4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page Chapter PRELIMINARIES OVERVIEW This chapter reviews the basic ideas you need to start calculus The topics include the real number system, Cartesian coordinates in the plane, straight lines, parabolas, circles, functions, and trigonometry We also discuss the use of graphing calculators and computer graphing software 1.1 Real Numbers and the Real Line This section reviews real numbers, inequalities, intervals, and absolute values Real Numbers Much of calculus is based on properties of the real number system Real numbers are numbers that can be expressed as decimals, such as - = - 0.75000 Á = 0.33333 Á 22 = 1.4142 Á The dots Á in each case indicate that the sequence of decimal digits goes on forever Every conceivable decimal expansion represents a real number, although some numbers have two representations For instance, the infinite decimals 999 Á and 1.000 Á represent the same real number A similar statement holds for any number with an infinite tail of 9’s The real numbers can be represented geometrically as points on a number line called the real line –2 –1 – ͙2 3 The symbol ޒdenotes either the real number system or, equivalently, the real line The properties of the real number system fall into three categories: algebraic properties, order properties, and completeness The algebraic properties say that the real numbers can be added, subtracted, multiplied, and divided (except by 0) to produce more real numbers under the usual rules of arithmetic You can never divide by www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 2 Chapter 1: Preliminaries The order properties of real numbers are given in Appendix The following useful rules can be derived from them, where the symbol Q means “implies.” Rules for Inequalities If a, b, and c are real numbers, then: a b Q a + c b + c a b Q a - c b - c a b and c Q ac bc a b and c Q bc ac Special case: a b Q -b - a a Q a If a and b are both positive or both negative, then a b Q 1 a b Notice the rules for multiplying an inequality by a number Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality Also, reciprocation reverses the inequality for numbers of the same sign For example, but - - and 1>2 1>5 The completeness property of the real number system is deeper and harder to define precisely However, the property is essential to the idea of a limit (Chapter 2) Roughly speaking, it says that there are enough real numbers to “complete” the real number line, in the sense that there are no “holes” or “gaps” in it Many theorems of calculus would fail if the real number system were not complete The topic is best saved for a more advanced course, but Appendix hints about what is involved and how the real numbers are constructed We distinguish three special subsets of real numbers The natural numbers, namely 1, 2, 3, 4, Á The integers, namely 0, ; 1, ;2, ; 3, Á The rational numbers, namely the numbers that can be expressed in the form of a fraction m>n, where m and n are integers and n Z Examples are , - -4 4 = = , 9 -9 200 , 13 and 57 = 57 The rational numbers are precisely the real numbers with decimal expansions that are either (a) terminating (ending in an infinite string of zeros), for example, = 0.75000 Á = 0.75 or (b) eventually repeating (ending with a block of digits that repeats over and over), for example 23 = 2.090909 Á = 2.09 11 www.elsolucionario.net The bar indicates the block of repeating digits Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1.1 Real Numbers and the Real Line A terminating decimal expansion is a special type of repeating decimal since the ending zeros repeat The set of rational numbers has all the algebraic and order properties of the real numbers but lacks the completeness property For example, there is no rational number whose square is 2; there is a “hole” in the rational line where 22 should be Real numbers that are not rational are called irrational numbers They are characterized by having nonterminating and nonrepeating decimal expansions Examples are p, 22, 5, and log10 Since every decimal expansion represents a real number, it should be clear that there are infinitely many irrational numbers Both rational and irrational numbers are found arbitrarily close to any point on the real line Set notation is very useful for specifying a particular subset of real numbers A set is a collection of objects, and these objects are the elements of the set If S is a set, the notation a H S means that a is an element of S, and a x S means that a is not an element of S If S and T are sets, then S ´ T is their union and consists of all elements belonging either to S or T (or to both S and T) The intersection S ă T consists of all elements belonging to both S and T The empty set Ô is the set that contains no elements For example, the intersection of the rational numbers and the irrational numbers is the empty set Some sets can be described by listing their elements in braces For instance, the set A consisting of the natural numbers (or positive integers) less than can be expressed as A = 51, 2, 3, 4, 56 The entire set of integers is written as 50, ;1, ; 2, ;3, Á Another way to describe a set is to enclose in braces a rule that generates all the elements of the set For instance, the set A = 5x ƒ x is an integer and x 66 is the set of positive integers less than Intervals A subset of the real line is called an interval if it contains at least two numbers and contains all the real numbers lying between any two of its elements For example, the set of all real numbers x such that x is an interval, as is the set of all x such that -2 … x … The set of all nonzero real numbers is not an interval; since is absent, the set fails to contain every real number between -1 and (for example) Geometrically, intervals correspond to rays and line segments on the real line, along with the real line itself Intervals of numbers corresponding to line segments are finite intervals; intervals corresponding to rays and the real line are infinite intervals A finite interval is said to be closed if it contains both of its endpoints, half-open if it contains one endpoint but not the other, and open if it contains neither endpoint The endpoints are also called boundary points; they make up the interval’s boundary The remaining points of the interval are interior points and together comprise the interval’s interior Infinite intervals are closed if they contain a finite endpoint, and open otherwise The entire real line ޒis an infinite interval that is both open and closed Solving Inequalities The process of finding the interval or intervals of numbers that satisfy an inequality in x is called solving the inequality www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 4 Chapter 1: Preliminaries TABLE 1.1 Types of intervals Finite: Notation Set description Type (a, b) 5x ƒ a x b6 Open 5x ƒ a … x … b6 Closed 5x ƒ a … x b6 Half-open 5x ƒ a x … b6 Half-open 5x ƒ x a6 Open [a, q d 5x ƒ x Ú a6 Closed s - q , bd 5x ƒ x b6 Open s - q , b] 5x ƒ x … b6 Closed [a, b] [a, b) (a, b] sa, q d Infinite: ( ޒset of all real numbers) s - q, q d EXAMPLE Picture a b a b a b a b a a b b Both open and closed Solve the following inequalities and show their solution sets on the real line (a) 2x - x + x –3 x (b) (c) Ú x - 2x - x + 2x x + x Add to both sides Subtract x from both sides The solution set is the open interval s - q , 4d (Figure 1.1a) x 11 x 2x + Solution (a) (a) (b) - (c) FIGURE 1.1 Solution sets for the inequalities in Example (b) - x 2x + -x 6x + 7x + -3 7x - x www.elsolucionario.net Multiply both sides by Add x to both sides Subtract from both sides Divide by Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1.1 Real Numbers and the Real Line The solution set is the open interval s - 3>7, q d (Figure 1.1b) (c) The inequality 6>sx - 1d Ú can hold only if x 1, because otherwise 6>sx - 1d is undefined or negative Therefore, sx - 1d is positive and the inequality will be preserved if we multiply both sides by sx - 1d, and we have Ú x - Ú 5x - 11 Ú 5x Multiply both sides by sx - 1d Add to both sides 11 Ú x Or x … 11 The solution set is the half-open interval (1, 11>5] (Figure 1.1c) Absolute Value The absolute value of a number x, denoted by ƒ x ƒ , is defined by the formula ƒxƒ = e EXAMPLE x, - x, x Ú x Finding Absolute Values ƒ ƒ = 3, ƒ -5 ƒ = - s -5d = 5, ƒ ƒ = 0, ƒ - ƒaƒƒ = ƒaƒ Geometrically, the absolute value of x is the distance from x to on the real number line Since distances are always positive or 0, we see that ƒ x ƒ Ú for every real number x, and ƒ x ƒ = if and only if x = Also, ͉– 5͉ ϭ –5 ͉3͉ ƒ x - y ƒ = the distance between x and y ͉4 Ϫ 1͉ ϭ ͉1 Ϫ 4͉ ϭ FIGURE 1.2 Absolute values give distances between points on the number line on the real line (Figure 1.2) Since the symbol 2a always denotes the nonnegative square root of a, an alternate definition of ƒ x ƒ is ƒ x ƒ = 2x It is important to remember that 2a = ƒ a ƒ Do not write 2a = a unless you already know that a Ú The absolute value has the following properties (You are asked to prove these properties in the exercises.) Absolute Value Properties ƒ -a ƒ = ƒ a ƒ ƒ ab ƒ = ƒ a ƒ ƒ b ƒ ƒaƒ a ` ` = b ƒbƒ ƒa + bƒ … ƒaƒ + ƒbƒ A number and its additive inverse or negative have the same absolute value The absolute value of a product is the product of the absolute values The absolute value of a quotient is the quotient of the absolute values The triangle inequality The absolute value of the sum of two numbers is less than or equal to the sum of their absolute values www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 6 Chapter 1: Preliminaries Note that ƒ -a ƒ Z - ƒ a ƒ For example, ƒ -3 ƒ = 3, whereas - ƒ ƒ = - If a and b differ in sign, then ƒ a + b ƒ is less than ƒ a ƒ + ƒ b ƒ In all other cases, ƒ a + b ƒ equals ƒ a ƒ + ƒ b ƒ Absolute value bars in expressions like ƒ -3 + ƒ work like parentheses: We the arithmetic inside before taking the absolute value a a –a x EXAMPLE Illustrating the Triangle Inequality a ƒ - + ƒ = ƒ ƒ = ƒ -3 ƒ + ƒ ƒ = ƒ3 + 5ƒ = ƒ8ƒ = ƒ3ƒ + ƒ5ƒ -3 - ƒ = ƒ - ƒ = = ƒ - ƒ + ƒ -5 ƒ ƒ ͉x͉ FIGURE 1.3 ƒ x ƒ a means x lies between -a and a The inequality ƒ x ƒ a says that the distance from x to is less than the positive number a This means that x must lie between -a and a, as we can see from Figure 1.3 The following statements are all consequences of the definition of absolute value and are often helpful when solving equations or inequalities involving absolute values Absolute Values and Intervals If a is any positive number, then ƒxƒ ƒxƒ ƒxƒ ƒxƒ ƒxƒ = … Ú a a a a a if and only if x = ; a if and only if -a x a if and only if x a or x - a if and only if -a … x … a if and only if x Ú a or x … - a The symbol is often used by mathematicians to denote the “if and only if ” logical relationship It also means “implies and is implied by.” EXAMPLE Solving an Equation with Absolute Values Solve the equation ƒ 2x - ƒ = Solution By Property 5, 2x - = ; 7, so there are two possibilities: 2x - = 2x = 10 x = 2x - = - 2x = - x = -2 Equivalent equations without absolute values Solve as usual The solutions of ƒ 2x - ƒ = are x = and x = - EXAMPLE Solving an Inequality Involving Absolute Values Solve the inequality ` - x ` www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1.1 Real Numbers and the Real Line Solution We have 2 ` - x ` -1 - x Property -6 - x -4 Subtract 33 x Multiply by - 1 x Take reciprocals Notice how the various rules for inequalities were used here Multiplying by a negative number reverses the inequality So does taking reciprocals in an inequality in which both sides are positive The original inequality holds if and only if s1>3d x s1>2d The solution set is the open interval (1>3, 1>2) EXAMPLE Solve the inequality and show the solution set on the real line: (a) ƒ 2x - ƒ … x (b) ƒ 2x - ƒ Ú Solution ƒ 2x - ƒ -1 … 2x - … 2x … x (a) (a) x (b) FIGURE 1.4 The solution sets (a) [1, 2] and (b) s - q , 1] ´ [2, q d in Example … … … … 1 Property Add Divide by The solution set is the closed interval [1, 2] (Figure 1.4a) (b) ƒ 2x - ƒ Ú 2x - Ú or 2x - … - x - Ú 2 or x Ú or x - … 2 Divide by x … Add The solution set is s - q , 1] ´ [2, q d (Figure 1.4b) www.elsolucionario.net Property Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 1.1 Real Numbers and the Real Line EXERCISES 1.1 Decimal Representations Inequalities Express 1>9 as a repeating decimal, using a bar to indicate the repeating digits What are the decimal representations of 2>9? 3>9? 8>9? 9>9? Express 1>11 as a repeating decimal, using a bar to indicate the repeating digits What are the decimal representations of 2>11? 3>11? 9>11? 11>11? If x 6 , which of the following statements about x are necessarily true, and which are not necessarily true? a x x c 6 e x b x - 1 d x 6 g - 6 - x h -6 - x - www.elsolucionario.net f ƒ x - ƒ Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL/Thomas_ch01p001-072 8/19/04 10:49 AM Page 8 Chapter 1: Preliminaries If - y - , which of the following statements about y are necessarily true, and which are not necessarily true? a y 6 b - 6 y - c y d y 6 y f 6 e y - 1 y g Solve the inequalities in Exercises 35–42 Express the solution sets as intervals or unions of intervals and show them on the real line Use the result 2a = ƒ a ƒ as appropriate 35 x 38 h ƒ y - ƒ 1 x2 41 x - x In Exercises 5–12, solve the inequalities and show the solution sets on the real line 36 … x 37 x 39 sx - 1d2 40 sx + 3d2 42 x - x - Ú Theory and Examples 43 Do not fall into the trap ƒ -a ƒ = a For what real numbers a is this equation true? For what real numbers is it false? - 2x - 3x Ú 5x - … - 3x 3s2 - xd 2s3 + xd 2x - Ú 7x + 11 sx - 2d sx - 6d Quadratic Inequalities - x 3x - 10 x + 12 + 3x … 12 44 Solve the equation ƒ x - ƒ = - x 45 A proof of the triangle inequality Give the reason justifying each of the numbered steps in the following proof of the triangle inequality ƒ a + b ƒ = sa + bd2 = a + 2ab + b Absolute Value … a2 + ƒ a ƒ ƒ b ƒ + b2 Solve the equations in Exercises 13–18 13 ƒ y ƒ = 14 ƒ y - ƒ = 15 ƒ 2t + ƒ = 16 ƒ - t ƒ = 17 ƒ - 3s ƒ = 18 ` 19 ƒ x ƒ 22 ƒ t + ƒ 20 ƒ x ƒ … 23 ƒ 3y - ƒ 21 ƒ t - ƒ … 24 ƒ 2y + ƒ 25 ` 26 ` 1 27 ` - x ` z - 1` … 2 28 ` x - ` 29 ƒ 2s ƒ Ú 30 ƒ s + ƒ Ú 31 ƒ - x ƒ 32 ƒ - 3x ƒ 33 ` 34 ` 3r - 1` 5 = ƒaƒ + 2ƒaƒ ƒbƒ + ƒbƒ = s ƒ a ƒ + ƒ b ƒ d2 ƒa + bƒ … ƒaƒ + ƒbƒ s - 1` = Solve the inequalities in Exercises 19–34, expressing the solution sets as intervals or unions of intervals Also, show each solution set on the real line z - 1` … (1) r + ` Ú (2) (3) (4) 46 Prove that ƒ ab ƒ = ƒ a ƒ ƒ b ƒ for any numbers a and b 47 If ƒ x ƒ … and x - 1>2 , what can you say about x? 48 Graph the inequality ƒ x ƒ + ƒ y ƒ … 49 Let ƒsxd = 2x + and let d be any positive number Prove that ƒ x - ƒ d implies ƒ ƒsxd - ƒs1d ƒ 2d Here the notation ƒ(a) means the value of the expression 2x + when x = a This function notation is explained in Section 1.3 50 Let ƒsxd = 2x + and let P be any positive number Prove P that ƒ ƒsxd - ƒs0d ƒ P whenever ƒ x - ƒ Here the no2 tation ƒ(a) means the value of the expression 2x + when x = a (See Section 1.3.) 51 For any number a, prove that ƒ -a ƒ = ƒ a ƒ 52 Let a be any positive number Prove that ƒ x ƒ a if and only if x a or x - a 53 a If b is any nonzero real number, prove that ƒ 1>b ƒ = 1> ƒ b ƒ ƒaƒ a for any numbers a and b Z b Prove that ` ` = b ƒbƒ 54 Using mathematical induction (see Appendix 1), prove that n n ƒ a ƒ = ƒ a ƒ for any number a and positive integer n www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 AP-22 8/27/04 6:50 AM Page 22 Appendices 18 Find the two square roots of i a Extend the results of Exercise 26 to show that ƒszd = ƒszd if 19 Find the three cube roots of -8i ƒszd = an z n + an - z n - + Á + a1 z + a0 20 Find the six sixth roots of 64 21 Find the four solutions of the equation z - 2z + = 22 Find the six solutions of the equation z + 2z + = 23 Find all solutions of the equation x + 4x + 16 = 24 Solve the equation x + = Theory and Examples 25 Complex numbers and vectors in the plane Show with an Argand diagram that the law for adding complex numbers is the same as the parallelogram law for adding vectors 26 Complex arithmetic with conjugates Show that the conjugate of the sum (product, or quotient) of two complex numbers, z1 and z2, is the same as the sum (product, or quotient) of their conjugates 27 Complex roots of polynomials with real coefficients come in complex-conjugate pairs is a polynomial with real coefficients a0, Á , an b If z is a root of the equation ƒszd = , where ƒ(z) is a polynomial with real coefficients as in part (a), show that the conjugate z is also a root of the equation (Hint: Let ƒszd = u + iy = ; then both u and y are zero Use the fact that ƒszd = ƒszd = u - iy ) 28 Absolute value of a conjugate Show that ƒ z ƒ = ƒ z ƒ 29 When z = z If z and z are equal, what can you say about the location of the point z in the complex plane? 30 Real and imaginary parts Let Re(z) denote the real part of z and Im(z) the imaginary part Show that the following relations hold for any complex numbers z, z1 , and z2 a z + z = 2Reszd b z - z = 2iImszd c ƒ Reszd ƒ … ƒ z ƒ d ƒ z1 + z2 ƒ = ƒ z1 ƒ + ƒ z2 ƒ + 2Resz1z2 d e ƒ z1 + z2 ƒ … ƒ z1 ƒ + ƒ z2 ƒ www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 AP-22 8/27/04 6:50 AM Page 22 Appendices A.6 The Distributive Law for Vector Cross Products In this appendix, we prove the Distributive Law u * sv + wd = u * v + u * w which is Property in Section 12.4 Proof To derive the Distributive Law, we construct u * v a new way We draw u and v from the common point O and construct a plane M perpendicular to u at O (Figure A.10) We then project v orthogonally onto M, yielding a vector v¿ with length ƒ v ƒ sin u We rotate v¿ 90° about u in the positive sense to produce a vector v– Finally, we multiply v– by the M' v u O M v' FIGURE A.10 v'' uϫv 90˚ As explained in the text, u * v = ƒ u ƒ v– www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 23 A.7 The Mixed Derivative Theorem and the Increment Theorem AP-23 length of u The resulting vector ƒ u ƒ v– is equal to u * v since v– has the same direction as u * v by its construction (Figure A.10) and ƒ u ƒ ƒ v– ƒ = ƒ u ƒ ƒ v¿ ƒ = ƒ u ƒ ƒ v ƒ sin u = ƒ u * v ƒ Now each of these three operations, namely, projection onto M rotation about u through 90° multiplication by the scalar ƒ u ƒ when applied to a triangle whose plane is not parallel to u, will produce another triangle If we start with the triangle whose sides are v, w, and v + w (Figure A.11) and apply these three steps, we successively obtain the following: A triangle whose sides are v¿, w¿, and sv + wd¿ satisfying the vector equation v¿ + w¿ = sv + wd¿ A triangle whose sides are v–, w– , and sv + wd– satisfying the vector equation v– + w– = sv + wd– (the double prime on each vector has the same meaning as in Figure A.10) w u vϩw w' M v v' (v ϩ w)' FIGURE A.11 The vectors, v, w, v + w , and their projections onto a plane perpendicular to u A triangle whose sides are ƒ u ƒ v–, ƒ u ƒ w– , and ƒ u ƒ sv + wd– satisfying the vector equation ƒ u ƒ v– + ƒ u ƒ w– = ƒ u ƒ sv + wd– Substituting ƒ u ƒ v– = u * v, ƒ u ƒ w– = u * w, and ƒ u ƒ sv + wd– = u * sv + wd from our discussion above into this last equation gives u * v + u * w = u * sv + wd, which is the law we wanted to establish www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 23 A.7 The Mixed Derivative Theorem and the Increment Theorem A.7 AP-23 The Mixed Derivative Theorem and the Increment Theorem This appendix derives the Mixed Derivative Theorem (Theorem 2, Section 14.3) and the Increment Theorem for Functions of Two Variables (Theorem 3, Section 14.3) Euler first published the Mixed Derivative Theorem in 1734, in a series of papers he wrote on hydrodynamics www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 AP-24 8/27/04 6:50 AM Page 24 Appendices THEOREM The Mixed Derivative Theorem If ƒ(x, y) and its partial derivatives ƒx , ƒy , ƒxy , and ƒyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b), then ƒxysa, bd = ƒyxsa, bd Proof The equality of ƒxysa, bd and ƒyxsa, bd can be established by four applications of the Mean Value Theorem (Theorem 4, Section 4.2) By hypothesis, the point (a, b) lies in the interior of a rectangle R in the xy-plane on which ƒ, ƒx , ƒy , ƒxy , and ƒyx are all defined We let h and k be the numbers such that the point sa + h, b + kd also lies in R, and we consider the difference ¢ = Fsa + hd - Fsad, (1) Fsxd = ƒsx, b + kd - ƒsx, bd (2) where We apply the Mean Value Theorem to F, which is continuous because it is differentiable Then Equation (1) becomes ¢ = hF¿sc1 d, (3) where c1 lies between a and a + h From Equation (2) F¿sxd = ƒxsx, b + kd - ƒxsx, bd , y so Equation (3) becomes ¢ = h[ƒxsc1, b + kd - ƒxsc1, bd] R k (a, b) (4) Now we apply the Mean Value Theorem to the function gs yd = fxsc1, yd and have R' gsb + kd - gsbd = kg¿sd1 d, h or x ƒxsc1, b + kd - ƒxsc1, bd = kƒxysc1, d1 d for some d1 between b and b + k By substituting this into Equation (4), we get FIGURE A.12 The key to proving ƒxysa, bd = ƒyxsa, bd is that no matter how small R¿ is, ƒxy and ƒyx take on equal values somewhere inside R¿ (although not necessarily at the same point) ¢ = hkƒxysc1, d1 d (5) for some point sc1, d1 d in the rectangle R¿ whose vertices are the four points (a, b), sa + h, bd, sa + h, b + kd, and sa, b + kd (See Figure A.12.) By substituting from Equation (2) into Equation (1), we may also write ¢ = ƒsa + h, b + kd - ƒsa + h, bd - ƒsa, b + kd + ƒsa, bd = [ƒsa + h, b + kd - ƒsa, b + kd] - [ƒsa + h, bd - ƒsa, bd] (6) = fsb + kd - fsbd, where fs yd = ƒsa + h, yd - ƒsa, yd (7) The Mean Value Theorem applied to Equation (6) now gives  = kfsd2 d www.elsolucionario.net Copyright â 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley (8) 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 25 A.7 The Mixed Derivative Theorem and the Increment Theorem AP-25 for some d2 between b and b + k By Equation (7), f¿s yd = ƒysa + h, yd - ƒysa, yd (9) Substituting from Equation (9) into Equation (8) gives ¢ = k[ƒysa + h, d2 d - ƒysa, d2 d] Finally, we apply the Mean Value Theorem to the expression in brackets and get ¢ = khƒyxsc2, d2 d (10) for some c2 between a and a + h Together, Equations (5) and (10) show that ƒxysc1, d1 d = ƒyxsc2, d2 d, (11) where sc1, d1 d and sc2, d2 d both lie in the rectangle R¿ (Figure A.12) Equation (11) is not quite the result we want, since it says only that ƒxy has the same value at sc1, d1 d that ƒyx has at sc2, d2 d The numbers h and k in our discussion, however, may be made as small as we wish The hypothesis that ƒxy and ƒyx are both continuous at (a, b) means that ƒxysc1, d1 d = ƒxysa, bd + P1 and ƒyxsc2, d2 d = ƒyxsa, bd + P2 , where each of P1, P2 : as both h, k : Hence, if we let h and k : 0, we have ƒxysa, bd = ƒyxsa, bd The equality of ƒxysa, bd and ƒyxsa, bd can be proved with hypotheses weaker than the ones we assumed For example, it is enough for ƒ, ƒx , and ƒy to exist in R and for ƒxy to be continuous at (a, b) Then ƒyx will exist at (a, b) and equal ƒxy at that point THEOREM The Increment Theorem for Functions of Two Variables Suppose that the first partial derivatives of z = ƒsx, yd are defined throughout an open region R containing the point sx0 , y0 d and that ƒx and ƒy are continuous at sx0 , y0 d Then the change ¢z = ƒsx0 + ¢x, y0 + ¢yd - ƒsx0 , y0 d in the value of ƒ that results from moving from sx0 , y0 d to another point sx0 + ¢x, y0 + ¢yd in R satisfies an equation of the form C(x0 ϩ , x, y0 ϩ ,y) ¢z = ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y + P1 ¢x + P2 ¢y, in which each of P1, P2 : as both ¢x, ¢y : A(x0, y0 ) Proof We work within a rectangle T centered at Asx0 , y0 d and lying within R, and we assume that ¢x and ¢y are already so small that the line segment joining A to Bsx0 + ¢x, y0 d and the line segment joining B to Csx0 + ¢x, y0 + ¢yd lie in the interior of T (Figure A.13) We may think of ¢z as the sum ¢z = ¢z1 + ¢z2 of two increments, where B(x0 ϩ , x, y0 ) T ¢z1 = ƒsx0 + ¢x, y0 d - ƒsx0 , y0 d FIGURE A.13 The rectangular region T in the proof of the Increment Theorem The figure is drawn for ¢x and ¢y positive, but either increment might be zero or negative is the change in the value of ƒ from A to B and ¢z2 = ƒsx0 + ¢x, y0 + ¢yd - ƒsx0 + ¢x, y0 d is the change in the value of ƒ from B to C (Figure A.14) www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 AP-26 8/27/04 6:50 AM Page 26 Appendices z S Q z ϭ f (x, y) P0 Q' , z2 P'' ,z P' , z1 A(x0 , y ) y0 y B x (x0 ϩ , x, y ) y ϩ ,y C(x0 ϩ , x, y ϩ ,y) FIGURE A.14 Part of the surface z = ƒsx, yd near P0sx0, y0, ƒsx0, y0 dd The points P0 , P¿, and P– have the same height z0 = ƒsx0, y0 d above the xy-plane The change in z is ¢z = P¿S The change ¢z1 = ƒsx0 + ¢x, y0 d - ƒsx0 , y0 d, shown as P–Q = P¿Q¿ , is caused by changing x from x0 to x0 + ¢x while holding y equal to y0 Then, with x held equal to x0 + ¢x , ¢z2 = ƒsx0 + ¢x, y0 + ¢yd - ƒsx0 + ¢x, y0 d is the change in z caused by changing y0 from y0 + ¢y , which is represented by Q¿S? The total change in z is the sum of ¢z1 and ¢z2 On the closed interval of x-values joining x0 to x0 + ¢x, the function Fsxd = ƒsx, y0 d is a differentiable (and hence continuous) function of x, with derivative F¿sxd = ƒxsx, y0 d By the Mean Value Theorem (Theorem 4, Section 4.2), there is an x-value c between x0 and x0 + ¢x at which Fsx0 + ¢xd - Fsx0 d = F¿scd¢x or ƒsx0 + ¢x, y0 d - ƒsx0, y0 d = ƒxsc, y0 d¢x or ¢z1 = ƒxsc, y0 d¢x (12) Similarly, Gs yd = ƒsx0 + ¢x, yd is a differentiable (and hence continuous) function of y on the closed y-interval joining y0 and y0 + ¢y, with derivative G¿s yd = ƒysx0 + ¢x, yd www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 27 A.7 The Mixed Derivative Theorem and the Increment Theorem AP-27 Hence, there is a y-value d between y0 and y0 + ¢y at which Gs y0 + ¢yd - Gs y0 d = G¿sdd¢y or ƒsx0 + ¢x, y0 + ¢yd - ƒsx0 + ¢x, yd = ƒysx0 + ¢x, dd¢y or ¢z2 = ƒysx0 + ¢x, dd¢y (13) Now, as both ¢x and ¢y : 0, we know that c : x0 and d : y0 Therefore, since ƒx and ƒy are continuous at sx0 , y0 d, the quantities P1 = ƒxsc, y0 d - ƒxsx0, y0 d, (14) P2 = ƒysx0 + ¢x, dd - ƒysx0, y0 d both approach zero as both ¢x and ¢y : Finally, ¢z = ¢z1 + ¢z2 = ƒxsc, y0 d¢x + ƒysx0 + ¢x, dd¢y From Equations (12) and (13) = [ƒxsx0 , y0 d + P1]¢x + [ƒysx0 , y0 d + P2]¢y From Equation (14) = ƒxsx0 , y0 d¢x + ƒysx0 , y0 d¢y + P1 ¢x + P2 ¢y, where both P1 and P2 : as both ¢x and ¢y : 0, which is what we set out to prove Analogous results hold for functions of any finite number of independent variables Suppose that the first partial derivatives of w = ƒsx, y, zd are defined throughout an open region containing the point sx0 , y0 , z0 d and that ƒx , ƒy , and ƒz are continuous at sx0 , y0 , z0 d Then ¢w = ƒsx0 + ¢x, y0 + ¢y, z0 + ¢zd - ƒsx0 , y0 , z0 d = ƒx ¢x + ƒy ¢y + ƒz ¢z + P1 ¢x + P2 ¢y + P3 ¢z, (15) where P1, P2, P3 : as ¢x, ¢y, and ¢z : The partial derivatives ƒx , ƒy , ƒz in Equation (15) are to be evaluated at the point sx0 , y0 , z0 d Equation (15) can be proved by treating ¢w as the sum of three increments, ¢w1 = ƒsx0 + ¢x, y0 , z0 d - ƒsx0 , y0 , z0 d (16) ¢w2 = ƒsx0 + ¢x, y0 + ¢y, z0 d - ƒsx0 + ¢x, y0 , z0 d (17) ¢w3 = ƒsx0 + ¢x, y0 + ¢y, z0 + ¢zd - ƒsx0 + ¢x, y0 + ¢y, z0 d, (18) and applying the Mean Value Theorem to each of these separately Two coordinates remain constant and only one varies in each of these partial increments ¢w1, ¢w2, ¢w3 In Equation (17), for example, only y varies, since x is held equal to x0 + ¢x and z is held equal to z0 Since ƒsx0 + ¢x, y, z0 d is a continuous function of y with a derivative ƒy , it is subject to the Mean Value Theorem, and we have ¢w2 = ƒysx0 + ¢x, y1, z0 d¢y for some y1 between y0 and y0 + Ây www.elsolucionario.net Copyright â 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 AP-28 8/27/04 6:50 AM Page 28 Appendices The Area of a Parallelogram’s Projection on a Plane A.8 This appendix proves the result needed in Section 16.5 that ƒ su * vd # p ƒ is the area of the projection of the parallelogram with sides determined by u and v onto any plane whose normal is p (See Figure A.15.) P u v Q THEOREM The area of the orthogonal projection of the parallelogram determined by two vectors u and v in space onto a plane with unit normal vector p is S R u' P' Area = ƒ su * vd # p ƒ p v' Q' S' R' FIGURE A.15 The parallelogram determined by two vectors u and v in space and the orthogonal projection of the parallelogram onto a plane The projection lines, orthogonal to the plane, lie parallel to the unit normal vector p Proof In the notation of Figure A.15, which shows a typical parallelogram determined by vectors u and v and its orthogonal projection onto a plane with unit normal vector p, § § u = PP ¿ + u¿ + Q¿Q § § § § sQ¿Q = - QQ ¿ d = u¿ + PP ¿ - QQ ¿ = u¿ + s p Similarly, (For some scalar s because § § (PP ¿ - QQ ¿ ) is parallel to p) v = v¿ + for some scalar t Hence, u * v = su¿ + spd * sv¿ + tpd = su¿ * v¿d + ssp * v¿d + tsu¿ * pd + stsp * pd (1) (')'* The vectors p * v¿ and u¿ * p are both orthogonal to p Hence, when we dot both sides of Equation (1) with p, the only nonzero term on the right is su¿ * v¿d # p That is, su * vd # p = su¿ * v¿d # p z In particular, P(0, 0, 3) ƒ su * vd # p ƒ = ƒ su¿ * v¿d # p ƒ Q(2, –1, 2) k S(1, 3, 2) (2) The absolute value on the right is the volume of the box determined by u¿, v¿ , and p The height of this particular box is ƒ p ƒ = 1, so the box’s volume is numerically the same as its base area, the area of parallelogram P¿Q¿R¿S¿ Combining this observation with Equation (2) gives Area of P¿Q¿R¿S¿ = ƒ su¿ * v¿d # p ƒ = ƒ su * vd # p ƒ , y R(3, 2, 1) which says that the area of the orthogonal projection of the parallelogram determined by u and v onto a plane with unit normal vector p is ƒ su * vd # p ƒ This is what we set out to prove x FIGURE A.16 Example calculates the area of the orthogonal projection of parallelogram PQRS on the xy-plane EXAMPLE Finding the Area of a Projection Find the area of the orthogonal projection onto the xy-plane of the parallelogram determined by the points P(0, 0, 3), Qs2, -1, 2d , R(3, 2, 1), and S(1, 3, 2) (Figure A.16) www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 29 A.9 Basic Algebra, Geometry, and Trigonometry Formulas Solution AP-29 With § u = PQ = 2i - j - k, § v = PS = i + 3j - k, and we have su * vd # p = -1 -1 -1 = ` 1 -1 ` = 7, so the area is ƒ su * vd # p ƒ = ƒ ƒ = www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley p = k, 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 29 A.9 Basic Algebra, Geometry, and Trigonometry Formulas A.9 AP-29 Basic Algebra, Geometry, and Trigonometry Formulas Algebra Arithmetic Operations ac a#c = b d bd asb + cd = ab + ac, a c ad + bc + = , b d bd a>b a d = #c b c>d Laws of Signs a a -a = - = b b -b -s - ad = a, Zero Division by zero is not defined If a Z 0: a = 0, a = 1, 0a = For any number a: a # = # a = Laws of Exponents a ma n = a m + n, sabdm = a mb m, sa m dn = a mn, n am>n = 2am = If a Z 0, am = a m - n, an a = 1, a -m = am The Binomial Theorem For any positive integer n, sa + bdn = an + nan - 1b + + nsn - 1d n - 2 a b 1#2 nsn - 1dsn - 2d n - 3 a b + Á + nabn - + bn 1#2#3 www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley n aBm A2 4100AWL/Thomas_APPp001-034 AP-30 8/27/04 6:50 AM Page 30 Appendices For instance, sa + bd2 = a2 + 2ab + b2, sa - bd2 = a2 - 2ab + b2 sa + bd3 = a3 + 3a2b + 3ab2 + b3, sa - bd3 = a2 - 3a2b + 3ab2 - b3 Factoring the Difference of Like Integer Powers, n>1 a n - b n = sa - bdsa n - + a n - 2b + a n - 3b + Á + ab n - + b n - d For instance, a2 - b2 = sa - bdsa + bd, a3 - b3 = sa - bdsa2 + ab + b2 d, a4 - b4 = sa - bdsa3 + a2b + ab2 + b3 d Completing the Square If a Z 0, b ax + bx + c = a ax + a xb + c b b2 b2 = a ax + a x + b + c 4a 4a b b2 b2 = a ax + a x + b + a ab + c 4a 4a b b2 b2 = a ax + a x + b + c 4a 4a ('''')''''* This is ax + b b 2a = au + C (')'* Call this part C su = x + sb>2add The Quadratic Formula If a Z and ax + bx + c = 0, then - b ; 2b - 4ac 2a x = Geometry Formulas for area, circumference, and volume: sA = area, B = area of base, C = circumference, S = lateral area or surface area, V = volumed Triangle Similar Triangles c h c' a' Pythagorean Theorem a c b b A ϭ bh b b' a' b' c' aϭbϭc www.elsolucionario.net a a2 ϩ b2 ϭ c2 Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100AWL/Thomas_APPp001-034 8/27/04 6:50 AM Page 31 AP-31 A.9 Basic Algebra, Geometry, and Trigonometry Formulas Parallelogram Trapezoid Circle a h h A ϭ r 2, C ϭ 2r r b b A ϭ bh A ϭ (a ϩ b)h Any Cylinder or Prism with Parallel Bases Right Circular Cylinder r h h h V ϭ Bh B B V ϭ r 2h S ϭ 2rh ϭ Area of side Any Cone or Pyramid Right Circular Cone Sphere h h h s r V ϭ Bh V ϭ r 2h S ϭ rs ϭ Area of side B B V ϭ r 3, S ϭ 4r Trigonometry Formulas y Definitions and Fundamental Identities Sine: Cosine: Tangent: P(x, y) y sin u = r = csc u x cos u = r = sec u y tan u = x = cot u www.elsolucionario.net O Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley r x y x 4100AWL/Thomas_APPp001-034 AP-32 8/27/04 6:50 AM Page 32 Appendices Identities sin s -ud = - sin u, cos s -ud = cos u sin2 u + cos2 u = 1, sec2 u = + tan2 u, cos 2u = cos2 u - sin2 u sin 2u = sin u cos u, cos2 u = csc2 u = + cot2 u + cos 2u , - cos 2u sin2 u = sin sA + Bd = sin A cos B + cos A sin B sin sA - Bd = sin A cos B - cos A sin B cos sA + Bd = cos A cos B - sin A sin B cos sA - Bd = cos A cos B + sin A sin B tan sA + Bd = tan A + tan B , - tan A tan B sin aA - p b = - cos A, sin aA + p b = cos A, sin A sin B = tan sA - Bd = cos aA cos aA + tan A - tan B + tan A tan B p b = sin A p b = - sin A 1 cos sA - Bd - cos sA + Bd 2 cos A cos B = 1 cos sA - Bd + cos sA + Bd 2 sin A cos B = 1 sin sA - Bd + sin sA + Bd 2 sin A + sin B = sin 1 sA + Bd cos sA - Bd 2 sin A - sin B = cos 1 sA + Bd sin sA - Bd 2 cos A + cos B = cos 1 sA + Bd cos sA - Bd 2 cos A - cos B = - sin 1 sA + Bd sin sA - Bd 2 www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 8/27/04 6:50 AM Page 33 A.9 Basic Algebra, Geometry, and Trigonometry Formulas AP-33 Trigonometric Functions Degrees Radian Measure Radians 45 ͙2 s θ C ir 45 90 r Un ͙2 it cir cl e 4100AWL/Thomas_APPp001-034 cle of diu sr 30 u s or r = = u 180° = p radians s u = r, ͙3 60 ͙3 90 The angles of two common triangles, in degrees and radians y y y ϭ sin x – – y ϭ cos x 3 2 x y ϭ sinx Domain: (–ϱ, ϱ) Range: [–1, 1] y – – 3 2 Domain: (–ϱ, ϱ) Range: [–1, 1] y y ϭ tan x y ϭ sec x – 3 – – 2 3 2 x Domain: All real numbers except odd integer multiples of /2 Range: (–ϱ, ϱ) y 3 2 x Ϯ , Ϯ 3 , 2 (–ϱ, –1] h [1, ϱ) Domain: x Range: y y ϭ csc x – – – 3 – – 2 y ϭ cot x 3 2 Domain: x 0, Ϯ, Ϯ2, Range: (–ϱ, –1] h [1, ϱ) www.elsolucionario.net x – – 3 2 Domain: x 0, Ϯ, Ϯ2, Range: (–ϱ, ϱ) Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley x x ... elements and topics of calculus, visit www.aw-bc.com /thomas www.elsolucionario.net Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 4100 AWL /Thomas_ ch01p001-072 8/19/04...4100 AWL /Thomas_ ch01p001-072 8/19/04 10:49 AM Page Chapter PRELIMINARIES OVERVIEW This chapter reviews the basic ideas you need to start calculus The topics include the... Addison-Wesley 4100 AWL /Thomas_ ch01p001-072 8/19/04 10:50 AM Page 19 1.3 Functions and Their Graphs 19 Functions and Their Graphs 1.3 Functions are the major objects we deal with in calculus because