Chapter 1 INTRODUCTION TO CALCULUS In the English language, the rules of grammar are used to speak and write effectively. Asking for a cookie at the age of ten was much easier than when you were first learning to speak. These rules developed over time. Calculus developed in a similar way. Sir Isaac Newton and Gottfried Wilhelm von Leibniz independently organized an assortment of ideas and methods that were circulating among the mathematicians of their time. As a tool in the service of science, calculus served its purpose very well. More than two centuries passed, however, before mathematicians had identified and agreed on its underlying principles—its grammar.In this chapter, you will see some of the ideas that were brought together to form the underlying principles of calculus. CHAPTER EXPECTATIONS In this chapter, you will • simplify radical expressions, Section 1.1 • use limits to determine the slope and the equation of the tangent to a graph, Section 1.2 • pose problems and formulate hypotheses regarding rates of change, Section 1.3, Career Link • calculate and interpret average and instantaneous rates of change and relate these values to slopes of secants and tangents, Section 1.3 • understand and evaluate limits using appropriate properties,Sections 1.4, 1.5 • examine continuous functions and use limits to explain why a function is discontinuous, Sections1.5, 1.6 NEL Review of Prerequisite Skills Before beginning this chapter, review the following concepts from previous courses: • determining the slope of a line: m (cid:1) ¢y ¢x • determining the equation of a line • using function notation for substituting into and evaluating functions • simplifying algebraic expressions • factoring expressions • finding the domain of functions • calculating average rate of change and slopes of secant lines • estimating instantaneous rate of change and slopes of tangent lines Exercise 1. Determine the slope of the line passing through each of the following pairs of points: a. 12, 52 and 16, (cid:2)72 d. 10, 02 and 1(cid:2)1, 42 b. 13, (cid:2)42 and 1(cid:2)1, 42 e. 1(cid:2)2.1, 4.412 and 1(cid:2)2, 42 3 1 7 1 c. 10, 02 and 11, 42 f. a , b and a , (cid:2) b 4 4 4 4 2. Determine the equation of a line for the given information. a. slope 4, y-intercept (cid:2)2 d. through 1(cid:2)2, 42 and 1(cid:2)6, 82 b. slope (cid:2)2, y-intercept 5 e. vertical, through 1(cid:2)3, 52 c. through 1(cid:2)1, 62 and 14, 122 f. horizontal, through 1(cid:2)3, 52 3. Evaluate for x (cid:1) 2. a. f1x2 (cid:1) (cid:2)3x (cid:3) 5 c. f1x2 (cid:1) (cid:2)3x2 (cid:3) 2x (cid:2) 1 b. f1x2 (cid:1) 14x (cid:2) 2213x (cid:2) 62 d. f1x2 (cid:1) 15x (cid:3) 222 4. For f1x2 (cid:1) x , determine each of the following values: x2(cid:3) 4 a. f1(cid:2)102 b. f1(cid:2)32 c. f102 d. f1102 (cid:1)3 (cid:2) x, if x 6 0 5. Consider the function f given by f1x2 (cid:1) e (cid:1)3 (cid:3) x, if x (cid:4) 0 Calculate each of the following: a. f1(cid:2)332 b. f102 c. f1782 d. f132 2 REVIEW OF PREREQUISITE SKILLS NEL 1 , i f (cid:2)3 6 t 6 0 t 6. A function s is defined for t 7 (cid:2)3 by s1t2 (cid:1) µ 5, if t (cid:1) 0 t3, i f t 7 0 Evaluate each of the following: a. s1(cid:2)22 b. s1(cid:2)12 c. s102 d. s112 e. s11002 7. Expand, simplify, and write each expression in standard form. a. 1x (cid:2) 621x (cid:3) 22 d. 1x (cid:2) 121x (cid:3) 32(cid:2)12x (cid:3) 521x (cid:2) 22 b. 15 (cid:2) x213 (cid:3) 4x2 e. 1a (cid:3) 223 c. x15x (cid:2) 32 (cid:2) 2x13x (cid:3) 22 f. 19a (cid:2) 523 8. Factor each of the following: a. x3 (cid:2) x c. 2x2 (cid:2) 7x (cid:3) 6 e. 27x3 (cid:2) 64 b. x2 (cid:3) x (cid:2) 6 d. x3 (cid:3) 2x2 (cid:3) x f. 2x3 (cid:2) x2 (cid:2) 7x (cid:3) 6 9. Determine the domain of each of the following: x2 (cid:3) 4 a. y (cid:1) Vx (cid:3) 5 d. h1x2 (cid:1) x 6x b. y (cid:1) x3 e. y (cid:1) 2x2 (cid:2) 5x (cid:2) 3 3 1x (cid:2) 321x (cid:3) 42 c. y (cid:1) f. y (cid:1) x (cid:2) 1 1x (cid:3) 221x (cid:2) 121x (cid:3) 52 10. The height of a model rocket in flight can be modelled by the equation h1t2 (cid:1) (cid:2)4.9t2 (cid:3) 25t (cid:3) 2, where h is the height in metres at t seconds. Determine the average rate of change in the model rocket’s height with respect to time during a. the first second b. the second second 11. Sacha drains the water from a hot tub. The hot tub holds 1600 L of water. It takes 2 h for the water to drain completely. The volume of water in the hot tub is modelled by V1t2 (cid:1) 1600 (cid:2) t2, where V is the volume in litres at 9 t minutes and 0 (cid:5) t (cid:5) 120. a. Determine the average rate of change in volume during the second hour. b. Estimate the instantaneous rate of change in volume after exactly 60 min. c. Explain why all estimates of the instantaneous rate of change in volume where 0 (cid:5) t (cid:5) 120 result in a negative value. 12. a. Sketch the graph of f1x2 (cid:1) (cid:2)21x (cid:2) 322 (cid:3) 4. b. Draw a tangent line at the point 15, f1522, and estimate its slope. c. Estimate the instantaneous rate of change in f1x2 when x (cid:1) 5. NEL CHAPTER 1 3 CAREER LINK Investigate CHAPTER 1: ASSESSING ATHLETIC PERFORMANCE Differential calculus is fundamentally about the idea of instantaneous rate of change. A familiar rate of change is heart rate. Elite athletes are keenly interested in the analysis of heart rates. Sporting performance is enhanced when an athlete is able to increase his or her heart rate at a slower pace (that is, to get tired less quickly). A heart rate is described for an instant in time. Time Number of Heart rate is the instantaneous rate of change in the total number (s) (min) Heartbeats of heartbeats with respect to time. When nurses and doctors count 10 0.17 9 heartbeats and then divide by the time elapsed, they are not 20 0.33 19 determining the instantaneous rate of change but are calculating the average heart rate over a period of time (usually 10 s). In this chapter, 30 0.50 31 the idea of the derivative will be developed, progressing from the 40 0.67 44 average rate of change calculated over smaller and smaller intervals 50 0.83 59 until a limiting value is reached at the instantaneous rate of change. 60 1.00 75 Case Study—Assessing Elite Athlete Performance The table shows the number of heartbeats of an athlete who is undergoing a cardiovascular fitness test. Complete the discussion questions to determine if this athlete is under his or her maximum desired heart rate of 65 beats per minute at precisely 30 s. DISCUSSION QUESTIONS 1. Graph the number of heartbeats versus time (in minutes) on graph paper, joining the points to make a smooth curve. Draw a second relationship on the same set of axes, showing the resting heart rate of 50 beats per minute. Use the slopes of the two relationships graphed to explain why the test results indicate that the person must be exercising. 2. Discuss how the average rate of change in the number of heartbeats over an interval of time could be calculated using this graph. Explain your reasoning. 3. Calculate the athlete’s average heart rate over the intervals of 30 s, 60 s4, 310 s, 50 s4, and 320 s, 40 s4. Show the progression of these average heart rate calculations on the graph as a series of secants. 4. Use the progression of these average heart-rate secants to make a graphical prediction of the instantaneous heart rate at t(cid:1)30 s. Is the athlete’s heart rate less than 65 beats per minute at t(cid:1)30 s? Estimate the heart rate at t(cid:1)60 s. 4 CAREER LINK NEL What Is Calculus? Two simple geometric problems originally led to the development of what is now called calculus. Both problems can be stated in terms of the graph of a function y (cid:1) f1x2. • The problem of tangents:What is the slope of the tangent to the graph of a function at a given point P? • The problem of areas:What is the area under a graph of a function y (cid:1) f1x2 between x (cid:1) a and x (cid:1) b? y slope = ? P y = f(x) area = ? x 0 a b Interest in the problem of tangents and the problem of areas dates back to scientists such as Archimedes of Syracuse (287–212 BCE),who used his vast ingenuity to solve special cases of these problems. Further progress was made in the seventeenth century,most notably by Pierre de Fermat (1601–1665) and Isaac Barrow (1630–1677),a professor of Sir Isaac Newton (1642–1727) at the University of Cambridge,England. Professor Barrow recognized that there was a close connection between the problem of tangents and the problem of areas. However,it took the genius of both Newton and Gottfried Wilhelm von Leibniz (1646–1716) to show the way to handle both problems. Using the analytic geometry of Rene Descartes (1596–1650),Newton and Leibniz showed independently how these two problems could be solved by means of new operations on functions,called differentiation and integration. Their discovery is considered to be one of the major advances in the history of mathematics. Further research by mathematicians from many countries using these operations has created a problem-solving tool of immense power and versatility,which is known as calculus. It is a powerful branch of mathematics, used in applied mathematics,science,engineering,and economics. We begin our study of calculus by discussing the meaning of a tangent and the related idea of rate of change. This leads us to the study of limits and, at the end of the chapter, to the concept of the derivative of a function. NEL CHAPTER 1 5 Section 1.1—Radical Expressions: Rationalizing Denominators Now that we have reviewed some concepts that will be needed before beginning the introduction to calculus, we have to consider simplifying expressions with radicals in the denominator of radical expressions. Recall that a rational number is a number that can be expressed as a fraction (quotient) containing integers. Sothe process of changing a denominator from a radical (square root) to a rational number (integer) is called rationalizing the denominator. The reason that we rationalize denominators is that dividing by an integer is preferable to dividing by a radical number. In certain situations, it is useful to rationalize the numerator. Practice with rationalizing the denominator prepares you for rationalizing the numerator. There are two situations that we need to consider: radical expressions with one-term denominators and those with two-term denominators. For both, the numerator and denominator will be multiplied by the same expression, which is the same as multiplying by one. EXAMPLE 1 Selecting a strategy to rationalize the denominator 3 Simplify by rationalizing the denominator. 4(cid:1)5 Solution 3 3 V5 (cid:1) (cid:6) (Multiply both the numerator 4V5 4V5 V5 and denominator by (cid:1)5) 3V5 (cid:1) (Simplify) 4 (cid:6) 5 3V5 (cid:1) 20 When the denominator of a radical fraction is a two-term expression, you can rationalize the denominator by multiplying by the conjugate. An expression such as (cid:1)a (cid:3) (cid:1)b has the conjugate Va (cid:2) Vb. Why are conjugates important? Recall that the linear terms are eliminated when expanding a difference of squares. For example, 1a (cid:2) b21a (cid:3) b2 (cid:1) a2 (cid:3) ab (cid:2) ab (cid:2) b2 (cid:1) a2 (cid:2) b2 6 1.1 RADICAL EXPRESSIONS: RATIONALIZING DENOMINATORS NEL If a and b were radicals, squaring them would rationalize them. Consider this product: QVm (cid:3) VnR QVm (cid:2) VnR, m, n rational (cid:1) Q(cid:1)mR2 (cid:2) (cid:1)mn (cid:3) (cid:1)mn (cid:2) Q(cid:1)nR2 (cid:1) m (cid:2) n Notice that the result is rational! EXAMPLE 2 Creating an equivalent expression by rationalizing the denominator 2 Simplify by rationalizing the denominator. (cid:1)6 (cid:3) (cid:1)3 Solution 2 2 V6 (cid:2) V3 (Multiply both the numerator and (cid:1) (cid:6) denominator by (cid:1)6(cid:2) (cid:1)3) V6 (cid:3) V3 V6 (cid:3) V3 V6 (cid:2) V3 2Q(cid:1)6 (cid:2) (cid:1)3R (cid:1) (Simplify) 6 (cid:2) 3 2Q(cid:1)6 (cid:2) (cid:1)3R (cid:1) 3 EXAMPLE 3 Selecting a strategy to rationalize the denominator 5 Simplify the radical expression by rationalizing the denominator. 2(cid:1)6 (cid:3) 3 Solution 5 5 2V6 (cid:2) 3 (cid:1) (cid:6) (The conjugate 2(cid:1)6(cid:3)(cid:1)3is 2(cid:1)6(cid:2)(cid:1)3) 2V6 (cid:3) 3 2V6 (cid:3) 3 2V6 (cid:2) 3 5Q2V6 (cid:2) 3R (cid:1) (Simplify) 4V36 (cid:2) 9 5Q2V6 (cid:2) 3R (cid:1) 24 (cid:2) 9 5Q2V6 (cid:2) 3R (cid:1) (Divide by the common factor of 5) 15 2V6 (cid:2) 3 (cid:1) 3 The numerator can also be rationalized in the same way as the denominator was in the previous expressions. NEL CHAPTER 1 7 EXAMPLE 4 Selecting a strategy to rationalize the numerator Rationalize the numerator of the expression (cid:1)7 (cid:2) (cid:1)3. 2 Solution V7 (cid:2) V3 V7 (cid:2) V3 V7 (cid:3) V3 (Multiply the numerator and (cid:1) (cid:6) 2 2 V7 (cid:3) V3 denominator by (cid:1)7(cid:3)(cid:1)3) 7 (cid:2) 3 (cid:1) (Simplify) 2Q(cid:1)7 (cid:3) (cid:1)3R 4 (cid:1) (Divide by the common factor of 2) 2Q(cid:1)7 (cid:3) (cid:1)3R 2 (cid:1) V7 (cid:3) V3 IN SUMMARY Key Ideas • To rewrite a radical expression with a one-term radical in the denominator, multiply the numerator and denominator by the one-term denominator. (cid:1)a (cid:1)a (cid:1)b (cid:1) (cid:6) (cid:1)b (cid:1)b (cid:1)b (cid:1)ab (cid:1) b • When the denominator of a radical expression is a two-term expression, rationalize the denominator by multiplying the numerator and denominator by the conjugate, and then simplify. 1 1 (cid:1)a(cid:3) (cid:1)b (cid:1) (cid:6) (cid:1)a(cid:2) (cid:1)b (cid:1)a(cid:2) (cid:1)b (cid:1)a(cid:3) (cid:1)b (cid:1)a(cid:3) (cid:1)b (cid:1) a(cid:2)b Need to Know • When you simplify a radical expression such as (cid:1)3, multiply the numerator 5(cid:1)2 and denominator by the radical only. (cid:1)3 (cid:1)2 (cid:1)6 (cid:6) (cid:1) 5(cid:1)2 (cid:1)2 5122 (cid:1)6 (cid:1) 10 • (cid:1)a(cid:3) (cid:1)b is the conjugate (cid:1)a (cid:2) (cid:1)b, and vice versa. 8 1.1 RADICAL EXPRESSIONS: RATIONALIZING DENOMINATORS NEL Exercise 1.1 PART A 1. Write the conjugate of each radical expression. a. 2(cid:1)3 (cid:2) 4 c. (cid:2)2V3 (cid:2) V2 e. V2 (cid:2) V5 b. V3 (cid:3) V2 d. 3V3 (cid:3) V2 f. (cid:2)V5 (cid:3) 2V2 2. Rationalize the denominator of each expression. Write your answer in simplest form. V3 (cid:3) V5 4V3 (cid:3) 3V2 a. c. V2 2V3 2V3 (cid:2) 3V2 3V5 (cid:2) V2 b. d. V2 2V2 PART B K 3. Rationalize each denominator. 3 V3 (cid:2) V2 2V3 (cid:2) V2 a. c. e. V5 (cid:2) V2 V3 (cid:3) V2 5V2 (cid:3) V3 2V5 2V5 (cid:2) 8 3V3 (cid:2) 2V2 b. d. f. 2V5 (cid:3) 3V2 2V5 (cid:3) 3 3V3 (cid:3) 2V2 4. Rationalize each numerator. V5 (cid:2) 1 2 (cid:2) 3V2 V5 (cid:3) 2 a. b. c. 4 2 2V5 (cid:2) 1 8(cid:1)2 C 5. a. Rationalize the denominator of . (cid:1)20 (cid:2) (cid:1)18 8(cid:1)2 b. Rationalize the denominator of . 2(cid:1)5 (cid:2) 3(cid:1)2 c. Why are your answers in parts a and b the same? Explain. 6. Rationalize each denominator. 2V2 2V2 3V5 a. c. e. 2V3 (cid:2) V8 V16 (cid:2) V12 4V3 (cid:2) 5V2 2V6 3V2 (cid:3) 2V3 V18 (cid:3) V12 b. d. f. 2V27 (cid:2) V8 V12 (cid:2) V8 V18 (cid:2) V12 7. Rationalize the numerator of each of the following expressions: A Va (cid:2) 2 Vx (cid:3) 4 (cid:2) 2 Vx (cid:3) h (cid:2) x a. b. c. a (cid:2) 4 x x NEL CHAPTER 1 9 Section 1.2—The Slope of a Tangent You are familiar with the concept of a tangentto a curve. What geometric interpretation can be given to a tangent to the graph of a function at a point P? A tangent is the straight line that most resembles the graph near a point. Its slope tells how steep the graph is at the point of tangency. In the figure below, four tangents have been drawn. T 2 T T 1 3 y = f(x) T 4 The goal of this section is to develop a method for determining the slope of a tangent at a given point on a curve. We begin with a brief review of lines and slopes. Lines and Slopes y P2 (x2, y2) l Dy P1 (x1, y1) Dx x 0 The slope m of the line joining points P 1x , y 2 and P 1x , y 2 is defined as 1 1 1 2 2 2 m (cid:1) ¢y (cid:1) y2 (cid:2) y1. ¢x x2 (cid:2) x1 The equation of the line lin point-slope form is y (cid:2) y1 (cid:1) mor y (cid:2) y (cid:1) m1x (cid:2) x 2. x (cid:2) x1 1 1 The equation in slope–y-intercept form is y (cid:1) mx (cid:3) b,where bis the y-intercept of the line. To determine the equation of a tangent to a curve at a given point, we first need to know the slope of the tangent. What can we do when we only have one point? We proceed as follows: y Q Q Q tangent at P secant P y = f(x) x 0 10 1.2 THE SLOPE OF A TANGENT NEL