric functions, hyperbolic functions and
inverse trigonometric or hyperbolic functions. Also, the reader should be comfortable with the idea of u-substitution both in indefinite integrals and with definite
integrals. At this point, evaluating definite integrals using the Fundamental Theorem of Calculus should seem more natural.
IDEA: The emphasis throughout this chapter will be to use tables, transformations and other techniques to evaluate integrals that do not qualify as any of
the “basic” integrals seen in Chapter 4.
A free CAS available online is
Maxima:
→ maxima.sourceforge.net
The emphasis in this chapter is going to be on using tables and transformation
methods. For example, appropriate transformation methods can simplify an integral so that the reader only has to use the basic integral rules that were introduced
in Chapter 4. Take some time now to locate the integral tables at the beginning and
end of the book. Before starting the chapter the reader may also want to identify a
CAS that they can use (for example, Maple, Mathematica and Derive are all good
choices). This is important for two reasons. First, using a CAS can be a valuable
tool for checking your answers as you progress through the material. On the other
hand, there are some problems that are intended to be done using a CAS.
Checklist of Key Ideas:
review basic integration techniques
technology; CAS
tables of integrals
transformation methods
143
7.2 Integration by Parts
PURPOSE: To develop a versatile integration technique that allows many integrals to be rewritten involving a simpler integrand.
Integration by parts can be thought of as the product rule for antiderivatives. In
fact, the product rule for derivatives gives a good way to remember this formula.
Here is the differential version of the product rule.
d(uv) = u dv + v du
Then the following is the integral version.
Z
(u dv + v du) =
Solve for
Z
Z
Z
u dv +
Z
v du = uv
u dv to find the integration by parts formula.
u dv = uv −
Z
v du
IDEA: Integration by parts is really just an antiderivative version of the product
rule.
This simple expression of the integration by parts formula can be deceptive, however. The variables u and v are actually supposed to represent functions of x. So
the differential form of the product rule could really be written as follows.
d u(x)v(x) = u(x) v′ (x)dx + v(x) u′ (x)dx
Then the integral version would be as shown below.
Z
u(x)v′ (x) dx +
Z
v(x)u′ (x) dx = u(x)v(x)
So the integration by parts formula is as follows.
Z
u(x)v′ (x) dx = u(x)v(x) −
Z
v(x)u′ (x) dx
CAUTION: Do not forget the constant of integration with indefinite integrals.
Another important thing is to not forget the constant of integration. As long as
there is still an indefinite integral involved, there should be a C as a part of the antiderivative. On the other hand, it is important to remember with definite integrals
that even the uv term needs to be evaluated at the limits of integration.
Zb
a
x=b Zb
u(x)v (x) dx = u(x)v(x) − v(x)u′ (x) dx
′
x=a
a
Note:
u = u(x)
dv = v′ (x) dx
144
IDEA: Integration by parts may allow a complicated integrand to be rewritten
as a more basic integrand.
R
The basic idea behind integration by parts is that the integral u dv may not be a
function that can be integrated by our basic techniques.
But through an appropriate
R
choice of u and dv, the resulting integral of v du may be easier to handle. The
toughest part about this method is picking u and dv. Finding du and v can be
accomplished by taking the derivative of u and integrating dv respectively.
IDEA: du is found by taking the derivative of u. v is found by integrating dv.
LIATE Method:
pick u towards the top
⇓
L - logarithmic
I - inverse trigonometric
A - algebraic, polynomial
T - trigonometric
E - exponential
⇓
pick dv towards the bottom
The LIATE method tries to make picking u and dv easier by following a particular
pattern. For example, the types of functions represented by LIATE (Logarithmic
– Inverse Trigonometric – Algebraic – Trigonometric – Exponential) can be used
as a guide for picking u and dv (see margin to the left). Algebraic functions really
stand for polynomials or rational functions.
tabular integration by parts
Tabular integration by parts can simplify the process of using integration by
parts. After u = u(x) and dv = v′ (x)dx are initially picked, tabular integration
by parts creates two columns. In the “u” column, derivatives of u are calculated.
In the “dv” column, dv terms are integrated over and over. This process should
simply be continued until the derivative of u either is zero or it repeats itself as the
original u. Then the appropriate terms should be matched from each column with
the appropriate signs and a constant
of integration added to the end. If the original
R
u and dv are repeated, then u dv will need to be solved for algebraically.
IDEA: Integration by parts can be tried on any integral such as
letting u = f (x) and dv = dx.
R
f (x) dx by
Another difficult thing to decide is whether or not to even use integration by parts.
It is a very versatile method and very often just trying it can lead to an integral
with a basic
formula. One choice for u and dv that can always be tried with the
R
integral f (x) dx is u = f (x) and dv = dx. On the other hand, there are times
when integration by parts must be used. Some obvious examples are when there
is a product of functions of different types that does not give rise to a basic integral
form.
Integration by parts also gives rise to the following useful trigonometric reduction
formulas.
Z
1
n−1
sinn x dx = − sinn−1 x cos x +
n
n
Z
sinn−2 x dx for n ≥ 2
n−1
1
sin x cosn−1 x +
cosn−2 x dx for n ≥ 2
n
n
These formulas are used in the next section on trigonometric integrals (see §7.3).
Z
cosn x dx =
Z
145
Checklist of Key Ideas:
reverse of product rule
guidelines for integration by parts; choosing u and dv
LIATE
tabular integration by parts
integration by parts and definite integrals
reduction formulas
7.3 Integrating Trigonometric Functions
PURPOSE: To develop integration techniques for integrands that
are comprised of powers or products of trigonometric functions.
Some integrals are comprised solely of trigonometric functions. These are integrals where powers of trigonometric functions occur either by themselves or as
products. Most often the techniques that are used here are reduction techniques.
The goal is to transform the integrand into a simpler expression for which the
basic methods from Chapter 4 can be applied. Sometimes this takes the form of
using reduction formulas or other times, an appropriate trigonometric identity can
be used with u-substitution to simplify the integrand.
reduction formulas
see §7.2
IDEA: The goal is to reduce or transform an integrand to the point where basic
antiderivative techniques may be used.
In many cases, trigonometric identities and definitions can be used to simplify a
given integral. For example, the integral
Z
sin2 x cos2 x dx
does not have a simple antiderivative and u-substitution does not work directly.
However, the Pythagorean identity and the cosine reduction formula can be used
on this integral. Using sin2 x = 1 − cos2 x leads to the following.
Z
(cos2 x − cos4 x) dx
Then the trigonometric reduction formulas (see §7.2) can be used to evaluate these
two integrals.
Z
cos2 x dx =
1
1
sin x cos x +
2
2
Z
1
dx = (sin x cos x + x) +C
2
Pythagorean Identity
sin2 u + cos2 u = 1
146
R
cos4 x dx=
=
3
1
sin x cos3 x +
4
4
Z
cos2 x dx
3
1
sin x cos3 x + (sin x cos x + x) +C
4
8
These are then combined to give the antiderivative below.
1
1
x
sin2 x cos2 x dx = − sin x cos3 x + sin x cos x + +C
4
8
8
Z
IDEA: Integrals involving trigonometric functions may often be evaluated in
more than one way.
There is often more than one way
Z to integrate trigonometric integrals. For example, consider again the integral
sin2 x cos2 x dx. By using a double-angle identity
for sine we have
sin2 x cos2 x = (sin x cos x)2 =
1 2
1
sin (2x) = (1 + cos (4x))
4
8
and then it is a simple matter to integrate as follows.
1
8
Note:
Many of these techniques make
use of basic trig definitions,
double angle-identities, and the
Pythagorean identity.
Z
(1 + cos (4x)) dx =
x sin (4x)
+
+C
8
32
General techniques are given for the following types of integrals in this section:
Z
sinm x cosn x dx
Z
tann x dx
Z
sin (mx) cos (nx) dx
Z
secn x dx
Z
tanm x secn x dx
where m and n are positive integers.
Checklist of Key Ideas:
reduction formulas
simplifying using trigonometric identities
products of sines and cosines to integer powers
integrating powers of tangent and secant
trigonometric identities and u-substitution
147
7.4 Trigonometric Substitutions
PURPOSE: To develop integration techniques to handle integrands that involve square roots.
Some integrals involve square root terms where u-substitution is ineffective. In
some of these cases, a special type of substitution called a trigonometric substitution may be used to simplify the square root so that the integral has a more
simple form. The idea with each trigonometric substitution is to make use of the
Pythagorean identity or one of its forms in order to simplify the radical.
Three Forms of
the Pythagorean identity:
sin2 u + cos2 u = 1
tan2 u + 1 = sec2 u
1 + cot2 u = csc2 u
IDEA: Trigonometric substitution involves u-substitution combined with the
Pythagorean identity.
√
For integrals that have a2 − x2 , a substitution of x = a sin u will often work. Notice that dx = a cos u du. This substitution can be remembered since this square
root looks similar to the one which is involved in the integral for sin−1 x. (Incidentally, a substitution of x = a cos u should also work.) The substitution will get
rid of the square root in the following way.
p
√
a2 − x2 = pa2 − a2 sin2 u
= a√ 1 − sin2 u
= a cos2 u
= ±a cos u
√
For integrals that have a2 + x2 or a2 + x2 , a substitution of x = a tan u will often work. Simplification will come from an alternative form of the Pythagorean
identity:
√
a2 − x2 → try x = a sin u
a2 + x2 → try x = a tan u
1 + tan2 x = sec2 x
This substitution x = a tan u can be remembered since the form a2 + x2 looks similar to the integral that is related to tan−1 x.
Completing the square on some quadratic terms can lead to terms that look like
a2 + (x − h)2 . The appropriate substitution should be x − h = a tan u and then
dx = a sec2 u du.
√
For integrals that have x2 − a2 , a substitution of x = a sec u will often work.
Simplification will come by rearranging the version of the Pythagorean identity
used above.
1 + tan2 x = sec2 x −→ sec2 x − 1 = tan2 x
This substitution x = a sec u can be remembered since the form
similar to the integral that relates to sec−1 x.
√
x2 − a2 looks
√
x2 − a2 → try x = a sec u
148
Checklist of Key Ideas:
method of trigonometric substitution
√
integrals involving a2 − x2
√
integrals involving a2 + x2
√
integrals involving x2 − a2
integrals involving irreducible quadratic denominators, a(x − h)2 + c2
7.5 Integrating Rational Functions by Partial Fractions
PURPOSE: To develop a technique to assist in integrating rational functions.
When a proper rational function is in an integral, it may sometimes be rewritten
using partial fraction decomposition. This amounts to taking a fraction and
writing it as a sum of fractions that do not have a common denominator. Often the
resulting pieces or partial fractions are simpler and easier to handle in an integral.
Partial Fraction Decomposition:
1. write as a proper rational function
2. factor the denominator
3. solve for constants in numerator
To be able to use partial fraction decomposition requires three things. First, the
rational function should be written as a proper rational function (degree of the
polynomial on the bottom is greater than the top). Second, the denominator of the
rational expressions should be factored. Last, after the partial fraction decomposition is performed, the values of the resulting constants in the numerators need to
be determined.
The quadratic equation is useful for
factoring.
If ax2 + bx + c = 0 then
√
−b ± b2 − 4ac
r1,2 =
2a
⇓
ax2 + bx + c = (x − r1 )(x − r2 )
There are several cases considered here. Each case depends upon the types of
factors in the denominator.
1. distinct linear factors
A
B
ax + b
=
+
(x − r1 )(x − r2 ) x − r1 x − r2
2. repeated linear factors
ax + b
A
B
=
+
(x − r)2
x − r (x − r)2
ax2 + bx + c
A
B
C
=
+
+
(x − r)3
x − r (x − r)2 (x − r)3
149
3. distinct quadratic factors
b3 x3 + b2 x2 + b1 x + b0
A1 x + A0
B1 x + B0
=
+
(a2 x2 + a1 x + a0 )(c2 x2 + c1 x + c0 ) a2 x2 + a1 x + a0 (c2 x2 + c1 x + c0 )
4. repeated quadratic factors
A1 x + A0
B1 x + B0
b3 x3 + b2 x2 + b1 x + b0
=
+
(a2 x2 + a1 x + a0 )2
a2 x2 + a1 x + a0 (a2 x2 + a1 x + a0 )2
There are two ways to evaluate the constants in the numerators after partial
fractions have been determined based upon the factors of the denominator. A new
numerator involving the new, unknown coefficients can be found by recombining
the partial fraction terms into one rational term. Then several equations for the
coefficients can be obtained by equating the numerators of the terms with the
full denominators. Then the coefficients can either be determined by trying to
match the coefficients of the terms with the corresponding powers of x or by
substituting different values for x into the equations.
Checklist of Key Ideas:
partial fraction decomposition
proper rational function
determining coefficients by matching terms
determining coefficients by substitution
linear factors
quadratic factors
quadratic formula
using long division (or synthetic division) when possible
7.6 Using Computer Algebra Systems and Tables of
Integrals
PURPOSE: To introduce alternative methods for evaluating integrals.
There are an infinite variety of integrals that can arise. It will not always be a
simple process to find a substitution or transformation that will allow a simple
antiderivative to be found. Using a table of integrals can shorten the effort required to evaluate a given integral. Likewise, a Computer Algebra System (CAS)
can greatly shorten the time required to evaluate an integral. It is still useful to
evaluating constants:
1. substitution
2. matching terms
150
understand how integration works, however, as sometimes the results that are returned by a table of integrals or CAS may seem more mysterious then the original
integral.
CAUTION: Sometimes no perfect match or no possible match for an integrand
can be found → u-substitution may help.
When using a table of integrals, sometimes a perfect match cannot be found for
the integral that is being considered. In some cases, an appropriate u-substitution
can be used to find a form in the tables that does match. One example of this is
integrals that involve fractional powers of x. In these cases, a good substitution to
try is u = x1/r where r is the LCD of all of the fractional powers of x. This can
have the effect of removing all of the fractional powers of x so that other methods
may then be employed.
Checklist of Key Ideas:
be familiar with what is in the table of integrals
perfect matches and imperfect matches
using substitution to create a match
integrals involving fractional powers of x
rational trigonometric functions
understanding and using integration with CAS
7.7 Numerical Integration; Simpson’s Rule
PURPOSE: To develop techniques for approximating the value of
a definite integral.
see also §4.4
midpoint method
Definite integrals will often arise that cannot be evaluated using the Fundamental
Theorem of Calculus since a simple antiderivative is not known for the integrand.
In these cases, numerical integration may be the only way to find a value for the integral. Some forms of numerical integration have already been encountered: left
endpoint approximations, right endpoint approximations and midpoint approximations. More sophisticated methods are introduced here.
The midpoint method has already been introduced (i.e., the rectangle method
with heights of rectangles determined at the midpoints of the intervals). This is
sometimes called the tangent line approximation because it is equivalent to drawing a tangent line to a curve at the midpoint of each subinterval. The trapezoidal
area underneath this tangent line is the same as the area of the rectangle determined by the midpoint method. This is due to the symmetry of the tangent line at
the midpoint of the subinterval.
151
The trapezoidal approximation is the average of the left and right endpoint approximations. In essence, each subinterval is approximated by the area of a trapezoid. The trapezoid is formed by connecting the function values at the left and
right endpoints of the subinterval with a straight line. Then the area of the resulting trapezoid is used to approximate the area under the curve on the subinterval.
trapezoidal method
Simpson’s rule is the result of drawing a quadratic function over the interval and
then finding the area under the quadratic function on the interval. The quadratic
function that is used is required to go through the function values at the left endpoint, right endpoint and the midpoint of the subinterval. As it turns out, the
following is true.
Simpson’s rule
1
1
Sn = (Ln/2 + 4Mn/2 + Rn/2 ) = (Tn/2 + 2Mn/2 )
6
3
Usually Simpson’s Rule is applied on an even number of intervals so that the
midpoints occur at the nodes of the partition that is being used and extra function
values do not need to be calculated. Thus, in the formula given above, n and n/2
should be integers.
Checklist of Key Ideas:
trapezoidal approximations
midpoint approximations; tangent line approximation
Simpson’s Rule
absolute errors; |EM |, |ET | and |ES |
overestimates and underestimates
error bounds
7.8 Improper Integrals
PURPOSE: To evaluate integrals with discontinuous integrands
or infinite limits of integration.
When an infinite discontinuity (such as a vertical asymptote) or an infinite limit of
integration appears within a definite integral then we have an improper integral.
In some of these cases, a value for the integral can be found. Then the definite
integral is said to converge. In cases where an improper integral does not evaluate
to some finite number, then the integral is said to diverge.
To evaluate improper integrals, the definite integral is rewritten using a dummy
variable such as b. There are two situations that may arise: if the limits of integration are infinite or if the integrand is discontinuous in some way (particularly
at the endpoints of the interval). In the case of infinite limits of integration, the
converge vs. diverge
Types of Improper Integrals:
1. infinite limit of integration
2. discontinuous integrand
152
limits are rewritten using a dummy variable (such as b). The integral is evaluated
with the dummy variable and then the limit is taken as the dummy variable (i.e.,
b) approaches the appropriate limiting value. For example, the following integral
has an infinite upper limit of integration.
Z b
0
e−x dx = lim
Z b
b→∞ 0
e−x dx
IDEA: First, evaluate the integral with a limit of integration of b. Then take the
limit involving b.
The definite integral is first evaluated using a limit of integration of b and then the
limit involving b is evaluated. If the limit exists, then the improper integral is said
to converge. Otherwise, the integral is said to diverge.
In the case where the integrand is discontinuous, the point of discontinuity may
occur either at an endpoint or in the middle of the interval. If the discontinuity occurs at an endpoint then a dummy variable is again used in place of the appropriate
limit of integration. The following integral has an integrand that is discontinuous
as it approaches the lower limit of integration.
x=1
Z 1
Z 1
√ 1/2 −1/2
−1/2
= lim 2 − 2 b = 2
x
dx = lim
x
dx = lim 2x b→0+ b
0
b→0+
x=b
b→0+
Again, the integral is evaluated first with the dummy variable, then the limit is
evaluated. If the limit exists and is finite then the improper integral converges,
otherwise it diverges.
IDEA: More than one dummy variable and more than one integral may be
required if
1. there are two infinite limits of integration.
2. there is a discontinuity in the middle of the interval.
discontinuity in middle of interval
If a discontinuity occurs in the middle of the interval then the integral needs to
be written as two integrals and two different limits taken. If either of the resulting
integrals diverges then the whole improper integral will diverge. For example,
Z 1
1
−1
x
dx = lim
Z a
1
a→0− −1
x
dx + lim
Z 1
1
b→0+ b
x
dx
which diverges because both integrals diverge. Notice that in writing two different
integrals, two different dummy variables are used.
A similar situation can also arise with infinite limits of integration. For example,
Z ∞
−∞
2
e−x dx = lim
Z 0
a→−∞ a
2
e−x dx + lim
Z b
b→∞ 0
2
e−x dx
Again, notice that two dummy variables are used, one for each infinite limit of
integration. In this example, the fact that x = 0 was chosen as the “middle” limit
of integration was arbitrary, although the same number had to be used in both
integrals.
153
Checklist of Key Ideas:
improper integrals; infinite discontinuities; infinite limits of integration
rewriting as a proper integral and taking a limit
convergence and divergence
applications to arc length and surface area
154
Chapter 7 Sample Tests
1
cos9 x +C
9
1
(b) − cos9 x +C
9
Section 7.1
1. Evaluate
Z
(a)
(8 − 2x)5 dx.
(8 − 2x)6
+C
6
−(8 − 2x)6
(b)
+C
6
−(8 − 2x)6
(c)
+C
12
−(8 − 2x)6
+C
(d)
3
Z √
2. Evaluate
4x + 9 dx.
(c) 9 cos9 x +C
(a)
1
√
+C
8 4x + 9
2
+C
(b) √
4x + 9
1
(c) (4x + 9)3/2 +C
6
1
(d) (4x + 9)3/2 +C
2
(d) −9 cos9 x +C
7.
Z
Z
(c)
(d) e
8.
Z
ex dx
5 + ex
x
+C
+C
=
5ex
+C
5 + ex
(b) 5 ln (5 + ex ) +C
(a)
9.
Z
(c) 2 ln |4 − x2 | +C
1 x2
e +C
2
cos x
sin x e
dx =
(a) cos x ecos x +C
(b) −ecos x +C
(c) − sin x
ecos x +C
(d) −xecos x +C
8
cos x sin x dx =
4x dx
=
4 − x2
(b) −2 ln |4 − x2 | +C
2
2
ex
+C
5 + ex
(a) 4 ln |4 − x4 | +C
x ex dx =
(c) ex +C
2
1
(d) x2 ex +C
2
6.
4
√
(d)
2
Z
e
(a) 3 cos (x2 ) +C
(b) 2ex +C
5.
3e x
+C
4x3/2
(c) ln (5 + ex ) +C
(a)
Z
(b)
3x sin (x2 ) dx =
(c) −6 cos (x2 ) +C
3
(d) − cos (x2 ) +C
2
4.
e x
+C
2x3/2
√
x
(b) 6 cos (x2 ) +C
Z
√
(a)
√
(a)
3.
√
e x
√ dx =
2 x
(d) 4 ln |4 − x2 | +C
10.
Z
4 sinh2 x cosh x dx =
(a)
4
sinh3 x +C
3
(b) 12 sinh3 x +C
(c) 4 sinh2 x +C
(d) 2 sinh2 x +C
11. Answer true or false. In evaluating
for u is u = x2 .
Z
12. Answer true or false. In evaluating
choice for u would be sin x.
2
x 7x dx a good choice
Z
cos6 x sin x dx a good
155
13. Answer true or false. In evaluating
choice for u would be ex + 8.
14. Answer true or false. In evaluating
for u would be sin x.
Z
15. Answer true or false. In evaluating
choice for u would be x4 .
Z
1
1
cos (2x) − x2 cos (2x) +C
2
4
1
1
(b) x sin (2x) + cos (2x) − x2 cos (2x) +C
4
2
1
(c) x sin (2x) + cos (2x) − x2 cos (2x) +C
2
1
(d) 2x sin (2x) + cos (2x) − x2 cos (2x) +C
2
ex (ex + 8) dx a good
(a) x sin (2x) +
4 sin x
dx a good choice
cos x
Z
x3 cos (x4 ) dx a good
6.
Z
Section 7.2
1.
Z
(a)
2.
Z
7.
(d)
8.
(d) 3x3 e3x +C
x cos (9x) dx =
x2 sin (9x)
+C
18
sin (9x)
(b)
+C
9
cos (9x) x sin (9x)
+
+C
(c)
81
9
cos (9x) x cos (9x)
(d)
+
+C
81
9
(a)
4.
Z
2x sin x dx =
9.
5e4x sin (3x) dx =
(a) 5e4x 3 sin (3x) + cos (3x) +C
(b) 5e4x 3 sin (3x) − cos (3x) +C
5
(c) e4x 3 sin (3x) − 3 cos (3x) +C
7
1
(d) e4x 4 sin (3x) − 3 cos (3x) +C
5
Z 1
4xe6x dx =
(b) 224.2
(c) 224.7
(d) 225.1
10.
Z 3
1
x2 ln x dx =
(a) 7.00
(b) 2 sin x − 2 cos x +C
(b) 7.03
(d) 2 cos x + 2x cos x +C
(d) 6.92
2x2 sin (2x) dx =
√
cos−1 (6x)
+C
6
(a) 223.9
(c) 6.96
(c) 2 cos x − 2x cos x +C
5.
Z
0
(a) 2 sin x − 2x cos x +C
Z
sin−1 (6x) dx =
1 − 36x2
+C
6
√
1 − 36x2
+C
(b) x cos−1 (6x) +
6
(c) 6 cos−1 (6x) +C
(c) 27e3x +C
3.
(d) x2 ln (2x) − x2 +C
(a) x sin−1 (6x) +
e3x
(9x2 − 6x + 2) +C
3
(b) xe3x +C
Z
Z
9x2 e3x dx =
(a)
1 2
x2
x ln (2x) − +C
4
4
5 2
5x2
(b) x ln (2x) −
+C
2
4
(c) 5x2 ln (2x) − 5x2 +C
(a)
x e6x dx =
e6x
(6x − 1) +C
6
e6x
(6x − 1) +C
(b)
36
1
(c) x2 e6x +C
2
e6x
(d)
+C
6
5x ln (2x) dx =
11.
Z 3
1
cos (ln x) dx =
156
1
3x
+ sin (2x) + sin 4x +C
2
8
3x 3
1
− sin (2x) + cos3 x sin x +C
(b)
2
4
8
3
3
(c) sin (2x) + sin x cos x +C
4
3x 3
+ cos (2x) + sin3 x cos x +C
(d)
2
4
(a) 1.57
(a)
(b) 1.52
(c) 1.48
(d) 1.42
12. Answer true or false.
Z π /4
0
x sin (2x) dx = 1/4.
13. Answer true or false.
Z 3π /4
14. Answer true or false.
Z 1
π /4
0
15. Answer true or false.
Z 1
0
x tan x dx = 1.
5.
Z
4xe−3x dx = 0.
1.
13
cos x sin x dx =
cos14 x
+C
14
cos14 x
+C
(b) −
14
cos14 x
(c)
+C
13
cos14 x
+C
(d) −
13
(a)
2.
Z
2 sin2 (5x) dx =
sin (10x)
(a) x −
+C
10
(b) 2x − 2 sin (5x) +C
6.
Z
ln |cos (9x)|
+C
9
ln |cos (9x)|
(b) −
+C
9
tan2 (9x)
+C
(c)
18
tan2 (9x)
+C
(d) −
18
7. Z
Answer true or false.
sin (6x) cos (4x) dx = −
8.
Z
− sin3 (2x) dx =
(a)
(b)
(c)
(d)
4.
Z
1
1
cos (2x) + sin3 (2x) +C
2
6
1
1
cos (2x) − cos3 (2x) +C
2
6
1
1
− cos (2x) + cos3 (2x) +C
2
6
1
1
− cos (2x) − sin3 (2x) +C
2
6
4 cos4 x dx =
sec2 (5x + 8) dx =
tan (5x + 8)
+C
5
tan (5x + 8)
+C
(b)
5x + 8
tan (5x + 8)
(c) −
+C
5
tan (5x + 8)
(d) −
+C
5x + 8
(a)
sin (5x)
+C
5
sin (5x)
(d) x −
+C
10
Z
tan (9x) dx =
(a)
(c) x −
3.
x sin (16x)
−
+C
8
128
x cos (16x)
(b) −
+C
8
128
x sin (8x)
(c) −
+C
8
128
x cos (8x)
(d) −
+C
8
128
(a)
ln (x2 + 20) dx = 1.
Section 7.3
Z
sin2 (4x) cos2 (4x) dx =
9.
Z
csc (6x) dx =
ln |tan (3x)|
+C
6
ln |tan (3x)|
(b) −
+C
6
ln |tan (6x)|
+C
(c)
6
ln |tan (6x)|
(d) −
+C
12
(a)
cos (2x) cos (10x)
−
+C
4
20
157
10. Answer true or false.
11.
Z
Z
tan11 x sec2 x dx =
tan12 x
+C
12
(b) 1.388
tan x sec5 x dx =
(c) 3.271
(a) sec5 x +C
(d) 1.398
sec6 x
(b)
+C
6
sec5 x
+C
(c)
5
(d) sec6 x +C
12. Answer
−
(a) 1.381
true
4.
Z 3
2
(−x)2
dx
p
=
(−x)2 − 1
(a) 14.941
(b) 0.077
(c) 0.093
or
Z
false.
cot6 (4x)
+C
24
13. Answer true or false.
Z π /4
Z π /4
0
15. Answer true or false.
5.
Z 2
4dx
√
=
4
x x2 + 5
(a) 2.003
1
0
14. Answer true or false.
(d) 17.01
cot5 (4x) csc2 (4x) dx =
Z π /3
tan2 (6x) dx = 1.00.
(b) 1.684
(c) 1.692
sec2 (x) dx = 0.
−π /3
(d) 1.698
tan (2x) dx = 0.
6.
Z 1
0
x3 dx
=
(9 + x2 )5/2
(a) 0.00086
Section 7.4
1.
Z p
(c) 0.00041
9 − x2 dx =
√
x
x 9 − x2
+ sin−1
+C
6
9
√
x 9 − x2 9 −1 x (b)
+ sin
+C
2
2
3
√
x
x 9 − x2
+ 2 sin−1
+C
(c)
2
3
√
x
x 9 − x2
(d)
+ 3 sin−1
+C
2
3
Z
dx
√
=
2.
5 − x2
√ sin−1 5 x
(a)
+C
5
−1
sin (5x)
(b)
+C
5
√ !
5x
+C
(c) sin−1
5
√
√ !
5 −1
5x
(d)
sin
+C
5
5
(a)
3.
(b) 0.00031
Z 0
−1
3ex
p
4 − 2e2x dx =
(d) 0.00082
7.
Z 4√ 2
3x − 2 dx
x
(a) 4.72
1
=
(b) 4.79
(c) 3.12
(d) 3.17
8.
Z 2π
π
cos θ d θ
p
=
4 − sin2 θ
(a) 1
(b) 0
(c) −1
(d) 4
9.
Z
dx
=
x2 + 3x + 9
√ !
√
√
2 3
−1 2 3 x + 3 3
tan
+C
(a)
9
9
2x + 3
2
+C
(b) tan−1
3
3
2
(c) tan−1 (2x + 3) +C
3
158
(d)
2
tan−1
11
2x + 3
11
10. Z
Answer true or false.
Z
dx
dx
√
= p
2
x − 4x + 2
(x − 2)2 − 2
11. Z
Answer true or false.
Z
Z
Z
dx
dx
dx
dx
=
−7
+
2
2
x
3
x − 7x + 3
x
Z 1
dx
√
=
12.
0 2 5x − x2
(a) 0.46
(b) 0.47
(c) 1
(d) 0
13.
Z 2
1
3
2
+
x−1 x
3. Which of the following is the partial fraction decomposition
3x2 − 2x + 2
of 2
?
(x + 2)(x − 1)
(d)
+C
dx
=
3x2 + 9x + 4
1
2
+
x2 + 2 x − 1
2
1
+
(b) 2
x +2 x−1
2x
1
(c) 2
+
x +2 x−1
x
2
(d) 2
+
x +2 x−1
Z
4x + 10
dx =
4.
x2 + 5x − 6
(a)
(a) 2 ln |x2 + 5x − 5| +C
(b) 2 ln |x2 + 5x − 6| +C
(a) 0.021
(c) 2 ln |x + 6| + ln |x − 1| +C
(b) 0.034
(d) 2 ln |x + 2| + ln |x + 3| +C
(c) 0.043
(d) 0.053
14. Answer true or false.
5.
Z 2
0
15. Answer true or false.
Z π
0
√
4x 16x − 1 dx = 90.9.
cos x sin x
1. Which of the following is the the partial fraction decomposi5x + 10
tion of
?
(x − 2)(x + 3)
(b) 2 ln |x2 + 1| + ln |x + 2| +C
(c) 2 tan−1 x + ln |x + 2| +C
(d) tan−1 x + ln |x + 2| +C
6. Answer true or false.
Z
ln |x + 2| + ln |x + 4| +C
7. Answer true or false.
ln |x − 2| +C
(a)
2. Which of the following is the partial fraction decomposition
5x − 2
?
of 2
x −x
1
2
(a)
+
x−1 x
1
2
+
(b)
x−1 x
3
2
(c)
+
x−1 x
x2 + 2x + 5
dx =
x3 + 2x2 + x + 2
(a) ln |4x2 + 1| + ln |2x + 4| +C
p
1 − sin2 x dx = 0.
Section 7.5
4
1
+
x+3 x−2
2
3
(b)
+
x+3 x−2
4
1
(c)
+
x+3 x−2
3
2
(d)
+
x+3 x−2
Z
8. Answer true or false.
ln |x2 + 2| +C
9.
Z
x2 + 4
dx =
(x − 1)3
(a) ln |x − 1| −
Z
x2 + 2x + 1
x3
dx =
+ x2 + x +
(x + 2)(x + 4)
3
Z
dx
= ln |x + 3| +
(x + 3)(x − 2)
2x3 + 4x2 + 2x + 2
dx = tan−1 x +
(x2 + 4)(x2 + 2)
5
2
+C
−
x − 1 2(x − 1)2
(b) ln |x − 1| +C
x3
+ 2x + ln3 |x − 1| +C
3
x3
1
(d)
+C
+ 2x −
3
2(x − 1)2
(c)
10. Z
Answer true or false.
x3 + x + 3
dx = ln |x| + ln |x + 5| +C
x(x + 5)
159
11. Answer true or false.
Z
12. Z
Answer true or false.
2x + 1
dx = ln |x2 + 2| + ln |x + 2| +C
(x2 + 2)(x + 2)
13. Z
Answer true or false.
x2 − 3x − 17
3
dx = ln |x + 7| − tan−1 (x/2) +C
2
(x + 7)(x2 + 4)
14. Z
Answer true or false.
dx
1
= tan−1 x + tan−1 (x/2) +C
2
(x2 + 1)(x2 + 4)
15. Answer true or false.
Z
x6 1
+ +C
6 x
1
ln (3x)
+C
−
(b) x6
6
36
dx
= ln3 |x − 6| +C
(x − 6)3
(a)
x6 ln (3x)
1
−
+C
6
36
x6 1
(d)
− +C
6
x
Z
√
5.
4 x ln x dx =
(c)
16
8x3/2
ln x − x3/2 +C
3
9
8 3/2
3/2
(b) 4x ln x − x +C
3
16
8x3/2
ln x −
+C
(c)
3
9
8x3/2
(d)
ln x +C
3
x dx
= ln |x + 3| +C
(x + 3)2
(a)
Section 7.6
1.
Z
6x
dx =
4x + 3
(a)
(b)
(c)
(d)
2.
Z
3 3
+ ln |4x + 3| +C
2 8
3 3
− ln |4x + 3| +C
2 8
x2
6 ln |4x + 3| + +C
2
3x 9
− ln |4x + 3| +C
2
8
(c)
(d)
Z
1
4
−
ln |4 − 5x| +C
25(4 − 5x) 25
2
ln |4 − 5x| +C
25
sin (3x) sin (7x) dx =
1
1
sin (4x) −
sin (10x) +C
8
20
1
1
(b) sin (7x) −
sin (3x) +C
8
20
1
1
(c) cos (7x) − sin (3x) +C
7
3
1
1
(d) cos (7x) + sin (3x) +C
7
3
(a)
4.
Z
Z
7.
Z
16x
dx =
(4 − 5x)2
2
(a) ln |4 − 5x| +C
5
64 + 16(4 − 5x) ln |4 − 5x|
+C
(b)
25(4 − 5x)
3.
6.
x5 ln (3x) dx =
e4x cos (2x) dx =
sin (2x)
4x cos (2x)
+C
+
(a) e
3
2
e4x
(b)
cos (2x) − sin (2x) +C
20
e4x
2 cos (2x) + sin (2x) +C
(c)
10
e4x
3 cos (2x) + 2 sin (2x) +C
(d)
6
e−3x sin (4x) dx =
e−3x
− 3 sin (4x) − 4 sin (4x) +C
5
e−3x
− 3 cos (4x) + 4 sin (4x) +C
(b)
5
e−3x
3 cos (4x) + 4 sin (4x) +C
(c)
25
−e−3x
3 sin (4x) + 4 cos (4x) +C
(d)
25
Z
dx
√
8.
=
2
x 2x2 + 6
−6x
(a) √
+C
2x2 + 6
6x
(b) √
+C
2x2 + 6
√
2x2 + 6
+C
(c)
6x
√
− 2x2 + 6
+C
(d)
6x
(a)
160
9.
Z
3. Use n = 4 subintervals with the midpoint rule to approximate
ln (4x + 2) dx =
the integral
2
(a) 2 ln (4x + 2) +C
0
(b) 2.5430
(c) 2.5621
ln2 (4x + 2)
(d)
+C
2
(d) 2.5745
10. Answer true or false. For
11. Answer true or false.
12. Answer true or false.
13.
√
Z
Z
Z
Z
x ln (6 − 2x2 ) dx a good choice
√ √ cos x dx = 2 sin x +C.
√
e
x
√
dx = e
x
√
( x + 1) +C
Z 1
0
|x − 1/3| dx.
(a) 0.2733
(b) 0.2708
(d) 0.2917
5. Use 2n = 6 subintervals with Simpson’s rule to approximate
the given integral:
Z 5
1
dx.
1 x
(a) 1.6131
2
(3x − 14)(x + 7)3/2 +C
15
(b) (x − 7)3/2 +C
2
(c) (x − 7)3/2 + x +C
3
5
2
(d) (x + 7)5/2 − (x − 7)3/2 +C
3
2
14. Answer true or false. The area enclosed by y =
π
y = 0, x = 0 and x = 4 is 128
3 .
15. Answer true or false.
4. Use n = 5 subintervals to approximate the following integral
using the trapezoidal rule:
(c) 0.2867
x x + 7 dx =
(a)
ln (x + 1) dx.
(a) 2.5257
(4x + 2) ln (4x + 2)
− x +C
4
(c) 4x ln (4x + 2) − 4x +C
(b)
for u is 6 − 2x2 .
Z 3
Z x
3
(b) 1.6094
(c) 1.7351
√
16 − x2 ,
1
dt
√
= + x.
3
t 3t − 5
(d) 1.6436
6. Use Simpson’s rule with 2n = 8 subintervals to approximate
the given integral:
Z 4
1
|3x − 4| dx.
(a) 10.875
(b) 10.833
Section 7.7
(c) 10.828
(d) 10.750
1. Use n = 10 to approximate the integral by the midpoint rule:
Z 1
2x3/2 dx
0
(a) 0.804
(b) 0.799
(c) 4.98
(d) 8.04
2. Use n = 10 to approximate the integral by the midpoint rule:
Z 1
0
(x3 + 1) dx
(a) 3.49
7. Use the midpoint rule with n = 5 subintervals to approximate
the given integral:
Z 1
0
(x3 + 4) dx.
(a) 4.24500
(b) 4.24875
(c) 4.24219
(d) 4.26000
8. Use n = 6 subintervals to approximate the integral by the
Z 3
dx
trapezoidal rule:
3
1 x +1
(a) 0.31546
(b) 1.249
(b) 0.32546
(c) 1.257
(c) 0.31398
(d) 3.51
(d) 0.31888
161
9. Use the midpoint rule with n = 5 subintervals to approximate
Z 1
dx
the integral:
.
−1 1 + x2
(a) 1.57746
(b) 1.55747
(c) 1.58118
Section 7.8
1. Answer true or false.
Z 8
dx
is an improper integral.
x−4
Z 4
dx
is an improper integral.
2. Answer true or false.
0 x−1
0
3. Answer true or false.
(d) 1.57542
4.
Z ∞
dx
x4
1
10. Use Simpson’s rule with 2n = 6 subintervals to approximate
Z 2p
x3 + 1 dx.
the integral:
Z 2
−∞
e5x dx is an improper integral.
=
(a) 1/3
0
(b) 1/6
(a) 3.23200
(c) 1/2
(b) 3.24594
(d) Diverges.
(c) 3.25988
5.
Z ∞
dx
√ =
x
6
(d) 3.24109
(a) 1/2
11. Use the trapezoidal rule with n = 6 subintervals to approximate the integral:
Z 5
2
(b) 1/6
x sin (1/x) dx.
(c) 2
(d) Diverges.
(a) 2.95032
(b) 2.95071
6.
(c) 2.94954
(c) 6
(d) Diverges.
x sin (1/x) dx.
7.
(a) 2.877695
dx
1 − x2
=
(b) −3/2
(c) 0
(d) 2.880575
(d) −π /2
13. Answer true or false. If the trapezoidal rule is used with
n = 38 subintervals to approximate the integral
| ≤ 1 × 10−3 .
Z 2
0
2
e−x dx,
8.
Z π /2
0
cot x dx =
(a) 0
If the midpoint rule is used with
n = 19 subintervals to approximate the integral
then |EM | ≤ 1 × 10−5 .
Z 2
0
2
e−x dx,
subintervals to approximate the integral
(b) 1
(c) −30.08
(d) Diverges.
15. Answer true or false. If Simpson’s rule is used with 2n = 8
|ES | ≤ 1 × 10−3 .
√
(a) 3/2
(c) 2.871765
14. Answer true or false.
Z 0
1
(b) 2.877027
then |ET
e6x dx =
(b) 1/6
12. Use Simpson’s Rule with 2n = 6 to approximate the integral:
1
−∞
(a) −1/6
(d) 2.94730
Z 4
Z 0
Z 2
0
2
e−x dx, then
9. Answer true or false.
Z 3
0
10. Answer true or false.
Z ∞
1
x−2 dx diverges.
x−2 dx diverges.
162
11. Answer true or false.
Z 1
0
12. Answer true or false.
ln (2x) dx diverges.
Z ∞
(d)
cos x dx diverges.
6. Answer true or false.
0
13. Answer true or false.
Z ∞
0
14. Answer true or false.
15. Answer true or false.
e−6x dx diverges.
Z 0
−∞
Z 4
−∞
7.
e−5x dx diverges.
Z
x−3 dx = 1 − lim
b→−∞
−2
= 1.
b2
sinh12 x cosh x dx =
(a)
Z
1
sinh13 x +C
13
choice for u is 4ex + 5.
4.
Z
x
1 p
x 9 − x2 + 9 sin−1
+C
2
9
x
9
1 p
+C
(b) x 9 − x2 + sin−1
2
2
3
p
1
x
+C
(c) x 9 − x2 − 9 sin−1
2
9
p
9
1
x
(d) x 9 − x2 − sin−1
+C
2
2
3
Z 1
2 dx
√
=
10.
0
6x − x2
(a) 1.68
(b) 1.66
(c) 1.72
(d) 1.84
Z
3x cos x dx =
x
x
e (4e + 5)dx, a good
2
1
+
is the partial fraction dex+4 x−8
3x − 12
.
composition of
(x + 4)(x − 8)
11. Answer true or false.
12.
Z
(b) 3 sin x + 3x sin x +C
(a) e3x 3 sin (6x) + 6 cos (6x) +C
(b) e3x 3 sin (6x) − 6 cos (6x) +C
(c) −e3x sin (6x) − 2 cos (6x) +C
dx =
(a) ln |9x2 + 1| + ln |2x + 4| +C
(c) 9 tan−1 x + 2 ln |x + 2| +C
(d) 3 cos x + 3x cos x +C
e3x sin (6x) dx =
2x2 + 9x + 20
x3 + 2x2 + x + 2
(b) 9 ln |x2 + 1| + 2 ln |x + 2| +C
(c) 3 cos x − 3x sin x +C
5.
tan (6x) dx =
1
ln |cos (6x)| +C
6
1
(b) − ln |cos (6x)| +C
6
1
(c)
tan2 (6x) +C
12
1
(d) − tan2 (6x) +C
12
8. Z
Answer true or false.
(a) 3 cos x + 3x sin x +C
Z
x cot x dx = 0.
(a)
2x dx
√
=
9 − x4
x
+C
(a) sin−1
3
2
x
(b) sin−1
+C
3
x
+C
(c) cos−1
3
2
x
(d) cos−1
+C
3
3. Answer true or false. In evaluating
5π /4
1
cos (14x)
sin (8x) cos (6x) dx = − cos (2x) −
+C
4
28
Z p
9 − x2 dx =
9.
(b) 13 sinh13 x +C
1
sinh11 x +C
(c)
11
(d) 11 sinh11 x +C
2.
Z
Z 7π /4
(a)
Chapter 7 Test
1.
e3x
sin (6x) − 2 cos (6x) +C
15
13.
Z
(d) 3 tan−1 x + 2 ln |x + 2| +C
x8 ln2 (x) dx =
x9
81 ln2 (x) − 18 ln (x) + 2 +C
729
1
x9 ln2 (x)
−
+C
(b)
9
81
(a)
163
(c)
x9 ln2 (x)
+C
9
17. Use 2n = 10 subdivisions to approximate the integral by
Simpson’s Rule:
x9 ln2 (x) 1
(d)
− +C
9
9
0
midpoint rule:
0
(cos x + 1) dx
(a) 4.1667
(c) 4.1995
(d) 4.2001
18.
Z ∞
0
e−5x dx =
(a) 1.8424
(a) 1/5
(b) 1.8422
(b) −1/5
(c) 5
(c) 1.8420
(d) Diverges.
(d) 1.8418
16. Use n = 10 subdivisions to approximate the integral by the
trapezoid rule:
Z 2
1
(a) 31.24
(b) 31.32
(c) 31.29
(d) 31.20
(x5 + 4) dx.
(b) 4.1892
14. Z
Answer true or false.
x2 x sin (4x) cos (4x)
x sin (4x) dx =
−
−
.
4
4
16
15. Use n = 10 subdivisions to approximate the integral by the
Z 1
Z 1
7
(x − 1) dx
19.
Z 3
0
√
dx
9 − x2
=
(a) π /2
(b) −π /2
(c) 0
(d) Diverges.
20. Answer true or false.
Z ∞
4
e2x dx diverges.
164
Chapter 7: Answers to Sample Tests
Section 7.1
1. c
9. b
2. c
10. a
3. d
11. true
4. a
12. false
5. b
13. true
6. b
14. false
7. d
15. true
8. c
2. a
10. a
3. c
11. b
4. a
12. true
5. c
13. false
6. b
14. false
7. a
15. false
8. d
2. a
10. true
3. b
11. c
4. a
12. true
5. a
13. false
6. b
14. false
7. true
15. true
8. a
2. c
10. true
3. c
11. false
4. b
12. a
5. d
13. c
6. a
14. true
7. a
15. true
8. b
2. c
10. false
3. c
11. false
4. b
12. false
5. c
13. true
6. false
14. false
7. false
15. false
8. false
2. b
10. true
3. a
11. false
4. b
12. false
5. a
13. a
6. c
14. false
7. d
15. false
8. d
2. b
10. d
3. c
11. c
4. c
12. b
5. a
13. true
6. d
14. false
7. a
15. true
8. b
2. true
10. false
3. true
11. false
4. a
12. true
5. d
13. false
6. b
14. true
7. d
15. false
8. d
2. b
10. a
18. a
3. true
11. true
19. a
4. a
12. c
20. true
5. d
13. a
6. false
14. false
7. b
15. d
8. true
16. c
Section 7.2
1. b
9. b
Section 7.3
1. b
9. a
Section 7.4
1. b
9. a
Section 7.5
1. a
9. a
Section 7.6
1. d
9. b
Section 7.7
1. b
9. a
Section 7.8
1. true
9. true
Chapter 7 Test
1. a
9. b
17. a
Chapter 8: Mathematical Modeling with
Differential Equations
Summary: This chapter brings together the two important ideas of differentiation and integration of functions. These ideas are related by looking at equations
that involve both a function, y(x) and some of its derivatives, (i.e., y′ (x), y′′ (x),
etc.) which may be solved in some cases using integration. Topics discussed
in this chapter include how to solve first order differential equations, how to use
differential equations to model various physical situations, and how to find the
unique solution of a differential equation that is paired with an initial condition.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Determine the order of a differential equation (§8.1).
2. Determine if a given function is a solution of a differential equation (§8.1).
3. Solve a first order separable differential equation (§8.2).
4. Draw a slope field or direction field for a differential equation (§8.3).
5. Use Euler’s method to approximate the solution of a differential equation
(§8.3).
6. Solve a first order linear differential equation (§8.4).
7. Model various applications using first order differential equations (§8.1,
§8.2, and §8.4).
8.1 Modeling With Differential Equations
PURPOSE: To define what a differential equation is and to solve
some first order differential equations.
165
166
This section introduces what differential equations are and what it means for a
function to solve a differential equation. Two types of differential equations are
discussed: first order linear equations and separable equations.
CAUTION: Differential equations will involve some unknown function y = y(t).
Usually this function is just written as y although it is still a function of some
independent variable.
A first-order differential equation is an equation that involves only y, its derivative,
y′ and an independent variable t. Sometimes, the independent variable will not
occur explicitly in the equation. For example, y′ = y2 , really could be written as
y′ (t) = y(t)2 .
solution
general solution
initial-value problem (IVP)
initial conditions
A function, y = φ (t), is a solution of the differential equation if an equality results
(i.e., both sides of the equation are equal) when the function is inserted into the
differential equation. A general solution of a differential equation involves a
constant of integration that allows for an infinite number of different functions to
solve the differential equation. An Initial-Value Problem (IVP) is a differential
equation that is paired with some initial condition. A solution of an IVP must solve
the differential equation and also satisfy the initial condition(s). Initial conditions
are substituted into the general solution in order to determine a unique value for
the constant of integration.
Checklist of Key Ideas:
differential equation
order
general solution
integral curves
initial-value problems
initial conditions
logistic growth
spread of disease
8.2 Separation of Variables
Purpose: To solve separable differential equations.
separable equation
Some first-order differential equations are separable. A separable equation can
be written in the form
H(y)y′ = F(t) −→ H(y)dy = F(t)dt.
167
Then to find y, just integrate both sides. The left side should be integrated with
respect to y and the right side should be integrated with respect to the independent
variable (t in this case). After doing this, it may or may not be possible to solve
explicitly for y.
CAUTION: Some separable differential equations do not appear to have an
independent variable. For example, y′ = y2 , really should be thought of as,
y′ (t) = y(t)2 .
Many applications can be modeled by first-order differential equations. Some of
the applications that are discussed in this section are (1) the exponential growth
or decay of some substance or population, (2) determining the amount of a drug
present in the bloodstream at some time, t, (3) the spread of disease over time,
and (4) continuous compounding. The equation that is most often used in these
models is the following initial-value problem:
y′ = ky, y(t0 ) = y0
This equation relates to exponential growth and decay which shows up in a surprising number of different applications.
In this differential equation, when k > 0 then it models growth and when k < 0
then it models decay. The general solution of the differential equation y′ = ky is
always of the form
y = Cekt .
The constant C is determined by the initial conditions at t = t0 . If t = t0 then the
solution becomes
y = y0 ekt
For all of the application problems that involve this differential equation, a key
item is being able to find the growth/decay constant, k. A typical problem usually
has the following format:
1. Given: initial condition at t0 .
2. Given: a value for y at some other time.
3. Find: the value of y at some t.
The method of solution typically has the following structure:
1. Use the initial condition to find C.
2. Find k using the other specified value for y.
3. Find the value of y at the time in question.
growth vs. decay
y′ = ky is solved either as a firstorder linear equation or as a separable equation
168
half-life
doubling time
Finding k often involves using what is called the half-life or the doubling time.
Each of these are simply ways of specifying a value for y at some given time. The
half-life gives the time after which only half of the original amount is left. Halflife occurs in decay problems (k < 0). The doubling time indicates how long it
takes for the amount present to become twice the original amount. Doubling time
occurs in growth problems (i.e., k > 0). Generally if t = T is the doubling time
then the growth constant k is given by
k=
ln 2
.
T
On the other hand, if t = T is the half-life, the decay constant k is given by
k=
ln (1/2)
ln 2
=−
.
T
T
IDEA: Half-life or doubling time are calculated the same way. For the differential equation y′ = ky if
1. y0 = y(t0 ) ← the original amount
2. ya = y(ta ) ← amount at some later time
then the growth/decay constant is k =
ln (ya /y0 )
.
ta
Checklist of Key Ideas:
first order separable equations
method of separation of variables
exponential growth or decay models
k, the growth or decay constant
general solution of y′ = ky
relative growth or decay rate
doubling time or half-life
radioactive decay
carbon-14 dating
8.3 Slope Fields; Euler’s Method
PURPOSE: To introduce a method for visualizing the behavior of
solutions to first-order differential equations.
169
A key feature of any differential equation is that it involves the derivative (or
derivatives) of some (unknown) function. The derivative itself gives a way for
finding the slope of a function at a point. A slope field uses a differential equation
to find the slopes of possible solutions to the differential equation at many different
points. This can often provide useful information about the differential equation
and how it will behave.
slope field
IDEA: The equation y′ = f (x, y) allows the slope of y to be calculated at some
point (x, y) by evaluating f (x, y) at the point.
Euler’s method uses the idea of the derivative representing the slope to approximate a solution of an initial-value problem that involves a first order differential
equation.
IDEA: Euler’s method for solving y′ = f (x, y) with y(x0 ) = y0 :
yn+1 = yn + ∆x f (xn , yn )
Euler’s method is closely related to the left rectangle method for approximating
the value of an integral. The fundamental theorem of calculus says
Z xn+1
xn
y′ dx = y(xn+1 ) − y(xn ).
Then Euler’s method makes two approximations: one for the left side of this
equation and one for the right. The integral is approximated by a left rectangle. The value of the derivative at the left end of the interval is f (xn , yn ). The
width of the interval is xn+1 − xn = ∆x. So the integral is approximately equal to
the area of the left rectangle which is given by ∆x · f (xn , yn ). Then the right hand
side of the equation is replaced with the approximate values of y. For example,
yn+1 − yn ≈ y(xn+1 ) − y(xn ). So the fundamental theorem of calculus becomes
∆x f (xn , yn ) = yn+1 − yn .
When this is solved for yn+1 , this gives Euler’s Method.
IDEA: Typically xi = x0 + i∆x and yi is the approximate value of y(xi ) in Euler’s
method.
Euler’s method is easily implemented. All that is needed is a starting value for
x = x0 and y = y0 which are given in an IVP. Then a value for ∆x is picked and the
next value y1 is approximated. Then x1 = x0 + ∆x and y1 are used to find y2 and
so forth. For example if y(0) = 1 then x0 = 0 and y0 = 1. Then to approximate
y(1), a choice for ∆x needs to be made. If ∆x = 0.5 then x1 = x0 + ∆x = 0.5 and
x2 = x1 + ∆x = 1 so y(1) would be approximated by y2 (i.e., y2 ≈ y(x2 ) = y(1)).
Checklist of Key Ideas:
slope field or direction field
Euler’s method
increment or step-size, ∆x
absolute and percentage error
Euler’s method
170
8.4 First-Order Differential Equations and Applications
PURPOSE: To solve linear first order differential equations and
their applications.
first-order linear
First-order linear differential equations are differential equations that can be
written in the following form.
y′ + p(t)y = g(t)
linear vs. nonlinear
Anything that cannot be written in this form is nonlinear. The equation, y′ = y2 ,
is first-order but it is nonlinear since it cannot be written in linear form.
method of integrating factor
First-order linear differential equations can be solved using the method of integrating factor. Here is an outline for this solution method:
1. write the D.E. as y′ + py = g
R
2. find the integrating factor µ (t) = e
p(t)dt
3. multiply by µ to obtain
d
y·µ = g·µ
dt
4. Integrate both sides (with respect to t) to obtain
y · µ = G(t) +C
R
where G(t) = g · µ dt
5. Solve for y by dividing by µ
mixing problems
Mixing problems always have the following format:
dy
= rate in − rate out,
dt
where y(t) represents the amount of some substance in a mixing tank. A common
mistake is to assume that y(t) represents the liquid in the tank or that y(t) is a
concentration.
CAUTION: The function y(t) represents an amount, not a concentration.
This is not the case. The function y(t) usually will have units that represent some
amount. The differential equation is often easier to write down by keeping track
of the units. All three terms, dy/dt, the “rate in”, and the “rate out” must have the
same units. For example,
rate in = (concentration in) × (flowrate in).
171
The velocity of a falling body can be modeled by the differential equation
dv c
+ v = −g,
dt m
falling body
v(0) = v0
where m is the mass of the object, g is the gravitational constant, and c is a constant of proportionality for the drag force due to wind resistance. Note that this
differential equation assumes that down is the negative direction. This is a linear
differential equation that can be solved by the method of integrating factor. After
solving the differential equation, it can be seen that the speed of the object after a
mg
.
long period of time will approach a terminal speed of vt =
c
Checklist of Key Ideas:
first order linear equations
integrating factor and the method of integrating factors
mixing problems
modeling free fall with air resistance
terminal or limiting velocity
terminal speed
172
Chapter 8 Sample Tests
Section 8.1
(a) y′ − 5y = −25t + 5
1. State the order of the differential equation 3y′′ + 7y = 0.
(a) 0
(c) y′ + 5y = 25t − 5
(d) y′ + 5y = 5t
(b) 1
9. Which of the following differential equations models logistic
growth?
(c) 2
(d) 3
2. State the order of the differential equation
(b) y′ − 5y = 5t
y′ − 5y2
(a) y′ = ky
= 0.
(a) 0
(b) 1
(c) 2
(d) 3
3. Answer true or false. The differential equation y′ − 5y = 0
has a general solution of y = Ce−5t .
dy
4. Answer true or false. The differential equation (3 + x) = 1
dx
is solved by y = ln |3 + x| +C when x ≥ 0.
5. Find the unique solution of the initial-value problem
dy
= t 3 , y(0) = 4.
dt
1
(a) y = t 4 + 4
4
(b) y = 4t 4 + 4
(c) y = 0
1
(d) y = t 2 + 4
2
6. Find the general solution of the differential equation
dy
= t 1/5 .
dt
(a) y = t 6/5 +C
(b) y = 2t 6/5 +C
6
(c) y = t 6/5 +C
5
5 6/5
(d) y = t +C
6
7. If y = Cet +5t is the general solution to a differential equation
with y(0) = 5 then what is the value of the constant C?
(a) 0
(b) −1
(c) 1
(b) y′ = ky2
y
y
(c) y′ = k 1 −
L
(d) y′ = ky(1 − y2 )
10. Find the unique solution of the differential equation y′′ = −3x
with y(0) = 2 and y′ (0) = −5.
1
(a) y = − x3 − 5x + 2
2
(b) y = −5x + 2
3
(c) y = − x2 − 5x + 2
2
(d) y = −3x2 − 5x + 2
Section 8.2
1. Find the general solution of the differential equation
dy
+ y3 = 0.
dt
(a) y = e3t +C
(b) y = et +C
(c) y = 31 t 1/3 +C
(d) y = ±(2t +C)−1/2
2. Find the unique solution of the initial-value problem
dy
= y2 , y(1) = −1.
dt
(a) 3y − y3 − 2 = 0
(b) y + y3 + 2 = 0
1
(c) y = t 1/3 + 2
3
1
(d) y = −
t
3. Find the general solution of the differential equation
(d) 5
8. The function y = Ce5t + 5t is the general solution to which of
the following differential equations?
(a) y = ±(2x2 +C)1/2
√
(b) y = ± 2x +C
2x
= y′ .
y
173
(c) k = −0.347 (g·days)−1
(c) y = Ce2x
(d)
y = e2Cx
4. Find the unique solution of the initial-value problem
x
= y′ ,
y
y(1) = 3.
(d) k = −0.0433 (g·days)−1
11. If y = yo ekt where k > 0 and yo > 0, then the quantity that is
being modeled
(a) is increasing.
(a) y = x + 2
(b) is decreasing.
(b) y = 2x + 1
√
(c) y = x2 + 8
(d) undetermined; more information is needed.
(d) y = 3ex−1
5. Answer true or false. Suppose that a quantity y = y(t)
changes in such a way that dy/dt = ky1/5 , where k > 0. It
can be said that the rate of change of the quantity is proportional to the fifth root of the amount present.
6. Answer true or false. Suppose that a quantity y = y(t)
√
changes in such a way that dy/dt = k y, where k > 0. It
can be said that y increases at a rate that is proportional to the
square root of the time, t.
7. If an initial population of Po bacteria is growing at a rate of
3% per hour, then the number of bacteria present t hours later
is
(a) Po (1 + 0.03t)
(c) is constant.
12. If a certain population, y, is growing exponentially according
to y = yo ekt then determine the value of k if the population
doubles in T = 10 months.
(a) k = 10 ln 2 (months)−1
yo ln 2
(months)−1
(b) k =
10
ln 2
(months)−1
(c) k =
10
(d) k = 10yo ln (1/2) (months)−1
13. Answer true or false. If y(0) = 20 and the substance represented increases at a rate of 8%, then y = 20(0.08)t .
14. Answer true or false. If y(0) = 40 and the substance that is
represented decreases at a rate of 16%, then y = 40(0.16)t .
(b) Po e0.03t
(c) Po e1.03t
(d) Po (1.03)t
8. A given radioactive substance has a half-life of 151 years.
Find a formula for the amount of substance, y, that is present
at time, t if 500 g of the substance are present initially.
(a) y = 500e−0.00459t
(b) y = 500e−0.693t
(c)
y = 500e0.717t
(d) y = 500e−0.011t
9. If 400 g of a radioactive substance decays to 60 g in 12 years,
then find the half-life of the substance.
Section 8.3
1. If y′ = −x + 5y, then the slope of the direction field at (1, 2)
is
(a) 9
(b) −9
(c) 1/9
(d) −1/9
2. If y′ = cos (xy), then the slope of the direction field at (0, 3)
is
(a) 1
(a) 7.06 years
(b) π
(b) 4.38 years
(c) −1
(c) 1.9 years
(d) 0.15 years
10. A particular radioactive substance is found to decay according to the equation y′ = ky2 . If 100 g of the substance are
initially present, then what is the value of k if 50 g are present
after 4 days?
(a) k = −0.173 (g·days)−1
(b) k = −0.0025 (g·days)−1
(d) 0
3. If y′ = cos (3xy), then the slope of the direction field at (5, 0)
is
(a) 1
(b) π
(c) −1
(d) 0
4. If y′ = x cos y, then the slope of the direction field at (7, 0) is
174
(a) 7
(b) −7
(c) 0
(d) 1
5. If y′ = yex , then the slope of the direction field at (6, 0) is
(a) 6e6
(b) e6
(c) 0
(d) 6
6. If y′ = 4x − 4y, then the slope of the direction field at (1, 1) is
11. Use Euler’s method with a step-size of ∆x = 0.2 to approximate the value of y(1.6) if y satisfies the initial value problem,
y′ = sin (x − y), y(1) = 1.
(a) y(1.6) ≈ 1
(b) y(1.6) ≈ 1.0397
(c) y(1.6) ≈ 1.1102
(d) y(1.6) ≈ 1.565
12. Suppose that Euler’s method is used to approximate the
value of y(t) if y satisfies the initial value problem,
y′ = xey , y(2) = 0. How big is each step-size, ∆t, if 12 steps
are used?
(a) ∆t = 2
(a) 4
(b) ∆t = 1/6
(b) −8
(c) ∆t = 1/12
(c) 8
(d) 0
x
7. If y =
, then what is the slope of the direction field at
3y
(5, 2)?
′
(a) 1/6
(d) ∆t = 1.2
13. Consider the initial value problem y′ = y + t, y(0) = 1. The
exact solution of this problem is y = −t − 1 + 2et . If the Euler method with a step-size of ∆t = 0.25 is used to approximate y(1) then what is the absolute error of this approximation?
(b) 0
(a) 1.56
(c) −1
(b) 0.554
(d) 5/6
8. If y′ = y cosh x, then what is the slope of the direction field at
(5, 0)?
(a) 5
(b) −5
(c) 1
(d) 0
9. If y′ = (sin x)(cos
x), then what is the slope of the direction
π π
field at
,
?
4 4
(a) 1
(b) 1/4
(c) 1/2
√
2
(d)
2
10. If y′ = 3 ln x − 2 ln y, then what is the slope of the direction
field at (1, 1)?
(a) 0
(b) 2
(c) 1
(d) 2e
(c) 0.4464
(d) 0.138
Section 8.4
1. When using the method of integrating factors to solve the
linear differential equation (t + 1)y′ + ty
R = 5, the correct integrating factor to multiply by is µ = e p(t)dt where p(t) is
(a) p(t) = t
(b) p(t) = (t + 1)
(c) p(t) =
(d) p(t) =
1
t+1
t
t+1
2. When using the method of integrating
factors to solve the
√
linear differential equation y′ + y t = g(t), the correct integrating factor to multiply by is
√
(a) µ = t
√
t
(b) µ = e
(c) µ = e(2/3)t
(d) µ =
3/2
3/2
t
e√
t
3. Solve the differential equation
dy
− 2y = 0.
dx
175
(a) y = Ce2x
(b)
y = Ce−x
(c) y = eCx
(d) y = e−Cx
4. Find the general solution of the differential equation
dy
− 3y = −2et .
dt
9. Find the general solution of the differential equation
y′ + y = sin (x).
(a) y = Ce−x − 21 cos (x) + 12 sin (x)
(b) y = Ce−x − 12 cos (x) + 12 sin (x)
(c) y = Cex − 12 cos (x) − 21 sin (x)
(d) y = Cex + 12 cos (x) − 21 sin (x)
(a) y = Ce−t
(b) y = Cet
(c) y = et +Ce3t
(d) y = − ln |t| +C
5. Find the unique solution of the initial-value problem
{y′ = 4y, y(1) = e4 }.
(a) y = e4t
(b) y = 4e4t
(c) y = e−4t
1
(d) y = −
t
6. Which of the following is the general solution of the differential equation y′ + 2y = e−2x ?
1
(a) y = Ce2x − e−2x
4
(b) y = Ce−2x + xe−2x
(c) y = Ce2x − 13 e−x
(d) y = Ce−2x + e−x
7. Find the unique solution of the initial-value problem
{y′ + 4y = t, y(0) = −2}.
1
31 −4t 1
e + t−
16
4
16
33
1
1
(b) y = e−4t + t −
16
4
16
1
1
1
(c) y = e−4t + t −
16
4
16
31
1
1
(d) y = − e4t − t −
16
4
16
8. Which of the following is the general solution of the differential equation ty′ + 5y = t?
(a) y = −
C t
−
t5 6
t
(b) y = Ct 5 −
4
t
(c) y = Ct 5 +
4
C t
(d) y = 5 +
6
t
(a) y =
Chapter 8 Test
1. The order of the differential equation 4xy′′ = 5y3 + t is
(a) 0
(b) 1
(c) 2
(d) 3
2. Answer true or false. The differential equation y′ − 5y = 0 is
solved by y = Ce5t .
3. Which of the following is the general solution of the differdy
− 9y = 0?
ential equation
dt
(a) y = Ce−3t
(b) y = Ce9t
(c) y = et + 9e9t +C
(d) y = c1 e−3t + c2 e3t
4. Which of the following is the general solution of the differential equation y′′ + 25y = 0
(a) y = c1 e−5t + c2 e5t
(b) y = c1 cos (5t) + c2 sin (5t)
(c) y = c1 e−5t + c2 te−5t
(d) y = c1 e5t + c2 te5t
5. Answer true or false. The function y = cos (3t) +C solves the
differential equation y′ + y = 0.
6. Answer true or false. The function y = Cet − 4 solves the
differential equation y′ − y = 4.
7. If y′ = x cos y, then the slope of the direction field at the point
(5, 0) is
(a) 0
(b) −5
(c) 5
(d) 1
8. If y′ = 4x + 6y then the slope of the direction field at the point
(1, 2) is
(a) 14
176
(b) 16
(a) y = 60e−0.0099t
(c) 10
(b) y = 60e−0.0116t
(d) 3
(c) y = 60(0.5)−t/70
9. If y′ = exy then the slope of the direction field at the point
(8, 0) is
(a) 0
(d) y = 60e−0.059t
15. If 480 g of a radioactive substance decays to 30 g in 17 years,
then the half-life of the substance is
(b) 1
(a) 9.07 years
(c) 8
(b) 4.25 years
(d) e8
(c) 6.06 years
10. If y′ = e3xy then the slope of the direction field at the point
(7, 0) is
(d) 2.77 years
(a) 0
16. Answer true or false. The differential equation y′′ = 25t is
solved by the function y = c1 cos (5t) + c2 sin (5t).
(b) 1
17. If y = yo ekt and k < 0 then the function being modeled is
(c) 7
(d) 21
11. Answer true or false. Suppose that a quantity y = y(t)
changes in such a way that dy/dt = kt 1/5 , where k > 0. It
can be said that the rate of change of the quantity is proportional to the fifth root of time.
12. Answer true or false. Suppose that √
a quantity y = y(t)
changes in such a way that dy/dt = k t, where k > 0. It
can be said that y increases at a rate that is proportional to the
square root of time.
13. Suppose that an initial population of 20, 000 bacteria is growing at a rate of 4% per hour, and that y = y(t) is the number
of bacteria present after t hours. Write an expression for y.
(a) y = 20, 000 + 800t
(b) y = 20, 000e0.04t
(c) y = 20, 000e1.04t
(d) y = 20, 000(1.04)t
14. Suppose that a radioactive substance decays with a half-life
of 70 days. Find a formula that relates the amount present to
t, if initially 60 g of the substance are present.
(a) increasing.
(b) decreasing.
(c) constant.
(d) undetermined; more information is needed.
18. Answer true or false. An exponential decay model y = yo ekt
used to find the half-life of a substance always uses −0.2 for
k.
19. Answer true or false. If y(0) = 20 and a substance grows at a
rate of 13%, this situation may be modeled by y = 20(0.13)t
20. Answer true or false. If y(0) = yo and a substance decreases at a rate of 8%, this situation may be modeled by
y = yo (1 − 0.08t).
21. The unique solution to the differential equation y′ + ty = 8t
with y(0) = 2 is
(a) y = 8 − 6e−t
(b) y = 4 − 2e−t
2
/2
2
(c) y = 8 − 6e−t/2
(d) y = 8 − 6et
2
/2
177
Chapter 8: Answers to Sample Tests
Section 8.1
1. c
9. c
2. b
10. a
3. false
4. true
5. a
6. d
7. d
8. a
2. d
10. b
3. a
11. a
4. c
12. c
5. true
13. false
6. false
14. false
7. d
8. a
2. a
10. a
3. a
11. c
4. a
12. b
5. c
13. b
6. d
7. d
8. d
2. c
3. a
4. c
5. a
6. b
7. a
8. d
2. true
10. b
18. false
3. b
11. true
19. false
4. b
12. true
20. false
5. false
13. d
21. a
6. true
14. a
7. c
15. b
8. b
16. false
Section 8.2
1. d
9. b
Section 8.3
1. a
9. c
Section 8.4
1. d
9. b
Chapter 8 Test
1. c
9. b
17. d
178
Chapter 9: Infinite Series
Summary: Three main ideas are discussed in this chapter: sequences of numbers, infinite series (a series being the sum of numbers) and the representation
of functions using a Taylor Series or a power series. The underlying idea in this
chapter is whether or not a given sequence, infinite series or Taylor series converges.
Sequences are discussed first. Then series are considered since they can be thought
of as a sum of the terms in a sequence. The chapter concludes by considering Taylor polynomials and how they may be used to approximate or represent a function
near a particular point. A Taylor series is an infinite sum of polynomial terms of
increasing order. Maclaurin polynomials/series, which are a special case of the
Taylor polynomials/series are also discussed.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Determine if a sequence is converging to a value (§9.1).
2. Identify whether a sequence is monotone or strictly monotone (§9.2).
3. Determine if a given series is convergent or not (§9.3, 9.4, 9.5, 9.6).
4. In some cases (such as geometric series) find the sum of the series if it
converges (§9.3, 9.6).
5. Find the n-th order Taylor polynomial for a function f (x) about a point
x = x0 (§9.7).
6. Find the interval and radius of convergence of a given power series in x (or
x − x0 ) (§9.8).
7. Represent/approximate a function f (x) by an appropriate Taylor series (§9.8, 9.9).
8. Understand the connection between Taylor series and power series in x (or
x − x0 ) (§9.10).
9. Differentiate or integrate a power series term-by-term (§9.10).
179
180
9.1 Sequences
PURPOSE: To define sequences and to discuss whether or not
they converge.
n-th term, an
Sequences are simply a list of numbers. In some cases, a formula may be found
for the n-th term in a sequence. When a formula is found for a general term
in a sequence, then the sequence may be thought of as a function and can even
be graphed. This function, however, only accepts nonnegative integer values as
inputs. In this case, the general terms are often given a subscript or an index to
indicate which order they come in the sequence and hence
∞ which integer they are
depending upon. Sequences are often written as an n=1 which stands for the
following list of numbers.
∞
an n=1 = a1 , a2 , a3 , . . .
IDEA: A sequence is a list of numbers. Two important things to know about
any sequence are
1. What does the n-th term look like?
2. Does the sequence approach a limit and converge?
Since sequences are simply lists of numbers, then different sequences can be combined in all of the usual arithmetic operations (i.e., addition, multiplication, etc.).
For example, if {cn } = {an } + {bn } then c5 = a5 + b5 . Or if {cn } = {an /bn } then
c5 = a5 /b5 .
limit of a sequence
converges vs. diverges
If the terms in a sequence eventually approach a particular finite value then that
value is said to be the limit of the sequence. If this number is a finite number then
the sequence is said to converge. Otherwise, the sequence diverges.
IDEA: An important concept that is central to this chapter is whether or not a
quantity converges or diverges.
Squeezing Theorem
if {an } and {bn } have the same limit
and an ≤ cn ≤ bn for all n “large
enough” then {cn } must also have
this limit.
This idea of convergence or divergence is one of the central themes of this chapter.
One important method for checking the convergence of a sequence and its limit
is by using the Squeezing Theorem. The main idea of the Squeezing Theorem
is to find two sequences that have known limits to act as outer boundaries for the
sequence that is being investigated. If these two boundary sequences have the
same limit then the limit in the middle must also have this limit.
CAUTION: The two bounding sequences only have to “squeeze in” on the third
sequence as they approach the limit (i.e., as n gets very large). For smaller
values of n they may not necessarily bound the third sequence.
recursive sequences
Sequences are sometimes defined recursively. Then the n-th term may be defined in using some combination of the previous n − 1 terms in the sequence. Any
recursive sequence must have a few terms that are called base terms from which
181
all other terms are defined. For example, consider the following sequence of numbers.
{1, 2, 3, 5, 8, 13, . . .}
At first this may seem to have no discernible pattern. However, this is actually
the Fibonacci sequence which arises in a variety of places in mathematics such
as number theory and finding roots of nonlinear equations. The terms of the Fibonacci sequence are defined in the following way.
a1 =1
a2 =2
a3 =a1 + a2 = 3
a4 =a2 + a3 = 5
..
.
an =an−1 + an−2
In this way, it can be found that the 8-th term of the sequence is a8 = 34, for
example.
Checklist of Key Ideas:
infinite sequence
terms of a sequence
general term, an , in a sequence; index notation
graph and limit of a sequence
convergence or divergence
arithmetic of sequences
Squeezing Theorem
behavior of |an | vs. an as n → ∞
recursion, recursion formulas
9.2 Monotone Sequences
PURPOSE: To determine whether a sequence is monotone and
how this affects the convergence of the sequence.
Some sequences can be described based upon the behavior of the general term,
an , in the sequence. Building upon the idea that a sequence can be graphed like
the functions that have been discussed earlier in this book, then sequences can
also be described as increasing or decreasing. Describing a sequence as either
increasing or decreasing is not, however, dependent upon having a formula for an .
182
If a formula is known for an then its derivative may help determine if the terms in
the sequence are increasing or not.
IDEA: If an = f (n) and f ′ (n) > 0 then {an } is an increasing sequence. If
f ′ (n) < 0 then {an } is a decreasing sequence.
CAUTION: Taking the derivative of an = f (n) does not make an a continuous
function. f (n) is still only defined at positive integer values.
increasing or decreasing
strictly
monotone
If the terms of a sequence are remaining constant or becoming more positive
(“growing larger”) then a sequence is called increasing. Likewise, if the terms
are remaining constant or becoming more negative (“growing smaller”) then a sequence is said to be decreasing. If the phrase “remaining constant” is removed
then a sequence will be strictly increasing or decreasing. Monotone is then
used to say that a sequence is either increasing or decreasing. Strictly monotone
would indicate either strictly increasing or decreasing.
{an } is monotone increasing
→ an ≥ an−1 for all n
IDEA: Monotone can describe both increasing or decreasing sequences.
Monotone says that a sequence may do one or the other (increase or decrease) but not both.
{an } is strictly increasing
→ an > an−1 for all n
IDEA: “Strictly” monotone means means that a sequence will not remain constant. A strictly monotone sequence may either increase or decrease but not
both and it may not remain constant.
Monotone sequences are important because if their terms have an appropriate
bound then they can be said to converge. For example, an increasing sequence
that is approaching some upper boundary which it cannot pass must converge to
a value that is not above this upper bound. Similar statements can be said about
decreasing sequences and lower bounds.
IDEA: Some monotone sequences must converge to a limit.
1. If {an } is monotone increasing and has an upper bound then it has a
limit.
2. If {an } is monotone decreasing and has a lower bound then it has a
limit.
Checklist of Key Ideas:
increasing/decreasing
strictly increasing/decreasing
monotone, monotonic
strictly monotone
“eventually” having a property
convergence of monotone sequences
upper/lower bound
183
9.3 Infinite Series
PURPOSE: To define infinite series and investigate the convergence and divergence of some specific types.
After considering sequences or lists of numbers, now the emphasis will be placed
on adding a list of numbers together or taking their sum. The idea of an infinite
sequence has already been introduced (i.e., a list having an infinite number of
terms). The next step will be to try and sum together an infinite number of terms.
This is the idea of an infinite series which is just that - the summation of infinitely
many terms.
As with sequences, a big concern is whether or not a given infinite series will
converge. Many methods will be introduced in this chapter to determine if any
given summation will converge or not. Even if an infinite series can be found to
converge, this is no guarantee that the sum it will converge to can be found.
∞
infinite series,
∑ uk
k=1
convergence or divergence
CAUTION: Even though a series may converge, it may not always be possible
to find the sum of the series.
In order to talk about convergence, infinite series can be thought of as sequences
where each term in the sequence is a sum of a finite number of terms. These terms
in this sequence are then called partial sums. For example, if an infinite series is
∞
given by
partial sums
∑ uk then the sequence of partial sums would be
k=1
s1 , s2 , s3 , . . .
where sk is defined as follows:
s1 = u1
s2 = u1 + u2
s3 = u1 + u2 + u3
..
.
n
sn =
∑ uk
k=1
Hence, sn is usually referred to as the n-th partial sum. Then the convergence
or divergence of an infinite series is equivalent to asking if the sequence sn
approaches a limit as n → ∞.
n-th partial sum, sn
∞
CAUTION: When talking about an infinite series,
∑ uk , the expressions uk
k=1
are called the terms of the series. Do not confuse the terms of the series with
the partial sums of the series.
Three types of special infinite series are identified in this section:
geometric series,
geometric series
if |r| < 1 then
∞
a
→ ∑ ark =
1
−
r
k=0
184
∞
∑ ark
k=0
telescoping sums
→ converge and possible to find the
sum
harmonic series
→ diverges
telescoping sums
∞ 1
1
∑ − k+a
k=1 k
and harmonic series. An actual value can be found for both geometric series and
telescoping sums if they are found to converge. On the other hand, a harmonic
series is deceptive:
∞
1
∑k
k=1
This infinite series is the best example of a series where the general term uk = 1/k
approaches zero as k → ∞ and yet the infinite series itself can be shown to diverge.
This is very good reason for why other convergence tests are needed.
IDEA: It is a good idea to remember the harmonic series since it shows that
even if the n-th term in the sum goes to zero, the overall sum may still diverge.
CAUTION: The geometric series is one of the series that begins with k = 0.
Rearranging the indices may be necessary to evaluate the sum.
Geometric series have to start with k = 0 in order to use the summation formula that is given here. To evaluate some geometric series, the indices have to
be shifted. Shifting the indices usually involves rewriting the term uk and then
factoring out any constants. Consider the following geometric sum that begins at
k = 3.
∞ k
∞
∞
3
=
u
=
k
∑
∑
∑ vj
j=0
k=3 4
k=3
In this case, u3 = v0 , so it follows that k = j + 3. To shift the indices, replace k
with j + 3 in the first series. Then the lower limit would be j = 0 instead of k = 3.
The result is
3 ∞ j
∞ k
∞ j+3
∞ 3 j
3
3
3
3
3
3
=∑
=
∑ 4 =∑ 4
∑ 4 .
4
4
4
j=0
j=0
j=0
k=3
Then the sum of the series would be
3
3
27
1
= .
·
4
1 − 3/4 16
185
Checklist of Key Ideas:
sum of infinitely many terms; infinite series
terms in sum; series
sum of a sequence
n-th partial sum, sk
sequence of partial sums
convergence/divergence of an infinite series
geometric series
telescoping series
harmonic series
9.4 Convergence Tests
PURPOSE: To introduce some convergence tests and some
general properties of convergent and divergent series.
∞
For the time being, infinite series will only be considered that have nonnegative
terms being summed. Because of the harmonic series mentioned in the previous
section, it is not enough to determine if the general term in an infinite series, uk ,
approaches zero or not. On the other hand, if uk does not approach zero then
certainly an infinite series will not converge. Equivalently, if an infinite series
does converge, then this would certainly indicate that uk does approach zero as
k → ∞.
∑ uk with uk ≥ 0
k=1
relationship between convergence
and the k-th term, uk
∞
IDEA: If a series,
∑ uk , converges then uk → 0. If uk 6→ 0 then series diverges.
k=1
CAUTION: Just because the k-th term approaches zero, does not guarantee
that the series will converge (i.e., harmonic series is divergent).
Two convergence tests are considered in this section. Others are introduced in
the following sections. The two mentioned here are the integral test and the pseries test. The integral test again uses the idea that the general term uk of the
terms being summed may be thought of as a function. And if the corresponding
integral of the function converges or diverges, then the infinite series will do the
∞
same. In other words, if f (k) = uk then we will compare
Z ∞
1
∑ uk with the integral
k=1
f (x) dx. If the integral converges or diverges, then the infinite series will also
converge or diverge.
integral test
186
p-series test
The p-series test is for a special infinite series of the form
∞
1
∑ kp
k=1
If p > 1 then the infinite series will converge. If 0 < p ≤ 1 then it will diverge.
Notice that p = 1 is simply a harmonic series which has already been shown to
diverge (see §9.3).
Checklist of Key Ideas:
Σ notation for series
general term, uk
Divergence Test and its converse
algebraic properties of infinite series
Integral Test
p-series Test
9.5 The Comparison, Ratio, and Root Tests
PURPOSE: To describe two comparison tests and the ratio and
root tests for the convergence of a series.
comparison test
limit comparison test
Three more types of important convergence tests are introduced in this section.
The first type is actually two different tests with the same idea: the comparison
test and the limit comparison test. Both of these compare the k-th general term
of an infinite series with the k-th term of another series. The limit comparison,
looks at the ratio between the two terms and is sometimes easier to use.
IDEA: All of the tests in this section involve using either the k-th term in the
series, uk , or a ratio of consecutive terms, uk+1 /uk .
CAUTION: Remember that if uk 6→ 0 then the series must diverge (see §9.4).
ratio test
root test
The next two types of tests, the ratio test and the root test, focus solely on the
terms in a given infinite series. The ratio test looks at the limit of consecutive
terms in the series. If their ratio is less than one in the limit then the series will
converge. If the ratio is equal to one, then no conclusion can be made. Otherwise,
the series will diverge. The root test on the other hand, only looks at the k-th root
of the k-th term and asks what this does in the limit as k → ∞. It has a similar
conclusion as the ratio test.
CAUTION: Both the ratio and the root test fail to offer any conclusion if they
result in a value of one. In these cases, another test should be tried.
187
The ratio test is the more versatile of the two and often is used when factorial
expressions occur in the general term of the infinite series. The root test is often
used when the k-th term involves something to the k-th power.
Checklist of Key Ideas:
Comparison Test; comparing terms of two sequences
handling dominant terms in inequalities
Limit Comparison Test
Ratio Test
Root Test
9.6 Alternating Series; Conditional Convergence
PURPOSE: To investigate the convergence or divergence of series that have both positive and negative terms.
Up until now, only infinite series with nonnegative terms have been considered.
Now we investigate series that may have negative terms. For example, the series
∞
∑ (−1)k+1 uk
k=1
will alternate between positive and negative values if uk > 0. Because things are
being added and then subtracted, it is sometimes possible to find a sum for the
series if it converges.
The alternating series test asks whether or not the absolute value of the general
terms, |uk |, are getting smaller or not. If this is the case, and the absolute value
of the terms are approaching zero, then the series will converge. The ratio test is
also used in this section except it is applied to the absolute value of the ratio of
two consecutive terms.
alternating series test
ratio test for absolute convergence
IDEA: If a series converges absolutely then the series will converge.
∞
A series
∞
∑ uk is said to converge absolutely if the related series ∑ |uk | con-
k=1
absolute convergence
k=1
verges. If a series is guaranteed to converge if it converges absolutely. On the
other hand, if the series converges but this absolute value series does not, then this
is called conditional convergence.
CAUTION: A series that converges does not have to converge absolutely.
conditional convergence
188
∞
For example, the alternating series
1
∑ (−1)k k converges and yet it does not con-
k=1
verge absolutely. When the absolute values are applied, the resulting series is the
harmonic series which diverges. So this series is conditionally convergent.
IDEA: The section in the book has a very nice summary of the convergence
tests used on series. The tests covered so far are the following:
1. divergence test (§9.4)
2. integral test (§9.4)
3. comparison test (§9.5)
4. limit comparison test (§9.5)
5. ratio test (§9.5) and ratio test for absolute convergence (§9.6)
6. root test (§9.5)
7. alternating series test (§9.6)
Checklist of Key Ideas:
alternating series
Alternating Series Test
approximating sums of alternating series
absolute convergence/divergence
conditional convergence
Ratio Test for absolute convergence
summary of convergence tests
9.7 Maclaurin and Taylor Polynomials
PURPOSE: To introduce Maclaurin and Taylor polynomial approximations of a function, f (x).
Maclaurin polynomial
Taylor polynomial
Maclaurin and Taylor Polynomials are simply an extension of finding the linear
or quadratic approximation of a function f (x) at a point x = x0 . The Maclaurin
polynomial is actually a special case of the Taylor polynomial in that it is always
approximating f (x) at x = 0. So an n-th order Taylor polynomial is simply an
approximation of f (x) near x = x0 that will match the function and its first n
derivative values at x = x0 .
189
IDEA: Maclaurin and Taylor polynomials are approximations of a function that
are designed to match the function value and the first n derivatives at a point
(this point is x = 0 for Maclaurin polynomials).
CAUTION: To be able to find an n-th order Taylor polynomial of a function, it
does have to have at least n derivatives that are defined a x = x0 .
Sigma notation can be used to concisely write out the n-th order Taylor polynomial of a function y = f (x) about the point x = x0 .
sigma notation
n
pn (x) =
∑ f (k) (x0 )(x − x0 )k /k!
k=0
The n-th remainder, Rn (x), of a Taylor polynomial which is defined by
Rn (x) = f (x) − pn (x)
can be used to find a particular approximation to a desired level of accuracy. For
example, the n-th remainder is bounded in the following way.
|Rn (x)| ≤
M
|x − x0 |n+1
(n + 1)!
Here M is a bound on the (n + 1)-st derivative of f (x) in the interval [x, x0 ]. Then
to achieve accuracy to m decimal places the following is required.
|Rn (x)| ≤ 5 × 10−(m+1)
One way to do this is to solve for n so that the following is true.
|Rn (x)| ≤
M
|x − x0 |n+1 ≤ 5 × 10−(m+1)
(n + 1)!
Since the bound on Rn (x) will be less than the desired error tolerance, then Rn (x)
will also be less than the desired error tolerance.
Checklist of Key Ideas:
local linear/quadratic approximations of f (x)
n-th Maclaurin polynomial
n-th Taylor polynomial for f (x) about x = x0
Σ notation for Maclaurin and Taylor polynomials
n-th remainder → error estimation
Taylor’s formula with remainder
approximating f (x) to m decimal places accurately
| f (n+1) (c)| ≤ M
where x ≤ c ≤ x0
190
9.8 Maclaurin and Taylor Series; Power Series
PURPOSE: To represent functions using Taylor Series and
Power series.
Taylor series
Since n-th order Taylor polynomials can be found for a function f (x), then the next
natural idea is to use an infinite number of terms or an infinite order polynomial.
The result is a Taylor series expansion of the function.
IDEA: Since a Taylor series depends upon x then where the series will converge will also depend upon x.
interval of convergence
radius of convergence
The only question then is whether or not this series will converge. This question is
answered by finding the interval and radius of convergence of the Taylor series.
Most useful in this case is to apply the absolute ratio test (see §9.6). Using this test
can help determine an interval x0 − R < x < x0 + R where the infinite series will
converge. Then this is called the interval of convergence and the value of R > 0
is called the radius of convergence. It is not clear what a power series will do at
x = x0 ± R. This needs to be determined on a case-by-case basis.
CAUTION: A Taylor Series may or may not converge at the endpoints of its
interval of convergence. Convergence at both x = x0 −R and x = x0 +R needs
to be determined for each individual function.
power series
Both Maclaurin and Taylor Series are examples of power series. The distinguishing factor for a power series is that it is not clear what function it is approximating
and so it may just be generically written as
∞
∑ ak (x − x0 )k .
k=1
Checklist of Key Ideas:
Maclaurin series; infinite order polynomial
Taylor series for f (x) about x = x0
power series in x (or x − x0 )
interval/radius of convergence
finding the interval of convergence
representing functions with power series
191
9.9 Convergence of Taylor Series
PURPOSE: To determine if f (x) is equal to its Taylor series representation on a given interval.
As mentioned in the previous section, the convergence of a Taylor series or a
power series is dependent upon the value of x. The region where a power series
does converge is called its interval of convergence. Even if a Taylor series does
converge within some interval, the next question is what does it converge to?
does Taylor series converge to f (x)?
IDEA: If Rn (x) → 0 then the Taylor series will converge to the function f (x).
A Taylor series for the function f (x) will only be equal to the function if the n-th
remainder term approaches zero for each x in an interval as n → ∞.
Several functions are approximated in this section including trigonometric functions, exponential functions, and logarithmic functions. In each case, there are
two main considerations. First, what point should the Taylor expansion be taken
about? Second, how many terms need to be used in order to ensure a certain decimal place accuracy in the approximation. Expansions should be taken about points
where the function and its derivatives are easily obtainable. The point should also
be close to the range of values for which approximations are sought.
two considerations:
1. point to expand about
2. achieving accuracy to m decimal places
IDEA: When approximating using Taylor series
1. Expand about a point where the function and its derivatives are obtainable.
2. Expand about a point near where the approximation is desired.
To achieve accuracy to m decimal places, use the bound on the n-th Remainder as
a guide to how many terms should be included. For example, choose n such that
|Rn (x)| ≤
M
|x − x0 |n+1 ≤ 5 × 10−(m+1) .
(n + 1)!
By choosing n so that this inequality is satisfied for the input value of x (or any
x within an interval), the truncation error due to only using n terms will be less
than the error tolerance. To ensure that the roundoff error does not ruin the
accuracy, all intermediate calculations should be done with m + 1 decimal places.
IDEA: Truncation error occurs when only n terms of a Taylor series are used.
Roundoff error occurs due to normal calculations that only use a finite number
of digits.
CAUTION: To ensure that a Taylor series approximates accurately to m decimal places, at least m + 1 decimal places should be used in all intermediate
calculations.
truncation error
roundoff error
192
Checklist of Key Ideas:
does f (x) equal its Taylor series?
convergence and the n-th remainder
picking the point x = x0 to expand about
how many terms are needed (for a desired accuracy)?
roundoff/truncation error
approximating functions near x = x0
binomial series
9.10 Differentiating and Integrating Power Series;
Modeling with Taylor Series
PURPOSE: To integrate and differentiate power series term-byterm.
New ideas that are presented in this section are those of integrating and differentiating power series term-by-term. This can be done with meaningful results so
long as it is done within the interval of convergence of the given power series.
This process can often be helpful for defining a function or integrating a function
that may not have an antiderivative in closed form.
For example, the function f (x) =
Z x
sint
dt does not have a closed form exprest
sion since sint/t does not have an antiderivative that can be expressed using common functions. But by dividing the power series for sint by t we obtain
0
t2 t4 t6
sint
= 1− + − +···
t
3! 5! 7!
which can be integrated term-by-term. Then the following would be a series for
f (x).
Z x
t2 t4 t6
f (x) =
1 − + − + · · · dt
3! 5! 7!
0
= x−
x3
x5
x7
+
−
+···
3 · 3! 5 · 5! 7 · 7!
IDEA: Term-by-term integration and differentiation of a power series is valid
within its interval of convergence.
Integration and differentiation of a power series or Taylor series term-by-term is
valid as long as it is performed within the radius of convergence of the original
193
series. Likewise, the resulting series will have the same radius of convergence as
the original.
Picking a point to expand about, x = x0 , is usually done by finding a value where
f (x) and its derivatives are conveniently obtained and where the point x = x0 is
close to the point where approximation is desired. This needs to be determined on
a case-by-case basis.
IDEA: The Taylor series and the power series for a function are equivalent.
The culmination of this chapter is that if a function is represented by a power
series, then that power series must be the Taylor series representation of the function. Similarly, if a function is represented by a Taylor series then that is the power
series representation of the function. The biggest implication here is that the coefficients of a power series representation of a function can now be directly related
to the derivatives of the function.
Checklist of Key Ideas:
differentiation of power series term-by-term
integration of power series term-by-term
relationship between power series and Taylor series
practical uses of Maclaurin/Taylor series
multiplying and dividing power series
194
Chapter 9 Sample Tests
Section 9.1
1. The general term for the sequence 3, 3/8, 1/9, 3/64, . . . is
(a) 3/n3
(b) 3/n2
(c) 1/n
∞
(d) n1/3
2. Write out the first five terms of
The sequence
2n
n+5
∞
cos (2nπ )
converges.
n=1
11. Answer true orfalse.
.
n=1
The sequence
(a) 2, 1, 2/3, 1/2, 2/5
∞
sin (nπ + π /2)
converges.
n=1
12. If the sequence converges, then find its limit. If it does not
converge, answer “diverges.”
∞
(−1)e2n
(b) 1/3, 4/7, 3/4, 8/9, 1
(c) 2/5, 1/3, 2/7, 1/4, 2/9
(d) 2/5, 2/3, 6/7, 1, 10/9
3. Write out the first five terms of
8. Answer true orfalse. 7n + 8 ∞
The sequence
converges.
2n + 5 n=1
9. Answer true orfalse. ∞
2
+
6
converges.
The sequence
n3
n=1
10. Answer true orfalse.
n=1
{3 sin (nπ )}∞
n=1 .
(a) 3π , 6π , 9π , 12π , 15π
(a) 0
(b) 1/e
(c) −e
(b) −3, 3, −3, 3, −3
(d) diverges
(c) 0, 0, 0, 0, 0
(d) 3, 0, −3, 0, 3
n
o∞
.
4. Write out the first five terms of (−1)n−1 n2
n=1
(a) 1, −4, 9, −16, 25
13. If the sequence converges, then find its limit. If it does not
converge, answer “diverges.”
∞
3n
8n n=1
(a) 3/4
(b) 3, 6, 9, 12, 15
(b) 3
(c) −3, −6, −9, −12, −15
(c) 0
(d) 1, 4, 9, 16, 25
(d) diverges
n
o∞
.
5. Write out the first five terms of (−1)n−1 3n
n=1
(a) −3, 6, −9, 12, −15
(b) 3, 6, 9, 12, 15
(c) −3, −6, −9, −12, −15
(d) 3, −6, 9, −12, 15
3 ∞
.
6. Write out the first five terms of 2 +
n n=1
(a) 5, 7/2, 3, 11/4, 13/5
(b) 4, 5/2, 2, 7/4, 8/5
(c) 5, 6, 7, 8, 9
(d) 4, 5, 6, 7, 8
7. Answer true or false.
3 ∞
3n
converges.
The sequence
n + 1 n=1
14. If the sequence converges, then find its limit. If it does not
converge, answer “diverges.”
1 1 1 1 1
,
,
,
,
,...
62 63 64 65 66
(a) 1/6
(b) 1/36
(c) 0
(d) diverges
15. If the sequence converges, then find its limit. If it does not
converge, answer “diverges.”
1, 3, 5, 7, 9, . . .
(a) 1
(b) 2
(c) 1/2
(d) diverges
195
Section 9.2
1. Determine which answer best describes the sequence:
4 ∞
3n2 n=1
6. Determine which answer best describes the sequence:
∞
sin (2π /n)n3n
n=1
(a) Strictly increasing
(b) Strictly decreasing
(a) Strictly increasing
(c) Increasing, but not strictly increasing
(b) Strictly decreasing
(d) Decreasing, but not strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
2. Determine which answer best describes the sequence:
∞
2en
n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
3. Determine which answer best describes the sequence:
∞
n−5
n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
4. Determine which answer best describes the sequence:
(n − 1)2 − n ∞
2
n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
5. Determine which answer best describes the sequence:
∞
e−3n
n=1
(a) Strictly increasing
(b) Strictly decreasing
7. Determine which answer best describes the sequence:
π ∞
3 cos
2n n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
8. Determine which answer best describes the sequence:
∞
2
2n − 2n
n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
9. Determine which answer best describes the sequence:
∞
6
n n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
10. Determine which answer best describes the sequence:
3 ∞
8− 3
n n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
11. Determine which answer best describes the sequence:
2 ∞
7+ 4
n n=1
(a) Strictly increasing
(c) Increasing, but not strictly increasing
(b) Strictly decreasing
(d) Decreasing, but not strictly decreasing
(c) Increasing, but not strictly increasing
196
(d) Decreasing, but not strictly decreasing
12. Determine which answer best describes the sequence:
∞
4
2
4n − 3n
n=1
(a) Strictly increasing
4. Answer true or false.
∞ k
7
The series ∑
converges.
k=1 3
5. Answer true or false.
∞
The series
1
∑ (k + 7)(k + 6) converges.
k=1
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
13. Determine which answer best describes the sequence:
∞
((n − 1)! − 1)en
n=1
6. Answer true or false.
∞
6
The series ∑ converges.
k=1 k
find its sum.
(a) 1/4
(a) Strictly increasing
(b) 6
(b) Strictly decreasing
(c) 4
(c) Increasing, but not strictly increasing
(d) diverges
(d) Decreasing, but not strictly decreasing
14. Determine which answer best describes the sequence:
∞
((n − 1)! − 1)(−e4n )
n=1
(a) Strictly increasing
8. Determine whether the following series converges or not. If
it converges, then find its sum. Otherwise, indicate that it
“diverges.”
∞ 4
1
∑ (k + 9)(k + 10) + k + 9
k=1
(b) Strictly decreasing
(a) 24/55
(c) Increasing, but not strictly increasing
(b) 1
(d) Decreasing, but not strictly decreasing
15. Determine which answer best describes the sequence:
∞
− n cos (2nπ + π )
n=1
(a) Strictly increasing
(b) Strictly decreasing
(c) Increasing, but not strictly increasing
(d) Decreasing, but not strictly decreasing
k
1
∑ 5 converges, and if so,
k=1
∞
7. Determine whether the series
(c) 2/45
(d) diverges
∞
9. Determine whether the series
∑ 10
k=1
so, find its sum.
k
4
converges, and if
3
(a) 16/3
(b) 10/3
(c) 2000/3
(d) diverges
∞
Section 9.3
1. Answer true or false.
The series 5 + 5/2 + 5/3 + · · · + 5/n converges.
2. Answer true or false.
n
7
1
7 7 7
converges.
+···+7
The series + + +
2 4 8 16
2
3. Answer true or false.
∞ k
1
The series ∑ 8
converges.
5
k=1
10. Determine whether the series
∑ 5k−3 9k converges, and if so,
k=1
find its sum.
(a) 9/25
(b) 1/45
(c) 45
(d) diverges
∞
11. Determine whether the series
k=1
if so, find its sum.
18
∑ (−1)k−1 5k
converges, and
197
(a) 18/5
(b) 3
(b) diverges
∞
3.
(c) 9/2
k=1
(a) converges
(d) diverges
∞ k
3
12. Determine whether the series ∑
converges, and if so,
k=1 4
find its sum.
1
∑ k+7
(b) diverges
∞
4.
4k2 + 7k + 2
3k2 − 6
k=1
∑
(a) 3/4
(a) converges
(b) 3
(b) diverges
(c) 5/4
∞
5.
(d) diverges
k=1
(a) converges
13. Write the number 0.39 as a fraction.
(b) diverges
(a) 39/100
(b) 13/33
∞
6.
(a) converges
(d) 3939/10000
(b) diverges
14. Write the number 0.17 as a fraction.
∞
7.
(a) converges
(b) diverges
(c) 17/100
∞
8.
15. Write the number 4.34 as a fraction.
1
∑ 5k + 2
k=1
(a) converges
(a) 430/99
(b) diverges
(b) 43/99
(c) 434/999
∑ 3k−3/4
k=1
(b) 171/999
(d) 171/1000
∑ 7k−5/2
k=1
(c) 393/1000
(a) 17/99
∑ 12 cos (kπ )
∞
9.
1
∑ k+2
k=1
(d) 217/50
(a) converges
(b) diverges
Section 9.4
∞
10.
k2 + 8
∑ k2 + 7
k=1
(a) converges
Determine whether each of the following series converges or diverges.
(b) diverges
∞
∞
1
1. ∑ 9
k=1 k
11.
∞
2.
1
∑ k5
k=1
(a) converges
k+4
k3
(a) converges
(a) converges
(b) diverges
∑
k=1
(b) diverges
∞
12.
1
∑ √k + 5
k=1
(a) converges
(b) diverges
198
∞
13.
1
(a) converges
∑ √2k + 7
3
(b) diverges
k=1
(a) converges
(c) convergence cannot be determined
(b) diverges
∞
14.
∑
k=1
∞
6.
3k
2
k3k
k=1 k!
∑
(a) converges
(a) converges
(b) diverges
(b) diverges
(c) convergence cannot be determined
∞
2
4
15. ∑ 4 + 3
k
k
k=1
∞
7.
k!
∑ k5k
k=1
(a) converges
(a) converges
(b) diverges
(b) diverges
(c) convergence cannot be determined
∞
Section 9.5
8.
1
∑ k3
k=1
(a) converges
For each of the following series, state whether it converges, diverges
or if convergence cannot be determined.
(b) diverges
(c) convergence cannot be determined
∞
1
1. ∑ 2
9k
− 2k
k=1
∞
9.
(a) converges
(a) converges
(b) diverges
(b) diverges
(c) convergence cannot be determined
∞
2.
1
∑ 7k3 + 3k
(c) convergence cannot be determined
∞
10.
k=1
k=1
(a) converges
(b) diverges
(b) diverges
(c) convergence cannot be determined
1
∑ k−5
(c) convergence cannot be determined
∞
11.
k=1
9k + 1
∑ 2k − 3
k=1
(a) converges
(a) converges
(b) diverges
(b) diverges
(c) convergence cannot be determined
(c) convergence cannot be determined
∞
sin2 k
4. ∑
k=1 k!
∞
12.
6
∑ ek
k=1
(a) converges
(a) converges
(b) diverges
(b) diverges
(c) convergence cannot be determined
∞
5.
k
∑ 7k
(a) converges
∞
3.
(6k)!
k
k=1 k
∑
k!
∑ k8
k=1
(c) convergence cannot be determined
∞
13.
1
∑ (7 ln (k + 2))k
k=1
199
(a) converges
(b) diverges
∞
(b) converges conditionally
1
∑ (k + 1)k
k=1
(a) converges
(b) diverges
(c) diverges
1 k
6. ∑ − 4
k
k=1
∞
(a) converges absolutely
(c) convergence cannot be determined
(b) converges conditionally
∞
15.
|sin (kx)|
3k
k=1
∑
(a) converges
∑
(a) converges absolutely
(c) convergence cannot be determined
14.
9 k
−
2
k=1
∞
5.
(c) diverges
∞
7.
k
∑ (−3)k
k=1
(b) diverges
(a) converges absolutely
(c) convergence cannot be determined
(b) converges conditionally
(c) diverges
Section 9.6
∞
8.
(−1)k
k=1 k − 6
∑
(a) converges absolutely
In questions #1 through #11, state whether the given series will converge absolutely, converge conditionally or diverge.
(b) converges conditionally
(c) diverges
∞
(−1)k
1. ∑
6k
+1
k=1
∞
9.
cos (π k)
4k
k=1
∑
(a) converges absolutely
(a) converges absolutely
(b) converges conditionally
(b) converges conditionally
(c) diverges
(c) diverges
∞
(−1)k
2. ∑
k=1 5k
∞
10.
∑
k=1
cos (π k + 2)
k
(a) converges absolutely
(a) converges absolutely
(b) converges conditionally
(b) converges conditionally
(c) diverges
k
∞
3
3. ∑ (−1)k
4
k=1
(a) converges absolutely
(b) converges conditionally
(c) diverges
∞
(−1)k (2k)!
4. ∑
2k2k
k=1
(a) converges absolutely
(c) diverges
∞
11.
4k
∑ cos (π k)
k=1
(a) converges absolutely
(b) converges conditionally
(c) diverges
∞
(−1)k
which of the following is equal to
k=1 k!
the fifth partial sum?
12. Given the series
(a) −0.6333
(b) converges conditionally
(b) 1.7167
(c) diverges
(c) −0.2
∑
200
(d) −0.0083
1 k
which of the following is equal
4
−
∑
2
k=1
to the fifth partial sum?
∞
13. Given the series
(a) 1.375
(b) −1.375
(c) 1.396
(d) 1 −
x2
2
4. Find the Maclaurin polynomial of order 2 for e7x .
49 2
x
2
49
(b) 1 − 7x + x2
2
(c) 1 + 7x + 49x2
(a) 1 + 7x +
(d) 1 − 7x + 49x2
(d) −1.396
∞
14. Given the series
k+1
∑ (−1)
k=1
4
which of the following is
2k
equal to the fifth partial sum?
5. Find the Maclaurin polynomial of order 2 for 2e−6x .
(a) 2 − 12x + 36x2
(b) 2 + 12x + 36x2
(c) 2 − 12x + 72x2
(a) 1.375
(d) 2 + 12x + 72x2
(b) 0.125
6. Find the Taylor polynomial for f (x) = ex of order 2 about the
point xo = 3.
(c) 3.875
(d) −3.875
15. Answer true or false.
∞
3
For the series ∑ (−1)k+1 √
the fourth partial sum is
k
+1
k=1
equal to 0.5476.
Section 9.7
1. Find the Maclaurin polynomial of order 2 for e3x .
9x2
2
(b) 1 − 3x + 3x2
(a) 1 + 3x +
(c) 1 + x + x2
(d) 1 + 3x + 9x2
2. Find the Maclaurin polynomial of order 2 for sin (x/2).
x
(a) 1 −
2
x
(b)
2
1
1
(c) 1 + x + x2
2
8
1
1
(d) 1 + x − x2
2
8
3. Find the Maclaurin polynomial of order 2 for cos (x).
(a) e3 − e3 (x − 3) + e3 (x − 3)2
e3
(x − 3)2
2
e3
(c) e3 + e3 (x − 3) + (x − 3)2
2
(d) e3 + e3 (x − 3) + e3 (x − 3)2
(b) e3 − e3 (x − 3) +
7. Find the Taylor polynomial for f (x) = e−3x of order 2 about
the point xo = 3.
9e−9
(x − 3)2
2
9e−9
(b) e−9 − 3e−9 (x − 3) +
(x − 3)2
2
(c) e−9 − 3e−9 (x − 3) + 9e−9 (x − 3)2
(a) e−9 + 3e−9 (x − 3) +
(d) e−9 + 3e−9 (x − 3) + 9e−9 (x − 3)2
8. Find the Taylor polynomial for f (x) = 3 ln x of order 2 about
the point xo = 2.
3
3
(a) 3 ln 2 + (x − 2) − (x − 2)2
2
4
3
3
(b) 3 ln 2 + (x − 2) − (x − 2)2
2
8
3
3
(c) 3 ln 2 + (x − 2) + (x − 2)2
2
4
3
3
(d) 3 ln 2 + (x − 2) + (x − 2)2
2
8
9. Find the Taylor polynomial for f (x) = 4 sin x of order 2 about
the point xo = π /2.
(a) 4 − 4x2
x2
2
(b) 1 + x2
(b) 4 + 4x2
(c) 1 − x2
(d) 4 + 2(x − π /2)2
(a) 1 +
(c) 4 − 2(x − π /2)2
201
10. Answer true or false. The Maclaurin polynomial of order 3
25
32
for e5x is 1 + 5x + x2 + x3 .
2
3
11. Answer true or false. The Maclaurin polynomial of order 3
ln 3 2 ln 3 3
for ln (3 + x) is ln 3 + ln 3x +
x +
x .
2
6
12. Answer true or false. The Maclaurin polynomial of order 3
for cosh x2 is cosh x2 + 2x2 sinh x2 + 2x4 cosh x2 + 2x6 sinh x2 .
13. Answer true or false. The Taylor polynomial of order 3 for ex
about the point xo = 3 is
e3 + e3 (x − 3) +
e3
2
(x − 3)2 +
e3
6
(x − 3)3 .
14. Answer true or false. The Taylor polynomial of order 2 for
π π 3
cos (xe−x ) about the point xo = π /2 is x −
.
− x−
2
2
15. Answer true or false. The Taylor polynomial of order 3 for
ln 4
ln x about the point xo = 4 is ln 4+ln 4(x−4)+
(x−4)2 +
2
ln 4
(x − 4)3 .
6
(c) 7
(d) ∞
4. Find the radius of convergence for the series
∞
3xk
∑ ln k
k=1
(a) 3
(b) 1
(c) 1/3
(d) ∞
5. Find the radius of convergence for the series
∞
xk
∑ (−1)k √x
k=1
(a) 1
(b) 2
(c) 1/2
(d) ∞
6. Find the interval of convergence for the series
∞
Section 9.8
xk
∑ (−1)k 4
k=1
1. Find the radius of convergence for the series
∞
2xk
∑ k+2
k=1
(a) 2
(b) 1
(a) (−1, 1)
(b) (−4, 4)
1 1
(c) − ,
4 4
(d) (−∞, ∞)
7. Find the interval of convergence for the series
(c) 1/2
∞
∑ (−1)k
(d) ∞
2. Find the radius of convergence for the series
∞
k k
∑8 x
k=1
k=1
(a) (−4, −2)
(b) (2, 4)
(c) (−1, 1)
(d) (−∞, ∞)
(a) 8
(b) 1
8. Find the interval of convergence for the series
(c) 1/8
∞
3. Find the radius of convergence for the series
(a)
∞
xk
∑ k+7
k=1
(a) 1/7
2(3x − 5)k
5k
k=1
10
− ,0
3
(−5, 5)
10
0,
3
(−∞, ∞)
∑
(d) ∞
(b) 1
(x − 3)k
3
(b)
(c)
(d)
202
(c) 0.30469
9. Answer true or false.
∞
6k xk
The interval of convergence for the series ∑
is (−1, 1).
k=1 k!
10. Answer true or false.
∞
6k xk+2
is
k=1 (3k)!
∑
The interval of convergence for the series
(−∞, ∞).
∞
3k (x − 3)k
is
k!
k=1
∑
(−∞, ∞).
(a) 0.30452
(b) 0.30456
∞
∑ (x − 8)k
The interval of convergence for the series
is
5. Estimate cosh (0.2) to 5 decimal-place accuracy.
(a) 1.01989
(7, 9).
13. Answer true or false.
∞
∑ (4x − 1)k
(c) 1.02019
(d) 1.02040
√
6. Estimate 5 e to 5 decimal-place accuracy.
k=1
The interval of convergence for the series
(d) 0.30462
(b) 1.02007
12. Answer true or false.
is
k=1
(0, 1).
(a) 1.22134
(b) 1.22140
(c) 1.22146
14. Answer true or false.
∞
The interval of convergence for the series
6k xk
is
k=1 k!
∑
(−2/3, 0).
(d) 1.22153
7. Estimate e−3 to 5 decimal-lace accuracy.
(a) 0.04959
15. Answer true or false.
∞
The interval of convergence for the series
4. Estimate sinh (0.3) to 5 decimal-place accuracy.
(c) 0.30459
11. Answer true or false.
The interval of convergence for the series
(d) 0.30462
4(x + 4)k
is
3k
k=1
∑
(−7, −1).
(b) 0.04965
(c) 0.04972
(d) 0.04979
8. Estimate sin (0.6) to 5 decimal-lace accuracy.
(a) 0.56453
Section 9.9
1. Estimate sin (5◦ ) to 5 decimal-place accuracy.
(a) 0.08721
(b) 0.08716
(c) 0.00720
(d) 0.00730
2. Estimate tan (9◦ ) to 5 decimal-place accuracy.
(a) 0.15827
(b) 0.15832
(c) 0.15838
(d) 0.15849
3. Estimate sin−1 (0.3) to 5 decimal-place accuracy.
(a) 0.30458
(b) 0.30459
(b) 0.56458
(c) 0.56464
(d) 0.56472
9. Answer true or false. cos (0.8) can be approximated to 4 decimal places to be 0.6967.
10. Answer true or false. ln 4 can be approximated to 3 decimal
places to be 1.388.
11. Answer true or false. e7 can be approximated to 3 decimal
places to be 1, 096.610.
12. Answer true or false. cosh (0.7) can be approximated to 3
decimal places to be 1.255.
13. Answer true or false. tanh−1 (0.14) can be approximated to 3
decimal places to be 0.141.
14. Answer true or false. sinh−1 (0.17) can be approximated to 3
decimal places to be 0.162.
15. Answer true or false. cosh−1 (1.19) can be approximated to
3 decimal places to be 1.421.
203
Section 9.10
Chapter 9 Test
1. Answer true or false. The Maclaurin series for e2x − e−x can
be obtained by subtracting the Maclaurin series for e−x from
the Maclaurin series for e2x .
2. Answer true or false. The Maclaurin series for x2 sin (x) can
be obtained by multiplying the Maclaurin series for sin (x) by
x2 .
3. Answer true or false. The Maclaurin series for sin2 x can be
obtained by multiplying the Maclaurin series for sin x by itself.
4. Answer true or false. The Maclaurin series for sin (2x) can
be obtained by multiplying the Maclaurin series for sin x by
itself.
5. Answer true or false. The Maclaurin series for 2 cosh x can be
obtained by multiplying the Maclaurin series for cosh x by 2.
6. Answer true or false. The Maclaurin series for tan x can be
obtained by dividing the Maclaurin series for sin x by the
Maclaurin series for cos x.
7. Answer true or false. The Maclaurin series for ex sinh x can
be obtained by multiplying the Maclaurin series for ex by the
Maclaurin series for sinh x.
ln (5 + x)
8. Answer true or false. The Maclaurin series for
can
1+x
be obtained by dividing the Maclaurin series for ln (5 + x) by
the Maclaurin series for 1 + x.
9. Answer true or false. The Maclaurin series for ln (3 + x) can
be differentiated term by term to determine that the derivative
1
.
of ln (3 + x) is
3+x
10. Answer true or false. The Maclaurin series for ln (6x + 2) can
be differentiated term by term to determine that the derivative
1
of ln (6x + 2) is .
x
11. Answer true or false. The Maclaurin series for sin (3x) can be
differentiated term by term to determine that the derivative of
sin (3x) is cos x.
12. Answer true or false. The Maclaurin series for sinh x can be
differentiated term by term to determine that the derivative of
sinh (7x) is cosh (7x).
ex
13. Answer true or false. The Maclaurin series for can be integrated term by term to determine that the integral of ex is
ex +C.
14. Answer true or false. The Maclaurin series for cos (4x) can
be integrated term by term to determine that the integral of
cos (4x) is −4 sin (4x) +C.
1
can be
15. Answer true or false. The Maclaurin series for
7+x
1
integrated term by term to determine that the integral of
7+x
is ln (7 + x) +C.
√
√
√
1. The general term for the sequence 2, 2 2, 2 3, 4, 2 5, . . .
is
√
(a) 2 n
√
(b) 2 n + 1
√
(c) 2 n − 1
√
(d) 2n n
−n − 5 ∞
2. Write out the first five terms of the sequence
.
n + 8 n=1
(a) −5/8, −2/3, −7/10, −8/11, −3/4
(b) −2/3, −7/10, −8/11, −3/4, −10/13
(c) −5/9, −3/5, −7/11, −2/3, −9/13
(d) −5/8, −5/8, −5/8, −5/8, −5/8
2
n +5 ∞
converges, then find its limit.
3. If the sequence
n5 − 8 n=1
Otherwise, indicate that it “diverges”.
(a) 0
(b) −5/7
(c) 1
(d) diverges
4. Determine which answer best describes the sequence
3 ∞
n
n4 + 5 n=1
(a) strictly increasing
(b) strictly decreasing
(c) increasing, but not strictly increasing
(d) decreasing, but not strictly decreasing
5. Determine which answer best describes the sequence
∞
(n − 1)!n7
n=1
(a) strictly increasing
(b) strictly decreasing
(c) increasing, but not strictly increasing
(d) decreasing, but not strictly decreasing
6. Answer true or false.
n
9 9 9
9
1
The series + + +
+···+9
converges.
2 4 8 16
2
∞ k
4
converges to 24.
7. Answer true or false. The series ∑ 6
5
k=1
204
∞
8. Write 2.19 as a fraction.
17.
(a) 73/33
(k + 1)!
k9k
k=1
∑
(b) 219/100
(a) converges
(c) 217/99
(b) diverges
(c) convergence cannot be determined
(d) 219/500
∞
9. Answer true or false. The series
1
∑ (k + 5)k converges.
k=1
∞
10. Answer true or false. The series
k3 + 2
∑ 5k3 + 2 converges.
k=1
11. Answer true or false. The Maclaurin polynomial of order 3
for e7x is 1 + 7x + 49x2 + 243x3 .
12. Answer true or false. The Maclaurin polynomial of order 3
x2 ln 5 x3 ln 5
+
.
for ln (x + 5) is ln 5 + x ln 5 +
2
6
13. Answer true or false. The Maclaurin polynomial of order 3
for cosh (x3 ) is
cosh (x3 ) + 3x2 sinh (x3 ) + 9x4 cosh (x3 ) + 27x6 sinh (x3 ).
14. Answer true or false. The Taylor polynomial of order 3 about
the point xo = 5 for ex is
e5 + e5 (x − 5) +
e5
2
(x − 5)2 +
e5
6
(x − 5)3 .
∞
1
15. ∑ 6
k
−7
k=1
(a) converges
(b) diverges
(c) convergence cannot be determined
∞
1
16. ∑ 4
5k
− k2
k=1
(a) converges
(b) diverges
(c) convergence cannot be determined
∞
18.
(−1)k
∑ 7k + 2
k=1
(a) converges absolutely
(b) converges conditionally
(c) diverges
∞
19. Find the radius of convergence for
∑ 6k xk .
k=1
(a) 6
(b) 1
(c) 1/6
(d) ∞
∞
20. Find the interval of convergence for
xk
∑ (−1)k 9
k=1
(a) (−1, 1)
(b) (−9, 9)
1 1
(c) − ,
9 9
(d) (−∞, ∞)
21. Estimate sin (11◦ ) to 5 decimal-place accuracy.
(a) 0.19081
(b) 0.19089
(c) 0.19094
(d) 0.19100
22. Answer true or false. The Maclaurin series for e−x + ln x can
be obtained by adding the Maclaurin series for e−x and ln x.
205
Chapter 9: Answers to Sample Tests
Section 9.1
1. a
9. true
2. b
10. true
3. c
11. false
4. a
12. d
5. d
13. c
6. a
14. c
7. false
15. d
8. true
2. a
10. a
3. a
11. b
4. c
12. a
5. b
13. c
6. c
14. d
7. a
15. a
8. b
2. true
10. d
3. true
11. b
4. false
12. b
5. true
13. b
6. false
14. a
7. a
15. a
8. d
2. a
10. b
3. b
11. a
4. b
12. b
5. b
13. b
6. a
14. b
7. b
15. a
8. b
2. a
10. a
3. b
11. b
4. a
12. a
5. b
13. a
6. b
14. a
7. a
15. a
8. a
2. b
10. b
3. a
11. c
4. a
12. a
5. c
13. b
6. a
14. a
7. a
15. true
8. b
2. b
10. false
3. d
11. false
4. a
12. false
5. a
13. true
6. c
14. false
7. b
15. false
8. b
2. c
10. true
3. b
11. true
4. b
12. true
5. a
13. false
6. a
14. false
7. b
15. true
8. c
2. c
10. false
3. c
11. false
4. a
12. true
5. b
13. true
6. b
14. false
7. d
15. false
8. c
2. true
10. false
3. true
11. false
4. false
12. false
5. true
13. true
6. true
14. false
7. true
15. true
8. true
2. b
10. false
18. b
3. a
11. false
19. c
4. b
12. false
20. a
5. a
13. false
21. a
6. true
14. true
22. true
7. true
15. a
8. c
16. a
Section 9.2
1. b
9. b
Section 9.3
1. false
9. d
Section 9.4
1. a
9. b
Section 9.5
1. a
9. b
Section 9.6
1. b
9. a
Section 9.7
1. a
9. c
Section 9.8
1. b
9. false
Section 9.9
1. b
9. true
Section 9.10
1. true
9. true
Chapter 9 Test
1. a
9. true
17. a
206
Chapter 10: Parametric And Polar Curves;
Conic Sections
Summary: This chapter begins by introducing the idea of representing curves
using parameters. These parametric equations of the curves can then be used
to graph, find tangent lines, and find arc lengths of the given curves. Polar coordinates and their relationship to Cartesian or rectangular coordinates are then
described. Polar coordinates may also be used as a particular parameterization of
curves. Then tangent lines, arc length, and area may be found using polar coordinates as the parameterization of the curve. The latter half of the chapter discusses
conic sections and how they may be graphed. Then the idea of conic sections is
extended by representing these curves using polar coordinates.x
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Represent a function as a parametric curve (§10.1).
2. Find tangent lines of curves that are either defined parametrically (§10.1) or
defined using polar coordinates (§10.3).
3. Find the arc length of a parametric curve (§10.1) or a polar curve (§10.3).
4. Convert between rectangular and polar coordinates (§10.2).
5. Graph functions in polar coordinates (§10.2).
6. Find the area between curves that have been defined in polar coordinates
(§10.3).
7. Identify different conic sections and their graphs (§10.4).
8. Rotate conic sections and determine their resulting quadratic equations (§10.5).
9. Describe conic sections and their graphs using polar coordinates (§10.6).
207
208
10.1 Parametric Equations; Tangent Lines And Arc
Length For Parametric Curves
PURPOSE: To discuss tangent lines and arc length of parametric
curves and curves in polar coordinates.
parameter
parametric curve
orientation
One of the main idea in this section is to come up with functions for both the x
and y coordinates of a curve. In this way, curves that are generally not considered
as functions may be graphed quite easily. For example, this gives a great way
to graph the inverse of a function. If x = f (t) and y = g(t) then the inverse can
be graphed by reversing the definitions. x = g(t) and y = f (t) would result in
the graph of the inverse. The parameter t used in the definitions of x and y also
results in the parametric curve having a definite orientation. If t is followed in
a positive direction, for example, then this will take you in one direction along a
curve while a negative direction for t would take you in the opposite direction.
Tangent lines were first introduced in Chapters 1 and 2 where the idea of a derivative was first developed. The same kinds of tangent lines to curves are discussed
in this section. The new approach taken here is to represent the curves parametrically by defining a function to each coordinate of a point on the curve. or as polar
curves.
To find a tangent line to a curve requires that the slope of the curve at a point be
found. This is done using the derivative of the curve. In other words, the slope of
a tangent line is equivalent to dy/dx. If x = f (t) and y = g(t) then this slope can
be found by dividing g′ (t) by f ′ (t). One way to remember this is to think of these
slopes as differentials and divide out some of the terms. For example,
dy dt
dy
g′ (t) dy/dt
=
=
·
=
f ′ (t) dx/dt
dt dx dx
which is the slope of the tangent line.
horizontal tangents
vertical tangents
singular points
Some special types of tangent lines are horizontal tangents and vertical tangents.
Horizontal tangent lines occur when dy/dt = 0 and dx/dt 6= 0. Vertical tangents occur at a point on the curve where dy/dt 6= 0 and dx/dt = 0. If both
dy/dt = 0 and dx/dt = 0 then this is a singular point on the curve that does not
have a tangent line.
CAUTION: To have a vertical tangent at a point does require that the curve be
defined at that point. For example, vertical asymptotes do not have vertical
tangents.
arc length
Calculating arc length for a parametric curve is done using the arc length formula
(see §6.4) but modified to include both x′ (t) and y′ (t) in the equation.
L=
Z t=b q
t=a
(x′ (t))2 + (y′ (t))2 dt
209
Checklist of Key Ideas:
tangent lines
infinite slope, vertical tangents
singular points
arc length of a parametric curve
parametric first and second derivatives
10.2 Polar Coordinates
PURPOSE: To introduce polar coordinates and to graph curves
that are represented using polar coordinates.
Until this point in the book, coordinates have been measured in a rectangular
fashion. So any given point is identified by a horizontal value, x, and a vertical
value, y. Polar coordinates offer a different way to identify each point. This new
identification can be seen by drawing a line from the origin (called the pole) out
to the point. The point is then identified by how far it is from the origin (this is the
r-coordinate) and the angle that the line makes with the positive x-axis (called the
polar axis). This angle is the θ -coordinate. Angles are measured using radian
measure.
CAUTION: The angle θ should be given in radian measure where
1 radian=180/π degrees.
Using trigonometric formulas it can be seen that if the same point has polar coordinates of (r, θ ) then the corresponding rectangular coordinates, (x, y), will be
determined by the equations
x = r cos θ and y = r sin θ .
Following similar reasoning, if a point is designated as (x, y) in rectangular coordinates then the point in polar coordinates (r, θ ) can be found by the following
equations:
r2 = x2 + y2 and tan θ = y/x.
CAUTION: Polar coordinates of a point are not unique.
Polar coordinates for a point are not unique. The reason for this is that there are
many values of θ that can be chosen so that tan (θ ) = y/x and so there are many
ways to choose θ . For example, the√
rectangular point
√ (2, 2) can be√identified by
any √
of the following polar points: (2 2, π /4), (−2 2, −3π /4), (2 2, 9π /4), or
(−2 2, 5π /4).
the pole, polar axis
r-coordinate
θ -coordinate
210
IDEA: Graphing polar curves involves always thinking of points by drawing a
line from the origin to the point.
Graphing in polar coordinates is not difficult although it is definitely different
than graphing in rectangular coordinates. Graphing involves the same process as
identifying points. Any point on a graph should be thought of using a straight
line to the origin. Then the length of the line and the angle of the line are used to
determine r and θ respectively.
Many polar graphs are symmetric about the x-axis, the y-axis or the origin. Symmetry in polar graphs often occurs because of the repetitive nature of angles. For
example, the graph of r = cos θ will be symmetric about the x-axis since cosine is
an even function (i.e., cos (θ ) = cos (−θ )). Generally, symmetry can be found by
replacing θ by either −θ , θ + π or θ − π in any given expression. If the resulting
expression is the same, then some type of symmetry exists.
IDEA: Symmetry of polar curves can be tested for in three ways:
1. symmetric about the x-axis if replacing θ with −θ gives an equivalent
equation
2. symmetric about the y-axis if replacing θ with θ + π gives an equivalent
equation
3. symmetric about the origin if replacing θ with θ − π (or replacing r with
−r) gives an equivalent equation
IDEA: When checking for symmetry, it can be helpful to remember two important properties of sine and cosine.
1. sine is odd → sin (−θ ) = − sin (θ )
2. cosine is even → cos (θ ) = cos (−θ )
The two trigonometric functions, sine and cosine, are odd and even functions
respectively. These can be useful ideas when testing to see whether a polar graph
will be symmetric or not as many of them contain either sine or cosine in some
fashion.
Many graphs of functions are introduced in this section. If recognizing all of the
different graphs is important then it may be helpful to write each different example
on note cards that can be quickly reviewed. Here are some of the main graphs and
their names.
1. lines through the pole: θ = θ0
2. circles: r = a, r = 2a cos θ , r = 2a sin θ
3. rose curves: r = a sin (nθ ) or r = a cos (nθ ) where n is a positive integer
4. lemniscates: r2 = a2 cos (2θ ) or r2 = a2 sin (2θ )
211
5. limaçons: r = a ± b sin θ or r = a ± b cos θ where a and b are positive. Here
are some types of limaçons:
→ with an inner loop: a/b < 1
→ a cardioid (special type of limaçon: a = b)
→ dimpled: 1 < a/b < 2
→ convex: a/b ≥ 2
6. spirals: there are several different types of these but they all have the common feature of circling around the pole in larger and larger paths. Here are
some common types with a and b positive.
→ Archimedean: r = aθ
√
→ parabolic: r = a θ
→ logarithmic: r = aebθ
√
→ Lituus: r = a/ θ
→ hyperbolic: r = a/θ
There are many varieties of polar curves that can be graphed. To get a handle
on what a graph may look like, some common points to try and find are where
the curve may intersect with the x- and y-axes. For example, the angle θ = 0
corresponds to the positive x-axis. When this is substituted into an expression, a
value for r may be found to determine where the curve touches the x-axis. Other
angles that may be checked easily in this manner are θ = π /2 (positive y-axis),
θ = π (negative x-axis), and θ = 3π /2 (negative y-axis).
Checklist of Key Ideas:
polar coordinate system
angular coordinate, radial coordinate
radian measure
converting between rectangular and polar coordinates
symmetry tests in polar coordinates
circles
roses
cardioids
lemniscates
limaçons
spirals
212
10.3 Tangent Lines, Arc Length, And Area For Polar Curves
PURPOSE: To find the area bounded by functions that are defined either parametrically or as polar curves.
If r = f (θ ) then
→ dy/d θ = f ′ (θ ) sin θ + f (θ ) cos θ
→ dx/d θ = f ′ (θ ) cos θ − f (θ ) sin θ
A similar argument can be followed for polar curves where r = f (θ ). For example,
since x = r cos θ = f (θ ) cos θ and y = r sin θ = f (θ ) sin θ then the slope at a point
can be found by
dy dy/d θ
=
dx dx/d θ
Both of these derivatives can be found using the product rule.
tangent at origin
A special situation occurs for tangent lines at the origin. If r = f (θ0 ) = 0 and
f ′ (θ0 ) 6= 0 then the line θ = θ0 is tangent to the curve at the origin. It is not
uncommon for a polar curve to have more than one tangent line at the origin.
CAUTION: At points where a polar curve intersects itself, it may have more
than one tangent line.
arc length
Calculating arc length for a polar curve is done using the arc length formula for
parameterized curves (see §10.1). For example, if a curve given by r = f (θ ) is
parameterized by x = r cos θ and y = r sin θ then x′ (t) and y′ (t) can be found using
the product rule. After squaring the terms and adding them together the resulting
arc length formula can be written in terms r = f (θ ):
L=
Z θ =b q
θ =a
( f (θ ))2 + ( f ′ (θ ))2 d θ
A similar expression may be obtained if f (θ ) is replaced by r and f ′ (θ ) with
dr/d θ .
CAUTION: To find arc length, the curve should not repeat itself on the interval
a < θ < b as given by the limits of integration and the derivative f ′ (θ ) needs
to be continuous over the same interval.
A couple of trouble spots for this formula come from the repetitive nature of polar
curves and the derivative f ′ (θ ). If the polar curve repeats itself for any values of
θ in the interval determined by the limits of integration, then this formula will not
give an accurate value. Likewise, if f (θ ) or f ′ (θ ) are undefined or discontinuous
then the arc length obtained may not be correct. In these situations, sometimes
the techniques of improper integrals (see §7.8) may be used to find the correct arc
length.
area between curves
Finding area between curves or in bounded regions has already been discussed
in rectangular coordinates. Now, the same idea will be applied to polar coordinates. One important difference between rectangular and polar coordinates are
the shapes that the area is broken into. With rectangular coordinates, an area was
213
broken into several vertical strips (when integrating with respect to dx) or horizontal strips (when integrating with respect to dy). In polar coordinates, the region to
be measured will be broken into small infinitesmal regions by rays that come out
from the origin. So the area will be broken into triangular wedges.
Each wedge is assumed to have area of a sector of a circle with central angle d θ .
Then to find the area formula, what is needed are the limits of integration and the
“radius” of each sector. The limits of integration are determined by the angles of
the starting and ending ray as determined by the boundaries of the area. Sometimes this is determined by where two curves intersect. The radius is determined
by the distance to the curve (usually r = f (θ )) that the ray is being drawn to.
infinitesmal wedges
area of a sector of a circle = 12 θ r2
IDEA: The area formula in polar coordinates is given by
A=
Z θ =β
1 2
r dθ
θ =α
2
where r = f (θ ).
Symmetry can often be use to simplify the limits of integration and then multiply
the resulting integral by an appropriate integer representing the symmetry. In a
rose with five petals, for example, one only needs to find the area of one petal and
then multiply by five. This can often reduce the complexity of the integration that
is taking place.
symmetry
IDEA: Symmetry of trigonometric functions can often be used with polar coordinates to simplify the limits of integration.
In the case where the area between two curves is being found, the problem should
be thought of as an inner area and outer area problem. The area that is bounded
by the inner function (the one closer to the origin) is subtracted from the area
bounded by the outer function (which includes the inner area).
area between two curves
CAUTION: When two curves are involved, the inner function is closer to the
origin. Which function is closer to the origin may change, however.
It is possible that over the interval of integration the inner function and outer function will switch places. In this case, either symmetry should be used if possible
to simplify the problem or multiple integrals for the different regions may need to
be evaluated. When multiple integrals need to be evaluated, the places where the
functions intersect should be considered as possible limits of integration. Finding
intersections of polar functions may be done by setting the two equations equal
and solving for θ . This may not give all of the places of intersection, however, so
graphing the function may be helpful as well.
CAUTION: Graphing two polar curves together can help find places where
they intersect that may not be found algebraically.
intersections of polar functions
214
Checklist of Key Ideas:
lines tangent at the origin
arc length of a polar curve
area
wedges
polar coordinate area formula
using symmetry to write limits of integration
intersections of polar graphs
10.4 Conic Sections in Calculus
PURPOSE: To define the various curves that are conic sections
and to graph these curves.
double-napped cone
Conic sections play an important role in calculus and arise naturally in many applications. Each conic section is actually defined to be the curve that results when
a plane intersects a double-napped cone (opening both upwards and downwards
around a central line). There are three main types of conic sections: parabolas,
ellipses, and hyperbolas. Of the three types, only the ellipse forms a closed curve.
The parabola is made up of one open curve and the hyperbola is made up of two
curves which open in opposite directions.
IDEA: The ellipse is the only conic section that is a closed curve. The hyperbola is the only conic section made up of two curves.
Each conic section is defined as a curve in such a way that any point on the curve
has some relationship in terms of distance to special points called the foci (or
focus if there is only one) of the curve.
IDEA: Each point on a conic section has some sort of relationship with a point
called the focus, or two points called the foci.
Some other common traits of the conic sections are listed in the following table:
215
ellipse
parabola
hyperbola
# of foci
2
1
2
# of vertices
2
1
2
center?
yes
no
yes
axis of symmetry?
no
yes
no
directrix
no*
yes
no*
asymptotes?
no
no
yes
focal/conjugate axis?
no
no
yes
minor/major axis?
yes
no
no
*all conic sections have a directrix (see §10.6)
All conic sections also have a similar relationships between the vertices, the foci
and the center of the curve. In each case, all of these points occur on the same line
(although the parabola does not have a center). This line is the axis of symmetry
for the parabola, the major axis of the ellipse and the focal axis of the hyperbola.
axis of symmetry
major axis
focal axis
Below are the standard equations of the conic sections.
parabola
(p > 0)
y2 = ±4p x
x2
= ±4p y
opens in ±x direction
opens in ±y direction
x2 /a2 + y2 /b2 = 1
major axis in x direction
(a > b > 0)
x2 /b2 + y2 /a2
=1
major axis in y direction
hyperbola
x2 /a2 − y2 /b2
=1
opens in x direction
y2 /a2 − x2 /b2 = 1
opens in y direction
ellipse
(a > 0, b > 0)
Also, each of the three conic sections has a special rectangle associated with it that
can aid in determining the location of the vertices, the center and other features.
The parabola’s rectangle has dimensions of 2p by 4p. The vertex is at the center
of the rectangle and the curve of the parabola goes through the corners of the
rectangle. One of the rectangle sides of length 4p borders the directrix line. In
the center of the opposite side is the focus of the parabola.
The rectangle for an ellipse has dimensions of 2a by 2b. The center of the ellipse
is at the center of the rectangle and the ellipse itself is contained within the rectangle. It should be noted that in the ellipse, a is always greater than b and both
numbers are positive. The value of a represents the distance between the center
of the ellipse and the vertices (which occur along the major axis). A circle is a
special case of an ellipse when a = b and then the radius is equal to a or b. For
x2
y2
example, the ellipse
+
= 1 has a value of a = 2 and b = 3/2. So the
4
(9/4)
ellipse is contained within a rectangle of dimensions 4 × 3.
CAUTION: In the hyperbola, a does not have to be bigger than b although a
will always measure the distance from the center to a vertex.
rectangle for a parabola
rectangle for an ellipse
216
rectangle for a hyperbola
asymptotes
branches
The rectangle for a hyperbola has dimensions of 2a by 2b, as for the ellipse,
although a is not always larger than b. In the case of the hyperbola, a is also the
distance between the center of the hyperbola and the vertices. The graph of the
hyperbola touches its rectangle only at the vertices. Otherwise, the curves of the
hyperbola occur entirely on the outside of the rectangle. If the diagonals of the
rectangle are extended as lines, then these are the lines that serve as asymptotes
for the branches of the hyperbola. The direction in which a hyperbola opens
can be found by looking for the x and y intercepts of the standard equation. For
example, in the case x2 − y2 = 1, the y-intercept would occur when x = 0 which
leaves the equation −y2 = 1. Since this equation has no real solution, the curve
does not ever touch the y-axis.
The ellipse and hyperbola share many similarities. First, the dimensions of both
rectangles are given by a and b. Both equations involve a squared x and y term.
Both ellipses and hyperbolas have a center, two foci and two vertices. The primary
difference between ellipses and hyperbolas is that an ellipse is a closed single
curve while a hyperbola is two distinct, non-closed curves.
IDEA: Conic sections may be shifted and translated like other curves. Replacing x with x − h translates the conic section horizontally and replacing y with
y − k shifts the conic section vertically.
Conic sections may be translated like other curves. By replacing x with x − h a
conic section will be translated horizontally |h| units (to the right if h > 0 or to the
left if h < 0). Replacing y with y − k will shift the conic section vertically by |k|
units (shifted up if k > 0 or down if k < 0).
Checklist of Key Ideas:
conic sections, plane intersecting a cone
degenerate conic sections
parabola
→ directrix, focus, axis of symmetry, vertex
ellipse
→ foci, center, minor/major axes, vertices
hyperbola
→ foci, branches, center, focal/conjugate axes, vertices
standard equations
translating curves
reflection properties of conic sections
217
10.5 Rotation of Axes; Second-Degree Equations
PURPOSE: To investigate conic sections and quadratic equations that are rotated.
In the previous section, all the conic sections were aligned along either the x- or
y-axes. That is, the line that contained the vertices and the foci was either parallel
to the x-axis or the y-axis. In this section, this generally is not the case.
A general quadratic equation is a way of representing a conic section that may not
be aligned along one of the coordinate axes. A general quadratic equation will
have the following form.
Ax2 + Bxy +Cy2 + Dx + Ey + F = 0
The term Bxy is called the cross-product term. In the case where B 6= 0 the
resulting conic section will not be aligned along the x- or y-axis.
cross-product term, Bxy
IDEA: A general quadratic equation where B 6= 0 is rotated by an angle of θ
from the xy-coordinate axes where cot (2θ ) = (A −C)/B.
When B 6= 0, the conic section will be aligned along a coordinate system with
(x′ , y′ ) coordinates. The new x′ -axis and y′ -axis are rotated from the x- and yaxes by an angle of θ . The quadratic equation may be represented in the new
coordinate system by first finding θ and then determining x′ and y′ in terms of x,
y, and θ .
(x′ , y′ ) coordinates
x′ = x cos θ + y sin θ
y′ = −x sin θ + y cos θ
IDEA: To eliminate the cross-product term, find θ and then rewrite x and y as
x′ and y′ .
IDEA: x′ and y′ can usually be found without calculating θ explicitly by writing
cos θ and sin θ in terms of cos (2θ ).
If it is not important to know the actual value of θ , then x′ and y′ may still be
calculated. Using the value of cot (2θ ), an appropriate right triangle for 2θ may
be constructed from which a value for cos (2θ ) may be obtained. Then cos θ and
sin θ may be calculated using cos (2θ ).
Checklist of Key Ideas:
quadratic equations in x and y
cross-product terms
rotating axes
rotation equations
rotated conic sections
cot 2θ =
A −C
with 0 < θ < π /2
B
218
10.6 Conic Sections in Polar Coordinates
PURPOSE: To use polar coordinates to describe conic sections
and to define the eccentricity of a conic section.
eccentricity
Conic sections may also be described in polar coordinates. The polar equations
ed
ed
and r =
where e
for conic sections have the form of r =
1 ± e cos θ
1 ± e sin θ
is the eccentricity of the conic section and d is the distance from the directrix of
the conic section to the pole (or the origin). Any point on a curve that is a conic
section is some distance from the focus and the directrix. The ratio of these two
distances is called the eccentricity.
IDEA: The eccentricity of a conic section is e = d1 /d2 where d1 is the distance
from a point to the focus and d2 is the distance from the same point to the
directrix. Eccentricity for conic sections is constant.
Knowing the eccentricity and the distance of the directrix from the pole, a conic
section will be in one of the following forms:
equation
ed
r=
1 + e cos θ
ed
r=
1 − e cos θ
ed
r=
1 + sin θ
ed
r=
1 − sin θ
directrix location (d > 0)
line at x = d
line at x = −d
line at y = d
line at y = −d
CAUTION: The focus that is nearer to the directrix is assumed to be at the
pole.
All conic sections can be represented using these equations. Which conic section
is represented depends upon the value of the eccentricity. If 0 < e < 1 then it is an
ellipse. If e = 1 then it is a parabola. Otherwise, if e > 1 then it is a hyperbola.
IDEA: The eccentricity of the conic sections obey the following relationships:
ellipse
→ 0<e<1
parabola →
e=1
hyperbola →
e>1
From e and d together with the location of the directrix, the rest of the important
information can be found about a conic section. For example, the line that contains
the vertices and the focus will be perpendicular to the directrix. In the case of the
parabola, the value p will be equal to d.
finding r0 and r1
For the ellipse and the hyperbola, the values r0 and r1 need to be determined in
order to find a, b and c. Recall that c is the distance from the center to the focus.
In the case of the ellipse, the focus that is closer to the directrix is in between the
219
center and one of the vertices. The distance from this focus to the closer vertex
is called r0 and the distance to the farther vertex is r1 . The values of r0 and r1
are determined in the same way for a hyperbola although it should be noted that
the closer vertex is in between the focus and the center.
IDEA: The values of r0 and r1 for the ellipse and the hyperbola are found by
finding θ so that (r, θ ) gives the location of the two vertices.
CAUTION: Finding values for r0 and r1 assumes that the focus closest to the
directrix is located at the pole.
Finding r0 and r1 means finding the value of θ so that the polar coordinate (r, θ )
is one of the vertices. When both vertices are found, r0 and r1 can be calculated.
The value of θ used will depend upon which line the center, the vertices and the
foci are located on. If the conic section is oriented along the x-axis then the values
of θ used would be θ = 0 and θ = π . Which point is the near vertex is dependent
upon where the directrix is located.
Then for an ellipse, a, b and c can be found using the following formulas:
a = (r1 + r0 )/2,
b=
√
r0 r1 ,
c = (r1 − r0 )/2.
For a hyperbola, a, b and c can be found using the following formulas.
a = (r1 − r0 )/2,
b=
√
r0 r1 ,
c = (r1 + r0 )/2.
Checklist of Key Ideas:
focus
directrix
eccentricity
using eccentricity to identify conic sections
polar equations of conic sections
Kepler’s Laws of orbits, areas and periods
220
Chapter 10 Sample Tests
Section 10.1
1. Find dy/dx if x and y are parameterized by
x = 3t 2 , y = 9t
3
2t
2t
(b)
3
(c) 6t
3t
(d)
2
2. If a curve is parameterized by x = sint and y = 2 cost then
find dy/dx.
(a)
(d) t = 0, π /2, 3π /2
8. Find the value(s) of
t for which the tangent line(s)to the curve
parameterized by x = 2et − 5, y = 7t 2 + 3t + 1 is/are horizontal.
(a) t = 1
(b) t = −3/14
(c) t = 3/2
(d) t = 1, −3/2
9. Find the value(s) oft for which the tangent line(s)
to the curve
parameterized by x = t 3 − t 2 − 5t, y = t 4 + 2 is/are horizontal.
(a) t = 0
(b) t = 5/2
(a) 2 tan (t)
(c) t = 5
(b) 2 cot (t)
(d) t = 0, 5/2
(c) −2 tan (t)
(d) −2 cos (t)
3. Find dy/dx if x = e2t and y = t.
(a)
(b)
(c)
(d)
e−2t
2
2et
t
22t
2te2t
4. Answer true or false. If x = t 3 and y = t 2 − 2, then
1
d2y
=− .
3t
dx2
5. Answer true or false. If x = sint and y = cost then
d2y
= cott.
dx2
6. Find the value oft for which the tangent line to the curve
parameterized by x = t 4 , y = 3t 2 − 2t + 6 is horizontal.
(a) t = 1/3
(b) t = 0
(c) t = 2/3
(d) t = 8
7. Find the value(s) of
t for which the tangent line(s)to the curve
parameterized by x = 3 sin (t) + 8, y = 5t 2 + 7 is/are horizontal.
(a) t = π /2, 3π /2
(b) t = 0
(c) t = −3/5
10. If 0 ≤ t < 2π , then find the value(s) oft for which the tangent
line(s) to the curve parameterized by x = 6t 3/2 , y = 4 sint
is/are horizontal.
(a) t = 0
(b) t = 0, π /2
(c) t = π /2, 3π /2
(d) t = 0, π /2, 3π /2
Section 10.2
1. Answer true or false. To plot (6, π /4) in polar coordinates, go
out 6 units from the pole to the right, then rotate π /4 clockwise.
2. Find the rectangular coordinates of the polar point (1, π /4).
√ √
(a) ( 2, 2)
√
√
(b) ( 2/2, 2/2)
√ √
(c) (2 2, 2 2)
√ √
(d) (4 2, 4 2)
3. Find the rectangular coordinates of the polar point (5, −π /2).
(a) (5, 0)
(b) (0, 5)
(c) (−5, 0)
(d) (0, −5)
4. Find the rectangular coordinates of the polar point
(−4, −π /4).
√
√
(a) (−2 2, −2 2)
221
√ √
(b) (2 2, 2 2)
√ √
(c) (−2 2, 2 2)
√
√
(d) (2 2, −2 2)
5. Use a calculating utility to approximate the polar coordinates
of the point (7, 3).
(a) (58, 1.1903)
(b) (7.6158, 0.4049)
(d) 6
12. Describe the curve r = 10 + 10 sin (θ ).
(a) limacon with inner loop
(b) cardioid
(c) dimpled limacon
(d) convex limacon
13. Describe the curve r = 4 + 6 cos (θ ).
(c) (58, 1.1659)
(a) limacon with inner loop
(d) (7.6158, 1.1659)
(b) cardioid
6. Use a calculating utility to approximate the polar coordinates
of the point (−2, −5).
(a) (29, 4.3319)
(c) dimpled limacon
(d) convex limacon
14. Describe the curve r = 2 + 8 cos (θ ).
(b) (5.3852, 4.3319)
(a) limacon with inner loop
(c) (58, 1.19031)
(b) cardioid
(d) (5.3319, 1.1903)
(c) dimpled limacon
7. Describe the curve θ = π /2.
(d) convex limacon
(a) a vertical line
(b) a horizontal line
15. Answer true or false. The graph of r = 7θ in polar coordinates gives an Archimedean spiral.
(c) a circle
(d) a semicircle
Section 10.3
8. Describe the curve r = 4 cos (θ ).
(a) a circle left of the origin
(b) a circle above the origin
(c) a circle right of the origin
(d) a circle below the origin
9. Describe the curve r = 8 sin (θ ).
(a) a circle left of the origin
(b) a circle above the origin
1. Find the approximate area of the region enclosed by r =
4 − 4 cos (θ ).
(a) 75.40
(b) 56.55
(c) 18.85
(d) 9.42
2. Find the approximate area of the region enclosed by r =
4 − 4 sin (θ ).
(c) a circle right of the origin
(a) 75.40
(d) a circle below the origin
(b) 56.55
10. What is the radius of the circle r = 8 sin (θ )?
(a) 8
(b) 16
(c) 4
(d) 1
11. How many petals does the rose r = 7 sin (3θ ) have?
(a) 1
(b) 4
(c) 3
(c) 18.85
(d) 9.42
3. Find the approximate area of the region enclosed by r =
2 − 6 cos (θ ) from θ = 0 to θ = π /2.
(a) 5.28
(b) 58.56
(c) 183.96
(d) 23.00
4. Find the approximate area of the region enclosed by r =
2 − 6 sin (θ ) from θ = 0 to θ = π /2.
222
(a) 183.96
(a) 12.6
(b) 58.56
(b) 29.3
(c) 5.28
(c) 117.1
(d) 23.00
(d) 58.6
5. Answer true or false. The approximate area of the region
bounded by the curve r = 3 cos (2θ ) from θ = 3π /2 to θ =
2π is 3.53.
15. Find the approximate area bounded by r = 6 − 3 sin (θ ) from
θ = π to θ = 3π /2.
(a) 12.1
6. Answer true or false. The area between the circle r = 5 and
the curve r = 2 + 2 cos (θ ) is π /2.
(b) 29.3
7. Answer true or false. The area of one petal of r = sin (2θ ) is
(d) 49.81
given by
Z π /2
0
0.5 sin (2θ ) d θ .
8. Answer true or false. The area in one petal of r = 2 sin (6θ )
is given by
Z π /3
0
2
2(sin (6θ )) d θ .
9. Answer true or false.
r = cos (6θ ) is given by
The area in all of the petals of
Z π /3
0
3(sin (6θ ))2 d θ .
10. Find the approximate area of the region bounded by r = 6θ
from θ = 0 to θ = π .
(a) 3.1
(b) 2.6
(c) 186.04
(d) 10.3
(c) 117.1
16. Answer true or false. If r = 7 sin (θ ), then the tangent line to
the curve at the origin is θ = 0.
17. Find the arc length of the spiral r = e5θ between θ = 0 and
θ = 1.
√
26
(e − 1)
(a)
5
2
(b) (e − 1)
5
√
26 5
(c)
(e − 1)
5
2
(d) (e5 − 1)
5
18. Find the arc length of the spiral r = sin θ between θ = 0 and
θ = 2π .
(a) 2
11. Find the approximate area of the region bounded by r = 10θ
from θ = 0 to θ = 2π .
(a) 41, 034
(b) 2π
√
(c) 2 2π
√
(d) 2 2
19. Answer true or false. The arc length of the curve r = sin (2θ )
between θ = 0 and θ = π is π .
(b) 2, 067
(c) 8, 268
20. Answer true or false. The arc length of the curve r = 8θ between θ = 0 and θ = π is 8π .
(d) 4, 134
12. Answer true or false.
The region bounded between
r = 2 cos (θ ) and r = 2 sin (θ ) is given by
Z π /4
0
2(sin (θ ) − cos (θ ))2 d θ .
13. Find the approximate area bounded by r = 6 − 4 cos (θ ) from
θ = π to θ = 3π /2.
Section 10.4
1. The vertex of the parabola y2 = 9x is located at
(a) (1, 9)
(b) (0, 0)
(a) 12.6
(c) (9, 1)
(b) 29.3
(d) (1, 1)
(c) 117.1
(d) 58.6
14. Find the approximate area of the region bounded by r =
6 − 4 cos (θ ) from θ = π /2 to θ = π .
2. The vertex of the parabola (y − 6)2 = 3(x − 2) is located at
(a) (−2, −6)
(b) (2, 6)
(c) (−3, −6)
223
Section 10.5
(d) (3, 3)
3. A parabola has a vertex at (4, 5) and a directrix of x = 0. The
focus of the parabola is located at
(a)
(b)
(c)
(d)
(a) parabola
(4, 0)
(4, 10)
(8, 5)
(8, 10)
(b) circle
(c) ellipse
(d) hyperbola
4. The graph of the parabola x = 9y2 + 2 opens along
(a)
(b)
(c)
(d)
2. The graph x2 + 6xy − 3y2 = 0 is what type of conic section?
the positive x-axis.
the negative x-axis.
the positive y-axis.
the negative y-axis.
(a) parabola
(b) circle
(c) hyperbola
(d) ellipse
5. What are the ends of the minor axis for the ellipse
(a)
(b)
(c)
(d)
3. The graph x2 + 5xy − 7y2 + 25 = 0 is what type of conic section?
x2 y2
+
= 1?
36
4
(6, 0) and (−6, 0)
(36, 0) and (−36, 0)
(0, 6) and (0, −6)
(0, 2) and (0, −2)
(a) parabola
(b) circle
(c) ellipse
x2 y2
6. Answer true or false. The foci of
+
= 1 are (7, 0) and
49 81
(−7, 0).
x2 y2
7. The foci of the ellipse
+
= 1 are located at
81 49
√
√
(a) (−4 2, 0) and (4 2, 0)
√
√
(b) (−2 65, 0) and (2 65, 0)
(c) (−9, 0) and (9, 0)
(d) (0, −7) and (0, 7)
8. Answer true or false. The foci of the ellipse
(8, 0) and (−8, 0).
1. What type of conic section is the graph of xy = 12?
x2 y2
+
= 1 are
64 36
x2
y2
−
=1
9. Answer true or false. The foci of the hyperbola
81 16
√
√
are (0, − 97) and (0, 97).
x2 y2
10. Answer true or false. The foci of the hyperbola
−
=1
64 36
are (10, 0) and (−10, 0).
11. Answer true or false. The hyperbola 7y2 − 12x2 = 84 opens
along the y-axis in both the positive and negative directions.
12. Answer true or false. 7x2 + y2 = 7 has a vertical major axis.
13. Answer true or false. The vertex of y = 2x2 + 8 is located at
(−2, 0).
14. Answer true or false. The vertex of y = x2 − 9 is located at
(0, −9).
15. Answer true or false. The vertex of x = 5y2 + 12 is located at
(5/12, 0).
(d) hyperbola
4. The graph of 3xy = 5 is a(n)
(a) parabola
(b) circle
(c) ellipse
(d) hyperbola
5. The graph 4x2 − 4y2 = 0 is best described as which of the
following?
(a) hyperbola
(b) circle
(c) line
(d) pair of intersecting lines
6. Answer true or false.
x2 + 7xy + 2y2 + 7x + 5y − 2 = 0 represents an ellipse.
7. Answer true or false.
x2 − 5xy + 6y2 + 5x − 3y + 4 = 0 represents a pair of intersecting lines.
8. Answer true or false.
6x2 − 6y2 + 8 = 0 represents a pair of intersecting lines.
9. Answer true or false.
x2 + 3xy + 2y2 = 0 represents a single point.
10. Answer true or false.
x2 + 3xy + 2y2 − 6 = 0 represents a circle.
224
Section 10.6
(c) circle
1. The eccentricity of r =
8
is
1 + 2 sin (θ )
(a) 4
7
is
6 + 4 cos (θ )
(b) ellipse
(c) 2
(c) circle
(d) 6
(d) hyperbola
2. The eccentricity of r =
8
is
4 + 8 cos (θ )
9. Answer true or false. The graph of r =
horizontally.
(a) 8
10. Answer true or false. The graph of r =
(b) 4
perbola that opens left and right.
(c) 2
(d) 1
1
orients
5 − cos (θ )
3
is a hy1 − 6 sin (θ )
11. Answer true or false. The graph of r =
3. The eccentricity of r =
3
is
5 + 15 sin (θ )
parabola that opens to the left.
12. Answer true or false. The graph of r =
(a) 12
6
is a
1 − sin (θ )
4
is a
1 + cos (θ )
parabola that opens to the left.
(b) 4
13. Answer true or false. The graph of r =
(c) 2
parabola that opens upward.
(d) 3
4. Answer true or false. r =
5
has its directrix left
1 − 3 cos (θ )
5. Write the equation of the ellipse that has e = 3 and directrix
x = 1.
3
(a) r =
1 + 3 cos (θ )
3
(b) r =
1 − 3 cos (θ )
3
(c) r =
1 + 3 sin (θ )
3
(d) r =
1 − 3 sin (θ )
6. The graph of r =
8. The graph of r =
(a) parabola
(b) 1
of the pole.
(d) hyperbola
6
is
5 − 2 sin (θ )
(a) parabola
(b) ellipse
(c) circle
(d) hyperbola
10
is
7. The graph of r =
2 + 4 cos (θ )
1
is a
6 − 3 sin (θ )
10
is a hy6 − 6 sin (θ )
perbola oriented along the vertical axis in both the positive
and negative directions.
14. Answer true or false. The graph of r =
15. A small planet is found 8 times as far from the sun as the
earth. What is its period?
(a) 22.6 years
(b) 64 years
(c) 32 years
(d) 4 years
Chapter 10 Test
1. Find the rectangular coordinates of the polar point (−4, π /4).
√ √
(a) (− 2, 2)
√
√
(b) (− 2/2, 2/2)
√
√
(c) (−2 2, −2 2)
√
√
(d) (−4 2, −4 2)
2. Use a calculating utility to approximate the polar coordinates
of the rectangular point (6, 8).
(a) (10, 0.9273)
(a) parabola
(b) (100, 0.9273)
(b) ellipse
(c) (10, 0.6435)
225
(d) (100, 0.6435)
3. Which of the following best describes the curve
r = 8 cos (θ )?
(a) a circle left of the origin
(b) a circle above the origin
(c) a circle right of the origin
(d) a circle below the origin
4. What is the radius of the circle r = 24 cos (θ )?
(a) 24
(b) 48
(b) t = 0
(c) t = −3/5
(d) t = 0, π /2, 3π /2
11. Answer true or false. If r = 12 sin (θ ), then the tangent line
to the curve at the origin is given by the equation θ = 0.
12. Answer true or false.
The arc length of the curve
r = 2 cos (3θ ) between θ = 0 and θ = π is 4π .
13. Find the approximate area of the region that is enclosed by
r = 4 − 4 sin (θ ).
(a) 24π
(b) 18π
(c) 12
(c) 6π
(d) 1
(d) 3π
5. How many petals does the rose r = 6 sin (3θ ) have?
(a) 1
(b) 4
(c) 3
(d) 6
6. Which of the following statements best describes the curve
r = 10 + 5 cos (θ )?
(a) a limacon with an inner loop
(b) a cardioid
(c) a dimpled limacon
(d) a convex limacon
7. Answer true or false. r = 7/θ is the graph of a hyperbolic
spiral.
8. Find dy/dx if a curve is parameterized by
x = −2 cos (t), y = −2 sin (t)
(a) tan (t)
(b) cot (t)
(c) − tan (t)
(d) − cot (t)
p
p
9. Answer true or false. If x = sin (t) and y = cos (t) then
d2y
= cot (t).
dx2
10. Suppose that a curve is parameterized by
x = sin (t), y = 3t 2 + 10 .
Find the value(s) of t for which the tangent line(s) to the curve
is/are horizontal.
(a) t = π /2, 3π /2
14. Answer true or false. The area of the region that is bounded
by the curve r = 3 sin (2θ ) from θ = 0 to θ = π is 4π .
15. Answer true or false. The area between the circle r = 12 and
the curve r = 5 + 5 cos (θ ) is π .
16. Answer true or false. The area in one petal of r = sin (4θ ) is
given by
Z 2π
0
0.5 sin (4θ ) d θ .
17. Find the area of the region that is bounded by
r = 6 − 4 cos (θ ).
(a) 12.6
(b) 29.3
(c) 117.1
(d) 138.23
18. The vertex of the parabola x2 = −3y is located at
(a) (0, 0)
(b) (3, 1)
(c) (1, 3)
(d) (1, 1)
19. The eccentricity of r =
6
is
1 + 3 cos (θ )
(a) 6
(b) 1
(c) 3
(d) 12
20. Answer true or false. The curve r =
rectrix to the left of the pole.
8
has its di1 − 2 cos (θ )
226
Chapter 10: Answers to Sample Tests
Section 10.1
1. a
9. a
2. c
10. c
3. a
4. false
5. false
6. a
7. b
8. b
2. b
10. c
3. d
11. c
4. c
12. b
5. b
13. a
6. b
14. a
7. a
15. true
8. c
2. a
10. c
18. b
3. a
11. d
19. false
4. c
12. false
20. false
5. true
13. d
6. false
14. d
7. false
15. d
8. false
16. true
2. b
10. true
3. c
11. true
4. a
12. true
5. d
13. false
6. false
14. true
7. a
15. false
8. false
2. c
10. false
3. d
4. d
5. d
6. false
7. false
8. false
2. c
10. false
3. d
11. false
4. true
12. true
5. a
13. false
6. b
14. false
7. d
15. a
8. b
2. a
10. b
18. a
3. c
11. true
19. c
4. c
12. false
20. true
5. c
13. a
6. d
14. false
7. true
15. false
8. d
16. false
Section 10.2
1. false
9. b
Section 10.3
1. a
9. false
17. c
Section 10.4
1. b
9. false
Section 10.5
1. d
9. false
Section 10.6
1. c
9. true
Chapter 10 Test
1. c
9. false
17. d
Chapter 11: Three Dimensional Space; Vectors
Summary: This chapter introduces three dimensional spaces and the concept of
a vector. Much of the material here is focused upon describing vectors, how they
interact with each other and their significance in both two and three dimensional
spaces. Dot products and cross products of vectors are two of the most significant ideas in the chapter. These two features of vectors allow many geometric
relationships to be built about planes, curves and later, vector-valued functions.
Using dot products, planes are defined in terms of normal vectors and both dot and
cross products give information about the angle between two vectors. Quadric surfaces are also discussed in the chapter. These are extensions of conic sections from
two dimensions to surfaces in three dimensions. Finally, cylindrical and spherical
coordinate systems are introduced as alternative ways of describing three dimensional spaces. Both cylindrical and spherical coordinates are similar in nature to
polar coordinates that were introduced in two dimensional space as an alternative
to rectangular coordinates.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Identify different surfaces in three dimensions such as
→ spheres (§11.1),
→ cylindrical surfaces (§11.1),
→ planes (§11.6),
→ quadric surfaces (§11.7).
2. Find a vector based upon two points (§11.1).
3. Calculate the norm of a vector, a scalar product and the sum of two vectors
(§11.2).
4. Normalize a vector (§11.2).
5. Find the angle between two vectors (§11.3, 11.4).
6. Calculate the dot product of two vectors (§11.3).
7. Calculate the cross product of two vectors (§11.4).
227
228
8. Calculate the scalar triple product of three vectors (§11.4).
9. Find the parametric equation of a line in the direction of a vector (§11.5).
10. Find the equation of a plane (§11.6).
11. Calculate various distances between planes, points and lines (§11.6).
12. Identify and sketch quadric surfaces (§11.7).
13. Convert between the rectangular, cylindrical and spherical coordinate systems (§11.8).
11.1 Rectangular Coordinates in 3-Space; Spheres;
Cylindrical Surfaces
PURPOSE: To extend rectangular coordinates to three dimensions and to introduce the equation of a sphere and some cylindrical surfaces that are extensions of curves in a plane.
3-space
xyz-coordinate system
octant
The main focus of this section is to extend the usual rectangular coordinate system
in the xy-plane into a 3 dimensional space or a 3-space. This is done by introducing a z-coordinate along a z-axis that is perpendicular to the xy-plane. Then this
new xyz-coordinate system is separated into 8 regions called octants where the
regions are separated by the xy-, xz- and yz-planes. If x > 0, y > 0 and z > 0 then
this is called the first octant.
right-handed system
The xyz-coordinate system is typically considered to be a right-handed system
(i.e., if the thumb on the right hand is pointing in the positive z direction then
the rest of the fingers curl from the positive x-axis towards the positive y-axis).
In other words, in a right handed system when one looks down at the xy-plane
from the positive z direction, the positive x-axis points to the right and the positive
y-axis points up.
distance formula
An important concept in 3-space is that of distance between two points. If two
points, P1 and P2 have the coordinates of (x1 , y1 , z1 ) and (x2 , y2 , z2 ) then the distance is given by the following formula.
q
d = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2
IDEA: The equation of a sphere is just the distance formula in the rectangular
coordinate system.
sphere
This is also the basis for the equation of a sphere. In a sphere, one point is the
center (x0 , y0 , z0 ) and then all points on the surface of the sphere, represented by
(x, y, z), are a fixed distance r from the center. Then the equation of a sphere
r2 = (x − x0 )2 + (y − y0 )2 + (z − z0 )2
229
is just the distance formula squared.
IDEA: An equation involving only two variables, represents a cylindrical surface in 3-space since the third variable can be any value. This extends (extrudes) the curve infinitely in the direction perpendicular to the plane of the
curve.
A special type of surface is also introduced in this section. Any curve in the xyplane can be extended to be a surface in 3-space by letting the surface consist of
all points that are directly above or below the curve in the xy-plane. For example,
the equation x2 + y2 = 1 would represent a circular cylinder. This is an infinite
circular tube that extends upwards and downwards in the z direction parallel to the
z-axis. In other words, the z-coordinate is not specified so any point whose x- and
y-coordinates satisfy the equation (with any z-coordinate) will be on the circular
cylinder. Surfaces of this type are called cylindrical surfaces. These types of
surfaces can also be generated by curves in the xz- or yz-planes. So long as the
equation only specifies two variables, then the third variable can be anything since
it is unspecified.
cylindrical surfaces
Checklist of Key Ideas:
1-space, 2-space and 3-space
rectangular or Cartesian coordinate system
left-handed or right-handed
coordinate planes
octants
distance formula in 3-space
equation of a sphere
cylindrical surfaces
11.2 Vectors
PURPOSE: To define vectors in two and three dimensions and
to discuss various mathematical operations on vectors such as
addition and scalar multiplication.
A vector is a way of representing a point in 2-space or 3-space relative to some
other point. For example, a vector indicates how far one point is from another and
the distance between the points. The two defining features of any vector are its
magnitude and its direction. The magnitude of a vector between two points (an
initial point and a terminal point) is equal to the distance between the two points.
The direction of a vector is described by saying it points from the initial point in
the direction of the terminal point.
magnitude
direction
230
IDEA: Vectors may be described or determined in several ways.
−→
v = PQ = hq1 − p1 , q2 − p2 , q3 − p3 i
1. Using an initial point and a terminal point.
v = kvk u where kuk = 1.
2. Using a magnitude and a unit vector in the same direction.
v = kvk hcos θ , sin θ i
3. Using a magnitude and an angle (in 2-space).
IDEA: The magnitude of a vector is equivalent to the norm of a vector or the
length of a vector. The norm of a vector is calculated by
√
also kvk = v · v
(see §11.3)
kvk =
q
v21 + v22 + v23
where v = hv1 , v2 , v3 i.
CAUTION: The norm of vector is a positive scalar value. When working with
vectors, it is important to know which operations result in vectors and which
result in scalar values.
All vectors may be described using components. A vector v from the point
P1 (x1 , y1 , z1 ) to the point P2 (x2 , y2 , z2 ) (in 3-space) has the component form of
−−→
v = P1 P2 = hx2 − x1 , y2 − y1 , z2 − z1 i
= (x2 − x1 )i + (y2 − y1 )j + (z2 − z1 )k
= v1 i + v2 j + v3 k.
In this case, P1 is the initial point and P2 is the terminal point. Notice that the
components v1 , v2 , and v3 simply represents the change in value of the x, y or z
coordinates respectively between the two points.
IDEA: Components of a vector are scalar values that determine the change in
coordinates from one point to another.
Once the components of a vector are known, the vector may be moved to a different initial point. The components will determine the change in the coordinate
from the new initial point to some new terminal point. For example, a vector can
generally be written as v = hv1 , v2 , v3 i. Then if the initial point of the vector is
(x, y, z) then the terminal point will be (x + v1 , y + v2 , z + v2 ).
IDEA: A vector may use any point as its initial point. The components determine what direction the vector points and how far.
vector addition
scalar multiplication
Many operations may be performed upon vectors using only addition and multiplication. Vectors may be added together by simply adding together their components. For example, v + w = hv1 + w1 , v2 + w2 , v3 + w3 i in 3-space. A scalar
multiple of a vector is simply the same vector but with components that have all
been multiplied by some scalar factor. The effect of multiplying a vector by k is
to multiply the magnitude of the vector by |k|. If k < 0 then this also reverses the
direction of the vector. Using vector addition (by adding components) and scalar
multiplication, then vector subtraction can be defined by just adding together a
vector v and a negative vector u. For example, v − u = v + (−u).
231
CAUTION: The operations of adding two vectors, v + w, and scalar multiplication, k v, both result in a new vector. When performing these operations, the
results must be vectors not scalars.
There are several special types of vectors and special ways of comparing vectors.
Two vectors are said to be equal only if each of their components are equal.
equal vectors
IDEA: If two vectors, v and w, are equal then
1. v − w = 0 (the zero vector),
2. kv − wk = 0 (this is a scalar value),
3. v1 = w1 , v2 = w2 , and v3 = w3 (the components are equal).
CAUTION: If kvk = kwk then it does not necessarily mean that v and w are
equivalent. This also does not mean that v and w are parallel.
Two vectors are said to be parallel if they have the same direction or exactly
opposite directions. If two vectors are not parallel, then the angle, θ , between
the two vectors can be found (see §11.3 dot product and §11.4 cross product). If
two vectors are orthogonal or perpendicular then the angle between them is a right
angle (see §11.3). A unit vector, u, is a vector that has a magnitude (or length) of
one (i.e., kuk = 1). A unit vector can be found that has the same direction as some
other vector, v. This process is called normalizing the vector v. The unit vector
u is found by dividing v by its magnitude. The vectors i = h1, 0, 0i, j = h0, 1, 0i,
and k = h0, 0, 1i are special unit vectors that lie along the positive x-, positive yand positive z-axes respectively.
IDEA: Normalizing a vector v is finding a unit vector u where u =
u has the same direction as v.
v
. Then
kvk
Vectors have many applications and often have some physical meaning. For example, a displacement vector will give the position of some object relative to the
origin. A velocity vector describes how fast an object is moving and in what direction. A force vector describes how much force is being applied to an object and
in what direction the force is pushing. Any physical quantity that can be described
in terms of a magnitude and a direction may be described using a vector.
IDEA: Any quantity that is described using magnitude and direction may be
represented by a vector.
Checklist of Key Ideas:
magnitude and direction
vector
displacement vector
initial point or tail
parallel vectors
angle between vectors
(see §11.3, 11.4)
unit vector
normalizing a vector
the unit vectors i, j and k
232
terminal point or head
scalars
equal vectors, zero vector, vector sum, scalar multiple
parallel vectors
components
norm of a vector
unit vectors
normalizing a vector
vectors determined by length and angle
11.3 Dot Products; Projections
PURPOSE: To define the dot product of two vectors and to describe how the dot product relates to the angle between two vectors, the orthogonal components of a vector and the projection of
a vector onto another vector.
There is more than one way to define a product involving a vector. In the previous
section, a scalar multiplication (or scalar product) was introduced, where a scalar
number is multiplied by each of the components of a vector. In this section, the
dot product of two vectors is defined. In the next section, the cross product and
scalar triple product will be defined.
CAUTION: Not all “products” involving vectors produce another vector.
dot product
As more products are defined for vectors, be careful to note what the result of
each product is. For example, a scalar product of a vector is a new vector with a
different length and either the same or opposite direction. A dot product results
in a scalar value.
IDEA: A dot product is the sum of the products of the components:
u · v = u1 v1 + u2 v2 + u3 v3
This is a scalar value, not a vector.
Some properties of u · v
u·v = v·u
u·(v + w) = u · v + u · w
0·v = 0
v · v = kvk2
There are several algebraic properties of the dot product of two vectors. All of
them may be verified by writing out the component definition of the dot product.
One of the more important algebraic properties of the dot product is its relationship
to the norm of a vector. The dot product of a vector with itself is the square of its
norm.
233
IDEA: The dot product and the vector norm are related by the following equations.
kvk2 = v · v
kvk =
or
dot product and norm
√
v·v
The dot product is related to the two important features of a vector: length and
direction. As seen above, the dot product can be used to find the magnitude or
length of a vector. The dot product may also be used to determine the angle
between two vectors.
IDEA: The dot product relates the angle θ between two vectors with the following equations.
cos θ =
u·v
kuk kvk
or
dot product and angle between
vectors
u · v = kuk kvk cos θ
Using this relationship with cos θ , the dot product may also be used to determine
the angle that a vector makes with one of the rectangular coordinate axes. These
angles are called direction angles. This can be done by taking the dot product of
a vector u with one of the unit vectors, i, j, or k, along the coordinate axes.
IDEA: If u · v = 0 then u ⊥ v. Then u and v are orthogonal vectors.
direction angles
orthogonal vectors
Since the dot product can be written in terms of cos θ , if two vectors are perpendicular to each other then cos θ = 0 and so u · v = 0. In other words, if the dot
product between two vectors is zero then the two vectors must be perpendicular
(or orthogonal). This is an important property of the dot product that may be one
of its most significant contributions to vector calculus.
IDEA: Given two unit vectors e1 and e2 where e1 ⊥ e2 , any vector v can be
broken into orthogonal components in the direction of e1 and e2 :
v = (v·e1 )e1 + (v·e2 )e2
Building upon the idea of orthogonal vectors, the dot product can be used to decompose a vector into orthogonal components. In other words, the vector can
be written as a sum of vectors where each of the vectors in the sum are mutually
orthogonal. This is related to the idea of finding the projection of one vector onto
another since the orthogonal component of v in the direction of e1 is actually just
the projection of v onto the vector e1 .
orthogonal components
IDEA: The projection of the vector v onto w is defined to be
projw v = (kvk cos θ )
v ·w
v·w
w
w.
=
w
=
kwk kwk2
w·w
Note: kvk cos θ =
v·w
.
kwk
So projections can be found using only dot products; the angle θ does not
need to be found.
The vector v can be projected onto any other vector w using the dot product. This
is done by finding a vector that is a scalar multiple of a unit vector in the same
projection of v onto w
234
Note that if w is already a unit vector
then w · w = kwk2 = 1.
direction as w. The unit vector is given by w/kwk. Then the length of this vector
should be equal to cos θ kvk where θ is the angle between v and w.
IDEA: Drawing a right triangle with v as the hypotenuse can be helpful. The
two legs of the triangle will be the orthogonal components of v. One of the
legs will the orthogonal projection of v onto w.
To remember this idea of projections, it may be helpful to draw a right triangle
with v as the hypotenuse. Then one of the legs of the right triangle will be parallel
to w and the other orthogonal to w. The vector that is parallel to w is projw v. By
vector addition, the other leg of the triangle (which is orthogonal to w) will be
v − projw v since the vector sum of the two legs should equal the hypotenuse.
Checklist of Key Ideas:
dot product
algebraic properties of the dot product
relationship between dot product and norm
angle between two vectors
the sign of the dot product
direction angles
orthogonal components of a vector
vector components, scalar components
orthogonal projections
work, displacement vector
11.4 Cross Product
PURPOSE: To define the cross product of two vectors, the scalar
triple product of three vectors and to describe the algebraic and
geometric properties of both of these vector products.
cross product
The cross product is another product that is defined between two vectors. Unlike
the dot product, the result of a cross product is another vector. Also, the cross
product is only defined for vectors in three space.
CAUTION: The cross product results in another vector. The cross product is
only defined for two vectors that are in 3-space.
The definition of the cross product is also more complicated than the definition
of the dot product. It may be helpful to remember that if w = u × v then the i-th
235
component of w does not depend upon the corresponding components of u or v.
For example, w1 = u2 v3 − u3 v2 does not depend upon either u1 or v1 .
IDEA: If w = u × v then wi does not depend upon either ui or vi .
Here is another way to define the cross product.
IDEA: If c = a × b then if i, j, k all have values of 1, 2 or 3 but i 6= j 6= k then
ci = a j bk − ak b j
where j < k.
For example, if c = hc1 , c2 , c3 i = a × b then consider c1 . In this case, i = 1 so j
and k must be 2 and 3. Also j < k so j = 2 and k = 3. Then
c1 = a2 b3 − a3 b2 .
CAUTION: Order is important when computing cross products. For example,
a × b = −(b × a). Because of this, the statement a × b × c is ambiguous and
should be avoided (i.e., it is unclear whether a × b or b × c should be calculated first).
An important feature of a cross product is that it will be orthogonal to both of
the vectors in the cross product. This is important in finding normal vectors to
planes (see §11.6) and also calculating the areas of parallelograms. In a righthanded system (see §11.1), the right hand may be used to visualize the direction
of the cross product vector. The right hand is held out in front of the body with the
thumb pointing up and the first finger pointing straight ahead. Then the second
finger should point to the left. The first finger represents the direction of the vector
u, the second finger represents the direction of v, and the thumb will represent the
direction of the cross product, u × v.
Another important feature of the cross product is its relationship with sin θ where
θ is the angle between two vectors a and b.
sin θ =
ka × bk
kak kbk
or
kak kbk sin θ = ka × bk
CAUTION: Using the inverse sine function with the cross product may not give
the correct angle between two vectors. To find the angle between two vectors
where 0 ≤ θ ≤ π , the dot product should be used with inverse cosine.
Although the cross product may be used to calculate sin θ , as a general rule it
should not be used to find the angle between two vectors. According to the cross
product sin θ is always positive (since norms are always nonnegative) and so the
inverse sine function would always predict that 0 ≤ θ ≤ π /2. The dot product is a
better choice for finding θ because it is related to cos θ . Then inverse cosine can
correctly predict if 0 ≤ θ ≤ π /2 or π /2 ≤ θ ≤ π .
IDEA: The cross product can be used to calculate the areas of parallelograms
and triangles.
cross product and orthogonality
(a × b) ⊥ a and (a × b) ⊥ b
236
As a result of this, the cross product may be used to find the areas of parallelograms and triangles. If two adjacent sides of a parallelogram are formed by the
vectors u and v (both starting from the same point) then the area of the parallelogram is ku × vk. On the other hand, a triangle that has two sides determined by
the vectors u and v may be thought of as half of a parallelogram. Consequently,
the area of the triangle would be ku × vk/2.
IDEA: The scalar triple product is a scalar value obtained using both the dot
and cross products.
The scalar triple product involves three vectors, a, b and c, and both the dot and
cross products. The result of the triple scalar product is a scalar value. Consider
the scalar triple product a·(b × c). The vector w = b × c is perpendicular to the
plane that contains b and c (as long as they are not parallel). Then a · w is related
to the angle between a and the plane formed by b and c.
IDEA: The scalar triple product may be used to find the volume of a parallelepiped that is formed by the three vectors a, b, and c which all have the
same initial point.
Checklist of Key Ideas:
determinant, cross product
algebraic properties of cross product
area of a parallelogram
sine of the angle between vectors
cross products and orthogonal vectors
cross products and parallel vectors
scalar triple products
volume of a parallelepiped
expressing dot and cross products using only angles and magnitudes
scalar moment or torque
vector moment or vector torque
11.5 Parametric Equations of Lines
PURPOSE: To use vectors to define a line parametrically in two
and three dimensions.
237
Vectors are an important part of describing lines in 2-space and 3-space. Recall
that if v = hv1 , v2 , v3 i then v1 , v2 , and v3 represent the change in x, y and z coordinates from some initial point to some terminal point (see §11.2). If the vector
between the points (x, y, z) and (x0 , y0 , z0 ) is parallel to v then
hx − x0 , y − y0 , z − z0 i = tv.
So given a starting point (in 3-space), of (x0 , y0 , z0 ) and a vector v = hv1 , v2 , v3 i,
the following represents a line through the given point that is parallel to the vector
v:
x = x0 + v1t,
y = y0 + v2t,
z = z0 + v3t.
Depending upon the value of t, the terminal point, (x, y, z), will be at different
points along the line in the direction determined by the vector v.
IDEA: If r0 = hx0 , y0 , z0 i with its initial point at the origin then the following is
the equation of a line through the point (x0 , y0 , z0 ) and parallel to the vector v:
r = r0 + tv
The coordinates of points on the line can be parameterized by the following functions.
x(t) = x0 + v1t,
y(t) = y0 + v2t,
z(t) = z0 + v3t.
If the vector r0 = hx0 , y0 , z0 i with its initial point at the origin then the line can be
written as a vector-valued function (see §12.1).
r(t) = hx(t), y(t), z(t)i = r0 + tv
So each component of the vector r(t) is a function of t.
IDEA: If the point (x1 , y1 , z1 ) is on the line through the point (x0 , y0 , z0 ) parallel
to the vector v then the same value of t must be used to determine x1 , y1 , and
z1 in each of the parametric equations of the line.
Different points on the line may be found by evaluating the parameterizations of
x(t), y(t) and z(t) at different values of t. Then to check if a point is on a line, the
same value of t should result in the values of x, y and z. For example, consider the
line
x = t,
y = 3 + t,
z = 2 − 5t.
If t = 2 then (x, y, z) = (2, 5, −8) which is a point on the line. On the other hand,
the point (1, 4, 2) is not on the line. This can be seen since x = 1 corresponds to
t = 1. The value of y = 4 also corresponds to t = 1. Then the value of z should be
z = 2 − 5(1) = −3 not z = 2. The values of x, y and z should all be obtained using
the same value of t.
CAUTION: Lines that are defined parametrically and intersect at a point, do
not have to both reach that point for the same parameter value.
238
Lines that are defined parametrically can still intersect although the same parameter value may not give the point of intersection for both lines. For example, the
lines {x = 1 +t1 , y = 2 −t1 , z = 4 + 4t1 } and {x = 3 − 2t2 , y = 1 +t2 , z = 6 − 2t2 }
both intersect at the point (1, 2, 4). However, the first line reaches this point at
t1 = 0 and the second line reaches this point at t2 = 1.
IDEA: To determine if two lines intersect, two of their coordinates should be
set equal to solve for the two parameters. Then the two parameter values
should give the same point if the lines intersect.
To determine if two lines intersect, set two of the coordinates equal to each other
and solve for the two different parameters. If values for t1 and t2 can be found, then
evaluate the third coordinate at these values. If both lines have the same coordinate
value then the lines intersect at the point determined by the two different parameter
values.
Checklist of Key Ideas:
determining a line with a point and a vector
parametric equation of a line
skew lines, parallel lines
line segments
vector equation of a line
11.6 Planes in 3-Space
PURPOSE: To discuss the relationship between planes and the
dot product and to investigate various distance problems involving planes, points and lines.
The equations of planes in 3-space are very closely related to vectors. The basic
idea is that the dot product of a vector normal to the plane and a vector within the
plane must be zero. In other words a normal vector is perpendicular to any vector
that is within the plane so their dot product must be zero (see §11.3).
Equations of planes:
point-normal form
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0
general form
ax + by + cz + d = 0
IDEA: The equation of a plane is given by
n·v = 0
where n = ha, b, ci is normal to the plane and v is in the plane.
If the point P(x0 , y0 , z0 ) is in the plane then the vector v = hx − x0 , y − y0 , z − z0 i
is in the plane for any generic point (x, y, z) in the plane. So if n = hn1 , n2 , n3 i is
normal to the plane then n · v = 0.
239
IDEA: The general form of the equation of a plane is ax + by + cz + d = 0. In
this case, the vector n = ha, b, ci is normal to the plane.
general form of plane
normal vector
A plane can be determined then by knowing a point in the plane and knowing a
vector that is normal to the plane. Any information that leads to these two things
is sufficient to determine a plane.
IDEA: A plane can be determined using any of the following information:
1. A normal vector and a point in the plane.
2. Three points in the plane.
3. A point in the plane and a known parallel plane.
4. A point in the plane and two non-parallel vectors in some parallel plane.
5. Three non-parallel lines that are contained in the plane.
Various combinations of information may be used to find a plane. The basic information needed to define a plane is a point in the plane and a normal vector.
IDEA: Given two vectors, a and b, in a plane, then the vector n = a × b is
normal to the plane.
If two non-parallel vectors are known that are contained within a plane then it is
a simple process to find a vector normal to the plane. For example, the vector
c = a × b is orthogonal to both a and b. The only way that this can happen if a
and b are not parallel is if the vector is normal to the plane that contains both a
and b. So c = a × b is a normal vector to the plane containing a and b.
IDEA: To find a point in the plane ax + by + cz + d = 0, specify values for x and
y and then solve the equation of the plane for z.
Once a normal vector is obtained using the cross product, all that remains is to
find a point in the plane. Usually this information has to either be given or enough
must be known so that a point in the plane can be found. For example, if a plane
is said to contain the point where two lines intersect then this point of intersection
must be found.
IDEA: Distance problems involving planes can usually be reduced to the problem of finding the distance between a point and a plane.
IDEA: Distance between a point and a plane can be found using the magnitude of a projection onto a normal vector.
Various questions involving distances between points, lines and planes can be
answered using the relationship of vectors and planes. The key is to try and phrase
each problem in terms of finding the distance between a point P(x0 , y0 , z0 ) and the
plane. First form a vector from a point in the plane to the point P. Then find the
projection of this vector onto the normal vector of the plane. The distance to be
found is the magnitude of this resulting projection onto the normal vector.
distance between point and a plane
240
IDEA: Finding the angle between two planes is equivalent to finding either the
angle θ between the two normal vectors n1 and n2 or its supplement, π − θ
(whichever is the acute angle).
angle between two planes
The angle between two planes is defined to be an acute angle. This can be found
using the dot product of the two normal vectors. For example, the angle between
the two normal vectors n1 and n2 can be found by
n1 ·n2
−1
.
θ = cos
kn1 k kn2 k
CAUTION: The angle between the two normal vectors may be obtuse. In this
case, the angle between the planes is the supplement of the angle between
the two vectors.
However, using the inverse cosine function results in an angle θ , that can be either
acute, 0 ≤ θ ≤ π /2, or obtuse, π /2 ≤ θ ≤ π . If θ is obtuse then the angle between
the two planes will be π − θ . This can also be accomplished by using the absolute
value of the dot product (i.e., |n1 ·n2 |) in the above formula for θ .
Checklist of Key Ideas:
using dot products to define a plane
point-normal form of equation of a plane
planes parallel to coordinate planes
intersecting planes
general form of equation of a plane
plane determined by two vectors
plane determined by three points
distance between a point and a plane
distance between two parallel planes
distance between two skew lines
11.7 Quadric Surfaces
PURPOSE: To define quadric surfaces in three dimensions to
describe how these surfaces may be graphed and visualized.
Quadratic equations in the xy plane such as
Ax2 + Bxy +Cy2 + Dx + Ey + F = 0
241
represents a cylinder in 3-space (see §12.1). The extension of quadratic cylinders
where all three coordinates appear in the equation are quadric surfaces which
have the general equation of
quadratic cylinders
quadric surfaces
Ax2 + By2 +Cz2 + Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0.
Graphs of quadric surfaces are best determined using traces of the surface in the
three coordinate planes. A trace is simply a curve in a given plane. For example,
to find the trace of a function in the xy-plane, then z should be set to zero.
trace of a surface
IDEA: The six quadric surfaces can be distinguished by the types of traces
that they make in the coordinate planes.
Any quadric surface may be determined by what types of traces occur in the coordinate planes. For example, an ellipsoid has a trace of an ellipse in each of the
coordinate planes. Six common types of paraboloids are given in the book. Pay
careful attention to Table 11.7.1 in the book which describes each of the types of
quadric surfaces and the types of traces that they make.
IDEA: The quadric surfaces of the form Ax2 + By2 + Cz2 = d can be distinguished by the values of A, B, C and d .
A quadric surface that is centered about the origin and oriented along the coordinate axes may have the form Ax2 + By2 +Cz2 = d where d = 1 or d = 0.
type
A, B,C
d
all positive
equals one
hyperboloid (one sheet)
one is negative
equals one
hyperboloid (two sheets)
two are negative
equals one
elliptic cone
two are negative
equals zero
ellipsoid
A quadric surface with one linear term and two quadratic terms such as z + Ax2 + By2 = 0
will be some type of paraboloid. If both A and B are negative then this is an elliptic
paraboloid. If A and B have opposite signs then this is a hyperbolic paraboloid.
Checklist of Key Ideas:
mesh lines, trace of a surface
quadric surfaces or quadrics
second-degree equation in x, y and z
ellipsoid
elliptic cone
elliptic paraboloid
hyperboloid of one or two sheets
hyperbolic paraboloid
identifying quadrics using coefficients
reflections and translations of surfaces
ellipsoid
hyperboloids (one or two sheets)
elliptic cone
elliptic paraboloid
hyperbolic paraboloid
242
11.8 Cylindrical and Spherical Coordinates
PURPOSE: To introduce cylindrical and spherical coordinate
systems in three dimensions as alternatives to rectangular coordinates and to describe how to convert equations of surfaces
between the three coordinate systems.
Previously in 2-space, polar coordinates were given as an alternative to rectangular
coordinates (see §10.1). There are two common ways to extend the idea of polar
coordinates to 3-space. One is called cylindrical coordinates and the second is
called spherical coordinates.
IDEA: Cylindrical coordinates combine polar coordinates in the xy-plane with
the z-coordinate from rectangular coordinates in 3-space.
cylindrical coordinates
Cylindrical coordinates essentially use polar coordinates to represent the x and
y coordinates and then use the same value of z as rectangular coordinates to represent the coordinate of a point in the z direction. So a cylindrical point will have
the coordinates of (r, θ , z). In polar coordinates, the value of r was the distance
from the origin out to any point. In cylindrical coordinates, the value of r will be
the distance of the point from the z-axis.
IDEA: Spherical coordinates are based upon a point being on a sphere of
some radius ρ . Two angles are given to determine where the point is located
on the sphere.
spherical coordinates
Spherical coordinates are based upon the idea that any point can be thought of
as being located on a sphere of some radius ρ . That is, the distance, ρ , from the
origin to any point can be found. Then two angles are given to determine where
the point is located on the sphere of radius ρ . Consider a line that is drawn from
the origin to the point. One angle, θ , is the angle that the line is rotated from the
positive x-axis. The other angle, φ , describes the angle made with the line and the
positive z-axis.
CAUTION: The variable r in cylindrical coordinates and ρ in spherical coordinates do not represent the same distance.
converting between coordinate systems
Converting between the three coordinate systems involves using trigonometric
relationships of two right triangles. Any point with coordinates in 3-space can
be written with the rectangular coordinates (x, y, z). The first right triangle that is
made in the xy-plane by drawing a line from the origin to the point (x, y, 0) and a
line from this same point perpendicular to the x-axis. The second right triangle is
formed by the z-axis, the line from the origin to the point (x, y, z), and the line from
(x, y, z) to (0, 0, z). Both of these right triangles can be used to relate rectangular,
cylindrical and spherical coordinates. When considering the formulas given in the
book, it may be helpful to also be using these triangles.
IDEA: Here are some helpful reminders when converting between coordinate
systems.
Let rectangular coordinates use (x, y, z), cylindrical coordinates use (r, θ , z) and
spherical coordinates use (ρ , θ , φ ). Then
243
1. z is the same in rectangular and cylindrical coordinates.
2. θ is the same in cylindrical and spherical coordinates.
3. r2 = x2 + y2 .
4. ρ 2 = x2 + y2 + z2 .
5. r = ρ sin φ
6. x = r cos θ , y = r sin θ , and z = ρ cos φ
Any other information may be determined from the above.
Checklist of Key Ideas:
cylindrical coordinates
spherical coordinates
surfaces determined by constant coordinates
converting between coordinate systems
surfaces in cylindrical and spherical coordinates
spherical coordinates in navigation
244
Chapter 11 Sample Tests
Section 11.1
(a) parallel to the x−axis
(b) parallel to the y−axis
1. Answer true or false. A box has a corner at the origin and
corners at (5, 0, 0), (0, 2, 0), and (0, 2, 1). If three of the
edges of the box lie on the axes, the point (5, 2, 1) is a corner
point of the box.
2. Answer true or false. (5, 7, 1), (−1, 11, 3), and (1, 5, 7)
are vertices of an equilateral triangle.
3. Find the distance from (1, 3, 2) to the xy−plane.
(c) parallel to the z−axis
(d) parallel to the line y = x
10. The equation for a circular cylinder with radius 5 oriented
symmetrically about the z−axis is
(a) x2 + y2 = 5.
(b) x2 + y2 = 25.
(a) 1
(c) z2 = 5.
(b) 2
(d) z2 = 25.
(c) 3
√
(d) 14
4. Find the distance from (1, −2, −3) to the origin.
11. Answer true or false. x2 + 2x + y2 + 2y + z2 + 2z = 16 describes a sphere of radius 4.
12. x2 + y2 + z2 = −1 graphs as
(a) a sphere.
(a) 1
(b) a point.
(b) 2
(c) nothing; there is no such graph.
(c) 3
√
(d) 14
(d) a cylinder.
5. The surface described by x2 + y2 + z2 = 15 is a(n)
(a) sphere.
13. Answer true or false. x2 + 7x + y2 + y + z2 + z = 1 describes
a sphere centered at the origin.
14. Find the distance the surface x2 +y2 +z2 = 1 is from the point
(0, 0, −3).
(b) cylinder.
(a) 1
(c) cone.
(b) 2
√
(c) 2
(d) ellipsoid.
6. The spherical surface (x − 2)2 + (y − 3)2 + (z + 4)2 = 5 is
centered at
(a) (4, 9, −16).
(b) (−4, −9, 16).
(c) (2, 3, −4).
(d) (−2, −3, 4).
7. Answer true or false. The sphere x2 + (y − 2)2 + z2 = 16 has
a radius of 4.
8. The graph of x2 + z2 = 5 is an infinitely long circular cylinder
whose central axis is the
(a) x−axis.
(b) y−axis.
(c) z−axis.
(d) line x = y.
9. z = sin y describes a surface. In what direction would it be
possible to traverse the surface in a straight line?
(d) 0
15. Answer true or false. (x − 3)2 + (y − 6)2 + (z + 2)2 = 25 represents a sphere centered at (−3, −6, 2).
Section 11.2
1. The vector with initial point P1 (1, 2) and terminal point
P2 (3, 10) is
(a) h2, 8i.
(b) h−2, −8i.
(c) h6, 10i.
(d) h−6, −10i.
2. The vector with initial point P1 (0, 1, 2) and terminal point
P2 (2, 3, 1) is
(a) h4, 6, 5i.
(b) h−4, −6, −5i.
(c) h2, 2, −1i.
245
(d) h−2, −2, 1i.
3. Find the terminal point of v = i + 4j + k if the initial point is
(0, 0, 1).
(a) (1, 2, 0)
(b) (1, 3, 1)
(c) (1, 4, 2)
(d) (1, 1, 1)
4. Let v = h3, −2i. Find the norm of v.
√
(a) − 5
√
(b) 5
√
(c) 13
√
(d) − 13
5. √
Answer true or false. Let u = 3i + 4j + 2k. The norm of u is
29.
6. Answer true or false. Let v = 5i − 5j. The norm of v is 0.
7. Add u = 2i + 2j + 3k to v = i + 2j + 2k.
(a) 3i + 4j + 5k
(b) i + 2j + 3k
(c) 2i + 3j + 4k
12. Answer true or false. If kvk = 16 and φ , the angle
the vector
makes
with the positive x−axis, is 45◦ , then
√
√
v = h 2/2, 2/2i.
13. Answer true or false. Two forces, one 60 N and the other 80
N, act at right angles. The resultant force has a magnitude of
100 N.
14. A particle is said to be in static equilibrium if the resultant
of all forces applied to it is zero. Find the force F that must
be applied to a particle to produce static equilibrium if there
are two forces, each of 20 N, applied so that one acts 60◦
above the positive x−axis and the other acts 60◦ below the
positive x−axis. Give the magnitude of the resultant acting
in the negative x direction.
(a) 20 N
(b) 40 N
√
(c) 40 2 N
√
(d) 80 2 N
15. Let u = h1, 1, 0i, v = h0, 1, 0i, and w = h0, 0, 2i. Find C1 ,
C2 , and C3 such that h6, 6, 4i = C1 u +C2 v +C3 w.
(a) (6, 0, 2)
(b) (6, 6, 4)
(c) (6, 6, 2)
(d) No such constraints exist.
(d) 5i + 6j + 7k
8. If u = 3i + 5j + k, 6u =
(a) 18i + 6j + 1k
(b) 8i + 10j + 6k
(c) 8i + 6j + k
(d) 18i + 30j + 6k
9. Let u = h2, 1i, v = h1, 3i, and w = h5, 1i. Find the vector x
that satisfies 2u + v − x = w + x.
(a) h0, 0i
(b) h0, −2i
(c) h5, 5i
(d) h0, 2i
10. Given that kvk = 2, find all values of k such that kkvk = 4.
(a) −2, 2
Section 11.3
1. Find the dot product h3, 1i · h5, 2i.
(a) 28
(b) 17
(c) 11
√
17
(d)
2. Find the dot product h0, 1, 1i · h2, −1, 1i.
(a) 0
(b) 2
(c) −2
(d) 4
3. Find the dot product h5, 3, 2i · h2, 7, 3i.
(b) 2
(a) 70
(c) −4, 4
(b) 48
(d) 4
11. Answer true or false. If kvk = 5 and φ , the angle
the vector makes
with the positive x−axis, is π /3, then
√
v = h5/2, 5 3/2i.
(c) 37
(d) 111
4. Answer true or false. 0 · v = 0.
5. Find the dot product u · v, where u = 4i + 3j and v = 2i − 2j.
246
(a) 8
(b) −8
(c) −2
(d) 2
6. Answer true or false. If u = 2i + 3j − k and v = 5i − 2j + 4k,
u · v = 0.
7. Find the angle between u = h4, 2i and v = h10, 4i.
(a) 4.21◦
(b) 4.76◦
(c) 5.12◦
(d) 8.13◦
8. Find the angle between u = 4i + 2k and v = 5i + 2k.
(a) 4.21◦
(b) 4.76◦
(c) 5.12◦
(d) 8.13◦
9. Let u = h4, 1i, v = h2, 8i, and w = h10, 3i. Find u · (w − 2v).
(a) −37
(b) 11
(c) −3
(d) −19
10. Answer true or false.
√
√ If u = i + j√+ 3k then the direction
11
11
3 11
cosines are cos α =
, cos β =
, and cos γ =
.
11
11
11
11. Answer true or false. u · u = 0.
12. Answer true or false. Let u = 3i + 5j + k. The direction
1
5
1
cosines are cos α = , cos β = , and cos γ = .
3
12
4
13. Answer true or false. u · v = 0 does not necessarily mean
u = 0.
14. Answer true or false. If v is a three-space vector that has di1
rection cosines α and β each equal to , then cos γ must be
3
0.
15. A box is pulled across a frictionless surface by applying a 5
N force. The force is applied by pulling on a rope at an angle
of 60◦ above the horizontal. If the box is moved by the force
a total of 60 m, how much work is done?
(a) 300 N·m
(b) 150 N·m
√
(c) 150 2 N·m
√
(d) 150 3 N·m
Section 11.4
1. Find (i + j + k) × −j.
(a) i − k
(b) −j
(c) −i − j
(d) j
2. If u = −2i − 3j − 4k and v = −i + 3j − 2k, u × v =
(a) 18i − 9k.
(b) 18i + 4j − 9k.
(c) 18i − 4j − 9k.
(d) 12i + 8j − 3k.
3. If u = h0, 2, 1i and v = h−1, −3, 0i, u × v =
(a) h3, −1, 2i.
(b) h−3, 1, −2i.
(c) h−3, −1, −2i.
(d) h−3, −1, 2i.
4. If a = 4i + k and b = −2j, find a × b.
(a) 0
(b) 2i + 8k
(c) −2i + 8k
(d) 2i − 8k
5. A parallelogram has u = −2j − k and v = i + 3j as adjacent
sides. The area of the parallelogram is
√
(a) 14.
√
(b) 5.
√
14
(c)
.
2
√
5
.
(d)
2
6. If u = i + 2j + k, v = 3i − 2j + k, and w = −3i − 4j + k, find
u · (w × v).
(a) 28
(b) −28
(c) 4
(d) 0
7. If u = h1, 2, 3i, v = h−1, −7, −2i, and w = h4, 1, 2i, find
u · (w × v).
(a) 57
(b) −57
(c) −81
247
(d) 81
8. Answer true or false. If u · (v × w) = 8, u · (w × v) = −8.
9. If u = i+j+k and v = −4i−3j, the area of the parallelogram
that has u and v as adjacent sides is
3. Consider the following lines:
L1 : x = 5 − 2t, y = 2 + t, z = 3 + t;
L2 : x = −7 + 2t, y = 8 − t, z = 9 − t.
These lines intersect at the point
(a) 6.
(a) (5, 2, 3).
(b) 74.
√
(c) 26.
(b) (−1, 5, 6).
(d) 14.
(d) (0, 0, 0).
10. If u = −i − j + k and v = −4i + 3j, the area of the parallelogram that has u and v as adjacent sides is
(a) 6.
(b) 74.
√
(c) 74.
(d) 14.
11. Calculate the triple scalar product of u = 2i + 3j + 4k,
v = i − j + k, and w = 3i − j + 5k.
(a) 6
(b) 28
(c) −28
(d) −6
12. Answer true or false. The volume of the parallelepiped that
has u, v, and w as adjacent edges, where u = −2i − j + k,
v = i − 4j + 3k, and w = 2i + 2j + 2k, is 42.
13. Answer true or false.
Consider the three vectors
u = h1, 1, 1i, v = h1, 0, 2i, and w = h3, 4, 9i. If u, v, and
w all have the same initial point then they lie in the same
plane.
14. Answer true or false.
Consider the three vectors
u = h1, 0, 0i, v = h0, 3, 4i, and w = h1, 6, 8i. If u, v, and
w all have the same initial point then they lie in the same
plane.
15. Answer true or false. A force of 50 N acts in the positive
z−direction at a point (2, 1, 3). If the object is free to rotate
about√the point (0, 0, 0), the scalar moment about (0, 0, 0)
is 50 5 N·m.
(c) (2, 2, 4).
4. Find the parametric equations for the line whose vector is
given by hx, yi = h7, 0i + th2, −3i.
(a) x = 7, y = 2t − 3
(b) x = 7 + 2t, y = 3
(c) x = 7 + 2t, y = −3t
2t
(d) x = , y = 0
7
5. Find the parametric equations for the line whose vector is
given by hx, y, zi = h−2, 1, 3i + th3, 2, 5i.
(a) x = −2 + 3t, y = 1 + 2t, z = 3 + 5t
(b) x = 2 + 3t, y = −1 + 2t, z = −3 + 5t
(c) x = −2 − 3t, y = 1 − 2t, z = 3 − 5t
(d) x = 2 − 3t, y = −1 − 2t, z = −3 − 5t
6. Express x = 5 + 3t, y = −3 + 4t in bracket notation.
(a) hx, yi = h5, 3i + th3, 4i
(b) hx, yi = th7, 1i
(c) hx, yi = th−3, 4i
(d) hx, yi = h5, −3i + th3, 4i
7. The lines given by x = −3 + 5t, y = 6 + 3t, z = 8 − 2t and
x = −5 + 5t, y = 2 + 3t, z = 7 − 2t are
(a) intersecting at one single point.
(b) skew.
(c) parallel.
(d) perpendicular.
8. The lines given by x = −2t, y = 9 + 2t, z = 5 − 6t and
x = 2t, y = 9 + 10t, z = 5 − 14t are
(a) intersecting at one single point.
(b) parallel.
Section 11.5
(c) skew.
(d) perpendicular.
1. Answer true or false. The parametric equations for the line
joining P1 (3, 5) and P2 (1, 1) are x = 3 − 2t, y = 5 − 4t.
2. Answer true or false. The parametric equations of the line
passing through (−2, 1, 6) and parallel to v = h2, 3, 5i are
x = −2 + 2t, y = 1 + 3t, z = 6 + 5t.
9. The lines x = −2t, y = −2t, z = −2t and x = 4t, y = 4t, z = 4t
are
(a) parallel.
(b) perpendicular.
248
(c) the same line.
(d) skew.
10. The lines x = 5 − t, y = 1 + 2t and x = 12 + t, y = −3 − 2t
are
(a) parallel.
(b) skew.
(c) the same line.
(d) perpendicular.
11. Where does the line x = 4−2t, y = 6+3t, z = 4−2t intersect
the yz−plane?
(a) (0, 12, 0)
Section 11.6
1. The equation of the plane that passes through P(8, 1, 3) and
has n = h1, 5, −2i as a normal vector is
(a) (x + 8) + 5(y + 1) − 2(z + 3) = 0.
(b) (x − 8) + 5(y − 1) − 2(z − 3) = 0.
(c) (x + 8) + (5y + 1) − (2z + 3) = 0.
(d) (x − 8) + (5y − 1) − (2z − 3) = 0.
2. The equation of the plane that passes through P(−5, −3, −1)
and has n = h3, 1, 2i as a normal vector is
(a) 3(x − 5) + (y − 3) + 2(z − 1) = 0.
(b) (3x − 5) + (y − 3) + (2z − 1) = 0.
(b) (2, 9, 2)
(c) (3x + 5) + (y + 3) + (2z + 1) = 0.
(c) (4, 6, 0)
(d) 3(x + 5) + (y + 3) + 2(z + 1) = 0.
(d) (4, 6, 4)
12. Where does the line x = 6−6t, y = 5+2t, z = 2−8t intersect
the yz−plane?
(a) (21, 0, 22)
(b) (0, 7, −6)
(c) (6, 5, 2)
(d) (−3, 1, −4)
13. Where does the line x = 5 − 4t, y = 7 + 3t, z = 2 + t intersect the plane parallel to the xy−plane that includes the point
(0, 0, 1)?
(a) (9, 4, 1)
(b) (6, 8, 3)
(c) (5, 7, 2)
(d) (−3, 4, 2)
14. Where does the line x = 6 − 6t, y = 3 − 3t intersect with the
line x = 0, y = 1 − t?
(a) (0, 0)
(b) (4, 11)
(c) (3, 14)
(d) An intersection point does not exist.
15. How far are the lines hx, y, zi = th1, 2, 4i and
hx, y, zi = h0, 3, 4i + th1, 2, 4i apart?
(a) 0
(b) 7
(c) 5
(d) 25
3. Find an equation of the plane that passes through P1 (2, 7, 1),
P2 (−1, −1, −3), and P3 (5, 2, 7).
(a) 68(x − 2) − 6(y − 7) + 39(z − 1) = 0
(b) −68(x − 2) + 6(y − 7) + 39(z − 1) = 0
(c) −68(x + 2) − 6(y + 7) + 39(z + 1) = 0
(d) 68(x + 2) + 6(y + 7) − 39(z + 1) = 0
4. Answer true or false.
The planes x − 2y + z = 5 and
4x − 8y + 4z = 5 are parallel.
5. Answer true or false.
The planes x − y + 3z = 6 and
−4x + 4y + 3z = 7 are parallel.
6. Answer true or false.
The planes x + 2y + z = 5 and
7
7
7
x + y + z = 1 are parallel.
4
2
4
7. Answer true or false. The line x = 9 + t, y = 7 − t, z = 3 − 3t
is parallel to the plane x − 2y + z = 5.
8. Answer true or false. The line x = 6 − t, y = 1 + 3t, z = 9 + 5t
is parallel to the plane x + y + z = 8.
9. Find the distance between the point (1, 2, 2) and the plane
2x + y + 2z + 1 = 0.
(a) −3
(b) −9
(c) 3
(d) 9
10. Find the distance between the point (0, 3, 4) and the plane
2x + 3y + 6z + 10 = 0.
(a) 7
1
(b)
7
43
(c)
7
249
31
7
11. Determine whether the planes 2x + 4y − 2z = 4 and
5x + 10y − 5z = 2 are parallel, perpendicular, or neither.
(d)
(c) Hyperboloid of two sheets
(d) Elliptic cone
4. Identify the quadratic surface defined by x2 + 3y2 + 8z2 = 1.
(a) parallel
(a) Ellipsoid
(b) perpendicular
(b) Hyperboloid of one sheet
(c) neither
(c) Hyperboloid of two sheets
12. Determine whether the planes 3x + 3y − 3z = 2
x + y − 2z = 3 are parallel, perpendicular, or neither.
and
(d) Elliptic cone
5. Identify the quadratic surface defined by z2 − 2x2 − 5y2 = 0.
(a) parallel
(a) Ellipsoid
(b) perpendicular
(b) Hyperboloid of one sheet
(c) neither
(c) Elliptic cone
13. Find the acute angle of intersection of 2x + 3y − z = 5 and
x + 3y + 4z = 2. (Round answer to nearest degree.)
(a) 60◦
(b) 63◦
(c) 68◦
(d) 71◦
14. Answer true or false. The equation of the plane passing
through the point (1, 2, 7) and perpendicular to n = h5, 2, 8i
is h5, 2, 8i · hx − 1, y − 2, z − 7i = 0.
15. Answer true or false. The equation of the plane passing
through the point (5, 2, 3) and perpendicular to n = h3, 3, 4i
is h3, 3, 4i · hx + 5, y + 2, z + 3i = 0.
(d) Elliptic paraboloid
6. Identify the trace of the surface 2x2 + 5y2 − z2 = 5 where
x = 1.
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
7. Identify the trace of the surface 3x2 + 5y2 + 5z2 = 10 where
x = 0.
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
Section 11.7
y2 z2
1. Identify the quadratic surface defined by x =
+ .
3
7
(a) Ellipsoid
(b) Elliptic cone
(c) Elliptic paraboloid
(d) Hyperbolic paraboloid
2. Identify the quadratic surface defined by 4x2 + 4y2 − z2 = 1.
(a) Sphere
(b) Ellipsoid
(c) Hyperboloid of one sheet
(d) Hyperboloid of two sheets
3. Identify the quadratic surface defined by z2 − 2x2 − 5y2 = 0.
(a) Ellipsoid
(b) Hyperboloid of one sheet
8. Identify the trace of the surface z = x2 − 3y2 where x = 5.
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
9. Identify the trace of the surface z = 3x2 + 2y2 where y = 2.
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
10. Identify the trace of the surface y = x2 − 2z2 where y = 7.
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
11. Identify the trace of the surface y = 4x2 − z2 where z = 1.
250
(a) Circle
(c) (81, 1.107, 0.730)
(b) Ellipse
(d) (9, 1.107, 0.730)
(c) Parabola
3. Convert (5, π /2, π /2) from spherical coordinates to rectangular coordinates.
(d) Hyperbola
12. Identify the trace of the surface 4x2 + 9y2 − 7z2 = 1 where
z = 0.
(a) (0, 0, 5)
(b) (5, 0, 0)
(a) Circle
(c) (0, 5, 0)
(b) Ellipse
(d) (5, 5, 5)
(c) Parabola
(d) Hyperbola
13. Identify the trace of the surface 2x2 +3y−z2 = 5 where x = 1.
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
14. Identify the trace of the surface
z = 0.
6x2 + 6y2 + 3z2
= 25 where
(a) Circle
(b) Ellipse
(c) Parabola
(d) Hyperbola
15. Identify the trace of the surface 4x2 − 4y2 − 4z2 = 0 where
x = 2.
(a) Circle
4. Convert (4, π /4, π /6) from spherical coordinates to rectangular coordinates.
√ √
√
(a) ( 2, 2, 2 3)
√
√
√
(b) (4 2, 4 2, 8 3)
√
√
√
(c) ( 10, 10, 2 15)
√
√
√
(d) ( 10, 10, 2 10)
5. Answer true or false. There is no way to convert from spherical coordinates directly to cylindrical coordinates.
6. Convert the equation ρ = 5 from spherical coordinates to
cylindrical coordinates.
(a) z2 = 25 − r2
(b) 5z2 = 1 − 5r2
(c) z2 = 5 − 5r2
(d) 5z2 = 1 − r2
7. Convert the equation ρ = 3 from spherical coordinates to
rectangular coordinates.
(b) Ellipse
(a) x2 + y2 + z2 = 9
(c) Parabola
(b) x2 + y2 + z2 = 3
(d) Hyperbola
(c) 9x2 + 9y2 + 9z2 = 1
(d) 9x2 + 9y2 + 9z2 = 9
Section 11.8
1. Convert (6, 8, 9) from rectangular coordinates to cylindrical
coordinates.
8. Convert the equation 9z = x2 + y2 from rectangular coordinates to cylindrical coordinates.
(a) 9z = r2
(b) z = 9r2
(a) (10, 0.927, 9)
(c) 3z = r
(b) (10, 0.644, 9)
(d) z = 3r
(c) (100, 0.927, 9)
(d) (100, 0.644, 9)
2. Convert (6, 3, 6) from rectangular coordinates to spherical
coordinates.
(a) (81, 0.464, 0.841)
(b) (9, 0.464, 0.841)
9. Answer true or false. (3, 0, 0) in rectangular coordinates and
(3, 0, 0) in cylindrical coordinates identify the same point.
10. Answer true or false. (0, 2, 0) in rectangular coordinates and
(0, 2, 0) in cylindrical coordinates identify the same point.
11. Answer true or false. The equation in rectangular coordinates, z = 5x2 + 5y2 , converts to the equation z = 5r2 in cylindrical coordinates.
251
12. Answer true or false. The equation z = 8ρ cos φ in spherical
8z
in rectangular
coordinates converts to cos φ = p
x2 + y2 + z2
coordinates.
13. Answer true or false. The equation z = 12 in cylindrical coordinates converts to z = 12 in rectangular coordinates.
p
14. Answer true or false. √
z = 6x2 + 6y2 in rectangular coordinates converts to z = 6r in cylindrical coordinates.
15. Answer true or false. The
pequation ρ = 3 in spherical coordinates converts to 3z = 3 x2 + y2 in rectangular coordinates.
7. Answer true or false. If kvk = 8 and φ , the angle√the vector
√
makes with the positive x−axis is 45◦ , then v = h4 2, 4 2i.
8. Find the dot product h3, 8i · h2, 5i.
(a) 18
(b) 46
(c) 31
(d) 77
9. Find the dot product h2, 4i · h3, 5i.
(a) 26
(b) 32
(c) 2
Chapter 11 Test
(d) 48
1. Find the distance from (5, 3, −6) to the xy−plane.
(a) 3
10. Let u = h5, 2i, v = h1, 3i, and w = h1, 2i.
Find (u · v) + (u · w).
(b) 5
(a) 26
(c) 6
√
(d) 2 15
(b) 22
(c) 20
2. The surface described by x2 + y2 + z2 = 10 is a(n)
(d) 48
11. Answer true or false. Let u = 5i + 3j + 4k. The direction
5
1
1
cosines are cos α = , cos β = , and cos γ = .
12
4
3
12. If u = 2i + 3j + 4k and v = i − 3j + 2k, u × v =
(a) sphere.
(b) cylinder.
(c) cone.
(d) ellipsoid.
(a) 18i − 9k.
3. Answer true or false.
√= 6 describes a sphere centered at (−3, 2, 3) with radius 6.
(b) 18i + 4j − 9k.
4. Find the norm of u = 2i + 4j + 3k.
(d) 12i + 8j − 3k.
(x + 3)2 + (y − 2)2 + (z − 3)2
(a) 29
√
29
(b)
(c) 18i − 4j − 9k.
13. If a = −2j and b = −4i − k, find a × b.
(a) 0
(c) 9
(b) 2i + 8k
(d) 3
(c) −2i − 8k
5. If u = 4i + 5j − 2k and v = i − j + 2k, u − v =
(a) 3i + 5j − 2k.
(b) 3i + 4j.
(c) 3i + 6j − 4k.
(d) 3i + 6j.
6. Let u = h1, 3i and v = h5, 9i. Find x that satisfies 2u + 2x =
4v − x.
(a) h6, 10i
(b) h12, 20i
(c) h18, 30i
(d) h3, 5i
(d) 2i − 8k
14. A parallelogram has u = 2j + k and v = i + 3j as adjacent
sides. The area of the parallelogram is
√
(a) 14.
√
(b) 5.
√
14
.
(c)
2
√
5
.
(d)
2
15. If u = h2, 3, 2i, v = h3, −2, 1i, and w = h3, 4, −1i, find
u · (v × w).
(a) 50
252
(b) 14
(c) 24
(d) 12
16. Answer true or false. The volume of the parallelepiped that
has u, v, and w as adjacent sides, where u = −i − j − k,
v = 2i + j − k, and w = i − 4j + 3k, is 21.
17. Answer true or false. The parametric equations to the line
passing through (−1, 3, 6) and parallel to v = h2, 4, 8i are
x = 2 − t, y = 4 + 3t, z = 8 + 6t.
18. Answer true or false. The parametric equations for the line
whose vector is given by hx, y, zi = h2, 1, 3i +th1, 4, 7i are
x = 1 + 2t, y = 4 + t, z = 7 + 3t.
19. The lines defined by hx, y, zi = h1, 4, 7i + th1, 1, 1i and
hx, y, zi = h3, 8, 1i + th1, 1, 1i are
(d) (x + 2) + (y − 7) + (z + 1) = 0
22. Answer true or false. The planes x + 3y + 2z = 6 and
−4x − 12y − 8z = 1 are parallel.
23. Identify the quadratic surface defined by x2 − 2y2 + 4z2 = 1.
(a) Sphere
(b) Ellipsoid
(c) Hyperboloid of one sheet
(d) Hyperboloid of two sheets
24. Identify the trace of the surface x2 − 2y2 + 3z2 = 1 where
y = 1.
(a) Circle
(b) Ellipse
(a) skew.
(c) Parabola
(b) perpendicular.
(d) Hyperbola
(c) parallel.
(d) the same line.
20. The equation for the plane that passes through P(2, 1, 4) and
has n = h4, 1, 5i as a normal vector is
(a) (4x − 2) + (y − 1) + (5z − 4) = 0.
(b) (4x + 2) + (y + 1) + (5z + 4) = 0.
(c) 4(x − 2) + (y − 1) + 5(z − 4) = 0.
(d) 4(x + 2) + (y + 2) + 5(z + 4) = 0.
21. Find the equation of the plane that passes through P1 (0, 0, 0),
P2 (−2, −1, −3), and P3 (−5, −2, −4).
25. Convert (4, 8, 8) from rectangular coordinates to spherical
coordinates.
(a) (144, 0.464, 0.841)
(b) (12, 1.107, 0.841)
(c) (144, 1.107, 0.730)
(d) (12, 0.464, 0.730)
p
26. Convert the equation x2 + y2 + z2 = 25 from rectangular
coordinates to spherical coordinates.
(a) ρ = 50
(a) 2x − 7y + z = 0
(b) ρ = 5
(c) −2x + 7y + z = 0
(d) ρ = 25
(b) (x − 2) + (y + 7) + (z − 1) = 0
(c) ρ = 10
253
Chapter 11: Answers to Sample Tests
Section 11.1
1. true
9. a
2. true
10. b
3. b
11. false
4. d
12. c
5. a
13. false
6. c
14. b
7. true
15. false
8. b
2. c
10. a
3. c
11. true
4. c
12. false
5. true
13. true
6. false
14. a
7. a
15. a
8. d
2. a
10. true
3. c
11. false
4. true
12. false
5. d
13. true
6. true
14. false
7. b
15. b
8. b
2. a
10. c
3. a
11. d
4. d
12. false
5. a
13. false
6. a
14. true
7. b
15. true
8. true
2. true
10. a
3. b
11. a
4. c
12. b
5. a
13. a
6. d
14. a
7. c
15. c
8. a
2. d
10. c
3. b
11. a
4. true
12. c
5. false
13. a
6. true
14. true
7. true
15. false
8. false
2. c
10. d
3. d
11. c
4. a
12. b
5. c
13. c
6. d
14. a
7. a
15. a
8. c
2. b
10. false
3. c
11. true
4. a
12. false
5. false
13. true
6. a
14. true
7. a
15. false
8. a
2. a
10. c
18. false
26. d
3. true
11. false
19. c
4. b
12. a
20. c
5. c
13. d
21. a
6. a
14. a
22. true
7. true
15. a
23. c
8. b
16. false
24. b
Section 11.2
1. a
9. d
Section 11.3
1. b
9. b
Section 11.4
1. a
9. c
Section 11.5
1. true
9. c
Section 11.6
1. b
9. c
Section 11.7
1. c
9. c
Section 11.8
1. a
9. true
Chapter 11 Test
1. c
9. a
17. false
25. b
254
Chapter 12: Vector-Valued Functions
Summary: Having defined vectors and their properties, now vector-valued
functions are considered. These are simply functions that have two or more outputs that are the components of a vector. As functions, the usual ideas of limits,
derivatives, and integration are applied to vector-valued functions. One natural
way to describe a vector-valued function is as a parameterization where each component is a function of a parameter such as time, t.
Parameterization and derivatives together lead to the ideas of unit tangent and unit
normal vectors and ways to describe the curvature of a given curve that can be
defined parametrically. The chapter finishes by discussing motion along a curve
and how this relates to Kepler’s Laws of Planetary Motion.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Identify the curve of a parametric equation (§12.1).
2. Represent a curve as a vector-valued function (§12.1).
3. Evaluate limits, derivatives, and integrals of vector-valued functions (§12.2).
4. Take derivatives of dot and cross products (§12.2).
5. Find an arc length parameterization of a curve (§12.3).
6. Use the chain rule to take the derivative of a vector-valued function (§12.3).
7. Find unit normal, unit tangent, and unit binormal vectors of a curve or
vector-valued function in 2-space or 3-space (§12.4).
8. Calculate the curvature of a parametric curve or vector-valued function at a
point (§12.5).
9. Relate position, velocity, and acceleration vectors of a particle traveling
along a curve (§12.6).
10. Find the normal and tangential components of acceleration (§12.6).
11. Understand the relationship between acceleration, velocity, and curvature
(§12.5, 12.6).
12. Understand and apply Kepler’s three laws of planetary motion (§12.7).
255
256
12.1 Introduction to Vector-Valued Functions
PURPOSE: To describe vector-valued functions which are themselves vectors and to introduce how vector-valued functions may
be defined parametrically.
scalar-valued function
vector-valued function
curves in 3-space
A function that has one input and only one output is called a scalar-valued function since the output is a scalar value. This section describes how curves can be
describe parametrically and how curves may represent functions that have vector
values or vector-valued functions.
In the last chapter, lines in 3-space were described using a parameterization for
each coordinate in terms of some parameter or variable such as t. Curves in 3space can also be parameterized in the same way. Each coordinate of a point on
the curve is determined by some parameter such as t.
x = f (t)
positive orientation
negative orientation
y = g(t)
z = h(t)
The orientation of a parametric curve describes in which direction the path of the
curve is being followed. The direction that is followed along the curve as the
parameter t increases is called the direction of positive orientation while if t
decreases the path would be along the direction of negative orientation. Often
the parameterizations for each coordinate are referred to as x(t), y(t), and z(t).
IDEA: The vector or parametric equation of a line is the simplest example of a
vector-valued function (see §11.5).
In the last chapter, the parametric equation of a line was introduced. This is the
simplest example of a vector-valued function. One parameter t determines each
of the coordinates, x, y, and z, for a point on the line. So the output for a single
input is a vector of values. For example, the vector-valued function representation
of a line in 3-space may be given by r(t) where
r(t) = r0 + tv = hx0 + v1t, y0 + v2t, z0 + v3ti
radius or position vector
for some given vector v = hv1 , v2 , v3 i. Here the vector r(t) is assumed to have
an initial point at the origin and the terminal point determined by the coordinate
functions. The vector-valued function r(t) is called the radius or position vector.
IDEA: A curve in 3-space is a representation of a vector-valued function.
As seen above, curves in 3-space may also be parameterized by defining a separate
function, x(t), y(t), and z(t), for each coordinate of a point on the curve. Each of
the coordinate functions depend upon some variable (or parameter). Then the
curve is a representation of a vector-valued function, r(t).
r(t) = hx(t), y(t), z(t)i
To find points on the curve, the coordinate or component functions are each evaluated at the same t-value.
257
IDEA: The curve defined by the intersection of two surfaces may be defined
parametrically by choosing one coordinate (either x, y or z) as the parameter
and then simultaneously solving the equations of the surfaces for the remaining variables in terms of the chosen parameter.
A curve that defines the intersection of two surfaces may often be described
parametrically. The two curves z = f (x, y) and z = g(x, y), may be parameterized by supposing that x = t is the parameter and then solving both equations
for y and z in terms of t. For example, suppose that one surface is the parabolic
cylinder x2 + y = 1 and the other surface is the plane z = x + y. Let x = t be the
chosen parameter. Then y and z need to be expressed in terms of t. From the
parabolic cylinder, y = 1 − t 2 . Then substituting this information into the plane
gives z = t + 1 − t 2 . So the parameterization of the curve would be as follows.
x=t
y = 1 − t2
z = 1 + t − t2
The point (1, 0, 1) is the point on this curve that corresponds to t = 1.
Checklist of Key Ideas:
parametric curves in 3-space
orientation
direction of increasing parameter
intersections of surfaces
vector-valued functions, components
radius vector or position vector
two-point vector form of a line (or line segment)
12.2 Calculus of Vector-Valued Functions
PURPOSE: To discuss limits, continuity, derivatives, and integrals for vector-valued functions and the meanings of these ideas
in two and three dimensions.
Earlier in this book (see Chapter 1), limits, derivatives, and integrals were defined
for scalar functions of one variable (scalar meaning one input gave only one output). The limits, derivatives, and integrals of vector-valued functions are found in
a similar fashion. In this section, the concepts of limits, derivatives, and integrals
are explained for vector-valued functions where the components are each functions of one variable (or parameter). The next chapter will take up a description of
these ideas if a vector-valued function is dependent upon more than one variable
or parameter (i.e., multi-variable functions).
intersection of two surfaces
258
IDEA: Limits, derivatives, and integrals for vector-valued functions of one
variable are calculated component-by-component where each component is
treated as a single variable, scalar function.
single-variable component
functions
If a vector-valued function is a function of just one variable then limits, derivatives
and integrals are defined in a similar fashion as for scalar-valued function. Since
the component functions of a vector-valued function are each scalar-valued
functions of one variable, then their limits, derivatives, and integrals may be
calculated in the usual manner.
IDEA: If r(t) is a vector-valued function then
1. lim r(t) is found by evaluating the corresponding limits of each of the
t→a
components.
2. r′ (t) is found by taking the derivative of each of the components.
Z
Z b
3.
r(t) dt or
r(t) dt are found by evaluating the integrals of each of
a
the components.
Each component of a vector-valued function is itself a scalar-valued function.
Limits, derivatives, and integrals of each component function may calculated in
the usual way. The result is a vector of values. To find the limit, derivative, or integral of a vector-valued function just means to find the corresponding quantities
for each of the components. The resulting vector is the desired limit, derivative,
or integral. For example, the derivative of r(t) is simply given by the following.
r′ (t) = hx′ (t), y′ (t), z′ (t)i
If r(t) = ht 2 , sint, e2t i then r′ (t) = h2t, cost, 2e2t i.
CAUTION: For limits, derivatives, and integrals to exist for a vector-valued
function, the corresponding items must exist for each component function.
vector-valued limits
A limit in the normal scalar-valued case will exist if the expression approaches a
single quantity for values arbitrarily close to the limiting value. In vector-valued
functions, the situation is slightly more complicated. For a limit to exist, for
example, each of the components must approach a limit. If even one of the com′
ponents
Z does not approach a limit then the limit will not exist. Similarly, for r (t)
and
r(t) dt to exist, the corresponding derivatives and integrals must exist for
each of the coefficient functions.
IDEA: The vector-valued function r(t) is continuous only if each of its components is continuous.
vector-valued continuity
Continuity and differentiability are ideas that extend from the ability to take limits
or to take derivatives. For functions with vector values, the function may only be
continuous if each of the component functions is continuous. This is because the
idea of continuity is related to a limit at a point being equal to the function value at
that point. For vector-valued functions, the limit being equal to the function value
must be true for each of the components.
259
IDEA: The vector-valued function r(t) is differentiable only if each of its components is differentiable.
Similarly, a vector-valued function may only be differentiable if each of the component functions is differentiable. For example, the function r(t) = ht 2 , et , |t|i is
not differentiable at t = 0. While the first two components of r(t) are differentiable
at t = 0, the third is not and so r(t) is not differentiable at t = 0.
IDEA: Taking derivatives of scalar multiples, dot products, cross products and
even scalar triple products may be found by first performing the appropriate
vector operations (i.e., scalar multiple, dot product, etc.) and then taking the
derivatives of each of the resulting scalar components.
IDEA: If a function is scalar-valued then its derivative should be scalar-valued.
If a function is vector-valued then its derivative should be vector-valued.
Once the idea of taking derivatives is understood to be a process done by differentiating each component, it is straightforward to describe the derivatives for scalar
multiplication, dot products, and cross products (or even scalar triple products).
In the case of dot products, the quantity r1 (t)·r2 (t) is a scalar quantity. So the
derivative of a dot product should also be a scalar value. One way to verify this is
to first calculate the dot product and then to take the derivative.
CAUTION: When taking derivatives using dot products, cross products, and
other vector operations that involve products of functions, some version of the
product rule will be used within each component.
Taking derivatives of scalar multiples, dot products, cross products, and scalar
triple products will all involve some use of the product rule. For example, in
taking the derivative of the cross product, this means that one should not simply
take the derivatives of all of the components separately, and then perform the cross
product. This is wrong. The vector operations should be performed first and then
the derivatives of the resulting components taken. Any components that contain a
product of functions will require the use of the product rule.
IDEA: The vector r′ (t0 ) is tangent to the curve described by
r = hx(t), y(t), z(t)i at t = t0 .
Checklist of Key Ideas:
limit of a function, limits of components
limits and continuity
continuous at a point
continuous on an interval
definition of a derivative
vector-valued derivative, r′ (t)
vector-valued differentiability
1. Perform vector operations.
2. Take derivatives
component-by-component.
260
geometric interpretation of the derivative
derivative of a function, derivatives of components
derivatives of constant vectors
derivatives of scalar multiples and vector sums
tangent line to the graph of a vector valued function
derivatives of dot and cross products
vector-valued functions with constant magnitude
definite integrals of vector-valued functions
antiderivatives of vector-valued functions
fundamental theorem of calculus for vector-valued functions
12.3 Change of Parameter; Arc Length
PURPOSE: To investigate different parameterizations of the
same curve and to describe how arc length of a curve may be
used as a parameter to define the curve.
smooth parameterization
continuously turning tangent vector
Curves that have smooth parameterizations are said to have a continuously
turning tangent vector. This means that for a curve r(t), the tangent vector r′ (t)
is never equal to the zero vector. In this section, only curves with smooth parameterizations are discussed. There are two main ideas introduced in this section. The
first is arc length and the second is a change of parameter.
arc length
Arc length is easy to express in terms of vectors. If a curve is parameterized by
the vector-valued function r(t) = hx(t), y(t), z(t)i then the arc length may be found
by integrating the norm of r′ (t) over the desired interval.
IDEA: The arc length of the curve r(t) = hx(t), y(t), z(t)i is given by
L=
Z b
a
kr′ (t)k dt.
IDEA: A change of parameters for a curve requires that the chain rule be used
to calculate derivatives. This means that the chain rule should be used on
each component. If t = g(τ ) then
dr dt
dr
=
.
dτ
dt d τ
261
Curves that are described parametrically may often use different parameters to
describe the curve. These different parameters will trace the same curve although
they may be at different points on the curve or have a different orientation for
different parameter values. The tangent vector of a curve may also change depending upon the parameter that is used to describe a curve. For example, dr/dt
and dr/d τ may not give the same tangent vector if t = g(τ ). To relate the two
tangent vectors requires the use of the chain rule. This simple means applying
the chain rule to the vector components when taking derivatives.
change of parameters
chain rule
IDEA: Arc length of a curve may be used to parameterize a curve. The arc
length parameter, s, may be defined by
s=
dr du.
du Z t
t0
The arc length parameter of a curve may be found by using the arc length formula starting at some reference point t = t0 for the position function r(t). When
using the arc length to parameterize a curve, there are a few nice results that relate
to tangent vectors.
IDEA: If a curve r(t) is parameterized using arc length, s, then kdr/dsk = 1.
In other words, the length of the tangent vector with respect to arc length is
always equal to 1.
For example, if s is the arc length parameter of a curve then kdr/dtk = ds/dt.
In particular, if the curve is parameterized using arc length then kdr/dsk = 1.
This means that when arc length is used as a parameter, then tangent vectors are
always unit vectors. This will become more important in the next section when
unit tangent vectors are discussed.
Checklist of Key Ideas:
smooth parameterizations
smooth function
continuously turning tangent vector
vector version of calculating arc length
arc length as a parameter
change of parameters and the chain rule
smooth change of parameters
positive or negative changes of parameter
smooth parameterizations and tangent vectors
finding an arc length parameterization
arc length parameter
262
12.4 Unit Tangent, Normal, and Binormal Vectors
PURPOSE: To introduce the unit tangent, unit normal, and unit
binormal vectors of a curve and to describe their geometric relationship to the curve and each other.
In this section, three special unit vectors are discussed: unit tangent vector, unit
normal vector, and unit binormal vector. These vectors are useful in providing a
frame of reference for an object at any given point since these vectors will be mutually orthogonal with each other. Then the motion of an object along a curve may
be described by considering the motion of the object in the component directions
along these three unit vectors.
The unit tangent vector, T(t), is probably the easiest to identify since it is obtained
by normalizing the tangent vector r′ (t). This results in a unit tangent vector. The
unit normal vector, N(t), is obtained from the unit tangent vector by taking its
derivative and normalizing. The unit binormal vector, B(t), is found by taking the
cross product of the unit tangent vector and the unit normal vector.
IDEA: Given a smooth curve r(t) then the unit tangent, normal, and binormal
vectors are defined as follows.
inward unit normal vector
r′ (t)
kr′ (t)k
unit tangent:
T(t) =
unit normal:
N(t) =
unit binormal:
B(t) = T(t)×N(t)
T′ (t)
kT′ (t)k
The unit tangent vector is tangent to the curve at a point. The unit normal vector
is orthogonal to the unit tangent vector and it points in the direction that the curve
is turning at a point. In 2-space, the unit normal vector is called the inward unit
normal vector. The binormal vector is only defined in 3-space since it requires
the cross product of T and N. Since it is defined as the cross product, it is a vector
that is orthogonal to both T and N. In 3-space, the binormal vector can be useful
in describing how a curve will bend and how its path will change.
CAUTION: A unit binormal vector requires the cross product which is only
defined for vectors in 3-space.
IDEA: The unit tangent, normal, and binormal vectors are mutually orthogonal
unit vectors that form a right-handed coordinate system.
CAUTION: The unit normal vector is not found by normalizing r′′ (t).
−→
T′ (t)
r′′ (t)
=
6
kT′ (t)k kr′′ (t)k
The vectors r′ (t) and r′′ (t) are usually not perpendicular. Then even if they
are normalized (which only changes their length), they still may not be perpendicular.
263
Computing T and N for curves that are parameterized by arc length is a much
simpler process. For example, if r(s) is a position vector that is parameterized
by arc length, s, then kr′ (s)k = 1. So T(s) = r′ (s). It follows that N(s) =
r′′ (s)/kr′′ (s)k.
curves parameterized by arc length
IDEA: If a curve is parameterized by arc length, s, (see §12.3) then the unit
tangent and unit normal vectors are related by
T(s) = r′ (s)
and
N(s) = r′′ (s)/kr′′ (s)k.
Checklist of Key Ideas:
unit tangent vectors
unit normal vectors
inward unit normal vectors in 2-space
curves parameterized by arc length
binormal vector
right-handed coordinate system
TNB coordinate system
rectifying plane
osculating plane
normal plane
expressing B in terms of r(t)
expressing B in terms of r(s)
12.5 Curvature
PURPOSE: To define the curvature of a curve and to give various
methods of computing the curvature of a curve at a point.
Curvature describes how a curve is changing direction. The idea is that at any
point on a curve there is some circle that is tangent to the curve at that point and
the curve and the circle have the same curvature at that point. Then the radius
of curvature is the radius of the tangent circle at that point. A smaller radius of
curvature means that the curve is bending around a smaller circle meaning that the
curve has to change direction more rapidly.
IDEA: Curvature describes how a curve is changing at a point. It is a measure
of how the unit tangent vector is changing at that point.
curvature, κ
radius of curvature, ρ = 1/κ
264
More generically, the curvature is defined by the magnitude of the change of the
unit tangent vector with respect to arc length. There are many equivalent formulas
for curvature and this section spends time describing many of them. Depending
upon what is given in a problem, one formula may be more useful than another.
Here is a summary of some curvature formulas.
κ (s) = kdT/dsk = kr′′ (s)k
N(s) =
r′′ (s)
κ (s)
κ (t) = kT′ (t)k/kr′ (t)k
κ (t) =
B(t) =
kr′ (t)×r′′ (t)k kv×ak
=
kr′ (t)k3
kvk3
r′ (t)×r′′ (t)
kr′ (t)×r′′ (t)k
Checklist of Key Ideas:
definition of curvature
various formulas for curvature
curvature in 2-space and curvature in 3-space
circle of curvature or osculating circle
concave and convex sides of a curve C in 2-space
angle between the positive x-axis and T
various formulas for T, N, B
12.6 Motion Along a Curve
PURPOSE: To investigate motion along a curve in terms of
speed, velocity and acceleration and to relate these to the unit
normal and unit tangent vectors.
So far in this chapter, vector-valued functions have been referred to as a radial
function r(t). This also can be thought of as a position function or displacement
function where the vector represents how an object is displaced from the origin at
a given time. Now in this section, r(t) is associated with the position of an object
that is moving along some curve. Then r′ (t) and r′′ (t) correspond to the velocity
and acceleration of the object at some point on the curve.
265
IDEA: If the function r(t) describes the position of some object along a curve,
C, then the velocity and acceleration of the object along the curve are given
by the derivatives of r(t).
velocity vector:
acceleration vector:
v(t) = r′ (t)
a(t) = r′′ (t) = v′ (t)
Notice that velocity and acceleration are vector quantities meaning that there are
velocity and acceleration components in all three coordinate directions (in 3space) that need to be accounted for. Sometimes it is necessary to know the speed
of an object. The speed is defined to be the norm or magnitude of the velocity
vector. So speed only describes how fast an object is moving and not the direction
in which it is moving.
IDEA: Velocity and acceleration are both vector quantities. Speed is defined
as kv(t)k which is a scalar quantity. Speed does not describe direction.
IDEA: Speed ds/dt is the same as the instantaneous rate of change of the
arc length s with respect to the parameter t .
Speed is important because it is related to the distance that an object travels. The
arc length formula that was given in terms of vectors in §12.3 says that
L=
Z b
a
kv(t)k dt
But this is just the integral of the speed over the time interval.
IDEA: Distance traveled over a period of time may be found by integrating the
speed of the object.
L=
Z b
a
s(t) dt =
Z b
a
kv(t)k dt.
Notice that distance traveled and displacement are very different. For example,
given the velocity vector,
CAUTION: Distance traveled is a scalar quantity. Displacement is a vector
quantity.
IDEA: The acceleration may be decomposed into the orthogonal components
along the unit tangent and unit normal vectors. These are called the tangential
and normal components of the acceleration.
The normal and tangential components of acceleration find the components of the
acceleration in the directions of the unit normal and unit tangent vectors. Also, the
velocity can be thought of as the vector in the same direction as the unit tangent
vector with the magnitude of speed. These two ideas together serve to relate many
concepts including arc length, speed, T, N, and curvature.
266
IDEA: Velocity is a scalar multiple of the unit tangent vector. It can be written
as v =
ds
T.
dt
IDEA: The acceleration vector can be decomposed into orthogonal components along the unit tangent and unit normal vectors.
a = aT T + aN N
where
aT =
d2s v · a
=
dt 2
kvk
and
aN = κ
ds
dt
2
=
kv × ak
.
kvk
Using the ideas of the dot product and projections from Chapter 11, the acceleration vector can be decomposed into orthogonal components along the unit tangent
vector and along the unit normal vector. These components relate to how the velocity is changing and how the curve is turning. For example, the tangential component is related to the how the speed is changing. If the particle has a constant
speed, then all of the acceleration will be in the direction of the normal component. In other words, accelerating in the normal direction will cause the particle
to turn although its speed will not change.
IDEA: If speed is constant then aT = 0. If curvature is zero (i.e., the path is
not curving) then aN = 0.
Similarly, if the particle is moving along a straight line and so it is not turning,
all of its acceleration will be along the tangential component. Consequently, the
curvature of the motion will be zero since κ is a multiple of aN .
A model for projectile motion can be found by integrating a given acceleration
vector assuming that the mass m is constant and that air resistance is not a factor.
In 2-space if a = −gj is the acceleration vector, then parametric equations of projectile motion can be shown to be r(t) = (−gt 2 /2 + s0 )j + tv0 by integrating. In
other words, gravity works in the j direction and position is linear in the other two
directions.
Checklist of Key Ideas:
position function or trajectory, r(t)
velocity and the unit tangent vector
acceleration
speed
displacement, distance traveled, and arc length
normal and tangential components of acceleration
scalar and vector components of acceleration
267
formulas for aT and aN
motion with constant speed
vector version of Newton’s 2nd law
parametric equations of projectile motion
12.7 Kepler’s Laws of Planetary Motion
PURPOSE: To use the ideas of vector-valued functions and functions defined using parameters as a tool to understanding Kepler’s Laws of Planetary Motion.
There are three laws which Kepler proposed about planetary motion. The first law
is called the law of orbits. It says that planets have elliptical orbits with the sun
being a focus of the elliptical path of the orbit. The second is called the law of
areas. It says that the line from the sun to a planet sweeps out an area at a constant
rate. The third law is called the law of periods. It says that the orbit time squared
is equal to the cube of the semimajor axis of the orbit.
Newton’s law of universal gravitation can be used to find the position function
of an object that is subjected to a central force. In other words, what is the position
function of an object that is orbiting another object (such as a planet orbiting the
sun)? If two objects have masses of M and m then the force between them can be
expressed as
kFk =
GMm
r2
where G is the universal gravitational constant and r is distance between the objects.
IDEA: The position of an object orbiting an object of mass M is given by
a=−
GM
r.
r3
The sum of the forces equals mass times acceleration. Then the force that is
applied to an object is a vector. This force vector is given by the mass times the
acceleration vector which is equal to the sum of all the force vectors. In other
words, the sum of the forces in any coordinate direction is equal to the mass times
that component of the acceleration. It then follows that the position vector is
proportional to the acceleration vector.
IDEA: If the object of mass M is at a focus of an ellipse then r is the equation
of an ellipse in polar coordinates:
r=
k
1 + e cos θ
where r is the distance between the objects of mass M and m.
law of orbits
law of areas
law of periods
Newton’s law of universal gravitation
268
escape velocity
Assuming an appropriate coordinate system (see the text), it can be shown that
the distance r is given by the polar equation of a conic section. In particular,
for an object in a repeating orbit, this is an elliptical path. In other words, the
eccentricity, e, of the orbiting object (see §10.6 for eccentricity) is between 0 and
1. Escape velocity for the object occurs when e = 1.
IDEA: Escape velocity occurs when e = 1 for the orbital path (meaning that its
path becomes parabolic instead of elliptical) and the velocity is given by
vesc =
r
2GM
r0
where r0 is the distance between the objects at t = 0.
aphelion and perihelion
apogee and perigee
For objects orbiting the sun, the time of aphelion and perihelion refer to the time
that the object is farthest from and closest to the sun respectively. The farthest
distance is called the apogee and the closest distance is called the perigee. These
distances occur at the vertices of the elliptical orbit.
Checklist of Key Ideas:
Kepler’s three laws
Law of Orbits
Law of Areas
Law of Periods
Newton’s Law of Universal Gravitation
universal gravitational constant, G
escape speed
aphelion and perihelion
apogee and perigee
269
Chapter 12 Sample Tests
Section 12.1
(c) Parabola
(d) Circle
1. Find the domain of r(t) = (5 + cost)i − 2tj.
(a) 0 ≤ t < ∞
11. Describe the graph of r(t) = 4ti + 8 sintj + 8 costk.
(b) −∞ < t < ∞
(a) Straight line
(c) 0 ≤ t ≤ 2π
(d) −π ≤ t ≤ π
(b) Spiral
√
2. Find the domain of r(t) = t − 4i + t 2 j − 3tk.
(c) Parabola
(a) 4 ≤ t < ∞
(d) Circle
(b) 0 ≤ t < ∞
12. Describe the graph of r(t) = t 3 i + t 2 j + tk.
(c) 5 ≤ t < ∞
(d) −4 ≤ t < ∞
3. Find the domain of r(t) = ht 2 , t − 2,
(a) Cubic
√
t + 1i.
(b) Twisted cubic
(a) 0 ≤ t < ∞
(c) Spiral
(b) 1 ≤ t < ∞
(c) −1 ≤ t < ∞
(d) Parabola
(d) 0 − ∞ < t < ∞
4. Answer true or false. r(t) = t 2 i + t 3 j can be expressed as a
parametric equation by x = sint, y = cost.
5. Answer true or false. r(t) = costi − sintj can be expressed as
a parametric equation by x2 − y2 = t.
= t2,
6. Answer true or false. The parametric equation x
y=t
can be expressed by the single vector equation r(t) = t 3 i+t 3 j.
7. Answer true or false. The parametric equation that is defined
by x = sint, y = 8t, z = t can be expressed by the single vector equation r(t) = sinti + 8tj + tk.
8. Describe the graph of r(t) = 6ti + 2tj + tk.
13. As t increases, the graph of r(t) = h4 sint, cost, 3ti sketches
(a) clockwise and up.
(b) counter-clockwise and up.
(c) clockwise and down.
(d) counter-clockwise and down.
14. As t increases, the graph of r(t) = hcost, 5 sint, −3ti
sketches
(a) Twisted cubic
(a) clockwise and up.
(b) Straight line
(b) counter-clockwise and up.
(c) Spiral
(d) Parabola
9. Describe the graph of r(t) = 4i + 2 costj + 2 sintk.
(a) Straight line
(b) Spiral
(c) Parabola
(c) clockwise and down.
(d) counter-clockwise and down.
15. As t increases, the graph of r(t) = h2 cost, 9 sint, 4ti
sketches
(a) clockwise and up.
(d) Circle
10. Describe the graph of r(t) = −2i + sintj − costk.
(a) Straight line
(b) Spiral
(b) counter-clockwise and up.
(c) clockwise and down.
(d) counter-clockwise and down.
270
Section 12.2
(a) 3ti − 4tj + k
(b) 6ti − 4tj + k
1. If r(t) = (6 − 2t)i + (t 2 − 3)j, find r′ (t).
(a) t 2 i +
(c) 6ti + 4j
t3
j
3
(d) 6ti + 4tj + k
t3
(b) (6t − t )i +
−3 j
3
2
(c) −2i + 2tj
10.
(b) (2t 2 +C)i + (6t +C)j
2. If r(t) = 4ti − 3tj + costk, find r′ (t).
(c) (2t 2 +C1 )i + (6t +C2 )j
(a) 4i − 3j − sintk
(d) 2t 2 + 6t +C
(b) 4i − 3j + sintk
(d) 7 sint
11.
hcost, sinti dt =
(a) h−1, −1i
(a) i
(b) h−1, 1i
(b) −i
(c) h1, 1i
(c) −j
(d) h1, −1i
(d) j
(a) 15i + 6j + k
Z π /2
0
3. Find r′ (π /2) if r(t) = costi + sintj + k.
4. Find r′ (0) if r(t) = 7t 4 i + 2t 3 j + tk.
(4ti + 6j) dt =
(a) 2t 2 i + 6tj +C
(d) −3t
(c) −7 sint
Z
12.
Z 3
0
ht, t 2 , 2i dt =
(b) 0
(a) h9/2, 9, 6i
(c) k
(b) h9/2, 9, 2i
(d) 5i + 2j + k
(c) h9, 9, 6i
5. lim (4t 2 i + 2tj) =
t→2
(d) h3, 9, 2i
(b) 16i + 4j
13. Answer true or false. If r(t) = t 3 i + 2tj, the tangent line at
t0 = 4 is given by r(t) = 3t 2 i + 2j.
(c) not defined
14. If y′ (t) = 8ti + 3t 2 j, y(0) = i + 2j, find y(t).
(a) 20
(d) 2
6. lim h2 sint, 3 cost, ti =
t→π
(a) h0, −3, π i
(b) h0, −3, 0i
(c) −π
(d) π
7. Answer true or false. r(t) = sinti + 5 costj is continuous at
t = 0.
(a) (12t 2 + 1)i + (12t 3 + 2)j
(b) (4t 2 + 1)i + (t 3 + 2)j
(c) 4t 2 i + 3t 3 j
(d) 12t 2 i + 12t 3 j
15. If y′ (t) = 9t 2 i + 2tj, y(1) = i + j, find y(t).
(a) (3t 3 − 2)i + t 2 j
(b) (3t 3 + 2)i + t 2 j
8. Answer true or false. r(t) = 4 lnti + costj − 4 lntk is continuous at t = 0.
(c) (3t 3 + 1)i + (t 2 + 1)j
9. r(t) = t 3 i + 2t 2 j + tk. Find r′′ (t).
(d) (3t 3 + 4)i + t 2 j
271
Section 12.3
1. Answer true or false. r(t) = 3t 2 i + 2t 3 j + sintk is a smooth
function of the parameter t.
2. Answer true or false. r(t) = 4t 2 i − t 3 j + sin (2t)k is a smooth
function of the parameter t.
√
3. Answer true or false. r(t) = 5 ti + 4t 2 j + 8t 3 k is a smooth
function of the parameter t.
4. Find the arc length of the graph of r(t) = 2ti + 4j + 3k for
4 ≤ t ≤ 7.
(a) 6
(b) −6
(c) 21
(d) −21
5. Find the arc length of the graph of r(t) = − sinti − costj + 9k
for 0 ≤ t ≤ π .
(a) 2
(b) 2π
(c) π
(d) 0
6. Find the arc length of the graph of r(t) = 2et i + 2et j + et k for
0 ≤ t ≤ 1.
(a) 3e
(b) 5e − 5
(c) 3e − 3
√
√
(d) (1 + 2 2)e − 1 − 2 2
7. What is the arc length of the parametric curve that is defined
by x = 3et , y = 4et , z = 2 for 0 ≤ t ≤ 1?
(a) 3e
(b) 5e − 5
(c) 3e − 3
√
√
(d) (1 + 2 2)e − 1 − 2 2
8. What is the arc length of the parametric curve that is defined
by x = − sint, y = −7, z = − cost for 0 ≤ t ≤ π ?
(a) 2
(b) 2π
(c) π
(d) 0
9. Find the arc length parameterization of the line x = 6t + 2,
y = 4t − 1 that has the same orientation as the given line and
uses (2, − 1) as a reference point.
s
s
(a) x = √ , y = √
2 13
13
2s
3s
(b) x = √ , y = √
13
13
s
s
(c) x = √ + 2, y = √ − 1
2 13
13
3s
2s
(d) x = √ + 2, y = √ − 1
13
13
10. Find the arc length parameterization of the line
x = 2 cost + 3, y = 2 sint − 4 that has the same orientation
as the given curve and uses (5, − 4) as a reference point.
s
s
(a) x = 2 cos
+ 3, y = 2 sin
−4
2
2
(b) x = cos s + 3, y = sin s − 4
s
s 3
+ , y = 2 sin
−2
(c) x = 2 cos
2
2
2
3
(d) x = cos s + , y = sin s − 2
2
11. Answer true or false. If r = 2ti + (−4t + 3)j, the arc length
parameterization of the curve relative
√ to the reference point
(0, 3) involves the parameter t = 2 5s.
12. Answer true or false. If r = (2t + 2)i + (6t + 2)j + (4t − 1)k,
the arc length parameterization of the curve relative to the refs
erence point (2, 2, − 1) involves the parameter t = √ .
2 14
13. Answer true or false. If r = 2 sinti − 2 costj + tk, the arc
length parameterization of the curve relative to the reference
s
point (0, − 2, 0) involves the parameter t = √ .
5
14. Answer true or false. If r = (2t − 2)i + (t − 3)j + 4tk, the arc
length parameterization of the curve relative to the reference
s
point (−2, − 3, 0) involves the parameter t = √ .
21
15. Answer true or false. If r = (3t − 1)i + (4t − 1)j + (2t + 1)k,
the arc length parameterization of the curve relative to the
s
reference point (−1, −1, 1) involves the parameter t = √ .
3
Section 12.4
1. r(t) = 4t 2 i + 8tj. Find T(t) for t = 2.
2
1
(a) √ i + √ j
5
5
1
(b) √ k
5
1
(c) − √ k
5
(d) i
2. Answer true or false. r(t) = 3t 2 i + 6tj. N(t) for t = 2 is
0.49i − 0.87j.
3. r(t) = 7t 2 i + 14tj. Find B(t) for t = 2.
272
(a) k
(b) 2i + j
(c) −k
1
2
(d) √ i − √ j
5
5
4. r(t) = 4(t 2 + 2)i + 4et j + 4et k. Find T(t) for t = 0.
1
1
(a) 2i + √ j + √ k
2
2
1
1
(b) √ j + √ k
2
2
1
1
(c) 2i + √ j + √ k
6
6
1
1
(d) √ j + √ k
3
3
5. r(t) = 4(t 2 + 2)i + 4et j + 4et k. Find N(t) for t = 0.
1
3
1
(d) √ i + √ j + √ k
11
11
11
10. r(t) = 9et i + 9e2t j + 9e3t k. Find T(t) for t = 0.
3
1
3
√ i− √ j+ √ k
7 7
7 7
7 7
(b) 6i − 6j + 2k
2
3
1
(c) √ i + √ j + √ k
14
14
14
1
4
9
(d) √ i + √ j + √ k
7 2
7 2
7 2
(a)
11. Answer true or false. r(t) = 9et i + 9e2t j + 9e3t k. When t = 0,
4
9
1
N(t) = √ i + √ j + √ k.
7 2
7 2
7 2
12. Answer true or false. r(t) = 9et i + 9e2t j + 9e3t k. When t = 0,
3
1
3
B(t) = √ i − √ j + √ k.
7 7
7 7
7 7
13. r(t) = 8ti + 8t 2 j + 8t 3 k. Find T(t) for t = 0.
(a) 0
(a) i
(b) −2ii
1
(b) √ j
10
3
1
(c) √ j + √ k
10
10
1
(d) − √ k
14
(c) i
(d) 1.334i + 1.334j + 1.334k
6. r(t) = 4(t 2 + 2)i + 4et j + 4et k. Find B(t) for t = 0.
(a) 2i + 0.707j + 0.707k
(b) 0.707j − 0.707k
(c) 2i + 0.662j + 0.662k
(d) 0.662j + 0.662k
7. r(t) = 6(t 2 + t)i + 6t 2 j + 18t 2 k. Find T(t) for t = 0.
14. Answer true or false. r(t) = 8ti + 8t 2 j + 8t 3 k. When t = 0,
1
N(t) = √ j.
10
15. Answer true or false. r(t) = 8ti + 8t 2 j + 8t 3 k. When t = 0,
1
B(t) = − √ k.
2 35
(a) i
(b) j
(c) k
(d) 0
8. r(t) = 6(t 2 + t)i + 6t 2 j + 18t 2 k. Find N(t) for t = 0.
(a) 6i
√
√
10
3 10
(b)
j+
k
10
10
√
√
3 10
10
j+
k
(c) −
10
10
(d) i + j + 3k
9. r(t) = 6(t 2 + t)i + 6t 2 j + 18t 2 k. Find B(t) for t = 0.
(a) i + j + 3k
(b) 0.95j + 0.316k
(c) 0.95j + 0.316k
Section 12.5
1. Find the curvature κ (t) for r(t) = − sinti − costj.
(a) 1
(b) −1
(c) 0
(d) sin2 t − cos2 t
2. Find the curvature κ (t) for r(t) = − costi − sintj − 6k.
√
(a) 11
(b) 1
(c) sin2 t − cos2 t
(d) sin2 t
3. Find the curvature κ (t) for r(t) = 2et i + 2et j + 6k.
(a) 0
273
2et
(b) √
2
3
(c) √
2et
1
(d) √
2
(c) 3
(d)
10. If y = − cos x + 4, find the curvature at x =
4. Find the curvature κ (t) for r(t) = 6i + 2tj + 3t 2 k at t = 0.
3
√
40 10
3
(b)
2
4
(c)
3
3
(d) √
2 10
(a)
π
.
2
(a) 0
(b) 1
(c) −1
(d)
1
√
2 2
11. If y = −7 + sin x, find the curvature at x =
π
.
2
(a) 0
5. Answer true or false. √
If r(t) = (t 3 +2)i+(t 4 −5)j+(t 5 +1)k,
the curvature κ (t) is 36t 2 + 144t 4 + 400t 6 .
6. Answer true or false. If r(t) =
9
κ (t) at t = 1 is √ .
13 26
2
3
ti + t 3 j + t 4 k,
the curvature
7. If r(t) = (2t 2 − 1)i + (t − 3)j + (4t + 8)k, find the curvature
κ (t) at t = 1.
√
2 17
(a) √
33
√
4 17
√
(b)
33 33
34
(c) √
33
√
34
(d) √
33
s s 8. If r(s) = 2 cos
i + 8 + 2 cos
j + 3k, find κ (s) at
2
2
s = 0.
√
2
(a)
2
1
(b)
4
(c) 2
1
(d)
2
s s 9. If r(s) = 6i + 3 cos
j + 2 + 3 cos
k, find κ (s) at
3
3
s = 0.
1
(a)
9
√
2
(b)
3
(b) 1
(c) −1
(d)
1
√
2 2
12. If x = t 3 + 8, y = t 2 − 1, then κ (t) at t = 1 is
(a)
6
√ .
13 13
6
(b) √ .
13
(c) 0.
(d)
18
√ .
13 13
13. Answer true or false. The curve y = 4x3 has a maximum curvature at x = 4.
14. At what points does 12x2 + 75y2 = 300 have maximum curvature?
(a) (0, −4), (0, 4)
(b) (−4, 0), (4, 0)
(c) (−5, 0), (5, 0)
(d) (0, −5), (0, 5)
15. At what points does 16x2 + 100y2 = 400 have minimum curvature?
(a) (−5, 0), (5, 0)
(b) (0, −5), (0, 5)
(c) (0, −2), (0, 2)
(d) (−2, 0), (2, 0)
274
Section 12.6
1. r(t) = (4t 3 + 7)i + (2t − 3)j is the position vector of a particle
moving in a plane. Find the velocity.
(a) 12t 2 i + 2j
(b) 12i
(c) 24ti + 2j
(d) 24ti
2. r(t) = (4t 3 + 8)i + (2t − 1)j is the position vector of a particle
moving in a plane. Find the acceleration.
(a) 12t 2 i + 2j
(b) 12i
(c) 24ti + 2j
(d) 24ti
3. r(t) = (4t 3 + 5)i + (2t − 2)j is the position vector of a particle
moving in a plane. Find the speed at t = 1.
√
(a) 2 37
(b) 12
(c) 24
(d) 0
4. Find the velocity of a particle moving along the curve
r(t) = (t 3 − 5)i + (4t − 6)j − (t 2 − 1)k at t = 1.
(a) 3i + 4j − 2k
(b) 6i − 2j
(c) 3i + 4j + 2k
(d) 0
5. Find the acceleration of a particle moving along the curve
r(t) = (t 3 + 3t)i + (4t − 2)j − (t 2 + t − 2)k at t = 1.
(a) 3i + 4j − 2k
(b) 6i − 2k
(c) 3i + 4j + 2k
(d) 0
6. Find the speed of a particle moving along the curve
r(t) = (t 3 + 6)i + (4t − 6)j − (t 2 − 4)k at t = 1.
√
(a) 29
√
(b) 21
√
(c) 4 2
√
(d) 2 10
7. Answer true or false. If a(t) = 2 sinti +tj, the velocity vector
t2
is v(t) = −2 costi + j, if v(0) = −j.
2
8. Answer true or false. If a(t) = sinti + tj, the position vector
3
t
is r(t) = − sinti +
+ 1 j, if v(0) = i and r(0) = i.
3
9. If v = i and a = i − 3j, find aT .
(a) 1
(b) 2
1
(c)
2
(d) 8
10. If v = 4i and a = i − 3j, find aN .
(a) 6
(b) −6
3
(c)
4
(d) 3
11. If v = 2j and a = j − 3k, find κ .
(a) 6
(b) −6
3
(c)
4
(d) 3
12. r(t) = 2t 3 i − 4tj; 1 ≤ t ≤ 2. Find the displacement.
(a)
(b)
(c)
(d)
14i − 4j
18i − 8j
14i + 4j
18i + 8j
13. r(t) = 2t 3 i − 4tj; 1 ≤ t ≤ 2. Find the distance.
√
(a) 2 53
√
(b) 6 5
√
(c) 2 85
(d) 18
14. v(t) = 2i + 3j. Find T(t).
3
2
(a) √ i + √ j
13
13
2
3
(b) √ i + √ j
5
5
1
1
(c) √ i + √ j
13
13
1
1
(d) √ i + √ j
5
5
15. Find aN if kak = 4 and θ = π /6.
(a)
(b)
(c)
(d)
2
√
√
1
2
3
275
Section 12.7
(d) 2.70 × 1024 km
1. Answer true or false. According to Kepler’s second law, a
planet moves fastest at a point on its semiminor axis.
2. If an object orbits the sun with rmax = 120, 000, 000 miles
and rmin = 110, 000, 000 miles, the orbit has the following
eccentricity.
(a) 23
1
(b)
23
(c) 20
1
(d)
20
3. If an object orbits the sun with rmax = 610, 000, 000 miles
and rmin = 600, 000, 000 miles, the orbit has the following
eccentricity.
(a) 121
1
(b)
121
(c) 120
1
(d)
120
4. Answer true or false. Object 1 has rmax = 110, 000, 000
miles and rmin = 100, 000, 000 miles.
Object 2 has
rmax = 320, 000, 000 miles and rmin = 310, 000, 000 miles.
Both elliptical orbits have the same eccentricity.
5. Find the speed of a particle in a circular orbit with radius
1028 m around an object of mass 1025 kg. (Recall that
G= 6.67 × 10−11 m/kg · s2 )
(a) 1.50 × 1013 m/s
(b) 6.67 × 10−14 m/s
(c) 3.87 × 106 m/s
(d) 2.58 × 10−7 m/s
6. Find the speed of a particle in a circular orbit with radius
1030 m around an object of mass 1027 kg. (Recall that
G= 6.67 × 10−11 m/kg · s2 )
(a)
(b)
(c)
1.50 × 1013
m/s
6.67 × 10−14
3.87 × 106
m/s
m/s
(d) 2.58 × 10−7 m/s
7. An object in orbit has rmax = 1025 km and e = 0.58.
Find rmin .
(a) 2.66 × 1023 km
(b) 2.70 × 1023 km
(c) 2.66 × 1024 km
8. An object in orbit has rmin = 1026 km and e = 0.58.
Find rmax .
(a) 3.76 × 1026 km
(b) 3.80 × 1026 km
(c) 3.80 × 1025 km
(d) 3.76 × 1024 km
9. An object in orbit has rmax = 1026 km and e = 0.52.
Find rmin .
(a) 3.16 × 1025 km
(b) 3.21 × 1025 km
(c) 3.24 × 1025 km
(d) 3.27 × 1025 km
10. An object in orbit has rmin = 1026 km and e = 0.52.
Find rmax .
(a) 3.15 × 1026 km
(b) 3.17 × 1026 km
(c) 3.19 × 1026 km
(d) 3.21 × 1026 km
11. If, for an elliptical orbit, rmin = 1026 km and e = 0.59, find a,
the semimajor axis.
(a) 2.40 × 1026 km
(b) 2.44 × 1026 km
(c) 2.47 × 1026 km
(d) 2.51 × 1026 km
12. If, for an elliptical orbit, rmax = 1026 km and e = 0.59, find
a, the semimajor axis.
(a) 6.25 × 1025 km
(b) 6.27 × 1025 km
(c) 6.29 × 1025 km
(d) 6.31 × 1025 km
13. If, for an elliptical orbit, rmin = 1027 km and e = 0.81, find a,
the semimajor axis.
(a) 5.23 × 1027 km
(b) 5.26 × 1027 km
(c) 5.29 × 1027 km
(d) 5.32 × 1027 km
14. If, for an elliptical orbit, rmax = 1027 km and e = 0.81, find
a, the semimajor axis.
(a) 5.41 × 1027 km
276
(b) 5.49 × 1027 km
(c) 5.44 × 1027 km
(d) 5.52 × 1026 km
15. Answer true or false. If a = 1.50 × 1012 km and e = 0.10,
rmax of an elliptical orbit is 1.65 × 1012 km, where a denotes
the semimajor axis.
(d)
8.
Z π /2
0
9 2
t + 3t +C
2
hcost, sint, 9 sinti dt =
(a) h1, 1, 9i
(b) h1, −1, −9i
(c) h−1, 1, 9i
(d) h−1, −1, −9i
Chapter 12 Test
√
1. Find the domain of r(t) = h t − 6, t 3 , t − 5i; t0 = 6.
(a) 0 ≤ t < ∞
(b) 6 ≤ t < ∞
(c) −6 ≤ t < ∞
(d) −∞ < t < ∞
2. Answer true or false. The vector equation r = costi + sintk
can be expressed in parametric form x = cost, y = 0, z = sint.
3. Describe the graph of r(t) = 15i + costj + sintk.
(a) Straight line
11. Find the arc length of the curve that is defined by
r(t) = − sinti + 6j − costk for 0 ≤ t ≤ π .
(a) 2
(b) 2π
(c) π
(d) 0
12. Find the arc length of the parametric curve x = sint, y = 10,
z = cost for 0 ≤ t ≤ π .
(a) 2
(c) Parabola
(b) 2π
(d) Circle
(c) π
(a) Cubic
(b) Twisted cubic
(c) Spiral
(d) Parabola
5. If r(t) = 2i + 5t 3 j + 3 costk, find r′ (t).
(a) 15t 2 j − 3 sintk
(b) 15t 2 j + 3 sintk
(c) ti + 15t 2 j − 3 sintk
(d) ti + 15t 2 j + 3 sintk
6. Answer true or false. r(t) = (t 2 + 4)i + 2t 4 j − (3t + 7)k is
continuous at t = 0.
Z
10. Answer true or false. r(t) = 6ti + 9 costj + 3t 5 k is a smooth
function of the parameter t.
(b) Spiral
4. Describe the graph of r(t) = ti + t 3 j + t 2 k.
7.
9. Answer true or false. If r(t) = sinti + 3 costj, the tangent line
at t0 = π is given by r(t) = −i.
(9ti + 3j) dt =
9 2
t i + 3tj +C
2
9 2
(b)
t +C i + (3t +C)j
2
9 2
t +C1 i + (3t +C2 )j
(c)
2
(a)
(d) 0
√
13. Answer true or false. If r = (t − 3)i + (4t + 2)j + ( 3 + 2t)k,
the arc length parameterization of the curve relative to the ref√
s
erence point (−3, 2, 3) involves the parameter t = √ .
19
2
14. r(t) = (t − 1)i + (2t − 5)j. Find T(2) =
1
2
(a) √ i + √ j
5
5
1
(b) √ k
5
1
(c) − √ k
5
(d) i
15. Answer true or false. r(t) = (t 2 − 1)i + (2t − 3)j, t = 2. N(t)
for the given value of t is i.
16. Answer true or false.
If r(t) = t 2 i + (2t − 1)j then
1
B(2) = − √ k.
5
17. Find the curvature κ (t) for r(t) = 2ti + 3t 2 j + 5k at t = 0.
3
√
40 10
3
(b)
2
(a)
277
3
16
(d) 2
(c)
18. Answer true or false. If r(t) = t 5 i + t 4 j + t 3 k, the curvature
κ (t) is 20t 6 − 30t 5 + 336t 4 .
19. Answer true or false. If r(t)√
= (t + 2)i + (2t 2 + 1)j + 4tk, the
4 17
curvature κ (t) at t = 1 is √ .
33 33
20. If an object orbits the sun with rmax = 290, 000, 000 miles and
rmin = 280, 000, 000 miles, the elliptical orbit has eccentricity
(a) 57.
1
(b)
.
57
(c) 56.
1
(d)
.
56
π
21. If y = 4 + sin x, find the curvature at x = .
2
(a) 0
(b) 1
(c) −1
1
(d) √
2 2
22. If x = t 3 + 2, y = t 2 − 1, then κ (t) at t = 1 is
6
√ .
13 13
6
(b) √ .
13
(a)
(c) 0.
(d)
180
√ .
13 13
23. r(t) = (4t 3 + 7)i + (2t − 3)j is the position vector of a particle
moving in a plane. Find the velocity.
(a) 12t 2 i + 2j
(b) 12i
(c) 24ti + 2j
(d) 24ti
24. r(t) = (4t 3 + 5)i + (6t − 9)j is the position vector of a particle
moving in a plane. Find the acceleration.
(a) 12t 2 i + 2j
(b) 12i
(c) 24ti + 2j
(d) 24ti
25. r(t) = (4t 3 + 3)i + (2t − 6)j is the position vector of a particle
moving in a plane. Find the speed at t = 1.
√
(a) 2 37
(b) 12
(c) 24
(d) 0
26. Answer true or false. If a(t) = sinti + 2tj + k, the position
vector is − sinti if v(0) = −i and r(0) = 0.
27. Answer true or false. Each planet moves in a circular orbit
with the sun at the center of the circle.
278
Chapter 12: Answers to Sample Tests
Section 12.1
1. b
9. d
2. a
10. d
3. c
11. b
4. false
12. b
5. false
13. a
6. false
14. d
7. true
15. b
8. b
2. a
10. c
3. b
11. c
4. c
12. a
5. b
13. false
6. a
14. b
7. true
15. a
8. false
2. true
10. a
3. false
11. false
4. a
12. true
5. c
13. true
6. c
14. true
7. b
15. false
8. c
2. false
10. c
3. c
11. false
4. b
12. false
5. c
13. a
6. b
14. false
7. a
15. false
8. b
2. b
10. a
3. a
11. b
4. b
12. a
5. false
13. false
6. true
14. c
7. b
15. c
8. a
2. d
10. d
3. a
11. c
4. a
12. a
5. b
13. a
6. a
14. a
7. false
15. a
8. false
2. b
10. b
3. b
11. b
4. false
12. c
5. d
13. b
6. d
14. d
7. c
15. true
8. a
2. true
10. true
18. false
26. false
3. d
11. c
19. true
27. false
4. b
12. c
20. b
5. a
13. false
21. b
6. true
14. a
22. a
7. c
15. false
23. a
8. a
16. false
24. d
Section 12.2
1. c
9. c
Section 12.3
1. true
9. d
Section 12.4
1. a
9. c
Section 12.5
1. a
9. b
Section 12.6
1. a
9. a
Section 12.7
1. true
9. a
Chapter 12 Test
1. b
9. false
17. b
25. a
Chapter 13: Partial Derivatives
Summary: While Chapter 12 introduced vector-valued functions, the components of these functions still only depended upon a single variable. In this chapter, functions are considered where there may be multiple independent variables.
Functions that have multiple inputs are often called multivariable functions.
The usual ideas about these functions are discussed. Limits are described as each
input approaches a particular value. Continuity is also discussed in a similar fashion. Derivatives are also discussed although they take on new meaning now.
Derivatives may be taken with respect to only one input and so they are called
partial derivatives. The idea of a directional derivative is a consequence of this
since partial derivatives can be thought of as derivatives along a particular line or
derivatives in a particular direction. This idea is extended to allow a derivative to
be taken in any direction.
The chain rule was an important concept for single variable functions and a corresponding idea is investigated for multivariable functions. With single variable
functions, derivatives lead to the ideas of tangent lines, local linearizations and
optimization problems. Similar topics are discussed for multivariable functions
such as tangent planes, gradients and finding the extreme values of a function of
two variables.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Define functions of two and three variables and understand what a level
curve is. (§13.1).
2. Graph a function of two or three variables (§13.1).
3. Evaluate limits along a curve and limits of a function with two or three
variables (§13.2).
4. Recognize when a function of two or three variables is continuous (§13.2).
5. Find partial derivatives of functions of two or three variables (§13.3).
6. Determine if a function is differentiable and find the local linearization of a
function of two or three variables at a point (§13.4).
279
280
7. Use the chain rule to find derivatives of a function of two or three variables
when the variables are represented using parameterizations (§13.5).
8. Find directional derivatives and the gradient of a function of two or three
variables at a point (§13.6).
9. Find tangent planes to surfaces and their normal vectors (§13.7).
10. Use gradients to find tangent lines to the intersections of surfaces (§13.7).
11. Find the absolute maximum and minimum of a function of two variables on
a bounded set (§13.8)
12. Use the method of Lagrange Multipliers to solve constrained optimization
problems for functions of two or three variables (§13.9).
13.1 Functions of Two or More Variables
PURPOSE: To describe functions of two or more variables and
to discuss level curves and level surfaces.
multivariable functions
Functions describe a rule in which some output value is determined by some input
value. To this point, only functions that have one input have been considered.
These are called single-variable functions. In the following sections, functions
that require two or more variables as input will be considered. These are called
multivariable functions.
IDEA: Functions in this section depend upon multiple variables (usually either
two or three) but still have only scalar values.
Functions that depend upon two more variables may have either a scalar output or
a vector output. For the time being, only functions with scalar output will be considered. An example of a scalar multivariable function might be the temperature
in a room. The temperature is measured by a single number at any given point.
However, the temperature may also change depending upon the point in the room
(in 3-space) it is begin measured.
IDEA: The graph of z = f (x, y) is a surface above the xy-plane.
the surface z = f (x, y)
level curves
contour plot
The function z = f (x, y) represents a surface above the xy-plane. The height of the
surface, z, at any given point is determined by the function f (x, y) which accepts
both x and y inputs. If the surface z = f (x, y) is intersected with the plane z = k
this results in a curve being traced where z = f (x, y) has a constant value k. These
are called level curves (since they are all at the same level). A plot of several level
curves together is called a contour plot.
IDEA: The curve determined by f (x, y) = k with some constant, k, is a curve
on the surface z = f (x, y). The curve traces the intersection of the plane z = k
with the surface z = f (x, y).
281
The function w = f (x, y, z) does not have an easy description in 3 − space. In fact,
it is difficult if not impossible to graph this function in 3 − space. The function
gives a different value at each point (x, y, z) in 3-space. Unfortunately a fourth
dimension would be required to show the “height”, w, of this function and this is
not easily visualized.
One approach that is taken is to consider the level surfaces of the function. These
are surfaces in 3-space where w has a constant value or f (x, y, z) = k.
level surfaces
IDEA: The graph of f (x, y, z) = k with some constant, k, is a level surface.
All of the values of f (x, y, z) on the level surface f (x, y, z) = k have the same
value.
To visualize the function it might be beneficial to create a movie with each frame
consisting of a level surface with a greater value of k. The movie when played
would show how the level surfaces change as k increases. This is not a method,
however, that is easily produced.
Checklist of Key Ideas:
function of two variables
function of three variables
domain
graph of z = f (x, y)
level curves
level surfaces
13.2 Limits and Continuity
PURPOSE: To define and investigate the concepts of limits and
continuity for functions of more than one variable.
Limits of single-variable functions were first discussed in Chapter 1. With the introduction of multivariable functions, new ideas for limits need to be discussed.
Many of these ideas are similar to those that have been defined for single variable
functions.
IDEA: Single-variable functions have limits from one side of a point or the
other. Multi-variable functions can have limits along any path that approaches
the point.
Perhaps one of the most important features of single-variable limits were onesided limits. For example, was a value approached from the positive or negative
side? For multivariable functions, there are many more directions than just the
limits of multivariable functions
282
limit along a path
positive or negative side. So when talking about multivariable limits, the limit
along a path approaching a point needs to be discussed.
IDEA: A multi-variable limit along a path that is parameterized by one variable,
can be thought of as a single variable limit.
If a curve approaching a point can be parameterized by a single variable, then a
multivariable limit along a path can be evaluated as a single variable limit by using
the parameterization to simplify the function so it only has one variable.
IDEA: For a limit to exist at a point then all the values within an open ball
around the point must be approaching the same value as they get arbitrarily
close to the point.
open circle or ball
In multivariable functions, the area around a particular point is described using
either an open circle or an open ball (an open interval was used for single variable
functions). For a limit to exist then all of the values of the function in an open
ball around the point must be approaching the same limit. In other words, if two
different limits can be found along two different paths approaching a point then a
limit will not exist. Note that it is nearly impossible to find a limit in this manner
since an infinite number of smooth curves would need to be checked.
IDEA: If a limit exists then the limit along every smooth curve to the point is
the same value. If two different paths have different limits then the limit does
not exist.
interior points
boundary points
Limits for single-variable functions are also talked about at endpoints of a limit.
To discuss this, an idea of interior and boundary points of a domain need to
be clarified. An open ball or circle around a point is essentially all of the points
within a circle around the point. If this circle can be drawn within the domain then
the point is an interior point. If this circle always lies both inside and outside the
domain then the point is a boundary point.
IDEA: An open circle or open ball around a point is best seen by drawing a
picture of the point. Then the open circle or ball is represented by a circle
drawn around the point.
IDEA: If the open circle about a point can be drawn completely in the domain
then the point is an interior point. If the open circle always lies both inside and
outside the domain then the point is a boundary point.
closed and open domains
In one dimension, intervals were described as open if they did not contain their
endpoints or as closed if they contained both endpoints. In two or more dimensions a similar idea holds. A domain will be closed if it contains all of its boundary
points. If a domain does not include any of its boundary points then it is considered to be open. It is possible for a domain to include some boundary points but
not all. In this case, the domain would not be described as open or closed.
IDEA: A domain is closed if it contains all of its boundary points. A domain is
open if it contains none of its boundary points.
283
CAUTION: A domain may contain some of its boundary points (but not all) and
therefore would be neither open nor closed.
Now that limits have been described for multivariable functions, the next important idea is that of continuity. As with single-variable functions, the continuity of
a multivariable function is related to its limit at a point. If the limiting value of
a function at a point is equal to the functions value at that point then the function
is continuous. This implies that the function must be at least defined at the point.
Normally, limits do not imply anything about the actual existence of a function at
a point. Continuity allows this connection between a limit and a function value to
be made.
continuity
IDEA: Continuity for multivariable functions is the same as for single-variable
functions: a function is continuous if its value at a point is equal to its limit at
that point.
IDEA: Continuity at boundary points is treated in the same way as continuity
at endpoints for single variable functions.
A limit at a boundary point may be described as coming from within the domain
or from the exterior of the domain. To describe continuity at a boundary point,
the function value at the boundary point must be equal to the limit of the function
as it comes from within the domain. This is the same as saying that that the
function value must be the same as the one-sided limit from within the interval for
single-variable functions.
Checklist of Key Ideas:
limit along a curve C
open and closed sets
limit of a function of two variables
limits and continuity
continuity of a function of two variables
continuity of compositions
limits at discontinuities
limit along curves from one side of a boundary
limit of a function of three variables
limit at a boundary point
continuity at the boundary
284
13.3 Partial Derivatives
PURPOSE: To define partial derivatives for functions of more
than one variable and to discuss the meaning of partial derivatives.
partial derivatives
Having discussed limits and continuity of a multivariable functions, this section
considers the idea of derivatives of multivariable functions. It is not possible to just
talk about “the derivative” of a multivariable function, however, since it depends
upon more than one variable. In fact, it is possible to take partial derivatives of
the function by assuming that all of the variables of the functions except one are
constant. The partial derivative of a function with respect to x would mean that x
is the variable that is not considered to be a constant.
IDEA: Partial derivatives are found using the same derivative rules as for the
derivatives of single-variable functions. All but one of the variables are treated
as constants when taking derivatives.
Since a partial derivative essentially is the derivative of just one variable, then it
may be calculated in a fashion similar to the derivative of a single variable function. All of rules that were introduced in chapter 2, for example, may be used to
find the partial derivative. The trick is simply to consider all of the other variables
of a function as constants. For example, if f (x, y, z) = x + 2y + 3z then ∂ f /∂ x = 1
since y and z are treated as constants. Likewise, ∂ f /∂ y = 2 (x and z are constant)
and ∂ f /∂ z = 3 (x and y are constant). Similarly if f (x, y, z) = x2 sin (yz) + y2 z2
then ∂ f /∂ x = 2x sin (yz).
notation for partial derivatives
There are several notations for the partial derivative of a function. Some of
the more common ones are given below. They are all equivalent expressions of
the same partial derivative. In words the partial derivative ∂ f /∂ y means to take
the derivative of the function f with respect to the variable y. This means that all
variables except y should be considered to be constants.
IDEA: There are many notations for partial derivatives. Here are some equivalent statements for the partial derivative of f with respect to the variable x:
Dx f =
∂f
∂
= fx =
( f ) = ∂x f
∂x
∂x
IDEA: A partial derivative describes the rate of change of the function
z = f (x, y) with respect to a particular variable.
Also, since the partial derivative may be found using the rules of derivatives
for single variable functions, it also has some similar meaning. For example,
the partial derivative with respect to x represents the change of the function in
the x-direction. Equivalently, for the function z = f (x, y), the partial derivative
∂ z/∂ x = fx would represent the slope of the tangent line to the surface in the xdirection. A similar statement can not be made about w = f (x, y, z) since it is not
a surface.
285
IDEA: If z = f (x, y) is intersected with a plane parallel to either the xz or yz
plane, then the the partial derivatives ∂ z/∂ x and ∂ z/∂ y describe the slopes
of the respective curves that result as the intersection of the surface and the
plane.
Another way of describing this change in a particular direction is to consider a
curve that results when a surface z = f (x, y) is intersected with a plane. If the
intersecting plane is parallel to the yz-plane (where y in particular is constant)
then the slope of the resulting curve is given by fx at some point.
It is helpful to think of higher order partial derivatives in a step-by-step process.
In particular, it matters in what order that derivatives are taken with respect to
∂
f means to take the derivative of f with respect
which variable. For example,
∂z
∂
it must then be differentiated
to z. But then if this quantity is “multiplied” by
∂x
∂2 f
. Notice
with respect to x. This higher order derivative would be written as
∂x ∂z
the order in which the differentiation took place: first with respect to z and then
with respect to x.
higher order partial derivatives
IDEA: Higher order partial derivatives may be thought of as multiplication of
derivative operations. For example,
respect to x. Then
∂2 f
=
∂ x2
∂
∂x
∂
∂x
f
and
∂
means to take the derivative with
∂x
∂2 f
=
∂y ∂x
∂
∂y
∂
∂x
f
or
fxy = ( fx )y =
∂
∂y
∂
∂x
f
In this step-by-step procedure of taking partial derivatives, a variable may be considered as a constant in one step and then as a variable in the next step. The steps
should not be mixed together but should be considered in order. In the example
∂2 f
above for a function w = f (x, y, z) when calculating
, z is first considered as
∂z ∂x
a constant as the derivative with respect to x is taken and then it is considered as a
variable when ∂ /∂ z is performed.
IDEA: Mixed partials or mixed partial derivatives are found by alternating
which variables are considered to be constants.
Unfortunately, not all notations seem to say the same thing when it comes to taking
∂2 f
and fxz refer to the same partial derivative
partial derivatives. For example,
∂z ∂x
but they seem to indicate that the derivatives should be taken in a different order.
This is not the case. Both representations indicate that differentiation should first
be done with respect to x and then with respect to z. Since the notation using
∂ /∂ z occurs to the left of f , it should be read right-to-left for order. On the other
mixed partial derivatives
286
hand, the subscript notation fxz occurs after the f and can be read as the normal
left-to-right.
∂2 f
should be taken right-to-left (first ∂ /∂ y
∂ x∂ y
then ∂ /∂ x). The derivatives of fxy should be taken left to right (first ∂ /∂ x then
∂ /∂ y).
CAUTION: The derivatives of
In most cases, the order of mixed partial derivatives will not matter. Care should
still be taken to ensure that derivatives are being taken in the correct order but
if the mixed partial derivatives with respect to the same variables are continuous
then they will be equal. For example, fxz = fzx , if they are both continuous. Note,
however, that if the derivatives are not with respect to the same variables then the
same assumptions may not be made. For example, fxz does not generally equal
fyx . One requires a derivative with respect to y while the other requires a derivative
with respect to z.
IDEA: Mixed partial derivatives written in different orders are equal if they are
continuous.
Checklist of Key Ideas:
partial derivative at a point
partial derivative functions
notation for partial derivatives
meaning of partial derivatives in terms of rates of change and slopes
taking partial derivatives implicitly
second-order partial derivatives and mixed partial derivatives
equality of second-order mixed partial derivatives
the wave equation
partial differential equation
verifying solution of a partial differential equation
13.4 Differentiability, Differentials, and Local Linearity
PURPOSE: To determine if functions of more than one variable
are differentiable and to look at the local behavior of multivariable
functions using differentials.
287
This section continues the progression of topics from limits and continuity to
derivatives and now to differentiability of a function. This is contrary to the
way these topics were discussed for single variable functions. For functions of
one variable, having a derivative is the same as being differentiable. This is not
the case for multivariable functions.
differentiability
Differentiability of a function z = f (x, y) is defined by a limit that involves the total
change of the function, ∆ f , and its partial derivatives. If this limit exists and is
equal to zero (see Definition 13.4.1 in the book) then it is said to be differentiable.
IDEA: A function f (x, y) is only differentiable if its first-order partial derivatives
exist and an appropriate limit is satisfied.
In words, the limit in the definition is saying that as small changes are made in the
variables, x, and y, that the actual change of the function value, ∆z, will approach
the value of the differential dz (see below) which approximates the change in the
function value.
Differentiability still implies continuity of a function. In other words, for small
changes in x and y there should be correspondingly small changes in the value of
the function f (x, y). From this it can be implied that near a point, the function
value must be equal to its limit which means that the function would be continuous.
differentiability and continuity
IDEA: Differentiability implies continuity. Continuity does not imply differentiability.
Remember that “the derivative” of z = f (x, y) is an ambiguous statement. Which
variable should the derivative be taken with respect to? So while differentiability
lead to an idea of what the derivative of a single variable function is, this is not
the case for a differentiable function f (x, y). The limit that is used to determine if
a function is differentiable will not give a “derivative” of f (x, y). It does relate the
total change of the function with the approximate change however.
total change
CAUTION: The limit evaluated to determine if f (x, y) is differentiable does not
readily determine the derivative of f (x, y) as was the case with single-variable
functions.
CAUTION: Even if fx and fy are both defined, a function may not necessarily
be differentiable.
A function z = f (x, y) may even have both partial derivatives, fx and fy . However,
unless they are both continuous, there is no guarantee that the function will be
differentiable. In other words, just because one can find the partial derivatives of a
function does not indicate whether or not the function is differentiable or not. But
the partial derivatives are still useful. If they can be found and they are continuous
then it can be concluded that the function is differentiable.
IDEA: The total differential, i.e., dz, of a function z = f (x, y) is an approximation of the change in the function z = f (x, y) for small changes in x and
y.
total differential, dz
288
∆z and dz
The real value of the limit in the definition of differentiability is that it says that
for small changes in the variables that ∆z and dz should be approximately equal.
If r = hx, yi and ∆r = h∆x, ∆yi, then another way to state the limit would be
∆z − dz
= 0.
k∆rk→0 k∆rk
lim
local linear approximation
Here we are representing x and y using the vector r and then ∆r corresponds
to a small changes in x and y (i.e., ∆x = dx and ∆y = dy). Consequently, this
means that if dz is approximately equal to ∆z then a local linear approximation
of z = f (x, y) can be made at a point. In the case of the function z = f (x, y) this
amounts to finding the tangent plane.
IDEA: Local linear approximations can be made to functions z = f (x, y) and
w = f (x, y, z) that are differentiable at a point.
The benefit of being able to find a local linear approximation is that an approximation of the total change of the function may be more easily found. For example,
if L(x, y) is the local linear approximation of f (x, y) then L(x, y) − f (x0 , y0 ) ≈ ∆ f .
On the other hand, if ∆x = dx and ∆y = dy then approximating the change in the
function z = f (x, y) using the local linear approximation is the same as approximating this change using the total differential dz.
IDEA: Using a local linear approximation to find the change in the value of a
function is the same as using a total differential.
tangent plane
IDEA: For a multivariable function with two inputs, z = f (x, y), the local linear
approximation is a plane that is tangent to the surface at a point (see §13.7).
Checklist of Key Ideas:
increment of f , ∆ f
differentiable at a point
differentiability and continuity
continuity and continuity of first order partial derivatives
relationship between differentials and partial derivatives
total differential approximately equal to ∆ f
local linear approximation at a point in 2-space and 3-space
13.5 The Chain Rule
PURPOSE: To use the chain rule to calculate derivatives of functions of more than one variable when the variables can be represented using a parameterization.
289
The main idea of this section comes about if the variables of a function can be
parameterized using a single variable. The result is that the multivariable function may be thought of as a single variable function. For example if x = g(t) and
y = h(t) then the function z = f (x, y) = f (x(t), h(t)) = F(t).
Now since z = f (x, y) can be thought of as z = F(t), it is possible to find the
derivative of z with respect to t. One way to to do this is to simply substitute the
parameterizations into the definition of z and take the derivative of the resulting
function using the usual single variable techniques. The other way is to use the
multivariable chain rule.
parameterizing variables
multivariable chain rule
IDEA: If a multivariable function may be parameterized so that it is instead
a function of one variable, then the chain rule may be used to evaluate its
derivative.
The concept is similar to the chain rule for single variable functions. The derivative of the inside function is taken and then this is multiplied by the derivative
of the parameterization. Since there are more than one variables, however, this
process needs to be done for each variable and then the results can be added to together. The process can be pictured using the tree diagrams in the book. Another
way to think of it is that the chain rule will have as many terms added together
as there are variables in the original function. For example, f (x, y) will have two
terms: one from x and one from y.
IDEA: For multivariable functions, the chain rule needs to be applied for each
variable of the function in order to find the derivative of the function with respect to a parameter.
If the variables of a function are parameterized using more than one parameter,
then the derivatives from the chain rule become partial derivatives with respect to
one parameter or the other. The process of finding these partial derivatives is the
same as if there were one parameter. Since these are partial derivatives, however,
the derivative should only be found with respect to one parameter at a time.
IDEA: If a multivariable function is parameterized by more than two parameters, then the chain rule produces partial derivatives with respect to the parameters.
Checklist of Key Ideas:
find derivatives using the chain rule for functions of two or three variables
(one parameter involved)
implicit differentiation of f (x, y) = c
related rates and the chain rule
find partial derivatives using the chain rule for multivariable functions (two
or more parameters involved)
more than one parameter
290
13.6 Directional Derivatives and Gradients
PURPOSE: To find directional derivatives in the direction of a
given vector and to discuss gradients and their relationship to
level curves.
directional derivative
gradient
This section introduces the directional derivative, Du f , which is a form of a
partial derivatives of a function in the direction of a given unit vector, u. While this
seems like a complicated idea since there are an infinite number of unit vectors that
can be used, there are some simple ways to remember directional derivatives using
the gradient, ∇ f . In fact, Du f is a linear combination of the partial derivatives of
f that can be written as simple dot product.
IDEA: A directional derivative is just a linear combination of the first order
partial derivatives of a function.
Du = ∇ f ·u = u1 fx + u2 fy
where ∇ f = h fx , fy i and u is a unit vector.
The meaning of the direction derivative is also straightforward. The directional
derivative Du f describes how the function f is changing in the direction of the
vector u. Specifically, Du f describes the instantaneous rate of change of f as the
input variables are changed in the direction of the vector u.
IDEA: The directional derivative Du f is the slope of the surface z = f (x, y) in
the direction of the unit vector u. It finds the instantaneous rate of change of
the function z = f (x, y) in the direction of u.
The function w = f (x, y, z) (and functions of more variables) also have directional
derivatives that may be defined by Du w = ∇w·n. However, in these cases, the
geometrical representation of Du w as the slope of a tangent line is no longer appropriate since these functions are not described by surfaces any longer. The directional derivative does, however, still describe the instantaneous rate of change
of the function in the direction of the vector u.
CAUTION: The directional derivative Du f is not a slope if w = f (x, y, z) since
the function is not a surface.
The gradient has already been described above as ∇ f = h fx , fy i. For functions of
more variables, this vector would include all of the first order partial derivatives
of the function. For example if w = f (x, y, z) then ∇w = h fx , fy , fz i.
IDEA: The gradient is a vector of the first-order partial derivatives of a function
f.
gradient and level curves
Aside from its connection to directional derivatives, the gradient also has some
other remarkable properties. In particular, for z = f (x, y), the gradient vector is
perpendicular to any level curves of f (x, y). This describes the direction that must
be moved from any level curve in order for the function to experience a maximum
291
change from that point. In other words, the biggest increase in a function will
occur in the direction of the gradient from that point and the biggest decrease will
occur in the opposite direction of the gradient.
IDEA: Gradients are perpendicular to level curves (for z = f (x, y)) and give
the direction from a level curve in which the function f (x, y) will experience
maximum increase.
For functions of more than two variables such as w = f (x, y, z), the gradient still
has meaning. For example, the gradient ∇w can be used to determine the plane
which is tangent to a level surface of w at a given point since it is perpendicular to
the level surface. This same idea can be used to find the tangent plane to a surface
z = f (x, y) at a point. More is said about this in the next section.
gradient and level surfaces
Checklist of Key Ideas:
directional derivative
slope of z = f (x, y) at a point in the direction of u
Du f (x0 , y0 )
relationship between differentiability and directional derivatives for a unit
vector u
the gradient, ∇ f
writing directional derivatives using the gradient
maximum value of Du f and the gradient
gradients are normal to level curves
finding a gradient path that seeks maximum change along a path
13.7 Tangent Planes and Normal Vectors
PURPOSE: To find tangent planes and normal vectors to surfaces represented by multivariable functions and to discuss their
relationship with total differentials.
Recall from Section 11.6 that planes may be defined using a dot product that involves a normal vector and two points in the plane (one being unknown). The
same idea can now be used with a gradient vector to determine local linear approximations and tangent planes to surfaces.
IDEA: Recall that the local linear approximation to z = f (x, y) at a point is a
tangent plane.
local linear approximation
tangent plane
292
the function G(x, y, z) = f (x, y) − z
With a little modification of the function z = f (x, y), a gradient vector can be
used to find the tangent plane to z = f (x, y) at a point. Define the function
G(x, y, z) = f (x, y) − z. Then G = 0 is just the surface z = f (x, y). On the other
hand, G(z, y, z) = 0 describes a level surface. The gradient of G, ∇G, will be perpendicular to the level surface. In other words, ∇G will be a normal vector to the
tangent plane of z = f (x, y).
IDEA: If G(x, y, z) = f (x, y) − z, then the gradient vector ∇G,
∇G = h fx , fy , −1i
gives the normal vector for a tangent plane to z = f (x, y).
So for ∆r = hx − x0 , y − y0 , z − z0 i = r − r0 , the tangent plane of z = f (x, y) is
defined by the dot product ∇G·∆r = 0. If the equation of this plane is solved for z
then this gives the local linear approximation.
total differential
The total differential dz = fx dx + fy dy is equivalent to using the local linear
approximation to approximate ∆z. For example, if dx = ∆x and dy = ∆y then
L(x, y) − f (x0 , y0 ) = dz ≈ ∆z.
IDEA: The total differential dz is change along a tangent plane which approximates the actual total change ∆z of a function for small changes in x and y
(see also §13.4).
If w = f (x, y, z) then the same process as above can be used to find a tangent plane
to any level surface w = k of the function.
Checklist of Key Ideas:
tangent planes to z = f (x, y)
tangent planes and local linear approximations
normal line to a surface at a point
tangent planes and differentials
relationship between dz and ∆z
tangent planes to level surfaces, F(x, y, z) = C
using gradients to find tangent planes
using gradients to find tangent lines of curves at the intersection of two
surfaces
293
13.8 Maxima and Minima of Functions of Two Variables
PURPOSE: To develop methods for finding the absolute extrema of a continuous function of two variables on a closed and
bounded region.
Recall that in chapter 3, one problem that was posed was finding the absolute
maximum and minimum values of a function y = f (x) on some closed interval. In this section, the same question is asked of a function z = f (x, y) on some
closed and bounded domain. As it turns out, if z = f (x, y) is continuous and the
domain is closed and bounded then the function will have an absolute maximum
and minimum value (this is similar to the guarantee made for y = f (x) in Chapter 3). Also similar to the situation for y = f (x) on a closed interval, these absolute
extrema must occur either at the boundaries of the domain or at critical points of
the function within the domain (at interior points).
absolute maximum and minimum
values
IDEA: Like single-variable functions that are continuous on a closed interval,
a multivariable function that is continuous on a closed and bounded domain
must have an absolute maximum and an absolute minimum.
IDEA: Finding absolute extrema is similar to the single-variable case. The
function values need to be checked at the boundary (although only at certain
points) and at interior points called critical points (found using the first-order
partial derivatives).
So the process here is to describe how to find critical points for f (x, y) and how
to find suitable points to check on the boundary. Similar to the situation with the
function y = f (x) critical points are defined using the derivatives of z = f (x, y).
Specifically, a critical point occurs where any of the first order partial derivatives
of z = f (x, y) are either equal to zero or undefined.
critical points
IDEA: A critical point for z = f (x, y) occurs if fx = 0, fy = 0 or either of these
are undefined. This is similar to the one dimensional case when f ′ (x) = 0 or
f ′ (x) is undefined determined a critical point.
The one dimensional case could make use of the second derivative test to determine if a relative extrema would occur at a point. A similar test can be used with
z = f (x, y) that involves the second order partial derivatives. The quantity that
should be computed is D = fxx fyy − ( fxy )2 and then Theorem 13.8.6 may be applied. As was the case for y = f (x) when y′′ = 0 using the second derivative test,
no conclusion may be drawn about a critical point if D = 0.
IDEA: The quantity D = fxx fyy − ( fxy )2 can be used to determine if a critical
point has relative extrema or a saddle point (see Theorem 13.8.6 in the book).
CAUTION: If the quantity D = fxx fyy − ( fxy )2 = 0 at a critical point then Theorem 13.8.6 can make no conclusion about that critical point.
the Second Partials Test
(Theorem 13.8.6)
294
While this quantity D is useful, it may only make conclusions about relative extrema when it is applied at a critical point. Otherwise, it does not indicate one way
or the other if there is a relative extrema. Also the function z = f (x, y) may have a
saddle point where no relative extrema may occur. This is similar to the situation
when y = f (x) had an inflection point at a critical point.
CAUTION: The Second Partials Test should only be applied at a critical point.
The quantity D does not provide information about relative extrema at other
points.
list of possible points
Now that critical points have been found using fx and fy , the idea is to find the
largest and smallest function values of z = f (x, y) among the list of possible
points, the boundary points, and the critical points. The boundary points may
be checked by parameterizing the boundary with a single variable and then finding the largest and smallest values on the boundary by using the techniques for a
single variable function in Chapter 3. Then these values would be compared with
those found at the critical points.
IDEA: Checking points on the boundary can be done as a single variable
extrema problem if the boundary can be parameterized with a one variable.
Again, similar to the situation for a single variable function y = f (x), if z = f (x, y)
is not continuous or if the domain is not closed or not bounded then there is no
longer any guarantee that an absolute maximum and minimum may be found. In
other words, these three things are essential: continuity of f (x, y), closed domain
and bounded domain. Otherwise, finding absolute maximum and minimum values
(if they even exist!) can be a longer process which is not discussed in this text.
IDEA: If a function is discontinuous, or if the domain is either open or unbounded then a function z = f (x, y) is not guaranteed to have absolute extrema.
Checklist of Key Ideas:
absolute and relative extrema of functions of two variables
bounded sets and unbounded sets
extreme value theorem for a function of two variables
definition of a critical point for a function of two variables
finding relative extrema for a function of two variables
saddle points
the second partials test
location of absolute extrema in interior of a domain
strategy for finding absolute extrema on a closed and bounded set
295
13.9 Lagrange Multipliers
PURPOSE: To use the approach of Lagrange multipliers to assist in finding the maximum or minimum value of a multivariable
function when it is subject to some constraint.
In some instances, a maximum or minimum value of a function z = f (x, y) is
sought not on some closed domain but rather on a curve defined by g(x, y) = 0
(similarly for z = f (x, y, z) and g(x, y, z) = 0).
IDEA: A constrained optimization problem can be thought of as trying to
minimize (or maximize) the function z = f (x, y) along the intersection of the
surface z = f (x, y) and the level curve g(x, y) = 0.
constrained optimization
g(x, y) = 0
In situations like this, there are two possible approaches to finding a maximum
(or minimum) value. The first is to substitute the relationship g(x, y) = 0 into the
function z = f (x, y) with the intent of reducing the number of variables. If this
is successful then the question may be asked of a single variable function. For
example, suppose that f (x, y) = x2 y + y2 and g(x, y) = 2x + y = 0. Then from the
g = 0 equation it can be seen that y = −2x. Substituting this into the equation
z = f (x, y) results in z = x2 (−2x) + (−2x)2 = −2x3 + 4x2 .
The other process of finding the sought after extrema of f (x, y) may be done using
Lagrange Multipliers. The basic idea is that any extrema that occurs will happen
at a place where the level curves of z = f (x, y) = c are tangent to g(x, y) = 0.
In other words, at this point, ∇ f = λ ∇g (their gradients are pointed in the same
direction).
tangent level curves
IDEA: The gradients ∇ f and ∇g are parallel at a constrained extrema of the
function z = f (x, y) along the curve g(x, y) = 0. That is to say ∇ f = λ ∇g.
So to find the extrema of z = f (x, y) subject to g(x, y) = 0 is equivalent to solving
the system of equations ∇ f − λ ∇g = 0 for (x, y, λ ). This system of equations will
possibly result in several sets of numbers (x, y, λ ). The value of λ is not important
but the extrema of z = f (x, y) must occur at one of these sets of (x, y) values. So
solving the system of equations ∇ f − λ ∇g = 0 results in a list of points where the
extrema might possibly occur.
IDEA: Find the points (x, y, λ ) such that ∇ f − λ ∇g = 0. Then the solution must
occur at one of these points (x, y). The value of λ is not directly important.
Checklist of Key Ideas:
constrained optimization problem
a constraint g(x, y) = 0 or g(x, y, z) = 0
intersection of f (x, y, z) = c and g(x, y, z) = 0
solve the system
∇ f − λ ∇g = 0
296
lagrange multipliers
constrained absolute and relative extrema
constrained extremum principle for functions of two variables subject to
one constraint
at a constrained extrema, gradient of f and gradient of g are parallel
297
Chapter 13 Sample Tests
Section 13.1
(a) 3x2 + y2 − z = 4
1. f (x, y, z) = x2 − yz. Find f (1, 2, 3).
(a) −4
(c) 3x2 + y2 − z = 2
(b) −5
(d) 3x2 + y2 − z = −2
(c) −7
(d) 5
2. f (x, y,
(b) 3x2 + y2 − z = 0
z) = 3exy + z.
Find f (1, 0, 6).
(a) 9
11. Let f (x, y, z) = 2x2 + y2 − z2 . Find an equation of a level
surface passing through (0, 0, 1).
(b) 3e + 6
(a) −z2 = 1
(c) 6e
(b) z2 = 1
(d) 9e
3. f (x, y, z) =
√
x + y + z. Find f (1, 2, 1).
(a) 4
(b) 0
(c) 2
(d) 1
4. Answer true or false. f (x, y) = 7 describes a plane parallel
to the xy−plane 7 units above it.
5. Answer true or false. f (x, y) = x2 + y2 graphs in 3-space
as a circle of radius 1 centered at (0, 0) and confined to the
xy−plane.
p
6. Answer true or false. f (x, y) = 3 x2 + y2 graphs as a semicircle.
p
7. Answer true or false. f (x, y) = x2 + y2 + 2 graphs as a
hemisphere.
8. The graph of z = 4x2 + 4y2 for z = 0 is
(a) a circle of radius 4.
(b) a circle of radius 2.
(c) a circle of radius 16.
(d) a point.
9. The graph of z = 4x2 − 2y2 for z = 0 includes the following
point.
(c) 2x2 + y2 − z2 = 1
(d) 2x2 + y2 − z2 = −1
12. f (x, y, z) = exyz . Find an equation of the level surface that
passes through (0, 1, 2).
(a) exyz = 2
(b) exyz = 3
(c) exyz = 1
(d) exyz = 0
13. Answer true or false. If V (x, y) is the voltage potential at a
point (x, y) in the xy−plane, then the level curve for V , called
2
, and it passes
the equipotential curve, is V (x, y) = p
2
x + y2
through (1, 0) when V (x, y) = 1.
14. Answer true or false. If V (x, y) is the voltage potential at a
point (x, y) in the xy−plane, then the level curve for V , called
2
the equipotential curve, is V (x, y) = p
, and it passes
x2 + y2
through (0, 1) when V (x, y) = 1.
15. What is/are the domain restriction(s) for f (x, y) = ln(xy2 )?
(a) (0, 0, 0)
(b) (1, 0, 0)
(a) x > 0, y 6= 0
(c) (0, 1, 1)
(b) x > 0, y > 0
(d) none of the above
10. Let f (x, y, z) = 3x2 + y2 − z. Find an equation of the level
surface passing through (1, 0, 1).
(c) x 6= 0, y 6= 0
(d) No restrictions exist
298
Section 13.2
1.
lim
(x,y)→(3,4)
(c) 0
(d) does not exist
(2x + y) =
11.
(a) 10
(a)
(c) 14
(d) does not exist
lim
(x,y)→(π ,0)
(c) 1
(1 + y) sin x =
(d) does not exist
12. Answer true or false. f (x, y, z) = 4x2 y2 z is continuous everywhere.
(a) 0
(b) 1
does not exist.
13. Answer true or false. f (x, y, z) = cos(xyz) is continuous everywhere.
2z
14. Answer true or false. f (x, y, z) =
is continuous evsin(xy)
erywhere.
does not exist.
15. Answer true or false. f (x, y, z) = yz ln |x| is continuous everywhere.
(c) 2
(d) does not exist
3. Answer true or false.
4. Answer true or false.
5. Answer true or false.
6. Answer true or false.
7.
lim
(x,y)→(1,1)
lim
lim
lim
lim
(x,y)→(0,0)
xy =
(x + y + 2) does not exist.
Section 13.3
1. f (x, y) = 6x4 y7 . Find fx (x, y).
(b) 168x3 y6
(c) 24x3 y7 + 42x4 y6
8e2xy =
(d) 42x2 y6
2. f (x, y) = ln(xy). Find fx (2, 3).
3
2
1
(b)
2
5
(c)
6
1
(d)
6
(a)
(a) 0
(b) 1
(c) 8
(d) does not exist
4 sin(x2 + y2 )
p
=
(x,y)→(0,0)
x2 + y2 + 1
lim
(a) 4
10.
does not exist.
(a) 24x3 y7
(d) does not exist
9.
y − 2x
(x,y)→(0,0) x2 + y2
(c) 2
lim
2
(x,y)→(0,0) 3x2 + y2
(b) 1
(x,y)→(0,0)
5
(x,y)→(0,0) x2 + 4y2
(a) 0
8.
x+1
=
y+2
1
2
(b) 0
(b) 7
2.
lim
(x,y)→(0,0)
3. z = e3xy . Find
(b) 0
(a) 3xe3xy
(c) 1
(b) 3e3xy
(d) does not exist
(c) 3ye3xy
lim
(x,y,z)→(1,1,2)
(a) 2
(b) 4
x2 yz =
∂z
.
∂y
(d) 3xye3xy
4. Answer true q
or false.
If f (x, y) =
4x3
x4 + 3y2 , fx (x, y) = p
.
2 x4 + 3y2
299
5. z = sin(x2 y4 ). Find
∂z
.
∂y
(c) 64xy2 e4xyz
(d) 4xy2 e4xyz
(a) (4x2 y3 + 2xy4 ) cos(x2 y4 )
(b) 8xy3 cos(x2 y4 )
15. f (x, y, z) = x sin(yz). fyz =
(c) 4y3 cos(x2 y4 )
(a) xyz sin(yz)
(d) 4x2 y3 cos(x2 y4 )
(b) x cos (yz) − xyz sin (yz)
6. f (x, y) = x4 y3 . fxx =
(a) 12x2 y3
(c) x sin(yz)
(b) 6x4 y
(d) −x sin(yz)
(c)
12x3 y2
(d) 12x2
7. Answer true or false. If f (x, y, z) = (x2 + y3 + z4 )1/3 − 2, then
6x
∂ f (x, y, z)
= 2
.
3
∂x
(x + y + z4 )2/3
8. f (x, y, z) = (x2 + y2 + z2 )1/4 . fx (1, 2, 3) =
1
√
4
2 143
1
(b) √
4
143
1
(c) √
4
4 143
7
(d) √
4
143
(a)
1. Answer true or false.
If f (x, y) = x3 + 2y2 , then d f (x, y) = 3x2 dx + 4ydy.
2. Answer true or false.
If f (x, y) = x4 y7 , then d f (x, y) = 4x3 dx + 7y6 dy.
3. For gas PV = nRT , where n and R are constants, estimate the
change in nRT as P goes from 1 atm to 1.001 atm and V goes
from 2 m3 to 2.002 m3 . Answer in atm·m3 .
(a) 0.004
(b) 0.003
9. f (x, y, z) = xeyz . fxz =
(a) xyeyz
(b)
Section 13.4
(c) 0.002
yeyz
(d) 0.0015
(c) 0
4. Use the total differential to approximate the change in
f (x, y) = x2 +3y3 as (x, y) varies from (3, 4) to (3.01, 3.98).
(d) y
10. f (x, y, z) = y2 exz . fzz =
(a) exz
(a) −0.01
(b) x2 exz
(b) −0.03
(c) y2 ez
(c) −2.82
(d) x2 y2 exz
11. Answer true or false. x sin x solves the wave equation.
12. Answer true or false. If z = sin x cos y,
∂z
∂z
=− .
∂x
∂y
13. Answer true or false. The tangent line to z = x2 y at (1, 1, 1)
in the y−direction has a slope of 2.
14. f (x, y, z) = e4xyz . fxyy =
(d) −0.78
5. Use the total differential to approximate the change in
f (x, y) = xy as (x, y) varies from (3, 4) to (3.01, 3.98).
(a) −0.05
(b) −1.10
(a) 4xyze4xyz
(c) −0.02
(b) (32xz2 + 64x2 yz3 )e4xyz
(d) −0.10
300
Section 13.5
(d) 8u − 2
1. w = r2 − 3s; r = 2x, s = x + 4y. Find
(a) 5
∂ w .
∂ x x=1,y=3
(b) 7
8. Answer true or false. If z = f (v) and v = g(x, y), then
∂ 2z
dz ∂ 2 v d 2 z ∂ 2 v
=
+
.
dv ∂ x2 dv2 ∂ x2
∂ x2
9. Answer true or false. If z = x1/3 y3 , fxy and fyx differ on the
xy−plane.
10. Answer true or false. If z = x7 y1/5 , fxy and fyx are equal
where y 6= 0.
(c) 4
(d) 3
2. w = 4x sin y; x = t 2 , y = 3t. Find
(a) 8π + 3
(b) 8π
∂ w .
∂ t π
11. A right triangle initially has legs of 1 m. If they are increasing, one by 3 m/s and the other by 4 m/s, how fast is the
hypotenuse increasing?
(a) 5 m/s
(b) 7 m/s
√
(c) 7 2 m/s
√
7 2
(d)
m/s
2
(c) −12π 2
(d) 5π
3. Let f (x, y) = xy8 . Find fxyx .
(a) 8y7
Section 13.6
(b) 0
(c) 1
1. z = 5x + 3y. Find ∇z.
(d) 8xy7
(a) 5i + 3j
4. Let f (x, y) = e2xy . Find fxyy .
(b) 5xi + 3yj
(a) 2xy2 e2xy
(c) xi + yj
(b) 4x2 ye2xy
(d) −5i − 3j
(c) 8x2 ye2xy
2. z = 3x2 + 4y2 . Find ∇z.
(d) 8xe2xy (xy + 1)
5.
z = 5x2 y3 ;
∂z
.
x = u + v, y = u − v. Find
∂u
(d) xi + yj
(b) 25u9
(c) 15(u + v)2 (u − v)2 + 10(u + v)(u − v)3
(d) 15u4
3
−u2 v
3
(c) (3u2 − 2uv)eu
3
(d) 2eu
(a) i + 2j
−u2 v
(a) 8u + 4
(b) 6
(c) 8u − 2 + 6v
(a) 12i + 2j
4. f (x, y) = xy. Find the gradient of f at (2, 1).
−u2 v
7. z = 4x − 2y; x = u2 , y = u − 3v. Find
3. f (x, y) = (x2 + y)3/2 . Find the gradient of f at (1, 3).
(b) 3i + j
1
(c) 3i + j
2
(d) 6i + 3j
∂z
.
∂u
(a) 2ue2u−v
(b) eu
(b) 6xi + 8yj
(c) 3i + 4j
(a) 15(u − v)2 + 10(u + v)
6. z = exy ; x = u2 , y = u − v. Find
(a) 3xi + 4yj
(b) 2i + j
∂z
.
∂u
(c) i + j
(d) 3i + 3j
3
2
5. f (x, y) = e4xy ; P = (1, 2); u = √ i + √ j. Find Du f at
13
13
P.
301
28e4
(a) √
13
28e8
(b) √
13
28
(c) √
13
(d) e8
3
2
6. f (x, y) = yex ; P = (0, 3); u = √ i + √ j. Find Du f at
13
13
P.
9
(a) √
13
(b) 0
14
(c) √
13
5
(d) √
13
7. Answer true or false. If f (x, y) = 5exy + 3x and a = 4i + 3j
is a vector, the direction derivative of f with respect to a at
4
(2, 3) is (15e6 + 3)i + 6e6 j.
5
8. Find the largest value among all possible directional derivatives of f (x, y) = 3x3 + 2y.
√
(a) 81x4 + 4
p
(b)
9x6 + 4y2
(c)
81x4 + 4
(d) 9x2 + 2
9. Find the smallest value among all possible directional derivatives of f (x, y) = 4x + y.
√
(a) − 5
√
(b) 5
√
(c) − 17
√
(d) 17
10. A particle is located at the point (3, 5) on a
metal surface whose temperature at a point (x, y) is
T (x, y) = 25 − 3x2 − 2y2 . Find the equation for the trajectory of a particle moving continuously in the direction of
maximum temperature increase. y =
(a) x2/3
5
(b) 2/3 x2/3
3
(3x)2/3
5
5 2/3
(d) x
3
(c)
11. A particle is located at the point (3, 5) on a
metal surface whose temperature at a point (x, y) is
T (x, y) = 16 − 2x2 − 3y2 . Find the equation for the trajectory of a particle moving continuously in the direction of
maximum temperature increase. y =
(a) x2/3
5
(b) 3/2 x3/2
3
(3x)2/3
(c)
5
5 2/3
(d) x
3
12. Answer true or false. z = 4x2 + y2 . k∇zk = 10 at (1, 1).
13. Answer true or false. The gradient of f (x, y) = 7x − 3y2 at
(1, 2) is 7i − 6j.
14. Answer true or false. The gradient of f (x, y) = 5x2 − 7y at
(2, 3) is 10i − 7j.
15. The gradient of f (x, y) = 5exy at (1, 0) is
(a) 5i
(b) 5j
(c) i
(d) j
16. f (x, y, z) = 5x + 3y2 + z3 . Find the gradient at (1, 2, 1).
(a) 9i + 2j + 3k
(b) 5i + 12j + 3k
(c) i + 2j + k
(d) 9i + 2j + k
17. f (x, y, z) = x2 yz. Find the gradient at (2, 3, 4).
(a) 24i + 16j + 6k
(b) 2i
(c) 48i + 16j + 12k
(d) k
18. f (x, y, z) = ln(xyz) and u = 2i + j − 4k. Find the directional
derivative of f at (1, 2, 3) in the direction of u.
1
√
6 21
1
(b)
6
1
1
2
(c) i + j − k
3
6
3
(d) 2 ln 6i + ln 6j − 4 ln 6k
(a)
19. f (x, y, z) = e−4xyz and u = 3i + 4j − k. Find the directional
derivative of f at (3, 2, 4) in the direction of u.
(a) 26i
302
1
(b) √ (3i + 4j − k)
26
3. Find an equation for the tangent plane to z = 4x7 y2 at
P(1, 2, 16).
(c) 48e8
(a) 8 + 56(x − 1) + 8(y − 2) = z
48e8
(d) √
26
(b) 112(x − 1) + 16(y − 2) + 16 = z
20. Answer true or false. f (x, y, z) = 4x2 + 3y2 + 7z2 . The directional derivative of f at (1, 1, 1) that has the largest value
is in the direction of 8i + 6j + 14k
z) = 4x2 + 3y2 + 7z2 .
21. Answer true or false. f (x, y,
The directional derivative of f at (1, 1, 1) that has the smallest value
is in the direction of −8i − 6j − 14k
22. Answer true or false. f (x, y, z) = 4x2 + 3y2 + 7z2 . The directional
derivative of f at (1, 1, 1) that has the largest value
√
is 296.
z) = 4x2 + 3y2 + 7z2 .
23. Answer true or false. f (x, y,
The directional
√ derivative of f at (1, 1, 1) that has the smallest value
is − 296.
24. The gradient of sin x + cos x + z2 at (0, 0, 2) is
(a) i + 4k.
(c) i + j + 4k.
(d) −i + 4k.
25. The gradient of sin x + y3 − cos z at (π /2, 2, π ) is
(a) 12j.
(b) i + 12j + k.
(c) −i + 12j + k.
(d) i + 12j − k.
4. For z = 4x7 y2 , find the parametric normal line to the surface
at P(1, 2, 16).
(a) x = 1 + 112t, y = 2 − 16t, z = 16 + t
(b) x = 1 + 112t, y = 2 + 16t, z = 16 − t
(c) x = 1 − 112t, y = 2 + 16t, z = 16 + t
(d) x = 1 + 112t, y = 2 − 16t, z = 16 + t
5. Find an equation for the tangent
sin(2x) cos(3y) + 2 at P(π , π , 2).
plane
to
z =
(a) −2(x − π ) − 3(y − π ) − (z − 2) = 0
(b) −2(x − π ) − 3(y − π ) + (z − 2) = 0
(d) 2(x − π ) + (z − 2) = 0
6. For z = sin(2x) cos(3y) + 2, find the parametric normal line
to the surface at P(π , π , 2).
(a) x = π + t, y = π , z = 2 − t
(b) x = π + 2t, y = π , z = 2 − t
(c) x = π − 2t, y = π , z = 2 + t
(d) x = π − 2t, y = π , z = 2 − t
7. Find an equation for the tangent plane to 3x2 + 4y2 + z2 = 19
at P(1, 0, −4).
(a) −6(x − 1) − z + 4 = 0
Section 13.7
1. Find an equation for the tangent plane to z = 5x2 y at
P(1, 2, 20).
(a) 20(x − 1) + 5(y − 2) + 20 = z
(b) 20(x − 1) − 5(y − 2) − 20 = z
(c) 4(x − 1) + (y − 2) + 4 = z
(d) 20x + 5y − z = 0
2.
(d) 16 + 112(x − 1) + 64(y − 2) = 0
(c) −2(x − π ) − (z − 2) = 0
(b) j + 4k.
For z = 5x2 y,
(c) 16 + 112(x − 1) + 64(y − 2) = 1
find the parametric normal line to the surface at
P(1, 2, 20).
(a) x = 1 + 20t, y = 2 − 5t, z = 20 + t
(b) x = 1 + 20t, y = 2 + 5t, z = 20 − t
(c) x = 1 − 20t, y = 2 + 5t, z = 20 − t
(d) x = 1 + 20t, y = 2 + 5t, z = 20 + t
(b) 6(x − 1) − 8(z + 4) = 0
(c) −6(x − 1) − 8y − 8(z + 4) = 0
(d) 6x + 8y − 2z = 0
8. For 3x2 + 4y2 + z2 = 8, find the parametric normal line to the
surface at P(1, 1, 1).
(a) x = 1 + 6t, y = 1 + 8t, z = 1 + 2t
(b) x = 1 + 6t, y = 1 + 8t, z = 1 − 2t
(c) x = 1 − 6t, y = 1 − 8t, z = 1 + 2t
(d) x = 1 + 6t, y = 1 − 8t, z = 1 + 2t
9. Find an equation for the tangent plane to 3x2 y = −12 at
P(2, −1, −21).
(a) −12(x − 2) + 12(y + 1) + z − 21 = 0
(b) −12(x − 2) − 12(y + 1) + z − 21 = 0
303
(c) −12(x − 2) + 12(y + 1) + z + 21 = 0
(d) −12x + 12y + z = 0
3x2 y + z3
10. For
= −4, find the parametric normal line to the
surface at P(2, −1, 2).
(a) x = 2 + 12t, y = −1 + 12t, z = 2 − 12t
(b) x = 2 + 12t, y = −1 + 12t, z = 2 + 12t
(c) x = 2 + t, y = −1 + t, z = 2 − t
(d) x = 12t, y = 12t, z = 12t
(a) a relative maximum.
(b) a relative minimum.
(c) a saddle point.
(d) cannot be determined.
8. f (x, y) = 4xy − 8x. (0, 2) is
(a) a relative maximum.
(b) a relative minimum.
(c) a saddle point.
(d) cannot be determined.
9. f (x, y) = xy2 + x2 y. (0, 0) is
Section 13.8
(a) a relative maximum.
1. f (x, y) = 2xy + 4x + 2y − 8. Find a critical point.
(a) (−2, −1)
(b) a relative minimum.
(c) a saddle point.
(b) (−1, −2)
(d) cannot be determined.
(c) (2, 1)
10. f (x, y) = x2 + 4xy + y2 + 8x + 2y.
(d) (1, 2)
2. f (x, y) = 5x2 + 2y2 − 9. Find a critical point.
(a) (10, 4)
is
(a) a relative maximum.
(b) a relative minimum.
(d) cannot be determined.
(c) (−10, −4)
11. f (x, y) = x2 y2 − x2 . (0, 1) is
(d) (0, 0)
3. f (x, y) = 6x2 − 2y2 + 11. Find a critical point.
(a) (6, −2)
(a) a relative maximum.
(b) a relative minimum.
(c) a saddle point.
(b) (−6, 2)
(d) cannot be determined.
(c) (−12, −4)
12. f (x, y) = x2 + 2x + y2 . (−1, 0) is
(d) (0, 0)
4. f (x, y) = exy + 4. Find a critical point.
(a) a relative maximum.
(b) a relative minimum.
(a) (0, 0)
(c) a saddle point.
(b) (4, 4)
(d) cannot be determined.
(c) (−4, −4)
(d) none exist
5. Answer true or false. f (x, y) = ex + ey − 3. There is no critical point.
6. f (x, y) = x3 − 12x − 9y. (2, 0) is
(a) a relative maximum.
13. Answer true or false. If f (x, y) has two critical points, one
must be a relative maximum and the other must be a relative
minimum.
14. Answer true or false. Every function f (x, y) has a relative
maximum.
15. f (x, y) = e3xy . (0, 0) is
(b) a relative minimum.
(a) a relative maximum.
(c) a saddle point.
(b) a relative minimum.
(d) cannot be determined.
(c) a saddle point.
7. f (x,
2 7
,−
3 3
(c) a saddle point.
(b) (5, 2)
y) = x5 − 80x + 3y.
(2, 0) is
(d) cannot be determined.
304
Section 13.9
Chapter 13 Test
1. 4xy subject to 2x + 2y = 20 is maximized at which point?
1. f (x, y, z) = 2x2 + yz. Find f (1, 2, 1).
(a) (2, 2)
(a) 4
(b) (4, 4)
(b) 0
(c) (5, 5)
(c) 1
(d) (0, 0)
(d) 5
2. xy subject to 4x + 2y = 8 is maximized at which point?
(a) (1, 2)
(b) (2, 1)
2. Answer true or false. The graph of f (x, y) =
a hemisphere.
p
x2 + y2 + 4 is
3. Let f (x, y, z) = 5xyz. Find an equation of the level surface
passing through (2, 3, 1).
(c) (8, 8)
(d) (2, 4)
(a) 5xyz = 6
3x2 y
(b) 5xyz = 30
3. Answer true or false.
To maximize
subject
to 4x − 2xy = 10, ∇ f (x, y) = λ ∇g(x, y) can be written as
6xi + 3x2 j = 4λ i − 2λ j.
4. Answer true or false. There are no relative extrema of
f (x, y, z) = x2 + (y + 3)2 + (z − 3)2 subject to the constraint
x2 + y2 + z2 = 1.
5. Answer true or false. 3xyz subject to x2 y + z2 = 6 has an extrema at (0, 0, 0).
6. Answer true or false. x2 yz2 subject to x + y + z = 5 has an
extrema at (0, 0, 0).
7. Answer true or false. x2 yz2 subject to x + y + z = 5 has an
extrema at (2, 1, 2).
8. Answer true or false. x2 yz2 subject to x + y + z = 5 has an
extrema at (1, 2, 1).
9. Answer true or false. To find an extrema subject to a constraint it is always necessary to find λ .
10. Answer true or false.
To find an extrema for
f (x, y, z) = (x − 4)2 + (y − 4)2 + (z − 4)2 subject to
x4 + y4 + z4 = 1, ∇ f (x, y, z) = λ ∇g(x, y, z) gives
x − 4 = 2x3 λ , y − 4 = 2y3 λ , z − 4 = 2z3 λ .
11. Answer true or false.
To find an extrema for
f (x, y, z) = (x − 4)2 + (y − 4)2 + (z − 4)2 subject to
1
1
1
+ − = 0, ∇ f (x, y, z) = λ ∇g(x, y, z) gives
x2 y2 z2
x − 4 = −xλ , y − 4 = −yλ , z − 4 = zλ .
12. Answer true or false. f (x, y, z) = (x − 4)2 + (y − 4)2 + (z − 4)2
9
9
9
subject to 2 + 2 + 2 = 1 has an extrema at (0, 0, 0).
x
y
z
13. Answer true or false. x3 + 2x2 y + y3 has an extrema when
subjected to x3 + y3 = 5 at (1, 1).
14. Answer true or false. x3 + 2x2 y + y3 has an extrema when
subjected to x3 + y3 = 5 at (−1, −1).
15. Answer true or false. x3 + 2x2 y + y3 has an extrema when
subjected to x3 + y3 = 2 at (1, 1).
(c) 5xyz = 0
(d) 5xyz = 5
4.
lim
(x,y)→(1,2)
(x − y) =
(a) −1
(b) 1
(c) 3
(d) does not exist
5. Answer true or false.
5
does not exist.
2
x
+
y2
(x,y)→(0,0)
lim
6. Answer true or false. f (x, y, z) = yz ln |x| is continuous everywhere.
7. z = e5xy . Find
∂z
.
∂y
(a) 5xe5xy
(b) 5e5xy
(c) 5ye5xy
(d) 5xye5xy
8. Answer true q
or false.
If f (x, y) =
10y4
x6 + 4y5 , fy (x, y) = p
.
x6 + 4y5
9. xy subject to 2x + 4y = 16 is maximized at which point?
(a) (4, 2)
(b) (2, 1)
(c) (8, 4)
(d) (0, 0)
10. z = 4x − 2y; x = u2 , y = u − 3v. Find
∂z
.
∂v
305
(a) −6uv
(b) 10i + j
(c) 6v
(d) 5i + 2j
(b) −6
(c) 10xi + 2yj
(d) 6
11. A right triangle initially has legs of 1 m. If they are increasing, one by 6 m/s and the other by 8 m/s, how fast is the
hypotenuse increasing?
(a) 10 m/s
(b) 14 m/s
√
(c) 14 2 m/s
√
(d) 7 2 m/s
12. Find an equation for the tangent plane to z = 4x7 y2 at P =
(1, 2, 16).
(a) 28x6 y2 (x − 1) + 8x7 (y − 2) − (z − 16) = 0
(b) 112(x − 1) + 16(y − 2) − (z − 16) = 0
(c) 28x6 y2 (x − 1) + 8x7 y(y − 2) + (z − 16) = 0
(d) 112(x − 1) + 16(y − 2) + (z − 16) = 0
(c) 7i + 4j + k
(b) x = 1 + 112t, y = 2 + 16t, z = 16 + t
(d) 2i + j + 2k
(c) x = 1 − 112t, y = 2 − 16t, z = 16 − t
20. f (x, y, z) = 9x2 − 3y2 − 3. There is a critical point at
(d) x = 1 − 112t, y = 2 + 16t, z = 16 + t
15. z = 5x2 + y2 . Find ∇z.
(a) 5i + j
f (x, y) = x4 y7 ,
(a) 7i + 8j + 12k
(b) 7i + 8j + 3k
(a) x = 1 − 112t, y = 2 − 16t, z = 16 + t
If
17. Answer true or false.If f (x, y) = 5exy + 3x and a = 8i + 6j
is a vector, the directional derivative of f with respect to a at
4
(2, 3) is (15e6 + 3)i + 6e6 j.
5
18. Answer true or false. f (x, y, z) = |y| − exyz is differentiable
everywhere.
19. f (x, y, z) = 7x + 4y2 + z3 . Find the gradient at (2, 1, 2).
13. For z = 4x7 y2 , find the parametric normal lines to the surface
at P(1, 2, 16).
14. Answer true or false.
d f = 4x3 dx + 7y6 dy.
2
3
16. f (x, y) = xey ; P = (3, 0); u = √ i + √ j. Find Du f at
13
13
P.
11
(a) √
13
(b) 0
4
(c) √
13
6
(d) √
13
then
(a) (9, −3).
(b) (−18, 9).
(c) (18, −9).
(d) (0, 0).
306
Chapter 13: Answers to Sample Tests
Section 13.1
1. b
9. a
2. a
10. c
3. c
11. d
4. true
12. c
5. false
13. true
6. false
14. false
7. false
15. a
8. d
2. a
10. a
3. true
11. a
4. true
12. true
5. true
13. true
6. false
14. false
7. b
15. false
8. c
2. b
10. d
3. a
11. false
4. true
12. false
5. d
13. false
6. a
14. b
7. false
15. b
8. a
2. false
3. a
4. c
5. c
2. c
10. true
3. b
11. d
4. d
5. c
6. c
7. d
8. false
2. b
10. b
18. a
3. d
11. b
19. d
4. a
12. false
20. true
5. b
13. false
21. true
6. a
14. false
22. true
7. false
15. b
23. true
8. a
16. b
24. a
2. b
10. a
3. b
4. b
5. c
6. d
7. b
8. a
2. d
10. c
3. d
11. d
4. a
12. b
5. true
13. false
6. d
14. false
7. d
15. c
8. c
2. a
10. true
3. false
11. false
4. true
12. false
5. false
13. false
6. false
14. false
7. true
15. false
8. false
2. false
10. d
18. false
3. b
11. d
19. a
4. a
12. b
20. d
5. true
13. b
6. false
14. false
7. a
15. c
8. true
16. a
Section 13.2
1. a
9. b
Section 13.3
1. a
9. b
Section 13.4
1. true
Section 13.5
1. a
9. false
Section 13.6
1. a
9. c
17. c
25. a
Section 13.7
1. a
9. a
Section 13.8
1. b
9. d
Section 13.9
1. c
9. false
Chapter 13 Test
1. a
9. a
17. false
Chapter 14: Multiple Integrals
Summary: The previous chapter focused upon taking derivatives and limits of
multivariable functions. In this chapter, the integration of multivariable functions
is considered. This naturally leads to integrating functions with respect to two or
three variables. The integration of these types of functions over their domains is
described best using double and triple integrals. Double integrals are useful for
calculating the area of a two dimensional region or its center of mass. They are
also useful for calculating the surface area of a surface determined by a function
z = f (x, y). Triple integrals allow the mass of an object to be determined and
consequently the location of its center of mass.
In many problems, situations in two and three dimensions are not easily described
using Cartesian or rectangular coordinates. In some cases, problems may be more
easily considered if the coordinates are given as polar coordinates in two dimensions or by cylindrical coordinates or spherical coordinates in three dimensions.
Just as u-substitution played an important role in the development of integration in
one dimension, using a change of variables is important in both double and triple
integral problems. In particular, when switching from rectangular coordinates to
either polar, cylindrical or spherical coordinates, a change of variables is being
performed to switch from one set of coordinates to another.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Set up double integrals and evaluate double integrals (§14.1).
2. Write double integrals with nonconstant limits of integration (§14.2).
3. Set up double integrals over nonrectangular regions (using rectangular, §14.2,
or polar coordinates, §14.3).
4. Find the volume under a surface using double integrals (§14.1 − 14.3).
5. Switch the order of integration in double integrals (§14.2, 14.3).
6. Change between rectangular and polar coordinates in double integrals (§14.3).
7. Calculate area using double integrals (§14.2, 14.3).
8. Define surfaces parametrically (§14.4).
307
308
9. Express a parametric surface in vector form (§14.4).
10. Find tangent planes to parametric surfaces and surface area of a parametric
surface (§14.4).
11. Calculate volume under a surface using a triple integral (§14.5).
12. Determine appropriate limits of integration for triple integrals (§14.5).
13. Set up and evaluate triple integrals using both cylindrical and spherical coordinates (§14.6).
14. Convert between different coordinate systems for triple integrals (§14.6).
15. Find the Jacobian and perform general changes of variables in double and
triple integrals (§14.7).
16. Find the centroid of a lamina using double integrals (§14.8).
17. Find the centroid of a solid using triple integrals (§14.8).
14.1 Double Integrals
PURPOSE: To introduce double integrals and to explain how they
are evaluated and some of their properties.
net signed volume
region of integration
Double integrals provide a way to integrate beneath a surface such as z = f (x, y)
that is defined in terms of two variables. Single integrals found the net signed
area beneath a function. Similarly, double integrals find the net signed volume
beneath a function z = f (x, y). The limits of integration have a similar meaning
as for single integrals. In the single integral case, the limits of integration defined
the interval over which the integration would occur. Double integrals describe a
region in a plane over which the integration occurs.
To make the transition from using single integrals to double integrals requires an
understanding of two concepts. These two concepts are then repeatedly applied
throughout the remainder of this chapter (and again with triple integrals). The first
idea is that a double integral is essentially just a set of nested single integrals. That
is, a double integral can be viewed as one integral which is inside of another (this
is referred to as iterated integrals). The other concept is that the limits of integration define a two-dimensional region in the xy-plane over which the integration is
taking place.
IDEA: Two important ideas related to double and triple integrals are that (i) the
limits of integration define the domain over the integration and that (ii) multiple
integrals should be evaluated from the inside out.
evaluating double integrals
To evaluate double integrals requires that the inside integral be evaluated first.
The result of this integration becomes the integrand of the outside integral which
309
can then be evaluated. Each integral that is evaluated is only integrated with respect
to one variable indicated by the order of the differentials. For example,
RR
f (x, y) dxdy indicates that the inside integral is done with respect to x and then
the outer integral is done with respect to y.
IDEA: Evaluating a double integral is done in the same way as a single integral
except that two integrals will need to be evaluated.
One interesting property of double integrals is that the order
of integration may
RR
be changed. A double integral is typically represented as
f (x, y) dA where dA
can either equal dxdy (integrate with respect to x first) or dydx (integrate with
respect to y first).
order of integration
IDEA: The dA in a double integral indicates in which order the integration
should be done.
The order of integration does not affect the region, R, that is being integrated over.
However, if the order of integration is changed, care should be taken in rewriting
the limits of integration since they may change. Limits of integration are talked
about more in the next section when nonrectangular regions are considered.
CAUTION: Changing the order of integration involves changing the limits of
integration. This is not necessarily done by just switching the limits of integration.
Double integrals may be used to find the volume beneath a surface defined by
z = f (x, y). They may also be used to find the area of a region. To find the
RR area,
the function in the integrand, f (x, y), should just be set equal to one. So R 1 dA
will calculate the area of the region R. Notice that this is written for a generic dA.
This same process will hold true with nonrectangular regions (see §14.2) and in
polar coordinates that will be introduced later (see §14.3).
Checklist of Key Ideas:
volume under a surface
infinitesimal rectangle
Riemann sum of infinitesimal rectangles
double integral of f (x, y) over a region R
net signed volume
partial integration
iterated integration and iterated integrals
order of integration
properties of double integrals
subdividing a region R into smaller subregions
calculating volume and area
310
14.2 Double Integrals over Nonrectangular Regions
PURPOSE: To describe how double integrals should be set up
over a nonrectangular region and how to evaluate double integrals over a nonrectangular region.
limits of integration
The main topic of discussion in this section regards the limits of integration of
double integrals. The limits of integration describe the region over which the
double integral is being evaluated. So there will be two x limits of integration and
two y limits of integration. These four things will determine what the region R
looks like. Remember that changing the limits of integration will not change the
region. On the other hand, changing the order of integration does not mean that
the limits of integration may simply be just interchanged.
CAUTION: When changing the order of integration, do not simply switch the
limits of integration. The limits of integration need to be picked carefully so
that they accurately describe the region R.
One basic idea to grasp onto is that when an integral is finished being evaluated
over a fixed region, R, the result should be a scalar value that does not involve
variables such as x or y. This does not mean that limits of integration have to
always be constant values however. On the other hand, the outer most limits of
integration should be constant values.
IDEA: Type I regions are defined by curves at the top and bottom and vertical
lines at the left and right. Type II regions are defined by curves at the left and
right and horizontal lines at the top and bottom.
type I and II regions
type I → dA = dydx
tpye II → dA = dxdy
Two types of nonrectangular regions are considered in this text: type I and type
II regions. Type I regions are bounded by curves on the top and bottom. The inner
most integration should be set RR
up between the curves. So for Type I regions, the
integral should have the order R f (x, y) dydx. The dy indicates that integration
in the y direction between the curves at the top and the bottom will occur first. For
Type II regions, the region R will be bounded on the left and right by curves and by
lines on the top and bottom. Again, the double integral should be set up so that the
inner most integral will integrate between the curves. In this case,
this is in the x
RR
direction (from left to right) so the integrals should be set up like R f (x, y) dxdy.
Sometimes an integral may be simpler to write in Type I fashion compared to
Type II fashion or vice versa. For example, consider the following double integral
written for a type II region (curves on the left and right):
y=3 x=(11−y)/2
Z
Z
y=1
f (x, y) dxdy
x=y
This may also be written in a type I fashion although it will require more integrals:
y=x
x=3 Z
Z
x=1 y=1
f (x, y) dydx +
y=3
x=4 Z
Z
x=3 y=1
f (x, y) dydx +
x=5 y=11−2x
Z
Z
x=4
y=1
f (x, y) dydx
311
Incidentally, the region in this example is the trapezoid with corners at the points
(1, 1), (3, 3), (4, 3), and (5, 1) in the xy-plane.
IDEA: To change limits of integration, be sure that the new limits of integration
describe the same region.
CAUTION: Sometimes changing the limits of integration may require that more
than one double integral be evaluated. Also, generally the limits of integration
may not simply be interchanged.
Checklist of Key Ideas:
nonconstant limits of integration
type I nonrectangular region
type II nonrectangular region
changing order of integration with nonconstant limits of integration
finding area of a region R
14.3 Double Integrals in Polar Coordinates
PURPOSE: To explain how double integrals can be written in
either rectangular or polar coordinates and to explain how double
integrals with polar coordinates may be set up and evaluated.
In this section, double integrals using polar coordinates are described. Recall that
the relationship between rectangular coordinates and polar coordinates are give
by the following equations:
polar coordinates
see §10.2
x = r cos θ , y = r sin θ
and
r2 = x2 + y2 , tan θ = y/x
The same things that are important for double integrals in rectangular coordinates
are important for double integrals in polar coordinates. In particular, this means
that the limits of integration indicate the region that is being integrated over and
the integral should be evaluated from the inside out.
For polar coordinates, only simple polar regions will be considered. These are
regions that will resemble portions of wedges coming from a circle. The regions
will be bounded on the left and right by rays from the origin. The other two
bounds will be inner and outer curves such as r = g(θ ) and r = h(θ ). Since polar
regions of this type are the only ones that are considered here, order of integration
will
RR always be in the r direction first. So double polar integrals will be set up as
R f (r, θ ) rdrd θ .
simple polar regions
dA = rdrd θ
312
RR
Any double integral will have the format R f dA where f is a function of the
coordinate variables. The term dA, however, will vary depending upon whether
rectangular coordinates or polar coordinates are used. In rectangular coordinates
dA = dxdy or dA = dydx which represents an infinitesimal rectangle of area dA. In
polar coordinates, dA = rdrd θ . This is the area of an infinitesimal polar wedge.
A description of its derivation is given in the book. For the purposes of setting
up double integrals it may be helpful to have what dA should be represented as
written down on a notecard as problems are attempted. In this way, it will become
second nature to use the appropriate term for dA in the appropriate type of double
integral.
converting between rectangular and
polar coordinates
Conversion between rectangular and polar coordinates is a straightforward
process. Three things need to be done whenever a change is made from one coordinate system to another. The limits of integration need to be changed, the
integrand needs to be rewritten in terms of the new variables and the appropriate
form of dA needs to be used. It is important to remember that changing coordinate
systems does not alter the region R that is being integrated over.
Checklist of Key Ideas:
simple polar region
inner and outer radial boundaries
polar rectangle
infinitesimal polar rectangle and polar Riemann sums
polar double integrals
finding area using double polar integrals
converting between rectangular and polar double integrals
use of symmetry in double polar integrals
14.4 Parametric Surfaces; Surface Area
PURPOSE: To describe surfaces parametrically and to use double integrals to find the area of a parametric surface.
multivariable vector-valued functions
This section introduces two important concepts. The first is a multivariable function that is vector-valued and the second is the calculation of the area of a surface. Vector-valued functions have already been seen in the text (see Chapter 12).
For example, the vector form of a parametric equation of a line is actually a vectorvalued function. Each output of the function for a given input determines one of
the coordinates of a point on the line. Another example of a vector-valued function is the gradient function of a multivariable function. The gradient vector of a
multivariable function is the vector-value of the gradient function at a particular
313
point. Multivariable functions have also been considered (see Chapter 13). This
section makes use of both the vector-valued and multivariable ideas.
Multivariable functions that are vector-valued are treated in much the same way as
regular vector-valued functions. For example, the continuity and derivatives of a
vector-valued function are investigated by looking at the component functions of
the function. Since the components are each multivariable functions themselves,
they will each have appropriate partial derivatives as well. Multivariable vectorvalued functions are introduced here as a quick and simple way of representing a
surface parametrically.
The basic idea is to attach a grid of some sort to the surface. This is done by
parameterizing coordinates of any given point on the surface. For example, if
each point on the surface can be parameterized so that x, y, and z are all functions
of the parameters u and v, then the partial derivatives of the coordinates with
respect to the parameters will indicate how the coordinates will be changing along
the uv-grid. So if r(u, v) = hx(u, v), y(u, v), z(u, v)i then ∂ r/∂ u and ∂ r/∂ v will
indicate how the uv grid is traced onto the surface. The vector ∂ r/∂ v would
indicate how the surface is changing when u is held constant. In other words,
∂ r/∂ v will be tangent to the u-curves where u is constant. More importantly,
however, is the idea that a “square” on the grid can be defined by these two partial
derivative vectors. The area of one “square” would be given by the magnitude of
their cross product.
parameterizing a surface
This leads directly to the idea of how to calculate the surface area. First, parameterize the coordinates of a point on the surface in terms of the parameters u and
v. Then integrate the magnitude of the cross product of the two partial derivative
vectors, ∂ r/∂ u and ∂ r/∂ v, over the entire surface.
calculating area
IDEA: Surface area of z = f (x, y) can be found by parameterizing the coordinates as x = u, y = v, and z = f (u, v). Then the surface area would be found
by
area =
ZZ ∂ r × ∂ r dudv
∂v
R̃ ∂ u
where r = hu, v, f (u, v)i and R̃ is the space R from the xy-plane as it is represented in uv-space.
Tangent planes to a parametric surface can also be determined using the vectors
∂ r/∂ u and ∂ r/∂ v. Since these two vectors must be tangent to the surface at a
point, then they both must lie within any tangent plane at that point. So the cross
product of these two vectors will be normal to the surface and to any tangent plane.
This normal vector can be used to define the equation of the tangent plane. If this
vector is normalized, then it is called the principle unit normal vector.
Checklist of Key Ideas:
parameterizing a curve using one parameter
parameterizing a surface using two parameters
tangent planes
principle unit normal vector
314
constant parameter curves or parametric mesh lines
lines of latitude or longitude
surfaces of revolution represented parametrically
vector form of a parametric surface
vector-valued function of two variables
graph of a vector-valued function of two variables
continuity of a vector-valued function of two variables
partial derivatives of multivariable vector-valued functions
tangent plane to a parametric surface
principal unit normal vector
smooth parametric surface
surface area of parametric surfaces
surface area of z = f (x, y)
14.5 Triple Integrals
PURPOSE: To introduce triple integrals and to explain how they
may be set up and evaluated over both rectangular and nonrectangular regions.
The natural way to extend double integrals is to talk about triple integrals. Single integrals integrated along an interval in one dimension, and double integrals
are evaluated over a planar region in two dimensions so triple integrals are correspondingly evaluated over some three dimensional region.
RRR
Typically a triple integral may be written as R f (x, y, z) dV where dV represents
some infinitesimal box. In rectangular coordinates this box has a volume give by
(dx) × (dy) × (dz). Then in rectangular coordinates, dV = dxdydz. As is the case
with double integrals, this ordering of the differentials indicates the variable of
integration for each integral. Also like double integrals, the order of integration
may be changed. The same cautionary statement from double integrals about the
limits of integration applies here. Although the order of integration may change,
this does not change the region over which the integration is changing. It will
generally change the limits of integration. They may not be simply interchanged
in most cases.
CAUTION: When the order of integration is changed, care must be taken to
use the appropriate limits of integration. Generally, the limits of integration
may not simply be interchanged. Instead they will need to be redefined.
315
Evaluating triple integrals proceeds in the same fashion as for double integrals.
The integral is considered to be three nested integrals and the order of evaluation
should be from the inside to the outside. As mentioned above, the variable of integration for each integral is determined by the ordering of the differentials given by
dV = dxdydz. As is the case with single and double integrals, each integral is evaluated by treating all variables except the variable of integration as constants. This
means that the limits of integration for the innermost integral could be surfaces
described by the other two variables. Then the middle integral to be evaluated
may be between two curves that are defined in terms of the variable of the outer
integration.
evaluating triple integrals
It is again important to realize that the limits of integration are chosen so that they
will describe the region over which the integration will occur. The region over
which integration is occurring is independent of the order of integration. In other
words, although changing the order of integration will almost certainly alter the
limits of integration (except in some very simple cases), this does not mean that
the region of integration is changing.
Double integrals could be used to calculate the area of the region of integration by
allowing the integrand to be equal to one. When this is done with triple integrals,
the result is to find the volume of the region of integration.
Checklist of Key Ideas:
finite solid
infinitesimal subboxes
triple integral of f (x, y, z) over a region G
three dimensional region G
subdividing the region G
projection of G onto the xy-plane
simple xy-, xz- or yz-solid
nonconstant limits of integration g1 (x, y) and g2 (x, y)
calculating volume of the region G
changing order of integration
14.6 Triple Integrals in Cylindrical and Spherical
Coordinates
PURPOSE: To set up triple integrals using both cylindrical and
spherical coordinates and to explain how to convert between the
different coordinate systems.
triple integrals and volume
316
cylindrical coordinates
spherical coordinates
see §11.8
In two dimensions, double integrals in rectangular coordinates may be converted
to double integrals in polar coordinates. This is often done to make a calculation
more straightforward particularly if a region is easier to describe in one coordinate system than another. In three dimensions, rectangular coordinates may be
changed to either cylindrical or spherical coordinates. The impact this has on
triple integrals is similar to the effect that polar coordinates may have on double
integrals. The bottom line, however, is that regardless of what coordinate system
is used, the region G in three dimensions is not being changed and integrals are
still evaluated by calculating the inner integrals first and working outward.
IDEA: Regardless of what coordinate system is used, the region of integration
will not change. On the other hand, the limits of integration describing the
region will most likely change when different coordinate systems are used.
The biggest differences when using cylindrical and spherical coordinates are that
the limits of integration must be chosen correctly, and the integrand f (x, y, z) and
dV must be expressed in terms of the appropriate coordinate
system. That is to
RRR
say, all triple integrals will have the general form of G f dV . How G, f , and dV
are represented in a particular coordinate system is the main issue.
IDEA: Refer to §11.8 for the appropriate formulas for converting variables between coordinate systems. Here are appropriate forms for dV in cylindrical
and spherical coordinates.
cylindrical → dV = r dz dr d θ
spherical → dV = ρ 2 sin φ d ρ d φ d θ
IDEA: The order of integration may also be switched in cylindrical and spherical coordinates. In each case, the appropriate change of limits of integration need to be made. However, the factor of r in cylindrical coordinates and
ρ 2 sin φ in spherical coordinates must remain with the integrand regardless of
the order of integration.
Checklist of Key Ideas:
cylindrical wedges
Riemann sum of infinitesimal cylindrical wedges
triple integral using cylindrical coordinates
dV in cylindrical coordinates
projection of the solid G onto the xy-plane
converting between rectangular and cylindrical coordinates
changing limits of integration when changing coordinate systems
spherical wedge
Riemann sum of infinitesimal spherical wedges
317
triple integral using spherical coordinates
dV in spherical coordinates
converting between rectangular and spherical coordinates
14.7 Change of Variables in Multiple Integrals;
Jacobians
PURPOSE: To discuss a change of variables in either double or
triple integrals and to explain how the change of variables procedure relates to the Jacobian of a function.
Change of variables in double and triple integrals are an extension of the idea
of change of variables or u-substitution in single integrals. Recall that when a
u-substitution was performed where u = g(x), then the new differential in the
integral, du, would replace the term g′ (x)dx. This process is described in greater
detail here in this section.
Now in multiple integrals, the integrands are very often multivariable functions so
finding “g′ (x)” is not so simple. The concept that is introduced here to allow a
change of variables is called the Jacobian. It is related to this factor of g′ (x) from
the single variable case.
The first idea that is mentioned is that of transformation. A transformation T is
simply a rule for converting from one coordinate system to the another. So if
a region is represented in the uv-plane, then the transformation (x, y) = T (u, v)
would indicate how this region would be represented in the xy-plane by taking
each point in the uv-plane and transforming it to a corresponding point in the
xy-plane.
Two important questions need to be considered. First, if x = g(u, v) and y = h(u, v)
then how can u and v be expressed in terms of x and y? Second, if an infinitesimal area such as dA is described in the xy-plane, what does it look like in the
uv-plane? The first issue is found by solving the parametric equations for x and y
for expressions for u and v.
CAUTION: Generally dxdy 6= dudv. To do a change of variables in either a
double or triple integral requires the use of the Jacobian.
The infinitesimal area dA is related by the Jacobian. In general, it will not be true
that dudv = dxdy. Instead, the Jacobian will act as a conversion fact between the
two coordinate systems.
IDEA:
In
two
dimensions
dxdy = |J(u, v)|dudv, where
∂x
J(u, v) = ∂ u
∂y
∂u
∂x
∂v
∂y
∂v
.
if
x = g(u, v)
and
y = h(u, v)
then
expressing u and v
in terms of x and y
318
Here is an example of a simple change of variables. Consider the double integral
below which will calculate the volume under the surface z = x + y and above a
trapezoidal region with corners in the xy-plane at (1, 1), (2, 3), (5, 3), and (4, 1).
y=3 x=(y+7)/2
Z
Z
(x + y) dxdy
y=1 x=(y+1)/2
To change this into uv-space requires that x and y be parameterized in terms of u
and v. Suppose that it is given that x = (u + v)/2 and y = v. Then the Jacobian
would be
∂x ∂x ∂ u ∂ v 1/2 1/2 J(u, v) = = 1/2.
=
∂y ∂y 0
1 ∂u ∂v
Also to change the limits of integration, requires them to be written in terms
of u and v. So y = 1 = v and y = 3 = v remain unchanged. On the other hand
x = (y + 7)/2 becomes (u + v)/2 = (v + 7)/2. Solving for u we find u = 7. Similarly, the lower limit of integration becomes u = 1. Finally the surface becomes
z = x + y = (u + v)/2 + v = (u + 3v)/2. So the new integral is
Z 3Z 7
(u + 3v)
1
· dudv = 30.
·
2
2
1 1
Checklist of Key Ideas:
change of variables in a single integral
the Jacobian
transformation from the xy-plane to the uv-plane
the transformation T (u, v) from (u, v) to (x, y) coordinates
the image of (u, v) in the xy-plane
the image of S under T
the inverse transformation T −1
constant u and v curves
the Jacobian of T (u, v)
“secant vectors” and tangent vectors
change of variables for double integrals
the transformation T (u, v, w) from (u, v, w) to (x, y, z) coordinates
change of variables for triple integrals
using change of variables as a guide for transforming between rectangular,
cylindrical and spherical coordinates
319
14.8 Centroid, Center of Gravity, Theorem of
Pappus
PURPOSE: To use double and triple integrals to find the center
of gravity for laminas and solids.
The focus of this section is to find the center of gravity (sometimes called the
center of mass) of two and three dimensional regions. The ideas all hinge upon
the units of the integrals that will be used. In finding centers of mass, the idea is
that a sum of moments divided by a sum of masses will yield a center of mass.
The term centroid simply refers to a center of gravity when the density function
δ is constant. These types of regions are called homogeneous.
center of gravity
center of mass
centroid
constant density
homogeneous region
The region that is being considered is essentially partitioned into much smaller
regions of size dA and dV for two and three dimensions respectively. Each of
these smaller portions has some mass given by δ (x, y)dA or δ (x, y, z)dV . In two
dimensions, the moments are defined by distance from a particular axis. So the
moment of one of these regions from the y-axis, My , would be xδ (x, y)dA and the
moment of one of these regions from the x-axis, Mx , would be yδ (x, y). If these
moments are divided by the mass then they give the corresponding x or y distance.
This is the main idea behind center of mass.
IDEA: The center of mass in two dimensions is described by the point (x̄, ȳ)
where
x̄ =
My
M
and
Mx
.
M
ȳ =
In the formulas above, My =
“center of mass equals
moment divided by mass”
RR
R xδ
dA, Mx =
RR
R yδ
dA and M =
RR
Rδ
dA.
It may be confusing to use the formulas above since the subscripts do not seem
to match the value being calculated. Instead, use the integrands to remember how
the calculation is done. For example, x̄ =“(xδ (x, y)dA)/(δ (x, y)dA)” which would
yield an x-value. The units that are being used should also serve to confirm which
integrals are to appear where in the formulas.
CAUTION: When calculating x̄ and ȳ, the integrals Mx , My , and M need to be
calculated separately. The integrands may not be combined.
The center of mass in three dimensions is very similar to the center of mass in
two dimensions. The only significant difference is that the moments are calculated
using distances from coordinate planes. So instead of Mx and My there are now
Mxy , Mxz and Myz . The term Mxy is the moment with respect to the xy-plane.
Again the subscripts may be confusing. The key here is the distance involved. For
Mxy , the distance to the xy-plane from a point is z (the distance will always be the
variable not found in the subscript).
center of mass in 3-space
320
IDEA: The center of mass in three dimensions (x̄, ȳ, z̄) is found by
x̄ =
Theorem of Pappus
Myz
,
M
whereRRR Mxy =
M = G δ dV .
ȳ =
RRR
G zδ
Mxz
,
M
dV ,
z̄ =
and
Mxz =
RRR
G yδ
Mxy
M
dV ,
Myz =
RRR
G xδ
dV ,
and
The Theorem of Pappus provides an interesting application of the center of mass
of a two dimensional region. If this region R with a center of mass located at (x̄, ȳ)
is rotated around a particular axis to generate a solid, then the volume will be
the area of the region times the circumference of the circle traveled by the center
of mass during the revolution. When rotated about the x-axis, for example, the
distance traveled by the center of mass would be 2π ȳ.
Checklist of Key Ideas:
lamina (planar region)
homogeneous or nonhomogeneous
density function δ (x, y) or δ (x, y, z)
mass of a lamina or solid
moment about the line x = a or y = c
lever arm
equilibrium of a system of masses
first moment about the x-, y-, or z-axis
moment about the xy-, xz-, or yz-planes
centroid of a planar region (a lamina) R
centroid of a solid region G
Theorem of Pappus
321
Chapter 14 Sample Tests
Section 14.1
1.
Z 1Z 3
0
0
(x + 4) dx dy =
33
2
27
(b)
2
9
(c)
2
11
(d)
2
(a)
2.
Z 1Z 3
0
0
4, −1 ≤ y ≤ 2} is
(c) 0
(d) 2
ex dx dy =
(a) e2 − 1
(b) e2 − 2
(c) 2e
(d) e2
5.
(a) 1
(c) 9
Z πZ π
0
π /2
cos x dx dy =
11. Answer true or false. The average value of the function
f (x, y) = sin x cos y over the rectangle [0, π ] × [0, 2π ] is
Z Z
1 π 2π
sin x cos y dy dx.
π 0 0
12. Answer true or false. The average value of the function f (x, y) = x2 y3 over the rectangle [0, 5] × [0, 2] is
Z Z
1 5 2 2 3
x y dy dx.
10 0 0
13. Answer true or false. The volume of the solid bounded by
z = x3 y and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
Z 1 Z 2
x3 y dy dx.
Z 1 Z 2
sin x cos y dy dx.
Z 1 Z 2
e3xy dy dx.
−1 −1
(a) π
14. Answer true or false. The volume of the solid bounded by
z = sin x cos y and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
(b) −π
π
(c)
2
−π
(d)
2
6. Evaluate
0
10. Find the volume of the solid bounded by z = 10 − 4x − 2y and
the rectangle R = [0, 2] × [0, 1].
(b) 10
0
x2 sin y dx dy.
1
(c) 20
(b) 6
0
Z 3 Z 2R
x2 sin y dA; R = {(x, y) : 0 ≤ x ≤
(d) 25
(d) 36
4.
ZZ
5xy3 dx dy.
(b) 12
(a) 5
Z 1Z 2
−2 −1
(a) 10
dx dy =
0
5xy3 dA; R = {(x, y) : −2 ≤ x ≤
9. Find the volume of the solid bounded by z = −2x − 2y and
the rectangle R = [0, 1] × [0, 3].
33
2
27
(b)
2
21
(c)
2
11
(d)
2
0
2, 1 ≤ y ≤ 3} is
ZZ
Z 4 ZR 2
8. Answer true or false.
(a)
3.
(b) 2
35
(c)
3
(d) 7
7. Answer true or false.
(x + 4) dy dx =
Z 2Z 3
(a) 4
−1 −1
ZZ
R
2xy2 dA; R = {(x, y) : −1 ≤ x ≤ 2, 1 ≤ y ≤ 2}.
15. Answer true or false. The volume of the solid bounded by
z = e3xy and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
−1 −1
322
Section 14.2
(c)
28
15
Z 2Z x
(d)
25 23
+
10
3
1.
0
0
xy dy dx =
(a) 1
8. Find the area of the region of the xy-plane that is enclosed by
y = x and y = x2 , for 0 ≤ x ≤ 2.
(b) 2
1
3
2
(b)
3
(c) 1
1
(d)
6
(c) 3
(a)
(d) 4
2.
Z π /2 Z cos x
0
0
dy dx =
(a) 1
(b) 0
9. Find the area of the the region in the xy-plane that is enclosed
by y = x and y = x2 , for 1 ≤ x ≤ 3.
(c) −1
(d) π
3.
Z 1Z x
0
0
(a) 4
14
(b)
3
16
(c)
3
(d) 6
ey dy dx =
(a) e − 2
(b) e + 1
(c) −e
(d) e
4.
10.
Z 1Z xp
0
1
x2 + 1 dy dx =
0
(b)
(c)
(d)
(b) 0.042
(c) 0.419
(d) 0.423
11. Answer true or false.
12.
ZZ
x4 dA, where R is the re-
R
gion bounded by y = x + 4, y = 2x, and x = 16, is
16
x4 dy dx.
x+4
6. Answer true or false.
ZZ
xy dA, where R is the region
R
bounded by y = x, y = 0, and x = 4, is
7.
Z 2 Z x2
0
y dy dx =
x
25
10
26
(b)
15
(a)
Z 1 Z x2
0
5. Answer true or false.
Z 28 Z 2x
(xy − 4) dy dx ≈
x
(a) 0.412
2 3/2
(2 + 1)
3
23/2 + 1
3
2 3/2
(2 − 1)
3
23/2 − 1
3
(a)
Z 2 Z x2
Z 4Z x
0
0
xy dy dx.
Z 0 Z x
−1 0
2
dy dx = − .
3
x
dy dx =
(a) −1
(b) 1
1
(c)
2
(d) −
1
2
13. Z
Answer
true or false.
Z 2
2
x
f (x, y) dy dx =
0
0
14. Z
Answer
true or false.
Z 2
3
0
x
0
f (x, y) dy dx =
Z 4 Z √y
0
0
Z 3 Z y2
0
0
15. Z
Answer
true or false.
Z 2Z
2 Z ln x
f (x, y) dy dx =
1
1
1
1
f (x, y) dx dy.
f (x, y) dx dy.
ey
f (x, y) dx dy.
323
Section 14.3
1.
0
Z π /2 Z 0
r cos θ dr d θ =
sin θ
0
(a) −
8. Answer true or false.
9. Answer true or false.
1
6
1
6
π
(c)
2
2.
r3 dr d θ ≈
−π /2 0
(b) −0.29
0
r2 dr d θ =
(a) 1
(b) −1
(c) 0
(d) 2
Z π Z cos 2θ
0
(c) 0
(d) 2
Z π /6 Z sin 3θ
0
dr d θ =
(a) 3
1
(b)
3
(c) −3
1
(d) −
3
6. Answer true or false.
Z 1 Z cos θ
0
0
0
0
Z 1Z x
0
0
(x2 + y2 ) dy dx =
dr d θ .
7. Answer true or false.
Z πZ 1
r dr d θ .
(a)
(b) −1
0
0
25π
3
25π
(b)
6
250π
(c)
3
125π
(d)
3
13. Find the area enclosed by the three-petaled rose r = 2 sin 3θ .
π
(a)
4
π
(b)
2
π
(c)
8
(d) π
dr d θ =
(a) 1
5.
0
r dr d θ .
(a)
(d) 0.33
Z π Z cos θ
0
Z πZ 2
0
25π
3
25π
(b)
6
250π
(c)
3
125π
(d)
3
12. Find the volume of the solid formed by the left hemisphere
x2 + y2 + z2 = 25.
(c) 0.29
4.
−1 0
dy dx =
0
8π
3
16π
(b)
3
32π
(c)
3
2π
(d)
3
11. Find the volume of the solid formed by the left hemisphere
r2 + z2 = 25.
π
2
Z π /2 Z sin θ
0
Z 0 Z √4−x2
Z π /2 Z 1
(a)
(a) −0.33
3.
0
dy dx =
10. Find the volume of the solid formed by the left hemisphere
r2 + z2 = 4.
(b)
(d) −
Z 1 Z √1−x2
Z 1 Z √1−x2
0
3
rer dr d θ .
0
14. Find the volume of the region enclosed by x2 + y2 = 4 and
x2 + z2 = 4 that is above the xy−plane.
8π
3
16π
(b)
3
(c) 8π
(a)
ex
2
+y2
dy dx =
324
(d) 4π
8. Answer true or false. To find the portion of the surface
z = x + y2 that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 3,
15. Find the region inside the circle r = 25 sin θ .
Z 2Z 3
evaluate
(a) 25
(c) 25π
(d) 5π
1
Section 14.4
1. The surface expressed parametrically by x = r cos θ , y =
r sin θ , z = 9 − r2 is
(a) a sphere.
(c) a paraboloid.
(d) a cone.
2. The surface√expressed parametrically by x = r cos θ , y =
r sin θ , z = 9 − r2 is
(a) a sphere.
2
1
11. Answer true or false. To find the portion of the surface
z = 5xy + 3 that lies above the rectangle 1 ≤ x ≤ 3, 2 ≤ y ≤ 5,
Z 5Z 3q
25x2 + 25y2 + 1 dx dy.
evaluate
0
(d) a cone.
3. Answer true or false. A parametric representation of the surface z + x2 + y2 = 5 in terms of the parameters u = x and v = y
is x = u, y = v, z = 5 − u2 − v2 .
4. Answer true or false.
The parametric equations for
x2 + y2 = 16 from the plane z = 1 to the plane z = 2 are
x = 4 cos v, y = 4 sin v, z = u with 0 ≤ v ≤ 2π and 1 ≤ u ≤ 2.
5. Answer true or false. Parametric equations for x2 + z2 = 25
from y = 0 to y = 1 are x = 5 cos u, y = v, z = 5 sin u with
0 ≤ u ≤ 2π and 0 ≤ v ≤ 1.
6. The cylindrical parameterization of z = xex
2
+y2
Z 4Z 2
2x dx dy.
0
14. Answer true or false. To find the portion of the surface
z = x2 − 2y that lies above the rectangle 1 ≤ x ≤ 2, 0 ≤ y ≤ 1,
Z 1Z 2p
4x2 − 1 dx dy.
evaluate
1
0
15. Answer true or false. To find the portion of the surface
z = x3 + y3 that lies above the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ 3,
Z 3Z 1q
evaluate
9x4 + 9y4 + 1 dx dy.
0
0
Section 14.5
1.
Z 1Z 2Z 3
0
is
0
0
x2 yz dx dy dz =
(a) 18
(a) x = r cos θ , y = r sin θ , z = er .
(b) 9
(b) x = r sin θ , y = r cos θ , z = er .
(c) 27
z = rer cos θ .
(d) x = r sin θ , y = r cos θ , z = rer cos θ .
0
13. Answer true or false. To find the portion of the surface
z = x2 − y that lies above the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 4,
0
(c) a paraboloid.
(d) 6
2.
Z 2 Z π /2 Z π /2
0
7. The equation of the tangent plane to x = u, y = v, z = u + v2
where u = 0 and v = 1 is
0
0
sin x sin y dx dy dz =
(a) −2
(b) 2
(a) x + 2(y − 2) + z − 1 = 0.
(c) 1
(b) x + 2(y − 2) − z + 1 = 0.
(d) x + 2y − 2 − z + 1 = 0.
1
12. Answer true or false.
To find the portion of the
surface z = x2 − x + y2 that lies above the rectangle
Z 3Z 4q
4x2 + 4y2 dx dy.
0 ≤ x ≤ 4, 0 ≤ y ≤ 3, evaluate
evaluate
(b) an ellipsoid.
0
10. Answer true or false.
To find the portion of the surface z = x2 + 3y2 + 4 that lies
above the rectangle 1 ≤ x ≤ 2, 2 ≤ y ≤ 4, evaluate
Z 4Z 2q
4x2 + 36y2 + 1 dx dy.
2
(b) an ellipsoid.
(c) x + 2y − 2 + z + 1 = 0.
(x + y2 ) dy dx.
9. Answer true or false.
To find the portion of the
surface z = 3x2 + 4y2 that lies above the rectangle
Z 3Z 2q
0 ≤ x ≤ 2, 1 ≤ y ≤ 3, evaluate
6x2 + 8y2 dx dy.
(b) 5
(c) x = r cos θ , y = r sin θ ,
0
0
(d) −1
3.
Z 2 Z z2 Z y
0
0
0
y dx dy dz =
325
(a) 4
128
(b)
21
(c) 8
8
(d)
5
4.
9.
0
0
0
(d) 4
3 dz dy dx =
10.
(c) 6
(d) 5
11.
x dx dy dz =
0
12. Answer true or false.
0
z3 dx dy dz =
13. Answer true or false.
14. Answer true or false.
1
3
0
Z 5Z zZ y
0
0
0
0
1.
Z π /2 Z π /2 Z 1
0
0
0
(c) 0
(d) 1
2.
0
dx dy dz = 25.
Z 4 Z z2 Z y3
0
0
dx dy dz = 45 .
(a) 2
(a)
0
8
3
(c) 6.095
8
3
(c) 2π 3
(d) 5
(d) π 3
(b) 6.082
0
Z π /2 Z π /2 Z 2
0
z3 y dx dy dz ≈
0
ρ 4 sin φ cos θ d ρ d φ d θ =
(a) 1
1
(b)
5
(c) −1
1
(d) −
5
sin y dx dy dz =
(b) 4
Z 2Z zZ y
0
dz dy dx = x2 .
Section 14.6
(a) 2
0
Z 5 Z x Z x+y
0
Z 5 Z π /2 Z cos y
0
sin x dx dy dz =
(d) 5
2
3
4
(d)
5
8.
−z2 0
(a) 8
14
(b)
3
(c) 6
(c)
1
Z 2 Z 0 Z 3π
1
5
1
(a)
3
7.
dx dy dz =
(b) 7
Z 1 Z zZ 2
(b) −
0
0
(a) 8
1
7
2
(b)
7
1
(c)
5
(d) 0
−1 0
Z 2 Z z2 Z 3
1
(a)
6.
yz dx dy dz =
(c) 2
Z 1 Z zZ y
−1 0
0
(b) 0
(a) 1
1
(b)
12
1
(c)
3
3
(d)
20
5.
0
(a) 0.29
Z 1 Z x Z y3
0
Z 1 Z z Z √y2 +5
(b) −
0
ρ 2 sin φ cos θ d ρ d φ d θ =
326
3. Answer true
√ or false.
36−r2
Z 2π Z π Z
0
4.
0
2r dz dr d θ =
0
Z 2π Z π Z 2
0
0
−2
√
12. Answer true or false.
16π 2 − 24π
3
0
13.
(a) 6π
(b) 4
(b) 9π
(c) −4
(c) 3
Z π Z π /2 Z 3
−π /2 1
14.
sin φ d ρ d θ d φ =
−2 0
δ (r, θ , z) dr d θ dz, where δ (r, θ , z) = r, is the
0
0
0
δ (r, θ , z) dr d θ dz, where δ (r, θ , z) = rz, is the
(a) 3π
9π
(b)
2
3
(c)
2
3π
(d)
2
ρ 3 sin φ cos θ d φ d θ d ρ =
15.
(a) 16
Z 1 Z 2π Z 3
0
0
0
δ (r, θ , z) dr d θ dz, where δ (r, θ , z) = z2 , is the
following.
(b) 0
(a) 6π
27π
(b)
2
(c) 3
(c) 8
(d) 4
Z 2π Z 6 Z 2
3
sin θ d ρ d φ d θ = 0
following.
(c) 4π
3π
(d)
2
Z 2 Z π /2 Z 2π
Z 1 Z 2π Z 3
0
(b) 3π
0
0
0
(d) 3π
(a) π
7.
0
0
following.
(a) 0
0
6.
Z 1 Z 2π Z 3
0
sin φ cos θ ρ d ρ d θ d φ =
(d) 6
5.
Z 2π Z 1−sin2 θ Z 1
1
dz dr d θ =
(d) 2π
(a) 6π
(b) 2π
Section 14.7
(c) 4π
(d) π
8.
Find the center of gravity of the sphere x2 +y2 +z2
= 4 where
δ (x, y, z) = 6x2 y2 z2 .
(a) (2, 2, 2)
1. Find
∂ (x, y)
, if x = u + 2v and y = 3u + v.
∂ (u, v)
(a) 5
(b) −5
(b) (4, 4, 4)
(c) 7
(c) (1, 1, 1)
(d) −7
(d) (0, 0, 0)
9. Answer true orpfalse. The center of
p gravity of the solid enclosed by z = x2 + y2 and z = − x2 + y2 , if the density is
δ (x, y, z) = x2 + y2 + z2 , is at the origin.
10. Answer true or false. The center of gravity of the solid enclosed by x2 + y2 = 1 and 2y2 + z2 = 1 is at the origin if
δ (x, y, z) is constant.
11. Answer true or false. The center of gravity of the solid enclosed by x2 + y2 = 1 and 3y2 + z2 = 1 is at the origin if
δ (x, y, z) = x2 + 1.
2. Find
∂ (x, y)
, if x = u2 and y = u − v.
∂ (u, v)
(a) 2u + 1
(b) 2u
(c) −2u
(d) −2u − 1
3. Find the Jacobian if x = eu and y = ev .
(a) 0
(b) euv
327
Section 14.8
(c) eu−v
(d) eu+v
4. Find the Jacobian if u = ex and v = yex .
ln v − v
(a)
u
ln v + v
(b)
u
1
(c) 2
u
1
(d) − 2
u
5. Find the Jacobian if x = 3u + w, y = vw, and z = u2 v.
3u2 v + 2uvw
3u2 v + 2uw2
6(u2 v + uw2 )
−6(u2 v + uw2 )
(a)
(b)
(c)
(d)
∂ (x, y) v u =
.
If x = uv, y = u + 2v, then
∂ (u, v) 1 2 8. Answer true or false.
∂ (x, y) 2u 1 2
2
If x = u , y = v , then
=
.
∂ (u, v) 1 2v 9. Answer true or false.
If x = u + 2v+ 3w, y = 7 + uv + 4v, z = eu + vw, then
1 2v 3 ∂ (x, y, z) 4 u .
= 7
∂ (u, v, w) u
e
w v 10. Answer true or
If x = eu , y = ev , z = ew , then
false.
eu 0
0 ∂ (x, y, z) v
0 .
= 0 e
∂ (u, v, w) 0 0 ew 0
0
(b) 18 kg
(c) 8 kg
(d) 10 kg
(a) 2
7. Answer true or false.
Z 1Z 1
(a) 14 kg
2. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its mass.
y
6. Find the Jacobian if u = x, v = , and w = x + z.
x
(a) −u
(b) −uvw
(c) u
(d) uvw
11. Answer true or false.
1. A uniform beam 10 m in length is supported at its center by a
fulcrum. A mass of 10 kg is placed at the left end, a mass of
4 kg is placed on the beam 4 m from the left end, and a third
mass is placed 2 m from the right end. What mass should the
third mass be to achieve equilibrium?
If u = x + 2y and v = 3x + v,
ex+2y e3x+y dx dy =
Z 3Z 4
0
eu ev du dv.
0
12. Answer true or false.
If u = 4x + y and v = x2 y,
Z 10 Z 4
Z 2Z 2
u
4x + y
dx dy =
dv du.
2
x
y
1 v
5
1 1
13. Answer true or false. If u = x + y and v = 2x − y,
Z 2Z 2
Z 4Z 2
(x + y)2
u2
dy dx =
− dv du.
v
1 1 2x − y
2 1
14. Answer true or false. If x = uvw, y = eu , and z = 2u, the
Jacobian is 0.
15. Answer true or false. If x = uvw, y = u − v2 w, and z = u, the
Jacobian has no dependence on v nor on w.
(b) 4
(c) 1
(d) 8
3. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its center of mass.
8 16
,
(a)
5 5
8 8
(b)
,
5 5
16 16
,
(c)
5 5
16 16
,
(d)
15 15
4. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its mass.
(a) 4
28
3
20
(c)
5
(d) 2
(b)
5. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its center of mass.
1 1
,
(a)
3 3
6 8
(b)
,
7 5
328
(c)
(d)
8 6
,
5 7
224
,8
15
6. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its moment of inertia about the
x−axis.
224
9
(b) 8
224
(c)
15
416
(d)
45
(a)
7. A lamina with density δ (x, y) = x2 +2y2 is bounded by x = y,
x = 0, y = 0, and y = 2. Find its moment of inertia about the
y−axis.
224
9
(b) 8
224
(c)
15
416
(d)
45
11. Answer true or false. The centroid given by z =
0.
12. The centroid of a rectangular solid in the first octant with vertices (0, 0, 0), (0, 0, 1), and (1, 1, 1) is the following.
1 1 1
(a)
, ,
2 2 2
(b) (1, 1, 1)
1 1 1
, ,
(c)
3 3 3
1 1 1
, ,
(d)
4 4 4
13. The centroid of a rectangular solid in the first octant with vertices (0, 0, 0), (0, 0, 2), and (2, 2, 2) is the following.
1 1 1
(a)
, ,
2 2 2
(b) (0, 1, 2)
(c) (1, 1, 1)
2 2 2
(d)
, ,
3 3 3
(a)
14. The centroid of the solid given by (x − 2)2 + y2 + (z + 3)2 = 9
is the following.
(a) (2, 0, 3)
(b) (−2, 0, 3)
(c) (0, 0, 0)
8. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its moment of inertia about the
x−axis.
(a)
(b)
(c)
(d)
16
5
32
5
8
3
16
3
9. A lamina with density δ (x, y) = xy is bounded by x = 2,
x = 0, y = x, and y = 0. Find its moment of inertia about the
y−axis.
16
5
32
(b)
5
8
(c)
3
16
(d)
3
(d) (2, 0, −3)
15. The centroid of the solid given by
1 is the following.
(x−3)2
4
+
(y−5)2
16
+
(a) (−3, −5, 2)
(b) (0, 0, 0)
(c) (2, 4, 3)
(d) (3, 5, −2)
Chapter 14 Test
1.
Z 4Z 2
0
(a)
10. Answer true or false. The moment of inertia about x = a,
where a is the x−coordinate of the center of mass, is 0.
p
x2 + y2 is
0
dx dy =
(a) 6
(b) 8
(c) 0
(d) 20
2.
ZZ
R
2yx2 dA; R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2} =
(a)
4
3
(z+2)2
9
=
329
8
3
(c) 0
1
(d)
3
3. Answer true or false. The volume of the solid bounded by
z = x7 y3 and R = {(x, y) : −1 ≤ x ≤ 1, −1 ≤ y ≤ 2} is
(b)
Z 1 Z 2
x7 y3 dy dx.
−1 −1
4. Answer true or false. The average value of the function f (x, y) = x3 y4 over the rectangle [0, 2] × [0, 3] is
Z Z
1 2 3 3 4
x y dy dx.
6 0 0
5.
Z π /2 Z cos x
0
0
dy dx =
(a) 1
(b) 0
(c) −1
(d) π
4π
3
8π
(b)
3
2π
(c)
3
16π
(d)
3
11. Find the area enclosed by one petal of the three-petaled rose
r = 2 sin 3θ .
π
(a)
12
π
(b)
6
π
(c)
24
π
(d)
3
12. The surface expressed
parametrically by x = r cos θ ,
√
y = r sin θ , and z = 16 − r2 is a(n)
(a)
(a) sphere.
6. Answer true or false.
ZZ
(b) ellipsoid.
sin x dA, where R is the
(c) paraboloid.
R
region bounded by y = x + 4, y = x, and x = 16 is
Z 28 Z 2x
16
7.
(a) x = r cos θ , y = r sin θ , z = er .
y dy dx =
(b) x = r sin θ , y = r cos θ , z = er .
35
(c) x = r cos θ , y = r sin θ , z = er cos θ .
10
35
(b)
5
35
−9
(c)
10
35
(d)
+9
10
(d) x = r sin θ , y = r cos θ , z = er cos θ .
(a)
8.
0
(d) cone.
13. The cylindrical parameterization of z = (x2 + y2 )ex is
x+4
Z 3 Z x2
0
sin x dy dx.
Z 0
Z sin θ
−π /2 0
(a) −
14. The equation of the tangent plane to x = u, y = v, z = u + v2
where u = 1 and v = 1 is
(a) x − 1 + 2(y − 2) + z − 2 = 0.
(b) x − 1 − 2(y − 1) − z + 2 = 0.
(c) x − 1 + 2y − 2 + z + 2 = 0.
(d) x − 1 + 2y − 2 − z + 2 = 0.
r cos θ dr d θ =
15. Answer true or false.
To find the portion of the
surface z = 3x2 + 4y2 that lies above the rectangle
Z 3Z 2q
0 ≤ x ≤ 2, 1 ≤ y ≤ 3, evaluate
36x2 + 64y2 + 1 dx dy.
1
6
1
1
6
π
(c)
2
(b)
π
(d) −
2
9. Answer true or false.
16.
Z 3 Z π /2 Z π /2
1
0
0
sin x sin y dx dy dz =
(a) −2
(b) 2
Z 3 Z √9−x2
−3 0
(c) 1
dx dy =
Z πZ 3
0
10. Find the volume of the solid left hemisphere if
(d) −1
r dr d θ
0
r 2 + z2
= 1.
17.
Z 5 Z π /2 Z sin y
1
0
0
0
cos y dx dy dz =
330
(a) 2
(b) 8π
(b) 4
(c) 0
(c) 0
(d) 2π
(d) 1
18. A lamina with density δ (x, y) = xy is bounded by x = 0,
x = y, y = 0, and y = 2. Find its mass.
(b) (6, 6, 6)
(b) 4
(c) (1, 1, 1)
(d) (0, 0, 0)
(d) 8
19. A lamina with density δ (x, y) = xy is bounded by x = 0,
x = y, y = 0, and y = 2. Find its center of mass.
8 16
(a)
,
5 5
8 8
,
(b)
5 5
16 16
,
(c)
5 5
16 16
,
(d)
15 15
20. The centroid of a rectangular solid in the first octant with vertices (0, 0, 0), (0, 1, 1), and (1, 0, 0) is
(a) (0, 1, 2).
1 1 1
(b)
, ,
.
2 2 2
(c) (1, 1, 1).
2 2 2
, ,
.
(d)
3 3 3
Z 2π Z π /2 Z 2
0
0
0
ρ 2 sin φ d ρ d φ d θ =
16π
(a)
3
16π
(b) −
3
3
(c) 4π
Z π /2 Z π Z 8
0
−π 4
(a) 4π
24. Answer true or false. The center of gravity of the solid enclosed by x2 + y2 = 1 and y2 + z2 = 1 is at the origin if
δ (x, y, z) = x + 1.
25. Answer true or false.
Z 2π Z 1−cos2 θ Z 1
0
26. Find
0
0
cos θ d ρ d φ d θ = 0
∂ (x, y)
, if x = 5u + 2v and y = 7u + v.
∂ (u, v)
(a) −9
(b) 9
(c) −19
(d) 19
27. Find the Jacobian if u = xy and v = 2x.
2u
v2
2u 1
(b) − 2 −
v
v
2u 1
(c) − 2 +
v
v
1
(d)
v
(a) −
28. Find the Jacobian if x = 4u + w, y = vw, and z = u2 v.
(a) 4u2 v + 2uvw
(b) 4u2 v + 2uw2
(c) 8(u2 v + uw2 )
(d) −8(u2 v + uw2 )
(d) −4π 3
22.
(a) (3, 3, 3)
(a) 2
(c) 1
21.
23. Find the center of gravity of the sphere x2 +y2 +z2 = 6 where
δ (x, y, z) = 6x2 y2 z2 .
sin φ d ρ d θ d φ =
29. Answer true or false.
Z 2Z 2
x+y 2
1
Z 4Z 2
2
1
−
u2
v2
1
2x − y
dx dy =
dv du, where u = x + y and v = 2x − y.
331
Chapter 14: Answers to Sample Tests
Section 14.1
1. a
9. b
2. b
10. b
3. b
11. false
4. a
12. true
5. b
13. true
6. d
14. true
7. false
15. true
8. true
2. a
10. b
3. a
11. false
4. d
12. d
5. false
13. false
6. true
14. false
7. c
15. false
8. c
2. c
10. b
3. c
11. c
4. c
12. c
5. b
13. d
6. false
14. b
7. false
15. c
8. true
2. a
10. true
3. true
11. true
4. true
12. false
5. true
13. false
6. c
14. false
7. d
15. true
8. false
2. b
10. b
3. b
11. b
4. d
12. false
5. d
13. false
6. d
14. false
7. a
8. c
2. a
10. true
3. false
11. false
4. a
12. true
5. c
13. b
6. b
14. b
7. a
15. d
8. d
2. c
10. true
3. d
11. false
4. c
12. false
5. a
13. false
6. c
14. true
7. true
15. false
8. false
2. a
10. false
3. a
11. false
4. b
12. a
5. b
13. c
6. a
14. d
7. d
15. d
8. c
2. a
10. c
18. a
26. a
3. true
11. d
19. d
27. d
4. true
12. a
20. b
28. a
5. a
13. c
21. a
29. false
6. false
14. d
22. b
7. a
15. true
23. d
8. b
16. b
24. false
Section 14.2
1. b
9. b
Section 14.3
1. a
9. false
Section 14.4
1. c
9. false
Section 14.5
1. b
9. a
Section 14.6
1. b
9. true
Section 14.7
1. b
9. false
Section 14.8
1. b
9. d
Chapter 14 Test
1. b
9. true
17. a
25. true
332
Chapter 15: Topics in Vector Calculus
Summary: This chapter is the culmination of the discussion of multivariable
scalar and vector-valued functions and the applications of double and triple integrals. The main focus in this chapter is to extend integration by parts to double
and triple integrals. The basic idea of integration by parts was to simplify the
integration of some single integrals by rewriting the integral as some terms evaluated at the boundary of the interval added together with a new integral. In double
and triple integrals, the boundaries are usually either curves in a plane or surfaces.
Line integrals and surface integrals are introduced to discuss boundary terms similar to those that show up when integration by parts is used.
The chapter begins by discussing vector fields. In particular, the divergence (a
scalar value) and curl (a vector) of a vector field are introduced. After the introduction of vector fields, the concept of line integrals and their relationship to such
quantities as work is described. In particular, so-called conservative vector fields
are found to obey the Fundamental Theorem of Line Integrals. Green’s Theorem
uses line integrals as a means to simplify the evaluation of some double integrals. For example, some work and area problems may be reduced using Green’s
Theorem to a problem of evaluating a line integral. Analogous to line integrals
along a curve, a surface integral may be evaluated when a surface is parameterized. This has applications relating to flow fields and flux of a vector field. The
chapter concludes by discussing the Divergence Theorem and Stokes’ Theorem.
The Divergence Theorem has some important applications involving flux and in
electromagnetic problems. Stokes’ Theorem is essentially an extension of Green’s
Theorem to higher dimensions. Both Green’s Theorem and Stokes’ Theorem are
closely related to integration by parts in one dimension.
OBJECTIVES: After reading and working through this chapter
you should be able to do the following:
1. Define a vector field (§15.1).
2. Find the divergence and curl of a vector field (§15.1).
3. Find the gradient field of a multivariable scalar function (§15.1).
4. Calculate line integrals on a smooth or piecewise smooth curve (§15.2).
5. Calculate the line integral of a vector field along a curve (§15.2).
333
334
6. Find work using line integrals (§15.2).
7. Determine if a vector field is conservative (§15.3).
8. Evaluate line integrals of conservative fields using the Fundamental Theorem of Line Integrals (§15.3).
9. Use Green’s Theorem to evaluate double integrals by evaluating line integrals (§15.4).
10. Define and evaluate surface integrals (§15.5).
11. Use surface integrals to determine the flux across a surface (§15.6).
12. Use the Divergence Theorem to evaluate surface integrals and to develop a
better understanding of divergence (§15.7).
13. Use Stokes’ Theorem to evaluate triple integrals through the use of surface
integrals (§15.8).
15.1 Vector Fields
PURPOSE: To describe vector fields and related concepts such
as divergence, curl, and the idea of a conservative field.
In this introductory section, the main topic is that of multivariable, vector-valued
functions. These have already been discussed to some extent (see §14.4). Multivariable, vector-valued functions are often called vector fields. A vector field is
just another way of saying that at any given point in the space (either 2-space or
3-space) there is a vector that is determined by the function at that point.
vector field
It is common to indicate a point (x, y, z) using the vector notation r = hx, y, zi.
Then a vector field can be written as F(r) = F(x, y, z). One type of common
vector field is a gradient field. This is the vector field that is determined by the
gradient of a given function at any given point. If a vector field can be shown to
be the gradient field of another function then it is called a conservative vector field
and the associated function is called a potential function.
IDEA: The gradient field ∇φ is a vector field. Additionally if F = ∇φ then F is
called a conservative vector field and φ (r) is called a potential function.
CAUTION: If φ (r) is a potential function, it is a multivariable function that has
scalar values. In other words, a potential function is not a vector field.
divergence and curl
Two common operators that are associated with a vector field F are the divergence
and the curl of F. To define these two operators, it is helpful
E del operD to use the
ator which can be thought of as a vector defined by ∇ = ∂∂x , ∂∂y , ∂∂z . Then the
divergence of F is equal “del dot F” or ∇·F and the curl of F is defined as “del
cross F” or ∇ × F.
335
IDEA: The del operator can be treated as a vector which then makes it easy
to define the divergence and the curl of F.
del operator, ∇
div F = ∇·F
curl F = ∇ × F
CAUTION: The divergence of F, ∇·F, is a scalar value. The curl of F, ∇ × F,
is a vector field.
If the vector field F(r) represents some sort of flow field, then divF describes
whether fluid is flowing toward or away from a given point. On the other hand,
curlF describes the rotation of a fluid flow about a point.
The Laplacian operator, △, can also be defined using the del operator. In this
case △ = ∇·∇ (sometimes written as ∇2 ). This vector notation is helpful because
it identifies that the Laplacian operator is a scalar multiplier. In other words, △φ
and △F may both be considered. Another way to think of the Laplacian operator
is as the divergence of the gradient of a function.
Checklist of Key Ideas:
vector field; gravitational field; velocity field
the notation F(r)
inverse square fields
gradient field of φ
conservative field
potential function
divergence of F, divF
curl of F
the “del” operator
the Laplacian operator
15.2 Line Integrals
PURPOSE: To introduce line integrals as a way of integrating
either scalar or vector valued functions along a curve that can be
parameterized.
A new type of integral is introduced in this section. So far, integrals have been
considered over intervals, planar regions, and over three dimensional regions. This
section introduces the idea of integrating a quantity along a curve. This is called
Laplacian operator
336
line integral
a line integral (even though the curve itself does not have to be linear). The key
to evaluating line integrals lies in being able to parameterize the curve in question
so that the integrand along the curve may be expressed in terms of the parameter.
IDEA: Being able to evaluate a line integral is dependent upon being able to
parameterize the curve along which the integration takes place. The resulting
integral is a single integral in terms of the parameter.
Once the curve is parameterized, the so-called line integral is rewritten as a simple
single integral with the variable of integration being the parameter that is used to
parameterize the curve. This means that the integrand needs to be rewritten in
terms of the parameter and the limits of integration will be written in terms of the
parameter. There are two common ways of parameterizing a curve: with respect
to arc length and with respect to the coordinate variables.
IDEA: There are two common ways for parameterizing a curve for calculating
line integrals: parameterizing with respect to arc length and parameterizing
with respect to coordinate variables.
Notation for line integralsRis perhaps the easiest place to become confused. For
example, the line integral C f (x, y) dx, does not mean integration with respect to
the variable x. What it really means is that the curve C should be parameterized
in terms of t. For example, let x = g(t), and y = h(t), then dx = x′ (t)dt. So when
this integral is written out it should really look like a single integral in terms of t:
Z
f (x, y) dx =
C
Z t1
t0
f (x(t), y(t))x′ (t)dt.
CAUTION: The notation for line integrals may easily cause a problem with
their evaluation. The key is to always remember to parameterize the curve
and then to express ds, dx, dy, or dz in terms of the parameter.
line integrals with respect to arc
length
Line integrals
that are calculated with respect to arc length typically have the
R
form C f (x, y, z) ds where s is the arc length parameter. This formula is rarely
used in this form. Instead, the curve C is parameterized so that r(t) = hx(t), y(t), z(t)i.
Then the integral is written as
Z
f (x, y, z) ds =
C
line integral with respect to x, y, or z
Z t1
t0
f (r(t))kr′ (t)kdt.
Similarly, if the line integral is being calculated with respect to a coordinate
variable such as x, y, or z, then given the parameterization r(t) the differentials
dx, dy, and dz should each be written in terms of t.
IDEA: If a curve, C, is parameterized by
r(t) = hx(t), y(t), z(t)i = x(t)i + y(t)j + z(t)k
then the differentials for line integrals are given as follows:
ds = kr′ (t)kdt, dx = x′ (t)dt, dy = y′ (t)dt, dz = z′ (t)dt.
337
In addition to calculating line integrals with respect to arc length, line integrals
are also calculated with
respect to the
coordinate variables.
This usually is written
R
R
R
in one of the forms C f (x, y, z)dx, C f (x, y, z)dy, or C f (x, y, z)dz. As mentioned
above, the key to evaluating these is to parameterize the curve and then rewrite the
variables and the differentials in terms of the parameter.
R
IDEA: The line integral CR f (x, y, z)ds means to calculate with respect to arc
length. The line integral C f (x, y, z)dx (or with dy or dz) means to calculate
with respect to the variable x.
Line integrals of vector fields may also be calculated. If the vector field is represented by F and a point represented using the position vector r, then the line
integral is just the integral of the dot product of F and dr. Again the key is to
parameterize the curve.
line integrals of vector fields
IDEA: The line integral of F along the curve C is given by
Z
F·dr =
C
t=b
Z
F(r(t))·r′ (t)dt.
t=a
One important application of the line integral of a vector field along a curve is
that of work. Work is the energy required to move a particle. The work can be
found by multiplying the force required times the distance the particle is moved.
The line integral of the force required (represented by the force field, F) would
represent the work done by the force field:
work =
Z
work and line integrals
F·dr.
C
IDEA: A curve may be broken into sections C = C1 +C2 + · · ·Cn and then line
integral over the whole curve may then be calculated by the sum of the line
integrals on the individual curves.
Line integrals are usually calculated on smooth curves where r′ (t) has some meaning. However, on curves that are not smooth, the line integral may still be calculated by simply dividing the curve into sections that are smooth. Then the individual line integrals may be added together. For example, it is possible to calculate
the line integral over a curve such as y = |x| by breaking the curve into two separate pieces for x ≥ 0 and x < 0.
Checklist of Key Ideas:
line integral of f with respect to s along a curve C
evaluating line integrals using parameters
oriented curves
line integrals with respect to x, y, or z along a curve
piecewise smooth curves
338
oriented in direction of increasing parameter
line integrals in vector form
integrating a vector field along a curve
calculating work using line integrals
work performed by a force field
piecewise smooth curves and line integrals
15.3 Independence of Path; Conservative Vector Fields
PURPOSE: To investigate conservative vector fields, how to
check if a vector field is conservative, and the property of independence of path.
potential functions
conservative vector field
In this section, the relationship between potential functions and their gradients
is explored in more detail. Recall that if a vector field F is the gradient of some
function φ then F is called a conservative vector field and φ is called a potential
function. In some sense, φ is similar to being an antiderivative of F = ∇φ . This
idea is made more clear when line integrals are considered.
The fundamental theorem of calculus says that if f = F ′ then the change in F
between two points is equal to the integral of f over that same interval.
Z b
a
f (x) dx = F(b) − F(a)
A similar relationship exists between a conservative vector field and its associated
potential function.
Fundamental Theorem of Line Integrals
IDEA: If F = ∇φ then
Z
C
F·dr =
Z t1
t0
F(r(t))·r′ (t)dt = φ (r(t1 )) − φ (r(t0 )).
Here r(t0 ) and r(t1 ) are the initial and final points of a curve that is parameterized by r(t).
path independent
Note that the change in φ between two points r0 and r1 does not depend upon
any particular path between the points. So the relationship that F = ∇φ implies
that the line integral of F between two points does not depend upon the path of
integration between the two points. In other words, conservative vector fields are
called path independent. As a result of this, if F is a conservative vector field
and the curve C starts and ends at the same point then the line integral of F along
C should be zero.
IDEA: If F = ∇φ then
I
C
F·dr = 0.
339
R
Recall the if F is an antiderivative of f that f (x) dx = F(x) +C. In other words,
any antiderivatives of f will differ by only a constant. A similar relationship holds
between a conservative vector field and its potential function. In fact, if φ is a
potential function of F then so is φ +C.
IDEA: If φ is a potential function of F then so is φ +C for any constant, C.
Checking to see whether a vector field is conservative is the same as taking the
mixed partial derivatives of the potential function φ . They must be equal. In the
two dimensional case this means that
h−
∂ ∂
, i·F = 0.
∂y ∂x
In the three dimensional case, this means that ∇ × F = 0. In other, words, vector
fields in 3-space that are conservative are said to be curl free.
Checklist of Key Ideas:
work integral
path of integration
independence of path
Fundamental Theorem of Line Integrals
conservative vector fields and line integrals
line integrals along closed curves or paths
connected domains
simple parametric curve
simply connected or multiply connected
Conservative Field Test
conservative fields and curl (in 3-space only)
conservation of energy principle
potential function; potential energy; kinetic energy
15.4 Green’s Theorem
PURPOSE: To introduce Green’s Theorem and describe its relationship with double integrals and line integrals.
Green’s Theorem is the subject of this section. Green’s Theorem is similar to the
idea of integration by parts that was used on single integrals. What the theorem
curl free
340
says is that a particular function that, g(x, y), is evaluated on a planar region may
be rewritten using a line integral around the boundary of the region.
Using the idea of determinants and the “del” operator, ∇, here is an interesting
way to remember Green’s Theorem. First, let F = hF1 , F2 i. Then consider the
following function defined using a determinant:
∂
∂ ∂
x
∂
y
= ∂ F2 − ∂ F1 .
g(x, y) = det ∂x
∂y
F1 F2 Notice that the first row of this determinant is ∇ and the second row is the vector
field F. Then Green’s Theorem says the following:
I
C
F·dr =
ZZ
g(x, y) dA
R
where g(x, y) is defined to be the determinant of ∇ and F.
This idea is consistent with what has already been said of conservative vector
fields. If F is conservative then on the one hand, a line integral along a closed path
will be zero (that is the left hand side of Green’s Theorem). On the other hand, if
g(x, y) is the determinant of ∇ and F then g(x, y) = 0. So the right hand side of the
equation would also be zero.
Green’s theorem is not incredibly useful if g(x, y) is known but F is not. In that
case, the components of F would not be known and it would be hard to make use
of the above relationship.
One point that is very important to understand is that when Green’s Theorem is
used to reduce a double integral to a line integral, the orientation of the curve
in the line integral is important. For the equality to hold, the curve needs to be
traversed in a direction so that the surface in question is always to the left of the
curve. In other words, the curve needs to be parameterized in such a way that the
line integral can be calculated while going in a counterclockwise fashion around
the region.
CAUTION: The line integral in Green’s theorem requires that the line around
the outside of a region should be traversed in a counterclockwise direction.
Interestingly enough, g(x, y) as defined above is related to the curl of F in 3-space.
Consequently, Stokes’ theorem at the end of this chapter can be seen as a direct
extension of Green’s Theorem from 2-space into 3-space.
Checklist of Key Ideas:
Green’s Theorem
notation for line integrals around simple closed curves
work and Green’s Theorem
finding area using line integrals
multiply connected regions
positive orientation
341
15.5 Surface Integrals
PURPOSE: To define surface integrals for parametric and nonparametric surfaces and to explain how they are evaluated.
Surface integrals are an extension of the idea from §14.4 of calculating the surface area of a given surface. The key to evaluate a function on a surface is to
parameterize the surface. This idea of parameterizing the surface is also what
allowed for the calculation of the surface area of a given surface. Surface integrals are also similar to line integrals. A line integral calculated an integral of a
function along a curve by parameterizing the curve. Surface integrals do the same
thing over a surface by parameterizing the surface.
surface integrals
Recall from section §14.4 that if a surface can be parameterized by
r(u, v) = hx(u, v), y(u, v), z(u, v)i
then the surface area could be found by
ZZ
R
Norm
∂r ∂r
×
dudv.
∂u ∂v
In this section, this same idea is used except that now, the integrand is multiplied
by some function f (x, y, z). The important thing to remember is that the surface is
parameterized.
This is not obvious from the notation used for a surface integral,
RR
σ f (x, y, z)ds. Again (as was the case in line integrals), ds and the variables x, y,
and z need to be rewritten in terms of the parameters u and v. Then the integral
can be written as follows:
ZZ
σ
f (x, y, z)ds =
ZZ
R
f (r(u, v)) kru × rv k dudv
IDEA: Calculating surface integrals is done in the same way that surface area
was calculated in §14.4. The surface is parameterized by two parameters u
and v. The resulting integral is the same as the surface area integral except
that the integrand is multiplied by some function f (x, y, z).
Since the integrand of the surface area integral is multiplied by some function
f (x, y, z), it is a simple matter to think of this function as a density function. Then
the mass of a curved lamina can be found by evaluating the appropriate surface
integral with a density function δ (x, y, z). Following this same line of reasoning,
the center of mass of a curved lamina can be found by using the formulas for
Mxy , Mxz , and Myz along with M to find (x̄, ȳ, z̄). In each case, the appropriate
differential notation to use should replace dV with ds where ds = kru × rv kdudv.
CAUTION: The center of mass of a curved lamina will not necessarily be on
the lamina itself.
density function
342
Checklist of Key Ideas:
curved lamina
surface integral
vector form of a surface integral
evaluating surface integrals
functions that are surfaces above different coordinate planes
15.6 Applications of Surface Integrals; Flux
PURPOSE: To describe flow fields, flux and how these concepts
relate to surface integrals over parametric, orientable surfaces.
orientation of a surface
In the previous section surface integrals were calculated with little regard for their
“orientation.” In this section, the orientation of a surface will be important since
the information that is sought involves the flux of a flow field through a given
surface. That is, if a particular vector field, F, is seen as a flow field (which
describes the velocity of some fluid at a given point) then the flux is the net amount
of the fluid to pass through the surface per unit time.
IDEA: The flux of a fluid is the net amount of the fluid to pass through a surface
per unit time.
principle unit normal vector
A surface may be oriented using the principle unit normal vector. If a surface
is parameterized by r(u, v) then the principle unit normal vector, n, is obtained by
normalizing the vector ru × rv .
IDEA: The principle unit normal vector, n, is defined as follows.
n=
ru × rv
kru × rv k
The direction that n points on a surface is called the positive orientation. The
direction −n would represent negative orientation.
positive orientation
flux through a surface
The principle unit normal vector, n, points in the direction of positive orientation of a surface. Depending upon how the parameters u and v are chosen, both
directions could be considered as positive orientation. The crucial element here is
to be consistent with how the parameters are defined so that only one direction is
associated with positive orientation of the surface.
IDEA: The flux through a surface is described by the surface integral
ZZ
ZZ
σ
F·nds =
R
F(r)· (ru × rv ) dudv.
343
Checklist of Key Ideas:
flow fields
streamlines, electric lines and tangent vectors
oriented surfaces; orientable and nonorientable
positive orientation and n
incompressible and compressible fluids
steady state
net volume of fluid to pass through a surface
flux of F across σ
flux integrals
orientation of nonparametric surfaces
15.7 The Divergence Theorem
PURPOSE: To state the Divergence Theorem (or Gauss’s Theorem) and to use its results to evaluate the flux of a flow field
across a surface that bounds a particular region of space.
The Divergence Theorem (also called Gauss’s Theorem) relates surface integrals
to triple integrals. It is similar in principle to both Green’s Theorem and the relationship between conservative fields and potential functions. In each of these
cases, two functions were related in such a way that one of the functions was
somehow obtained using the derivatives of the other.
CAUTION: The principle unit normal vector from the previous section and the
unit outward normal vector may not necessarily have the same direction. This
will depend upon how the surface is parameterized using u and v.
To be able to used the Divergence Theorem requires that the unit outward normal vector, n, of a given surface be known. This is the unit vector that points
away from the interior of the three dimensional domain, G. Then the Divergence
Theorem states
ZZ
σ
F · nds =
ZZZ
∇·F dV
G
where the first integral is a surface integral over some closed surface σ and the
second is a triple integral over the region G that is enclosed by σ . Often the triple
integral is easier to evaluate than the surface integral. In other words, the process
of finding the outward flux of a fluid flow from some solid region may be obtained
by evaluating a triple integral instead of a surface integral.
unit outward normal vector
344
IDEA: The Divergence Theorem relates the divergence of a vector field, F, to
the outward flux from a region. In other words, the integral of the divergence
is equal to the outward flux of the fluid.
Checklist of Key Ideas:
closed surfaces; smooth surfaces; piecewise surfaces
inward and outward orientation of a surface
Divergence Theorem (or Gauss’s Theorem)
outward flux
using the integral of the divergence to find flux across a surface
outward flux density of F across G
describing divergence at a point in terms of flux density
sources and sinks and divergence
continuity equation for incompressible fluids
Gauss’s Law for Inverse-Square Fields
Gauss’s Law for Electric Fields
15.8 Stokes’ Theorem
PURPOSE: To state Stokes’ Theorem and to describe how it relates surface integrals with line integrals.
Recall that Green’s Theorem allowed a double integral over some planar region
to be written instead using a related line integral around the boundary of the region. Stokes’ Theorem does essentially the same thing in 3-space. Stokes’ Theorem then can allow a general surface integral to be evaluated using a line integral
around the boundary of the surface.
Recall the determinant that was used in Green’s Theorem to write the integral
over the planar region. Now in 3-space the idea of this determinant needs to be
extended since ∇ = h∂x , ∂y , ∂z i and F will have three components as well. The
result is to consider the cross product of the “del” operator and F on the surface.
But this is just the curl of F, ∇ × F.
So Stokes’ Theorem says
I
C
oriented surface
F · Tds =
ZZ
σ
(∇ × F)·n dS.
where C is a curve that bounds the oriented surface σ . In this case, the surface
345
σ has some orientation as determined by the principle unit normal vector, n (see
§14.5). Also, the curve C must be traversed in the positive direction. This means
that if T and N are the normal and tangential components of the curve C then B,
the binormal vector should point in “same direction” as n.
Checklist of Key Ideas:
oriented surfaces bounded by a simple closed parametric curve
positive and negative orientation of a curve around a surface
Stokes’ Theorem
calculating work with Stokes’ Theorem
different surfaces with the same boundary curve
curl and circulation
circulation of F around a curve
circulation density of F around a curve
direction of maximum circulation density
irrotational fields
346
Chapter 15 Sample Tests
Section 15.1
1. Answer true or false. φ (x, y) = x2 y is the potential function
for F(x, y) = 2xi + j.
2. Answer true or false. φ (x, y) = sin x + cos y is the potential
function for F(x, y) = cos xi − sin yj.
3. F(x, y, z) = x3 i + 3j + xzk. Find divF.
(a) 3x2 + x
(c) ex i + yex j + p
(d) ex i − yex j + p
2x
x2 + y2
x
x2 + y2
k
k
9. Answer true or false. ∇2 φ = 6xy2 + 2x3 , if φ = x3 y2 .
10. Answer true or false.
∇2 φ = cos x − sin(xy), if φ = sin x + cos(xy).
11. Answer true or false. ∇2 φ = ex + 2xyexy , if φ = exy .
(b) 3x2 i + xk
12. Answer true or false. ∇2 φ = (xy + yz + xz)2 exyz , if φ = exyz .
(c)
3x3
13. Answer true or false. ∇2 φ = 0, if φ = 7x + 3y + 2z.
(d)
3x2 − x
14. Answer true or false. ∇2 φ = 6x, if φ = x3 + y + z.
4. F(x, y, z) = x3 i + 3j + xzk. Find curlF.
(a) zj
15. Answer true or false.
∇2 φ = − sin x − sin y − sin z, if φ = sin x + sin y + sin z.
(b) −zj
(c) z
Section 15.2
(d) −z
5. F(x, y, z) = xyzi + yj + xk. Find divF.
and y = t, (0 ≤ t ≤ 1).
(a) yz + 1
(1 + x2 y) ds, where x = 3t
Z
(1 + x2 y) dy, where x = 3t
Z
(1 + x2 y) dx, where x = 3t
C
(b) 10.28
(c) xy − 2 − xz
(c) 13
(d) xy − 2 + xz
6. F(x, y, z) = xyzi + yj + xk. Find curlF.
(d) 2
2. Find the approximate value of
(a) xzi + j
and y = t, (0 ≤ t ≤ 1).
(b) xzi − xj
(c) (xy − 1)j − xzk
2y
x2 + y2
2x
(c) ex + yex + p
x2 + y2
x
(d) ex + yex + p
2
x + y2
p
8. F(x, y, z) = ex i + x2 + y2 j + yex k. Find curlF.
(a) e i + p
(b) ex i + p
2y
x2 + y2
y
x2 + y2
(c) 9.75
(d) 2.3
and y = t, (0 ≤ t ≤ 1).
x2 + y2
x
(b) 3.25
3. Find the approximate value of
y
x
C
(a) 6.75
(d) (xy − 1)j + xzk
p
7. F(x, y, z) = ex i + x2 + y2 j + yex k. Find divF.
(b) e + p
Z
(a) 3.25
(b) yz − 1
(a) ex + p
1. Find the approximate value of
j
j
C
(a) 6.75
(b) 3.25
(c) 9.75
(d) 2.3
4. Find the approximate value of
Z
y = 2t, and z = −2t, (0 ≤ t ≤ 1).
(a) −1
(b) −2
C
(xy + z) ds, where x = t,
347
13. Find the work done by F(x, y) = xi + xyj along the curve
y = x2 from (0, 0) to (1, 1).
(c) 0
(d) 0.05
Z
5. Find the value of
(a) 0.75
(xy + z) dy, where x = t, y = 2t, and
(b) 0.9
C
z = −2t, (0 ≤ t ≤ 1).
(c) 1
(a) −1
(d) 2
(b) 1
2
(c)
3
(d) −
14. Answer true or false. The work done by F(x, y) = xi + yex j
along the curve y = x2 from (−1, −1) to (0, 0) is the same
as the work done in moving the same particle along the same
curve from (0, 0) to (1, 1).
2
3
Z
6. Find the value of
(xy + z) dz, where x = t, y = 2t, and
C
z = −2t, (0 ≤ t ≤ 1).
(a) −1
(b) 1
2
(c)
3
(d) −
15. Answer true or false. The work done by F(x, y) = xi + yex j
along the curve y = x3 from (0, 0) to (1, 1) is the same as the
work done in moving the same particle along the same curve
from (−1, −1) to (0, 0).
Section 15.3
1. Answer true or false. F(x, y) = 2xi + 3yj is a conservative
vector field.
2
3
7. Find the value of
Z
3xy2 dx + 5xy2 dy along the curve x = y2
C
from (0, 0) to (1, 1).
(a) 2
(b) 7
2. Answer true or false. F(x, y) = xyi + xyj is a conservative
vector field.
x3
3. Answer true or false. F(x, y) = x2 yi + j is a conservative
3
vector field.
4. Answer true or false. F(x, y) = sin xi + sin yj is a conservative
vector field.
(c) −2
(d) −7
Z
8. Find the value of
xy ds, where x = cost, y = sint,
C
0 ≤ t ≤ π.
5. Answer true or false. F(x, y) = sin yi + sin xj is a conservative
vector field.
6.
(a) 0
(d) 4
9. Answer true or false. If x = t, y = 5t, and (0 ≤ t ≤ 1) then
C
Z 1
5t 2 dt.
0
10. Answer true or false.
(0 ≤ t ≤ 2π ), then
Z
C
If x = cost and y = sint, with
(x − 2y) ds =
Z 2π
0
7.
(cost − 2 sint) dt.
C
(xy − z) ds = 2
Z 2π
0
(− sint cost − 3t) dt.
12. Answer true or false. If x = et , y = 3et , and z = et with
Z
Z 1 √
(0 ≤ t ≤ 2π ), then (x + y + z) ds =
5 11et dt.
C
0
Z (1,3)
(0,1)
x2 y dx +
x3
dy =
3
(a) 1
9
(b)
2
(c) 2
7
(d)
2
11. Answer true or false. If x = cost, y = − sint, and z = 3t with
(0 ≤ t ≤ 2π ), then
Z
2x dx + 3y dy =
27
2
39
(b)
2
33
(c)
2
(d) 16
(c) 2
xy ds =
(1,0)
(a)
(b) 1
Z
Z (2,3)
8.
Z (2,5)
(1,2)
3x2 y dx + x3 dy =
348
(a) 76
(c) 5
(b) 38
(d) 0
(c) 80
(d) 40
9.
Z (π ,π )
(0,0)
sin x dx + sin y dy =
(c) −11.
(b) 2
(c) −4
(d) 0
10.
(0,0)
(d) 0
11. Answer true or false. If F(x, y) = 5yi + 5xj, then φ = 5.
12. Find the work done by the force field F(x, y) = x2 y3 i + x3 y2 j
on a particle moving along an arbitrary smooth curve from
P(1, 1) to Q(0, 0).
1
2
(b) 1
(a)
5xy dx + 7xy dy, where C is the rectangle
(a) 13
45
(b)
95
47
(c)
97
51
(d)
101
2. Find the area enclosed in the ellipse
x2 y2
+
= 1.
9
4
(a) 36
(b) 36π
1
3
2
3
x3
13. Find the work done by the force field F(x, y) = x2 yi + j
3
on a particle moving along an arbitrary smooth curve from
P(0, 0) to Q(1, 2).
2
(a)
3
5
(b) −
6
1
(c)
3
1
(d) −
3
14. Find the work done by the force field F(x, y) = i + j on a particle moving along an arbitrary smooth curve from P(0, 0) to
Q(2, 3).
(b) 3
I
C
(c) −4
(a) 2
1. Evaluate
0 ≤ x ≤ 1, and 1 ≤ y ≤ 3.
(b) 2
(c) −
(d) −7.
Section 15.4
sin x dx − sin y dy =
(a) 4
(d)
(a) 11.
(b) 7.
(a) 4
Z (π ,π )
15. Find the work done by the force field F(x, y) = x2 i + yj on a
particle moving along an arbitrary smooth curve from P(0, 0)
to Q(3, 2) is
(c) 6
(d) 6π
3. Find the area enclosed in the ellipse
(x − 2)2 (y + 3)2
+
= 1.
25
9
(a) 225
(b) 225π
(c) 15
(d) 15π
4. Find the work done by the field F(x, y) = y3 i + (y − x3 )j on
a particle that travels once around a unit circle x2 + y2 = 1 in
a counterclockwise direction.
3π
(a)
4
3π
(b)
2
π
(c)
4
(d) π
5. Find the work done by the field F(x, y) = (x2 + y3 )i + (sin y − x3 )j
on a particle that travels once around a unit circle x2 + y2 = 1
in a counterclockwise direction.
349
3π
4
3π
(b)
2
π
(c)
4
(d) π
11. Evaluate
(a)
4x2 + 9y2 = 36.
(a) 6π
(b) 6
(c) 0
12. Use a line integral to find the area of the triangle with vertices
(0, 0), (a, 0), and (a, b).
(a)
7. Find the work done by the field F(x, y) = hx2 + y3 , sin y − x3 i
on a particle that travels once around a circle x2 + y2 = 4 in a
counterclockwise direction.
(a) −24π
3π
(b)
2
π
(c)
4
(d) π
I
(d) 12
1
2
(b) ab
ab
(c)
2
(d) 2ab
(a) −24π
3π
(b)
2
π
(c)
4
(d) π
13. I
Answer true or false.
(ey + 2x) dx + (3x4 + 2y) dy; C = x2 + y2 = 1, is 2π .
C
14. I
Answer true or false.
sin x dx + cos y dy; C = x2 + y2 = 1, is π .
C
15. I
Answer true or false.
sin y dx + cos x dy; C = x2 + y2 = 1, is 2π .
C
(x2 + y) dx + (y2 + x) dy, where C is x2 + y2 = 9.
C
Section 15.5
1. Evaluate
(a) 6π
ZZ
xz dS, where σ is the part of the plane
σ
(b) 9π
(c) 18π
(d) 0
9. Evaluate
(ex + 2y) dx + (3y4 + 2x) dy, where C is
C
6. Find the work done by the field F(x, y) = y3 i + (y − x3 )j on
a particle that travels once around a circle x2 + y2 = 4 in a
counterclockwise direction.
8. Evaluate
I
I
(ex + 2y) dx + (3y4 + 2x) dy, where C is
C
x2 + y2 = 1.
(a) π
x + y + 2z = 1 in the first octant.
√
6
(a)
96
√
3
(b)
12
√
5 6
(c)
16
√
(d) 3
(b) 2π
2. Answer true or false.
(c) 4π
part of the plane x + 2y + z = 1 in the first octant is
I
C
x2 y2
+
= 1.
9
4
(a) 6π
(ex + 2y) dx + (3y4 + 2x) dy, where C is
Z 1 Z 1/2−x
0
0
(x − x2 − xy) dy dx.
3. Find the surface area of the cone z =
the plane z = 4.
(a) 16π
(b) 6
√
(b) 16π 2
(c) 0
(c) 8π
(d) 12
xz dS, where σ is the
σ
(d) 0
10. Evaluate
ZZ
√
(d) 8π 2
p
x2 + y2 that lies below
350
4. Find the surface area of the cone z =
tween the planes z = 3 and z = 4.
p
x2 + y2 that lies be-
(a) 7π
√
(b) 7π 2
13. Answer true or false. If σ is the part of x + y + z = 6 that lies
inZZthe first octant above
the region R then
√ ZZ y 6−x−y
x y
e e
dA.
e e dS = 3
σ
(c) 8π
√
(d) 8π 2
5. Find the surface area of x2 + y2 + (z − 2)2 = 4 that lies below
z = 2.
(a) 8π
R
14. Answer true or false. If σ is the part of x + 2y + z = 5 that
lies
above the region R then
ZZ in the first octant
√ ZZ 2
2
zx y dS = 6
x y(5 − x − 2y) dA.
σ
R
15. Answer true or false. If σ is the part of x + 2y + z = 5 that
lies in the first octant above
R then
the region
ZZ
√ ZZ 2 5 − x − z
2
dA.
zx dS = 6
zx
2
σ
(b) 16π
R
(c) 32π
Section 15.6
(d) 64π
6. Find the surface area of x2 + y2 + (z − 2)2 = 4 that lies below
z = 4.
1. Find the flux of the vector field F(x, y, z) = 2zk across the
sphere x2 + y2 + z2 = 9 oriented outward.
(a) 8π
(a) 72π
(b) 32π
(b) 36π
(c) 16π
(c) 0
(d) 108π
(d) 64π
7. Answer true or false. If σ is the part of x + y +
ZZz = 2 that
xz dS =
lies in the first octant above the region R then
√ Z 1 Z 1−z
3
0
0
σ
(1 − y − z)z dy dz.
8. Answer true or false. If σ is the part of sin x + cos y +
z ZZ
= 0 that lies inZZ
the first octant above the region R then
√
2
x z dS = 2
x2 (− sin x − cos y) dy dz.
σ
R
9. Answer true or false. If σ is the part of x + y − 2z = 5 that
lies
above the region R then
ZZ in the first octant
√ ZZ
x sin y dS = 6
sin y(5 − y + 2z) dA.
σ
R
10. Answer true or false. If σ is the part of x + y + 3z = 8 that
lies
octant
ZZ in the first √
ZZabove the region R then
y
xe dS = 11
ey (8 − 3z − y) dA.
σ
R
x2 + y2 + 4
11. Answer true or false. If σ is the part of z =
lies
aboveq
the region R then
ZZ
ZZ in the first octant
y2 z2 dS =
σ
(x4 + y2
that
4x2 + 4y2 ) dA.
R
12. Answer true or false. If σ is the part of x + y + z = 6 that lies
inZZthe first octant above
the region R then
√ ZZ x 6−y−x
x y
e e dS = 3
e e
dA.
σ
R
2. Find the flux of the vector field F(x, y, z) = 5zk across the
sphere x2 + y2 + z2 = 1 oriented outward.
20π
3
500π
(b)
3
(c) 0
4π
(d)
3
3. Answer true or false. If σ is the portion of the surface
z = 4 − x2 − y2 that lies above the xy−plane, and σ is oriented up, the flux of the vector field F(x, y, z) = xi + yj + zk
(a)
across σ is Φ =
Z 2π Z 1
0
0
(x2 + y2 + 4) dA.
4. Answer true or false. If σ is the portion of the surface
z = 1 − x2 − y2 that lies above the xy−plane, and σ is oriented up, the fluxZZ
of the vector field F(x, y, z) = x2 i + yj + zk
across σ is Φ =
R
(x3 + y2 − x2 + 2) dA.
5. Answer true or false. Let F(x, y, z) = xi + yj + 3k.
The flux outward through the surface x2 + y2 + z2 = 1 is
Z 2π Z π
0
0
(2 sin2 φ cos θ sin θ + 3 sin φ cos φ ) d φ d θ .
6. Let F(x, y, z) = xi + yj + zk and σ be the portion of the surface z = 5 − x2 − y2 that lies above the xy−plane. Find the
upward flux of the vector field across σ .
(a)
75π
2
351
5π
2
15π
(c)
2
(d) 0
Section 15.7
(b)
7. Answer true or false. If F(x, y, z) = 2xi+yk, the flux through
the xy−plane,
the portion of the surface σ that lies above
ZZ
where σ is defined by z = 6 − x2 − y2 , is
(x2 + y2 + 6) dA.
R
8. Answer true or false. If F(x, y, z) = 2xi + yk, the flux
through the portion of the surface σ that lies above the
xy−plane,
where σ is defined by z = 3x2 + 3y2 + 1, is
ZZ
(3x2 + 3y2 + z) dA.
R
9. If F(x, y, z) = xi + yk, the magnitude of the flux through the
portion of the surface σ that lies to the right of the xz−plane,
where σ is defined by y = 1 − x2 − z2 , is
5π
.
2
15π
.
(b)
2
(c) π .
(a)
(d) 0.
10. If F(x, y, z) = yj + zk, the magnitude of the flux through the
portion of the surface σ that lies in front of the yz−plane,
where σ is defined by x = 1 − y2 − z2 , is
5π
.
2
15π
.
(b)
2
(c) π .
(a)
(d) 0.
11. Answer true or false. The surface z = −x3 − y2 + 5 has a
−3x2 i − 2yj + k
normal vector n = p
.
9x2 + 4y2 + 1
12. Answer true or false. The surface y = −x2 − z2 + 8 has a
−2xi − 2zj + k
normal vector n = √
.
4x2 + 4z2 + 1
13. Answer true or false. The surface z = −2x3 − 2y3 + 7 has a
−6xi − 6yj + k
.
normal vector n = p
36x2 + 36y2 + 1
14. Answer true or false. The surface z = 4x + 2y has a normal
−4i − 2j + k
√
vector n =
.
21
1. Find the outward flux of the vector field F(x, y, z) = xi across
the sphere x2 + y2 + z2 = 4.
32π
(a)
3
256π
(b)
3
(c) 0
16π
(d)
3
x
2. Find the outward flux of the vector field F(x, y, z) = i
2
across the sphere x2 + y2 + z2 = 4.
32π
(a)
3
256π
(b)
3
(c) 0
16π
(d)
3
3. Find the outward flux of the vector field F(x, y, z) = 3zk
across the sphere x2 + y2 + z2 = 4.
32π
(a)
3
(b) 32π
256π
(c)
3
(d) 256π
5xi + 5yj + 5zk
and let σ be a closed, ori4. Let F(x, y, z) = 2
(x + y2 + z2 )3/2
entable surface that surrounds the origin. Find Φ.
(a)
(b)
(c)
(d)
−25π
0
100π
25π
8xi + 8yj + 8zk
and let σ be a closed,
(4x2 + 4y2 + 4z2 )3/2
orientable surface that surrounds the origin. Find Φ.
5. Let F(x, y, z) =
(a)
(b)
(c)
(d)
0
4π
8π
16π
x2 i + y2 j + z2 k
and
(x4 + y4 + z4 )3/2
let σ be a closed, orientable surface that surrounds the origin.
Then Φ = π .
7. Find the outward flux of F(x, y, z) = xi + (2y + 3)j + 6z2 k
across the unit cube in the first octant that has a vertex at the
origin.
6. Answer true or false. Let F(x, y, z) =
352
(a) 1
(b) 0
(c) 9
(d) 8
8. Find the outward flux of F(x, y, z) = xi + yj + (z − 2)k across
the rectangle with vertices (0, 0, 0), (0, 0, 1), (2, 1, 0), and
(2, 1, 1).
(a) 6
(b) 8
(c) 0
(d) 1
9. Find the outward flux of F(x, y, z) = (x − 1)i + (y − 3)j +
zk across the rectangle with vertices (0, 0, 0), (0, 0, 1),
(2, 1, 0), and (2, 1, 1).
(a) 3
(b) 6
(c) 0
(d) 1
10. Find the outward flux of F(x, y, z) = 2x2 i + 3yj + zk across
the rectangle with vertices (0, 0, 0), (0, 0, 1), (2, 1, 0), and
(2, 1, 1).
(a) 12
(b) 20
(c) 30
(d) 7
11. Answer true or false. The outward flux of the vector field
x3
y3
z3 − 1
F(x, y, z) = i + j +
k across the surface of the
3
3
3
p
region that is enclosed by the hemisphere z = 16 − x2 − y2
2π
and the plane z = 0 is
.
5
12. Answer true or false. The outward flux of the vector field
F(x, y, z) = x2 i + 2yj + zk across the cube bounded by the
axes and the planes x = 2, y = 3, and z = 4 is 240.
13. Answer true or false. The outward flux of the vector field
F(x, y, z) = 2x2 i + 2y2 j + 2z2 k across the cube bounded by
the axes and the planes x = 2, y = 3, and z = 4 is 58.
14. Answer true or false. The outward flux of the vector field
F(x, y, z) = (2x2 + 3)i + (y2 + 2)j + 2z2 k across the cube
bounded by the axes and the planes x = 2, y = 3, and z = 4 is
178.
15. Answer true or false. The outward flux of the vector field
F(x, y, z) = x3 i + yj + zk across the cube bounded by the
axes and the planes x = 2, y = 3, and z = 4 is 192.
Section 15.8
1. Answer true or false. If σ is the surface z = −x2 − y2 + 4 and
F(x,
ZZzk, then
ZZ y, z) = 3xi + 4yj +
σ
(curl F) · n dS =
(6x + 8y + 1) dA.
R
2. Answer true or false. If σ is the surface z = −x2 − y2 + 4 and
2
F(x,
ZZz k, then
ZZ y, z) = 5xi + 2yj +
σ
(curl F) · n dS =
(10x + 4y + 2z) dA.
R
3. Answer true or false. If σ is the surface z = −x2 − y2 + 4 and
F(x,
+ (3x + 5y)j + xk, then
ZZ y, z) = (4x + 2y)iZZ
σ
(curl F) · n dS =
(12x + 8y + z) dA.
R
2
2
4. Answer true or false. If σ is the surface
ZZ z = −x − y + 4 and
F(x, y, z) = 6xi + 2yj + 3zk, then
σ
(curl F) · n dS = 0.
2
2
5. Answer true or false. If σ is the surface
ZZz = −x − y + 4
and F(x, y, z) = xyzi + xzj + x2 yzk, then
σ
(curl F) · n dS is
equivalent
to
ZZ
R
(2x2 (xz − 1) + 2y(xyz − 2xz) + z(1 − x)) dA.
6. Answer true or false. The amount of work needed to move a
particle around the rectangle (0, 0, 0), (0, 2, 3), (1, 2, 3),
(1, 0, 0) and
back to (0, 0, 0) byF(x, y, z) = xyzi+yzj+xyk
Z 2Z 1
2(−y + xy)
is
− x − 2y dy dx.
3
0 0
7. Answer true or false.
The work done by the force
field F(x, y, z) = xi + yj + zk to move a particle completely
around the rectangle (0, 0, 0), (0, 0, 2), (1, 0, 2), and
(1, 0, 0) is 0.
8. Answer true or false. The work done by the force field
F(x, y, z) = x2 i + y3 j + zk to move a particle completely
around the quadrangle (0, 0, 0), (0, 1, 3), (2, 1, 3), and
(2, 0, 0) is 0.
9. Answer true or false. If F(x, y, z) = xi + 2yj + z2 k and
2
2
σZZis a surface such that
ZZ z = −x − y + 2, where z ≥ 0,
σ
(curl F) · n dS =
R
(−2xi − 9yj + 2zk) dA.
10. Answer true or false. If F(x, y, z) = yi + zj + xk and
2
2
σZZis a surface such that
ZZ z = −x − y + 2, where z ≥ 0,
σ
(curl F) · n dS =
(−2x + 2y + 1) dA.
R
11. Answer true or false. If F(x, y, z) = yi + zj + xk and σ
y2
x2
+ 1, where z ≥ 0,
is a surface such that z = − −
2
2
ZZ
ZZ
(curl F) · n dS =
(−x + y + 1) dA.
σ
R
353
12. Answer true or false. If F(x, y, z) = yi + zj + xk and
σZZis a surface such ZZ
that z = 2x + 3y + 4, where z ≥ 0,
σ
(curl F) · n dS =
(2x + 3y + z) dA.
R
13. Answer true or false. F(x, y, z) = x2 i + y3 j + x2 zk. The
work needed to move a particle around a triangle (0, 0, 0),
(0, 2, 3), (0, 0, 3) is 0.
14. Answer true or false. F(x, y, z) = sin xi + ey j + z4 k. The
work needed to move a particle around a triangle (0, 0, 0),
(0, 2, 3), (0, 0, 3) is 0.
15. Answer true or false. F(x, y, z) = sin yi + ex j + z4 k. The
work needed to move a particle around a triangle (0, 0, 0),
(0, 2, 3), (0, 0, 3) is 0.
7. Answer true or false. If x = sint, y = cost, (0 ≤ t ≤ 2π ),
Z
C
(x − 2y) ds =
Z 2π
0
(sint − 2 cost) dt.
8. Answer true or false. The work done by F(x, y) = xi + yex j
along the curve y = x2 from (0, 0) to (1, 1) is the same as
the work done moving the same particle along the same curve
from (1, 1) to (4, 2).
9. Answer true or false. F(x, y) = 6xi + 2yj is a conservative
vector field.
10. Answer true or false. F(x, y) = 3x2 yi + x3 j is a conservative
vector field.
11.
Z (2,4)
(0,1)
6x dx + 2y dy =
(a) 12
(b) 27
Chapter 15 Test
1. Answer true or false. φ (x, y) = xey is the potential function
for F(x, y) = ey i + xey j.
2. F(x, y, z) = zi + yj + xyzk. Find divF.
(a) 1 + xy
(b) 1 − xy
(c) xz + 1 − yz
(d) xz + 1 − yz
3. F(x, y, z) = x3 i + 3j + xzk. Find curlF.
(c) 28
(d) 11
12. F(x, y) = 6xi + yj the work done by the force field on a particle moving along an arbitrary smooth curve from P(0, 0) to
Q(1, 2) is
(a) 5
7
(b)
2
(c) 10
7
(d) −
2
13. Find the area enclosed in the ellipse
(a) j + xyk
(a) 225
(b) j − zk
(b) 225π
(d) xzi − (1 − yz)j
(d) 15π
(c) −zj
4. Answer true or false. ∇2 φ , if φ = x4 y, is 12x2 y + 4x3 .
5. Evaluate
Z
2
C
(x y − 1) ds, where x = 3t and y = t, (0 ≤ t ≤ 1).
(a) 3.95.
(b) 1.75.
(c) 7.
(d) 1.
6. Evaluate
Z
C
6xy2 dx + 15xy2 dy along the curve x = y2 from
(0, 0) to (1, 1).
(c) 15
14. Find the work done by the field F(x, y) = hsin x + y3 , −x3 + ey cos y)i
on a particle that travels once around a unit circle x2 + y2 + 1
in a counterclockwise direction.
3π
(a)
4
3π
(b) −
2
π
(c)
4
(d) π
15. Evaluate
I
(a) 5.
4y2 = 36.
(b) 21.
(a) π
(c) −6.
(b) 12π
(d) −2.
x2 y2
+
= 1.
25
9
(c) 4π
C
(ex + 2y) dx + (3y4 + 2x) dy, where C is 9x2 +
354
(d) 0
16. Evaluate
ZZ
xz dS where σ is the part of the plane
σ
x+y+
across σ is Φ =
z
= 1 in the first octant.
2
(b) 0.121
(c) 0.131
(d) 0.142
σ
0
0
18. Find the flux of the vector field F(x, y, z) = 4zk across the
sphere x2 + y2 + z2 = 9 oriented outward.
64π
3
(c) 0
(d) 144π
0
0
(x2 + y2 + 5) dA.
32π
3
256π
(b)
3
(c) 0
16π
(d)
3
(a)
17. Answer true or false. If σ is the part of x+y+z = 2 that lies in
ZZ
√ Z 1 Z 1−x
the first octant,
x(1 − x − z) dz dx.
xy dS = 3
(b)
Z 2π Z 5
20. Answer true or false. The surface z = 4x + 2y has a normal
4i + 2j + k
√
vector n =
.
21
21. Find the outward flux of the vector field F(x, y, z) = yj across
the sphere x2 + y2 + z2 = 4.
(a) 0.204
16π
(a)
3
19. Answer true or false. If σ is the portion of the surface
z = 5 − x2 − y2 that lies above the xy−plane, and σ is oriented up, the flux of the vector field F(x, y, z) = xi + yj + zk
3xi + 3yj + 3zk
(9x2 + 9y2 + 9z2 )3/2
and σ be a closed orientable surface that surrounds the origin.
Then Φ = 4π .
22. Answer true or false. Let F(x, y, z) =
23. Answer true or false. If σ is the surface z = −x2 − y2 − 4 and
F(x,
ZZzk, then
ZZ y, z) = 6xi + 2yj +
σ
(curl F) · n dS =
(12x + 4y + 1) dA.
R
355
Chapter 15: Answers to Sample Tests
Section 15.1
1. false
9. true
2. true
10. false
3. a
11. false
4. b
12. false
5. a
13. true
6. c
14. true
7. b
15. true
8. d
2. c
10. true
3. b
11. false
4. a
12. false
5. d
13. b
6. c
14. false
7. a
15. false
8. a
2. false
10. d
3. true
11. false
4. true
12. c
5. false
13. a
6. d
14. c
7. a
15. a
8. b
2. d
10. c
3. d
11. c
4. b
12. c
5. b
13. false
6. a
14. false
7. a
15. false
8. d
2. false
10. true
3. b
11. false
4. b
12. false
5. a
13. false
6. c
14. true
7. false
15. false
8. false
2. a
10. c
3. false
11. false
4. false
12. true
5. false
13. false
6. a
14. true
7. false
8. false
2. d
10. a
3. b
11. false
4. b
12. true
5. a
13. false
6. false
14. false
7. c
15. false
8. b
2. false
10. false
3. false
11. false
4. true
12. false
5. false
13. false
6. false
14. true
7. true
15. false
8. true
2. a
10. true
18. d
3. c
11. b
19. false
4. false
12. a
20. false
5. a
13. d
21. a
6. a
14. b
22. false
7. true
15. d
23. c
8. false
16. a
24. false
Section 15.2
1. b
9. false
Section 15.3
1. true
9. a
Section 15.4
1. a
9. d
Section 15.5
1. a
9. true
Section 15.6
1. a
9. c
Section 15.7
1. a
9. b
Section 15.8
1. false
9. false
Chapter 15 Test
1. true
9. true
17. false