Calculus I, II, III PDF

CALCULUS Table of Contents Calculus I, First Semester Chapter 1. Rates of Change, Tangent Lines and Differentiation 1 1.1. Newton’s Calculus 1 1.2. Liebniz’ Calculus of Differentials...

CALCULUS Table of Contents Calculus I, First Semester Chapter 1. Rates of Change, Tangent Lines and Differentiation 1 1.1. Newton’s Calculus 1 1.2. Liebniz’ Calculus of Differentials 13 1.3. The Chain Rule 14 1.4. Trigonometric Functions 16 1.5. Implicit Differentiation and Related Rates 19 Chapter 2. Theoretical Considerations 24 2.1. Limit Operations 24 2.2. Limits at Infinity 29 2.3. Some Basic Theorems 34 2.4. l’Hôpitals Rule 36 Chapter 3. Extrema, Concavity and Graphs 40 3.1. Monotonicity and the First Derivative 40 3.2. Optimization 42 3.3. Concavity and the Second Derivative 45 3.4. Graphing Functions 47 Rational Functions 47 Other Sketches 51 Chapter 4. Integration: A Differential Equations Approach 55 4.1. Antiderivatives 55 4.2. Area 59 4.3 Separation of Variables 64 4.4. The Exponential Function 69 4.5 The Logarithm 75 4.6 Growth and Decay 79 Inhibited Growth 80 Chapter 5. Integration: The Accumulation Method 86 5.1. The Definite Integral he Fundamental Theorem of the Calculus 86 5.2. Summation and the Definition of Area 92 5.3. Volume 95 Volumes of Revolution 99 5.4. Arc Length 104 5.5. Work 109 5.6 Mass and Moments 113 Centroids 118 i Calculus II, Second Semester Chapter 6. Transcendental Functions 122 6.1. Inverse Functions 122 6.2. The Inverse Trigonometric Functions 127 6.3 First Order Differential Equations 130 Chapter 7. Techniques of Integration 136 7.1. Substitution 136 7.2. Integration by Parts 139 7.3. Partial Fractions 143 7.4. Trigonometric Methods 149 Chapter 8. Indeterminate Forms and Improper Integrals 153 8.1. L’Hôpital’s Rule 153 8.2 Other Indeterminate Forms 156 8.3 Improper Integrals: Infinite Intervals 158 8.4 Improper Integrals: Finite Asymptotes 161 Chapter 9. Sequences and Series 164 9.1. Sequences 164 9.2. Series 134 9.3. Tests for Convergence 175 9.4. Power Series 180 9.5. Taylor Series 184 Chapter 10 Numerical Methods 191 10.1. Taylor Approximation 191 10.2. Newton’s Method 195 10.3. Numerical Integration 198 Chapter 11. Conics and Polar Coordinates 203 11.1. Quadratic Relations 203 11.2. Eccentricity and Foci 210 11.3. String and Optical Properties of the Conics 215 11.4. Polar Coordinates 219 11.5. Calculus in Polar Corrdinates 225 Chapter 12. Second Order Linear Differential Equations 228 12.1. Homogeneous Equations 228 12.2. Behavior of the Solutions 233 12.3. Applications 235 12.4. The Inhomogeneous equation 238 ii Calculus III, Third Semester Chapter 13. Vector Algebra 241 13.1 Basic Concepts 241 13. 2. Vectors in the Plane 243 13.3. Vectors in Space 253 13.4. Lines and Planes in Space 259 Chapter 14. Particles in Motion; Kepler’s Laws 265 14.1 Vector Functions 265 14.2 Planar Particle Motion 269 14.3 Particle Motion in Space 273 14.4 Derivation of Kepler’s Laws of Planetary Motion from Newton’s Laws 276 Chapter 15. Coordinates and Surfaces 281 15.1 Change of Coordinates in Two Dimensions 281 15.2 Special Coordinate Systems 287 15.3 Surfaces; Graphs and Level Curves 292 15.4 Cylinders and Surfaces of Revolution 295 15.5. Quadric Surfaces 296 Chapter 16. Differentiable Functions of Several Variables 302 16.1 The Differential and Partial Derivatives 302 16.2 Gradients and Vector Methods 309 16. 3 Theoretical Considerations 315 16.4 Optimization 317 The Method of Lagrange Multipliers 320 Chapter 17. Multiple Integration 324 17.1 Integration on Planar Regions 324 17.2. Applications 331 17.3. Theoretical Considerations 334 17.4. Integration in Other Coordinates 337 17.5 Triple Interals 347 Integration in Other Coordinates 350 Chapter 18. Vector Calculus 354 18.1 Vector Fields 354 18.2 Line Integrals and Work 360 18.3 Independence of Path 364 18.4. Green’s Theorem in the Plane 367 18.5 Stokes’ and Gauss’ Theorems in Three Dimensions 370 iii Preface As I neared the end of a second decade of teaching Calculus at the University of Utah, I became aware of a student type that persisted in all my classes, comprising between 10 and 20 percent of the class. These students came to class irregularly, often just to ask a question about one of the more demanding problems. They took all the exams, and, almost uniformly ended up with grades in the top quartile. Looking at their transcripts, and talking with several of them, I discovered that they were among the better prepared of the Calculus students, having taken Calculus in high school, either in an AP class, or from another country. Now, the University of Utah has an AP Calculus course for such students, but as that course emphasizes theory, these students preferred to be in the regular engineering sequence. As these students were well grounded in Algebra, and had already seen some of the basics of Calculus, I concluded that what was appropriate for them was a course in Calculus that emphasized a deep intuitive understanding of Calculus and problems sets that depended on, and extended that understanding. So, with the collusion of my chair, I started an intensive summer calculus course: all three semesters in ten weeks. After all, if this population was ten percent of the total Calculus cohort, and if half of those took the summer course, that would provide a class of about 50 students enough to make the experiment economically feasible. We met in this class for five hours a day, five days a week. For about half the time the instructor developed the basic ideas, illustrating them through problems, and for the rest of the time, the students do their homework in the presence of graduate assistants. This worked, and that course continues still with about the same numbers of students. I taught that course for three years, and then turned to developing an online course, based on the same approach. The first few semesters were difficult: there was not a good fit between the text and the online problem sets I had created. I spent many hours responding to email inquiries, started to write and post supplementary notes, and decided that it was a good idea to keep past years’ problems with solutions posted. As new email inquiries and responses were integrated into the supplementary notes, it occurred to me that I had almost a complete first draft of a text on Calculus. I spent about a semester organizing the material and completing it to a full text that still exists as a supplementary text for the online course. That is the genesis of this text. As a result, it is thin on drill exercises, informal and intuitive on theoretical issues, approaches the ideas of the subject in the context of the problems they solve, and is rich in examples that illustrate those ideas. As such, it has served well that class of students with sufficient technical competence to follow the ideas in an argument. Conceptual Apporach of the Text There is no chapter 0: survey of algebra, trigonometry and pre-calculus. It is assumed that students have sufficient grasp of the concept of function to be able to get right into that which the Calculus is about. Ideas and techniques from the pre-calculus are reviewed in context as the need for them arises. The approach in Chapter 1 is intuitive and informal: the fundamental issue is to see how un- derstanding of rules of change leads to qualitative comprehension of processes and predictions of future behavior. The first two sections start with the Newton and Liebniz approaches to Differen- tial Calculus. The Newtonian approach is presented as one focusing on rates of change of functions of a given independent variable (usually time), while that of Liebniz deals with variables and how iv they change with respect to each other. In this way, the two threads of derivatives (Newton) and differentials (Liebniz) are introduced and used throughout the text to develop the various ways of looking at a particular process. I’ve always been of two minds concerning theoretical issues. I feel that they are too deep for the first calculus course, but find it difficult to be vague and intuitive about them. So, the approach I’ve adopted in this text is that of Newton: position and velocity are measurable attributes of moving bodies, and the limit idea of calculus is the tool for solving problems about movement. Similarly, area is a measurable attribute of planar figures, and the idea of accumulation of the Calculus is the tool for calculating area from the algebraic expressions delimiting the figure. The problems of existence of limits and area are thus avoided. On the other hand, there are real issues in relating velocity with change in position, and in defining area, and I can’t allow myself to sneak through the calculus without pointing it out. These discussions are collected in sections entitled something like ”theoretical considerations.” As a result this text has no preliminary section on limits. I feel that this would be misplaced: students get the erroneous impression that a limit is calculated by substitution if the function is given by a formula, and otherwise one should look at the graph. So, at the beginning, the calculus of polynomials is developed, and the calculation of the derivative is done algebraically. Derivation of the trigonometric functions is presented using Pascal’s idea. Here the issue of existence of limits is joined: we must know that lim sin x/x = 1 as x approaches zero. This is discussed in chapter 2, in which a wide variety of questions of limits is discussed, including l’Hôpital’s rule, which is usually postponed to second semester. Similarly, integration is introduced as the problem of finding functions with given derivatives using the idea of differentials. It is essential that students understand that it is a differential, not a function, that is integrated. By showing that the differential of the area underneath a curve y = f (x) is f (x)dx, we see that that area is given by integration. The technigue of integration by substitution is a direct consequence of the chain rule. From there, we move directly into solving separable differential equations (of the type f (x)dx = g(y)dy) and from there to the introduction of the exponential function as the solution of dy/y = rdx. Then, in the subsequent chapter the issue of existence of area is taken up; we turn to the integral as a limit of an accumulation process and the Fundamental Theorem of the Calculus is now seen as the binding together of the Newton and Liebniz approaches. New techniques and ideas are introduced (where ever possible) in the context of a problem to be solved. Thus the chapter on sequences and series is viewed as developing a technique for approximating values of transcendental functions. Conics are introduced as a tool for understanding quadratic relations, and this leads directly into the study of second order equations. The Chapter on Vector Calculus is based on intuition provided by fluid flows. The motivation is that it is better to emphasize the understanding of the physical content of the fundamental theorems of vector calculus, rather than providing proofs of these theorems. After all, the correcrt proofs of these theorems reduce them to the one-variable fundamental theorem of the calculus by judicious choice of coordinates. The hard part, best deferred to the course in Advanced Calculus, is to show that the statement of the theorems is independent of the coordinates in which they are calculated. v Outline of the Text First Semester In Chapter 1 the basics of the differential calculus are introduced, mostly in the context of poly- nomial functions and relations. In section four the derivatives of the trigonometric functions are found using the differential triangle of Pascal. Chapter 2 discusses the concept of limit, techniques for calculating limits, including l’Hôpital’s rule. We show that the sine function is differentiable (a fact postponed from chapter 1). Limits at infinity and asymptotes are discussed. The point of the last section is the affirmation that a function with derivative 0 everywhere in an interval is constant. This is always a difficult passage in the first calculus course, for the students see this as the tautology that a function with no change throughout an interval ends up at the same value with which it started. So, the instructor has to first convince the student that there is a real issue here. In class, I’ll do that by exhibiting the Cantor function, which is almost always constant, but somehow gets from 0 to 1. Then the instructor must confess that the issue will not be joined, and instead this fundamental fact will be shown to follow from statements (intermediate value, existence of maxima for continuous functions) that, for some reason, are more intuitively obvious. I leave the strategy for doing this to the instructor, and instead just lay down the minimal exposition. Chapter 3 covers the geometric significance of the first and second derivatives, optimization and graph sketching, Chapter 4 introduces integration as the process reverse to differentiation, and extends this to separable differential equations. This leads to the introduction of the exponential function as the solution of dy/y = rdx. I do not feel that it is necessary to discuss the general concept of inverse functions to get from the exponential to the logarithm; the logarithm is simply what we use to find a if we know ea. Chapter 5 turns to Liebniz’ concept of integration as a method of accumulation, and shows (through the Fundamental theorem) that the Calculi of Newton and Liebniz are the same subject. This is followed by the application of integration to a variety of accumulation processes. Covering these first 5 chapters in one semester is a tall order. It may be that the course will end with the fundamental theorem, and that the second semester starts with the applications of integration. Alternatively, some sections of these first five chapters may be omitted. Second Semester We are now in a position to abstractly discuss inverse functions and the fact that a function with nonzero derivative in an interval has an inverse, leading to the introduction of many useful tran- scendental functions. This, together with the solution of general first order differential equations, is covered in Chapter 6. Chapter 7 covers the techniques of integration necessary to make use of a table of integrals: substi- tution, integration by parts and partial fractions. It is the intent of this text that from this point on, students will use a table of integrals. vi Chapter 8 begins with a brief review of l’Hôpital’s rule, as some instructors may prefer to cover this in the second semester. Then Improper Integrals are covered as this material will be necessary for the next chapter. We begin Chapter 9 (Sequences and Series) with a more formal discussion of the Principle of Mathematical Induction (although it has already been used informally in several places in the text), in connection with the definition of sequences by recursion. As the point of this chapter is to develop ways to approximate values of transcendental functions, the focus is to get to Taylor series quickly. It also works to cover section 10.1 (on the Taylor formula with estimate) before Chapter 9, to better motivate the discussion of series. The important thing is to emphasize that a calculation is not an approximation without some estimate of the error. Chapter 10 is a brief introduction to the simplest numerical methods; enough to give an idea of how the basic computer algorithms work. Chapter 11 covers materials on the conics and polar coordinates. Although this material is tradi- tionally pre-calculus, it is necessary to review it before launching into mulit-variate calculus. The string properties of the conics are derived, and their optical properties are derived as differential versions of the string properties. Chapter 12 covers the usual material on second order differential equations and basic harmon- ics. Many students will have already covered this in the physics classes, and in many cases this is done at the beginning of the course on differential equations. Thus, this chapter may be omitted, particularly if Calculus is three quarters rather than three semesters. Third Semester The introduction to Vector Algebra (chapter 13) is motivated by geometry, the point being to characterize linear subspaces in two and three dimensions in a variety of ways. One of the central themes of this text is the study of particles in motion; we discuss the calculus of vector-valued functions of a single variable (Chapter 14)in this context. Sections 2 and three cover the differential geometry of curves in two and three dimensions, and section 4 provides a derivation of Kepler’s Laws of Planetary Motion from Newton’s Laws of Motion. Chapter 15 (Coordinates and Surfaces) covers the basic geometric tools of the several variable calculus: change of coordinates in two and three variables, surfaces, and in particular, the normal forms of quadrics. Chapters 16 and 17 cover differentiation and integration in several variables. The differential of a function is introduced as the linear function which best fits the given function near a base point. Applications of integrals are discussed in detail in two dimensions. Chapter 18 (Vector Calculus) is motivated as the study of fluids in motion, paralleling the theme of particles in motion for the single variable calculus. The basic forms (curl and divergence) are introduced in this context. This leads to an interpretation that makes the fundamental theorems (Green, Stokes, Gauss) plausible. Here the emphasis is on the examples which show how to use and apply these theorems. vii  CALCULUS I, First Semester I. Rate of Change, Tangent Line and Differentiation 1.1 Newton’s Calculus Early in his career, Isaac Newton wrote, but did not publish, a paper referred to as the tract of october 1666. This was his sole work purely on mathematics, and contained the fundamental ideas and techniques of the calculus. While writing (1684-87) the Principia Mathematica, the fundamental exposition of his mathematical physics of “the system of the world”, he reworked and expanded these ideas and included them as part of this treatise. Newton’s central conception was that of objects in motion. To Newton motion is described by the position and velocity of the particle relative to a fixed coordinate system, as functions of time. These then are the fundamental variables: x, y, z, etc., and their velocities are denoted by ẋ, ẏ, ż, etc. Now, velocity is a measure of the rate of change of position (and acceleration, denoted ẍ, etc., is a rate of change of velocity), and what was needed was a means to express this relationship, and a process of deriving relations among the various velocities and accelerations from given relations among the variables. This is the Calculus of Newton. Calculus, as it is presented today starts in the context of two variables, or measurable quantities, x, y, which are related in the sense that values of one of the variables determine values of the other. A function y = f (x) is a rule which specifies this relation between the input or independent variable x and the output or dependent variable y. This may be given by a formula, a table, or a computer algorithm; in fact, any set of rules which uniquely determine outputs for given inputs. The set of allowable values for the input variable is called the domain of the function, and the set of outputs, the range. In our context both the domain and the range are sets of real numbers. An interval is the set of all real numbers between specified real numbers c and d. This is denoted by (1.1) (c, d) = {x : c < x < d} if neither endpoint is included, and by (1.2) [c, d] = {x : c ≤ x ≤ d} if both are included. We shall say “I is an interval about a” to mean that a is between the endpoints of the interval I. Now, suppose that y = f (x) is a function defined for all x in an interval I = (c, d). The graph of f is the set of all points (x, y) in the plane where x is in the interval (c, d) and y = f (x). Calculus studies the behavior of y as a function of x in terms of the way y changes as x changes. For x0 , x1 points in the interval, and y0 = f (x0 ), y1 = f (x1 ), the ratio ∆y y 1 − y0 (1.3) = ∆x x1 − x0 is called the average rate of change of y with respect to x in the interval between x0 and x1. This is the ratio of the change in y (denoted ∆y) with the change in x (denoted ∆x). 1 Linear Functions If the ratio (1.3) is the same for all points (x0 , y0 ), (x1 , y1 ) on the graph, we say that y = f (x) is a linear function of x.This is because that condition is equivalent to saying that the graph of y = f (x) is a straight line (which is easy to see using similar triangles; see figure (1.1)). Figure 1.1 x3 y3
 x1 y1
 y3 y2  x2 y2
 x3  x2 y1 y0 x0 y0 
 x1  x0 y1 y0 y3 y2   x1  x0  x3  x2 For y = f (x) a linear function, the ratio (1.3) is called the slope of the line, denoted m. Then another point (x, y) is on the line if and only if the calculation of (1.3) gives the slope. Thus, the condition for (x, y) to be on the line is y − y0 (1.4) =m or y − y0 = m(x − x0 ). x − x0 This is called the point-slope equation of the line. If we wish to find the equation of the line through two points (x0 , y0 ) and (x1 , y1 ), we use those points to find the slope and then use the above equation. Thus, the condition for (x, y) to be on the line through these points is y − y0 y1 − y 0 (1.5) =. x − x0 x1 − x0 This is the two-point equation for the line. Example 1.1. Find the equation of the line through the points (2,-1), (3,7). Then find the line through (6,-1) of the same slope. Using (1.5) with (x0 , y0 ) = (2, −1) and (x1 , y1 ) = (3, 7), we have that (x, y) is on the line when y − (−1) 7 − (−1) y+1 8 = or = , x−2 3−2 x−2 1 giving us the equation y = 8x − 17. This line has slope m = 8. Thus the line through (6,-1) of the same slope has the equation y − (−1) = 8 or y = 8x − 49. x−6 2 Example 1.2. Is P (5,12) on the line joining Q(2,7) and R(8,15)? The slope of the line through Q and R is (15 − 7)/(8 − 2) = 4/3. The slope of the line through P and Q is (12 − 7)/(5 − 2) = 5/3. Since these two lines do not have the same slope, they cannot be the same line. Thus P is not on the line through Q and R. Here are some facts about lines which will be useful when studying more general curves. a. If L is a line of slope m, then (1.6) m = tan θ where θ is the angle that L makes with a horizontal line. If the line is vertical, then L has infinite slope. Suppose we are given two lines: L1 of slope m1 , and L2 of slope m2. Then b. L1 and L2 are parallel if and only if m1 = m2. c. L1 and L2 are perpendicular if and only if m1 m2 = −1. d. The length of of the line segment between two points P (x1 , y1 ) and Q(x2 , y2 ) (the distance between the two points) is p (1.7) |P Q| = (x1 − x2 )2 + (y1 − y2 )2. Example 1.3 Find the equation of the line through (2,3) which is perpendicular to the line L : 2x + 3y = 11. The line L has slope m = −2/3. Thus the line perpendicular to L has slope −1/(−2/3) = 3/2. Thus the equation of the line we seek is y−3 3 3 = or y = x. x−2 2 2 Polynomial functions For the general curve given by the equation y = f (x), the ratio (1.3): ∆y/∆x is the slope of the line joining the two points (x0 , y0 ) and (x1 , y1 ) on the graph of f. But now, if the graph is not a line, this ratio changes as the point (x1 , y1 ) moves. As x1 approaches x0 , this ratio may approach a specific number. If it does, this number is called the derivative of y with respect to x, evaluated at the point x0. It is the instantaneous rate of change of y with respect to x at x0 , and also the slope of the line which best approximates the curve at (x0 , y0 ), called the tangent line to the curve (see Figure 1.2). 3  Figure 1.2 x1 y1
 x1 y1
 Tangent x0 y0 x1 y1 Line 

In this chapter we shall concentrate on finding the derivative of functions given by a formula; this process is called differentiation. It turns out to be quite simple for polynomial functions. But first, we make these definitions explicit. Definition 1.1. Let y = f (x) be a function defined for all values of x in an interval about the point a. If the difference quotient f (x) − f (a) x−a approaches a specific number L, then we say that f is differentiable at a, and the number L is called the derivative of f at a, denoted f 0 (a). It is the slope of the tangent line of y = f (x) at a. For now, we will rely on an intuitive understanding of the phrase “approaches a specific number L”; the meaning will be made clear through the examples. After these examples, we introduce an explicit definition of the idea of limit, and in the next chapter we will discuss this definition in greater depth. Example 1.4. Consider f (x) = x2 , Find the tangent line to this curve at the point (a, f (a)). We take a point (x, f (x)) near (a, f (a)) and calculate the difference quotient f (x) − f (a) x2 − a2 (x − a)(x + a) = = = x + a.) x−a x−a x−a Clearly, as x approaches a, x + a approaches 2a, so we get f 0 (a) = 2a. Since this is true for any value a, we can conclude that if f (x) = x2 , its derivative is f 0 (x) = 2x. Example 1.5. Find the equation of the line tangent to the curve y = x2 at the point (3,9). For x = 3, the derivative is f 0 (3) = 2(3) = 6. Thus the tangent line has the equation y−9 =6 or y = 6x − 9. x−3 4 The ease with which we calculated the derivative for y = x2 followed from simple algebraic facts. We shall see that this works in general for polynomials; but first, one more example: Example 1.6. If f (x) = x3 , f 0 (x) = 3x2. Fix a point (a, a3 ) on the graph, and let (x, x3 ) be a nearby point. We look at the slope of the line joining these points: ∆y x3 − a3 =. ∆x x−a Since the quotient of x3 − a3 by x − a is x2 + ax + a2 this can be rewritten as x3 − a3 (x − a)(x2 + ax + a2 ) = = x2 + ax + a2 , x−a x−a and evaluating this at a, we get f 0 (a) = 3a2. Now, for any polynomial y = f (x), this process will work: divide f (x) − f (a) by x − a, and evaluate the quotient at x = a to calculate the derivative. Let’s spell this out, starting with the division theorem of algebra: Theorem 1.2. Let f be a polynomial of degree d. Then, for any number a when we divide f (x) by x − a, we get a quotient which is a polynomial of degree d − 1 and a remainder of f (a): f (x) f (a) (1.8) = q(x) +. x−a x−a Now, to apply this to the calculation of instantaneous rate of change, move the second term on the right to the left: f (x) − f (a) = q(x). x−a Aas we let x approach a, the difference quotient q(x) approaches q(a), so the polynomial is differ- entiable at a, and its derivative is q 0 (a). Theorem 1.3. A polynomial y = f (x) is everywhere differentiable. Its derivative at x = a is q(a), where q is the quotient of f (x) − f (a) by x − a. Newton realized that using long division would be a tedious way to calculate derivatives, and with them instantaneous rates of change, so he had the genius to take a slightly more abstract approach to lead to an automatic way of calculating derivatives. First, we must make explicit what we mean by the phrase ”the expression approaches a specific number” by introducing the notion of limit. Suppose that y = g(x) defines a function in an interval about x0. We say that g has the limit L as x approaches x0 if we can make the difference |g(x) − L| as small as we need it to be by taking x as close to x0 as we have to. More precisely, but less intuitively, Definition 1.4. limx→x0 g(x) = L if, for any  > 0 there is a δ > 0 such that |x − x0 | < δ implies |g(x) − L| < . We now observe that limits behave well under algebraic operations. 5 Proposition 1.5. Suppose that f and g are functions defined in an interval about x0 and that lim f (x) = L , lim g(x) = M. x→x0 x→x0 Then a) lim (f (x) + g(x)) = L + M x→x0 b) lim (f (x) · g(x)) = L · M x→x0 c) If M 6= 0, then f (x) L lim =. x→x0 g(x) M Applying this proposition to the calculation of derivatives, we see how differentiation behaves under algebraic operations: Proposition 1.6. Suppose that f and g are functions defined and differentiable in an interval I. Then a) f + g is differentiable in I, and (f + g)0 = f 0 + g 0. b) f g is differentiable in I, and (f g)0 = f 0 g + f g 0. We give a brief justification of these rules, which follow from the corresponding rules for limits (proposition 1.5). This is straighforward for part a), but for the product, the argument requires some preliminary algebraic manipulation. Suppose then, that f and g are differentiable at a, and let h = f g. Then to see that h is differentiable, we must take the limit, as x approaches a of h(x) − h(a) f (x)g(x) − f (a)g(a) =.) x−a x−a The product rule for limits does not apply directly, for this is not a product. However, if we add and multiply f (a)g(x), we get f (x)g(x) − f (a)g(a) = [(f (x) − f (a))g(x)] + [f (a)(g(x) − g(a))] , which leads to a sum of products f (x)g(x) − f (a)g(a) f (x) − f (a) g(x) − g(a) = g(x) + f (a). x−a x−a x−a Now, we can take the limits using proposition 1.5. We have to note that, since g is differentiable, we also have limx→a g(x) = g(a). This brings us to the rule for differentiating polynomials. 6 Proposition 1.7. a) If f is constant, then f 0 = 0. b) If f (x) = axn for some positive integer n, then f 0 (x) = anxn−1. c) A polynomial is differentiated term by term, using b) for each term. To verify a), we only have to note that a constant function is unchanging; its graph is a horizontal line, so has slope 0. c) follows from the fact that the limit of a sum is the sum of the limits. b) follows by a bootstrap method. We have already seen this for n = 0, 1, 2, 3. To proceed, we use the product rule. For example, take n = 4: (x4 )0 = (x3 x)0 = (x3 )0 x + x3 x0 = (3x2 )x + x3 (1) = 4x3. If we have the proposition for all integers up to n − 1, then we have it for n by the same method: (xn )0 = (xn−1 x)0 = ((n − 1)xn−2 )x + xn−1 (1) = nxn−1. Example 1.7. Let f (x) = 2x2 − 3x + 3. Find f 0 (x). What is the equation of the line tangent to the curve given by y = f (x), at the point (1,2)? Using proposition 1.7, we have f 0 (x) = 2(2x) − 3(1) + 0 = 4x − 3. This gives the slope of the tangent line at (1,2) by evaluating at x = 1 : f 0 (1) = 4(1) − 3 = 1. Thus the equation of the tangent line is y−2 = 1 or y =x+1 x−1 Example 1.8. If f (x) = 2x5 − x4 + 8x3 + 2x − 5 , then f 0 (x) = 2(5x4 ) − 4x3 + 8(3x2 ) + 2x0 − 0 = 10x4 − 4x3 + 24x2 + 2. If a function f is differentiable at every point of an interval I, then the derivative is defined at every point in the interval I, and thus is a function on I. This function, denoted f 0 , is defined by the rule: for all x in I, f (x + h) − f (x) (1.9) f 0 (x) = lim. h→0 h In particular, since f 0 is now a function on I, it too may be differentiable. If so, its derivative is denoted f 00 (x), and is called the second derivative of f with respect to x. Proceeding, we can define third and fourth derivatives and so forth. 7 So far, we have been interpreting the derivative as the instantaneous rate of change of y with respect to x, or the slope of the tangent line. Another fundamental interpretation is in terms of motion. Consider an object moving along a straight line. Let the variable t represent time, and x the displacement of the object from a fixed point, 0, on the line. Then the position of the object at time t is given by a function x = x(t). The velocity (denoted at time t as v(t)) of the object is the instantaneous rate of change of x with respect to t. The acceleration of this object (denoted a(t)) is the instantaneous rate of change of v with respect to t. Thus, if v(t) is the velocity of the object at time t, and a(t) its acceleration, we have (1.10) v(t) = x0 (t) , a(t) = v 0 (t) = x00 (t). Example 1.9 Suppose an object is moving in a straight line so that its displacement at time t is given by x(t) = 4t2 + 12t. Find the formulas for the velocity and acceleration of this object. What are the velocity and acceleration at time t = 5? Differentiating, we find that v(t) = x0 (t) = 8t + 12, a(t) = x00 (t) = 8. Thus the velocity at time t = 5 is v(5) = 8(5) + 12 = 52, and the acceleration is a(5) = 8. In many dynamical problems, an object is moving at constant acceleration. For example, an object falls near the surface of the earth at an acceleration of 32 ft/sec2 downward (or 9.8 m/sec2 downward, in the metric system). If the acceleration is constant, that tells us that ratio of the change in velocity over the change in time is constant: that is, the velocity is a linear function of time. Similarly, since the velocity is a linear function, the distance travelled must be given by a quadratic function; all we have to do is to use the given data to find the coefficients. We conclude: Proposition 1.8. Suppose an object moves at constant acceleration a. If at time t = 0 the object is at position x0 and has velocity v0 , then at any time we have a 2 (1.11) x(t) = t + v0 t + x0 , v(t) = at + v0. 2 It is easy to check that these functions do have the desired properties, that is, that x0 (t) = v(t) and v 0 (t) = a, and that their values at t = 0 are as given. Furthermore, we can argue intuitively, as we have done above, that these are the precise formulas for distance and velocity. For, since the rate of change of velocity is constant, velocity must be linear, and since the rate of change of distance is linear, it must be quadratic. This was the way Newton argued; but there are loose ends as was pointed out very articulately by Newton’s contemporary, George Berkeley. Why indeed, are these the only formulas with the desired properties? How do we know that there does not exist some as yet unknown mysterious functions which have the same values at t = 0, and the given acceleration? The third book of Newton’s Principia gives formidable evidence that no such mysteries exist, and that work, together with much subsequent experimental evidence, carried the day. But Berkeley’s objections were valid on logical grounds, and the issue was not satisfactorily resolved until the nineteenth century. Example 1.10. An object is projected upward at an initial velocity of 48 ft/sec. How high does it go? We measure distance upward from the starting point, so that x0 = 0 and v0 = 48. The acceleration due to gravity is a = -32 ft/sec, so (by proposition 1.8), the equations of motion (1.11) are x(t) = −16t2 + 48t , v(t) = −32t + 48. 8 If we complete the square for x(t), we have x(t) = −16(t − 3/2)2 + 36. Thus the greatest value of x is achieved at t = 3/2, and is x(3/2) = 36 feet. Note that at this highest point, v(3/2) = 0, confirming our intuition that at the moment the object turns around its velocity must be zero. Example 1.11. An automobile is traveling at 60 mph. At what rate must it decelerate so as to stop in 100 yards? Converting everything to feet and seconds, we have an initial velocity of 88 ft/sec, and we can take s(0) = 0. At some future time T , we have s(T ) = 300 feet, v(T ) = 0. Call the rate of deceleration a. The equations of motion (1.11) are a x(t) = − t2 + 88t , v(t) = −at + 88. 2 At time T we have 300 = −(a/2)T 2 + 88T, 0 = −aT + 88. From the second we get T = 88/a; putting that in the first we get a 88 2 88 1 1 882 300 = − + 88 or 300 = 882 (− + ) or 300 = , 2 a a 2a a 2a so a = 12.91 ft/sec2. More rules of differentiation Eventually we will develop a full set of rules for finding the derivative of any function given by a formula. We turn now to the quotient rule to handle quotients of polynomials (called rational functions). Proposition 1.9. Suppose that f and g are differentiable at a point a, and g(a) 6= 0. Then 1/g and h = f /g are differentiable at a, and 1
0 g0 gf 0 − f g 0 (1.12) = − 2, h0 =. g g g2 To show that 1/g is differentiable, we must calculate the limit as x → a of 1 1 g(x) − g(a).) x−a Once again, a little algebra helps us. Simplifying the compound fraction, we get 1 g(a) − g(x) −1 g(x) − g(a) · = , x−a g(x)g(a) g(x)g(a) x−a which has as its limit 1
0 g 0 (a) (a) = −. g (g(a))2 9 Now the second equation of (1.38) follows from this and the product rule appied to f /g considered as f · (1/g). f
0 1
0 1
−g 0
0 1
gf 0 − f g 0 =f + f0 =f + f =. g g g g2 g g2 In particular, we have d 1 1 ( )=− 2. dx x x Proposition 1.10. Let n be any integer, positive, zero, or negative. Then (1.13) for f (x) = xn we have f 0 (x) = nxn−1. By proposition 1.7b, this is true for n positive or zero. For negative exponents, we apply the quotient rule to f (x) = 1/xn with n positive: nxn−1 f 0 (x) = − = (−n)x−n−1 , (xn )2 which is just (1.13) for the negative exponent −n. Example 1.12. Find the derivative of 3 5 f (x) = x2 − 2x + −. x x2 Rewrite the function in exponential notation: f (x) = x2 − 2x + 3x−1 − 5x−2. Now use (1.13): f 0 (x) = 2x − 2 + 3(−x−2 ) − 5(−2x−3 ), which can be rewritten as 3 10 f (x) = 2x − 2 − 2 + 3. x x Example 1.13. Let f (x) = 30x + 2x−1. For what value of x is f 0 (x) = 0? Differentiate: f 0 (x) = 30 − 2x−2. Now solve f 0 (x) = 0: 2 0 = 30 − so that x2 = 15 x2 √ and the answer is x = ± 15. As a last observation, we return to the definition of the derivative to differentiate the square root function: √ √ Proposition 1.11. If f (x) = x for x > 0, then f 0 (x) = 1/(2 x). √ √ √ √ Here we use the fact that x − a = ( x − a)( x + a). Thus √ √ √ √ x− a x− a 1 1 = √ √ √ √ =√ √ → √ x−a ( x − a)( x + a) x+ a 2 a 10 as x → a, for a 6= 0. Problems 1.1 1. Find the equation of the line which goes through the point (2,-1) and is perpendicular to the line given by the equation 2x − y = 1. 2. a) Let f (x) = x2 +3x−1. Find the slope of the secant line joining the points (2, 9) and (x, f (x)). b) Find the slope of the tangent line to the curve y = f (x) at the point (2, 9). c) What is the equation of this tangent line? 3. Let 1 f (x) =. x2 a) Find the slope of the secant line through the points (x, x12 ) and (x + h, (x+h) 1 2 ). b) Find the slope of the line tangent to the graph of y = f (x) at the point (3, 91 ). 4. Find the derivatives of the following functions: a) f (x) = x3 − x2 + 1 1 b) g(x) = x2 + x3 1 c) h(x) = (x2 + )(x3 − x2 + 1) x3 5. Find the derivatives of the given functions: a) f (x) = 3x−1 + x3 1 b) f (x) = (x2 + )(x3 − x2 + 1) x3 6. Find the derivative of the given functions: x2 + 1 a) f (x) = x+1 1 b) f (x) = x2 + x3 7. Find the derivative of x2 + 1 f (x) = x+1 11  1 − t2 8. Differentiate : h(t) = 1 + t3 10. Sketch the graph of a function with these properties: a) f (0) = 2 and f (1) = 0; b) f 0 (x) < 0 for 0 < x < 2; c) f 0 (x) > 0 for x < 0 or x > 2. 11. Sketch the graph of a function with these properties: a) f (0) = 1 and f 0 (0) = 0; b) f (−1) = 0, f (1) = 0, c) f 0 (x) < 0 for 0x < −1 and 0 < x < 1, c) otherwise, f 0 (x) > 0. 12. Find the value of x where the graphs of these two functions have parallel tangent lines: f (x) = x2 − 3x + 2 , g(x) = 5x2 − 11x − 17. 13. Find the points on the curve y = 3x2 − 3x + 1 whose tangent line is perpendicular to the line x + 2y = 7. 14. Let C1 and C2 be curves given by the equations C1 : y = x3 + x2 , C2 : y = x2 + x. For what values of x do these curves have parallel tangent lines? 15. Find the derivative of f (x) = (x + 1)( x1 + 1). 16. Find the slope of the line tangent to the curve y = x2 − 3x + 1/x at the point (3,1/3). 17. Let y = x3 − 48x + 1. Find the x coordinate of the points at which the graph has a horizontal tangent line. 18. A man standing at the edge of the roof of a building 120 feet high throws a ball directly upwards at a velocity of 48 ft/sec. a) How high does the ball go? b) Assuming that it proceeds to fall along the side of the building, how long does it take to hit ground level? 19. Another man standing on ground level throws the ball back to his friend on the roof. At what initial velocity must he throw it in order to reach the roof? 20. On the planet Garbanzo in the Weirdoxus solar system, the equation of motion of a falling body is s = s0 + v0 t − 10t3 where s0 is the initial height above ground level and v0 is the initial velocity. Distance is mea- sured in garbanzofeet. If a ball is thrown upwards from ground level at an initial velocity of 120 garbanzofeet/second, how high does the ball rise? 12 1.2 Liebniz’ Calculus of Differentials Up to this point we have been following the development of the Calculus according to Newton. We have been considering variables y, z, u, v, etc. as functions of a particular variable (called the “independent variable”) x, and discovering how to find rates of change of the dependent variables relative to the independent variable. The ideas of Liebniz follow a different, but equivalent, set of ideas. Liebniz is concerned with a collection of variables x, y, z, u, v, etc. and their “infinitesimal increments”. This is a hard concept to get a hold on, but we can think of it this way. When we actually make measurements, we always have in mind, even if unspecified, an “error bar”; that is, a largest allowable error. Thus, our calculators display numbers to 8 decimal points, allowing for a “negligible” error of at most 10−8. A more efficient computer has a smaller error bar, perhaps 10−32 , or 2−128. Instruments of measurement, no matter how delicate, have to allow for such an error bar. So, if u is a measurable variable, it comes equipped with an error bar: an allowable increment in a measurement which does not change the accepted value of the measurement. It is this which we should call the “infintesimal increment” in u, called by Liebniz the differential of u, denoted du. However, the important feature of this concept is that it is not tied down to the level of accuracy of today’s instruments, but it represents the error bar for all time: du stands for the smallest measurable increment for all ways of measuring ever to come. We get a more concrete interpretation of the differential by relating it to the linear approximation to the variables. More precisely, suppose the variables x, y are related by y = f (x). The tangent line at a point (x0 , y0 ) is the line which best approximates the curve. We have used the symbols ∆x, ∆y to represent changes in the variables x and y along the curve; now we let dx, dy represent changes in the variables along the tangent line. Since the slope of the tangent line at (x0 , y0 ) is f 0 (x0 ), we obtain this important relationship between the differentials: at the point (x0 , f (x0 )), dy ∆y (1.14) = lim = f 0 (x0 ) , dx ∆x→0 ∆x or, without specifying the particular point, dy = f 0 (x)dx. This we can interpret as the equation of the tangent line by replacing dx and dy by x − x0 , y − y0. Example 1.14. Find the equation of the tangent line to the curve y = x3 − 2x + 5 at the point (2, 9). First we calculate the relation between the differentials: dy = (3x2 − 2)dx. At x = 2, this gives dy = 10dx. Now we interpret this as the equation of the tangent line by replacing dy by y − 9 and dx by x − 2. The equation of the tangent line is thus y − 9 = 10(x − 2) or y = 10x − 11. Finally, considering the equation dy = f 0 (x)dx as the linear approximation to the equation y = f (x) (at a particular point), we can make preliminary estimates of the change in y, given a change in x. Example 1.15. The volume of a sphere of radius r is V = (4/3)πr3. Suppose the surface of a sphere of radius 6 feet is covered by a 1 inch coat of paint. About how much paint will be needed? 13 From the defining equation we have dV = 4πr2 dr; so letting r = 6 feet and dr = 1/12 feet, we can estimate the change in volume to be 1 dV = 4π(6)2 ( ) = 37.7 cu. ft. 12 Thinking of the derivative as the ratio of two quantities which eventually become zero has its philosophical problems and was also subjected to the scathing criticism of Berkeley. This concept of “evanescent quantities” (as Berkeley sarcastically identified them) was controversial in the days of Newton and Liebniz and remained so for the following 200 years. Note, by the way, that one can object to Newton’s methods on the same ground: when we write f (x) − f (a) x2 − a2 (x − a)(x + a) = = =x+a , x−a x−a x−a by what right are we now able to let x become a? If we did so one step sooner, we’d be dividing by zero, which is forbidden. So, in this set of equations x cannot be a. But in the next line we say, “let x be a”! These philosophical obstacles were eventually overcome; we shall proceed without resolution, as did Newton, Liebniz and their successors to enormous effect. Suffice it to say that this can all be put on a logical footing, while at the same time, the concept of differential as “smallest possible increment” is a powerful intuitive tool throughout mathematics and its applications. To illustrate, in the next section, we shall give a heuristic derivation of the law of differentiation for composite functions. Problems 1.2 x 1. Let y=. x2 +1 Find the equation of the tangent line to the graph at the point (2,0.4). 2. Find the equation of the tangent line to y = x2 (x3 − 1) at (2,28). √ 3. Find the equation of the tangent line to the curve y = x cos x at (π/4, π 2/8). 4. Find the the equation of the tangent line to the curve y = x − x−2 at (2,7/4). 5. Let y = x + 25x−1. Find an approximate value of y when x = 3.2. 1.3 The Chain Rule Suppose that y, u and x are variables such that u is a function of x: u = f (x), and y is a function of u: y = g(u). Then y can be viewed as a function of x by writing y = h(x) = f (g(x)). How do we find the rate of change of y with respect to x? Using differentials, we have: dy = f 0 (u)du, and du = g 0 (x)dx, so that dy = f 0 (u)g 0 (x)dx, it being understood that in this formula u is to be expressed in terms of x. A shorthand for this is dy dy du = ·. dx du dx 14 Example 1.16. Let y = (4x + 1)3. We introduce the intermediate variable u = 4x + 1, so that y = u3. Then dy = 3u2 du, and du = 4dx, so that dy = 3u2 (4) = 12(4x + 1)2. dx Example 1.17. If y = x−n with n positive , we introduce u = xn , so that y = u−1. Then d −n dy dy du 1 nxn−1 (x ) = = = − 2 nxn−1 = − n 2 = −nx−n−1 , dx dx du dx u (x ) giving another derivation of proposition 1.11. Example 1.18. Let y = ((2x + 1)3 + 5)2. Here we need to use the chain rule more than once. We think of y = v 2 , where v = u3 + 5, and u = 2x + 1. Then dy dy dv du = = 2v(3u2 )(2). dx dv du dx Now replace v and u by their expressions in x: dy == 12((2x + 1)3 + 5)(2x + 1)2. dx The statement of the chain rule is as follows. Proposition 1.12. Suppose that g is differentiable at the point a, and f is differentiable at g(a). Then the composed function h(x) = f (g(x)) is differentiable at a and (1.15) h0 (a) = f 0 (g(a))g 0 (a). In particular, Proposition 1.13. If f is differentiable at a, and n is any positive or negative integer, h(x) = (f (x))n is also differentiable at a and (1.16) h0 (x) = n(f (x))n−1 f 0 (x). Of course, the Liebniz formulation (1.55) is easier to remember and apply than proposition 1.12. For that reason we shall begin to adopt the Liebniz notation for differentiation: if y = f (x) is differentiable in an interval I, we write dy (1.17) f 0 (x) = dx and use f 0 (x) and dy/dx interchangeably. The notation for higher derivatives is: d2 y d3 y (1.18) f 00 (x) = , f 000 (x) = , etc. dx2 dx3 15 In this notation, Proposition I.13 becomes simply d n dy (y ) = ny n−1. dx dx Problems 1.3 1. Find the derivative of g(x) = (x3 + 1)4. √ 2. Find the first and second derivatives of f (x) = x 1 − x2 √ 3. Differentiate: f (x) = 2x2 − 3x + 1. (x + 1)2 4. Find f 0 (x) : f (x) = (x − 1)2 5. Find g 0 (x), g 00 (x) : g(x) = (x3 + 1)4. 1.4 Trigonometric Functions Consider a particle moving in the counterclockwise direction around the circle of radius 1 with constant angular velocity of 1 radian/second such that at time t = 0 it is at the point (1, 0). Then its position at time t is (cos t, sin t). These functions are defined for all values of t, and are periodic of period 2π since in time 2π the particle will make one full circuit of the circle. There are four other trigonometric functions defined by the equations sin t cos t 1 1 tan t = , cot t = , sec t = , csc t =. cos t sin t cos t sin t Assuming that these functions are differentiable, we can calculate the derivatives by an argument using differentials due to Blaise Pascal. Figure 1.3: Pascal’s derivation. dy dt y cost dx   dt I
t x sint  In figure 1.3 we have located the moving point at P = (cos t, sin t) at time t, and its position Q after an infinitesimal increment dt. Since we are on the unit circle t also measures arc length 16 along the circle. The triangle with sides dx, dy, dt is called the “differential triangle”. It may be of concern that dt represents an arc of the circle, but, remember - at the differential level an arc and a straight line are indistinguishable. Since the tangent line to the circle is perpendicular to the radius at the point P , the differential triangle is similar to the triangle (I). Thus −dx dt dy dt = , = , sin t 1 cos t 1 so dx dy = − sin t , = cos t. dt dt Since x = cos t, y = sin t, we obtain the first part of the following. Proposition 1.14. d d a) (sin t) = cos t , (cos t) = − sin t , dt dt d d b) (tan t) = sec2 t , (cot t) = − csc2 t , dt dt d d c) (sec t) = sec t tan t , (csc t) = − csc t cot t. dt dt b) anc c) follow from the quotient rule. For example, b): d d sin t cos t cos t − sin t(− sin t) 1 (tan t) = = 2 = = sec2 t. dt dt cos t cos t cos2 t Remember: in the the above discussion we have assumed that the trigonometric functions were differentiable, and it was that assumption that allowed us to consider an arc of a (differential) circle as a straight line. These formulae imply the following limit results, just by the definition of the derivative. Proposition 1.15. sin x cos x − 1 (1.19) lim =1, lim =0. x→0 x x→0 x We remind the reader that we started out assuming that the functions sin x and cos x are differen- tiable, so this problem tells us what the derivatives are at x = 0 under those assumptions. In the next chapter we wil prove proposition 1.15 directly by geometric methods, from which we conclude that the sine and cosine functions are indeed differentiable. Problems 1.4 1. From a point 1000 feet away from the base of a building, the angle of elevation of its roof is 17 degrees. How tall is the building? 17 2. A marker is rotating counterclockwise around a circle of radius 4 centered at the origin at the rate of 7 revolutions per minute. a) What is its position after 2.3 minutes? b) How soon after 2.3 minutes will it cross the x-axis again? √ 3. If tan α = − 3, what are the possible values of sin α? 4. Express as a function of 2x: sin x − cos x sin x + cos x 5. Find the derivative: f (x) = sin x cos x 6. Find the derivatives of the following functions: a) f (x) = cos2 x sin2 x b) g(x) = cos x 7. Find the derivative: sin x g(x) = cos x 8. Differentiate: y = (x2 − 1) sin(x2 + 1). 9. Find the derivative: h(x) = (cos(2x) + 1) sin(3x) 10. Differentiate: g(x) = (sin(3x) + 1)3. 11. A point moves around the unit circle so that the angle it makes with the x-axis at time t is θ(t) = (t2 + t)π. Let (x(t), y(t)) be the cartesian coordinates of the point at time t. What is dy/dt when t = 3? 12. Let f (x) = x sin x. Find the equation of the tangent line to the graph y = f (x) at the points x = (2n + 1/2)π for any integer n. 13. Find the derivatives of these functions: (a) h(x) = (cos(2x) + 1) sin(3x) b) g(x) = (tan(3x) − 1)2 14. Consider the curves C1 : y = sin x and C2 : y = cos x. a) At which points x between −π/2 and π/2 do the curves have parallel tangent lines? b) At which such points do they have perpendicular tangent lines? 15. Differentiate: 1 + tan x f (x) = 1 − tan x 18 1.5 Implicit Differentiation and Related Rates Suppose that x and y are variables which are related by a functional equation: F (x, y) = c, a constant. We say that this relation defines y implicitly as a function of x. For, in principle, given a value of x, say x = a, we can solve the equation F (a, y) = c for y, giving the “rule” defining y in terms of x. However, to find dy/dx we need not solve this equation. When y and x are so related, their differentials are related as well, and the chain rule can be used to find that relationship, as a function of both x and y. The idea is to think of z = F (x, y) as another variable which, because of the relation F (x, y) = c is constant, so dz = 0. Now, apply the chain rule to the expression for z. Example 1.19. Suppose that the variables x and y satisfy the relation (1.20) x2 − xy + 2y 2 = 4. Letting z represent this defining relation, we have dz = 0. Now, using (1.15) and the rules for differentiation, dz = 2xdx − (xdy + ydx) + 4ydy = 0 , giving us (1.21) (−x + 4y)dy = (−2x + y)dx , leading to this expression for the derivative: dy 2x − y =. dx 4y − x Example 1.20. What is the equation of the tangent line to the curve given by (1.68) at the point (2,1)? We find the slope by substituting the values x = 2, y = 1 in equation (1.21): dy 2(2) − 1 3 m= = =. dx 4(1) − 2 2 Then the equation of the line is 3 3 y−1= (x − 2) or y= x−2. 2 2 Notice that, if we substitute x = 2, y = 1, dy = y − 1, dx = x − 2 into equation (1.21) we get the same result. That is because equation (1.21) is the linear approximation to the relation between x and y, which is of course the same as the equation of the tangent line. Example 1.21. Find the equation of the tangent line to the curve y 3 + 2 cos2 x = 0 at the point (π/4, −1). We differentiate implicitly: 3y 2 dy − 2 cos x sin xdx = 0. √ Now, at the point (π/4, −1), y 2 = 1, cos x = sin x = 2/2, so this becomes 3dy − dx = 0. Replacing dx by x − π/4 and dy by y − (−1) gives the equation of the tangent line: π π 3(y + 1) − (x − )=0, or 3y − x = −3. 4 4 19 Related Rates Suppose we are in a situation where one or more variables are related, and the variables are functions of time. For example, if a spherical balloon is being inflated, then during this process the volume (V ), area (A) and radius (r) are increasing with time. Since these are all related, we are able, by differentiation to relate the rates of growth. For example, suppose the balloon is being inflated by putting gas in at a steady rate of 3 cc/sec. We may ask “at what rate is the radius changing?” We start with the formula relating volume with radius: V = (4/3)πr3. V and r are functions of time, so, differentiating with respect to time we obtain dV 4 dr dr = π(3r2 ) = 4πr2. dt 3 dt dt Putting in the datum dV /dt = 3 , we find dr 3 = cm/sec , dt 4πr2 so depends upon the radius at the time. Here is a protocol for attacking such problems. Step 1. Draw a picture (if appropriate), and identify the relevant variables: those things which can change. State the problem in terms of the variables. Step 2. Find a relationship among the variables. Step 3. Differentiate, to obtain a relationship among the variables and their rates of change. Step 4. Put in the values of the variables at the time in question, and solve the resulting equation. Example 1.22. Suppose as above a balloon is being inflated with gas at a rate of 3 cc/sec. At what rate is the area increasing when the radius is 14 cm? First, we identify the variables as volume: V , area: A, and radius: r. Now, these are related by the equations 4 V = πr3 , A = 4πr2. 3 Now, we differentiate these equations: dV dr dA dr = 4πr2 , = 8πr. dt dt dt dt Now, at the specific time of interest, r = 14 cm, and dV /dt = 3 cc/sec. Substituting these values, we have: dr dA dr 3 = 4π(14)2 , = 8π(14). dt dt dt Then dr/dt = 3/[4(14)2 π], so the second equation gives dA 3 3 = 8π(14) 2 = cm2 /sec. dt 4(14) π 7 20 Example 1.23. A ship leaves port at noon heading north at 25 knots (nautical miles per hour), and 2 hours later another ship leaves heading west at 30 knots. Assuming the ships travel in straight lines, at what rate is the distance between the ships increasing after an additional 3 hours? First, the variables are: t, the time elapsed since noon, N , the distance travelled in that time by the ship heading north, W , the distance travelled by the ship heading west, and Z, the distance between them. The relations among these variables are: Z2 = N 2 + W 2 , from the Pythagorean theorem. In t hours after noon, the first ship has travelled 25t nautical miles: N = 25t, and, since the second ship started two hours later, it has travelled 30(t − 2) nautical miles (notice, we are assuming that t ≥ 2). Now since we have been given the rates of change of N and W , and want to find that of Z, we differentiate the first equation with respect to t to relate these rates: dZ dN dW 2Z = 2N + 2W. dt dt dt Now, at t = 5, we have N = 125, W = 75, dN/dt = 25, dW/dt = 30, giving dZ Z = 125(25) + 75(30) = 5375. dt √ We find Z by the first relation Z 2 = 1252 + 752 = (25)2 (34), so Z = 25 34. Finally, dZ 5375 = √ = 36.87 nautical miles/hour. dt 25 34 Example 1.24. Suppose that x and y are functions of t which satisfy the relation x3 y 2 + 2y = 8. Suppose that at the point (1, 2), the velocity of x is 3 in/sec. What is the velocity of y? Differentiating the relation implicitly, we get dx 2 dy dy 3x2 y + x3 (2y ) + 2 =0. dt dt dt Now substituting x = 1, y = 2, dx/dt = 3, in this equation: dy dy 3(1)2 (3)(22 ) + (1)3 (2(2) )+2 =0. dt dt Solving for dy/dt, we find 36 + 6dy/dt = 0, or, the velocity of y is -6 in/sec. Definition 1.16. For integers p and q, the function (1.23) y = xp/q is defined, for all positive x, as the positive solution of the equation y q = xp. √ So, for example, x can be written as x1/2 , the cube root of x as x1/3 , etc. Using implicit differentiation we verify: 21 Proposition 1.17. d n (1.24) x = nxn−1 f or all rational numbers n. dx A rational number is a quotient p/q of integers. Differentiate the equation y q = xp implicitly: dy p xp−1 qy q−1 dy = pxp−1 dx or =.) dx q y q−1 Replacing y by xp/q , we get dy p xp−1 p p = p−p/q = xp−1−p+p/q = x(p/q)−1 dx qx q q which is the desired result, since n = p/q. Problems 1.5 1. A curve is given by the equation x2 − xy + y 2 = 7. Find the equation of the line tangent to this curve at the point (2,-1). 2. Find the slope of the curve defined by the relation 4(x2 + xy) = 2y 3 − y 2 at the point (1, 2). 3. Variables x and y are related by the formula x sin y + y sin x = π. If dy/dt = 3 when x = 3π/2 and y = π/2, what is dx/dt? 4. The relation cos y + x = sin y determines a curve in the x-y plane. Find the slope of the line tangent to the curve at the point (1, π/2). 5. Consider the curve given by the equation: y 2 + xy + x2 = 1. At what points does this curve have a horizontal tangent line? 6. Consider the curve given by the equation: x2 y − y 3 = 1. At what points does this curve have a vertical tangent line? 7. A ship is travelling in a circle of radius 6 nautical miles around an island at a speed of 10 knots (nautical miles per hour). A lighthouse is 10 miles due east of the island. At what rate is the distance between ship and lighthouse increasing when the ship is exactly due north of the island? 8. A new stadium, built like a cylinder capped with a hemispherical dome is proposed to have a diameter of 500 feet. To include another 2000 seats, the diameter must be increased by 10 feet. By approximately how much will the area of the dome be increased? (Note: the area of a sphere of radius r is 4πr2.) 22 9. A cat is walking toward a telelphone pole of height 30 feet. She is walking at a steady rate of 4 ft/sec. A bird is perched on top of the tellephone pole. When the cat is 45 feet from the base of the pole, at what rate is the distance between bird and cat decreasing? 10. Water is flowing into a conical tank through an opening at the vertex at the top at the rate of 12 cu. ft./min. The base of the tank is a circle of radius 12 ft. and the height of the cone is 20 ft. At what rate is the water level rising when the water level is 4 ft. from the top? The formula for the volume of a cone of base radius r and height h is V = (1/3)πr2 h. 11. Let P be an upward-opening parabola whose axis is the y-axis and whose vertex is the origin. Suppose the line y = C intersects the parabola in two points. Show that the tangent lines at these points intersect on the y-axis of the parabola. 12. Suppose that a point moves along the x-axis according to the formula x(t) = 1/(t2 + 1). Let A(t) be the area of the circle with diameter joining the origin to the point x(t). Find A0 (t) when t = 3. 23  II. Theoretical Considerations 2.1 Limit Operations In this section we shall go more deeply into the concept of limits than we did in chapter 1. Suppose that y = f (x) is a function defined in an interval about the point x0. Each value of x determines a value y using the rule represented by the function f. We say that y approaches a number L as x approaches x0 if we can be sure that y is as close as we please to L just by taking x close enough to x0. A little more precisely, if we allow an error  > 0 in the calculation of L, we can find an error δ > 0 for x0 such that if x is within δ of x0 , then y is within  of L. If the limit L is the number y0 calculated from x0 by f , then we say that f is continuous at x0. That is the content of the following two definitions. Definition 2.1. limx→x0 f (x) = L if, for any  > 0 there is a δ > 0 such that |x − x0 | < δ implies |f (x) − L| < . If the limit L is f (x0 ), then we say that f is continuous at x0 : Definition 2.2. A function f , defined in an interval about x0 is continuous at x0 if limx→x0 f (x) = f (x0 ). A function is said to be continuous if it is continuous at every point where it is defined. Example 2.1. Let f (x) = x/|x|. Then, f (0) = 0, for x > 0, f (x) = 1, and for x < 0, f (x) = −1. Thus, in any interval about 0, there are values of x for which f (x) = 1 and other values of x for which f (x) = −1. There is thus no number L such that both 1 and -1 are within.5 of L, so there can be no limx→0 f (x). Example 2.2. Let f (x) = cos(1/x) for x 6= 0. There is no value to assign to f (0) to make this function continuous. For if x = (2πn)−1 , f (x) = 1 for n even, and f (x) = −1 for n odd, so we are in the same situation as that of example 1. However, for the function g(x) = x cos(1/x), we can define g(0) = 0 to get a continuous function. For |g(x)| ≤ |x| for every x, since the cosine is bounded by 1. Thus, for any  > 0, if |x| < , we also have |g(x)| < . Now we state the basic facts describing how limits behave under algebraic operations. Proposition 2.1. Suppose that f and g are functions defined in an interval about x0 and that lim f (x) = L , lim g(x) = M. x→x0 x→x0 Then a) lim (f (x) + g(x)) = L + M x→x0 b) lim (f (x) · g(x)) = L · M x→x0 c) If M 6= 0, then f (x) L lim =. x→x0 g(x) M 24 This proposition then tells us the following about continuous functions: Proposition 2.2 Suppose that f and g are defined in an interval about x0 , and are continuous at x0. Then the sum f + g and product f · g are also continuous at x0. If g(x0 ) 6= 0, the quotient f /g is also continuous at x0. Now it is clear that a constant function is continuous: if f (x) = C for all x, then the difference |f (x) − C| = 0 no matter what x is. Also, the function f (x) = x is continuous everywhere: we can make |f (x) − f (x0 )| <  just by taking |x − x0 | < . Thus, by proposition 2.2, any function formed from constants and tne function f (x) = x by taking products and sums is continuous. But these are the polynomials. Proposition 2.3. All polynomials are continuous everywhere. A rational function (that is, a quotient of polynomials) is continuous everywhere where its denominator is non-zero. Example 2.3. That is not to say that a rational function is not continuous where the denominator is zero; perhaps it can be defined at those points so as to be continuous, For example, consider x2 − 4x − 5 f (x) =. x−5 Since we cannot divide by zero, f (x) is not defined for x = 5. But, can we define f (5) so that the function is continuous? Noting that x2 − 4x − 5 = (x − 5)(x + 1), we see that for x 6= 5, f (x) = x + 1. Thus by defining f (5) = 6, we get a continuous function. Suppose now that g is defined in an interval around x0 and f is a function defined on the range (set of values) of g. Then we can form the composition of the two functions, f ◦ g, just by applying the rule defining f to the value of g : f ◦ g(x) = f (g(x)). Proposition 2.4. Suppose that g is defined in an interval about the point x0 , g(x0 ) = y0 and f is defined in an interval about y0. If g is continuous at x0 , and f is continuous at y0 , then h = f ◦ g is also continuous at x0. To show this, we have to show that we can insure that h(x) is within  of h(x0 ) by taking x close enough to x0. By the continuity of f we can be sure that f (y) is within  of f (y0 ) by taking y within some small number, η of y0. But then, by the continuity of g, there is a δ such that, if x is within δ of x0 , g(x) is within η of g(x0 ) = y0 , and finally, f (g(x)) is within  of f (y0 ) = f (g(x0 )). A useful technique is what is called the squeeze theorem. Suppose, in some interval containing the point a, the values of f lie between those of two other functions g and h. Suppose also that g and h have the same limit as x approaches a, then f also has that limit. Proposition 2.5 (Squeeze Theorem). Suppose that f, g, h are defined in a interval containing a and that g(x) ≤ f (x) ≤ h(x). If lim g(x) = lim h(x) = L , x→a x→a we also have limx→a f (x) = L. 25 Suppose an allowable error  > 0 is specified. From the hypothesis, we know that there is a δ1 > 0 such that if x is within δ1 of x0 , then g(x) ≥ L − , and there is a δ2 > 0 such that if x is within δ2 of x0 , then h(x) ≤ L + . Then, so long as δ is less than both δ1 and δ2 , we have L −  ≤ g(x) ≤ f (x) ≤ h(x) ≤ L +  which is to say that f (x) is within  of L. Now suppose again that f is defined in a neighborhood of x0 and continuous there. We now turn to the question of the differentiability of f at x0. Definition 2.3. Let f be defined in a neighborhood of x0. If the limit f (x) − f (x0 ) lim x→x0 x − x0 exists, it is denoted f 0 (x0 ), and is called the derivative of f at x0. f is said to be differentiable at x0. Proposition 2.1c suggests that the limit does not exist since the denominator approaches 0. But we have to be careful: the numerator is also going to zero (as in example 2.3). In fact, as we saw by the division theorem of chapter 1, If f is a polynomial, then so is this difference quotient, and the limit is the value of that quotient at x0. In fact, in general it is a necessary condition for differentiability that the limit of the numerator is zero - a fact we already used several times in chapter 1. Proposition 2.6. Let f be defined in a neighborhood of x0. If f is differentiable at x0 , then it is continuous at x0. Let L = f 0 (x0 ). The hypothesis tells us that we can be sure the difference quotient is within  of L by taking x close enough to x0. So, taking, for example,  = 1, then if x is close enough to x0 , f (x) − f (x0 ) −1 < − L < +1, x − x0 from which we conclude that (L − 1)(x − x0 ) < f (x) − f (x0 ) < (L + 1)(x − x0 ). Now, the left and right hand sides tend to 0 as x approaches x0 , so, by the squeeze theorem, lim(f (x) − f (x0 )) = 0. But that is the same as lim f (x) = f (x0 ); that is, f is continuous at x0. Now, in section 1 of chapter 1, no problems arose in calculating limits, since we were there dealing with polynomials (even in calculating derivatives). However, more generally questions about limits can become real issues. For example, when we turned to trigonometric functions and the square root function, we tacitly assumed their continuity. Since the continuity is intuitively clear (if we envision the graph of these functions), this was not an obstacle to finding derivatives. However, in more general contexts, the continuity is not at all clear. As preparation for this, we shall here reconsider the assumptions of continuity made in chapter 1. First, the square root. 26 Example 2.4. For a ≥ 0, √ √ lim x= a. x→a We have to distinguish the cases a 6= 0 and a = 0. First, the case a 6= 0. We start with the identity √ √ √ √ ( x − a)( x + a) = x − a , which, for our purposes should be written as √ √ x−a x− a= √ √ , x+ a since √ it is the expression on the left we need to make small. Given  > 0, choose δ > 0 so that δ/ a < . Then if |x − a| < δ, √ √ |x − a| |x − a| δ | x − a| = √ √ < √ < √ M. √ √ Now, given  > 0, choose the integer n so that n > 1/2. Then n > 1/, so 1/ n <  which is what we need. For x < 1/n, √ 1 x< √ x0 , we call the limit the limit from the right, denoted limx→x+ f (x), and if we restrict to those x < x0 we call it the limit from the 0 left, denoted limx→x− f (x). Suppose that f is defined in an interval about x0 , and both the limit 0 from the left and the limit from the right exist. Show that if they are both equal to L, then lim f (x) = L. x→x0 2. Show that if f and g are functions defined in an interval near x0 and lim f (x) = L , lim g(x) = M , x→x0 x→x0 then lim f (x)g(x) = LM. x→x0 Hint: Write f (x) = L + a(x), g(x) = M + b(x), noting that by the hypothesis we can ensure that a(x), b(x) can be made as small as we please by taking x sufficiently close to x0. x2 − 4 3. lim = x→2 x2 − 3x + 2 28 Hint: Factor numerator and denominator. cos x − 1 4. lim = x→0 x sin x Hint: multiply and divide by cos x + 1. 5. Suppose that f is defined in an interval about 0, and that |f (x)| ≤ |x|2 in that interval. Show that f is differentiable at 0 and f 0 (0) = 0. 6. Show that the function f defined by f (0) = 0 and for x 6= 0, f (x) = x2 sin(1/x) is differentiable at 0 and has derivative zero. 7. Show that the function f defined by f (0) = 0 and for x 6= 0, f (x) = x sin(1/x) is not differentiable at 0. 2.2 Limits at Infinity Suppose that f is defined for all positive numbers. We say that f has the limit L as x → +∞ if we can make f as close as we please to L by taking x large enough. For example 1 lim =0, x→+∞ x since we can make 1/x <  just by taking x > 1/. Definition 2.4. Suppose that f (x) is defined for all x > M0. We say that lim f (x) = L x→+∞ if, for every  > 0, we can find an M ≥ M0 such that if x > M , then |f (x) − L| < . Suppose that f (x) is defined for all x < M0. We say that lim f (x) = L x→−∞ if, for every  > 0, we can find an M ≤ M0 such that if x < M , then |f (x) − L| < . Example 2.7. x lim =1. x→+∞ x + 1 For, given  > 0, choose M = 1/. Then, for x > M , we have x x − (x + 1) 1 1 1 −1 = = < < =. x+1 x+1 x+1 x M Now, we define what it means to have ±∞ as a limit. Definition 2.5. Let f be defined for all x in an interval about a, except perhaps at a. We write lim f (x) = +∞ x→a 29 if, for any M > 0, there is an  > 0 such that for |x − a| < , we have f (x) > M. Similarly, lim f (x) = −∞ x→a if, for any M > 0, there is an  > 0 such that for |x − a| < , we have f (x) < −M. We will also say that limx→+∞ f (x) = +∞ if we can make f (x) as large as we please by taking x sufficiently large, and similarly, we define limx→+∞ f (x) = −∞, limx→−∞ f (x) = +∞, and so forth. Proposition 2.9. Let p be a polynomial of degree n > 1, with leading coefficient 1. a) If n is even, lim p(x) = +∞. x→±∞ b) If n is odd