Derivatives and Rates of Change

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1 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

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 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

3 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 1 =1/ = Recll, tt te point-slope eqution of line is (x 2 +x+ 2 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 [(+) (+)] [ ] [ ] [ ] ( ) 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 = ( ) = 2 1 =1/ = f(x)=x x [(+) (+)] [ ] ( x 1 ) ( ) = 2 [((+) )((+) 2 +(+)+ 2 )] [(+) ] [(+) 2 +(+)+ 2 1] ((+) 2 +(+)+ 2 1) = = 2 1 =1/ = nd te sme result follows. ( ) = 2

4 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

5 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 (x 2 2 ) 4.9()(x+) 4.9(x+) = 4.9(+) = 4.9(2) = 9.8 or v() f(+) f() 4.9(+) ( ) ( ) 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) = = 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 = = t 1 = s Te velocity of te bll s it its te ground is terefore 450 v(t 1 ) = 9.8t 1 = m/s 4.9 5

6 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

7 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 x x 2 2 8x+8 ()(x+) 8() ()[(x+) 8] [(x+) 8] or f () f(+) f() [(+) 2 8(+)+9] [ 2 8+9] (2+ 8) (2+ 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

8 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 Interpret nd estimte te vlue of D (1990). t D(t)

9 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 Interpret nd estimte te vlue of D (1990). t D(t) 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 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 From tis tble we see tt D (1990) lies somewere between nd 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) = billion dollrs per yer Anoter metod would be to plot te debt function nd estimte te slope of te tngent line wen t =

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