3.3. The Derivative as a Rate of Change. Instantaneous Rates of Change. DEFINITION Instantaneous Rate of Change

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1 3.3 The Derivative a a Rate of Change The Derivative a a Rate of Change In Section 2.1, we initiated the tudy of average and intantaneou rate of change. In thi ection, we continue our invetigation of application in which derivative are ued to model the rate at which thing change in the world around u. We reviit the tudy of motion along a line and examine other application. It i natural to think of change a change with repect to time, but other variable can be treated in the ame way. For example, a phyician may want to know how change in doage affect the body repone to a drug. An economit may want to tudy how the cot of producing teel varie with the number of ton produced. Intantaneou Rate of Change If we interpret the difference quotient ƒx + hd - ƒxdd>h a the average rate of change in ƒ over the interval from x to x + h, we can interpret it limit a h : a the rate at which ƒ i changing at the point x. DEFINITION Intantaneou Rate of Change The intantaneou rate of change of ƒ with repect to x at ƒx + hd - ƒx d ƒ x d = lim, h: h provided the limit exit. x i the derivative Thu, intantaneou rate are limit of average rate. It i conventional to ue the word intantaneou even when x doe not repreent time. The word i, however, frequently omitted. When we ay rate of change, we mean intantaneou rate of change.

2 172 Chapter 3: Differentiation EXAMPLE 1 How a Circle Area Change with It Diameter The area A of a circle i related to it diameter by the equation A = p 4 D 2. How fat doe the area change with repect to the diameter when the diameter i 1 m? Solution The rate of change of the area with repect to the diameter i da dd = p 4 # 2D = pd 2. When D = 1 m, the area i changing at rate p>2d1 = 5p m 2 >m. Poition at time t and at time t t f(t) f(t t) FIGURE 3.12 The poition of a body moving along a coordinate line at time t and hortly later at time t + t. Motion Along a Line: Diplacement, Velocity, Speed, Acceleration, and Jerk Suppoe that an object i moving along a coordinate line (ay an -axi) o that we know it poition on that line a a function of time t: The diplacement of the object over the time interval from t to t + t(figure 3.12) i and the average velocity of the object over that time interval i y ay = diplacement travel time = ƒtd. = ƒt + td - ƒtd, = t ƒt + td-ƒtd =. t To find the body velocity at the exact intant t, we take the limit of the average velocity over the interval from t to t + ta t hrink to zero. Thi limit i the derivative of ƒ with repect to t. DEFINITION Velocity Velocity (intantaneou velocity) i the derivative of poition with repect to time. If a body poition at time t i = ƒtd, then the body velocity at time t i ytd = d dt = lim t: ƒt + td-ƒtd. t EXAMPLE 2 Finding the Velocity of a Race Car Figure 3.13 how the time-to-ditance graph of a 1996 Riley & Scott Mk III-Old WSC race car. The lope of the ecant PQ i the average velocity for the 3-ec interval from t = 2 to t = 5 ec; in thi cae, it i about 1 ft> ec or 68 mph. The lope of the tangent at P i the peedometer reading at t = 2 ec, about 57 ft> ec or 39 mph. The acceleration for the period hown i a nearly contant 28.5 ft>ec 2 during

3 3.3 The Derivative a a Rate of Change Ditance (ft) 5 Secant lope i average velocity 4 for interval from Q t 2 to t 5. Tangent lope 3 i peedometer 2 reading at t 2 (intantaneou 1 velocity). P t Elaped time (ec) FIGURE 3.13 The time-to-ditance graph for Example 2. The lope of the tangent line at P i the intantaneou velocity at t = 2 ec. each econd, which i about.89g, where g i the acceleration due to gravity. The race car top peed i an etimated 19 mph. (Source: Road and Track, March 1997.) Beide telling how fat an object i moving, it velocity tell the direction of motion. When the object i moving forward ( increaing), the velocity i poitive; when the body i moving backward ( decreaing), the velocity i negative (Figure 3.14). f(t) f (t) d dt d dt increaing: poitive lope o moving forward t decreaing: negative lope o moving backward t FIGURE 3.14 For motion = ƒtd along a traight line, y = d/dt i poitive when increae and negative when decreae. If we drive to a friend houe and back at 3 mph, ay, the peedometer will how 3 on the way over but it will not how -3 on the way back, even though our ditance from home i decreaing. The peedometer alway how peed, which i the abolute value of velocity. Speed meaure the rate of progre regardle of direction.

4 174 Chapter 3: Differentiation DEFINITION Speed Speed i the abolute value of velocity. d Speed = ƒ ytd ƒ = ` dt ` EXAMPLE 3 Horizontal Motion Figure 3.15 how the velocity y = ƒ td of a particle moving on a coordinate line. The particle move forward for the firt 3 ec, move backward for the next 2 ec, tand till for a econd, and move forward again. The particle achieve it greatet peed at time t = 4, while moving backward. y MOVES FORWARD (y ) y f'(t) FORWARD AGAIN (y ) Speed up Steady (y cont) Slow down Speed up Stand till (y ) t (ec) Greatet peed Speed up Slow down MOVES BACKWARD (y ) FIGURE 3.15 The velocity graph for Example 3. HISTORICAL BIOGRAPHY Bernard Bolzano ( ) The rate at which a body velocity change i the body acceleration. The acceleration meaure how quickly the body pick up or loe peed. A udden change in acceleration i called a jerk. When a ride in a car or a bu i jerky, it i not that the acceleration involved are necearily large but that the change in acceleration are abrupt.

5 3.3 The Derivative a a Rate of Change 175 DEFINITIONS Acceleration, Jerk Acceleration i the derivative of velocity with repect to time. If a body poition at time t i = ƒtd, then the body acceleration at time t i atd = dy = d 2 dt dt 2. Jerk i the derivative of acceleration with repect to time: jtd = da dt = d 3 dt 3. t (econd) t t 1 t 2 t 3 (meter) Near the urface of the Earth all bodie fall with the ame contant acceleration. Galileo experiment with free fall (Example 1, Section 2.1) lead to the equation where i ditance and g i the acceleration due to Earth gravity. Thi equation hold in a vacuum, where there i no air reitance, and cloely model the fall of dene, heavy object, uch a rock or teel tool, for the firt few econd of their fall, before air reitance tart to low them down. The value of g in the equation = 1>2dgt 2 depend on the unit ued to meaure t and. With t in econd (the uual unit), the value of g determined by meaurement at ea level i approximately 32 ft>ec 2 (feet per econd quared) in Englih unit, and g = 9.8 m>ec 2 (meter per econd quared) in metric unit. (Thee gravitational contant depend on the ditance from Earth center of ma, and are lightly lower on top of Mt. Everet, for example.) The jerk of the contant acceleration of gravity g = 32 ft>ec 2 d i zero: An object doe not exhibit jerkine during free fall. EXAMPLE 4 Modeling Free Fall Figure 3.16 how the free fall of a heavy ball bearing releaed from ret at time t = ec. (a) How many meter doe the ball fall in the firt 2 ec? (b) What i it velocity, peed, and acceleration then? Solution = 1 2 gt 2, j = d gd =. dt (a) The metric free-fall equation i = 4.9t 2. During the firt 2 ec, the ball fall 2d = 4.92d 2 = 19.6 m. (b) At any time t, velocity i the derivative of poition: FIGURE 3.16 A ball bearing falling from ret (Example 4). ytd = td = d dt 4.9t 2 d = 9.8t.

6 176 Chapter 3: Differentiation Height (ft) max y 256 t? At t = 2, the velocity i y2d = 19.6 m>ec in the downward (increaing ) direction. The peed at t = 2 i Speed = ƒ y2d ƒ = 19.6 m>ec. The acceleration at any time t i atd = y td = td = 9.8 m>ec 2. At t = 2, the acceleration i 9.8 m>ec 2. EXAMPLE 5 Modeling Vertical Motion 4, y (a) 16t 16t 2 A dynamite blat blow a heavy rock traight up with a launch velocity of 16 ft> ec (about 19 mph) (Figure 3.17a). It reache a height of = 16t - 16t 2 ft after t ec. (a) How high doe the rock go? (b) What are the velocity and peed of the rock when it i 256 ft above the ground on the way up? On the way down? (c) What i the acceleration of the rock at any time t during it flight (after the blat)? (d) When doe the rock hit the ground again? y d dt t (b) FIGURE 3.17 (a) The rock in Example 5. (b) The graph of and y a function of time; i larget when y = d/dt =. The graph of i not the path of the rock: It i a plot of height veru time. The lope of the plot i the rock velocity, graphed here a a traight line. t Solution (a) In the coordinate ytem we have choen, meaure height from the ground up, o the velocity i poitive on the way up and negative on the way down. The intant the rock i at it highet point i the one intant during the flight when the velocity i. To find the maximum height, all we need to do i to find when y = and evaluate at thi time. At any time t, the velocity i The velocity i zero when The rock height at t = 5 ec i max = 5d = 165d - 165d 2 = 8-4 = 4 ft. See Figure 3.17b. (b) To find the rock velocity at 256 ft on the way up and again on the way down, we firt find the two value of t for which To olve thi equation, we write y = d dt = d dt 16t - 16t 2 d = 16-32t ft>ec t = or t = 5 ec. td = 16t - 16t 2 = t 2-16t = 16t 2-1t + 16d = t - 2dt - 8d = t = 2 ec, t = 8 ec.

7 3.3 The Derivative a a Rate of Change 177 y (dollar) Slope marginal cot x (ton/week) x h y c(x) FIGURE 3.18 Weekly teel production: c(x) i the cot of producing x ton per week. The cot of producing an additional h ton i cx + hd - cxd. x The rock i 256 ft above the ground 2 ec after the exploion and again 8 ec after the exploion. The rock velocitie at thee time are y2d = d = = 96 ft>ec. y8d = d = = -96 ft>ec. At both intant, the rock peed i 96 ft> ec. Since y2d 7, the rock i moving upward ( i increaing) at t = 2 ec; it i moving downward ( i decreaing) at t = 8 becaue y8d 6. (c) At any time during it flight following the exploion, the rock acceleration i a contant a = dy dt = d dt 16-32td = -32 ft>ec2. The acceleration i alway downward. A the rock rie, it low down; a it fall, it peed up. (d) The rock hit the ground at the poitive time t for which =. The equation 16t - 16t 2 = factor to give 16t1 - td =, o it ha olution t = and t = 1. At t =, the blat occurred and the rock wa thrown upward. It returned to the ground 1 ec later. y x x 1 x 1 c y c(x) dc dx FIGURE 3.19 The marginal cot dc>dx i approximately the extra cot c of producing x = 1 more unit. x Derivative in Economic Engineer ue the term velocity and acceleration to refer to the derivative of function decribing motion. Economit, too, have a pecialized vocabulary for rate of change and derivative. They call them marginal. In a manufacturing operation, the cot of production c(x) i a function of x, the number of unit produced. The marginal cot of production i the rate of change of cot with repect to level of production, o it i dc>dx. Suppoe that c(x) repreent the dollar needed to produce x ton of teel in one week. It cot more to produce x + h unit per week, and the cot difference, divided by h, i the average cot of producing each additional ton: cx + hd - cxd h = The limit of thi ratio a h : i the marginal cot of producing more teel per week when the current weekly production i x ton (Figure 3.18). dc dx = lim cx + hd - cxd h: h Sometime the marginal cot of production i looely defined to be the extra cot of producing one unit: c x average cot of each of the additional h ton of teel produced. = marginal cot of production. cx + 1d - cxd =, 1 which i approximated by the value of dc>dx at x. Thi approximation i acceptable if the lope of the graph of c doe not change quickly near x. Then the difference quotient will be cloe to it limit dc>dx, which i the rie in the tangent line if x = 1 (Figure 3.19). The approximation work bet for large value of x.

8 178 Chapter 3: Differentiation Economit often repreent a total cot function by a cubic polynomial cxd = ax 3 + bx 2 + gx + d where d repreent fixed cot uch a rent, heat, equipment capitalization, and management cot. The other term repreent variable cot uch a the cot of raw material, taxe, and labor. Fixed cot are independent of the number of unit produced, wherea variable cot depend on the quantity produced. A cubic polynomial i uually complicated enough to capture the cot behavior on a relevant quantity interval. EXAMPLE 6 Suppoe that it cot Marginal Cot and Marginal Revenue cxd = x 3-6x x dollar to produce x radiator when 8 to 3 radiator are produced and that rxd = x 3-3x x give the dollar revenue from elling x radiator. Your hop currently produce 1 radiator a day. About how much extra will it cot to produce one more radiator a day, and what i your etimated increae in revenue for elling 11 radiator a day? Solution c 1d: The cot of producing one more radiator a day when 1 are produced i about c xd = d dx Ax3-6x xB = 3x 2-12x + 15 c 1d = 31d - 121d + 15 = 195. The additional cot will be about $195. The marginal revenue i r xd = d dx Ax3-3x xB = 3x 2-6x The marginal revenue function etimate the increae in revenue that will reult from elling one additional unit. If you currently ell 1 radiator a day, you can expect your revenue to increae by about if you increae ale to 11 radiator a day. r 1d = 31d - 61d + 12 = $252 EXAMPLE 7 Marginal Tax Rate To get ome feel for the language of marginal rate, conider marginal tax rate. If your marginal income tax rate i 28% and your income increae by $1, you can expect to pay an extra $28 in taxe. Thi doe not mean that you pay 28% of your entire income in taxe. It jut mean that at your current income level I, the rate of increae of taxe T with repect to income i dt>di =.28. You will pay $.28 out of every extra dollar you earn in taxe. Of coure, if you earn a lot more, you may land in a higher tax bracket and your marginal rate will increae.

9 3.3 The Derivative a a Rate of Change 179 Senitivity to Change When a mall change in x produce a large change in the value of a function ƒ(x), we ay that the function i relatively enitive to change in x. The derivative ƒ xd i a meaure of thi enitivity. EXAMPLE 8 Genetic Data and Senitivity to Change The Autrian monk Gregor Johann Mendel ( ), working with garden pea and other plant, provided the firt cientific explanation of hybridization. Hi careful record howed that if p (a number between and 1) i the frequency of the gene for mooth kin in pea (dominant) and 1 - pd i the frequency of the gene for wrinkled kin in pea, then the proportion of mooth-kinned pea in the next generation will be y = 2p1 - pd + p 2 = 2p - p 2. The graph of y veru p in Figure 3.2a ugget that the value of y i more enitive to a change in p when p i mall than when p i large. Indeed, thi fact i borne out by the derivative graph in Figure 3.2b, which how that dy>dp i cloe to 2 when p i near and cloe to when p i near 1. dy/dp 2 1 y dy dp 2 2p y 2p p 2 1 p 1 p (a) (b) FIGURE 3.2 (a) The graph of y = 2p - p 2, decribing the proportion of mooth-kinned pea. (b) The graph of dy>dp (Example 8). The implication for genetic i that introducing a few more dominant gene into a highly receive population (where the frequency of wrinkled kin pea i mall) will have a more dramatic effect on later generation than will a imilar increae in a highly dominant population.