Math 23 Module PDF

Preview

Summary

University of the Philippines DilimanMATHEMATICS 23Elementary Analysis IIICourse ModuleInstitute of Mathematicsiv➞2019 by the Institute of Mathematics, University of the Philippines Diliman. All rights reserved. No part of this document may be distributed in any way, shape, or form without prior wri...

Description

University of the Philippines Diliman

MATHEMATICS 23 Elementary Analysis III Course Module

Institute of Mathematics

MATHEMATICS 23 Elementary Analysis III Course Module

Institute of Mathematics University of the Philippines Diliman

iv 2019 by the Institute of Mathematics, University of the Philippines Diliman. All rights reserved. No part of this document may be distributed in any way, shape, or form without prior written permission from the Institute of Mathematics, University of the Philippines Diliman.

Mathematics 23 Module Committee: Julius Caesar C. Agapito Richell O. Celeste Aaron J. Ramos Louie John D. Vallejo

Special Contributors: Jeanine Concepcion H. Arias-Ona Mark Philip F. Ona Renier G. Mendoza Victoria May Paguio Mendoza Rolando B. Perez III

Reviewer: Marrick C. Neri

Contents 1 Functions of More Than One Variable 1 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.4 1.5 1.6 1.7

Differentiability, Differentials and Local Linear Approximation . . . . . . . . . . . . 22 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Directional Derivatives and Gradients . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Relative Extrema of Functions of Two Variables . . . . . . . . . . . . . . . . . . . . 44

1.8 1.9

Absolute Extrema of Functions of More than One Variable . . . . . . . . . . . . . . 48 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

2 Multiple Integration 63 2.1 Double Integrals over Rectangular Regions . . . . . . . . . . . . . . . . . . . . . . . . 63 2.2 2.3 2.4 2.5

Double Integrals Over General regions . . . . . . . . . . . . . . . . . . . . . . . . . . Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.6 2.7 2.8

Triple Integrals in Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 101 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 106 Applications of Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

3 Vector Calculus

67 77 83 93

117

3.1 3.2 3.3 3.4

Scalar and Vector Fields . . . Divergence and Curl . . . . . Conservative Vector Fields . . Line Integrals of Scalar Fields

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

117 121 124 130

3.5 3.6 3.7 3.8

Line Integrals of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Fundamental Theorem of Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 141 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 Surface Integrals of Scalar Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

3.9 Surface Integrals of Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 3.10 Stokes’ Theorem and Gauss’s Divergence Theorem . . . . . . . . . . . . . . . . . . . 166 v

Chapter 1

Functions of More Than One Variable 1.1

Introduction

Recall that a function f of a single variable x is a correspondence from a set D ⊆ R to R such that each element x ∈ D is associated a unique real number f (x). Definition. Let D ⊆ R2 . A function f of two variables x and y is a correspondence from the set D to R such that each element (x, y) ∈ D is associated to a unique real number f (x, y). Example 1.1.1. Suppose a particle, with mass m, moves with acceleration a. Newton’s second law of motion tells us that the net force F acting upon the particle can be computed using m and a. Hence, F is a function of m and a. In particular, the formula is F = f (m, a) = ma.



The idea of functions of two variables can be extended further to functions depending on three or more variables. Definition. Let D ⊆ Rn , where Rn = {(x1 , x2 , . . . , xn ) : x1 , x2 , . . . , xn ∈ R}. A function f of n variables x1 , x2 , . . . , xn−1 and xn , is a correspondence from the set D to R such that each point (x1 , x2 , . . . , xn ) ∈ D is associated to a unique real number f (x1 , x2 , . . . , xn ). The set D is called the domain of f . Definition. Let D1 , D2 ⊆ Rn . Two functions f : D1 → R and g : D2 → R are equal if and only if

D1 = D2 and for all (x1 , . . . , xn ) ∈ D1 , f (x1 , . . . , xn ) = g(x1 , . . . , xn ).

Remark. If no restrictions are stated explicitly, then it is understood that the domain comprises all points (x1 , . . . , xn ) for which f (x1 , . . . , xn ) is a unique real number. Example 1.1.2. If a rectangular box has length l, width w and height h units, then its volume V can be computed using these three variables. We say that V is a function of l, w and h, and write this as V = f (l, w, h). In particular, the formula is given by V = f (l, w, h) = lwh. 1

2

CHAPTER 1. FUNCTIONS OF MORE THAN ONE VARIABLE A physical restriction to these variables is that they should be positive. Hence, the domain is {(l, w, h) ∈ R3 : l > 0, w > 0, h > 0}.

Example 1.1.3. Let f (x, y) = x2 +



√ 3 xy. Find f (2, −4), f (1, 0), f (t, t2 ) and f (2y2 , 4y).

Solution: By substitution, we get

f (2, −4) = 22 + f (1, 0) = 12 +

p 3

p 3

f (t, t2 ) = t2 +

2(−4) = 4 − 2 = 2

1(0) = 1

Note that the domain of f is the entire xy-plane.

p 3

t(t2 ) = t2 + t

 2 p f (2y2 , 4y) = 2y2 + 3 2y2 (4y) = 4y4 +2y

1 . x2 − y Solution: For the expression to be a real number, the radicand in the denominator must be positive. That is, x2 > y. Hence, the domain is Example 1.1.4. Identify and sketch the domain of f (x, y) = p {(x, y) ∈ R2 : y < x2 }. The parabola y = x2 divides the plane into two, all points on or above it, and all points below it. The point (0, −1), which is a point below the parabola, satisfies y < x2 . Hence, the domain

includes all points below the parabola. The parabola being a dashed curve indicates that points on it are not part of the domain.

Figure 1.1 

Example 1.1.5. Identify and sketch the domain of f (x, y) = sin−1 (x − 1). Solution: The inverse sine function is defined only for values in the interval [−1, 1]. Thus, −1 6 x − 1 6 1, or that 0 6 x 6 2. Hence, the domain is

{(x, y) ∈ R2 : 0 6 x 6 2}.

3

1.1. INTRODUCTION

Graphically, the domain consists of points between the lines x = 0 and



x = 2, including these lines.

Figure 1.2: Domain of f (x, y) = sin−1 (x − 1) Example 1.1.6. Identify the domain of g(x, y, z) = ln(1 − x2 − y2 − z 2 ). Solution: The natural logarithm is defined only for positive values. Thus, 1 − x2 − y2 − z 2 > 0 or that x2 + y2 + z 2 < 1. Hence, the domain is {(x, y, z) ∈ R3 : x2 + y2 + z 2 < 1}. Graphically, the domain is the set of points inside the unit sphere centered at the origin, excluding the sphere itself.

Graphs of Functions of Two Variables Suppose f is a function of two variables x and y. The graph of f is the set of all points (x, y, z ) in the three-dimensional space such that (x, y) ∈ dom f and z = f (x, y). Example 1.1.7. Sketch the graph of the following functions. 1. f (x, y) = 2 − 2x − y. Solution: The graph of f is the graph of the equation z = 2 − 2x − y, which is a plane with normal vector h2, 1, 1i.

Figure 1.3: Graph of f (x, y) = 2 − 2x − y

4

CHAPTER 1. FUNCTIONS OF MORE THAN ONE VARIABLE 2. g(x, y) =

p

x2 + y 2 .

Solution: The graph of g is the graph of z =

p

x2 + y2 . Squaring both sides, we have

z 2 = x2 + y 2 , which represents a circular cone. But the formula for g imposes that z ≥ 0. Thus, the graph of g is the upper nappe of the cone.

Figure 1.4: Graph of g(x, y) =

p

x2 + y 2 

Level Curves Let f be a function of two variables x and y, say f (x, y) = 4x2 + y2 . Let us cut across the graph of f by a horizontal plane z = k, for several values of k ∈ R. k = 4 : 4x2 + y2 = 4 k = 3 : 4x2 + y2 = 3 k = 2 : 4x2 + y2 = 2 k = 1 : 4x2 + y2 = 1 k = 0 : {(0, 0)} k 0, there is a corresponding small number δ > 0 such that |f (x, y) − L| < ε whenever (x, y) ∈ D and 0 <

p

(x − a)2 + (y − b)2 < δ.

Remark. In the definition, by “D contains points arbitrarily close to (a, b)”, we mean for any p δ > 0, there is at least one point (x, y) ∈ D such that 0 < (x − a)2 + (y − b)2 < δ .

Example 1.2.1. Prove that

lim

(3x + 2y) = 1.

(x,y )→(1,−1)

Solution: Let f (x, y) = 3x + 2y, and L = 1. For any small number ε > 0 we choose, we want to find a small number δ > 0, such that |f (x, y) − L| = |3x + 2y − 1| < ε whenever the distance between (x, y) and (1, −1) is less than δ . Note that |3x + 2y − 1| = |3(x − 1) + 2(y + 1)| 6 3|x − 1| + 2|y + 1|, and we want this to be less

than ε. Also, notice that

|x − 1| = Similarly,

p

(x − 1)2 ≤

p

(x − 1)2 + (y + 1)2 < δ.

|y + 1| < δ. Thus, |3x + 2y − 1| ≤ 3|x − 1| + 2|y + 1| < 3δ + 2δ = 5δ. ε Hence, by taking δ = , we have 5 ε |3x + 2y − 1| < 5δ = 5 = ε. 5 Therefore,

lim

(x,y)→(1,−1)

(3x + 2y) = 1.

Remark. If the limit of f (x, y) as (x, y ) approaches (a, b) exists, then that limit is unique. Recall that on the real number line, one can approach a number from two directions, from the right and from the left. On the xy-plane, there are infinitely many ways one can approach a point (a, b). Hence, we extend the notion of one-sided limits for functions of one variable. Instead, we

10

CHAPTER 1. FUNCTIONS OF MORE THAN ONE VARIABLE

Figure 1.12: The point (a, b) can be approached in infinitely many ways. shall have the limit of the function along a curve, as (x, y) approaches a point. Recall that for functions of one variable, if the limit from the left is not equal to the limit from the right, then the limit does not exist. The case for functions of two variables is analogous. Definition. Let C be a smooth curve that passes through the point (x0 , y0 ). We define the limit of f (x, y) as (x, y) approaches (x0 , y0 ) along C as follows: If C has equation y = g(x), then lim

f (x, y) = lim f (x, g (x)). x→x0

(x,y)→(x0 ,y 0 ) along C

If C has equation x = g(y), then lim

(x,y )→(x0 ,y 0 ) along C

f (x, y) = lim f (g(y), y ). y→y0

Remark. If the limits of f (x, y) as (x, y) approaches a point along two different curves are not equal, or if the limit along a curve does not exist (provided that the curve lies in the domain of f ), then the limit of f does not exist. Example 1.2.2. Determine whether

lim

(x,y)→(0,0) x2

1 exists or not. + y2

Solution: Consider the curve C to be the line y = x. Along C, we have 1 1 1 = 2 = . x2 + y 2 x + x2 2x2 Thus, lim

2 (x,y )→(0, 0) x along C

and the limit does not exist.

1 1 = +∞, = lim 2 x→0 2x2 +y

11

1.2. LIMITS AND CONTINUITY Example 1.2.3. Show that

x2 − y 2 does not exist. (x,y)→(0,0) x2 + y 2 lim

Solution: Let us consider the limit of f (x, y) =

x2 − y 2 along the x-axis (y = 0), which passes x2 + y 2

through the origin. The limit is x2 − y 2 x2 − 0 2 x2 = lim = lim = 1. (x,y )→(0, 0)x2 + y 2 x→0 x2 + 02 x→0 x2 lim

along x-axis

Also, let us consider the limit of f (x, y) along the y-axis (x = 0). The limit is x2 − y 2 02 − y 2 −y2 = lim 2 = lim 2 = −1. 2 2 2 (x,y )→(0, 0)x + y y→0 y y→0 0 + y lim

along y-axis

Since the limits along different curves are not equal, then lim

(x,y)→(0,0)

Example 1.2.4. Show that

x2 − y 2 x2 + y 2

does not exist.

xy 2 does not exist. (x,y)→(0,0) x2 + y 4 lim

Solution: Let C1 be the line y = x. Then lim

(x,y)→(0, 0) along C1

x xy 2 x(x)2 = lim = 0. = lim 2 4 2 4 x→0 x + x x→0 1 + x2 x +y

Let C2 be the parabola x = y2 . Then xy 2 y2 (y2 ) y4 1 = lim = lim = . 2 4 y→0 y 4 + y 4 y→0 2y 4 2 (x,y )→(0, 0) x + y lim

along C2

Since the limits along different curves are distinct,

Example 1.2.5. Determine

lim

lim

xy 2 does not exist. + y4

(x,y)→(0,0) x2

3x2 y . + y2

(x,y)→(0,0) x2

Solution: Let C1 be a non-vertical line through the origin. That is, C1 : y = mx, for some m ∈ R. Then 3mx 3x2 y 3x2 (mx) lim = lim = 0. = lim 2 2 2 2 2 (x,y )→(0, 0) x + y x→0 x + m x x→0 1 + m2 along

C1

Along the curves C2 : y = x2 , C3 : x = y2 , C4 : y = x3 , C5 : x = y3 , it can easily be verified that the limits of the function as (x, y) → (0, 0) are also zero. The above computations seem to indicate that the limit is zero. However, this is not enough to say that the limit is zero. We need to prove that it is so by definition.

12

CHAPTER 1. FUNCTIONS OF MORE THAN ONE VARIABLE Let ε > 0. We find a δ > 0 such that      3x2 y   3x2 y       x2 + y 2 − 0  = x2 + y 2  < ε

whenever

0<

p

x2 + y2 < δ.

  2  3x2 y   = 3x |y| and that x2 ≤ x2 + y2 . Thus  Note that  2 x + y 2  x2 + y 2

p p 3x2 |y| 3x2 |y| ≤ = 3|y| = 3 y2 ≤ 3 x2 + y2 2 2 2 x +y x

ε which we want to be less than ε. Thus, if we take δ = , then 3   ε p   3x2 y 2 2    x2 + y2 − 0  ≤ 3 x + y < 3δ = 3 3 = ε. Therefore, the limit is indeed zero.

Continuity Definition. Let D ⊆ R2 and (a, b) ∈ D. A function f : D → R is said to be continuous at (a, b) if

1. f (a, b) is defined, 2. 3.

lim

f (x, y) exists, and

lim

f (x, y) = f (a, b).

(x,y )→(a,b) (x,y )→(a,b)

Remarks. 1. The function f is said to be continuous on the set D ⊆ R2 if it is continuous at all points on D. 2. If f is continuous at all points on the xy-plane, we say that f is continuous everywhere. 3. If a function is not continuous at a point, then we say it is discontinuous at that point. Moreover, if the limit exists, the discontinuity is removable. Otherwise, the discontinuity is essential. The following theorem illustrates ways of producing continuous functions from known continuous functions.

1.2. LIMITS AND CONTINUITY

13

Theorem 1.2.6. Let h : R2 → R and g : R → R. Fix (a, b) ∈ R2 . 1. A sum, difference or product of continuous functions is continuous. 2. A quotient of continuous functions is continuous, except at the points where the denominator is zero. 3. If h is continuous at (a, b) and g is continuous at h(a, b), then f (x, y) = g(h(x, y)) is continuous at (a, b). Example 1.2.7. Use the above theorem to show that the following functions are continuous everywhere. 1. f (x, y) = xexy + y 2/3 2. g(x, y) = cos(x3 + y3 ) − |x3 + y3 | 3. h(x, y) =

xy 1 + x2 + y 2

Solution: The polynomials x and xy are continuous. Also, ex and y2/3 are continuous. By part (3.) of the theorem, the composition exy is continuous. By part (1.), xe xy is continuous since it is a product of continuous functions. Therefore, by part (1.) again, f (x, y) is continuous since it is a sum of continuous functions. Likewise, g is continuous since it is a difference and composition of continuous functions cos x, x + y3 , and |x|. Finally, h is continuous being a quotient of continuous functions 1 +x2 + y2 , which is never zero, and xy . 3

Example 1.2.8. Evaluate the following limits. 1. 2.

lim

(x,y)→(−1,2)

lim

(x,y)→(2,2)

xy x2 + y 2 x3 y 2 1 − xy

Solution: xy is continuous at (−1, 2), being a quotient of continuous + y2 2 functions at (−1, 2). Since the denominator is non-zero, the limit of f (x, y) is f (−1, 2) = − . 5

1. The function f (x, y) =

x2

x3 y 2 is continuous at (2, 2), being a quotient of continuous functions 1 − xy 32 at (2, 2). Since the denominator is non-zero, the limit is f (2, 2) = − . 3

2. Similarly, f (x, y) =

14

CHAPTER 1. FUNCTIONS OF MORE THAN ONE VARIABLE

Example 1.2.9. Analyze the continuity at the origin of x2 − xy f (x, y) = √ √ . x− y Solution: Its domain is the set of points in the first quadrant, including the positive x- and y-axes, excluding the line y = x. For all points in the domain, we obtain an equivalent fraction with a limit that’s easily computed. That is, if y 6= x, we have √ √ (x2 − xy)( x + y) √ √ √ √ ( x − y)( x + y) √ √ x(x − y)( x + y) = x−y √ √ = x( x + y).

x2 − xy √ √ x− y

=

Thus, √ x2 − xy √ √ x( x + y) = 0. lim √ = x − y (x,y)→(0,0) (x,y )→(0,0) lim

x2 − xy Moreover, we say that f (x, y) = √ √ has a removable discontinuity at (0, 0). x− y Example 1.2.10. Find

Solution: Consider

lim

(x,y )→(1,0)

sin−1



 x2 + xy − x . x2 + 2xy + y2 − 1

x2 + xy − x . x2 + 2xy + y2 − 1

As (x, y) → (1, 0), the limit is of the form

0 . 0

However, if y 6= 1 − x, we obtain the equivalent fraction x2

x2 + xy − x + 2xy + y2 − 1

=

x(x + y − 1) x = (x + y − 1)(x + y + 1) x+y+1

which is continuous at (1, 0). Thus, as (x, y) → (1, 0), 1 x → . x+y+1 2 Since the inverse sine is continuous at lim

(x,y)→(1,0)

−1

sin



1 , 2

then

x2 + xy − x x2 + 2xy + y2 − 1



−1

= sin

  1 π = . 6 2

15

1.3. PARTIAL DERIVATIVES EXERCISES 1.2 I. Show that the following limits do not exist.

1.

lim

2xy + y2

4.

(x,y )→(0,0) x2

2

2. 3.

lim

3x y x4 + y 2

lim

2xy 2 5x2 + y4

(x,y)→(0,0)

(x,y)→(0,0)

lim

(x,y)→(0,0)

3xy p 3

x9 − 2y3

ln(1 − x2 ) + y4 (x,y)→(0,0) x2 + y 4 sin(x − y) 6. lim (x,y)→(1,1) 1 − xy

5.

lim

II. Determine whether the discontinuity of the following functions at the point P is removable or essential.