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Course Workbook

Understanding Calculus II: Problems, Solutions, and Tips

Professor Bruce H. EdwardsUniversity of Florida

Professor Bruce H. Edwards is Professor of Mathematics at the University of Florida, where he has won a host of teaching awards. Professor Edwards received his B.S. in Mathematics from Stanford University and his Ph.D. in Mathematics from Dartmouth College. Between receiving the two degrees, he taught mathematics at a university near Bogotá, Colombia, as a Peace Corps volunteer. His current research focuses on the CORDIC algorithm, used in computers and graphing calculators to calculate function values. With coauthor Ron Larson, Professor Edwards has published leading texts in calculus, algebra, trigonometry, and other mathematical disciplines.

Understanding C

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i

Bruce H. Edwards, Ph.D.Professor of Mathematics

University of Florida

Professor Bruce H. Edwards has been a Professor of Mathematics at the University of Florida since 1976. He received his B.S. in Mathematics from Stanford University in 1968 and his Ph.D. in Mathematics from

Dartmouth College in 1976. From 1968 to 1972, he was a Peace Corps volunteer in Colombia, where he taught mathematics (in Spanish) at Universidad Pedagógica y Tecnológica de Colombia.

Professor Edwards’s early research interests were in the broad area of pure mathematics called algebra. His dissertation in quadratic forms was titled “Induction Techniques and Periodicity in Clifford Algebras.” Beginning in 1978, Professor Edwards became interested in applied mathematics while working summers for NASA at the Langley Research Center in Virginia. This work led to his research in numerical analysis and the solution of differential equations. During his sabbatical year, 1984 to 1985, he worked on two-point boundary value problems with Professor Leo Xanthis at the Polytechnic of Central London. Professor Edwards’s current research is focused on the algorithm called CORDIC that is used in computers and graphing calculators for calculating function values.

Professor Edwards has coauthored a number of mathematics textbooks with Professor Ron Larson of Penn State Erie, The Behrend College. Together, they have published leading texts in calculus, applied calculus, linear algebra, finite mathematics, algebra, trigonometry, and precalculus.

Over the years, Professor Edwards has received many teaching awards at the University of Florida. He was named Teacher of the Year in the College of Liberal Arts and Sciences in 1979, 1981, and 1990. In addition, he was named the College of Liberal Arts and Sciences Student Council Teacher of the Year and the University of Florida Honors Program Teacher of the Year in 1990. He also served as the Distinguished Alumni Professor for the UF Alumni Association from 1991 to 1993. The winners of this two-year award are selected by graduates of the university. The Florida Section of the Mathematical Association of America awarded Professor Edwards the Distinguished Service Award in 1995 for his work in mathematics education for the state of Florida. His textbooks have been honored with various awards from the Text and Academic Authors Association.

Professor Edwards has taught a wide range of mathematics courses at the University of Florida, from first-year calculus to graduate-level classes in algebra and numerical analysis. He particularly enjoys teaching calculus to freshman because of the beauty of the subject and the enthusiasm of the students.

Professor Edwards has been a frequent speaker at both research conferences and meetings of the National Council of Teachers of Mathematics. He has spoken on issues relating to the Advanced Placement calculus examination, especially on the use of graphing calculators.

Professor Edwards has taught three other Great Courses: Understanding Calculus: Problems, Solutions, and Tips; Mathematics Describing the Real World: Precalculus and Trigonometry; and Prove It: The Art of Mathematical Argument. ■

Table of Contents

ii

Professor Biography ��iScope ��1

LEsson 1Basic Functions of Calculus and Limits ��3

LEsson 2Differentiation Warm-Up ��7

LEsson 3Integration Warm-Up ��12

LEsson 4Differential Equations—Growth and Decay ��16

LEsson 5Applications of Differential Equations ��19

LEsson 6Linear Differential Equations ��23

LEsson 7Areas and Volumes ��26

LEsson 8Arc Length, Surface Area, and Work ��31

LEsson 9Moments, Centers of Mass, and Centroids ��35

LEsson 10Integration by Parts ��40

LEsson 11Trigonometric Integrals ��43

LEsson 12Integration by Trigonometric Substitution ��47

LEsson 13Integration by Partial Fractions ��52

LEsson 14Indeterminate Forms and L’Hôpital’s Rule ��56

LEsson 15Improper Integrals ��60

LEsson GUIDEs

InTRoDUCTIon

iii

Table of Contents

LEsson 16Sequences and Limits ��64

LEsson 17Infinite Series—Geometric Series ��68

LEsson 18Series, Divergence, and the Cantor Set ��72

LEsson 19Integral Test—Harmonic Series, p-Series ��76

LEsson 20The Comparison Tests ��80

LEsson 21Alternating Series ��83

LEsson 22The Ratio and Root Tests ��87

LEsson 23Taylor Polynomials and Approximations ��91

LEsson 24Power Series and Intervals of Convergence ��96

LEsson 25Representation of Functions by Power Series �� 100

LEsson 26Taylor and Maclaurin Series �� 104

LEsson 27Parabolas, Ellipses, and Hyperbolas �� 109

LEsson 28Parametric Equations and the Cycloid ��113

LEsson 29Polar Coordinates and the Cardioid ��118

LEsson 30Area and Arc Length in Polar Coordinates �� 123

LEsson 31Vectors in the Plane �� 127

LEsson 32The Dot Product of Two Vectors �� 131

LEsson 33Vector-Valued Functions �� 136

Table of Contents

iv

LEsson 34Velocity and Acceleration �� 140

LEsson 35Acceleration’s Tangent and Normal Vectors �� 144

LEsson 36Curvature and the Maximum Bend of a Curve �� 148

sUPPLEmEnTaL maTERIaL

Solutions �� 152Glossary �� 201Summary of Differentiation Formulas �� 209Summary of Integration Formulas �� 210Power Series for Elementary Functions ��211Summary of Tests for Convergence and Divergence �� 212Bibliography �� 213

1

Understanding Calculus II: Problems, solutions, and Tips

Scope:

The goal of this course is to further your understanding and appreciation of calculus. Just as in Understanding Calculus: Problems, Solutions, and Tips, you will see how calculus plays a fundamental role in all of science and engineering.

In the first third of the course, you’ll use the tools of derivatives and integrals that you learned in calculus I to solve some of the great detective stories of mathematics—differential equations. The middle portion of the course will take you to the beautiful world of infinite series and their connection to the functions you have learned about in your studies of precalculus and calculus. Finally, the third part of the course will lead to a solid understanding of key concepts from physics, including particle motion, velocity, and acceleration.

Calculus is often described as the mathematics of change. The concepts of calculus—including velocities, accelerations, tangent lines, slopes, areas, volumes, arc lengths, centroids, curvatures, and work—have enabled scientists, engineers, and economists to model a host of real-life situations.

For example, a physicist might need to know the work required for a rocket to escape Earth’s gravitational field. You will see how calculus allows the calculation of this quantity. An engineer might need to know the balancing point, or center of mass, of a planar object. The integral calculus is needed to compute this balancing point. A biologist might want to calculate the growth rate of a population of bacteria, or a geologist might want to estimate the age of a fossil using carbon dating. In each of these cases, calculus is needed to solve the problem.

Although precalculus mathematics (geometry, algebra, and trigonometry) also deals with velocities, accelerations, tangent lines, slopes, and so on, there is a fundamental difference between precalculus mathematics and calculus. Precalculus mathematics is more static, whereas calculus is more dynamic. The following are some examples.

• An object traveling at a constant velocity can be analyzed with precalculus mathematics. To analyze the velocity of an accelerating object, you need calculus. In fact, vectors and vector-valued functions will be developed in this course in order to analyze motion in the plane.

• The slope of a line can be analyzed with precalculus mathematics. To analyze the slope of a curve, you need calculus. Furthermore, the amount that a curve bends (its curvature) will be studied in this course.

• Finite summations can be analyzed with precalculus mathematics. However, calculus is needed to study infinite summations, or series. These series can be used to represent all the functions you have studied in precalculus and elementary calculus.

This course is presented in the same order as a university-level calculus course. The material is based on the 10th edition of the bestselling textbook Calculus by Ron Larson and Bruce H. Edwards (Brooks/Cole, 2013). However, any standard calculus textbook can be used for reference and support throughout the course.

As you progress through the course, most concepts will be introduced using illustrative examples. You will encounter all the important theoretical ideas and theorems but not dwell on their technical proofs. You will find that it is easy to understand and apply calculus to real-world problems without knowing these theoretical intricacies.

2

Graphing calculators and computers are playing an increasing role in the mathematics classroom. Without a doubt, graphing technology can enhance the understanding of calculus, so some instances where graphing calculators are used to verify and confirm calculus results have been included.

This course begins with a brief review of key ideas from precalculus and first-semester calculus, including pitfalls to avoid. You will then dive immediately into the vast field of differential equations, including their applications to science and engineering, focusing on topics covered in the Advanced Placement Program.

Next, you will deepen your understanding of two other topics from elementary calculus. First, you will learn more applications of integration, including centers of mass and work. These applications follow the general principles of finding areas and volumes of geometric objects. Second, you will learn more advanced integration techniques, including integration by parts, trigonometric integrals, trigonometric substitution, and partial fractions. A particularly interesting topic involves the calculation of limits by L’Hôpital’s rule and the study of integrals defined on infinite intervals.

The course then turns to the intriguing topic of infinite series. How can you add up an infinite number of numbers? Do infinite summations make sense? You will see that this is indeed possible. You might have seen the following geometric series.

1 1 11 2.2 4 8

+ + + + =

You will also see that most of your favorite functions can be represented by infinite summations. For example, the exponential function has the following infinite series representation.

2 3 4

1 .2 6 24

= + + + + +x x x xe x

One of the surprising consequences of your study of series will be the derivation of the beautiful formula

1,ie π = - where i is the imaginary number 1 .i = -

There are two main themes of the third part of this course. The first is the study of particle motion, including velocity and acceleration. The second is the development of key ideas that will be needed in third-semester calculus—the extension of calculus from two to three dimensions. Hence, this third part of the course covers conics, parametric equations, polar coordinates, vectors, and vector-valued functions.

By the end of the course, you will have covered all the main topics of second-semester university calculus, including those covered in the Advanced Placement Calculus BC course. In particular, you will be ready to study the calculus of three dimensions, which completes a three-semester university calculus sequence.

Students are encouraged to use all course materials to their maximum benefit, including the video lessons, which you can review as many times as you wish; the individual lesson summaries and accompanying problems in the workbook; and the supporting materials in the back of the workbook, including the solutions to all problems and various review items.

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3

Basic Functions of Calculus and LimitsLesson 1

Topics

• Introduction.

• Linear functions.

• Polynomial and radical functions.

• Trigonometric functions.

• Logarithmic and exponential functions.

• Limits.

Definitions and Theorems

• The line y mx b= + has slope m and y-intercept ( )0,b .

• The functions f and g are inverses of each other if ( ( )) , ( ( ))f g x x g f x x= = for all values of x in the appropriate domains.

• Key limits: 0

sinlim 1x

xx→

= , 0

cos 1lim 0x

xx→

-= ,

0

1lim 1x

xe

x→

+ =

.

• Key logarithmic properties:

( )ln ln ln .

ln ln ln .

ln ln .b

ab a b

a a bb

a b a

= +

= -

=

Summary

Welcome to Understanding Calculus II: Problems, Solutions, and Tips! In our first lesson, we review our basic list of functions and their graphs. Without a doubt, confidence in precalculus is crucial for success in calculus. Hence, we open with a list of the top 10 pitfalls in precalculus. The subsequent review of functions will be brief, and you should not hesitate to consult textbooks on algebra and trigonometry to fill in the blanks. Along the way, we will indicate some of the new topics that are coming in later lessons. We will also recall some important limits and their meanings. The lesson closes by discussing how to find the slope of the sine curve at a given point.

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Example 1: The Equation of a Line

The equation 3 5y x= - represents a line with slope 3 and y-intercept ( )0, 5- . The line 3 1y x= + is parallel to the given line, but with y-intercept ( )0,1 . The slope of a line perpendicular to these lines would be the negative reciprocal of 3—that is, 1

3- .

Example 2: Inverse Functions

The functions 3( )f x x= and 13 3( )g x x x= = are inverses of each other, which means that their graphs are

symmetric across the line .y x=

Example 3: The Absolute Value Function

The absolute value function ( )f x x= is continuous, but not differentiable, at 0.x = Its graph appears in the shape of the letter V.

Example 4: Inverse Trigonometric Functions

The inverse sine function is defined by first restricting the domain of the sine function to the closed interval

,2 2π π -

. Then, we have if and only if sin ,x y= where 2 2yπ π

- ≤ ≤ and 1 1x- ≤ ≤ . A similar argument

applies to the definition of the arctangent function.

Example 5: Graphs of Inverses

The functions ( ) lnf x x= and ( ) xg x e= are inverses of each other. Their graphs are symmetric across the line .y x=

Study Tips

• Horizontal lines have equations of the form y c= , a constant. Slope is not defined for vertical lines. Their equations are of the form x b= , where b is a constant.

• The graphs of inverse functions are symmetric across the line .y x=

• You have to restrict the domains of the trigonometric functions in order to define their inverses.

• The natural logarithmic and exponential functions are inverses of each other—that is, ln xe x= and ln xe x= .

5

Pitfalls

The following are the top 10 precalculus pitfalls.

10. Make sure that your graphing calculator is in the correct mode—radian or degree.

9. The notation ( )1f x- does not mean ( )1

f x. For example, ( )1 1sin

sinx

x- ≠ , but rather, 1sin arcsinx x- = .

In general, this is the notation for the inverse function of .f

8. Infinity is not a number. For example, never write [ ]0,∞ , but rather, [ )0,∞ .

7. You can’t divide by zero in mathematics. For example, 0

sinlim 1x

xx→

= .

6. Answers can appear in different forms. For example, ln sec ln cosθ θ= - .

5. Be careful with radical expressions. For example, ( )( ) ( )( )4 4 1 4 1 2 2i i= - - ≠ - - = = - .

4. Be careful with exponents. For example, 3 4 122 2 2≠ , but instead, 3 4 3 4 72 2 2 2+= = .

3. Be careful with logarithms. Remember that for , 0,a b > ln ln lna b ab+ = . The logarithm of a sum is not the sum of the logarithms.

2. You can’t split up denominators. For example, 1 1 1

3 cos 3 cosx x≠ -

- .

1. Logarithms can be confusing. Remember that ln 1e = and ln1 0,= whereas ln 0 is undefined.

Problems

1. Find the slope and y-intercept of the line 2 3 9x y- = .

2. Find the equation of the line passing through the point ( )2,1 and perpendicular to the line in Exercise 1.

3. How does the graph of ( ) 2 3g x x= - + differ from that of ( )f x x= ?

4. Show that ( ) 5 1f x x= + and 1( )5

xg x -= are inverses of each other.

6

5. Find the inverse function of ( ) 2.f x x= -

6. Find the horizontal and vertical asymptotes of the graph of 2

2

2 1( )1

xf xx

+=

-.

7. How would you restrict the domain of the cosine function in order to define its inverse function?

8. Use the formula for the sine of a sum to prove the formula sin 2 2sin cosu u u= .

9. Use the formula for the cosine of a sum to prove the formula 2cos 2 1 2sinu u= - .

10. Calculate the limit 20

1 coslimx

xx→

- .

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Differentiation Warm-UpLesson 2

Topics

• Tangent lines to curves.

• Derivative formulas.

• Derivatives of trigonometric functions.

• Derivatives of logarithmic and exponential functions.

• Implicit differentiation and inverse functions.

• Derivatives and graphs.


• The derivative of a constant function is 0.

• For polynomial and radical functions, you have ( ) 1n nd x nxdx

-= .

• The derivatives of the 6 trigonometric functions:

2 2

(sin ) cos , (cos ) sin .(tan ) sec , (cot ) csc .(sec ) sec tan , (csc ) csc cot .

x x x xx x x xx x x x x x

′ = ′ = -

′ = ′ = -′ = ′ = -

• The derivative of the logarithmic function: [ ] 1lnd xdx x

= .

• The derivative of the exponential function: (sin ) cos , (cos ) sinx x x x′ = ′ = - .

• The derivatives of the 3 most important inverse trigonometric functions:

[ ]

[ ]

[ ]

2

2

2

1arcsin .11arctan .

11arcsec .

1

d xdx xd xdx xd xdx x x

=−

=+

=−

• The chain rule: If ( )y f u= is a differentiable function of u and ( )u g x= is a differentiable function

of x, then ( )( )y f g x= is a differentiable function of x and dy dy dudx du dx

= ⋅ , or equivalently,

( )( ) ( )( ) ( )d f g x f g x g xdx

′ ′= .

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Summary

In this second review lesson, we recall the main concepts of differentiation. We first use the formal definition of derivative to show that the slope of the tangent to the curve siny x= at any point ( ),x y equals cos .x We then review the basic formulas for calculating derivatives. We recall the derivatives of trigonometric, logarithmic, and exponential functions. We then apply our knowledge of derivatives to the analysis of a graph. Finally, we close by reversing the problem: Given the derivative of a function, what is the original function?

Example 1: The Slope of the Tangent Line to the Sine Function

Find the slope of the tangent line to the graph of the sine function at any point ( ),x y .

Solution

Let ( )( ),sinx x x x+ ∆ + ∆ be a point near ( ),x y on the sine graph. Then, the slope of the line joining these

points is ( )( )

( )sin sin sin sinx x x x x xm

x x x x+ ∆ - + ∆ -

≈ =+ ∆ - ∆

.

The exact slope is obtained by taking the limit as 0x∆ → :

( )

( ) ( )0 0

0 0

sin sin sin cos( ) cos sin( ) sinlim lim

sin cos 1lim cos lim sin cos (1) sin (0) cos .

x x

x x

x x x x x x x xmx x

x xx x x x x

x x

∆ → ∆ →

∆ → ∆ →

+ ∆ - ∆ + ∆ -= =

∆ ∆∆ ∆ -

= + = + =∆ ∆

Example 2: The Derivative of a Function

Calculate the derivative of the function 6 4 2( ) 3f x x x xx

π= + - + - .

Solution

We first rewrite the function and then differentiate, as follows.

( )16 4 6 4 12

15 3 2 5 322

23 3 2

1 1 26 12 2 6 12 .2 2

d dx x x x x x xdx x dx

x x x x x xxx

π π-

- -

+ - + - = + - + -

= + - - = + - -

Example 3: The Derivative of a Logarithmic Function

Calculate the derivative of ( ) ln(4 )f x x= .

Solution

We can use the chain rule, or properties of the logarithmic function.

( )

( ) ( )0 0

0 0

sin sinlim lim

sin cos 1lim cos lim sin cos (1) sin (0) cos .

x x

x x

x x x x x x x xmx x

x xx x x x x

x x

∆ → ∆ →

∆ → ∆ →

+ ∆ − ∆ + ∆ −= =

∆ ∆∆ ∆ −

= + = + =∆ ∆

9

( ) ( )

( ) ( )

1 1ln 4 4 .4

1ln 4 ln 4 ln .

xx x

x xx

′ = =

′ ′= + =

Example 4: The Derivative of an Exponential Function

Use the quotient rule to calculate the derivative of 2

( )xef x

x= .

Solution

( ) ( )( )( ).

2 22 22

2 2 2

( ) 2 1 2 1 2 1x xx x

xx e ed e ee x

dx x xx x x

− = = − = −

Example 5: Implicit Differentiation

Use implicit differentiation to calculate dydx

if sin .x y=

Solution

We implicitly differentiate both sides to obtain 11 cos .cos

dy dyydx dx y

= ⇒ =

Next, we use the fundamental

trigonometric identity to express our answer in terms of .x

2 2

2

1 1cos 1 sin 1cos 1

dyy y xdx y x

= - = - ⇒ = =-

.

Notice that we are assuming that 2 2

yπ π- ≤ ≤ , which ensures that cos 0y ≥ .

Example 6: Derivatives and Graphs

Analyze the graph of the function 5 4

3 3( ) 2 5f x x x= - .

Solution

We first calculate the derivatives ( ) ( )1 13 310 2

3f x x x′ = - and ( )

( )13

23

20 1.

9

xf x

x

-′′ = Solving ( ) 0f x = , we

obtain the intercepts ( ) 1250,0 and ,08

. Setting the first derivative equal to zero, we obtain the critical

numbers 0x = and 8.x = By analyzing the sign of the first derivative, we see that the graph is increasing on

( ) ( ),0 and 8,-∞ ∞ and decreasing on ( )0,8 . Analyzing the second derivative, we see that the graph is concave downward on ( ),0-∞ and ( )0,1 and concave upward on ( )1,∞ . There is an inflection point at 1.x = Finally, we see that there is a relative minimum at ( )8, 16- and a relative maximum at ( )0,0 . Try graphing the function with your graphing utility to verify these results.

10

( )0,0

( )8, 16-

y x x= −2 55

34

3

1258

1 3,−( )

Study Tips

• In general, the rules for differentiation are straightforward. Many computers and graphing calculators have built-in capabilities for calculating derivatives.

• Keep in mind that your answer might not look like the answer in the textbook.

• For example, your answer for the derivative of sin cosy x x= might be 2 2cos sinx x- , whereas the textbook might have the equivalent answer cos 2x .

• In Example 5, we actually calculated the derivative of the inverse sine function.

• In Example 1, some textbooks use the notation h instead of x∆ .

Pitfalls

• The derivative of a product is not the product of the derivatives. In fact,

( ) ( ) ( ) ( ) ( ) ( ).d f x g x f x g x g x f xdx

′ ′= +

• The derivative of a quotient is not the quotient of the derivatives. In fact,

( )( )

( ) ( ) ( ) ( ) ( ) 2

f x g x f x f x g xddx g x g x

′ ′−=

.

• Remember to use the chain rule. The derivative of sin 6y x= is not cos 6 ,y x′ = but rather, 6cos 6 .y x′ =

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1. Find the derivative of 43

1 3( ) 2 17.g x xxx

= - + +

2. Find the equation of the tangent line to cosy x= at the point ( )0,1 .

3. Find the derivative of 3 ln 4xy e x= .

4. Find the slope of the graph of sin 2( )cos3

xf xx

= at the point ( ),0π .

5. Use implicit differentiation to find dydx

if tan .x y= What is the derivative of the arctangent function?

6. Find the critical numbers of 2 2 1( )

1x xf x

x- +

=+

. Determine the open intervals on which the graph is increasing or decreasing.

7. Find the critical numbers of ( )2

3( ) 2g x x= + . Determine the open intervals on which the graph is increasing or decreasing.

8. Determine the intervals on which the graph of 2

2

4( )4xf x

x+

=-

is concave upward or concave downward.

9. Find all relative extrema of the function ( ) cos sin .f x x x= +

10. What is the function f if 6 2 21( ) sec ,0 ?2

xf x x x e xx

π′ = + + + < <

Problems

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Integration Warm-UpLesson 3

Topics

• Antiderivatives.

• Integration by substitution.

• The definite integral and area.

• The fundamental theorem of calculus.

• The second fundamental theorem of calculus.


• Let f be a continuous function on the closed interval [ ],a b and let F be an antiderivative of .f The

fundamental theorem of calculus says that ( ) ( ) ( )b

af x dx F b F a= -∫ .

• Let f be continuous on the open interval I containing .a Then, for every x in the interval, the second

fundamental theorem of calculus says that ( ) ( )x

a

d f t dt f xdx

= ∫ .

Summary

In this third review lesson, we recall the basic facts about integration. We first discuss how to find antiderivatives using our knowledge of derivatives. After a simple example of integration by substitution, we turn to definite integrals and the area problem. The fundamental theorem of calculus is the major theorem here, permitting us to evaluate definite integrals using antiderivatives. This theorem and the second fundamental theorem of calculus show how integration and differentiation are basically inverse operations. We end the lesson by solving a differential equation.

Example 1: Finding Antiderivatives

a. ( )5

4 cos sin5

x xxx x e dx x e C+ + = + + +∫ .

b. 1 lndx x Cx

= +∫ .

Example 2: Integration by Substitution

Calculate the integral 2sinx x dx∫ .

13

Solution

We use integration by substitution by letting 2u x= and 2du x dx= . Hence, we have

( ) ( )2 2 21 1 1sin 2 cos cos2 2 2

x x dx x C x C.= − + = − +∫

Example 3: Definite Integrals

Set up the definite integral for the area under the semicircle 225y x= − .

Solution

The limits of integration are 5x = - and 5.x = Hence, the area is given by the integral

( )5 22

5

1 2525 52 2

x dx ππ−

− = =∫ .

225y x= − .

Example 4: The Fundamental Theorem of Calculus

Use the fundamental theorem of calculus to evaluate the integral ( )4 2

416 x dx

--∫ .

Solution

We find an antiderivative for the integrand and then evaluate it at the two endpoints.

( )43 3 34 2

44

4 ( 4) 128 25616 16 16(4) 16( 4) 1283 3 3 3 3xx dx x

--

-- = - = - - - - = - =

∫ .

14

Example 5: The Second Fundamental Theorem of Calculus

Verify the second fundamental theorem of calculus for the derivative 1

cosxd t dt

dx ∫ .

Solution

We first integrate, then differentiate, as follows

[ ] [ ]11cos sin sin sin1 cos

x xd d dt dt t x xdx dx dx

= = - = ∫ .

Study Tips

• You can check your work to an integration problem by differentiating the answer. For instance, in

Example 2, ( )( )2 2 21 1cos sin 2 sin2 2

x x x x x ′- = - - =

, which is the original integrand.

• It is often helpful to take advantage of symmetry. For instance, in Example 4, the integral is equivalent

to ( ) ( )4 42 2

4 016 2 16x dx x dx

-- = -∫ ∫ . The lower limit of 0 will make the integration much easier.

• When using the fundamental theorem of calculus, you don’t need to add a constant of integration. In other words, any antiderivative will do.

• In general, integration is more difficult than differentiation. Fortunately, computers and calculators can find many antiderivatives. Furthermore, there are many numerical techniques for definite integrals, such as the trapezoidal rule and Simpson’s rule.

• Many integration formulas come directly from corresponding derivative formulas. For example, because

we know that 2

1arctan1

d xdx x

=+

, 2

1 arctan1

dx x Cx

= ++∫ .

• Keep in mind that you need to have a constant of integration for indefinite integrals (antiderivatives), but not for definite integrals.

Pitfalls

• You cannot move variables outside the integral sign. For example, 1 13 3

0 0cos cosx x dx x x dx≠∫ ∫ .

• When using the second fundamental theorem of calculus, make sure that the constant a is the lower

limit of integration and that x is the upper limit. For example, ( ) ( )a x

x af t dt f t dt= -∫ ∫ , whereas the

integral 2

( )x

af t dt∫ also requires the chain rule.

• The integral of ( ) 1f x x= is ln .x C+ You can omit the absolute value sign if the variable x is positive.Le

sson

3: I

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n W

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-Up

15

Problems

1. Find the antiderivative: ( )xx e dx-+∫ .

2. Set up the definite integral for the area bounded by the line 2y x= and the x-axis, between 1x = and 5.x =

3. Use the fundamental theorem of calculus to evaluate ( )0 1 23 3

1t t dt

--∫ .

4. Use the fundamental theorem of calculus to evaluate 2

0

2cos3x dx

π ∫ .

5. Use the fundamental theorem of calculus to evaluate 1

1

12 3

dxx- +∫ .

6. Find the area bounded by 2 2, 0, 0xy e y x-= + = = , and 2.x =

7. Solve the differential equation ln , (1) 2dy x ydx x

= = - .

8. Solve the differential equation 2 1 1( ) sin , (0) , (0)4 2

xf x x e f f′′ ′= + = = .

9. Find ( )F x′ if 2

0( ) sin

xF x t dt= ∫ .

10. Use the substitution 6u x= + to calculate 6x x dx+∫ .

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Differential Equations—Growth and DecayLesson 4

Topics

• Verifying solutions to differential equations.

• Separation of variables.

• Euler’s method.

• Growth and decay models.

• Applications.


• A differential equation in x and y is an equation that involves x, y, and derivatives of y.

• A function ( )f x is a solution to a differential equation if the equation is satisfied when y and its derivatives are replaced by ( )f x and its derivatives.

• Euler’s method for approximating the solution to 0 0( , ), ( )y F x y y x y′ = = is given by

( )1 1 1 1, ,n n n n n nx x h y y hF x y- - - -= + = + . Here, ( ),F x y is a convenient notation to indicate that y′

equals an expression involving both x and y .

• The solution to the growth and decay model dy kydt

= is .kty Ce=

Summary

In this lesson, we review the basic concepts of differential equations. We show how to use the technique of separation of variables to solve separable equations. We develop Euler’s method, a simple numerical algorithm for approximating solutions to differential equations. Then, we look at growth and decay models and present two applications.

Example 1: Verifying a Solution to a Differential Equation

Verify that the function 3y Cx= is a solution to the differential equation ' 3 0.xy y- =

Solution

We have 3,y Cx= which implies that 23 .y Cx′ = Hence, ( ) ( )2 3' 3 3 3 0,xy y x Cx Cx- = - = which verifies that 3y Cx= is a solution to the differential equation.

17

Example 2: Separation of Variables

Use the technique of separation of variables to solve the differential equation 2 .xyy

′ =

Solution

We bring all the terms involving x to one side and those involving y to the other side. Then, we

integrate both sides to obtain the solution: 2' 2 .dy xy y dy x dxdx y

= = ⇒ = Integrating both sides,

22

12 .2yy dy x dx x C= ⇒ = +∫ ∫ Simplifying, we obtain the solution 2 22y x C- = .

Example 3: Using Euler’s Method

Use Euler’s method with 0.1h = to approximate the solution to the differential equation ' , (0) 1y x y y= - = at the points 0.1 and 0.2x = .

Solution

Euler’s method in this case is ( ) 0 0' , , 0, 1, 0.1y F x y x y x y h= = - = = = . Then, we have

( ) ( )( )1 1 0 0 0 0 00.1 and , 1 0.1 1 (0.1)(0 1) 0.9.x y y hF x y x y= = + = + - = + - = The next step is

( ) ( )( )2 2 1 1 1 1 10.2 and , 0.9 0.1 0.9 (0.1)(0.1 0.9) 0.82x y y hF x y x y= = + = + - = + - = .

Example 4: An Application to Carbon Dating

Carbon 14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true, then the amount of carbon 14 absorbed by a tree that grew several centuries ago would be the same as the amount absorbed by a tree today. A piece of ancient charcoal contains only 15% as much as a piece of modern charcoal. How long ago was the tree burned? (The half-life of carbon 14 is 5715 years.)

Solution

Using the growth and decay model kty Ce= , the initial amount of carbon 14 is .C Hence,

( ) ( )5715 57151 1 .2 2

k kC Ce e= ⇒ = Using the properties of logarithms, we have ( ) 15715 ln 0.0001213.2

k k= ⇒ ≈ -

Hence, 0.00012130.15 0.15kt tC Ce e-= ⇒ = . Solving for t, we obtain ln 0.15 0.0001213 15,640 yearst t= − ⇒ ≈ .

18

Study Tips

• The general solution to a differential equation will contain one or more arbitrary constants. You need to have appropriate initial conditions to determine these constants.

• You can always check your solution to a differential equation by substituting it back into the original equation.

• Euler’s method is the simplest of all the numerical methods for solving differential equations. You can obtain more accurate approximations by using a smaller step size, or a more accurate method, such as the Runge-Kutta method.

• For the growth and decay model kty Ce= , there is growth when 0k > and decay when 0.k <

• Logarithmic manipulations play an important role in solving growth and decay applications. For

example, the following is how to solve for an exponent .k

Finally, 1ln ln 22

5715 5715k -

= = .

Pitfalls

• Be careful when working with logarithms. For example, in the computation

( )32 ln 0.42367k k- = ⇒ ≈ , the final answer is positive because ( )3ln 07 < .

• Unfortunately, not all differential equations can be solved by separation of variables. To use this method, you have to be able to move all of the terms containing x to one side and all of the terms containing y to the other side.

Problems

1. Verify that 4xy Ce= is a solution to the differential equation 4 .y y′ =

2. Verify that 2 22 lny y x- = is a solution to the differential equation ( )2 1 .y xy y′ = -

3. Verify that cos xy e-= satisfies the differential equation ( )sin , 2 1.y y x y π′ = =

4. Verify that 1 2sin 3 cos3y C x C x= + satisfies the differential equation 9 0.y y′′+ = Then, find the particular solution satisfying ( )6 2y π = and ( 6) 1.y π′ =

5. Use integration to find the general solution of the differential equation 2

.xy xe′ =

6. Solve the differential equation 5dy xdx y

= .

7. Solve the differential equation (1 ).y x y′ = +

8. Find the equation of the graph that passes through the point ( )9,1 and has slope2yyx

′ = .

9. Use Euler’s method with 0.1h = to approximate (0.2)y for the differential equation , (0) 2.y x y y′ = + =

10. Radioactive radium has a half-life of approximately 1599 years. What percent of a 20-gram amount remains after 1000 years?

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e e kk k5715 571512

12 5715 1

2( ) = ⇒ = ⇒ =ln ln ( ) ln .( )

19

applications of Differential EquationsLesson 5

Topics

• More on separation of variables.

• Orthogonal trajectories.

• The logistic differential equation.

• An application of the logistic differential equation.


• The orthogonal trajectories of a given family of curves are another family of curves, each of which is orthogonal to every curve in the given family.

• The logistic differential equation is 1 ; , 0dy yky k Ldt L

= - >

, where k is the rate of growth (or decay)

and L is a limit on the growth (or decay). Its solution is 1 kt

Lybe-=

+.

Summary

In this lesson, we look at some important applications of differential equations. After reviewing the technique of separation of variables, we study orthogonal trajectories, which consist of a family of curves, each of which is orthogonal (perpendicular) to every curve in a given family. For example, in thermodynamics, the flow of heat across a plane surface is orthogonal to the isothermal curves (curves of constant temperature). Then, we develop the famous logistic differential equation and use it in an application involving animal populations. This model was first proposed by the Dutch mathematical biologist Pierre-Francois Verhulst in the 1840s.

Example 1: Separation of Variables

Solve the differential equation ( )2 4 dyx xydx

+ = .

Solution

Move all of the terms involving x to one side of the equation and all of the terms involving y to the other side. Then, integrate both sides.

( )212 2

1ln ln 44 4 2

dy x dy xdx dx y x Cy x y x

= ⇒ = ⇒ = + ++ +∫ ∫ .

The calculus portion of the problem is finished. Next, we need to use our precalculus skills to solve for .y

1 1 12 2 2ln ln 4 ln ln 4 4C C Cy x e e x y e x = + + = + ⇒ = + . Hence, the final answer is 2 4y C x= + .

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Example 2: Orthogonal Trajectories

Find the orthogonal trajectories for the family of hyperbolas given by .Cyx

=

Solution

Rearrange the equation so that the constant C is on one side, and implicitly differentiate to find :dydx

Cy yx Cx

= ⇒ = . Hence, 0dy dy yy xdx dx x

+ = ⇒ = - . Because the slope at any point ( ),x y is ,yx

- the

slope for the orthogonal family is .dy xdx y

= Next, we separate variables to find the orthogonal trajectories:

2 2

1.2 2y xy dy x dx y dy x dx C= ⇒ = ⇒ = +∫ ∫ Finally, we see that the curves are hyperbolas, 2 2y x K- = .

xy C=2 2y x K- =

Example 3: An Application of the Logistic Differential Equation

A state game commission releases 40 elk into a game refuge. After 5 years, the elk population is 104. The commission believes that the environment can support no more than 4000 elk. Use the logistic model to estimate the elk population after 15 years.

Solution

The carrying capacity is 4000,L = so the solution becomes 40001 kty

be-=+

. At time 0,t = there are 40 elk, so we

21

can find ,b as follows. (0)

4000 4000(0) 40 40 40 40 4000 3960 / 40 991 1ky b b

be b-= ⇒ = = ⇒ + = ⇒ = =+ +

. At time

5, 104,t y= = so we can find .k (5)

4000104 0.1941 99 k k

e-

= ⇒ ≈ + . Our model is complete:

0.194

4000 .1 99 ty

e-=+

At

time 0.194(15)

400015, 626 elk1 99

t ye−

= = ≈+

.

Study Tips

• Keep in mind that you can always check your solution to a differential equation by substituting it, and its derivatives, into the original differential equation.

• In the growth and decay model ,kty Ce= growth (or decay) is unlimited. In the logistic model, there is a limit on the population, called the carrying capacity. In other words, y L= is a horizontal asymptote of the graph of the solution.

• Two trivial solutions to the logistic differential equation 1dy ykydt L

= -

are 0y = and y L= .

Pitfalls

• Although sometimes true, the function 0y = is not always a solution to a separable differential equation.

For example, 0y = is not a solution to the equation dy xdx y

= .

• Scientists use many different symbols for the independent and dependent variables. For example, you will often see t or θ instead of the independent variable .x In addition, the dependent variable might be P or N instead of y .

• Don’t forget to include the absolute value sign when integrating the function 1 ,x

1 lndx x Cx

= +∫ .

Problems

1. Solve the differential equation dy x ydx

= .

2. Solve the differential equation ln 0y x xy′- = .

3. Find the particular solution of the differential equation 2sin , (0) 1.du uv v udv

= =

4. Find the orthogonal trajectories of the family .xy Ce=

22

5. Find the orthogonal trajectories of the family 2 .x Cy=

6. Verify that 1 kt

Lybe-=

+ satisfies the logistic differential equation 1 .dy yky

dt L = -

7. Solve the logistic differential equation 1 , (0) 4.36

dy yy ydt

= - =

8. Solve the logistic differential equation 24 , (0) 8.

5 150dy y y ydt

= - =

9. At time 0t = , a bacterial culture weighs 1 gram. Two hours later, the culture weighs 4 grams. The maximum weight of the culture is 20 grams. Write a logistic equation that models the weight of the bacterial culture. Then, use your model to find the weight after 5 hours.

10. Write the differential equation that models the following verbal statement: The rate of change of y with respect to x is proportional to the difference between y and 4.

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Linear Differential EquationsLesson 6

Topics

• Linear differential equations.

• Integrating factors.

• Applications.


• A first-order linear differential equation is an equation that can be written in the standard form

( ) ( )dy P x y Q xdx

+ = . Here, ( )P x and ( )Q x are continuous functions of the independent variable .x

• The integrating factor for a first-order linear differential equation in the standard form is ( )P x dxu e∫= .

Summary

In this lesson, we study linear differential equations, which typically cannot be solved by separation of variables. The key to their solution is the use of an integrating factor. By multiplying both sides of the equation by the integrating factor, the left-hand side becomes the derivative of a product involving the solution .y Hence, the solution is obtained by integrating both sides and solving for .y We present some applications of linear differential equations and finally return to an equation we considered early in our development of Euler’s method.

Example 1: Using an Integrating Factor

Multiply both sides of the differential equation xy y e′+ = by the integrating factor xe and solve the resulting equation.

Solution

Multiplying through by the integrating factor xe produces the product of a derivative on the left-hand side of the

original differential equation: 2x x x x x xy e ye e e ye e′′ + = ⇒ = . Next, integrate both sides and solve for y:

2 21 12 2

x x x x xye e dx e C y e Ce-= = + ⇒ = +∫ .

Example 2: Solving a Linear Differential Equation

Solve the differential equation 2 , 0y y x xx

′ - = > .

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Solution

We first calculate the integrating factor: 22( ) 2 ln lnP x dx dx x xx

= - = - = -∫ ∫ , which implies that the

integrating factor is 2

2

( ) ln2ln

1 1( )P x dx x

xu x e e

xe-∫= = = = . Next, we multiply the original differential equation by

the integrating factor. 2 2 2 2 3 2

2 1 2 1 1 1 2 1 1 1y y x y y x y y yx x x x x x x x x x

′′ ′ ′- = ⇒ - = ⇒ - = ⇒ =

. The left-hand

side is a derivative, so we then integrate both sides: 2 22

1 1 ln lny dx x C y x x Cxx x

= = + ⇒ = +∫ .

Example 3: An Application to Falling Bodies with Air Resistance

A calculus textbook of mass m is dropped from a hovering helicopter. Find its velocity at time t if the air resistance is proportional to the book’s velocity.

Solution

The downward force on the book is given by F mg kv= - , where g is the gravitational constant, v is the velocity of the textbook, and k is a constant of proportionality. By Newton’s second law,

dvF ma m mg kvdt

= = = - . This simplifies to the linear differential equation dv k v gdt m

+ = . Solving this equation

gives ( )1kt

mmgv ek

-= - .

Study Tips

• The separable differential equation ( )650dN k Ndt

= - can be rewritten in the form of a linear

differential equation 650dN kN kdt

+ = .

• The differential equation 2y x y x′+ = is not linear due to the square root term.

• When computing an integrating factor, you do not need a constant of integration.

• Many textbooks emphasize the formula for solving linear differential equations. Instead, just memorize

the formula for the integrating factor, ( )P x dxu e∫= .

• Notice in Example 3 that the velocity approaches /mg k as t increases. If there were no air resistance, the velocity would increase without bound.

Pitfalls

• Although there are exceptions, separation of variables cannot be used for linear differential equations.

25

• A differential equation might not seem linear at first glance. However, you might be able to rearrange the equation into the standard form of a linear equation, as indicated in the following examples.

1. ( ) 0.x xy xy e y y x e y′ ′+ = ⇒ + - =

2. 2 22 .xy y x y y xx

′ ′- = ⇒ - =

• Besides using an integrating factor to solve a linear differential equation, you might need to use advanced integration techniques, such as integration by parts (Lesson 10) to complete the solution.

Problems

1. Determine whether the differential equation 3 1xx y xy e′+ = + is linear.

2. Determine whether the differential equation 2siny y x xy′ - = is linear.

3. Verify that the function 12

x xy e Ce-= + is the solution to xy y e′+ = .

4. Find the integrating factor for the differential equation 12 6dy y xdx x

+ = - + .

5. Find the integrating factor for the differential equation 2 10dy xy xdx

+ = .

6. Solve the linear differential equation 16y y′ - = .

7. Solve the linear differential equation ( 1)siny x dx dy- = .

8. Find the particular solution of the linear differential equation tan sec cos , (0) 1.y y x x x y′+ = + =

9. Solve the falling object equation dv kv gdt m

+ = for ,v where andk m are constants.

10. Solve the weight loss equation 17.53500 3500

dw C wdt

= - for ,w where C is a constant.

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areas and VolumesLesson 7

Topics

• Areas of planar regions.

• Volumes by the disk method.

• Volumes by the shell method.

• Applications to ellipses.


• If ( ) ( )f x g x≥ on the interval a x b≤ ≤ , then the area bounded by the graphs of f and g between the

vertical lines x a= and x b= is [ ]( ) ( )b

aA f x g x dx= -∫ .

• For a horizontal axis of revolution, the volume of a solid with the disk method is given by the integral

( )2b

aV R x dxπ= ∫ . Similarly, for a vertical axis, the volume is ( )2d

cV R y dyπ= ∫ .

• For a vertical axis of revolution, the volume of a solid of revolution with the shell method is given by the integral

2 ( ) ( )b

aV p x h x dxπ= ∫ , where ( )p x is the radius and ( )h x is the height of the representative rectangle.

Summary

In this lesson, we study some fundamental applications of integration: the calculation of areas and volumes. We first review how to find the area of a planar region bounded by two curves. Then, we turn to volumes and review the disk and shell methods. We will look closely at ellipses as well as solids obtained by rotating ellipses about an axis. We apply our knowledge of ellipsoids to the shape of the planet Saturn.

Example 1: Finding the Area of a Region Bounded by Two Curves

Find the area bounded by the graphs of the functions 4 2( ) 2f x x x= - and 2( ) 2g x x= .

Solution

The most important part of this problem is the setup of the area integral. You have to first determine the region under consideration by finding where the two graphs intersect.

Setting ( ) ( ),f x g x= you obtain 4 2 2 4 2 22 2 4 0 ( 2)( 2) 0x x x x x x x x- = ⇒ - = ⇒ - + = . Thus, 0, 2, and 2,x = - and the points of intersection are (0,0), (2,8), and ( 2,8).-

27

g x= 2 2

f x x= −4 22

x x f x4 22− = ( )

2 2x g x= ( )

2 8,( )−( )2 8,

0 0,( )

Taking advantage of symmetry, you see from the graph that the area is given by the integral

( )2

2 4 2

0

2 2 2A x x x dx = - - ∫ . This integral is easy to evaluate:

22 3 5

2 4

0 0

4 32 32 1282 4 2 23 5 3 5 15x xA x x dx

= - = - = - = ∫ .

Example 2: Volumes by the Disk Method

Consider the region bounded by , 0, and 3.y x y x= = = Find the volume of the solid when the region is revolved about (a.) the x-axis, (b.) the y-axis, and (c.) the line 3.x =

Solution

a. The disk method gives ( )33 3 22

0 0 0

92 2xV x dx x dx ππ π π

= = = =

∫ ∫ .

b. For this one, we use a technique related to the disk method, the so-called washer method. We will

integrate with respect to y and use 2y x x y= ⇒ = . Thus, we have

( ) ( )3 3

22 2 4

0 0

36 33 95

V y dy y dy ππ π = - = - = ∫ ∫ .

c. In this case, the radius is 23 y- , so the volume becomes

( ) ( )3 3

22 2 4

0 0

24 33 9 65

V y dy y y dy ππ π= − = − + =∫ ∫ .

28

Example 3: The Shell Method

Use the shell method to find the volume in Example 2b.

Solution

The shell method results in the same answer. 3

3 3 3 5 52 2 2

0 00

2 2 36 32 2 2 2 35 5 5

V x x dx x dx x ππ π π π = = = = = ∫ ∫ .

Example 4: The Volume of a Football

The ellipse 2 2

125 9x y

+ = is revolved about the x-axis. Calculate the volume of the resulting football-shaped solid.

2 2

125 9x y

+ =

Solution

We first solve for 2 :y

( )2 2 2 2 2

2 2 225 91 1 9 2525 9 9 25 25 25x y y x xy y x

-+ = ⇒ = - ⇒ = ⇒ = -

.

Hence, the equation for the upper half of the ellipse is 23 25 .5

y x= - The volume is therefore

( )2

5 52 2

5 5

3 925 25 605 25

V x dx x dxπ π π- -

= - = - = ∫ ∫ . Note that if the ellipse had been rotated about the y-axis,

then the volume of the resulting “M&M” would be 100π .

Study Tips

• The setup of an area or volume integral is more important than the actual integrations. These integrations can often be done with graphing calculators or computers.

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• Take advantage of symmetry when evaluating areas and volumes. For instance, in Example 1, we noted

that ( ) ( )2 2

2 4 2 2 4 2

2 0

2 2 2 2 2A x x x dx x x x dx-

= - - = - - ∫ ∫ .

• The area bounded by the ellipse 2 2

2 2 1x ya b

+ = is .abπ

Pitfalls

• Drawing appropriate graphs is very important in area and volume problems. The graphs help determine

which curve is on the top and which is on the bottom. For example, the area bounded by the graphs of

y x= and 3y x= is not given by the integral ( )1 3

1x x dx

--∫ , but rather, ( )1 3

012 .2x x dx- =∫

• Be careful when solving equations for an unknown. In Example 1, you cannot cancel the common

factor 2x or you will lose the solution 0.x = In other words, the solutions to 4 24 0x x- = are not the

same as the solutions to 2 4 0x - = .

Problems

1. Find the area of the region bounded by the graphs of 2 2y x x= + and 2.y x= +

2. Find the area of the region bounded by the graphs of (2 ) and .x y y x y= - = -

3. Find the area of the region bounded by sin and cos 2 , .2 6

y x y x xπ π= = - ≤ ≤

4. Find b such that the line y b= divides the region bounded by 29y x= - and 0y = into regions of equal area.

5. Let , 0.a b > Show that the area bounded by the ellipse 2 2

2 2 1x ya b

+ = is .abπ

6. Find the volume of the solid generated by revolving the region bounded by the curves 2y x= and 24y x x= - about the x-axis.

7. Find the volume of the solid generated by revolving the region bounded by the curves 2y x= and 24y x x= - about the line 6.y =

8. Use the disk method to verify that the volume of a right circular cone is 21 ,3

r hπ where r is the radius of the

base and h is the height.

30

9. The region bounded by 2 2and 4y x y x x= = - is revolved about the line 4.x = Use the shell method to find the volume of the resulting solid.

10. Consider a sphere of radius r cut by a plane, thus forming a segment of height .h Show that the volume of

this segment is ( )21 33

h r hπ - .

11. The ellipse 2 2

125 9x y

+ = is rotated about the y-axis. Show that the resulting volume is 100 .π

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arc Length, surface area, and WorkLesson 8

Topics

• The arc length of a curve.

• The area of surfaces of revolution.

• Work.

• Applications.


• If ( )f x is a smooth curve on the interval a x b≤ ≤ , then the arc length of f is

( ) [ ]2 21 1b b

a as f x dx y dx′ ′= + = + ∫ ∫ .

• Consider the surface of revolution obtained by revolving the graph of the smooth function ( ) , ,f x a x b≤ ≤

about the x-axis. The surface area is given by the integral ( ) 22 ( ) 1

b

aS f x f x dxπ ′= + ∫ . Similarly, if

rotated about the y-axis, the surface area is ( ) 22 1

b

aS x f x dxπ ′= + ∫ .

• Suppose a constant force F moves an object a distance D . The work W done by the force is W FD= .

• If the force is variable, given by ( )f x , then the work W done by moving the object from x a= to x b=

is ( )b

aW F x dx= ∫ .

• Hooke’s law says that the force F required to compress or stretch a spring (within its elastic limits)

is proportional to the distance d that the spring is compressed or stretched from its original length:

F kd= . The proportionality constant k is the spring constant and depends on the nature of the spring.

• Newton’s law of universal gravitation says that the force F of attraction between two particles of masses

1m and 2m is proportional to the product of the masses and inversely proportional to the square of the

distance x between them: 1 22

m mF kx

= . Here, k is the gravitational constant, often written as .G

Summary

In this lesson, we apply calculus to three applications. First, we review arc length computations. Then, we move to the calculation of surface area of a surface of revolution. Finally, we develop the concept of work as force times distance. We close with some applications of work to physics and engineering.

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Example 1: Finding Arc Length

Find the arc length of the graph of the function 3

22 1,0 13

y x x= + ≤ ≤ .

Solution

The derivative of the function is 1

2y x x′ = = , so the arc length between 0 and 1 is

( ) ( ) ( ) ( )1

21 1 32 20 0

0

2 21 1 1 1 8 1 1.2193 3

b

as y dx x dx x dx x′= + = + = + = + = - ≈∫ ∫ ∫ .

Example 2: Finding Surface Area

Find the area of the surface obtained by revolving 31 ,0 3,3

y x x= ≤ ≤ about the x-axis.

Solution

The derivative is 2y x′ = . Hence, the surface area is given by

( ) ( ) ( )3 13 32 4 4 32

0 02 ( ) 1 2 1 1 4

3 6b

a

xS f x f x dx x dx x x dxππ π′= + = + = + ∫ ∫ ∫ .

Evaluating this integral by substitution, we have ( )

( ) ( ) ( )33

4 2 33 34 2 2

00

11 82 1 258.85

36 9 92

xS xπ π π

+ = = + = - ≈

.

33

Example 3: Hooke’s Law and Work

A force of 750 pounds compresses a spring 3 inches from its natural length of 15 inches. Find the work done in compressing the spring from 0 to 3 inches.

Solution

By Hooke’s law, ( )F x kx= , so (3) 750 (3) 250, ( ) 250 .F k k F x x= = ⇒ = = Then, the work done is

3 32

00( ) 250 125 1125 inch-pounds

b

aW F x dx x dx x = = = = ∫ ∫ .

Study Tips

• The integrals for arc length and surface area can be especially difficult to evaluate by hand. Often, these integrals don’t have elementary antiderivatives. Of course, you can always resort to numerical methods.

• Some graphing utilities have built-in arc length capabilities. If you have such a calculator, check that your answer makes sense.

• The concept of work is the motivation of the line integral in calculus of three dimensions.

Pitfalls

• Notice that in both Examples 1 and 2, the lower limit of integration was 0. However, in these integrals, this endpoint did not make the problem easier. We still had to substitute 0 into the antiderivative.

• In Example 3, the amount of work in compressing the spring from 3 to 6 inches is not the same as the work in compressing the spring from 0 to 3 inches.

• In the U.S. measurement system, force is frequently measured in pounds or tons. In the metric system, kilograms and metric tons are always units of mass, while force is measured in dynes (grams × cm/sec2) or newtons (kilograms × m/sec2).

• Suppose that you hold a heavy textbook in the air for 2 hours. How much work did you do? The answer is 0 because the textbook was not moved.

Problems

1. Find the arc length of the graph of 3 1 1, 2.6 2 2xy x

x= + ≤ ≤

2. Find the arc length of the graph of ( )ln cos ,0 .4

y x x π= ≤ ≤

3. Set up the definite integral that represents the arc length of the graph of lny x= , 1 5.x≤ ≤ Use a graphing utility to approximate the arc length.

34

4. Find the area of the surface formed by revolving the graph of 2 ,0 9y x x= ≤ ≤ about the x-axis.

5. Set up the definite integral that represents the surface area formed by revolving the graph of ln ,1y x x e= ≤ ≤ about the x-axis. Use a graphing utility to approximate the surface area.

6. Find the area of the surface formed by revolving the graph of 29 ,0 3y x x= - ≤ ≤ about the y-axis.

7. Determine the work done by lifting a 100-pound bag of sugar 20 feet.

8. A force of 750 pounds compresses a spring 3 inches from its natural length of 15 inches. Use Hooke’s law to determine how much work is done in compressing the spring an additional 3 inches.

9. A force of 250 newtons stretches a spring 30 centimeters. Use Hooke’s law to determine how much work is done in stretching the spring from 20 centimeters to 50 centimeters.

10. A lunar module weighs 12 tons on the surface of Earth. How much work is done in propelling the module from the surface of the moon to a height of 50 miles? Consider the radius of the moon to be 1100 miles and its force of gravity to be 1

6 that of Earth.

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moments, Centers of mass, and CentroidsLesson 9

Topics

• Mass.

• Moments about a point.

• Two-dimensional systems.

• Centers of mass and centroids.

• The theorem of Pappus.


• Mass is the measure of a body’s resistance to change in motion and is independent of the gravitational field.

• If a mass m is concentrated at a point, and if x is the distance between the mass and another point ,P then the moment of m about P is .mx

• Consider the masses 1,..., nm m located at positions 1,..., nx x along the x-axis. The moment about the

origin is 0 1 1 ... n nM m x m x= + + . The center of mass is 0Mx

m= , where 1 2 nm m m m= + + + .

• For a flat plate of uniform density 1 (planar lamina) bounded by the graphs of

( )f x and ( ), ,g x a x b≤ ≤ the moments about the axes are given by

( ) [ ]( ) ( ) ( ) ( ) and ( ) ( )2

b b

x ya a

f x g xM f x g x dx M x f x g x dx+ = - = - ∫ ∫ .

• The center of mass (or centroid) is ,y xM Mx y

m m= = , where m is the mass of the lamina.

• Let R be a planar region to the right of the y-axis. Let x be the distance from the center of mass of the region to the y-axis, and let A be the area of the region (that is, its mass). If the region is rotated about the y-axis, then the theorem of Pappus says that the volume of the resulting solid of revolution is 2V xAπ= .

Summary

In this lesson, we study moments and centers of mass. In particular, we develop formulas for finding the balancing point of a planar area, or lamina. First, we study one- and two-dimensional examples and then develop the formulas for arbitrary planar regions. We assume that the region is of uniform density 1. Hence, the mass of the lamina will be the same as its area. Finally, we present the famous theorem of Pappus for the volume formed by revolving a planar region about an axis and use it to calculate the volume of a torus.

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Example 1: The Center of Mass of a Linear System

Four masses of 10, 15, 5, and 10 are located on the x-axis at positions 5,0,4- and 7. Find the center of mass of the system.

Solution

The mass of the system is 10 15 5 10 40.m = + + + = The moment about the origin is

1 1 2 2 3 3 4 4 10( 5) 15(0) 5(4) 10(7) 40oM m x m x m x m x= + + + = − + + + = .

Hence, the center of mass is 40 1.40

oMx

m= = = If you imagine the fulcrum at the origin, then the system is not

in equilibrium.

Example 2: The Center of Mass of a Two-Dimensional System

Find the center of mass of a system of point masses—m1 = 6, m2 = 3, m3 = 2, and m4 = 9—located at

( )3, 2 , (0,0), ( 5,3),- - and ( )4,2 .

Solution

The total mass is 6 3 2 9 20.m = + + + = The moments about the axes are

6(3) 3(0) 2( 5) 9(4) 44.

6( 2) 3(0) 2(3) 9(2) 12.y

x

MM

= + + - + =

= - + + + =

Hence, 44 11 12 3and .20 5 20 5

y xM Mx y

m m= = = = = = The center of mass is

11 3, .5 5

Example 3: The Center of Mass of a Planar Lamina

Find the center of mass of the planar lamina of uniform density 1 bounded by the parabola 216y x= - and the x-axis.

Solution

In this example, 2( ) 16f x x= - and ( ) 0.g x = The mass is given by the area of the region,

[ ]4 2

4

256( ) ( ) (16 )3

b

am f x g x dx x dx

-= - = - =∫ ∫ . The moments are

37

( ) ( )

[ ] ( )

24 2

4

4 2

4

( ) ( ) 16 8192( ) ( ) 16 546.13.2 2 15

( ) ( ) 16 0.

b

x a

b

y a

f x g x xM f x g x dx x dx

M x f x g x dx x x dx

-

-

+ - = - = - = ≈

= - = - =

∫ ∫

∫ ∫

The center of mass is ( ) 32, , 0,5

y xM Mx y

m m = =

. By symmetry, the center of mass lies on the y-axis.

4-4

216y x= -

6 4 325

. =

Example 4: Using the Theorem of Pappus

Find the volume of the torus formed by revolving the circular region ( )2 22 1x y- + = about the y-axis.

38

Solution

The area of the region is π, and the center of mass x is 2 units from the axis of revolution. Hence, by the

theorem of Pappus, 22 2 (2)( ) 4V xAπ π π π= = = . This problem would be more difficult to solve using the disk or shell method.

Study Tips

• For a region of uniform density, the center of mass is often called the centroid of the region.

• The moment of a point is large if either the distance from the other point P is large or if the mass m is large.

• The center of mass for the system in Example 1 is not at the origin. The system would be in equilibrium if the fulcrum were located at 1.x =

• The same principles apply to the study of centers of mass of planar regions of nonuniform density.

• Consider a segment of a plane curve C that is revolved about an axis that does not intersect the curve (except possibly at the endpoints). The second theorem of Pappus says that the area of the resulting surface of revolution is equal to the product of the length of C times the distance d traveled by the centroid of C .

Pitfalls

• Notice in Example 1 that one of the points is at the origin, and hence, its moment is 0. That said, if you remove the point, then the resulting system of three masses has a different center of mass (1.6).

• For two-dimensional systems, notice that the formula for yM involves the x values, while the formula

for xM involves the y values.

• The center of mass of a planar region need not lie in the region at all. For example, consider the annular region between the circles of radius 1 and 2 centered at the origin. The center of mass is the origin, which is outside the region.

Problems

1. A 15-kilogram child is on a seesaw, 3 meters from the fulcrum. Where should the 20-kilogram friend sit in order to balance the seesaw?

2. Find the center of mass of the point masses lying on the x-axis if 1 2 37, 3, 5m m m= = = and

1 2 35, 0, 3x x x= - = = .

3. Find the center of mass of the system of point masses: ( )1 5 at 2,2m = , ( )2 1at 3,1m = - and ( )3 3 at 1, 4 .m = -

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4. Find ( ), , and ,x yM M x y for the lamina of uniform density 1 bounded by the graphs of

3, 0, and 0.y x y x= - + = =

5. Find ( ), , and ,x yM M x y for the lamina of uniform density 1 bounded by the graphs of

, 0, and 4.y x y x= = =

6. Find ( ), , and ,x yM M x y for the lamina of uniform density 1 bounded by the graphs of 22 and 0.x y y x= - =

7. Find the centroid of the region bounded by the semicircle 2 2 and 0.y r x y= - =

8. Use the integration capabilities of a graphing utility to approximate the centroid of the region bounded by

the graphs of the equations 310 125 and 0.y x x y= - =

9. A torus is formed by revolving the circle ( )2 25 16x y- + = about the y-axis. Use the theorem of Pappus to find the volume of the resulting solid of revolution.

10. A solid is formed by revolving the region bounded by the graphs of , 4, and 0y x y x= = = about the x-axis. Use the theorem of Pappus to find the volume of the resulting solid of revolution.

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Integration by PartsLesson 10

Topics

• Review of integration formulas.

• Integration by parts.

• Applications.


• integration by parts: u dv uv v du= -∫ ∫ .

• The integral of the logarithmic function is ln lnx dx x x x C= - +∫ .

Summary

We begin a series of lessons on techniques of integration. In general, integration is more difficult than differentiation. Fortunately, there are many techniques, as well as computer algorithms, for calculating antiderivatives. After recalling some basic formulas from calculus I (see the appendix containing a summary of integration formulas), we develop the technique called integration by parts. This important method is based on the product rule for derivatives. We will present examples and then applications to centers of mass and area.

Example 1: Evaluating Integrals

Although the following two integrals seem similar, their solutions are quite different. The first is solved by substitution, whereas the second is a formula involving the inverse sine function.

a. ( ) ( )( )

12 2

12 22

2

11 11 2 112 21 2

xx dx x x dx C x Cx

− − = − − − = − + = − − + −

∫ ∫ .

b. 2

1 arcsin1

dx x Cx

= +−

∫ .

Example 2: Integration by Parts

Use integration by parts to evaluate the integral sinx x dx∫ .

Solution

We let , , sin , and cos .u x du dx dv x dx v x= = = = - Then, we have sin ( cos ) ( cos ) cos sinx x dx uv v du x x x dx x x x C= - = - - - = - + +∫ ∫ ∫ .

41

Example 3: Repeated Integration by Parts

Calculate 2 .xx e dx∫

Solution

We perform integration by parts twice. First, let 2 , 2 , , and .x xu x du x dx dv e dx v e= = = = Then,

( )2 2 22 2x x x x xx e dx x e e x dx x e xe dx= - = -∫ ∫ ∫ . We then use integration by parts a second

time to evaluate the remaining integral. Let , , , and .x xu x du dx dv e dx v e= = = = Finally,

2 2 22 2 2x x x x x x xx e dx x e xe e dx x e xe e C = - - = - + + ∫ ∫ . You can check this answer by differentiation.

Example 4: An Application to Area

Find the area bounded by the graphs of ( ) ln , 0f x x y= = and .x e=

Solution

The area is given by the integral 1

lne

A x dx= ∫ , which requires integration by parts. Let

1ln , , , and .u x dv dx du dx v xx

= = = = The antiderivative of the logarithm function is

therefore 1ln (ln ) ln lnx dx x x x dx x x dx x x x Cx

= - = - = - +∫ ∫ ∫ . Finally, the area is

[ ]11ln ln ( ln ) (ln1 1) 1 1

e eA x dx x x x e e e e e= = - = - - - = - + =∫ .

1 e

y x= ln

42

Study Tips

• A key skill for integration by parts is identifying u and dv . In general, let u be a function whose derivative is simple, and let dv be a differential that can be integrated.

• You do not need a constant of integration when integrating dv because any antiderivative will be sufficient.

• You can always check your answers by differentiation.

• Numerical algorithms in modern computers and calculators have reduced the need to dwell on manual techniques for complicated integrals.

Pitfalls

• Although you do not need a constant of integration when integrating dv in an intermediate step, you do need a constant of integration in your final answer.

• Keep in mind that answers to integration problems can take many forms. For example, the following expressions are equivalent: 1andx C xe C e+ .

Problems

1. Find the indefinite integral 32 3 1 .t t dt-∫

2. Find the indefinite integral sin .cos

x dxx∫

3. Solve the differential equation ( )25 .xy e′ = +

4. Evaluate the definite integral 1

1 ln .e x dx

x-

∫

5. Find the area bounded by the graphs of 2

51

yx

=+

and 0,y = between 4x = - and 4.x =

6. Use integration by parts to find the indefinite integral sin 3 .x x dx∫7. Use integration by parts to find the indefinite integral 4arccos .x dx∫8. Use integration by parts to find the indefinite integral 2 cos .x x dx∫9. Evaluate the integral cos x dx∫

by first using substitution, then integration by parts.

10. Find the centroid of the region bounded by the graphs of arcsin , 0,y x x= = and .2y π=

11. Use integration by parts twice to find the indefinite integral sin .xe x dx∫

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Trigonometric IntegralsLesson 11

Topics

• Integrals with products of sines and cosines.

• Integrals with products of tangents and secants.

• The difficult cases.

• Applications.


• fundamental identities: 2 2 2 2sin cos 1, tan 1 secx x x x+ = + = .

• half-angle formulas: 2 21 cos 2 1 cos 2sin ;cos2 2

x xx x- += = .

• sec ln sec tanx dx x x C= + +∫ .

• 3 1 1sec sec tan ln sec tan2 2

x dx x x x x C= + + +∫ .

Summary

In this lesson, we look at integrals of trigonometric functions. Our first examples deal with products and other

products of sines and cosines: sin cosm nx x dx∫ . In general, these integrals will be easy to evaluate if either sine or cosine occur to an odd power. If both m and n are even, then we will need to use half-angle trigonometric formulas. Next, we look at products and other products of tangents and secants. Similarly, these integrals will be easy, except in the case where tangent occurs to an even power and secant to an odd power. In all cases, we use the fundamental trigonometric identities to simplify the integrands.

Example 1: Products of Sines and Cosines

Evaluate the integral 3 2sin cosx xdx∫ .

Solution

The key idea is to convert the integrand to cosines, keeping one sine function on the right as the negative derivative of cosine.

( )

( ) ( )

3 2 2 2 2 2

5 32 4 4 2

sin cos sin cos sin 1 cos cos sin

cos coscos cos sin cos cos ( sin ) .5 3

x x dx x x x dx x x x dx

x xx x x dx x x x dx C

= = -

= - = - - = - +

∫ ∫ ∫

∫ ∫

44

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Example 2: Even Powers of Sine and Cosine

One arch of the sine curve is revolved about the x-axis. Find the resulting volume.

Solution

The disk method says that the volume is 2

0sinV x dx

ππ= ∫ . Because this is an integral without odd powers of

sine or cosine, we have to use a half-angle formula.

[ ]

[ ]

2

0 0 0

2

0

1 cos 2sin 1 cos 22 2

1 sin 2 0 .2 2 2 2

xV x dx dx x dx

x x

π π π

π

ππ π

π π ππ

- = = = -

= - = - =

∫ ∫ ∫

Example 3: Products of Tangents and Secants

Evaluate the integral 4 3sec tanx x dx∫ .

Solution

Because secant occurs to an even power, we can convert two of the secants to tangents, reserving 2sec x as the derivative of tangent

( )4 3 2 3 2 2 3 2

6 45 3 2

sec tan sec tan sec tan 1 tan sec

tan tantan tan sec .6 4

x x dx x x x dx x x x dx

x xx x x dx C

= = +

= + = + +

∫ ∫ ∫

∫

Because tangent occurs to an odd power, we could have also solved the problem as follows.

( )

( )

4 3 3 2 3 2

6 45 3

1

sec tan sec tan (sec tan ) sec sec 1 sec tan

sec secsec sec sec tan .6 4

x x dx x x x x dx x x x x dx

x xx x x x dx C

= = -

= - = - +

∫ ∫ ∫

∫

These two answers are equivalent, because they differ by a constant.

Example 4: The Integral of the Secant Function

Evaluate the integral sec x dx∫ .

Solution

Because secant occurs to an odd power, we cannot use our fundamental identities. However, we can use a simple trick, as follows.

45

2sec tan sec sec tansec sec ln sec tansec tan sec tan

x x x x xx dx x dx dx x x Cx x x x+ +

= = = + ++ +∫ ∫ ∫ .

Example 5: Which Integral Is Easier to Solve?

Suppose that you were taking a test and you had to solve one, and only one, of the following integrals. Which would you choose?

a. 365sin cosx x dx∫ .

b. 4 4sin cosx x dx∫ .

Solution

Although the first integral looks more difficult at first glance, it is simply 366

365 sinsin cos366

xx x dx C= +∫ .

The second integral can be done using half-angle formulas, but it would take a long time.

Study Tips

• An integral like 4 6sin cosx x dx∫ is difficult because both powers are even. You need the half-angle formulas to solve it.

• An integral like 3 4sec tanx x dx∫ is difficult because the secant power is odd and the tangent power is

even. Usually, you have to convert to sines and cosines.

• Sometimes an integral can be solved in different ways, leading to equivalent antiderivatives. For example, the following answers are all correct.

2 2

1 2 3sin cos cos 2sin cos

2 2 4x x xx x dx C C C= + = − + = − +∫ .

Pitfall

• Be aware of the notation for powers of trigonometric functions. For example, the expression 2sin x

means 2(sin )x . On a calculator, you will need to place parentheses around the sine function before you square it.

46

Problems

1. Evaluate the integral 2 3sin cos .x x dx∫

2. Evaluate the integral 2cos 3 .x dx∫

3. Evaluate the integral 5 4sin 5 cos 5 .x x dx∫

4. Evaluate the integral 4sec tan .x x dx∫

5. Evaluate the integral 5 3tan 2 sec 2 .x x dx∫

6. Find the centroid of the region bounded by the graphs of cos , 0, 0y x y x= = = , and .2x π=

7. Evaluate the integral csc .x dx∫

8. Verify that [ ] 2

seccsc .tan

d xx Cdx x

- + =

9. Evaluate the integral 4sin .x dx∫

10. Use integration by parts twice to evaluate the difficult integral 3sec .x dx∫

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Integration by Trigonometric substitutionLesson 12

Topics

• Integrals involving terms like 2 2 2 2 2 2, , , and 0.a x a x x a a- + - >

• Definite integrals and conversion of limits of integration.


• inverse trigonometric formulas: Let 0.a >

2 2

2 2

2 2

arcsin .

1 arctan .

1 arcsec .

dx x Caa x

dx x Ca x a a

xdx Ca ax a

= +-

= ++

= +-

∫

∫

∫

• sine substitution: sinx a θ= for integrands involving 2 2a x- .

• tangent substitution: tanx a θ= for integrands involving 2 2a x+ .

• secant substitution: secx a θ= for integrands involving 2 2x a- .

Summary

Trigonometric substitution is a technique for converting integrands to trigonometric integrals. For example, we

can use the substitution sinx a θ= for integrands containing radicals such as 2 2a x- . These substitutions convert an integral in the original variable x to an integral in the new variable .θ This new integral will involve trigonometric functions. After evaluating the new integral, you convert your answer back to the original variable. You can conveniently represent these substitutions by right triangles, as shown in the first three examples. For definite integrals, you can also convert the limits of integration to the new variable θ , thus avoiding the need to return to the original variable.

Example 1: A Sine Substitution

Evaluate the integral 2 29

dxx x-∫ .

Solution

We begin by making the substitution 3sin .x θ= Then, we have 2 29sinx θ= and

48

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2 2 2 23cos , 9 9 9sin 3 1 sin 3 cos 3cosdx d xθ θ θ θ θ θ= - = - = - = = .

Next, go from the variable x to the variable θ , as follows.

( )( )

2222 2

3cos 1 1 1 1csc cot9 9 9sin9sin 3cos9

ddx d d Cx x

θ θ θ θ θ θθθ θ

= = = = − +−

∫ ∫ ∫ ∫ .

Finally, return to the original variable 21 1 9: cot

9 9xx C

xθ -

- = - + . That is,

2

2 2

999

dx x C.xx x−

= − +−

∫

You can summarize this substitution by the following useful right triangle.

3x

9 2- x

Example 2: A Tangent Substitution

Evaluate the integral( )

32 21

dx

x +∫ .

Solution

We begin with the substitution tan tanx a θ θ= = . Then, we have

( )3

2 2 2 2 2 32sec , 1 tan 1 sec , 1 sec .dx d x xθ θ θ θ θ= + = + = + =

Moving to the new variable, ( )

2

3 32 2

sec cos sinsecsec1

ddx d d Cx

θ θ θ θ θ θθθ

= = = = ++

∫ ∫ ∫ ∫ . Converting back to

the original variable, 2

sin1

xC Cx

θ + = ++

. In other words, ( )

3 22 2 11

dx x Cxx

= +++

∫ .

49

x2 1+

1

x

Example 3: A Secant Substitution

Evaluate the integral 2 25

dxx -

∫ .

Solution

We begin with the substitution 5sec .x θ= Then, we have

2 2 25sec tan , 25 25sec 25 5 sec 1 5 tandx d xθ θ θ θ θ θ= - = - = - = .

With this substitution, we have

2

2

5sec tan 25sec ln sec tan ln5 tan 5 525

ddx x xd C Cx

θ θ θ θ θ θ θθ

-= = = + + = + +

-∫ ∫ ∫ .

x2 5-

5

x

Example 4: An Application of Arc Length

A thin wire is in the shape of the parabola 21 ,0 1.2

y x x= ≤ ≤ What is the length of the wire?

50

Solution

The derivative is ,y x′ = and hence, the formula for arc length gives

( )1 12 2

0 01 1s y dx x dx′= + = +∫ ∫ .

We then use the trigonometric substitution 2tan , sec ,x dx dθ θ θ= = and hence, 2 21 1 tan secx θ θ+ = + = .

The indefinite integral becomes

2 2 31 sec (sec ) secx dx d dθ θ θ θ θ+ = =∫ ∫ ∫ .

Rather than return to the variable x , we can use the equation tanx θ= to convert the limits of integration:

When 0, 0,x θ= = and when 1, .4x πθ= = Hence, we have the interesting conclusion that

1 42 3

0 01 secs x dx d

πθ θ= + =∫ ∫ .

Using integration by parts twice, you obtain the formula for the right-hand integral, and we have the final answer:

( )40

1 1sec tan ln sec tan 2 ln 2 1 1.1482 2

sπ

θ θ θ θ = + + = + + ≈ .

Study Tips

• It is often helpful to use a right triangle to represent the trigonometric substitution.

• Remember that you can always check your answer by differentiation.

• You can do definite integrals two ways: (1) First, solve the indefinite integral as in the examples, then apply the fundamental theorem of calculus using the original limits of integration; or (2) use the substitution to change the original limits of integration to θ - limits, which is how we did Example 4.

Pitfalls

• Sometimes, your answer might look different than the answer from a calculator. In Example 3, for

instance, a calculator could give 2ln 25 ,x x+ - which is equivalent to the answer we found,

2 25ln

5 5x x C-

+ + .

• For indefinite integrals, don’t forget to convert back to the original variable.

• Be aware that the integrals arising from arc length and surface area applications can be quite challenging, as in Example 4. The curve was quite simple, a parabola, but the integration required a trigonometric substitution.

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Problems

1. Find the integral 2

12 .1 9

dxx+∫


1 .9

dxx-

∫


1 .4 1

dxx x -∫

4. Use the substitution 4sinx θ= to find the indefinite integral ( )3

2 2

1 .16

dxx-

∫

5. Use the substitution 5secx θ= to find the indefinite integral 2 25 .x dxx-

∫

6. Use the substitution tanx θ= to find the indefinite integral ( )

2

22.

1

x dxx+

∫

7. Evaluate the definite integral ( )

232

30 2 2.

1

x dxx-

∫

8. Solve the differential equation 2 4 1, 2, (0) 4.dyx x ydx

+ = ≥ - =

9. Find the arc length of the curve 21 ,0 4.2

y x x= ≤ ≤

52

Less

on 1

3: In

tegr

atio

n by

Par

tial F

ract

ions

Integration by Partial FractionsLesson 13

Topics

• Integrals with distinct linear terms.

• Integrals with repeated linear terms.

• Integrals with irreducible quadratic terms.

• Application to the logistic differential equation.


• linear factors: If the linear factor ( )px q+ occurs to the mth power, then use

1 22 ...

( ) ( )m

m

AA Apx q px q px q

+ + ++ + +

.

• quadratic factors: If the irreducible quadratic factor 2( )ax bx c+ + occurs to the nth power, then use

1 1 2 22 2 2 2...

( ) ( )n n

n

B x CB x C B x Cax bx c ax bx c ax bx c

++ ++ + +

+ + + + + +.

Summary

The technique of partial fractions is really an algebraic skill. It consists of splitting up complicated algebraic expressions, in particular rational functions, into a sum of simpler functions. These can then be integrated easily using our previous techniques. We will study how to deal with linear terms, repeated linear terms, and irreducible quadratic terms. Finally, we will apply these skills to the solution of the logistic differential equation.

Example 1: Distinct Linear Factors

Evaluate the integral 2 5 6

dxx x- +∫ .

Solution

We begin by factoring the denominator and splitting up the integrand into its two partial fractions.

2

1 1( 3)( 2) 3 25 6

A Bx x x xx x

= = +- - - -- +

.

To find the unknown constants A and B, we obtain a common denominator.

53

2

1 ( 2) ( 3)3 2 ( 3)( 2)5 6

A B A x B xx x x xx x

- + -= + =

- - - -- +.

By equating the numerators, we obtain the basic equation 1 ( 2) ( 3)A x B x= - + - . This equation holds for all values of the variable ,x so by letting x equal key values, we can determine the constants.

3 :1 (3 2) 1.2 :1 (2 3) ( 1) 1.

x A A Ax B B B= = - = ⇒ == = - = - ⇒ = -

Hence, the original integral can be written in a much simpler form and easily evaluated.

2

1 1 ln 3 ln 23 2 3 25 6

dx A B dx dx x x Cx x x xx x

- = + = + = - - - + - - - -- + ∫ ∫ ∫ .

Example 2: Repeated Linear Factors

Determine the partial fraction decomposition for the integral 2

3 2

5 20 62

x x dxx x x

+ ++ +∫ .

Solution

We factor the denominator, ( ) ( )23 2 22 2 1 1x x x x x x x x+ + = + + = + , and set up the decomposition. Notice how

we handle the repeated linear factor ( )

2

3 2 2

5 20 612 1

x x A B Cx xx x x x

+ += + +

++ + +.

To find the constants A, B, and C, we get a common denominator and form the basic equation.

Letting 0, 1,x = - and 1, we obtain 6, 1,A B= = - and 9.C = Thus, the integral has been split up into simpler pieces.

( )

2

3 2 2

5 20 6 6 1 912 1

x x dx dxx xx x x x

+ + = - + ++ + +

∫ ∫ .

This integral is easy to evaluate, and the final answer is 96ln ln 1

1x x C

x- + - +

+.

Example 3: A Quadratic Factor

Set up the partial fraction decomposition for the integral ( )( )

3

2

2 4 81 4

x x dxx x x

- -- +∫ .

( )( )

( )22

223 2 2

1 ( 1)5 20 6 5 20 6 1 ( 1)2 1

A x Bx x Cxx x x x A x Bx x Cxx x x x x

+ + + ++ += ⇒ + + = + + + +

+ + +.

54

Solution

We split up the expression into its partial fractions. Notice how we handle the irreducible quadratic factor.

( )( ) ( )3

2 2

2 4 811 4 4

x x A B Cx Dx xx x x x

- - += + +

-- + +.

Obtaining a common denominator,

( )( ) ( )3 2 2

2 2

2 4 8 ( 1)( 4) ( 4) ( ) ( 1)1 4 ( 1) 4

x x A x x Bx x Cx D x xx x x x x x

- - - + + + + + -=

- + - +.

We have the basic equation

3 2 22 4 8 ( 1)( 4) ( 4) ( ) ( 1)x x A x x Bx x Cx D x x- - = - + + + + + - .

As before, we solve for the constants and obtain 2, 2, 2,A B C= = - = and 4.D =

Example 4: The Logistic Differential Equation

The partial fraction decomposition for the logistic differential equation is

( )1 1dy L L

y L y y L y= +

- -.

Study Tips

• Keep in mind that finding the partial fraction decomposition of a rational function is a precalculus topic.

• Many calculators and computers can perform the partial fraction decomposition automatically.

• If the degree of the numerator is greater than or equal to the degree of the denominator, then use long division to simplify. For example,

2 2555 5

x dx x dxx x

= + + - - ∫ ∫ .

Pitfalls

• Repeated linear and quadratic factors require extra terms for the partial fraction decomposition to work.

For example, five terms are needed for

( ) ( )2 2 3 2 23 2 2

2 311 1

x A B C Dx E Fx Gdx dxx x x xx x x

- + + = + + + + ++ +

∫ ∫ .

Less

on 1

3: In

tegr

atio

n by

Par

tial F

ract

ions

55

• Don’t be upset if your answer looks different from a calculator answer. For instance, in Example 1, a

calculator could give the answer as 2ln3

xx

- - -

. By logarithmic properties, this is equivalent to our

answer. Note that calculators don’t include the constant of integration.

Problems

1. Evaluate the integral 2

19

dxx -∫ .


5 .3 4

dxx x+ -∫


3 2

4 2 1x x dxx x+ -+∫ .


3

1x dxx x

-+∫ .

5. Set up the partial fraction decomposition for the integral ( ) ( )

2

232 2

3

2 9

x x dxx x x

+ -

- +∫ . Do not solve for

the constants.

6. Use properties of logarithms to verify that 2

ln 3 ln 2 ln .3

xx x

x -

- - - = - -

7. Evaluate the integral ( )( )

3

2

2 4 8 .1 4

x x dxx x x

- -- +∫


2

4 3 .2 1x x dxx x

+ ++ -∫

9. Find the area of the region bounded by the graphs of 2

157 12

yx x

=+ +

, 0, 0,y x= = and 2.x =

10. Show that the solution to the equation ( )1 ln ln ky L y C tL L

- - + = is 1 kt

Lybe-=

+.

56

Less

on 1

4: In

dete

rmin

ate

Form

s an

d L’

Hôp

ital’s

Rul

e

Indeterminate Forms and L’Hôpital’s RuleLesson 14

Topics

• L’Hôpital’s rule.

• Limits of the form 0 .0

• Limits of the form .∞∞

• Other indeterminate forms.

• The number e.


• L’Hôpital’s rule: Let f and g be functions that are differentiable on an open interval ( ),a b containing

,c except possibly at c itself. Assume that ( ) 0g x′ ≠ for all x in ( ),a b , except possibly at c itself. If the

limit of ( ) ( )f x g x as x approaches c produces the indeterminate form 0 0, then ( )( )

( )( )

lim limx c x c

f x f xg x g x→ →

′=

′,

provided that the limit on the right exists (or is infinite).

• This result also applies if the limit of ( ) ( )f x g x as x approaches c produces any one of the

indeterminate forms ( ) ( ), ,∞ ∞ ∞ -∞ -∞ ∞ , or ( ) ( )-∞ -∞ . Furthermore, L’Hôpital’s rule applies to

one-sided limits.

Summary

Limits of the form 00 or ∞∞

are called indeterminate because they do not guarantee that a limit exists, nor

do they indicate what the limit is, if one does exist. The classic example is 0

sinlimx

xx→

, which we know equals 1

from elementary calculus. In this lesson, we present the famous L’Hôpital’s rule for evaluating such indeterminate forms. We apply this valuable theorem to a variety of examples.

57

Example 1: The Indeterminate Form 00

Evaluate the limit 2

0

1limx

x

ex→

- .

Solution

This limit is indeterminate because both numerator and denominator approach 0 as x approaches 0. We apply L’Hôpital’s rule by differentiating numerator and denominator separately.

2 2

0 0

1 2lim lim 21

x x

x x

e ex→ →

-= = .

Example 2: The Indeterminate Form ∞∞

Evaluate the limit lnlimx

xx→∞

.

Solution

Both numerator and denominator tend to infinity as x approaches infinity.

1ln 1lim lim lim 01x x x

x xx x→∞ →∞ →∞

= = = .

You see that the rate of growth of the logarithmic function is smaller than that of the function ( ) .f x x=

Example 3: A Polynomial Approximation for the Cosine Function

Evaluate the limit 20

1 coslimx

xx→

- .

Solution

We use L’Hôpital’s rule twice, as follows.

20 0 0

1 cos sin cos 1lim lim lim2 2 2x x x

x x xx x→ → →

-= = = .

This means that near 0, 2

1 cos 12

xx

-≈ . By solving for the cosine, you have 21cos 1

2x x≈ - , which is a polynomial

approximation of the cosine function. We will see more such approximations when we study infinite series.

Example 4: The Indeterminate Form 0 ∞

Evaluate the limit lim x

xx e-

→∞.

58

Solution

We can convert this to an indeterminate form, ,∞ ∞ and then use L’Hôpital’s rule.

112lim lim lim lim 0

2x

x x xx x x x

x xx ee e x e

-

→∞ →∞ →∞ →∞= = = = .

Example 5: The Number e

Evaluate the limit 1lim 1x

x x→∞

+

.

Solution

This is an example of the indeterminate form 1∞ . We can evaluate it by using logarithms, as follows.

Let 11x

yx

= +

. Then, ( )1ln 11 1ln ln 1 ln 1

1

xxy x

x x x

+ = + = + =

.

As x tends to infinity, this is an indeterminate form, 00 . By L’Hôpital’s rule,

( ) ( )2

2

1 11 1ln 1 1 1lim ln lim lim lim 1

1 1 11x x x x

xx xy

x xx→∞ →∞ →∞ →∞

- + + = = = =

- + .

Because ln y tends to 1, y tends to .e That is, 1lim 1 .x

xe

x→∞

+ =

Study Tips

• As seen in Examples 4 and 5, there are many types of indeterminate forms. These include the forms 0 00 , 1 , , 0 ,∞⋅∞ ∞ and .∞ -∞

• Sometimes you have to use L’Hôpital’s rule more than once. For example,

3 20 0 0

1 1lim lim lim63

x x x

x x x

e x e exx x+ + +→ → →

- - -= = = ∞ .

• Sometimes L’Hôpital’s rule doesn’t work at all.

( )

2

12 2 2

1 1lim lim lim11 1 (2 )2

x x x

x xxx x x

-→∞ →∞ →∞

+= =

+ +.

If you continue to apply L’Hôpital’s rule, you will return to the original limit.

Less

on 1

4: In

dete

rmin

ate

Form

s an

d L’

Hôp

ital’s

Rul

e

59

Pitfalls

• Make sure that the hypotheses of L’Hôpital’s rule are verified. For example, you cannot use it for the

limit 2

0

1limx

xx

ee→

- because the denominator is not tending to 0. This limit actually equals 0.

• One-sided limits can be tricky, as in the following example.

3 20 0 0

1 1lim lim lim63

x x x

x x x

e x e exx x+ + +→ → →

- - -= = = ∞ .

If the limit had been from the left, the answer would have been .-∞

Problems

1. Evaluate the limit 0

sin 4lim .sin 3x

xx→


3

2 3lim .3x

x xx→

- --


arcsinlim .x

xx→

4. Evaluate the limit 3lim .x

x

ex→∞


21

lnlim .1x

xx→ -

6. Evaluate the limit 1lim sin .x

xx→∞


lim xx

x→∞

.

8. Evaluate the limit ( )1

lim 1 .xx

x→∞

+


8lim .4 2x

xx x+→

- - -

10. Try to use L’Hôpital’s rule to evaluate the limit 2

tanlimsecx

xxπ -→

. What do you observe? What is the limit?

60

Less

on 1

5: Im

prop

er In

tegr

als

Improper IntegralsLesson 15

Topics

• Improper integrals with infinite limits of integration.

• Improper integrals with infinite discontinuities.

• A special type of improper integral.


• If f is continuous on the interval [ ), ,a ∞ then ( ) ( )lim

b

a abf x dx f x dx

∞

→∞=∫ ∫ .

• If f is continuous on the interval ( ],b-∞ , then ( ) ( )limb b

aaf x dx f x dx

-∞ →-∞=∫ ∫ .

• If f is continuous on the interval ( ),-∞ ∞ , then ( ) ( ) ( )c

cf x dx f x dx f x dx

∞ ∞

-∞ -∞= +∫ ∫ ∫ .

• If f is continuous on the interval [ ),a b and has an infinite discontinuity at ,b then

( ) ( )limb c

a ac bf x dx f x dx

-→=∫ ∫ . There are similar definitions for discontinuities at the left-hand endpoint

and in the interior.

• Theorem: 1

1 1 , 11p dx p

px∞

= >-∫ . The integral diverges if 1.p ≤

Summary

Improper integrals appear to be outside the hypotheses of the fundamental theorem of calculus, which requires that the integrand be continuous on a closed interval [ ], .a b In this lesson, we consider how to evaluate such improper integrals, where one of the limits of integration is ∞ or -∞ , or those that are not continuous on the closed interval [ ],a b . We analyze these integrals by considering a related “normal” integral and then taking an appropriate limit. If the limit exists, the improper integral converges. Otherwise, it diverges. We also prove an important theorem that will be used in our later study of infinite series.

Example 1: A Divergent Improper Integral

Evaluate the integral 1

1 dxx

∞

∫ .

Solution

We consider the appropriate standard integral and take the limit.

61

[ ] [ ]11 1

1 1lim lim ln lim ln ln1b b

b b bdx dx x b

x x∞

→∞ →∞ →∞= = = − = ∞∫ ∫ .

Because the limit is infinity, we say that the integral diverges.

Example 2: A Convergent Improper Integral


1 dxx

∞

∫ .

Solution

We proceed as in the previous example, but this time, the integral converges. We say that the area under this infinitely long region is finite.

22 21 1 1

1

1 1 1 1 1lim lim lim lim 11

bb b

b b b bdx dx x dx

x bx x∞ -

→∞ →∞ →∞ →∞

- - - = = = = - = ∫ ∫ ∫ .

Example 3: An Infinite Discontinuity at an Endpoint


30

1 dxx∫ .

Solution

The integrand has an infinite discontinuity at the endpoint 0.x =

( )1

1 1 1 2 23 3 3

30 0 0 0

1 3 3 3lim lim lim 12 2 2bb b bb

dx x dx x bx + + +

−

→ → →

= = = − = ∫ ∫ .

Example 4: A Special Improper Integral

Show that 1

1 1 , 11p dx p

px∞

= >-∫ . If 1p ≤ , the integral diverges.

Solution

If 1,p = Example 1 verifies that the integral diverges. If 1,p ≠

1

1 11 111

1 1 1 1 1 11 1 1

b bpb b pp p p

xdx x dxp p px x b

− +−

− −

= = = = − − + − − ∫ ∫ .

Next, take the limit as .b →∞ If 1,p > 1

1 11p dx

px∞

=-∫ . Otherwise, the integral diverges. This integral will

play an important role in our study of infinite series.

62

Study Tips

• Improper integrals are evaluated by considering an appropriate “normal integral” and then taking a limit.

• Improper integrals can have surprising properties, especially when interpreted as areas, volumes, and surface areas. For example, the area under the curve 1y x= , 1 x≤ , is infinite. However, the volume obtained by rotating this region about the x-axis is finite. This solid is often called Gabriel’s horn. Despite finite volume, the surface area of Gabriel’s horn is infinite.

• Sometimes, an integral is “doubly improper.” For example, to analyze the following integral, you have to split it into two pieces.

( ) ( ) ( ) ( ) ( )

1 1

0 0 1 10lim lim

1 1 1 1 1

c

b cb

dx dx dx dx dxx x x x x x x x x x+

∞ ∞

→∞→= + = +

+ + + + +∫ ∫ ∫ ∫ ∫ .

Pitfalls

• Sometimes, an integral has an interior discontinuity. Without proper analysis, you might obtain an erroneous answer. For example, the following computation is incorrect.

22 2 3

3 21 11

1 1 1 1 38 2 82

dx x dxx x

-

- --

- - = = = + = ∫ ∫ .

In fact, the integral diverges. You can see this by splitting up the integral into two improper integrals,

2 0 2

3 3 31 1 0

1 1 1dx dx dxx x x- -

= +∫ ∫ ∫ . Both of these integrals diverge.

• It’s worth repeating: Make sure to check for discontinuities when evaluating definite integrals. Using a graphing utility to graph the integrand can often help in identifying points of discontinuity.

• Don’t forget that infinity is not a number. For example, the infinite interval [ )2,∞ should not be written as [ ]2, .∞

Problems

1. Determine whether the improper integral 4

2x dx

∞ -∫ converges or diverges. Evaluate the integral if it converges.


51

x dx∞ -

∫ converges or diverges. Evaluate the integral if it converges.


0

xe dx∞ -∫ converges or diverges. Evaluate the integral if it converges.


2

11

dxx-∞ +∫ converges or diverges. Evaluate the integral if it converges.


0

1 dxx∫ converges or diverges. Evaluate the integral if it converges.Le

sson

15:

Impr

oper

Inte

gral

s

63


0

10 dxx∫ converges or diverges. Evaluate the integral if it converges.


30

11

dxx -∫ converges or diverges. Evaluate the integral if it converges.


31

1 dxx-∫ converges or diverges. Evaluate the integral if it converges.

9. Determine whether the improper integral ( )0

11

dxx x

∞

+∫ converges or diverges. Evaluate the integral if it converges.

10. Show that the surface area of Gabriel’s horn is 41

1 12 1 .S dxx x

π∞

= +∫ Then, prove that this integral diverges by

comparing it to the divergent integral 1

12 .dxx

π∞

∫

64

Less

on 1

6: s

eque

nces

and

Lim

its

sequences and LimitsLesson 16

Topics

• Sequences.

• Limits of sequences.

• Monotonic sequences.

• Bounded sequences.

• The sequence of partial sums.


• A sequence { } { }1 2, ,...na a a= is a function whose domain is the set of positive integers. That is, a

sequence is an infinite list of real numbers.

• Let L be a real number. The limit of the sequence { }na is L if for all 0,ε > there exists 0M > such

that na L ε- < whenever .n M> We say that the sequence converges to L and write lim nn

a L→∞

= . If the limit does not exist, then the sequence diverges.

• Theorem: If 1, then lim 0n

nr r

→∞< = .

• A sequence is monotonic if its terms are nondecreasing, 1 2 3 ...a a a≤ ≤ ≤ , or nonincreasing, 1 2 3 ...a a a≥ ≥ ≥ .

• A sequence is bounded above if there is a real number M such that na M≤ for all .n M is an upper bound of the sequence.

• A sequence is bounded below if there is a real number N such that na N≥ for all .n N is a lower bound of the sequence.

• A sequence is bounded if it is both bounded above and bounded below.

• Theorem: If a sequence is bounded and monotonic, then it converges.

Summary

In this lesson, we begin our study of infinite series, perhaps the most important topic in calculus II. The concept of infinite series is based on sequences, which is the topic of this lesson. Informally, a sequence is an infinite list of real numbers. In the Fibonacci sequence, 1, 1, 2, 3, 5, 8, … , the next term is the sum of the previous two terms. The definition of the limit of a sequence is complicated, but basically means that the terms of the sequence

approach a fixed real number. For example, the limit of the sequence 1 1 11, , , ,2 3 4

is 0 because the terms get

65

closer and closer to 0: 1lim 0n n→∞

= . Finally, we show how an infinite series can be analyzed as a sequence of

partial sums.

Example 1: The Terms of a Sequence

Write out the first five terms of the sequences { }1n

nan

= + and { } ( ){ }3 1 n

nb = + - .

Solution

We let 1, 2, 3, 4, and 5n = in each sequence.

( ){ } { }1 2 3 4 5, , , , ,... ; 3 1 2,4,2,4,2,...

1 2 3 4 5 6nn

n = + - = +

.

Example 2: Limits of Sequences

a. ( )( )lim 3 1 n

n→∞+ - does not exist.

b. 2

1lim 5 5n n→∞

- =

.

c. 1lim sin 0n n→∞

= .

d. 0

1sin1 sinlim sin lim lim 11n n x

xnnn xn

→∞ →∞ →= = = .

Example 3: Bounded and Monotonic Sequences

a. The sequence 1 2 3 4, , , ,2 3 4 5

is bounded and monotonic.

b. The sequence 2, 4, 2, 4, 2, … is bounded but not monotonic.

c. The Fibonacci sequence is monotonic but not bounded.

d. Quotients of successive terms of the Fibonacci sequence produce a sequence that is bounded but not monotonic.

Example 4: The Partial Sums of an Infinite Series

Write out the first three partial sums of the infinite series 1 1 1 1 .2 4 8 16

+ + + +

66

Solution

The first three partial sums are 1 2 31 1 1 3 1 1 1 7, , and .2 2 4 4 2 4 8 8

S S S= = + = = + + =

If you continue forming this sequence of partial sums, you obtain

1 3 7 15 31, , , , ,2 4 8 16 32

.

Notice that the limit of this sequence of partial sums is 1. Hence, we say that the infinite series

1 1 1 12 4 8 16

+ + + + converges to 1.

Study Tips

• Loosely speaking, L is the limit of a given sequence if you can get as close to L as you want by going out far enough in the sequence. That is, there exists M such that all the terms past the Mth term are close to .L

• If a sequence has a limit, then that limit is unique.

• Many calculators and computers can evaluate limits of sequences. You can also get a good idea about the behavior of a sequence by writing out the first few terms. It also helps to analyze the terms for large values.

Pitfalls

• You can’t tell what the rule is for a sequence by just observing the first few terms. For example, the

sequence 1 1 1, , ,...2 4 8

could have the following rules: ( )( )2

1 6or .2 1 6n n n n

+ - +

• It is possible for a sequence to not be monotonic yet still converge. For example, the quotients of successive terms of the Fibonacci sequence converge to the golden ratio.

1 1 2 3 5 8 1 5, , , , , 1.618.1 1 2 3 5 2

n

n

aa

+ + = → ≈

Problems

1. Write the first five terms of the sequence 3 .!

n

nan

=

2. Write the first five terms of the sequence ( ) 1 21 .nna

n+= -

Less

on 1

6: s

eque

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and

Lim

its

67

3. Write the first five terms of the sequence sin .2n

na π=

4. Find the limit (if possible) of the sequence 3

3

8 .2n

nan

=-

5. Find the limit (if possible) of the sequence 2

2 .1

nna

n=

+

6. Find the limit (if possible) of the sequence ( )1 1 .nna = + -

7. Find the limit (if possible) of the sequence ( )1 1.

n

nan

+ -=

8. Find the limit (if possible) of the sequence 31 .n

nan

= +

9. Determine whether the sequence 14nan

= - is monotonic or whether it is bounded.

10. Determine whether the sequence 23

n

na = -

is monotonic or whether it is bounded.

11. Write the first five terms of the sequence nna n= and find its limit.

68

Less

on 1

7: In

finite

Ser

ies—

Geo

met

ric S

erie

s

Infinite Series—Geometric SeriesLesson 17

Topics

• Infinite series.

• The sequence of partial sums.

• Convergent and divergent series.

• Telescoping series.

• Geometric series.

• Repeating decimals.


• For the infinite series 1 2 31

...nn

a a a a∞

=

= + + +∑ , the nth partial sum is given by 1 2 2n nS a a a a= + + + + .

• If the sequence of partial sums converges to S, then the series converges to S. The limit S is called the

sum of the series, 1

limn nnna S S

∞

→∞=

= =∑ . If the sequence of partial sums diverges, then the series diverges.

• Let a be a nonzero constant. The geometric series with common ratio r is 2 3

0...k

kar a ar ar ar

∞

=

= + + + +∑ .

The series converges to 1

ar-

if 1.r < The geometric series diverges if 1.r ≥

Summary

Fundamental to all of calculus is the notion of a “limit machine” as we express what happens as we go to infinity. We define infinite series in terms of a sequence of partial sums of the series. If the sequence converges, then the series converges. On the other hand, if the sequence of partial sums diverges, then so does the series. After presenting some elementary examples, we look at an example of a telescoping series, where intermediate terms of the partial sums drop out. Then, we study geometric series, in which each term in the summation is a fixed multiple of the previous term, and prove an important convergence theorem. Finally, we apply our knowledge of geometric series to repeating decimals.

Example 1: A Convergent Series

Show that the series 1

1 1 1 1 12 4 8 162n

n

∞

=

= + + + +∑

converges.

Solution

The sequence of partial sums is 1 21 3 2 1, ,...,2 4 2

n

n nS S S -= = = , which converges to 1. That is,

69

1

1 lim 12 nn nn

S∞

→∞=

= =∑ , and the series converges to 1.

Example 2: A Divergent Series

Show that the series 11 1 1 1

n

∞

=

= + + +∑

diverges.

Solution

The sequence of partial sums is 1 2 31, 1 1 2, 1 1 1 3,..., nS S S S n= = + = = + + = = , which diverges. Hence,

11 lim limnn nn

S n∞

→∞ →∞=

= = = ∞∑ , and the series diverges.

Example 3: A Telescoping Series

Determine the convergence or divergence of the telescoping series 1

1( 1)n n n

∞

= +∑ .

Solution

Using partial fractions, we can rewrite the series: 1 1

1 1 1( 1) 1n nn n n n

∞ ∞

= =

= - + + ∑ ∑ .

This is called a telescoping series because the nth partial sum is

1 1 1 1 1 1 1 1 1... 11 2 2 3 3 4 1 1nS

n n n = - + - + - + + - = - + +

.

All the terms have canceled except the first and last terms. Finally,

1 1

1 1 1 1lim lim 1 1( 1) 1 1nn nn n

Sn n n n n

∞ ∞

→∞ →∞= =

= − = = − = + + + ∑ ∑ .

.

Example 4: A Geometric Series

Determine the convergence or divergence of the geometric series 0 0

3 1322

n

nn n

∞ ∞

= =

=

∑ ∑ .

70

Solution

For this geometric series, 3a = and 12r = . Hence, the series converges, and its sum is 3 6

11 1 2

aSr

= = =- -

.

Example 5: A Repeating Decimal

Use a geometric series to write the repeating decimal 0.08 0.080808...= as a fraction.

Solution

We have 2 4 6 2 2

0

8 8 8 8 10.08 0.080808... ...10 10 10 10 10

n

n

∞

=

= = + + + =

∑ , which is a geometric series with

2 2

8 1and10 10

a r= = . Hence, we obtain the following.

2 2

2 22

2

8 88 810 100.08

11 9910 1 10 11 1010

ar

= = = = =- - --

.

Study Tips

• There are two key questions about an infinite series: (1) Does the series converge? (2) If so, what is the limit? In general, the second question is more difficult.

• It is often helpful to write out the first few terms of an infinite series to get a feel for how it behaves.

• However, an infinite series can begin at any value of .n The convergence or divergence does not depend on the first few terms but, rather, on the nature of the terms at the “end” of the series.

• Any variable can be used as the summation variable. For example, the following two series are the

same: 0 0

1 13 3k n

k n

∞ ∞

= =

=∑ ∑ .

• You should memorize the formula for the sum of a geometric series: 2 3

0... , 1.

1k

k

aar a ar ar ar rr

∞

=

= + + + + = <-∑

• A geometric series diverges if 1.r ≥ For example, 2

32

n

n

∞

=

∑ diverges.

• All repeating or terminating decimals can be written as rational numbers (fractions).

• Some rational numbers have two decimal representations. For example, 0.9999... 1= .

Pitfalls

• A common error is to confuse sequences and series. A sequence is an infinite list of numbers. A series is an infinite summation.Le

sson

17:

Infin

ite S

erie

s—G

eom

etric

Ser

ies

71

• It might seem that the infinite repeating decimal 0.999999... is less than 1, but this decimal can be expressed as a geometric series and can be shown to equal 1.

• It is worth repeating: We define a series by its sequence of partial sums, but try not to confuse the two. A sequence is an ordered list of numbers: 1 2 3, , ,...b b b . A series is an infinite sum of

numbers: 1 2

1n

na a a

∞

=

= + +∑

.

Problems

1. Find the first five terms of the sequence of partial sums for the series 1

1

3 .2n

n

∞

-=∑

2. Find the first five terms of the sequence of partial sums for the series ( )0

1.

!

n

n n

∞

=

-∑

3. Determine whether { } 1n

nan+ =

and

1 1

1n

n n

nan

∞ ∞

= =

+=∑ ∑ are convergent or divergent.

4. Determine the convergence or divergence of the series 0

75

n

n

∞

=

∑ .


4( 2)n n n

∞

=

+

∑ .


2

11n n

∞

=

-

∑ .

7. Find the sum of the convergent series 0

13

n

n

∞

=

-

∑ .


34

n

n

∞

=

∑ .

9. Write the repeating decimal 0.075 as a rational number (fraction).

10. Find all values of x for which the series ( )1

3 n

nx

∞

=∑ converges.

72

Less

on 1

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erie

s, D

iver

genc

e, a

nd th

e C

anto

r set

series, Divergence, and the Cantor setLesson 18

Topics

• Review of infinite series.

• The nth term test for divergence.

• Operations with series.

• Application: the bouncing ball problem.

• The Cantor set.


• Theorem: If the series 1 2 31

...nn

a a a a∞

=

= + + +∑ converges, then lim 0nna

→∞= .

• The nth term test for divergence: If the terms of an infinite series do not tend to zero, then the series diverges.

• Algebraic properties of convergent series: If ,n na A b B= =∑ ∑ are convergent series, then

( ) andn n n na b A B ca c a cA+ = + = =∑ ∑ ∑ .

Summary

We begin this lesson with a brief review of infinite series. Then, we present an important test for divergence: If the terms of a series do not tend to zero, then the series diverges. Equivalently, if a series converges, then its terms tend to zero. We discuss algebraic manipulations of convergent series. As an application of geometric series, we solve a bouncing ball problem. Finally, we present the fascinating Cantor set.

Example 1: The nth Term Test for Divergence

The following series diverge because their terms do not approach zero.

a. ( )0

1 1 1 1 1 1 1 ...n

n

∞

=

- = - + - + - +∑ .

b. 1

7 .6

n

n

∞

=

∑

c. 1

1 2 3 .1 2 3 4n

nn

∞

=

= + + ++∑

73

Example 2: The Sum of Two Geometric Series

Find the sum of the series 0

1 32 5n n

n

∞

=

-

∑ .

Solution

Here we have the sum of two convergent geometric series. Their sum is

0 0 0

1 3 1 1 1 3 15 73 21 12 5 4 42 5 1 12 5

n n

n nn n n

∞ ∞ ∞

= = =

- = - = - = - = - - -

∑ ∑ ∑ .

Example 3: The Bouncing Ball Problem

A ball is dropped from a height of 6 feet and begins bouncing. The height of each bounce is 34

the height of the previous bounce. Find the total vertical distance traveled by the ball.

Solution

The ball first drops 6 feet to the ground. On each subsequent bounce, the distances traveled are

2

1 23 3 3 3 3 3 3 36 6 12 , 6 6 12 ,...4 4 4 4 4 4 4 4

D D = + = = + =

So, the total distance is

2

0

3 3 3 3 16 12 12 ... 6 12 6 9 6 36 42 feet34 4 4 4 1 4

n

nD

∞

=

= + + + = + = + = + = - ∑ .

Example 4: The Cantor Set

The famous Cantor set is constructed as follows. Starting with the unit interval [ ]0,1 , remove the middle third, 1 2,3 3

, thus leaving two closed intervals, 10,3

and 2 ,13

. Next, remove the middle of each of these two

intervals, leaving four closed intervals, 1 2 1 2 7 80, , , , , and ,1 .9 9 3 3 9 9

If you continue this construction

forever, the points on the unit interval that remain form the Cantor set. This set has an infinite number of points,

because all the endpoints of the closed intervals are in it. However, the total length of the segments removed can

be calculated as a geometric series:

2

0

11 1 1 1 1 2 1 2 1 2 31 2 4 ... ... 123 9 27 3 3 3 3 3 3 3 1 3

n

n

∞

=

+ + + = + + + = = = -

∑ .

There are infinitely many points in the Cantor set, yet the total length of the segments removed is 1.

74

Study Tips

• It is very helpful to use your calculator or computer to add up some partial sums in a series. For example,

the geometric series ( )0

cos1 n

n

∞

=∑ converges because 0 cos1 0.5403 1< ≈ < . Moreover, the sum is

1 1 2.17531 cos1 1 0.5403

≈ ≈- -

. You can confirm this by computing an appropriate partial sum. For

example, ( )100

0cos1 2.1753.n

n=

≈∑• Memorize the harmonic series

1

1 1 1 112 3 4n n

∞

=

= + + + +∑ , which we will study more closely in the

next lesson.

Pitfalls

• The nth term test for divergence says that if the terms of a series do not approach zero, then the series diverges. However, you cannot conclude anything from this test if the terms do approach zero. That is, the series might converge or it might diverge.

• Be very careful with notation. For example, the following are three similar-looking series.

1 1 1

a. b. c.n k kn k n

a a a∞ ∞ ∞

= = =∑ ∑ ∑

• The first two are the same, whereas the third one is just the sum k k ka a a+ + + .

• Once again, just because the terms of a series get smaller and smaller, you cannot conclude that the series converges. We will see appropriate examples in the next lesson.

Problems

1. Verify that the series 0

53

n

n

∞

=

-

∑ diverges.

2. Verify that the series 0 2 3n

nn

∞

=

+

∑ diverges.

3. Verify that the series 0

!2n

n

n∞

=

∑ diverges.

4. Find the sum of the convergent series ( ) ( )1

0.7 0.9n n

n

∞

=

- ∑ .


1 12 3n n

n

∞

=

-

∑ .

Less

on 1

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erie

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iver

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e, a

nd th

e C

anto

r set

75

6. Find the sum of the convergent series 1 0.1 0.01 0.001+ + + + .

7. Find the sum of the convergent series ( )1

sin1 n

n

∞

=∑ .


1lnn n

∞

=∑ .

9. Find all values of x for which the series ( )1

1 n

nx

∞

=

-∑ converges.

10. A ball is dropped from a height of 16 feet. Each time it drops h feet, it rebounds 0.81h feet. Find the total distance traveled by the ball.

76

Less

on 1

9: In

tegr

al T

est—

Har

mon

ic s

erie

s, p

-ser

ies

Integral Test—Harmonic series, p-seriesLesson 19

Topics

• The divergence of the harmonic series.

• The integral test.

• p-series.

• The Euler-Mascheroni constant.


• The integral test: Let f be a positive, continuous, and decreasing function for 0,x ≥ and ( )na f n= .

Then, 1

1and ( )n

na f x dx

∞ ∞

=∑ ∫

either both converge or both diverge.

• The p-series 1

1 1 1 11 2 3p p p p

n n

∞

=

= + + +∑

converges if 1p > and diverges if 0 1p< ≤ .

• The harmonic series 1

1 1 1 1 ...1 2 3n n

∞

=

= + + +∑ is a p-series with p = 1, and hence, the harmonic series diverges.

• Let 1 1 1 1...1 2 3nS

n= + + + +

be the nth partial sum of the harmonic series. The Euler-Mascheroni constant

is [ ]lim ln 0.577216nnS n γ

→∞- = ≈ .

Summary

We begin by looking closely at the harmonic series and showing why it diverges. We then present our first

major theorem on the convergence of infinite series: the integral test, which compares a given series with a corresponding improper integral. If the integral converges, then so does the series. Conversely, if the integral diverges, then so does the series. We define an important class of series called p-series and prove a theorem

concerning their convergence. In particular, the harmonic series 11 1

1 1 1 112 3n nn n

∞ ∞

= =

= = + + +∑ ∑ is a p-series with

p = 1. Finally, we discuss the fascinating Euler-Mascheroni constant, which is not known to be rational or irrational.

77

Example 1: Using the Integral Test

Use the integral test to determine the convergence or divergence of the series 21

11n n

∞

= +∑ .

Solution

The function 2

1( )1

f xx

=+

is positive, continuous, and decreasing for 0.x ≥ Furthermore, the improper

integral converges: [ ] [ ]2 2 11 1

1 1lim lim arctan lim arctan arctan12 4 41 1

b b

b b bdx dx x b

x xπ π π∞

→∞ →∞ →∞= = = - = - =

+ +∫ ∫ .

Hence, the series converges, too.


Determine the convergence or divergence of the series 31

1 1 118 27n n

∞

=

= + + +∑ .

Solution

The function 31( )f x x= is positive, continuous, and decreasing for 0.x > Furthermore,

33 2 21 1

1 1

1 1 1 1 1lim lim2 22 2

b b

b bdx x dx

x x b∞ ∞ -

→∞ →∞

- - - = = = - = ∫ ∫ . Hence, the original series converges. However, it

does not converge to 12

. In fact, nobody knows the exact value of this sum of the reciprocals of cubes.



1 1 112 3n n

∞

=

= + + +∑ .

Solution

The function 1( )f xx

= is positive, continuous, and decreasing for 0.x > Furthermore,

1 12 2

1 1 11

1 lim 2 lim 2 2b b

b bdx x dx x b

x∞ ∞ -

→∞ →∞

= = = - ∫ ∫ , which diverges. Hence, the original series diverges.

Example 4: Comments on p-Series

A p-series converges for 1p > and diverges for 0 1p< ≤ . For example,

78

a.

31

1 1 118 27n n

∞

=

= + + +∑

converges (Example 2).

b.

121 1

1 1 1 112 3n nn n

∞ ∞

= =

= = + + +∑ ∑

diverges (Example 3).

c. 1

1 1 1 1 ...1 2 3n n

∞

=

= + + +∑ diverges (harmonic series).

Study Tips

• Remember: The harmonic series 1

1 1 1 1 ...1 2 3n n

∞

=

= + + +∑ diverges even though its terms are getting

smaller and smaller.

• The integral test does not have to begin at the number 1. It can start at 2 or 5, or any other positive integer.

• The integral test can reveal convergence for a series whose actual sum is unknown. For example, the

series 3

1

1n n

∞

=∑ converges, but its exact sum is unknown.

• It is unknown whether the Euler-Mascheroni constant (written γ ) is rational or irrational.

Pitfalls

• In Example 1, the improper integral converges to 4

π . However, that does not mean that the series also

converges to the same value 4

π . It probably converges to a different sum.

• You must verify all three requirements in order to use the integral test: positive, continuous, and decreasing. For example, the integral test cannot be used for a series with negative terms, such as

( ) 1

1

1 1 1 11 12 3 4

n

n n

∞+

=

- = - + - +∑

.

• The nth term test for divergence says that if the terms of a series don’t tend to zero, then the series must diverge. However, the converse is false, and the key example is the harmonic series. Its terms tend to zero, but the series “adds up to infinity.”

Less

on 1

9: In

tegr

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mon

ic s

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-ser

ies

79

Problems

1. Use the integral test to determine the convergence or divergence of the series 1

1 .3n n

∞

= +∑

2. Use the integral test to determine the convergence or divergence of the series 13 .n

n

∞-

=∑

3. Use the integral test to determine the convergence or divergence of the series ( )1

1 .1n n n

∞

= +∑

4. Use the integral test to determine the convergence or divergence of the series 2

1 .lnn n n

∞

=∑


3 .n n

∞

=∑


31

3 .n n

∞

=∑


2 .5 1n

nn

∞

=

++∑

8. Determine the convergence or divergence of the series ( )2

2

1 .lnn n n

∞

=∑


ln .n

nn

∞

=∑

10. Explain why the integral test does not apply to the series 2

1

sin .n

nn

∞

=

∑

80

Less

on 2

0: T

he C

ompa

rison

Tes

ts

The Comparison TestsLesson 20

Topics

• The direct comparison test.

• The limit comparison test.


• The direct comparison test: Let 0 .n na b< ≤ If 1

nn

b∞

=∑ converges, then so does

1n

na

∞

=∑ . Similarly, if

1n

na

∞

=∑

diverges, then so does

1n

nb

∞

=∑ .

• The limit comparison test: Let 0, 0.n na b> > Assume that lim n

nn

aL

b→∞

=

, where L is finite and

positive. Then, the two series andn na b∑ ∑ either both converge or both diverge.

Summary

In this lesson, we develop more convergence tests. The direct comparison test for positive-term series compares a given series with a known series. The limit comparison test is similar, but more powerful. It allows us to analyze the convergence of a series without having a term-by-term comparison with a known series, as in the direct comparison test.

Example 1: Using the Direct Comparison Test

Use the direct comparison test to show that the series 2

1

1 1 1 13 6 112n n

∞

=

= + + ++∑

converges.

Solution

Note that 2 2

1 12n n

<+

for all n and 2

1

1 convergesn n

∞

=∑ . Hence,

21

1 converges2n n

∞

= +∑ .

Example 2: Using the Limit Comparison Test


11n n

∞

= -∑ .

81

Solution

We compare the series to the convergent p-series 32

1n n

∞

=∑ .

( )

33

33

11lim lim lim 1

11

n

n n nn

a nn Lb n

n→∞ →∞ →∞

−

= = = = −

.

Hence, the series converges. Notice that it would be difficult to use the direct comparison test for this example.

Example 3: Using the Limit Comparison Test


24 1

n

n

nn

∞

= +∑ .

Solution

We compare the series to the divergent series 21

2n

n n

∞

=∑ . This series diverges because of the nth term test for

divergence. That is, the numerator is greater than the denominator for 4.n > By the limit comparison test,

we have

( )

3 32

2 3 3

3

22 4 1 4 1 1lim lim lim lim 4 4

2 24 1

nn

n nn n n n

n nnn n n n n

n→∞ →∞ →∞ →∞

+ + = = = + =

+

.

Hence, the original series diverges.

Study Tips

• Loosely speaking, the direct comparison test says that if the “larger” series converges, then the “smaller” series also converges. Equivalently, if the “smaller” series diverges, then the “larger” series diverges.

• Given a series whose behavior you wish to analyze, the secret in using the limit comparison test is to find a similar series whose convergence properties you know. For example, to analyze the convergence

or divergence of the series 1

1sinn n

∞

=∑ , you can compare it to the divergent harmonic series

1

1n n

∞

=∑ . Then,

you have ( )1

1 0

sinsinlim lim 1n

n xn

xx→∞ →

= = , which shows that the original series diverges.

• For a series that you might think is convergent, try to compare it with a convergent p-series or a convergent geometric series.

• Similarly, if you think that a given series is divergent, try comparing it to a divergent p-series or a divergent geometric series.

82

Pitfalls

• The comparison tests are for positive-term series only. You cannot use the test on a series like ( )

21

1 n

n n

∞

=

-∑ ,

which has negative terms.

• For the limit comparison test, the limit must be a real number L greater than zero. In particular, the limit cannot be infinity. Remember, infinity is not a number.

Problems

1. Use the direct comparison test to determine the convergence or divergence of the series 21

1 .3 2n n

∞

= +∑


1 .5 3n n

∞

= -∑


ln .1n

nn

∞

= +∑

4. Use the limit comparison test to determine the convergence or divergence of the series 21

.1n

nn

∞

= +∑

5. Use the limit comparison test to determine the convergence or divergence of the series ( )21

1 .3n n n

∞

= +∑

6. Use the limit comparison test to determine the convergence or divergence of the series 1

5 .4 1n

n

∞

= +∑


2 .3 1n

nn

∞

= -∑


1tan .n n

∞

=∑


2 1.5 1

n

nn

∞

=

++∑


51

2 1 .3 2 1n

nn n

∞

=

-+ +∑

Less

on 2

0: T

he C

ompa

rison

Tes

ts

83

alternating seriesLesson 21

Topics

• Alternating series.

• The alternating series test.

• Conditional and absolute convergence.

• Approximations.


• Let 0.na > The following are alternating series:

•

( )

( )

1 2 31

11 2 3

1

1 .

1 .

nn

n

nn

n

a a a a

a a a a

∞

=

∞+

=

- = - + - +

- = - + -

∑

∑

Let 0.na > The alternating series test says that the alternating series ( ) ( ) 1

1 11 and 1n n

n nn n

a a∞ ∞

+

= =

- -∑ ∑converge if 1lim 0 and , for all nn n nn

a a a+→∞= ≤ .

• The series na∑ is conditionally convergent if the series converges, but the series na∑ diverges.

• The series na∑ is absolutely convergent if the series na∑

converges.

• Theorem: If converges, then converges.n na a∑ ∑• Given a convergent alternating series with sum S , the absolute value of the remainder satisfies

1N N NR S S a += - ≤ .

Summary

Having developed tests for positive-term series, we turn to series having terms that alternate between positive and negative. The alternating series test is the main theorem of this lesson. We then use absolute value to look at the concepts of conditional and absolute convergence for series with positive and negative terms. Finally, we study the error in approximating an alternating series by its nth partial sum.

84

Example 1: Using the Alternating Series Test

Show that the series ( ) 1

1

1 1 11 ...2 3

n

n n

+∞

=

-= - + -∑

converges.

Solution

In this example, we have 1 .nan

= We see that 1lim lim 0nn na

n→∞ →∞= = and 1

1 11n na a

n n+ = < =+

for all n . Thus, the

series converges by the alternating series test.

Example 2: Absolute Convergence

The series ( )

31

1 n

n n

∞

=

-∑ is absolutely convergent because the series 3

1

1n n

∞

=∑

converges (p-series, 3 1p = > ).

Example 3: Conditional Convergence

The series ( )

1

1 n

n n

∞

=

-∑ is conditionally convergent because the series converges (by the alternating series test), but

the series 1

1n n

∞

=∑ diverges (p-series, 1 12p = < ).

Example 4: Approximation and Errors

Approximate the error in using the first 1000 terms of the convergent alternating series

( ) 1

1

1 1 11 ...2 3

n

n n

+∞

=

-= - + -∑

to approximate the sum of the series.

Solution

Summing the first 1000 terms, you have ( ) 11000

10001

10.69264743.

n

nS

n

+

=

-= ≈∑ The error is bounded by the absolute

value of the first omitted term, 1000 1000 10011

1001R S S a= - ≤ = .

It can be shown that the infinite series sums to ln 2 0.69314718S = ≈ . Hence, the real error,

100010.00049975 0.000999 0.001

1001S S- ≈ < ≈ ≈ , is well within the bound given by the approximation theorem.

Less

on 2

1: a

ltern

atin

g se

ries

85

Study Tips

• To apply the alternating series test, you must verify two things: (1) The terms tend to zero, and (2) the terms (in absolute value) are nonincreasing.

• There are three possibilities for an arbitrary infinite series: (1) converges absolutely, (2) converges conditionally, or (3) diverges.

• The alternating harmonic series ( ) 1

1

1 1 1 1 11 ...2 3 4 5

n

n n

+∞

=

-= - + - + -∑ is a classic example of a

conditionally convergent series.

Pitfalls

• You cannot arbitrarily rearrange the terms of an alternating series and expect to obtain the same sum. This can only be done if the series is absolutely convergent.

• Make sure you verify both hypotheses of the alternating series test. For example, the terms of the series

( )1

1

( 1) 1 2 3 4 5 ...1 2 3 4

n

n

nn

+∞

=

- += - + - +∑

do not tend to zero. In fact, the series diverges by the nth term test

for divergence (the terms to not tend to zero).

Problems

1. Determine the convergence or divergence of the series ( ) 1

1

12

n

n n

+∞

=

-+∑ .

2. Determine the convergence or divergence of the series ( )1

1

15

n

nn

∞

+=

-∑ .

3. Determine the convergence or divergence of the series ( ) 1

1

1!

n

n n

+∞

=

-∑ .

4. Determine whether the series ( )1

1!

n

n n

∞

=

-∑ converges conditionally or absolutely, or diverges.

5. Determine whether the series ( ) 1

2

1ln

n

n n n

+∞

=


6. Determine whether the series ( )3

3

15

n

n

nn

∞

=

--∑ converges conditionally or absolutely, or diverges.

7. Determine whether the series 1( 1) arctann

nn

∞

=


86

8. Is the series 21

sinn

nn

∞

=∑ an alternating series? Determine whether the series converges conditionally or

absolutely, or diverges.

9. It can be shown that ( )( )

1

1

1 11 !

n

n n e

+∞

=

-=

-∑ . Use the fourth partial sum to approximate 1 .e Use the remainder

theorem to estimate the error in this approximation.

10. Is the following statement true or false? If the series na∑ diverges, then the series na∑ also diverges.

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87

The Ratio and Root TestsLesson 22

Topics

• The ratio test.

• The root test.

• Functions represented by series.


• The ratio test: Let na∑ be a series with nonzero terms.

• If 1lim 1n

nn

aa

+

→∞< , then the series converges absolutely.

If 1lim 1n

nn

aa

+

→∞> , then the series diverges.

The test is inconclusive if 1lim 1n

nn

aa

+

→∞= .

• The root test: Let na∑ be a series.

na∑ converges absolutely if lim 1n

nna

→∞< .

na∑ diverges if lim 1n

nna

→∞> .

The test is inconclusive if lim 1nnn

a→∞

= .

Summary

In this lesson, we look at two more powerful convergence tests: the ratio test and the root test. The ratio test is particularly useful for series having factorials, whereas the root test is useful for series involving nth roots. We close by looking ahead at an infinite series involving a variable .x

Example 1: Using the Ratio Test


2!

n

n n

∞

=∑ .

Solution

We form the absolute value of the quotient of successive terms and take the limit.

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( )

1

11

21 ! 2 ! 2lim lim lim lim 0

( 1)! ( 1)2 2!

n

nn

n nn n n nn

na na n n

n

+

++

→∞ →∞ →∞ →∞

+= = = =

+ + .

Because the limit is 0 1,L = < the series converges by the ratio test.

Example 2: Using the Ratio Test

Determine the convergence or divergence of the series 1 !

n

n

nn

∞

=∑ .

Solution

We take the limit of the quotient of successive terms.

( )( ) ( )

( )( )

1

11

11 ! 1 1

lim lim lim lim1

!

n

n nn

n n nn n n nn

nn n na

a n n n nn

+

+

+

→∞ →∞ →∞ →∞

++ + +

= = =+

.

To evaluate this limit, we do the following.

( )1 1 1lim lim lim 1 1

n n n

nn n n

n n en nn→∞ →∞ →∞

+ + = = + = >

.

Because 1L e= > , the series diverges by the ratio test.

Example 3: Using the Root Test


1

n

nn

en

∞

=∑ .

Solution

We form the nth root of the absolute value of the nth term and take the limit.

1 22 2 2

lim lim lim lim lim 0nn n n n

n nn n n nn n n n nn

e e e eann n n→∞ →∞ →∞ →∞ →∞

= = = = = .

Because 0 1,L = < the series converges by the root test.

89

Example 4: A Series with a Variable

Find all values of x for which the series 2 3

01 ...

! 2! 3!

n

n

x x xxn

∞

=

= + + + +∑ converges.

Solution

We consider x fixed and use the ratio test, as follows.

( )1

11 1 ! !lim lim lim lim 0

( 1)! ( 1)!

nn

nn nn n n n

n

xna x n x

xa n x nn

+

++

→∞ →∞ →∞ →∞

+= = = =

+ +

.

Hence, the series converges for all values of .x Furthermore, this series equals its derivative, and hence, the

series represents the exponential function, ( ) .xf x e=

Study Tips

• Notice the absolute value signs in the ratio and root tests. This means that the tests determine if a given series is absolutely convergent.

• In using the ratio test, we often have to simplify factorials. For example, ( )! ! 11 ! ( 1) ! 1

n nn n n n

= =+ + +

and 0! 1= .

• The ratio test is generally preferred if the series has factorials. The root test is useful for series containing nth roots.

• It is interesting to note the relative rates of growth of the expressions we have studied in these lessons. For 4,n > you have 2ln 2 !n nn n n n n< < < < < .

Pitfalls

• The ratio test and root test are inconclusive for p-series. Fortunately, we know that the p-series

1

1p

n n

∞

=∑

converges for 1.p >

• The ratio test and root test are both inconclusive when the corresponding limit equals 1. In these situations, you will need to analyze the series using a different test.

90

Problems

1. Use the ratio test to determine the convergence or divergence of the series 1

1 .2n

n

∞

=∑


3 .!

n

n n

∞

=∑


6 .5

n

nn

∞

=

∑


!.nn

nn

∞

=∑

5. Use the root test to determine the convergence or divergence of the series 1

1 .7n

n

∞

=∑


2 1 .1

n

n

nn

∞

=

+ -

∑


ln .n

n

nn

∞

=

∑


7 .!

n

n

nn

∞

=∑

9. Find all values of x for which the series 12

3

n

n

x∞

=

∑ converges.

10. Show that the root test is inconclusive for p-series.

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91

Taylor Polynomials and approximations Lesson 23

Topics

• Maclaurin and Taylor polynomials.

• Taylor’s theorem with remainder.


• Let f have n derivatives at 0. The nth-degree Maclaurin polynomial for f is

( )2(0) (0)( ) (0) (0)

2! !

nn

nf fP x f f x x x

n′′

′= + + + + .

• Let f have n derivatives at .c The nth-degree Taylor polynomial for f at c is ( )

2( ) ( )( ) ( ) ( )( ) ( ) ( )2! !

nn

nf c f cP x f c f c x c x c x c

n′′

′= + - + - + + -

.

• Taylor’s theorem: Let f be differentiable through order 1n + on an interval

containing c . Then, there exists a number z between x and c such that ( )

2( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )2! !

nn

nf c f cf x f c f c x c x c x c R x

n′′

′= + - + - + + - +

,

where ( ) ( )

( ) ( )1

1( )1 !

nn

n

f zR x x c

n

++= -

+. That is,

( )( )

11( ) max ( )

1 !

nn

n

x cR x f z

n

++-

≤+

.

Summary

In this lesson, we show how to approximate a function with a polynomial. The first example shows how to construct the first-degree Maclaurin polynomial for the exponential function. This polynomial agrees with the exponential function and its first derivative at the point 0.x = More generally, the Maclaurin polynomial

of degree n for a function ( )f x agrees with the given function and its first n derivatives at the point 0.x = Taylor polynomials generalize this by requiring that the function and its n derivatives agree with the

polynomial and its n derivatives at a point .x c= Finally, we show how Taylor’s theorem can be used to estimate the accuracy of a Taylor polynomial.

Example 1: Finding a Maclaurin Polynomial

Find the first-degree Maclaurin polynomial for ( ) .xf x e=

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Solution

The polynomial ( )1 0 1P x a a x= + satisfies 00 (0) 1a f e= = = and ( ) 0

1 0 1.a f e′= = = Hence, ( )1 1 .P x x= +

This polynomial agrees with ( ) xf x e= at the point ( )0,1

and agrees with the derivative at that point. Notice

that this polynomial is nothing more than the tangent line to the curve at the point 0.x =

( )0,1

y ex=

y a a x= +0 1

Example 2: Finding a Taylor Polynomial

Find the third-degree Taylor polynomial for ( ) lnf x x= centered at 1.c =

Solution

We evaluate the function and its derivatives at 1.

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

Next, we apply the definition of the Taylor polynomial.

2 3

3

2 3

2 3

( ) ( )( ) ( ) ( )( ) ( ) ( )2! 3!

(1) (1)(1) (1)( 1) ( 1) ( 1)2! 3!

1 1( 1) ( 1) ( 1) .2 3

f c f cP x f c f c x c x c x c

f ff f x x x

x x x

′′ ′′′′= + - + - + -

′′ ′′′′= + - + - + -

= - - - + -

2 33

2 3

2 3

( ) ( )( ) ( ) ( )( ) ( ) ( )2! 3!

(1) (1)(1) (1)( 1) ( 1) ( 1)2! 3!

1 1( 1) ( 1) ( 1) .2 3

f c f cP x f c f c x c x c x c

f ff f x x x

x x x

′′ ′′′′= + - + - + -

′′ ′′′′= + - + - + -

= - - - + -

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

2

3

( ) ln . (1) 0.1( ) . (1) 1.

1( ) . (1) 1.

2( ) . (1) 2.

f x x f

f x fx

f x fx

f x fx

= =

′ ′= =

′′ ′′= - = -

′′′ ′′′= =

93

From the graph, you can see that the polynomial is a good approximation of the function near the point 1.c =

1 0,( )

P x x x x32 31 1

21 1

31( ) = −( ) − −( ) + −( )

( ) lnf x x=

Example 3: Using Taylor’s Theorem

The third-degree Maclaurin polynomial for ( ) sinf x x= is

3

3 ( )3!xP x x= - . Use this polynomial to approximate

( )sin 0.1 . Then, use Taylor’s theorem to estimate the accuracy of this approximation.

Solution

Substituting 0.1x = into the polynomial, the approximation is ( ) ( ) ( )3

3

0.1sin 0.1 0.1 0.1 0.0998333

3!P≈ = - ≈ .

Taylor’s theorem says that ( )43 3

43

( )sin ( )3! 3! 4!x x f zx x R x x x= - + = - + , where 0 0.1z< < . Because sin 1z ≤ ,

and ( )4

4 4( ) sin4! 4!

f z zx x= , we have the following error bound:

You can verify that the actual error is 3 (0.1) sin 0.1 0.00000042P - ≈ .

( ) ( )4 43

sin 1 0.0001(0.1) 0.1 0.1 0.0000044! 4! 4!

zR = ≤ = ≈ .

94

Study Tips

• In general, a Taylor or Maclaurin polynomial will be a better approximation at values of x near the expansion point c (where 0c = for a Maclaurin polynomial).

• In general, higher-degree polynomial approximations are more accurate.

• The first-degree polynomial calculated in Example 1 is nothing more than the tangent line to the curve at the point ( )0,1 .

• It is usually difficult to actually determine the value of z in the error formula of Taylor’s theorem, but we are often able to find an upper bound, as in Example 3.

• Many graphing utilities and computers have built-in programs for calculating Maclaurin and Taylor polynomials.

• Given a polynomial f of degree ,n the Maclaurin polynomial of degree n for f is itself. For example, the third-degree Maclaurin polynomial for 3( ) 2 1f x x x= + - is the same polynomial. That is, the best approximation of a polynomial of degree n is itself.

Pitfall

• The Maclaurin polynomial for the function lny x= does not exist because ln 0 is not defined.

Problems

1. Find the second-degree Maclaurin polynomial for ( ) .xf x e=

2. Find the first-degree Taylor polynomial for ( ) tan at .4

f x x c π= =

3. Find the second-degree Maclaurin polynomial for ( ) sec .f x x=

4. Find the third-degree Maclaurin polynomial for 3( ) 2 1.f x x x= + -

5. Find the third-degree Maclaurin polynomial for ( ) sin .f x x=

6. Find the third-degree Taylor polynomial for 2( ) at 1.f x cx

= =

7. Find the fourth-degree Taylor polynomial for ( ) ln at 2.f x x c= =

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95

8. Use the fourth-degree Maclaurin polynomial for the cosine function to approximate cos(0.3). Then, use Taylor’s theorem to obtain an upper bound for the error of this approximation.

9. Use the fifth-degree Maclaurin polynomial for the exponential function to approximate .e Then, use Taylor’s theorem to obtain an upper bound for the error of this approximation.

10. Differentiate the fifth-degree Maclaurin polynomial for the sine function to obtain the fourth-degree polynomial for the cosine function.

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Power series and Intervals of Convergence Lesson 24

Topics

• Power series.

• The radius of convergence of a power series.

• The interval of convergence of a power series.

• Differentiation and integration of power series.


• If x is a variable, then an infinite series of the forms below are power series.

2

0 1 20

20 1 2

0

.

( ) ( ) ( ) .

nn

n

nn

n

b x b b x b x

b x c b b x c b x c

∞

=

∞

=

= + + +

- = + - + - +

∑

∑

The first series is centered at 0, and the second is centered at .c

• For a power series centered at c , one of the following is true.

a. The series converges only at .c The radius of convergence is 0.R =

b. The series converges absolutely for all .x The radius of convergence is .R = ∞

c. There exists a real number 0R > , the radius of convergence, such that the series converges

absolutely for x c R- < and diverges for x c R- > .

Summary

Loosely speaking, a power series is an infinite polynomial. In other words, it is an infinite series in the variable .x

One familiar example is the geometric series, 2

01n

nx x x

∞

=

= + + +∑ . Given a power series, the key question is to

find its interval of convergence—that is, to find the points where the series converges. In general, this will be either

a point, an interval, or perhaps the entire real line. We also discuss differentiation and integration of power series.

Example 1: Some Power Series Examples

2 3

01

! 2 3!

n

n

x x xxn

∞

=

= + + + +∑

is a power series centered at 0.

97

( ) 2

0( 1) 1 1 ( 1) ( 1)nn

nx x x

∞

=

- + = - + + + -∑

is a power series centered at 1- .

Example 2: The Interval of Convergence

Find the interval of convergence of the power series 2 3

01 ... .

! 2! 3! !

n n

n

x x x xxn n

∞

=

= + + + + + +∑

Solution

We compute the limit of the quotient of successive terms.

( )1

1 1 !lim lim lim 0

1!

n

nnn n n

n

xna x

a nxn

+

+

→∞ →∞ →∞

+= = =

+ . The ratio test says that the series converges if this limit is less than 1.

Hence, the power series converges for all .x The radius of convergence is R = ∞ . The interval of convergence is ( ),-∞ ∞ .

Example 3: Geometric Series

The geometric series 2 3

0

11 ...1

n

nx x x x

x

∞

=

= + + + + =-∑ is a power series centered at 0. Use the ratio test to

confirm that this series converges for 1.x <

Solution

We compute the limit of the quotient of successive terms.

11lim lim lim

nn

nn n nn

a x x xa x

++

→∞ →∞ →∞= = = .

The ratio test says that the series converges if this limit is less than 1. That is, the geometric series converges for 1x < . The radius of convergence of the series is 1. Furthermore, the series diverges at 1x = and 1x = - , so its

interval of convergence is 1 1x- < < .

10-1R =1

98

Example 4: The Interval of Convergence of a Power Series

Find the interval of convergence of the power series 2 3

0! 1 2 6 ...n

nn x x x x

∞

=

= + + + +∑ .

Solution

We use the ratio test: ( ) 1

1 1 !lim lim lim ( 1)

!

nn

nn n nn

n xan x

a n x

++

→∞ →∞ →∞

+= = + = ∞ .

Hence, the series only converges at 0. The radius of convergence is 0.

Example 5: Differentiation of a Power Series

The interval of convergence of the power series 2 3

1( ) ...

2 3

n

n

x x xf x xn

∞

=

= = + + +∑ is the half-open interval [ )1,1- ,

whereas the interval of convergence of its derivative, 1

1 2

1 1( ) 1 ...

nn

n n

nxf x x x xn

-∞ ∞-

= =

′ = = = + + +∑ ∑ , is ( )1,1- .

Notice how we lost an endpoint.

Study Tips

• You can use the ratio test to find the radius of convergence of a power series, but it might take more work to find the interval of convergence because you have to analyze the series at the endpoints.

• In general, you might lose endpoints when you differentiate a power series, and you might gain endpoints when you integrate.

Pitfalls

• Be careful when using the ratio test for a series like ( )2 1

0

( 1)2 1 !

n n

n

xn

+∞

=

-+∑ . The nth term is clearly ( )

2 1( 1)2 1 !

n nxn

+-+

,

whereas the next term does not involve ( )2 2n + but, rather, ( )2 3n + . That is, the next term is

( ) ( )1 2( 1) 1 1 2 3( 1) ( 1)

2( 1) 1 ! 2 3 !

n n n nx xn n

+ + + + +- -=

+ + +.

• Remember that the interval of convergence can be open, closed, or neither, so make sure that you use the proper notation. For example, the interval of convergence for the series in Example 5 is denoted [ )1,1- .

Problems

1. Find the radius of convergence of the power series ( )0

4 n

nx

∞

=∑ .

2. Find the radius of convergence of the power series ( )2

0 2 !

n

n

xn

∞

=∑ .

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99

3. Find the radius of convergence of the power series ( ) 2

0

2 !!

n

n

n xn

∞

=∑ .

4. Find the interval of convergence of the power series 0 7

n

n

x∞

=

∑ .

5. Find the interval of convergence of the power series ( )1

1 n n

n

xn

∞

=

-∑ .

6. Find the interval of convergence of the power series ( ) 1

1

1 ( 4)9

n n

nn

xn

+∞

=

- -∑ .

7. Find the interval of convergence of the power series ( )0

2 !3

n

n

xn∞

=

∑ .

8. Find a power series that has ( )1,0- as its interval of convergence.

9. Let ( ) ( )1 1

1

1 1( ) .

1

n n

n

xf x

n

+ +∞

=

- -=

+∑ Find the intervals of convergence of the functions ( )( ), ,f x f x′ and

( ) .f x dx∫10. Verify that the interval of convergence of

1

1 ( 1)

n

n

xn n

+∞

= +∑ is the closed interval [ ]1,1-

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Representation of Functions by Power seriesLesson 25

Topics

• Representing functions by power series.

• A formula for π using the arctangent function.


• If x is a variable, then an infinite series of the forms below are power series.

2

0 1 20

20 1 2

0

.

( ) ( ) ( ) .

nn

n

nn

n

b x b b x b x

b x c b b x c b x c

∞

=

∞

=

= + + +

- = + - + - +

∑

∑

The first series is centered at 0, and the second is centered at .c

• For a power series centered at c , one of the following is true.

a. The series converges only at .c

b. The series converges absolutely for all .x

c. There exists a real number 0R > , the radius of convergence, such that the series converges

absolutely for x c R- < and diverges for x c R- > .

Summary

The geometric series 2 3

0

11 ... , 11

n

nx x x x x

x

∞

=

= + + + + = <-∑ represents the function 1

1 x- on its

interval of convergence. By appropriate manipulations of the series, you can generate power series for other functions. These manipulations include differentiation and integration of known series. At the end of the lesson, we will present some beautiful series formulas for π .

Example 1: Representing a Function by a Power Series

Find a power series centered at 1c = for the function ( ) 1 .f xx

=

Solution

We use a geometric series as follows.

101

( ) ( ) [ ] [ ]

0 0 0

1 1 1 1 ( 1) ( 1) ( 1)1 1 1 1

n n n n

n n nx x x

x x x

∞ ∞ ∞

= = =

= = = - + = - - = - -- - - - + ∑ ∑ ∑ .

Because this is a geometric series, the interval of convergence is 1 1 0 2.x x- < ⇒ < <

Example 2: Integrating a Series to Find a Power Series

Integrate the series in Example 1 to find a power series for ln x.

Solution

We integrate the series term by term.

( ) 1

0 0

11 ( 1) ( 1) ln ( 1)1

nn n n

n n

xdx x dx x C

x n

+∞ ∞

= =

−= − − ⇒ + = −

+∑ ∑∫ ∫ .

Letting 1x = in this equation shows that 0C = , and the power series is

( ) ( ) ( )1 2 3

0

1 1 11ln ( 1) ...1 1 2 3

nn

n

x x xxxn

+∞

=

- - --= - = - + -

+∑ .

Using the ratio test, we conclude that the radius of convergence is 1.R = Checking the endpoints, we conclude

that the interval of convergence is ( ]0,2 . Try graphing the logarithmic function together with the first few terms of its series.

2 30-1

Interval (0,2]

Example 3: A Power Series for the Arctangent Function

Find a power series for the arctangent function.

Solution

Our goal is to obtain the arctangent function as the integral of the power series for 2

1 .1 x+

To this end, begin by

substituting 2x for x in the series for 11 x+

.

( ) ( ) 22

0 0 0

1 1 1( ) 1 11 1 ( ) 1

n nn n n

n n nx x x

x x x

∞ ∞ ∞

= = =

= = - = - ⇒ = -+ - - +∑ ∑ ∑ .

Next, integrate this series term by term.

( ) ( )2 1

22

0 0

1 1 arctan 12 11

nn nn

n n

xdx x dx x Cnx

+∞ ∞

= =

= - ⇒ = - +++ ∑ ∑∫ ∫ .

102

Letting 0x = shows that 0.C = Hence, we have

( )2 1 3 5 7

0arctan 1 ...

2 1 3 5 7

nn

n

x x x xx xn

+∞

=

= - = - + - ++∑ .

This series converges on [ ]1,1- . In fact, it converges to the arctangent function on this interval. Letting 1x = , we obtain a beautiful series for π .

1 1 1 1 1 1arctan1 1 ... 4 1 ...4 3 5 7 3 5 7π π = = - + - + ⇒ = - + - +

.

Study Tips

• The key to using a geometric series is to use algebra, differentiation, and integration to manipulate the function into the appropriate form. For instance, in Example 1, we used only algebra.

( ) ( )1 1 1 1

1 1 1 1 1x x x r= = =

- - - - + - .

• We can use our skills from the previous lesson to determine the interval of convergence of a power series. However, it is more difficult to actually prove that the series converges to the function in question. This is a subtle point that will be discussed in the next lesson.

• In general, the radius of convergence does not change when you differentiate or integrate a power series. However, you might lose endpoints under differentiation and gain endpoints under integration.

• There are many infinite series approximations for π , and some converge much more quickly than the arctangent series in Example 3. For instance, try writing out the first few terms of the expression

1 14arctan arctan4 5 239π= - .

• In general, a series with factorials in the denominator will give a more accurate approximation than one without factorials.

Pitfall

When manipulating the geometric series, make sure that the new series is precisely in the form 1 .1 r-

For example, let ( ) 1 .3

f xx

=-

a. Centered at 0 :c = ( ) 0

11 133 3 31 3

n

n

xxx

∞

=

= = - -∑ .

b. Centered at 1c = : ( )11 1 2

13 2 1 12

xx x= =

-- - - -

.

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103

Problems

1. Find a geometric power series for the function 1( )2

f xx

=+

.

2. Find a geometric power series for the function 2( )5

f xx

=-

.

3. Find a geometric power series for the function 2

1( )1

f xx

=-

.

4. Find a power series for the function 1( )3

f xx

=-

centered at 1.c =

5. Use the power series ( )0

1 11

n n

nx

x

∞

=

= -+ ∑ to determine a power series for ( )ln 1x +

centered at 0.

6. Use the power series ( )0

1 11

n n

nx

x

∞

=

= -+ ∑ to determine a power series centered at 0 for 2

2 1 11 1 1x x x

-= +

- + -.

7. Find a series representation for ( )2

1( )1

f xx

=-

and determine its interval of convergence.

8. Find the interval of convergence for the series ( )2 1

01

2 1

nn

n

xn

+∞

=

-+∑ .

9. The radius of convergence of the series 0

nn

na x

∞

=∑ is 3. What is the radius of convergence of the series

1

1

nn

nna x

∞-

=∑ ?

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Taylor and maclaurin series Lesson 26

Topics

• Taylor and Maclaurin series.

• An application to integration.

• The famous formula 1ie π = - derived from familiar infinite series.


• Theorem: If the function f is represented by the power series ( )0

( ) nn

nf x a x c

∞

=

= -∑ ,

then the

coefficients are given by ( ) ( )

!

n

n

f ca

n= .

• The Taylor series for f at c is

( ) ( ) ( )2

0

( ) ( )( ) ( ) ( )( ) ( ) ( )! 2! !

nnn n

n

f cf c f cx c f c f c x c x c x cn n

∞

=

′′′- = + - + - + + - +∑

.

• The series is a Maclaurin series if 0c = :

( ) ( ) ( )2

0

0(0) ( )(0) (0)! 2! !

nnn n

n

ff f xx f f x x xn n

∞

=

′′′= + + + + +∑

.

Summary

In this lesson, we explore the general technique to represent a function ( )y f x= as an infinite series. We saw some

examples in the previous lesson, such as lny x= , 1yx

= , and arctany x= . The basic idea is to extend Maclaurin

and Taylor polynomials to series. For example, the nth-degree Maclaurin polynomial for the exponential function 2 3

12! 3! !

nx x xxn

+ + + + +

generalizes to the power series 2 3

01

! 2! 3! !

n n

n

x x x xxn n

∞

=

= + + + + + +∑

. We show how

you can approximate a definite integral with an infinite series. Finally, we close with the derivation of a very

famous and beautiful formula connecting , ,e π and 1.i = -

Example 1: The Maclaurin for the Sine Function

Find the Maclaurin series for ( ) sinf x x= .

105

Solution

We need to calculate the derivatives of the sine function at 0.x =

( )4

( ) sin .( ) cos .( ) sin .( ) cos .

( ) sin .

f x xf x xf x xf x x

f x x

=′ =′′ = -′′′ = -

=

( )4

(0) sin 0 0.(0) cos 0 1.

(0) sin 0 0.(0) 1.

(0) 0.

ffff

f

= =′ = =′′ = - =′′′ = -

=

The pattern for the coefficients is clear, so we have

( ) ( ) ( )2

03 5 7

2 3 4 5

0(0) (0)(0) (0)! 2! !

0 ( 1) 0 10 (1) ... .2! 3! 4! 5! 3! 5! 7!

nnn n

n

ff fx f f x x xn n

x x xx x x x x x

∞

=

′′′= + + + + +

-= + + + + + + = - + - +

∑

By the ratio test, this series converges for all .x In fact, it can be shown that the series converges to the original

function ( ) sinf x x= .

Example 2: Manipulating a Maclaurin Series

Use the series for the cosine function to find the Maclaurin series for ( ) cosf x x= .

Solution

We substitute x for x in the cosine series.

2 4 6 2 3

cos 1 ... cos 1 ...2! 4! 6! 2! 4! 6!x x x x x xx x= - + - ⇒ = - + - .

The series converges for 0x ≥ .

Example 3: An Application to Integration

Use four terms of the Maclaurin series for 2xe- to approximate 21

0

xe dx-∫ .

( ) ( ) ( )2

03 5 7

2 3 4 5

0(0) (0)(0) (0)! 2! !

0 ( 1) 0 10 (1) ... .2! 3! 4! 5! 3! 5! 7!

nnn n

n

ff fx f f x x xn n

x x xx x x x x x

∞

=

′′′= + + + + +

-= + + + + + + = - + - +

∑

106

Solution

Notice first that 2xe- does not have an elementary antiderivative, so we can’t just use the fundamental theorem

of calculus. We derive the series for 2xe- from the series for .xe

2

2 3

04 6 8

2

1 .! 2! 3! !

1 .2! 3! 4!

n nx

n

x

x x x xe xn n

x x xe x

∞

=

-

= = + + + + + +

= - + - + +

∑

Next, integrate the first four terms.

2

14 6 3 5 71 1 2

0 00

1 0.7428572! 3! 3 5(2!) 7(3!)

x x x x x xe dx x dx x- ≈ - + - = - + - ≈

∫ ∫ .

This is close to the calculator answer, 0.746824. A more accurate answer can be obtained by using more terms of the series.

Example 4: A Famous Formula

Consider the series for the exponential function, the sine function, and the cosine function.

( )( )

( )( )

2 1 2

0 0 0

1 1, sin , cos

! 2 1 ! 2 !

n nn nnx

n n n

x xxe x xn n n

+∞ ∞ ∞

= = =

- -= = =

+∑ ∑ ∑ .

Substitute ix for x in the exponential series, where 1i = - .

That is, cos sinixe x i x= + . Letting and cos sin ,ix e iππ π π= = + we have the beautiful formula 1ie π = - .

Study Tips

• If you differentiate the Maclaurin series for the exponential function, your answer will be the same exponential function.

• You can obtain the Maclaurin series for the cosine function by directly applying the definition, or just by differentiating the series for the sine function.

• Notice that the series for the sine function only contains odd powers of .x This is to be expected because the sine function is odd. Similarly, the series for the cosine function contains only even powers of .x

( ) ( ) ( ) ( ) ( )2 3 4

0

2 4 3 5

1 ...! 2! 3! 4!

1 ... ... .2! 4! 3! 5!

nix

n

ix ix ix ixe ix

nx x x xi x

∞

=

= = + + + + +

= - + - + - + -

∑

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( ) ( ) ( ) ( ) ( )2 3 4

0

2 4 3 5

1 ...! 2! 3! 4!

1 ... ... .2! 4! 3! 5!

nix

n

ix ix ix ixe ix

nx x x xi x

∞

=

= = + + + + +

= - + - + - + -

∑

107

• You can calculate series representations in many ways. For example, the Maclaurin series for

2siny x= could be obtained by term-by-term multiplication of the series for the sine function times

itself, or by using the trigonometric identity 2 1 cos 2sin2

xx -= and the series for cos 2x .

• Fortunately, the series representations of the most important functions that we use in calculus always converge to that function.

Pitfalls

• Keep in mind that the square root of 1- , denoted by i , is not a real number. That is, 1i = - , while

2 1.i = - Note that ( )23i i i i= = - (also not a real number), while 4 1.i =

• There exist functions f for which the Taylor series does not converge to the function itself. For

example, the series for 2

1, 0( )

0, 0

xe xf xx

- ≠= =

is equal to 0, not f . If you graph ,f you will see that it is

extremely flat near the origin.

-1

2xy e=

0�1

-0�1

-1 1

108

Problems

1. Show that the interval of convergence for 0 !

n

n

xn

∞

=∑ is ( ), .-∞ ∞

2. Show that the interval of convergence for ( )( )

2 1

0

12 1 !

n n

n

xn

+∞

=

-+∑ is ( ), .-∞ ∞

3. Use the definition of a Maclaurin series to derive the series for the cosine function.

4. Use the definition of a Taylor series to find the series for 1( )f xx

= centered at 1.c =

5. Use the definition of a Taylor series to find the series for ( ) xf x e= centered at 1.c =

6. Find the Maclaurin series for the function 3( ) xf x e-= .

7. Find the Maclaurin series for the function ( ) sin 3 .f x x=

8. Use the identity 2 1 cos 2sin2

xx -= to find the first two nonzero terms of the Maclaurin series for 2sin x

9. Find the first four nonzero terms of the Maclaurin series for ( ) sinxf x e x= .

10. Use the first three terms of the Maclaurin series for ( ) cosf x x= to approximate the integral 1

0.5cos x dx∫ .

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109

Parabolas, Ellipses, and Hyperbolas Lesson 27

Topics

• Parabolas.

• Ellipses and eccentricity.

• Hyperbolas.


• A parabola is the set of all points ( ),x y that are equidistant from a fixed line (the directrix) and a fixed point (the focus).

• An ellipse is the set of all points ( ),x y the sum of whose distances from two fixed points (the foci)

is constant.

• The standard equations of an ellipse with center ( ),h k and major and minor axes of lengths 2a and 2b , ,a b> are as follows.

( ) ( ) ( )

( ) ( ) ( )

2 2

2 2

2 2

2 2

1 horizontal major axis .

1 vertical major axis .

x h y ka b

x h y kb a

- -+ =

- -+ =

• The eccentricity of an ellipse is cea

= .

• A hyperbola is the set of all points ( ),x y for which the absolute value of the difference between the distances between two fixed points (the foci) is constant.

• The standard equations of a hyperbola are as follows.

( ) ( )

( ) ( )

2 2

2 2

2 2

2 2

1 (horizontal transverse axis).

1 (vertical transverse axis).

x h y ka b

y k x ha b

- -- =

- -- =

110

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Summary

In this lesson, we review conic sections: parabolas, ellipses, and hyperbolas. This is probably a familiar topic, details of which are best left for a precalculus course, so we will mainly emphasize the calculus aspects. We first look at parabolas and an arc length computation. We then turn to ellipses, their formulas, and the concept of eccentricity. Finally, we briefly discuss hyperbolas. The conics appear briefly in second-semester calculus, but their three-dimensional generalizations play an important role in third-semester calculus.

Example 1: The Arc Length of a Parabola

Set up the integral for the arc length of the parabola 2 , 0 4.y x x= ≤ ≤

Solution

We calculate the derivative 2y x′ = and use the formula for arc length.

( ) ( )( )42 2

0

ln 65 8 8 651 1 2 16.8186

4b

as y dx x dx

+ +′= + = + = ≈∫ ∫ .

Example 2: The Equation of an Ellipse

Find the center, vertices, and foci of the ellipse with equation 2 24 8 4 8 0x y x y+ - + - = .

Solution

We first complete the square to find the standard form of the ellipse.

( ) ( )( ) ( )

2 2

2 2

2 2

4( 2 1) ( 4 4) 8 4 4

4 1 2 16

1 21.

4 16

x x y y

x y

x y

- + + + + = + +

- + + =

- ++ =

Next, we see that the center is ( )1, 2 ;- the axis of the ellipse is vertical; and 4, 2,a b= = and

2 2 16 4 12 2 3c a b= - = - = = . Hence, the vertices are ( ) ( )1, 6 , 1,2- , and the foci are

( ) ( )1, 2 2 3 , 1, 2 2 3- - - + .

111

Vertex (1,2)

Vertex (1,-6)

(1,-2)(-1,-2) (3,-2)

4

Focus

Focusx y−( )+

+( )=

14

216

12 2

Example 3: The Area Bounded by an Ellipse

Find the area bounded by the ellipse 2 2

14 1x y

+ = .

Solution

We will find the area in the first quadrant and multiply by 4. To this end, solve for y ::

2 2 222 211 1 1 444 1 4 2

x y x xy y x+ = ⇒ = − ⇒ = − = − .

The total area is 2 2

0

14 4 22

A x dx π= - =∫ .

Study Tips

• The integral in Example 1 could be evaluated using the trigonometric substitution 2 tanx θ= .

• Parabolas and ellipses have reflective properties. • Because 0 c a< < for an ellipse, the eccentricity satisfies 0 1e< < . If the eccentricity is near 0, then the

ellipse is nearly circular. On the other hand, if the eccentricity is near 1, then the ellipse will appear elongated.

• In general, the area bounded by the ellipse 2 2

2 2 1x ya b

+ = is .abπ On the other hand, the integral for the

arc length of an ellipse is difficult to evaluate.

• Because the formula for arc length, ( )21b

as y dx′= +∫ , appears often in calculus, it is worth memorizing.

112

Pitfall

• The integral for the arc length of an ellipse does not have an elementary antiderivative and leads to the study of so-called elliptic integrals.

Problems

1. Set up the integral for the arc length of the parabola 24 0x y- = for 0 4y≤ ≤ .

2. Use trigonometric substitution to evaluate the integral 4 2

01 4x dx+∫ .

3. Find the center, vertices, and foci of the ellipse 2 216 16x y+ = .

4. Find the center, vertices, and foci of the ellipse 2 29 4 36 24 36 0x y x y+ + - + = .

5. Find the eccentricity of the ellipse in Exercise 3.

6. Find the area enclosed by the ellipse in Exercise 3.

7. Find an equation of the ellipse with center ( )0,0 , focus ( )5,0 , and vertex ( )6,0 .

8. Find the equations of the tangent lines to the hyperbola 2

2 19x y- = at 6.x =

9. Earth moves in an elliptical orbit with the Sun at one of the foci. The length of half of the major axis is 149,598,000 kilometers, and the eccentricity is 0.0167. Find the minimum distance (perihelion) and maximum distance (aphelion) of Earth from the Sun.

10. Use a graphing utility to approximate the arc length of the ellipse 2

2 14x y+ = .

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113

Parametric Equations and the Cycloid Lesson 28

Topics

• Parametric equations.

• Slope in parametric equations.

• The cycloid.

• Arc length in parametric equations.


• Definition of parametric curve: Let f and g be continuous functions of t on an interval .I The

equations ( ) ( ), x f t y g t= = are parametric equations of the parameter .t The curve C given by

these equations is smooth if the derivatives of f and g are continuous and not both zero on .I

• Given the parametric equations ( ), ( )x f t y g t= = for the smooth curve C , the slope at the point ( ),x y

on C is , 0dy

dy dxdtdxdx dt

dt= ≠ .

• A cycloid is defined as the curve traced out by a point on the circumference of a circle rolling along a

line. The parametric equations of a cycloid are ( sin ), (1 cos ).x a y aθ θ θ= - = -

• The arc length of a parametric curve is given by

( ) ( )2 2

2 2b b

a a

dx dys dt f t g t dtdt dt

′ ′= + = + ∫ ∫ .

Summary

In this lesson, we introduce parametric equations. The idea is to consider x and y as functions of a third variable (“parameter”) .t This allows us to add an orientation to the graph of the parametric curve. After a brief example of how to analyze a parametric curve, we look at the calculus concept of slope in parametric equations. We will look closely at the equation of the cycloid, its derivative, and its arc length.

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Example 1: Sketching a Parametric Curve

Sketch the curve described by the parametric equations 2 4, , 2 32tx t y t= - = - ≤ ≤ .

Solution

By letting the parameter t take on values on the interval [ ]2,3- , you can generate a table of values for the

points ( ),x y . For example, when 2,t = - ( )22 4 0x = - - = and 2 1.

2y -= = -

t −2 −1 0 1 2 3

x 0 −3 −4 −3 0 5

y −1 12- 0 1

2 1 32

Next, plot these points in the xy-plane and connect them with a smooth curve. You will see that you have a piece of a parabola opening to the right. In fact, you can eliminate the parameter in this example to obtain 2 4.x 4y= −

5 32

,

(0, 1)

(0, -1)

5(-4, 0)

−

3 1

2,

t = 1

t = -1 t = -2

t = 0

t = 3

t = 2

x t

y t

t

= −

=

− ≤ ≤

2 4

22 3− −

3 1

2,

Example 2: The Slope of a Parametric Curve

The slope of the curve 3cos , 4sinx t y t= = is 4cos3sin

dydy tdt

dxdx tdt

= =-

.

Example 3: Analyzing a Cycloid

Find the points on the cycloid ( sin ), (1 cos ),x a y aθ θ θ= - = - 0,θ ≥ where the curve has a horizontal tangent.

115

Solution

We first calculate the derivative.

sin sincos 1 cos

dydy ad

dxdx a ad

θ θθθ θ

θ= = =

- -.

The graph has a horizontal tangent when the numerator is 0 (and the denominator is not 0). These are the points , 3 , 5 ,...θ π π π= .

0 0,( )θ = 0

πa a, 2( ) 3 2πa a,( )

2 0πa,( ) 4 0πa,( )θ π= 2

θ π=

Example 4: Finding Arc Length of a Parametric Curve

Set up the integral for the arc length of one arch of the cycloid 2( sin )x t t= - , 2(1 cos )y t= - .

Solution

The derivatives are 2 2cos , 2sin .dx dyt tdt dt

= - = Hence, the arc length is given by

( ) ( )2 2

2 2

02 2 2cos 2sin

b

a

dx dys dt t t dtdt dt

π = + = - + ∫ ∫ .

This integral can be evaluated using standard techniques, and the final length is 16.

Study Tips

• Most graphing utilities have built-in features for graphing parametric curves. Try to verify the examples in the lesson and workbook with your calculator.

116

• You can eliminate the parameter in Example 2. Use the fundamental trigonometric identity as follows:

cos , sin ,3 4x yt t= = which implies that

2 22 2cos sin 1

9 16x yt t+ = + = , an ellipse.

• Notice that we lose the orientation of a parametric curve if we eliminate the parameter.

• The cycloid is closely related to the famous brachistochrone problem, which consists of finding the shape of the curve that an object will take when sliding from point A down to point B in the least amount of time. The resulting curve is an inverted cycloid.

Pitfalls

• When finding the points where a parametric curve has vertical tangents, be careful to check if both the denominator and numerator vanish at the point. In this case, you probably have a cusp or node. A good

example is 3 2,x t y t= = , for which both derivatives vanish at 0.t = Notice that if you eliminate the

parameter ( )21 22 3 3 ,y t x x= = = you obtain a function with a cusp at the origin.

• You will encounter many different variables for the parameter besides .t

• Remember that parametric equations include orientation. The two sets of parametric equations cos , sinx t y t= = and sin , cosx t y t= = have the same graph (a circle of radius 1). However, the

equations are different because the orientation in the first set is counterclockwise and clockwise for the second set.

• Algebraic manipulation can mislead you into thinking that two sets of parametric equations describe the same curve when they do not. If you eliminate the parameter in the two sets of parametric equations

, x t y t= = and 2 2, ,x t y t= = you obtain .y x= However, these equations do not describe the same curve because the second set is the restricted line , 0.y x x= ≥

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117

Problems

1. Identify the curve given by the parametric equations 2 3, 3 1.x t y t= - = +

2. Identify the curve given by the parametric equations 2sec , 3 tan .x t y t= =

3. Identify the curve given by the parametric equations 2 3cos , 4 3sin .x t y t= + = - +

4. Find the slope of the tangent line to the curve , 3 1x t y t= = - at 1t = .

5. Find the slope of the tangent line to the curve 3 3cos , sinx yθ θ= = at 4πθ = .

6. Find all points (if any) of horizontal and vertical tangency to the parametric curve 21, 3x t y t t= + = + .

7. Find all points (if any) of horizontal and vertical tangency to the parametric curve 5 3cos , 2 sinx yθ θ= + = - + .

8. Find the arc length of the curve 3 5, 7 2 , 1 3x t y t t= + = - - ≤ ≤ .

9. Find the arc length of the curve 2 36 , 2 , 1 4x t y t t= = ≤ ≤ .

10. Find the equations of the two tangent lines at the point where the parametric curve 2sin 2 , 3sinx t y t= = crosses itself.

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Polar Coordinates and the Cardioid Lesson 29

Topics

• Definition of polar coordinates.

• Conversion formulas.

• Graphs of polar equations.

• Graph of the cardioid.

• The slope of a polar graph.

• Horizontal and vertical tangents.


• Let ( , )P x y= be a point in the plane. Let r be the distance from P to the origin, and let θ be the angle

that the segment OP makes with the positive x-axis. Then, ( ),r θ is a set of polar coordinates for the

point .P

• conversion formulas: 2 2 2cos , sin ; tan , .yx r y r r x yx

θ θ θ= = = = +

• Let ( )r f θ= be a polar curve. The slope of the graph of the curve is given by

( ) ( )( ) ( )

cos sinsin cos

dy f fdy ddxdx f f

d

θ θ θ θθθ θ θ θ

θ

′+= =

′- +.

Summary

You are probably familiar with polar coordinates. Even so, this lesson reviews their main properties and graphs. One example we consider is the cardioid. We then look at the derivative of a function in polar coordinates and study where the graph has horizontal and vertical tangents.

Example 1: Converting from Polar Coordinates to Rectangular Coordinates

Find the rectangular coordinates of the point ( ) ( ), 3, 6r πθ = .

119

Solution

We use the conversion formulas from polar to rectangular, as follows.

( )

3 3cos 3 cos 3 .6 2 2

31sin 3 sin 3 .6 2 2

x r

y r

πθ

πθ

= = = =

= = = =

Hence, rectangular coordinates are ( ) 33, ,2 2x y =

.

Example 2: Converting from Rectangular Coordinates to Polar Coordinates

Find two sets of polar coordinates for the point ( ) ( ), 1, 3x y = - - .

Solution

We use the conversion formulas from rectangular to polar. Notice that there are an infinite number of solutions.

( ) ( )222 1 3 4 2,2.

3 4tan 3 , .1 3 3

r r

yx

π πθ

= - + - = ⇒ = -

-= = = ⇒

-

Hence, two sets of polar coordinates are ( ) ( ) ( )4, 2, , 2,3 3r π πθ = - .

Example 3: Graphs in Polar Coordinates

a. The graph of 2r = is the circle 2 2 2 4x y r+ = = .

2r =

x y2 2 4+ =

120

The graph of 3πθ =

is a line: tan tan 3 3

3y y xx

πθ = = = ⇒ = .

π3

3πθ =

y x= 3

b. The graph of 2(1 cos )r θ= - is a cardioid.

2(1 cos )r θ= - θπ

=43

θπ

=23

θπ

=2

θπ

=3

θπ

=53

θ = 0

θ π=

-4

Example 4: Horizontal and Vertical Tangents

Find the horizontal and vertical tangents for the polar graph sin , 0r θ θ π= ≤ ≤ .

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Solution

We have 2cos sin cos , sin sin sin sinx r y rθ θ θ θ θ θ θ= = = = = . Then, the vertical tangents

occur when 2 2 3cos sin cos 2 0 ,4 4

dxd

π πθ θ θ θθ= - = = ⇒ = . The horizontal tangents occur when

2sin cos sin 2 0 0,2

dyd

πθ θ θ θθ= = = ⇒ = .

Study Tips

• In general, converting from polar coordinates to rectangular coordinates is easier than converting from rectangular to polar. Furthermore, you can always convert back to the original coordinates to verify your answer.

• Most graphing utilities have built-in features for converting between polar coordinates and rectangular coordinates.

• Try using your graphing utility to graph the polar curves in this lesson. Make sure you set the calculator to polar mode.

• The curve in Example 4 is actually a circle. To see this, convert to rectangular coordinates: 2 2 2sin sinr r r x y yθ θ= ⇒ = ⇒ + = . Next, complete the square:

2 2

2 2 21 1 1 14 4 2 2

x y y x y + - + = ⇒ + - =

.

• Do not memorize the formula for the derivative in polar coordinates,

( ) ( )( ) ( )

cos sinsin cos

dy f fdy ddxdx f fd

θ θ θ θθθ θ θ θθ

′+= =

′- +.

Instead, just remember that cosx r θ= and sin .y r θ= From these equations and the product rule, you

can easily derive the formula for dydx

.

Pitfalls

• The polar coordinates of a point are not unique. For example, the point ( ) ( ), 0, 2x y = has an infinite

number of polar representations, such as ( ) ( ) ( )5, 2, 2,2 2r π πθ = = .

• In polar coordinates, the r value can be negative. For example, if ( ) ( ), 0, 2x y = , then

( ) ( ) ( )3, 2, 2,2 2r π πθ -= - = - are both polar representations of the point.

• It is helpful to identify the coordinate system under consideration. For example, to avoid confusion,

write ( ) ( ), 1,r θ π= instead of just ( )1, .π

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• If both derivatives equal zero at a point, then you might have a cusp there. For example, the cardioid

2(1 cos )r θ= - has a cusp at the origin.

Problems

1. Convert the point ( ) ( ), 8, 2r πθ = to rectangular coordinates.

2. Convert the point ( ) ( )5, 2, 3r πθ = - to rectangular coordinates.

3. Convert the point ( ) ( ), 2, 2x y = to polar coordinates. Give two sets of polar coordinates.

4. Convert the point ( ) ( ), 0, 6x y = - to polar coordinates. Give two sets of polar coordinates.

5. Convert the rectangular equation 2 2 9x y+ = to polar form. Describe the curve.

6. Convert the rectangular equation 12x = to polar form. Describe the curve.

7. Convert the polar equation 5cosr θ= to rectangular form. Describe the curve.

8. Find the points of horizontal tangency to the polar curve 1 sinr θ= - .

9. Find the points of vertical tangency to the polar curve 2(1 cos )r θ= - .

10. Find the points of intersection of the graphs 1 cosr θ= + and 1 cosr θ= - .

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area and arc Length in Polar Coordinates Lesson 30

Topics

• Area of a circular sector.

• Area in polar coordinates.

• Arc length in polar coordinates.


• The area of a circular sector of radius r and angle θ is 212

A rθ= .

• Consider a region in the plane bounded by the polar equation ( )r f θ= , where f is

continuous and nonnegative on the interval [ ], , 0 2 .α β β α π< - < The area of the region is

( ) 2 21 12 2

A f d r dβ β

α αθ θ θ= = ∫ ∫ .

• arc length in polar coordinates: ( ) ( ) ( )22 2 2 drs f f d r ddβ β

α αθ θ θ θθ′= + = + ∫ ∫ .

Summary

We continue with our study of polar coordinates, focusing on applications involving integration. First, we develop the polar equation for the area bounded by a polar curve. The formula is based on the area of a circular sector. After some examples, we turn to arc length in polar coordinates. Here, the formula is similar to that for parametric equations.

Example 1: Area in Polar Coordinates

Find the area enclosed by the cardioid 2(1 cos ), 0 2 .r θ θ π= - ≤ ≤

Solution

The area is [ ]2 22

0

1 1 2(1 cos )2 2

A r d dβ π

αθ θ θ= = -∫ ∫ . To evaluate this integral, we expand the quadratic and use

the half-angle trigonometric formula.

2 22

0 0

2

0

1 cos 22 (1 2cos cos ) 2 (1 2cos )2

1 12 2sin sin 2 2(2 ) 6 .2 4

A d d Aπ π

π

θθ θ θ θ θ

θ θ θ θ π π π

+= - + = - +

= - + + = + =

∫ ∫2 22

0 0

2

0

1 cos 22 (1 2cos cos ) 2 (1 2cos )2

1 12 2sin sin 2 2(2 ) 6 .2 4

A d d Aπ π

π

θθ θ θ θ θ

θ θ θ θ π π π

+= - + = - +

= - + + = + =

∫ ∫

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2(1 cos )r θ= -cardioid

Example 2: Area in Polar Coordinates

Set up the integral for the area of the inner loop of the region 1 2sinr θ= + .

Solution

The key issue is establishing the limits of integration. By graphing the curve, you see that the inner loop is

traced out on the interval 7 116 6π πθ≤ ≤ . Thus, the area is given by [ ]

11 267

6

1 1 2sin2

A dπ

πθ θ= +∫ .

1 2sinr θ= +

Example 3: Arc Length in Polar Coordinates

Set up the integral for the length of the cardioid 2(1 cos ), 0 2 .r θ θ π= - ≤ ≤

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Solution

The derivative is 2sindrd

θθ

= . Hence, the area integral is

( ) ( )2 2 22 2

0(2 2cos ) 2sindrs r d dd

β π

αθ θ θ θθ= + = - +∫ ∫ .

This integral requires some trigonometric formulas to evaluate, and the final answer is 16.

Study Tips

• You can take advantage of symmetry in Example 1. That is,

[ ] [ ]2 2 2

0 0

1 12(1 cos ) (2) 2(1 cos )2 2

A d dπ π

θ θ θ θ= - = -∫ ∫ .

• The integral for the arc length of an ellipse leads to the interesting and difficult subject of elliptic integrals. For example, a calculator approximation for the circumference of the ellipse

4 , 0 22 cos

r θ πθ

= ≤ ≤+

is 16.65293.

• You have probably noticed that the formula for arc length in polar coordinates is similar to the formula in parametric equations.

• Some graphing utilities have the ability to compute the arc length of a curve. For example, a calculator’s approximation for Example 3 could be 16.0.

Pitfalls

• In area problems, make sure that the curve is traced out exactly once on the interval α θ β≤ ≤ . For

example, the area bounded by the curve 8sinr θ= is given by the integral [ ]2

0

1 8sin2

A dπ

θ θ= ∫ , where

the θ-interval is 0 θ π≤ ≤ , not 0 2θ π≤ ≤ . The graph of this equation is a circle of radius 4.

• There can be more than one polar equation for the same polar graph. For example, sin 2 , sin 2 ,r rθ θ= = -

and ( )sin 2r θ= - yield the same polar graph.

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Problems

1. Find the area of the region bounded by 3cosr θ= .

2. Verify that the area bounded by one petal of the rose curve 3cos3r θ= is 34

π .

3. Find the area of one petal of the rose curve 4sin 3r θ= .

4. Set up the integral for the area of the inner loop of 1 2cosr θ= + .

5. Set up the integral for the area between the loops of 3 6sinr θ= - .

6. Find the length of the polar curve 8, 0 2 .r θ π= ≤ ≤

7. Find the length of the polar curve 4cos , 0 .r θ θ π= ≤ ≤

8. Set up the integral for the length of the polar curve 2 , 0 2r θ θ π= ≤ ≤ .

9. Use a graphing utility to approximate the arc length of the ellipse 63 sin

rθ

=-

.

10. What conic does the polar equation 2sin 3cosr θ θ= + represent?

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Vectors in the Plane Lesson 31

Topics

• Vectors in the plane.

• Equivalent vectors.

• The component form of a vector.

• Vector operations and properties.

• Unit vectors and tangent unit vectors.

• Applications.


• The directed line segment PQ

has initial point P and terminal point Q . The length or magnitude

of the directed line segment is denoted . Directed line segments having the same length and

direction are said to be equivalent vectors and are often denoted as PQ=v

.

P

Q

Directed Line Segment

PQ

= u

• A vector is in standard position if its initial point is the origin ( )0,0 . If a vector is in standard

position and its terminal point is ( )1 2,v v , then the component form of the vector is 1 2,v v=v ,

where 1 2andv v are the components. In particular, the zero vector is 0,0=0 . A vector of length

one is a unit vector.

• Addition and scalar multiplication of vectors: Given 1 2 1 2, , ,u u v v= =u v ,

1 1 2 2 1 2 1 2, and , ,u v u v c c u u cu cu+ = + + = =u v u.

1 2 1 1 2 2( 1) , ; ( ) ,u u u v u v- = - = - - - = + - = - -u u u v u v .

• The standard unit vectors are 1,0 and 0,1= =i j . Consider the unit vector

cos ,sin cos sinθ θ θ θ= = +u i j . Any vector v can be written as

cos ,sin cos sinθ θ θ θ= = = +v v u v v i v j .

PQ

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Summary

The next few lessons concern vectors in the plane. In this lesson, we define vectors and their properties. We show how to express an arbitrary vector in terms of the standard unit vectors i and j . Finally, we apply our knowledge of vectors to a problem involving force.

Example 1: Vectors

Find the component form and length of the vector having initial point ( )3, 7- and terminal point ( )2,5- .

Solution

We have 1(3, 7), ( 2,5) 2 3 5P Q v- - ⇒ = - - = - and 2 5 ( 7) 12v = - - = . Hence, 5,12= -v and

2 2( 5) 12 169 13= - + = =v .

Example 2: Vector Operations

You can add vectors together and multiply them by scalars (real numbers). For example, if

2,5 and 3,4 ,= - =v w then 2 3,5 4 1,9 ;+ = - + + =v w 3 3 2,5 6,15 ;= - = -v and

3,4 3,4 0,0- = - = =w w 0 , the zero vector. There is a useful geometric interpretation of vector addition.

The vectors form two adjacent sides of a parallelogram, and the diagonal is their sum, as indicated in the figure.

(1,9)

(-2,5)

u + w

u

w

(3,4)

Example 3: Unit Vectors

Find a unit vector in the direction of the vector 3, 4= -v .

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Solution

We first find the length of the given vector, 2 23 ( 4) 9 16 5= + - = + =v . The unit vector is

1 1 3 43, 4 ,5 5 5

= = - = -u vv

.

Example 4: An Application of Vectors

Two tugboats are pushing a cruise ship. Each boat is exerting a force of 400 pounds at angles of 20° and 20 .- ° What is the resultant force on the cruise ship?

Solution

The two forces are as follows.

1

2

400cos 20 400sin 20 .400cos( 20 ) 400sin( 20 )400cos(20 ) 400sin(20 ) .

= ° + °= - ° + - °= ° - °

F i jF i j

i j

The resultant force is 2 800cos 20 752 pounds= + = ° ≈1F F F i i .

Study Tips

• Two vectors are considered equivalent if they have the same length and direction. For example, if

( ) ( ) ( )0,0 , 3, 2 , 1, 2 ,P Q R and ( )4,4S are four points in the plane, then the vectors andPQ RS

are

equivalent. They have the same length and point in the same direction.

P

QR

S

(0,0)

(1,2)

(3,2)

(4,4)

• The set of vectors in the plane, together with the operations of addition and scalar multiplication, satisfy the axioms of a vector space.

• Any vector can be written as a linear combination of the standard unit vectors. For example, ( )4, 6 4 1,0 6 0,1 4 6- = + - = -i j .

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Pitfalls

• Keep in mind that the notation for vectors and their lengths varies widely in textbooks. For example, you might see ( )1 2 1 2, ,PQ v v v v= = = =v v

.

• Don’t confuse the standard unit vector 1,0=i with the complex number 1i = - . When writing

notes, underline the vector i with a little squiggle.

Problems

1. The vector u has initial point ( )3,2 and terminal point ( )5,6 . The vector v has initial point ( )1,4 and terminal point ( )3,8 . Show that the two vectors are equivalent.

2. The vector u has initial point ( )0,3 and terminal point ( )6, 2- . The vector v has initial point ( )3,10 and terminal point ( )9,5 . Show that the two vectors are equivalent.

3. The vector u has initial point ( )2,0 and terminal point ( )5,5 . Write the vector in component form and as a linear combination of the standard unit vectors andi j .

4. The vector u has initial point ( )0, 4- and terminal point ( )5, 1- - . Write the vector in component form and as a linear combination of the standard unit vectors andi j .

5. Let 4,9 and 2, 5= = -u v . Find 2, , , and 5 3 .3

- -u u v u u v

6. Let 3, 4 and 3,0= - =u v . Find , 3 , , and 6 .- - -u u v u u v

7. Find the unit vector in the direction of 5,12 .

8. Find the vector v if 2=v and it makes an angle of 150° with the positive x-axis.

9. Two tugboats are pushing a cruise ship. Each boat is exerting a force of 400 pounds at angles of 10° and 10 .- ° Find the resultant force on the cruise ship.

10. Find a unit vector parallel to the graph of 3( )f x x= at the point ( )1,1 .

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The Dot Product of Two Vectors Lesson 32

Topics

• The dot product of two vectors.

• The angle between two vectors.

• Orthogonal vectors.

• Projections.

• Force and work.


• The dot product of the vectors 1 2 1 2, and ,u u v v= =u v is 1 1 2 2u v u v= +u • v .

• Theorem: If θ is the angle between the two nonzero vectors u and v , then

cosθ =u vu v .

• The vectors u and v are orthogonal (perpendicular) if 0=u v .

• The projection of the vector u onto the vector v is proj =

v 2

u vu vv .

• If 1 proj= vw u is the projection of u onto v , then 2 1= -w u w is called the vector component of u orthogonal to v .

Summary

The dot product of two vectors is a real number, not a vector. This product provides a method for determining the angle between two nonzero vectors. In particular, two vectors are orthogonal (perpendicular) if their dot product is zero. We then turn to projections of one vector onto another. We close with some typical applications of dot product and projection involving force and work.

Example 1: The Dot Product of Two Vectors

Find the dot product of the vectors 2,3 and 1, 4= = -u v .

Solution

2,3 1, 4 2(1) 3( 4) 2 12 10= - = + - = - = -u v .

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Example 2: The Angle between Two Vectors

Find the angle between the vectors 3,1 and 2, 1= = -u v .

Solution

We use the formula for the cosine of the angle between two vectors.

3,1 2, 1 3(2) 1( 1) 5 5 2cos23,1 2, 1 10 5 50 5 2

θ- + -

= = = = = =-

u vu v

.

Hence, the angle is 4πθ = .

u

v

(3,1)

(2,-1)

Example 3: Orthogonal Vectors

The vectors 1,1 and 2, 2= = -u v are orthogonal because 0=u v .

Example 4: The Projection of One Vector onto Another

Find the projection of the vector 5,10=u onto the vector 4,3=v .

Solution

Draw a sketch of these vectors and the corresponding projection.

2

5,10 4,3 20 30proj 4,3 4,3 2 4,3 8,6254,3

+ = = = = = v 2

u vu vv

.

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u

v

(5,10)

(4,3)

1

2

4

6

8

10

2 3 4 5

Example 5: An Application to Work

A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a 20° angle with the horizontal. Find the work done in pulling the wagon 50 feet.

F

20°

50i

Solution

We define the force vector, ( )25 cos 20 sin 20= ° + °F i j , and the vector giving the distance traveled,

50PQ= =v i

. Finally, the work is (50)(25)cos 20 1250cos 20 1174.6 ft-lbW = = ° = ° ≈F v .

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Study Tips

• The dot product of vectors is a real number (scalar) and can be positive, negative, or zero.

• The length of a vector can be expressed using the dot product.

22 2 2 21 2 1 2 1 2 1 2 1 2, , ,v v v v v v v v v v= ⇒ = + ⇒ = + = •v v v .

• The zero vector is orthogonal to every vector.

• You can rewrite the formula for the cosine of the angle between two vectors as follows:

cos cosθ θ= ⇒ ⋅ =u v u v u vu v . This shows that and cosθ⋅u v have the same sign.

• If the projection of u onto v equals u , then the vectors are parallel. Similarly, if the projection of u

onto v is the zero vector, then the vectors are orthogonal.

Pitfalls

• Keep in mind that the words “orthogonal,” “perpendicular,” and “normal” often mean the same thing in mathematics.

• Be careful when writing expressions involving vector operations. For example, the expression +u v w makes no sense because you can’t add the scalar u v to the vector .w

• When calculating the magnitude of a vector, don’t just stop once you’ve found the dot product; remember that magnitude is the square root of the dot product of the vector with itself.

Problems

1. Find the dot product of the vectors 3, 4 and 1,5= = -u v .

2. Find the dot product of the vectors 6, 4 and 3,2= - = -u v .

3. Find the angle between the vectors 3, 4 and 8,6= = -u v .

4. Use a calculator to approximate the angle between the vectors 3,1=u and 2,4= -v .

5. Find the projection of 6,7 onto 1,4= =u v .

6. Find the projection of 2 3 onto 3 2 .= - = +u i j v i j

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7. Find the projection of 2 3 onto 5= + = +u i j v i j and the vector component of u orthogonal to v .

8. A 48,000-pound truck is parked on a 10° slope. Find the force required to keep the truck from rolling down the hill.

9. A car is towed using a force of 1600 newtons. The chain used to pull the car makes a 25° angle with the horizontal. Find the work done in towing the car 2 kilometers.

10. What is known about an angle between two vectors if their dot product is negative?

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Vector-Valued FunctionsLesson 33

Topics

• Vector-valued functions.

• Limits of vector-valued functions.

• Derivatives and integrals of vector-valued functions.

• Position, velocity, and acceleration.


• A function of the form ( ) ( ) ( )t f t g t= +r i j is a vector-valued function; f and g are the component functions, and t is the parameter.

• The derivative of the vector-valued function ( )tr is ( )0

( ) ( )limt

t t ttt∆ →

+ ∆ -′ =∆

r rr , providing that this limit exists.

r r( ) ( )t t t+ ∆ −

( )tr

′r ( )t

r( )t t+ ∆

• Application to motion: If ( )tr is the position of a particle, then the velocity is ( ) ( )t t′=v r , the speed

is ( )tv , and the acceleration is ( ) ( ) ( )t t t′ ′′= =a v r .

• Theorem: If ( ) ( ) constant, then ( ) ( ) 0t t t t′= =r r r r .

Summary

In this lesson, we use our knowledge of vectors to study vector-valued functions, which are functions whose values are vectors. The domain of a vector-valued function consists of real numbers, and the output is a vector. As will become clear, this topic is very similar to parametric equations. We will see how to apply calculus to the study of these functions. In particular, the derivative of a vector-valued function is a vector tangent to the graph and points in the direction of motion. Perhaps the most important application is using vector-valued functions to describe the motion of a particle.

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Example 1: A Vector-Valued Function

Describe the graph of the vector-valued function ( ) 2cos 3sin , 0 2 .t t t t π= - ≤ ≤r i j

Solution

By letting 2cosx t= and 3siny t= - , we can eliminate the parameter

2 2 2 2

2 2 2 2cos , sin cos sin 12 3 2 3x y x yt t t t = = ⇒ + = + = - -

.

The graph is the ellipse 2 2

14 9x y

+ = oriented clockwise. Some of the values are

(0) 2 , ( ) 3 , ( ) 2 .2π π= = - = -r i r j r i

2

3

1

-1

-2

-2 -1 1 2

-3

-3j

-2i

2i

t = 0

t = π2

t = π

Example 2: Limits of Vector-Valued FunctionsYou can evaluate limits componentwise. For example,

0

21

sina. lim .

1 1b. lim does not exist.1

t

t

t

tet

t t

→

→

+ = + + -

i j i j

i j

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Example 3: Derivatives of Vector-Valued Functions

You can evaluate derivatives componentwise. For example,

( ) ( )( ) ( ) ( )

2( 2) 2 .

( sin ) (1 cos ) 1 cos sin .

t t t t t

t t t t t t t

′= + + ⇒ = +

′= - + - ⇒ = - +

r i j r i j

r i j r i j

Example 4: Integrals of Vector-Valued Functions

You can evaluate integrals componentwise. For example,

( )2 21cos sin2

t tt e dt t e+ = + +∫ i j i j C .

Note that C is a constant vector, not a scalar.

Study Tips

• It is very important to visualize vector-valued functions. We often graph just the endpoint of a vector, whereas at other times, we indicate the entire vector coming out from the origin.

• The vector-valued function ( ) ( ) ( )t f t g t= +r i j is often written with brackets, ( ) ( ), ( )t f t g t=r .

• The derivative of a vector-valued function is a tangent vector to the graph pointing in the direction of motion. When graphing a vector-valued function, it is helpful to place the tangent vector at the endpoint of the position vector.

• The usual properties of derivatives hold for vector-valued functions.

• There is a natural orientation of a vector-valued function. For example, the parabola 2 1y x= + can

be described by the vector-valued function 2( ) ( 1)t t t= + +r i j or by the oppositely oriented function 2( ) ( 1)t t t= - + +r i j .

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Pitfalls

• You will often see other parameters besides t . For example, θ is very common in textbooks.

• The graph of a vector-valued function might not be smooth at points where the derivative vanishes. For

example, the cycloid given by ( ) ( ) ( )( sin ) (1 cos ) 1 cos sint t t t t t t′= - + - ⇒ = - +r i j r i j has cusps at

0, 2 , 4 ,...t π π= .

• Watch out for C appearing as a vector, instead of the more familiar scalar, as in Example 4:

( )2 21cos sin2

t tt e dt t e+ = + +∫ i j i j C .

Problems

1. Find the domain of the vector-valued function ( ) 1 ln2

t tt

= +-

r i j .

2. Evaluate the vector-valued function ( ) 2 ( 1)t t t= + -r i j at the points 1 and 2t t= = - .

3. Describe the curve represented by the vector-valued function ( ) ( 1)t t t= + -r i j and indicate the orientation.

4. Describe the curve represented by the vector-valued function ( ) cos 2sint t t= +r i j and indicate the orientation.

5. Describe the curve represented by the vector-valued function ( ) 3sec 2 tan .t θ θ= +r i j

6. Represent the plane curve 2 2 25x y+ = by a vector-valued function.

7. Find the limit: 2

2

2lim1

t

t

tet

-

→∞

+ +

i j .

8. Find the first and second derivatives of the vector-valued function ( ) 3 3t t t= -r i j .

9. Find the first derivative of the vector-valued function ( ) 4cos 4sint t t= +r i j .

10. Evaluate the integral: 21 sec t dtt

+ ∫ i j .

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Velocity and acceleration Lesson 34

Topics

• Motion in the plane.

• The velocity and acceleration vectors.

• Speed.

• Application to projectile motion.


• Motion in the plane:

[ ] [ ]2 2

( ) ( ) ( ) (position function).( ) ( ) ( ) ( ) (velocity).( ) ( ) ( ) ( ) (acceleration).

( ) ( ) ( ) ( ) (speed).

t x t y tt t x t y tt t x t y t

t t x t y t

= +′ ′ ′= = +′′ ′′ ′′= = +

′ ′ ′= = +

r i jv r i ja r i j

v r

• Projectile motion: 20 0

1( )2

t gt t= - + +r j v r . Here, 0v is the initial velocity and 0r is the initial position.

• Projectile motion: ( ) 20 0

1( ) cos ( sin )2

t v t h v t gtθ θ = + + - r i j . Here, h is the initial height, 0v is the

initial speed, and θ is the angle with the horizontal.

Summary

In this lesson, we combine parametric equations, curves, vectors, and vector-valued functions to form a model for motion in the plane. We derive equations for the motion of a projectile subject to gravity.

Example 1: Motion in the Plane

Find the velocity, speed, and acceleration of a particle moving along the plane curve given by

( ) 2sin 2cos2 2t tt = +r i j .

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Solution

The velocity is the derivative of the position function, so we have the following: ( ) ( ) cos sin2 2t tt t′= = -v r i j .

The acceleration is the rate of change of the velocity, or the second derivative of the position,

1 1( ) ( ) sin cos2 2 2 2

t tt t′′= = - -a r i j . Finally, the speed is the magnitude of the velocity,

2 2( ) cos sin 12 2t tt′ = + =r .

-2

-2

2

2r π( )

v π( )

a π( )

Example 2: Describing the Path of a Particle

Describe the path of the particle moving according to ( )2( ) 4t t t= - +r i j .

Solution

By letting y t= , you see that the movement is along the parabola 2 4x y= - . We also have ( ) ( ) 2t t t′= = +v r i j

and ( ) ( ) 2t t′′= =a r i . Notice that the speed is not constant: 2 2( ) (2 ) 1 4 1t t t′ = + = +r .

Example 3: An Application to Projectile Motion

A baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45° with respect to the ground. Find the maximum height of the baseball. Will it clear the 10-foot fence located 300 feet from home plate?

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1045°

Solution

We use the position function ( ) 20 0

1( ) cos ( sin )2

t v t h v t gtθ θ = + + - r i j with 03, 100,h v= = and 45 .θ = °

Hence,

( )

2

2

1( ) 100cos 3 (100sin ) 324 4 2

50 2 3 50 2 16 .

t t t t

t t t

π π = + + - = + + -

r i j

i j

The maximum height is attained when the y-component of the derivative

( )( ) ( ) 50 2 50 2 32t t t ′ = = + - r v i j is zero.

50 2 25 250 2 32 0 2.21 sec32 16

t t- = ⇒ = = ≈ .

For this time, the maximum height is

2

25 2 25 2 6493 50 2 16 81 feet16 16 8

y = + - = ≈

.

Next, we find the time when the ball is 300 feet from home plate.

300 6300 50 2 3 2 4.24 sec50 2 2

t t= ⇒ = = = ≈ .

At that time, the ball is ( ) ( )23 50 2 3 2 16 3 2 303 288 15 feety = + - = - =

high and, hence, clears the fence.

Study Tips

• Given a position function, you can calculate the velocity, acceleration, and speed. In particular, the velocity vector is tangent to the path of the particle and points in the direction of motion.

• To get a sense of what is going on in a problem, look at a particular point on the curve and plot the velocity and acceleration vectors at that point.

• Notice in Example 1 that the speed is constant, but the velocity is not because the motion is circular. Furthermore, the acceleration vector points toward the center of the circle. In general, if an object is traveling at constant speed, then the velocity and acceleration vectors are perpendicular.

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Pitfalls

• Keep in mind that speed is a scalar, whereas velocity is a vector.

• The acceleration of an object is not the derivative of the speed but, rather, the derivative of the velocity.

Problems

1. Find the velocity and acceleration for the position function ( ) t tt e e-= +r i j .

2. Find the velocity and acceleration for the position function 4( ) lnt t t= +r i j .

3. Find the velocity, acceleration, and speed at the point ( )3,0 for the position function ( ) 3 ( 1)t t t= + -r i j .

4. Find the velocity, acceleration, and speed at the point ( )3,0 for the position function ( ) 3cos 2sint t t= +r i j .

5. Find the velocity, acceleration, and speed at the point ( )1,1 for the position function 2 3( )t t t= +r i j .

6. Use the acceleration function ( )t = +a i j to find the position function if (0) =v 0 and (0) =r 0 .

7. Use the acceleration function ( ) 2 3t = +a i j to find the position function if (0) 4=v j and (0) =r 0 .

8. Find the vector-valued function for the path of a projectile launched at a height of 10 feet above the ground with an initial velocity of 88 feet per second and at an angle of 30° above the horizontal.

9. A baseball is hit from a height of 2.5 feet above the ground with an initial velocity of 140 feet per second and at an angle of 22° above the horizontal. Will the ball clear a 10-foot-high fence located 375 feet from home plate?

10. Determine the maximum height of a projectile fired at a height of 1.5 meters above the ground with an initial velocity of 100 meters per second and at an angle of 30° above the horizontal.

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acceleration’s Tangent and normal VectorsLesson 35

Topics

• The unit tangent vector.

• The unit normal vector.

• The tangential and normal components of acceleration.


• For the position vector ( ) ( ) ( )t x t y t= +r i j , the unit tangent vector is ( )( )( )ttt

′=

′rTr

.

• The unit normal vector (or principle unit normal) is ( )( )( )ttt

′=

′TNT

.

• The acceleration vector can be decomposed into its tangential and normal components:

( ) ( ) ( )t a t a t= +T Na T N .

• tangential component: dadt

= = = Tv av a Tv

.

• normal component: 2 2a a′= = = -N Tv T a N a .

Summary

Acceleration is the rate of change of velocity and has two components: the rate of change in speed and the rate of change in direction. To analyze this phenomenon, we first present the unit tangent vector. This unit vector points in the direction of motion. The unit normal vector is orthogonal to the unit tangent vector and points in the direction the particle is turning. Then, we show how the acceleration vector can be decomposed into its tangential and normal components.

Example 1: A Unit Tangent Vector

Find the unit tangent vector if 2( )t t t= +r i j .

Solution

The derivative is ( ) 2t t′ = +r i j , and its magnitude is 2( ) 1 4 .t t′ = +r Hence, the unit tangent vector is

( )2

( ) 1( ) 2( ) 1 4

tt tt t

′= = +

′ +

rT i jr

.

145

2( )t t t= +r i j y x= 2

T(1)

(1,1)

Example 2: A Unit Normal Vector

Find the unit normal vector if ( ) 3cos 3sint t t= +r i j .

Solution

The derivative of the function is ( ) 3sin 3cost t t′ =- +r i j , and its magnitude is

( ) ( )2 2( ) 3sin 3cos 3.t t t′ = - + =r Hence, the unit tangent vector is

3sin 3cos( )( ) sin cos( ) 3

t ttt t tt

′ - += = = - +

′i jrT i j

r.

The derivative of the unit tangent vector is ( ) cos sint t t′ = - -T i j , and its magnitude is ( ) 1t′ =T . Finally, the

unit normal vector is ( )( ) cos sin( )tt t tt

′= = - -

′TN i jT

.

Example 3: The Tangential and Normal Components of Acceleration

Find the tangential and normal components of acceleration for the curve given by 1( )t tt

= +r i j at the point

1.t =

146

Solution

We have 2

1(1) , ( ) ( ) , (1)t tt-′ ′= + = = + = -r i j r v i j r i j and 3

2( ) ( ) , (1) 2 .t tt

′′ ′′= = =r a j r j

The unit tangent vector is ( ) ( )2

4

1 1( ) , (1) .21

t tt

= - = -+

T i j T i j

The unit normal vector is ( ) ( )2

4

1 1( ) , (1) .21

t tt

= + = ++

N i j N i j

The tangential and normal components of acceleration are

( )

( )

1 22 2.2 21 22 2.2 2

a

a

= = - = - = -

= = + = =

T

N

a T j i j

a N j i j

Study Tips

• The tangential component of acceleration, aT , is equal to the rate of change of the speed. If the tangential component of acceleration is zero, then the speed is constant.

• When the normal component of acceleration is zero, then the motion is in a straight line.

• For an object traveling at constant speed, the velocity and acceleration vectors are orthogonal to each other.

• If the path of a particle is a straight line, then the unit normal does not exist. For example, if

( ) 3 4t t t= +r i j , then 3 4( )5 5

t = +T i j and ( )tN does not exist.

Pitfalls

• There are two vectors orthogonal to the unit tangent vector. The unit normal vector is the one that points in the direction that the curve is bending.

T(t)

r(t)

N(t)

• As in Example 3, finding the unit normal can be difficult. A graphing utility might make the derivative computations easier.

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Problems

1. Find the unit tangent vector to the curve 2( ) 2t t t= +r i j at 1.t =

2. Find the unit tangent vector to the curve ( ) 4cos 4sint t t= +r i j at .4t π=

3. Find the unit tangent vector to the curve ( ) 3 lnt t t= -r i j at .t e=

4. Find the unit normal vector to the curve 21( )2

t t t= +r i j at 2.t =

5. Find the unit normal vector to the curve ( ) cos sint t tπ π= +r i j at .6t π=

6. Find the unit tangent vector to the curve ( ) 4t t=r i . Does the unit normal exist? Describe the form of the path.

7. Find the tangential and normal components of acceleration for the plane curve 2( ) 2t t t= +r i j at 1.t =

8. Find the tangential and normal components of acceleration for the plane curve ( ) 2cos3 4sin 3t t t= +r i j at 0.t =

9. Prove the formula: a =Tv av .

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Curvature and the maximum Bend of a CurveLesson 36

Topics

• The definition of curvature.

• The curvature for a circle.

• Curvature and the tangential and normal components of acceleration.

• The maximum curvature of the exponential function.


• The curvature of the graph of the function ( ) ( ) ( )t x t y t= +r i j is ( )( )t

Kt′

=′

Tr

.

• In rectangular coordinates, the curvature of ( )y f x= is ( )

32 21

yK

y

′′= ′+

.

• The tangential and normal components of acceleration can be described in terms of the curvature and

arc length: 22

2( ) d s dst Kdtdt

= +

a T N . Here, dsdt

is the rate of change of arc length, which is equivalent

to the speed.

Summary

In our final lesson, we develop the concept of curvature: How much does a curve bend? How do you measure the curviness of a planar graph? After illustrating this concept with some examples, we show how curvature helps in decomposing the acceleration vector into its tangential and normal components. We then use the formula for curvature for functions of the form ( )y f x= to calculate the curvature of the graph of

the exponential function ( ) xf x e= and determine the point of maximum curvature. We close with a brief discussion of future areas of study.

Example 1: The Curvature of a Circle

Find the curvature of the circle of radius r given by ( ) cos sint r t r t= +r i j .

Solution

The unit tangent vector is ( ) sin cost t t= - +T i j . The derivative is ( ) cos sin ,t t t′ = - -T i j and ( ) 1t′ =T .

149

Hence, the curvature is ( ) 1( )t

Kt r′

= =′

Tr

.

Example 2: The Curvature of the Exponential Function

Find the curvature of the exponential function xy e= .

Solution

The derivatives are all the same: xy y e′ ′′= = . The curvature is, therefore,

( ) ( )3 3 3

2 2 22 2 211 1

x x

xx

ey eKey e

′′= = = +′+ +

.

Example 3: The Maximum Curvature of the Exponential Function

Find the point on the graph of xy e= where the curvature is a maximum.

Solution

We use our skills from basic calculus to maximize the curvature.

( ) ( ) ( )

( )( )

( )

3 12 2 22 2 2

2 53 2 22 2

31 1 2 1 2211

x x x x x x x

xx

e e e e e e edKdx ee

+ - + -= =

++

.

Setting 0dKdx

= , we have 2 2 12 12

x xe e⇒ = ⇒ = . Using logarithms, we have

1 12 ln ln 2 ln 2 0.34662 2

x x= = - ⇒ = - ≈ - . By the first derivative test, this value of x is a maximum. At that

point, the maximum curvature is 1 2 3ln 2 0.3852 9

K - = ≈

.

150

-5

0�007 0�304 0�354

0�112

0�002

-1 1 3

− ≈ −( )12

2 0 347ln .

*

0�385

y ex=

Study Tips

• Curvature is measured by the rate of change of the unit tangent vector with respect to the rate of change of the position function.

• The curvature of a circle is the reciprocal of its radius. Hence, small circles have large curvature, and large circles have small curvature.

• The curvature of a line is zero. For example, for an object falling from the sky, all acceleration is

in the tangential direction (which does not depend on curvature). For example, if ( ) 4 2t t t= -r i j ,

then ( ) 4 2t′ = -r i j and ( )22( ) 4 2 20 2 5.t′ = + - = =r Thus, ( ) 4 2( )( ) 2 5ttt

′ -= =

′r i jTr

and

( ) 0( ) 0 0( ) 2 5t

t Kt′

′ = ⇒ = = =′

TT

r.

• If you are in a car turning a corner at constant speed, all of the acceleration is in the normal direction. The normal component of acceleration depends on both curvature and speed, and hence, you can be pushed against the car door by more speed or by more curvature.

• The curvature of the exponential function tends to zero as x tends to or .∞ -∞

Pitfalls

• The tangential component of acceleration is the rate of change of speed, which can be positive, negative, or zero.

• On the other hand, the normal component of acceleration depends on both the curvature and the speed.

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Problems

1. Find the curvature of the plane curve 2( )t t= +r i j at 2.t =

2. Find the curvature of the plane curve 1( )t tt

= +r i j at 1.t =

3. Find the curvature of the plane curve ( ) sint t t= +r i j at .2t π=

4. Find the curvature of the plane curve ( ) 5 2t t t= - +r i j . What is this curve?

5. Find the curvature of the plane curve 2( ) 2 3f x x= + at the point 1.x = -

6. Find the curvature of the plane curve ( ) cos 2f x x= at the point 2 .x π=

7. Verify that the maximum curvature of the exponential function is 2 3 .9

8. Find the maximum and minimum curvature for the ellipse 2 2

1.9 25x y

+ =

9. Find all points on the graph of 31y x= - where the curvature is 0.

152

solu

tions

solutions

Lesson 1

Rewrite the equation as 2 33

y x= - . The slope is 23

, and the y-intercept is the point ( )0, 3- .

The slope is 3 .2

- Hence, ( )31 22

y x- = - - or 3 42

y x= - + .

The graph of g is a vertical shift upward of 3 units and a horizontal shift to the right of 2 units.

We verify that the two compositions are the identity:

( )

1 1( ( )) 5 1 ( 1) 1 .5 5

(5 1) 1 5( ( )) 5 1 .5 5

x xf g x f x x

x xg f x g x x

- - = = + = - + =

+ -= + = = =

2 22 2 2y x y x x y= - ⇒ = - ⇒ = + . We now reverse the roles of x and y to obtain the inverse function

( )1 2 2, 0.f x x x- = + ≥

The horizontal asymptote is 2y = , and the vertical asymptotes are 1x = and 1.x = -

One way is to restrict the domain to the closed interval [ ]0,π .

Using the formula sin( ) sin cos sin cosu v u v v u+ = + , let u v= to obtain

sin 2 sin( ) sin cos sin cos 2sin cosu u u u u u u u u= + = + = .

Using the formula cos( ) cos cos sin sinu v u v u v+ = - , let u v= to obtain

2 2 2cos 2 cos( ) cos cos sin sin (1 sin ) sin 1 2sin .u u u u u u u u u u= + = - = - - = -

The limit is 12

:

2

2 2 20 0 0

22

20 0 0

1 cos 1 cos 1 cos 1 coslim lim lim1 cos (1 cos )

sin sin 1 1 1lim lim lim 1 .(1 cos ) 1 cos 2 2

x x x

x x x

x x x xx x x x x

x xx x x x

→ → →

→ → →

- - + -= ⋅ =

+ +

= = = = + +

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

153

Lesson 2

We first rewrite the function and then calculate the derivative.

314 3 3 4 32 23 4

2

1 1 9( ) 2 3 17 ( ) 8 9 82 2

g x x x x g x x x x xxx

−− − −′= − + + ⇒ = + − = + − .

( ) ( ) ( )1 1

2 21cos cos cos sin2

y x x y x x-′= = ⇒ = -

At 0, 0.x y′= = The equation of the tangent line is 1 0( 0)y x- = - , or 1.y =

3 3 31 1(4) 3 ln 4 3ln 44

x x xy e e x e xx x

′ = + = +

.

We use the quotient rule.

2 2

cos3 (2cos 2 ) sin 2 ( 3sin 3 ) 2cos 2 cos3 3sin 2 sin 3( )cos 3 cos 3

x x x x x x x xf xx x

− − +′ = = .

Hence, we have ( ) 2(1)( 1) 0 21

f π - +′ = = - .

22

1tan 1 secsec

dy dyx y ydx dx y

= ⇒ = ⇒ =

.

Next, replace 2sec y with 2 2tan 1 1y x+ = + to obtain 2

11

dydx x

=+

. This is the derivative of the

arctangent function.

Using the quotient rule, ( )

( )( )( )

2

2 2

1 32 3( )1 1

x xx xf xx x

- ++ -′ = =+ +

. Analyzing the sign of the derivative in the

three intervals determined by the critical numbers 3, 1x = - and the vertical asymptote 1,x = - you see

that f is increasing on the intervals ( ) ( ), 3 and 1,-∞ - ∞ and decreasing on ( ) ( )3, 1 and 1,1- - - .

( )( )

13

13

2 2( ) 23 3 2

g x xx

−′ = + =

+.

The only critical number is 2.x = - The graph is increasing on ( )2,- ∞ and decreasing on ( ), 2-∞ - .

The first and second derivatives can be calculated using the quotient rule.

( )( ) ( )

( ) ( )( ) ( )

2

2 2 3 3

16 3 416 ,2 2 2 2

xxf x f xx x x x

− +′ ′′= =

− + − +.

Analyzing the sign of the second derivative and the intervals determined by 2x = ± , you see that the graph is concave upward on ( )2,2- and concave downward on ( ), 2-∞ - and ( )2,∞ .

1.

2.

3.

4.

5.

6.

7.

8.

154

( ) sin cos 0 sin cos tan 1f x x x x x x′ = - + = ⇒ = ⇒ = . The critical numbers are 4

x nπ π= + . By the first

derivative test, the function has relative maxima of 2 at 24

x nπ π= + and relative minima of 2- at

(2 1)4

x nπ π= + - , where n is an integer.

We find the antiderivative: 7

21( ) tan ln7 2

xxf x x x e C= + + + + .

Lesson 3

( ) ( ) 312 22

3x x xx e dx x e dx x e C− − −+ = + = − +∫ ∫ .

5

12A x dx= ∫ .

( )0

50 1 2 43 3 3 3

11

3 3 3 3 2704 5 4 5 20

t t dt t t−

−

− = − = − + = − ∫ .

22 2

0 00

2 3 2 2 3 2 3 3 3 3cos cos sin3 2 3 3 2 3 2 2 4x x xdx dx

ππ π = = = = ∫ ∫ .

( ) ( )1 1 1

11 1

1 1 1 1 1 12 ln 2 3 ln 5 ln1 ln 52 3 2 2 3 2 2 2

dx dx xx x −− −

= = + = − = + +∫ ∫ .

22 2 2 4 4

00

1 1 1 9 1( 2) 2 42 2 2 2 2

x xA e dx e x e e− − − − = + = − + = − + + = − ∫ .

( )2lnln .2xxy dx C

x= = +∫

The initial condition says that (1) 2 0y C= - = + , which implies that 2.C = - The solution is

( )21 ln 22

y x= − .

We need to integrate twice and apply the initial conditions.

2 2

2 21

1 1 1( ) cos ; (0) ( ) cos 1.2 2 2

1 1 1( ) sin ; (0) ( ) sin .4 4 4

x x

x x

f x x e C f f x x e

f x x e x C f f x x e x

′ ′ ′= - + + = ⇒ = - + +

= - + + + = ⇒ = - + +

Note that 2sin t does not have an elementary antiderivative. From the second fundamental theorem of

calculus, 2( ) sinF x x′ = .

6, , 6u x du dx x u= + = = - . ( ) ( )3 5 312 2 2 226 6 6 4

5x x dx u u du u u du u u C+ = - = - = - +∫ ∫ ∫ . Next, convert back

to the original variable ( ) ( )5 3

2 226 6 4 65

x x dx x x C+ = + - + +∫ .

9.

10.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

solu

tions

155

Lesson 4

4 44 .x xy Ce y Ce′= ⇒ = Hence, ( )4 44 4 4 .x xy Ce Ce y′ = = =

2 2 22 ln 2 2 .y y x yy y xy

′ ′- = ⇒ - = Hence, 1 ,y y xy

′- =

which implies that 21 1x xyy

yy y′ = =

--.

( )cos sin sin .xy e x y x-′ = = Finally, we verify the initial condition ( ) ( )cos 022 1.y e eπ

π-

= = =

1 23 cos3 3 sin 3y C x C x′ = - and 1 29 sin 3 9 cos3y C x C x′′ = - - . Thus, we have

( ) ( )1 2 1 29 9 sin 3 9 cos3 9 sin 3 cos3 0.y y C x C x C x C x′′ + = - - + + = We then apply the initial conditions:

( ) 1 2 12 6 sin cos 22 2

y C C Cπ ππ= = + ⇒ = and ( ) 1 2 2 211 6 3 cos 3 sin 3 .

2 2 3y C C C Cπ ππ′= = - = - ⇒ = -

The particular solution is 12sin 3 cos3 .3

y x x= -

( )2 2 21 122 2

x x xy xe dx e x dx e C= = = +∫ ∫ .

5 5 5 .dyy x y dy x dx y dy x dxdx

= ⇒ = ⇒ =∫ ∫ Hence, the solution is 2 2

2 21

5 5 .2 2y x C y x C= + ⇒ - =

( )1 .1 1

dy dy dyx y x dx x dxdx y y

= + ⇒ = ⇒ =+ +∫ ∫ Hence, the solution is

( )2 2 21

22 2 2

1ln 1 1 1.2

x C x xxy C y e Ce y Ce+

+ = + ⇒ + = = ⇒ = -

2 2 .2

dy y dy dx dy dxdx x y x y x

= ⇒ = ⇒ =∫ ∫ Hence, the solution is

2 212 ln ln ln ln ln ln .y x C x C y Cx y Cx= + = + ⇒ = ⇒ =

The initial condition (9,1) implies 11 9 , .9

C C= = The particular solution is 2 1 ,9

y x= or 29 0.y x- =

We have ( ) ( )( )1 0 0 0, 2 0.1 0 2 2.2y y hF x y= + = + + = and

( ) ( )( )2 1 1 1, 2.2 0.1 0.1 2.2 2.43y y hF x y= + = + + = , the approximation of (0.2).y

Because the initial quantity is 20 grams, 20 .kty e= Because the half-life is 1599 years,

15991 1 110 20 1599 ln ln 2.2 2 1599

kt ke e k k= ⇒ = ⇒ = ⇒ = - Therefore, we have [ ]ln 2/159920 ,ty e -= and at

time [ ]ln 2/1599 (1000)1000, 20 12.96t y e -= = ≈ grams.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

156

Lesson 5

.dy dyx dx x dxy y= ⇒ =∫ ∫ So, we have ( ) ( )3 3

2 21

2 233 32

12ln .3

x C xy x C y e Ce+= + ⇒ = =

ln lnln .dy x dy x dyy x x dx dxdx x y x y

= ⇒ = ⇒ =∫ ∫ Hence, we have

( ) ( )( ) ( )2 2

112 ln ln /22

11 ln ln .2

x C xx C y y e Ce++ = ⇒ = = Note that 0.x >

( )2cos /22 21

1sin ln cos .2

vdu v v dv u v C u Ceu

-= ⇒ = - + ⇒ =∫ ∫ The initial condition gives

11/2 2(0) 1 .u Ce C e-= = ⇒ = The particular solution is, therefore,

( ) ( )2 21 cos /2 1 cos /22 v vu e e e- -= = .

The given family of exponential functions is .x xy Ce y Ce y′= ⇒ = = The orthogonal trajectories satisfy 2

11 1 .

2dy yy y dy dx x K

y dx y′ = - ⇒ = - ⇒ = - ⇒ = - +∫ ∫ Hence, the orthogonal trajectories are the family

of parabolas 2 2y x K= - + .

The given family of parabolas is 2 2x Cy x Cy′= ⇒ = , and we can solve for y′,

as follows: 22 2 2 .x x yy

xC xy

′ = = = The orthogonal trajectories satisfy the equation 2

212 .

2 2dy x xy dy dx y Kdx y

= - ⇒ = - ⇒ = - +∫ ∫ Hence, the orthogonal trajectories are the family of

ellipses 2 22x y K+ = .

( ) ( ) ( )1 21 1

1kt kt kt

kt

Ly L be y L be kbebe

- -- - --

′= = + ⇒ = - + -+

. This can be rearranged as follows:

( ) ( )

( )

( )

21

11 1

1 11

111

1 1 .1

kt kt

ktkt kt

kt

kt

kt

kt

y L be kbe

Lk bebe be

bekybe

kybe

L yky kyLL be

-- -

-- -

-

-

-

-

′ = - + -

= + + + -

= + = - + = - = - +

1.

2.

3.

4.

5.

6.

solu

tions

157

For this equation, 1 and 36.k L= = Therefore, the solution is 36 .1 1kt t

Lybe be- -= =

+ + We determine

b by

using the initial condition:

364 8.1

bb

= ⇒ =+

Hence, the solution is 36 .1 8 ty

e-=+

We can rewrite the equation as 4 1 .5 120

dy yydt

= -

For this equation, 4 0.8 and 120.5

k L= = = Therefore,

the solution is 0.8

120 .1 1kt t

Lybe be- -= =

+ +

We determine b by using the initial condition:

1208 14.1

bb

= ⇒ =+

Hence, the solution is 0.8

120 .1 14 ty

e-=+

The model is , 20, (0) 1, (2) 4.1 kt

Ly L y ybe-= = = =

+ Hence, 20

1 ktybe-=

+.

2

20 20(0) 1 1 19. (2) 4 4 .1 1 19 ky b y

b e-= ⇒ = ⇒ = = ⇒ =+ +

Solving for ,k

2 2 4 4 1 191 19 5 2 ln ln 0.7791.19 19 2 4

k ke e k k- -+ = ⇒ = ⇒ - = ⇒ = ≈ The logistic model is

0.7791

201 19 ty

e-=+

. At 5, 14.43t y= ≈ grams.

( )4 .dy k ydx

= -

Lesson 6

This equation is linear because it can be written as ( )2 3

1 1 1xy y ex x

′+ = + .

This equation is not linear because of the 2y term.

1 1 1 .2 2 2

x x x x x x xy e Ce y y e Ce e Ce e- - - ′= + ⇒ + = - + + =

The differential equation can be rewritten as 1 6 2dy y xdx x

+ = - . Hence, the integrating factor is

1 ln( )dx xxu x e e x∫= = = .

The integrating factor is 22

( ) .xdx xu x e e∫= =

7.

8.

9.

10.

1.

2.

3.

4.

5.

158

The integrating factor is ( 1)

( ) .dx xu x e e- -∫= = Multiplying by the integrating factor

gives ( )16 16 .x x x x xe y e y e ye e- - - - -′′ - = ⇒ = Integrating both sides, we have

16 16 16x x x xye e dx e C y Ce- - -= = - + ⇒ = - +∫ .

First, rewrite the equation as (sin ) sin .y x y x′ - = - The integrating factor is ( sin ) cos( ) .

x dx xu x e e-∫= =

Multiplying by the integrating factor gives ( )cos cos cos cos cos(sin ) ( sin ) ( sin ) .x x x x xe y e x y x e ye x e′′ - = - ⇒ = -

Integrating both sides, we have cos cos cos cos( sin ) 1x x x xye x e dx e C y Ce-= - = + ⇒ = +∫ .

The integrating factor is tan ln sec( ) sec .

xdx xu x e e x∫= = = Multiplying by the integrating factor gives

( ) 2(sec ) sec (tan ) sec (sec cos ) sec sec 1.x y x x y x x x y x x′′ + = + ⇒ = + Integrating both sides, we have

2sec (sec 1) tan sin cos cosy x x dx x x C y x x x C x= + = + + ⇒ = + +∫ .

Because ( )0 1, 1y C= = and sin cos cos .y x x x x= + +

The equation is linear, but we can also solve it by separating variables. Let b k m= and write as

.dv dvg bv dtdt g bv

= - ⇒ =-

Integrating both sides, we have 11 ln .dv dt g bv t C

g bv b= ⇒ - - = +

-∫ ∫ Next,

solve for v as follows: 11ln .bt bC btg bv bt bC g bv e Ce- - -- = - - ⇒ - = = Because the object is dropped,

0v = when 0.t = So, g C= , and hence, we have the following.

( )/1bt

bt kt mg ge mgbv g ge v eb k

-- --

- = - + ⇒ = = - .

Although the equation is separable, we will solve it as a linear equation. Rewriting, we have

0.005 .3500

dw Cwdt

+ = The integrating factor is 0.005 0.005( ) .dt tu t e e∫= = Multiplying by the integrating factor

gives ( )0.005 0.005 0.005 0.005 0.005(0.005) .3500 3500

t t t t tC Ce w e w e we e′′ + = ⇒ = Integrating both sides, we have

0.005 0.005 0.005 0.005 .3500 17.5 17.5

t t t tC C Cwe e dt e K w Ke-= = + ⇒ = +∫ When 00, (0)t w w= = , and we have the

final solution: 0.005 /2000 017.5 17.5 17.5 17.5

t tC C C Cw w e w e- - = + - = + -

.

6.

7.

8.

9.

10.

solu

tions

159

Lesson 7

The points of intersection are given by 2 2 2x x x+ = + . Solving this

equation, we have 2 2 0 ( 2)( 1) 0 2,1x x x x x+ - = ⇒ + - = ⇒ = - . The area is

13 21 2

22

1 1 8 9( 2) ( 2 ) 2 2 2 43 2 3 2 3 2x xA x x x dx x

−−

− = + − + = − − + = − + − − − = ∫ .

The points of intersection are given by 22y y y- = - . Solving this equation, we have

( 3) 0 0,3y y y- = ⇒ = . The area is

( )32 33 2

00

3 27 27 92 ( )2 3 2 3 2y yA y y y dy

= − − − = − = − = ∫ .

( )66

22

1 3 3 3 3cos 2 sin sin 2 cos 02 4 2 4

A x x dx x xπ

π

π π−−

= − = + = + − = ∫ .

The area under the curve is ( )3 2

39 36.A x dx

-= - =∫ The line y b= intersects

the parabola at ( )9 ,b b- - and ( )9 ,b b- . So, we need to find b such that

( ) ( )9 92 2

9 09 2 9 18

b b

bx b dx x b dx

- -

- - - - = - - = ∫ ∫ . Evaluating this integral, we have

( ) ( ) ( )39 93 3 23 3

2 2

0 0

9 29 (9 ) 9 9 93 3 3 3

b b bx xx bx b x b b- -

- - - = - - = - - = - =

, which implies that

( )3

23 3

27 9 99 9 9 3.3302 4 4

b b b− = ⇒ − = ⇒ = − ≈ .

In the first quadrant, 2

21 xy ba

= - , so the area bounded by the ellipse is

22 2

20 0

44 1a ax bA b dx a x dx

a a= - = -∫ ∫ . This integral represents the area of 1

4 of a circle. Hence,

244

b aA aba

π π

= =

.

1.

2.

3.

4.

5.

160

The curves intersect at the points ( ) ( )0,0 and 2,4 . The volume is

( ) ( )2

2 222 4 2 3 3 4

0 00

16 324 16 8 23 3

V x x x dx x x dx x x ππ π π = − − = − = − = ∫ ∫ .

The curves intersect at the points ( ) ( )0,0 and 2,4 . The volume is

( ) ( ) ( )2 22 22 2 3 2

0 0

646 6 4 8 5 63

V x x x dx x x x dx ππ π = − − − + = − + = ∫ ∫ .

We revolve around the x-axis the triangle in the first quadrant formed by the line ,0 .ry x x hh

= ≤ ≤

The volume is

The curves intersect at the points ( ) ( )0,0 and 2,4 . The volume is, therefore,

( )( ) ( )242 22 3 2 3 2

0 00

2 4 4 2 4 6 8 4 2 4 164xV x x x dx x x x dx x xπ π π π

= − − = − + = − + =

∫ ∫ .

We use the disk method. The curve in the first quadrant has the equation 2 2( )f y r y= - , and we

integrate from to y r h y r= - = . The volume is

( ) ( ) ( )33 32 2 2 3 2

3 3 3

rr

r hr h

r hy rV r y dy r y r r r hπ π π−

−

− = − = − = − − − −

∫ .

This simplifies to ( )21 33

h r hπ - .

We use the disk method. The curve to the right of the y-axis has the equation

25( ) 9 , 3 3.3

f y y y= - - ≤ ≤

( )32 33 32 2

3 00

5 25 509 (2) 9 9 1003 9 9 3

yV y dy y dy yπ ππ π−

= − = − = − = ∫ ∫ .

2 2 32

200

13 3

hh r r xV x dx r h

h hπ π π

= = = ∫ .

6.

7.

8.

9.

10.

11.

solu

tions

161

Lesson 8

22

2 2

1 1 12 2 2

dy x xdx x x

= - = -

. The arc length is given by

222 22 4

2 21 12 2

232 221

2 12

1 1 1 11 1 22 4

1 1 1 1 1 8 1 1 332 .2 2 3 2 3 2 24 16

b

a

dys dx x dx x dxdx x x

xx dxx x

= + = + - = + +

= + = - = - - - =

∫ ∫ ∫

∫

sin tancos

dy x xdx x

= - = - . The arc length is given by

[ ] ( ) ( )

24 42 2

0 0

4 400

1 1 tan sec

sec ln sec tan ln 2 1 ln1 ln 2 1 0.881.

b

a

dys dx x dx x dxdx

x dx x x

π π

π π

= + = + =

= = + = + - = + ≈

∫ ∫ ∫

∫

1dydx x

= . Hence, 2

5

21

11 1 4.367b

a

dys dx dxdx x

= + = + ≈ ∫ ∫ .

Because 1yx

′ = , we have

( ) ( )9

9 9 33 12 22

4 44

1 8 82 2 1 4 1 1 10 5 171.263 3

S x dx x dx xx

ππ π π = + = + = + = − ≈ ∫ ∫ .

Because 1y x′ = , we have

21

12 ln 1 7.055e

S x dxx

π= + ≈∫ .

Because 2y x′ = - , we have

( ) ( ) ( )333 3 32 2 2 2 2

0 00

2 1 4 1 4 8 1 4 37 1 117.324 6 6

S x x dx x x dx xπ π ππ = + = + = + = − ≈ ∫ ∫ .

( )( )100 20 2000 ft-lbW Fd= = = .

1.

2.

3.

4.

5.

6.

7.

[ ] ( ) ( )

24 42 2

0 0

4 400

1 1 tan sec

sec ln sec tan ln 2 1 ln1 ln 2 1 0.881.

b

a

dys dx x dx x dxdx

x dx x x

π π

π π

= + = + =

= = + = + - = + ≈

∫ ∫ ∫

∫

162

( ) (3) 750 (3) 250F x kx F k k= ⇒ = = ⇒ = . So, the work done is

6 62

33( ) 250 125 4500 1125 3375 in-lb.

b

aW F x dx x dx x = = = = - = ∫ ∫

25( ) 250 (30) 3F x kx k k= ⇒ = ⇒ = . So, the work done is50250

2020

25 25 31250 5000 8750 n-cm.3 6 3 3

xW x dx

= = = - =

∫

The weight on the surface of the moon is ( )1 12 26

= tons. The weight varies inversely as the square of the

distance from the center of the moon. Hence,

( )( )

622 2 2.42 10

1100k kF x kx

= ⇒ = ⇒ = × .

11506 61150 621100

1100

2.42 10 2.42 10 1 12.42 10 95.652mi-ton1100 1150

W dxx x

× − × = = = × − ≈ ∫ .

Lesson 9

In order to balance, you must have 15(3) 20( ) 2.25x x= ⇒ = feet from the fulcrum.

7( 5) 3(0) 5(3) 20 47 3 5 15 3

x − + + −= = = −

+ +.

5(2) 1( 3) 3(1) 105 1 3 9

x + - += =

+ + and 5(2) 1(1) 3( 4) 1

5 1 3 9y + + -= = -

+ +. The center of mass is ( ) 10 1, , .

9 9x y = -

The mass is ( )323

00

9 93 3 92 2 2xm x dx x

= - + = - + = - + =

∫ .

( ) ( )

( )

333 32 2 2

0 00

33 23

00

1 1 1 93 6 9 3 9 .2 2 2 3 2

3 93 .3 2 2

x

y

xM x dx x x dx x x

x xM x x dx

= - + = - + = - + =

= - + = - + =

∫ ∫

∫

Hence, ( ) ( )9 9

2 21, 1 and , 1,19 92 2

y xM Mx y x ym m

= = = = = = = .

8.

9.

10.

1.

2.

3.

4.

solu

tions

163

The mass is

4324

00

2 163 3xm x dx

= = =

∫ .

424

0045

24

00

4.2 4

2 64 .5 5

x

y

x xM x dx

xM x x dx

= = =

= = =

∫

∫

Hence, ( )64 3 12 3 3 12 3, 4 and , ,5 16 5 16 4 5 4

y xM Mx y x ym m

= = = = = = =

.

The mass is ( )232 2 2

00

423 3ym y y dy y

= - = - =

∫ .

( ) ( )

( )

22 3 52 2 4

00

23 42 2

00

2 1 4 82 .2 2 3 5 15

2 42 .3 4 3

y

x

y y y yM y y dy y

y yM y y y dy

- = - = - + =

= - = - =

∫

∫

Hence, ( )8 3 2 4 3 2, 1 and , ,115 4 5 3 4 5

y xM Mx y x ym m

= = = = = = =

.

By symmetry, 0.x = The area (mass) is 21 .2

A rπ= Hence, we have

( )3 32

2 2 22 2 2

2 1 1 1 4 42 3 3 3

rr

rr

x r ry r x dx r xr r rπ π π π-

-

= - = - = =

∫ .

The centroid is ( ) 4, 0,3

rx yπ

=

.

The mass is 5 3

010 125 1033.0m x x dx= - ≈∫ .

( )35 3

0

5 2 3

0

10 125 10 125 130,208.3.2

10 125 3105.6.

x

y

x xM x x dx

M x x dx

-= - ≈

= - ≈

∫

∫

Hence, ( ) ( )3105.6 1302083.0, 126.0 and , 3.0,126.01033.0 1033.0

y xM Mx y x ym m

= ≈ ≈ = ≈ ≈ ≈ .

5.

6.

7.

8.

164

The distance from the center of the circle and the y-axis is 5r = . The area of the circle is 16 .A π=

So, ( )( ) 22 2 5 16 160V rAπ π π π= = = .

The area of the region is ( )( )1 4 4 82

A = = . We need to calculate y :

( )( )434

00

1 1 1 84 4 168 2 16 3 3

xy x x dx x = + − = − =

∫ .

So, 83

r y= = . ( )8 1282 2 83 3

V rA ππ π = = =

.

Lesson 10

Let 3 21, 3u t du t dt= - = . Hence, we have

( ) ( ) ( ) ( )4 4

3 33 31

32 3 3 231 11 11 1 3 43 3 43

t tt t dt t t dt C C

− −− = − = + = +∫ ∫ .

Let cos , sin .u x du x dx= = - Hence, we have

( )1

2sin cos ( sin ) 2 coscos

x dx x x dx x Cx

−= − − = − +∫ ∫ .

( )2 25 10 25x x xy e e e′ = + = + + . So, the solution is

( )2 2110 25 10 252

x x x xy e e dx e e x C= + + = + + +∫ .

Let 11 ln ,u x du dxx-

= - = . Hence, we have

( ) ( ) ( ) ( )2

1 11

1 ln 1 1 1 1 11 ln 1 ln 0 1 .2 2 2 2

ee ex dx x dx x

x x- - = - - = - - = - + = ∫ ∫

[ ]4 4 4

2 2 04 0

5 12(5) 10 arctan 10arctan 41 1

A dx dx xx x−

= = = =+ +∫ ∫ .

Let 1, , sin 3 , cos33

u x du dx dv x dx v x= = = = - . Then, we have

1 1 1sin 3 cos3 cos3 cos3 sin 3 .3 3 3 9

xx x dx uv v du x x x dx x x C = - = - - - = - + + ∫ ∫ ∫

Let 2

1arccos , , ,1

u x du dx dv dx v xx

-= = = =

-. Then, we have

( )2

24arccos 4 4 arccos 4 arccos 1 .

1xx dx uv v du x x dx x x x C

x

= - = + = - - + - ∫ ∫ ∫

9.

10.

1.

2.

3.

4.

5.

6.

7.

solu

tions

165

We do integration by parts twice. Let 2 , 2 , cos , sinu x du x dx dv x dx v x= = = = .

So, we have 2 2cos sin 2 sinx x dx x x x x dx= -∫ ∫ .

For the right-hand integral, let 2 , 2 , sin , cosu x du dx dv x dx v x= = = = - . Then,

2 2 2cos sin 2 cos 2cos sin 2 cos 2sinx x dx x x x x x dx x x x x x C = − − + = + − + ∫ ∫ .

Let 2, , 2w x w x wdw dx= = = . We have a new integral that can be solved by integration by parts:

cos (cos ) (2 ) 2 cosx dx w w dw w wdw= =∫ ∫ ∫ . Let

, , cos , sinu w du dw dv wdw v w= = = = . Then, we have

( )( ) ( )

cos 2 cos 2 sin sin

2 sin cos 2 sin cos .

x dx w wdw w w wdw

w w w C x x x C

= = -

= + + = + +

∫ ∫ ∫

The area is 1

1 2

00

arcsin arcsin 1 12 2

A x dx x x x xπ π = - = - - - = ∫ .

( )1

0

1

0

arcsin2 arcsin 1.2 2

arcsin .2 8

x

y

xM x dx

M x x dx

π π

π π

+ = - =

= - =

∫

∫

Hence, ( ), ,18

x y π =

.

You can do this problem two ways. In each case, the desired integral “magically” appears on both sides of an equation.

Method 1: Let Then, you have

For the integral on the right, let So, you have

8.

9.

10.

11.

u e du e dx dv x dx v xx x= = = = −, , sin , cos .

e x dx e x x e dx e x e x dxx x x x xsin cos cos cos cos .∫ ∫ ∫= −( ) − −( ) = − +

u e du e dx dv x dx v xx x= = = =, , cos , sin .

e x dx e x e x dx e x e x x e dx

e

x x x x x xsin cos cos cos sin sin∫ ∫ ∫= − + = − + − ( )( )2 xx x x

xx x

x dx e x e x

e x dx e x e x C

sin sin cos

sin sin cos .

∫

∫

= −

=−

+2

Method 2: Let Then, you have

For the integral on the right, let So, you have

u x du x dx dv e dx v ex x= = = =sin , cos , , .

e x dx e x e x dxx x xsin sin cos .∫ ∫= −

u x du x dx dv e dx v ex x= = − = =cos , sin , , .

e x dx e x e x dx e x e x e x dx

e x d

x x x x x x

x

sin sin cos sin cos sin

sin∫ ∫ ∫= − = − −

2 xx e x e x

e x dx e x e x C

x x

xx x

∫

∫

= −

=−

+

sin cos

sin sin cos .2

166

Lesson 11

We convert two of the cosines to sines, as follows:

( ) ( )3 5

2 3 2 2 2 4 sin sinsin cos sin 1 sin cos sin sin cos .3 5

x xx x dx x x x dx x x x dx C= - = - = - +∫ ∫ ∫

This integral requires the half-angle formula:

2 1 cos 6 1 1 sin 6cos 3 sin 62 2 6 2 12

x x xx dx dx x x C C+ = = + + = + + ∫ ∫ .

We convert four of the sines to cosines, as follows:

( )( )( )

( )

5 4 2 2 4

4 6 8

4 6 8

5 7 9

sin 5 cos 5 1 cos 5 1 cos 5 cos 5 sin 5

cos 5 2cos 5 cos 5 sin 5

1 cos 5 2cos 5 cos 5 ( 5sin 5 )5cos 5 2cos 5 cos 5 .

25 35 45

x x dx x x x x dx

x x x x dx

x x x x dx

x x x C

= - -

= - +

-= - + -

= - + - +

∫ ∫∫

∫

We convert two secants to tangents, as follows:

( )

( )

4 2 2

2 43 2

sec tan 1 tan tan sec

tan tantan tan sec .2 4

x x dx x x x dx

x xx x x dx C

= +

= + = + +

∫ ∫

∫Note: An equivalent answer is 4

11 sec4

x C+ .

We convert four tangents to secants, as follows:

( )( ) ( )

( )( )

5 3 2 2 2

6 4 2

7 5 3

tan 2 sec 2 sec 2 1 sec 2 1 sec 2 sec 2 tan 2

1 sec 2 2sec 2 sec 2 2sec 2 tan 22sec 2 sec 2 sec 2 .

14 5 6

x x dx x x x x x dx

x x x x x dx

x x x C

= - -

= - +

= - + +

∫ ∫

∫

The area of the region is [ ]2 200

cos sin 1A x dx xπ π

= = =∫ . Next, use integration by parts:

, , cos , sinu x du dx dv x dx v x= = = = .

[ ] [ ]

( )

2 22 20 00 0

22 22

0 00

cos sin sin sin cos 1.2

1 1 1 1cos 1 cos 2 sin 2 .2 4 4 2 8

x x x dx x x x dx x x x

y x dx x dx x x

π ππ π

ππ π

π

π

= = - = + = -

= = + = + =

∫ ∫

∫ ∫

So, the centroid is ( ), 1,2 8

x y π π = -

.

1.

2.

3.

4.

5.

6.

solu

tions

167

csc cotcsc csc ln csc cotcsc cot

x xx dx x dx x x Cx x

+= = - + +

+∫ ∫ .

[ ]2

2 22

cos1 cos seccoscsc csc cot .

sinsin sin tancos

xd x xxx C x x

xdx x x xx

- + = = = =

We use the half-angle formulas:

( )

4

2

1 cos 2 1 cos 2sin2 2

1 1 2cos 2 cos 241 1 cos 41 2cos 24 21 1 sin 4sin 24 2 83 1 1sin 2 sin 4 .8 4 32

x xx dx dx

x x dx

xx dx

xx x x C

x x x C

- - =

= - +

+ = - + = - + + +

= - + +

∫ ∫

∫

∫

Let 2sec , sec tan , sec , tanu x du x x dx dv x dx v x= = = = . Then,

( )

3 2

2

3

sec sec tan sec tan

sec tan sec sec 1

sec tan sec sec .

x dx x x x x dx

x x x x dx

x x x dx x dx

= -

= - -

= - +

∫ ∫∫∫ ∫

Hence, we can solve for the integral of secant cubed:

3

3

2 sec sec tan sec

1 1sec sec tan ln sec tan .2 2

x dx x x x dx

x dx x x x x C

= +

⇒ = + + +

∫ ∫

∫

Lesson 12

Let 3 , 3 .u x du dx= = ( )22

12 34 4arctan 31 9 1 3

dxdx x Cx x

= = ++ +∫ ∫ .

2

1 arcsin39xdx C

x= +

−∫ .

Let 2 , 2 .u x du dx= = 2

1 arcsec 24 1

dx x Cx x

= +-

∫ .

7.

8.

9.

10.

1.

2.

3.

168

Let 24sin , 4cos , 16 4cosx dx d xθ θ θ θ= = - = . Then, we have

( ) ( )2

3 32 2

2

1 4cos 1 sec164cos16

1 1 sin 1tan .16 16 cos 16 16

dx d dx

xC C Cx

θ θ θ θθ

θθθ

= =-

= + = + = +-

∫ ∫ ∫

Let 25sec , 5sec tan , 25 5 tanx dx d xθ θ θ θ θ= = - = . Then, we have

( ) ( )

22

2 2

25 5 tan 5sec tan 5 tan5sec

5 sec 1 5 tan 25 5arcsec .5

x dx d dx

xd C x C

θ θ θ θ θ θθ

θ θ θ θ

-= =

= - = - + = - - +

∫ ∫ ∫

∫

Let 2 2tan , sec , 1 secx dx d xθ θ θ θ= = + = . Then, we have

( )( )

2 2 22

2 42

22 2

tan sec 1sin 1 cos 2sec 21

1 sin 2 1 1 sin cos2 2 2 2

1 1 1arctan arctan .2 2 11 1

x ddx d dx

C C

x xx C x Cxx x

θ θ θ θ θ θ θθ

θθ θ θ θ

= = = -+

= - + = - + = - + = - + + + +

∫ ∫ ∫ ∫

Let 2 2sin , cos ,1 cosx dx d xθ θ θ θ= = - = . When 0, 0,x θ= = and when 3 , .2 3x πθ= = Hence, we have

2 2

1 14 4

dy y dxdx x x

= ⇒ =+ +

∫ . Let 2tan , secx dx dθ θ θ= = and 2 24 4secx θ+ = . Then, we have

(0) 4 4 ln 2 4 ln 2y C C= ⇒ = + ⇒ = - . Hence, the solution is

( )[ ]

2 232 3 3 2

3 30 0 02 2

3 2 300

sin cos tancos1

(sec 1) tan 3 .3

x ddx dx

d

π π

π π

θ θ θ θ θθ

πθ θ θ θ

= =-

= - = - = -

∫ ∫ ∫

∫

2

1

22

1

2sec sec ln sec tan2sec

4ln ln 4 .2 2

dy d C

x x C x x C

θ θ θ θ θ θθ

= = = + +

+= + + = + + +

∫ ∫

2ln 4 4 ln 2y x x= + + + − .

4.

5.

6.

7.

8.

solu

tions

( )[ ]

2 232 3 3 2

3 30 0 02 2

3 2 300

sin cos tancos1

(sec 1) tan 3 .3

x ddx dx

d

π π

π π

θ θ θ θ θθ

πθ θ θ θ

= =-

= - = - = -

∫ ∫ ∫

∫

2

1

22

1

2sec sec ln sec tan2sec

4ln ln 4 .2 2

dy d C

x x C x x C

θ θ θ θ θ θθ

= = = + +

+= + + = + + +

∫ ∫

169

( )22 21 , ,1 12

y x y x y x′ ′= = + = + . So, the arc length is 4 2

01s x dx= +∫ . Using the substitution tan ,x θ=

you obtain ( ) ( )4

4 2 2 2

00

1 11 1 ln 1 4 17 ln 4 17 9.29362 2

s x dx x x x x = + = + + + + = + + ≈ ∫ .

Lesson 13

( )( )2

1 19 3 3 3 3

A Bx x x x x

= = +- - + + -

. The basic equation is, therefore, 1 ( 3) ( 3)A x B x= - + + .

We next find the constants andA B .

1 13 1 6 ; 3 1 66 6

x B B x A A= ⇒ = ⇒ = = - ⇒ = - ⇒ = -

2

1 1 1 1 1 1 1 1 3ln 3 ln 3 ln .9 6 3 6 3 6 6 6 3

xdx dx dx x x C Cx x x x

-= - + = - + + - + = +

- + - +∫ ∫ ∫

( )( )2

5 53 4 4 1 4 1

A Bx x x x x x

= = ++ - + - + -

. The basic equation is, therefore, 5 ( 1) ( 4)A x B x= - + + .

We next find the constants

1 5 5 1; 4 5 5 1x B B x A A= ⇒ = ⇒ = = - ⇒ = - ⇒ = -

2

5 1 1 1ln 4 ln 1 ln .3 4 4 1 4

xdx dx dx x x C Cx x x x x

- -= + = - + + - + = +

+ - + - +∫ ∫ ∫

( )2

2 2

4 2 11 1

x x A B Cx x x x x+ -

= + ++ +

. The basic equation is, therefore,

2 24 2 1 ( 1) ( 1)x x Ax x B x Cx+ - = + + + + . We next find the three constants.

0 1; 1 1; 1 3x B x C x A= ⇒ = − ⇒ = − ⇒ = = ⇒ = .

2

3 2 2

4 2 1 3 1 1 13ln ln 11

x x dx dx x x Kx x xx x x

+ − = − + = + + + + ++ ∫ ∫ .

andA B.

9.

1.

2.

3.

170

( )2

22

111

x A Bx Cx xx x

- += +

++. ( ) ( )2 21 1x A x Bx C x- = + + + .

0 1; 1 0 2 ; 1 0 2x A x B C x B C= ⇒ = - = ⇒ = - + + = - ⇒ = - + - .

Solving these equations, you obtain 1, 2, 0.A B C= - = = The integral becomes

2 22

3 2

1 1 2 1ln ln 1 ln1

x x xdx dx dx x x K Kx xx x x

− += − + = − + + + = +

+ +∫ ∫ ∫ .

Notice how the repeated factors are handled.

( ) ( ) ( ) ( ) ( )2

2 2 3 22 232 2 2

32 92 22 9 9

x x A B C D E Fx G Hx Kx x x xx xx x x x

+ − + += + + + + + +

− +− −− + +.

( ) 2ln 3 ln 2 ln 2 ln 3 ln .

3x

x x x xx

-- - - = - - - - = - -

( )( )3

22

2 4 81 41 4

x x A B Cx Dx x xx x x

- - += + +

- +- +. The basic equation is, therefore,

( )( ) ( ) ( )3 2 22 4 8 1 4 4 ( 1)x x A x x Bx x Cx D x x- - = - + + + + + - . Solving for the constants, you obtain

2, 2, 2, 4.A B C D= = - = = The integral becomes

( )( )3

2 22

2

2 4 8 2 2 2 41 4 41 4

2ln 2ln 1 ln 4 2arctan .2

x x xdx dxx x x xx x x

xx x x C

- - = - + + - + +- +

= - - + + + +

∫ ∫

We first divide: 2

2 2

4 3 52 .2 1 2 1x x xx x x x+ + -

= ++ - + -

We determine the partial fraction decomposition in the usual

way to obtain 2

5 3 22 2 .2 1 2 1 1

xx x x x

-+ = + -

+ - - +

Hence, the integral is 2

2

4 3 3 2 32 2 ln 2 1 2ln 1 .2 1 2 1 1 2x x dx dx x x x Cx x x x

+ + = + - = + - - + + + - - + ∫ ∫

4.

5.

6.

7.

8.

solu

tions

171

( )( )2

15 157 12 3 4 3 4

A Bx x x x x x

= = ++ + + + + +

. The basic equation is 15 ( 4) ( 3)A x B x= + + + .

We determine the constants as follows:

4 15 15; 3 15x B B x A= − ⇒ = − ⇒ = − = − ⇒ = .

We solve for y as follows:

( )

ln ln ln .

.

1 .1 1 1

kt CL kt CL kt CL

kt CLkt CL kt CL

kt CL kt CL kt

yy L y kt CLL y

y e y Le yeL y

Le L Ly e Le ye e be

- - -

-- -

- - + -

- - = = --

= ⇒ = --

+ = ⇒ = = =+ + +

Lesson 14

0 0

sin 4 4cos 4 4lim lim .sin 3 3cos3 3x x

x xx x→ →

= =

2

3 3

2 3 2 2lim lim 4.3 1x x

x x xx→ →

- - -= =

-

2

0 0

1arcsin 1lim lim 1.

1x x

x xx→ →

-= =

3 2lim lim lim lim3 6 6

x x x x

x x x x

e e e ex x x→∞ →∞ →∞ →∞= = = = ∞ .

2

2 21 1 1

2ln 2lnlim lim lim 11 1 2x x x

x x xx x x→ → →

= = =− −

.

This is the indeterminate form ( )( )0 .∞

( ) ( )( ) ( )2

2

1 11 cossin1 1lim sin lim lim lim cos 11 1x x x x

xx xx xx x x→∞ →∞ →∞ →∞

− = = = = −

.

[ ]

2 2 2

2 00 0

15 15 15 15 ln 3 ln 47 12 3 4

5(4) 1015 ln 5 ln 6 ln 3 ln 4 15ln 15ln .6(3) 9

A dx dx x xx x x x

= = - = + - + + + + +

= - - + = =

∫ ∫

9.

10.

1.

2.

3.

4.

5.

6.

[ ]

2 2 2

2 00 0

15 15 15 15 ln 3 ln 47 12 3 4

5(4) 1015 ln 5 ln 6 ln 3 ln 4 15ln 15ln .6(3) 9

A dx dx x xx x x x

= = - = + - + + + + +

= - - + = =

∫ ∫

172

This is the indeterminate form 0∞ . Let 1 1 lnln lnx x xy x y x

x= ⇒ = = .

11lnlim ln lim lim 0 lim 11

xx x x x

x xy xx→∞ →∞ →∞ →∞

= = = ⇒ = .

This is the form 1∞. Let ( ) ( )1 1 ln(1 )1 ln ln 1x x xy x y x

x+

= + ⇒ = + = .

1

0 0 0 0

1ln(1 ) (1 )lim ln lim lim 1 lim(1 )

1x

x x x x

x xy x ex+ + + +→ → → →

+ += = = ⇒ + = .

This is the indeterminate form ∞-∞ . We rewrite as follows:

2

2 2 22 2 2

2

8 8 ( 2) 8 2lim lim lim4 2 4 4

2 2 6 3lim .2 4 2

x x x

x

x x x x xx x x x

xx

+ + +

+

→ → →

→

- + - - - = = - - - - - - - = = = -

The limit cycles back and forth.

2

22 2 2 2 2

tan sec sec sec tan tanlim lim lim lim limsec sec tan tan sec secx x x x x

x x x x x xx x x x x xπ π π π π- - - - -→ → → → →

= = = = .

The limit can be calculated as follows: 2 2 2

sintan coslim lim lim sin 11sec cosx x x

xx x xx x

π π π- - -→ → →= = = .

Lesson 15

The integral converges to 1 :24

34 4

32 22

1 1 1lim lim lim3 3 24 24

bb

b b b

xx dx x dxb

-∞ - -

→∞ →∞ →∞

- = = = + = - ∫ ∫ .

The integral diverges: 4

51 1 45 5 5

1 1

1

5lim lim lim 14 45

b

b

b b b

xx dx x dx b∞ - -

→∞ →∞ →∞

= = = - = ∞

∫ ∫ .

The integral converges to 1 :4

4 4 4 4

0 00

1 1 1 1lim lim lim4 4 4 4

bbx x x b

b b be dx e dx e e

∞ - - - -

→∞ →∞ →∞

- = = - = + = ∫ ∫ .

The integral converges to :2π

[ ] [ ]

0 0 0

2 2

1 1lim lim arctan lim 0 arctan2 21 1 bbb b b

dx dx x bx x

π π−∞ →−∞ →−∞ →−∞

= = = − = − − = + + ∫ ∫ .

7.

8.

9.

10.

1.

2.

3.

4.

solu

tions

173

The integral converges to 4: 44 4 1

20 0 0 0

1 lim lim 2 lim 4 2 4b bb b b

dx x dx x bx + + +

-

→ → → = = = - = ∫ ∫ .

The integral diverges:

[ ] [ ]5 5 5

0 0 0 0

10 10lim lim 10ln lim 10ln 5 10lnbbb b b

dx dx x bx x+ + +→ → →

= = = − = ∞∫ ∫ .

The integral is undefined at 1 and converges to 0:

The integral is undefined at 0. 2 0 2

3 3 31 1 0

1 1 1dx dx dxx x x- -

= +∫ ∫ ∫ . The right-hand integral diverges (in

fact, they both diverge), so the original integral diverges.

The integral is doubly improper, so we split this integral at a convenient point—for example, at 1.x =

We have ( ) 2

1 1( ) , .f x f xx x

-′= = The surface area is:

[ ]241 1

1 12 ( ) 1 ( ) 2 1S f x f x dx dxx x

π π∞ ∞

′= + = +∫ ∫ .

On the interval [ )1,∞ , 4

1 1 11x x x

+ > , and the integral 1

12 dxx

π∞

∫ diverges. Hence, the integral

41

1 12 1 dxx x

π∞

+∫ also diverges.

( ) ( )

( ) ( )

2 1 2

3 3 30 0 1

21 13 3

01 12

2 23 3

1 10

1 1 11 1 1

lim 1 lim 1

3 3 3 3lim 1 lim 1 0.2 2 2 2

b

bb bb

b b b

dx dx dxx x x

x dx x dx

x x

- +

- +

- -

→ →

→ →

= +- - -

= - + -

= - + - = - + =

∫ ∫ ∫

∫ ∫

( ) ( ) ( )1

0 0 1

1

10

1 1 11 1 1

lim 2arctan lim 2arctan

2 0 2 2 .4 2 4

c

cbb

dx dx dxx x x x x x

x x

π π π π

+

∞ ∞

→∞→

= ++ + +

= +

= - + - =

∫ ∫ ∫

5.

6.

7.

8.

9.

10.

( ) ( )

( ) ( )

2 1 2

3 3 30 0 1

21 13 3

01 12

2 23 3

1 10

1 1 11 1 1

lim 1 lim 1

3 3 3 3lim 1 lim 1 0.2 2 2 2

b

bb bb

b b b

dx dx dxx x x

x dx x dx

x x

- +

- +

- -

→ →

→ →

= +- - -

= - + -

= - + - = - + =

∫ ∫ ∫

∫ ∫( ) ( )

( ) ( )

2 1 2

3 3 30 0 1

21 13 3

01 12

2 23 3

1 10

1 1 11 1 1

lim 1 lim 1

3 3 3 3lim 1 lim 1 0.2 2 2 2

b

bb bb

b b b

dx dx dxx x x

x dx x dx

x x

- +

- +

- -

→ →

→ →

= +- - -

= - + -

= - + - = - + =

∫ ∫ ∫

∫ ∫

( ) ( ) ( )1

0 0 1

1

10

1 1 11 1 1


2 0 2 2 .4 2 4

c

cbb

dx dx dxx x x x x x

x x

π π π π

+

∞ ∞

→∞→

= ++ + +

= +

= - + - =

∫ ∫ ∫( ) ( ) ( )1

0 0 1

1

10

1 1 11 1 1


2 0 2 2 .4 2 4

c

cbb

dx dx dxx x x x x x

x x

π π π π

+

∞ ∞

→∞→

= ++ + +

= +

= - + - =

∫ ∫ ∫

174

Lesson 16

2 3 4 5

1 2 3 4 53 3 9 3 27 9 3 81 27 3 243 813, , , ,1! 2! 2 3! 6 2 4! 24 8 5! 120 40

a a a a a= = = = = = = = = = = = = .

1 2 3 4 52 2 2 2 1 22, 1, , ,1 2 3 4 2 5

a a a a a= = = − = − = = − = − = .

1 2 3 4 52 3 4 5sin 1, sin 0, sin 1, sin 0, sin 1

2 2 2 2 2a a a a aπ π π π π= = = = = = − = = = = .

3 3 3

3 33 3

18 8 8lim lim lim 81 22 2 1n n n

n n nn n

n n→∞ →∞ →∞

= ⋅ = = −− − −

.

22 2

2 2 2lim lim lim 21 11 1 1n n n

n nn n n n

→∞ →∞ →∞= = =

+ + +.

The sequence is 0, 2,0, 2,0... and, hence, diverges.

The sequence is 1 10,1,0, ,0, ,0,...2 3

and converges to 0.

We know that 1lim 1 ,n

ne

n→∞

+ =

so ( )33

33 3lim 1 lim 1nn

n ne

n n→∞ →∞

+ = + =

.

The sequence is monotonic because 11 14 4

1n na an n += - < - =

+. The sequence is bounded because 4na < .

The terms are 2 4 8, , ,...3 9 27

- - , so the sequence is not monotonic. It is bounded because 23na ≤ .

The first five terms are

1 2 43 51 1, 2 1.414, 3 1.442, 4 1.414, 5 1.380= ≈ ≈ ≈ ≈ .

The limit is 1: Let 1

1lnlim ln lim lim 01

xx x x

x xy x yx→∞ →∞ →∞

= ⇒ = = = . Hence, we have

1lim ln lim ln 0 lim lim 1n nn x n n

y y n n→∞ →∞ →∞ →∞

= = ⇒ = = .

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

solu

tions

175

Lesson 17

The partial sums are as follows.

1

2

3

4

5

3.33 4.5.23 33 5.25.2 43 3 33 5.625.2 4 83 3 3 33 5.8125.2 4 8 16

S

S

S

S

S

=

= + =

= + + =

= + + + =

= + + + + =

The partial sums are as follows.

( )0

1

2

3

4

5

11.

0!11 0.

1!11 1 0.5.2!1 11 1 0.3333.2! 3!1 1 11 1 0.375.2! 3! 4!

S

S

S

S

S

-= =

-= + =

= - + =

= - + - ≈

= - + - + =

The sequence { } 1n

nan+ =

converges to 1: 1lim 1.

n

nn→∞

+= The series

1 1

1 2 3 4 ...1 2 3n

n n

nan

∞ ∞

= =

+= = + + +∑ ∑ diverges because the partial sums diverge.

This geometric series diverges because the ratio is 7 1.5≥

This telescoping series converges because the nth partial sum is

The sequence of partial sums converges to 2 2 31 2

+ = .

4 2 2( 2) 2

2 2 2 2 2 2 2 21 3 2 4 3 5 2

2 2 2 2 .1 2 1 2

nSn n n n

n n

n n

= = -+ +

= - + - + - + + - +

= + - -+ +

1.

2.

3.

4.

5.

4 2 2( 2) 2

2 2 2 2 2 2 2 21 3 2 4 3 5 2

2 2 2 2 .1 2 1 2

nSn n n n

n n

n n

= = -+ +

= - + - + - + + - +

= + - -+ +

4 2 2( 2) 2

2 2 2 2 2 2 2 21 3 2 4 3 5 2

2 2 2 2 .1 2 1 2

nSn n n n

n n

n n

= = -+ +

= - + - + - + + - +

= + - -+ +

176

This telescoping series converges because the nth partial sum is

The sequence of partial sums converges to 1 1 312 2 4 + =

.

The geometric series with 13r -= converges to

( )1 3

1 41 3=

--.

The geometric series begins at 2n = and can be rewritten as follows.

2

2 0

3 3 3 9 1 934 4 4 16 41 4

n n

n n

∞ ∞

= =

= = ⋅ = −

∑ ∑ .

0.075 10 0.75,1000 75.75.x x x= ⇒ = = Subtracting, 990 75x = , which gives 75 5990 66

x = = .

The series is geometric with 3 .r x= Hence, it converges for 3 1x < , which is equivalent to 1 13 3

x- < < .

Lesson 18

The geometric series diverges because the ratio satisfies 5 13-

≥ .

The series diverges because the terms do not tend to 0: 1lim 02 3 2n

nn→∞

= ≠+

.

The terms do not tend to 0: !lim 02nn

n→∞

= ∞ ≠ .

This is the difference of two geometric series. The sum is

( ) ( ) ( ) ( )

( ) ( )

1 00.7 0.9 0.7 0.9

1 1 10 20107 9 3 31 110 10

n n n n

n n

∞ ∞

= =

− = −

= − = − = −− −

∑ ∑

.

2

1 11 1 1 12 21 1 1 2 1 1

1 1 1 1 1 1 1 112 3 2 4 3 5 1 11 1 1 11 .2 2 1

nSn n n n n

n n

n n

= = - = - - - + - + = - + - + - + + - - + = + - - +

6.

7.

8.

9.

10.

1.

2.

3.

4.

solu

tions

2

1 11 1 1 12 21 1 1 2 1 1

1 1 1 1 1 1 1 112 3 2 4 3 5 1 11 1 1 11 .2 2 1

nSn n n n n

n n

n n

= = - = - - - + - + = - + - + - + + - - + = + - - +

2

1 11 1 1 12 21 1 1 2 1 1

1 1 1 1 1 1 1 112 3 2 4 3 5 1 11 1 1 11 .2 2 1

nSn n n n n

n n

n n

= = - = - - - + - + = - + - + - + + - - + = + - - +

( ) ( ) ( ) ( )

( ) ( )

1 00.7 0.9 0.7 0.9

1 1 10 20107 9 3 31 110 10

n n n n

n n

∞ ∞

= =

− = −

= − = − = −− −

∑ ∑

.

177

0 0 0

1 1 1 1 1 1 3 121 12 3 2 3 2 21 12 3n n n n

n n n

∞ ∞ ∞

= = =

- = - = - = - = - -

∑ ∑ ∑ .

The series is geometric: ( )0

1 1 10110 91 10

n

n

∞

=

= = -

∑ .

Because sin1 1< , the geometric series ( ) ( )1 0

sin1 sin(1) sin1n n

n n

∞ ∞

= =

=∑ ∑ converges to sin1 5.3080

1 sin1≈

-.

The series diverges because the kth term ( )ln k- does not tend to 0.

1 1

1ln ln 0 ln 2 ln 3 lnk k

kn n

S n kn= =

= = − = − − − −∑ ∑ � .

The series is geometric and converges for 1 1x - < 0 2x⇒ < < .

The ball first drops 16 feet. Then, it goes up and down, as follows:

( ) ( ) ( )

( ) ( )

( ) ( )

1

22 2 2

3

2

1 0

16.0.81(16) 0.81(16) 32(0.81).

16 0.81 16 0.81 32 0.81 .

16 32 0.81 32 0.81

16 32 0.81 16 32 0.81

3216 152.42 feet.1 0.81

n n

n n

DD

D

D∞ ∞

= =

== + =

= + =

= + + +

= + = - +

= - + ≈-

∑ ∑

Lesson 19

Let 1( ) .3

f xx

=+

This function is positive, continuous, and decreasing for 1.x ≥

[ ]11

1 ln( 3)3

dx xx

∞ ∞= + = ∞+∫ . Hence, the series diverges.

Let ( ) 3 .xf x -= This function is positive, continuous, and decreasing for 1.x ≥

( )11

1 13ln 3 3 3ln 3

xxdx

∞∞ - -

= =

∫ . Hence, the series converges.

5.

6.

7.

8.

9.

10.

1.

2.

178

Let ( )

1( ) .1

f xx x

=+

This function is positive, continuous, and decreasing for 1.x ≥

( )1 1

1 2 ln( 1)1

dx xx x

∞∞ = + = ∞ +∫ . The series diverges.

Let 1( ) .ln

f xx x

= This function is positive, continuous, and decreasing for 2.x ≥

[ ]22

1 ln(ln )ln

dx xx x

∞ ∞= = ∞∫ . Hence, the series diverges.

This is a convergent p-series, 5 1.3

p = >

This is a divergent p-series, 2 1.3

p = <

This series diverges because 2 1lim 05 1 5n

nn→∞

+= ≠

+.

Let ( )2

1( ) .ln

f xx x

= This function is positive, continuous, and decreasing for 2.x ≥

( ) ( )222

1 1 1ln 2ln ln

dxx x x

∞∞ -

= =

∫ . The series converges.

Let 3

ln( ) .xf xx

= This function is positive, continuous, and decreasing for 2.x ≥ Use integration by parts

3( ln , )u x dv x dx-= = to obtain the following:

( )3 22

2

2 ln 1ln 2ln 2 14 l6

xx dxx x

∞∞ + +

= - =

∫ . Hence, the series converges.

The function 2sin( ) xf x

x =

is not decreasing for 1.x ≥

Lesson 20

2 2

1 13 2 3n n

<+

. Hence, the series 21

13 2n n

∞

= +∑ converges by comparison to the convergent p-series 21

1 1 .3 n n

∞

=∑

1 15 3 5n n

>-

. Hence, the series 1

15 3n n

∞

= -∑ diverges by comparison to the divergent p-series

(harmonic) 1

1 1.5 n n

∞

=∑

3.

4.

5.

6.

7.

8.

9.

10.

1.

2.

solu

tions

179

For ln 13,1 1nn

n n≥ >

+ +. Hence, the series diverges by comparison to the divergent series

1

1 .1n n

∞

= +∑

(Note: this series is divergent by the integral test.)

The series diverges by a limit comparison with the divergent harmonic series:

( )2 2

2

1lim lim 11 1n n

nn n

nn→∞ →∞

+= =

+.

The series converges by a limit comparison with the convergent p-series 3

1

1n n

∞

=∑ :

( )2 3

23

13

lim lim 11 ( 3)n n

n n nn n

n→∞ →∞

+= =

+.

The series converges by a limit comparison with the convergent geometric series 1

14

n

n

∞

=

∑ :

( )5

4 1 5(4 )lim lim 51 4 14

n n

nn nn

→∞ →∞

+= =

+.

This series diverges because 2 2lim 0.3 1 3n

nn→∞

= ≠-

The series diverges by a limit comparison with the divergent harmonic series:

( ) ( ) ( )( ) ( )

22

2

2

1 11 sectan1lim lim lim sec 11 1n n n

nn nn

n n→∞ →∞ →∞

−= = =

−.

The series converges by a limit comparison with the convergent geometric series 1

25

n

n

∞

=

∑ :

( )( )

( )

2 15 1 2 1 5lim lim 1

5 1 225

n

n n n

n n nn n→∞ →∞

++ +

= ⋅ =+

.

The series converges by a limit comparison with the convergent p-series 3

1

1n n

∞

=∑ :

( )( )

2

5 5 3

53

2 13 2 1 2 2lim lim1 3 2 1 3n n

nn n n n

n nn

→∞ →∞

−+ + −

= =+ +

.

3.

4.

5.

6.

7.

8.

9.

10.

180

Lesson 21

Because 11 1

3 2n na an n+ = < =

+ + and 1lim lim 0

2nn na

n→∞ →∞= =

+, the series converges by the alternating series test.

Because 1 2 1

1 15 5n nn na a+ + += < = and

1

1lim lim 05n nn n

a +→∞ →∞= = , the series converges by the alternating series test.

Because 11 1

( 1)! !n na an n+ = < =

+ and 1lim lim 0

!nn na

n→∞ →∞= = , the series converges by the alternating series test.

For 2

1 14,!

nn n

≥ < and 2

1

1n n

∞

=∑ is a convergent p-series. Hence, the series

1

1!n n

∞

=∑ converges by the direct

comparison test. Therefore, the series ( )1

1!

n

n n

∞

=

-∑

converges absolutely.

The series converges by the alternating series test. The series 2

1lnn n n

∞

=∑ diverges by the integral test.

Hence, the series ( ) 1

2

1ln

n

n n n

+∞

=

-∑

converges conditionally.

The series 3

3 5n

nn

∞

= -∑ converges by the limit comparison test with the p-series 23

1n n

∞

=∑ . Therefore, the series

( )3

3

15

n

n

nn

∞

=

--∑ converges absolutely.

The series diverges because lim arctan 02n

n π→∞

= ≠ .

The first few terms are 21

sin sin1 sin 2 sin 3 ...1 4 9n

nn

∞

=

= + + +∑ . Some of these terms are positive and some are

negative, but it is not an alternating series. However, the series converges absolutely because 2 2

sin 1nn n

≤

and 2

1

1n n

∞

=∑ is a convergent p-series.

The fourth partial sum is

( )( )

14

41

1 1 1 11 11 ! 2 6 3

n

nS

n

+

=

−= = − + − =

−∑ .

The remainder is 41 14! 24

R = = , and 4 41 0.0416724

S S R- = = ≈ . Note that 41 0.03455Se- ≈ .

True. If the series na∑ converged, then so would na∑ .

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

solu

tions

181

Lesson 22

11

1

1 2 12lim lim lim 11 2 22

nnn

nn n nn n

aa

++

+→∞ →∞ →∞= = = < . Therefore, the series converges by the ratio test.

( )( )

1

1

31 ! 3( !) 3lim lim lim lim 0 1

3 1 ! 1!

n

nnn n n n

n

na na n n

n

+

+

→∞ →∞ →∞ →∞

+= = = = <

+ +. Therefore, the series converges by the ratio test.

( )( )( )

1

161 1 6 65lim lim lim 1

5 565

n

nnn n n

n

na na nn

+

+

→∞ →∞ →∞

+ + = = = > . Therefore, the series diverges by the ratio test.

The series converges by the ratio test.

( )( ) ( )

( )

( )

11

1

1 !1 1

lim lim lim! 1

1 1lim lim lim 1.11 11

n nn

nn n nn n

nn

n nn n n

nn n na

na nn

n nn en

n

+

++→∞ →∞ →∞

→∞ →∞ →∞

++ +

= =+

= = = = < + + +

11 1lim lim 17 7

nn

n nn na

→∞ →∞

= = < . Therefore, the series converges by the root test.

2 1 2 1lim lim lim 2 11 1

n

n nnn n n

n nan n→∞ →∞ →∞

+ + = = = > - - . Therefore, the series diverges by the root test.

1

ln lnlim lim lim 0 1n n

nnn n n

n nan n→∞ →∞ →∞

= = = <

. Therefore, the series converges by the root test.

( )( ) ( )

( )

1

1

1 71 ! 1 7( !) 7lim lim lim lim 0 1

7 1 !!

n

nnn n n n

n

nn n na

na n n nn

+

+

→∞ →∞ →∞ →∞

++ +

= = = = <+

. Therefore, the series converges by the

ratio test.

( )( )

1

12 3lim lim lim

3 32 3

n

nnn n n

n

xa x xa x

+

+

→∞ →∞ →∞= = = . For the series to converge, we need 1 3 3 3.

3x x x< ⇒ < ⇒ - < <

The series also diverges at 3.x = ± Therefore, the series converges on the open interval ( )3,3 .-

Note that this could have been answered using our knowledge of geometric series.

1 1lim lim lim 1n nn ppn n n n

an n→∞ →∞ →∞

= = = . Hence, the root test is inconclusive.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

182

Lesson 23

We calculate the derivatives, as follows.

( )( ) ( )

( )

( ) , 0 1.

, 0 1.

( ) , 0 1.

x

x

x

f x e f

f x e f

f x e f

= =

′ ′= =

′′ ′′= =

The second-degree polynomial is ( ) ( ) ( ) ( ) 22

2

00 0 1

2! 2f xP x f f x x x′′

′= + + = + + .


( )( ) ( )2

( ) tan , 1.4

sec , 2.4

f x x f

f x x f

π

π

= =

′ ′= =

The first-degree polynomial is ( ) ( ) ( )( )1 1 24 4 4 4P x f f x x ππ π π ′= + - = + -

.


( )( ) ( )

( )3 2

( ) sec , 0 1.

sec tan , 0 0.

( ) sec sec tan , 0 1.

f x x f

f x x x f

f x x x x f

= =

′ ′= =

′′ ′′= + =

The second-degree polynomial is ( ) ( ) ( ) ( ) 22

2

00 0 1

2! 2f xP x f f x x′′

′= + + = + .

Because this is already a third-degree polynomial, it equals its Maclaurin polynomial.


( )( ) ( )

( )( )

( ) sin , 0 0.

cos , 0 1.

( ) sin , 0 0.

( ) cos , 0 1.

f x x f

f x x f

f x x f

f x x f

= =

′ ′= =

′′ ′′= - =

′′′ ′′′= - = -

The third-degree polynomial is ( ) ( ) ( ) ( ) ( ) 3 32 3

4

0 00 0

2! 3! 3! 6f f x xP x f f x x x x x′′ ′′′

′= + + + = - = - .

1.

2.

3.

4.

5.

solu

tions

183


( )

( ) ( )( )

( )

1

2

3

4

2( ) 2 , 1 2.

2 , 1 2.

( ) 4 , 1 4.

( ) 12 , 1 12.

f x x fx

f x x f

f x x f

f x x f

-

-

-

-

= = =

′ ′= - = -

′′ ′′= =

′′′ ′′′= - = -

The third-degree Taylor polynomial is

( ) ( ) ( ) ( ) ( )2 3 2 33

4 122 2( 1) 1 1 2 2( 1) 2 1 2 12! 3!

P x x x x x x x= - - + - - - = - - + - - - .


( )

( ) ( )

( )

( )

( ) ( ) ( )

1

2

3

4 44

( ) ln , 2 ln 2.1 1, 2 .

21( ) , 2 .4

1( ) 2 , 2 .4

3( ) 6 , 2 .8

f x x f

f x x fx

f x x f

f x x f

f x x f

-

-

-

-

= =

′ ′= = =

′′ ′′= - = -

′′′ ′′′= =

= - = -

The fourth-degree Taylor polynomial is ( ) ( ) ( ) ( )

( ) ( ) ( )

2 3 44

2 3 4

31 11 84 4ln 2 ( 2) 2 2 22 2! 3! 4!

1 1 1 1ln 2 ( 2) 2 2 2 .2 8 24 64

P x x x x x

x x x x

= + - - - + - - -

= + - - - + - - -

The fourth-degree polynomial evaluated at 0.3x = gives ( ) ( ) ( )2 40.3 0.3cos 0.3 1 0.9553375

2! 4!≈ - + ≈ .

The fifth derivative ( )5 ( ) sinf x x= - is bounded by 1. Hence, ( ) ( )5 54

1 0.3 2.025 105!

R x -≤ = × . This is

much larger than the exact error.

The fifth-degree polynomial evaluated at 1x = gives 1 1 1 1 11 1 2.716672! 3! 4! 5!

e e= ≈ + + + + + ≈ .

The sixth derivative ( )6 ( ) xf x e= is bounded by 3 on the interval [ ]0,1 . Hence,

( ) ( )65

3 1 0.0041676!

R x ≤ ≈ .

3 5 2 4 2 43 51 13! 5! 3! 5! 2! 4!x x x x x xx

′ - + = - + = - +

.

6.

7.

8.

9.

10.

( ) ( ) ( ) ( )

( ) ( ) ( )

2 3 44

2 3 4

31 11 84 4ln 2 ( 2) 2 2 22 2! 3! 4!

1 1 1 1ln 2 ( 2) 2 2 2 .2 8 24 64

P x x x x x

x x x x

= + - - - + - - -

= + - - - + - - -

184

Lesson 24

The series is geometric and converges for 14 1 .4

x x< ⇒ < The radius of convergence is 14

R = .

Using the ratio test, we see that the radius of convergence is R = ∞ .

( )

( )( ) ( )

2 22

12

2 2 !lim lim lim 0

2 2 2 12 !

n

nnn n n

n

xna xL

xa n nn

+

+

→∞ →∞ →∞

+= = = =

+ +.

Using the ratio test, we see that the radius of convergence is 0R = .

( )( )

( )( )( )

2 2

21

2

2 2 !1 ! 2 2 2 1

lim lim lim2 ! 1

!

n

nnn n n

n

n xn n n xaL

n xa nn

+

+

→∞ →∞ →∞

++ + +

= = = = ∞+

.

Because the series is geometric, it converges for 1 7 77x x< ⇒ − < < .

( )( )

( )

1 1

1

11

lim lim lim11

n n

nn nn n n

n

xna nxL x

a nxn

+ +

+

→∞ →∞ →∞

-+

= = = =+-

. The series converges for 1.x < At 1x = , the

alternating series ( ) ( )1 1

1 1 1n nn

n nn n

∞ ∞

= =

- -=∑ ∑ converges. At 1x = - , the series ( )

1 1

1 ( 1) 1n n

n nn n

∞ ∞

= =

- -=∑ ∑ diverges.

The interval of convergence is ( ]1,1- .

( ) ( )( )

( ) ( ) ( )

2 1

11

1

1 41 9 ( 4) 1lim lim lim 4

1 9 91 49

n n

nn

n nn n nn

n

xna n xL x

a nxn

+ +

++

+→∞ →∞ →∞

- -+ -

= = = = -+- -

.

The series converges for 4 9,x - < or 5 13.x- < < At 13x = , the alternating series

( ) ( )1 1

1 1

1 9 19

n nn

nn nn n

+ +∞ ∞

= =

- -=∑ ∑ converges. At 5x = - , the series ( ) 1

1 1

1 ( 9) 19

n n

nn nn n

+∞ ∞

= =

- - -=∑ ∑ diverges. The

interval of convergence is ( ]5,13- .

1.

2.

3.

4.

5.

6.

solu

tions

185

( ) ( )( ) ( )

( )( )1

12 2 ! 2 2 2 13lim lim lim

32 ! 3

n

nnn n n

n

xn n n xaLa xn

+

+

→∞ →∞ →∞

+ + += = = = ∞

. The radius of convergence is 0R = ,

and the series converges only for 0.x =

There are many possible answers. The series ( )1

2 1 n

nx

∞

=

+∑ is geometric and converges for

2 1 1 1 2 1 1 1 0x x x+ < ⇔ - < + < ⇔ - < < .

Using the ratio test, you will see that the intervals of convergence are as follows.

( ) ( ) ( ]

( ) ( ) ( )

( ) ( )( )( ) [ ]

1 1

1

1

11 2

1

1 1( ) , 0, 2 .

1

( ) 1 1 , 0,2 .

1 1( ) , 0, 2 .

1 2

n n

n

n n

nn n

n

xf x

n

f x x

xf x dx C

n n

+ +∞

=

∞+

=

+ +∞

=

- -=

+

′ = - -

- -= +

+ +

∑

∑

∑∫

We have ( )( )2

11

1 2lim lim lim

2( 1)

n

nnn n n

n

xn na nx x

xa nn n

+

++→∞ →∞ →∞

+ += = =

++

. So, the radius of convergence is 1. For 1,x =

the series 1

1 1

1 1( 1) ( 1)

n

n nn n n n

+∞ ∞

= =

=+ +∑ ∑ converges by limit comparison with 2

1

1n n

∞

=∑ . For

1

1

( 1)1,( 1)

n

nx

n n

+∞

=

-= -

+∑

converges by the alternating series test. Hence, the interval of convergence is [ ]1,1- .

Lesson 25

( )( )

10 0

1 11 122 2 2 21 2

nn n

nn n

xxxx

∞ ∞

+= =

−− = = = −+ −∑ ∑ .

The series converges on the open interval ( )2,2- .

( ) 10 0

22 2 25 ;( 5,5)5 5 5 51 5

n n

nn n

x xxx

∞ ∞

+= =

= = = − − −∑ ∑ .

( )2 22

0 0

1 ;( 1,1)1

nn

n nx x

x

∞ ∞

= =

= = −− ∑ ∑ .

7.

8.

9.

10.

1.

2.

3.

186

( )1

0 0

1 11 1 1 1213 2 ( 1) 2 2 21

2

nn

nn n

xxxx x

∞ ∞

+= =

−− = = = = −− − − −

∑ ∑ .

The series converges on the interval ( )1,3- .

( ) ( ) 1

0 0

11ln( 1) 1 , 1 11 1

n nn n

n n

xx dx x dx C x

x n

+∞ ∞

= =

- + = = - = + - < ≤ + + ∑ ∑∫ ∫ .

To solve for the constant ,C let 0x = and conclude that 0.C = Therefore,

( ) ( ]1

0

1ln( 1) ; 1,1

1

n n

n

xx

n

+∞

=

-+ = -

+∑ .

We combine the two geometric series as follows.

( ) ( ) 22

0 0 0

2 1 1 1 1 1 2 2 ...; 1 1.1 1 1

n nn n n

n n nx x x x x

x x x

∞ ∞ ∞

= = =

- = + = - + = - + = + + - < < - + - ∑ ∑ ∑

( )1

20 1

1 111

n n

n n

d d x nxdx x dxx

∞ ∞-

= =

= = = - - ∑ ∑ .

We next use the ratio test to determine the radius of convergence.

11

( 1) 1lim lim limn

nnn n n

n

a n x n x xa nnx

+-→∞ →∞ →∞

+ += = = . So, the radius is 1. Checking the endpoints, the series

diverges at both 1x = and 1x = - . The interval of convergence is the open interval ( )1,1- .

1 2 32 21

2 1

( 1) / (2 3) 2 1lim lim lim2 3( 1) / (2 1)

n nn

n nn n nn

a x n n x xa nx n

+ ++

+→∞ →∞ →∞

- + += = =

+- +. So, the radius is 1. At 1,x =

( ) ( )2 1

0 0

11 12 1 2 1

nn n

n n

xn n

+∞ ∞

= =

- = -+ +∑ ∑ , which converges (alternating series test). At 1,x = -

( ) ( ) ( )2 1

0 0

11 1

2 1 2 1

nn n

n n

xn n

+∞ ∞

= =

-- = -

+ +∑ ∑ , which also converges. Hence, the interval of convergence is the closed

interval [ ]1,1- .

Because the second series is the derivative of the first series, the radius of convergence is the same: 3.

4.

5.

6.

7.

8.

9.

solu

tions

187

Lesson 26

11 / ( 1)!lim lim lim 0

1/ !

nn

nn n nn

a x n xa nx n

++

→∞ →∞ →∞

+= = =

+. Hence, the series converges for all values of .x

1 2 3 21

2 1

( 1) / (2 3)!lim lim lim 0(2 3)(2 2)( 1) / (2 1)!

n nn

n nn n nn

a x n xa n nx n

+ ++

+→∞ →∞ →∞

- += = =

+ +- +. Hence, the series converges for all

values of .x

We calculate the derivatives at 0.

( ) cos , (0) 1.( ) sin , (0) 0.( ) cos , (0) 1.( ) sin , (0) 0.

.....

f x x ff x x ff x x ff x x f

= =′ ′= - =′′ ′′= - = -′′′ ′′′= =

The Maclaurin series is ( )

( )2 4 2

0 0

(0) ( 1)1 ...! 2! 4! 2 !

n n nn

n n

f x x xxn n

∞ ∞

= =

-= - + - =∑ ∑ .

We calculate the derivatives at 1.

1

22

33

44

1( ) , (1) 1.

1( ) , (1) 1.

2( ) 2 , (1) 2.

6( ) 6 , (1) 6.

.....

f x x fxf x x fxf x x fxf x x fx

−

−

−

−

= = =

′ ′= − = − = −

′′ ′′= = =

′′′ ′′′= − = − = −

The Taylor series is ( )

( ) ( ) ( )2

0 0

( ) 1 ( 1) ( 1) ... 1 1!

nn n n

n n

f c x c x x xn

∞ ∞

= =

- = - - + - - = - -∑ ∑ .

All of the derivatives are equal to e at 1x = .

( ) ( )( ) , (1) , ( ) , (1) , , (1)n nx x xf x e f e f x e f e f e f e′ ′= = = = = = .

The Taylor series is ( )

( ) ( ) ( ) ( )2 3

0 0

1 1 1(1) 1 1 ( 1) ... ...! 2! 3! !

nnn

n n

x x xf x e x en n

∞ ∞

= =

- - -- = + - + + + =

∑ ∑ .

Substitute 3x- for x in the series for the exponential function.

( ) ( ) 2 3

0 0

3 1 3 9 271 3! ! 2 3!

n n n n

n n

x x x xxn n

∞ ∞

= =

- -= = - + - +∑ ∑ .

1.

2.

3.

4.

5.

6.

188

Substitute 3x for x in the series for the sine function.

( ) ( )( )

( ) ( )2 1 3 5

0

1 3 3 33

2 1 ! 3! 5!

n n

n

x x xx

n

+∞

=

−= − + −

+∑ �.

2 1 cos 2sin .2 2

xx = - The series for cos 2x is obtained from the cosine series:

2 42 4 2 42 2 21 1 2

2! 4! 3x x x x- + - = - + -

. Hence, the series for 2sin x is

2 4 2 41 1 1 1 2 1cos 2 1 22 2 2 2 3 3

x x x x x - = - - + - = - +

.

Instead of using the definition of the Maclaurin series, we can multiply the terms of the two series, as follows.

2 3 4 3 5 3 521

2 6 24 6 120 3 30x x x x x x xx x x x

+ + + + + - + - = + + - -

.

12 2 31 1

0.5 0.50.5

55 253 187cos 1 0.324652 24 4 72 72 576 576x x x xx dx dx x

≈ - + = - + = - = ≈

∫ ∫ .

Note: Using a calculator, 1

0.5cos 0.32433x dx ≈∫ .

Lesson 27

On the given interval, the equation is equivalent to 2 ,0 4.y x x= ≤ ≤ Hence, 1yx

′ = , and the arc

length is 2

4 4

0 0

1 11 5.916xs dx dxxx

+= + = ≈

∫ ∫ .

Let 2 2 22 tan ,2 sec ,1 4 secx dx d xθ θ θ θ= = + = . Then, we have

22 3sec 1 11 4 sec sec sec tan ln sec tan .

2 2 4x dx d d Cθθ θ θ θ θ θ θ θ

+ = = = + + +

∫ ∫ ∫ Hence, the

definite integral is ( )44 2 2 2

0 0

1 11 4 1 4 (2 ) ln 1 4 2 8 65 ln 65 84 4

x dx x x x x + = + + + + = + + ∫ .

7.

8.

9.

10.

1.

2.

solu

tions

189

Rewrite as 2 2

1. 4, 1, 16 1 151 16x y a b c+ = = = = - = . The center is (0,0) , the vertices are ( )0, 4± , and

the foci are ( )0, 15± .

We first complete the square to obtain ( ) ( )2 22 3

14 9

x y+ -+ = . Hence, 3, 2, 9 4 5a b c= = = - = . The

center is ( )2,3- , the vertices are ( ) ( )2,6 , 2,0- - , and the foci are ( )2,3 5- ± .

Eccentricity =154

cea

= = .

Area = (4)(1) 4abπ π π= = .

We have 6, 5, 36 25 11a c b= = = - = . The major axis is horizontal, and the equation is 2 2

136 11x y

+ = .

Differentiating implicitly, 2 2 0 .9 9x xyy y

y′ ′- = ⇒ = At 6, 3x y= = ± and

6 2 399 3

y ± ±′ = = . At the point ( ) ( )2 36, 3 , 3 6 2 3 3 39

y x x y- = - ⇒ - = . At the point

( ) ( )2 36, 3 , 3 6 2 3 3 39

y x x y-- + = - ⇒ + = .

Eccentricity = 0.0167 2,498,286.6149,598,000

c c ca

= = ⇒ ≈ . The least distance (perihelion) is

147,099,713.4 kma c- ≈ . The greatest distance (aphelion) is 152,096,286.6 kma c+ ≈ .

We find the arc length in the first quadrant and multiply by 4. Solving for

( ) ( )2 2

2

22

4, , ,2 4 42 4

x x xy y y yxx

- -′ ′= = =--

. The arc length of the ellipse is given by

( )2 2

04 1 4(2.422112) 9.68845y dx′+ ≈ ≈∫ . Note that this is an improper integral and that the integrand

does not have an elementary antiderivative.

3.

4.

5.

6.

7.

8.

9.

10.

190

Lesson 28

Eliminating the parameter, 3 3 3 11, 3 12 2 2 2

x xt y x+ + = = + = +

. This is a line in the plane.

We use a trigonometric identity to verify that the curve is a hyperbola.

2 2 2 2

2 2tan 1 sec 1 13 2 4 9y x x yt t + = ⇒ + = ⇒ − =

.

This is a circle of radius 3 and center ( )2, 4 .-

( ) ( )2 2

2 22 22 4 cos sin 1 2 4 93 3

x y t t x y− + + = + = ⇒ − + + =

.

( )3 6 .

12

dydy dt t

dxdxdt t

= = = At 1, 6.dytdx

= =

2

2

3sin cos tan3cos sin

dydy d

dxdx d

θ θθ θθ θθ

= = = --

. At , tan 1.4 4

dydx

π πθ = = - = -

Horizontal tangents: 32 3 02

dy t tdt

= + = ⇒ = - . The corresponding point is ( ) 1 9, , .2 4

x y = - -

There are

no vertical tangents because 1 0.dxdt

= ≠

Horizontal tangents: 3cos 0 , .2 2

dydt

π πθ θ= = ⇒ = The corresponding points are ( ) ( )5, 1 , 5, 3 .- - Vertical

tangents: 3sin 0 0,dxdt

θ θ π= - = ⇒ = . The corresponding points are ( ) ( )8, 2 , 2, 2- - . Note that this curve

is an ellipse.

3, 2dx dydt dt

= = - . The arc length is given by the integral

2 233

1 19 4 13 4 13

b

a

dx dys dt dt tdt dt - -

= + = + = = ∫ ∫ . Note that this is a line.

212 , 6dx dyt tdt dt

= = . The arc length is given by the integral

( )

( )

2 2 434 42 4 2 2 2

1 11

3 12 2

144 36 6 4 2 4

2 20 5 70 5.

b

a

dx dys dt t t dt t t dt tdt dt

= + = + = + = +

= - =

∫ ∫ ∫

1.

2.

3.

4.

5.

6.

7.

8.

9.

solu

tions

191

The curve crosses itself at the origin. At this point, 0t = or .t π= 3cos4cos 2

dy tdx t

= . At 30, ,4

dytdx

= = and

the tangent line is 3 .4

y x= At 3, ,4

dytdx

π= = - and the tangent line is 3 .4

y x= -

Lesson 29

8cos 0, 8sin 8.2 2

x yπ π= = = = ( ) ( ), 0,8x y = .

5 52cos 1, 2sin 3.3 3

x yπ π= - = - = - =

( ) ( ), 1, 3 .x y = -

2 2 2 52 2 2 2; tan 1 ,2 4 4

r π πθ θ= ± + = ± = = ⇒ = . There are many possible answers:

( ) 5, 2 2, , 2 2,4 4

r π πθ = -

.

36; tan undefinied ,2 2

r π πθ θ= ± ⇒ = . There are many possible answers:

( ) 3, 6, , 6,2 2

r π πθ = - .

2 9, 3r r= = . This is a circle of radius 3 centered at the origin.

cos 12 12sec .x r rθ θ= = ⇒ = This is a vertical line.

2 2 25 cos 5r r x y xθ= ⇒ + = . Next, complete the square to obtain the equation 2 2

25 52 2

x y - + =

.

This is a circle of radius 52

centered at ( )5 ,02.

We have ( )sin 1 sin siny r θ θ θ= = - , which implies that

( ) ( )1 sin cos cos sin cos 1 2sin 0dyd

θ θ θ θ θ θθ= - - = - = . cos 0θ = or 1sin .2θ = Hence,

3 5, , , .2 2 6 6π π π πθ = These values correspond to the three points ( ) 3 1 1 5, 2, , , , ,

2 2 6 2 6r π π πθ =

.

Note that there is not a horizontal tangent at the origin ( )2πθ = because 0dx

dθ= at that point as well.

10.

1.

2.

3.

4.

5.

6.

7.

8.

192

We have ( ) 2cos 2 1 cos cos 2cos 2cosx r θ θ θ θ θ= = - = - . Hence,

( )2sin 4cos sin 2sin 2cos 1 0dxd

θ θ θ θ θθ= - + = - = . The solutions are 50, , ,3 3

π πθ π= . For 0θ = , dydθ

is also zero, so that point is a cusp. The three vertical tangents occur at ( ) ( ) ( ) ( )5, 1, , 1, , 4,3 3r π πθ π= .

Setting the equations equal to each other, we obtain 1 cos 1 cos cos 0.θ θ θ+ = - ⇒ = This gives two

points of intersection: ( ) 3, 1, , 1, .2 2

r π πθ =

However, the graphs indicate that there is a third point, the

origin ( )0,0 .

Lesson 30

[ ]2

0 00

1 9 1 cos 2 9 sin 2 93cos2 2 2 4 2 4

A d dπ

π π θ θ πθ θ θ θ+ = = = + = ∫ ∫ . Note that the region is a circle of

radius 32

.

The petal is traced out on the interval 6 6π πθ- ≤ ≤ . The area is

[ ]626 6

6 66

1 9 1 cos 6 9 sin 6 9 33cos32 2 2 4 6 4 6 6 4

A d dπ

π π

π ππ

θ θ π π πθ θ θ θ- -

-

+ = = = + = + = ∫ ∫ .

The petal is traced out on the interval 0 3πθ≤ ≤ . The area is

[ ]323 3

0 00

1 1 cos 6 sin 6 44sin 3 8 4 42 2 6 3 3

A d dπ

π π θ θ π πθ θ θ θ- = = = - = = ∫ ∫ .

Half of the inner loop is traced out on the interval 23

π θ π≤ ≤ . The total area is given by

22

3

12 (1 2cos )2

A dπ

πθ θ= +∫ . Note: 2 3 3

2A π -

= .

The area inside the outer loop is 3

2 21 5

6

12 (3 6sin )2

A dπ

πθ θ= -∫ . The area inside the inner loop is

2 22

6

12 (3 6sin )2

A dπ

πθ θ= -∫ . The area between the loops is the difference, 1 2A A- . Note: This difference

equals 9 27 3π + .

8, 0drrdθ

= = . [ ]2

2 2200

64 0 8 16drs r d dd

β π π

αθ θ θ π

θ = + = + = = ∫ ∫ .

Note that this is the circumference of a circle of radius 8.

9.

10.

1.

2.

3.

4.

5.

6.

solu

tions

193

You can check that this is a circle of radius 2. Using the formula, 4sindrd

θθ= - .

[ ]2

2 2 200

16cos 16sin 4 4 .drs r d dd

β π π

αθ θ θ θ θ π

θ = + = + = = ∫ ∫

2drdθ

= . 2

22 2

04 4drs r d d

dβ π

αθ θ θ

θ = + = + ∫ ∫ ( )42.5≈ .

Using a graphing utility, you obtain 13.736s ≈ .

You can see that this represents a circle by multiplying both sides by r , converting to Cartesian coordinates, and completing the square.

( )2

22 2 2 3 132 sin 3 cos 2 3 12 4

r r r x y y x x yθ θ = + ⇒ + = + ⇒ − + − =

.

Lesson 31

5 3,6 2 2,4 , 3 1,8 4 2,4= - - = = - - =u v . The vectors are equivalent.

6 0, 2 3 6, 5 , 9 3,5 10 6, 5= - - - = - = - - = -u v . The vectors are equivalent.

5 2,5 0 3,5 3 5= - - = = +u i j .

5 0, 1 ( 4) 5,3 5 3= - - - - - = - = - +u i j .

2 2 2 2 84 9 97, 4,9 ,6 ,3 3 3

= + = = =u u

2, 5 4,9 2, 14 ,5 3 5 4,9 3 2, 5 14,60− = − − = − − − = − − =v u u v .

( )2 23 4 25 5, 3 3 3,4 9, 12 ,= - + = = - = - - = -u u

3,0 3,4 6, 4 ,6 6 3,4 3,0 21,24− = − − = − − = − − = −v u u v .

2 25,12 5 12 169 13.= + = = Unit vector: 1 5 125,12 , .13 13 13

=

( ) ( )2 cos150 sin150 3 3,1= ° + ° = - + = - v i j i j .

7.

8.

9.

10.

1.

2.

3.

4.

5.

6.

7.

8.

194

We have the following two forces.

( ) ( ) ( ) ( )1

2

400cos10 400sin10 .400cos 10 400sin 10 400cos 10 400sin 10 .

= ° + °

= - ° + - ° = ° - °

F i jF i j i j

The resultant force is 1 2= + =F F F 800cos10 788 pounds.° ≈i i

( ) ( )23 1 3.f x x f′ ′= ⇒ = Let 1,3 , 1 9 10= = + =w w . Then, the two unit vectors are

1 1,310

± = ±ww

.

Lesson 32

( ) ( )3,4 1,5 3 1 4 5 17= − = − + =u vi i .

( ) ( )( )6, 4 3,2 6 3 4 2 26= - - = - + - = -u v .

( ) ( )3 8 4 6 0cos 05(10)9 16 64 36

θ- +

= = = =+ +

u vu v . The vectors are orthogonal, and the angle is 90°.

( ) ( )3 2 1 4 2 1cos9 1 4 16 200 50

θ- + -

= = = = -+ +

u vu v . Using the inverse cosine function,

1arccos 98.150

θ = - ≈ °

.

( ) ( )( )2 2 2

6 1 7 4 34proj 2 2,8171 4

+ = = = = = +

vu vu v v v vv .

( ) ( )( )( )2 2 2

2 3 3 2 0proj 0,0133 2

+ - = = = = = +

vu vu v v v 0v .

( ) ( )( )( )1 2 2 2

2 5 3 1 13 1 5 1=proj ,26 2 2 25 1

+ = = = = = +

vu vw u v v v vv . The vector component normal to v is

2 15 1 1 52,3 , ,2 2 2 2

= - = - = -w u w .

9.

10.

1.

2.

3.

4.

5.

6.

7.

solu

tions

195

The gravitational force is 48,000 . cos10 sin10= - = ° + °F j v i j . We have

( ) ( ) ( ) ( )( )48,000 sin10 8335.1 cos10 sin10= = − ° ≈ − ° + °2

F vw v = F v v v i jvi i .

8335.1 pounds≈w .

( )1600 cos 25 sin 25 , 2000= ° + ° =F i j v i . The work done is

( ) ( )1600 2000 cos 25 2900.2 km-N 2,900,184.9 Joules .W = = ° ≈ ≈F v

If the dot product is negative, then so is the cosine of the angle, so the angle must be obtuse.

Lesson 33

The domain is all values of 0, 2.t t> ≠

( ) 21 1 (1 1) ;= + - =r i j i ( ) ( )22 2 ( 2 1) 4 3- = - + - - = -r i j i j .

, 1 1.x t y t x= = - = - The curve is the line 1,y x= - oriented left to right.

22 2 2cos , sin ,cos sin 1

2 4y yx t t t t x= = + = + = . The curve is an ellipse, oriented counterclockwise.

2 22 2sec , tan ,sec tan 1

3 2 9 4x y x yθ θ θ θ= = - = - = . The curve is a hyperbola.

Let 5cos , 5sinx t y t= = . One representation is ( ) 5cos 5sint t t= +r i j .

2

2

2lim 21

t

t

tet

-

→∞

+ = +

i j j because the limit of the first term is zero.

( ) ( )23 3 , 6t t t t′ ′′= - =r i j r i .

( ) 4sin 4cost t t′ = - +r i j .

21 sec ln tant dt t tt

+ = + + ∫ i j i j C .

8.

9.

10.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

196

Lesson 34

( ) ( ) , ( ) ( )t t t tt t e e t t e e- -′ ′′= = - + = = +v r i j a r i j .

3 22

1 1( ) ( ) 4 , ( ) ( ) 12t t t t t tt t

-′ ′′= = + = = +v r i j a r i j .

( ) ( ) 3 , ( ) ( ) 0 0t t t t′ ′′= = + = = + =v r i j a r i j 0 . Speed = ( ) 9 1 10.t = + =v At the point ( )3,0 , 1t = and

(1) 3 , (1)= + =v i j a 0 , and the speed is 10.

( ) ( ) 3sin 2cos , ( ) ( ) 3cos 2sint t t t t t t t′ ′′= = - + = = - -v r i j a r i j . Speed = 2 2( ) 9sin 4cost t t= +v . At the

point ( )3,0 , 0t = and (0) 2 , (0) 3= = -v j a i , and the speed is ( ) 0 4 2t = + =v .

2( ) ( ) 2 3 , ( ) ( ) 2 6t t t t t t t′ ′′= = + = = +v r i j a r i j . Speed = 2 4( ) 4 9t t t= +v . At the point ( )1,1 , 1t = and

(1) 2 3 , (1) 2 6= + = +v i j a i j , and the speed is (1) 4 9 13= + =v .

We integrate the acceleration function as follows.

( )

( )2 2

1 1

( ) ( ) . (0) .

( ) ( ) . (0) .2 2

t t dt dt t t

t tt t dt t t dt

= = + = + + = ⇒ =

= = + = + + = ⇒ =

∫ ∫

∫ ∫

v a i j i j C v 0 C 0

r v i j i j C r 0 C 0

Hence, the position function is 2 2

( )2 2t tt = +r i j .

We integrate the acceleration function as follows.

( )

( )( )2

21 1

( ) ( ) 2 3 2 3 . (0) 4 4 .

3( ) ( ) 2 3 4 4 . (0) .2

t t dt dt t t

tt t dt t t dt t t

= = + = + + = ⇒ =

= = + + = + + + = ⇒ =

∫ ∫

∫ ∫

v a i j i j C v j C j

r v i j i j C r 0 C 0

Hence, the position function is 2

2 3( ) 42tt t t

= + +

r i j .

We use the formula for the position function.

( ) ( ) ( )2 2( ) 88cos30 10 88sin 30 16 44 3 10 44 16t t t t t t t = ° + + ° − = + + − r i j i j.

1.

2.

3.

4.

5.

6.

7.

8.

solu

tions

197

The position function is ( ) ( ) 2( ) 140cos 22 2.5 140sin 22 16t t t t = ° + + ° - r i j .

When 375, 2.889x t= ≈ seconds. For this value of , 20.5t y ≈ feet. So, the ball clears the 10-foot fence.

The position function now uses 4.9- instead of 16,- due to the units.

( ) ( ) 2( ) 100cos30 1.5 100sin 30 4.9t t t t = ° + + ° - r i j . The maximum height occurs

when 0dydt

= : 100sin 30 9.8 5.102t t° = ⇒ ≈ seconds. The maximum height is

( )21.5 100sin 30 (5.102) 4.9 5.102 129.1y = + ° - ≈ meters.

Lesson 35

We calculate the derivative and use the formula for the unit tangent.

( )2 2 2 2(1) 2 2( ) 2 2 . (1) 2 2 , (1)

2 21 4 4 2 2t t

′ + +′ ′= + = + = = = = +′ +

i j i jrr i j r i j T i jr

.

We calculate the derivative and use the formula for the unit tangent.

( )

( ) 4sin 4cos . ( ) 2 2 2 2 .4( ) 2 2 2 2 2 2 2 2 2 24( ) .4 4 2 28 8

4

t t t π

ππ

π

′ ′= - + = - +

′ - + - += = = = - +

+′

r i j r i j

r i j i jT i jr

( ) ( )( ) 2

2

131 1 3( ) 3 , 3 , ( )1 9 19

e eet e et e e e

e

−′ −′ ′= − = − = = =′ ++

i jr i jr i j r i j Tr

.

We differentiate the unit tangent vector.

( )( )

( )( ) ( )

( )( ) ( )

( ) ( )( ) ( )

2

3 3 3 32 22 2 2 2

( ) , ( ) .1

1 2 1, 2 .5 51 1

2 12 2 .2 5

t tt t tt t

ttt t

′ +′ = + = =′ +

- -′ ′= + = ++ +

′= = - +

′

r i jr i j Tr

T i j T i j

TN i j

T

9.

10.

1.

2.

3.

4.

198

We differentiate the unit tangent vector.

( )( )

( ) ( ) ( )

( ) ( )( )

sin cos( ) sin cos , ( ) , ( ) sin cos .

3 1cos sin , 1. .6 2 2

3 16 .6 2 26

t t tt t t t t t tt

t t t t

π ππ π ππ

π

ππ

π

′ - +′ ′= - + = = = = - +′

′ ′ ′= - - = = - -

′= = - -

′

r i jr i j r T i jr

T i j T T i j

TN i j

T

( ) ( )( )

44 , ( ) , ( )4

tt t t

t′

′ ′= = = = =′

r ir i T i T 0r

. The unit normal is not defined. The path is a line (the x-axis).

We calculate the unit tangent, unit normal, and acceleration vectors.

( ) ( ) ( ) ( ) ( )( )( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

2 2

3 3 3 32 22 2 2 2

2

2 2 , 1 2 2 , 2 1 .

1 1 1( ) 2 2 , (1) .24 4 1

1 1 1 1 1( ) , (1) .2 2 2 22 21 1

( ) 1 2 2, 1 .( ) 2 21

t t t t

tt t t

t t ttt

t t

tt tt t

′= = + = + = =

= = + = + = ++ +- -′ ′= + = + = -

+ +

′= = - = -

′ +

v r i j v i j a i a

vT i j i j T i j

v

T i j T i j i j

TN i j N i jT

The tangential and normal components of acceleration at 1t = are 2, 2= = = =T Na a T a a N .

We will use the fact that the unit normal points in the direction that the curve is bending. The curve is

1 ,y x= oriented left to right.

( ) ( ) ( ) ( ) ( ) ( )

( )( ) ( )

( ) ( ) ( )

2 4 3

22

2 24 4

4

1 1 2, 1 , 1 , , 1 2 .

1 1 1 1( ) .1 1 11

1 1(1) 1 .2 2

t t t tt t t

t tt tt tt t t

t

′= = - = + = - = =

= = - = - = - + ++

= - ⇒ = +

v r i j v v i j a j a j

vT i j i j i j

v

T i j N i j

The tangential and normal components are 2, 2= = - = =T Na a T a a N .

.a = = = =Tv v aa T T a av v

5.

6.

7.

8.

9.

solu

tions

199

Lesson 36

For 0,t > ( ) ( ) ( ) ( )( )( )

2 , 2 , , , 0.t

t t t t t t Kt

′′ ′ ′= = = = = =

′

Tr i r T i T 0

r

This seems reasonable because the path is a straight line.

We use the results of Problem 8 in Lesson 35. Recall that

( ) ( ) ( )2

1 , 1 , 1 2tt

′ ′ ′= - = - =r i j r i j r . Furthermore, ( )2

4

1( )1

t tt

= -+

T i j , which implies that

( ) ( )

3

3 34 42 2

2 2( )1 1

t ttt t

′ = ++ +

T i j and 3 3

2 2

2 2 1 1(1)2 22 2

′ = + = +T i j i j .

Finally, ( )( )1 1(1) 1 and1 2

K′

′ = = =′

TT

r.

( ) ( )2( ) cos , ( ) 1 cos , , 12 2t t t t π π′ ′ ′ ′= + = + = =r i j r r i r .

( ) ( ) ( ) ( )2

1sin , 1 , cos2 1 cost t t t

tπ= − = − = +

+a j a j T i j .

( )( ) ( )

( )3 32 22 2

sin cos sin , 21 cos 1 cos

t t ttt t

π′ ′= - = -+ +

T i j T j . Hence, the curvature is ( ) ( )( )

12 12 12

Kπ

ππ

′= = =

′

T

r.

( ) ( )1( ) 5 2 , ( ) 5 2 , 029

t t t K′ ′= - + = - + = ⇒ =r i j T i j T 0 . The curve is a straight line, and hence, the

curvature is zero.

( )22 3, 4 , 1 4, 4.y x y x y y′ ′ ′′= + = - = - = The formula for curvature in rectangular coordinates gives

( ) ( )3 3 3

2 22 2 2

4 4 0.057171 1 4

yK

y

′′= = = ≈ ′+ + -

.

( ) ( )cos 2 , 2sin 2 , 2 0, 4cos 2 , 2 4y x y x y y x yπ π′ ′ ′′ ′′= = - = = - = - .

The curvature is ( )

3 32 22 2

4 41 01

yK

y

′′= = = +′+

.

1.

2.

3.

4.

5.

6.

200

We have seen in the lesson that the curvature for the exponential function is

( )3

2 2.

1

x

x

eKe

=+

Differentiating, ( )

( )

2

52 2

1 2

1

x x

x

e eK

e

-′ =

+. Let the derivative equal zero:

12 2 21 1 1 1 11 2 0 2 ln ln ln 2 ln 22 2 2 2 2

x xe e x x-

- = ⇒ = ⇒ = ⇒ = = - = . By the first derivative

test, this is a maximum. For this x-value, 12 12

2xe

-= =

and 2 12xe -= = 1 .2

Hence,

( ) ( ) ( )3 3 3

2 2 2 2

11 2 2 32 1 .

2 93 3311 1 2 2

x

x

eKe

= = = = =+ +

The position function for the ellipse is ( ) 3cos 5sin .t t t= +r i j We have

2 2( ) 3sin 5cos , ( ) 9sin 25cost t t t t t′ ′= - + = +r i j r and ( )2 2

3sin 5cos .9sin 25cos

t ttt t

- +=

+

i jT

Differentiating, ( )( )

32 2 2

75cos 45sin .9sin 25cos

t ttt t

- -′ =+

i jT So, the curvature is

When 3,2 2t π π= the maximum is 59

, and when 0,2t π= the minimum is 325 .

3 21 , 3 , 6 .y x y x y x′ ′′= - = - = - ( )

3 32 42 2

6

1 91

y xK

xy

′′ -= = +′+

. So, the curvature is zero when 0.x =

Try graphing the function 31y x= - to verify this answer.

( )( )

( )( )

( ) ( )

2 2 2 2

32 2 2

12 2 2

2 2

2 32 2 2 2 2

75 cos 45 sin

9sin 25cos

9sin 25cos

15 25cos 9sin 15 .9sin 25cos 9sin 25cos

t t

t ttK

t t t

t t

t t t t

+

+′ = =′ +

+= =

+ +

Tr

7.

8.

9.

solu

tions

( )( )

( )( )

( ) ( )

2 2 2 2

32 2 2

12 2 2

2 2

2 32 2 2 2 2

75 cos 45 sin

9sin 25cos

9sin 25cos

15 25cos 9sin 15 .9sin 25cos 9sin 25cos

t t

t ttK

t t t

t t

t t t t

+

+′ = =′ +

+= =

+ +

Tr

201

Glossary

Note: The number in parentheses indicates the lesson in which the concept or term is introduced.

absolute value function (1): The absolute value function is defined by , 0

( ), 0

x xf x x

x x≥

= = - <. It is continuous, but

not differentiable, at 0.x = Its graph appears in the shape of the letter V.

acceleration (34–35): In calculus, acceleration is the rate of change of velocity and has two components: the rate of change in speed and the rate of change in direction.

alternating series (21): The terms of an alternating series alternate in sign. For 0na > , the following are alternating series:

( )

( )

1 2 31

11 2 3

1

1 ....

1 ....

nn

n

nn

n

a a a a

a a a a

∞

=

∞+

=

- = - + - +

- = - + -

∑

∑

arc length (8): If ( )f x is a smooth curve on the interval a x b≤ ≤ , then the arc length of f is

( ) [ ]2 21 1b b

a as f x dx y dx′ ′= + = + ∫ ∫ . The differential of arc length is 21 ( )ds f x dx′= + .

area of a region in the plane (7): Let f be continuous and nonnegative on the interval [ ],a b . Partition the

interval into n equal subintervals of length 0 1 2 1, , , , , ,-

-∆ = = = n n

b ax x a x x x x bn

. The area of the region bounded

by f , the x-axis, and the vertical lines and x a x b= = is 11

lim ( ) , -→∞=

= ∆ ≤ ≤∑n

i i i in iA f c x x c x , provided that this

limit exists.

axis of revolution (30): If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution.

bounded (16): When a sequence is both bounded above and bounded below.

cardioid (29): Polar equations of the form ( )1 cosr a= + θ and ( )1 sinr a= + θ are called cardioids.

center of mass (9): Calculated by dividing moment about the origin by the total mass, 0Mx

m= , where

1 2 nm m m m= + + +

and the moment about the origin is given by 0 1 1 ... n nM m x m x= + + . For a region of

uniform density, the center of mass is often called the centroid of the region. See moment.

centroid (9): The geometric center of a planar lamina or higher-dimensional object. When mass is uniformly distributed, the centroid is equivalent to the center of mass.

compound interest formula: Let P be the amount of a deposit at an annual interest rate of r (as a decimal)

202

compounded n times per year. The amount after t years is 1ntrA P

n = +

. If the interest is compounded

continuously, the amount is rtA Pe= . See Understanding Calculus: Problems, Solutions, and Tips, Lesson 27.

concavity (2): Let f be differentiable on an open interval I . The graph of f is concave upward on I if f ′ is increasing on I and concave downward on I if f ′ is decreasing on .I A graph is concave upward if the graph is above its tangent lines and concave downward if the graph is below its tangent lines.

continuous function (1): A function f is continuous at c if the following three conditions are met: ( )f c is

defined, lim ( )x c

f x→

exists, and lim ( ) ( )x c

f x f c→

= .

convergence (17–24): If the sequence of partial sums of an infinite series converges to a limit L, then we say that the infinite series converges to L. The interval of convergence of a series is the set of all real numbers for which the series converges. For a series centered at c, the interval of convergence satisfies one of the following: The series converges only at c; there exists a real number R > 0 (the radius of convergence) such that the series

converges absolutely for x c R- < and diverges for x c R- > ; or the series converges for all real numbers.

curvature (36): A measure of how much a curve bends, ( )( )t

Kt′

=′

Tr

. The curvature of ( )y f x= is ( )

32 21

yK

y

′′= ′+

.

cycloid (28): The curve traced out by a point on the circumference of a circle rolling along a line.

definite integral (3): Let f be defined on the interval [ ],a b . Partition the interval into n equal subintervals

of length 0 1 2 1, , , , , , -

-∆ = = = n n

b ax x a x x x x bn

. Assume that the following limit exists: 1

lim ( ) ,n

in if c x

→∞=

∆∑ where

1i i ix c x- ≤ ≤ . Then, this limit is the definite integral of f from a to b and is denoted ( )b

af x dx∫ .

∆x (2, 7): x∆ (read “delta x”) denotes a (small) change in x. Some textbooks use h instead of .x∆

derivative (2): The derivative of f at x is given by the following limit, if it exists:

0

( ) ( )( ) limx

f x x f xf xx∆ →

+ ∆ -′ =∆ .

Notations for the derivative of ( ) :y f x=

[ ] [ ]( ), , , ( ) , ′ ′dy df x y f x D y

dx dx

The definitions of slope and the derivative are based on the difference quotient for slope:

change inslopechange in

y yx x

∆= =

∆

differential (2): Let ( )y f x= be a differentiable function. Then, dx x= ∆ is called the differential of x . The differential of y is ( )dy f x dx′= .

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differential equation (4): A differential equation in x and y is an equation that involves x, y, and derivatives of y. The order of a differential equation is determined by the highest-order derivative in the equation. A first-order

linear differential equation (6) can be written in the standard form ( ) ( )dy P x y Q xdx

+ = .

dot product (32): The dot product of two vectors 1 2 1 2, , ,u u v v= =u v is 1 1 2 2u v u v⋅ = +u v . The dot product of two vectors is a real number, not a vector, and provides a method for determining the angle between two nonzero vectors.

Euler’s method (5): A simple numerical algorithm for approximating solutions to differential equations. An

approximate solution to 0 0( , ), ( )y F x y y x y′ = = is given by ( )1 1 1 1, ,n n n n n nx x h y y hF x y- - - -= + = + .

Euler-Mascheroni constant (19): Denoted by gamma (γ ), where ( )lim lnnnS n

→∞γ = - , and 1 1 11

2 3nSn

= + + + +

. It is not known if this limit is rational or irrational.

exponential function (1): The inverse of the natural logarithmic function lny x= is the exponential function xy e= . The exponential function is equal to its derivative, x xd e e

dx = . The exponential function to base , 0,a a >

is defined by ( )ln a xxa e=

first derivative test (2): Let c be a critical number of .f If f ′ changes from positive to negative at ,c then f has

a relative maximum at ( )( ), .c f c If f ′ changes from negative to positive at ,c then f has a relative minimum at

( )( ), .c f c See second derivative test.

fundamental theorem of calculus (3): If f is a continuous function on the closed interval [ ],a b and F is an antiderivative of ,f then ( ) ( ) ( )

b

af x dx F b F a= -∫ . This theorem and the second fundamental theorem of

calculus show how integration and differentiation are basically inverse operations. If f is continuous on an open interval I containing a , the second fundamental theorem of calculus says that for any x in the interval,

( ) ( )x

a

d f t dt f xdx

= ∫ .

geometric series (17): A series in which each term in the summation is a fixed multiple of the previous term.

The geometric series with common ratio r is 2 3

0...k

kar a ar ar ar

∞

=

= + + + +∑ . The series converges to 1

ar-

if

1.r < The geometric series diverges if 1.r ≥

growth and decay model (5): The solution to the growth and decay model dy kydt

= is kty Ce= .

half-angle formulas (11): 2 21 cos 2 1 cos 2sin ; cos2 2

x xx x- += = . Used when exponents m and n are both even in

the integral sin cosm nx x dx∫ .

harmonic series (19): 1

1 1 112 3n n

∞

=

= + + +∑ diverges even though its terms tend to zero.

.

204

Hooke’s law (8): The force F required to compress or stretch a spring (within its elastic limits) is proportional to the distance d that the spring is compressed or stretched from its original length: F kd= . The proportionality constant k is the spring constant and depends on the nature of the spring.

horizontal asymptote (5): The line y L= is a horizontal asymptote of the graph of f if lim ( ) or lim ( ) .x x

f x L f x L→∞ →-∞

= =

implicit differentiation (2): A technique used when it is difficult to express y as a function of x explicitly. The steps are as follows: Differentiate both sides with respect to x, collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side, factor dy/dx out of the left side, and solve for dy/dx.

improper integral (15): An integral where one of the limits of integration is ∞ or -∞ , of the form

( ) lim ( )b

a abf x dx f x dx

∞

→∞=∫ ∫ , or those that are not continuous on the closed interval [ ],a b .

infinite series (17): An infinite series is defined in terms of a sequence of partial sums. See series.

integrating factor (6): For a linear differential equation, the integrating factor is ( )P x dxu e∫= .

integration by partial fractions (13): An algebraic technique for splitting up complicated algebraic expressions—in particular, rational functions—into a sum of simpler functions, which can then be integrated easily using other techniques.

integration by parts (10): u dv uv v du= -∫ ∫ .

integration by substitution (3): Let F be an antiderivative of f . If ( )u g x= , then ( )du g x dx′= , so we have

( ( )) ( ) ( ( )) because ( ) ( ) .f g x g x dx F g x C f u du F u C′ = + = +∫ ∫

inverse functions (1): Those whose graphs are symmetric across the line .y x=

A function g is the inverse function of the function f if ( ( ))f g x x= for all x in the domain of g and ( ( ))g f x x= for all x in the domain of f . The inverse of f is denoted 1f - .

inverse trigonometric functions (1): These inverse functions are defined by restricting the domain of the original function, as follows:

1arcsin sin sin , for 1 1 and 2 2

y x x y x x y- π π= = ⇔ = - ≤ ≤ - ≤ ≤

1arccos cos cos , for 1 1 and 0y x x y x x y-= = ⇔ = - ≤ ≤ ≤ ≤ π

1arctan tan tan , for and 2 2

y x x y x x y- π π= = ⇔ = -∞ < < ∞ - < <

1arcsec sec sec , for 1, 0 , and 2

y x x y x x y y- π= = ⇔ = ≥ ≤ ≤ π ≠

L’Hôpital’s rule (14): A technique for evaluating indeterminate forms for limits such as 00 or ∞∞ , where no

guaranteed limit exists.

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limit (1): Defined informally, if ( )f x becomes arbitrarily close to a single number L as x approaches c from

either side, we say that the limit of ( )f x as x approaches c is L , which we write as lim ( )x c

f x L→

= . Also, the

equation lim ( )x c

f x→

= ∞ means that ( )f x increases without bound as x approaches c . More formally: Let f be a

function defined on an open interval containing c (except possibly at c), and let L be a real number. The statement

lim ( )x c

f x L→

= means that for each ε > 0, there exists a δ > 0 such that if 0 ,x c< - < δ then ( )f x L- < ε .

limit of a sequence (16): When the terms of a sequence approach a fixed real number. When the limit exists, the sequence converges to L , and we write lim nn

a L→∞

= . If the limit does not exist, then the sequence diverges.

log rule for integration (3): 1 lndx x Cx

= +∫ . For a logarithmic function to base a, when 0 and 1,a a> ≠ 1log ln

lna x xa

= . See Understanding Calculus: Problems, Solutions, and Tips, Lesson 27.

logistic differential equation (5): Defined as 1 ; , 0,dy yky k Ldt L

= - >

where k is the rate of growth (or

decay) and L is a limit on the growth (or decay). Its solution is 1 kt

Lybe-=

+.

moment (9): Related to the turning force of a mass around a pivot or fulcrum. More precisely, if a mass m is

concentrated at a point, and if x is the distance between the mass and another point ,P then the moment of m about P is mx.

monotonic (16): A sequence is monotonic if its terms are nondecreasing, 1 2 3 ...a a a≤ ≤ ≤ , or nonincreasing,

1 2 3 ...a a a≥ ≥ ≥ .

natural logarithmic function (3): The natural logarithmic function is defined by the definite integral

1

1ln , 0= >∫x

x dt xt

.

normal (35): Perpendicular or orthogonal; the normal component of acceleration is the direction of the acceleration

and is given by 2 2a a′= = = -N Tv T a N a .

one-sided limits (1): The limit from the right means that x approaches c from values greater than .c The

notation is lim ( )x c

f x L+→

= . Similarly, the limit from the left means that x approaches c from values less than

, notated lim ( )x c

c f x L-→

= . See Understanding Calculus: Problems, Solutions, and Tips, Lesson 5.

orthogonal (5): The orthogonal trajectories of a given family of curves are another family of curves, each of which is orthogonal (perpendicular) to every curve in the given family. See normal and dot product.

p-series (19): 1

1 1 1 1 ...1 2 3p p p p

n n

∞

=

= + + +∑ is equal to the harmonic series when 1p = . The p-series test states that

a p-series converges if 1p > and diverges if 0 1p< ≤ .

206

parametric equation (28): Consider x and y as functions of a third variable (“parameter”) .t The curve traced out by the parametric equations ( ), ( )x f t y g t= = induces an orientation to the graph of the curve.

planar lamina (9): A flat plate of uniform density.

polar coordinates (29): An alternative to rectangular (Cartesian) coordinates of ( , )P x y= , with each point

instead given by ( ),r θ , where r is the distance from P to the origin and θ is the angle the segment OP makes with the positive x-axis.

power series (24): An infinite series in the variable x; loosely speaking, an infinite polynomial. One familiar example is the geometric series, 2

01n

nx x x

∞

=

= + + +∑

.

radians (1): Calculus uses radian measure. If a problem is stated in degree measure, you must convert to radians: 360° is 2π radians; 180° is π radians.

second derivative test (2): Let ( ) 0f c′ = ( c is a critical number of f ). If ( ) 0f c′′ > , then f has a relative

minimum at c . If ( ) 0f c′′ < , then f has a relative maximum at c .

sequence (16): An infinite list of real numbers, written { } { }1 2, ,...na a a= . That is, a sequence is a function whose domain is the set of positive integers.

series (17): A series is written using sigma notation for sums (See Understanding Calculus: Problems,

Solutions, and Tips, Lesson 19); for an infinite series 1 2 31

...nn

a a a a∞

=

= + + +∑ , the nth partial sum is given by

1 2 2n nS a a a a= + + + + .

Note that it is common to say simply “series” when referring to an infinite series.

slope field (4): Given a differential equation in x and y, ( , )y F x y′ = , the equation determines the derivative at

each point ( ),x y . If you draw short line segments with slope ( , )F x y at selected points ( ),x y , then these line segments form a slope field. See Understanding Calculus: Problems, Solutions, and Tips, Lesson 35.

solid of revolution (7): If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution. When the plane is a circle, the resulting solid is a torus.

solution curves (5): The general solution of a first-order differential equation represents a family of curves known as solution curves, one for each value of the arbitrary constant. See Understanding Calculus: Problems, Solutions, and Tips, Lesson 35.

summation formulas:

1.

n

ic c c c cn

=

= + + + =∑

1

( 1).2

n

i

n ni=

+=∑

2

1

( 1)(2 1).6

n

i

n n ni=

+ +=∑

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207

Taylor polynomial (23): The Taylor polynomial of degree n for f at c is

( ) ( ) ( )2

0

( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )! 2! !

nnnn n

nn

f cf c f cP x x c f c f c x c x c x cn n=

′′′= - = + - + - + + -∑ .

Taylor series (26): A power series representation of a function. The Taylor series for f at c is

( ) ( ) ( )2

0

( ) ( )( ) ( ) ( )( ) ( ) ( )! 2! !

nnn n

n

f cf c f cx c f c f c x c x c x cn n

∞

=

′′′- = + - + - + + - +∑

.

Taylor’s theorem (23): Can be used to estimate the accuracy of a Taylor polynomial. Let f be differentiable through order 1n + on an interval containing c . Then, there exists a number z between x and c such that

( )

2( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )2! !

nn

nf c f cf x f c f c x c x c x c R x

n′′

′= + - + - + + - + ,

where ( ) ( )

( ) ( )1

1( )1 !

nn

n

f zR x x c

n

++= -

+. That is,

( )( )

11( ) max ( )

1 !

nn

n

x cR x f z

n

++-

≤+

.

telescoping series (17): A series in which intermediate terms of the partial sums cancel one another and drop out, typically leaving only the first and last terms.

torus (31): A surface or solid shaped like a tire or doughnut and formed by revolving the region bounded by the circle 2 2 2x y r+ = about the line ( )x R r R= < .

theorem of Pappus (9): If a region is rotated about the y-axis, then the volume of the resulting solid of revolution is 2V xAπ= .

trigonometric identities (2): Trigonometric identities are trigonometric equations that are valid for all values of the variable (typically x or θ) and offer an important technique for simplifying differentiation and integration problems. In addition to those described under trigonometric functions, some of the most useful are as follows:

2 2 1.sin x cos x+ =

2 2tan sec 1.x x+ =

2 2

2

2

cos2 cos sin .sin 2 2sin cos .

1 cos2cos .2

1 cos2sin .2

x x xx x x

xx

xx

= -=

+=

-=

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trigonometric functions (1): The right triangle definition of the trigonometric functions uses the following right triangle:

sin .

cos .

sintan .cos

acbc

ab

θ =

θ =

θθ = =

θ

For a point ( ),x y on the unit circle 2 2 1x y+ = , the unit circle definition of the trigonometric functions is

sinsin , cos , tan .cos

1 1 1csc , sec , cot .sin cos tan

yy xx

xy

θθ = θ = θ = =

θ

θ = θ = θ = =θ θ θ

trigonometric substitution (12): A technique for converting integrands to trigonometric integrals.

vector (31): For quantities that have both magnitude and direction, such as velocity, acceleration, and force; contrasted with scalars, which have only magnitude, such as speed, mass, volume, and time. Usually appear in lowercase, bold letters.

vector-valued functions (33): Functions that use vectors, instead of algebraic variables, to define their outputs,

usually of the form ( ) ( ) ( )t f t g t= +r i j , where f and g are the component functions and t is the parameter.

velocity (34): The derivative of the position function. The velocity vector is tangent to the path of a particle and points in the direction of motion.

work (8): Force times distance: W F D= . If the force is variable, given by ( )f x , then the work W done by

moving the object from x a= to x b= is ( )b

aW F x dx= ∫ .

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Summary of Differentiation Formulas

1. Constant multiple rule: [ ]d cu cudx

= ′ .

2. Sum or difference rule: [ ]d u v u vdx

± = ′ ± ′ .

3. Product rule: [ ]d uv uv vudx

= ′ + ′ .

4. Quotient rule: 2

d u vu uvdx v v

′ - ′ = .

5. Constant rule: [ ] 0d cdx

= .

6. Chain rule: [ ( )] ( )d f u f u udx

= ′ ′ .

7. General power rule: 1[ ]n nd u nu udx

-= ′ .

8. [ ] 1d xdx

= .

9. 1[ln ]d xdx x

= .

10. [ ]x xd e edx

= .

11. 1[log ](ln )a

d xdx a x

= .

12. [ ] (ln )x xd a a adx

= .

13. [sin ] cosd x xdx

= .

14. [cos ] sind x xdx

= - .

15. 2[tan ] secd x xdx

= .

16. 2[cot ] cscd x xdx

= - .

17. [sec ] sec tand x x xdx

= .

18. [csc ] csc cotd x x xdx

= - .

19. 2

1[arcsin ]1

d xdx x

=-

.

20. 2

1[arctan ]1

d xdx x

=+

.

21. 2

1[arcsec ]1

d xdx x x

=-

.

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Summary of Integration Formulas

1. ( ) ( )kf x dx k f x dx=∫ ∫ .

2. [ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx± = ±∫ ∫ ∫ .

3. ( ) d x x C= +∫ .

4. Power rule for integration: 1

, for 11

nn xx dx C n

n

+

= + ≠ -+∫ .

5. Log rule for integration: lndx x Cx

= +∫ .

6. x xe dx e C= +∫ .

7. 1 ln

x xa dx a Ca

= + ∫ .

8. sin cosxdx x C= - +∫ .

9. cos sinxdx x C= +∫ .

10. tan ln cosxdx x C= - +∫ .

11. cot ln sinxdx x C= +∫ .

12. sec ln sec tanxdx x x C= + +∫ .

13. csc ln csc cotxdx x x C= - + +∫ .

14. 2sec tanxdx x C= +∫ .

15. 2csc cotxdx x C= - +∫ .

16. sec tan secx xdx x C= +∫ .

17. csc cot cscx xdx x C= - +∫ .

18. 2 2 arcsindx x C

aa x= +

-∫ .

19. 2 2

1 arctandx x Ca x a a

= ++∫ .

20. 2 2

1 arcsecxdx C

a ax x a= +

-∫ .

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Power Series for Elementary Functions

1. ( ) ( ) ( ) ( ) ( )2 31 1 1 1 1 1 1 0 2n nx x x x xx= - - + - - - + + - - + < ≤

2. ( )2 31 1 1 1 11

n nx x x x xx= - + - + + - + - < <

+

3. 2 31 1 1 11

nx x x x xx= + + + + + + - < <

-

4. ( ) ( ) ( ) ( ) ( ) ( )2 3 411 1 1 1

ln 1 1 0 22 3 4

nnx x x x

x x xn

-- - - -= - - + - + + - + < ≤

5. 2 3 4

12! 3! 4! !

nx x x x xe x x

n= + + + + + + + -∞ < < ∞

6. ( ) ( )3 5 7 2 1

sin 13! 5! 7! 2 1 !

nnx x x xx x x

n

+

= - + - + + - + -∞ < < ∞+

7. ( ) ( )2 4 6 2

cos 1 12! 4! 6! 2 !

nnx x x xx x

n= - + - + + - + -∞ < < ∞

8. ( )3 5 7 2 1

arctan 1 1 13 5 7 2 1

nnx x x xx x x

n

+

= - + - + + - + - ≤ ≤+

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

( ) ( ) ( ) ( )2


tπ= − = − = +

+a j a j T i j .

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Summary of Tests for Convergence and Divergence

• If the nth term of the series does not approach 0, then the series diverges.

• Try to compare with a known series. Remember: The harmonic series diverges.

• Is the series geometric? 0

n

nar

∞

=∑ converges for 1r < and diverges for 1.r ≥

• Is the series a p-series? The series 1

1p

n n

∞

=∑ converges for 1p > and diverges for 0 1.p< ≤

• Is the series a telescoping series? If so, use cancelation to analyze the nth partial sum.

• Is the series alternating? If so, try to use the alternating series test for convergence.

• Try to use the integral test by comparing the series to an appropriate improper integral.

• The ratio test is especially useful for series containing factorials n! or powers xn.

• The root test is especially useful for series containing radicals.

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Bibliography

Larson, Ron, and Bruce H. Edwards. Calculus. 10th ed. Florence, KY: Brooks/Cole, 2013. This text covers an entire three-semester calculus sequence.

———. Calculus: Early Transcendental Functions. 5th ed. Florence, KY: Brooks/Cole, 2010. This text offers a different approach to the exponential and logarithmic functions.

Larson, Ron, Robert Hostetler, and Bruce H. Edwards. Precalculus Functions and Graphs: A Graphing Approach. 5th ed. Boston: Houghton Mifflin, 2008. An excellent source for precalculus, algebra, and trigonometry.

Saxon, John H., Jr., and Frank Y. H. Wang. Calculus with Trigonometry and Analytic Geometry. 2nd ed. Edited by Bret L. Crock and James A. Sellers. Wilmington, MA: Saxon Publishers, 2002. This is another good choice among calculus texts and is edited by James A. Sellers, who produced Algebra I, Algebra II, and Mastering the Fundamentals of Mathematics with The Great Courses.

Simmons, George. Calculus with Analytic Geometry. 2nd ed. New York: McGraw-Hill, 1996. A somewhat different approach to calculus.

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