KEY CONCEPT: THE DERIVATIVE

Transcription

1 Capter Two KEY CONCEPT: THE DERIVATIVE We begin tis capter by investigating te problem of speed: How can we measure te speed of a moving object at a given instant in time? Or, more fundamentally, wat do we mean by te term speed? We give a definition of speed tat as wide-ranging implications not just for te speed problem, but for measuring te rate of cange of any quantity. Our journey will lead us to te key concept of derivative, wic forms te basis for our study of calculus. Te derivative can be interpreted geometrically as te slope of a curve, and pysically as a rate of cange. Because derivatives can be used to represent everyting from fluctuations in interest rates to te rates at wic fis populations vary and gas molecules move, tey ave applications trougout te sciences.

2 66 Capter Two KEY CONCEPT: THE DERIVATIVE 2. HOW DO WE MEASURE SPEED? Te speed of an object at an instant in time is surprisingly difficult to define precisely. Consider te statement At te instant it crossed te finis line, te orse was traveling at 42 mp. How can suc a claim be substantiated? A potograp taken at tat instant will sow te orse motionless it is no elp at all. Tere is some parado in trying to study te orse s motion at a particular instant in time, since by focusing on a single instant we stop te motion! Problems of motion were of central concern to Zeno and oter pilosopers as early as te fift century B.C. Te modern approac, made famous by Newton s calculus, is to stop looking for a simple notion of speed at an instant, and instead to look at speed over small time intervals containing te instant. Tis metod sidesteps te pilosopical problems mentioned earlier but introduces new ones of its own. We illustrate te ideas discussed above by an idealized eample, called a tougt eperiment. It is idealized in te sense tat we assume tat we can make measurements of distance and time as accurately as we wis. A Tougt Eperiment: Average and Instantaneous Velocity We look at te speed of a small object (say, a grapefruit) tat is trown straigt upward into te air at t = 0 seconds. Te grapefruit leaves te trower s and at ig speed, slows down until it reaces its maimum eigt, and ten speeds up in te downward direction and finally, Splat! (See Figure 2..) Suppose tat we want to determine te speed, say, at t = second. Table 2. gives te eigt, y, of te grapefruit above te ground as a function of time. During te first second te grapefruit travels 90 6 = 84 feet, and during te second second it travels only = 52 feet. Hence te grapefruit traveled faster over te first interval, 0 t, tan te second interval, t 2. Velocity positive Start Splat! Velocity negative Ground Figure 2.: Te grapefruit s pat is straigt up and down Table 2. Heigt of te grapefruit above te ground t (sec) y (feet) Velocity versus Speed From now on, we will distinguis between velocity and speed. Suppose an object moves along a line. We pick one direction to be positive and say tat te velocity is positive if it is in te same direction, and negative if it is in te opposite direction. For te grapefruit, upward is positive and downward is negative. (See Figure 2..) Speed is te magnitude of te velocity and so is always positive or zero. If s(t) is te position of an object at time t, ten te average velocity of te object over te interval a t b is Average velocity = Cange in position Cange in time = s(b) s(a). b a In words, te average velocity of an object over an interval is te net cange in position during te interval divided by te cange in time.

3 2. HOW DO WE MEASURE SPEED? 67 Eample Compute te average velocity of te grapefruit over te interval 4 t 5. Wat is te significance of te sign of your answer? During tis interval, te grapefruit moves (06 50) = 44 feet. Terefore te average velocity is 44 ft/sec. Te negative sign means te eigt is decreasing and te grapefruit is moving downward. Eample 2 Compute te average velocity of te grapefruit over te interval t 3. Average velocity = (62 90)/(3 ) = 72/2 = 36 ft/sec. Te average velocity is a useful concept since it gives a roug idea of te beavior of te grapefruit: if two grapefruits are urled into te air, and one as an average velocity of 0 ft/sec over te interval 0 t wile te second as an average velocity of 00 ft/sec over te same interval, te second one is moving faster. But average velocity over an interval does not solve te problem of measuring te velocity of te grapefruit at eactly t = second. To get closer to an answer to tat question, we ave to look at wat appens near t = in more detail. Te data in Figure 2.2 sows te average velocity over small intervals on eiter side of t =. Notice tat te average velocity before t = is sligtly more tan te average velocity after t =. We epect to define te velocity at t = to be between tese two average velocities. As te size of te interval srinks, te values of te velocity before t = and te velocity after t = get closer togeter. In te smallest interval in Figure 2.2, bot velocities are 68.0 ft/sec (to one decimal place), so we define te velocity at t = to be 68.0 ft/sec (to one decimal place). Average velocity 66.4 ft/sec Average velocity 69.6 ft/sec t =. y = Average velocity 67.8 ft/sec Average velocity 68.2 ft/sec t = 0.9 y = t =.0 y = t = 0.99 y = Average velocity 68.0 ft/sec Average velocity 68.0 ft/sec t =.00 y = t = y = Figure 2.2: Average velocities over intervals on eiter side of t = : Sowing successively smaller intervals Of course, if we calculate to more decimal places, te average velocities before and after t = would no longer agree. To calculate te velocity at t = to more decimal places of accuracy, we take smaller and smaller intervals on eiter side of t = until te average velocities agree to te number of decimal places we want. In tis way, we can estimate te velocity at t = to any accuracy. Te data is in fact calculated from te formula y = t 6t 2.

4 68 Capter Two KEY CONCEPT: THE DERIVATIVE Defining Instantaneous Velocity Using Limit Notation Wen we take smaller and smaller intervals, it turns out tat te average velocities get closer and closer to 68 ft/sec. It seems natural, ten, to define instantaneous velocity at te instant t = to be 68 ft/sec. Its definition depends on our being convinced tat smaller and smaller intervals give average speeds tat come arbitrarily close to 68; tat is, te average speeds approac 68 as a limit. Notice ow we ave replaced te original difficulty of computing velocity at a point by a searc for an argument to convince ourselves tat te average velocities approac a limit as te time intervals srink in size. Sowing tat te limit is eactly 68 requires te precise definition of limit given in Section.8. To define instantaneous velocity at an arbitrary point t = a, we use te same metod as for t =. On small intervals of size around t = a, we calculate Average velocity = s(a + ) s(a). Te instantaneous velocity is te number tat te average velocities approac as te intervals decrease in size, tat is, as becomes smaller. So we make te following definition: Let s(t) be te position at time t. Ten te instantaneous velocity at t = a is defined as Instantaneous velocity at t = a s(a + ) s(a) = lim. 0 In words, te instantaneous velocity of an object at time t = a is given by te limit of te average velocity over an interval, as te interval srinks around a. Tis limit refers to te number tat te average velocities approac as te intervals srink. To estimate te limit, we look at intervals of smaller and smaller, but never zero, lengt. Visualizing Velocity: Slope of Curve Now we visualize velocity using a grap of eigt. Te cornerstone of te idea is te fact tat, on a very small scale, most functions look almost like straigt lines. Imagine taking te grap of a function near a point and zooming in to get a close-up view. (See Figure 2.3.) Te more we zoom in, te more te curve appears to be a straigt line. We call te slope of tis line te slope of te curve at te point. Curve More linear Almost completely linear P P P Slope of line = Slope of curve at P Figure 2.3: Estimating te slope of te curve at te point by zooming in To visualize te instantaneous velocity, we tink about ow we calculated it. We took average velocities over small intervals containing at t =. Two suc velocities are represented by te slopes

5 2. HOW DO WE MEASURE SPEED? 69 of te lines in Figure 2.4. As te lengt of te interval srinks, te slope of te line gets closer to te slope of te curve at t =. Slope of curve = Instantaneous velocity at t = y y = s(t) Slope = Average velocity over t 3.5 Slope = Average velocity over t t Figure 2.4: Average velocities over small intervals Te instantaneous velocity is te slope of te curve at a point. Let s go back to te grapefruit. Figure 2.5 sows te eigt of te grapefruit plotted against time. (Note tat tis is not a picture of te grapefruit s pat, wic is straigt up and down.) How can we visualize te average velocity on tis grap? Suppose y = s(t). We consider te interval t 2 and te epression Average velocity = Cange in position Cange in time = s(2) s() 2 = = 52 ft/sec. Now s(2) s() is te cange in position over te interval, and it is marked vertically in Figure 2.5. Te in te denominator is te time elapsed and is marked orizontally in Figure 2.5. Terefore, Average velocity = Cange in position Cange in time (See Figure 2.5.) A similar argument sows te following: = Slope of line joining B and C. Te average velocity over any time interval a t b is te slope of te line joining te points on te grap of s(t) corresponding to t = a and t = b. Figure 2.5 sows ow te grapefruit s velocity varies during its journey. At points A and B te curve as a large positive slope, indicating tat te grapefruit is traveling up rapidly. Point D y (eigt) 42 Grapefruit moving fast (upward) A 90 B Velocity zero D E C y = s(t) s(2) s() F Grapefruit moving fast G (downward) t (time) Figure 2.5: Te eigt, y, of te grapefruit at time t

6 70 Capter Two KEY CONCEPT: THE DERIVATIVE is almost at te top: te grapefruit is slowing down. At te peak, te slope of te curve is zero: te fruit as slowed to zero velocity for an instant in preparation for its return to eart. At point E te curve as a small negative slope, indicating a slow velocity of descent. Finally, te slope of te curve at point G is large and negative, indicating a large downward velocity tat is responsible for te Splat. Using Limits to Compute te Instantaneous Velocity Suppose we want to calculate te instantaneous velocity for s(t) = t 2 at t = 3. We must find: We sow two possible approaces. s(3 + ) s(3) (3 + ) 2 9 lim = lim. 0 0 Eample 3 (3 + ) 2 9 Estimate lim numerically. 0 Te limit is te value approaced by tis epression as approaces 0. Te values in Table 2.2 seem to be converging to 6 as 0. So it is a reasonable guess tat (3 + ) 2 9 lim = 6. 0 However, we cannot be sure tat te limit is eactly 6 by looking at te table. To calculate te limit eactly requires algebra. Table 2.2 Values of ( (3 + ) 2 9 ) / near = ( (3 + ) 2 9 ) / Eample 4 (3 + ) 2 9 Use algebra to find lim. 0 Epanding te numerator gives (3 + ) 2 9 = = Since taking te limit as 0 means looking at values of near, but not equal, to 0, we can cancel, giving (3 + ) 2 9 lim = lim (6 + ). 0 0 As approaces 0, te values of (6 + ) approac 6, so (3 + ) 2 9 lim = lim (6 + ) =

7 2. HOW DO WE MEASURE SPEED? 7 Eercises and Problems for Section 2. Eercises. Te distance, s, a car as traveled on a trip is sown in te table as a function of te time, t, since te trip started. Find te average velocity between t = 2 and t = 5. t (ours) s (km) Slope 3 0 /2 2 Point A B C D E F 2. In a time of t seconds, a particle moves a distance of s meters from its starting point, were s = 3t 2. (a) Find te average velocity between t = and t = + if: (i) = 0., (ii) = 0.0, (iii) = (b) Use your answers to part (a) to estimate te instantaneous velocity of te particle at time t =. 3. In a time of t seconds, a particle moves a distance of s meters from its starting point, were s = 4t Figure For te function sown in Figure 2.7, at wat labeled points is te slope of te grap positive? Negative? At wic labeled point does te grap ave te greatest (i.e., most positive) slope? Te least slope (i.e., negative and wit te largest magnitude)? E (a) Find te average velocity between t = and t = + if: (i) = 0., (ii) = 0.0, (iii) = (b) Use your answers to part (a) to estimate te instantaneous velocity of te particle at time t =. B CD A Figure 2.7 F 4. In a time of t seconds, a particle moves a distance of s meters from its starting point, were s = sin(2t). (a) Find te average velocity between t = and t = + if: (i) = 0., (ii) = 0.0, (iii) = (b) Use your answers to part (a) to estimate te instantaneous velocity of te particle at time t =. 5. Matc te points labeled on te curve in Figure 2.6 wit te given slopes. Estimate te limits in Eercises 7 0 by substituting smaller and smaller values of. For trigonometric functions, use radians. Give answers to one decimal place. 7. lim 0 (3 + ) lim lim 0 cos 0. lim 0 e + e Problems. A car is driven at a constant speed. Sketc a grap of te distance te car as traveled as a function of time. 2. A car is driven at an increasing speed. Sketc a grap of te distance te car as traveled as a function of time. 3. A car starts at a ig speed, and its speed ten decreases slowly. Sketc a grap of te distance te car as traveled as a function of time. 4. For te grap y = f() in Figure 2.8, arrange te following numbers from smallest to largest: Te slope of te grap at A. Te slope of te grap at B. Te slope of te grap at C. Te slope of te line AB. Te number 0. Te number. y A B y = Figure 2.8 C y = f()

8 72 Capter Two KEY CONCEPT: THE DERIVATIVE 5. Te grap of f(t) in Figure 2.9 gives te position of a particle at time t. List te following quantities in order, smallest to largest. A, average velocity between t = and t = 3, B, average velocity between t = 5 and t = 6, C, instantaneous velocity at t =, D, instantaneous velocity at t = 3, E, instantaneous velocity at t = 5, F, instantaneous velocity at t = 6. as a function of time, t. Sketc a possible grap for f if te average velocity of te particle between t = 2 and t = 6 is te same as te instantaneous velocity at t = Find te average velocity over te interval 0 t 0.2, and estimate te velocity at t = 0.2 of a car wose position, s, is given by te following table. t (sec) s (ft) f(t) Figure A particle moves at varying velocity along a line and s = f(t) represents te particle s distance from a point 2.2 THE DERIVATIVE AT A POINT t Use algebra to evaluate te limits in Problems lim 0 (2 + ) lim 0 3(2 + ) lim 0 (3 + ) 2 (3 ) lim 0 ( + ) 3 Average Rate of Cange In Section 2., we looked at te cange in eigt divided by te cange in time, wic tells us Average rate of cange of eigt wit respect to time = s(a + ) s(a). Tis ratio is called te difference quotient. Now we apply te same analysis to any function f, not necessarily a function of time. We say: Average rate of cange of f over te interval from a to a + = f(a + ) f(a). Te numerator, f(a + ) f(a), measures te cange in te value of f over te interval from a to a +. So, te difference quotient is te cange in f divided by te cange in. See Figure 2.0. Slope = Average rate of cange = f(a+) f(a) A B f() f(a + ) f(a) a a + Figure 2.0: Visualizing te average rate of cange of f

9 2.2 THE DERIVATIVE AT A POINT 73 Altoug te interval is no longer necessarily a time interval, we still talk about te average rate of cange of f over te interval. If we want to empasize te independent variable, we talk about te average rate of cange of f wit respect to. Average Rate of Cange versus Absolute Cange Te average rate of cange of a function over an interval is not te same as te absolute cange. Absolute cange is just te difference in te values of f at te ends of te interval: f(a + ) f(a). Te average rate of cange is te absolute cange divided by te size of te interval: f(a + ) f(a). Te average rate of cange tells ow quickly (or slowly) te function canges from one end of te interval to te oter, relative to te size of te interval. It is often more useful to know te rate of cange tan te absolute cange. For eample, if someone offers you a $00 salary, you will want to know ow long to work to make tat money. Just knowing te absolute cange in your money, $00, is not enoug, but knowing te rate of cange (i.e., $00 divided by te time it takes to make it) elps you decide weter or not to accept te salary. Blowing Up a Balloon Consider te function wic gives te radius of a spere in terms of its volume. For eample, tink of blowing air into a balloon. You ve probably noticed tat a balloon seems to blow up faster at te start and ten slows down as you blow more air into it. Wat you re seeing is variation in te rate of cange of te radius wit respect to volume. Eample Te volume, V, of a spere of radius r is given by V = 4πr 3 /3. Solving for r in terms of V gives r = f(v ) = ( ) /3 3V. 4π Calculate te average rate of cange of r wit respect to V over te intervals 0.5 V and V.5. Using te formula for te average rate of cange gives Average rate of cange of radius for 0.5 V Average rate of cange of radius for V.5 = = f() f(0.5) 0.5 f(.5) f() 0.5 = 2 = 2 ( ( ) /3 3 4π ( (4.5 4π ( ) ) / π ) /3 ( ) ) / π So we see tat te rate decreases as te volume increases. Instantaneous Rate of Cange: Te Derivative We define te instantaneous rate of cange of a function at a point in te same way tat we defined instantaneous velocity: we look at te average rate of cange over smaller and smaller intervals. Tis instantaneous rate of cange is called te derivative of f at a, denoted by f (a).

10 74 Capter Two KEY CONCEPT: THE DERIVATIVE Te derivative of f at a, written f (a), is defined as Rate of cange of f at a = f f(a + ) f(a) (a) = lim. 0 If te limit eists, ten f is said to be differentiable at a. To empasize tat f (a) is te rate of cange of f() as te variable canges, we call f (a) te derivative of f wit respect to at = a. Wen te function y = s(t) represents te position of an object, te derivative s (t) is te velocity. Eample 2 By coosing small values for, estimate te instantaneous rate of cange of te radius, r, of a spere wit respect to cange in volume at V =. Te formula for r = f(v ) is given in Eample. Wit = 0.0 and = 0.0, we ave te difference quotients f(.0) f() and f(0.99) f() Wit = 0.00 and = 0.00, f(.00) f() and f(0.999) f() Te values of tese difference quotients suggest tat te limit is between and We conclude tat te value is about 0.207; taking smaller values confirms tis. So we say f () = Instantaneous rate of cange of radius wit respect to volume at V = In tis eample we found an approimation to te instantaneous rate of cange, or derivative, by substituting in smaller and smaller values of. Now we see ow to visualize te derivative. Visualizing te Derivative: Slope of Curve and Slope of Tangent As wit velocity, we can visualize te derivative f (a) as te slope of te grap of f at = a. In addition, tere is anoter way to tink of f (a). Consider te difference quotient (f(a + ) f(a))/. Te numerator, f(a + ) f(a), is te vertical distance marked in Figure 2. and is te orizontal distance, so Average rate of cange of f = f(a + ) f(a) = Slope of line AB. As becomes smaller, te line AB approaces te tangent line to te curve at A. (See Figure 2.2.) We say Instantaneous rate of cange of f at a f(a + ) f(a) = lim = Slope of tangent at A. 0

11 2.2 THE DERIVATIVE AT A POINT 75 Slope = Average rate of cange = f(a+) f(a) A a B a + f() f(a + ) f(a) Figure 2.: Visualizing te average rate of cange of f A a B B B B f() Slope = Derivative = f (a) Figure 2.2: Visualizing te instantaneous rate of cange of f Te derivative at point A can be interpreted as: Te slope of te curve at A. Te slope of te tangent line to te curve at A. Te slope interpretation is often useful in gaining roug information about te derivative, as te following eamples sow. Eample 3 Is te derivative of sin at = π positive or negative? Looking at a grap of sin in Figure 2.3 (remember, is in radians), we see tat a tangent line drawn at = π as negative slope. So te derivative at tis point is negative. Negative slope f() = sin π 2π 3π 4π Figure 2.3: Tangent line to sin at = π Recall tat if we zoom in on te grap of a function y = f() at te point = a, we usually find tat te grap looks like a straigt line wit slope f (a). Eample 4 By zooming in on te point (0, 0) on te grap of te sine function, estimate te value of te derivative of sin at = 0, wit in radians. Figure 2.4 sows graps of sin wit smaller and smaller scales. On te interval 0. 0., te grap looks like a straigt line of slope. Tus, te derivative of sin at = 0 is about f() = sin f() = sin f() = sin Figure 2.4: Zooming in on te grap of sin near = 0 sows te derivative is about at = 0

12 76 Capter Two KEY CONCEPT: THE DERIVATIVE Later we will sow tat te derivative of sin at = 0 is eactly. (See page 36 in Section 3.5.) From now on we will assume tat tis is so. Eample 5 Use te tangent line at = 0 to estimate values of sin near = 0. In te previous eample we see tat near = 0, te grap of y = sin looks like te straigt line y = ; we can use tis line to estimate values of sin wen is close to 0. For eample, te point on te straigt line y = wit coordinate 0.32 is (0.32, 0.32). Since te line is close to te grap of y = sin, we estimate tat sin (See Figure 2.5.) Cecking on a calculator, we find tat sin , so our estimate is quite close. Notice tat te grap suggests tat te real value of sin 0.32 is sligtly less tan y y = y = sin 0.32 Figure 2.5: Approimating y = sin by te line y = Wy Do We Use Radians and Not Degrees? After Eample 4 we stated tat te derivative of sin at = 0 is, wen is in radians. Tis is te reason we coose to use radians. If we ad done Eample 4 in degrees, te derivative of sin would ave turned out to be a muc messier number. (See Problem 20, page 80.) Estimating te Derivative of an Eponential Function Eample 6 Estimate te value of te derivative of f() = 2 at = 0 grapically and numerically. Grapically: Figure 2.6 indicates tat te grap is concave up. Assuming tis, te slope at A is between te slope of BA and te slope of AC. Since Slope of line BA = (20 2 ) (0 ( )) = 2 and Slope of line AC = (2 2 0 ) ( 0) =, we know tat at = 0 te derivative of 2 is between /2 and. Numerically: To estimate te derivative at = 0, we look at values of te difference quotient f(0 + ) f(0) = = 2 for small. Table 2.3 sows some values of 2 togeter wit values of te difference quotients. (See Problem 26 on page 80 for wat appens for very small values of.)

13 2.2 THE DERIVATIVE AT A POINT 77 B Slope = 2 2 A Slope = f() = 2 C Tangent line Slope = f (0) Figure 2.6: Grap of y = 2 sowing te derivative at = 0 Table 2.3 at = 0 Numerical values for difference quotient of 2 2 Difference quotient: Te concavity of te curve tells us tat difference quotients calculated wit negative s are smaller tan te derivative, and tose calculated wit positive s are larger. From Table 2.3 we see tat te derivative is between and To tree decimal places, f (0) = Eample 7 Find an approimate equation for te tangent line to f() = 2 at = 0. From te previous eample, we know te slope of te tangent line is about Since te tangent line as y-intercept, its equation is y = Computing te Derivative of / Te grap of f() = / in Figure 2.7 leads us to epect tat f (2) is negative. To compute f (2) eactly, we use algebra. f() = / Slope = f (2) 2 3 Figure 2.7: Tangent line to f() = / at = 2 Eample 8 Find te derivative of f() = / at te point = 2. Te derivative is te limit of te difference quotient, so we look at f f(2 + ) f(2) (2) = lim. 0 Using te formula for f and simplifying gives ( f (2) = lim ) = lim 2 0 ( ) 2 (2 + ) = lim 2(2 + ) 0 2(2 + ). Since te limit only eamines values of close to, but not equal to, zero, we can cancel. We get f (2) = lim 0 2(2 + ) = 4. Tus, f (2) = /4. Te slope of te tangent line in Figure 2.7 is /4.

14 78 Capter Two KEY CONCEPT: THE DERIVATIVE Eercises and Problems for Section 2.2 Eercises. Te table sows values of f() = 3 near = 2 (to tree decimal places). Use it to estimate f (2) a b c d e f f () (a) Make a table of values rounded to two decimal places for te function f() = e for =,.5, 2, 2.5, and 3. Ten use te table to answer parts (b) and (c). (b) Find te average rate of cange of f() between = and = 3. (c) Use average rates of cange to approimate te instantaneous rate of cange of f() at = (a) Make a table of values, rounded to two decimal places, for f() = log (tat is, log base 0) wit =,.5, 2, 2.5, 3. Ten use tis table to answer parts (b) and (c). (b) Find te average rate of cange of f() between = and = 3. (c) Use average rates of cange to approimate te instantaneous rate of cange of f() at = (a) Let f() = 2. Eplain wat Table 2.4 tells us about f (). (b) Find f () eactly. (c) If canges by 0. near =, wat does f () tell us about ow f() canges? Illustrate your answer wit a sketc. Table 2.4 Difference in 2 successive 2 values Grap f() = sin, and use te grap to decide weter te derivative of f() at = 3π is positive or negative. 6. For te function f() = log, estimate f (). From te grap of f(), would you epect your estimate to be greater tan or less tan f ()? 7. Estimate f (2) for f() = 3. Eplain your reasoning. 8. Figure 2.8 sows te grap of f. Matc te derivatives in te table wit te points a, b, c, d, e. Figure Label points A, B, C, D, E, and F on te grap of y = f() in Figure 2.9. (a) Point A is a point on te curve were te derivative is negative. (b) Point B is a point on te curve were te value of te function is negative. (c) Point C is a point on te curve were te derivative is largest. (d) Point D is a point on te curve were te derivative is zero. (e) Points E and F are different points on te curve were te derivative is about te same. y Figure 2.9 y = f() 0. Te grap of y = f() is sown in Figure Wic is larger in eac of te following pairs? y Figure 2.20 y = f() (a) Average rate of cange: Between = and = 3? Or between = 3 and = 5? (b) f(2) or f(5)? (c) f () or f (4)?

15 2.2 THE DERIVATIVE AT A POINT 79 Problems. Suppose tat f() is a function wit f(00) = 35 and f (00) = 3. Estimate f(02). 2. Te function in Figure 2.2 as f(4) = 25 and f (4) =.5. Find te coordinates of te points A, B, C. C A B Figure 2.2 Tangent line f() 3. Use Figure 2.22 to fill in te blanks in te following statements about te function g at point B. (a) g( ) = (b) g ( ) = 6. Wit te function f given by Figure 2.23, arrange te following quantities in ascending order: 0, f (2), f (3), f(3) f(2) 7. Suppose y = f() graped in Figure 2.23 represents te cost of manufacturing kilograms of a cemical. Ten f()/ represents te average cost of producing kilogram wen kilograms are made. Tis problem asks you to visualize tese averages grapically. (a) Sow ow to represent f(4)/4 as te slope of a line. (b) Wic is larger, f(3)/3 or f(4)/4? 8. On a copy of Figure 2.24, mark lengts tat represent te quantities in parts (a) (d). (Pick any convenient, and assume > 0.) (a) f() (b) f( + ) (c) f( + ) f() (d) (e) Using your answers to parts (a) (d), sow ow te f( + ) f() quantity can be represented as te slope of a line on te grap. y (.95, 5.02) B (2, 5) Figure 2.22 g() Tangent line 4. Sow ow to represent te following on Figure (a) f(4) (b) f(4) f(2) (c) f(5) f(2) 5 2 (d) f (3) f() Figure For eac of te following pairs of numbers, use Figure 2.23 to decide wic is larger. Eplain your answer. (a) f(3) or f(4)? (b) f(3) f(2) or f(2) f()? f(2) f() (c) or 2 (d) f () or f (4)? f(3) f()? 3 y = f() Figure Consider te function sown in Figure (a) Write an epression involving f for te slope of te line joining A and B. (b) Draw te tangent line at C. Compare its slope to te slope of te line in part (a). (c) Are tere any oter points on te curve at wic te slope of te tangent line is te same as te slope of te tangent line at C? If so, mark tem on te grap. If not, wy not? A C f a c b Figure 2.25 B

16 80 Capter Two KEY CONCEPT: THE DERIVATIVE 20. (a) Estimate f (0) if f() = sin, wit in degrees. (b) In Eample 4 on page 75, we found tat te derivative of sin at = 0 was. Wy do we get a different result ere? (Tis problem sows wy radians are almost always used in calculus.) 2. Estimate te derivative of f() = at = For y = f() = 3 3/2, use your calculator to construct a grap of y = f(), for 0 2. From your grap, estimate f (0) and f (). 23. Let f() = ln(cos ). Use your calculator to approimate te instantaneous rate of cange of f at te point =. Do te same ting for = π/4. (Note: Be sure tat your calculator is set in radians.) 24. Te population, P, of Cina, in billions, can be approimated by te function P =.5(.04) t, were t is te number of years since te start of 993. According to tis model, ow fast is te population growing at te start of 993 and at te start of 995? Give your answers in millions of people per year. 25. (a) Grap f() = 2 2 and g() = f() + 3 on te same set of aes. Wat can you say about te slopes of te tangent lines to te two graps at te point = 0? = 2? Any point = 0? (b) Eplain wy adding a constant value, C, to any function does not cange te value of te slope of its grap at any point. [Hint: Let g() = f() + C, and calculate te difference quotients for f and g.] 26. Suppose Table 2.3 on page 77 is continued wit smaller values of. A particular calculator gives te results in Table 2.5. (Your calculator may give sligtly different results.) Comment on te values of te difference quotient in Table 2.5. In particular, wy is te last value of (2 )/ zero? Wat do you epect te calculated value of (2 )/ to be wen = 0 20? Table 2.5 Questionable values of difference quotients of 2 near = 0 Difference quotient: (2 )/ Use algebra to evaluate te limits in Problems lim 0 ( 3 + ) lim 0 /( + ) 28. lim 0 (2 ) lim 0 /( + ) lim [Hint: Multiply by in numerator and 0 denominator.] / 32. lim /2 Find te derivatives in Problems algebraically. 33. f() = 5 2 at = f() = 3 at = g(t) = t 2 + t at t = 36. f() = at = 37. g() = / at = g(z) = z 2, find g (2) For Problems 39 42, find te equation of te line tangent to te function at te given point. 39. f() = 5 2 at = f() = 3 at = 2 4. f() = at = f() = / 2 at (, ) 2.3 THE DERIVATIVE FUNCTION In te previous section we looked at te derivative of a function at a fied point. Now we consider wat appens at a variety of points. Te derivative generally takes on different values at different points and is itself a function. First, remember tat te derivative of a function at a point tells us te rate at wic te value of te function is canging at tat point. Geometrically, we can tink of te derivative as te slope of te curve or of te tangent line at te point.

17 2.3 THE DERIVATIVE FUNCTION 8 Eample Estimate te derivative of te function f() graped in Figure 2.26 at = 2,, 0,, 2, 3, 4, Slope of tangent = f ( ) Figure 2.26: Estimating te derivative grapically as te slope of te tangent line From te grap we estimate te derivative at any point by placing a straigtedge so tat it forms te tangent line at tat point, and ten using te grid squares to estimate te slope of te straigtedge. For eample, te tangent at = is drawn in Figure 2.26, and as a slope of about 2, so f ( ) 2. Notice tat te slope at = 2 is positive and fairly large; te slope at = is positive but smaller. At = 0, te slope is negative, by = it as become more negative, and so on. Some estimates of te derivative are listed in Table 2.6. You sould ceck tese values. Are tey reasonable? Is te derivative positive were you epect? Negative? Table 2.6 Estimated values of derivative of function in Figure f () Notice tat for every -value, tere s a corresponding value of te derivative. Terefore, te derivative is itself a function of. For any function f, we define te derivative function, f, by f f( + ) f() () = Rate of cange of f at = lim. 0 For every -value for wic tis limit eists, we say f is differentiable at tat -value. If te limit eists for all in te domain of f, we say f is differentiable everywere. Most functions we meet are differentiable at every point in teir domain, ecept peraps for a few isolated points. Te Derivative Function: Grapically Eample 2 Sketc te grap of te derivative of te function sown in Figure We plot te values of tis derivative given in Table 2.6. We obtain Figure 2.27, wic sows a grap of te derivative (te black curve), along wit te original function (color).

18 82 Capter Two KEY CONCEPT: THE DERIVATIVE 5 4 f () 3 f() Figure 2.27: Function (colored) and derivative (black) from Eample You sould ceck tat tis grap of f makes sense. Were te values of f are positive, f is increasing ( < 0.3 or > 3.8) and were f is negative, f is decreasing. Notice tat at te points were f as large positive slope, suc as = 2, te grap of te derivative is far above te -ais, as it sould be, since te value of te derivative is large tere. At points were te slope is gentler, suc as =, te grap of f is closer to te -ais, since te derivative is smaller. Wat Does te Derivative Tell Us Grapically? Were f is positive, te tangent line to f is sloping up; were f is negative, te tangent line to f is sloping down. If f = 0 everywere, ten te tangent line to f is orizontal everywere, and f is constant. We see tat te sign of f tells us weter f is increasing or decreasing. If f > 0 on an interval, ten f is increasing over tat interval. If f < 0 on an interval, ten f is decreasing over tat interval. Moreover, te magnitude of te derivative gives us te magnitude of te rate of cange; so if f is large (positive or negative), ten te grap of f is steep (up or down), wereas if f is small te grap of f slopes gently. Wit tis in mind, we can learn about te beavior of a function from te beavior of its derivative. Te Derivative Function: Numerically If we are given values of a function instead of its grap, we can estimate values of te derivative. Eample 3 Table 2.7 Table 2.7 gives values of c(t), te concentration (µg/cm 3 ) of a drug in te bloodstream at time t (min). Construct a table of estimated values for c (t), te rate of cange of c(t) wit respect to time. Concentration as a function of time t (min) c(t) (µg/cm 3 )

19 2.3 THE DERIVATIVE FUNCTION 83 We estimate values of c using te values in te table. To do tis, we ave to assume tat te data points are close enoug togeter tat te concentration does not cange wildly between tem. From te table, we see tat te concentration is increasing between t = 0 and t = 0.4, so we epect a positive derivative tere. However, te increase is quite slow, so we epect te derivative to be small. Te concentration does not cange between 0.4 and 0.5, so we epect te derivative to be rougly 0 tere. From t = 0.5 to t =.0, te concentration starts to decrease, and te rate of decrease gets larger and larger, so we epect te derivative to be negative and of greater and greater magnitude. Using te data in te table, we estimate te derivative using te difference quotient: c c(t + ) c(t) (t). Since te data points are 0. apart, we use = 0., giving c c(0.) c(0) (0) = = 0.5 µg/cm 3 /min c c(0.2) c(0.) (0.) = = 0.5 µg/cm 3 /min c c(0.3) c(0.2) (0.2) = = 0.4 µg/cm 3 /min c c(0.4) c(0.3) (0.3) = = 0.2 µg/cm 3 /min c c(0.5) c(0.4) (0.4) = = 0.0 µg/cm 3 /min and so on. Tese values are tabulated in Table 2.8. Notice tat te derivative as small positive values up until t = 0.4, were it is rougly 0, and ten it gets more and more negative, as we epected. Te slopes are sown on te grap of c(t) in Figure Table 2.8 Estimated derivative of concentration t c (t) Slope = 0.5 Slope = 0 c(t) Slope = Figure 2.28: Grap of concentration as a function of time t Improving Numerical Estimates for te Derivative In te previous eample, te estimate for te derivative at 0.2 used te interval to te rigt; we found te average rate of cange between t = 0.2 and t = 0.3. However, we could equally well ave gone to te left and used te rate of cange between t = 0. and t = 0.2 to approimate te derivative at 0.2. For a more accurate result, we could average tese slopes and say c (0.2) ( Slope to left + 2 of 0.2 Slope to rigt of 0.2 ) = In general, averaging te slopes leads to a more accurate answer = 0.45.

20 84 Capter Two KEY CONCEPT: THE DERIVATIVE Derivative Function: From a Formula If we are given a formula for f, can we come up wit a formula for f? We often can, as sown in te net eample. Indeed, muc of te power of calculus depends on our ability to find formulas for te derivatives of all te functions we described earlier. Tis is done systematically in Capter 3. Derivative of a Constant Function Te grap of a constant function f() = k is a orizontal line, wit a slope of 0 everywere. Terefore, its derivative is 0 everywere. (See Figure 2.29.) If f() = k, ten f () = 0. f() = k Slope = 0 Figure 2.29: A constant function Derivative of a Linear Function We already know tat te slope of a straigt line is constant. Tis tells us tat te derivative of a linear function is constant. If f() = b + m, ten f () = Slope = m. Derivative of a Power Function Eample 4 Find a formula for te derivative of f() = 2. Before computing te formula for f () algebraically, let s try to guess te formula by looking for a pattern in te values of f (). Table 2.9 contains values of f() = 2 (rounded to tree decimals), wic we can use to estimate te values of f (), f (2), and f (3). Table 2.9 Values of f() = 2 near =, = 2, = 3 (rounded to tree decimals) Near =, te value of 2 increases by about eac time increases by 0.00, so f () = 2.

21 2.3 THE DERIVATIVE FUNCTION 85 Similarly, near = 2 and = 3, te value of 2 increases by about and 0.006, respectively, wen increases by So f (2) = 4 and f (3) = 6. Knowing te value of f at specific points can never tell us te formula for f, but it certainly can be suggestive: Knowing f () 2, f (2) 4, f (3) 6 suggests tat f () = 2. Te derivative is calculated by forming te difference quotient and taking te limit as goes to zero. Te difference quotient is f( + ) f() = ( + )2 2 = = Since never actually reaces zero, we can cancel it in te last epression to get 2 +. Te limit of tis as goes to zero is 2, so f () = lim 0 (2 + ) = 2. Eample 5 Calculate f () if f() = 3. We look at te difference quotient f( + ) f() = ( + )3 3. Multiplying out gives ( + ) 3 = , so f () = lim = lim. 0 0 Since in taking te limit as 0, we consider values of near, but not equal to, zero, we can cancel giving f () = lim = lim ( ). 0 0 As 0, te value of (3 + 2 ) 0 so f () = lim 0 ( ) = 3 2. Te previous two eamples sow ow to compute te derivatives of power functions of te form f() = n, wen n is 2 or 3. We can use te Binomial Teorem to sow te power rule for a positive integer n: If f() = n ten f () = n n. Tis result is in fact valid for any real value of n.

22 86 Capter Two KEY CONCEPT: THE DERIVATIVE Eercises and Problems for Section 2.3 Eercises For Eercises 9, sketc a grap of te derivative function of eac of te given functions.. y y For f() = ln, construct tables, rounded to four decimals, near =, = 2, = 5, and = 0. Use te tables to estimate f (), f (2), f (5), and f (0). Ten guess a general formula for f ().. (a) Estimate f (2) using te values of f in te table. (b) For wat values of does f () appear to be positive? Negative? 3. 4 y y f() Find approimate values for f () at eac of te -values given in te following table y y f() In Eercises 3 4, find a formula for te derivative using te power rule. Confirm it using difference quotients. 7. y 8. 4 y 4 3. k() = / 4. l() = / Find a formula for te derivatives of te functions in Eercises 5 6 using difference quotients g() = m() = /( + ) 9. y For Eercises 7 20, sketc te grap of f(), and use tis grap to sketc te grap of f (). 7. f() = 2 8. f() = ( ) 9. f() = cos 20. f() = log Problems 2. Given te numerical values sown, find approimate values for te derivative of f() at eac of te -values given. Were is te rate of cange of f() positive? Were is it negative? Were does te rate of cange of f() seem to be greatest? f() Values of and g() are given in te table. For wat value of is g () closest to 3? g()

23 2.3 THE DERIVATIVE FUNCTION 87 For Problems 23 32, sketc te grap of f () f() f() 34. Figure 2.3 sows a grap of voltage across an electrical capacitor as a function of time. Te current is proportional to te derivative of te voltage; te constant of proportionality is positive. Sketc a grap of te current as a function of time. voltage 25. f() f() Figure 2.3 time f() 28. f() 35. A veicle moving along a straigt road as distance f(t) from its starting point at time t. Wic of te graps in Figure 2.32 could be f (t) for te following scenarios? (Assume te scales on te vertical aes are all te same.) (a) A bus on a popular route, wit no traffic (b) A car wit no traffic and all green ligts (c) A car in eavy traffic conditions f() f() 2 3 (I) t (II) t (III) 3. f() 32. f() t Figure In te grap of f in Figure 2.30, at wic of te labeled -values is (a) f() greatest? (b) f() least? (c) f () greatest? (d) f () least? 36. Te derivative of f is te spike function in Figure Wat can you say about te grap of f? f (t) t f() Figure 2.30 Figure 2.33

24 88 Capter Two KEY CONCEPT: THE DERIVATIVE 37. A cild inflates a balloon, admires it for a wile and ten lets te air out at a constant rate. If V (t) gives te volume of te balloon at time t, ten Figure 2.34 sows V (t) as a function of t. At wat time does te cild: (a) Begin to inflate te balloon? (b) Finis inflating te balloon? (c) Begin to let te air out? (d) Wat would te grap of V (t) look like if te cild ad alternated between pincing and releasing te open end of te balloon, instead of letting te air out at a constant rate? debt (trillions of dollars) year Figure V (t) Figure Te population of a erd of deer is modeled by ( P (t) = sin 2πt π ) 2 t 40. Draw te grap of a continuous function y = f() tat satisfies te following tree conditions. f () > 0 for < 2, f () < 0 for 2 < < 2, f () = 0 for > Draw te grap of a continuous function y = f() tat satisfies te following tree conditions. f () > 0 for π 2 < < π 2, f () < 0 for π < < π 2 and π 2 < < π, f () = 0 at = π 2 and = π Figure 2.36 is te grap of f, te derivative of a function f. On wat interval(s) is te function f (a) Increasing? (b) Decreasing? were t is measured in years from January. (a) How does tis population vary wit time? Sketc a grap of P (t) for one year. (b) Use te grap to decide wen in te year te population is a maimum. Wat is tat maimum? Is tere a minimum? If so, wen? (c) Use te grap to decide wen te population is growing fastest. Wen is it decreasing fastest? (d) Estimate rougly ow fast te population is canging on te first of July. 39. Te grap in Figure 2.35 sows te accumulated federal debt since 970. Sketc te derivative of tis function. Wat does it represent? f Figure 2.36: Grap of f, not f 43. Using a grap, eplain wy if f() is an even function, ten f () is odd. 44. Using a grap, eplain wy if g() is an odd function, ten g () is even. 2.4 INTERPRETATIONS OF THE DERIVATIVE We ave seen te derivative interpreted as a slope and as a rate of cange. In tis section, we see oter interpretations. Te purpose of tese eamples is not to make a catalog of interpretations but to illustrate te process of obtaining tem. An Alternative Notation for te Derivative So far we ave used te notation f to stand for te derivative of te function f. An alternative notation for derivatives was introduced by te German matematician Wilelm Gottfried Leibniz

25 (646 76). If te variable y depends on te variable, tat is, if ten e wrote dy/d for te derivative, so y = f(), 2.4 INTERPRETATIONS OF THE DERIVATIVE 89 dy d = f (). Leibniz s notation is quite suggestive if we tink of te letter d in dy/d as standing for small difference in.... Te notation dy/d reminds us tat te derivative is a limit of ratios of te form Difference in y-values Difference in -values. Te notation dy/d suggests te units for te derivative: te units for y divided by te units for. Te separate entities dy and d officially ave no independent meaning: tey are all part of one notation. In fact, a good way to view te notation dy/d is to tink of d/d as a single symbol meaning te derivative wit respect to of.... So dy/d can be viewed as d d (y), meaning te derivative wit respect to of y. On te oter and, many scientists and matematicians tink of dy and d as separate entities representing infinitesimally small differences in y and, even toug it is difficult to say eactly ow small infinitesimal is. Altoug not formally correct, it can be elpful to tink of dy/d as a small cange in y divided by a small cange in. For eample, recall tat if s = f(t) is te position of a moving object at time t, ten v = f (t) is te velocity of te object at time t. Writing v = ds dt reminds us tat v is a velocity, since te notation suggests a distance, ds, over a time, dt, and we know tat distance over time is velocity. Similarly, we recognize dy d = f () as te slope of te grap of y = f() since slope is vertical rise, dy, over orizontal run, d. Te disadvantage of Leibniz s notation is tat it is awkward to specify te -value at wic we are evaluating te derivative. To specify f (2), for eample, we ave to write dy d. =2 Using Units to Interpret te Derivative Te following eamples illustrate ow useful units can be in suggesting interpretations of te derivative. We use te fact tat te units of te instantaneous and te average rate of cange are te same. For eample, suppose s = f(t) gives te distance, in meters, of a body from a fied point as a function of time, t, in seconds. Ten knowing tat ds dt = f (2) = 0 meters/sec t=2 tells us tat wen t = 2 seconds, te body is moving at an instantaneous velocity of 0 meters/sec. Tis means tat if te body continued to move at tis speed for a wole second, it would move 0 meters. In practice, owever, te velocity of te body may not remain 0 meters/sec for long. Notice tat te units of instantaneous velocity and of average velocity are te same. Eample Te cost C (in dollars) of building a ouse A square feet in area is given by te function C = f(a). Wat is te practical interpretation of te function f (A)?

26 90 Capter Two KEY CONCEPT: THE DERIVATIVE In te alternative notation, f (A) = dc da. Tis is a cost divided by an area, so it is measured in dollars per square foot. You can tink of dc as te etra cost of building an etra da square feet of ouse. Ten you can tink of dc/da as te additional cost per square foot. So if you are planning to build a ouse rougly A square feet in area, f (A) is te cost per square foot of te etra area involved in building a sligtly larger ouse, and is called te marginal cost. Te marginal cost is probably smaller tan te average cost per square foot for te entire ouse, since once you are already set up to build a large ouse, te cost of adding a few square feet is likely to be small. Eample 2 Te cost of etracting T tons of ore from a copper mine is C = f(t ) dollars. Wat does it mean to say tat f (2000) = 00? In te alternative notation, f (2000) = dc dt = 00. T =2000 Since C is measured in dollars and T is measured in tons, dc/dt must be measured in dollars per ton. So te statement f (2000) = 00 says tat wen 2000 tons of ore ave already been etracted from te mine, te cost of etracting te net ton is approimately $00. Eample 3 If q = f(p) gives te number of pounds of sugar produced wen te price per pound is p dollars, ten wat are te units and te meaning of te statement f (3) = 50? Since f (3) is te limit as 0 of te difference quotient f(3 + ) f(3), te units of f (3) and te difference quotient are te same. Since f(3 + ) f(3) is in pounds and is in dollars, te units of te difference quotient and f (3) are pounds/dollar. Te statement f (3) = 50 pounds/dollar tells us tat te instantaneous rate of cange of q wit respect to p is 50 wen p = 3. In oter words, wen te price is $3, te quantity produced is increasing at 50 pounds/dollar. Tus, if te price increased by a dollar, te quantity produced would increase by approimately 50 pounds. Eample 4 Eample 5 You are told tat water is flowing troug a pipe at a constant rate of 0 cubic feet per second. Interpret tis rate as te derivative of some function. You migt tink at first tat te statement as someting to do wit te velocity of te water, but in fact a flow rate of 0 cubic feet per second could be acieved eiter wit very slowly moving water troug a large pipe, or wit very rapidly moving water troug a narrow pipe. If we look at te units cubic feet per second we realize tat we are being given te rate of cange of a quantity measured in cubic feet. But a cubic foot is a measure of volume, so we are being told te rate of cange of a volume. One way to visualize tis is to imagine all te water tat is flowing troug te pipe ending up in a tank somewere. Let V (t) be te volume of water in te tank at time t. Ten we are being told tat te rate of cange of V (t) is 0, or V (t) = dv dt = 0. Suppose P = f(t) is te population of Meico in millions, were t is te number of years since 980. Eplain te meaning of te statements: (a) f (6) = 2 (b) f (95.5) = 6 (c) (f ) (95.5) = 0.46