Differentiation. introduction to limits

Transcription

1 9 9A Introduction to limits 9B Limits o discontinuous, rational and brid unctions 9C Dierentiation using i rst principles 9D Finding derivatives b rule 9E Antidierentiation 9F Deriving te original unction rom te gradient unction Dierentiation areas of stud Tis area o stud will include: Te derivative as te gradient o te grap o a unction at a point and its representation b a gradient unction Notation or derivatives D ( ), d d, (), d( ( )) d First principles approac to inding te gradient unction or () = n, n Z and simple polnomial unctions Derivatives o simple power unctions and polnomial unctions Antidierentiation as te reverse process o dierentiation and identiication o amilies o curves wit te same gradient unction. ebookplus 9a introduction to limits Digital doc Quick Questions In matematics it is important to understand te concept o a limit. Tis concept is especiall important in te stud o calculus. In everda lie we use te term limit to describe a restriction put on a quantit. For eample, te legal blood alcool concentration limit or a driver is normall.5 g/ ml. As te number o standard alcoolic drinks consumed in our approaces, te average adult s blood alcool concentration approaces.5. Likewise, some time ater a celebration, a person wo as been drinking eavil at an earlier time ma ave a blood alcool concentration wic is approacing te legal limit o.5 rom a iger level, as te number o drinks not et metabolised b teir bod approaces. We could sa tat as te number o standard drinks remaining in te bod approaces, te blood alcool concentration approaces.5. In essence te blood alcool concentration is a unction, sa (), o te number o drinks,, remaining in te bod. mats Quest matematical metods Cas or te Casio Classpad

2 Worked Eample Add te ollowing series o numbers and state wat value it is approacing Tink write Add te irst terms. Sum o irst terms is (=.75). Add te irst terms. Sum o irst terms is 7 (=.875). 8 Add te irst terms. Sum o irst terms is 5 (.98). 6 Add te irst 5 terms. Sum o irst 5 terms is (.969). 5 Add te irst 6 terms. Sum o irst 6 terms is 6 (.98). 6 6 Give te upper limit. Te sum is approacing. Worked Eample Epressing limits in matematical language, we sa tat a limit can be used to describe te beaviour o a unction, (), as te independent variable,, approaces a certain value, sa a. In some cases te unction will not be deined at a. Using te correct notation or te eample on blood alcool concentration described earlier, we would write: lim ( ) = 5. Tis is read as te limit o () as approaces is equal to.5. B investigating te beaviour o te unction () = + in te vicinit o =, sow tat lim ( ) = 5: a b and b using a CAS calculator. Tink a Create a table o values or and () in te vicinit o =. Consider te values taken b () as approaces. a Write/displa () As approaces rom te let and te rigt, () approaces a value o 5. So lim ( ) = 5. 5 () b On te Spreadseet screen, label column b A, and enter te -values. To write a ormula to calculate values or () = + using te -values in column A, label column B (), and in cell B tpe: = A + Fill down to B8 b igligting B:B8; ten tap: Edit Fill Range OK Capter 9 Dierentiation

3 Find te let- and rigt-and limits as approaces. Te let- and rigt-and limits are equal, so wen () = +, lim ( ) = 5. Limits o continuous unctions A continuous unction as a grap tat orms a continuous line; tat is, it as no breaks. I a unction is continuous at te point were a limit is being ound, ten te limit alwas eists and can easil be ound b direct substitution. Worked Eample Find lim( + ): a b and b using a CAS calculator. Tink Write/displa a To consider weter te unction is continuous, sketc te grap o = + in te vicinit o =. a 6 () As te grap is continuous, substitute = (rom lim ) into + to evaluate te limit. b On te Main screen, tap: b ) - Complete te entr line as: lim ( + ) Ten press E. Te unction is continuous at =. lim ( + ) = + = 6 Write te answer. lim ( + ) = 6 In general i a unction, (), is continuous wen = a, ten lim ( ) = ( a ). a Teorems on limits Teorem For te constant unction were () = c, ten lim ( ) = c. a I lim ( ) = A and lim g ( ) = B, ten te ollowing teorems appl. a a Mats Quest Matematical Metods CAS or te Casio ClassPad

4 Teorem For eample: Teorem For eample: Teorem For eample: lim[ ( ) ± g ( )] = lim ( ) ± lim g ( ) = A ± B a a a lim( + ) = lim + lim = + = lim[ ( ) g ( )] = lim ( ) lim g ( ) = A B a a a lim [ ( )] = lim ( ) lim ( ) = - = - lim ( ) a g ( ) = lim ( ) a A = i B lim g ( ) B a + lim( + ) lim 5 + = 5 lim( + ) 5 = 9 6 Tese teorems on limits can be epressed in word orm:. Te limit o a sum equals te sum o te limits.. Te limit o a dierence equals te dierence o te limits.. Te limit o a product equals te product o te limits.. Te limit o a quotient equals te quotient o te limits. remember. I a unction, (), is continuous wen = a, ten lim ( ) = ( a ). a. For te constant unction were () = c, lim ( ) = c. a. lim[ ( ) ± g ( )] = lim ( ) ± lim g ( ) a a a. lim[ ( ) g ( )] = lim ( ) lim g ( ) a a a lim ( ) 5. a g ( ) = lim ( ) a lim g ( ) a eercise 9a introduction to limits We Add te ollowing series o numbers and state wat value it is approacing Te diagram at rigt sows regular polgons wit, and 5 sides. As te number o sides gets ver large ( ) wat sape emerges? n n n 5 Capter 9 Dierentiation

5 mc I n represents te number o sides o a regular polgon, ten wic o te ollowing correctl describes te situation given in question above? A n B lim n a Find te value o as n gets ininitel large. n b Write tis using limit notation. C lim D n E n 5 n 5 a I S = and n represents te number o terms to be summed in 6 te series, cop and complete te ollowing table: b n 5 6 S mc Wic o te ollowing is equal to lim n S? A.75 B.95 C D E 6 We B investigating te beaviour o te unction () = + 5 in te vicinit o =, sow tat lim ( ) = 8. ebookplus Digital doc SkillSHEET 9. Substituting into a unction 7 mc From te grap at rigt it can be seen tat te lim ( ) is equal to: A B - C D E () 8 For eac o te unctions graped below, ind lim ( ). a b () c () () d e () 5 () 5 () 9 Wea Evaluate te limits below. a lim ( + 5 ) b lim ( k) c lim( 9a + ) k a d lim ( + ) e lim( + ) lim ( 8 5) 5 g lim ( p p+ ) lim ( + ) p mats Quest matematical metods Cas or te Casio Classpad

6 ebookplus Digital doc Investigation Sneaking up on a limit 9B mc Web I () = - ten te value o lim ( ) is equal to: A B C D - E mc Te lim ( 5 ) is equal to: A B - 5 C 9 D - E Find te value o te ollowing limits. + a lim 6 c lim b lim + + d lim limits o discontinuous, rational and brid unctions limits o discontinuous unctions I a unction is discontinuous at te point were te limit is being investigated ten te limit will eist onl i te unction is approacing te same value rom te let as rom te rigt. Consider te discontinuous unctions graped below.. From te let, lim ( ). (Te smbol - = indicates tat we are letting approac rom te let side.) From te rigt, lim ( ). = + Let limit = rigt limit. Tereore, lim ( ) =.. From te let, lim ( ). = = + From te rigt, lim ( ). Let limit rigt limit. Tereore, lim ( ) does not eist. limits o rational unctions Finding te limit o a rational unction involves simpliing te unction beore direct substitution can take place and recognising an values o or wic it is discontinuous. () WorkeD eample a Find lim : i b and ii using a CAS calculator. b Sketc te grap o ( ) =, stating te value o or wic it is discontinuous. Capter 9 Dierentiation 5

7 Tink a i We cannot substitute = directl as we will get, wic is undeined, so we actorise te numerator. ii Cancel ( ) rom te numerator and denominator. Now substitute = (rom lim ) and evaluate. On te Main screen, tap: ) - Complete te entr line as: lim Ten press E. a WriTe/D ispla i lim = lim ( ) = ii = lim, Write te answer. lim = b Te grap o ( )= is te same as te grap o () =, ecept were te point (, ) does not eist. b () Te unction is discontinuous at =. WorkeD eample 5 a B irst actorising te numerator, simpli te rational unction ( )=, stating te value or wic te unction does not eist + (tat is, is discontinuous). b Find lim ( ), were a is te value at wic () is discontinuous. a ebookplus Tutorial int- Worked eample 5 Tink WriTe a Factorise te numerator a ( )= + = ( + )( + ) ( + ) 6 mats Quest matematical metods Cas or te Casio Classpad

8 Cancel ( + ) rom te numerator and denominator. ( ) = +, so () is discontinuous at = b Write an epression or lim ( ) a evaluate b substituting = -. and b lim + = + = - Limits o brid unctions Hbrid unctions are unctions tat ave dierent rules or dierent parts o te domain. Worked Eample 6, (, ] a Sketc te grap o te brid unction ( ) =., (, ) b Find i lim ( ) ii lim ( ) + iii lim ( ) i it eists. c Find solutions to parts a and b using a CAS calculator. Tink Write/Displa a Sketc () = over te domain ( -, ]. Sketc (on te same aes) () = - over te domain (, ). a () b i Substitute = into () =. b i lim ( ) = lim ii Substitute = into () = -. ii lim ( ) = lim ( ) iii Are tese limits equal? iii lim ( ) does not eist c On te Grap & Tab screen, complete te entr lines as: = = > Tick te and boes and tap!. + = = + (as let limit rigt limit). Capter 9 Dierentiation 7

9 On te Main screen, complete te entr lines as: lim ( ) lim ( ) + Press E ater eac entr. Write te answers. c i lim ( ( ) ) = ii lim ( ( ) ) = + iii lim( ( ) ) does not eist as te let limit does not equal te rigt limit. REMEMBER. I a unction is discontinuous at te -value were te limit is being investigated ten te limit will eist onl i te unction is approacing te same value rom te let as rom te rigt.. Finding te limit o a rational unction involves simpliing te unction beore direct substitution can take place and recognising an values o or wic it is discontinuous.. Hbrid unctions are unctions tat ave dierent rules or dierent parts o te domain. Eercise 9B Limits o discontinuous, rational and brid unctions Wic o te ollowing graps are discontinuous? a b c 8 Mats Quest Matematical Metods CAS or te Casio ClassPad

10 d e For eac discontinuous unction above, state te value o or wic it is discontinuous. WE a Find lim. b Sketc te grap o ( ) =, stating te value o or wic it is discontinuous. + a Evaluate () wen = i ( ) =. Comment on tis result. b For wat value o is () discontinuous? c Factorise te numerator o (). d Now simpli (). e Sketc te grap o (). Evaluate lim ( ) i it eists. 5 WE 5a B irst actorising te numerator, simpli te ollowing rational unctions, stating te value or wic te unction does not eist (is discontinuous) a ( )= b ( )= c ( )= d ( )= e ( )= 6 + g ( )= + 8 ( )= + 7 ( )= 6 WE 5b For eac rational unction in question 5 above ind lim ( ), were a is te a value at wic () is discontinuous. 7 WE 6a Sketc te graps o te ollowing brid unctions. +, (, ) a ( ) =, [, ), (, ) c ( ) =, [, ), (, ] b g ( ) = +, (, ) +, (, ] d p ( ) = +, (, ) 8 WE 6b For eac o te corresponding unctions in question 7 above evaluate te ollowing. a i lim ( ) b i lim g ( ) c i lim ( ) d i lim p ( ) ii lim ( ) + ii lim g ( ) + ii lim ( ) + ii lim p ( ) + iii lim ( ) iii lim g ( ) iii lim ( ) iii lim p ( ) Capter 9 Dierentiation 9

11 9 WE 6c Investigate weter te ollowing limits eist. For tose tat do eist, state te limit., ( +, ), (, ] a lim b lim, [, ), (, ) 5, (, ), (, ] c lim d lim +, [, ) +, (, ), (, ], (, ) e lim lim +, (, ) +, [, ) 9C Evaluate te ollowing. a lim ( ) b lim + c lim, ( 9 +, ) d lim + +, [, ) e lim ( +, (, ] ) lim +, (, ) + + g lim + i lim ( + )( ) k lim + lim( + 5) j lim ( + )( ) l lim Dierentiation using irst principles Te gradient unction is te rule or te instantaneous rate o cange o a given unc tion at an point. Te gradient at an point (, ) can be ound b substitution into te gradient unction. Consider te secant PQ drawn to te curve () at rigt. Te () coordinates o P are [, ()] and te coordi nates o Q Q are [ +, ( + )]. So te gradient o te secant ( ) PQ = rise run Tangent at P = ( ) ( () P + ) + = ( + ) ( ) As approaces zero, tat is, as Q draws as close as possible to P along te curve, PQ eectivel becomes a tangent to te curve at P. We can tereore sa tat te gradient o te tangent at P is Q lim ( + ) ( ) or ( ) lim ( + ) = ( ),, were () denotes te gradient o a tan gent at an point,, on te grap o (). P Q Q Q moves closer to P as approaces. Mats Quest Matematical Metods CAS or te Casio ClassPad

12 Worked Eample 7 Tat is, () is te gradient unction o (). Te process o inding te gradient unction lim ( + ) ( ) is called dierentiation rom irst principles. Two dierent orms o notation are commonl used to represent a unction and its derivative.. Te European notation o Leibniz is:. Te alternative notation is: (a) or te unction (a) () or te unction (b) d or te derivative. d (b) Find te derivative o - rom irst principles: a b and b using a CAS calculator. () or d d [ ()] or D ( ) or te derivative. Tink Write/displa a Deine (). a () = - Te derivative is equal to: lim ( + ) ( ). ( ) lim ( + ) = ( ) Find () and ( + ). () = ( + ) = ( + ) ( + ) Simpli te numerator ( + ) - (). ( + ) - () = ( + ) - ( + ) - ( - ) = = Factorise te numerator ( + ) - (). = ( + - ) 6 Simpli lim ( + ) ( ) b cancelling te common actor o. lim ( + ) ( ) = lim ( + ) = lim ( + ), 7 Evaluate te limit b substituting =. = -. b On te Main screen, complete te entr line as: b Deine () = ; ten press E. To evaluate te gradient o te secant, complete te entr line as: ( ( + ) ( )) ; ten press E. To simpli te epression: igligt it, cop it and tap: Action Transormation simpli Ten paste te epression and press E. To evaluate te derivative, complete te entr line as: lim ( + ); ten press E. Capter 9 Dierentiation

13 Write te answer. + ( ) = lim ( ) ( ) ( ) = WorkeD eample 8 I g () = + 5 -, ind: a g () using irst principles b te value(s) o were te gradient equals. Tink WriTe ebookplus Tutorial int-5 Worked eample 8 a Let g () = a g() = Te derivative is equal to: g lim ( + ) g ( ). g g () = lim ( + ) g ( ) Find g() and g( + ). g() = + 5 g( + ) = ( + ) + 5( + ) Simpli te numerator g( + ) - g(). 5 Factorise. = ( + + 5) g 6 Simpli lim ( + ) g ( ) b cancelling te common actor o. 7 Evaluate te limit b substituting =. g( + ) - g() = ( + ) + 5( + ) - - ( ) = ( + + ) = = g lim ( + ) g ( ) = lim ( ) = lim ( ), = + 5 So g () = + 5. b Solve g () =. b g () = + 5 = = - 5 = 5 So te gradient equals wen = 5. Note: For an polnomial unction, (), wen te epression ( + ) - () is simpli ied, all o its terms ave as a actor. remember. Te process o inding te gradient unction lim ( + ) ( ) is called dierentiation rom i rst principles.. Dierentiating gives d d.. Dierentiating () gives (). mats Quest matematical metods Cas or te Casio Classpad

14 eercise 9C 9D Dierentiation using irst principles We7a Find te derivative o te ollowing rom irst principles. a 5-7 b + c - 8 d + We7b Use irst principles to ind d d. a = + b = - + c = d = 9 - e = 6 - = We8 I g() = - 6 ind: a g () using irst principles. b te value(s) o were te gradient equals. a I () = - 8, ind () using irst principles. b Hence, determine te value(s) o were te gradient unction is equal to. 5 B irst deriving te gradient unction (), evaluate () wen () is equal to: a b + c - + d mc Wic o te ollowing do not denote te gradient at an point on a grap? (One or more answers.) A () B lim ( + ) ( ) C lim ( + ) ( ) D d E ( + ) ( ) d 7 mc Te most accurate metod or inding te gradient wen = or te unction () = + is b: A sketcing te grap and drawing a tangent at = to ind te gradient B inding te gradient o te secant to te curve joining te points were = and =. C inding () using irst principles and evaluating () D guessing E inding te gradient o te line rom te origin to te point (, 5). 8 mc Given tat () = i () = and g () = + i g() = +, ten te deriv ative o + + must be equal to: A + + B + C + D 5 + E 5 + Finding derivatives b rule ebookplus Digital docs SkillSHEET 9. Dierentiating rom irst principles WorkSHEET 9. Fortunatel, te tedious process o inding derivatives rom irst principles need not be applied once rules are establised. For polnomial unctions, te ollowing rules appl. Rule. I () = n, ten () = n n -. Rule. I () = a n, ten () = na n -. Rule. I () = c, ten () = (were c is constant). Rule. I () = g() + (), ten () = g () + (). WorkeD eample 9 Dierentiate eac o te ollowing: a = 8 b = c = 7+ d 5 = Capter 9 Dierentiation

15 Tink WriTe a Write te epression or. Appl rule to ind te derivative. a = 8 d = 8 d = 8 b Appl rule. b = c Appl rules, and. Remember tat =. d Dierentiate te terms separatel (tat is, appl rules and ). d d 8 7 = ( ) = 6 c = 7 + d = 7 + d = 7 = 7 d = d d = 5 ( ) + ( ) ( 6 ) 5 5 = = + 6 WorkeD eample Find () i () = ( - ). Tink WriTe Write down (). () = ( - ) Epand te brackets. () = - 6 Dierentiate b rule. () = 6-6 WorkeD eample I g( ) = +, ind g (): a b and b using a CAS calculator. ebookplus Tutorial int-988 Worked eample Tink WriTe/Displa a Factorise te numerator because at tis stage we can onl dierentiate a constant denominator. + a g ( ) = ( + ) = Simpli g(). = ( + ), Epand te brackets. = + Dierentiate g() b rule. g () = 8 + mats Quest matematical metods Cas or te Casio Classpad

16 b On te Main screen, complete te entr b line as: + Deine g ( )= Ten press E. To ind g (), tap: ) - Complete te entr line as: d d g Ten press E. Write te answer. Given g() = +, g () = 8 + Worked Eample Dierentiate eac o te ollowing. Epress all answers wit a positive inde. a ( )= b ( ) = c ( )= d ( ) =. 7 Tink Write a Write down (). a () = () = Dierentiate b rule. = - - Epress answer wit a positive inde. = b Write down (). b () = 7 Bring te term to te numerator using te inde laws, as we can onl dierentiate a constant denominator. = - 7 Dierentiate b rule. () = - 7( ) = Epress answer wit a positive inde. = c Write down (). c () = Dierentiate b rule. () = ( ) Epress answer wit a positive inde. = 7 8 = Capter 9 Dierentiation 5

17 d Write down (). d () = Convert to inde orm. ( )= Bring te term to te numerator using te inde laws. () = Dierentiate b rule. () = 5 Epress wit a positive inde. = ( ) = 6 Epress te power o back in surd orm. = Worked Eample To evaluate te gradient o a curve at a given point, substitute te given value o into te gradient unction or derivative. For te unction (), te gradient at te point (a, (a)), is ound b evaluating (a). Tis gives te gradient o te tangent at = a, wic equals te gradient o te curve onl at tat point. For te unction () = , evaluate ( - ), () and (). Tink Write te original unction and ten ind te gradient unction, (). write/display () = 5 + '() = Substitute = - into te gradient unction. '( - ) = ( - ) - ( - ) - 5 = = - '( - ) = - Substitute = into te gradient unction. '() = () - () - 5 = - 5 '() = - 5 On te Main screen, complete te entr line as: Deine () = Ten press E. To evaluate (), complete te entr line as: d d ( ( )) = Ten press E. 5 Write te answer. () = 6 Mats Quest Matematical Metods CAS or te Casio ClassPad

18 Tangents and normals As mentioned earlier in tis capter te derivative () is actuall te gradient unction. Tis means tat te value o te gradient at an particular point on a curve is equal to te numerical value o te derivative at tat point. Recall tat i te gradient o a tangent to a curve at point P is m T, ten te normal, m N, is a straigt line perpendicular (at rigt angles) to te tangent suc tat m N = and passing troug te point P as sown at rigt. m T Also recall tat te equation o a straigt line is given b - = m( - ) were (, ) is te point P, above, and m is te gradient. Normal P Tangent Worked Eample a Find te equation o te tangent to te curve () = at te point were te gradient as a value o 8. b Hence, ind te equation o te normal at tis point. c Find te equation o te tangent at tis point using a CAS calculator. Tink a Find te gradient unction o te curve, (). Find, te value o were () = 8; tat is, solve + 6 = 8. Alternativel, on te Main screen, complete te entr line as: solve d d ( + 6 8) = 8, Ten press E. Write/DISPLAY a () = () = + 6 For gradient = = 8 = = So =. Find ( ) to determine te value o. Simpli te equation - = m T ( - ) to ind te equation o te tangent. b Find te gradient o te normal using b m N = mn =. m T = ( ) = () = () + 6() - 8 = - Te equation o te tangent at te point (, - ) is - - = 8( - ) + = 8-8 = Capter 9 Dierentiation 7

19 Simpli te equation - = m N ( - ) to ind te equation o te normal. c Te gradient is 8 at te point (, ). c Substitute tese values into te equation. On te Main screen, complete te entr line as: solve( - - = 8( - ), ) Ten press E. Te equation o te normal at te point (, - ) is - - = - ( - ) 8 + = ( ) = = Write te answer. Te equation o te tangent to te curve () = + 6-8, at te point (, - ) wit a gradient o 8 is: T = 8-9. WorkeD eample 5 For te unction () = - +, ind te coordinates o te points were te gradient is perpendicular to te line = - +. Hence, ind te equations o te tangents and normals at tese points. Tink WriTe Find te derivative. () = - + () = -. m = Find te gradient o te tangent. As te tangent is perpendicular to te line = - +, te product o teir gradients is -, i.e. m m = -. Let m equal te gradient o te tangent and let m equal te gradient o te line = - +. = m were m = - = m tangent = ( ) ebookplus Tutorial int-7 Worked eample 5 Put () = and solve or. - = - - = ( + )( - ) = = or = Substitute = and = into () to ind () = () - () + = teir -coordinates and state te points. + = 7 (, ) and, are te coordinates o te points. 7 8 mats Quest matematical metods Cas or te Casio Classpad

20 5 Find te equations o te tangent lines. Equation o tangent at te point (, ) - = ( - ) = - + = Equation o tangent at te point, 7 7 = = = = 6 Find te equations o te normal lines. Use m normal = to ind te gradient o m tangent te normal. m normal = = - Equation o normal at te point (, ) - = - ( - ) = = - + Equation o normal at te point, 7 7 = = = = WorkeD eample 6 a Grap te unction () = - and () on te same set o aes. b How can () be used to predict te location and nature o te stationar points o ()? a Tink To grap () = - and () on te same set o aes, on te Grap & Tab screen complete te entr line as: = - = d d ( - ) Tick te and boes and tap!. a WriTe/Displa ebookplus Tutorial int-989 Worked eample 6 Capter 9 Dierentiation 9

21 b Te -intercepts o () predict te location and nature o te stationar points o (). b At = -.86, () goes rom positive to negative (let to rigt), indicating a maimum turning point or () = -. At =.86, () goes rom negative to positive (let to rigt) indicating a minimum turning point or (). REMEMBER. Rule : I () = n, ten () = n n -.. Rule : I () = a n, ten () = na n -.. Rule : I () = c, ten () = (were c is constant).. Rule : I () = g() + (), ten () = g () + (). 5. Equation o tangent: - = m T ( - ) 6. Equation o normal: - = m N ( - ), were m N = m T Eercise 9D Finding derivatives b rule WE 9 Dierentiate eac o te ollowing. a = 6 b = 7 c = 5 + d = Find d i is: d a b 8 7 c 5 5 d - 6 e Matc te correct derivative rom te set A to G below to eac o te ollowing. a 8 b + c d e g p A B + C p p - D 8 7 E 6 - F + 6 G + 6 Dierentiate te ollowing. a = b = c = d = e = 6 = Find () i () is: a b 5 8 c d + - e WE Find () i () is: a () = ( + ) b () = ( - 5) c () = ( + ) d () = 9(8 - ) e () = ( + ) () = ( - 5). Mats Quest Matematical Metods CAS or te Casio ClassPad

22 7 We Find g () b irst simpliing g (). ebookplus Digital docs Spreadseet 7 Gradient at a point Spreadseet 9 Tangent and normal a g ( )= c g ( )= b g ( )= d g ( )= 8 We Dierentiate eac o te ollowing. Epress all answers wit a positive inde. a - b - 7 c - d 5-8 e g 9 i 5 j 6 k l m n 5 o p q r s 9 We a Evaluate i () ii ( - ) and iii () or eac o te ollowing. a () = b () = + - c () = d () = We b a Find te -intercepts o te parabola = b Find te gradient o te parabola at te points were it crosses te -ais. c Determine te value o or wic te gradient o te parabola is: i ii 7 iii -. a Find te -intercepts o te curve = b Find te gradient o te curve at tese points. c Find te coordinates o te point were te gradient is. Find te coordinate(s) o te points on te curve - were te tangent: a is parallel to te -ais b is parallel to te line = - +. We Find te equations o te tangent and normal or eac o te ollowing curves. a () = at te point (, 7) b () = + - at = - c () = at = d () = at te point (, - 8) e () = + at = a () = - at te point (a, a - 8a) We a Find te equation o te tangent at te point on te curve + -, were te gradient is 6. b Hence, ind te equation o te normal at tis point. 5 Find te equation o te normal to te curve, = - + 5, at te point were te curve crosses te -ais. 6 Find te equation o te normal to te curve = - + at: a = b =. 7 Find te equation o te normal to te grap o = + + at =. 8 I te equation o te normal to te curve = is = - + c, ind te: a coordinates at te point o tangenc b value o c. Capter 9 Dierentiation

23 9 We 5 Find te equations o te tangents to te curve () = wic are parallel to te line - =. Find te coordinates o te points on te curve () = were te tangents are parallel to te line =. Hence, ind te equation o te tangents at tese points. 9e Find te equation o te tangent to te curve () = - + wic is perpen dicular to te line = - +. We 5 Find te coordinates o te points were te tangents to te curve () = - - are perpendicular to te line + =. Hence ind te equations o te normals at tese points. Te tangent to te curve = a + b + c at te point ( -, - 8) is parallel to te line =. I te curve also passes troug te point ( -, -.5), ind te values o a, b and c. Hence, ind te equation o te tangent at te point ( -, - 8). a Find te equation o te tangent to te curve = - at = a. b Hence ind te -intercept o te tangent line in terms o a. c A straigt line wit equation = - passes troug te -intercept o te tangent line. Find te value o a. d Wat is te equation o te tangent line? 5 We 6 Sketc te graps o: a ( )= b ( )= + c ( )= + + and te grap o () or eac on te same set o aes. How can () be used to predict te location and nature o te stationar points o () in eac case? antidierentiation From te previous investigation ou ma ave observed a pattern to obtain te original unction rom te derivative. For eample, i () = ten () = = For eac term, add to te power and divide b te resulting power. Tis process is just te inverse o te one used to dierentiate b rule. Rule. I () = n, ten te antiderivative is n+ ( )= + c, n - were c is a constant. n + Rule. I () = a n, ten te antiderivative is ebookplus Interactivit int-68 Antidierentiation an+ ( )= c n + +, n - were c is a constant. Rule. I () = g () + (), ten te antiderivative is () = g() + () + c Note: I ou dierentiate () above, () will be obtained (te derivative o c is ). For eample:. te derivative o is. te derivative o + is also. te derivative o - is also. te derivative o + c is also (were c is a constant). mats Quest matematical metods Cas or te Casio Classpad

24 Wen inding te antiderivative, or indeinite integral as it is also known, tere are an ininite number o possibilities or c.. Te antiderivative o () is denoted b F().. Tis can be epressed as te indeinite integral and is written as: F ( ) = ( ) d. We read tis as te integral o () wit respect to.. Te antiderivative o d is + c. d Worked Eample 7 Find te antiderivative o +. Tink Write Antidierentiate eac term b rule and add a constant. Te antiderivative is + + c. Worked Eample 8 I d d = + 5-7, ind te antiderivative,. Tink Write down d d. Antidierentiate eac term b rule and add a constant. Write d d = = ( + 5 7) d 5 = + 7+ c Worked Eample 9 Find () i () = ( + ). Tink Write Write down (). ( ) = ( + ) Epand te brackets. ( )= Antidierentiate eac term b rule and add a constant. ( ) = ( ) d = c Capter 9 Dierentiation

25 WorkeD eample Find ( + 5 ) d. Tink Epand te brackets to epress in basic polnomial orm so it can be antidierentiated. WriTe ( + 5 ) d = ( + 5 d ) Antidierentiate eac term b rule and add a constant. 5 = + + c WorkeD eample Find an antiderivative o a ( ) = and b ( ) = ( + ). Tink WriTe/Displa ebookplus Tutorial int-99 Worked eample a To antidierentiate () = + 5-7, on a te Main screen tap: ) - Complete te entr line as: ( + 5 7)d Ten press E. Write te answer. Note: Te calculator does not include te constant c. It is important tat ou include it in our answer. 5 ( + 5 7) d = + 7+ c b To antidierentiate () = ( + ), b complete te entr line as: ( + ) d Ten press E. To epand te brackets, igligt and cop te answer; ten tap: Action Transormation epand Paste te answer and press E. mats Quest matematical metods Cas or te Casio Classpad

26 Note: Replace 9 wit te constant c, indicating te original equation can ave an constant not just 9 until more inormation is known. Write te answer. ( ) 6 + d = c REMEMBER. I () = n n+, ten te antiderivative is () = ( )= + c, n -, were c is a constant. n +. I () = a n an+, ten te antiderivative is () = ( )= c n + +, n -, were c is a constant.. I () = g () + (), ten te antiderivative is () = g() + () + c.. Te antiderivative o () is denoted b F(). 5. F ( ) = ( ) d d 6. Te antiderivative o is + c, were c is a constant. d Eercise 9E Antidierentiation a Cop and complete te ollowing table. Function () Derivative (gradient unction) () i ii + iii iv v vi vii 6 viii i 9 5 i ii 6 7 iii iv 7 (continued) Capter 9 Dierentiation 5

27 Function () Derivative (gradient unction) () v 7 vi 8 + vii + 7 viii 6 b Eplain in words ow to ind te original unction rom te derivative (gradient unction). We7 Find te antiderivative o eac o te ollowing. a 9 b c + - d e g We8 I d d We9 Find () i () is: = 5, ind te antiderivative,. a 5( + 7) b ( - ) c - ( + 5) d ( + 7)( + 5) e ( - )( + ) ( - )( + ) g ( + ) ( - ) i ( - ) j ( + )( - + ) 5 mc I () = ( - 5) ten te antiderivative () is: A ( - 5) + c B c C - D E mc I d = ( + )( 5, ) ten te antiderivative is: d A c B - C c D c ebookplus Digital doc WorkSHEET 9. E c 7 We Find te ollowing. a ( + 7) d b ( 6 + ) d c ( 6 5 ) d d ( )( + ) d e + 7 d ( + + ) d 8 mc I () = 6 - +, wic one o te ollowing cannot be ()? A B C D E We Find te antiderivative o a ( - ) b ( - 5) 6 mats Quest matematical metods Cas or te Casio Classpad

28 9 Deriving te original unction rom te gradient unction Te process o antidierentiation identiies a amil () o original unctions rom a given gradient unction. () For eample i () =, ten te antiderivative () would be given as () = + c. Tereore, () () could be, +, +, +, -, or () -, and so on. () Tat is, te constant, c, could take an real value. In act we ave an ininite number o poss ibilities wic orm a amil o straigt lines translated up or down, depending on te value taken b c. Consider te eample o () =. In tis case te antiderivative would be given as: F () = + c F() could be, +, +, -, -, and so on. Again, tere are an ininite number o possibilities wic orm a amil o parabolas translated up or down, depending on te value taken b c. () () () () () Worked Eample I () =, ten one possible grap or () is sown at rigt. On te same set o aes, sketc tree more possibilities or te grap o (). () () Tink write () is te antiderivative o (). ( )= d = + c Te unction () is a straigt line wit c being te -intercept. Coosing dierent values or c will provide more possibilities or (). Wen: c =, () = +. c =, () = +. c = -, () = -. () c = c = c = c = In te eamples above it is possible to ind one particular member o te amil o curves rom te gradient unction wen special conditions called boundar conditions are given. Tese elp us to identi wic member o te amil we are dealing wit b providing clues about te original unction. Capter 9 Dierentiation 7

29 WorkeD eample Te grap o a gradient unction is sown at rigt: a From te grap, write down: i te value o were te gradient is ii te sign o te gradient to te let o tis point iii te sign o te gradient to te rigt o tis point. b i State te sape o te grap o (). ii I () =, sketc te grap o (). '() ebookplus Tutorial int- Worked eample Tink WriTe a b i Te gradient is were te gradient unction grap crosses te -ais. ii To te let, te gradient unction is below te -ais. iii To te rigt, te gradient unction is above te -ais. i I te grap o te derivative is linear, ten its antiderivative, (), will be quadratic. ii Since () =, te grap o () passes troug te point (, ). a i Te gradient is wen =. b ii I <, te gradient is negative. iii I >, te gradient is positive. ii i Te grap o () will be a parabola. Negative gradient () Positive gradient Zero gradient WorkeD eample I d = and = wen =, determine te rule or ; a b and b using a CAS calculator. d Tink a Antidierentiate d b rule to d obtain an epression or, remembering to add a constant. Substitute = and = into te equation. a WriTe/Displa d = d = + c Substituting (, ) into te equation: = () + c Solve te equation or c. = + c c = - c = 5 Write te rule or. = mats Quest matematical metods Cas or te Casio Classpad

30 b To antidierentiate, on te Main b screen tap: ) - Complete te entr line as: ()d Ten press E. Deine () = + c b completing te entr line as: Deine () = + c Ten press E. To determine c, complete te entr line as: solve (() =, c) Ten press E. Write te answer. I d = and = wen =, ten d = + 5. REMEMBER. Antidierentiation o te gradient unction, (), gives a amil o unctions () + c, wic can be sketced as a amil o curves.. It is possible to ind one particular member o te amil o curves rom te gradient unction wen special conditions called boundar conditions are given. Eercise 9F Deriving te original unction rom te gradient unction WE I () =, one possible grap or () is sown at rigt. On te same set o aes, sow tree more possibilities or (). () Capter 9 Dierentiation 9

31 An original unction or te gradient unction d = 8 is sown at rigt. On te same aes, sow our oters. d WE Te grap o a gradient unction is sown at rigt: a From te grap, write down: i te value o were te gradient is ii te sign o te gradient to te let o tis point iii te sign o te gradient to te rigt o tis point. b i State te sape o te grap o (). ii I () =, sketc te grap o (). '() On te same aes sketc our curves wit a gradient described as () = +. 5 For eac gradient unction grap below give: i te value o were te gradient is ii te sign o te gradient (positive or negative) let o tis point iii te sign o te gradient rigt o tis point. a Gradient b unction Gradient unction c Gradient unction d Gradient unction 6 MC Eamine te gradient unction, '(), sketced at rigt. Ten coose te correct answer to te ollowing questions about its original unction. a () as a gradient o wen equals: A B - and C and D and E b () as a negative gradient wen: A < B < < C > D < E > c () as a positive gradient wen: A < and > B > and < C > and > D = and = E = 7 For te gradient unction sketced at rigt, state all values o were te gradient is: a zero b negative c positive. Gradient unction '() 6 Mats Quest Matematical Metods CAS or te Casio ClassPad

32 8 Questions a to g relate to te ollowing igures, A to G. A b '() g'() c '() d '() e g'() '() g '() a Eamine te gradient unction () in igure A. Sketc te grap o (), given tat () =. b Eamine te grap o g () in igure b. Sketc te grap o g(), given tat g() =. c Eamine te grap o () in igure c. Sketc () i () = -. d Eamine te grap o () in igure d. Sketc () i () =. e For g () graped in igure e, sketc g() i g () =. I () is graped as in igure, sketc te unction (), given tat ( - ) = - 9, () = and () = 9. g I () is graped as in igure g, sketc (), given tat ( - ) = 5, () = and () = I d = - 5 and = wen =, ind te rule or. d WE a Find te equation o te curve wit a gradient unction - and passing troug te point (, - ). WE b I d d = + and te point (, ) belongs to te curve, ind te equation or. I () = ( + ) and F() =, ind F(). Capter 9 Dierentiation

33 Summar Introduction to limits I a unction, (), is continuous wen = a, ten lim ( ) = ( a ). a For te constant unction were () = c, lim ( ) = c. a lim[ ( ) ± g ( )] = lim ( ) ± lim g ( ). a a a lim[ ( ) g ( )] = lim ( ) lim g ( ). a a a lim ( ) lim ( ) a a g ( ) = lim g ( ). a Limits o discontinuous, rational and brid unctions I a unction is discontinuous at te -value were te limit is being investigated, ten te limit will eist onl i te unction is approacing te same value rom te let as rom te rigt. Finding te limit o a rational unction involves simpliing te unction beore direct substitution can take place and recognising an values o or wic it is discontinuous. Hbrid unctions are unctions tat ave dierent rules or dierent parts o te domain. Deriving te gradient unction ( ) () Q () P Tangent at P Gradient o a secant = rise run ( + ) ( ) = () is te gradient unction o (). () = lim ( + ) ( ), Dierentiation using irst principles Q Q Q moves closer to P as approaces. P Q Mats Quest Matematical Metods CAS or te Casio ClassPad

34 Te process o inding te gradient unction lim ( + ) ( ) Dierentiating gives d d. Dierentiating () gives (). is called dierentiation rom irst principles. Finding derivatives b rule I () = n, ten () = n n -. I () = a n, ten () = na n -. I () = c, ten () = (were c is constant). I () = g() + (), ten () = g () + (). Equation o tangent: - = m T ( - ) Equation o normal: - = m N ( - ), were m N = m T Antidierentiation Antidierentiation, or integration, is te reverse process o dierentiation. I () = n n+, ten te antiderivative is ( )= + c, n -, were c is a constant. n + I () = a n an+, ten te antiderivative is ( )= c n + +, n -, were c is a constant. I () = g () + (), ten te antiderivative is () = g() + () + c. Te antiderivative o () is denoted b F(). F ( ) = ( ) d d Te antiderivative o is + c, were c is a constant. d Deriving te original unction rom te gradient unction Antidierentiation o te gradient unction, (), gives a amil o unctions () + c, wic can be sketced as a amil o curves. For eample, i () =, ten te antiderivative is () = + c. Tis produces a amil o curves as sown. () () () () () () It is possible to ind one particular member o te amil o curves rom te gradient unction wen special conditions, called boundar conditions, are given. Capter 9 Dierentiation

35 capter review Sort answer Find lim ( ). I () = - +, ind lim ( ). a B irst actorising te numerator, simpli te + 7+ rational unction ( )=, stating + te value or wic te unction does not eist (tat is, is discontinuous). b Find lim ( ), were a is te value at wic a () is discontinuous. a Sketc te grap o te brid unction +, (, ) ( ) =, [, ) b Find lim ( ). 5 Find lim ( + ) ( ) ; tat is, ind () or () = +. 6 Dierentiate () = using irst principles. 7 I g() = : a ind g () b evaluate: i g () ii g ( - ) c ind te coordinates wen te gradient is. 8 Find te coordinates were te gradient o te tangent to = is: a parallel to te -ais b parallel to te line - c equal to te gradient o te unction g () = - or te same value(s) o. 9 Find te equations o te tangent and normal or () = - + at te point (, ). Find te antiderivative o eac o te ollowing using te rule. a 6 b 5 + c d e ( - )( + 7) Find te ollowing: a ( + 8 7) d b ( 6 + ) d c ( ) d d ( 7)( + 7) d Find te rule o te original unction i its gradient unction is d = ( + )( ) and = 6 d wen =. Te grap o g () is sown below. g'() I g () = 6 and g () =, sketc te grap o g(). Multiple coice Te lim( 7 ) equals: 5 a - 7 b is undeined c 5 d - e 8 Te lim ( + 5 ) is equal to: a b is undeined c 5 d 8 e + Te lim is equal to: a b 7 c d - e 5 Questions to 6 reer to te unction () graped below. () Te lim ( ) equals: a b c d e - 5 Te lim ( ) is: a b c undeind d e - 6 Te lim ( ) equals: a b c undeind d e - 7 I () = -, ten lim ( + ) ( ) equals: a - b + c d - e Mats Quest Matematical Metods CAS or te Casio ClassPad

36 8 I () = + 7, ten () is equal to: ( + ) ( ) A B lim ( + ) ( ) ( + ) ( ) C D lim ( + ) E lim ( ) 9 Te gradient o te tangent to te curve () at = 5 is: A lim ( + ) () 5 B lim () 5 C lim ( 5 + ) ( ) D lim ( 5+ ) () 5 E lim ( ) I g() = , ten g () equals: a 5 - b c - d 8 e - - I = , ten d is equal to: d A B C D E - 7 Te unction () = ( - )( + 5), so () equals: a + 5 b + c - d e + 5 I = -, ten d d equals: a - b - c d - e - Te derivative o - is: a 6 b 6 c d e Wen dierentiated, 5 is equal to: a 5 b 5 d e 5 c 5 6 Given tat () = + 7 -, ( - ) equals: a 7 b - c - d - e 7 Te gradient o = equals wen is: a 8 b 5 c - d e Te gradient o te normal to te curve = at = is: a b - c d e 9 I te tangent to te curve = is parallel to te line = +, ten te coordinate at te point o tangenc is: a b c - d e I d d = - + 7, ten could be: A 6 - B - C D E + 7 Te antiderivative o is: a 9 + c b + c c 9 d e Te antiderivative o ( - )( + 5) is: A c B ( + 5) + c C + c E c D c Te epression + 6 d is equal to: A c C c E + + c B c D c Te unction g() graped below as a negative gradient wen: g() A < < B - < < C < - and > D < E > - 5 From te gradient unction () graped at rigt, it can be deduced tat () as a gradient o wen: A = B > C < - D = - and = E = - 6 Te grap o g () is sown below. g'() '() Capter 9 Dierentiation 5

37 I g() =, ten te grap o g() is best represented b: A B g() Wic one o te ollowing could be te grap o = '()? A B C D g() C D E g() g() E g() 7 At =, te grap o te unction, wit rule () = ( + ) ( - ) + as a: A -ais intercept B -ais intercept C local minimum D local maimum E point o inlection wit zero gradient [VCAA ] 8 I = + +, te rate o cange o wit respect to at = k is: A k + B k + 6 C k + k D k + k + E k + k + k [VCAA ] 9 Te grap o te unction wit rule = () is sown below. [VCAA ] For te grap o = () sown, '() is negative wen: (, 5) 5 (, ) A < < B C < - or > D - or E - 5 < < or > [VCAA 6] For te grap = + 7 +, te subset R or wic te gradient is negative is given b te interval: A (.5, 5.) B ( -.99,.5) C ( -, ) D (- 5, ) E (.5, ) [VCAA 7] Etended Response On te grap o te unction () = -, a secant is drawn rom te point A (, ) to a point B were = + a. a Find te coordinates o te point B in terms o a. b Write an epression or te gradient o te secant. c Find te limit, k, or te above epression as a. d Write te derivative unction (). e Sow tat () = k. For te unction () = : a write an epression or te gradient o te secant QR connecting te points Q(, ()) and R( +, ( + )) 6 Mats Quest Matematical Metods CAS or te Casio ClassPad

38 b write an epression or te gradient o te secant PQ connecting te points P( -, ( - )) and Q(, ()) c write an epression or te gradient o te secant PR connecting te points P( -, ( - )) and R( +, ( + )) d sow tat gradient PR = gradient PQ + gradientqr e state te relationsip between te gradient o te secant PR and te gradient at te point Q(, ()). Find te coordinates o te point o intersection o te tangents to te curve = at te points were = and =. A giant park slide is made rom a straigt section o metal seet joined to a curved section o metal seet suc tat its proile is modelled b te brid unction, [, ) ( ) = ( 6), [, 5] were is te orizontal distance in metres rom te start o te slide. a Find te eigt o te slide at te start. b Find te eigt o te slide at te end. c Sketc te grap o (). d Given tat te two sections are joined at a orizontal distance o metres rom te start o te slide, ind te eact lengt o te straigt section. e Sow tat te slide is smootl joined so tat cildren will not eperience a bump. 5 A mountain trail can be modelled b te curve wit equation = , were and are, respectivel, te orizontal and vertical distances measured in kilometres, < <. a Find te gradient at te beginning and end o te trail. b Calculate te point were te gradient is. c Veri tat tis point represents te maimum gradient witin te given domain b tracing te unction wit a CAS calculator. d Hence, state te maimum eigt o te pat. e Find te point were te pat is lowest and, ence, state te minimum eigt. 6 Te position o a particle at an time, t, is represented b te antiderivative o its velocit, v, wit respect to time; tat is, = vd t. I its velocit is given b te rule v = t - 5, and te initial position o te particle is cm let o te origin, ind: a te rule or its position,, at an time, t b te position o te particle ater seconds. For te net tree questions use te ollowing series epansions: e = + + +! +! + 5 5!... sin () = -! + 5 5! - 7 7! + 9 9!... cos () = - +! - 6 6! + 8 8!... Remember 5! = 5. 7 a Find () i () = e. b Wat do ou notice? 8 a Find g () i g() = sin(). b Wat do ou notice? 9 a Find () i () = cos(). b Wat do ou notice? ebookplus Digital doc Test Yoursel Capter 9 Capter 9 Dierentiation 7

39 ebookplus activities Capter opener Digital doc Quick Questions: Warm-up wit ten quick questions on dierentiation (page ) 9A Introduction to limits Digital docs SkillSHEET 9.: Practise substituting values into a unction (page ) Investigation: Investigate evaluating and approacing a limit (page 5) 9B Limits o discontinuous, rational and brid unctions Tutorial We5 int-: Watc a tutorial on simpliing a rational unction and ten calculating te limit o te unction at te point or wic it is discontinuous (page 6) 9C Dierentiation using irst principles Tutorial We8 int-5: Watc a tutorial on ow to ind te gradient equation using irst principles and te values or wic te gradient equation equals zero (page ) Digital docs SkillSHEET 9.: Practise inding te derivative using irst principles (page ) WorkSHEET 9.: Evaluate limits and determine derivates o polnomials using irst principles (page ) 9D Finding derivatives b rule Tutorials We int- 988 : Watc a tutorial on dierentiating a rational unction ater simpliing it (page ) We5 int-7: Watc a tutorial on ow to ind te equations o te tangent and normal at t particular point on a cubic (page 8) We6 int- 989: Watc a tutorial on graping a unction and its derivative on te same set o aes (page 9) Digital docs Spreadseet 7: Investigate te gradient at a point using a spreadseet (page ) Spreadseet 9: Investigate a tangent and normal using a spreadseet (page ) 9E Antidierentiation Interactivit Antidierentiation int-68: Consolidate our understanding o antidierentiating polnomials (page ) Tutorial We int- 99 : Watc a tutorial on inding an antiderivative (page ) Digital doc WorkSHEET 9.: Use irst principles and te rule to ind te derivative, calculate gradients at dierent -values and determine te derivative (page 6) 9F Deriving te original unction rom te gradient unction Tutorial We int-: Watc a tutorial on sketcing te grap o a unction given te grap o its derivative (page 8) Capter review Digital doc Test Yoursel: Take te end-o-capter test to test our progress (page 7) To access ebookplus activities, log on to 8 mats Quest matematical metods Cas or te Casio Classpad