Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk Derivtives nd Rtes of Cnge Te Tngent Problem EXAMPLE: Grp te prbol y = x 2 nd te tngent line t te point P(1,1). Solution: We ve: DEFINITION: Te tngent line to te curve y = f(x) t te point P(,f()) is te line troug P wit slope m (1) provided tt tis limit exists. Tere is noter (equivlent) expression for te slope of te tngent line: m f(+) f() (2) 1
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk EXAMPLE: Find n eqution of te tngent line to te yperbol y = /x t te point (,1). Solution 1: Let f(x) = /x. Ten te slope of te tngent line t (,1) is f(x)=/x m x ( x x ) x () x x x x() x x() ( x) x() () x() x = = = 2 = 1 2 Recll, tt te point-slope eqution of line is y y 0 = m(x x 0 ) Terefore, n eqution of te tngent line t te point (,1) is y 1 = 1 (x ) wic simplifies to x+y 6 = 0 Solution 2: Equivlently, if we use formul (2), we get m f(+) f() f(x)=/x + ( (+) + ) (+) (+) + (+) (+) (+) (+) (+) (+) nd te sme result follows. (+) = (+0) = = = 2 = 1 2 EXAMPLE: Find n eqution of te tngent line to y = x x t te point (1/,f(1/)). 2
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk EXAMPLE: Find n eqution of te tngent line to te yperbol y = x x t te point (1/,f(1/)). Solution 1: Let f(x) = x x. Ten te slope of te tngent line t (1/,f(1/)) is m x x+ ()(x 2 +x+ 2 1) f(x)=x x (x x) ( ) (x ) () = 2 + 2 + 2 1 = 2 1 =1/ = Recll, tt te point-slope eqution of line is (x 2 +x+ 2 1) ( ) 2 1 1 = 2 y y 0 = m(x x 0 ) x + Terefore, n eqution of te tngent line t te point (1/,f(1/)) is y y 0 = m(x x 0 ) = y f(1/) = 2 ( x 1 ) wic simplifies to 18x+27y +2 = 0 Solution 2: Equivlently, if we use formul (2), we get m f(+) f() f(x)=x x [(+) (+)] [ ] [ + 2 + 2 + ] [ ] ( 2 ++ 2 1) nd te sme result follows. Solution : Similrly, if we use formul (2), we get m f(+) f() [(+) ] [(+) ] ((+) 2 +(+)+ 2 ) ()(x 2 +x+ 2 ) () 1 = y + 8 27 = 2 2 + 2 + ( 2 ++ 2 1) = 2 1 =1/ = f(x)=x x [(+) (+)] [ ] ( x 1 ) ( ) 2 1 1 = 2 [((+) )((+) 2 +(+)+ 2 )] [(+) ] [(+) 2 +(+)+ 2 1] ((+) 2 +(+)+ 2 1) = 2 + 2 + 2 1 = 2 1 =1/ = nd te sme result follows. ( ) 2 1 1 = 2
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk Te Velocity Problem Suppose n object moves long strigt line ccording to n eqution of motion s = f(t), were s is te displcement (directed distnce) of te object from te origin t time t. Te function f tt describes te motion is clled te position function of te object. In te time intervl from t = to t = + te cnge in position is f(+) f(). Te verge velocity over tis time intervl is verge velocity = displcement time = f(+) f() wic is te sme s te slope of te secnt line PQ in te second figure. Now suppose we compute te verge velocities over sorter nd sorter time intervls [, +]. In oter words, we let pproc 0. We define velocity (or instntneous velocity) v() t time t = to be te limit of tese verge velocities: v() f(+) f() REMARK: Equivlently, v() EXAMPLE: Suppose tt bll is dropped from te upper observtion deck of te CN Tower, 450 m bove te ground. () Wt is te velocity of te bll fter 5 seconds? (b) How fst is te bll trveling wen it its te ground? 4
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk EXAMPLE: Suppose tt bll is dropped from te upper observtion deck of te CN Tower, 450 m bove te ground. () Wt is te velocity of te bll fter 5 seconds? (b) How fst is te bll trveling wen it its te ground? Solution: () We use te eqution of motion s = f(t) = 4.9t 2 were t is time (in seconds) nd s is te displcement (in meters) to find te velocity v() fter seconds: v() 4.9x 2 4.9 2 4.9(x 2 2 ) 4.9()(x+) 4.9(x+) = 4.9(+) = 4.9(2) = 9.8 or v() f(+) f() 4.9(+) 2 4.9 2 4.9( 2 +2+ 2 ) 4.9 2 4.9( 2 +2+ 2 2 ) 4.9(2+ 2 ) 4.9(2+) 4.9(2+) = 4.9(2+0) = 4.9(2) = 9.8 Terefore te velocity fter 5 seconds is v(5) = 9.8 5 = 49 m/s. (b) Since te observtion deck is 450 m bove te ground, te bll will it te ground t te time t 1 wen s(t 1 ) = 450, tt is, 4.9t 2 1 = 450 = t 2 1 = 450 450 = t 1 = 4.9 4.9 9.6 s Te velocity of te bll s it its te ground is terefore 450 v(t 1 ) = 9.8t 1 = 9.8 94 m/s 4.9 5
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk Derivtives We ve seen tt limits of te form lim or lim f(+) f() rise in finding te slope of tngent line or te velocity of n object. Moreover, te sme type of limit rises wenever we clculte rte of cnge in ny of te sciences or engineering, suc s rte of rection in cemistry or mrginl cost in economics. Since tis type of limit occurs so widely, it is given specil nme nd nottion. DEFINITION: Te derivtive of function f t number, denoted by f (), is if tis limit exists. f () f(+) f() REMARK: Equivlently, f () EXAMPLE: Find te derivtive of te function y = x 2 8x+9 t te number. Ten find n eqution of te tngent line to te prbol y = x 2 8x+9 t te point (, 6). 6
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk EXAMPLE: Find te derivtive of te function y = x 2 8x+9 t te number. Ten find n eqution of te tngent line to te prbol y = x 2 8x+9 t te point (, 6). Solution: We ve f () (x 2 8x+9) ( 2 8+9) x 2 8x+9 2 +8 9 x 2 8 2 +8 x 2 2 8x+8 ()(x+) 8() ()[(x+) 8] [(x+) 8] or f () f(+) f() [(+) 2 8(+)+9] [ 2 8+9] 2 +2+ 2 8 8+9 2 +8 9 2+ 2 8 (2+ 8) (2+ 8) = 2+0 8 = 2 8 so = 2 8 f () = 2 8 To find n eqution of te tngent line to te prbol y = x 2 8x+9 t te point (, 6) we will use te point-slope eqution of line: y y 0 = m(x x 0 ) Since (x 0,y 0 ) = (, 6) nd m = f () = 2 8 = 2, we obtin y ( 6) = 2(x ) or y = 2x 7
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk Rtes of Cnge Suppose y is quntity tt depends on noter quntity x. Tus y is function of x nd we write y = f(x). If x cnges from x 1 to x 2, ten te cnge in x (lso clled te increment of x) is x = x 2 x 1 nd te corresponding cnge 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 clled te verge rte of cnge of y wit respect to x over te intervl [x 1,x 2 ] nd cn be interpreted s te slope of te secnt line PQ in te figure bove. By nlogy wit velocity, we consider te verge rte of cnge over smller nd smller intervls by letting x 2 pproc x 1 nd terefore letting x pproc 0. Te limit of tese verge rtes of cnge is clled te (instntneous) rte of cnge of y wit respect to x t x = x 1, wic is interpreted s te slope of te tngent curve y = f(x) t (P(x 1,f(x 1 )): y instntneous rte of cnge 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 ), we ve second interprettion of te rte of cnge: x 2 x 1 x 2 x 1 Te derivtive f () is te instntneous rte of cnge of y = f(x) wit respect to x wen x = EXAMPLE: Let D(t) be te US ntionl debt t time t. Te tble below gives pproximte vlues of tis function by providing end of yer estimtes, in billions of dollrs, from 1980 to 2000. Interpret nd estimte te vlue of D (1990). t D(t) 1980 90.2 1985 1945.9 1990 2. 1995 4974.0 2000 5674.2 8
Section 2.1 Derivtives nd Rtes of Cnge 2010 Kiryl Tsiscnk EXAMPLE: Let D(t) be te US ntionl debt t time t. Te tble below gives pproximte vlues of tis function by providing end of yer estimtes, in billions of dollrs, from 1980 to 2000. Interpret nd estimte te vlue of D (1990). t D(t) 1980 90.2 1985 1945.9 1990 2. 1995 4974.0 2000 5674.2 Solution: Te derivtive D (1990) mens te rte of cnge of D wit respect to t wen t = 1990, tt is, te rte of increse of te ntionl debt in 1990. We know tt D D(t) D(1990) (1990) t 1990 t 1990 So we compute vlues of te difference quotient (te verge rtes of cnge) s follows: D(t) D(1990) t t 1990 1980 20.1 1985 257.48 1995 48.14 2000 244.09 From tis tble we see tt D (1990) lies somewere between 257.48 nd 48.14 billion dollrs per yer. [Here we re mking te resonble ssumption tt te debt didn t fluctute wildly between 1980 nd 2000.] We estimte tt te rte of increse of te ntionl debt of te United Sttes in 1990 ws te verge of tese two numbers, nmely D (1990) 257.48+48.14 2 = 02.81 0 billion dollrs per yer Anoter metod would be to plot te debt function nd estimte te slope of te tngent line wen t = 1990. 9