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Vector

Calculus

Michael Corral

About the author : Michael Corral is an Adjunct Faculty member of the Department of Mathematics at Schoolcraft College. He received a B.A. in Mathematics from the University of California at Berkeley, and received an M.A. in Mathematics and an M.S. in Industrial & Operations Engineering from the University of Michigan.

This text was typeset in LATEX 2 ε with the KOMA-Script bundle, using the GNU Emacs text

editor on a Fedora Linux system. The graphics were created using MetaPost, PGF, and Gnuplot.

Copyright © 2008 Michael Corral. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled “GNU Free Documentation License”.

Preface

This book covers calculus in two and three variables. It is suitable for a one-semester course, normally known as “Vector Calculus”, “Multivariable Calculus”, or simply “Calculus III”. The prerequisites are the standard courses in single-variable calculus (a.k.a. Calculus I and II). I have tried to be somewhat rigorous about proving results. But while it is important for students to see full-blown proofs - since that is how mathematics works - too much rigor and emphasis on proofs can impede the flow of learning for the vast majority of the audience at this level. If I were to rate the level of rigor in the book on a scale of 1 to 10, with 1 being completely informal and 10 being completely rigorous, I would rate it as a 5. There are 420 exercises throughout the text, which in my experience are more than enough for a semester course in this subject. There are exercises at the end of each sec- tion, divided into three categories: A, B and C. The A exercises are mostly of a routine computational nature, the B exercises are slightly more involved, and the C exercises usu- ally require some effort or insight to solve. A crude way of describing A, B and C would be “Easy”, “Moderate” and “Challenging”, respectively. However, many of the B exercises are easy and not all the C exercises are difficult. There are a few exercises that require the student to write his or her own computer pro- gram to solve some numerical approximation problems (e.g. the Monte Carlo method for approximating multiple integrals, in Section 3.4). The code samples in the text are in the Java programming language, hopefully with enough comments so that the reader can figure out what is being done even without knowing Java. Those exercises do not mandate the use of Java, so students are free to implement the solutions using the language of their choice. While it would have been simple to use a scripting language like Python, and perhaps even easier with a functional programming language (such as Haskell or Scheme), Java was cho- sen due to its ubiquity, relatively clear syntax, and easy availability for multiple platforms. Answers and hints to most odd-numbered and some even-numbered exercises are pro- vided in Appendix A. Appendix B contains a proof of the right-hand rule for the cross prod- uct, which seems to have virtually disappeared from calculus texts over the last few decades. Appendix C contains a brief tutorial on Gnuplot for graphing functions of two variables. This book is released under the GNU Free Documentation License (GFDL), which allows others to not only copy and distribute the book but also to modify it. For more details, see the included copy of the GFDL. So that there is no ambiguity on this matter, anyone can make as many copies of this book as desired and distribute it as desired, without needing my permission. The PDF version will always be freely available to the public at no cost (go to http://www.mecmath.net). Feel free to contact me at [email protected] for

iii

Contents

• 1 Vectors in Euclidean Space Preface iii
  • 1.1 Introduction
  • 1.2 Vector Algebra
  • 1.3 Dot Product
  • 1.4 Cross Product
  • 1.5 Lines and Planes
  • 1.6 Surfaces
  • 1.7 Curvilinear Coordinates
  • 1.8 Vector-Valued Functions
  • 1.9 Arc Length
• 2 Functions of Several Variables
  • 2.1 Functions of Two or Three Variables
  • 2.2 Partial Derivatives
  • 2.3 Tangent Plane to a Surface
  • 2.4 Directional Derivatives and the Gradient
  • 2.5 Maxima and Minima
  • 2.6 Unconstrained Optimization: Numerical Methods
  • 2.7 Constrained Optimization: Lagrange Multipliers
• 3 Multiple Integrals
  • 3.1 Double Integrals
  • 3.2 Double Integrals Over a General Region
  • 3.3 Triple Integrals
  • 3.4 Numerical Approximation of Multiple Integrals
  • 3.5 Change of Variables in Multiple Integrals
  • 3.6 Application: Center of Mass
  • 3.7 Application: Probability and Expected Value
• 4 Line and Surface Integrals
  • 4.1 Line Integrals
  • 4.2 Properties of Line Integrals
  • 4.3 Green’s Theorem
  • 4.4 Surface Integrals and the Divergence Theorem vi Contents
  • 4.5 Stokes’ Theorem
  • 4.6 Gradient, Divergence, Curl and Laplacian
• Bibliography
• Appendix A: Answers and Hints to Selected Exercises
• Appendix B: Proof of the Right-Hand Rule for the Cross Product
• Appendix C: 3D Graphing with Gnuplot
• GNU Free Documentation License
• History
• Index

2 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE

The coordinate system shown in Figure 1.1.1 is known as a right-handed coordinate system , because it is possible, using the right hand, to point the index finger in the positive direction of the x -axis, the middle finger in the positive direction of the y -axis, and the thumb in the positive direction of the z -axis, as in Figure 1.1.3.

x

z

y

Figure 1.1.3 Right-handed coordinate system

An equivalent way of defining a right-handed system is if you can point your thumb up- wards in the positive z -axis direction while using the remaining four fingers to rotate the x -axis towards the y -axis. Doing the same thing with the left hand is what defines a left- handed coordinate system. Notice that switching the x - and y -axes in a right-handed system results in a left-handed system, and that rotating either type of system does not change its “handedness”. Throughout the book we will use a right-handed system. For functions of three variables, the graphs exist in 4-dimensional space (i.e. R^4 ), which we can not see in our 3-dimensional space, let alone simulate in 2-dimensional space. So we can only think of 4-dimensional space abstractly. For an entertaining discussion of this subject, see the book by ABBOTT.^1 So far, we have discussed the position of an object in 2-dimensional or 3-dimensional space. But what about something such as the velocity of the object, or its acceleration? Or the gravitational force acting on the object? These phenomena all seem to involve motion and direction in some way. This is where the idea of a vector comes in.

(^1) One thing you will learn is why a 4-dimensional creature would be able to reach inside an egg and remove the yolk without cracking the shell!

1.1 Introduction 3

You have already dealt with velocity and acceleration in single-variable calculus. For example, for motion along a straight line, if y = f ( t ) gives the displacement of an object after time t , then d y / dt = f ′( t ) is the velocity of the object at time t. The derivative f ′( t ) is just a number, which is positive if the object is moving in an agreed-upon “positive” direction, and negative if it moves in the opposite of that direction. So you can think of that number, which was called the velocity of the object, as having two components: a magnitude , indicated by a nonnegative number, preceded by a direction , indicated by a plus or minus symbol (representing motion in the positive direction or the negative direction, respectively), i.e. f ′( t ) = ± a for some number a ≥ 0. Then a is the magnitude of the velocity (normally called the speed of the object), and the ± represents the direction of the velocity (though the + is usually omitted for the positive direction). For motion along a straight line, i.e. in a 1-dimensional space, the velocities are also con- tained in that 1-dimensional space, since they are just numbers. For general motion along a curve in 2- or 3-dimensional space, however, velocity will need to be represented by a multi- dimensional object which should have both a magnitude and a direction. A geometric object which has those features is an arrow, which in elementary geometry is called a “directed line segment”. This is the motivation for how we will define a vector.

Definition 1.1. A (nonzero) vector is a directed line segment drawn from a point P (called its initial point ) to a point Q (called its terminal point ), with P and Q being distinct points. The vector is denoted by

PQ. Its magnitude is the length of the line segment, denoted by

PQ

, and its direction is the same as that of the directed line segment. The zero vector is just a point, and it is denoted by 0.

To indicate the direction of a vector, we draw an arrow from its initial point to its terminal point. We will often denote a vector by a single bold-faced letter (e.g. v ) and use the terms “magnitude” and “length” interchangeably. Note that our definition could apply to systems with any number of dimensions (see Figure 1.1.4 (a)-(c)).

S 0 R P Q x

−−→^ − PQ −→ RS

(a) One dimension

x

y

0

P

Q

R

S

−−→ PQ

−−→ RS

v

(b) Two dimensions

x

y

z

0 P

Q

R

S −−→ PQ

−−→ RS

v

(c) Three dimensions Figure 1.1.4 Vectors in different dimensions

1.1 Introduction 5

Unless otherwise indicated, when speaking of “the vector” with a given magnitude and direction, we will mean the one whose initial point is at the origin of the coordinate system.

Thinking of vectors as starting from the origin provides a way of dealing with vectors in a standard way, since every coordinate system has an origin. But there will be times when it is convenient to consider a different initial point for a vector (for example, when adding vectors, which we will do in the next section). Another advantage of using the origin as the initial point is that it provides an easy cor- respondence between a vector and its terminal point.

Example 1.1. Let v be the vector in R^3 whose initial point is at the origin and whose ter- minal point is (3, 4 , 5). Though the point (3, 4 , 5) and the vector v are different objects, it is convenient to write v = (3, 4 , 5). When doing this, it is understood that the initial point of v is at the origin (0, 0 , 0) and the terminal point is (3, 4 , 5).

x

y

z

0

P (3, 4, 5)

(a) The point (3,4,5)

x

y

z

0

v = (3, 4, 5)

(b) The vector (3,4,5) Figure 1.1.6 Correspondence between points and vectors

Unless otherwise stated, when we refer to vectors as v = ( a , b ) in R^2 or v = ( a , b , c ) in R^3 , we mean vectors in Cartesian coordinates starting at the origin. Also, we will write the zero vector 0 in R^2 and R^3 as (0, 0) and (0, 0 , 0), respectively. The point-vector correspondence provides an easy way to check if two vectors are equal, without having to determine their magnitude and direction. Similar to seeing if two points are the same, you are now seeing if the terminal points of vectors starting at the origin are the same. For each vector, find the (unique!) vector it equals whose initial point is the origin. Then compare the coordinates of the terminal points of these “new” vectors: if those coordinates are the same, then the original vectors are equal. To get the “new” vectors starting at the origin, you translate each vector to start at the origin by subtracting the coordinates of the original initial point from the original terminal point. The resulting point will be the terminal point of the “new” vector whose initial point is the origin. Do this for each original vector then compare.

6 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE

Example 1.2. Consider the vectors

PQ and

RS in R^3 , where P = (2, 1 , 5), Q = (3, 5 , 7), R = (1, − 3 , −2) and S = (2, 1 , 0). Does

PQ =

RS?

Solution: The vector

PQ is equal to the vector v with initial point (0, 0 , 0) and terminal point Q − P = (3, 5 , 7) − (2, 1 , 5) = (3 − 2 , 5 − 1 , 7 − 5) = (1, 4 , 2). Similarly,

RS is equal to the vector w with initial point (0, 0 , 0) and terminal point S − R = (2, 1 , 0) − (1, − 3 , −2) = (2 − 1 , 1 − (−3), 0 − (−2)) = (1, 4 , 2). So

PQ = v = (1, 4 , 2) and

RS = w = (1, 4 , 2).

PQ =

RS

y

z

x

0

−− PQ →

−− RS →

Translate −−→ PQ to v

Translate − RS −→ to w

P (2, 1, 5)

Q (3, 5, 7)

R (1, −3, −2)

S (2, 1, 0)

(1, 4, 2)

v = w

Figure 1.1.

Recall the distance formula for points in the Euclidean plane:

For points P = ( x 1 , y 1 ), Q = ( x 2 , y 2 ) in R^2 , the distance d between P and Q is:

d =

( x 2 − x 1 )^2 + ( y 2 − y 1 )^2 (1.1)

By this formula, we have the following result:

For a vector

PQ in R^2 with initial point P = ( x 1 , y 1 ) and terminal point Q = ( x 2 , y 2 ), the magnitude of

PQ is: ∥∥−−→ PQ

( x 2 − x 1 )^2 + ( y 2 − y 1 )^2 (1.2)

8 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE

Example 1.3. Calculate the following:

(a) The magnitude of the vector

PQ in R^2 with P = (− 1 , 2) and Q = (5, 5). Solution: By formula (1.2),

PQ

(5 − (−1))^2 + (5 − 2)^2 =

p 36 + 9 =

p 45 = 3

p

(b) The magnitude of the vector v = (8, 3) in R^2. Solution: By formula (1.3), ‖ v ‖ =

p 82 + 32 =

p

(c) The distance between the points P = (2, − 1 , 4) and Q = (4, 2 , −3) in R^2. Solution: By formula (1.4), the distance d =

p (4^ −^ 2)^2 +^ (2^ −^ (−1))^2 +^ (−^3 −^ 4)^2 = 4 + 9 + 49 =

p

(d) The magnitude of the vector v = (5, 8 , −2) in R^3. Solution: By formula (1.5), ‖ v ‖ =

52 + 82 + (−2)^2 =

p 25 + 64 + 4 =

p

Exercises

A

1. Calculate the magnitudes of the following vectors: (a) v = (2, −1) (b) v = (2, − 1 , 0) (c) v = (3, 2 , −2) (d) v = (0, 0 , 1) (e) v = (6, 4 , −4) 2. For the points P = (1, − 1 , 1), Q = (2, − 2 , 2), R = (2, 0 , 1), S = (3, − 1 , 2), does

PQ =

RS?

3. For the points P = (0, 0 , 0), Q = (1, 3 , 2), R = (1, 0 , 1), S = (2, 3 , 4), does

PQ =

RS?

B

4. Let v = (1, 0 , 0) and w = ( a , 0 , 0) be vectors in R^3. Show that ‖ w ‖ = | a | ‖ v ‖. 5. Let v = ( a , b , c ) and w = (3 a , 3 b , 3 c ) be vectors in R^3. Show that ‖ w ‖ = 3 ‖ v ‖.

C

x

y

z

0

P ( x 1 , y 1 , z 1 )

Q ( x 2 , y 2 , z 2 )

R ( x 2 , y 2 , z 1 )

S ( x 1 , y 1 , 0) T ( x 2 , y 2 , 0) U ( x 2 , y 1 , 0) Figure 1.1.

6. Though we will see a simple proof of Theorem 1. in the next section, it is possible to prove it using methods similar to those in the proof of Theorem 1.2. Prove the special case of Theorem 1.1 where the points P = ( x 1 , y 1 , z 1 ) and Q = ( x 2 , y 2 , z 2 ) satisfy the fol- lowing conditions: x 2 > x 1 > 0, y 2 > y 1 > 0, and z 2 > z 1 > 0. ( Hint: Think of Case 4 in the proof of Theorem 1.2, and consider Figure 1.1.9. )

1.2 Vector Algebra 9

1.2 Vector Algebra

Now that we know what vectors are, we can start to perform some of the usual algebraic operations on them (e.g. addition, subtraction). Before doing that, we will introduce the notion of a scalar.

Definition 1.3. A scalar is a quantity that can be represented by a single number.

For our purposes, scalars will always be real numbers.^3 Examples of scalar quantities are mass, electric charge, and speed (not velocity).^4 We can now define scalar multiplication of a vector.

Definition 1.4. For a scalar k and a nonzero vector v , the scalar multiple of v by k , denoted by k v , is the vector whose magnitude is | k | ‖ v ‖, points in the same direction as v if k > 0, points in the opposite direction as v if k < 0, and is the zero vector 0 if k = 0. For the zero vector 0 , we define k 0 = 0 for any scalar k.

Two vectors v and w are parallel (denoted by v ∥ w ) if one is a scalar multiple of the other. You can think of scalar multiplication of a vector as stretching or shrinking the vector, and as flipping the vector in the opposite direction if the scalar is a negative number (see Figure 1.2.1).

v (^) 2 v 3 v 0.5 v − v (^) − 2 v

Figure 1.2. Recall that translating a nonzero vector means that the initial point of the vector is changed but the magnitude and direction are preserved. We are now ready to define the sum of two vectors.

Definition 1.5. The sum of vectors v and w , denoted by v + w , is obtained by translating w so that its initial point is at the terminal point of v ; the initial point of v + w is the initial point of v , and its terminal point is the new terminal point of w.

(^3) The term scalar was invented by 19th (^) century Irish mathematician, physicist and astronomer William Rowan Hamilton, to convey the sense of something that could be represented by a point on a scale or graduated ruler. The word vector comes from Latin, where it means “carrier”. (^4) An alternate definition of scalars and vectors, used in physics, is that under certain types of coordinate trans- formations (e.g. rotations), a quantity that is not affected is a scalar, while a quantity that is affected (in a certain way) is a vector. See MARION for details.

1.2 Vector Algebra 11

Theorem 1.3. Let v = ( v 1 , v 2 ), w = ( w 1 , w 2 ) be vectors in R^2 , and let k be a scalar. Then (a) k v = ( kv 1 , kv 2 )

(b) v + w = ( v 1 + w 1 , v 2 + w 2 )

Proof: (a) Without loss of generality, we assume that v 1 , v 2 > 0 (the other possibilities are handled in a similar manner). If k = 0 then k v = 0 v = 0 = (0, 0) = (0 v 1 , 0 v 2 ) = ( kv 1 , kv 2 ), which is what we needed to show. If k 6 = 0, then ( kv 1 , kv 2 ) lies on a line with slope kv kv^21 = v v^21 , which is the same as the slope of the line on which v (and hence k v ) lies, and ( kv 1 , kv 2 ) points in the same direction on that line as k v. Also, by formula (1.3) the magnitude of ( kv 1 , kv 2 ) is √ ( kv 1 )^2 + ( kv 2 )^2 =

k^2 v^21 + k^2 v^22 =

k^2 ( v^21 + v^22 ) = | k |

v^21 + v^22 = | k | ‖ v ‖. So k v and ( kv 1 , kv 2 ) have the same magnitude and direction. This proves (a).

x

y

0

w 2 v 2

w 1 v 1 v 1 +^ w 1

v 2 + w 2 w 2

v^ w^1

v

w

w v + w

Figure 1.2.

(b) Without loss of generality, we assume that v 1 , v 2 , w 1 , w 2 > 0 (the other possibilities are han- dled in a similar manner). From Figure 1.2.5, we see that when translating w to start at the end of v , the new terminal point of w is ( v 1 + w 1 , v 2 + w 2 ), so by the definition of v + w this must be the ter- minal point of v + w. This proves (b). QED

Theorem 1.4. Let v = ( v 1 , v 2 , v 3 ), w = ( w 1 , w 2 , w 3 ) be vectors in R^3 , let k be a scalar. Then (a) k v = ( kv 1 , kv 2 , kv 3 )

(b) v + w = ( v 1 + w 1 , v 2 + w 2 , v 3 + w 3 )

The following theorem summarizes the basic laws of vector algebra.

Theorem 1.5. For any vectors u , v , w , and scalars k , l , we have (a) v + w = w + v Commutative Law

(b) u + ( v + w ) = ( u + v ) + w Associative Law

(c) v + 0 = v = 0 + v Additive Identity

(d) v + (− v ) = 0 Additive Inverse

(e) k ( l v ) = ( kl ) v Associative Law

(f) k ( v + w ) = k v + k w Distributive Law

(g) ( k + l ) v = k v + l v Distributive Law

Proof: (a) We already presented a geometric proof of this in Figure 1.2.4(a). (b) To illustrate the difference between analytic proofs and geometric proofs in vector alge- bra, we will present both types here. For the analytic proof, we will use vectors in R^3 (the proof for R^2 is similar).

12 CHAPTER 1. VECTORS IN EUCLIDEAN SPACE

Let u = ( u 1 , u 2 , u 3 ), v = ( v 1 , v 2 , v 3 ), w = ( w 1 , w 2 , w 3 ) be vectors in R^3. Then

u + ( v + w ) = ( u 1 , u 2 , u 3 ) + (( v 1 , v 2 , v 3 ) + ( w 1 , w 2 , w 3 )) = ( u 1 , u 2 , u 3 ) + ( v 1 + w 1 , v 2 + w 2 , v 3 + w 3 ) by Theorem 1.4(b) = ( u 1 + ( v 1 + w 1 ), u 2 + ( v 2 + w 2 ), u 3 + ( v 3 + w 3 )) by Theorem 1.4(b) = (( u 1 + v 1 ) + w 1 , ( u 2 + v 2 ) + w 2 , ( u 3 + v 3 ) + w 3 ) by properties of real numbers = ( u 1 + v 1 , u 2 + v 2 , u 3 + v 3 ) + ( w 1 , w 2 , w 3 ) by Theorem 1.4(b) = ( u + v ) + w

This completes the analytic proof of (b). Figure 1.2.6 provides the geometric proof.

u

v

w u + v

v + w

u + ( v + w ) = ( u + v ) + w

Figure 1.2.6 Associative Law for vector addition

(c) We already discussed this on p.10. (d) We already discussed this on p.10. (e) We will prove this for a vector v = ( v 1 , v 2 , v 3 ) in R^3 (the proof for R^2 is similar):

k ( l v ) = k ( lv 1 , lv 2 , lv 3 ) by Theorem 1.4(a) = ( klv 1 , klv 2 , klv 3 ) by Theorem 1.4(a) = ( kl )( v 1 , v 2 , v 3 ) by Theorem 1.4(a) = ( kl ) v

(f) and (g): Left as exercises for the reader. QED

A unit vector is a vector with magnitude 1. Notice that for any nonzero vector v , the vector (^) ‖ vv ‖ is a unit vector which points in the same direction as v , since (^) ‖^1 v ‖ > 0 and

∥∥ v ‖ v ‖

‖ v ‖ ‖ v ‖ =^ 1. Dividing a nonzero vector^ v^ by^ ‖ v ‖^ is often called^ normalizing^ v. There are specific unit vectors which we will often use, called the basis vectors : i = (1, 0 , 0), j = (0, 1 , 0), and k = (0, 0 , 1) in R^3 ; i = (1, 0) and j = (0, 1) in R^2. These are useful for several reasons: they are mutually perpendicular, since they lie on distinct coordinate axes; they are all unit vectors: ‖ i ‖ = ‖ j ‖ = ‖ k ‖ = 1; every vector can be written as a unique scalar combination of the basis vectors: v = ( a , b ) = a i + b j in R^2 , v = ( a , b , c ) = a i + b j + c k in R^3. See Figure 1.2.7.