Calculus. Sungpyo Hong. Pohang University of Science and Technology

Transcription

1 Calculus Sungpyo Hong Pohang University of Science and Technology POSTECH 6

2 Preface Calculus is the most fundamental course in the study of science and engineering, whose main contents include the differentiation and the integration of functions. Current cutting edge technology would not be possible without the notions of differentiation and integration. Most universities in Korea are using Calculus text books from the US, which begin with the most basics of differentiation and integration and some applications. However, these are already the contents of high school courses for those students of science and engineering in Korea. The faculty members at POSTECH felt that it is necessary to challenge the students with more advanced contents, and they replaced the basics of differentiation and integration with their applications to ordinary differential equations. At first part of the course, the students feel a little bit difficulties, but at the later part of the course, they feel more challenging and get better motivations. We are confident that the students at POSTECH have enough ability to absorb this much. At this point, we felt that we need our own style text book of Calculus. This lecture note is outgrown from the lectures delivered to freshmen in POSTECH for many years with this in mind. This book includes the techniques of integration as an appendix. Those students who feel uncomfortable with integrations may refer this as they need. We also think that one of the most prominent applications of the calculus is to the highly sophisticated theory of Gamma function, which plays quite important roles not only in science and engineering but also in mathematical theories. Hence, we include the theory of the Gamma function in an appendix, so that one can refer this part whenever they need. We are very grateful to many colleagues in the mathematics department at Pohang University of Science and Technology (POSTECH). We would also like to acknowledge the invaluable assistance we have received from the teaching assistants who have helped to make the problems and exercises in this book. v

3 vi Preface We have endeavored to eliminate typos and mistakes, but there may still be many such things. We will appreciate it very much if anyone points out such mistakes or misprints. Sungpyo Hong, & Young Sun Shim & July 6, in Pohang, Korea

4 Contents Preface v Preliminaries. Functions Limits of Function Values One-sided Limits and Limits at Infinity Exercises Continuity of Functions Exercises Derivatives Exercises Differentials Parametric equations Implicit differentiation Applications of derivatives Exercises Indeterminate Forms Exercises Antiderivatives Definite integrals Integrals by substitution Exercises Transcendental Functions 37. Inverse Functions Trigonometric Functions Inverse trigonometric functions Derivatives of the inverse trig functions Exercises vii

5 viii CONTENTS.3 The Natural Logarithm The Exponential Function e x The Functions a x The Functions y = log a x Order and Oh-Notation Applications Hyperbolic Functions The inverse hyperbolic functions Techniques of integration Basic integration formulas Integration by parts Integration by partial fractions Trigonometric substitutions Improper integrals Infinite Sequences and Series Sequences Infinite Series Tests for Convergence of Series The integral test The comparison test The ratio and root tests The alternating series test Power Series Term-by-term differentiation and integration Multiplication of power series Taylor and Maclaurin Series Applications of Power Series Fourier Series Convergence of Fourier series First Order Differential Equations 5 4. First order Linear Differential Equations Separable Equations Population models Brachistochrone problem Exact Equations Existence and Uniqueness Theorem

6 CONTENTS ix 5 Second Order Differential Equations Second Order Linear Differential Equations H O-LDE with Constant Coefficients Nonhomogeneous Equations Variation of parameters Method of undetermined coefficients Applications to Mechanical Vibrations Series Solutions Singular points Regular singular points, method of Frobenius Laplace Transforms Properties of Laplace transforms Discontinuous non-homogeneous functions The Dirac delta function The Convolution Integral Vectors in the Space 6. Cartesian Coordinates The Cross Product Lines and Planes in Space Cylinders and Quadric Surfaces Vector-valued functions 7. Vector Functions Projectile Motion Arc Length Kepler s Laws Functions of several variables Limits and Continuity Partial Derivatives Differentiability Directional Derivatives Derivatives and Chain Rule Taylor s Polynomial Extreme Values Lagrange Multipliers

7 x CONTENTS 9 Multiple Integrals 6 9. Double Integrals Center of Mass Double Integrals in Polar Form Triple Integrals Triple Integrals in Cylinder and Spherical Forms Coordinate Transforms Vector Fields 83. Line Integrals Vector Fields Potential Functions Green s Theorem Surfaces in the Space Stokes Theorem The Divergence Theorem Appendix A 39. The Gamma Function The definition of the gamma function Uniqueness of the gamma function Differentiability of the gamma function The Euler s first integral Γ(x) for large x Stirling s formula The connection with sin x Applications to definite integrals Convex Functions Index 345

8 Chapter Preliminaries. Functions Functions are the major objects we deal with in mathematics because they relate the inputs and the outputs in most phenomenon of the real world. Definition.. A function from a set D to a set R is a rule f that assigns a unique element y R to each element x D. In this case, we use the following notation: We say f is a function from D into R, denoted by f : D R, if x D, y R y = f(x). Note that a function is denoted by f, while a functional value y of f at an element x is denoted by f(x). The set D of all possible input x, called the independent variable, is called the domain of f, and the set of all values y = f(x), the dependent variable, is called the range, or image, of f. The range needs not be the whole set R. When a functional value y = f(x) is given by a formula of real numbers without mentioning the domain, the domain is assumed to be the largest set of x-values for which the formula gives real y-values: For instance, for y = x, the domain is R and the range is R + = {y R y }. For y = x, the domain is R {}. Definition.. A function f : D R is one-to-one (or, or injective) on the domain D if f(x ) f(x ) whenever x x in D. A function f : D R is onto (or surjective to) the range R if, for any y R, there is an element x D such that y = f(x).

9 Chapter. Preliminaries In mathematical notations: f is if f(x ) = f(x ) implies x = x in D. f is onto if y R, x D y = f(x). Definition..3 For a function f : D R, the inverse of f is a function g : R D defined by g(y) = x for y = f(x). It is quite easy to see that if such a function g exists, then f has to be and onto, and conversely. Moreover, such a function g is unique and so denoted by g = f. A function f is said to be invertible if the inverse exists. The following is easy consequence of the definition: Theorem.. A function f : D R is invertible if and only if (abbreviated by iff ) there is a function g : R Dsuch that f g = Id and g f = Id, iff f is and onto.. Limits of Function Values Let f(x) be a function defined on an open interval about x, except possibly at x. A usual intuitive definition of the concept of the limit of f can be read as follows: A number L is the limit of f as x approaches x, if f(x) gets arbitrarily close to a number L for all x sufficiently close to x. However, this definition is informal because the meaning of arbitrarily close and sufficiently close depends on the context. To a machinist manufacturing a clock, close may mean within a few thousandths of a centimeter, while to an astronomer observing distant galaxies, close may mean within a few thousand light-years. Hence, one has to make the meaning of arbitrarily close and sufficiently close more precisely: Definition.. Let f(x) be a function defined on an open interval about x, except possibly at x. We say that f has the limit L as x approaches x, and write lim x x f(x) = L, if, for every number ε >, there exists a corresponding number δ > such that f(x) L < ε, (f(x) is arbitrarily close to L) for all x that satisfies < x x < δ (x is sufficiently close to x ).

10 .. Limits of function values 3 Example.. Show that lim x (5x 3) =. Solution: Given any ε >, we have to find a δ > such that (5x 3) < ε whenever < x < δ.. For x, solve (5x 3) < ε about x = : a simple computation shows (5x 3) = 5x 5 = 5 x < ε, x < ε 5.. Thus, if we choose a value δ = ε 5 δ = ε 5, we have >, then, for x with < x < (5x 3) = 5x 5 = 5 x < 5 ε 5 = ε, which shows that lim x (5x 3) =. In fact any number δ > such that δ ε 5 will do the job. That is, we can choose sufficiently small δ that will work. Example.. For given ε =, find a δ > such that, for all x 5 with < x 5 < δ, the inequality x < which will imply lim x 5 x =. Solution: We find a necessary condition for x < : x < < x < < x < 3 < x < 9 < x <, which means that the inequality holds for all x in (, ). From the given point x = 5, the shortest distance is 5 = 3 < 5 = 5. Thus if we take < δ 3 then the interval ( < 5 δ, 5 + δ < 8) lies in (, ). That is, if x satisfies < x 5 < δ, then clearly < x < so that x < holds. Example..3 For f(x) = x, show that lim x f(x) = 4. Solution: Given any ε >, we want to find a δ > such that f(x) 4 < ε whenever < x < δ.. Solve f(x) 4 < ε about x = : For x, x 4 < ε ε < x 4 < ε, 4 ε < x < 4 + ε 4 ε < x < 4 + ε 4 ε < x < 4 + ε.

11 4 Chapter. Preliminaries. Choose a value δ > such that ( δ, + δ) ( 4 ε, 4 + ε). For this, one can take δ min{ 4 ε, 4 + ε }. Then clearly < x < δ implies that 4 ε < x < 4 + ε, or f(x) 4 < ε. The next theorem tells how to calculate the limits of functions that are arithmetic combinations of functions whose limits are known. Theorem.. Suppose that lim x a f(x) = L and lim x c g(x) = M. Then () lim x c (f(x) ± g(x)) = L ± M. () lim x c (f(x) g(x)) = L M. (3) lim x c (kf(x)) = kl. f(x) (4) lim x c g(x) = L M, provided M. (5) lim x c (f(x)) r/s = L r/s. Proof: We prove (), and leave the others to the readers. For a given ε >, we want to find a δ > such that f(x)+g(x) (L+M) < ε if < x c < δ holds. f(x) + g(x) (L + M) = (f(x) L) + (g(x) M) f(x) L + g(x) M. Since lim x a f(x) = L, δ > f(x) L < ε whenever < x c < δ. Similarly, lim x c g(x) = M implies that δ > g(x) M < ε whenever < x c < δ. Choose δ = min{δ, δ } >. Then < x c < δ implies that f(x) L < ε and g(x) M < ε. Thus, f(x) + g(x) (L + M) f(x) L + g(x) M < ε + ε = ε. The following is a direct consequence of Theorem... Theorem.. If P (x) = a n x n + a n x n + + a and Q(x) are polynomials, then () lim x c P (x) = P (c). () lim x c P (x) Q(x) = P (c) Q(c), provided Q(c).

12 .. Limits of function values 5 Theorem..3 If f(x) g(x) for all x in an open interval containing c, except possibly at x = c, and the limits of both f and g exist as x approaches c, then lim x c f(x) lim x c g(x). Proof: Let lim x c f(x) = L and lim x c g(x) = M. Suppose that L > M. From () in Theorem.., we have lim x c (g(x) f(x)) = M L. Thus, for any ε >, there is a δ > such that (g(x) f(x)) (M L) < ε whenever < x c < δ. Since L M >, take ε = L M. Then for this ε, there exists a δ > such that, for x with < x c < δ, (g(x) f(x)) (M L) < ε = L M holds. This implies g(x) < f(x) for x close to c, which is a contradiction. Corollary..4 (Sandwich Theorem) Suppose that g(x) f(x) h(x) on some open interval about c. If then lim x c f(x) = L. lim g(x) = lim h(x) = L, x c x c.. One-sided Limits and Limits at Infinity () If f(x) is defined on an interval (c, b) with c < b, then the right-hand limit L of f at c is defined as lim f(x) = L : x c + if ε > δ > f(x) L < ε if < x c < δ. () If f(x) is defined on an interval (a, c) with a < c, then the left-hand limit K of f at c is defined as lim f(x) = M : x c if ε > δ > f(x) M < ε if < c x < δ.

13 6 Chapter. Preliminaries (3) If f(x) is defined on an interval (a, b) with c (a, b), then f has a limit at c iff it has both left-hand and right-hand limits and these one-sided limits are equal. (4) By lim x f(x) = L we mean that ε > N > f(x) L < ε if x > N. (5) By lim x f(x) = L we mean that ε > N > f(x) L < ε if x < N. The line y = L is called a horizontal asymptote. For those cases, similar rules like Theorem.. hold. The meaning of the followings are meant by the notations themselves. (6) By lim x c f(x) = we mean that M > δ > f(x) > M if < x c < δ. (7) By lim x c f(x) = we mean that M > δ > f(x) < M if < x c < δ. In the case of 6 and 7, the vertical line x = c is called a vertical asymptote of f. Example..4 Show that sin θ () lim sin θ =, () lim cos θ =, (3) lim θ θ θ θ =. Solution: () From the definition of sin θ, we have θ sin θ θ for all θ. Since lim θ ( θ ) = lim θ ( θ ) =, by Corollary..4, we have lim θ sin θ =. O θ cos θ T P θ tan θ sin θ Q A y = θ y = sin θ y = θ y = θ y = cos θ

14 .. Limits of function values 7 () From the definition of cos θ, we have cos θ θ for all θ. Since lim θ ( θ ) =, again by Corollary..4, we have lim θ cos θ =. (3) We prove the right-hand limit first. Assume that < θ < π. Then from the figure above, we have sin θ < θ < tan θ. Thus < θ sin θ < sin θ, or > cos θ θ > cos θ. Since lim θ cos θ =, by Corollary..4 again, lim sin θ θ + θ =. Since sin θ θ is an even function, we also have lim sin θ θ θ =. Example..5 Prove that lim x + x =. Solution: Let ε > be given. Here we have x = and L =. x = x < ε < x < ε < x < ε. Hence, if we choose < δ ε, then for x < δleε we have < x < ε < x < ε. Example..6 Prove that lim x x =. Solution: Let ε > be given. To find an M such that x = x < ε for all x with x > M, we simply take M ε so that lim x x =. This also shows that the x-axis is a horizontal asymptote of f(x) = x on the right. Example..7 Prove that lim x x =. Solution: Let B > be given. W want to find a δ > such that > B x for all x with < x < δ. Note that x > B x < B x < B. Thus, we simply take δ B. Then for x with x < δ, we have > x δ B. This also shows that the y-axis is a vertical asymptote of f(x) = at x x =.

15 8 Chapter. Preliminaries.. Exercises. For the functions f(x) graphed here, find the left-hand, right-hand, and the limits or explain why they do not exist at the points x =,, 3: 3 3. If lim x f(x) = 5, can we say anything about the value of f at x =? Explain. 3. Prove that lim x a c = c and lim x a x = a. Use this to find the following limits: () lim x x 3 + 4x 3 x + x x, () lim 4x 3, (3) lim + 5 x x x x. 4. Find the following limits: () lim x x + x, () lim x 4 4 x x For given ε > and x, find δ > such that f(x) L < ε for all x with < x x < δ. () f(x) = x, L =, x =, ε =.. () f(x) = x 5, L =, x = 9, ε =. (3) f(x) = x, L = 4, x = 4, ε =.5. (4) f(x) = x 5, L =, x = 4, ε = Prove the following limit statements: () lim x 4 x =. () lim x f(x) = 4 for f(x) = (3) lim x x sin x =. { x, x x =. 7. Prove or disprove the following statements: () The number L is the limit of f(x) as x approaches x if f(x) gets closer to L as x approaches x.

16 .3. CONTINUITY OF FUNCTIONS 9 () The number L is the limit of f(x) as x approaches x if, given any ε >, there exists a value of x for which f(x) L < ε. 8. Find the limit and the aymptote of the following functions: () lim x 5x + 8x 3 3x, () lim + x 3y 9. Find the asymptotes of the following functions: (x 3). () y = + sin x x, () y = x + 3 x +, (3) y = 8 x 4, (4) y = x 3 x 4..3 Continuity of Functions Definition.3. A function f : (a, b) R defined on an interval (a, b) is continuous at a point c (a, b) if lim f(x) = f(c). x c i.e., if (i) lim x c f(x) exists, (ii) f(c) is defined, (iii) lim x c f(x) = f(c). Continuity of f at the end points a and b are defined by the right-hand and left-hand limits of f at a and b. A function is continuous on an interval if it is continuous at every point of the interval. Theorem.3. Let f and g be continuous functions at c. Then the following algebraic combinations of them are also continuous at c: f ± g, f g, kf, f r/s, f g provided g(c). This theorem can be proved from the limit rules in Theorem... For instance, the first part can be seen by: lim x c (f ± g)(x) = lim(f(x) ± g(x)) = lim x c x c = f(c) ± g(c) = (f ± g)(c). f(x) ± lim x c g(x) Thus the polynomials and rational functions are all continuous wherever they are defined at c. The functions y = sin x and y = cos x are continuous at x = by Example..4. In fact, they are continuous everywhere (see problem 5 in Exercise.3.). Hence, all the six trigonometric functions are continuous where they are defined by Theorem.7..

17 Chapter. Preliminaries Example.3. Prove that the function f(x) = x is continuous everywhere in R. { x, x > Solution: Note that f(x) = x = which consists of two x, x < polynomials on R {}. At the origin, lim x x = =. Thus the function is continuous on R. Theorem.3. If f is continuous at c and g is continuous at f(c), the the composition g f is continuous at c. If a function f is not defined at c so that f(c) does not exist, but lim x c f(x) = L exists, then one can define a new function F (x) by the rule: { f(x) if x is in the domain of f F (x) = L if x = c. The function F is continuous at x = c, called the continuous extension of f to x = c. Example.3. The domain of the function f(x) = sin x x is R {} on which it is continuous everywhere. However, since lim x f(x) =, we can extend the domain of f to the whole real line R by { sin x F (x) = x, x, x =. which is continuous on R. Example.3.3 The domain of the function f(x) = x +x 6 is R {±} on x 4 which it is continuous everywhere. Can f(x) be extended to bigger set? Solution: Note that f(x) = x + x 6 x 4 = (x + 3)(x ) (x + 3) = (x + )(x ) (x + ), so that lim x f(x) = 5 4. Thus f(x) can be extended to R { } by F (x) = { x +x 6 x 4, x ± 5 4, x =.

18 .3. Continuity of functions However, since lim x f(x) = ±, it cannot be extended to R. Continuous functions have a very useful property, called intermediate value property, saying that, whenever it takes on two values in range, it takes on all the values in between: Theorem.3.3 If f : [a, b] R is continuous on the closed interval [a, b], then f takes its maximum and minimum at some points c, d [a, b]. Theorem.3.4 (Intermediate Value Theorem) Let f : [a, b] R be a continuous function on [a, b]. Then, for any value y between f(a) and f(b), there is a point c [a, b] such that f(c) = y. The proof of this theorem depends on the completeness property of R, and may found in an advanced calculus..3. Exercises. Find the points where the following functions fail to be continuous. At which points, if any, are the discontinuities removable? Not removable? Give reasons for your answers. () y = (x ) + 4. () y = x + x 4x + 3. (3) y = cos x x. (4) y = x4 + + sin x.. For what value of a is f(x) = { x, x < 3 ax, x Suppose that a function f is continuous on the closed interval [, ], and f(x) for every x [, ]. Show that there must exist a number c in [, ] such that f(c) = c (c is called a fixed point of f). 4. Suppose that a function f defined on an interval (a, b) is continuous at some c (a, b) and f(c) >. Prove that there is an interval (c δ, c + δ) about c such that f(x) > for all x (c δ, c + δ). 5. Prove that f is continuous at a if and only if lim h f(a + h) = f(a). Use this fact to prove that y = sin x and y = cos x are continuous at every point x = a.

19 Chapter. Limits and Continuity.4 Derivatives A newly designed car by Hyun Dai Motors has been tested and its fuel economy (Kilometer/litter) depending on the speed is shown in the following diagram: Fuel Economy (km/litter) A B C Speed (km/hr) At the speed A, the fuel economy is still improving, and at the speed C it is getting worse as the sped increases, since the curve is rising and decaying at those points. However, at the speed B it is instantaneously staying still, which means the curve reached to the pick point. Thus the speed B is the most economic speed of the car. Now, problem is how can we find the point B? In the figure, we draw lines tangent to the graph at several points and compared the slope of the lines, and picked up point B since the line is horizontal, that is, instantaneously the curve is not increasing nor decreasing there. Then, how can we find the lines tangent to the curve, and how can we compute the slopes of the lines? To find the tangent line at a point P of the curve, we pick up a point on the curve nearby P and draw the secant line P Q joining P and Q, and then move the point Q toward P along the curve. Q P Q

20 .4. Derivatives 3 We do this in both sides of P : from the left and from the right sides of P. If the two limit lines, if exist, as Q approaches infinitely close to P coincide, it is called the tangent line to the curve at P. Now, the slope of the tangent line will be the limit of the slopes of the secant lines P Q: Suppose the points are denoted by P = (x, f(x )) and Q = (x, f(x)). Then the slope of the secant line P Q is f(x) f(x ) x x. Thus the slope of the tangent line is f(x) f(x ) lim, x x x x provided the limit exists. Historically, the slope of the secant line was called the difference quotient of f at x, and the limit, as x approaches x, of the difference quotient was called the derivative of f at x, and denoted by f (x ) = df(x ) dx = dy dx (x ) = y (x ). In this case, f is said to be differentiable, or has a derivative, at x. In practical meaning, it represents the rate of change of f(x ) as x moves nearby x. Now, the point x can be any point in the domain of f considered as a variable in the domain, and so replace it by x: Definition.4. The derivative of f is the function f whose value at x is f f(x + h) f(x) (x) = lim. h h If f exists at every point in the domain I of f, we call f differentiable on I. There are many notations for the same derivative of f: f (x) = f = y = dy dx = df dx = d f(x) = D(f)(x). dx Example.4. Find the derivative of y = x, and the tangent line at x = 4. Solution: A direct computation from the definition goes as follow: f f(z) f(x) z x (x) = lim = lim z x z x z x z x z x = lim z x ( z x)( z + x) = lim z x ( z + x) = x.

21 4 Chapter. Limits and Continuity At x = 4, f () = = 4 4. Thus the tangent line is y = 4 (x 4), or y = 4 x +. Example.4. Find the derivative of y = x at x =. Solution: The right-hand and left-hand derivatives of x at x = are f ( + ) = + h h lim = lim x + h x + h = lim h x + h =, f ( ) = + h h lim = lim x h x h = lim h x h =, which are different. Thus x is not differentiable at x =. For the example given at the outset of this section, we are looking for a point B where f (x) =. Theorem.4. (First derivative test) If a differentiable function f has a local extremum at an interior point c of its domain, then f (c) =. Theorem.4. If f has a derivative at x = c, then f is continuous at x = c. Proof: We need to show that lim x c f(x) = f(c), or lim h f(c+h) = f(c). Since f (c) exists, lim h f(c + h) = lim[f(c) + (f(c + h) f(c))] h = f(c + h) f(c) lim f(c) + lim h h h h = f(c) + f (c) = f(c). Theorem.4.3 (Intermediate value theorem of derivatives) If f is differentiable on an interval I = [a, b], then for any value z between f (a) and f (b), there is a point c [a, b] such that f (c) = z.

22 .4. Derivatives 5 Proof: Suppose that f (a) z f (b). Consider a function g(x) = f(x) zx, which is continuous on I. Then g (a + ) = f (a + ) z < implies g(a) > g(x ) at some x (a, b). Similarly, g (b ) = f (b ) z > implies g(x ) < g(b) at some x (a, b). Thus the minimum of g(x) is at some point c in (a, b). Since g is differentiable at c (a, b), by Theorem.4. we must have = g (c) = f (c) z, or f (c) = z. Theorem.4.4 (Differentiation rules) Let f and g be differentiable functions on an interval, and k a constant. () (kf + g) = kf + g. () (f g) = f g + f g. (3) ( f g ) = f g f g g. (4) (x n ) = nx n, for any integer n (x if n < ). Proof: () is clear and left for the reader. () (Product rule) d (f(x)g(x)) = lim dx h = lim h (3) (Quotient rule) f(x + h)g(x + h) f(x + h)g(x) + f(x + h)g(x) f(x)g(x) [ h ] g(x + h) g(x) f(x + h) f(x) f(x + h) + g(x) h h g(x + h) g(x) f(x + h) f(x) + g(x) lim h h h h = lim f(x + h) lim h = f(x)g (x) + g(x)f (x). d dx ( ) f(x) g(x) = lim h f(x+h) g(x+h) f(x) g(x) h = lim h g(x)f(x + h) f(x)g(x + h) hg(x + h)g(x) = lim h f(x + h)g(x) f(x)g(x) + f(x)g(x) f(x)g(x + h) hg(x + h)g(x) g(x) f(x+h) f(x) h h = lim g(x + h)g(x) = g(x)f (x) f(x)g (x) g(x). f(x) g(x+h) g(x) h

23 6 Chapter. Limits and Continuity (4) For a positive integer n, use induction on n: For n =, x = x and d (x + h) x h (x) = lim = lim dx h h h h = = x = x. Assume for n. Then, by the product rule, d dx (xn ) = d dx (x xn ) = x n + x (n )x n = nx n. For a negative integer n = m, f(x) = x n = x m. Thus, by the quotient rule, d dx (xn ) = d dx x m = xm (mx m ) x m = mx m = nx n. by If f is derivative, then its derivative is the second derivative of f, denoted d dx ( ) dy = dy dx dx = y = f (x) = d y dx = D (f)(x). The higher order derivatives are also defined in the same way. Example.4.3 Show that d d dx (sin x) = cos x and dx (cos x) = sin x. Solution: From the usual identities of the trigonometric functions, d sin(x + h) sin x (sin x cos h + cos x sin h) sin x (sin x) = lim = lim dx h h h h sin x(cos h ) + cos x sin h = lim h h cos h sin h = sin x lim + cos x lim h h h h = sin x + cos x = cos x. Similarly, one can find the derivative of cos x.

24 .5. DIFFERENTIALS 7.4. Exercises. Prove the following formulas: (tan x) = sec x, (sec x) = sec x tan x, (cot x) = csc x, (cscx) = cscx cot x.. Find the first and second derivatives of the functions: () y = 3x ( x. () y = (x + ) x ). ( ) x x + 3x (3) y =. (4) y = (3 x). x 3z.5 Differentials Let y = f(x) be a differentiable function on an interval I, and a I. When y = f(a) is known, f(x), for x I close to a, can be approximated by the value on the tangent line to the graph at a as follows: The tangent line to y = f(x) at x = a is L(x) = f(a) + f (a)(x a). Thus the increment y = f(x) f(a) of f at x I can be approximated by the increment L = L(x) f(a) = f (a)(x a) of the tangent line at x: i.e., y = f(x) f(a) L(x) f(a) = f (a) x, x = x a, or f(x) L(x) = f(a) + f (a)(x a), which is called the linear approximation, or linearization of f at x. f(x) L(x) f(a) L y a x

25 8 Chapter. Preliminaries Originally, the notation dy dx for the derivative of y with respect to x does not represent the ratio of dy and dx, but simply represents the slope of the graph of y = f(x). However, the formula of the linear approximation shows that the new variables dy and dx can be separated and the derivative dy dx is a true ratio of them: From the first equation above, by setting dx = x and x = a, the differential dy of f is defined as dy = f (a)dx. Geometrically, the differential dy is the change L(x) = L(x) L(a) in the linearization of f given as f (x)dx when x = a changes by an amount dx = x. If dx, then the quotient of the differential dy by the differential dx is equal to the derivative f (x) because dy dx = f (x)dx dx = f (x) = dy dx. Sometimes, we write df = f (x)dx in place of dy = f (x)dx, calling the differential of f. Example.5. Find the linear approximation of f(x) = + x at x = 3.. Solution: Let a = 3. Then f(3) =, f (3) = ( + x) / x=3 = 4, and so the linear approximation is L(3.) = + 4 (x 3) x=3. = = The true change y = f(x) f(a) of y = f(x) approximated by the differential estimate dy = f (a)dx when x changes by dx = x = x a with an error y dy = y f (a)dx = f(x) f(a) f (a) x ( ) f(a + x) f(a) = f (a) x x = ε x. Note that ε = f(a + x) f(a) x f (a), x.

26 .5. Differentials 9 Hence, the approximation error ε x is small when x is small. Therefore, we have y = f (a) x + ε x, where ε as x. Note that this equation is sometimes taken as the definition for the differentiabilty of f at x = a. Theorem.5. (Chain rule) Suppose that y = g(x) is differentiable at x, and z = f(y) is also differentiable at y = g(x). Then the composition z = f g(x) = f(g(x)) = f(y) is differentiable at x and (f g) (x) = f (g(x)) g (x), or dz dx = dz dy dy dx. Proof: Let x be an increment in x and let y and z be the corresponding increments in y and z. Then, from the differential equation, y = g (x) x + ε x, ε as x, z = f (y) y + ε y, ε as y = (f (y) + ε )(g (x) + ε ) x. z x = f (y)g (x) + g (x)ε + f (y)ε + ε ε, Since y as x, ε and ε as x. Thus dz dx = lim z x x = f (y)g (x) = f (g(x)) g (x). If y = f(x) is a differentiable function of x and x = g(t) is a differentiable function of t, then dy dt = dy dx dx dt, or d dt f(x) = f (x) dx dt. For example, if x is a differentiable function of t, then the power chain rule says that d dt xn n dx = nx dt.

27 Chapter. Preliminaries Example.5. d dt tan(5 sin t) = sec (5 sin t) d (5 sin t) dt = cos t sec (5 sin t). ( ) d = d (3x ) dt 3x dt = (3x ) d dt (3x ) = 3 (3x ). Example.5.3 Note that, when we take the derivatives of the trigonometric functions, the variable x is assumed to be measured in radian, not degree: 8 = π radians, or x = π 8x radians. The chain rule says that d dx sin(x ) = d πx sin( dx 8 ) = π πx cos 8 8 = π 8 cos(x )..5. Parametric equations If a car is moving on the xy-plane, the position (x, y) of the car in time may be represented as (x = f(t), y = g(t)) over a time interval t (a, b). In this case, the path in which the car moves is called a parametric curve, and the equations are called parametric equations for the curve in the parameter t. Thus the position of the car may be written as a vector valued function in t: α(t) = (f(t), g(t)) R, t [a, b]. For example, the parametric curve α(t) = (cos t, sin t), t [, π] represents the unit circle in the xy-plane tracing the entire circle once counterclockwise. If the two functions f(t) and g(t) are differentiable, we say the parametric curve α is differentiable and its derivative is denoted by α (t) = (f (t), g (t)). Since, in α (t) = dα dt = lim t α(t + t) α(t), t the numerator is a vector from the position α(t) to α(t + t) and the denominator t represents the time interval, α (t) geometrically represents the infinitesimal direction of motion, called the velocity vector of the car, and the magnitude α (t) = distance time is called the speed of the car.

28 .5. Differentials y α(t) α (t) α(t + t) x If y is considered as a differentiable function of x in the parametric curve, then the derivative dy dx can be obtained by the chain rule: dy dt = dy dx dx dt, dy or dx = dy/dt dx/dt, provided that dx. If the curve α(t) = (f(t), g(t)) R, t [a, b], defines y as twice differentiable functions of x, then for dy dx = y d y dx = d dx y = dy /dt dx/dt. Example.5.4 The ellipse x + y = can be parametrized as α(t) = a a (a cos t, b sin t) R, t [a, b]. Find the line tangent to the curve at the point ( a b, ) when t = π 4. Solution: The slope of the tangent line at t = π 4 is dy = dy/dt = b cos t dx dx/dt a sin t t= π 4 The tangent line is now t= π 4 t= π 4 = b/ a/ = b a. y = b/ b a (x a/ ) = b a x + b. Example.5.5 If an aircraft releases a bomb toward a ground target, the bomb moves along the path α(t) = (x(t), y(t)) = (t, 6t + 5), t. When does the bomb hit the target? How far did it fly in horizontal distance? Find the Cartesian equation of the path of the bomb, and the rate of descent relative to its forward motion when it hit the target.

29 Chapter. Preliminaries Solution: The bomb hit the target when y =. Thus 5 6t + 5 = t = 6 = 5 5 >. The horizontal distance it flew is x( ) = 5 = 3 5 meter. The Cartesian equation is obtained by eliminating t: Plug t = y = 6t + 5 = 6( x ) + 5 = 9 x + 5. x in The rate of descent relative to its forward motion when t = 5 5 is dy = dy/dt = 3t = dx dx/dt 3 t= 5 5 t= 5 5 t= 5 5 Thus it falls about.5 m per unit meter of forward motion when it hits the target..5. Implicit differentiation Sometimes, a function y = f(x) may be given by an implicit equation F (x, y) =, e.g., x + y 6 =. In some cases we may be able to solve such an equation for y as an explicit function y = f(x) in x. In some cases it is not easy to solve such an equation for y as an explicit function y = f(x) in x. In those cases we may find dy dx by implicit differentiation: Take differentiation both sides of the equation in x and then solve the resulting equation for y. Example.5.6 Find dy dx for y = x + sin xy. Solution: y dy dx (y x cos xy) dy d dx (y ) = d dx (x ) + d (sin xy) dx dx dy dx = x + cos xy(y + x dy dx ) = x + y cos xy = x + y cos xy y x cos xy.

30 .5. Differentials 3 Theorem.5. (Power rule) For a rational number p q, d dx x p q = p q x p q. Proof: For y = x p q, y q = x p. Thus by the implicit differentiation, qy q y = px p. y = p q xp y (q ) = p q xp x p q (q ) = p q xp x p+ p q = p q x p q..5.3 Applications of derivatives Many important theorems in mathematics require the following Extreme value theorem, whose proof requires a detailed knowledge of real number system. Hence we just state the theorem without proof. Theorem.5.3 (Extreme Value Theorem) If f is a continuous function on a closed interval [a, b], then there are numbers x and x [a, b] such that f(x ) f(x) f(x ) for all x [a, b]. The numbers f(x ) = m and f(x ) = M are called the absolute minimum and absolute maximum of f. A function has a local maximum (or, local minimum) at a point c in the domain if f(x) f(c) (or, f(x) f(c), respectively) for all x in some (half-)open interval containing c when c is an (end point) interior point of its domain. Theorem.5.4 (First Derivative Test) If f has a local maximum or local maximum at an interior point c of its domain and f (c) exists, then f (c) =. Proof: Suppose that f has a local maximum at x = c so that f(x) f(c) for all x near c. Then f(x) f(c) lim = f f(x) f(c) (c) = lim, x c x c x c + x c which implies f (c) =. Similarly, the same holds at a local minimum.

31 4 Chapter. Preliminaries Example.5.7 Find the absolute maximum and minimum values of y = f(x) = x /3 on [, 3]. Solution: The first derivative of f is f (x) = 3 x /3 = 3 3 x = { <, if x < > if x >, for x. f () is not defined. Since f() =, f( ) = 3 4, and f(3) = 3 9, the absolute minimum is f() = and the absolute maximum is f(3) = 3 9. Theorem.5.5 (Rolle s Theorem) Suppose that f : [a, b] R is continuous on [a, b] and differentiable on (a, b). If f(a) = f(b), then there is at least one point c (a, b) such that f (c) =. Proof: If f is a constant function with f(x) = f(a) = f(b) for all x [a, b], then f (x) = for all x [a, b]. If f is not a constant function with f(a) = f(b) and is differentiable on (a, b), then there is a point c where f take its extremum value by Theorem.5.3. By Theorem.5.4, f (c) =. Theorem.5.6 (The Mean Value Theorem) Suppose that f : [a, b] R is continuous on [a, b] and differentiable on (a, b). Then there is at least one point c (a, b) such that f(b) f(a) b a = f (c). Proof: The equation of the line through (a, f(a)) and (b, f(b)) is g(x) = f(a) + f(b) f(a) (x a). b a The vertical distance between the graphs of f and g is h(x) = f(x) g(x) = f(x) f(a) f(b) f(a) (x a) b a which satisfies the hypothesis of Rolle s Theorem on [a, b]: it is continuous on [a, b] and differentiable on (a, b), and h(a) = h(b) =. Hence h (c) = at some point c (a, b). But = h (c) = f (c) f(b) f(a). b a

32 .5. Differentials 5 Corollary.5.7 If f (x) = at each point x in an open interval (a, b), then f is a constant function on the interval: i.e, f(x) = C for all x (a, b). Proof: For any two points x x in (a, b), by Theorem.5.6, there is a point c (x, x ) such that f(x ) f(x ) x x = f (c) =. Thus, f(x ) f(x ) =, or f(x ) = f(x ). Corollary.5.8 If f (x) = g (x) at each point x in an open interval (a, b), then f g is a constant function on the interval: i.e, f(x) = g(x) + C for all x (a, b). Proof: For h(x) = f(x) g(x), h (x) = f (x) g (x) = for all x (a, b). Thus h(x) = f(x) g(x) = C on (a, b) by Corollary Exercises. Find the differential dy for y = x x. At x = what is the value of dy for dx =.?. Find the linearization L(x) of f(x) at x = a. () f(x) = x + x, a =. () f(x) = cos x, a = π. (3) f(x) = ( + x) k, a =. (4) f(x) = +x, a =. 3. Estimate (.) 5 and 3.9 by using the linearization. 4. Estimate the change in the volume V = 4 3 πr3 of the sphere when the radius changes from to.. 5. Estimate the change in the surface area S = 6x of a cube when the edge lengths changes from to.. 6. Find the derivatives dy dx of the following functions. () y = sin x sin x. () y = sin(x + x + ). + x (3) x y =. (4) y = x.

33 6 Chapter. Limits and Continuity 7. () Express the lateral surface area S of the right circular cone in terms of the base radius r and the height h. () Find ds dt (3) Find ds dt (4) Find ds dt dr in terms of dt dh in terms of dt dr in terms of dt when h is constant. when r is constant. and dh dt when neither r nor h is constant. 8. A particle moves along the curve y = x 3/ in the first quadrant in such a way that its distance from the origin increases at the rate of unit per second. when t = 3. Find dx dt.6 Indeterminate Forms Let f(x) and g(x) be both continuous functions such that f(a) = = g(a). Then f(x) lim x a g(x) = f(a) g(a) = is called an indeterminate form. However, from the definition of the derivative: f (a) g (a) = lim f(x) f(a) x a x a f(x) f(a) = lim x a g(x) g(a) = lim f(x) x a g(x). g(x) g(a) lim x a x a Theorem.6. (L Hôpital s Rule) Suppose that f(a) = = g(a), f (a) and g (a) exist, and that g (a). Then f(x) lim x a g(x) = f (a) g (a). 3x sin x Example.6. () lim x x = 3 cos x x= =. () lim +x x x = +x =. x= Theorem.6. (Stronger form of L Hôpital s Rule) Suppose that f(a) = = g(a), f and g are both differentiable on an open interval I containing a, and that g (x) on I if x a. Then f(x) lim x a g(x) = lim f (x) x a g (x).

34 .6. Indeterminate Forms 7 Example.6. x sin x cos x () lim x x 3 = lim x 3x + x x/ () lim x x = lim x /4( + x) 3/ sin x = lim x 6x = lim cos x = x 6 6. = lim x /( + x) / / x = 8. Theorem.6. is based on the following Cauchy s Mean Value Theorem: Theorem.6.3 (Cauchy s Mean Value Theorem) Let f(x) and g(x) be both continuous functions on [a, b] and differentiable on (a, b), and g (x) on (a, b). Then there is a number c (a, b) such that f (c) g (c) = f(b) f(a) g(b) g(a). Proof: Since g (x) on (a, b), we have g(b) g(a). Consider a function F (x) = f(x) f(a) f(b) f(a) (g(x) g(a)), g(b) g(a) which is differentiable where f and g are, and F (b) = F (a) =. Thus, by the Mean Value Theorem, there is c [a, b] such that F (c) = : i.e., F (c) = f (c) f(b) f(a) g(b) g(a) g (c), or f (c) g (c) = f(b) f(a) g(b) g(a). Notice that the Mean Value Theorem.5.6 is a special case of the Cauchy s Mean Value Theorem with g(x) = x. Moreover, a geometric interpretation of Theorem.6.3 can be seen as follows: For a curve C defined by the parametric equations α(t) = (x, y) = (g(t), f(t)), t [a, b], the slope of the curve at t is given by dy/dt dx/dt = f (t) g (t). Thus, f (c) g (c) is the slope of the tangent to the curve C at t = c. The secant line joining the two points (g(a), f(a)) and (g(b), f(b)) has the slope f(b) f(a) g(b) g(a).

35 8 Chapter. Limits and Continuity (g(c), f(c)) (g(b), f(b)) (g(a), f(a)) Proof: [Proof for Theorem.6.] For a < x, g (x). Theorem.6.3, then c [a, x] such that Set b = x in f (c) g (c) since f(a) = = g(a). Thus = f(x) f(a) g(x) g(a) = f(x) g(x), f(x) lim x a + g(x) = lim f (c) c a + g (c) = lim f (x) x a + g (x). The same holds as x a. The following examples show Theorem.6. can also be used to deal with the cases like,,, as x a. Example.6.3 sec x () lim x (π/) + tan x = lim sec x tan x x (π/) sec = lim sin x =. x x (π/) () lim x x x 3x + 5x = lim 4x x 6x + 5 = 3. (3) lim (x sin x x ) = lim h + h (4) lim x ( = lim x sin x x ) = lim x cos h sin h = lim =. h + x sin x x sin x = lim x cos x sin x + x cos x sin x cos x x sin x =.

36 .7. ANTIDERIVATIVES 9.6. Exercises. Evaluate the limits. cos x sin(a + h) sin a () lim x x. () lim h. h (3) lim x x tan x. (4) lim x x x + x x sec x (5) lim x + (6) lim x (π/) sin x tan x..7 Antiderivatives Many real problems require a function F from its known derivative f. If such a function F exists, it is called an antiderivative of f. Definition.7. A function F is an antiderivative of f on an interval I if F (x) = f(x) for all x I. Corollary.5.8 says that if F (x) and G(x) are both antiderivatives of f(x), then G(x) = F (x)+c for some constant C. Thus, F (x)+c represents all the antiderivatives of f(x). Definition.7. The set of all antiderivatives of f is the indefinite integral of f with respect to x, defined by f(x)dx = F (x) + C, or F (x)dx = F (x) + C, where F (x) is an antiderivative of f(x), is called an integral sign, f is the integrand of the integral, and x is the variable of integration. For example, (x + cos x)dx = x + sin x + C, and Function f(x) Antiderivative F (x) = f(x)dx x n xn+ + C, n, n rational n+ sin kx cos kx k + C, k a constant, k sin kx 3 cos kx k + C, k a constant, k 4 sec x tan x + C 5 csc x cot x + C 6 sec x tan x sec x + C 7 csc x cot x csc x + C

37 3 Chapter. Limits and Continuity Finding an antiderivative for a function f(x) is the same problem as finding a function y(x) that satisfies the equation dy dx = f(x). In general, an equation involving derivatives of a function y = f(x) is called a differential equation. Solving a differential equation means finding a function y = f(x) which satisfies the given differential equation..7. Definite integrals Let y = f(x) be an arbitrary function defined on a closed interval [a, b]. Partition [a, b] into n subintervals by choosing n points between a and b: P = {a = x < x < < x n = b}. Let x k = x k x k denote the width of the subinterval [x k, x k ]. This is the change of x from x k to x k and approximated by the differential dx. Let c k be any point in [x k, x k ]. Then f(c k ) x k is ± the area of the vertical rectangle over the subinterval [x k, x k ] with height f(c k ) and ± depending on the sign of f(c k ). A Riemann sum for f on [a, b] is the sum n S P = f(c k ) x k. k= We now can make various partitions of the interval [a, b], and many choices of the points c k. et P = max{ x k x k P }, called the norm of P. The limit of S P as P, if it exists, is called the definite integral of f over [a, b]: Definition.7.3 For a function f(x) on [a, b], the definite integral of f over [a, b] is denoted by lim P k= n f(c k ) x k = I = b a f(x)dx,

38 .7. Antiderivatives 3 provided the limit exists: i.e., for any ε > there is a number δ > such that for every partition P of [a, b] with P < δ and any choice of c k in [x k, x k ], n f(c k ) x k I < ε. k= In this case f is said to be integrable over [a, b]. Theorem.7. A continuous function is always integrable. Example.7. The function f(x) = is not integrable over [, ]. {, if x is rational, if x is irrational Theorem.7. Let f and g be integrable functions over [a, b]. The following rules hold: () a b f(x)dx = b a f(x)dx. () a a f(x)dx =. (3) b a kf(x)dx = k b a f(x)dx, b a f(x)dx = b a f(x)dx. (4) b a (f(x) ± g(x))dx = b a f(x)dx ± b a g(x)dx. (5) c a f(x) + b c f(x)dx = b a f(x)dx. (6) If M = max f(x) and m = min f(x) over [a, b], then m(b a) b a f(x)dx M(b a). (7) If f(x) on [a, b], then b a f(x)dx. If f(x) g(x) on [a, b], then b a f(x)dx b a g(x)dx. Definition.7.4 If f is integrable functions over [a, b], the average value of f on [a, b], called its mean value, is av(f) = b a b a f(x)dx.

39 3 Chapter. Limits and Continuity Theorem.7.3 (The Mean Value Theorem) If f is continuous on [a, b], then there is a point c [a, b] such that f(c) = b a b a f(x)dx. Proof: By the Extreme Value Theorem.5.3, f assumes the absolute maximum and minimum on [a, b]. Thus, min f b a b a f(x)dx max f. Since f is continuous, by the Intermediate Value Theorem for continuous functions, there exists a number c [a, b] such that f(c) = b b a a f(x)dx. Example.7. Suppose that f is continuous on [a, b], a b, and if b then f(c) = at some c [a, b], since min f av(f) = b a a b a f(x)dx =, f(x)dx = = max f, b a and so one can use the intermediate value theorem of continuous functions for the existence of c in [a, b]. Let f be a continuous functions over I = [a, b]. For x I, define a new function F (x) = x a f(t)dt. If f is nonnegative on [a, b], F (x) is the area under the graph of f from a to x. Let M = max f on [a, b]. For any ε >, take < δ ε M. Then, for any h < δ, (x + h) x = h < δ. By Theorem.7.3, there is a c [x, x + h] such that x+h F (x + h) F (x) = f(t)dt = hf(c) < Mδ M ε M = ε, x which shows that F is also continuous at each point x in [a, b].

40 .7. Antiderivatives 33 Theorem.7.4 (The First Fundamental Theorem of Calculus) If f is continuous on [a, b], then F (x) = x a f(t)dt is continuous on [a, b] and differentiable on (a, b) and its derivative is f(x): F (x) = d dx x a f(t)dt = f(x). Proof: The continuity of F is already shown. Note that F (x + h) F (x) h = h x+h x f(t)dt = f(c), where the last equality holds from the mean value theorem for some c [x, x + h]. As h, x + h x, and so c x. Since f is continuous at x, f(c) f(x): Thus F F (x + h) F (x) (x) = lim = lim f(c) = f(x). h h h Theorem.7.5 (The Second Fundamental Theorem of Calculus) If f is continuous on [a, b] and F is an antiderivative of f on [a.b], then b a f(x)dx = F (b) F (a). Proof: Note that, by the first fundamental theorem, F (x) = x a f(t)dt is an antiderivative of f. If G(x) is any other antiderivative of f, then G(x) = F (x) + C for some constant C. Then G(b) G(a) = F (b) F (a) = b a f(x)dx a a f(x)dx = b a f(x)dx. This theorem has a significant meaning in mathematics: the left side is computing the area under the graph which seems to be noting to do with the differentiation. But it turns out that it is computed from an antiderivative evaluated just at the two end points regardless of its values at points in the interval.

41 34 Chapter. Limits and Continuity.7. Integrals by substitution Note that a definite integral of a continuous function is a number as the limit of the Riemann sums, and the fundamental theorem says that it can be computed easily if we can find an antiderivative of the function. Finding antiderivatives is generally more difficult than fining derivatives. Here we present some techniques for computing them. Note that, for a differentiable function z = F (u), F (u) + C = F (u)du. Suppose also that u = g(x) is differentiable function. Then, by the Chain rule, d dx F (g(x)) = F (g(x))g (x) = f(g(x))g (x), where F (u) = f(u). Hence, d f(g(x))g (x)dx = F (g(x))g (x)dx = F (g(x))dx = F (g(x)) + C dx = F (u) + C = F (u)du = f(u)du. Theorem.7.6 (The Substitution Rule) If f is a continuous function in u and u = g(x) is a differentiable function, then f(u)du = f(g(x))g (x)dx. Example.7.3 4x u du dx = 4x 4dx = 4 4 dx dx = 4 u +/ Example.7.4 x 3 x + dx = u / du = 4 / + + C = 6 u3/ + C = 6 (4x )3/ + C. u /3 du = u/3 /3 + C = 3 u3/ + C = 3 (x + ) /3 + C. If u = g(x) has continuous derivative on [a, b] and z = f(u) is continuous on the range of g, then, for a anti derivative F (u) of f, b a f(g(x))g (x)dx = F (g(x)) x=b x=a = F (g(b)) F (g(a)) g(b) = F (u) u=g(b) u=g(a) = f(u)du. g(a)

42 .7. Antiderivatives 35 Example.7.5 3x x 3 + dx = udu = 3 u3/ = 3 3/ = Exercises. Find the indefinite integrals. 6dx () (x ). () sec s 5 ds. (3) cos θ + cos 4t 5 dθ. (4) dt. cscθ sin t (5) dθ. (6) cscθ sin θ t cos 3 t dt.. Evaluate the following definite integrals. () (3) (5) 4 9 π/ π x u dx. () π/3 cos θ 5 dθ. (4) (cos θ + cos θ )dθ. (6) π/ sec sds. + cos t dt. t + t t dt. 3. Find the limit: lim x x x 3 t t 4 + dt. 4. Suppose that z = f(u) is a continuous function on [a, b], and u(x), v(x) are differentiable functions of x whose values lie in [a, b]. Prove that d dx u(x) v(x) 5. Find the linearization of f(x) = f(t)dt = f(u(x)) du dv f(v(x)) dx dx. x+ 9 dt, at x =. + t