Chapter Primer on Differentiation

Transcription

1 Capter 0.01 Primer on Differentiation After reaing tis capter, you soul be able to: 1. unerstan te basics of ifferentiation,. relate te slopes of te secant line an tangent line to te erivative of a function,. fin erivatives of polynomial, trigonometric an transcenental functions, 4. use rules of ifferentiation to ifferentiate functions, 5. fin maima an minima of a function, an 6. apply concepts of ifferentiation to real worl problems. In tis primer, we will review te concepts of ifferentiation you learne in calculus. Mostly tose concepts are reviewe tat are applicable in learning about numerical metos. Tese inclue te concepts of te secant line to learn about numerical ifferentiation of functions, te slope of a tangent line as a backgroun to solving nonlinear equations using te Newton-Rapson meto, fining maima an minima of functions as a means of optimization, te use of te Taylor series to approimate functions, etc. Introuction Te erivative of a function represents te rate of cange of a variable wit respect to anoter variable. For eample, te velocity of a boy is efine as te rate of cange of te location of te boy wit respect to time. Te location is te epenent variable wile time is te inepenent variable. Now if we measure te rate of cange of velocity wit respect to time, we get te acceleration of te boy. In tis case, te velocity is te epenent variable wile time is te inepenent variable. Wenever ifferentiation is introuce to a stuent, two concepts of te secant line an tangent line (Figure 1) are revisite

2 0.01. Capter 0.01 f( secant line Q P tangent line Figure 1 Function curve wit tangent an secant lines. Let P an Q be two points on te curve as sown in Figure 1. Te secant line is te straigt line rawn troug P an Q. f ( Q P Figure Calculation of te secant line. a a+ Te slope of te secant line (Figure ) is ten given as f ( a + ) f ( a) m PQ, secant ( a + ) a

3 Primer on Differentiation f ( a + ) f ( a) As Q moves closer an closer to P, te limiting portion is calle te tangent line. Te slope of te tangent line m PQ, tangent ten is te limiting value of m PQ, secant as 0. f ( a + ) f ( a) mpq,tangent 0 Eample 1 Fin te slope of te secant line of te curve y 4 between points (,6) an (5,100) f( (5,100) 50 (,6) Figure Calculation of te secant line for te function Te slope of te secant line between (,6) an (5,100) is f (5) f () m y 4. Eample Fin te slope of te tangent line of te curve Te slope of te tangent line at (,6) is y 4 at point (,6).

4 Capter 0.01 f ( + ) f () m 0 4( + ) 4() 0 4( ) (4 + 4) 0 (4 + 4) f( Figure 4 Calculation of te tangent line in te function y 4. Te slope of te tangent line is f ( + ) f () m 0 4( + ) 4()

5 Primer on Differentiation (4 + 4) 0 (4 + 4) 0 4 Derivative of a Function Recall from calculus, te erivative of a function f ( at a is efine as f ( a + ) f ( a) f ( a) 0 Eample Fin f () if Eample 4 f ( 4. f ( + ) f () f () 0 4( + ) 4() 0 4( ) (4 + 4) 0 (4 + 4) 0 4 π Fin f if f ( sin( 4 π π f + f π f 4 4 lim 4 0 π π sin + sin 4 4 0

6 Capter 0.01 π π sin + sin lim 0 π π π sin cos() + cos sin() sin lim 0 cos() cos() from knowing tat 1 cos( ) lim 0 0 Secon Definition of Derivatives Tere is anoter form of te efinition of te erivative of a function. Te erivative of te function f ( at a is efine as f ( f ( a) f ( a) a a As a, te efinition is noting but te slope of te tangent line at P. f ( (, f ( ) P ( a, f ( a)) Q a f ( f ( a) a Figure 5 Grap sowing te secon efinition of te erivative. Eample 5 Fin f () if f ( 4 by using te form

7 Primer on Differentiation f ( f ( a) f ( a) a a of te efinition of a erivative. f ( f () f () 4 4() 4 6 4( 9) 4( )( + ) 4( + ) 4( + ) 4 Fining equations of a tangent line One of te numerical metos use to solve a nonlinear equation is calle te Newton- Rapson meto. Tis meto is base on te knowlege of fining te tangent line to a curve at a point. Let us look at an eample to illustrate fining te equation of te tangent line to a curve. Eample 6 Fin te equation of te line tangent to te function 4 f ( at Te line tangent is a straigt line of te form y m + c To fin te equation of te tangent line, let us first fin te slope m of te straigt line. f ( f (0.05) (0.05) m To fin te value of te y -intercept c of te straigt line, we first fin te value of te function at f (0.05) (0.05) 0.165(0.05)

8 Capter 0.01 Te tangent line passes troug te point ( 0.05, ), so m(0.05) + c (0.05) + c c f( Figure 6 Grap of function f( an te tangent line at Hence, y m + c is te equation of te tangent line. Oter Notations of Derivatives Derivates can be enote in several ways. For te first erivative, te notations are y f (, f (, y, an For te secon erivative, te notations are y f (, f (, y, an t For te n erivative, te notations are

9 Primer on Differentiation f ( n) (, n n f (, y ( n), n y n Teorems of Differentiation Several teorems of ifferentiation are given to sow ow one can fin te erivative of ifferent functions. Teorem 1 Te erivative of a constant is zero. If f ( k, were k is a constant, f ( 0. Eample 7 Fin te erivative of f ( 6. f ( 6 f ( 0 Teorem Te erivative of f n (, were 0 n is n 1 f ( n. Eample 8 Fin te erivative of 6 f ( 6 f ( f (. Eample 9 Fin te erivative of f ( 6 f ( f (. Teorem Te erivative of f ( kg(, were k is a constant is f ( kg (.

10 Capter 0.01 Eample 10 6 Fin te erivative of f ( f ( 10 6 f ( (10 ) (6 ) 5 60 Teorem 4 Te erivative of f ( u( ± v( is f ( u ( ± v (. Eample 11 Fin te erivative of f ( + 8. f ( + 8 f ( ( + 8) ( ) + (8) ( ) + 0 ( ) 9 Teorem 5 Te erivative of f ( u( v( is f ( u( v( + v( u(. (Prouct Rule) Eample 1 Fin te erivative of f ( ( 6)( + 8)

11 Primer on Differentiation Using te prouct rule as given by Teorem 5 were, f ( u( v( f ( u( v( + v( u( f ( ( 6)( + 8) u ( 6 v ( + 8 Taking te erivative of u (, u ( 6) ( ) (6) ( ) 0 ( 4 Taking te erivative of v (, v ( + 8) ( ) + (8) ( ) + 0 ( ) 9 Using te formula for te prouct rule f ( u( v( + v( u( ( 6)(9 ) + ( + 8)( Teorem 6 Te erivative of u( f ( v( is

12 Capter 0.01 v( u( u( v( f ( (Quotient Rule) ( v( ) Eample 1 ( 6) Fin te erivative of f (. ( + 8) Use te quotient rule of Teorem 6, if u( f ( v( ten v( u( u( v( f ( ( v( ) From ( 6) f ( ( + 8) we ave u ( 6 v ( + 8 Taking te erivative of u (, u ( 6) ( ) (6) ( ) 0 ( 4 Taking te erivative of v (, v ( + 8) ( ) + (8) ( ) + 0 ( )

13 Primer on Differentiation Using te formula for te quotient rule, ( + 8)(4 ( 6)(9 ) f ( ( + 8) Table of Derivatives f ( f ( n, n 0 n n 1 n k, n 0 n 1 kn sin ( cos( cos ( sin( tan ( sec ( sin ( cos( cos ( sin( tan ( 1 tan ( sin 1 ( 1 1 cos 1 ( tan 1 ( 1+ csc ( csc( cot( sec ( sec( tan( cot ( csc ( csc ( cot( csc( sec ( tan( sec(

14 Capter 0.01 cot ( 1 cot ( csc 1 ( 1 sec 1 ( 1 cot 1 ( 1 1+ ln ( a) a a ln( 1 log a ( 1 ln ( a) e e Cain Rule of Differentiation Sometimes functions tat nee to be ifferentiate o not fall in te form of simple functions or te forms escribe previously. Suc functions can be ifferentiate using te cain rule if tey are of te form f ( g( ). Te cain rule states ( f ( g( ) f ( g( ) g ( 4 For eample, to fin f ( of f ( (, one coul use te cain rule. g( ( g ( 6 f ( g( ) 4( g( ) 4 (( ) 4( (6 ) Implicit Differentiation Sometimes, te function to be ifferentiate is not given eplicitly as an epression of te inepenent variable. In suc cases, ow o we fin te erivatives? We will iscuss tis via eamples. Eample 14 y Fin if y + y

15 Primer on Differentiation y y ( + y ) (y) ( ) + ( y ) (y) y y + y + y y y y y y (y y y y y y 1 Eample 15 If y + y 5, fin te value of y. y + y 5 ( y + y ) (5) ( ) ( y) + ( y ) 0 y y y + y 0 y ( + y) + y y y y y y y Higer orer erivatives So far, we ave limite our iscussion to calculating first erivative, f ( of a function f (. Wat if we are aske to calculate iger orer erivatives of f (. A simple eample of tis is fining acceleration of a boy from a function tat gives te location of te boy as a function of time. Te erivative of te location wit respect to time

16 Capter 0.01 is te velocity of te boy, followe by te erivative of velocity wit respect to time being te acceleration. Hence, te secon erivative of te location function gives te acceleration function of te boy. Eample 16 Given f ( 7, fin te secon erivative, f ( ) an te tir erivative, f (. Given f ( 7 we ave f ( ( ) 9 f ( ( f ( ) (9 ) 9( 18 f ( ( f ( ) (18 18 Eample 17 If y + y 5, fin te value of y. From Eample 15 we obtain y y, y ( y y y (( y y ) ( y ( y ( y ) + y (y ( y) ( y ( y + y (y 1) y y y y y

17 Primer on Differentiation After substitution of y, y y y y y y 6( y y + ) (y Fining maimum an minimum of a function Te knowlege of first erivative an secon erivative of a function is use to fin te minimum an maimum of a function. First, let us efine wat te maimum an minimum of a function are. Let f ( be a function in omain D, ten f (a) is te maimum of te function if f ( a) f ( for all values of in te omain D. f (a) is te minimum of te function if f ( a) f ( for all values of in te omain D. Te minimum an maimum of a function are also te critical values of a function. An etreme value can occur in te interval [ c, ] at en points c,. a point in [ c, ] were f ( 0. a point in [ c, ] were f ( oes not eist. Tese critical points can be te local maimas an minimas of te function (See Figure 8). Eample 18 Fin te minimum an maimum value of f ( 5 in te interval [ 0,5]. maimum Domain [c,] minimum c Figure 7 Grap illustrating te concepts of maimum an minimum.

18 Capter 0.01 f( Absolute Maimum Local Maimum Local Maimum (f ( oes not eist) Local Minimum Local Minimum Absolute Minimum c Figure 8 Te plot sows critical points of f ( in [, ] c. f ( 5 f ( f ( 0 at 1. f ( eists everywere in [ 0,5]. So te critical points are 0, 1, 5. f (0) (0) (0) 5 5 f (1) (1) (1) 5 6 f (5) (5) (5) 5 10 Hence, te minimum value of f ( occurs at 1, an te maimum value occurs at 5.

19 Primer on Differentiation maimum 6 4 f( minimum Figure 9 Maimum an minimum values of f ( 5 over interval [0,5]. Figure 10 sows an eample of a function tat as no minimum or maimum value in te omain ( 0, ). f ( f ( 1/ Figure 10 Function tat as no maimum or minimum. Figure 11 sows te maimum of te function occurring at a singular point. Te function f ( as a sarp corner at a.

20 Capter 0.01 f ( f (a) a Figure 11 Grap emonstrates te concept of a singular point wit iscontinuous slope at a Eample 19 Fin te maimum an minimum of f ( in te interval [ 0,5]. f ( f ( f ( 0 on [ 0,5]. So te critical points are 0 an 5. f ( f (0) (0) 0 f (5) (5) 10 So te minimum value of f ( is at 0, an te maimum value is at 5. Te point(s) were te secon erivative of a function becomes zero is a way to know weter te critical point foun in te first erivative test is a local minimum or maimum. Let f ( be a function in te interval ( c, ) an f ( a) 0. f (a) is a local maimum of te function if f ( a) < 0. f (a) is a local minimum of te function if f ( a) > 0. If f ( a) 0, ten te secon erivative oes not offer any insigt into te local maima or minima. Eample 0 Remember Eample 18 were we foun f '( 0 at 1 for f ( 5 in te interval [ 0,5]. Is 1 a local maima or minima of te function?

21 Primer on Differentiation f ( 5 f ( f ( 0 at 1 f ( f (1) > 0 So te f (1) is te local minimum of te function. Applications of Derivatives Below are some eamples to sow real-life applications of ifferentiation. Eample 1 A rain gutter cross-section is sown below. A E D θ B C θ Figure 1 Gutter imensions for Eample 1. Wat angle of θ woul make te cross-sectional area of ABCD maimum? Note tat common sense or intuition may lea us to believe tat θ π / 4 woul maimize te crosssectional area of ABCD. Question your intuition. 1 Area ( BC + AD) CE CE CDsin(θ ) sin( θ ) BC AD BC + CDcos( θ ) + AB cos( θ ) AD + cos( θ ) + cos( θ ) AD + 6cos( θ ) 1 Area ( + + 6cos( θ ))(sin( θ ))

22 0.01. Capter sin ( θ ) + 9sin( θ ) cos( θ ) 9 9sin ( θ ) + sin(θ ) A 9 9cos( θ ) + cos(θ ) θ 9cos ( θ ) + 9cos(θ ) Wen is A 0? θ 9 cos ( θ ) + 9cos(θ ) 0 π θ Te angle at wic te area is maimum is θ 60. π π 9 π Area 9sin + sin For te interval of θ [ 0, π ], te area at te en points is Area(0) 0 Area( π ) 0 Eample A classic eample of te application of ifferentiation is to fin te imensions of a circular cyliner for a specific volume but wic uses te least amount of material. Do tis classic problem for a volume of 9m. Te total surface area, A of te cyliner is A top surface + sie surface + bottom surface π r + πr + πr πr + πr Te volume, V of te cyliner is V πr since V 9m. We can write

23 Primer on Differentiation πr 9 πr Tis gives te surface area just in terms of r as A π r + πr 9 πr 18 πr + r 1 πr + 18r V 9 m r Figure 1 Cyliner rawing for Eample 0. To fin te minimum, take te first erivative of A wit respect to r as A 4π r + 18( 1) r r 18 4π r r Solving for A 0, r 18 4πr 0 r 4πr r 4π

24 Capter r 4π m Since 9 πr, 9 π (1.175).5450 m But oes tis value of r correspon to a minimum? A 4π 18( ) r r 6 4π + r 6 4π A Tis value > 0 for r 1.175m. As per te secon erivative test, r 1.175m r correspons to a minimum. DIFFERENTIATION Topic Primer on Differentiation Summary Tese are tetbook notes of a primer on ifferentiation Major General Engineering Autors Autar Kaw, Luke Snyer Date July 17, 008 Web Site ttp://numericalmetos.eng.usf.eu