Contents CHAPTER 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant Circular Motion A Review of Trigonometry A Thousand Points of Light Computing in Calculus CHAPTER 2 Derivatives The Derivative of a Function Powers and Polynomials The Slope and the Tangent Line Derivative of the Sine and Cosine The Product and Quotient and Power Rules Limits Continuous Functions CHAPTER 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 Applications of the Derivative Linear Approximation Maximum and Minimum Problems Second Derivatives: Minimum vs. Maximum Graphs Ellipses, Parabolas, and Hyperbolas , Iterations x,+ = F(x,) Newton's Method and Chaos The Mean Value Theorem and l'H8pital's Rule Contents CHAPTER 4 4.1 4.2 4.3 4.4 The Chain Rule Derivatives by the Chain Rule Implicit Differentiation and Related Rates Inverse Functions and Their Derivatives Inverses of Trigonometric Functions CHAPTER 5 Integrals 5.1 The Idea of the Integral 177 5.2 Antiderivatives 182 5.3 Summation vs. Integration 187 5.4 Indefinite Integrals and Substitutions 195 5.5 The Definite Integral 201 5.6 Properties of the Integral and the Average Value 206 5.7 The Fundamental Theorem and Its Consequences 213 5.8 Numerical Integration 220 CHAPTER 6 Exponentials and Logarithms 6.1 An Overview 228 6.2 The Exponential ex 236 6.3 Growth and Decay in Science and Economics 242 6.4 Logarithms 252 6.5 Separable Equations Including the Logistic Equation 259 6.6 Powers Instead of Exponentials 267 6.7 Hyperbolic Functions 277 CHAPTER 7 7.1 7.2 7.3 7.4 7.5 Techniques of Integration Integration by Parts Trigonometric Integrals Trigonometric Substitutions Partial Fractions Improper Integrals CHAPTER 8 8.1 8.2 8.3 8.4 8.5 8.6 Applications of the Integral Areas and Volumes by Slices Length of a Plane Curve Area of a Surface of Revolution Probability and Calculus Masses and Moments Force, Work, and Energy Contents CHAPTER 9 Polar Coordinates and Complex Numbers 9.1 Polar Coordinates 348 9.2 Polar Equations and Graphs 351 9.3 Slope, Length, and Area for Polar Curves 356 9.4 Complex Numbers 360 CHAPTER 10 10.1 10.2 10.3 10.4 10.5 Infinite Series The Geometric Series Convergence Tests: Positive Series Convergence Tests: All Series The Taylor Series for ex, sin x, and cos x Power Series CHAPTER 11 11.1 11.2 11.3 11.4 11.5 Vectors and Matrices Vectors and Dot Products Planes and Projections Cross Products and Determinants Matrices and Linear Equations Linear Algebra in Three Dimensions CHAPTER 12 Motion along a Curve 12.1 The Position Vector 446 12.2 Plane Motion: Projectiles and Cycloids 453 12.3 Tangent Vector and Normal Vector 459 12.4 Polar Coordinates and Planetary Motion 464 CHAPTER 13 Partial Derivatives 13.1 Surfaces and Level Curves 472 13.2 Partial Derivatives 475 13.3 Tangent Planes and Linear Approximations 480 13.4 Directional Derivatives and Gradients 490 13.5 The Chain Rule 497 13.6 Maxima, Minima, and Saddle Points 504 13.7 Constraints and Lagrange Multipliers 514 Contents CHAPTER 14 14.1 14.2 14.3 14.4 Multiple Integrals Double Integrals Changing to Better Coordinates Triple Integrals Cylindrical and Spherical Coordinates CHAPTER 15 15.1 15.2 15.3 15.4 15.5 15.6 Vector Calculus Vector Fields Line Integrals Green's Theorem Surface Integrals The Divergence Theorem Stokes' Theorem and the Curl of F CHAPTER 16 16.1 16.2 16.3 Mathematics after Calculus Linear Algebra Differential Equations Discrete Mathematics Study Guide For Chapter 1 Answers to Odd-Numbered Problems Index Table of Integrals Contents CHAPTER 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant Circular Motion A Review of Trigonometry A Thousand Points of Light Computing in Calculus CHAPTER 2 Derivatives The Derivative of a Function Powers and Polynomials The Slope and the Tangent Line Derivative of the Sine and Cosine The Product and Quotient and Power Rules Limits Continuous Functions CHAPTER 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 Applications of the Derivative Linear Approximation Maximum and Minimum Problems Second Derivatives: Minimum vs. Maximum Graphs Ellipses, Parabolas, and Hyperbolas , Iterations x,+ = F(x,) Newton's Method and Chaos The Mean Value Theorem and l'H8pital's Rule CHAPTER 1 Introduction to Calculus 1.4 Velocity and Distance The right way to begin a calculus book is with calculus. This chapter will jump directly into the two problems that the subject was invented to solve. You will see what the questions are, and you will see an important part of the answer. There are plenty of good things left for the other chapters, so why not get started? The book begins with an example that is familiar to everybody who drives a car. It is calculus in action-the driver sees it happening. The example is the relation between the speedometer and the odometer. One measures the speed (or velocity); the other measures the distance traveled. We will write v for the velocity, and f for how far the car has gone. The two instruments sit together on the dashboard: Fig. 1.1 Velocity v and total distance f (at one instant of time). Notice that the units of measurement are different for v and f.The distance f is measured in kilometers or miles (it is easier to say miles). The velocity v is measured in km/hr or miles per hour. A unit of time enters the velocity but not the distance. Every formula to compute v from f will have f divided by time. The central question of calculus is the relation between v and f. 1 Introduction to Calculus Can you find v if you know f , and vice versa, and how? If we know the velocity over the whole history of the car, we should be able to compute the total distance traveled. In other words, if the speedometer record is complete but the odometer is missing, its information could be recovered. One way to do it (without calculus) is to put in a new odometer and drive the car all over again at the right speeds. That seems like a hard way; calculus may be easier. But the point is that the information is there. If we know everything about v, there must be a method to find f . What happens in the opposite direction, when f is known? If you have a complete record of distance,could you recover the complete velocity?In principle you could drive the car, repeat the history, and read off the speed. Again there must be a better way. The whole subject of calculus is built on the relation between u and f . The question we are raising here is not some kind of joke, after which the book will get serious and the mathematics will get started. On the contrary, I am serious now-and the mathematics has already started. We need to know how to find the velocity from a record of the distance. (That is called &@erentiation, and it is the central idea of dflerential calculus.) We also want to compute the distance from a history of the velocity. (That is integration, and it is the goal of integral calculus.) Differentiation goes from f to v; integration goes from v to f . We look first at examples in which these pairs can be computed and understood. CONSTANT VELOCITY Suppose the velocity is fixed at v = 60 (miles per hour). Then f increases at this constant rate. After two hours the distance is f = 120 (miles). After four hours f = 240 and after t hours f = 60t. We say that f increases linearly with time-its graph is a straight line. 4 velocity v ( t ) v Area 240 : time t 4 distancef ( t ) 2 4 0 ~ ~ s 1 ~ =4 " = 6 0 I time t Fig. 1.2 Constant velocity v = 60 and linearly increasing distance f= 60t. Notice that this example starts the car at full velocity. No time is spent picking up speed. (The velocity is a "step function.") Notice also that the distance starts at zero; the car is new. Those decisions make the graphs of v and f as neat as possible. One is the horizontal line v = 60. The other is the sloping line f = 60t. This v, f , t relation needs algebra but not calculus: if v is constant andf starts at zero then f = vt. The opposite is also true. When f increases linearly, v is constant. The division by time gives the slope. The distance is fl = 120 miles when the time is t 1 = 2 hours. Later f' = 240 at t , = 4. At both points, the ratio f / t is 60 miles/hour. Geometrically, the velocity is the slope of the distance graph: slope = change in distance change in time -- -vt t - - v . 1.1 Velocity and Distance + Fig. 1.3 Straight lines f = 20 60t (slope 60) and f = -30t (slope -30). The slope of thef-graph gives the v-graph. Figure 1.3 shows two more possibilities: 1. The distance starts at 20 instead of 0. The distance formula changes from 60t + to 20 60t. The number 20 cancels when we compute change in distance-so the slope is still 60. 2. When v is negative, the graph off goes downward. The car goes backward and the slope o f f = - 30t is v = - 30. I don't think speedometers go below zero. But driving backwards, it's not that safe to watch. If you go fast enough, Toyota says they measure "absolute valuesw-the + speedometer reads 30 when the velocity is - 30. For the odometer, as far as I know it just stops. It should go backward.? VELOCITY vs. DISTANCE: SLOPE vs. AREA How do you compute f' from v? The point of the question is to see f = ut on the graphs. We want to start with the graph of v and discover the graph off. Amazingly, the opposite of slope is area. The distance f is the area under the v-graph. When v is constant, the region under the graph is a rectangle. Its height is v, its width is t , and its area is v times t. This is integration, to go from v to f by computing the area. We are glimpsing two of the central facts of calculus. 1A The slope of the f-graph gives the velocity v. The area under the v-graph gives the distance f. That is certainly not obvious, and I hesitated a long time before I wrote it down in this first section. The best way to understand it is to look first at more examples. The whole point of calculus is to deal with velocities that are not constant, and from now on v has several values. EXAMPLE (Forward and back) There is a motion that you will understand right away. The car goes forward with velocity V, and comes back at the same speed. To say it more correctly, the velocity in the second part is - V. If the forward part lasts until t = 3, and the backward part continues to t = 6, the car will come back where it started. The total distance after both parts will be f = 0. -- +This actually happened in Ferris Bueller's Day 08,when the hero borrowed his father's sports car and ran up the mileage. At home he raised the car and drove in reverse. I forget if it worked. 1 Introductionto Calculus 1u(r)= slope off ( t ) + Fig. 1.4 Velocities V and - V give motion forward and back, ending at f (6)=0. + The v-graph shows velocities + V and - V. The distance starts up with slope V and reaches f = 3V. Then the car starts backward. The distance goes down with slope - V and returns to f = 0 at t = 6. Notice what that means. The total area "under" the v-graph is zero! A negative velocity makes the distance graph go downward (negative slope). The car is moving backward. Area below the axis in the v-graph is counted as negative. FUNCTIONS This forward-back example gives practice with a crucially important idea-the concept of a "jiunction." We seize this golden opportunity to explain functions: The number v(t) is the value of the function t. at the time t. The time t is the input to the function. The velocity v(t) at that time is the output. Most people say "v oft" when they read v(t). The number "v of 2" is the velocity + when t = 2. The forward-back example has v(2) = V and v(4) = - V. The function contains the whole history, like a memory bank that has a record of v at each t. It is simple to convert forward-back motion into a formula. Here is v(t): The ,right side contains the instructions for finding v(t). The input t is converted into + the output V or - V. The velocity v(t) depends on t. In this case the function is "di~continuo~sb,e~c'ause the needle jumps at t = 3. The velocity is not dejined at that instant. There is no v(3). (You might argue that v is zero at the jump, but that leads to trouble.) The graph off' has a corner, and we can't give its slope. The problem also involves a second function, namely the distance. The principle behind f(t) is the same: f (t) is the distance at time t. It is the net distance forward, and again the instructions change at t = 3. In the forward motion, f(t) equals Vt as before. In the backward half, a calculation is built into the formula for f(t): At the switching time the right side gives two instructions (one on each line). This would be bad except that they agree: f (3)= 3 V . v h e distance function is "con- ?A function is only allowed one ~:alue,f'(orr) ~ ( ta)t each time r 1.1 Velocity and Distance tinuous." There is no jump in f, even when there is a jump in v. After t = 3 the distance decreases because of - Vt. At t = 6 the second instruction correctly gives f (6)= 0. Notice something more. The functions were given by graphs before they were given by formulas. The graphs tell you f and v at every time t-sometimes more clearly than the formulas. The values f (t) and v(t) can also be given by tables or equations or a set of instructions. (In some way all functions are instructions-the function tells how to find f at time t.) Part of knowing f is knowing all its inputs and outputs-its domain and range: The domain of a function is the set of inputs. The range is the set of outputs. The domain of f consists of all times 0 < t < 6. The range consists of all distances + 0 0). Then draw (b) U(t) + 2 ( 4 U(t + 2) ( 4 3UW (e) U(3t). + 45 (a) Draw the graph of f (t) = t 1 for -1 Q t 6 1. Find the domain, range, slope, and formula for (b) 2f ( 0 ( 4 f (t - 3) (d) -f (0 (el f k t ) . 46 If f (t) = t - 1 what are 2f (3t) and f (1 -t) and f (t - I)? 47 In the forward-back example find f (* T )and f (3T). Verify that those agree with the areas "under" the v-graph in Figure 1.4. 48 Find formulas for the outputs fl(t) and fi(t) which come from the input t: (1) inside = input * 3 + output =inside 3 (2) inside +input + 6 output t inside* 3 Note BASIC and FORTRAN (and calculus itself) use = instead of t.But the symbol t or := is in some ways better. + The instruction t + t 6 produces a new t equal to the old t + plus six. The equation t = t 6 is not intended. 49 Your computer can add and multiply. Starting with the number 1 and the input called t, give a list of instructions to lead to these outputs: f1(t)=t2+t f2(t)=fdfdt)) f3(t)=f1(t+l)50 In fifty words or less explain what a function is. The last questions are challenging but possible. 51 If f (t) = 3t - 1 for 0 6 t Q 2 give formulas (with domain) and find the slopes of these six functions: (a) f (t + 2) (b) f(t) + 2 ( 4 2f ( 0 ( 4 f (2t) (e) f (- t) (f) f ( f (t)). + 52 For f (t) = ut C find the formulas and slopes of (a) 3f(0 + 1 (b) f (3t + 1) (c) 2f(4t) (dl f (- t) (el f ( 0 -f (0) (f) f ( f (t)). 53 (hardest) The forward-back function is f (t) = 2t for O < t ~ 3f (,t ) = 12-2t for 3 6 t d 6 . Graph f(f(t)) and find its four-part formula. First try t = 1.5 and 3. 54 (a) Why is the letter X not the graph of a function? (b) Which capital letters are the graphs of functions? (c) Draw graphs of their slopes. 1.2 Calculus Without Limits The next page is going to reveal one of the key ideas behind calculus. The discussion is just about numbers-functions and slopes can wait. The numbers are not even special, they can be any numbers. The crucial point is to look at their differences: Suppose the numbers are f = 0 2 6 7 4 9 Their differences are v = 2 4 1-35 The differences are printed in between, to show 2 -0 = 2 and 6 - 2 = 4 and 7 - 6 = 1. 1.2 Calculus Without Limits Notice how 4 - 7 gives a negative answer -3. The numbers in f can go up or down, the differences in v can be positive or negative. The idea behind calculus comes when you add up those differences: 2+4+1-3+5=9 The sum of differences is 9. This is the last number on the top line (in f). Is this an accident, or is this always true? If we stop earlier, after 2 + 4 + 1, we get the 7 in f. Test any prediction on a second example: Suppose the numbers are f= 1 3 7 8 5 10 Their differences are v = 2 4 1 -3 5 The f's are increased by 1. The differences are exactly the same-no change. The sum of differences is still 9. But the last f is now 10. That prediction is not right, we don't always get the last f. The first f is now 1. The answer 9 (the sum of differences) is 10 - 1, the last f minus the first f. What happens when we change the f's in the middle? Suppose the numbers are f= 1 5 12 7 10 Their differences are v = 4 7 -5 3 The differences add to 4 + 7 - 5+ 3 = 9. This is still 10 - 1. No matter what f's we choose or how many, the sum of differences is controlled by the first f and last f. If this is always true, there must be a clear reason why the middle f's cancel out. The sum of differences is (5- 1)+ (12 - 5)+ (7- 12) + (10 - 7)= 10 - 1. The 5's cancel, the 12's cancel, and the 7's cancel. It is only 10 - 1 that doesn't cancel. This is the key to calculus! EXAMPLE 1 The numbers grow linearly: f= 2 3 4 5 6 7 Their differences are constant: v = 1 1 1 1 1 The sum of differences is certainly 5. This agrees with 7 - 2 =fast -ffirst. The numbers in v remind us of constant velocity. The numbers in f remind us of a straight line f= vt + C. This example has v = 1 and the f's start at 2. The straight line would come from f= t + 2. EXAMPLE 2 The numbers are squares: f= 0 1 4 9 16 Their differences grow linearly: v = 1 3 5 7 1+ 3 + 5 + 7 agrees with 42 = 16. It is a beautiful fact that the first jodd numbers always add up to j2. The v's are the odd numbers, the f's are perfect squares. Note The letter j is sometimes useful to tell which number in f we are looking at. For this example the zeroth number is fo = 0 and the jth number is fj =j2. This is a part of algebra, to give a formula for the f's instead of a list of numbers. We can also use j to tell which difference we are looking at. The first v is the first odd number v,= 1. The jth difference is the jth odd number vj = 2j- 1.(Thus v4 is 8 - I = 7.) It is better to start the differences with j= 1, since there is no zeroth odd number vo. With this notation the jth difference is vj =fj -f -1.Sooner or later you will get comfortable with subscripts like j and j - 1,but it can be later. The important point is that the sum of the v's equals flast -first. We now connect the v's to slopes and the f's to areas. 10 0~~~~~~~ v44 = 7 1 Introduction to Calculus 1 nrdcin oCluu f4= 1 v3 = 5 v2 = 3 f 3 =9 1 =I t 12 3 4 f2=4 f, = 1 1 2 t 3 4 Fig. 1.7 Linear increase in v = 1, 3, 5, 7. Squares in the distances f= 0, 1,4, 9, 16. Figure 1.7 shows a natural way to graph Example 2, with the odd numbers in v and the squares in f. Notice an important difference between the v-graph and the f-graph. The graph of f is "piecewise linear." We plotted the numbers in f and connected them by straight lines. The graph of v is "piecewise constant." We plotted the differences as constant over each piece. This reminds us of the distance-velocity graphs, when the distance f(t) is a straight line and the velocity v(t) is a horizontal line. Now make the connection to slopes: The slope of the f-graph is distance up ddisistatannccee across ccchhhaaannngggeee iiinnn f t Over each piece, the change in t (across) is 1.The change in f (upward) is the difference that we are calling v. The ratio is the slope v/1l or just v. The slope makes a sudden change at the breakpoints t = 1,2, 3,.... At those special points the slope of the f-graph is not defined-we connected the v's by vertical lines but this is very debatable. The main idea is that between the breakpoints, the slope of f(t) is v(t). Now make the connection to areas: The total area under the v-graph is flast -ffirst This area, underneath the staircase in Figure 1.7, is composed of rectangles. The base of every rectangle is 1. The heights of the rectangles are the v's. So the areas also equal the v's, and the total area is the sum of the v's. This area is flast -first. Even more is true. We could start at any time and end at any later time -not necessarily at the special times t = 0, 1,2, 3,4. Suppose we stop at t = 3.5. Only half of the last rectangular area (under v = 7) will be counted. The total area is 1+ 3 + 5 + 2(7) = 12.5. This still agrees with flast -first = 12.5 - 0. At this new ending time t = 3.5, we are only halfway up the last step in the f-graph. Halfway between 9 and 16 is 12.5. This is nothing less than the Fundamental Theorem of Calculus. But we have only used algebra (no curved graphs and no calculations involving limits). For now the Theorem is restricted to piecewise linear f(t) and piecewise constant v(t). In Chapter 5 that restriction will be overcome. Notice that a proof of 1+ 3 + 5 + 7 = 42 is suggested by Figure 1.7a. The triangle under the dotted line has the same area as the four rectangles under the staircase. The area of the triangle is ½.base . height = -4 8, which is the perfect 9quare 42 When there are j rectangles instead of 4, we get .j. 2j =j2 for the area. 1.2 Calculus Wnhout Limits The next examples show other patterns, where f and v increase exponentially or oscillate around zero. I hope you like them but I don't think you have to learn them. They are like the special functions 2' and sin t and cos t-except they go in steps. You get a first look at the important functions of calculus, but you only need algebra. Calculus is neededfor a steadily changing velocity, when the graph off is curved. The last example will be income tax-which really does go. in steps. Then Section 1.3 will introduce the slope of a curve. The crucial step for curves is working with limits. That will take us from algebra to calculus. EXPONENTIAL VELOCITY AND DISTANCE Start with the numbers f = 1,2,4,8, 16. These are "powers of 2." They start with the zeroth power, which is 2' = 1. The exponential starts at 1 and not 0. Afterj steps there are j factors of 2, and & equals 2j. Please recognize the diflerence between 2j and j2 and 2j. The numbers 2j grow linearly, the numbersj2grow quadratically, the numbers 2' grow exponentially. At j = 10 these are 20 and 100 and 1024. The exponential 2' quickly becomes much larger than the others. The differences off = 1,2,4,8, 16 are exactly v = 1,2,4,8..We get the same beautiful numbers. When the f's are powers of 2, so are the v's. The formula vj = 2"-' is slightly different from & = 2j, because the first v is numbered v,. (Then v, = 2' = 1. The zeroth power of every number is 1,except that 0' is meaningless.)The two graphs in Figure 1.8 use the same numbers but they look different, because f is piecewise linear and v is piecewise constant. 1234 1234 Fig. 1.8 The velocity and distance grow exponentially (powers of 2). Where will calculus come in? It works with the smooth curve f (t)= 2'. This exponential growth is critically important for population and money in a bank and the national debt. You can spot it by the following test: v(t) is proportional to f (t). Remark The function 2' is trickier than t2. For f = t2 the slope is v = 2t. It is proportional to t and not t2. For f = 2' the slope is v = c2', and we won't find the constant c = .693 ... until Chapter 6. (The number c is the natural logarithm of 2.) Problem 37 estimates c with a calculator-the important thing is that it's constant. OSCILLATING VELOCITY AND DISTANCE We have seen a forward-back motion, velocity V followed by - V. That is oscillation of the simplest kind. The graph o f f goes linearly up and linearly down. Figure 1.9 shows another oscillation that returns to zero, but the path is more interesting. The numbers in f are now 0, 1, 1,0, -1, -l,O. Since f6 = 0 the motion brings us back to the start. The whole oscillation can be repeated. 1 lnhoductlonto Calculus The differences in v are 1,0, -1, -1,0, 1. They add up to zero, which agrees with Jast -Airst. It is the same oscillation as in f (and also repeatable),but shifted in time. The f-graph resembles (roughly) a sine curve. The v-graph resembles (even more roughly) a cosine curve. The waveforms in nature are smooth curves, while these are "digitized"-the way a digital watch goes forward in jumps. You recognize that the change from analog to digital brought the computer revolution. The same revolution is coming in CD players. Digital signals (off or on, 0 or 1 ) seem to win every time. The piecewise v and f start again at t = 6. The ordinary sine and cosine repeat at t = 2n. A repeating motion isperiodic-here the "period" is 6 or 2n. (With t in degrees the period is 360-a full circle. The period becomes 2n when angles are measured in radians. We virtually always use radians-which are degrees times 2n/360.) A watch has a period of 12 hours. If the dial shows AM and PM, the period is . Fig. 1.9 Piecewise constant "cosine" and piecewise linear "sine."They both repeat. A SHORT BURST O F SPEED The next example is a car that is driven fast for a short time. The speed is V until the distance reaches f = 1, when the car suddenly stops. The graph of f goes up linearly with slope V , and then across with slope zero: V upto t=T v(t) = 0 after t = T Vt up to t = T f (0= 1 after t = T This is another example of "function notation." Notice the general time t and the particular stopping time T. The distance is f (t). The domain off (the inputs) includes all times t 3 0. The range of f (the outputs) includes all distances 0 ff < 1. Figure 1.10 allows us to compare three cars-a Jeep and a Corvette and a Maserati. They have different speeds but they all reach f = 1. So the areas under the v-graphs are all 1. The rectangles have height V and base T = 1/ V. v~ EQUAL AREAS Maserati vc - - - - - 7 I Corvette v~ I I Jeep I EQUAL DISTANCES 1 II deltaII function I I II steD T~ T~ Fig. 1.10 Bursts of speed with V, TM= VcTc = 'V,T,= 1. Step function has infinite slope. Optional remark It is natural to think about faster and faster speeds, which means steeper slopes. The f-graph reaches 1 in shorter times. The extreme case is a step function, when the graph of f goes straight up. This is the unit step U(t),which is zero up to t = 0 and jumps immediately to U = 1 for t >0. 1.2 Calculus Without Limits 13 What is the slope of the step function? It is zero except at the jump. At that moment, which is t = 0, the slope is infinite. We don't have an ordinary velocity v(t)-instead we have an impulse that makes the car jump. The graph is a spike over the single point t = 0, and it is often denoted by 6-so the slope of the step function is called a "deltafunction." The area under the infinite spike is 1. You are absolutely not responsible for the theory of delta functions! Calculus is about curves, not jumps. Our last example is a real-world application of slopes ands rates-to explain "how taxes work." Note especially the difference between tax rates and tax brackets and total tax. The rates are v, the brackets are on x, the total tax is f. EXAMPLE 3 Income tax is piecewise linear. The slopes are the tax rates .15,.28,.31. Suppose you are single with taxable income of x dollars (Form 1040, line 37-after all deductions). These are the 1991 instructions from the Internal Revenue Service: If x is not over $20,350, the tax is 15% of x. If $20,350 < x < $49,300, the tax is $3052.50 + 28% of the amount over $20,350. If x is over $49,300, the tax is $11,158.50 + 31% of the amount over $49,300. The first bracket is 0 < x < $20,350. (The IRS never uses this symbol <, but I think it is OK here. We know what it means.) The second bracket is $20,350 < x < $49,300. The top bracket x > $49,300 pays tax at the top rate of 31%. But only the income in that bracket is taxed at that rate. Figure 1.11 shows the rates and the brackets and the tax due. Those are not average rates, they are marginal rates. Total tax divided by total income would be the average rate. The marginal rate of.28 or .31 gives the tax on each additionaldollar of incomeit is the slope at the point x. Tax is like area or distance-itadds up. Tax rate is like slope or velocity-it depends where you are. This is often unclear in the news media. v2= 60 ov = 20 A• 1'I.OUon - acros1s8s0u3p =slope60 ktax to payf(x) 11,158- tsaloxpreat.e2=8 31% 2 5 Fig. 1.11 f(2)= 40 S• slpe 20 2 5 3,052- 15% I 20,350 taxable income I Y 49,300 The tax rate is v,the total tax is f. Tax brackets end at breakpoints. Question What is the equation for the straight line in the top bracket? Answer The bracket begins at x = $49,300 when the tax is f(x) = $11,158.50. The slope of the line is the tax rate .31. When we know a point on the line and the slope, we know the equation. This is important enough to be highlighted. Section 2.3 presents this "point-slope equation" for any straight line. Here you see it for one specific example. Where does the number $11,158.50 come from? It is the tax at the end of the middle bracket, so it is the tax at the start of the top bracket. 1 Introduction to Calculus Figure 1.11 also shows a distance-velocity example. The distance at t = 2 is f (2)= 40 miles. After that time the velocity is 60 miles per hour. So the line with slope 60 on the f-graph has the equation + + f (t) = starting distance extra distance =40 60(t - 2). The starting point is (2'40). The new speed 60 multiplies the extra time t - 2. The point-slope equation makes sense. We now review this section, with comments. Central idea Start with any numbers in f. Their differences go in v. Then the sum of those differences is ha,,-ffirst. Subscript notation The numbers are f,, fl, ... and the first difference is v, =fl -f,. A typical number is fi and the jth difference is vj =fi -fi- . When those differences are added, all f's in the middle (like f,) cancel out: fi Examples =j or j2or 2'. Then vj = 1 (constant) or 2j - 1 (odd numbers) or 2'- '. Functions Connect the f's to be piecewise linear. Then the slope v is piecewise constant. The area under the v-graph from any t,,,,, to any ten,equals f (ten,)-f (t,,,,,). Units Distance in miles and velocity in miles per hour. Tax in dollars and tax rate in (dollars paid)/(dollars earned). Tax rate is a percentage like .28, with no units. 1.2 EXERCISES Read-through questions Problems 1-4 are about numbers f and differences v. Start with the numbers f = 1,6,2,5. Their differences are v = a .The sum of those differences is b .This is equal + + to f,,,, minus c . The numbers 6 and 2 have no effect on this answer, because in (6 - 1) (2 -6) (5 - 2) the numbers 6 and 2 d . The slope of the line between f(0) = 1 and f (1) = 6 is e . The equation of that line is f (t) = f . With distances 1, 5, 25 at unit times, the velocities are g . These are the h of the f-graph. The slope of the tax graph is the tax i . If f(t) is the postage cost for t ounces or t grams, the slope is the i per k . For distances 0, 1,4,9 the velocities are I . The sum of the first j odd numbers is fi = m . Then flo is n and the velocity ulo is 0 . The piecewise linear sine has slopes P . Those form a piecewise q cosine. Both functions have r equal to + 6, which means that f (t 6) = s for every t. The veloci- ties v = 1,2,4,8, ... have vj = t . In that case fo = 1 and jj.= u . The sum of 1,2,4,8, 16 is v . The difference 2J-2'- ' equals w . After a burst of speed V to time T, the distance is x . If f(T) = 1 and V increases, the burst lasts only to T = Y . When V approaches infinity, f (t) approaches a function. The velocities approach a A function, which is concentrated at t = 0 but has area B under its graph. The slope of a step function is c . 1 From the numbers f =0,2,7,10 find the differences u and the sum of the three v's. Write down another f that leads to the same v's. For f = 0,3,12,10 the sum of the u's is still . 2 Starting from f = 1,3,2,4draw the f-graph (linear pieces) and the v-graph. What are the areas "under" the u-graph that add to 4 - l? If the next number in f is 11, what is the area under the next v? 3 From v = 1,2, 1'0, -1 find the f's starting at fo = 3. Graph v and f. The maximum value of f occurs when v= .Where is the maximum f when u = 1,2,1, -l? 4 For f = 1, b, c, 7 find the differences vl ,u2, v, and add them up. Do the same for f = a, b, c, 7. Do the same for f =a, b, c, d. Problems 5-11 are about linear functions and constant slopes. 5 Write down the slopes of these linear functions: (a) f ( t ) = 1.lt (b) f ( t ) = 1 -2t (c) f ( t ) = 4 + 5(t -6). Compute f (6) and f (7) for each function and confirm that f (7) -f (6) equals the slope. 6 If f (t)= 5 + 3(t - 1) and g(t) = 1.5 + 2S(t - 1) what is h(t) =f (t)-g(t)? Find the slopes of f, g, and h. I .2 CalculusWithout Llmits Suppose ~ ( t=) 2 for t < 5 and v(t)= 3 for t > 5. (a) If f (0)=0 find a two-part formula for f (t). (b) Check that f (10) equals the area under the graph of v(t) (two rectangles) up to t = 10. Suppose u(t)= 10 for t < 1/10, v(t)=0 for t > 1/10. Starting from f (0)= 1 find f (t) in two pieces. + 9 Suppose g(t)= 2t 1 and f (t)=4t. Find g(3) and f (g(3)) and f (g(t)).How is the slope of f (g(t)) related to the slopes of f and g? 10 For the same functions, what are f (3) and g(f (3)) and g(f (t))?When t is changed to 4t, distance increases times as fast and the velocity is'multiplied by . 11 Compute f (6) and f (8) for the functions in Problem 5. Confirm that the slopes v agree with slope = f (8)-f 8 -6 (6) -- change change in in f t ' Problems 12-18 are based on Example 3 about income taxes. 12 What are the income taxes on x=$10,000 and x =$30,000 and x = $50,000? 13 What is the equation for income tax f(x) in the second bracket $20,350 1, define f beyond t = 1 so it is (a) continuous (b) discontinuous. (c) Define a tax function f(x) with rates .15 and .28 so you would lose by earning an extra dollar beyond the breakpoint. 17 The difference between a tax credit and a deduction from income is the difference between f (x)-c and f (x - d). Which is more desirable, a credit of c = $1000 or a deduction of d = $1000, and why? Sketch the tax graphs when f (x)= .15x. 18 The average tax rate on the taxable income x is a(x) = f (x)/x.This is the slope between (0,O)and the point (x,f (x)). Draw a rough graph of a(x).The average rate a is below the marginal rate v because . . Problems 19-30 involve numbers fo, f,,f2, ...and their differ- ences vj =& -&-, They give practice with subscripts 0, ...,j. 19 Find the velocities v,, v2, v3 and formulas for vj and &: (a) f = l , 3 , 5 , 7 ... (b) f=0,1,0,1, ... (c) f=O,$,$,i ,... fi 20 Find f,, f2, f3 and a formula for with fo =0: (a) v=l,2,4,8,... (b) u = - l , l , - l , l , ... 21 The areas of these nested squares are 12,22,32,.... What are the areas of the L-shaped bands (the differences between squares)?How does the figure show that I + 3 + 5 + 7 =42? 22 From the area under the staircase (by rectangles and then by triangles) show that the first j whole numbers 1 to j add up to G2+&. Find 1+2 + .-.+ 100. 23 If v=1,3,5,... then&=j2. If v = I, 1, 1,... then & = . Add those to find the sum of 2,4,6, ...,2j. Divide by 2 to find the sum of 1,2,3, ...,j. (Compare Problem 22.) 24 True (with reason) or false (with example). (a) When the f's are increasing so are the 0's. (b) When the v's are increasing so are the f's. (c) When the f's are periodic so are the 0's. (d) When the v's are periodic so are the f 's. 25 If f (t)= t2, compute f (99) and f (101). Between those times, what is the increase in f divided by the increase in t? + 26 If f (t)= t2 t, compute f (99) and f (101). Between those times, what is the increase in f divided by the increase in t? + + 27 If & =j2 j 1 find a formula for vj. 28 Suppose the 0's increase by 4 at every step. Show by , example and then by algebra that the "second difference" &+ -2&+&- equals 4. 29 Suppose fo = 0 and the v's are 1, 3, 4, $, 4, 4, 4, .... For which j does & = 5? , , , 30 Show that aj=&+ -2fj+fj- always equals vj+ -vj. If v is velocity then a stands for . Problems 31-34 involve periodic f's and v's (like sin t and cos t). 31 For the discrete sine f=O, 1, 1,0, -1, -1,O find the + second differencesal =f2 -2f1 +.fo and a2=f, - 2f2 fl and a3. Compare aj with &. + 32 If the sequence v,, v2, ... has period 6 and wl, w2, ... has period 10, what is the period of v, w,, v2 + w2, ...? 33 Draw the graph of f (t) starting from fo = 0 when v = 1, -1, -1, 1. If v has period 4 find f(12), f(l3), f(lOO.l). 16 1 lntroductlonto Calculus 34 Graph f(t) from f o = O to f 4 = 4 when v = 1,2, l,O. If v has period 4, find f (12) and f (14) and f (16). Why doesn't f have period 4? Problems 35-42 are about exponential v's and f 's. 35 Find the v's for f = 1,3,9,27. Predict v, and vj. Algebra gives 3j - 3j- = (3- 1)3j- '. 36 Find 1 + 2 + 4 + +32 and also 1+ j + d + +&. - a - 37 Estimate the slope of f (t)= 2' at t = 0. Use a calculator to compute (increase in f )/(increase in t) when t is small: f (t) -f t (0) - 2 1 1 and - 2.l - 1 and .I - 2.O' - 1 and .o1 2.0°1 - 1 - .001 . , 38 Suppose fo = I and vj = 2fi- . Find f,. 39 (a) From f = 1, j , b , find v,, v,, v, and predict vj. + + (b) Check f3 -fo = v, v2 v3 and fi-A- = vj. ' 40 Suppose vj = rj. Show that fi = (rj' - l)/(r- 1) starts from fo = 1 and has fj -fi-, = uj. (Then this is the correct fi = 1+ r + + rj = sum of a geometric series.) 41 From fi =(- 1)' compute vj. What is v, + v2 + + vj? 42 Estimate the slope of f (t) = et at t = 0. Use a calculator that knows e (or else take e = 2.78) to compute f(t)-f(0) t - e1 1 and e.' - 1 - .I and e-O1- 1 - .01 - Problems 43-47 are about U(t)= step from 0 to 1 at t = 0. 43 Graph the four functions U(t - 1) and U(t) - 2 and U(3t) and 4U(t). Then graph f (t) =4U(3t - 1)- 2. 44 Graph the square wave U(t) - U(t - 1).If this is the velocity v(t), graph the distance f(t). If this is the distance f (t), graph the velocity. 45 Two bursts of speed lead to the same distance f = 10: v= tot=.001 v = v t o t = . As V+ co the limit of the f (t)'s is + + 46 Draw the staircase function U(t) U(t - 1) U(t - 2). Its slope is a sum of three functions. 47 Which capital letters like L are the graphs of functions when steps are allowed? The slope of L is minus a delta function. Graph the slopes of the others. 48 Write a subroutine FINDV whose input is a sequence fo, f,, ...,f, and whose output is v,, v,, ...,v,. Include graphical output if possible. Test on fi = 2j and j2 and 2j. 49 Write a subroutine FINDF whose input is v,, ...,v, and fo, and whose output is fo, f,, ...,f,. The default value of fo is zero. Include graphical output if possible. Test vj =j. 50 If FINDV is applied to the output of FINDF, what sequence is returned? If FINDF is applied to the output of FINDV, what sequence is returned? Watch fo. 4 51 Arrange 2j and j2and 2' and in increasing order (a) when j is large:j = 9 (b) when j is small: j = &. 52 The average age of your family since 1970 is a piecewise linear function A(t). Is it continuous or does it jump? What is its slope? Graph it the best you can. 1.3 The Velocity at an Instant We have arrived at the central problems that calculus was invented to solve. There are two questions, in opposite directions, and I hope you could see them coming. 1. If the velocity is changing, how can you compute the distance traveled? 2. If the graph of f(t) is not a straight line, what is its slope? Find the distance from the velocity, find the velocity from the distance. Our goal is to do both-but not in one section. Calculus may be a good course, but it is not magic. The first step is to let the velocity change in the steadiest possible way. Question 1 Suppose the velocity at each time t is v(t) = 2t. Find f (t). With zr= 2t, a physicist would say that the acceleration is constant (it equals 2). The driver steps on the gas, the car accelerates, and the speedometer goes steadily up. The distance goes up too-faster and faster. If we measure t in seconds and v in feet per second, the distance f comes out in feet. After 10 seconds the speed is 20 feet per second. After 44 seconds the speed is 88 feetlsecond (which is 60 miles/hour). The acceleration is clear, but how far has the car gone? 1.3 The Velocity at an Instant Question 2 The distance traveled by time t is f ( t )= t2. Find the velocity v(t). The graph off ( t )= t 2 is on the right of Figure 1.12. It is a parabola. The curve starts at zero, when the car is new. At t = 5 the distance is f = 25. By t = 10, f reaches 100. Velocity is distance divided by time, but what happens when the speed is changing? Dividing f = 100 by t = 10 gives v = 10-the average veEocity over the first ten seconds. Dividing f = 121 by t = 11 gives the average speed over 11 seconds. But how do we find the instantaneous velocity-the reading on the speedometer at the exact instant when t = lo? change in distance (t + h)2 - time t t t+h t Fig. 1.12 The velocity v = 2t is linear. The distance f = t2 is quadratic. I hope you see the problem. As the car goes faster, the graph of t 2 gets steeperbecause more distance is covered in each second. The average velocity between t = 10 and t = 11 is a good approximation-but only an approximation-to the speed at the moment t = 10. Averages are easy to find: average velocity is f (11)-f (10) - 121- 100 = 21. 11- 10 1 The car covered 21 feet in that 1 second. Its average speed was 21 feetlsecond. Since it was gaining speed, the velocity at the beginning of that second was below 21. Geometrically, what is the average? It is a slope, but not the slope of the curve. The average velocity is the slope of a straight line. The line goes between two points on the curve in Figure 1.12. When we compute an average, we pretend the velocity is constant-so we go back to the easiest case. It only requires a division of distance by time: change in f average velocity = change in t ' Calculus and the Law You enter a highway at 1:00. If you exit 150 miles away at 3 :00, your average speed is 75 miles per hour. I'm not sure if the police can give you a ticket. You could say to the judge, "When was I doing 75?" The police would have 1 Introductionto Calculus to admit that they have no idea-but they would have a definite feeling that you must have been doing 75 sometime.? We return to the central problem-computing v(10) at the instant t = 10. The average velocity over the next second is 21. We can also find the average over the half-second between t = 10.0and t = 10.5. Divide the change in distance by the change in time: f (10.5) -f (10.0) -- (10.5)2- (10.0)2-- 110.25 - 100 = 20.5. 10.5 - 10.0 .5 .5 That average of 20.5 is closer to the speed at t = 10. It is still not exact. The way to find v(10) is to keep reducing the time interval. This is the basis for Chapter 2, and the key to differential calculus. Find the slope between points that are closer and closer on the curve. The "limit" is the slope at a single point. + Algebra gives the average velocity between t = 10 and any later time t = 10+ h. The distance increases from lo2 to (10 h)l. The change in time is h. So divide: This formula fits our previous calculations. The interval from t = 10 to t = 11 had + h = 1, and the average was 20 h = 21. When the time step was h = i,the average 4+ was 20 = 20.5. Over a millionth of a second the average will be 20 plus 1/1,000,000-which is very near 20. Conclusion: The velocity at t = 10 is v = 20. That is the slope of the curve. It agrees with the v-graph on the left side of Figure 1.12, which also has v(10) = 20. We now show that the two graphs match at all times. If f (t) = t 2 then v(t)= 2t. You are seeing the key computation of calculus, and we can put it into words before + equations. Compute the distance at time t h, subtract the distance at time t, and divide by h. That gives the average velocity: + This fits the previous calculation, where t was 10. The average was 20 h. Now the + average is 2t h. It depends on the time step h, because the velocity is changing. But we can see what happens as h approaches zero. The average is closer and closer to the speedometer reading of 2t, at the exact moment when the clock shows time t: I + I 1E As h approaches zero, the average velooity 2t h approaches v(t) = 2t. Note The computation (3) shows how calculus needs algebra. If we want the whole v-graph, we have to let time be a "variable." It is represented by the letter t. Numbers are enough at the specific time t = 10 and the specific step h = 1-but algebra gets beyond that. The average between any t and any t + h is 2t + h. Please don't hesitate to put back numbers for the letters-that checks the algebra. +This is our first encounter with the much despised "Mean Value Theorem." If the judge can prove the theorem, you are dead. A few u-graphs and f-graphs will confuse the situation (possibly also a delta function). 1.3 The VelocHy at an Instant There is also a step beyond algebra! Calculus requires the limit of the average. As h shrinks to zero, the points on the graph come closer. "Average over an interval" becomes "velocity at an instant.'' The general theory of limits is not particularly + simple, but here we don't need it. (It isn't particularly hard either.) In this example the limiting value is easy to identify. The average 2t h approaches 2t, as h -,0. What remains to do in this section? We answered Question 2-to find velocity from distance. We have not answered Question 1. If v(t) = 2t increases linearly with time, what is the distance? This goes in the opposite direction (it is integration). The Fundamental Theorem of Calculus says that no new work is necessary. Zfthe slope o f f (t) leads to v(t), then the area under that v-graph leads back to the f-graph. The odometer readings f = t2 produced speedometer readings v = 2t. By the Fundamental Theorem, the area under 2t should be t2. But we have certainly not proved any fundamental theorems, so it is better to be safe-by actually computing the area. Fortunately, it is the area of a triangle. The base of the triangle is t and the height is v = 2t. The area agrees with f (t): area =i(base)(height)= f(t)(2t)= t2. (4) EXAMPLE 1 The graphs are shifted in time. The car doesn't start until t = 1. Therefore v = 0 and f = O up to that time. After the car starts we have v = 2(t - 1) and f = (t - You see how the time delay of 1 enters the formulas. Figure 1.13 shows how it affects the graphs. + + Fig. 1.13 Delayed velocity and distance. The pairs v = at b and f =$at2 bt. EXAMPLE 2 The acceleration changes from 2 to another constant a. The velocity changes from v = 2t to v = at. The acceleration is the slope ofthe velocity curve! The distance is also proportional to a, but notice the factor 3: acceleration a 9 velocity v = at 9 distance f = fat2. If a equals 1, then v = t and f = f t2. That is one of the most famous pairs in calculus. If a equals the gravitational constant g, then v = gt is the velocity of a falling body. The speed doesn't depend on the mass (tested by Galileo at the Leaning Tower of Pisa). Maybe he saw the distance f = >2more easily than the speed v =gt. Anyway, this is the most famous pair in physics. 1 Introductionto Calculus + + EXAMPLE 3 Suppose f (t) = 3t t2. The average velocity from t to t h is + + + f (t Vave = h) -f (t) -- 3(t + h) h (t h)2 - 3t - t2 h + The change in distance has an extra 3h (coming from 3(t h) minus 3t). The velocity contains an additional 3 (coming from 3h divided by h). When 3t is added to the distance, 3 is added to the velocity. If Galileo had thrown a weight instead of dropping it, the starting velocity vo would have added vot to the distance. FUNCTIONS ACROSS TIME The idea of slope is not difficult-for one straight line. Divide the change in f by the change in t. In Chapter 2, divide the change in y by the change in x. Experience shows that the hard part is to see what happens to the slope as the line moves. Figure 1.l4a shows the line between points A and B on the curve. This is a "secant line." Its slope is an average velocity. What calculus does is to bring that point B down the curve toward A. 1 speed Fig. 1.14 Slope of line, slope of curve. Two velocity graphs. Which is which? . Question I What happens to the "change in f "-the height of B above A? Answer The change in f decreases to zero. So does the change in t. Question 2 As B approaches A, does the slope of the line increase or decrease? Answer I am not going to answer that question. It is too important. Draw another secant line with B closer to A. Compare the slopes. This question was created by Steve Monk at the University of Washington-where 57% of the class gave the right answer. Probably 97% would have found the right slope from a formula. Figure 1.14b shows the opposite problem. We know the velocity, not the distance. But calculus answers questions about both functions. Question 3 Which car is going faster at time t = 3/4? Answer Car C has higher speed. Car D has greater acceleration. Question 4 If the cars start together, is D catching up to C at the end? Between t = $ and t = 1, do the cars get closer or further apart? Answer This time more than half the class got it wrong. You won't but you can see why they did. You have to look at the speed graph and imagine the distance graph. When car C is going faster, the distance between them . 1.3 The VelocHy at an Instant To repeat: The cars start together, but they don't finish together. They reach the same speed at t = 1,not the same distance. Car C went faster. You really should draw their distance graphs, to see how they bend. These problems help to emphasize one more point. Finding the speed (or slope)is entirely different from finding the distance (or area): 1. To find the slope of the f-graph at a'particular time t, you don't have to know the whole history. 2. To find the area under the v-graph up to a particular time t, you do have to know the whole history. A short record of distance is enough to recover v(t). Point B moves toward point A. The problem of slope is local-the speed is completely decided by f (t) near point A. In contrast, a short record of speed is not enough to recover the total distance. We have to know what the mileage was earlier. Otherwise we can only know the increase in mileage, not the total. 1.3 EXERCISES Read-through questions Between the distances f (2)= 100 and f (6)= 200, the average velocity is a . If f(t) =i t 2 then f (6)= b and f(8) = c . The average velocity in between is d . The instantaneous velocities at t = 6 and t = 8 are e and f. + The average velocity is computed from f (t)and f (t h) by uave= g . If f ( t ) = t 2 then o,,,= h . From t = l to t = 1.1 the average is 1 . The instantaneous velocity is the I of u,,,. If the distance is f (t)=+at2 then the velocity is u(t) = k and the acceleration is 1 . 4 For the same f (t) = t2 + t, find the average speed between (a) t = O a n d l (b) t = O a n d + (c) t = O a n d h . 5 In the answer to 3(c), find the limit as h +0. What does that limit tell us? 6 Set h =0 in your answer to 4(c). Draw the graph of + f (t)= t2 t and show its slope at t = 0. + 7 Draw the graph of v(t) = 1 2t. From geometry find the area under it from 0 to t. Find the slope of that area function f (t). 8 Draw the graphs of v(t)= 3 - 2t and the area f(t). On the graph of f(t), the average velocity between A and B is the slope of m . The velocity at A is found by n . The velocity at B is found by 0 . When the velocity is positive, the distance is P . When the velocity is increasing, the car is q . 9 True or false (a) If the distance f (t) is positive, so is v(t). (b) If the distance f (t) is increasing, so is u(t). (c) If f (t) is positive, v(t) is increasing. (d) If v(t) is positive, f (t) is increasing. 1 Compute the average velocity between t = 5 and t = 8: (a) f (0= 6t + (b) f (t)= 6t 2 (c) f(t) =+at2 (d) f(t)='t-t2 (4 f(t)=6 (f) u(t)=2t + 2 For the same functions compute [f (t h) -f (t)]/h. This depends on t and h. Find the limit as h -,0. + 3 If the odometer reads f (t)= t2 t (f in miles or kilo- meters, t in hours), find the average speed between (a) t = l and t = 2 (b) t = 1 and t = 1.1 (c) t = l a n d t = l + h (d) t = 1 and t = .9 (note h = - .l) 10 If f(t) = 6t2 find the slope of the f-graph and also the v-graph. The slope of the u-graph is the 11 Iff (t) = t 2 what is the average velocity between t = .9 and + t = 1.1? What is the average between t - h and t h? 12 (a) Show that for f (t)=*at2 the average velocity between t - h and t +h' is exactly the velocity at t. + (b) The area under v(t)= at from t - h to t h is exactly the base 2h times 13 Find f (t)from u(t) = 20t iff (0)= 12. Also if f (1)= 12. 14 True or false, for any distance curves. (a) The slope of the line from A to B is the average velocity between those points. 22 1 lntroductlonto Calculus (b) Secant lines have smaller slopes than the curve. (c) If f (t) and F(t) start together and finish together, the average velocities are equal. (d) If v(t) and V(t) start together and finish together, the increases in distance are equal. 15 When you jump up and fall back your height is y = 2t - t2 in the right units. (a) Graph this parabola and its slope. (b) Find the time in the air and maximum height. (c) Prove: Half the time you are above y = 2. Basketball players "hang" in the air partly because of (c). 16 Graph f (t)= t2 and g(t) =f (t) - 2 and h(t) =f (2t), all from t =0 to t = 1. Find the velocities. 17 (Recommended)An up and down velocity is v(t) =2t for t < 3, v(t)= 12- 2t for t 2 3. Draw the piecewise parabola f (t). Check that f (6)=area under the graph of u(t). 18 Suppose v(t)= t for t < 2 and v(t)= 2 for t 2 2. Draw the graph off (t) out to t = 3. 19 Draw f (t) up to t =4 when u(t) increases linearly from (a) 0 to 2 (b) - I t 0 1 (c) -2 to 0. 20 (Recommended) Suppose v(t) is the piecewise linear sine function of Section 1.2. (In Figure 1.8 it was the distance.) Find the area under u(t) between t = 0 and t = 1,2,3,4,5,6. Plot those points f (1),...,f (6) and draw the complete piece- wise parabola f (t). 21 Draw the graph of f (t) = (1-t2( for 0 < t <2. Find a three-part formula for u(t). 22 Draw the graphs of f (t) for these velocities (to t = 2): (a) v(t)= 1- t (b) ~ ( t=) 11- tl (c) ~ ( t=) (1 -t) + 11- t1. 23 When does f (t) = t2- 3t reach lo? Find the average velocity up to that time and the instantaneous velocity at that time. + + 24 If f (t)=*at2 bt c, what is v(t)? What is the slope of v(t)? When does f (t) equal 41, if a = b = c = I? 25 If f (t)= t2 then v(t)= 2t. Does the speeded-up function f(4t) have velocity v(4t) or 4u(t) or 4v(4t)? 26 If f (t)= t - t2 find v(t) and f (3t). Does the slope of f (3t) equal v(3t) or 3v(t) or 3v(3t)? 27 For f (t) = t Z find vaVe(tb)etween 0 and t. Graph vave(t) and v(t). 28 If you know the average velocity uaVe(t)h,ow can you find the distance f (t)? Start from f (0)= 0. 1.4 Circular Motion This section introduces completely new distances and velocities-the sines and cosines from trigonometry. As I write that last word, I ask myself how much trigonometry it is essential to know. There will be the basic picture of a right triangle, with sides cos t + and sin t and 1. There will also be the crucial equation (cos t)2 (sin t)2= 1, which + is Pythagoras' law a' b2 = c2. The squares of two sides add to the square of the hypotenuse (and the 1 is really 12). Nothing else is needed immediately. If you don't know trigonometry, don't stop-an important part can be learned now. You will recognize the wavy graphs of the sine and cosine. W e intend to Jind the + slopes of those graphs. That can be done without using the formulas for sin(x y) and cos (x + y)-which later give the same slopes in a more algebraic way. Here it is only basic things that are needed.? And anyway, how complicated can a triangle be? Remark You might think trigonometry is only for surveyors and navigators (people with triangles). Not at all! By far the biggest applications are to rotation and vibration and oscillation. It is fantastic that sines and cosines are so perfect for "repeating motionw-around a circle or up and down. ?Sines and cosines are so important that I added a review of trigonometry in Section 1.5. But the concepts in this section can be more valuable than formulas. 1.4 Circular Motion 1 f = sin t 1 sin t -1 COS t Fig. 1.15 As the angle t changes, the graphs show the sides of the right triangle. Our underlying goal is to offer one more example in which the velocity can be computed by common sense. Calculus is mainly an extension of common sense, but here that extension is not needed. We will find the slope of the sine curve. The straight line f = vt was easy and the parabola f = +at2 was harder. The new example also involves realistic motion, seen every day. We start with circular motion, in which the position is given and the velocity will be found. A ball goes around a circle of radius one. The center is at x = 0, y = 0 (the origin). + The x and y coordinates satisfy x 2 y2 = 12, to keep the ball on the circle. We specify its position in Figure 1.16a by giving its angle with the horizontal. And we make the ball travel with constant speed, by requiring that the angle is equal to the time t. The ball goes counterclockwise. At time 1 it reaches the point where the angle equals 1. The angle is measured in radians rather than degrees, so a full circle is completed at t = 271 instead of t = 360. The ball starts on the x axis, where the angle is zero. Now find it at time t: The ball is at the point where x = cos t and y = sin t. This is where trigonometry is useful. The cosine oscillates between 1 and - 1, as the ball goes from far right to far left and back again. The sine also oscillates between 1 and - 1, starting from sin 0 = 0. At time 7112 the sine (the height) increases to one. The cosine is zero and the ball reaches the top point x = 0, y = 1. At time 71 the cosine is - 1 and the sine is back to zero-the coordinates are (- 1,O). At t = 271 the circle is complete (the angle is also 271), and x = cos 27~= 1, y = sin 271 = 0. vertical distance vertical velocity Fig. 1.16 Circular motion with speed 1, angle t, height sin t, upward velocity cos t. I Introduction to Calculus Important point: The distance around the circle (its circumference) is 2nr = 2n, because the radius is 1. The ball travels a distance 2n in a time 2n. The speed equals 1. It remains to find the velocity, which involves not only speed but direction. Degrees vs. radians A full circle is 360 degrees and 271 radians. Therefore = 1 radian = 36012~degrees 57.3 degrees = 1 degree = 2711360 radians .01745 radians Radians were invented to avoid those numbers! The speed is exactly 1, reaching t radians at time t. The speed would be .01745, if the ball only reached t degrees. The ball would complete the circle at time T = 360. We cannot accept the division of the circle into 360 pieces (by whom?), which produces these numbers. = To check degree mode vs. radian mode, verify that sin l oz .017 and sin 1 34. VELOCITY OF THE BALL At time t, which direction is the ball going? Calculus watches the motion between t and t + h. For a ball on a string, we don't need calculus-just let go. The direction of motion is tangent to the circle. With no force to keep it on the circle, the ball goes oflon a tangent. If the ball is the moon, the force is gravity. If it is a hammer swinging around on a chain, the force is from the center. When the thrower lets go, the hammer takes off-and it is an art to pick the right moment. (I once saw a friend hit by a + hammer at MIT. He survived, but the thrower quit track.) Calculus will find that same tangent direction, when the points at t and t h come close. The "velocity triangle" is in Figure 1.16b. It is the same as the position triangle, but rotated through 90". The hypotenuse is tangent to the circle, in the direction the ball is moving. Its length equals 1 (the speed). The angle t still appears, but now it is the angle with the vertical. The upward component of velocity is cos t, when the upward component of position is sin t. That is our common sense calculation, based on a figure rather than a formula. The rest of this section depends on it-and we check v = cos t at special points. At the starting time t = 0, the movement is all upward. The height is sin 0 = 0 and the upward velocity is cos 0 = 1. At time ~ 1 2t,he ball reaches the top. The height is sin 4 2 = 1 and the upward velocity is cos n/2 = 0. At that instant the ball is not moving up or down. The horizontal velocity contains a minus sign. At first the ball travels to the left. The value of x is cos t, but the speed in the x direction is -sin t. Half of trigonometry + is in that figure (the good half), and you see how sin2t cos2t = 1 is so basic. That equation applies to position and velocity, at every time. Application of plane geometry: The right triangles in Figure 1.16 are the same size and shape. They look congruent and they are-the angle t above the ball equals the angle t at the center. That is because the three angles at the ball add to 180". OSCILLATION: UP AND DOWN MOTION We now use circular motion to study straight-line motion. That line will be the y axis. Instead of a ball going around a circle, a mass will move up and down. It oscillates between y = 1 and y = - 1. The mass is the "shadow of the ball," as we explain in a moment. 1.4 Circular Motion There is a jumpy oscillation that we do not want, with v = 1 and v = -1. That "bang-bang" velocity is like a billiard ball, bouncing between two walls without slowing down. If the distance between the walls is 2, then at t = 4 the ball is back to + the start. The distance graph is a zigzag (or sawtooth) from Section 1.2. We prefer a smoother motion. Instead of velocities that jump between 1and -1, a real oscillation slows down to zero and gradually builds up speed again. The mass is on a spring, which pulls it back. The velocity drops to zero as the spring is fully stretched. Then v is negative, as the mass goes the same distance in the opposite direction. Simple harmonic motion is the most important back and forth motion, while f = vt and f = f at2 are the most important one-way motions. (.p=mst;///J ) fup = sin t turn UP down turn Fig. 1.17 Circular motion of the ball and harmonic motion of the mass (its shadow). How do we describe this oscillation?The best way is to match it with the ball on the circle. The height of the ball will be the height of the mass. The "shadow of the ball" goes up and down, level with the ball. As the ball passes the top of the circle, the mass stops at the top and starts down. As the ball goes around the bottom, the mass stops and turns back up the y axis. Halfway up (or down), the speed is 1. Figure 1.17a shows the mass at a typical time t. The height is y =f (t)= sin t, level with the ball. This height oscillates between f = 1 and f = -1. But the mass does not move with constant speed. The speed of the mass is changing although the speed of the ball is always 1 . The time for a full cycle is still 2n, but within that cycle the mass speeds up and slows down. The problem is to find the changing velocity u. Since the distance is f = sin t, the velocity will be the slope of the sine curve. THE SLOPE OF THE SINE CURVE At the top and bottom (t = n/2 and t = 3~12t)he ball changes direction and v = 0. The slope at the top and bottom of the sine curve is zero.? At time zero, when the ball is going straight up, the slope of the sine curve is v = 1. At t = n, when the ball and mass and f-graph are going down, the velocity is v = -1. The mass goes fastest at the center. The mass goes slowest (in fact it stops)when the height reaches a maximum or minimum. The velocity triangle yields v at every time t. To find the upward velocity of the mass, look at the upward velocity of the ball. Those velocities are the same! The mass and ball stay level, and we know v from circular motion: The upward velocity is v = cos t. ?That looks easy but you will see later that it is extremely important. At a maximum or minimum the slope is zero. The curve levels off. 1 Introductionto Calculus Figure 1.18 shows the result we want. On the right, f = sin t gives the height. On the left is the velocity v = cos t. That velocity is the slope of the f-curve. The height and velocity (red lines) are oscillating together, but they are out of phase-just as the position triangle and velocity triangle were at right angles. This is absolutely fantastic, that in calculus the two most famous functions of trigonometry form a pair: The slope of the sine curve is given by the cosine curve. When the distance is f (t)= sin t, the velocity is v(t)= cos t . Admission of guilt: The slope of sin t was not computed in the standard way. Previously we compared (t + h)' with t2,and divided that distance by h. This average velocity approached the slope 2t as h became small. For sin t we could have done the same: + average velocity = change in sin t - sin (t h) - sin t change in t h (1) + This is where we need the formula for sin (t h), coming soon. Somehow the ratio in (1) should approach cosmats h -,0. (It d,oes.)The sine and cosine fit the same pattern as t2 and 2 t o u r shortcut was to watch the shadow of motion around a circle. Fig. 1.I 8 v =cos t when f = sin t (red); v = -sin t when f =cos t (black). Question 1 What if the ball goes twice as fast, to reach angle 2t at time t? Answer The speed is now 2. The time for a full circle is only n. The ball's position is x = cos 2t and y = sin 2t. The velocity is still tangent to the circle-but the tangent is at angle 2t where the ball is. Therefore cos 2t enters the upward velocity and -sin 2t enters the horizontal velocity. The difference is that the velocity triangle is twice as big. The upward velocity is not cos 2t but 2 cos 2t. The horizontal velocity is -2 sin 2t. Notice these 2's! Question 2 What is the area under the cosine curve from t = 0 to t = n/2? You can answer that, if you accept the Fundamental Theorem of Calculuscomputing areas is the opposite of computing slopes. The slope of sin t is cos t, so the area under cos t is the increase in sin t. No reason to believe that yet, but we use it anyway. From sin 0 = 0 to sin n/2 = 1, the increase is 1. Please realize the power of calculus. No other method could compute the area under a cosine curve so fast. 1.4 Circular Motion THE SLOPE OF THE COSINE,CURVE I cannot resist uncovering another distance and velocity (another f-v pair) with no extra work. This time f is the cosine. The time clock starts at the top of the circle. The old time t = n/2is now t = 0.The dotted lines in Figure 1.18 show the new start. But the shadow has exactly the same motion-the ball keeps going around the circle, and the mass follows it up and down. The f-graph and v-graph are still correct, both with a time shift of 4 2 . The new f-graph is the cosine. The new v-graph is minus the sine. The slope of the cosine curve follows the negative of the sine curve. That is another famous pair, twins of the first: When the distance is f (t)= cos t, the velocity is v(t)= - sin t. You could see that coming, by watching the ball go left and right (instead of up and down). Its distance across is f = cos t. Its velocity across is v = -sin t. That twjn pair completes the calculus in Chapter 1 (trigonometry to come). We review the ideas: v is the velocity the slope of the distance curve the limit of average velocity over a short time the derivative of f. f is the distance the area under the velocity curve the limit of total distance over many short times the integral of v. Differential calculus: Compute v from f . Integral calculus: Compute f from v. With constant velocity, f equals vt. With constant acceleration, v = at and f =t a t 2. In harmonic motion, v = cos t and f = sin t . One part of our goal is to extend that list-for which we need the tools of calculus. Another and more important part is to put these ideas to use. Before the chapter ends, may I add a note about the book and the course? The book is more personal than usual, and I hope readers will approve. What I write is very close to what I would say, if you were in this room. The sentences are spoken before they are written.? Calculus is alive and moving forward-it needs to be taught that way. One new part of the subject has come with the computer. It works with a finite step h, not an "infinitesimal" limit. What it can do, it does quickly-even if it cannot find exact slopes or areas. The result is an overwhelming growth in the range of problems that can be solved. We landed on the moon because f and v were so accurate. (The moon's orbit has sines and cosines, the spacecraft starts with v = at and f = )at2. Only the computer can account for the atmosphere and the sun's gravity and the changing mass of the spacecraft.) Modern mathematics is a combination of exact formulas and approximate computations. Neither part can be ignored, and I hope you will see numerically what we derive algebraically. The exercises are to help you master both parts. t o n television you know immediately when the words are live. The same with writing. I lntroductlon to Calculus The course has made a quick start-not with an abstract discussion of sets or functions or limits, but with the concrete questions that led to those ideas. You have seen a distance function f and a limit v of average velocities. We will meet more functions and more limits (and their definitions!) but it is crucial to study important examples early. There is a lot to do, but the course has definitely begun. 1.4 EXERCISES Read-through questions A ball at angle t on the unit circle has coordinates x = a and y = b . It completes a full circle at t = c .Its speed is d . Its velocity points in the direction of the e , which is f to the radius coming out from the center. The upward velocity is g and the horizontal velocity is h . A mass going up and down level with the ball has height ' f(t)= i .This is called simple i motion. The velocity is u(t) = k . When t = n/2 the height is f = I and the velocity is v = m . If a speeded-up mass reaches f = sin 2t at time t, its velocity is v = n . A shadow traveling under the ball has f= cos t and v = o . When f is distance = area = integral, v is P = q = r . 1 For a ball going around a unit circle with speed 1, (a) how long does it take for 5 revolutions? (b) at time t = 3n/2 where is the ball? (c) at t = 22 where is the ball (approximately)? 7 A mass moves on the x axis under or over the original ball (on the unit circle with speed 1). What is the position x =f (t)? Find x and v at t = 4 4 . Plot x and v up to t =n. 8 Does the new mass (under or over the ball) meet the old mass (level with the ball)? What is the distance between the masses at time t? 9 Draw graphs of f(t) =cos 3t and cos 2nt and 271 cos t, marking the time axes. How long until each f repeats? + + 10 Draw graphs of f = sin(t n) and v =cos (t n). This oscillation stays level with what ball? 11 Draw graphs of f = sin ( 4 2 - t) and v = -cos (n/2 - t). This oscillation stays level with a ball going which way starting where? + 12 Draw a graph of f(t)= sin t cos t. Estimate its greatest height (maximum f ) and the time it reaches that height. By computing f check your estimate. 2 For the same motion find the exact x and y coordinates at t = 2x13. At what time would the ball hit the x axis, if it goes off on the tangent at t = 2n/3? 3 A ball goes around a circle of radius 4. At time t (when it reaches angle t) find (a) its x and y coordinates (b) the speed and the distance traveled (c) the vertical and horizontal velocity. 13 How fast should you run across the circle to meet the ball again? It travels at speed 1. 14 A mass falls from the top of the unit circle when the ball of speed 1 passes by. What acceleration a is necessary to meet the ball at the bottom? Find the area under v = cos t from the change in f = sin t: 4 O n a circle of radius R find the x and y coordinates at time t (and angle t). Draw the velocity triangle and find the x and y velocities. 15 from t = O to t = n 17 from t = O to t = 2 n j6 from t = 0 to t = n/6 18 from t = n/2 to t = 3x12. 5 A ball travels around a unit circle (raalus 1)with speed 3, starting from angle zero. At time t, (a) what angle does it reach? (b) what are its x and y coordinates? (c) what are its x and y velocities? This part is harder. 6 If another ball stays n/2 radians ahead of the ball with speed 3, find its angle, its x and y coordinates, and its vertical velocity at time t. 19 The distance curve f = sin 4t yields the velocity curve v = 4 cos 4t. Explain both 4's. 20 The distance curve f = 2 cos 3t yields the velocity curve v = - 6 sin 3t. Explain the -6. 21 The velocity curve v = cos 4t yields the distance curve f = $ sin 4t. Explain the i. 22 The velocity v = 5 sin 5t yields what distance? 23 Find the slope of the sine curve at t =4 3 from v =cos t. Then find an average slope by dividing sin n/2 -sin 4 3 by the time difference 4 2 - 4 3 . 24 The slope of f = sin t at t = 0 is cos 0 = 1. Compute average slopes (sin t)/t for t = 1, . l , .01, .001. The ball at x = cos t, y = sin t circles (1) counterclockwise (2)with radius 1 (3)starting from x = 1, y = 0 (4)at speed 1. Find (1)(2)(3)(4) for the motions 25-30. 25 x=cos3t, y=-sin3t 26 x = 3 cos 4t, y = 3 sin 4t 27 x = 5 sin 2t, y = 5 cos 2t 30 x =cos (- t), y =sin(- t) The oscillation x = 0, y = sin t goes (1)up and down (2)between -1 and 1 (3) starting from x = 0, y = 0 (4) at velocity v = cos t. Find (1)(2)(3)(4) for the oscillations 31-36. 31 x=cost, y=O 32 x = 0, y = sin 5t 33 x=O, y=2sin(t+O) 34 x=cost, y = c o s t 35 x=O, y = - 2 c o s i t 36 x=cos2t, y=sin2t 37 If the ball on the unit circle reaches t degrees at time t, find its position and speed and upward velocity. 38 Choose the number k so that x = cos kt, y = sin kt completes a rotation at t = 1. Find the speed and upward velocity. 39 If a pitcher doesn't pause before starting to throw, a balk is called. The American League decided mathematically that there is always a stop between backward and forward motion, even if the time is too short to see it. (Therefore no balk.) Is that true? 1.5 A Review of Trigonometry Trigonometry begins with a right triangle. The size of the triangle is not as important as the angles. We focus on one particular angle-call it 8-and on the ratios between the three sides x, y, r. The ratios don't change if the triangle is scaled to another size. Three sides give six ratios, which are the basic functions of trigonometry: RnIy cos 8 = x -r = near side hypo tenuse sin 8 = y - = opposite side r hypotenuse set 8 = -r = - 1 x cos 8 csc 8 = -r = - 1 y sin 8 X Fig. 1.19 tan 8 = y - x = opposite side near side cot x g=-=- y 1 tan 8 Of course those six ratios are not independent. The three on the right come directly from the three on the left. And the tangent is the sine divided by the cosine: Note that "tangent of an angle" and "tangent to a circle" and "tangent line to a graph" are different uses of the same word. As the cosine of 8 goes to zero, the tangent of 8 goes to infinity. The side x becomes zero, 8 approaches 90", and the triangle is infinitely steep. The sine of 90" is y/r = 1. Triangles have a serious limitation. They are excellent for angles up to 90°, and they are OK up to 180", but after that they fail. We cannot put a 240" angle into a triangle. Therefore we change now to a circle. 1 Introduction to Calculus Fig. 1.20 Trigonometry on a circle. Compare 2 sin 8 with sin 28 and tan 8 (periods 2n,n,n). Angles are measured from the positive x axis (counterclockwise). Thus 90" is straight up, 180" is to the left, and 360" is in the same direction as 0". (Then 450" is the same as 90°.) Each angle yields a point on the circle of radius r. The coordinates x and y of that point can be negative (but never r). As the point goes around the circle, the six ratios cos 8, sin 9, tan 8, ... trace out six graphs. The cosine waveform is the same as the sine waveform-just shifted by 90". One more change comes with the move to a circle. Degrees are out. Radians are in. The distance around the whole circle is 2nr. The distance around to other points is Or. We measure the angle by that multiple 8. For a half-circle the distance is m, so the angle is n radians-which is 180". A quarter-circle is 4 2 radians or 90". The distance around to angle 8 is r times 8. When r = 1 this is the ultimate in simplicity: The distance is 8. A 45" angle is Q of a circle and 27118 radians-and the length of the circular arc is 27~18.Similarly for 1": 360" = 2n radians 1" = 27~1360radians 1 radian = 3601271 degrees. 4 An angle going clockwise is negative. The angle -n / 3 is - 60" and takes us of the wrong way around the circle. What is the effect on the six functions? Certainly the radius r is not changed when we go to - 8. Also x is not changed (see Figure 1.20a). But y reverses sign, because - 8 is below the axis when + 8 is above. This change in y affects y/r and y / x but not xlr: The cosine is even (no change). The sine and tangent are odd (change sign). The same point is 2 of the right way around. Therefore 2 of 2n radians (or 300") gives the same direction as -n / 3 radians or -60". A diflerence of 2n makes no di$erence to x, y, r. Thus sin 8 and cos 8 and the other four functions have period 27~. We can go five times or a hundred times around the circle, adding 10n or 200n to the angle, and the six functions repeat themselves. EXAMPLE Evaluate the six trigonometric functions at 8 = 2n/3 (or 8 = -4 4 3 ) . This angle is shown in Figure 1.20a (where r = 1). The ratios are cos 8 = x/r = - 1/2 sin 8 = y/r = &/2 & tan 8 = y / x = - sec e = - 2 csc e = 2/& cot e = -i/d Those numbers illustrate basic facts about the sizes of four functions: The tangent and cotangent can fall anywhere, as long as cot 8 = l/tan 8. 1.5 A Review of Ttlgonometry 3 The numbers reveal more. The tangent - is the ratio of sine to cosine. The secant -2 is l/cos 8. Their squares are 3 and 4 (differing by 1). That may not seem remarkable, but it is. There are three relationships in the squares of those six numbers, and they are the key identities of trigonometry: + Everything flows fvom the Pythagoras formula x2 y2 = r2. Dividing by r2 gives + + ( ~ / r ) (~y/r)2= 1. That is cos28 sin28= 1. Dividing by x2gives the second identity, + which is 1 ( y / ~=) (~r / ~ )D~i.viding by y2 gives the third. All three will be needed throughout the book-and the first one has to be unforgettable. DISTANCES AND ADDITION FORMULAS To compute the distance between points we stay with Pythagoras. The points are in Figure 1.21a. They are known by their x and y coordinates, and d is the distance between them. The third point completes a right triangle. For the x distance along the bottom we don't need help. It is x, - xl (or Ix2 - x1I since distances can't be negative). The distance up the side is ly2- y, 1. Pythagoras immediately gives the distance d: distance between points = d = J(x2 - x , ) ~+ (y2- y1)'. (1) x=coss y = sin s Fig. 1.21 Distance between points and equal distances in two circles. By applying this distance formula in two identical circles, we discover the cosine of s - t. (Subtracting angles is important.) In Figure 1.21b, the distance squared is + d2= (change in x ) ~ (change in y)* + = (COSs - cos t)* (sin s - sin t)2. (2) Figure 1 . 2 1 ~shows the same circle and triangle (but rotated). The same distance squared is d2= (cos(s - t) - + (sin(s - t))2. (3) Now multiply out the squares in equations (2) and (3). Whenever ( c o ~ i n e+) ~(sine)2 appears, replace it by 1. The distances are the same, so (2)= (3): (2)= 1 + 1- 2 cos s cos t - 2 sin s sin t 1 Introduction to Calculus + After canceling 1 1 and then -2, we have the "additionformula" for cos (s - t): + The cosine of s - t equals cos s cos t sin s sin t. (4) + The cosine of s t equals cos s cos t - sin s sin t. (5) The easiest is t = 0. Then cos t = 1and sin t = 0. The equations reduce to cos s = cos s. To go from (4) to (5) in all cases, replace t by - t. No change in cos t, but a "minus" appears with the sine. In the special case s = t, we have cos(t + t ) = (COSt)(cos t) - (sin t)(sin t). This is a much-used formula for cos 2t: Double angle: cos 2t = cos2t - sin2t = 2 cos2t - 1 = 1 - 2 sin2t. (6) + I am constantly using cos2t sin2t = 1, to switch between sines and cosines. We also need addition formulas and double-angle formulas for the sine of s - t + and s t and 2t. For that we connect sine to cosine, rather than (sine)2to (co~ine)~. The connection goes back to the ratio y/r in our original triangle. This is the sine of the angle 0 and also the cosine of the complementary angle 7112 - 0: sin 0 = cos (7112 - 0) and cos 0 = sin (7112 - 0). (7) The complementary angle is 7112 - 0 because the two angles add to 7112 (a right angle). By making this connection in Problem 19, formulas (4-5-6) move from cosines to sines: sin (s - t) =sin s cos t - cos s sin t (8) + sin(s t) = sin s cos t + cos s sin t (9) + sin 2t = sin(t t) = 2 sin t cos t (10) I want to stop with these ten formulas, even if more are possible. Trigonometry is full of identities that connect its six functions-basically because all those functions come from a single right triangle. The x, y, r ratios and the equation x2 + y2 = r2 can be rewritten in many ways. But you have now seen the formulas that are needed by ca1culus.t They give derivatives in Chapter 2 and integrals in Chapter 5. And it is typical of our subject to add something of its own-a limit in which an angle approaches zero. The essence of calculus is in that limit. Review of the ten formulas Figure 1.22 shows d2= (0 - $)2+ (1 - -12)~. + cos 71 - = cos 71 - cos 71 - sin 71 - sin 71 - 6 2 3 2 3 (s - t) sin 71 - = sin 71 - cos 71 - - cos 71 - sin 71 - 6 2 3 2 3 cos 571 -= cos 71 - cos 71 - - sin 71 - sin 71 - 6 2 3 2 3 (s+ t) + sin 571 -= sin 71 - cos 71 - cos 71 - sin 71 - 6 2 3 2 3 (2t) sin 2 71 - 3 = 2 sin 71 - 3 cos 71 -3 cos 2 6 = sin 71 -3 = -12 (4-0) sin -71 6 = cos 71 -3 = 112 + tcalculus turns (6) around to cos2t = i(1 cos 2t) and sin2t =i(1 - cos 2t). A Review of Ttlgonometry Fig. 1.22 1.5 EXERCISES Read-through questions Starting with a a triangle, the six basic functions are the b of the sides. Two ratios (the cosine x/r and the c ) are below 1. Two ratios (the secant r/x and the d ) are above 1. Two ratios (the e and the f ) can take any value. The six functions are defined for all angles 8, by chang- ing from a triangle to a g . The angle 8 is measured in h . A full circle is 8 = i , when the distance around is 2nr. The distance to angle 8 is I . All six functions have period k . Going clockwise changes the sign of 8 and I and m . Since cos(- 9) = cos 8, the cosine is n . + Coming from x2 y2=r2 are the three identities + sin28 cos28 = 1 and 0 and P . (Divide by r2 and q and r .) The distance from (2, 5) to (3, 4) is d = s . The distance from (1, 0) to (cos(s -t), sin(s -t)) leads to the addition formula cos(s -t) = t . Changing + the sign of t gives cos(s t) = u . Choosing s = t gives cos 2t = v or w . Therefore i ( l + cos 2t) = x , a formula needed in calculus. 9 Find the distance d from (1,O) to (4, &/2) and show on a circle why 6d is less than 2n. 10 In Figure 1.22compute d2and (with calculator)12d. Why is 12d close to and below 2n? 11 Decide whether these equations are true or false: (a) -1--sc-ion-s8--8 = -1+s-icno-8s-8 (b) sec tan 8 e + csc +cot 8 e = sin 8 + cos 8 (c) cos 8 -sec 8 = sin 0 tan 8 (d) sin (2n -8) =sin 8 + 12 Simplify sin (n - O), cos(n-8), sin(n/2 + 8), cos (n/2 8). + 13 From the formula for cos(2t t) find cos 3t in terms of cos t. + 14 From the formula for sin (2t t) find sin 3t in terms of sin t. 1 In a 60-60-60 triangle show why sin 30" =3. 2 Convert x, 371, -7114 to degrees and 60°, 90°, 270" to radians. What angles between 0 and 2n correspond to 8 = 480" and 8 = -I0? 3 Draw graphs of tan 8and cot 8 from 0to 2n. What is their (shortest) period? 4 Show that cos 28 and cos28 have period n and draw them on the same graph. 5 At 8= 3n/2 compute the six basic functions and check cos28 +sin28, sec20 -tan28, csc28 -cot28. 6 Prepare a table showing the values of the six basic functions at 8 =0, 7114, n/3, ~ / 2n,. 7 The area of a circle is nr2. What is the area of the sector that has angle 8? It is a fraction of the whole area. 8 Find the distance from (1, 0) to (0, 1)along (a) a straight line (b) a quarter-circle (c) a semicircle centered at (3,i). + 15 By averaging cos (s -t) and cos (s t) in (4-5) find a for- mula for cos s cos t. Find a similar formula for sin s sin t. + + 16 Show that (cos t i sin t)2=cos 2t i sin 2t, if i2= -1. 17 Draw cos 8 and sec 8 on the same graph. Find all points where cos B = sec 8. + 18 Find all angles s and t between 0 and 2n where sin (s t) = sin s + sin t. - 19 Complementary angles have sin 8 = cos (n/2 8). Write + sin@ t) as cos(n/2 -s -t) and apply formula (4) with n/2 -s instead of s. In this way derive the addition formula (9). 20 If formula (9) is true, how do you prove (8)? 21 Check the addition formulas (4-5) and (8-9) for s = t = n/4. 22 Use (5) and (9) to find a formula for tan (s + t). 34 1 Introduction to Calculus In 23-28 find every 8 that satisfies the equation. 23 sin 8 = -1 24 sec 8 = -2 25 sin 8 = cos 8 26 sin 8 = 8 27 sec28 + csc28 = 1 28 tan 8 = 0 f i + 29 Rewrite cos 8 +sin 0 as sin(8 4) by choosing the correct "phase angle" 4. (Make the equation correct at 8 =0. Square both sides to check.) + + 30 Match a sin x b cos x with A sin (x 4). From equation (9)show that a =A cos 4 and b =A sin 4. Square and add to find A = . Divide to find tan 4 = bla. 31 Draw the base of a triangle from the origin 0 = (0'0) to P =(a, 0). The third corner is at Q =(bcos 8, b sin 8). What are the side lengths OP and OQ? From the distance formula (1) show that the side PQ has length + d2=a2 b2-2ab cos 8 (law of cosines). 32 Extend the same!riangle to a parallelogram with its fourth + corner at R =(a b cos 0, b sin 8).Find the length squared of the other diagonal OR. Draw graphs for equations 33-36, and mark three points. 33 y = sin 2x 35 y =3 cos 2xx 34 y = 2 sin xx 36 y=sin x+cos x 37 Which of the six trigonometric functions are infinite at what angles? 38 Draw rough graphs or computer graphs of t sin t and sin 4t sin t from 0 to 2n. -1 1.6 A Thousand Points of Light - 1 The graphs on the back cover of the book show y = sin n. This is very different from y = sin x. The graph of sin x is one continuous curve. By the time it reaches x = 10,000, the curve has gone up and down 10,000/27rtimes. Those 1591 oscillations would be so crowded that you couldn't see anything. The graph of sin n has picked 10,000points from the curve-and for some reason those points seem to lie on more than 40 separate sine curves. The second graph shows the first 1000points. They don't seem to lie on sine curves. Most people see hexagons. But they are the same thousand points! It is hard to believe that the graphs are the same, but I have learned what to do. Tilt the second graph and look from the side at a narrow angle. Now the first graph appears. You see "diamonds." The narrow angle compresses the x axis-back to the scale of the first graph. The effect of scale is something we don't think of. We understand it for maps. Computers can zoom in or zoom out-those are changes of scale. What our eyes see 1.6 A Thousand Points of Light depends on what is "close." We think we see sine curves in the 10,000 point graph, and they raise several questions: 1. Which points are near (0, O)? 2. How many sine curves are there? 3. Where does the middle curve, going upward from (0, 0), come back to zero? A point near (0,O) really means that sin n is close to zero. That is certainly not true of sin 1 (1 is one radian!). In fact sin 1 is up the axis at .84, at the start of the seventh sine curve. Similarly sin 2 is .91 and sin 3 is .14. (The numbers 3 and .14 make us think of n. The sine of 3 equals the sine of n - 3. Then sin .l4 is near .14.) Similarly sin 4, sin 5, ...,sin 21 are not especially close to zero. The first point to come close is sin 22. This is because 2217 is near n. Then 22 is close to 771, whose sine is zero: sin 22 = sin (7n - 22) z sin (- .01) z - .01. That is the first point to the right of (0,O) and slightly below. You can see it on graph 1, and more clearly on graph 2. It begins a curve downward. The next point to come close is sin 44. This is because 44 is just past 14n. 44 z 14n + .02 so sin 44 z sin .02 z .02. This point (44, sin 44) starts the middle sine curve. Next is (88, sin 88). Now we know something. There are 44 curves. They begin near the heights sin 0, sin 1, ..., sin 43. Of these 44 curves, 22 start upward and 22 start downward. I was confused at first, because I could only find 42 curves. The reason is that sin 11 equals -0.99999 and sin 33 equals .9999. Those are so close to the bottom and top that you can't see their curves. The sine of 11 is near - 1 because sin 22 is near zero. It is almost impossible to follow a single curve past the top-coming back down it is not the curve you think it is. The points on the middle curve are at n = 0 and 44 and 88 and every number 44N. Where does that curve come back to zero? In other words, when does 44N come + very close to a multiple of n? We know that 44 is 14n .02. More exactly 44 is 14n + .0177. So we multiply .0177 until we reach n: if N=n/.0177 then 44N=(14n+.0177)N3 14nN+n. This gives N = 177.5. At that point 44N = 7810. This is half the period of the sine curve. The sine of 7810 is very near zero. If you follow the middle sine curve, you will see it come back to zero above 7810. The actual points on that curve have n = 44 177 and n = 44 178, with sines just above and below zero. Halfway between is n = 7810. The equationfor the middle sine curve is y = sin (nx/78lO). Its period is 15,620-beyond our graph. Question The fourth point on that middle curve looks the same as the fourth point coming down from sin 3. What is this "double point?" Answer 4 times 44 is 176. On the curve going up, the point is (176, sin 176).On the curve coming down it is (179, sin 179). The sines of 176 and 179 difler only by .00003. The second graph spreads out this double point. Look above 176 and 179, at the center of a hexagon. You can follow the sine curve all the way across graph 2. Only a little question remains. Why does graph 2 have hexagons? I don't know. The problem is with your eyes. To understand the hexagons, Doug Hardin plotted points on straight lines as well as sine curves. Graph 3 shows y = fractional part of n/2x. Then he made a second copy, turned it over, and placed it on top. That produced graph 4-with hexagons. Graphs 3 and 4 are on the next page. 36 1 Introduction to Calculus This is called a Moivt pattevn. If you can get a transparent copy of graph 3, and turn it slowly over the original, you will see fantastic hexagons. They come from interference between periodic patterns-in our case 4417 and 2514 and 1913 are near 271. This interference is an enemy of printers, when color screens don't line up. It can cause vertical lines on a TV. Also in making cloth, operators get dizzy from seeing Moire patterns move. There are good applications in engineering and optics-but we have to get back to calculus. 1.7 Computing in Calculus Software is available for calculus courses-a lot of it. The packages keep getting better. Which program to use (if any) depends on cost and convenience and purpose. How to use it is a much harder question. These pages identify some of the goals, and also particular packages and calculators. Then we make a beginning (this is still Chapter 1) on the connection of computing to calculus. The discussion will be informal. It makes no sense to copy the manual. Our aim is to support, with examples and information, the effort to use computing to help learning. For calculus, the gveatest advantage of the computev is to o$er graphics. You see the function, not just the formula. As you watch, f ( x ) reaches a maximum or a minimum or zero. A separate graph shows its derivative. Those statements are not 100% true, as everybody learns right away-as soon as a few functions are typed in. But the power to see this subject is enormous, because it is adjustable. If we don't like the picture we change to a new viewing window. This is computer-based graphics. It combines numerical computation with gvaphical computation. You get pictures as well as numbers-a powerful combination. The computer offers the experience of actually working with a function. The domain and range are not just abstract ideas. You choose them. May I give a few examples. EXAMPLE I Certainly x3 equals 3" when x = 3. Do those graphs ever meet again'? At this point we don't know the full meaning of 3", except when x is a nice number. (Neither does the computer.) Checking at x = 2 and 4, the function x 3 is smaller both times: 23 is below 3* and 43 = 64 is below 34 = 81. If x3 is always less than 3" we ought to know-these are among the basic functions of mathematics. 1.7 Computing in Calculus The computer will answer numerically or graphically. At our command, it solves x3 = 3X.At another command, it plots both functions-this shows more. The screen proves a point of logic (or mathematics) that escaped us. If the graphs cross once, they must cross again-because 3" is higher at 2 and 4. A crossing point near 2.5 is seen by zooming in. I am less interested in the exact number than its position-it comes before x = 3 rather than after. A few conclusions from such a basic example: 1. A supercomputer is not necessary. 2. High-level programming is not necessary. 3. We can do mathematics without completely understanding it. The third point doesn't sound so good. Write it differently: We can learn mathematics while doing it. The hardest part of teaching calculus is to turn it from a spectator sport into a workout. The computer makes that possible. EXAMPLE 2 (mental computer) Compare x2 with 2X.The functions meet at x = 2. Where do they meet again? Is it before or after 2? That is mental computing because the answer happens to be a whole number (4). Now we are on a different track. Does an accident like Z4 = 42 ever happen again? Can the machine tell us about integers? Perhaps it can plot the solutions of xb= bx. I asked Mathernatica for a formula, hoping to discover x as a function of b-but the program just gave back the equation. For once the machine typed HELP icstead of the user. Well, mathematics is not helpless. I am proud of calculus. There is a new exercise at the end of Section 6.4, to show that we never see whole numbers again. EXAMPLE3 Find the number b for which xb= bx has only one solution(at x = b). When b is 3, the second solution is below 3. When b is 2, the second solution (4) is above 2. If we move b from 2 to 3, there must be a special "double point"-where the graphs barely touch but don't cross. For that particular b-and only for that one value-the curve xb never goes above bx. This special point b can be found with computer-based graphics. In many ways it is the "center point of calculus." Since the curves touch but don't cross, they are tangent. They have the same slope at the double point. Calculus was created to work with slopes, and we already know the slope of x2. Soon comes xb. Eventually we discover the slope of bx, and identify the most important number in calculus. The point is that this number can be discovered first by experiment. EXAMPLE4 Graph y(x)= ex- xe. Locate its minimum. + The next example was proposed by Don Small. Solve x4 - 1 l x 3 5x - 2 = 0.The first tool is algebra-try to factor the polynomial. That succeeds for quadratics, and then gets extremely hard. Even if the computer can do algebra better than we can, factoring is seldom the way to go. In reality we have two good choices: 1. (Mathematics)Use the derivative. Solve by Newton's method. 2. (Graphics)Plot thefunction and zoom in. Both will be done by the computer. Both have potential problems! Newton's method is fast, but that means it can fail fast. (It is usually terrific.) Plotting the graph is also fast-but solutions can be outside the viewing window. This particular function is 1 lntroductlonto Calculus zero only once, in the standard window from -10 to 10. The graph seems to be leaving zero, but mathematics again predicts a second crossing point. So we zoom out before we zoom in. The use of the zoom is the best part of graphing. Not only do we choose the domain and range, we change them. The viewing window is controlled by four numbers. They can be the limits A running to request input. Execute with P E N T E R after selecting the P R G M ( E X E C menu. Answer ? with 4 and E N T E R. After completion, rerun by pressing E N T E R again. The function is y = 14 - x if x < 7, y = x if x > 7. PrgmP: PIECES :Di s p " x = " :Input X :14-X-+Y :If 7 of tests ( 3 < X ( X < 7 1 evaluates to 1 if all true and to 0 if any false. TRACE and ZOOM The best feature is graphing. But a whole graph can be like a whole book-too much at once. You want to focus on one part. A computer or calculator will trace along the graph, stop at a point, and zoom in. There is also Z 0 0 M 0 U T, to widen the ranges and see more. Our eyes work the same way-they put together information on different scales. Looking around the room uses an amazingly large part of the human brain. With a big enough computer we can try to imitate the eyes-this is a key problem in artificial intelligence. With a small computer and a zoom feature, we can use our eyes to understand functions. Press T R A C E to locate a point on the graph. A blinking cursor appears. Move left or right-the cursor stays on the graph. Its coordinates appear at the bottom of the screen. When x changes by a pixel, the calculator evaluates y(x). To solve y(x) = 0, read off x at the point when y is nearest to zero. To minimize or maximize y(x), read off the smallest and largest y. In all these problems, zoom in for more accuracy. To blow up a figure we can choose new ranges. The fast way is to use a Z 0 0 M command. Forapresetrange,use Z O O M S t a n d a r d or Z O O M T r ig.Toshrink or stretch by X F a c t or Y F a c t (default values 4), use Z 0 0 M In or Z 0 0 M 0 u t . Choose the center point and press E N T E R. The new graph appears. Change those scaling factors with Z 0 0 M S e t F a c t o r s . Best of all, create your own viewing window. Press Z 0 0 M B o x . To draw the box, move the cursor to one corner. Press E N T E R and this point is a small square. The same keys move a second (blinking) square to the opposite corner-the box grows as you move. Press E N T E R, and the box is the new viewing window. The graphs show the same function with a change of scale. Section 3.4 will discuss the mathematics-here we concentrate on the graphics. EXAMPLE9 Place : Y l = X s i n ( 1 / X I intheY=editscreen.PressZOOM T r i g for a first graph. Set X F a c t = 1 and Y F a c t = 2.5. Press Z 0 0 M In with center at (O,O).Toseealargerpicture,use X F a c t = 10and Y F a c t = 1.Then Zoom O u t again. As X gets large, the function X sin (l/X) approaches . Now return to Z 0 0 M T r i g . Z o o m In with the factors set to 4 (default). Zoom again by pressing E N T E R. With the center and the factors fixed, this is faster than drawing a zoom box. 1.7 Computing in Calculus EXAMPLE 10 Repeat for the more erraticfunction Y = sin(l/X). After Z0 0 M T r ig, create a box to see this function near X = .01. The Y range is now Scaling is crucial. For a new function it can be tedious. A formula for y(x) does not easily reveal the range of y's, when A < x ( K,L , which increases K by 1 and skips a line if the new K exceeds L. Otherwise the command G o t o 1 restarts the loop. The screen shows the short form on the left. Example: Y l =x3+10x2-7x+42 with range Xrnin=-12 and Xrnax=lO. Set tick spacing X s c l = 4 and Y s c l = 2 5 0 . Execute with PRGM (EXEC) A E NT E R. For this program we also list menu locations and comments. PrgmA :AUTOSCL :All-Off :Xmin+A :19+L :(Xmax-A) / L + H :A+X :Y1 + C :C+D :I+ K :Lbl I :AtKH + X :Y1 + Y : I F Y (K,L) :Goto 1 :YI-On :C+Ymin :D+Ymax :DispGraph -Menu (Submenu) Comment Y V A R S ( 0 F F Turn off functions V A R S (RNG) Store X m i n using ST0 Store number of evaluations (19) Spacing between evaluations -Start at x = A Y V A R S ( Y ) Evaluate the function Start C and D with this value Initialize counter K = 1 PR G M ( C TL ) Mark loop start Calculate next x Evaluate function at x PGRM (CTL) New minimum? Update C PRGM (CTL) New maximum? Update D PRGM (CTL) Add 1 to K, skip G o t o if > L PRGM (CTL) Loop return to L b l 1 Y - V A R S ( O N ) Turnon Y1 ST0 V A R S ( R N G ) Set Y m i n = C ST0 V A R S (RNG) Set Ymax=D PR G M ( I/ 0 1 Generate graph Contents CHAPTER 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 Introduction to Calculus Velocity and Distance Calculus Without Limits The Velocity at an Instant Circular Motion A Review of Trigonometry A Thousand Points of Light Computing in Calculus CHAPTER 2 Derivatives The Derivative of a Function Powers and Polynomials The Slope and the Tangent Line Derivative of the Sine and Cosine The Product and Quotient and Power Rules Limits Continuous Functions CHAPTER 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 Applications of the Derivative Linear Approximation Maximum and Minimum Problems Second Derivatives: Minimum vs. Maximum Graphs Ellipses, Parabolas, and Hyperbolas , Iterations x,+ = F(x,) Newton's Method and Chaos The Mean Value Theorem and l'H8pital's Rule CHAPTER 2 Derivatives 2.1 The Derivative of a Function This chapter begins with the definition of the derivative. Two examples were in Chapter 1. When the distance is t 2 , the velocity is 2t. When f ( t ) = sin t we found v(t)= cos t. The velocity is now called the derivative o f f (t). As we move to a more formal definition and new examples, we use new symbols f' and dfldt for the derivative. 2A At time t , the derivativef' ( t )or df /dt or v(t) is f'(t)= lim fCt -t At) -f (0 At+O At (1) The ratio on the right is the average velocity over a short time At. The derivative, on the left side, is its limit as the step At (delta t ) approaches zero. + + G o slowly and look at each piece. The distance at time t At is f (t At). The distance at time t is f(t). Subtraction gives the change in distance, between those + times. We often write Af for this difference: Af =f (t At) -f (t).The average velocity is the ratio AflAt-change in distance divided by change in time. The limit of the average velocity is the derivative, if this limit exists: -df -- lim -A.f dt A t - 0 At This is the neat notation that Leibniz invented: Af/At approaches df /dt. Behind the innocent word "limit" is a process that this course will help you understand. Note that Af is not A times f ! It is the change in f . Similarly At is not A times t. It is the time step, positive or negative and eventually small. To have a one-letter symbol we replace At by h. The right sides of (1) and (2)contain average speeds. On the graph of f ( t ) , the distance up is divided by the distance across. That gives the average slope Af /At. The left sides of ( 1 ) and (2)are instantaneous speeds dfldt. They give the slope at the instant t. This is the derivative dfldt (when At and Af shrink to zero). Look again 2.1 The Derivative of a Function at the calculation for f(t) = t2: + -Af-- f(t+At)-f(t) -- t2+2tAt+(At)'-t2 = 2t At. At At At Important point: Those steps are taken before At goes to zero. If we set At = 0 too soon, we learn nothing. The ratio Af/At becomes 010 (which is meaningless). The numbers Af and At must approach zero together, not separately. Here their ratio is + 2t At, the average speed. To repeat: Success came by writing out (t + At)2 and subtracting t2 and dividing by At. Then and only then can we approach At = 0. The limit is the derivative 2t. There are several new things in formulas (1)and (2). Some are easy but important, others are more profound. The idea of a function we will come back to, and the definition of a limit. But the notations can be discussed right away. They are used constantly and you also need to know how to read them aloud: f (t) = "f of t" = the value of the function f at time t At = "delta t" = the time step forward or backward from t + + f (t At) = "f of t plus delta t" = the value off at time t At + Af = "delta f" = the change f (t At) -f (t) Af/At = "delta f over delta t" = the average velocity ff(t)= "f prime of t" = the value of the derivative at time t df /dt = "d f d t" = the same as f ' (the instantaneous velocity) lim = "limit as delta t goes to zero" = the process that starts with At+O numbers Af /At and produces the number df /dt. From those last words you see what lies behind the notation dfldt. The symbol At indicates a nonzero (usually short) length of time. The symbol dt indicates an infinitesimal (even shorter) length of time. Some mathematicians work separately with df and dt, and df/dt is their ratio. For us dfldt is a single notation (don't cancel d and don't cancel A). The derivative dfldt is the limit of AflAt. When that notation dfldt is awkward, use f ' or v. Remark The notation hides one thing we should mention. The time step can be negative just as easily as positive. We can compute the average Af/At over a time interval before the time t, instead of after. This ratio also approaches dfldt. The notation also hides another thing: The derivative might not exist. The averages AflAt might not approach a limit (it has to be the same limit going forward and backward from time t). In that case ft(t) is not defined. At that instant there is no clear reading on the speedometer. This will happen in Example 2. + + EXAMPLE 1 (Constant velocity V = 2) The distance f is V times t. The distance at time t At is V times t At. The diference Af is V times At: -Af----VAt - V so the limit is - df = V. At At dt The derivative of Vt is V. The derivative of 2t is 2. The averages AflAt are always V = 2, in this exceptional case of a constant velocity. 2 Derivatives EXAMPLE 2 Constant velocity 2 up to time t = 3, then stop. For small times we still have f ( t )= 2t. But after the stopping time, the distance is + fixed at f ( t )= 6 . The graph is flat beyond time 3. Then f (t At) =f ( t ) and Af = 0 and the derivative of a constant function is zero: t > 3: f ' ( t )= lim f ( t + At) - f A~+O At (0= lim at-o 0 - At = 0. In this example the derivative is not defined at the instant when t = 3. The velocity falls suddenly from 2 to zero. The ratio A f / A t depends, at that special moment, on whether At is positive or negative. The average velocity after time t = 3 is zero. The average velocity before that time is 2. When the graph o f f has a corner, the graph of v has a jump. It is a step function. One new part of that example is the notation (dfldt or f' instead of v). Please look also at the third figure. It shows how the function takes t (on the left)to f ( t ) .Especially it shows At and A f . At the start, A f / A t is 2. After the stop at t = 3, all t's go to the same f ( t )= 6 . So Af = 0 and df /dt = 0. u =df/dt=f' time distance f'(3) not defined slope undefined slope 2 t 3 3 Fig. 2.1 The derivative is 2 then 0.It does not exist at t = 3. THE DERIVATIVE OF 111 Here is a completely different slope, for the "demand function" f ( t )= lit. The demand is l / t when the price is t . A high price t means a low demand l l t . Increasing the price reduces the demand. The calculus question is: How quickly does l / t change when t changes? The "marginal demand" is the slope of the demand curve. The big thing is to find the derivative of l / t once and for all. It is - l / t 2 . + 1 1 EXAMPLE3 f ( t ) = - h a s A f = - - - 1 . This equals t - (t At) -- -At + + t t+At t t(t At) t(t At) ' + Divide by At and let At -,0: - Af -- - - 1 approaches - df - -- 1 At t(t At) dt t2 ' Line 1 is algebra, line 2 is calculus. The first step in line 1 subtracts f ( t ) from + f(t At). The difference is l / ( t+ At) minus l / t . The common denominator is t times +t At-this makes the algebra possible. We can't set At = 0 in line 2, until we have divided by At. + The average is A f / A t = - l / t ( t At). Now set At = 0. The derivative is - l / t 2 . Section 2.4 will discuss the first of many cases when substituting At = 0 is not possible, and the idea of a limit has to be made clearer. 2.1 The Derhrathre of a Function Fig. 2.2 Average slope is -&,true slope is -4. Increase in t produces decrease in f. Check the algebra at t = 2 and t + At = 3. The demand llt drops from 112 to 113. The difference is Af = - 116, which agrees with -1/(2)(3)in line 1. As the steps Af and At get smaller, their ratio approaches -1/(2)(2)= -114. This derivative is negative. The function llt is decreasing, and Af is below zero. The graph is going downward in Figure 2.2, and its slope is negative: An increasing f (t) has positive slope. A decreasing f (t) has negative slope. The slope -l/t2 is very negative for small t. A price increase severely cuts demand. The next figure makes a small but important point. There is nothing sacred about t. Other letters can be used-especially x. A quantity can depend on position instead of time. The height changes as we go west. The area of a square changes as the side changes. Those are not affected by the passage of time, and there is no reason to use t. You will often see y =f (x), with x across and y up-connected by a function f . Similarly, f is not the only possibility. Not every function is named f! That letter is useful because it stands for the word function-but we are perfectly entitled to write y(x) or y(t) instead off (x) or f (t). The distance up is a function of the distance across. This relationship "y of x" is all-important to mathematics. The slope is also a function. Calculus is about two functions, y(x) and dyldx. Question If we add 1 to y(x), what happens to the slope? Answer Nothing. Question If we add 1 to the slope, what happens to the height? Answer The symbols t and x represent independent variables-they take any value they want to (in the domain). Once they are set, f (t) and y(x) are determined. Thus f and y represent dependent variables-they depend on t and x. A change At produces a 1 2 Fig. 2.3 The derivative of l/t is -l/t2. The slope of l/x is -1/x2. 2 Derivatives change Af. A change Ax produces Ay. The independent variable goes inside the parentheses in f ( t )and y(x). It is not the letter that matters, it is the idea: independent variable t or x dependent variable f or g or y or z or u derivative dfldt or dfldx or dyldx or --• The derivative dyldx comes from [change in y] divided by [change in x ] . The time step becomes a space step, forward or backward. The slope is the rate at which y changes with x. The derivative of a function is its "rate of change." I mention that physics books use x(t) for distance. Darn it. To emphasize the definition of a derivative, here it is again with y and x: + -Ay-- y(x Ax) - y(x) -- distance up Ax Ax distance across - d y = lim - AY = yl(x). dx Ax AX+O The notation yl(x)pins down the point x where the slope is computed. In dyldx that extra precision is omitted. This book will try for a reasonable compromise between logical perfection and ordinary simplicity. The notation dy/dx(x)is not good; yl(x)is better; when x is understood it need not be written in parentheses. You are allowed to say that the function is y = x2 and the derivative is y' = 2xeven if the strict notation requires y(x)= x2 and yl(x)= 2x. You can even say that the function is x2 and its derivative is 2x and its second derivative is 2-provided everybody knows what you mean. Here is an example. It is a little early and optional but terrific. You get excellent practice with letters and symbols, and out come new derivatives. EXAMPLE 4 If u(x)has slope duldx, what is the slope off ( x )= ( ~ ( x ) ) ~ ? From the derivative of x2 this will give the derivative of x4. In that case u = x2 and f = x4. First point: The derivative of u2is not ( d ~ l d x W) ~e.do not square the derivative + 2x. To find the "square rule" we start as we have to-with Af =f ( x Ax) -f (x): + + + + Af = (U(X AX))^ - ( u ( x ) =) ~[u(x A X ) u(x)][U(X A X )- ~ ( x ) ] . + This algebra puts Af in a convenient form. We factored a' - b2 into [a b] times [a - b]. Notice that we don't have (AM)"We have A f , the change in u2.Now divide by Ax and take the limit: + I + + -Af-- [u(x Ax) u(x)][ X k~- U ( X ) approaches 2u(x)- du . Ax dx (5) This is the square rule: The derivative of (u(x))' is 2u(x) times duldx. From the derivatives of x2 and l / x and sin x (all known) the examples give new derivatives. EXAMPLE 5 (u= x 2 )The derivative of x4 is 2u duldx = 2(x2)(2x=) 4x3. EXAMPLE6 (u = l / x )The derivative of 1/x2is 2u duldx = (2/x)(-1 / x 2 )= -2/x3. EXAMPLE 7 (u = sin x, duldx = cos x ) The derivative of u2 = sin2x is 2 sin x cos x. Mathematics is really about ideas. The notation is created to express those ideas. Newton and Leibniz invented calculus independently, and Newton's friends spent a lot of time proving that he was first. He was, but it was Leibniz who thought of 2.1 The Derivative of a Function writing dyldx-which caught on. It is the perfect way to suggest the limit of AylAx. Newton was one of the great scientists of all time, and calculus was one of the great inventions of all time-but the notation must help. You now can write and speak about the derivative. What is needed is a longer list of functions and derivatives. Read-through questions + 9 Find Ay/Ax for y(x) = x x2. Then find dyldx. + + The derivative is the a of Af /At as At approaches b . 10 Find Ay/Ax and dy/dx for y(x) = 1 2x 3x2. Here Af equals c . The step At can be positive or d . + The derivative is written v or e or 1 . Iff (x) = 2x 3 and A x = 4 then Af= g . If A x = - 1 then Af= h . If Ax = 0 then Af= 1 . The slope is not 010 but dfldx = j . + 11 When f (t) = 4/t, simplify the difference f (t At) -f (t), divide by At, and set At = 0. The result is f '(t). + 12 Find the derivative of 1/t2 from Af (t) = l/(t At)2- 1/t2. + Write Af as a fraction with the denominator t2(t At)2. The derivative does not exist where f(t) has a k and v(t) has a I . For f (t)= l / t the derivative is m . The Divide the numerator by At to find Af/At. Set At = 0. + 13 Suppose f (t)= 7t to t = 1. Afterwards f (t) = 7 9(t - 1). slope of y = 4/x is dyldx = n . A decreasing function has (a) Find df /dt at t =3 and t =.; a o derivative. The P variable is t or x and the q variable is f or y. The slope of y2 (is) (is not) ( d ~ / d x ) ~ . (b) Why doesn't f (t) have a derivative at t = l? The slope of ( ~ ( x )is) ~ r by the square rule. The slope of (2x + 3)2is s . 14 Find the derivative of the derivative (the second derivative) of y = 3x2. What is the third derivative? 1 Which of the following numbers (as is) gives df /dt at time t? If in doubt test on f (t)= t2. (b) )m f (t + 2h) -f (t) 0 -+ 2 h (c) lim at-o f (t - At) -f -At (t) (d) lim t-10 f (t + At) -f At (t) 2 Suppose f (x) = x2. Compute each ratio and set h = 0: + + + 3 For f (x)= 3x and g(x) = 1 3x, find f (4 h) and g(4 h) and f1(4) and g1(4).Sketch the graphs of f and g-why do they have the same slope? 4 Find three functions with the same slope as f (x)= x2. + + 5 For f (x)= l/x, sketch the graphs off (x) 1 and f (x 1). Which one has the derivative -1/x2? 6 Choose c so that the line y = x is tangent to the parabola + y = x2 C. They have the same slope where they touch. 15 Find numbers A and B so that the straight line y = x fits + + smoothly with the curve Y= A Bx x2 at x = 1. Smoothly means that y = Y and dyldx = dY/dx at x = 1. 16 Find numbers A and B so that the horizontal line y = 4 + + fits smoothly with the curve y = A Bx x2 at the point x = 2. 17 True (with reason) or false (with example): (a) If f(t) < 0 then df /dt < 0. (b) The derivative of (f (t))2is 2 df /dt. (c) The derivative of 2f (t) is 2 df /dt. (d) The derivative is the limit of Af divided by the limit of At. + 18 For f (x) = l/x the centered diflerencef (x h) -f (x - h) is + l/(x h) - l/(x - h). Subtract by using the common denomi+ nator (x h)(x - h). Then divide by 2h and set h = 0. Why divide by 2h to obtain the correct derivative? + + 19 Suppose y = mx b for negative x and y = Mx B for x 3 0. The graphs meet if . The two slopes are . The slope at x = 0 is (what is possible?). 20 The slope of y = l / x at x = 114 is y' = -1/x2 = -16. At h = 1/12, which of these ratios is closest to -16? 7 Sketch the curve y(x) = 1 - x2 and compute its slope at x=3. 8 Iff (t) = l/t, what is the average velocity between t = 3 and t = 2? What is the average between t = 3 and t = l? What is the average (to one decimal place) between t = 3 and t = 101/200? ~(x+h)-y(x) y(x)-y(x-h) h h y(x+h)-y(x-h) 2 h 21 Find the average slope of y = x2 between x = x, and x = x2.What does this average approach as x2approaches x,? 22 Redraw Figure 2.1 when f(t) = 3 - 2t for t < 2 and f (t) = -1 for t > 2. Include df /dt. 50 2 Derivatives 23 Redraw Figure 2.3 for the function y(x)= 1- ( l / x ) . Include dyldx. 24 The limit of O/At as At 0 -+ is not 010. Explain. 25 Guess the limits by an informal working rule. Set At = 0.1 and -0.1 and imagine At becoming smaller: *26 Suppose f ( x ) / x-+ 7 as x -+ 0. Deduce that f (0)= 0 and f '(0)= 7. Give an example other than f ( x )= 7x. 33 The right figure shows f ( x )and Ax. Find Af /Ax and f '(2). 34 Draw f ( x )and Ax so that Af /Ax = 0 but f ' ( x )# 0. 35 If f = u2 then df/dx = 2u duldx. If g =f then dg/dx = 2f df /dx. Together those give g = u4 and dgldx = 36 True or false, assuming f (0)= 0: (a) If f ( x )6 x for all x, then df /dx 6 1. (b) If df /dx 6 1 for all x, then f ( x )6 x. + 37 The graphs show Af and Af /h for f ( x )= x2.Why is 2x h the equation for Aflh? If h is cut in half, draw in the new graphs. 27 What is lim ( 3 + X , - f ( 3 ) if it exists? What if x -+ l? x-0 Problems 28-31 use the square rule: d(u2)/dx= 2u(duldx). 28 Take u = x and find the derivative of x2 (a new way). 29 Take u = x 4 and find the derivative of x8 (using du/dx =4x3). 30 If u = 1 then u2= 1. Then d l / d x is 2 times d lldx. How is this possible? &. 31 Take u = The derivative of u2= x is 1 = 2u(du/dx).So what is duldx, the derivative of &? + 32 The left figure shows f ( t )= t2.Indicate distances f (t At) and At and Af. Draw lines that have slope Af /At and f '(t). 38 Draw the corresponding graphs for f ( x )= jx. + 39 Draw l l x and l / ( x h) and Aflh-either by hand with h = 5 or by computer to show h -+ 0. 40 For y = ex, show on computer graphs that dyldx = y. 41 Explain the derivative in your own words. 2.2 Powers and Polynomials - This section has two main goals. One is to find the derivatives of f (x)= x3 and x4 and x5 (and more generally f (x)= xn).The power or exponent n is at first a positive integer. Later we allow x" and x2s2and every xn. The other goal is different. While computing these derivatives, we look ahead to their applications. In using calculus, we meet equations with derivatives in them"diflerentialequations." It is too early to solve those equations. But it is not too early to see the purpose of what we are doing. Our examples come from economics and biology. 2.2 Powers and Polynomials 51 With n = 2, the derivative of x2 is 2x. With n = - 1, the slope of x-' is - 1xp2. Those are two pieces in a beautiful pattern, which it will be a pleasure to discover. We begin with x3 and its derivative 3x2, before jumping to xn. EXAMPLE 1 If f (x) = x3 then Af = (x + h)3- x3 = (x3+ 3x2h + 3xh2 + h3)- x3. Step 1: Cancel x3. Step 2: Divide by h. Step 3: h goes to zero. + + - Af = 3x2 3xh h2 approaches df - = 3x2. h dx + + + That is straightforward, and you see the crucial step. The power (x + h)3 yields four separate terms x3 3x2h 3xh2 h3. (Notice 1, 3, 3, 1.) After x3 is subtracted, we can divide by h. At the limit (h = 0) we have 3x2. For f(x) = xn the plan is the same. A step of size h leads to f(x + h) = (x + h)". + One reason for algebra is to calculate powers like (x h)", and if you have forgotten the binomial formula we can recapture its main point. Start with n = 4: Multiplying the four x's gives x4. Multiplying the four h's gives h4. These are the easy + terms, but not the crucial ones. The subtraction (x h)4- x4 will remove x4, and the limiting step h -,0 will wipe out h4 (even after division by h). Theproducts that matter + are those with exactly one h. In Example 1 with (x h)3, this key term was 3x2h. Division by h left 3x2. With only one h, there are n places it can come from. Equation (1) has four h's in parentheses, and four ways to produce x3h. Therefore the key term is 4x3h. (Division ' by h leaves 4x3.)In general there are n parentheses and n ways to produce xn- h, so ' the binomialformula contains nxn- h: Subtract xnfrom (2). Divide by h. The key term is nxn-'. The rest disappears as h + 0: + + + -Af-- (X Ax h)" - xn - nxn-' h ..- h h hn SO -dd=xf n x n - l . The terms replaced by the dots involve h2 and h3 and higher powers. After dividing by h, they still have at least one factor h. All those terms vanish as h approaches zero. + + + + + EXAMPLE 2 (x h)4= x4 4x3h 6x2h2 4xh3 h4. This is n = 4 in detail. Subtract x4, divide by h, let h + 0. The derivative is 4x3. The coefficients 1,4, 6, 4, 1 are in Pascal's triangle below. For (x + h)5 the next row is 1, 5, 10, 2. Remark The missing terms in the binomial formula (replaced by the dots) contain all the products xn-jhj. An x or an h comes from each parenthesis. The binomial coefficient "n choose j" is the number of ways to choose j h's out of n parentheses. It involves n factorial, which is n(n - 1) ... (1). Thus 5! = 5 4 3 2 1= 120. 2 Derivatives These are numbers that gamblers know and love: (;) b L nc/zoose j*'= n! = - j!(n -j)! 1 Pascal's 1 1 triangle 121 1 3 3 1 n=3 1 4 6 4 1 n=4 In the last row, the coefficient of x3h is 4 ! / 1 ! 3 ! = 4 * 3 * 2 * 1 / 1 * 3 * 2 - 1 = 4F.or the x2h2 term, with j = 2, there are 4 3 2 112 1 2 1 = 6 ways to choose two h's. Notice that 1 + 4 + 6 + 4 + 1 equals 16, which is z4. Each row of Pascal's triangle adds to a power of 2. Choosing 6 numbers out of 49 in a lottery, the odds are 49 48 47 46 45 44/6! to 1. That number is N = "49 choose 6" = 13,983,816. It is the coefficient of ~~~h~ + in (x h)49.If itimes N tickets are bought, the expected number of winners is A. The chance of no winner is e-'. The chance of one winner is Ae-'. See Section 8.4. Florida's lottery in September 1990 (these rules) had six winners out of 109,163,978 tickets. DERIVATIVES OF POLYNOMIALS Now we have an infinite list of functions and their derivatives: x x2 x3 x4 x5 ..- 1 2.x 3x2 4x3 5x4 ... The derivative of xn is n times the next lower power x n - l . That rule extends beyond these integers 1, 2, 3, 4, 5 to all powers: f = 1/x has f ' = - 1/x2: Example 3 of Section 2.1 (n = - 1) f = l/x2 has f ' = - 2/x3: Example 6 of Section 2.1 (n = - 2) & f = has f ' = + x L i 2 : true but not yet checked (n = i) Remember that - Y - ~ means l/x2 and x-112 means l/&. Negative powers lead to decreasing functions, approaching zero as x gets large. Their slopes have minus signs. Question What are the derivatives of x10 and x ~an.d .~-Ii2? + Answer lox9 and 2 . 2 ~ ' .a~nd - i x P 3 l 2 . Maybe (x h)2.2 is a little unusual. Pascal's triangle can't deal with this fractional power, but the formula stays firm: Afier .u2.2comes 2 . 2 ~ ' . ~ Thh. e complete binomial formula is in Section 10.5. That list is a good start, but plenty of functions are left. What comes next is really simple. A tremendous number of new functions are "linear combinations" like What are their derivatives? The answers are known for x3 and x2, and we want to multiply by 6 or divide by 2 or add or subtract. Do the same to the derivatices: 2C The derivative of c times f (x)is c times f '(x). + 20 The derivative of f (x) g(x) is f '(x) + gf(x). The number c can be any constant. We can add (or subtract) any functions. The rules allow any combination of f and g : The derivative of 9f (x)- 7g(x) is 9f '(x)- 7g1(x). 2.2 Powers and Polynomials 53 + The reasoning is direct. When f (x) is multiplied by c, so is f (x h). The difference Af is also multiplied by c. All averages Af /h contain c, so their limit is cf '. The only incomplete step is the last one (the limit). We still have to say what "limit" means. Rule 2D is similar. Adding f + g means adding Af + Ag. Now divide by h. In the + limit as h + 0 we reach f ' g'-because a limit of sums is a sum of limits. Any example is easy and so is the proof-it is the definition of limit that needs care (Section 2.6). You can now find the derivative of every polynomial. A "polynomial" is a combina- + tion of 1,x, x2, ...,xn-for example 9 2x - x5.That particular polynomial has slope 2 - 5x4. Note that the derivative of 9 is zero! A constant just raises or lowers the graph, without changing its slope. It alters the mileage before starting the car. The disappearance of constants is one of the nice things in differential calculus. The reappearance of those constants is one of the headaches in integral calculus. When you find v from f , the starting mileage doesn't matter. The constant in f has no effect on v. (Af is measured by a trip meter; At comesfrom a stopwatch.) To find distance from velocity, you need to know the mileage at the start. A LOOK AT DIFFERENTIAL EQUATIONS (FIND y FROM dyldx) We know that y = x3 has the derivative dyldx = 3x2. Starting with the function, we found its slope. Now reverse that process. Start with the slope andfind the function. This is what science does all the time-and it seems only reasonable to say so. Begin with dyldx = 3x2. The slope is given, the function y is not given. Question Can you go backward to reach y = x3? Answer Almost but not quite. You are only entitled to say that y = x3 + C. The constant C is the starting value of y (when x = 0). Then the dzrerential equation dyldx = 3x2 is solved. Every time you find a derivative, you can go backward to solve a differential + equation. The function y = x2 x has the slope dyldx = 2x + 1. In reverse, the slope 2x + 1 produces x2 + x-and all the other functions x2 + x + C, shifted up and down. + After going from distance f to velocity v, we return to f C. But there is a lot more to differential equations. Here are two crucial points: 1. We reach dyldx by way of AylAx, but we have no system to go backward. With dyldx = (sin x)/x we are lost. What function has this derivative? 2. Many equations have the same solution y = x3. Economics has dyldx = 3ylx. Geometry has dyldx = 3y213T. hese equations involve y as well as dyldx. Function and slope are mixed together! This is typical of differential equations. To summarize: Chapters 2-4 compute and use derivatives. Chapter 5 goes in reverse. Integral calculus discovers the function from its slope. Given dyldx we find y(x). Then Chapter 6 solves the differential equation dyldt = y, function mixed with slope. Calculus moves from derivatives to integrals to diferential equations. This discussion of the purpose of calculus should mention a sp~cificexample. 4: Differential equations are applied to an epidemic (like AIDS). In most epi emics the number of cases grows exponentially. The peak is quickly reached by e , and the epidemic dies down. Amazingly, exponential growth is not happening witb AIDSthe best fit to the data through 1988 is a cubic polynomial (Los Alamos Sciehce, 1989): + The number of cases fits a cubic within 2%: y = 174.6(t - 1981.2)3 340. 2 Derivatives This is dramatically different from other epidemics. Instead of dyldt = y we have dyldt = 3y/t. Before this book is printed, we may know what has been preventing d (fortunately). Eventually the curve will turn away from a cubic-I hope that mathematical models will lead to knowledge that saves lives. Added in proofi In 1989 the curve for the U.S. dropped from t to t'. MARGINAL COST AND ELASTICITY IN ECONOMICS First point about economics: The marginal cost and marginal income are crucially important. The average cost of making automobiles may be $10,000. But it is the $8000 cost of the next car that decides whether Ford makes it. "The average describes the past, the marginal predicts thefuture." For bank deposits or work hours or wheat, which come in smaller units, the amounts are continuous variables. Then the word "marginal" says one thing: Take the derivative.? The average pay over all the hours we ever worked may be low. We wouldn't work another hour for that! This average is rising, but the pay for each additional hour rises faster-possibly it jumps. When $10/hour increases to $15/hour after a 40-hour week, a 50-hour week pays $550. The average income is $ll/hour. The marginal income is $15/hour-the overtime rate. Concentrate next on cost. Let y(x) be the cost of producing x tons of steel. The cost of x + Ax tons is y(x + Ax). The extra cost is the difference Ay. Divide by Ax, the number of extra tons. The ratio Ay/Ax is the average cost per extra ton. When Ax is an ounce instead of a ton, we are near the marginal cost dyldx. Example: When the cost is x2, the average cost is x2/x= x. The marginal cost is 2x. Figure 2.4 has increasing slope-an example of "diminishing returns to scale." I I fixed supply any price --I E = O I any supply fixed price E= .. x quantity equilibrium price price Fig. 2.4 Marginal exceeds average. Constant elasticity E = +I. Perfectly elastic to perfectly inelastic (rcurve). This raises another point about economics. The units are arbitrary. In yen per kilogram the numbers look different. The way to correct for arbitrary units is to work with percentage change or relative change. An increase of Ax tons is a relative increase of Axlx. A cost increase Ay is a relative increase of Ayly. Those are dimensionless, the same in tons/tons or dollars/dollars or yen/yen. A third example is the demand y at price x. Now dyldx is negative. But again the units are arbitrary. The demand is in liters or gallons, the price is in dollars or pesos. ?These paragraphs show how calculus applies to economics. You do not have to be an economist to understand them. Certainly the author is not, probably the instructor is not, possibly the student is not. We can all use dyldx. 2.2 Powen and Polynomials Relative changes are better. When the price goes up by lo%, the demand may drop by 5%. If that ratio stays the same for small increases, the elasticity of demand is f. Actually this number should be - f.The price rose, the demand dropped. In our 4. definition, the elasticity will be - In conversation between economists the minus sign is left out (I hope not forgotten). DEFINITION The elasticity of the demand function y(x) is E(x) = lim AX-o - AY/Y Axlx -- -d.yldx Y/X Elasticity is "marginal" dividedby "average." E(x) is also relative change in y divided by relative change in x . Sometimes E(x) is the same at all prices-this important case is discussed below. EXAMPLE 1 Suppose the demand is y = c / x when the price is x. The derivative dy/dx = -c/x2comes from calculus. The division y/x = c / x 2is only algebra. The ratio is E = - 1 : For the demand y = c / x , the elasticity is (- c / x 2 ) / ( c / x 2=) -1 . All demand curves are compared with this one. The demand is inelastic when 1El < 1 . It is elastic when IEl > 1. The demand 20/& is inelastic ( E = - f), while x - i~s elastic (E = - 3). The power y = cxn, whose derivative we know, is the function with constant elasticity n: if y = cxn then dyldx = cnxn-' and E = cnxn-l/(cxn/x)= n. It is because y = cxn sets the standard that we could come so early to economics. In the special case when y = clx, consumers spend the same at all prices. Price x times quantity y remains constant at xy = c . EXAMPLE 2 The supply curve has E > 0-supply increases with price. Now the baseline case is y = cx. The slope is c and the average is y / x = c. The elasticity is E =c/c= 1. Compare E = 1 with E = 0 and E = CQ. A constant supply is "perfectly inelastic." The power n is zero and the slope is zero: y = c . No more is available when the harvest is over. Whatever the price, the farmer cannot suddenly grow more wheat. Lack of elasticity makes farm economics difficult. The other extreme E = a~is "perfectly elastic." The supply is unlimited at a fixed price x. Once this seemed true of water and timber. In reality the steep curve x = constant is leveling off to a flat curve y = constant. Fixed price is changing to fixed supply, E = CQ is becoming E = 0, and the supply of water follows a "gamma curve" shaped like T. EXAMPLE 3 Demand is an increasing function of income-more income, more demand. The income elasticity is E(I)= ( d y / d I ) / ( y / I )A. luxury has E > 1 (elastic). Doubling your income more than doubles the demand for caviar. A necessity has E < 1 (inelastic). The demand for bread does not double. Please recognize how the central ideas of calculus provide a language for the central ideas of economics. Important note on supply = demand This is the basic equation of microeconomics. Where the supply curve meets the demand curve, the economy finds the equilibrium price. Supply = demand assumes perfect competition. With many suppliers, no one can raise the price. If someone tries, the customers go elsewhere. 2 Derivatives The opposite case is a monopoly-no competition. Instead of many small producers of wheat, there is one producer of electricity. An airport is a monopolist (and maybe the National Football League). If the price is raised, some demand remains. Price fixing occurs when several producers act like a monopoly-which antitrust laws try to prevent. The price is not set by supply = demand. The calculus problem is different-to maximize profit. Section 3.2 locates the maximum where the marginal profit (the slope!) is zero. Question on income elasticity From an income of $10,000 you save $500. The income elasticity of savings is E = 2. Out of the next dollar what fraction do you save? Answer The savings is y = c x 2 because E = 2. The number c must give 500 = ~(10,000)s~o, c is 5 Then the slope dyldx is 2cx = 10 lo4 = &. This is the marginal savings, ten cents on the dollar. Average savings is 5%, marginal savings is lo%, and E = 2. 2.2 EXERCISES Read-through questions The derivative of f = x4 is f ' = a . That comes from expanding (x + h)4 into the five terms b . Subtracting x4 and dividing by h leaves the four terms c . This is Af /h, and its limit is d . + The derivative o f f = xn is f ' = e . Now (x h)" comes from the f theorem. The terms to look for are x n - ' h, containing only one g . There are h of those terms, + + + so (x h)" = .un i . After subtracting i and dividing by h, the limit of Aflh is k . The coefficient of .un-JhJ,not needed here, is " n choose j" = I , where n! means m . The derivative of x - i~s n . The derivative of x1I2 is + o . The derivative of 3.u (llx) is P , which uses the following rules: The derivative of 3f (.u)is CI and the deriv- + ative off (.u) g(x)is r . Integral calculus recovers s from dy/d.u. If dy1d.u = .u4 then y(.u) = t . + + + 12 Find the mistake: x2 is x x 0 . . x (with x terms). Its + + + derivative is 1 1 .-. 1 (also x terms). So the derivative of x2 seems to be x. 13 What are the derivatives of 3x'I3 and -3x-'I3 and (3x'I3)- '? + 14 The slope of .u ( 1 1 ~i)s zero when x = . What does the graph do at that point? 15 Draw a graph of y = x3 - x. Where is the slope zero? 16 If df /dx is negative, is f (x) always negative? Is f (x) negative for large x? If you think otherwise, give examples. 17 A rock thrown upward with velocity 16ft/sec reaches height f = 16t - 16t2 at time t. (a) Find its average speed Af /At from t = 0 to t = $. 4 (b) Find its average speed Af /At from t = to t = 1. (c) What is df /dt at t = i? 1 Starting with f = .u6, write down f ' and then f ". (This is "f double prime," the derivative off '.) After deriva- tives of x6 you reach a constant. What constant? 2 Find a function that has .u6 as its derivative. Find the derivatives of the functions in 3-10. Even if n is nega- '. tive or a fraction, the derivative of xn is nxn- 18 When f is in feet and t is in seconds, what are the units of f ' and its derivative f "? In f = 16t - 16t2, the first 16 is ft/sec but the second 16 is . + 19 Graph y = x3 x2 - x from x = - 2 to x = 2 and estimate where it is decreasing. Check the transition points by solving dyldx = 0. 20 At a point where dyldx = 0, what is special about the & - graph of y(x)? Test case: y = x2. 21 Find the slope of y = by algebra (then h 0): - A - JFG-J; -- J T h - J ; h h h JJzzi +iJ+; J;. 11 Name two functions with df/dx = 1/x2. 22 Imitate Problem 21 to find the slope of y = I/&. 2.2 Powers and Polynomials 57 23 Complete Pascal's triangle for n = 5 and n = 6. Why do the numbers across each row add to 2"? (:) (:) (i)? + + 24 Complete (x h)5=x5 mial coefficients and and . What are the bino- + 25 Compute (x h)3-(x - h)3, divide by 2h, and set h = 0. Why divide by 2h to Jind this slope? 26 Solve the differential equation y" = x to find y(x). + + 27 For f (x)= x2 x3, write out f (x Ax) and Af /Ax. What is the limit at Ax = 0 and what rule about sums is confirmed? 28 The derivative of ( ~ ( x )is) ~ from Section 2.1. Test this rule on u = xn. + + 29 What are the derivatives of x7 1 and (x Shift the graph of x7. 30 If df /dx is v(x), what functions have these derivatives? (a) 4+) (c) v(x + 1) (b) + 1 + (d) v(x) v'(x). 31 What function f(x) has fourth derivative equal to l? 32 What function f (x) has nth derivative equal to l? + 33 Suppose df /dx = 1 x + x2 + x3. Find f (x). 34 Suppose df /dx = x- - x- 3. Find f (x). 35 f (x) can be its own derivative. In the infinite polynomial f = 1+ x + 5x2+ &x3+ ,what numbers multiply x4 and x5 if df /dx equals f ? 36 Write down a differential equation dy/dx = that is solved by y = x2. Make the right side involve y (notjust 2x). 37 True or false: (a) The derivative of x" is nx". (b) The derivative of axn/bxnis a/b. (c) If df /dx = x4 and dgldx = x4 then f (x)=g(x). (d) (f (x)-f (a))/(x- a) approaches f '(a) as x -+ a. (e) The slope of y = (x - is y' = 3(x - Problems 38-44 are about calculus in economics. + 38 When the cost is y = yo cx, find E(x) = (dy/dx)/(y/x). It approaches for large x. 39 From an income of x = $10,000 you spend y = $1200 on your car. If E = 3,what fraction of your next dollar will be spent on the car? Compare dy/dx (marginal) with y/x (average). 40 Name a product whose price elasticity is (a) high (b) low (c) negative (?) 41 The demand y = c/x has dyldx = - y/x. Show that Ay/Ax is not -y/x. (Use numbers or algebra.) Finite steps miss the special feature of infinitesimal steps. 42 The demand y = xn has E = . The revenue xy (price times demand) has elasticity E = . + 43 y = 2x 3 grows with marginal cost 2 from the fixed cost 3. Draw the graph of E(x). 44 From an income I we save S(I). The marginal propensity to save is . Elasticity is not needed because S and I have the same . Applied to the whole economy this is (microeconomics) (macroeconomics). 45 2' is doubled when t increases by . t3 is doubled when t increases to t. The doubling time for AIDS is proportional to t. 46 Biology also leads to dyly = n dxlx, for the relative growth of the head (dyly) and the body (dxlx). Is n > 1 or n < 1 for a child? 47 What functions have df/dx = x9 and df/dx = xn? Why does n = -1 give trouble? 48 The slope of y = x3 comes from this identity: + + (x + h)3- x3 h = ( x h)2+ ( x h)x + x 2 . (a) Check the algebra. Find dyldx as h -+0. (b) Write a similar identity for y = x4. 49 (Computer graphing) Find all the points where y = + + x4 2x3- 7x2 3 = 0 and where dy/dx = 0. + 50 The graphs of y,(x) = x4 x3 and y,(x) = 7x - 5 touch at the point where y3(x)= = 0. Plot y3(x)to see what is special. What does the graph of y(x) do at a point where y = y' = O? 51 In the Massachusetts lottery you choose 6 numbers out of 36. What is your chance to win? 52 In what circumstances would it pay to buy a lottery ticket for every possible combination, so one of the tickets would win? -58 2 Derivatives 2.3 The Slope and the Tangent Line - Chapter 1 started with straight line graphs. The velocity was constant (at least piecewise). The distance function was linear. Now we are facing polynomials like x3- 2 + or x4 - x2 3, with other functions to come soon. Their graphs are definitely curved. Most functions are not close to linear-except if you focus all your attention near a single point. That is what we will do. Over a very short range a curve looks straight. Look through a microscope, or zoom in with a computer, and there is no doubt. The graph of distance versus time becomes nearly linear. Its slope is the velocity at that moment. We want to find the line that the graph stays closest to-the "tangent linew-before it curves away. The tangent line is easy to describe. We are at a particular point on the graph of y =f (x). At that point x equals a and y equals f (a) and the slope equals f '(a). The tangent line goes through that point x = a, y =f (a) with that slope m = fl(a). Figure 2.5 shows the line more clearly than any equation, but we have to turn the geometry into algebra. We need the equation of the line. + EXAMPLE1 Suppose y = x4 - x2 3. At the point x = a = 1, the height is y =f (a)= 3. The slope is dyldx = 4x3 - 2x. At x = 1 the slope is 4 - 2 = 2. That is fl(a): The numbers x = 1, y = 3, dyldx = 2 determine the tangent line. The equation of the tangent line is y - 3 = 2(x - l), and this section explains why. Fig. 2.5 The tangent line has the same slope 2 as the curve (especially after zoom). THE EQUATION OF A LINE A straight line is determined by two conditions. We know the line if we know two of its points. (We still have to write down the equation.) Also, if we know one point and the slope, the line is set. That is the situation for the tangent line, which has a known slope at a known point: 1. The equation of a line has the form y = mx + b 2. The number m is the slope of the line, because dyldx = m 3. The number b adjusts the line to go through the required point. I will take those one at a time-first y = mx + b, then m, then b. + 1. The graph of y = mx b is not curved. How do we know? For the specificexample + y = 2x 1, take two points whose coordinates x, y satisfy the equation: x=O, y = 1 and x = 4 , y = 9 both satisfy y = 2 x + 1. 2.3 The Slope and the Tangent Line Those points (0, 1) and (4,9)lie on the graph. The point halfway between has x = 2 + and y = 5. That point also satisfies y = 2x 1. The halfway point is on the graph. If we subdivide again, the midpoint between (0, 1) and (2, 5) is (1, 3). This also has y = 2x + 1. The graph contains all halfway points and must be straight. 2. What is the correct slope m for the tangent line? In our example it is m =f '(a)= 2. The curve and its tangent line have the same slope at the crucial point: dyldx = 2. + Allow me to say in another way why the line y = mx b has slope m. At x = 0 its + height is y = b. At x = 1 its height is y = m b. The graph has gone one unit across (0 to 1) and m units up (b to m + b). The whole idea is slope = distance up - - m distance across 1 ' Each unit across means m units up, to 2m + b or 3m + b. A straight line keeps a + constant slope, whereas the slope of y = x4 - x2 3 equals 2 only at x = 1. + 3. Finally we decide on b. The tangent line y = 2x b must go through x = 1 , y = 3. Therefore b = 1. With letters instead of numbers, y = mx + b leads to f (a)= ma + b. So we know b: 2E The equation of the tangent line has b =f (a)- ma: y=mx+f(a)-ma or y-f(a)=m(x-a). I(2) That last form is the best. You see immediately what happens at x = a. The factor x - a is zero. Therefore y =f (a)as required. This is the point-slope form of the equa- tion, and we use it constantly: y-3=2(x-1) or y -3 - -- distance up = sbpe 2. x - 1 distance across EXAMPLE 2 The curve y = x3 - 2 goes through y = 6 when x = 2. At that point dyldx = 3x2 = 12. The point-slope equation of the tangent line uses 2 and 6 and 12: y - 6 = 12(x-2), which is also y = 12x- 18. There is another important line. It is perpendicular to the tangent line and perpendicular to the curve. This is the normal line in Figure 2.6. Its new feature is its slope. When the tangent line has slope m, the normal line has slope - llm. (Rule: Slopes of tangent line: / distance track A :a' + 4 .*' your speed is V 4 T Fig. 2.6 Tangent line y - yo = m(x - x,). Normal line y - yo = - 1 - (x - x,). Leaving a roller- coaster and catching up to a car. m 2 Derivatives perpendicular lines multiply to give - 1.) Example 2 has m = 12, so the normal line has slope - 1/ 12: tangent line: y - 6 = 12(x- 2) normal line: y - 6 = - & ( x - 2). Light rays travel in the normal direction. So do brush fires-they move perpendicular to the fire line. Use the point-slope form! The tangent is y = 12x - 18, the normal is not y = - &x - 18. EXAMPLE 3 You are on a roller-coaster whose track follows y = x 2 + 4. You see a friend at (0,O)and want to get there quickly. Where do you step off? Solution Your path will be the tangent line (at high speed). The problem is to choose x = a so the tangent line passes through x = 0 , y = 0. When you step off at x = a, + the height is y = a2 4 and the slope is 2a + the equation of the tangent line is y - (a2 4) = 2a(x - a) + this line goes through (0,O)if - (a2+ 4 )= - 2a2 or a = 2. The same problem is solved by spacecraft controllers and baseball pitchers. Releasing a ball at the right time to hit a target 60 feet away is an amazing display of calculus. Quarterbacks with a moving target should read Chapter 4 on related rates. Here is a better example than a roller-coaster. Stopping at a red light wastes gas. It is smarter to slow down early, and then accelerate. When a car is waiting in front of you, the timing needs calculus: EXAMPLE 4 How much must you slow down when a red light is 7 2 meters away? In 4 seconds it will be green. The waiting car will accelerate at 3 meters/sec2. You cannot pass the car. Strategy Slow down immediately to the speed V at which you will just catch that car. (If you wait and brake later, your speed will have to go below V.) At the catchup time T , the cars have the same speed and same distance. Two conditions, so the distance functions in Figure 2.6d are tangent. Solution At time T, the other car's speed is 3 ( T - 4). That shows the delay of 4 + seconds. Speeds are equal when 3 ( T - 4 ) = V or T = V 4. Now require equal distances. Your distance is V times T . The other car's distance is 72 + $at2: 7 2 + 5 3 ( ~ - 4 ) ~ = V Tbecomes 7 2 + f - f - v 2 = V ( 3 V + 4 ) . The solution is V = 12 meters/second. This is 43 km/hr or 27 miles per hour. Without the other car, you only slow down to V = 7214 = 18 meters/second. As the light turns green, you go through at 65 km/hr or 40 miles per hour. Try it. THE SECANT LINE CONNECTING TWO POINTS O N A CURVE Instead of the tangent line through one point, consider the secant line through two points. For the tangent line the points came together. Now spread them apart. The point-slope form of a linear equation is replaced by the two-point form. The equation of the curve is still y =f (x).The first point remains at x = a, y =f (a). The other point is at x = c, y =f (c). The secant line goes between them. and we want its equation. This time we don't start with the slope-but rn is easy to find. 2.3 The Slope and the Tangent Line EXAMPLE 5 The curve y = x3- 2 goes through x = 2, y = 6. It also goes through x = 3, y = 25. The slope between those points is m = change in change in y x ---2-5 - 6 3-2 - 19. The point-slope form (at the first point) is y - 6 = 19(x- 2). This line automatically goes through the second point (3,25). Check: 25 - 6 equals 19(3- 2). The secant has the right slope 19 to reach the second point. It is the average slope AylAx. A look ahead The second point is going to approach the first point. The secant slope AylAx will approach the tangent slope dyldx. We discover the derivative (in the limit). That is the main point now-but not forever. Soon you will be fast at derivatives. The exact dyldx will be much easier than AylAx. The situation is turned around as soon as you know that x9 has slope 9x8. Near x = 1, the distance up is about 9 times the distance across. To find Ay = l.0019- 19,just multiply Ax = .001 by 9. The quick approximation is .009, the calculator gives Ay = .009036. It is easier to follow the tangent line than the curve. Come back to the secant line, and change numbers to letters. What line connects x = a, y =f (a) to x = c, y =f (c)?A mathematician puts formulas ahead of numbers, and reasoning ahead of formulas, and ideas ahead of reasoning: (1) The slope is m = distance up distance across - f (c)-f (a) c -a (2) The height is y =f (a) at x = a (3) The height is y =f (c) at x = c (automatic with correct slope). - - The t f ~ v a ruses the slope between the f4d -f@ c-a (3) At x = a the right side is zero. So y =f (a) on the left side. At x = c the right side has two factors c - a. They cancel to leave y =f (c). With equation (2)for the tangent line and equation (3) for the secant line, we are ready for the moment of truth. THE SECANT LlNE APPROACHES THE TANGENT LlNE What comes now is pretty basic. It matches what we did with velocities: + average velocity = A distance A time - f (t At) -f (t) At The limit is df /dt. We now do exactly the same thing with slopes. The secant tine turns into the tangent line as c approaches a: slope of secant line: - A f -- f ( 4 -f@) Ax c - a slope of tangent line: - df dx = limit of -A.f Ax There stands the fundamental idea of differential calculus! You have to imagine more secant lines than I can draw in Figure 2.7, as c comes close to a. Everybody recognizes c - a as Ax. Do you recognize f (c)-f (a) as f (x + Ax) -f (x)? It is Af, the change in height. All lines go through x = a, y =f (a). Their limit is the tangent line. secant secant tangent secant y -f (a) = c - a tangent y - f(a)= f'(a)(x- a) a ccc Fig.2.7 Secants approach tangent as their slopes Af/Ax approach df /dx. Intuitively, the limit is pretty clear. The two points come together, and the tangent line touches the curve at one point. (It could touch again at faraway points.) Mathematically this limit can be tricky-it takes us from algebra to calculus. Algebra stays away from 010, but calculus gets as close as it can. The new limit for df /dx looks different, but it is the same as before: f'(a) = lim f ( 4 -f (a) c+a C- 9 EXAMPLE 6 Find the secant lines and tangent line for y =f (x) = sin x at x = 0. The starting point is x = 0, y = sin 0. This is the origin (0,O). The ratio of distance up to distance across is (sin c)/c: secant equation y = - sin c x tangent equation y = lx. C As c approaches zero, the secant line becomes the tangent line. The limit of (sin c)/c is not 010, which is meaningless, but 1, which is dyldx. & EXAMPLE 7 The gold you own will be worth million dollars in t years. When & does the rate of increase drop to 10% of the current value, so you should sell the gold and buy a bond? At t = 25, how far does that put you ahead of = 5? &, Solution The rate of increase is the derivative of which is 1/2&. That is 10% & of the current value when 1/2& = &/lo. Therefore 2t = 10or t = 5. At that time you sell the gold, leave the curve, and go onto the tangent line: y-fi=$(t-5) becomes y - f i = 2 f i at t=25. With straight interest on the bond, not compounded, you have reached y = 3 f i = 6.7 million dollars. The gold is worth a measly five million. 2.3 EXERCISES Read-through questions A straight line is determined by a points, or one point and the b .The slope of the tangent line equals the slope of the c . The point-slope form of the tangent equation isy-f(a)= d . + The tangent line to y =x3 x at x = 1has slope . Its 2.3 The Slope and the Tangent Line 63 equation is f . It crosses the y axis at g and the x axis at h . The normal line at this point (1, 2) has slope i . Its equation is y - 2 = j . The secant line from (1, 2) to (2, k ) has slope I . Its equation is y-2= m . The point (c, f (c)) is on the line y -f (a) = m(x - a) pro- vided m = n . As c approaches a, the slope m approaches 0 . The secant line approaches the p line. 1 (a) Find the slope of y = 12/x. (b) Find the equation of the tangent line at (2, 6). (c) Find the equation of the normal line at (2, 6). (d) Find the equation of the secant line to (4, 3). + 2 For y = x2 x find equations for (a) the tangent line and normal line at (1, 2); + + + + (b) the secant line to x = 1 h, y =(1 h)2 (1 h). 3 A line goes through (1, -1) and (4, 8). Write its equation + in point-slope form. Then write it as y = mx b. + 4 The tangent line to y = x3 6x at the origin is Y=- . Does it cross the curve again? + 5 The tangent line to y = x3 - 3x2 x at the origin is Y=- . It is also the secant line to the point . 6 Find the tangent line to x = y2 at x = 4, y = 2. 7 For y = x2 the secant line from (a, a 2 ) to (c, c2) has the equation . Do the division by c - a to find the tan- gent line as c approaches a. 8 Construct a function that has the same slope at x = 1 and x = 2. Then find two points where y = x4 - 2x2 has the same tangent line (draw the graph). 9 Find a curve that is tangent to y = 2x - 3 at x = 5. Find the normal line to that curve at (5, 7). 10 For y = llx the secant line from (a, lla) to (c, llc) has the equation . Simplify its slope and find the limit as c approaches a. 11 What are the equations of the tangent line and normal line to y = sin x at x = n/2? 12 If c and a both approach an in-between value x = b, then the secant slope (f(c)-f (a))/(c- a) approaches . 13 At x = a on the graph of y = l/x, compute (a) the equation of the tangent line (b) the points where that line crosses the axes. The triangle between the tangent line and the axes always has area . + 14 Suppose g(x) =f (x) 7. The tangent lines to f and g at x =4 are . True orfalse: The distance between those lines is 7. + 15 Choose c so that y = 4x is tangent to y = x2 c. Match heights as well as slopes. + 16 Choose c so that y = 5x - 7 is tangent to y = x2 cx. 17 For y = x3 + 4x2- 3x + 1, find all points where the tan- gent is horizontal. 18 y = 4x can't be tangent to y = cx2. Try to match heights and slopes, or draw the curves. 19 Determine c so that the straight line joining (0, 3) and + (5, -2) is tangent to the curve y = c/(x 1). + + 20 Choose b, c, d so that the two parabolas y = x2 bx c and y = dx - x2 are tangent to each other at x = 1,y = 0. 21 The graph o f f (x)= x3 goes through (1, 1). + (a) Another point is x = c = 1 h, y =f (c)= . (b) The change in f is Af = . (c) The slope of the secant is m = (d) As h goes to zero, m approaches 22 Construct a function y =f (x) whose tangent line at x = 1 is the same as the secant that meets the curve again at x = 3. 23 Draw two curves bending away from each other. Mark the points P and Q where the curves are closest. At those points, the tangent lines are and the normal lines are . + '24 If the parabolas y = x2 1 and y = x - x2 come closest at + (a, a 2 1) and (c, c - c2), set up two equations for a and c. 25 A light ray comes down the line x = a. It hits the parabolic reflector y = x2 at P = (a, a2). (a) Find the tangent line at P. Locate the point Q where that line crosses the y axis. (b) Check that P and Q are the same distance from the focus at F = (0, $). (c) Show from (b) that the figure has equal angles. (d) What law of physics makes every ray reflect off the parabola to the focus at F? vertical ray 26 In a bad reflector y = 2/x, a ray down one special line x = a is reflected horizontally. What is a? 64 2 Derivatives 27 For the parabola 4py = x2, where is the slope equal to l? At that point a vertical ray will reflect horizontally. So the focus is at (0, 1. 28 Why are these statements wrong? Make them right. (a) If y = 2x is the tangent line at (1, 2) then y = - i x is the normal line. (b) As c approaches a, the secant slope (f (c)-f (a))& - a) approaches (f (a)-f (a))/(a-a). (c) The line through (2, 3) with slope 4 is y - 2 =4(x - 3). 29 A ball goes around a circle: x =cos t, y = sin t. At t = 3 4 4 the ball flies offon the tangent line. Find the equation of that line and the point where the ball hits the ground (y = 0). 30 If the tangent line to y =f(x) at x = a is the same as the tangent line to y =g(x) at x = b, find two equations that must be satisfied by a and b. 31 Draw a circle of radius 1 resting in the parabola y = x2. At the touching point (a, a2),the equation of the normal line is . That line has x = 0 when y = . The dis- tance to (a, a2)equals the radius 1 when a = . This locates the touching point. 32 Follow Problem 31 for the flatter parabola y =3x2 and explain where the circle rests. 33 You are applying for a $1000 scholarship and your time is worth $10 a hour. If the chance of success is 1-(l/x)from x hours of writing, when should you stop? 34 Suppose If (c)-f (a)l< Ic - a1 for every pair of points a and c. Prove that Idf /dxl< 1. 35 From which point x = a does the tangent line to y = 1/x2 hit the x axis at x = 3? 36 If u(x)/v(x)= 7 find u'(x)/v'(x). Also find (u(x)/v(x))'. 37 Find f(c) = l.OO110 in two ways-by calculator and by .-f(.c.)-f- (a) . . xf-'(a. ).(.c - a). Choose a = 1 and -f(.x),= xlO. 38 At a distance Ax from x = 1, how far is the curve y = l/x above its tangent line? 39 At a distance Ax from x = 2, how far is the curve y = x3 above its tangent line? 40 Based on Problem 38 or 39, the distance between curve and tangent line grows like what power (Ax)P? 41 The tangent line to f (x) = x2- 1 at x, = 2 crosses the x axis at xl = . The tangent line at x, crosses the x axis at x2= . Draw the curve and the two lines, which are the beginning of Newton's method to solve f (x) = 0. + 42 (Puzzle) The equation y = mx b requires two numbers, the point-slope form y -f (a)=f '(a)(x -a) requires three, and the two-point form requires four: a, f (a),c, f (c). How can this be? 43 Find the time T at the tangent point in Example 4, when you catch the car in front. 44 If the waiting car only accelerates at 2 meters/sec2,what speed V must you slow down to? 45 A thief 40 meters away runs toward you at 8 meters per second. What is the smallest acceleration so that v = at keeps you in front? 46 With 8 meters to go in a relay race, you slow down badly + (f= - 8 6t -$t2). How fast should the next runner start (choose u in f = vt) so you can just pass the baton? This section does two things. One is to compute the derivatives of sin x and cos x. The other is to explain why these functions are so important. They describeoscillation, which will be expressed in words and equations. You will see a "di~erentiael quation." It involves the derivative of an unknown function y(x). The differential equation will say that the second derivative-the derivative of the derivative-is equal and opposite to y. In symbols this is y" = - y. Distance in one direction leads to acceleration in the other direction. That makes y and y' and y" all oscillate. The solutions to y" = - y are sin x and cos x and all their combinations. We begin with the slope. The derivative of y = sin x is y' = cos x. There is no reason for that to be a mystery, but I still find it beautiful. Chapter 1 followed a ball around a circle; the shadow went up and down. Its height was sin t and its velocity was cos t . 2.4 The Derhrutii of the Sine and Cosine We now find that derivative by the standard method of limits, when y(x) = sin x: + - dy = limit of - AY= lim sin (x h) - sin x dx Ax h + o h + The sine is harder to work with than x2 or x3. Where we had (x h)2 or (x + h)3, we + now have sin(x h). This calls for one of the basic "addition formulas" from trigo- nometry, reviewed in Section 1.5: + + sin (x h) = sin x cos h cos x sin h (2) + cos(x h) = cos x cos h - sin x sin h. (3) + Equation (2) puts Ay = sin (x h) - sin x in a new form: ( ) (T). + -Ay-- sin x cos h Ax cos x sin h - sin x = sin x h cos h - 1 + cos x sin h (41 The ratio splits into two simpler pieces on the right. Algebra and trigonometry got us this far, and now comes the calculus problem. What happens as h +O? It is no longer easy to divide by h. (I will not even mention the unspeakable crime of writing (sin h)/h = sin.) There are two critically important limits-the first is zero and the second is one: cos h - 1 lim h-0 h =0 and sin h lim h-0 - h - 1. The careful reader will object that limits have not been defined! You may further object to computing these limits separately, before combining them into equation (4). Nevertheless-following the principle of ideas now, rigor later-I would like to proceed. It is entirely true that the limit of (4) comes from the two limits in (5): + + -dy-- (sin x)(first limit) (cos x)(second limit) = 0 cos x. dx (6) The secant slope Ay/Ax has approached the tangent slope dyldx. .995 r-995 cOs We cannot pass over the crucial step-the two limits in (5). They contain the real ideas. Both ratios become 010 i f wejust substitute h = 0. Remember that the cosine of a zero angle is 1, and the sine of a zero angle is 0. Figure 2.8a shows a small angle h (as near to zero as we could reasonably draw). The edge of length sin h is close to zero, and the edge of length cos h is near 1. Figure 2.8b shows how the ratio of sin h to h (both headed for zero) gives the slope of the sine curve at the start. When two functions approach zero, their ratio might do anything. We might have Lh .1 sinh No clue comes from 010. What matters is whether the top or bottom goes to zero more quickly. Roughly speaking, we want to show that (cos h - l)/h is like h2/h and (sin h)/h is like hlh. .loo.. . + Time out The graph of sin x is in Figure 2.9 (in black). The graph of sin(x Ax) Fig. 2.8 sits just beside it (in red). The height difference is Af when the shift distance is Ax. sin h sin (x + h) + Fig. 2.9 sin ( x h) with h = 10"= 11/18 radians. Af/Ax is close to cos x. Now divide by that small number Ax (or h). The second figure shows Af /Ax. It is close to cos x. (Look how it starts-it is not quite cos x.) Mathematics will prove that the limit is cos x exactly, when Ax -,0. Curiously, the reasoning concentrates on only one point (x = 0). The slope at that point is cos 0= 1. We now prove this: sin Ax divided by Ax goes to 1. The sine curve starts with + slope 1. By the addition formula for sin (x h), this answer at one point will lead to the slope cos x at all points. + Question Why does the graph of f (x Ax) shift left from f (x) when Ax >O? Answer When x = 0, the shifted graph is already showing f (Ax). In Figure 2.9a, the red graph is shifted left from the black graph. The red graph shows sin h when the black graph shows sin 0. THE LIMIT OF (sin h ) / h IS 4 There are several ways to find this limit. The direct approach is to let a computer draw a graph. Figure 2.10a is very convincing. Thefunction (sin h)/h approaches 1at the key point h = 0. So does (tan h)/h. In practice, the only danger is that you might get a message like "undefined function" and no graph. (The machine may refuse to divide by zero at h = 0. Probably you can get around that.) Because of the importance of this limit, I want to give a mathematical proof that it equals 1. sin h -n/2 h=O n/2 Fig.2.40 (sin h)/hsqueezed between cos x and 1; (tan h)/h decreases to 1. Figure 2.10b indicates, but still only graphically, that sin h stays below h. (The first graph shows that too; (sin h)/h is below 1.) We also see that tan h stays above h. Remember that the tangent is the ratio of sine to cosine. Dividing by the cosine is enough to push the tangent above h. The crucial inequalities (to be proved when h is small and positive) are s i n h < h and t a n h > h . (7) 2.4 The Derlvcrthre of the Sine and Cosine Since tan h = (sin h)/(cos h), those are the same as sin h < 1 and - sin h > cos h. h h What happens as h goes to zero? The ratio (sin h)/h is squeezed between cos h and 1. But cos h is approaching I! The squeeze as h +0 leaves only one possibility for (sin h)/h, which is caught in between: The ratio (sin h)/h approaches 1. Figure 2.10 shows that "squeeze play." lf twofunctions approach the same limit, so does any function caught in between. This is proved at the end of Section 2.6. For negative values of h, which are absolutely allowed, the result is the same. To the left of zero, h reverses sign and sin h reverses sign. The ratio (sin h)/h is unchanged. (The sine is an odd function: sin (- h) = - sin h.) The ratio is an even function, symmetric around zero and approaching 1 from both sides. The proof depends on sin h < h < tan h, which is displayed by the graph but not explained. We go back to right triangles. Fig. 2.11 Line shorter than arc: 2 sin h < 2h. Areas give h f h. The inequalities sin h < h < tan h are now proved. The squeeze in equation (8) produces (sin h)/h -,1. Q.E.D. Problem 13shows how to prove sin h < h from areas. Note All angles x and h are being measured in radians. In degrees, cos x is not the derivative of sin x. A degree is much less than a radian, and dyldx is reduced by the factor 2~1360. THE LIMIT OF (COS h - 1 ) / h IS 0 This second limit is different. We will show that 1 - cos h shrinks to zero more quickly + than h. Cosines are connected to sines by (sin h)2 (cos h)2= 1. We start from the +If we try to prove that, we will be here all night. Accept it as true. 2 Derivatives known fact sin h < h and work it into a form involving cosines: + (1 - cos h)(l cos h) = 1 - (cos h)2 = (sin h)2 < h2. (9) + Note that everything is positive. Divide through by h and also by 1 cos h: + o < 1 - cos h < h h 1 cos h ' Our ratio is caught in the middle. The right side goes to zero because h + 0. This is another "squeezew-there is no escape. Our ratio goes to zero. For cos h - 1 or for negative h, the signs change but minus zero is still zero. This confirms equation (6). The slope of sin x is cos x. Remark Equation (10) also shows that 1 - cos h is approximately i h 2 . The 2 comes + from 1 cos h. This is a basic purpose of calculus-to find simple approximations like $h2. A "tangent parabola" 1 - $h2 is close to the top of the cosine curve. THE DERIVATIVE OF THE COSINE This will be easy. The quick way to differentiate cos x is to shift the sine curve by xl2.That yields the cosine curve (solid line in Figure 2.12b).The derivative also shifts by 4 2 (dotted line). The derivative of cos x is - sin x. Notice how the dotted line (the slope) goes below zero when the solid line turns downward. The slope equals zero when the solid line is level. Increasing functions have positive slopes. Decreasing functions have negative slopes. That is important, and we return to it. There is more information in dyldx than "function rising" or "function falling." The slope tells how quickly the function goes up or down. It gives the rate of change. The slope of y = cos x can be computed in the normal way, as the limit of AylAx: .( Ay - cos(x + h)- cos x -- Ax h =cos cos h - 1 )-sinx(y) - dy - - (COS x)(O)- (sin \-)(I)= - sin u. d.u (11) + The first line came from formula (3) for cos(x h). The second line took limits, reaching 0 and 1 as before. This confirms the graphical proof that the slope of cos x is - sin x. --.. / Y = sin .\- is increasing p> v =,sin r bends down v' = - sin .\- is negative 1 ' = cos t decrease; y" = - sin t is negative Fig. 2.12 y(s) increases where y' is positive. y(s) bends up where jl"is positive. THE SECOND DERIVATIVES OF THE SINE AND COSINE We now introduce the derivative of the derivative. That is the second derivative of the original function. It tells how fast the slope is changing, not how fast y itself is 2.4 The Derivative of the Sine and Cosine changing. The second derivative is the "rate of change of the velocity." A straight line has constant slope (constant velocity), so its second derivative is zero: f (t) = 5t has df /dt = 5 and d2f /dt2 = 0. The parabola y = x2 has slope 2x (linear) which has slope 2 (constant). Similarly f ( t ) = r a t 2 has df/dt=at and d2f/dt2=a. There stands the notation d2f/dt2 (or d2y/dx2)for the second derivative. A short form is f " or y". (This is pronounced f double prime or y double prime). Example: The second derivative of y = x3 is y" = 6x. In the distance-velocity problem, f " is acceleration. It tells how fast v is changing, while v tells how fast f is changing. Where df/dt was distanceltime, the second derivative is di~tance/(time)T~.he acceleration due to gravity is about 32 ft/sec2 or 9.8 m/sec2,which means that v increases by 32 ftlsec in one second. It does not mean that the distance increases by 32 feet! The graph of y = sin t increases at the start. Its derivative cos t is positive. However the second derivative is -sin t. The curve is bending down while going up. The arch is "concave down" because y" = - sin t is negative. At t = n the curve reaches zero and goes negative. The second derivative becomes positive. Now the curve bends upward. The lower arch is "concave up." y" > 0 means that y' increases so y bends upward (concave up) y" < 0 means that y' decreases so y bends down (concave down). Chapter 3 studies these things properly-here we get an advance look for sin t. The remarkable fact about the sine and cosine is that y" = -y. That is unusual and special: acceleration = -distance. The greater the distance, the greater the force pulling back: + y = sin t has dy/dt = cos t and d2y/dt2= - sin t = - y. y = cos t has dy/dt = - sin t and d y/dt2= - cos t = - y. Question Does d2y/dt2< 0 mean that the distance y(t) is decreasing? Answer No. Absolutely not! It means that dy/dt is decreasing, not necessarily y. At the start of the sine curve, y is still increasing but y" < 0. Sines and cosines give simple harmonic motion-up and down, forward and back, out and in, tension and compression. Stretch a spring, and the restoring force pulls it back. Push a swing up, and gravity brings it down. These motions are controlled by a diyerential equation: + All solutions are combinations of the sine and cosine: y = A sin t B cos t. This is not a course on differential equations. But you have to see the purpose of calculus. It models events by equations. It models oscillation by equation (12). Your heart fills and empties. Balls bounce. Current alternates. The economy goes up and down: high prices -+ high production -,low prices -, -.. We can't live without oscillations (or differential equations). 70 2 ~erhrcrthres 2.4 EXERCISES Read-through questions 11 Find by calculator or calculus: The derivative of y = sin x is y' = a . The second derivative (the b of the derivative) is y" = c . The fourth derivative is y"" = d . Thus y = sin x satisfies the differential equations y" = e and y"" = f . So does y = cos x, whose second derivative is g . a sin 3h 2 1-cos 2h (b) rli-m+o 1-cos h ' 12 Compute the slope at x = 0 directly from limits: (a) y = tan x (b) y =sin (- x) All these derivatives come from one basic limit: (sin h)/h approaches h . The sine of .O1 radians is very close to i . So is the i of .01. The cosine of .O1 is not .99, because 1-cos h is much k than h. The ratio (1 -cos h)/h2 approaches I . Therefore cos h is close to 1-i h 2 and cos .Ol x m . We can replace h by x. The differential equation y" = -y leads to n . When y is positive, y" is o . Therefore y' is P . Eventually y goes below zero and y" becomes q . Then y' is r . Examples of oscillation in real life are s and t . 13 The unmarked points in Figure 2.11 are P and S. Find the height PS and the area of triangle OPR. Prove by areas that sin h < h. 14 The slopes of cos x and 1-i x 2are -sin x and . The slopes of sin x and are cos x and 1-3x2. 15 Chapter 10 gives an infinite series for sin x: From the derivative find the series for cos x. Then take its 1 Which of these ratios approach 1 as h -,O? derivative to get back to -sin x. 16 A centered diference for f (x)=sin x is zih (a) sin2h (b) zzi sin h 7 sin(- h) (a + f (x h) -f (x -h) - sin (x + h) -sin(x -h) = ? 2 h 2 h 2 (Calculator) Find (sin h)/h at h = 0.5 and 0.1 and .01. Where does (sin h)/h go above .99? Use the addition formula (2). Then let h -* 0. 3 Find the limits as h -,0 of + Repeat Problem 16to find the slope of cos x. Use formula to simplify cos(x h) -cos(x -h). sin2h (a) sin 5h (b) sin 5h (c) sin h (dl Find the tangent line to y = sin x at (a) x = 0 (b) x = 11 (c) x = 1114 4 Where does tan h = 1.01h? Where does tan h = h? + Where does y =sin x cos x have zero slope? + 5 y =sin x has period 211, which means that sin x = . The limit of (sin(211 h) -sin 2z)lh is 1 because . This gives dyldx at x = + 6 Draw cos (x Ax) next to cos x. Mark the height differ- ence Ay. Then draw AylAx as in Figure 2.9. 7 The key to trigonometry is cos20 = 1-sin20. Set sin 0 x 0 to find cos20x 1- 02. The square root is cos 0 x 1-30'. Reason: Squaring gives cos20 x and the correction term is very small near 0 = 0. + Find the derivative of sin (x 1) in two ways: + (a) Expand to sin x cos 1 cos x sin 1. Compute dyldx. + (b) Divide Ay = sin (x + 1 Ax) -sin (x + 1)by Ax. Write X instead of x + 1. Let Ax go to zero. Show that (tan h)/h is squeezed between 1 and l/cos h. As h -,0 the limit is . 22 For y = sin 2x, the ratio Aylh is + sin 2(x + h) -sin 2x sin 2x(cos 2h - 1) cos 2x sin 2h 8 (Calculator) Compare cos 0 with 1-302 for (a) 0 = 0.1 (b) 0 = 0.5 (c) 0 = 30" (d) 0 = 3". Explain why the limit dyldx is 2 cos 2x. 9 Trigonometry gives cos 0 = 1-2 sin2$0. The approxima- tion sin 30 x leads directly to cos 0 x 1-)02. 10 Find the limits as h -,0: 23 Draw the graph of y = sin ix. State its slope at x =0, 1112, 11, and 211. Does 3 sin x have the same slopes? f i + 24 Draw the graph of y =sin x cos x. Its maximum value is y = at x = .The slope at that point is . 25 By combining sin x and cos x, find a combination that starts at x =0 from y = 2 with slope 1. This combination also solves y" = . 2.5 The Product and Quotient and Power Rules 71 26 True or false, with reason: (a) The derivative of sin2x is cos2x (b) The derivative of cos (- x) is sin x (c) A positive function has a negative second derivative. (d) If y' is increasing then y" is positive. 27 Find solutions to dyldx =sin 3x and dyldx =cos 3x. 28 If y =sin 5x then y' = 5 cos 5x and y" = -25 sin 5x. So this function satisfies the differential equation y" = 29 If h is measured in degrees, find lim,,, (sin h)/h. You could set your calculator in degree mode. 30 Write down a ratio that approaches dyldx at x =z. For y = sin x and Ax = .O1 compute that ratio. 31 By the square rule, the derivative of ( ~ ( x )is) ~2u duldx. + Take the derivative of each term in sin2x cos2x = 1. 32 Give an example of oscillation that does not come from physics. Is it simple harmonic motion (one frequency only)? 33 Explain the second derivative in your own words. What are the derivatives of x + sin x and x sin x and l/sin x and xlsin x and sinnx? Those are made up from the familiar pieces x and sin x, but we need new rules. Fortunately they are rules that apply to every function, so they can be established once and for all. If we know the separate derivatives of two functions u and v, then the derivatives of u + v and uu and llv and u/u and un are immediately available. This is a straightforwardsection, with those five rules to learn. It is also an important section, containing most of the working tools of differential calculus. But I am afraid that five rules and thirteen examples (which we need-the eyes glaze over with formulas alone) make a long list. At least the easiest rule comes first. When we add functions, we add their derivatives. Sum Rule The derivative of the sum u(x) + v(x) is - d (u + v) = du - + -d.v dx dx dx + EXAMPLE 1 The derivative of x sin x is 1 + cos x. That is tremendously simple, but it is fundamental. The interpretation for distances may be more confusing (and more interesting) than the rule itself: Suppose a train moves with velocity 1. The distance at time t is t. On the train a professor paces back and forth (in simple harmonic motion). His distance from his seat is sin t. Then the total distance from his starting point is t + sin t, and + his velocity (train speed plus walking speed) is 1 cos t. If you add distances, you add velocities. Actually that example is ridiculous, because the professor's maximum speed equals the train speed (= 1). He is running like mad, not pacing. Occasionally he is standing still with respect to the ground. The sum rule is a special case of a bigger rule called "linearity." It applies when we add or subtract functions and multiply them by constants-as in 3x - 4 sin x. By linearity the derivative is 3 - 4 cos x. The rule works for all functions u(x) and v(x). + A linear combination is y(x) = au(x) bv(x), where a and b are any real numbers. Then AylAx is 2 Derivatives The limit on the left is dyldx. The limit on the right is a duJdx+ b dvldx. We are allowed to take limits separately and add. The result is what we hope for: Rule of Linearity The derivative of au(x) + bv(x) is - d (au + bu) = a du - + b -d.v dx dx dx The prorluct rule comes next. It can't be so simple-products are not linear. The sum rule is what you would have done anyway, but products give something new. The krivative of u times v is not duldx times dvldx. Example: The derivative of x5 is 5x4. Don't multiply the derivatives of x3 and x2. (3x2 times 2x is not 5x4.) For a product of two functions, the derivative has two terms. Product Rule (the key to this section) The derivative of u(x)v(x) is - d (uu) = u dv - + v -d.u dx dx dx EXAMPLE 2 u = x3 times v = x2 is uv = x5. The product rule leads to 5x4: EXAMPLE 3 In the slope of x sin x, I don't write dxldx = 1 but it's there: + d - dx (x sin x) = x cos x sin x. EXAMPLE 4 If u = sin x and v = sin x then uv = sin2x. We get two equal terms: + sin x - ddx (sin x) sin x - ddx (sin x) = 2 sin x cos x. This confirms the "square rule" 2u duldx, when u is the same as v. Similarly the slope of cos2x is -2 cos x sin x (minus sign from the slope of the cosine). Question Those answers for sin2x and cos2x have opposite signs, so the derivative + of sin2x cos2x is zero (sum rule). How do you see that more quickly? + + EXAMPLE 5 The derivative of uvw is uvw' uv'w u'vw-one derivative at a time. The derivative of xxx is xx + xx + xx. + + + Fig. 2.13 Change in length =Au Av. Change in area = u Av v Au Au Av. 2.5 The Product and Quotient and Power Rules After those examples we prove the product rule. Figure 2.13 explains it best. The area of the big rectangle is uv. The important changes in area are the two strips u Av and v Au. The corner area Au Av is much smaller. When we divide by Ax, the strips give u Av/Ax and v AulAx. The corner gives Au AvlAx, which approaches zero. Notice how the sum rule is in one dimension and the product rule is in two dimensions. The rule for uvw would be in three dimensions. The extra area comes from the whole top strip plus the side strip. By algebra, + + This increase is u(x h)Av v(x)Au-top plus side. Now divide by h (or Ax) and let h +0. The left side of equation (4) becomes the derivative of u(x)v(x).The right side becomes u(x) times dvldx-we can multiply the two limits-plus v(x) times duldx. That proves the product rule-definitely useful. We could go immediately to the quotient rule for u(x)/v(x). But start with u = 1. The derivative of l/x is - 1/x2(known). What is the derivative of l/v(x)? Reciprocal Rule The derivative of 1 ---- is - dvldx --- 44 u2 - The proof starts with (v)(l/v)= 1. The derivative of 1 is 0. Apply the product rule: "(A)=- ( ddx- )v1 + v1= ddxvO sothat dx v -dvv2ldx ' It is worth checking the units-in the reciprocal rule and others. A test of dimensions is automatic in science and engineering, and a good idea in mathematics. The test ignores constants and plus or minus signs, but it prevents bad errors. If v is in dollars and x is in hours, dv/dx is in dollars per hour. Then dimensions agree: hour and also - - dvvldx w dollars/hour dollar^)^ From this test, the derivative of l/v cannot be l/(dv/dx). A similar test shows that Einstein's formula e = mc2 is dimensionally possible. The theory of relativity might be correct! Both sides have the dimension of (mas~)(distance)~/(timew)h~e,n mass is converted to energy.? EXAMPLE6 The derivatives ofx-', x - ~ x, -" are -1xP2, - Z X - ~ ,-nx-"-I. Those come from the reciprocal rule with v = x and x2 and any xn: The beautiful thing is that this answer -nx-"-' fits into the same pattern as xn. Multiply by the exponent and reduce it by one. For negative and positive exponents the derivative of xn is nxn-l. (7) +But only Einstein knew that the constant is 1. 1 1 -Av A1Reciprocal - v + Au - -v = v(v + Av) Quotient u+Au -- -u -- vAu-uAv Av v + A v v v(v+ Av) AD v Fig. 2.14 Reciprocal rule from (- Av)/v2.Quotient rule from (v Au -u Av)/v2. EXAMPLE 7 The derivatives of - co1s x and - sin1 x are - +cossi n2xx and -.- c o s x sin2 x Those come directly from the reciprocal rule. In trigonometry, l/cos x is the secant of the angle x, and l/sin x is the cosecant of x. Now we have their derivatives: -d(set dx x)= - sin x -- cos2x - co1s x - csoins xx - sec x tan x. -d(CSCX)=--=-c-o-=s -x dx sin2 x 1 cos x sin x sin x csc x cot x. Those formulas are often seen in calculus. If you have a good memory they are worth storing. Like most mathematicians, I have to check them every time before using them (maybe once a year). It is really the rules that are basic, not the formulas. The next rule applies to the quotient u(x)/v(x).That is u times llv. Combining the product rule and reciprocal rule gives something new and important: Quotient Rule The derivative of u(x) - is 1 - du - - u - dvldx -- v duldx - u dvldx u(x) vdx v2 v2 You must memorize that last formula. The v2 is familiar. The rest is new, but not very new. If v = 1 the result is duldx (of course). For u = 1 we have the reciprocal + + rule. Figure 2.14b shows the difference (u Au)/(v Av) - (ulv). The denominator + V(V Av) is responsible for v2. EXAMPLE8 (only practice) If u/v = x5/x3(which is x2) the quotient rule gives 2x: EXAMPLE9 (important) For u = sin x and v = cos x, the quotient is sin xlcos x = + tan x. The derivative of tan x is sec2x. Use the quotient rule and cos2x sin2x = 1: cos x(cos x) - sin x(- sin x) ----1 - sec2x. c0s2X c0s2X (11) Again to memorize: (tan x)' = sec2x. At x = 0, this slope is 1. The graphs of sin x and x and tan x all start with this slope (then they separate). At x = n/2 the sine curve is flat (cos x = 0) and the tangent curve is vertical (sec2x = co). The slope generally blows up faster than the function. We divide by cos x, once for the tangent and twice for its slope. The slope of l/x is -l/x2. The slope is more sensitive than the function, because of the square in the denominator. EXAMPLE 10 -d -sin-x x cos x - sin x dx(x)- x2 2.5 The Product and -dent and Power Rules That one I hesitate to touch at x = 0. Formally it becomes 010. In reality it is more like 03/02,and the true derivativeis zero. Figure 2.10 showed graphically that (sin x)/x is flat at the center point. The function is even (symmetric across the y axis) so its derivative can only be zero. This section is full of rules, and I hope you will allow one more. It goes beyond xn to (u(x)r. A power of x changes to a power of u(x)-as in (sin x ) ~or (tan x)' or + (x2 I)*. The derivative contains nun-' (copying nxn-'), but there is an extra factor + duldx. Watch that factor in 6(sin x)' cos x and 7(tan x ) s~ec2x and 8(x2 l)'(2x): Power Rule The derivative of [u(x)In is n[~(x)]~-' ; dui; For n = 1 this reduces to du/dx = duldx. For n = 2 we get the square rule 2u duldx. Next comes u3.The best approach is to use mathematical induction, which goes from + each n to the next power n 1 by the product rule: + That is exactly equation (12)for the power n 1. We get all positive powers this way, going up from n = 1-then the negative powers come from the reciprocal rule. Figure 2.15 shows the power rule for n = 1,2,3. The cube makes the point best. The three thin slabs are u by u by Au. The change in volume is essentially 3u2Au. From multiplying out ( ~ + A u ) ~th,e exact change in volume is 3u2Au + ~ u ( A u+) ~(A~)~-whichalso accounts for three narrow boxes and a midget cube in the corner. This is the binomial formula in a picture. U(AU)* 3 bricks u Au u Au u Au Fig. 2.15 Length change =Au; area change x 21.4Au; volume change x 3u2Au. u2 AU 3 slabs ' EXAMPLE 11 -d (sin x)" = n(sin x)"- cos x. The extra factor cos x is duldx. dx Our last step finally escapes from a very undesirable restriction-that n must be ' a whole number. We want to allow fractional powers n = p/q, and keep the same formula. The derivative of xn is still nxnTo deal with square roots I can write (&)' = x. Its derivative is 2&(&)' = 1. Therefore (&)' is 1/2& which fits the formula when n = f. Now try n = p/q: 2 Derivatives Fractional powers Write u = xPIqas uq= xP. Take derivatives, assuming they exist: ' qU4-1 - du = pxpdx (power rule on both sides) -du---px-' dx qu-' (cancel xP with uq) du - = n x n - 1 dx (replace plq by n and u by xn) EXAMPLE 12 The slope of x'I3 is ~ x - ~ ITh~e. slope is infinite at x = 0 and zero at x = a.But the curve in Figure 2.16 keeps climbing. It doesn't stay below an "asymptote." 1;s 1 118 I Fig. 2.16 Infinite slope of xn versus zero slope: the difference between 0 < n < 1 and n > 1. EXAMPLE 13 The slope of x4I3is 4x'I3. The slope is zero at x = 0 and infinite at x = co.The graph climbs faster than a line and slower than a parabola (4 is between 1 and 2). Its slope follows the cube root curve (times j). WE STOP NOW! I am sorry there were so many rules. A computer can memorize them all, but it doesn't know what they mean and you do. Together with the chain rule that dominatesChapter 4, they achievevirtually all the derivativesever computed by mankind. We list them in one place for convenience. Rule of Linearity Product Rule Reciprocal Rule Quotient Rule Power Rule + + (au bv)' = au' bv' + (uv)' = ud VU' (Ilv)' = - v'/v2 (ulv)' = (vu' - uv')/v2 (un)'= nu''-'u' ', The power rule applies when n is negative, or a fraction, or any real number. The derivative of x" is zx"- according to Chapter 6. The derivative of (sin x)" is . And the derivatives of all six trigonometric functions are now established: (sin x)' = cos x (tan x)' = sec2x (sec x)' = sec x tan x (COSx)' = - sin x (cot x)' = - csc2x (csc x)' = - csc x cot x . 2.5 The Product and Quotient and Pwer Rules 77 2.5 EXERCISES Read-through questions The derivatives of sin x cos x and l/cos x and sin x/cos x and tan3x come from the a rule, b rule, c rule, and d rule. The product of sin x times cos x has + (uv)' = uv' e = 1 . The derivative of l / v is g , so the slope of sec x is h . The derivative of u/v is 1 , so the slope of tan x is I . The derivative of tan3x is k . The slope of xn is I and the slope of (~(x))i"s m . With n = -1 the derivative of (cos x)-' is n , which agrees with the rule for sec x. Even simpler is the rule of 0 , which applies to + au(x)+ bv(x). The derivative is P . The slope of 3 sin x + 4 cos x is q . The derivative of (3 sin x 4 cos x ) ~is r . The derivative of s is 4 sin3x cos x. Find the derivatives of the functions in 1-26. (X- 1)(x-2)(x - 3) + x2 cos x 2x sin x x3+ 1 +x 1 + csoins xx x1I2 sin2x + (sin x)'I2 + x4cos x x C O S x~ 3x2sinx-xcosx+sinx sec2x -tan2x 6 (X- 1 ) 2 (-~2)2 + 8 x'I2(x sin x) x2+1 sinx lo-+- x2- 1 COS X 12 x3I2sin3x + (sin x ) ~ / ~ + + 14 &(& l)(& 2) 16 ( ~ - 6 ) ' ~ + s i n ' ~ x 18 csc2x- cot2x sin x - cos x 20 sin x + cos x 30 A cylinder has radius r = 312 -and height h = - t 1 +t3I2 1+ t ' (a) What is the rate of change of its volume? (b) What is the rate of change of its surface area (including top and base)? 31 The height of a model rocket is f (t) = t3/(l + t). (a) What is the velocity v(t)? (b) What is the acceleration duldt? 32 Apply the product rule to u(x)u2(x)to find the power rule for u3(x). 33 Find the second derivative of the product u(x)v(x). Find the third derivative. Test your formulas on u = u =x. 34 Find functions y(x) whose derivatives are (a) x3 (b) l/x3 (c) (1 -x ) ~ (d~) co~s2x sin x. 35 Find the distances f (t), starting from f (0)=0, to match these velocities: (a) v(t) =cos t sin t (c) v(t)=Jl+t (b) v(t)=tan t sec2t 36 Apply the quotient rule to ( ~ ( x ) ) ~ / ( u ( xa)n)d~ -u'/v2. The latter gives the second derivative of -. 37 Draw a figure like 2.13 to explain the square rule. 38 Give an example where u(x)/u(x)is increasing but du/dx = dvldx = 1. 39 True orfalse, with a good reason: (a) The derivative of x2" is 2nx2"-'. + (b) By linearity the derivative of a(x)u(x) b(x)u(x) is a(x)du/dx.+ b(x)dvldx. (c) The derivative of 1xI3 is 31xI2. (d) tan2x and sec2x have the same derivative. (e) (uv)' = u'u' is true when u(x) = 1. --1 - 1 tan x cot x 26 x sin x + cos x + A growing box has length t, width 1/(1 t), and height COS t. (a) What is the rate of change of the volume? (b) What is the rate of change of the surface area? 28 With two applications of the product rule show that the derivative of uvw is uvw' + uv'w + u'uw. When a box with sides u, v, wgrows by Au, Av, Aw, three slabs are added with volume uu Aw and and . 29 Find the velocity if the distance is f (t)= 5t2 for t < 10, 500 + loo,/= for t 2 10. 40 The cost of u shares of stock at v dollars per share is uv dollars. Check dimensions of d(uv)/dtand u dv/dt and v duldt. 41 If u(x)/v(x)is a ratio of polynomials of degree n, what are the degrees for its derivative? + 42 For y = 5x 3, is ( d y / d ~t)h~e same as d 2 y / d ~ 2 ? 43 If you change from f (t)= t cos t to its tangent line at t = 7112, find the two-part function df /dt. 44 Explain in your own words why the derivative of u(x)v(x) has two terms. 45 A plane starts its descent from height y = h at x = -L to land at (0,O). Choose a, b, c, d so its landing path + + + y =ax3 bx2 cx d is smooth. With dx/dt = V =constant, find dyldt and d2y/dt2 at x = 0 and x = -L. (To keep d2y/dt2 small, a coast-to-coast plane starts down L > 100 miles from the airport.) You have seen enough limits to be ready for a definition. It is true that we have survived this far without one, and we could continue. But this seems a reasonable time to define limits more carefully. The goal is to achieve rigor without rigor mortis. First you should know that limits of Ay/Ax are by no means the only limits in mathematics. Here are five completely different examples. They involve n + a,not Ax +0: 1. a, = (n - 3)/(n + 3) (for large n, ignore the 3's and find a, + 1) 2. a, = )a,-, + 4 (start with any a, and always a, + 8) 3. an= probability of living to year n (unfortunately an+0) 4. a, = fraction of zeros among the first n digits of n (an+h?) 5. a, = .4, a2= .49, a, = .493, .... No matter what the remaining decimals are, the a's converge to a limit. Possibly a, + .493000 ...,but not likely. The problem is to say what the limit symbol + really means. A good starting point is to ask about convergence to zero. When does a sequence of positive numbers approach zero? What does it mean to write an+O? The numbers a,, a,, a,, ..., must become "small," but that is too vague. We will propose four definitions of convergence to zero, and I hope the right one will be clear. 1. All the numbers a, are below 10- lo. That may be enough for practical purposes, but it certainly doesn't make the a, approach zero. 2. The sequence is getting closer to zero-each a,, is smaller than the preceding a,. This test is met by 1.1, 1.01, 1.001, ... which converges to 1 instead of 0. 3. For any small number you think of, at least one of the an's is smaller. That pushes something toward zero, but not necessarily the whole sequence. The condition would be satisfied by 1, ), 1,f, 1, i,...,which does not approach zero. 4. For any small number you think of, the an's eventually go below that number and stay below. This is the correct definition. I want to repeat that. To test for convergence to zero, start with a small number- say 10-lo. The an's must go below that number. They may come back up and go below again-the first million terms make absolutely no difference. Neither do the next billion, but eventually all terms must go below lo-''. After waiting longer (possibly a lot longer), all terms drop below The tail end of the sequence decides everything. Question 1 Doesthesequence lo-,, 1 0 - ~ , 1 0 - ~ , ...approacho? Answer Yes. These up and down numbers eventually stay below any E. a,<~ifn>3 a,,< E if n > 6 non-convergence Fig. 2.17 Convergence means: Only a finite number of a's are outside any strip around L. 2.6 Limits Question 2 Does lo-', lo-*, lo-',, 10-lo, ... approach zero? Answer No. This sequence goes below but does not stay below. There is a recognized symbol for "an arbitrarily small positive number." By worldwide agreement, it is the Greek letter E (epsilon). Convergence to zero means that the sequence eventuallygoes below E and stays there. The smaller the E,the tougher the test and the longer we wait. Think of E as the tolerance, and keep reducing it. To emphasize that E comes from outside, Socrates can choose it. Whatever E he proposes, the a's must eventually be smaller. After some a,, all the a's are below the tolerance E. Here is the exact statement: for any E there is an N such that a, < E if n > N. Once you see that idea, the rest is easy. Figure 2.17 has N = 3 and then N = 6. EXAMPLE I The sequence f ,$, 8, ... starts upward but goes to zero. Notice that 1,4,9, ...,100, ...are squares, and 2,4, 8, ...,1024, ... are powers of 2. Eventually 2" grows faster than n2, as in alo = 100/1024. The ratio goes below any E. EXAMPLE 2 1, 0, f ,0, f ,0, ... approaches zero. These a's do not decrease steadily (the mathematical word for steadily is monotonica ally") but still their limit is zero. The choice E = 1 / 1 0 produces the right response: Beyond azoolall terms are below 1/1000. So N = 2001 for that E. The sequence 1,f ,f ,4,f ,f,...is much slower-but it also converges to zero. Next we allow the numbers a, to be negative as well as positive. They can converge upward toward zero, or they can come in from both sides. The test still requires the a, to go inside any strip near zero (and stay there). But now the strip starts at -E. The distancefrom zero is the absolute value la,l. Therefore a, -,0 means lanl+0. The previous test can be applied to lanl: for any E there is an N such that la,l < E if n > N. EXAMPLE 3 1, - f ,f ,- 4,...convergesto zerobecause 1,f ,f ,$, ...convergesto zero. It is a short step to limits other than zero. The limit is L if the numbers a, - L converge to Zero. Our final test applies to the absolute value la, - LI: for any E there is an N such that (a, - L(< E if n > N. This is the definition of convergence! Only a finite number of a's are outside any strip around L (Figure 2.18). We write a, -,L or lim -a,= L or limn,, a, = L. Fig. 2.18 a, -,0in Example 3;a, -* 1 in Example 4;a, -,rn in Example 5 (but a,,, -a, -,0). EXAMPLE 4 The numbers 3, 2 , g, ... converge to L = 1. After subtracting 1 the k, differences 3, f , ... converge to zero. Those difference are la, - LI. The distance between terms is getting smaller. But those numbers a,, a,, a3,a,, ... go + 4, past any proposed limit L. The second term is 15. The fourth term adds on 3 & + + so a, goes past 2. The eighth term has four new fractions 4 + f $, totaling more than $ + $ + $ + & = 3. Therefore a, exceeds 23. Eight more terms will add more than 8 times &, so a,, is beyond 3. The lines in Figure 2 . 1 8 ~are infinitely long, not stopping at any L. In the language of Chapter 10, the harmonic series 1 + 3 + 3 + does not converge. The sum is infinite, because the "partial sums" a, go beyond every limit L (a,,,, is past L = 9). We will come back to infinite series, but this example makes a subtle point: The steps between the a, can go to zero while still a, -,a. Thus the condition a,+, - a, -,0 is not suficient for convergence. However this condition is necessary. If we do have convergence, then a,,, - a, -,0. That is a good exercise in the logic of convergence, emphasizing the difference between "sufficient" and "necessary." We discuss this logic below, after proving that [statement A ] implies [statement B]: , If [a, converges to L] then [a,+ - a, converges to zero]. (1) Proof Because the a, converge, there is a number N beyond which (a, - L( < s and , also la, + - LI < E. Since a, +, - a, is the sum of a, +, - L and L - a,, its absolute + , value cannot exceed E E = 2s. Therefore a,+ - a, approaches zero. , Objection by Socrates: We only got below 2s and he asked for s. Our reply: If he particularly wants la, + - a, 1 < 1/10, we start with s = 1/20. Then 2s = 1/10. But this juggling is not necessary. To stay below 2s is just as convincing as to stay below s. THE LOGIC OF "IF" AND "ONLY IF" The following page is inserted to help with the language of mathematics. In ordinary language we might say "I will come if you call." Or we might say "I will come only if you call." That is different! A mathematician might even say "I will come if and only if you call." Our goal is to think through the logic, because it is important and not so fami1iar.t Statement A above implies statement B. Statement A is a, -,L; statement B is a,+, - a, -,0. Mathematics has at least five ways of writing down A => B, and I though you might like to see them together. It seems excessive to have so many expressions for the same idea, but authors get desperate for a little variety. Here are the five ways that come to mind: A implies B if A then B A is a suflcient condition for B B is true if A is true ?Logical thinking is much more important than E and 6. EXAMPLES If [positive numbers are decreasing] then [they converge to a limit]. + If [sequences a, and b, converge] then [the sequence a, b, converges]. If [f (x) is the integral of v(x)] then [v(x) is the derivative of f (x)]. Those are all true, but not proved. A is the hypothesis, B is the conclusion. Now we go in the other direction. (It is called the "converse," not the inverse.) We exchange A and B. Of course stating the converse does not make it true! B might imply A, or it might not. In the first two examples the converse was false-the a, can converge without decreasing, and a, + b, can converge when the separate sequences do not. The converse of the third statement is true-and there are five more ways to state it: A* B A is implied by B i f B then A A is a necessary condition for B B is true only i f A is true Those words "necessary" and "sufficient" are not always easy to master. The same is true of the deceptively short phrase "if and only if." The two statements A* B and A e B are completely different and they both require proof. That means two separate proofs. But they can be stated together for convenience (when both are true): A-B A implies B and B implies A A is equivalent to B A is a necessary and suficient condition for B - - - EXAMPLES [a, +L] A is true if and only i f B is true [2an-,2L] [a, + 1 + L + 11 [a, - L+ 01. RULES FOR LIMITS Calculus needs a definition of limits, to define dyldx. That derivative contains two limits: Ax +0 and AylAx + dyldx. Calculus also needs rulesfor limits, to prove the sum rule and product rule for derivatives. We started on the definition, and now we start on the rules. Given two convergent sequences, a, + L and b, + M, other sequences also converge: + Addition: a, + b, + L M Subtraction: a, - b, -,L - M Multiplication: a,b, -,LM Division: a,/b, + LIM (provided M # 0) We check the multiplication rule, which uses a convenient identity: a,b, - LM = (a, - L)(b, - M) + M(a, - L) + L(b, - M). (2) Suppose Jan- LJ< E beyond some point N, and 1b, - MI < E beyond some other point N'. Then beyond the larger.of N and N', the right side of (2) is small. It is less than + + E E ME LE.This proves that (2) gives a,b, +LM. An important special case is can-,cL. (The sequence of b's is c, c, c, c, ....) Thus a constant can be brought "outside" the limit, to give lim can= c lim a,. THE LIMIT OF f ( x ) AS x -,a The final step is to replace sequences by functions. Instead of a,, a2, ... there is a continuum of values f(x). The limit is taken as x approaches a specified point a (instead of n -,co). Example: As x approaches a = 0, the function f (x) = 4 - x2 approaches L = 4. As x approaches a = 2, the function 5x approaches L = 10. Those statements are fairly obvious, but we have to say what they mean. Somehow it must be this: i f x is close to a thenf (x) is close to L. If x - a is small, then f (x) - L should be small. As before, the word small does not say everything. We really mean "arbitrarily small," or "below any E." The difference f(x) - L must become as small as anyone wants, when x gets near a. In that case lim,,, f (x)= L. Or we write f (x)-,L as x -,a. The statement is awkward because it involves two limits. The limit x +a is forcing f (x) + L. (Previously n + co forced a, + L.) But it is wrong to expect the same E in both limits. We do not and cannot require that Jx- a1 < E produces )f (x)- LI < E. It may be necessary to push x extremely close to a (closer than E).We must guarantee that if x is close enough to a, then If (x)- LI < E. We have come to the "epsilon-deltadefinition" of limits. First, Socrates chooses E. He has to be shown that f (x) is within E of L, for every x near a. Then somebody else (maybe Plato) replies with a number 6. That gives the meaning of "near a." Plato's goal is to get f(x) within E of L, by keeping x within 6 of a: if 0< lx - a1 < S then (f(x)- LI < E. (3) The input tolerance is 6 (delta), the output tolerance is E. When Plato can find a 6 for every E,Socrates concedes that the limit is L. EXAMPLE Prove that lim 5x = 10. In this case a = 2 and L = 10. x+2 Socrates asks for 15x - 101< E. Plato responds by requiring Ix - 21 < 6. What 6 should he choose? In this case 15x - 101is exactly 5 times Jx- 21. So Plato picks 6 below ~ / 5 (a smaller 6 is always OK). Whenever J x- 21 < 45, multiplication by 5 shows that 15x - 101< E. Remark 1 In Figure 2.19, Socrates chooses the height of the box. It extends above and below L, by the small number E. Second, Plato chooses the width. He must make the box narrow enough for the graph to go out the sides. Then If (x)- Ll< E. 1 limit L is notf ( o ) f ( x ) = step function I I Fig. 2.19 S chooses height 2.5, then P chooses width 26. Graph must go out the sides. When f(x) has a jump, the box can't hold it. A step function has no limit as x approaches thejump, because the graph goes through the top or bottom-no matter how thin the box. Remark 2 The second figure has f (x)+L, because in taking limits we ignore the Jinalpoint x = a. The value f (a) can be anything, with no effect on L. The first figure has more: f (a)equals L. Then a special name applies- f is continuous.The left figure shows a continuous function, the other figures do not. We soon come back to continuous functions. Remark 3 In the example with f = 5x and 6 = 45, the number 5 was the slope. That choice barely kept the graph in the box-it goes out the corners. A little narrower, say 6 = ~110,and the graph goes safely out the sides. A reasonable choice is to divide E by 21ff(a)l. (We double the slope for safety.) I want to say why this 6 works-even if the E-6 test is seldom used in practice. The ratio off (x) - L to x - a is distance up over distance across. This is Af/Ax, close to the slope f'(a). When the distance across is 6, the distance up or down is near 61ff(a)l. That equals ~ / f2or our "reasonable choice" of 6-so we are safely below E. This choice solves most exercises. But Example 7 shows that a limit might exist even when the slope is infinite. EXAMPLE 7 lim ,/x - 1= 0 (a one-sided limit). x+1+ Notice the plus sign in the symbol x + 1+. The number x approaches a = 1 onlyfrom above. An ordinary limit x + 1 requires us to accept x on both sides of 1 (the exact value x = 1 is not considered). Since negative numbers are not allowed by the square root, we have a one-sided limit. It is L = 0. Suppose E is 1/10. Then the response could be 6 = 1/100. A number below 1/100 has a square root below 1/10. In this case the box must be made extremely narrow, 6 much smaller than E,because the square root starts with infinite slope. Those examples show the point of the 6-6 definition. (Given E, look for 6. This came from Cauchy in France, not Socrates in Greece.) We also see its bad feature: The test is not convenient. Mathematicians do not go around proposing 8's and replying with 8's. We may live a strange life, but not that strange. It is easier to establish once and for all that 5x approaches its obvious limit 5a. ' The same is true for other familiar functions: xn+an and sin x +sin a and (1 - x)- -t (1 - a)- '-except at a = 1. The correct limit L comes by substituting x = a into thefunction. This is exactly the property of a "continuousfunction." Before the section on continuous functions, we prove the Squeeze Theorem using E and 6. Proof g(x)is squeezed between f (x)and h(x). After subtracting L, g(x)- L is between f (x)- L and h(x)- L. Therefore Ig(x) - LI < E if If(x) - L(< E and Ih(x)- LJ< E. For any E,the last two inequalities hold in some region 0 < Jx- a1 < 6. So the first one also holds. This proves that g(x) +L. Values at x = a are not involved-until we get to continuous functions. 84 2 Derivatives 2.6 EXERCISES Read-through questions The limit of a, = (sin n)/n is a . The limit of a, = n4/2" is b . The limit of a, = (- I)"is c . The meaning of a, -+ 0 is: Only d of the numbers la,/ can be e . The meaning of a, -+ L is: For every f there is an g such that h ifn> i .Thesequencel,l+$,l+$+~,...isnot i because eventually those sums go past k . "5 If the sequence a,, a,, a,, ... approaches zero, prove that we can put those numbers in any order and the new sequence still approaches zero. + + *6 Suppose f (x) -+ L and g(x) -,M as x -t a. Prove from the definitions that f (x) g(x)-,L M as x -,a. Find the limits 7-24 if they exist. An E-6 test is not required. The limit of f ( x ) = sin x as x a -+ is I . The limit of f(x)=x/lxlasx-+-2is m ,but thelimitasx+Odoes not n . This function only has o -sided limits. The meaning of lirn,,, f (x)= L is: For every E there is a 6 such that I f (x)- LI < E whenever P . 7 lirn t+2 - t + 3 t2-2 9 lim f (X+ h) -f (4 X-~O h Two rules for limits, when a, -+ L and b, -+ M, are + u, h, -+ q and a,b, -+ r . The corresponding rules for functions, when f(x) -+ L and g(x)-+ M as x -+a, are s and t . In all limits, la, - LI or If (x)- LI must eventually go below and u any positive v . - A * B means that A is a w condition for B. Then B is true x A is true. A B means that A is a Y condition for B. Then B is true z A is true. 1 What is u, and what is the limit L? After which N is la, - LI < &?(Calculator allowed) $+a+&, (a) -1, + f , -f , ... (b) 4,++$, ... (c) i,$, i,... a n = n / 2 " (d) 1.1, 1.11, 1.111, ... r (e) a,, =/ ; n (f) ~ , = , / ' ~ - n + (g) 1 1, (1 +4I2, (1 + f ) 3 , ... 2 Show by example that these statements are false: (a) If a, -,L and h, -+ L then a,/b, 1 -+ (b) u, -+ L if and only if a: -+ L~ (c) If u, < 0 and a, -+ L then L < 0 (d) If infinitely many an's are inside every strip around zero then a, -+ 0. 3 Which of these statements are equivalent to B = A? (a) If A is true so is B (b) A is true if and only if B is true (c) B is a sufficient condition for A (d) A is a necessary condition for B. 4 Decide whether A B or B * A or neither or both: (a) A = [a, -+ 11 B = [-a, -+ - 11 (b) A =[a, -+0] B = [a,-a,-, -01 (c) A = [a, < n] B = [a, = n] (d) A = [a, -,O] B = [sin a, -+ 0) (e) A = [a, -+ 01 B = [lla, fails to converge] (f) A = [a, < n] B = [a,/n converges] sin2h cos2h 11 lirn h+O h2 13 lim+ x+o 1x1 - x (one-sided) sin x 15 lirn x-+l - x x2 + 25 17 lirn x-+5 -x--- 5 - 2x tan x 12 lirn ---- X + O sin x 14 lirn I x I - (one-sided) x-0- X JI+x-1 J4-x 19 lim (test x = .01) 20 lim ---- x+o Y x42 21 lim [f(x)-f(a)](?) x-+a 22 lim (sec x - tan x) x+42 sin x 23 lirn - X + O sin x/2 24 lim sin (x - 1) x-tl x2-1 25 Choose 6 so that If(.x)l < Aif 0 < x < 6. 26 Which does the definition of a limit require? - (1) I f ( x - ) - L l < ~ = O < I x - a ( < 6 (2) I f ( x ) - L l < ~ = O r l x - a l < G (3) If(x)- LI < E 0~I.x-a1< 6 27 The definition of "f(x) -+ L as x -+ x" is this: For any E there is an X such that < E if x > X. Give an example in which f (x)3 4 as x -+ rrc . 28 Give a correct definition of ''f(.x) -+ 0 as x -,- x'." 29 The limit of f(x) =(sin x)/x as x -+x is E = .O1 find a point X beyond which If(x)l < E. . For + 30 The limit of f (x)= 2x/(l x) as x -+ rx is L = 2. For t: = .O1 find a point X beyond which If ( x )- 21 < E. 31 The limit of , f ( s )= sin s as s -+ r_ does not exist. Explain why not. 2.7 Continuous Functions 85 ( :r+ 32 (Calculator) Estimate the limit of 1 - as x + a. + 33 For the polynomial f (x)=2x - 5x2 7x3 find (c) lirn x-im f- x(34 (d) lirn x4-00 - fx(34 34 For f (x)=6x3+ l00Ox find (a) lirn - f (x) x+m X (c) lirn x-rm - fx(44 + (d) lirn x4m - f ( 4 x3 1 Important rule As x + co the ratio of polynomials f(x)/g(x) has the same limit as the ratio of their leading terms. f (x)= + + + x3-x 2 has leading term x3 and g(x)= 5x6 x 1 has leading term 5x6.Therefore f (x)/g(x)behaves like x3/5x6+0, g(x)/f (x) behaves like 5x6/x3+ a,(f ( x ) ) ~ / ~ (bxeh) aves like x6/5x6 115. 38 If a, -+ L prove that there is a number N with this property: If n > N and m > N then (a, -a,( < 2 ~T.his is Cauchy's test for convergence. 39 No matter what decimals come later, a l = .4, a2= .49, a, = .493, ... approaches a limit L. How do we know (when we can't know L)? Cauchy's test is passed: the a's get closer to each other. (a) From a, onwards we have la, - aml< (b) After which a, is lam- a,l < 40 Choose decimals in Problem 39 so the limit is L = .494. Choose decimals so that your professor can't find L. 41 If every decimal in .abcde-.. is picked at random from 0, 1, ...,9, what is the "average" limit L? 42 If every decimal is 0 or 1(at random), what is the average limit L? + 43 Suppose a, =$an- 4 and start from al = 10. Find a2 and a, and a connection between a, - 8 and a,-, - 8. Deduce that a, -,8. 35 Find the limit as x + co if it exists: + 3x2+ 2 x + 1 x4 x2 1000 3 + 2 x + x 2 x3+x2 x3-1000 x sin -1. x 44 "For every 6 there is an E such that If (x)] N, then b,+ L. 37 The sum of 1+ r +r2 + ..- + r"-' is a, =(1 -r")/(l -r). What is the limit of a, as n -,co? For which r does the limit exist? 46 Explain in 110 words the difference between "we will get there if you hurry" and "we will get there only if you hurry" and "we will get there if and only if you hurry." 1 2.7 Continuous Functions - 1 This will be a brief section. It was originally included with limits, but the combination was too long. We are still concerned with the limit off (x) as x -,a, but a new number is involved. That number is f (a), the value off at x = a. For a "limit," x approached a but never reached it-so f(a) was ignored. For a "continuous function," this final number f (a) must be right. May I summarize the usual (good) situation as x approaches a? 1. The number f (a) exists (f is defined at a) 2. The limit of f (x) exists (it was called L) 3. The limit L equals f (a) (f (a) is the right value) In such a case, f (x) is continuous at x = a. These requirements are often written in a single line: f (x) +f (a) as x -,a. By way of contrast, start with four functions that are not continuous at x = 0. Fig. 2.20 Four types of discontinuity (others are possible) at x =0. In Figure 2.20, the first function would be continuous if it had f (0)= 0. But it has f (0) = 1.After changingf (0) to the right value, the problem is gone. The discontinuity is removable. Examples 2, 3, 4 are more important and more serious. There is no "correct" value for f (0): 2. f (x) = step function (jump from 0 to 1 at x = 0) 3. f (x)= 1/x2 (infinite limit as x +0) 4. f (x)= sin (1/x) (infinite oscillation as x +0). The graphs show how the limit fails to exist. The step function has a jump discontinuity. It has one-sided limits, from the left and right. It does not have an ordinary (twosided) limit. The limit from the left (x +0-) is 0. The limit from the right (x +0') is 1. Another step function is x/lxl, which jumps from -1 to 1. In the graph of l/x2, the only reasonable limit is L= + co. I cannot go on record as saying that this limit exists. Officially, it doesn't-but we often write it anyway: l/x2+ m as x +0. This means that l/x2 goes (and stays) above every L as x +0. In the same unofficial way we write one-sided limits for f (x)= l/x: + From the left, lim x+o- 1 -x = - co. From the right, lim 1 -= x+o+ X oo. (1) Remark l/x has a "pole" at x = 0. So has l/x2 (a double pole). The function l/(x2- X)has poles at x = 0 and x = 1. In each case the denominator goes to zero and the function goes to + oo or - oo. Similarly llsin x has a pole at every multiple of n (where sin x is zero). Except for l/x2these poles are "simplew-the functions are completely smooth at x = 0 when we multiply them by x: (x)(!-) =1 and (x) (A) 1 and ( ) are continuousat x =0. l/x2 has a double pole, since it needs multiplication by x2 (not just x). A ratio of + + polynomials P(x)/Q(x) has poles where Q = 0, provided any common factors like (X 1)/(x 1)are removed first. Jumps and poles are the most basic discontinuities, but others can occur. The fourth graph shows that sin(l/x) has no limit as x +0. This function does not blow up; the sine never exceeds 1. At x = 4 and $ and & it equals sin 3 and sin 4 and sin 1000. Those numbers are positive and negative and (?). As x gets small and l/x gets large, the sine oscillates faster and faster. Its graph won't stay in a small box of height E, no matter how narrow the box. CONTINUOUS FUNCTIONS DEFINITION f is "continuous at x = a" if f (a) is defined and f (x) 4f (a) as x -,a. Iff is continuous at every point where it is defined, it is a continuousfunction. 2.7 Continuous FuncHons 87 Objection The definition makes f(x)= 1/x a continuous function! It is not defined at x = 0, so its continuity can't fail. The logic requires us to accept this, but we don't have to like it. Certainly there is no f(0) that would make 1lx continuous at x = 0. It is amazing but true that the definition of "continuous function" is still debated (Mathematics Teacher, May 1989). You see the reason-we speak about a discontinuity of l/x, and at the same time call it a continuous function. The definition misses the difference between 1/x and (sin x)/x. The function f(x) = (sin x)/x can be made continuousat all x. Just set f(0) = 1. We call a function "continuable'iif its definition can be extended to all x in a way that makes it continuous. Thus (sin x)/x and \/; are continuable. The functions l/x and tan x are not continuable. This suggestion may not end the debate, but I hope it is helpful. EXAMPLE sin x and cos x and all polynomials P(x) are continuous functions. EXAMPL2E The absolute value Ixl is continuous. Its slope jumps (not continuable). EXAMPLE3 Any rational function P(x)/Q(x) is continuous except where Q = 0. EXAMPLE4 The function that jumps between 1 at fractions and 0 at non-fractions is discontinuous everywhere. There is a fraction between every pair of non-fractions and vice versa. (Somehow there are many more non-fractions.) EXAMPLE5 The function 02 is zero for every x, except that 00 is not defined. So define it as zero and this function is continuous. But see the next paragraph where 00 has to be 1. We could fill the book with proofs of continuity, but usually the situation is clear. "A function is continuous if you can draw its graph without lifting up your pen." At a jump, or an infinite limit, or an infinite oscillation, there is no way across the discontinuity except to start again on the other side. The function x" is continuous for n > 0. It is not continuable for n < 0. The function x0 equals 1 for every x, except that 00 is not defined. This time continuity requires 00 = 1. The interesting examples are the close ones-we have seen two of them: sin x 1 -cos x EXAMPL6E and are both continuable at x = 0. x x Those were crucial for the slope of sin x. The first approaches 1 and the second approaches 0. Strictly speaking we must give these functions the correct values (1 and 0) at the limiting point x = O-which of course we do. It is important to know what happens when the denominators change to x2. EXAMPLE7 sin x blows up but 1 -cos x has the limit 1 at x = 0. X2 2 Since (sin x)/x approaches 1, dividing by another x gives a function like 1lx. There is a simple pole. It is an example of 0/0, in which the zero from x2 is reached more quickly than the zero from sin x. The "race to zero" produces almost all interesting problems about limits. _ I·_I _ _·___ __ · _ �_·_I__ ____ 2 Derivatives For 1 - cos x and x2 the race is almost even. Their ratio is 1 to 2: 1 - cos x - 1 - cos2x ---s.in2x 1 1 -+- as x -+ 0. x2 x2(1+c0sx) x2 ~ + C O S X 1 + 1 This answer will be found again (more easily) by "1'HBpital's rule." Here I emphasize not the answer but the problem. A central question of differential calculus is to know how fast the limit is approached. The speed of approach is exactly the information in the derivative. These three examples are all continuous at x = 0. The race is controlled by the slope-because f (x) -f (0) is nearly f '(0) times x: - derivative of sin x is 1 - derivative of sin2x is 0 - derivative of xli3 is CQ sin x decreases like x sin2x decreases faster than x x1I3decreases more slowly than x. DIFFERENTIABLE FUNCTIONS The absolute value 1x1 is continuous at x = 0 but has no derivative. The same is true for x113. Asking for a derivative is more than asking for continuity. The reason is fundamental, and carries us back to the key definitions: Continuous at x: Derivative at x: + f (x Ax) -f ( x ) -+ 0 as Ax -+ 0 + f (x A.u) -f ( x ) Ax f -+ "(x) as Ax -+ 0. In the first case, Af goes to zero (maybe slowly). In the second case, Af goes to zero as fast as Ax (because AflAx has a limit). That requirement is stronger: 21 At a point where f(x) has a derivative, the function must be continuous. But f (x) can be continuous with no derivative. + Proof The limit of Af = (Ax)(Af/Ax) is (O)(df/dx)= 0. So f (x Ax) -f (x) -+ 0. The continuous function x113 has no derivative at x = 0, because +xw2I3blows up. The absolute value 1x1 has no derivative because its slope jumps. The remarkable 4 + + function cos 3x cos 9x is continuous at all points and has a derivative at no points. You can draw its graph without lifting your pen (but not easily-it turns at every point). To most people, it belongs with space-filling curves and unmeasurable areas-in a box of curiosities. Fractals used to go into the same box! They are beautiful shapes, with boundaries that have no tangents. The theory of fractals is very alive, for good mathematical reasons, and we touch on it in Section 3.7. I hope you have a clear idea of these basic definitions of calculus: 1 Limit ( n -+ ,xor s -+a ) 2 Continuity (at x = a) 3 Derivative (at x = a). Those go back to E and 6, but it is seldom necessary to follow them so far. In the same way that economics describes many transactions, or history describes many events, a function comes from many values f (x). A few points may be special, like market crashes or wars or discontinuities. At other points dfldx is the best guide to the function. 2.7 Continuous Functions This chapter ends with two essential facts about a continuousfunction on a closed interval. The interval is a 6 x < b, written simply as [a, b1.t At the endpoints a and b we require f (x) to approach f (a) and f (b). Extreme Value Property A continuous function on the finite interval [a, b] has a maximum value M and a minimum value m. There are points x,, and x, in [a, b] where it reaches those values: f(xmax)=M 3 f(x) 3 f(xmin)=m for all x in [a, b]. Intermediate Value Property If the number F is between f(a) and f(b), there is a point c between a and b where f (c) = F. Thus if F is between the minimum m and the maximum M, there is a point c between xminand x,, where f (c)= F. Examples show why we require closed intervals and continuous functions. For 0 < x < 1 the function f (x)= x never reaches its minimum (zero). If we close the interval by definingf (0) = 3 (discontinuous)the minimum is still not reached. Because of the jump, the intermediate value F = 2 is also not reached. The idea of continuity was inescapable, after Cauchy defined the idea of a limit. 2.7 EXERCISES Read-through questions Continuity requires the a of f (x) to exist as x -,a and to agree with b . The reason that x/lxl is not continuous at x = 0 is c . This function does have d limits. The reason that l/cos x is discontinuous at e is f . The reason that cos(l/x) is discontinuous at x =0 is g . The function f (x)= h has a simple pole at x = 3, where f has a i pole. The power xnis continuous at all x provided n is i . It has no derivative at x =0 when n is k . f (x)=sin (-x)/x approaches I as x -,0, so this is a m function pro- vided we define f (0)= n . A "continuous function" must be continuous at all 0 . A ','continuable function" can be extended to every point x so that P . Iff has a derivative at x =a then f is necessarily q at x =a. The derivative controls the speed at which f(x) approaches r . On a closed interval [a, b], a continuous f has the s value property and the t value property. It reaches its t~ M and its v m, and it takes on every value w . (sin x)/x2 x # 0 9 f( 4= c x#4 11 f(x)= 112 ~ = 4 x+c xdc lo f(x)= 1 x>c c xQO 12 f(x)= sec x x 2 0 In Problems 1-20, find the numbers c that make f(x) into (A) a continuous function and (B) a differentiable function. In one case f (x)-,f (a) at every point, in the other case Af /Ax has a limit at every point. isin x x < 1 1 f (4= c x2l icos3x X # 7 r 2 f (x)= C x=n {(tan x)/x x # 0 x2 x d c 15 f(x)= 16 f(x)= c x=o 2x x > c +The interval [a, b] is closed (endpoints included). The interval (a, b) is open (a and b left out). The infinite interval [0, ao) contains all x 3 0. i(sin x -x)/xc x # 0 19 f(x) = O x=O 20 f(x)=Ix2+c21 Construct your own f (x) with these discontinuities at x = 1. Removable discontinuity Infinite oscillation Limit for x -+ 1+,no limit for x + 1- A double pole + lirn f(x)=4 lim+f(x) x+1- x+ 1 lim f (x)= GO but lim (x - 1)f (x)=0 x+ 1 x-r 1 lim (X- 1)f (x)= 5 x-r 1 The statement "3x + 7 as x -+ 1" is false. Choose an E for which no 6 can be found. The statement "3x -* 3 as x -,1" is true. For E =4 choose a suitable 6. 29 How many derivatives f ', f ", ... are continuable functions? (a) f =x3I2 (b) f =x3I2sin x (c) f =(sin x)'I2 30 Find one-sided limits at points where there is no twosided limit. Give a 3-part formula for function (c). (b) sin 1x1 31 Let f (1)= 1 and f (- 1)= 1 and f (x)=(x2-x)/(x2- 1) otherwise. Decide whether f is continuous at (a) x = 1 (b) x = 0 (c) x=-1. '32 Let f (x)=x2sin l/x for x # 0 and f (0)=0. If the limits exist, find (a) f ( 4 (b) df /dx at x = 0 (c) Xli+mO f '(x). 33 If f(0) = 0 and f'(0) = 3, rank these functions from smallest to largest as x decreases to zero: 34 Create a discontinuous function f(x) for which f 2(x)is continuous. 35 True or false, with an example to illustrate: (a) If f(x) is continuous at all x, it has a maximum value M. (b) Iff (x)< 7 for all x, then f reaches its maximum. (c) If f (1)= 1 and f (2)= -2, then somewhere f (x)=0. (d) If f (1)= 1 and f (2) = -2 and f is continuous on [I, 21, then somewhere on that interval f (x) =0. 36 The functions cos x and 2x are continuous. Show from the property that cos x = 2x at some point between 0 and 1. 37 Show by example that these statements are false: (a) If a function reaches its maximum and minimum then the function is continuous. (b) If f(x) reaches its maximum and minimum and all values between f (0) and f (1), it is continuous at x =0. (c) (mostly for instructors) If f(x) has the intermediate value property between all points a and b, it must be continuous. 38 Explain with words and a graph why f(x) =x sin (llx) is continuous but has no derivative at x =0. SetflO)=0. 39 Which of these functions are continuable, and why? sin x x c 0 f l ( ~=) cos x x > 1 sin llx x < O f 2 ( 4 = cos l/x x > 1 X f3(x)= - sin x when sin x # 0 f4(x)=x0 + 0"' 40 Explain the difference between a continuous function and a continuable function. Are continuous functions always continuable? "41 f (x) is any continuous function with f (0)=f (1). + (a) Draw a typical f (x). Mark where f (x)=f (x 4). (b) Explain why g(x)=f (x +3)-f (x) has g(4)= -g(0). (c) Deduce from (b)that (a)is always possible: There must + be a point where g(x)=0 and f (x)=f (x 4). 42 Create an f (x) that is continuous only at x = 0. 43 If f (x) is continuous and 0