Derivatives and Rates of Change
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1 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Derivatives and Rates of Cange Measuring te Rate of Increase of Blood Alcool Concentration Biomedical scientists ave studied te cemical and pysiological canges in te body tat result from alcool consumption. Te reaction in te uman body occurs in two stages: a fairly rapid process of absorption and a more gradual one of metabolism. To predict te effect of alcool consumption, one needs to know te rate at wic alcool is absorbed and metabolized. Medical researcers measured te blood alcool concentration(bac) of eigt fasting adult male subjects after rapid consumption of 15 ml of etanol corresponding to one alcoolic drink). Te data tey obtained were modeled by te concentration function C(t) = te t were t is measured in minutes after consumption and C is measured in mg/ml Te grap of C is sown in te Figure below. EXAMPLE: How quickly is te BAC(given by te Equation above) increasing after 10 minutes? Solution: We are asked to find te rate of cange of C wit respect to t wen t = 10. Te difficulty is tat we are dealing wit a single instant of time (t = 10 min) and so no time interval is involved. However, we can approximate te desired quantity by calculating te average rate of cange of C wit respect to t in te time interval from t = 10 to t = 11: average rate of cange = cange in C cange in t = C(11) C(10) = (mg/ml)/min Te following table sows te results of similar calculations of te average rates of cange [in (mg/ml)/ min] over successively smaller time periods. It appears tat as we sorten te time period, te average rate of cange is becoming closer and closer to a number between and (mg/ml)/min. Te instantaneous rate of cange at t = 10 is defined to be te limiting value of tese average rates of cange over sorter and sorter time periods tat start or end at t = 10. So we estimate tat te BAC increased at a rate of about (mg/ml)/min. 1
2 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Te Tangent Problem EXAMPLE: Grap te parabola y = x 2 and te tangent line at te point P(1,1). Solution: We ave: DEFINITION: Te tangent line to te curve y = f(x) at te point P(a,f(a)) is te line troug P wit slope m (1) provided tat tis limit exists. Tere is anoter (equivalent) expression for te slope of te tangent line: m f(a+) f(a) (2) 2
3 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka EXAMPLE: Find an equation of te tangent line to te yperbola y = /x at te point (,1). Solution 1: Let f(x) = /x. Ten te slope of te tangent line at (,1) is f(x)=/x m x ( xa a x ) a xa () xa x xa a xa() a x xa() (a x) xa() () xa() xa = a= = a 2 = 1 2 Recall, tat te point-slope equation of a line is y y 0 = m(x x 0 ) Terefore, an equation of te tangent line at te point (,1) is y 1 = 1 (x ) wic simplifies to x+y 6 = 0 Solution 2: Equivalently, if we use formula (2), we get m f(a+) f(a) f(x)=/x a+ a ( (a+)a a+ ) a (a+)a (a+)a a+ (a+)a a (a+)a a (a+) (a+)a a a (a+)a (a+)a and te same result follows. (a+)a = (a+0)a = a= = a 2 = 1 2
4 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Derivatives We ave seen tat limits of te form lim or lim f(a+) f(a) arise wenever we calculate a rate of cange in any of te sciences, suc as a rate of growt in biology or a rate of reaction in cemistry. Since tis type of limit occurs so widely, it is given a special name and notation. DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. f (a) f(a+) f(a) REMARK: Equivalently, f (a) EXAMPLE: Find te derivative of te function y = x 2 8x+9 at te number a. Ten find an equation of te tangent line to te parabola y = x 2 8x+9 at te point (, 6). 4
5 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka EXAMPLE: Find te derivative of te function y = x 2 8x+9 at te number a. Ten find an equation of te tangent line to te parabola y = x 2 8x+9 at te point (, 6). Solution: We ave f (a) (x 2 8x+9) (a 2 8a+9) x 2 8x+9 a 2 +8a 9 x a x 2 a 2 8x+8a ()(x+a) 8() ()[(x+a) 8] [(x+a) 8] or f (a) f(a+) f(a) [(a+) 2 8(a+)+9] [a 2 8a+9] a 2 +2a+ 2 8a 8+9 a 2 +8a 9 2a+ 2 8 (2a+ 8) (2a+ 8) = 2a+0 8 = 2a 8 so = 2a 8 f (a) = 2a 8 To find an equation of te tangent line to te parabola y = x 2 8x+9 at te point (, 6) we will use te point-slope equation of a line: y y 0 = m(x x 0 ) Since (x 0,y 0 ) = (, 6) and m = f () = 2 8 = 2, we obtain y ( 6) = 2(x ) or y = 2x 5
6 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Rates of Cange Suppose y is a quantity tat depends on anoter quantity x. Tus y is a function of x and we write y = f(x). If x canges from x 1 to x 2, ten te cange in x (also called te increment of x) is x = x 2 x 1 and te corresponding cange in y is Te difference quotient y = f(x 2 ) f(x 1 ) y x = f(x 2) f(x 1 ) x 2 x 1 is called te average rate of cange of y wit respect to x over te interval [x 1,x 2 ] and can be interpreted as te slope of te secant line PQ in te figure below. By analogy wit velocity, we consider te average rate of cange over smaller and smaller intervals by letting x 2 approac x 1 and terefore letting x approac 0. Te limit of tese average rates of cange is called te (instantaneous) rate of cange of y wit respect to x at x = x 1, wic is interpreted as te slope of te tangent curve y = f(x) at (P(x 1,f(x 1 )): y instantaneous rate of cange x 0 x f(x 2 ) f(x 1 ) x 2 x 1 x 2 x 1 Since f f(x 2 ) f(x 1 ) (x 1 ) x 2 x 1 x 2 x 1 we ave a second interpretation of te rate of cange: Te derivative f (a) is te instantaneous rate of cange of y = f(x) wit respect to x wen x = a 6
7 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka EXAMPLE: Draw te tangent line to te BAC curve in Example 1 at t = 10 and interpret its slope. Solution: In Example 1 we estimated tat te rate of increase of te blood alcool concentration wen t = 10 is about (mg/ml)/min. Te equation of te curve is C(t) = te t wic gives C(10) So, using te point-slope equation of a line, we get tat an approximate equation of te tangent line at t = 10 is C = (t 10) = C = t Te concentration curve and its tangent line are graped in te Figure below and te slope of te tangent line is te rate of increase of BAC wen t = 10. EXAMPLE: Te table at te rigt, supplied by Andrew Read, sows experimental data involving malarial parasites. Te time t is measured in days and N is te number of parasites per microliter of blood. (a) Find te average rates of cange of N wit respect to t over te intervals [1,],[2,],[,4], and [,5]. Solution: Te average rate of cange over [1,] is N() N(1) = 12, = 6261 (parasites/µl)/day 1 2 Similar calculations give te average rates of cange in te following table: (b) Te derivative N () means te rate of cange of N wit respect to t wen t = days. We ave N N(t) N() () t t Te difference quotients in tis expression (for various values of t) are just te rates of cange in te table in part (a). So N () lies somewere 10,9 and 1,911 (parasites/µl)/day. We estimate tat te rate of increase of te parasite population on day was approximately te average of tese two numbers, namely N () 12,152 (parasites/µl)/day 7
8 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Te Velocity Problem Suppose an object moves along a straigt line according to an equation of motion s = f(t), were s is te displacement (directed distance) of te object from te origin at time t. Te function f tat describes te motion is called te position function of te object. In te time interval from t = a to t = a+ te cange in position is f(a+) f(a). Te average velocity over tis time interval is average velocity = displacement time = f(a+) f(a) wic is te same as te slope of te secant line PQ in te second figure. Now suppose we compute te average velocities over sorter and sorter time intervals [a, a+]. In oter words, we let approac 0. We define velocity (or instantaneous velocity) v(a) at time t = a to be te limit of tese average velocities: v(a) f(a+) f(a) REMARK: Equivalently, v(a) EXAMPLE: Suppose tat a ball is dropped from te upper observation deck of te CN Tower, 450 m above te ground. (a) Wat is te velocity of te ball after 5 seconds? (b) How fast is te ball traveling wen it its te ground? 8
9 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka EXAMPLE: Suppose tat a ball is dropped from te upper observation deck of te CN Tower, 450 m above te ground. (a) Wat is te velocity of te ball after 5 seconds? (b) How fast is te ball traveling wen it its te ground? Solution: (a) We use te equation of motion s = f(t) = 4.9t 2 were t is time (in seconds) and s is te displacement (in meters) to find te velocity v(a) after a seconds: v(a) 4.9x 2 4.9a 2 4.9(x 2 a 2 ) 4.9()(x+a) 4.9(x+a) = 4.9(a+a) = 4.9(2a) = 9.8a or v(a) f(a+) f(a) 4.9(a+) 2 4.9a 2 4.9(a 2 +2a+ 2 ) 4.9a 2 4.9(a 2 +2a+ 2 a 2 ) 4.9(2a+ 2 ) 4.9(2a+) 4.9(2a+) = 4.9(2a+0) = 4.9(2a) = 9.8a Terefore te velocity after 5 seconds is v(5) = = 49 m/s. (b) Since te observation deck is 450 m above te ground, te ball will it te ground at te time t 1 wen s(t 1 ) = 450, tat is, 4.9t 2 1 = 450 = t 2 1 = = t 1 = s Te velocity of te ball as it its te ground is terefore 450 v(t 1 ) = 9.8t 1 = m/s 4.9 9
10 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Appendix EXAMPLE: Find an equation of te tangent line to te yperbola y = x x at te point (1/,f(1/)). Solution 1: Let f(x) = x x. Ten te slope of te tangent line at (1/,f(1/)) is m x a x+a ()(x 2 +xa+a 2 1) f(x)=x x (x x) (a a) (x a ) () = a 2 +a 2 +a 2 1 = a 2 1 a=1/ = Recall, tat te point-slope equation of a line is (x 2 +xa+a 2 1) ( ) = 2 y y 0 = m(x x 0 ) x +a Terefore, an equation of te tangent line at te point (1/,f(1/)) is y y 0 = m(x x 0 ) = y f(1/) = 2 ( x 1 ) ()(x 2 +xa+a 2 ) () 1 = y = 2 ( x 1 ) wic simplifies to 18x+27y +2 = 0 10
11 Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Solution 2: Equivalently, if we use formula (2), we get m f(a+) f(a) f(x)=x x [(a+) (a+)] [a a] [a +a 2 +a 2 + a ] [a a] a 2 +a 2 + (a 2 +a+ 2 1) (a 2 +a+ 2 1) = a 2 1 ( ) 2 a=1/ 1 = 1 and te same result follows. = 2 Solution : Similarly, if we use formula (2), we get m f(a+) f(a) f(x)=x x [(a+) (a+)] [a a] [(a+) a ] [(a+) a] [((a+) a)((a+) 2 +(a+)a+a 2 )] [(a+) a] ((a+) 2 +(a+)a+a 2 ) [(a+) 2 +(a+)a+a 2 1] ((a+) 2 +(a+)a+a 2 1) = a 2 +a 2 +a 2 1 = a 2 1 ( ) 2 a=1/ 1 = 1 and te same result follows. = 2 11
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