LEVEL III Module 16: GAC016 Mathematics III: Calculus & Advanced Applications

Transcription

1 GLOBAL ASSESSMENT CERTIFICATE FACILITATOR GUIDE LEVEL III Module 6: GAC06 Mathematics III: Calculus & Advanced Applications Date of Issue: May 0 Version: 7.. All rights reserved. The material printed herein remains the property of ACT Education Solutions, Limited and cannot be reproduced without prior permission. The authors and publisher have made every attempt to ensure that the information contained in this book is complete, accurate and true at the time of printing. You are invited to provide feedback of any errors, omissions and suggestions for improvement. Every attempt has been made to acknowledge copyright. However should any infringement have occurred, the publisher invites copyright owners to contact the address below. ACT Education Solutions, Limited Suite 0, Level, 75 Alfred Street, North Sydney NSW 060 AUSTRALIA

2 Table of Contents INTRODUCTION... I MODULE OVERVIEW... I LEARNING OUTCOMES... I BEFORE YOU BEGIN... II UNIT BREAKDOWN... III ASSESSMENT EVENTS... III SUGGESTED DELIVERY SCHEDULE... IV ICONS... V UNIT : DIFFERENTIATION... PART A UNIT INTRODUCTION... PART B TERMINOLOGY INTRODUCED... PART C DIFFERENTIATION CONTINUITY AND LIMITS... PART D THE DERIVATIVE GRADIENTS OF SECANTS AND TANGENTS... 8 PART E DIFFERENTIATION OF FUNCTIONS... PART F RATES OF CHANGE... 8 PART G THE PRODUCT, QUOTIENT AND CHAIN RULES... 0 PART H DIFFERENTIATION EXPONENTIALS AND LOGARITHMS... 0 PART I APPLICATIONS OF THE DERIVATIVE... 7 PART J DIFFERENTIATION CIRCULAR FUNCTIONS... PART K HIGHER DERIVATIVES... 7 PART L CURVE SKETCHING... 5 UNIT : INTEGRATION... 6 PART A UNIT INTRODUCTION... 6 PART B TERMINOLOGY INTRODUCED... 6 PART C THE PRIMITIVE FUNCTION AND INDEFINITE INTEGRALS PART D DEFINITE INTEGRALS AND THE AREA UNDER A CURVE... 7 PART E THE VOLUME OF ROTATION UNIT : ADVANCED APPLICATIONS... 0 PART A UNIT INTRODUCTION... 0 PART B TERMINOLOGY INTRODUCED... 0 PART C MOTION IN A STRAIGHT LINE (RECTILINEAR MOTION)... 0 PART D APPROXIMATION METHODS IN INTEGRATION PART E PROBLEMS INVOLVING MAXIMA AND MINIMA... PART F EXPONENTIAL GROWTH AND DECAY... 7 PART G VERHULST-PEARL LOGISTIC FUNCTION THE NATURAL LAW OF GROWTH & DECAY.. 9 APPENDIX: FORMULAE...

3 Introduction Introduction Module Overview Welcome to GAC06: Mathematics III: Calculus & Advanced Applications. Calculus is one of the most important and useful topics in mathematics as it can be applied to many fields of study, particularly science, engineering and computing. This module, GAC06: Mathematics III: Calculus and Advanced Applications is an important introduction to this field of mathematics for those students wishing to study any of these fields at a tertiary level. Those universities that require students to have a competent understanding of mathematics as a prerequisite epect that calculus has been covered to a similar depth as it is in this module. The module will introduce students to some important techniques of differentiation and integration, what they represent for any given function, and provide an insight into their applications to solving problems. Learning Outcomes By the end of this module students should be able to:. Determine the derivative (if it eists) of most mathematical functions, and use the derivative to analyse functional behaviour.. Use techniques of integration to find indefinite and definite integrals.. Apply differentiation and integration techniques in a variety of practical problems. Page I Version 7. May 0

4 Introduction Before You Begin In the previous mathematics modules, students maintained a Mathematical Terminology Logbook. This form of assessment will continue in a similar way as before. All the words written in bold print within the Student Manual should be added to the logbook along with the definition written in students own words. As facilitator, you are required to check the logbooks three times. Various sections of the module can be allocated as Independent Study or homework. Allocate this work as determined by the needs of your class. Students will need to have each of the following items for the mathematics modules: a silent, non-programmable scientific calculator that has statistical functions for use in the statistics section mathematical drawing and measuring tools: ruler, compass and protractor a notebook/a folder to write notes and complete eercises in for regular marking a pocket-sized notebook for the Mathematical Terminology Logbook access to a computer for spreadsheet/eploratory applications throughout the module, and to undertake the two projects now included in this module. The volume of work covered in this module is quite large. It is therefore important that all students keep up to date with their homework. Advise students to see you if they are having difficulty keeping up with the course work so that assistance may be organised. Page II May 0 Version 7.

5 Introduction Unit Breakdown The following is a list of units to be covered in the module GAC06: Mathematics III: Calculus and Advanced Applications. Unit Unit Unit Differentiation Integration Advanced Applications Assessment Events No. Assessment Event Weight In-class Test: Units 0% Project : Unit Differentiation Project : Unit Advanced Applications 0% Eamination: Units 50% Course Work: Includes Mathematical Terminology Logbook and In-class Tasks. 0% Note: All details of Assessments can be found in the GAC06 Assessment Folder. It is your responsibility to ensure that the students are fully prepared for the assessments and that the assessment tools (test/eamination papers, record sheets, etc.) are photocopied and distributed to the students according to the guidelines in the Assessment Folder. You will need to liaise with the GAC Director of Studies. Page III Version 7. May 0

6 Introduction Suggested Delivery Schedule Week Week Week Week Week 5 Week 6 Week 7 Week 8 Week 9 Week 0 Week Week Assessment Event : Project Mathematical Terminology Logbook due Unit : Integration Unit : Integration Unit : Integration Unit : Integration Assessment Event : In-class Test Mathematical Terminology Logbook due Unit : Advanced Applications Assessment Event : Project Unit : Advanced Applications Unit : Advanced Applications Assessment Event : Eamination Assessment Event : Course Work including Mathematical Terminology Logbook Page IV May 0 Version 7.

7 Introduction Icons The following icons will be used as a visual aid throughout the Student Manual and : Icon Meaning Information Task Demonstration Review Independent Study Including Use of Spreadsheets Assessment Events Language Focus Hints and Cautions Page V Version 7. May 0

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9 Part A Part B Part C Part D Part E Part F Part G Part H Part I Part J Part K Part L Unit Introduction Terminology Introduced Differentiation Continuity and Limits The Derivative Gradients of Secants and Tangents Differentiation of Functions Rates of Change The Product, Quotient and Chain Rules Differentiation Eponentials and Logarithms Applications of the Derivative Differentiation Circular Functions Higher Derivatives Curve Sketching Part A Unit Introduction Overview In this unit, students will learn to apply appropriate methods of differential calculus to solving various algebraic problems. In this unit, students will learn to: differentiate various types of functions from first principles apply the basic formula for differentiation using the product, quotient and chain rules where appropriate find the first and second derivatives sketch the curve of any function over a given domain. In Unit, the techniques learnt here will be applied to various problems of quantitative analysis. This unit includes a series of tasks that students will work through to practise the course material. They will be epected to complete some work in their own time. You will guide them through the unit. Page Version 7. May 0

10 Eploring Calculus Calculus is about graphs of functions, so this section links very closely with the sections about graphing functions in GAC00. There will be many times when ideas from that module are used to help students understand what is happening in this module. You should feel free to encourage students to use the spreadsheets about graphing functions from GAC00. Assessment Events,, & Provide students with an overview of the assessment events for this module. Highlight the timing and nature of projects, the scope of the review test required for this unit (Test at end of Unit ), and the scope and nature of the final eamination. Assessment Event Provide students with an overview of the Mathematical Terminology Logbook. Students should be familiar with the requirements of this logbook from previous maths modules. This is due to be collected at the end of each unit. Page May 0 Version 7.

11 Part B Terminology Introduced By the end of this unit, students need to know the meanings of these terms. The terms are bolded in the first instance they occur. Please make sure students understand the definitions of new terminology and include these words in their Mathematical Terminology Logbook. Summary of Terms limit continuity derivative quotient rule turning point maimum absolute maimum approaches monotonic point tangent differentiation differentiate chain rule stationary point minimum absolute minimum nature relative maimum secant first principles product rule second derivative inflection point boundary values normal etremities relative minimum Page Version 7. May 0

12 Part C Eploring Differentiation Continuity Limits Determining Limits Differentiation Continuity and Limits This section is about continuity. This idea is important because of the idea of the gradient of the tangent to come. The gradient of the tangent is the limit of the gradient of the secant (the line joining two points on the curve) as the two points get closer together. Now if the curve happens to have no values, or suddenly changes direction, then the limit idea breaks down. Students refer to spreadsheet Many functions and Student Manual (pp 6). Continuity at a Point A function f() is defined to be continuous at a point ( = a ) if the following three conditions are satisfied:. f(a) eists. The limit of f() as approaches a eists ( eists ). f(a) is equal to this limit (ie. f(a) = ) If one or more of the conditions in the definition above fails to be true, then f() is said to be discontinuous at = a. Task. Continuity and Limits. Write down whether each of the following graphs of functions are continuous or discontinuous. a) b) c) d). Find: lim 5 a) y y b) lim y y Page May 0 Version 7.

13 c) lim 0 d) h lim h h 6 h e) lim 7 f) m lim m0 9m m m g) hy lim h0 h y hy 5h h h) a 5a 6 lim a a a 6 i) t 5t 7 lim t 5t t c j) lim c c. For what values of is the function discontinuous? y 6 6. Determine whether the function: 8 f () 9 is continuous at =. = Task. Solutions. a) continuous b) discontinuous c) continuous d) continuous. a) lim (5 - ) = 5() - = b) lim c) 8 lim 0 ( ) lim 0 = ( ) = 0 Page 5 Version 7. May 0

14 d ) e ) (h )(h + ) lim h (h + ) (h ) = lim h = - ( )( 9) lim = =7 lim ( 9) f) lim m0 m m ( 9 ) m m = lim m0 ( m 9m) = - lim h0 g) h y hy y h ( 5) = lim h0 y hy y 5 = y +y-5 h) lim a ( a)( a) ( a)( a) = lim a 5 ( a ) ( a ) i) lim t t 5t 7 t t t 5t t t t t Page 6 May 0 Version 7.

15 5 7 lim t t t 5 t t 5 j) ( c) lim c ( c)( c c ) lim c ( c c ). c 6 y ( )( ) and - discontinuous at = or = -. f = 9 ( 6)( ) 6 lim 9 eists and f () = 9 f () lim f () eists therefore f() is continuous at =. Page 7 Version 7. May 0

16 Part D Eploring the Derivative The Derivative Gradients of Secants and Tangents The idea of the derivative is the value of the gradient of a curve at a point. It is the same as the gradient of a tangent to the curve at that point. The spreadsheet Gradients close up demonstrates this. For any curve, the tangent to the curve is drawn, and as you move in closer to the curve (as with a magnifying glass) you see that the tangent gradient is just the same as the curve gradient. Here it is demonstrated for y =, at =. We are looking at the tangent drawn from 0.0 under to 0.0 over. The curve and the tangent have the same gradient. The Gradient of a Secant Gradient of a Tangent The gradient of the tangent is the limit of the gradient of the secant (the line joining two points on the curve) as the two points get closer together. The method of finding the gradient algebraically, known as first principles is equivalent to this. You epress the gradient as rise (the difference between two epressions) divided by the run, h. Then as h approaches 0, the gradient epression approaches the gradient function. Gradient of a tangent from first principles: = Page 8 May 0 Version 7.

17 Task. Gradient of a Tangent from First Principles Find the gradient of the tangent for each of the following functions by differentiating from first principles.. f at the point (, 5). y 6 at the point (, 0) f 5 at the point (0, -5).. y at the point where = 5. g ( ) where = 6. y at the point (9, ) 7. y at the points where = and = 8. f at the point where = Task. Solutions. f () = = = lim h0 lim h0 lim h0 h 6h h hh ( 6) h [( h) ( h) ] [ () ] = 6. dy [( h) ( h) 6] [( ) ( ) 6] = lim d h0 h h(h8) lim h0 = h (h 8) lim h0 = = -8. (0 h) (0 h) 5 0 (0) 5 f () = lim h0 h h(h ) = lim h 0 h = lim h = - h 0 h Page 9 Version 7. May 0

18 . dy d = = = lim h0 lim h0 = h h( 6 h h ) h [( h) ( h)] [ ] lim h 6h h0 5. g () = lim h 0 h 0 lim h0 [ ( h) ( h) ] [ () ] h( h) h lim h = h 6. dy d = lim h0 lim h 0 lim h 0 / / (9 h) 9 h / (9 h) X h h ( 9 h) h 9 6 lim h 0 ( 9 h ) / (9 h) / (9 h) 7. dy = lim d h0 h h = lim h0 ( h) = lim h0 dy at =, d at =, dy d Page 0 May 0 Version 7.

19 8. dy [( h) ( h) ( h)] [( ) ( ) ] = lim d h0 h h h = lim h0 h h h = lim h0 = 0 Page Version 7. May 0

20 Part E Eploring Differentiation of Functions Differentiation of Functions The spreadsheet Gradient functions aims to make the ideas here as clear as possible. There are three steps in the eplanation. See how the small tangent changes in gradient as changes. See the actual value of the gradient plotted on the graph. See the gradient values joined into a smooth curve. Below is the screen for the function y =. It shows that the gradient at = is. This is also the gradient of the pink gradient function. As this goes through the origin the gradient function is y =. By changing the -values you can see that the gradient is always double the -value. For any given function y = n The first derivative, of the function y = n is given by: d d ( )n = n n- where n is a real number. The resultant epression is the equation for the gradient of any tangent to the curve y = n. or The resultant epression indicates the rate of change of y = n for any given value of. Page May 0 Version 7.

21 Eploring Gradients of Rational Functions Rational functions is a name given to functions that have the variable in the denominator. For y = the graph is an hyperbola. For y = the curve looks like the trunk of a tree. Using the spreadsheet Gradients of reciprocal functions: the screen below shows the graph of y = and shows that the gradient at = is. This is a particular value for the gradient function for which the formula is y =. There is no gradient at = 0 as the function is discontinuous at that point. If you use y = as your function you will see the gradient as the function y =. Eploring Tangent and Normal to a Curve Theorems of Derivatives Refer to the Student Manual (p. 7) for further information on looking at the gradient function for any polynomial term and page 8 for information on eploring tangent and normal to a curve.. The derivative of a constant = 0. Polynomials are differentiated term by term.. The gradient of a tangent to a curve is determined using the first derivative. The normal to a curve is the line perpendicular to the tangent at the point of contact. m n = m t Page Version 7. May 0

22 Task. Differentiation of y = n, Tangents and Normals. Differentiate each of the following with respect to : a) y b) y c) 5 y 8 d) y 7 e) y 6 f) y 5 5 g) f 8 h) i) f 6 g j) f h l) 5 k) m) y 6 y 6 n) y 5 6 o) y p) y. Find the equation of the gradient and the normal to the curve y at the point (, 0). 5. Show that the gradient of y is always negative.. Find the equation of the tangent at the point (-, 6) on the curve y. If the tangent cuts the y ais at A and the ais at B, find the coordinates of A and B. 5. Find the point on the curve 5 f where the normal is parallel to the line y The curve y a b c passes through the points A(, 0) and B(, 0). The slope of the normal at A is and the gradient of the tangent at B is 9. Find the values of a, b and c. Task. Solutions. a) b) c) d) e) dy = d dy = 9 d dy d = 0 dy = d dy = - d Page May 0 Version 7.

23 f) dy = d g) f ' () = h) g'() = 0. 5 i) f() = + f j) k) f( ) = - 5 f = h'( ) l) m) dy = - - d y 6 dy d n) y = = dy.5 d 5 6 o) y dy = + - d p) y = = = dy = d Page 5 Version 7. May 0

24 . dy d = +. When =-, dy d = -5. Using y = m + c at (-, 0) where m = -5, equation of the tangent is y = -5-0 For equation of the normal m = /5 at (-, 0). Using y=m + c, equation of normal is y = /5 + /5. dy 5 = d which is always negative for all values of ecept zero. dy d = +. When =-, dy d = -. Using m = - at (-,6), equation of the tangent is y = - + Tangent cuts y ais at =0, coordinate of A is (0, ) Tangent cuts -ais at y=0, coordinate of B is (½,0 ) 5. f() = f '() = + 9. Gradient of tangent is + 9 and the gradient of the normal is. 9 This is parallel to a line with gradient m = The two gradients are equal. 9 Solving, + 9 = - gives = -. Point on the curve is (-, -) 6. Substituting (-,0) and (,0) in the equation y = a + b + c, we obtain 0 = a b + c and 0 = 9a + b + c dy = a + b. d At A(-, 0), m = -a + b = - [A](normal). Tangent at B(, 0) = 6a + b = 9 [B] Solving [A] and [B] gives a=, b=. Substituting in one of the above equations, c= a =, b =, c = Page 6 May 0 Version 7.

25 Independent Study. Work through Eample using a) Substitution/Elimination methods b) Matri Algebra methods to confirm the given values of a, b and c.. Using the values given for a, b and c show the steps necessary to obtain the given equation for the curve. Independent Study Solutions Summary of Information: y a b c, and point P (,7) has a tangent with gradient m. Point Q is on the graph at, and the tangent rises at 5 o A tangent with a slope of 5 o has a gradient m, o tan 5, or a. from b. from 5 o right angle triangle ratio, :: Given a b c () ab () ab (). a) Using Simultaneous equations substitution/elimination methods to solve: () () ab ab a Substitute () into () b b () (5) b Check answer by substituting () and (5) into L.H.S of ()? answer R. H. S Substitute () and (5) into () c 7 c 7 c 7 Thus a, b and c 7 which confirms the values given.. The general equation for the curve is y a b c now, given the values of a, b and c 7 the equation for this curve is y 7 Page 7 Version 7. May 0

26 Part F Eploring Rates of Change Distance & Speed Rates of Change Rate of Change and Gradients Rate of Change and Derivatives Rates of Change It is very useful to relate all this material with abstract functions to something real that is also understood intuitively by your students. Motion is a good eample of this. At a later point this will etend to uniformly accelerated motion, but at this stage we will just see that for any linear distance function the gradient is the speed. The reason for using the comple spreadsheet at this stage is to prepare students for the more comple situations later. Below is a screenshot of the spreadsheet Motion. The symbols are t (time), d (distance or displacement), v (speed or velocity) and a (acceleration). The initial speed (when t = 0) is u and c is the distance from the origin at t = 0. We may use any time and distance units (and hence speed). For the first eposure to this graph the Student Manual uses km and h. However you could use m and sec if you wish. This graph is interpreted this way. The horizontal ais shows the passage of time. The red line is the distance. It shows that the motion starts units behind 0, and its gradient is ( units across and units up). So d = t. The green line shows the speed (or velocity), and it is v =. The table shows that the distance increases constantly over each 0. time interval, and the speed is always. Because there is no change in speed, a = 0. Task.. A car moves along a straight road such that the distance s kilometres from a starting point O at a time t hours is given by the formula s t 6t 9t. Find the car s rate of change in distance. Page 8 May 0 Version 7.

27 . If the cost of supply of raw fish to a fish monger is C 50 0 f 0.0 f, calculate the rate of change of cost of supplying 8 tonnes of fish. Solutions Task.. Let f ( t) S t 6t 9t ds Rate of change, or f '( S ) dt d n n Using n d Then, d dt t t t 6 9 t 6 t 9t t t 9 rate of change in distance is given by t t 9. Let C f f ; 8 f tonnes. dc df d df f Cost 50 0 f 0.0 f f ; rate of change of cost When 8 $9.0 Page 9 Version 7. May 0

28 Part G Eploring the Product Rule The Product, Quotient and Chain Rules The reason we need a product rule is that some functions cannot be simply epanded. An eample would be y = e. However the best way to eplain and justify the rule is to use eamples where we can actually check that the rule works products of two polynomial functions that can be multiplied to form the product function. Here is the second eample, where u = +, and v = +. The product values ( + ) are in the column called uv, and the gradient values are calculated as the increase in uv divide by the common difference in (0.). Now u = and v =, so the product rule calculates the gradient as uv + vu = ( + ) + ( + ) = +. The green columns show that these values are the same as the gradient values of the product function. Refer to spreadsheet Product rule below. The Product Rule The product rule is used when differentiating the product of two functions (symbolised by u and v). If y = uv, then d du (uv) = v d d or = vu + uv + u dv d Eploring the Quotient Rule The quotient rule is more comple than the product rule. We can treat it the same way, using polynomials that we can simplify to understand how it works. In the spreadsheet Quotient rule, the first tab Quotient of two functions aims to show both functions and their quotient. The graph for u is blue, the graph for v is red and the graph for the quotient (u v) is black. The value of (u v) for any value of is found simply by dividing the value for u by the value for v. So if v has zero values they will produce an asymptote in the quotient function. Page 0 May 0 Version 7.

29 The second tab Quotient rule shows the gradient of the quotient function. Here is the second eample from the Student Manual. Note: You can use this spreadsheet to remind students about sin/cos = tan or get the gradients of the reciprocal trig functions. The Quotient Rule u The quotient of two numbers can be differentiated using the v product rule by writing the epression in the form uv. However, differentiating a quotient of two functions requires the use of a special rule. If y =, then d d ቀu v ቁ = u dv d = or v du d v Page Version 7. May 0

30 Eploring The Chain Rule Many students struggle with the idea of the chain in the chain rule the function of a function. Use the tab Function of another function in spreadsheet Chain rule to eplore this first. Here is the first function of a function: square of ( + ) that is mentioned in the Student Manual. The blue line shows the first function u = +, and it is those values that are squared to give the red parabola. For the tab Chain rule (below) the red parabola is still there but the green line is its gradient. The table shows that v is the function ( + ), and its gradient values are twice the values of u. The gradient of u = + is just (du/d), and the product of du/d by dv/du is the same as the gradient of v (see the two green columns). Page May 0 Version 7.

31 The Chain Rule The chain rule is sometimes also referred to as the function of a function rule. If where, then = or Task.5 Product, Quotient and Chain Rules Differentiate each of the following using the appropriate rule: Find the gradient to the tangent where = on the following functions: a) f b) f c) f 6. Find the point where the curve g the gradient of the tangent at this point. cuts the -ais and find Page Version 7. May 0

32 7. If the point (,) lies on the curve y, find the equations of 8 the tangent and the normal at this point. 8. Find the equations of the tangent and the normal to the curve y at the point (-, -). If the tangent meets the -ais at P and the normal meets the -ais at Q, find the distance of the line PQ (leave the answer in surd form). Solutions Task.5. Use Chain Rule : u =, y = u 5 dy dy du = d du. d dy du = du 5u, = d = 0( ). Chain Rule : u = +, y u dy dy du = d du. d dy 0.5 du u, du u d =. Chain Rule : u= - 5, y = u 6, dy dy du = d du. d dy du = du 6u5, = 8 d = 8( - 5) 5. Chain Rule : u = +, y= u, dy dy du = d du. d du dy = +, = u d du. = ( + )( + ) 5. Chain Rule : u= 6, dy dy du = d du. d dy du = u = ( 6 ) y u, du = - d Page May 0 Version 7.

33 6. Chain Rule: u = 5 +, y = u 6 dy dy du = d du d dy du = du 6u5, = 5 9 d =6( 5 + ) 5 (5 9 ) 7. Product Rule : D(uv) = udv + vdu u = v = + du = dv = d(uv) = udv + vdu dy = 9 + d 8. Product Rule : D(uv) = udv + vdu u = + v = + du = dv = D(uv) = Product Rule : D(uv) = udv + vdu u = + v = / + 5 du = dv= D(uv) = Product Rule : D(uv) = udv + vdu u = v= ( + 5) du = dv = 8( + 5) simplify above to obtain D(uv) = 8( )( + 5) + ( + 5) ( ). Product Rule : D(uv) = udv + vdu u v du = dv = ( + ) D(uv) = d uv 6. Product Rule : D(uv) = udv + vdu Page 5 Version 7. May 0

34 u = (5+) 0.5 v 5 du = 5(5 + ) dv = ( 5) -.5 5(5) (5) D(uv) = ( 5) ( 5) D(uv) = (5+) (5 5) ( 5).5 u vdu udv. Quotient Rule : D( ) = v v u = + v = 6, du = dv = 6 u 6 D( ) = v (6 ) u vdu udv. Quotient Rule : D( ) = v v u = v= + 5 du = dv = u D( ) = v ( + 5) u vdu udv 5. Quotient Rule : D( ) = v v u = v = du = dv = u D( ) = v u vdu udv 6. Quotient Rule : D( ) = v v u = v= du = dv = u - D( ) = v ( ) u vdu udv 7. Quotient Rule : D( ) = v v u = 5, v= du =, dv = u D( ) = v ( ) 8. Product Rule : D(uv) = udv + vdu u v ( ) du 6 dv 8 ( ) d( uv) 8 6 d( uv) 8 6 Page 6 May 0 Version 7.

35 9. Product Rule : D(uv) = udv + vdu u = - v = ( ) - du = - -5 dv = -8 ( ) - 5 D(uv) = ( - ) OR 5 ( ) u vdu udv 0. Quotient Rule : D( ) = v v u du = ( ) -0.5 dv = v= + u D( ) = v ( ) u vdu udv. Quotient Rule: d v v u = v= ( ) du = 8 dv = u ( ) ( 8 ) ( ) ( ) D( ) = v ( ) = 8 u vdu udv. Quotient Rule : D( ) = v v u =, v= 6 5 du = 0 dv = 5 (6 ) 5 6- D( v u ) = = 5 (6 - ).5 (6.. Product Rule : d(uv) = udv + vdu.5 ) u = (6 + ) v = (6 ) du (6 ) dv 6 du ( v) = (6 ) (6 ) (- ) (6 ) (6 ) / -/ Page 7 Version 7. May 0

36 u vdu udv. Quotient Rule :D( ) = v v 5 u v du = ( + ) (6 + ) dv = ( 5) ( 5) u ( 5 ) ( ) (6 ) ( ) ( 5 ) ( 5) d( )= v 8 ( 5 ) = ( ) (6 ) ( ) ( 5) ( 5 ) ( 5 ) 5 5. (a) f '() = ( ) ( ) ( ) f '() = 5 = (b) f '() = ( ) ( ) ( ) f '() = 8 5 (c) f '() = 6 f '() = (, 0) Curve cuts -ais where y = 0 at Gradient of tangent g '() = 9( ) at = is f () = /, so at (, ) the equation of the tangent line is y = For the normal line, m = -/ So the equation of the normal line is y = Equation of tangent : y = + 8 equation of normal : y = 7 Tangent cuts the -ais at : -6 P,0 and the normal cuts the -ais at : Page 8 May 0 Version 7.

37 Q -7,0 Distance between P and Q = = = Page 9 Version 7. May 0

38 Part H Eploring Differentiation of Eponential Functions Differentiation Eponentials and Logarithms The fundamental idea is that the gradient of y = e at any value of is e. Here is the second demonstration described in the Student Manual. Differentiation of Eponential Functions The basic rule for the differentiation of an eponential function is as follows: If y = e, then d d e = e If the power of the eponential is itself a function, ie f y e Then using the chain rule: dy dy du d du d f u Let y e = e, where u f then dy d dy du du d d e u f d f f ' e = = Page 0 May 0 Version 7.

39 Therefore, the rules of differentiation for eponential functions are:. If, then. If, then Differentiation of Logarithmic Functions From Student Manual (pages ) Therefore the rules for differentiation of logarithms are:. If, then. If, then Eploring Differentiation of Logarithmic Functions Students will have noticed the gap in the pattern of derivatives. function y = derivative y = 0 Where is the function that has as its derivative? This is it! You can check by asking the computer to draw y = see both branches. when you will Page Version 7. May 0

40 Two Special Results From Student Manual, pages.. The Derivative of y = a If e y a log y log e a ln y ln a ln y ln a d dy ln a y dy y ln a d = a ln a If, then. The Derivative of y = log a If y log a log e Using the change of base law, y log e a dy d log a e If, then Task.6 Differentiation of Eponentials and Logarithms Differentiate each of the following:.. y e e. y y e. 5. y y e e 6. y y log e 8. y ln 9. y log e 0. y lne. y ln. e y log e Page May 0 Version 7.

41 . Find the gradients of the tangent and the normal to the curve y e at the point =.. Find the equations of the tangent and the normal to the curve f log at the point where =. e 5. Find the equations of the tangents to the curve y e where = and =. Hence, find the point T where the two tangents intersect each other. Solutions Task.6. Chain Rule : u=, y = e u ; dy u du = e, = du d dy dy du = d du d = e. Quotient Rule : u =, v = e, dv = e, du = 0 u vdu udv D v v dy = -e - d. Chain Rule : u =, y = e u, dy d dy = du dy 6 d = dy du u e, e du d du = 6 d Page Version 7. May 0

42 . Chain Rule : u= 7, y = e u, dy dy du = d du d dy du = e u, = - du d dy = -e 7 d 5. Product Rule : u =, v = e du =, dv = e dy = e ( + ) d 6. Quotient Rule : u = e, v = 9 + u vdu udv D v v du = e, dv = dy e ( 7 ) = d (9 ) 7. Chain Rule : u = +, y = ln u ; dy dy du = d du d dy du = /u, = du d dy = d 8. Chain Rule : u =, y = ln u ; dy dy du = d du d dy du = /u, = - du d dy = d, Page May 0 Version 7.

43 9. Chain Rule : u =, y = ln( u), dy dy du = d du d dy = du u, du = d dy = d - 0. Chain Rule : u = e +, y = ln u ; dy dy du = d du d dy = / u, du = e du d dy e = d e. Product Rule : u =, v = ln ; u d( ) udv vdu v du =, dv = / dy = ln d. y = ln ( ) ln ( ) dy = d. dy = 6e. d The gradient of tangent at = is 6e 6 Hence the gradient of normal : = e. f '() = Gradient of tangent = -. Equation of tangent at (-, 0) using y = m + c is y = - Gradient of normal =. Equation of normal at (-, 0) is y Page 5 Version 7. May 0

44 5. Equation for the gradient of tangent is dy = e - d Equation of tangent using y = m + c at (-, e ) with m = e =7.89 is y = e + e + Equation of tangent using y = m + c at (, e - ) with m= e - is y = e + e point of intersection : solve e + e + = e + e e e, e y e e e Co-ordinates are ( -.075,.9) Page 6 May 0 Version 7.

45 Part I Applications of the Derivative Significance of the First Derivative Stationary Points Maimum Turning Point Refer to the Student Manual (pp 5 8). Significance of the First Derivative f() dy measures the rate of change of f() or y in relation to. f ' or d > 0 function is increasing Eploring Horizontal Point of Inflection Therefore the gradient of the tangent is always positive < 0 function is decreasing Therefore the gradient of the tangent is always negative f ' = 0 function is stationary Therefore the tangent is parallel to the -ais with a gradient of 0. For Eample: The following sketch of a function shows where it is increasing, decreasing or stationary (constant). increasing decreasing increasing constant 0 Page 7 Version 7. May 0

46 Ask your students to interpret the gradient of the cubic in terms of the appearance of the parabola. Look at when the parabola is zero (zero gradient), when it is positive and when it is negative. Look at the special point when the parabola has a minimum and the gradient switches from getting more negative to slowly becoming more positive the point of inflection. Then turn off the function and see if they can reconstruct the black curve having only the gradient curve. Note that the shape of the curve can be obtained from the gradient function, but not its vertical position, as the addition of a constant to the curve makes no difference to the gradient. A Stationary Point dy A stationary point is defined as where = 0. d There are two types of stationary points:. Turning points (relative maimum and relative minimum): relative maimum relative maimum relative maimum relative minimum relative minimum boundary/end point boundary/end point The boundary points at either end of the diagram are also relative maimum/minimum values. You will look at this later on. Page 8 May 0 Version 7.

47 Horizontal point of inflection (there are other points of inflection where the tangents are not horizontal) Maimum turning point (curve is concave down). f ' = 0. f ' > 0 before the point where. f ' < 0 after the point where y f ' = 0 f ' = 0 concave down An eample of this kind of curve would be y Minimum turning point (curve is concave up). f ' = 0. f ' < 0 before the point where. f ' > 0 after the point where f ' = 0 f ' = 0 y concave up An eample of this kind of curve would be y 8 Horizontal Point of Inflection (Monotonic Point). f ' = 0 dy has the same sign over an interval, then. f ' or d is said to be monotonic over that interval. y f Page 9 Version 7. May 0

48 Eploring Distance, Speed & Acceleration The same ideas of differentiation help us understand the equations of motion. The first screen shows the car increasing speed: a = m/s. The speed is v = t and distance is d = 0.5t. The second screen shows the car decreasing speed: a = m/s. The initial speed is so v = t and distance is d = 0.5t t. Page 0 May 0 Version 7.

49 Task.7. Determine whether the curve y 5 is increasing, horizontal or decreasing at =.. Find any turning points on the curve y and determine if they are maimum, minimum or neither.. Find the stationary points on the curve f 0 6 and determine whether they are maimum or minimum turning points.. If the curve a f has a stationary point where =, find the value of a and hence determine the type of turning point. f 5 has a monotonic point of inflection, determine 5. If the coordinates for the point of inflection and describe the shape of the graph of the function. Solutions Task.7. dy = 6. d f ' (-) = 9. Increasing at = -. dy = d = 6( + )( + ) = 0 = -, = - Turning points at (-, -6) and (-, -6) dy In the neighborhood of = -, is 0 + d minimum at (-, -6) dy In the neighborhood of = -, is + 0 d maimum at (-, -6). f ' () = 0 = ( 5)( + ) = 0 = 0, 5, - Turning points at (0, 6), ( 5, -69), (-, -6) dy In the neighborhood of = -, is 0 + d minimum Page Version 7. May 0

50 dy In the neighborhood of = 0, is + 0 d maimum dy In the neighborhood of = 5, is 0 + d minimum. f ' () = a = 0 a = when = Turning point occurs at (, - ) dy In the neighborhood of =, is 0 + d minimum 5. At horizontal point of inflection, first and second derivative both equal to zero f ' 0 f '' f( ) 5 Coordinates are 0,5 Page May 0 Version 7.

51 Part J Differentiation Circular Functions Eploring Differentiation of Circular Functions The Student Manual uses the length of day (sunrise to sunset) as an eample of a sine function (refer to Page ). The graph below, however, showing day length in Beijing, looks like a negative cosine function! Clearly the function depends on the time of year that we call zero time. To get a sine function we should start at March. Eploring Derivatives of Sin, Sin n, Cos, Cos n Differentiation of Circular Functions The surprise for many students meeting this for the first time is that the gradients are so closely related to the function itself. This is partly due to the fact that angles are now being measured in radians and is the basic reason for the use of radians. Shown below is the gradient of sin. The function sin has been squeezed so that two waves fit into π. As a result its gradient is twice as steep as the regular function, so it has a gradient of at = 0. The gradient function is f = cos. Page Version 7. May 0

52 Task.8. Find the derivative with respect to of each of the following: a. sin 6 b. cos5 c. tan 7 d. cos f. cos e. sin. Find f ' n for each of the following: a. f cos sin b. cos c. f tan sin f sin d. f sin ; Solutions Task.8. a) f ' sin 6 6cos 6 b) f ' cos 5 5sin 5 c) f ' tan 7 7sec 7 d) Let y cos u, so y cosu and dy dy du. d du d Chain Rule So, dy sinu du du d dy sin u. d dy sinu d Substitute u and dy sin d e) This requires a two step approach ysin Firstly, u sin, so y u Page May 0 Version 7.

53 So, Now, Net need to solve du du dv. d dv d ; where v dv d du dv du d du cos v cos cos v v d du cos d Chain Rule dy u du du cos Chain Rule d dy u.cos d Substituteu sin and dy sin cos d f) Let y cos u cos, so y u and dy dy du. Chain Rule d du d dy du u sin du d dy u sin u sin d Substitute u cos and dy cos d sin Page 5 Version 7. May 0

54 . a. f cos sin f a b f ' f ' a f ' b f ' sin cos b. f cos sin f ' f ' a f ' b Let u cos so y u du dy sin u d du dy u. sin u sin d Substitute u cos So f ' a cossin f ' b cos f ' cos cossin c. f tan sin f ' sec cos sin du dv v u d u f '( ) d d d v v d. f Quotient Rule Let u sin and v du cos d dv d du dv v u d u f '( ) d d d v v cos sin Page 6 May 0 Version 7.

55 Part K Higher Derivatives Significance of the Second Derivative By applying the same method of differentiation to the first derivative of a given function, you obtain the second derivative of that function. For any given function y = f() The first derivative of the function y = f() is:, y, f() The second derivative of the function y = f() is obtained by finding the derivative of the first derivative function. This is signified by:, y, f () The resultant epression is the equation for the rate of change of the gradient. or The resultant epression indicates the rate of change of f() for any given value of. Task.9 dy. Find the second derivative and determine whether is d increasing, decreasing or neither, for each of the following functions at the point where =. a) y b) y c) y d) y 8 e) y 6 f) y g) 5 y ( ) h) y i) y log j) k) log e y e y e l) y e. For what values of is the curve y concave up?. Show that a point of inflection eists where = on the curve y 70. Page 7 Version 7. May 0

56 Solutions Task.9. Graphs of f() are shown (a) d y d = f ''() ( = 0) is > 0 ( = 0), hence f ' () is increasing at = b) d y d = f ''() ( = ) is > 0, hence f ' () is increasing at = Page 8 May 0 Version 7.

57 c) d y d = 6 f ''() ( = -) is < 0, hence f ' () is decreasing at = d) d y d = 8 f ''( ) ( = -6) is < 0, hence f ' () is decreasing at = Page 9 Version 7. May 0

58 e) d y d = f ''() ( = 0) is = 0, hence f ' () is a point of infleion at = f) d y ( )[( ) ( )] = d ( ) f ''() ( = 0.5) ) is > 0, hence f ' () is increasing at = Page 50 May 0 Version 7.

59 g) d y = 0 ( ) d f ''() is = 0, hence f ' () is a point of infleion at = h) d y d = ( ) Since f ( )<0 at, the function is decreasing. Page 5 Version 7. May 0

60 i) d y d = f '' () (= -) is < 0, hence f ' () is decreasing at = j) d y = e e d f '' () ( = 6.) is > 0, hence f ' () is increasing at = k) Since is not in the domain of f, nothing happens at. Page 5 May 0 Version 7.

61 l) d y 9 = d e f ''() (= 0.) is > 0, hence f ' () is increasing at = d y. For a graph to be concave up, > 0. d d y = 6 6 d for >, the curve is concave up.. f ( ) 8 and f () 0, hence point of infleion eists where =. Page 5 Version 7. May 0

62 Part L Eploring Curve Sketching Fundamentals of Curve Sketching Curve Sketching Common Conventions (from Student Manual p. 6). All aes must be labelled, where possible.. Scales should be appropriately selected for each ais.. The size of the graph should be sufficient to reasonably convey, in a clear and concise manner, all relevant information.. The graph of a function should be labelled with its equation. 5. The coordinates of turning points, points of inflection, asymptotes, points of intersection, boundary points and any other important information should be clearly identified. 6. Where there are more than one function being charted on the same set of aes, arrows may be required to clearly identify relevant information. 7. All graphs should be drawn with a reasonable attempt to accurately represent the shape and features of the graph. Task.0 Sketch the graph for each of the following functions on a separate number plane. Show and label: - stationary points - inflection points - boundary values (if given a set domain) - absolute maimum and minimum (if given a set domain). 9 y. y y y y y y 6 8. y ; 5 9. y ; y. y 6sin(0 ), 6 6. y ln( ) Challenge Question. y cos(0 ), 6 6 Page 5 May 0 Version 7.

63 Solutions Task.0. Graph of y = 9 + maimum minimum point of inflection at (, -0) maimum at (-, 6) minimum at (, -6) boundary values for the given domain are (-, 6) and (6, 55) absolute maimum is 55 and absolute minimum is -6. Graph of y = 9 7 maimum minimum Point of infleion at ( ½, -56½ ) Minimum at (, -9) Maimum at (-, 6) Page 55 Version 7. May 0

64 . Graph of y = 5 + Point of inflection at ( 0,5) Maimum at (,) Minimum at ( -,-). Graph of y = maimum 0 0 Series minimum point of inflection at (, ) 7 maimum at (0,0) minimum at (, ) 7 boundary values for given domain are (, ) and (, 8) absolute maimum is 8 and absolute minimum is - Page 56 May 0 Version 7.

65 5. Graph of y = + maimum : not found minimum at (-.5, ) points of inflection at (0, -) and (-, -) 6. Graph of y = ( ) maimum point of inflection at (0,0) and (, 6) maimum at (, 7) absolute maimum is 7 and absolute minimum is -5 boundary values for given domain are (-, -8) and (5, -5) Page 57 Version 7. May 0

66 7. Graph of y = 6 maimum minimum minimum points of inflection at (, -8) and (-, -8) maimum at ( 0, -) minimum at (, -) and ( -, -) 8. Graph of y = absolute maimum is - absolute minimum is - relative maimum at (-, -) boundary values for given domain are (-5, -6.5) and (-, ) Page 58 May 0 Version 7.

67 9. Graph of y = no relative maima, relative minima, absolute maima and absolute minima. Boundary values for given domain are (-5, 0.5) and (5, ) 0. y relative maima occurs at (0, -) and relative minima occurs at (, ) No absolute maima or minima. Page 59 Version 7. May 0

68 . y 6sin(0 ), 6 6, 8 6 maimum minimum local minimum at, 6, local maimum at,6, point of inflection at 0,0, Boundary points 6,0, 6,0. y ln( ) asymptote at Page 60 May 0 Version 7.

69 Challenge Question. y cos(0 ), maimum Boundary Points 6,, 6, Local maimum at 0, Points of inflection at,,, Assessment Event Project : Project should be undertaken now. Consider allocating a small amount of class time to ensure that students understand the requirements and also to ensure that the work undertaken is each student s own work. Collect the Project in one week s time. Assessment Event Collect Mathematical Terminology Logbooks from students for marking. Page 6 Version 7. May 0

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71 Unit : Integration Unit : Integration Part A Part B Part C Part D Part E Unit Introduction Terminology Introduced The Primitive Function and Indefinite Integrals Definite Integrals and the Area under a Curve The Volume of Rotation Part A Overview Unit Introduction In this unit, students will learn to apply appropriate methods of integral calculus to solving various numerical and graphical problems. In this unit, students will learn to: select appropriate methods of integral calculus calculate the area enclosed by a given curve calculate the length of a curve calculate the volume enclosed by a given surface. This unit includes a series of tasks that students will work through to practise the course material. They will be epected to complete some work in their own time. You will guide them through the unit. Assessment Event In-class Test should be administered at the end of Unit. Note that this is an in-class test under test conditions. Assessment Event Assessment Reminder: You must collect students Mathematical Terminology Logbooks at the end of each unit. Page 6 Version 7. May 0

72 Unit : Integration Part B Terminology Introduced By the end of this unit, students need to know the meanings of these terms. The terms are bolded in the first instance they occur. Please make sure students understand the definitions of new terminology and include these words in their Mathematical Terminology Logbook. Summary of terms primitive function lower limit constant of integration integration upper limit bounded volume of rotation trapezoidal rule Simpson s rule intervals subintervals indefinite integral definite integral fundamental theorem anti-derivative integrand variable of integration numerical methods approimation Page 6 May 0 Version 7.

73 Unit : Integration Part C The Primitive Function and Indefinite Integrals Integration An anti-derivative of a function f() is another function F() such that: F() = f() or Any two anti-derivatives of a function differ only by a constant. Anti- Derivatives The Primitive Function (Refer to Student Manual p. 58) The primitive of a function is the name usually given to the antiderivative. It is the original function from which the derivative was found. So far you have only looked at the concept of what the primitive function is in relation to the derivative. Within the eamples on the previous page you may have noticed a pattern/rule emerging when finding the primitive from the function of the derivative. Rules: If then + C, where k is any constant If then, n If then, n Task.. Find the primitive function of: a) b) c) 9 d) 5 e) f). Find f () if: g) 6 h) i) a) f () = 5 9 b) f () = 6 c) f () = 8 d) f () = e) f () = f) f () = Page 65 Version 7. May 0

74 Unit : Integration. Epress y in terms of if: dy a) 5 d c) dy d 5 dy b) d 9 d) dy dy dy e) f) d d d. Given that f () = 6 and f () = 9, find an epression for f(). 5. Given that f (t) =t and f (6) =, find f (). 6. If y = ( ) and = 7 when y = 0, find when y =. 7. Given that f () = 5, f (0) = and f ( ) =, find f (). 8. If the gradient of the tangent to a curve is given by and the curve passes through the point (, 0), find the equation of the curve. 9. The rate of change of V with respect to t is given by dv t dt If V = 5 when t = 05, find V when t = 5. Solutions Task.. Using Rules : if f '() = k then f () = k : if f '() = n then f () = n n + C : if f '() = n ( a b) then f () = n ( a b) a( n ) + C a) + C b) + C c) + C d) C Page 66 May 0 Version 7.

75 Unit : Integration e) f) + C 5 C g) ( 6) 5 5 C h) f ( ), then f ( ) C i) 7 / 7 + C. Using Rules : if f ' () = k then f() = k : if f ' () = n then f() = n n + C : if f ' () = n a b then f () = n ( a b) a( n ) + C a) f() = C b) f() = C c) f() = ( 8) C d) f ' () = f() = C e) f() =.5 + C f) f ' () = / then f() = 5/ + C 5 Page 67 Version 7. May 0

76 Unit : Integration. a) y = C b) y = + + C c) y = C d) dy d.5 y = e) f ' () = C y = + C f ) f ' () = + y = + C. f() = + + C, =, y = 9 c = - f() = + 5. f (t) = ( t ) + C t = 6, y = f(t) = ( t ) + 9 f() = 0 Page 68 May 0 Version 7.

77 Unit : Integration 6. y = C substitute =7, y = 0 C = 6 y = ( ) 6 When y =, = f '() = 5 + f () = C = -, y= gives C = 6 5 Thus f () = and f () f '() = +. Curve passes through ( -, 0). Substitute = - and y= 0 in f() = + + C C = -6 Hence equation of the curve is y = dv = (t ). dt V = (t ) 6 + C. V = 5, t = 0.5 gives C = 5. (t ) Substitute t = 5 in V = V = 6.5 Page 69 Version 7. May 0

78 Unit : Integration The Indefinite Integral (from Student Manual p. 6) Notation where F() is the primitive function of f() General Rules, where k is a constant, where k is a constant Task. Find the indefinite integral of:. d. 9 d. d 5. d d 6. t 6 t d 8. y 7 y 5 d 0. 5d 5 d. d. 6 dt y dy Page 70 May 0 Version 7.

79 Unit : Integration Solutions Task. Using rules of integration: f () d = F() + C k d = k + C n d = n n + C ( a b ) n d = n ( a b) a( n ) + C, obtain C. ( ) 9 + C C C 5. Note = y = C 6. Epand the bracket to obtain (t t 6t + )dt t y = t 8t + t + C 7. Epand the bracket to obtain ( + ) d y = C Page 7 Version 7. May 0

80 Unit : Integration 8. y = y y = y (7y y y ) dy y 8y + 8y + C 9. y = (5 9) 0 + C 0. ( 5d = ( 5) d y = 5 C. y = 9( ) + C. ( 6 5 ) d then y = = (6 5) 7 (6 ) C d = 6 7 Page 7 May 0 Version 7.

81 Unit : Integration Part D Eploring Definite Integrals and Area under Curve Definite Integrals and the Area under a Curve The magic of integral calculus is that the area under a curve between two endpoints is the same as the difference between the anti-derivative, evaluated at the two endpoints. It is hard to appreciate the power of this until you see it and verify that it works for a few cases where you can actually calculate the area and compare it with difference between the evaluated antiderivatives. Students should be encouraged to understand how and why it operates in this way. Here is the screen for the fourth dot point in the Student Manual: Spreadsheet Adding areas (trapezoidal) showing the area under y = +. The trapezium area is the average of the parallel sides times the distance between them. The second idea with this spreadsheet is to show that increasing the number of trapezia (trapezoids) approaches the area under the curve. This is done with curves and you simply use the slider to get more trapezia (trapezoids). Page 7 Version 7. May 0

82 Unit : Integration Definite Integrals Definite integrals are different to the integrals performed so far because they have limits that determine the value of the integral. The rules of integration applied earlier to find the primitive function are the same here; however, no constant of integration is added to the result. [From Student Manual, page 6.] General Rule If the primitive function of f() is F(), then: = F(b) F(a). where b is called the upper limit and a is called the lower limit. To find the value of a definite integral:. Calculate the primitive function and simplify where possible.. Substitute the upper and lower limits into the primitive function.. Find the difference between the results. Eploring Distance, Speed & Acceleration Just as derivatives (gradients) were useful to interpret the graphs of motion, so are integrals (the areas under the curves). Here is the situation as described in the Student Manual (pp. 6 65). Page 7 May 0 Version 7.

83 Unit : Integration Task. Solutions Task.. Find d. Find 0 d. Find d 0. Find 5. Find y dy y d 6. Find 6 7. Evaluate 6 0 d d 8. Evaluate d 9. Evaluate a b c 0. Find 0 5 d d. Find d 8. Find 5 d. Evaluate d a. Find a a d c 5. Evaluate c if 0 d 6. Find the value of a if a. ( ) d. = 8 d ( ) ( ) d 0 8 (. ) ( ) = 6 0 Page 75 Version 7. May 0

84 Unit : Integration. + d (9 6 9 = [ 6 ( + ) ] 0 = 6 ቀ9 ቁ.9 ) ( ) y y ( y y )dy (6 ) (6) ( ). (6 ) d () 0 ( ) d a b 9b c 9a c 5 5 =.85 0 Page 76 May 0 Version 7.

85 Unit : Integration.. ( ) d a a (a.05 d ( ) ( 8) )d a a a a a ( a) a a (5 ) a c 5. 0 c c = c c + = 0 6. No Solution! (Draw a graph for students to show why this is so ) a 8 a( 6) ( 6) a a = 6 Page 77 Version 7. May 0

86 Unit : Integration Eploring Integrations Involving Key Functions Integrations Involving Eponential, Reciprocal (to Logarithmic) and Circular Functions We have found that the anti-derivative primitive function actually describes the area between the function and the -ais, between the left and right boundaries you choose. The spreadsheet Area functions is designed to show how this area function relates to a wide range of functions. You set the left boundary and the values of the area function are the areas calculated as definite integrals from that left boundary. So the area function always starts at 0 at the left boundary. Shown below is a slightly more comple function and its interpretation. The function is y =. Starting at 0, the area is under the ais so the area function goes negative, but as the function approaches 0 the areas being added lessen, so that at = the area function has a minimum. Then the areas increase to the point, at =, when the total area under and over the ais is 0 a zero value for the area function. It then continues to increase in a non-linear way, as more is being added in each step. Integrations involving Eponentials and Logarithms Eponentials General Rule * special result: Logarithms Page 78 May 0 Version 7.

87 Unit : Integration General Rule Circular Functions General Rule Task. Integration of Eponentials, Logarithms and Circular Functions Evaluate the following indefinite integrals:. e d. e d e e. e d. e e 5. d 6. e d e 5 7. d d 8. d d 0. d 8. d. 6 d Evaluate the following definite integrals:. 5. e d. 0 5 d e d e 6. e d Page 79 Version 7. May 0

88 Unit : Integration 6 7. e t5 d 8. t e e d d. d Integrate the following: 5. sin d dt d. d ln e. 0 e d sin d 7. cos sin d 8. cos d sin cos d cos d Solutions Task..... e d e e e d C d e e d e e e e d e e C e d C d ln( e e ) C Page 80 May 0 Version 7.

89 Unit : Integration d C ln e e d e e e e C 5 5 d 5 d ln( ) C d ln( ) C 7 6 ( 7) ln( 7) 6 d d C 6 + d = (+) d... = ln( + 8) + C d ln( 6) C 6 d d ln( ) C 0 e d e 0 0 e e ( ) e 0.6 Page 8 Version 7. May 0

90 Unit : Integration. e e e e e e e e 6 5 (e t 5 ( t e ) dt ) (e t = t e = 7.6 ln(5 9) 5 ln 9 ln ) ln( ) ln(6 ) ln( ) ln 7780 ln e 0 ( e e ) Page 8 May 0 Version 7.

91 Unit : Integration... ln( ) = ( ln) ( ln 0) = Undefined ln( ) - ln - ln 0 ln( e ln( e ) ln ln 0 ln ln ) ln( e 0 ) d 0 0 sin. cos 0.9 n nsin k ad cos( k a) C k sin d d cos sin d cos d sin d sin cos sin cos cos d sin sin sin ( ).707 Page 8 Version 7. May 0

92 Unit : Integration sin cos d sin d cos d cos cos 0 sin sin cos d cos d sin C sin C Page 8 May 0 Version 7.

93 Unit : Integration Area Under Eponential Function Area Under the Eponential Function y = a e k You may wish to eperiment with the left interval for the spreadsheet Area functions a e^k. Shown below is the screen for the area under the green curve y = e. If the area starts at 0 for =, then the constant is 0. Area Under Reciprocal Functions Area Under the Reciprocal Function y = k The area under y = is shown below. Note that it is not possible to find the area from 0, as the curve never reaches = 0. It is an asymptote. So a suitably small value is chosen. The constant has been adjusted so that the log curve passes through (, 0), and the arbitrary constant is therefore 0. Page 85 Version 7. May 0

94 Unit : Integration Area Under Circular Functions Here is the screen for the area under the function: y = sin 0.5. Because the curve is twice as wide, the areas are twice as great. Note that the turning point happens when the function starts going negative. Substitution Rule. Find d We can write the function as ( ) d. The anti-derivative techniques we have covered so far are not sufficient to evaluate such an integral. In such cases we try to simplify the function by changing the variable to a new variable. If we assume u to be what s inside the root sign, u, the differential of u is du du d d. Then our integration will be: du u u d ( ) u u du C C ( ) In general, this method works when an integral is of the form u ( ) f u( ) ( ) f, as in the previous eample above. The Substitution Rule u ( ) f u( ) d f u( ) du where f is a continuous function and u () is a differentiable function. Page 86 May 0 Version 7.

95 Unit : Integration Task.5 Integration by Substitution Determine the following integrals:. ( ) 99 d, Let u. d, Let u. d, Let u e d, Let 5 5. d ln 5 u, Let u ln 6. ln d 5, Let u ln 7. e d, Let u e e 5 ( ) 8. d, Let u e 9. d, Let e u e ln( 0. ) d, Let u ln Page 87 Version 7. May 0

96 Unit : Integration Solutions Task.5. Let u du d du d 99 u du = 00 u c 00 ( ) C. Let u u u u du = 5 u u C 5 5 ( ) 5 + ( ) C.. Let u 6 du 6 d du 5 d u du C 6 5 5u C 5 5( 6) C Let u du 5 d 5 du u d e du 5 5 e C 5 5 Page 88 May 0 Version 7.

97 Unit : Integration 5. Let u ln d du 6. d du u c ln ln u ln(ln ) C ln ln d d 5 5 Let u ln du d du d (ln ) d u du u C (ln ) C 5 5 ln C Let u e du e d du d u du u c e e C e e d e C 0.5 Let u ( ) du 0.5 d du d du u 6 5 u du 6 6 C = + + then du e d C Page 89 Version 7. May 0

98 Unit : Integration 9. Let u e du d e du ln uc u ln e C 0. Let u ln( ) Then du d d du udu u C ln C ln C Eploring Area under a Curve Definite integrals are the area under a curve between two boundaries. (Refer to Student Manual pp 7 75). Area under a Curve Eploring the Fundamental Theorem of Calculus Or, more formally: The Fundamental Theorem of Calculus For function, continuous over the interval where on Page 90 May 0 Version 7.

99 Unit : Integration Eploring the Fundamental Theorem and Motion Area Between Two Curves Distance, Speed and Acceleration The Fundamental Theorem brings it all together. To consolidate it, show how the theorem applies to motion using spreadsheet Motion. The screen is shown below. Task.6 Area under a Curve For each of the following area questions, first sketch the graph and shade the area that you are calculating.. Use integration to calculate the area bounded by the graph of the straight line y, the -ais and the ordinates = and = 5.. Calculate the area bounded by the curve y, the -ais and the ordinates = and =.. Calculate the area of the region bounded by the -ais and the graph of the function f 6.. Calculate the size of the area bounded by the ordinates = and =, the -ais and the curve y Find the area of the region bounded by the curve y ( ) and the -ais. 6. Calculate the area of the region bounded by the curve y-ais, the ordinates y = and y = 5. y 9, 7. Calculate the area bounded by the y-ais, the ordinates y = and y 6. y = 5 and the curve Page 9 Version 7. May 0

100 Unit : Integration 8. Find the area of the region (to decimal places) bounded by the curve y e and: a) the ordinates = and = and the -ais. b) the ordinates y = and y = and the y-ais. 9. Find the area of the shaded regions in each of the diagrams below. a) y = + y = b) bounded by the y-ais y y = - + y = c) y = 8 y = y = d) y y = y = - Page 9 May 0 Version 7.

101 Unit : Integration e) y y = y = = f) g) y y = + y = Page 9 Version 7. May 0

102 Unit : Integration Solutions Task.6. Area 5 ( ) d 75 ( 0) (.5 ) 5. y= + ( ) d 6 Area = 6 Page 9 May 0 Version 7.

103 Unit : Integration. f()=6+ d 9 8 (8 9) ( ) (6 ) 6 = Area = y= ( 7 6) d ( 6) ( 56 ) Area.5 Page 95 Version 7. May 0

104 Unit : Integration 5. y= (-) d ( ) 0 6 Area 0 6. y= 9.5 (9 y ) ( 9 y ) dy y Area =. 6.5 Page 96 May 0 Version 7.

105 Unit : Integration 7. y=-( -6) 5 (6 y) (6 y) dy (8) Area 5 8. y=e e d e e e Area = a) d ( ) Page 97 Version 7. May 0

106 Unit : Integration b) c) d) e) (f) ( ) d y y ( y y ) dy 0 0 ( ) ( d d 0 ( 5 ) 0. d ln ln ln.6 Area= ( ) ( ) ( ) d d d 0 (g) 0 0 ( ) d d [( ) ] d 0 8 Page 98 May 0 Version 7.

107 Unit : Integration Part E The Volume of Rotation Refer to Student Manual (pp 85 87). Task.7 Volumes of Rotation. Find the volume formed by rotating the curve y = between = 0 and = around the -ais.. Calculate the volume formed when the curve y = + between = 0 and = is rotated around the y-ais.. Find the volume created when the area bounded by the curve y = and the -ais is rotated around: (a) the -ais. (b) the y-ais.. Calculate the volume of revolution resulting from rotating the curve y = e between = 0 and = around the -ais. 5. Find the volume formed from rotating the curve y = ( + ) between = 0 and = around the -ais. Solutions for Task.7. V = π 6 d 0 = π [ 7 7 ] 0 = 8 7 π 5. V = π ( y ) dy = π [ y y] 5 = 8π. (a) V = π ( ) d = π [ ] = 5 5 π (b) V = π ( y) dy 0 = π [y y ] 0 = 8π. V = π e d 0 = π [ e ] 0 = π ( e ) 5. V = π + d 0 0 = π udu (Substituting u = + ) = π 0 [ u ] = π (0 0 ) Page 99 Version 7. May 0

108 Unit : Integration Assessment Event The In-class Test should be administered at the end of Unit. Note that this is an in-class test conducted under test conditions. Assessment Event Collect the Mathematical Terminology Logbooks from students for marking. Page 00 May 0 Version 7.

109 Unit : Advanced Applications Unit : Advanced Applications Part A Part B Part C Part D Part E Part F Part G Unit Introduction Terminology Introduced Motion in a Straight Line (Rectilinear Motion) Approimation Methods in Integration Problems involving Maima and Minima Eponential Growth and Decay Verhulst-Pearl Logistic Function The Natural Law of Growth and Decay Part A Unit Introduction Overview In this unit, students will learn to apply differentiation and integration techniques to solve a variety of problems. In this unit, students will learn to: apply their knowledge of calculus to a variety of practical situations; learn some applications of differentiation and integration using techniques learnt in the previous two units learn how to solve different problems that involve maimising or minimising a quantity. This unit includes a series of tasks that students will work through to practise the course material. They will be epected to complete some work in their own time. You will guide them through the unit. Assessment Events, and Project to be undertaken by students following completion of Part C Motion in a Straight Line Rectilinear Motion in this Unit. The Eamination is to be conducted at the end of Unit. Students are also required to submit their Mathematical Terminology Logbooks at the end of Unit. Remind students of assessment due dates and requirements. Page 0 Version 7. May 0

110 Unit : Advanced Applications Part B Terminology Introduced By the end of this unit, students need to know the meanings of these terms. The terms are bolded in the first instance they occur. Please make sure students understand the definitions of new terminology and include these words in their Mathematical Terminology Logbook. Summary of Terms averages marginal total cost revenue demand function variation quantity marginal revenue marginal cost trapezoidal rule Simpson s rule numerical method marginal revenue product consumer s surplus producer s surplus equilibrium point logistic function Verhulst-Pearl logistic maimising disease minimising subintervals approimation intervals Page 0 May 0 Version 7.

111 Unit : Advanced Applications Part C Motion in a Straight Line Important Motion in a Straight Line (Rectilinear Motion) Refer to Student Manual (pp. 9). Important Terms to Know for This Topic: i) INITIALLY: At the Beginning (t = 0) ii) AT REST: Velocity = 0 (V = 0) iii) AT THE ORIGIN: Displacement = 0 ( S 0 ) Note: S does not measure the distance travelled but measures how far the particle is from the origin, S measures displacement. Integration in Motion Equations Since the acceleration is the derivative of the velocity and the velocity is the derivative of the displacement therefore velocity is the Primitive Function of acceleration and displacement is the Primitive Function of velocity. if a f (t) V adt Note: Do NOT forget the constant when finding these Primitive Functions Task.. The equation for displacement of a particle is given by S t 0t i) Find its initial velocity and acceleration. ii) Find when the particle is at rest iii) Find the displacement and acceleration after seconds.. The equation for displacement of a particle is given by S e t i) Find the initial velocity ii) Find the acceleration after seconds iii) Show that the acceleration is always double the velocity. The acceleration of a particle is given by a sin t if initially the particle is at rest at the origin find the eact displacement after seconds. t. The velocity of a particle is given by v if the particle t is initially at the origin find the displacement after seconds. Page 0 Version 7. May 0

112 Unit : Advanced Applications Solutions Task.. S t 0t V 8t 0 a 8 (i.) initial velocity (t = 0) = 0m / s initial acceleration = 8 m / s (ii.) at rest: V 0 8t 0 0 t (iii.) t, S () 0() 7m a 8m / s. S e t t V e a e t (i.) Initial velocity (t = 0) = e 0 m / s (ii.) When t =, a = e (iii.) a e t V e t a V. a sin t V cos t C 6 cos t C When t 0, V 0 0 6cos 0 C 6 C V 6cos t 6 S sin t 6t C t 0, S C C 0 S sin t 6t When t, S sin 6 Page 0 May 0 Version 7.

113 Unit : Advanced Applications. V t t t S dt t t dt t S ln( t ) C t 0, S = 0 0 = ln(0 + ) + C C = ln = ln When t : S() = ln( + ) ln = ln ቀ 0 ቁ = ln 5 Assessment Event Project is to be undertaken in the students own time. Consider allocating a small amount of class time to ensure that students understand the requirements and to ensure that the work undertaken is each group s own work. The students have one week to complete the project. Page 05 Version 7. May 0

114 Unit : Advanced Applications Part D Approimation Methods in Integration [From Student Manual pages ] Some functions have primitive functions that are quite difficult to obtain and sometimes it is not possible to evaluate the integral by algebraic means. So there are numerical methods that can be used to give a close approimation of the area under such difficult functions: Trapezoidal Rule. Trapezoidal Rule. Simpson s Rule. Substitution Rule Therefore the General Trapezoidal Rule is: ò b a f ( h éë f a ( ) + f ( a + h) + f ( a + h) f ( b) ù û where: b a h = width of each interval n n = the number of subintervals. Simpson s Rule For n (n even) equal subintervals (n + function values) Simpson s Rule states: b a h f ( ) d f a0 f an f a f a... f a f a... h [(sum of the end values) + (odds) + (evens)] b a where h = n Task. Numerical Methods of Integration. Calculate the area between the curve y, the -ais and the bounds = and = using the trapezoidal rule with subintervals ( function values).. A fisherman decided to measure the depth of a river every 0 metres. The results of his measurements where as follows: Distance across the river Depth in metres 0m 0m 0m 0m 0m 50m 0m 6m 8m m m 0m Use the trapezoidal rule to estimate the area of the cross-section of that part of the river. Page 06 May 0 Version 7.

115 Unit : Advanced Applications. Use the trapezoidal rule, with three function values to estimate: 5 d. Use the trapezoidal rule with four subintervals to approimate the value of 0 d Evaluate 5 d using Simpson s rule with 5 function values. 6. Estimate the value of rule with subintervals. 7. The curve y d using Simpson s between = and = is rotated about the -ais. Estimate the volume of the solid formed using Simpson s rule with four subintervals. 8. Estimate the area between the curve y and the line 5 y using Simpson s rule with 7 function values. Solutions Task.. f () =, f () = 9, f () = 8 ( ) d () (9) 8 = units squared Page 07 Version 7. May 0

116 Unit : Advanced Applications. Area = (0) [(0 + 0) + ( )] =8m. f ()= = 7, f () = = 8, f (5) = 5 = 5 d = () [ (8) ] = 6 unit. Page 08 May 0 Version 7.

117 Unit : Advanced Applications f ()=.5, f (.5) =.06, f () =.7, f (0.5) =.50, f (0) =. 0 d 0.5 = [ ( ] =.68 =.6 units 5. f(5)=0, f()=, f()=, f()= =.6, f()= = d = [ (0 +.9) + (.6 + ) + ( ) ] =. 6. Page 09 Version 7. May 0

118 Unit : Advanced Applications f () = 65, f (0) = 8, f (-) =, f (-) =, f (-) = 8 ( ) d = [ (8 + ) + () ] = f() =, f(.5) = 8 5, f() =, f(.5) =, f() = π ( ) d = (0.5)( ) [ ( + 0) + ( ) + ( )] = π(0.5) ( 8 ) [ ] = 77 π Page 0 May 0 Version 7.

119 Unit : Advanced Applications 5 Solve y = and y = and obtain = 0 or = [( ) ( )] d gives the required area. 0.5 [.5] d f (0)= 0, f (-0.5) = -0.5, f (-0.5) = -0.5, f (-0.75) = , f (-) = -0.5, f (-.5) = -0.5, f (-.5) = [.5] d 0.5 = [ 0+0+( ) + ( )] (0.5) (-6.75) Page Version 7. May 0

120 Unit : Advanced Applications Part E Problems involving Maima and Minima Eamples Involving Minima and Maima Problems Steps in Solving Applied Maimum and Minimum Problems Refer to Student Manual (pp 0 0). Task. Maima and Minima Problems. Find the two numbers whose sum is 50 and whose product is maimum.. A manufacturer knows that the total cost c of producing q units is given by c 0.0q q 800. Find the value of q which minimises the average cost.. Suppose the demand function for a certain product is p 00 q, where p is the price (in dollars) per unit (q) and the 00 average cost function (in dollars) is c. Find the value q of q which maimises the profit?. A hotel has 80 rooms that can be rented for $00 per month. However, for each $0 per month increase, there will be two vacancies with no possibilities of filling them. What rent per room will maimise the monthly revenue? 5. Find the maimum area of a rectangular that can be inscribed in a circle of radius A rectangle of cardboard is 6 cm by 8 cm. In order to form a bo without a cover, identical small squares are cut from each corner and the remaining cardboard section is folded. Find the dimensions of the bo that maimises the volume and find that volume. 7. A power station is on one side of a river, which is km away from the beach and a house is km downstream on the other side of the river. It costs $ per metre to run the power line on the land and $8 per metre to run it water. How we should connect the power lines to minimise the cost? 8. The demand function for a product is p 50 q and the average cost function is c 8 00, where q is number of units, q and both p and c are epressed in dollars per unit. a. Determine the level of output at which profit is maimised. b. Determine the price at which maimum profit occurs. c. Determine the maimum profit. d. If, as a regulatory device, the government imposes a ta of $6 per unit on the monopolist, what is the new price for profit maimisation? Page May 0 Version 7.

121 Unit : Advanced Applications Solutions Task.. Let the numbers be and y. Then y 50 y 50 Product y (50 ) p( ) 50 5, and y The numbers are 5 and 5.. c 0.0 q 800 c q q c q q Average cost q 800 q c ( q) q 0 q. which minimises the average cost.. p 00 q c c q q c q Revenue pq 00q q Profit = revenue cost 00q q (q 00) q 98q 00 Ma Profit when 8q 98 0 Maimum profit when q.75. p = price, q = apartment. Hotel has 80 rooms, rent = $00 per month. Revenue = pq = $000 $0 increase per month gives vacancies. Then, Revenue $(00 0), rooms 80 If represents the number of $0 increases R (00 0 )(80 ) R( ) Revenue is ma when =0; Revenue = $ $600 per room will maimise profits. Page Version 7. May 0

122 Unit : Advanced Applications 5. Triangle: Length = 8 Rectangle : Length = 8 Width = Area of rectangle = l w ( 8 ) 8 56 A( ) This is a square of size.units Maimum area =.. = 8. unit Alternatively, use angle in a semicircle (consider diameter instead) y 6 y 6 Area y 6 A gives the same maimum area as above. Page May 0 Version 7.

123 Unit : Advanced Applications 6. length = 6, width = 8, height = Vol (8 )(6 ) 8 8 V ( ) = 6. or.69(ma) height=.69, width=.6, length =.6 Ma Volume = 98.5 cm 7. BX = ( ) Total cost=c() 6 8 8( 8 0) C( ) 8( )( 8 0) ( 8) ( 8) 80 0 =.85 (or 5.5 unacceptable) 8 0 Hence power line on land is.85km and.km under water. The minimum cost is $9.88. Page 5 Version 7. May 0

124 Unit : Advanced Applications 8. p = 50 q 00 c 8 q c cq c = 8q + 00 revenue = pq = (50 q) (q) = 50q q profit = revenue cost = (50q q ) (8q + 00) = - q + q 00 (a) p' (q) = -8q + q = 7 the profit is maimised (b) price = 50 q = 50 7 = $79 Maimum profit occurs when price is $79 per item (c) maimum profit= - q + q 00 = 7 + ( 7 ) 00 = $60.5 is the maimum profit (d) new price= $79 + $6 = $95 per unit Page 6 May 0 Version 7.

125 Unit : Advanced Applications Part F Eponential Growth and Decay Task. Eponential Growth and Decay Refer to Student Manual (pages 06 08). Eponential Growth and Decay Definition: Any quantity is said to have an eponential growth/decay model if at each instant in time, its rate of increase/decrease is proportional to the amount of the quantity present. dp ) Show that if P=A+Be Kt then K P A ) If 0.0P 00 dt dp dt Write and equation for P in terms of t if t=0, P=500 Hence find P when t=5 ) Assume Newton s Law of Cooling for the following problem: dt i.e. K T To dt An oven has been heated to 00 o C. An object with temperature of 5 o C is placed in the oven and after 0 minutes its temperature is 80 o C. i) What will be the temperature of the object after 0 minutes? ii) How long will it take the object to reach a temperature of 80 o C? Solutions Task. kt. P A Be dp B e kt K dt kt KBe But Be kt ( P A) dp k( P A) dt dp. 0.0( P 00) dt 0.0t P 00 Be Be 00 B 0.0t P 00 00e t 5, P e = Page 7 Version 7. May 0