level iii module 16: gac016 mathematics iii: calculus &...

GLOBAL ASSESSMENT CERTIFICATE

FACILITATOR GUIDE

LEVEL III Module 16: GAC016

Mathematics III: Calculus & Advanced Applications

Date of Issue: May 2013 Version: 7.3

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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 1: DIFFERENTIATION ....................................................................................................... 1

PART A UNIT INTRODUCTION .......................................................................................................... 1 PART B TERMINOLOGY INTRODUCED .............................................................................................. 3 PART C DIFFERENTIATION – CONTINUITY AND LIMITS .................................................................... 4 PART D THE DERIVATIVE – GRADIENTS OF SECANTS AND TANGENTS ............................................ 8 PART E DIFFERENTIATION OF FUNCTIONS ..................................................................................... 12 PART F RATES OF CHANGE ........................................................................................................... 18 PART G THE PRODUCT, QUOTIENT AND CHAIN RULES .................................................................. 20 PART H DIFFERENTIATION – EXPONENTIALS AND LOGARITHMS ................................................... 30 PART I APPLICATIONS OF THE DERIVATIVE .................................................................................. 37 PART J DIFFERENTIATION – CIRCULAR FUNCTIONS ...................................................................... 43 PART K HIGHER DERIVATIVES ...................................................................................................... 47 PART L CURVE SKETCHING ........................................................................................................... 54

UNIT 2: INTEGRATION ............................................................................................................. 63

PART A UNIT INTRODUCTION ........................................................................................................ 63 PART B TERMINOLOGY INTRODUCED ............................................................................................ 64 PART C THE PRIMITIVE FUNCTION AND INDEFINITE INTEGRALS ................................................... 65 PART D DEFINITE INTEGRALS AND THE AREA UNDER A CURVE .................................................... 73 PART E THE VOLUME OF ROTATION ............................................................................................. 99

UNIT 3: ADVANCED APPLICATIONS .................................................................................. 101

PART A UNIT INTRODUCTION ...................................................................................................... 101 PART B TERMINOLOGY INTRODUCED .......................................................................................... 102 PART C MOTION IN A STRAIGHT LINE (RECTILINEAR MOTION) .................................................. 103 PART D APPROXIMATION METHODS IN INTEGRATION ................................................................. 106 PART E PROBLEMS INVOLVING MAXIMA AND MINIMA ............................................................... 112 PART F EXPONENTIAL GROWTH AND DECAY .............................................................................. 117 PART G VERHULST-PEARL LOGISTIC FUNCTION – THE NATURAL LAW OF GROWTH & DECAY .. 119

APPENDIX: FORMULAE ........................................................................................................ 121

GAC016 Mathematics III: Calculus & Advanced Applications Facilitator Guide

Introduction

Page I ©ACT Education Solutions, Limited Version 7.3 May 2013

Introduction

Module Overview

Welcome to GAC016: 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,

GAC016: 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 expect 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:

1. Determine the derivative (if it exists) of most mathematical functions, and use the

derivative to analyse functional behaviour.

2. Use techniques of integration to find indefinite and definite integrals.

3. Apply differentiation and integration techniques in a variety of practical problems.

Facilitator Guide GAC016 Mathematics III: Calculus & Advanced Applications

Introduction

©ACT Education Solutions, Limited Page II May 2013 Version 7.3

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/A4 folder to write notes and complete exercises in for regular marking

a pocket-sized notebook for the Mathematical Terminology Logbook

access to a computer for spreadsheet/exploratory 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.


Introduction

Page III ©ACT Education Solutions, Limited Version 7.3 May 2013

Unit Breakdown

The following is a list of units to be covered in the module GAC016: Mathematics III:

Calculus and Advanced Applications.

Unit 1 Differentiation

Unit 2 Integration

Unit 3 Advanced Applications

Assessment Events

No. Assessment Event Weight

1 In-class Test: Units 1 – 2 20%

2 Project 1: Unit 1 Differentiation

Project 2: Unit 3 Advanced Applications

20%

3 Examination: Units 1 – 3 50%

4 Course Work: Includes Mathematical Terminology

Logbook and In-class Tasks.

10%

Note: All details of Assessments can be found in the GAC016 Assessment Folder. It is your

responsibility to ensure that the students are fully prepared for the assessments and that the

assessment tools (test/examination 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.


Introduction

©ACT Education Solutions, Limited Page IV May 2013 Version 7.3

Suggested Delivery Schedule

Week 1

Unit 1: Differentiation

Week 2 Unit 1: Differentiation



Assessment Event 2: Project 1

Mathematical Terminology Logbook due

Week 5 Unit 2: Integration




Assessment Event 1: In-class Test

Mathematical Terminology Logbook due

Week 9 Unit 3: Advanced Applications

Assessment Event 2: Project 2



Week 12 Assessment Event 3: Examination

Assessment Event 4: Course Work including Mathematical

Terminology Logbook


Introduction

Page V ©ACT Education Solutions, Limited Version 7.3 May 2013

Icons

The following icons will be used as a visual aid throughout the Student Manual and

Facilitator Guide:

Icon Meaning

Information

Task

Demonstration

Review

Independent Study – Including

Use of Spreadsheets

Assessment Events

Language Focus

Hints and Cautions



Page 1 ©ACT Education Solutions, Limited Version 7.3 May 2013


Part A Unit Introduction

Part B Terminology Introduced

Part C Differentiation – Continuity and Limits

Part D The Derivative – Gradients of Secants and Tangents

Part E Differentiation of Functions

Part F Rates of Change

Part G The Product, Quotient and Chain Rules

Part H Differentiation – Exponentials and Logarithms

Part I Applications of the Derivative

Part J Differentiation – Circular Functions

Part K Higher Derivatives

Part L Curve Sketching


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 3, 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 expected to complete

some work in their own time. You will guide them through the unit.



©ACT Education Solutions, Limited Page 2 May 2013 Version 7.3

Exploring Calculus

Calculus is about graphs of functions, so this section links very

closely with the sections about graphing functions in GAC004.

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 GAC004.

Assessment Events 1, 2, 3 & 4

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 2), and the

scope and nature of the final examination.

Assessment Event 4

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.





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

maximum

absolute maximum

approaches

monotonic point

tangent

differentiation

differentiate

chain rule

stationary point

minimum

absolute minimum

nature

relative maximum

secant

first principles

product rule

second derivative

inflection point

boundary values

normal

extremities

relative minimum




Part C Differentiation – Continuity and Limits

Exploring Differentiation

Continuity

Limits

Determining 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 4 – 6).

Continuity at a Point

If one or more of the conditions in the definition above fails to be

true, then f(x) is said to be discontinuous at x = a.

Task 1.1

Continuity and Limits 1. Write down whether each of the following graphs of functions

are continuous or discontinuous.

a) b)

c) d)

2. Find:

a) 45lim3

xx

b) 43lim 3

4

xx

x

-4 -2 2 4 6

-3

-2

-1

1

2

x

y

-4 -2 2 4 6

-8

-6

-4

-2

2

4

6

x

y

-4 -2 2 4 6

-8

-6

-4

-2

2

4

6

x

y

-4 -2 2 4 6

-8

-6

-4

-2

2

4

6

x

y

A function f(x) is defined to be continuous at a point (x = a ) if

the following three conditions are satisfied:

1. f(a) exists

2. The limit of f(x) as x approaches a exists

( exists )

3. f(a) is equal to this limit

(ie. f(a) = )




c)

xx

x

x 2lim

20 d)

2

6lim

2

2 h

hh

h

e)

3

27lim

3

3 x

x

x f)

m

mmm

m

139lim

23

0

g)

h

hhyyhhy

h

532lim

223

0 h)

6

65lim

2

2

2 aa

aa

a

i)

245

752lim

2

2

tt

tt

t j)

33lim

cx

cx

cx

3. For what values of x is the function

2611

62

xx

xy

discontinuous?

4. Determine whether the function:

f (x)

9

3

1832

x

xx

is continuous at x = 3.

Task 1.1 Solutions

1. a) continuous

b) discontinuous

c) continuous

d) continuous

2. a)

3

limx (5x - 4)

= 5(3) -4 = 11

b) 3

4lim 3 4x

x x

34 3 4 4

48

c)

0

limx

( 2)

x

x x

= 0

limx

1

( 2)x

= 20

1

1

2

x 3

x = 3




d )

limℎ→2

(ℎ − 3)(ℎ + 2)

(ℎ + 2)

= limℎ→2

(ℎ − 3)

1

= -1

e ) 3

limx

2( 3)( 3 9)

3

x x x

x

= 3

limx

2( 3 9)

1

x x

=27

f)

0

limm

2( 9 13)m m m

m

= 0

limm

2( 9 13)

1

m m

= - 13

g) 0

limh

3 2(2 3 5)h y hy y

h

= 0

limh

3 22 3 5

1

y hy y

= 2y3 +3y-5

h) 2

lima

( 3)( 2)

( 3)( 2)

a a

a a

= 2

lima

( 3)

( 3)

a

a

1

5

i) t

lim

2

2 2 2

2

2 2 2

2 5 7

5 4 2

t t

t t t

t t

t t t




2

2

5 72

lim4 2

5t

t t

t t

2

5

j) cx

lim

2 2

( )

( )( )

x c

x c x xc c

2 2

1lim

( )x c x xc c

2

1

3c

3.

6

( 13)( 2)

xy

x x

x ≠ 13 and x ≠ -2

discontinuous at x = 13 or x = -2

4.

( 6)( 3)

3

x xf x

x

= 9

3

6 3lim 9

3x

x x

x

exists and

f (3) = 9 exists

)3(lim)3(3

ffx

therefore f(x) is continuous at x = 3.




Part D The Derivative – Gradients of Secants and Tangents

Exploring the Derivative

The Gradient of a Secant

Gradient of a Tangent

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 = x3, at x = 1. We are looking at the tangent

drawn from 0.01 under 1 to 0.01 over 1. The curve and the tangent

have the same gradient.

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 express the gradient as rise

(the difference between two expressions) divided by the run, h.

Then as h approaches 0, the gradient expression approaches the

gradient function.

Gradient of a tangent from first principles:

=




Task 1.2 Gradient of a Tangent from First Principles Find the gradient of the tangent for each of the following functions

by differentiating from first principles.

1. 322 xxxf at the point (2 , 5)

2. 642 2 xxy at the point (1 , 0)

3. 153 xxxf at the point (0 , -5)

4. xxy 3at the point where x = 2

5. 23)( 2 xxxg where x = 1

6. xy at the point (9 , 3)

7. 12 xy at the points where x = 1 and x = 1

8. xxxxf 23 at the point where x =

3

1

Task 1.2 Solutions

1.

f (x) = 0

limh

2 2[(2 ) 2(2 ) 3] [2 (2 2) 3]h h

h

=

0

limh

2 6h h

h

=

0

limh

( 6)h h

h

= 6

2.

dx

dy =

0limh

2 2[2( 1 ) 4( 1 ) 6] [2( 1) 4( 1) 6]h h

h

= 0

limh

(2 8)h h

h

= 0

limh

(2 8)

1

h

= -8

3.

f (x) =0

limh

2 23(0 ) 2(0 ) 5 3 0 2(0) 5h h

h

=

0limh

(3 2)h h

h

=

0limh

3 2h

= -2




4.

dx

dy =

0

limh

3 3[(2 ) (2 )] [2 2]h h

h

=

0

limh

2(11 6 )h h h

h

=

0

limh

2 6 11h h

= 11

5.

g (x) =

0

limh

2 2[ (1 ) 3(1 ) 2] [ 1 3(1) 2]h h

h

0

limh

( 1)h h

h

0

limh

1 h

= 1

6.

dx

dy =

0limh

1/2 1/2(9 ) 9h

h

0

limh

1/2(9 ) 3h

h

X 1/2

1/2

(9 ) 3

(9 ) 3

h

h

0

limh

( 9 3)

h

h h

0

limh

1

( 9 3)1h

1

9 3

1

6

7.

dx

dy =

0

limh

2 2

x h x

= 0

limh

2

( )

h

x x h

= 0

limh

2

x x

at x = 1 , 2

21 1

dy

dx

at x = 1, 2

21 1

dy

dx




8.

dx

dy=

0limh

3 2 3 21 1 1 1 1 1[( ) ( ) ( )] [( ) ( ) ]

3 3 3 3 3 3h h h

h

= 0

limh

2 32h h

h

= 0

limh

22

1

h h

= 0




Part E Differentiation of Functions

Exploring Differentiation of Functions

The spreadsheet Gradient functions aims to make the ideas here

as clear as possible. There are three steps in the explanation.

See how the small tangent changes in gradient as x 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 = x2. It shows that the

gradient at x = 1 is 2. This is also the gradient of the pink gradient

function. As this goes through the origin the gradient function is

y´=2x. By changing the x-values you can see that the gradient is

always double the x-value.

For any given function y = xn

The first derivative, of the function y = xn is

given by:

𝑑

𝑑𝑥 ( x )n = nx n-1

where n is a real number.

The resultant expression is the equation for the

gradient of any tangent to the curve y = xn.

or

The resultant expression indicates the rate of

change of y = xn for any given value of x.




Exploring Gradients of Rational Functions

‘Rational functions’ is a name given to functions that have the

variable in the denominator. For y = 1

x the graph is an hyperbola.

For y = 1

x2 the curve looks like the trunk of a tree. Using the

spreadsheet Gradients of reciprocal functions: the screen below

shows the graph of y = 1

x and shows that the gradient at x = 1 is –1.

This is a particular value for the gradient function for which the

formula is y´ = –1

x2 . There is no gradient at x = 0 as the function is

discontinuous at that point.

If you use y = 1

x2 as your function you will see the gradient as the

function y´ = –2

x3 .

Exploring Tangent and Normal to a Curve

Refer to the Student Manual (p. 17) for further information on

looking at the gradient function for any polynomial term and page

18 for information on exploring tangent and normal to a curve.

Theorems of Derivatives

1. The derivative of a constant = 0

2. Polynomials are differentiated term by term.

3. 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.

mn = tm

1




Task 1.3 Differentiation of y = xn, Tangents and Normals

1. Differentiate each of the following with respect to x:

a) 4xy b)

33xy

c) 85 2 xy d) 722 xxy

e) 23 2xxy f) xxxxy 245 346

g) 2518 xxf h) 22

1

3 xxxg

i) x

xxxf

32 63 j) 12

4

1 24 xxxf

k) 3

5

3

xxh l) 321 462 xxxy

m) 6 xy n) xxxy

o) 2

25

2

362

x

xxy

p)

x

xxy

32

2. Find the equation of the gradient and the normal to the curve

232 2 xxy at the point (2 , 0).

3. Show that the gradient of x

y5

is always negative.

4. Find the equation of the tangent at the point (-1 , 6) on the curve

232 23 xxxy . If the tangent cuts the y axis at A and

the x axis at B, find the coordinates of A and B.

5. Find the point on the curve 512 xxxf where the

normal is parallel to the line .0123 xy

6. The curve cbxaxy 2 passes through the points A(2 , 0)

and B(3 , 20). The slope of the normal at A is 1 and the gradient

of the tangent at B is 9. Find the values of a , b and c .

Task 1.3 Solutions

1. a) dx

dy = 4x3

b) dx

dy = 9x2

c) dx

dy = 10x

d) dx

dy = 2x – 2

e) dx

dy = 3x2 - 4x




f) dx

dy = 6x5 – 20x3 – 12x2 – 2

g) f ' (x) = 10x0.25

h) g'(x) = 5.02

1

x +

3

6

x

i)

2 33 6f(x) = +

f 3 12

x x

x x

x x

j)

6 4

5 3

1f( ) = - 3

4

3f = 12

2

x x x

x x x

k)

8

3'( ) 5 h x x

l) dx

dy = -2x-2 + 12x-3 – 12x-4

m)

1

6 y x

5

61

6

dyx

dx

n)

1 3

1 2 2 = = y x x x x x

1 1.5dy

xdx

o) y

5 2

2 2 2

2 6 3

2 2 2

x x

x x x

23 3

32

xx

dx

dy = 3x2 + 3x -3

p) y = x

xx 32

= x

x

x

x 32

= x 2

1

2

3

3x

dx

dy =

xx

2

3

2

3




2. dx

dy = 4x + 3. When x=-2,

dy

dx= -5 .

Using y = mx + c at (-2, 0) where m1 = -5, equation of the tangent

is y = -5x - 10

For equation of the normal m2 = 1/5 at (-2, 0).

Using y=mx + c, equation of normal is y = 1/5 x + 2/5

3. dx

dy =

2

5

x

which is always negative for all values of x except zero

4. dx

dy = 3x2 + 4x – 3. When x=-1,

dy

dx= -4.

Using m1 = -4 at (-1,6), equation of the tangent is y = -4x + 2

Tangent cuts y axis at x=0, coordinate of A is (0, 2)

Tangent cuts x-axis at y=0, coordinate of B is (½,0 )

5. f(x) = 2x 2 + 9x – 5 . f '(x) = 4x + 9 .

Gradient of tangent is 4x + 9 and the gradient of the normal is

94

1

x

.

This is parallel to a line with gradient m = 1

3

The two gradients are equal.

1 1

4 9 3x

Solving, 4x + 9 = -3 gives x= -3 .

Point on the curve is (-3, -14)

6. Substituting (-2,0) and (3,20) in the equation

y = ax2 + bx + c , we obtain 0 = 4a – 2b + c and

20 = 9a + 3b + c

dx

dy= 2ax + b.

At A(-2, 0), m1 = -4a + b = -1 [A](normal).

Tangent at B(3, 20) = 6a + b = 9 [B]

Solving [A] and [B] gives a=1, b= 3.

Substituting in one of the above equations, c=2

a = 1, b = 3, c = 2




Independent Study

1. Work through Example 4 using

a) Substitution/Elimination methods

b) Matrix Algebra methods

to confirm the given values of a, b and c.

2. 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:

2 ,y ax bx c and point (2,7)P has a tangent with gradient

2m . Point Q is on the graph at 3

2x , and the tangent rises at

45o

A tangent with a slope of 45o has a gradient 1m ,

a. from tan 45 1o , or

b. from 45o right angle triangle ratio, 1:1: 2

Given

4 2 7a b c ------------- (1)

4 2a b ------------- (2)

3 1a b ------------- (3)

1. a) Using Simultaneous equations substitution/elimination

methods to solve:

(2) – (3)

4 2

3 1

1

a b

a b

a

------------- (4)

Substitute (4) into (2)

4 1 2

2 4

2

b

b

b

------------- (5)

Check answer by substituting (4) and (5) into L.H.S of (3)

3 1 2 ?

3 2 1

. .answer R H S

Substitute (4) and (5) into (1)

4 1 2 2 7

4 4 7

7

c

c

c

Thus 1,a 2b and 7c which confirms the values given.

2. The general equation for the curve is 2y ax bx c

now, given the values of 1,a 2b and 7c the equation

for this curve is2 2 7y x x




Part F Rates of Change

Exploring Rates of Change

Distance & Speed

Rates of Change

Rate of Change and Gradients

Rate of Change and Derivatives

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 example of this. At a later point this will extend 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 complex spreadsheet at this stage is to prepare students

for the more complex 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 exposure 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 axis shows the

passage of time. The red line is the distance. It shows that the

motion starts 4 units behind 0, and its gradient is 2 (2 units across

and 4 units up). So d = 2t – 4. The green line shows the speed (or

velocity), and it is v = 2.

The table shows that the distance increases constantly over each 0.1

time interval, and the speed is always 2. Because there is no change

in speed, a = 0.

Task 1.4 1. 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 3 26 9s t t t .

Find the car’s rate of change in distance.




2. If the cost of supply of raw fish to a fish monger is

250 40 0.01C f f , calculate the rate of change of cost of

supplying 48 tonnes of fish.

Solutions Task 1.4

1. Let 3 2( ) 6 9f t S t t t

Rate of change ,ds

dt or '( )f S

Using 1n ndx nx

dx

Then,

3 2

6 9d

t t tdt

3 1 2 1 1 13 6 2 9t t t

23 12 9t t

rate of change in distance is given by 23 12 9t t

2. Let 250 40 0.01 ;C f f 48f tonnes.

Cost 250 40 0.01dC d

f fdf df

40 0.02 f

When 48f ; rate of change of cost 40 0.02 48

$39.04




Part G The Product, Quotient and Chain Rules

Exploring the Product Rule

The Product Rule

The reason we need a product rule is that some functions cannot be

simply expanded. An example would be y = x2 ex. However the best

way to explain and justify the rule is to use examples 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 example, where u = x + 2, and v = x + 2. The

product values (x + 2)2 are in the column called uv, and the gradient

values are calculated as the increase in uv divide by the common

difference in x (0.2).

Now u´ = 1 and v´ = 1, so the product rule calculates the gradient as

uv´ + vu´ = (x + 2) + (x + 2) = 2x + 4. 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 is used when differentiating the product of two

functions (symbolised by u and v).

Exploring the Quotient Rule

The quotient rule is more complex 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 x 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.

If y = uv, then 𝑑

𝑑𝑥(𝑢𝑣) = 𝑣

𝑑𝑢

𝑑𝑥+ 𝑢

𝑑𝑣

𝑑𝑥

or

= vu + uv




The Quotient Rule

The second tab Quotient rule shows the gradient of the quotient

function. Here is the second example 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 of two numbers v

u can be differentiated using the

product rule by writing the expression in the form 1uv . However,

differentiating a quotient of two functions requires the use of a

special rule.

If y = , then

𝑑

𝑑𝑥ቀ

𝑢

𝑣ቁ =

𝑢𝑑𝑣𝑑𝑥

− 𝑣𝑑𝑢𝑑𝑥

𝑣2

or

=




Exploring 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

explore this first. Here is the first function of a function: square of (x + 1)

that is mentioned in the Student Manual. The blue line shows the first

function u = x + 1, 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 (x + 1)2, and its gradient values are

twice the values of u.

The gradient of u = x + 1 is just 1 (du/dx), and the product of du/dx by

dv/du is the same as the gradient of v (see the two green columns).




The Chain Rule

The chain rule is sometimes also referred to as the function of a function

rule.

Task 1.5 Product, Quotient and Chain Rules

Differentiate each of the following using the appropriate rule:

1. 542 x 2. 23 x

3. 62 54 x 4. 42 3xx

5. 3 26 x 6. 635 43 xx

7. 232 xx 8. 2342 2 xx

9.

5

212

2xx 10. 43 524 xx

11. 423 xx 12. 2

13

515

xx

13. 26

4

x

x 14.

5

32 x

x

15. 42

3

x

x 16.

xx

x

22

17. 4

52

2

x

x 18. 4223 243 xxx

19. 324 233 xx 20.

32

22

x

x

21. 1

42

23

xx

xx 22.

4 526

1

x

23. 4 2416.36 xx 24.

42

33

5

2

xx

xx

25. Find the gradient to the tangent where x = 2 on the following functions:

a) 1

33

x

xxf b)

1

12

2

x

xxf c) xxxf 23 2

26. Find the point where the curve 323 xxg cuts the x-axis and find

the gradient of the tangent at this point.

If where , then

or

=




27. If the point (2,1) lies on the curve x

xy

28

2

, find the equations of

the tangent and the normal at this point.

28. Find the equations of the tangent and the normal to the curve

2

12

xxy at the point (-2 , -3).

If the tangent meets the x-axis at P and the normal meets the x-axis at Q,

find the distance of the line PQ (leave the answer in surd form).

Solutions Task 1.5

1. Use Chain Rule : u = 2x – 4, y = u 5

dx

dy =

.du

dy dx

du

dy

du= 5u4 ,

dx

du = 2

= 10(2x – 4)4

2. Chain Rule : u = 3x + 2,

1

2y u

dx

dy =

.du

dy dx

du

0.51 1

2 2

dyu

du u

, 3du

dx

= 232

3

x

3. Chain Rule : u= 4x2 - 5, y = u6 ,

dx

dy =

.du

dy dx

du

dy

du = 6u5 ,

dx

du = 8x

= 48x(4x2 - 5)5

4. Chain Rule : u = x2 + 3x , y= u 4 ,

dx

dy =

.du

dy dx

du

dx

du= 2x + 3,

.du

dy = 4u3

= 4(2x + 3)(x2 + 3x)3

5. Chain Rule : u= 6 – 2x ,

1

3y u ,

dx

dy =

.du

dy dx

du

dy

du =

1

3u 3

2

dx

du = -2

=3 2)26(3

2

x




6. Chain Rule: u = x5 – 3x3 + 4 , y = u 6

dx

dy =

dy

du dx

du

dy

du = 6u5 ,

dx

du= 5x4 – 9x2

=6(x5 – 3x3 + 4)5(5x4 – 9x2 )

7. Product Rule : D(uv) = udv + vdu

u = x2 v = 3x + 2

du = 2x dv = 3

d(uv) = udv + vdu

dx

dy = 9x2 + 4x


u = 2x2 + 4 v = 3x + 2

du = 4x dv = 3

D(uv) = 18x2 + 8x + 12


u = 2x + 1 v = x2/2 + 5

du = 2 dv=x

D(uv) = 3x2 + x + 10


u = 4x3 v= (2x + 5)4

du = 12x2 dv = 8(2x + 5)3

simplify above to obtain

D(uv) = 8(4x3)(2x + 5)3 + (2x + 5)4(12x2)


3u x 2 4v x

du = 3x2 dv = 1

2(2x + 4) 2

1

D(uv) = 1

2

0.5 0.53 2x 2 4 3 2 4x x x

32

3 2

23

3 2

3 2 42 2 4

3 2 4 2 2 4

12 2 4 2 2 4

6 2 4

2 2 4 2 2 4

13 24

2 2 4

xd uv x x

x

x x x x

x x

x xx

x x

x x

x





u = (5x+1)3 0.5

5v x

du = 15(5x + 1)2 dv = 1

2 (x – 5) -1.5

D(uv) =

2 3

0.5 1.5

15(5 1) (5 1)0.5

( 5) ( 5)

x x

x x

D(uv) = (5𝑥+1)2(25𝑥−151)

2(𝑥−5)1.5

13. Quotient Rule : D(v

u) =

2v

udvvdu

u = x + 4 v = 6x – 2,

du = 1 dv = 6

D(v

u) =

2)26(

26

x


u) =

2v

udvvdu

u = 3x v= x2 + 5

du = 3 dv = 2x

D(v

u) =

2

2

2

-3x + 15

(x + 5)


u) =

2v

udvvdu

u = x3 v = x2 – 4

du = 3x2 dv = 2x

D(v

u) =

2 2

22

12

4

x x

x


u) =

2v

udvvdu

u = x v= 2x2 – x

du = 1 dv = 4x – 1

D(v

u) =

2

2 2

-2

(2 )

x

x x


u) =

2v

udvvdu

u = x2 – 5, v= x2 – 4

du = 2x, dv = 2x

D(v

u) =

2 2

2

( 4)

x

x


3 2 2 4

2 2 3

3 4 ( 2)

3 6 8 ( 2 )

u x x v x

du x x dv x x

3 4

3 2 2 2 2( ) 3 4 8 2 3 6 2d uv x x x x x x x

3

2 3 2 2 2( ) 2 8 3 4 2 3 6d uv x x x x x x x





u = 3x -4 v = (3x2 – 2) -3

du = -12x -5 dv = -18x ( 3x 2 – 2 )-4

D(uv) = 3 2 3

2 2

12 54

3 2

x (3x - 2 )

x x

OR 3 2 3 2 2

1 12 54

(3 2) 3 2x x x x


u) =

2v

udvvdu

2 2u x v= 2x + 3

du = x (x2 – 2 )-0.5 dv = 2

D(v

u) =

2 2

3 4

(2 3) 2

x

x x

21. Quotient Rule: 2

u vdu udvd

v v

u = x3 – 4 x 2 v= (x2 –x – 1)

du = 3x2 – 8x dv = 2x – 1

D(v

u) =

2 2 3 2

2 2

( 1) (3 8 ) ( 4 ) (2 1)

( 1)

x x x x x x x

x x

=

3 2

22

2 8

1

x x x x

x x


u) =

2v

udvvdu

u = 1, v= 5

4 6 2x

du = 0 dv = 4

1

)26(2

5x

D(

v

u) =

1

4

52

4

5 6-2x

2(6 - 2x)

= 25.2)2.6(

5.2

x

2.25

5

2 6 2x

23. Product Rule : d(uv) = udv + vdu

𝑢 = (6𝑥 + 3)4 𝑣 = (16 − 𝑥2)1

4

324(6 3)du x

32 4

116

2dv x x

3 2 1/ 4 4 2 -3/ 41

24(6 3) (16 ) (- ) (6 3)( ) (16 )= 2

x x x xv xd u




24. Quotient Rule :D(v

u) =

2v

udvvdu

3 4

3 22 5u x x v x x

du = 3(2x3 + x)2 (6x2 + 1) dv = 4(x2 – 5x)3 (2x – 5)

d(v

u)=

2 4 3 2 2 3 3 2 3

2 8

( 5 ) 3(2 ) (6 1) (2 ) 4( 5 ) (2 5)

( 5 )

x x x x x x x x x x

x x

= 3 2 2 3 3

2 4 2 5

3(2 ) (6 1) 4(2 ) (2 5)

( 5 ) ( 5 )

x x x x x x

x x x x

25. (a) f '(x) = 23

23

)1(

)3(3)1(3

x

xxx

f '(2) =

49

51 = -1.0408

(b) f '(x) = 2 2

2 2

2 ( 1) 2 ( 1)

( 1)

x x x x

x

f '(2) =

25

8

(c) f '(x) =

xx

x

232

26

2

f '(2) = 5

8

26. Curve cuts x-axis where y = 0 at

( 2

3

, 0)

Gradient of tangent

g '(x) = 9(3x – 2)2 at 𝑥 =

2

3 is 0.

27. f ‘(2) = 3/2, so at (2, 1) the equation of the tangent line is

𝑦 =3

2𝑥 − 2

For the normal line, m2 = -2/3

So the equation of the normal line is 𝑦 = −2

3𝑥 +

7

3

28. Equation of tangent : y =

2

11 x + 8

equation of normal :

𝑦 = −

2

11𝑥 −

37

11

Tangent cuts the x-axis at :

,0-16

11P

and the normal cuts the x-axis at :




,0-37

2Q

Distance between P and Q = 2

-16 -37

11 2

= 2

375

22

= 375

22




Part H Differentiation – Exponentials and Logarithms

Exploring Differentiation of Exponential Functions

Differentiation of Exponential Functions

The fundamental idea is that the gradient of y = ex at any value of x

is ex. Here is the second demonstration described in the Student

Manual.

The basic rule for the differentiation of an exponential function is

as follows:

If y = ex , then 𝑑

𝑑𝑥 𝑒𝑥 = 𝑒𝑥

If the power of the exponential is itself a function, ie xfey

Then using the chain rule:

dx

du

du

dy

dx

dy

Let xfey =

ue , where xfu

then dx

du

du

dy

dx

dy

= xfdx

deu

= xfexf '




Therefore, the rules of differentiation for exponential functions are:

Differentiation of Logarithmic Functions

Exploring Differentiation of Logarithmic Functions

From Student Manual (pages 31 – 32)

Therefore the rules for differentiation of logarithms are:

Students will have noticed the gap in the pattern of derivatives.

function y = x3 x2 x 1 x–1 x–2

derivative y´ = 3x2 2x 1 0 –x–2 –2x–3

Where is the function that has x–1 as its derivative? This is it!

You can check by asking the computer to draw y = 1

x when you will

see both branches.

1. If , then

2. If , then

1. If , then

2. If , then




Two Special Results

From Student Manual, pages 33 – 34.

1. The Derivative of y = ax

If xay

x

ee ay loglog

axy lnln

a

yx

ln

ln

yady

dx 1

ln

1

aydx

dyln

= aa x ln

2. The Derivative of y = logax

If xy alog

Using the change of base law, a

xy

e

e

log

log

xadx

dy

e

1

log

1

Task 1.6

Differentiation of Exponentials and Logarithms

Differentiate each of the following:

1. xey 4 2.

xey

2

1

3. 13 2 xey 4.

xey 27

5. 2

32 xexy 6. 2

3

9 x

ey

x

7. 13log xy e 8. 221ln xy

9. 3log xxy e 10. 4ln 3 xey

11.

2

ln2

xy x 12.

4

3log

2x

xy e

If , then

If , then




13. Find the gradients of the tangent and the normal to the curve

xey 32 at the point x = 2.

14. Find the equations of the tangent and the normal to the curve

3log xxf e at the point where x = 1.

15. Find the equations of the tangents to the curve xey 2 where

x = 2 and x = 2. Hence, find the point T where the two tangents

intersect each other.

Solutions Task 1.6

1. Chain Rule : u= 4x , y = eu ;

du

dy =

ue , dx

du = 4

dx

dy =

du

dy

dx

du

= 4e4x

2. Quotient Rule : u = 1, v = e2x ,

dv = 2e 2x , du = 0

D2v

udvvdu

v

u

dx

dy = -2e-2x

3. Chain Rule : u = 3x2 – 1, y = eu ,

dx

dy =

du

dy

dx

du

du

dy =

ue , dx

du= 6x

23 16 xdy

xedx




4. Chain Rule : u= 7 – 2x, y = eu ,

dx

dy =

du

dy

dx

du

du

dy = eu ,

dx

du= -2

dx

dy = -2e7 – 2x

5. Product Rule : u = 2x – 3, v = ex2 ,

du = 2, dv = 2xe2x

dx

dy = 2ex

2

(2x2 – 3x + 1)

6. Quotient Rule : u = e3x , v = 9 + x2

D2v

udvvdu

v

u

du = 3e3x , dv = 2x

dx

dy =

3 2

2 2

( 27 3 2 )

(9 )

xe x x

x

7. Chain Rule : u = 3x + 1, y = ln u ;

dx

dy =

du

dy

dx

du

du

dy = 1/u ,

dx

du = 3

dx

dy =

13

3

x

8. Chain Rule : u = 1 – 2x2 , y = ln u ;

dx

dy =

du

dy

dx

du

du

dy = 1/u ,

dx

du= -4x

dx

dy =

221

4

x

x




9. Chain Rule : u = x – x3 , y = ln( – u) ,

dx

dy =

du

dy

dx

du

du

dy =

1

u ,

dx

du= 1 – 3x2

dx

dy =

2

3

1 3

-

x

x x

10. Chain Rule : u = e3x + 4, y = ln u ;

dx

dy =

du

dy

dx

du

du

dy = 1/ u,

dx

du =

33 xe

dx

dy =

3

3

3

4

x

x

e

e

11. Product Rule : u =

2

2

x , v = ln x ;

( )u

d udv vduv

du = x, dv = 1/x

dx

dy = ln

2

xx x

12. y = ln (x – 3) – ln (x2 – 4 )

dx

dy =

3

1

x –

4

22 x

x

13. dx

dy = 6e3x .

The gradient of tangent at x=2 is 6e6

Hence the gradient of normal : – 66

1

e = - 0.0004

14. f '(x) = x

3

Gradient of tangent = -3.

Equation of tangent at (-1, 0)

using y = mx + c is y = -3x – 3

Gradient of normal = 1

3.

Equation of normal at (-1, 0) is

1 1

3 3y x




15. Equation for the gradient of tangent is

dx

dy = e-x

Equation of tangent using y = mx + c at (-2, 2 – e2 ) with m = e2

=7.389 is y = xe2 + e2 + 2

Equation of tangent using y = mx + c at (2, 2 – e-2) with m= e-2 is

y = 2

1

ex –

2

3

e + 2

point of intersection : solve xe2 + e2 + 2

= 2

1

ex –

2

3

e + 2

4

4

3

1

ex

e

,

4

4 2 2

3 1 32

1

ey

e e e

Co-ordinates are ( -1.075, 1.449)




Part I Applications of the Derivative

Significance of the First Derivative Stationary Points Maximum Turning Point Exploring Horizontal Point of Inflection

Refer to the Student Manual (pp 35 – 38).

Significance of the First Derivative f(x)

xf ' or dx

dy measures the rate of change of f(x) or y in relation to x.

Therefore the gradient of the tangent is always positive

Therefore the gradient of the tangent is always negative

Therefore the tangent is parallel to the x-axis with a gradient of 0.

For Example: The following sketch of a function shows where it is

increasing, decreasing or stationary (constant).

> 0

function is increasing

< 0

function is decreasing

xf ' = 0

function is stationary

increasing constant increasing decreasing

0 2 4 x




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

A stationary point is defined as where dx

dy= 0.

There are two types of stationary points:

1. Turning points (relative maximum and relative minimum):

The boundary points at either end of the diagram are also relative

maximum/minimum values. You will look at this later on.

relative

maximum relative

maximum

relative

maximum

relative

minimum

relative

minimum

boundary/end

point

boundary/end

point




Horizontal point of inflection (there are other points of inflection

where the tangents are not horizontal)

Maximum turning point (curve is concave down)

1. xf ' = 0

2. xf ' > 0 before the point where xf ' = 0

3. xf ' < 0 after the point where xf ' = 0

An example of this kind of curve would be 2 4y x

Minimum turning point (curve is concave up)

1. xf ' = 0

2. xf ' < 0 before the point where xf ' = 0

3. xf ' > 0 after the point where xf ' = 0

An example of this kind of curve would be 2 8y x

Horizontal Point of Inflection (Monotonic Point)

1. xf ' = 0

2. xf ' or dx

dy has the same sign over an interval, then xfy

is said to be monotonic over that interval.

x

y

concave

down

x

y

concave

up




Exploring Distance, Speed & Acceleration

The same ideas of differentiation help us understand the equations

of motion.

The first screen shows the car increasing speed: a = 1 m/s2.

The speed is v = t and distance is d = 0.5t2.

The second screen shows the car decreasing speed:

a = –1 m/s2. The initial speed is 4 so v = 4 – t and distance is

d = 0.5t2 –4t.




Task 1.7 1. Determine whether the curve 53 23 xxy is increasing,

horizontal or decreasing at x = 1.

2. Find any turning points on the curve

3536152 23 xxxy and determine if they are

maximum, minimum or neither.

3. Find the stationary points on the curve

6204 234 xxxxf and determine whether they are

maximum or minimum turning points.

4. If the curve 232 xaxxf has a stationary point where x

= 2, find the value of a and hence determine the type of turning

point.

5. If 3 5f x x has a monotonic point of inflection, determine

the coordinates for the point of inflection and describe the shape

of the graph of the function.

Solutions Task 1.7

1. dx

dy = 3x2 – 6x .

f ' (-1) = 9.

Increasing at x = -1

2. dx

dy = 6x2 + 30x + 36

= 6(x + 3 )(x + 2) = 0

x = -3, x = -2

Turning points at (-3, -62) and (-2, -63)

In the neighborhood of x = -2, dx

dy is – 0 +

minimum at (-2, -63)


dy is + 0 –

maximum at (-3, -62)

3. f ' (x) = 4x3 – 12x2 – 40x

= 4x( x – 5)(x + 2) = 0

x = 0, 5, -2

Turning points at (0 , 6), ( 5, -369), (-2, -26)


dy is – 0 +

minimum




In the neighborhood of x = 0, dx

dy is + 0 –

maximum


dy is – 0 +

minimum

4. f ' (x) = 2ax – 3 = 0

a = 3

4 when x = 2

Turning point occurs at (2, -1 )


dy is – 0 +

minimum

5. At horizontal point of inflection,

first and second derivative both equal to zero

' 0f x

23 0x

0

( ) 5

x

f x

'' 0f x

6 0x

Coordinates are 0,5




Part J Differentiation – Circular Functions

Exploring Differentiation of Circular Functions

The Student Manual uses the length of day (sunrise to sunset) as

an example of a sine function (refer to Page 41). 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 21.

Exploring Derivatives of Sin x, Sin nx, Cos x, Cos nx

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 2x.

The function sin 2x has been squeezed so that two waves fit into

2π. As a result its gradient is twice as steep as the regular function,

so it has a gradient of 2 at x = 0.

The gradient function is f´ = 2cos2x.




Task 1.8 1. Find the derivative with respect to x of each of the following:

2. Find 'f n for each of the following:

a. 3cos 2sin 2f x x x b. 2cos sinf x x x

c. tan sin 4f x x x d.

sin

2

xf x

x

; 1x

a. sin 6x b. cos5x c. tan 7x

d. 2cos 4x e.

2sin4

x

f.

2cos x

Solutions Task 1.8

1.

a) ' sin 6 6cos 6f x x

b) ' cos 5 5sin 5f x x

2c) ' tan 7 7sec 7f x x

d) Let 2cos 4y x

2 4u x , so

cosy u and

.dy dy du

dx du dx

Chain Rule

So,

sin

2

sin . 2

2 sin

dyu

du

dux

dx

dyu x

dx

dyx u

dx

Substitute 2 4u x and

22 sin 4dy

x xdx

e) This requires a two step approach

Firstly,

2sin4

y x

sin4

u x

, so

2y u




Next need to solve

.du du dv

dx dv dx

; where

4v x

So,

1

cos

1 cos

cos

cos4

dv

dx

duv

dv

duv

dx

duv

dx

dux

dx

Chain Rule

Now,

2

cos4

2 .cos4

dyu

du

dux

dx

dyu x

dx

Chain Rule

Substitute sin4

u x

and

2sin cos4 4

dyx x

dx

f) Let 2cosy x

cosu x , so 2y u and

.dy dy du

dx du dx Chain Rule

2 sin

2 sin 2 sin

dy duu x

du dx

dyu x u x

dx

Substitute cosu x and

2cos sindy

x xdx




2.

a. 3cos 2sin 2f x x x

' ' '

' 3sin 4cos 2

f x a b

f x f a f b

f x x x

b. 2cos sinf x x x

' ' 'f x f a f b

Let cosu x so 2y u

sin 2

2 . sin 2 sin

du dyx u

dx du

dyu x u x

dx

Substitute cosu x

So ' 2cos sinf a x x

' cosf b x

' cos 2cos sinf x x x x

c. tan sin 4f x x x

2' sec 4cos 4f x x x

d. sin

2

xf x

x

2'( )

du dvv u

d u dx dxf xdx v v

Quotient Rule

Let sinu x and 2v x

cos

1

dux

dx

dv

dx

2'( )

du dvv u

d u dx dxf xdx v v

2

2 cos sin

2

x x x

x




Part K Higher Derivatives

Significance of the Second Derivative

Task 1.9

By applying the same method of differentiation to the first derivative of

a given function, you obtain the second derivative of that function.

1. Find the second derivative and determine whether dx

dy is

increasing, decreasing or neither, for each of the following

functions at the point where x = 1.

a) xxxy 22 23 b) 2243 23 xxxy

c) 31

xx

y d) 34 8xxy

e) 633

1 23 xxxy f)

1

2

x

xy

g) 5)1( xy h) 1 xy

i) xy elog j) 2xey

k) 2log 2 xy e l) xe

y32

1

2. For what values of x is the curve 243 23 xxxy concave

up?

3. Show that a point of inflection exists where x = 2 on the curve

7024 24 xxy .

For any given function y = f(x)

The first derivative of the function y = f(x) is:

, y , f(x)

The second derivative of the function y = f(x) is obtained

by finding the derivative of the first derivative function.

This is signified by:

, y , f (x)

The resultant expression is the equation for the

rate of change of the gradient.

or

The resultant expression indicates the rate of

change of f(x) for any given value of x.




Solutions Task 1.9

1. Graphs of f(x) are shown

(a) xd

yd2

2

= 12x – 2

f ''(1) ( = 10) is > 0 ( = 10), hence f ' (x) is

increasing at x=1

b) xd

yd2

2

= 6x + 6

f ''(1) ( = 12) is > 0, hence f ' (x) is increasing at x=1




c) xd

yd2

2

= 3

2

x – 6x

f ''(1) ( = -4) is < 0, hence f ' (x) is decreasing at x=1

d) xd

yd2

2

= 12x 2 – 48x

f ''(1 ) ( = -36) is < 0, hence f ' (x) is decreasing at x=1




e) xd

yd2

2

= 2x – 2

f ''(1) ( = 0) is = 0,

hence f ' (x) is a point of inflexion at x=1

f)

xd

yd2

2

=

2 2

4

2( 1)[( 1) ( 2 )]

( 1)

x x x x

x

f ''(1) ( = 0.25) ) is > 0,

hence f ' (x) is increasing at x=1




g) xd

yd2

2

= 20 ( x – 1) 3

f ''(1) is = 0,

hence f ' (x) is a point of inflexion at x=1

h)

xd

yd2

2

= 3)1(4

1

x

Since ( )<0f x at 1x , the function is decreasing.




i) xd

yd2

2

= 2

1

x

f '' (1) (= -1) is < 0, hence f ' (x) is decreasing at x=1

j)

2 2

22

2= 4 2x xd y

x e ed x

f '' (1) ( = 16.3) is > 0, hence f ' (x) is increasing at x=1

k) Since 1 is not in the domain of f, nothing happens at 1.




l) xd

yd2

2

= 3

9

2 xe

f ''(1) (= 0.224) is > 0,

hence f ' (x) is increasing at x=1

2. For a graph to be concave up, xd

yd2

2

> 0.

xd

yd2

2

= 6x – 6

for x > 1, the curve is concave up.

3. 2( ) 12 48 f x x and (2) 0f , hence point of inflexion

exists where x = 2.




Part L Curve Sketching

Exploring Curve Sketching

Fundament-als of Curve Sketching

Task 1.10

Common Conventions (from Student Manual p. 46) 1. All axes must be labelled, where possible.

2. Scales should be appropriately selected for each axis.

3. The size of the graph should be sufficient to reasonably convey,

in a clear and concise manner, all relevant information.

4. 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 axes, 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.

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 maximum and minimum (if given a set domain)

1. 193 23 xxxy 2. 72492 23 xxxy

3 x 6

3. 3125 xxy 4.

23 xxy

2 x 4

5. 12 34 xxy 6. xxy 43

2 x 5

7. 36 24 xxy 8. 1

2

x

xy ; 5 x 1

9.3

3

x

xy ; 5 x 5 10.

1

12

x

xxy

11. 6sin(30 ), 6 6y x x

12. 24ln( )y x

Challenge Question

13. cos(30 ) 2y x , 6 6x




Solutions Task 1.10

1. Graph of y = x3 – 3x2 – 9x + 1

point of inflection at (1, -10)

maximum at (-1, 6)

minimum at (3, -26)

boundary values for the given domain are (-3, 26) and (6, 55)

absolute maximum is 55 and absolute minimum is -26

2. Graph of y = 2x3 – 9x2 – 24x – 7

Point of inflexion at ( 1½ , -56½ )

Minimum at (4, -119)

Maximum at (-1, 6)

maximum

minimum

-30

-25

-20

-15

-10

-5

0

5

10

15

20

-4 -2 0 2 4 6 8 Series1

minimum

maximum




3. Graph of y = 5 + 12x – x3

Point of inflection at ( 0,5)

Maximum at ( 2,21)

Minimum at ( -2,-11)

4. Graph of y = x3 – x2

point of inflection at (1

3 ,

27

2 )

maximum at (0,0)

minimum at (2

3,

27

4)

boundary values for given domain are (–2, –12) and (4, 48)


-20

-10

0

10

20

30

40

50

60

-4 -2 0 2 4 6

Series1

minimum

maximum




5. Graph of y = x4 + 2x3 – 1

maximum : not found

minimum at (-1.5, -2.6875)

points of inflection at (0, -1) and (-1, -2)

6. Graph of y = x3 ( 4 – x )

point of inflection at (0,0) and (2, 16)

maximum at (3 , 27)


boundary values for given domain are (-2, -48) and (5, -125)

maximum




7. Graph of y = x4 – 6x2 – 3

points of inflection at ( 1, -8) and (-1, -8)

maximum at ( 0, -3)

minimum at (√3 , -12) and ( - √3, -12)

8. Graph of y = 1

2

x

x

absolute maximum is -4

absolute minimum is -

relative maximum at (-2, -4)

boundary values for given domain are (-5, -6.25) and (-1, ∞)

minimum minimum

maximum




9. Graph of y = 3

3

x

x

no relative maxima, relative minima, absolute maxima and absolute

minima.

Boundary values for given domain are (-5, 0.25) and (5, 4)

10. 1

12

x

xxy

relative maxima occurs at (0, -1) and relative minima occurs at (2, 3)

No absolute maxima or minima.




11. 6sin(30 ), 6 6y x x ,

local minimum at 3, 6 , local maximum at 3,6 , point of

inflection at 0,0 , Boundary points 6,0 , 6,0

12. 24ln( )y x

asymptote at 1x

-8

-6

-4

-2

0

2

4

6

8

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

0

1

2

3

4

5

6

7

8

9

maximum

minimum




Challenge Question

13. cos(30 ) 2y x , 6 6x

Boundary Points 6,1 , 6,1

Local maximum at 0,3

Points of inflection at 3,2 , 3,2

Assessment Event 2

Project 1: Project 1 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 4

Collect Mathematical Terminology Logbooks from students for

marking.

0

0.5

1

1.5

2

2.5

3

3.5

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

maximum


Unit 2: Integration


Unit 2: Integration



Part C The Primitive Function and Indefinite Integrals

Part D Definite Integrals and the Area under a Curve

Part E The Volume of Rotation


Overview In this unit, students will learn to apply appropriate methods of integral

calculus to solving various numerical and graphical problems.


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 expected to complete some

work in their own time. You will guide them through the unit.

Assessment Event 1

In-class Test should be administered at the end of Unit 2. Note that this

is an in-class test under test conditions.

Assessment Event 4

Assessment Reminder: You must collect students’ Mathematical

Terminology Logbooks at the end of each unit.


Unit 2: Integration






and include these words in their Mathematical Terminology Logbook.

Summary of terms

primitive function

lower limit

volume of rotation

intervals

definite integral

integrand

approximation

constant of integration

integration

trapezoidal rule

subintervals

fundamental theorem

variable of integration

upper limit

bounded

Simpson’s rule

indefinite integral

anti-derivative

numerical methods


Unit 2: Integration


Part C The Primitive Function and Indefinite Integrals

Integration

Anti-Derivatives

Task 2.1

The Primitive Function (Refer to Student Manual p. 58)

The primitive of a function is the name usually given to the anti-

derivative. 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 examples on the

previous page you may have noticed a pattern/rule emerging when

finding the primitive from the function of the derivative.

1. Find the primitive function of:

a) 3 b) 2x c) 9x2

d) 54 x e) 243 3 xx f) 24 310 xx

g) 463 x h)

x

1 i) 7

5

x

2. Find f (x) if:

a) f (x) = xx 29 b) f (x) = 346 35 xx

c) f (x) = 483 x d) f (x) = 43 xx

e) f (x) = 2

1

3x f) f (x) = 3 2x

An anti-derivative of a function f(x) is another function F(x)

such that:

F(x) = f(x)

or

Any two anti-derivatives of a function differ only by a constant.

Rules:

If then + C, where k is any constant

If then , n 1

If then , n 1


Unit 2: Integration


3. Express y in terms of x if:

a) 54 3 xdx

dy b)

24

21

xxdx

dy

c) 9

2

5x

x

dx

dy d) xx

dx

dy 3

e) 3

321

x

x

dx

dy f)

2

123 23

x

xxx

dx

dy

4. Given that f (x) = 116 x and f (3) = 19, find an expression for

f(x).

5. Given that f (t) = 24t and f (6) = 22, find f (1).

6. If y = (x 3)2 and x = 7 when y = 0, find x when y = 4.

7. Given that f (x) = 5x4, f (0) = 3 and f (–1) = 1, find f (2).

8. If the gradient of the tangent to a curve is given by 12 x and the

curve passes through the point (3, 0), find the equation of the

curve.

9. The rate of change of V with respect to t is given by

212 tdt

dV

If V = 5 when t = 05, find V when t = 5.

Solutions Task 2.1

1. Using Rules

: if f '(x) = k then f (x) = kx

: if f '(x) = xn then f (x) = 1

1

n

xn

+ C

: if f '(x) = nbax )( then f (x) =

)1(

)( 1

na

bax n

+ C

a) 3x + C

b) x2 + C

c) 3x3 + C

d) 2x2 + 5x + C


Unit 2: Integration


e) 4 23

+ 2 2 4

x x x C

f) 5 32 x x C

g) 51

(3 6) 15

x C

h)

1

2( ) , then ( ) 2 f x x f x x C

i) 12

7 7/12x + C

2. Using Rules

: if f ' (x) = k then f(x) = kx

: if f ' (x) = xn then f(x) = 1

1

n

xn

+ C

: if f ' (x) = n

ax b then f (x) = )1(

)( 1

na

bax n

+ C

a) f(x) = 3 21

3 2

x x C

b) f(x) = x6 – x 4 + 3x + C

c) f(x) = 15

)83( 5x + C

d) f ' (x) = x 2 – x – 12

f(x) = 3 21 1

12 3 2

x x x C

e) f(x) = 2x1.5 + C

f) f ' (x) = x 2/3

then f(x) = 5

3x5/3 + C


Unit 2: Integration


3. a) y = x4 + 5x + C

b) y = 33

1

x

+

x

2 + C

c) y = 15

3x +

10

10x + C

d) dx

dy

1

3 3.52 x x x

y = 9

2 5.4x + C

e) f ' (x) = 3

1

x –

3

32

x

x

y = 22

1

x

– 2x + C

f ) f ' (x) = 2

33

x

x –

2

22

x

x +

2

1

x

x

y = 2

3x2 – 2x –

22

1

x + C

4. f(x) = 3x2 + 11x + C ,

x = 3, y = 19 → c = -41

f(x) = 3x2 + 11x – 41

5. f (t) = 3

)4( 3t + C

t = 6, y = 22 f(t) = 3

)4( 3t + 19

1

3

f(1) = 1

103


Unit 2: Integration


6. y =

33

3

xC

substitute x=7 , y = 0 C = 3

64

y = 3

)3( 3x –

64

3

When y = 4, x = 7.24

7. f '(x) = x5 + 3

f (x) = 6

6x + 3x + C

x = -1, y= 1 gives C = 36

5

Thus f (x) = 6

6x + 3x + 3

6

5

and (2) 20.5f

8. f '(x) = 2x + 1.

Curve passes through ( -3, 0).

Substitute x= -3 and y= 0 in f(x) = x2 + x + C

C = -6

Hence equation of the curve is y = x2 + x – 6

9. dt

dV = (2t – 1 ) 2 .

V = 6

)12( 3t + C .

V = 5, t = 0.5 gives C = 5 .

Substitute t = 5 in V = 6

)12( 3t + 5

V = 126.5


Unit 2: Integration


The Indefinite Integral (from Student Manual p. 61)

Task 2.2 Find the indefinite integral of:

1. dxxx 123 2 2. dxx

243

3. dxxxx 34912 23 4. dxxx 435

5.

dxxxxx 24

1

3 3 6. dttt 2162

7. dxxx 21 2 8.

dyyy

y

47

3

4

3

9. dxx3

95 10. dxx 52

11.

dxx4

132 12. dxx5

26

Notation

where F(x) is the primitive function of f(x)

General Rules

, where k is a constant

, where k is a constant


Unit 2: Integration


Solutions Task 2.2

Using rules of integration:

)(xf dx = F(x) + C

kx dx = kx + C

x n dx = 1

1

n

xn

+ C

nbax )( dx = )1(

)( 1

na

bax n

+ C , obtain

1. x3 - x2 + x + C

2. 9

)43( 3x + C

3. 3x4 + 3x3 + 2x2 – 3x + C

4. 6

6x +

4

4x + 4x + C

5. Note x =

1

2 x

y =22

1

x

+

3

22

3

x –

5

412

5

x +

31

3x + C

6. Expand the bracket to obtain

(t3 – 2t2 – 16t + 32 )dt

y = 4

4t –

2

3t3 – 8t2 + 32t + C

7. Expand the bracket to obtain

( 2 +2 x2 – x – x3 ) dx

y =

43 22 1

2 4 3 2

xx x x C


Unit 2: Integration


8. 3 y =

1

3y

1

2= y y →

3 1 1

4 3 2(7 4 )y y y dy

41 34

3 28

4

yx y +

3

28

3

y + C

9. y = 20

)95( 4x + C

10. dxx 52( =

1

2(2 5) x dx

y =

32 5

3

xC

11. y = 3)13(9

2

x + C

12. 5)26( x dx

=

5

2 (6 5) x dx

then y =

7

2(6 2)

76

2

xC

=

76 2

21

x


Unit 2: Integration


Part D Definite Integrals and the Area under a Curve

Exploring Definite Integrals and Area under 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 anti-

derivatives. 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 =

x + 1. 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).


Unit 2: 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 64.] To find the value of a definite integral:

1. Calculate the primitive function and simplify where possible.

2. Substitute the upper and lower limits into the primitive

function.

3. Find the difference between the results.

Exploring 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. 64 – 65).

General Rule If the primitive function of f(x) is F(x), then:

= F(b) F(a).

where b is called the upper limit and a is called the

lower limit.


Unit 2: Integration


Task 2.3 1. Find

2

1

2 1 dxx

2. Find 2

0

312 dxx

3. Find 4

034 dxx

4. Find 3

1

2 1 dxxx

5. Find 3

1 3

2dy

y

y

6. Find 2

1

23 63 dxxxx

7. Evaluate

2

0 46

2dx

x

8. Evaluate 1

1

2 1 dxxx

9. Evaluate 3

0

2 dxcbxax

10. Find 4

1

5

3

dxx

11. Find

2

1

4

23

21dxx

xx

12. Find

8

1

23

2

45 dxxx

13. Evaluate

21

41 32

11dx

xx

14. Find a

adxxa

2

15. Evaluate c if 2

02 4

c

x dx

16. Find the value of a if 2

228 dxxa

Solutions Task 2.3

1. 1. 2

1

2 )1( dxx

=

2

1

3

3

x

x

3322

3

11

3

= 6

2.

22

3

00

4(2 1)

8(2 1)x

xdx

4 4(2.2 1) ( 1)

8 8

10


Unit 2: Integration


3. ∫ √4𝑥 + 3𝑑𝑥4

0= [

1

6(4𝑥 + 3)

3

2]0

4

=1

6ቀ19

3

2 − 33

2ቁ ≅ 12.94

4.

3

26

)2

1

3

11()3

2

99(

x2

x

3

x3

1

23

5.

3

1

2132

2

y2

1

ydy)y2y(

9

2

9

1

3

1

6.

2

1

234

x62

x3

3

x

4

x

16 8 1 1 36 12 6

4 3 4 3 2

312

4

7.

12

2

02 (6 4)x dx

333.13

4

)24(3

2

)4()16(3

2

)4x6(3

2

2

1

2

1

2

0

2

1

8.

14 31

3 2

11

( ) 0.6674 3

x xx x dx

9.

33 2

0

99 3

3 2 2

ax bx bcx a c

10.

42

5

1

2

5

x

= 1.853


Unit 2: Integration


11.

22

3 2 4

2 31

1

1 2 1 1( 2 )

2 3 3x x x dx

x x x

12.

35

13

1

5 413.05

5 1

3

x x

13. 1

22 3

21

4

1

21

4

1 1

2x x dx

x x

4

)84()22(

14.

a2

a

2

3

2

1

2

1

a2

a2

1

2

3

xxadx)xa(

33 3 3 22 2 2

3

2

3

2

2 22 (2 )

3 3

4 2 22 1

3 3

(5 4 2)

3

aa a a

a

a

15. 42

xx2

c2

0

2

∴ 2𝑐 − 2𝑐2 = 4 ⟹ 𝑐2 − 𝑐 + 2 = 0

No Solution! (Draw a graph for students to show why

this is so…)

16.

232

2)162

4()16

2

4(

282

2

2

2

a

a

xx

a

a = 16

1


Unit 2: Integration


Exploring Integrations Involving Key Functions

Integrations Involving Exponential, Reciprocal (to Logarithmic) and Circular Functions

We have found that the anti-derivative ‘primitive function’ actually

describes the area between the function and the x-axis, 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 complex function and its interpretation.

The function is y = x – 1. Starting at 0, the area is under the axis so

the area function goes negative, but as the function approaches 0 the

areas being added lessen, so that at x = 1 the area function has a

minimum.

Then the areas increase to the point, at x = 2, when the total area

under and over the axis 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 Exponentials and Logarithms

Exponentials

Logarithms

General Rule

* special result:


Unit 2: Integration


Circular Functions

Task 2.4 Integration of Exponentials, Logarithms and Circular Functions

Evaluate the following indefinite integrals:

1. dxex x43 2.

dxex x322

3. dxe x 4.

dx

ee

eexx

xx

5. dxx4 6.

dxe

ex

x1

7. dx

x

x

2

55

4

8.

dx

xx

x

3

322

9. dx

x

x

72 3

2

10.

dx

xx

x

82

12

11.

dx

xx

x

62

142

12. dx

x

x3

2

1

4

Evaluate the following definite integrals:

13.

1

0

2

dxxe x 14.

3

2

2 dxe

x

15.

3

2

532 dxe x 16.

3

1

223 dxex x

General Rule

General Rule


Unit 2: Integration


17. 6

5

3 42 dxxe x 18.

1

0

52 dtet t

19.

4

1 95

5dx

x 20.

6

2 5

4

4dx

x

x

21.

1

0

22 3

dxex x 22.

3

2

2

1

1dx

xx

23.

1

1 2 1

4dx

x

x 24.

2ln

0 1dx

e

ex

x

Integrate the following:

25. 4

0sin xdx

26. 1

sin2

dx

27. cos 4 sin 4x x dx 28. 4

2

cos d

29. 2

0sin cos d

30. 1

cos 22 3

x dx

Solutions Task 2.4

1.

4

4

4

3

314

4

1

4

x

x

x

x e dx

x e dx

e C

2.

3

3

3

2 2

22 21 1

66 6

x

xx

x e dx

ex e dx C

3.

2

2

12

2

2

x

x

x

e dx

e dx

e C

4.

ln( )x x

x x

x x

e edx e e C

e e


Unit 2: Integration


5.

4

4ln 4

xx dx C

6.

1 1x x

x x x

x

e edx dx

e e e

e x C

7.

45

5

5ln( 2)

2

xdx x C

x

8.

2

2

2 3ln( 3 )

3

xdx x x C

x x

9.

2 2

3 3

3

1 6

2 7 6 (2 7)

1ln(2 7)

6

x xdx dx

x x

x C

10.

∫𝑥+1

𝑥2+2𝑥−8𝑑𝑥 =

1

2∫

2(𝑥+1)

𝑥2+2𝑥−8𝑑𝑥

=1

2ln(𝑥2 + 2𝑥 − 8) + 𝐶

11.

2

2

4 1ln(2 6)

2 6

xdx x x C

x x

12.

2 2

3 3

3

4 4 3

1 3 1

4ln(1 )

3

x xdx dx

x x

x C

13.

21

0

xxe dx

2 1

0

1 0

1

2

1

2

1 1( 1)

2

0.316

xe

e e

e


Unit 2: Integration


14.

3

2

22

x

e

3

22 2

3.5268

e e

15.

3

2

5x3

3

e2

761820

e3

2e

3

2 1114

16.

3

23

1

172.022

xex

17.

6523x x4xe

13.5115

)2025e()2436e( 89

18.

1

0

52 )( dtet t

=1

0

522

22

tet

= 473.61

19.

4

1ln(5 9)

ln 29 ln14

0.73

x

20.

65 5 5

2

1 1 1ln( 4) ln(6 4) ln(2 4)

5 5 5

1ln 7780 ln 36 1.075

5

x

21.

3 12

0

1 2

1

3

1( )

3

0.0775

xe

e e


Unit 2: Integration


22.

2

3

1

ln( 1)3

xx

= )0ln

3

1()1ln

3

2(

3

= Undefined

23.

12

-12ln( 1)

2 ln 2 - 2 ln 2

0

x

24.

4055.0

2ln3ln

)1ln()1ln(

)1ln(

02ln

2ln

0

ee

e x

25.

4 400

sin . cos 0.293x dx x

sin cos( )n

n kx a dx kx a Ck

26.

1sin 0.00873

2

0.00873

dx dx

x

27.

cos 4 sin 4 cos 4 sin 4

1 1sin 4 cos 4

4 4

1sin 4 cos 4

4

x x dx x dx x dx

x x

x x

28.

4 4

22

cos sin

sin sin4 2

0.707 ( 1)

1.707

d


Unit 2: Integration


29.

2 2 2

0 0 0sin cos sin cos

cos cos 0 sin sin 02 2

0 1 1 0

2

d d d

30.

1 1cos 2 cos 2

2 3 2 3

1 1sin 2

2 2 3

1sin 2

4 3

x dx x dx

x C

x C


Unit 2: Integration


Area Under Exponential Function

Area Under the Exponential Function y = a ekx

You may wish to experiment with the left interval for the

spreadsheet Area functions – a e^kx.

Shown below is the screen for the area under the green curve

y = e2x. If the area starts at 0 for x = –2, then the constant is 0.

Area Under Reciprocal Functions

Area Under the Reciprocal Function y = kx

The area under y = 2

x is shown below.

Note that it is not possible to find the area from 0, as the curve

never reaches x = 0. It is an asymptote. So a suitably small value is

chosen. The constant has been adjusted so that the log curve passes

through (1, 0), and the ‘arbitrary constant’ is therefore 0.


Unit 2: Integration


Area Under Circular Functions

Here is the screen for the area under the function: y = sin –0.5x.

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

1. Find dxxx 32 2

We can write the function as dxxx 2

1

2 )3(2 . 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 x to a new variable. If we assume u to be

what’s inside the root sign, 32 xu , the differential of u is

x

dudxdxxdu

22 . Then our integration will be:

CxCu

duux

duxudxux 2

3

22

3

2

1

2

1

2

1

)3(3

2

2

322)(2

In general, this method works when an integral is of the form

)()( xufxu , as in the previous example 3)3( 22 xfx

above.

The Substitution Rule

duxufdxxufxu )()()(

where f is a continuous function and )(xu is a differentiable

function.


Unit 2: Integration


Task 2.5

Integration by Substitution

Determine the following integrals:

1. dxxx 992 )3( , Let 32 xu

2. dxxx 1 , Let 1 xu

3. dx

x

x

5 3

2

162

2, Let 162 3 xu

4. dxex x

5342 , Let

53xu

5. dxxx ln

1, Let xu ln

6. dxx

x 5

ln4

, Let xu ln

7.

dx

e

ex

x

x

4

3

3

32

, Let 43

xeu

8. dxx

x

5)3(, Let xu 3

9. dxe

ex

x

2

, Let xeu 2

10. dxxx

xxx

3

33)3ln(

3

2

3, Let xxu 3ln 3


Unit 2: Integration


Solutions Task 2.5

1. 2Let 3u x

→ 2 du x dx

→ 2

dudx

x

→ duu

2

99

=

100

2 100

1

2 100

( 3)

200

uc

xC

2. Let 1u x

→ 1 x u

1

2

5 3

2 2

1

22

5 3

u u du

uuC

=

5

22( 1)

5

x +

3

22( 1)

3

xC

3. 3Let 2 16u x

→ 2 6 du x dx

→

1

52

4

5

1

6 3

1 5

3 4

dudx u du C

x

uC

4

3 55(2 16)

12

xC

4. 5Let 3u x

→ 4 15 du x dx

→ 4

2

15 15

ududx e du

x

→

532

15

xeC


Unit 2: Integration


5. Let lnu x

→ dx x du

Cx

cuduu

dxxx

)ln(ln

ln1

ln

1

6.

44 lnln

5 5

xxdx dx

x x

Let ln u x

1

1

du

dx x

du dxx

duudxx

x 44

5

1

5

)(ln

Cx

Cx

Cu

5

5

5

ln25

1

)(ln25

1

55

1

7. 3

Let 4xu e

→ 323 xdu x e dx

→ 3

1

2 23 x

dudx u du u c

x e

3

2 4xe C

→

3

3

3

232 4

4

xx

x

x edx e C

e

8.

1

0.5 2Let (3 ) 3 0.5u x x du x dx

1

21

2

22

dudx x du

x

652 2

6

uu du C

61

= 3 + x + 3

C

then xdu e dx


Unit 2: Integration


Exploring Area under a Curve

Area under a Curve

Definite integrals are the area under a curve between two

boundaries. (Refer to Student Manual pp 73 – 75).

Exploring the Fundamental Theorem of Calculus

Or, more formally:

9. Let 2 xu e

xe

dudx

ln

ln 2 x

duu C

u

e C

10. 3Let ln( 3 ) u x x

2

3

3 3Then

3

xdu dx

x x

→

3

2

3

3 3

x xdx du

x

udu Cu

2

2

23

3

1ln 3

2

ln 3

x x C

x x C

The Fundamental Theorem of Calculus

For function , continuous over the interval

where on


Unit 2: Integration


Exploring 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 2.6 Area under a Curve

For each of the following area questions, first sketch the graph and

shade the area that you are calculating.

1. Use integration to calculate the area bounded by the graph of the

straight line 23 xy , the x-axis and the ordinates x = 1 and x =

5.

2. Calculate the area bounded by the curve 12 xy , the x-axis and

the ordinates x = 2 and x = 1.

3. Calculate the area of the region bounded by the x-axis and the graph

of the function 26 xxxf .

4. Calculate the size of the area bounded by the ordinates x = 1 and

x = 4, the x-axis and the curve 672 xxy .

5. Find the area of the region bounded by the curve )2(2 xxy and

the x-axis.

6. Calculate the area of the region bounded by the curve 29y x ,

y-axis, the ordinates y = 1 and y = 15.

7. Calculate the area bounded by the y-axis, the ordinates y = 2 and

y = 5 and the curve 62 xy .


Unit 2: Integration


8. Find the area of the region (to 2 decimal places) bounded by the

curve xey and:

a) the ordinates x = 1 and x = 3 and the x-axis.

b) the ordinates y = 1 and y = 4 and the y-axis.

9. Find the area of the shaded regions in each of the diagrams below.

a)

b) bounded by the y-axis

c)

d)

y = x + 2

y = x2

2 y = - x2 + 2

x

y

y = x

y = x3 y = x

y = 8

x

y

y = -x

y =

2


Unit 2: Integration


e)

f)

g)

x

y

2 y = 2

x = 2

y =

x

y

y = x

y = x2 + 1

–1

1

2


Unit 2: Integration


Solutions Task 2.6

1.

Area

52

5

11

75( 10) (1.5 2)

2

4

3(3 2) 2

2

4

xx dx x

2.

13

12

22

(

1 21 4

3 3

6

1)3

xx dx x

Area = 6

y=x2+1


Unit 2: Integration


3.

32 3

32

22

9 8(18 9) ( 12 2 )

2

(6 ) 6

3

2 3

x xx x dx x

= 206

5

Area = 206

5

4.

1

3 21

2

44

1 7 64( 6) ( 56 24)

3 2

7( 7 6) 6

3 2

3

Area 13.5

x xx x dx x

f(x)=6+x–x2

y=x2 + 7x + 6


Unit 2: Integration


5.

2

3 42

2 3

00

2(2 )

3

164

3

1Area 1

3

4

x xx x dx

6.

1.53

1 2 21.52 2

1

1

17.54 22

(9 )(

.63

4.5 3

3.6

Are

9 )

a 3.

3

= 6

yy dy

y

y=xx(2-x)

y=29 x


Unit 2: Integration


7.

53

1 252

2

2

2(6 )(6 )

3

2 2(8)

3 3

2Area 4

3

yy dy

8.

3

1

133

1 eeedxe xx

Area = 17.37

9. a) 2

3 22

2

11

( 2) 23 2

14

2

x xx x dx x

y=-(x2-6)

y=ex


Unit 2: Integration


b) 1

3 21

2

00

( 2 ) 23 2

11

6

x xx x dx x

c) 8

41 2 383

1

1

( )42

3

120

4

y yy y dy

d)

023

222 0

2 2

2 2

2(2 )( 2

3

3

1 2

10 ( 5 ) 0 2 .31

3

x xxdx xdx

x

e)

2 2

11

22

12 2 lndx x x

x

1

4 ln 2 1 ln2

1.63

Area=1.63

(f)

2 3 2

2 2

1 2 1

2 3 22 3 3 2

2

1 2 1

( 1) ( 1) ( 2) 2

3 22 3 3 2

1 2 14 1 4

2 3 2

210

3

x dx x x x dx x x dx

x x x xx x x x

(g) 0 0 2

2 2

1 1 0( 1) [( 1) ]x dx xdx x x dx

1 1 81

3 2 3

14

2


Unit 2: Integration


Part E The Volume of Rotation

Refer to Student Manual (pp 85 – 87).

Task 2.7 Volumes of Rotation

1. Find the volume formed by rotating the curve 𝑦 = 𝑥3 between x =

0 and x = 2 around the x-axis.

2. Calculate the volume formed when the curve 𝑦 = 𝑥2 + 1 between x

= 0 and x = 2 is rotated around the y-axis.

3. Find the volume created when the area bounded by the curve 𝑦 =

4 − 𝑥2 and the x-axis is rotated around:

(a) the x-axis.

(b) the y-axis.

4. Calculate the volume of revolution resulting from rotating the curve

𝑦 = 𝑒−𝑥 between x = 0 and x = 1 around the x-axis.

5. Find the volume formed from rotating the curve 𝑦 = √𝑥2(𝑥2 + 1)4

between x = 0 and x = 3 around the x-axis.

Solutions for Task 2.7

1. 𝑉 = 𝜋 ∫ 𝑥6𝑑𝑥2

0= 𝜋 [

1

7𝑥7]

0

2=

128

7𝜋

2. 𝑉 = 𝜋 ∫ (√𝑦 − 1)2

𝑑𝑦5

1= 𝜋 [

1

2𝑦2 − 𝑦]

1

5= 8𝜋

3. (a) 𝑉 = 𝜋 ∫ (4 − 𝑥2)2𝑑𝑥2

−2= 𝜋 [16𝑥 −

8

3𝑥3 +

1

5𝑥5]

−2

2=

512

15𝜋

(b) 𝑉 = 𝜋 ∫ (√4 − 𝑦)2

𝑑𝑦4

0= 𝜋 [4𝑦 −

1

2𝑦2]

0

4= 8𝜋

4. 𝑉 = 𝜋 ∫ 𝑒−2𝑥𝑑𝑥1

0= 𝜋 [−

1

2𝑒−2𝑥]

0

1=

𝜋

2(1 − 𝑒−2)

5. 𝑉 = 𝜋 ∫ 𝑥√𝑥2 + 1𝑑𝑥3

0

= 𝜋 ∙1

2∫ √𝑢𝑑𝑢

10

1 (Substituting 𝑢 = 𝑥2 + 1)

=𝜋

2[

2

3𝑢

3

2]1

10

=𝜋

3(10√10 − 1)


Unit 2: Integration


Assessment Event 1

The In-class Test should be administered at the end of Unit 2. Note that

this is an in-class test conducted under test conditions.

Assessment Event 4

Collect the Mathematical Terminology Logbooks from students for

marking.


Unit 3: Advanced Applications





Part C Motion in a Straight Line (Rectilinear Motion)

Part D Approximation Methods in Integration

Part E Problems involving Maxima and Minima

Part F Exponential Growth and Decay

Part G Verhulst-Pearl Logistic Function – The Natural Law of Growth and

Decay


Overview In this unit, students will learn to apply differentiation and

integration techniques to solve a variety of problems.


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

maximising or minimising a quantity.

This unit includes a series of tasks that students will work through

to practise the course material. They will be expected to complete

some work in their own time. You will guide them through the

unit.

Assessment

Events 2, 3

and 4

Project 2 to be undertaken by students following completion of

Part C – Motion in a Straight Line – Rectilinear Motion – in this

Unit.

The Examination is to be conducted at the end of Unit 3.

Students are also required to submit their Mathematical

Terminology Logbooks at the end of Unit 3.

Remind students of assessment due dates and requirements.








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

maximising

disease

minimising

subintervals

approximation

intervals




Part C Motion in a Straight Line (Rectilinear Motion)

Motion in a Straight Line

Important

Integration in Motion Equations

Task 3.1

Refer to Student Manual (pp. 91).

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 ( 0S )

Note: ‘S’ does not measure the distance travelled but measures how far the

particle is from the origin, ‘S’ measures displacement.

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 )(tfa

adtV

1. The equation for displacement of a particle is given by

3104 2 ttS

i) Find its initial velocity and acceleration.

ii) Find when the particle is at rest

iii) Find the displacement and acceleration after 4

seconds.

2. The equation for displacement of a particle is given by

32 teS

i) Find the initial velocity

ii) Find the acceleration after 2 seconds

iii) Show that the acceleration is always double the

velocity

3. The acceleration of a particle is given by 12sin 2a t if

initially the particle is at rest at the origin find the exact

displacement after 4

seconds.

4. The velocity of a particle is given by2 4

tv

t

if the particle

is initially at the origin find the displacement after 4 seconds.

Note: Do NOT forget the constant

when finding these Primitive

Functions




Solutions Task 3.1

1. 3104 2 ttS

8

108

a

tV

(i.) initial velocity (t = 0) = sm /10

initial acceleration = 82/ sm

(ii.) at rest: 0V

4

11

0108

t

t

(iii.) 4t , 3)4(10)4(4 2 S

m27 2/8 sma

2. 32 teS

t

t

ea

eV

2

2

4

2

(i.) Initial velocity (t = 0) = sme /22 0

(ii.) When t = 2, 𝑎 = 4𝑒4

(iii.) tea 24

Va

eV t

2

2 2

3. ta 2sin12

CtV

2cos

2

112

Ct 2cos6

When 0t , 0V

CttS

tV

C

C

62sin3

62cos6

6

0cos60

0t , 0S

ttS

C

C

62sin3

0

000

When 4

t ,

46

2sin3

S

2

33




4. 42

t

tV

dtt

tS

42

dtt

t

4

2

2

12

CtS )4ln(2

1 2

0t , S = 0

∴ 0 =1

2ln(02 + 4) + 𝐶

𝐶 = −1

2ln 4 = − ln 2

When 4t :

𝑆(4) =1

2ln(42 + 4) − ln 2 = ln ቀ

√20

2ቁ =

1

2ln 5

Assessment Event 2

Project 2 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.




Part D Approximation Methods in Integration

Trapezoidal Rule Simpson’s

Rule

[From Student Manual pages 97 – 99.] 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 approximation

of the area under such difficult functions:

1. Trapezoidal Rule

2. Simpson’s Rule

3. Substitution Rule

Task 3.2 Numerical Methods of Integration

1. Calculate the area between the curve 13 xy , the x-axis and

the bounds x = 1 and x = 3 using the trapezoidal rule with 2

subintervals (3 function values).

2. A fisherman decided to measure the depth of a river every 10

metres. The results of his measurements where as follows:

Use the trapezoidal rule to estimate the area of the cross-section of

that part of the river.

Distance

across the

river

0m

10m

20m

30m

40m

50m

Depth in

metres 0m 26m 48m 32m 42m 0m

Therefore the General Trapezoidal Rule is:

f x( )dxa

b

ò @h

2f a( ) + 2 f a+h( ) +2 f a+ 2h( ) +...+ f b( )éë ùû

where: n

abh

= width of each interval

n = the number of subintervals.

For n (n even) equal subintervals (n + 1 function values) Simpson’s

Rule states:

b

an afafafafafaf

hdxxf ...2...4

3)( 42310

3

h [(sum of the end values) + 4(odds) + 2(evens)]

where h = n

ab




3. Use the trapezoidal rule, with three function values to estimate:

5

33 dxx

4. Use the trapezoidal rule with four subintervals to approximate the

value of 2

0

22 dxx .

5. Evaluate 5

1

225 dxx using Simpson’s rule with 5

function values.

6. Estimate the value of 1 4

32 3x dx

using Simpson’s

rule with 4 subintervals.

7. The curve 23 xxy between x = 1 and x = 3 is rotated about

the x-axis. Estimate the volume of the solid formed using

Simpson’s rule with four subintervals.

8. Estimate the area between the curve 2xxy and the line

xy2

5 using Simpson’s rule with 7 function values.

Solutions Task 3.2

1.

f (1) =2, f (2) = 9, f (3) = 28

3

3

1

1( 1) (1) 2 2(9) 28

2x dx

= 24 units squared




2.

Area =

2

1(10) [(0 + 0) + 2(2.6+4.8+3.2+4.2)]

2

=148m

3.

f (3)=33 = 27, f (4) = 34 = 81 , f (5) = 35 = 243

5

33 dxx

1

= (1) [ 27 + 243 + 2(81) ] 2

= 216 unit2

4.




f (2)= 2.45, f (1.5) = 2.06, f (1) = 1.73, f (0.5) = 1.50, f (0) = 1.41

2

0

22 dxx

2

0.5= [2.45 + 1.41 + 2(2.06 + 1.73 + 1.50 ]

2

=3.6128

= 3.62 units

5.

f(5)=0, f(4)=3, f(3)=4, f(2)= 21 =4.6,

f(1)= 24 =4.9

5

2

125 x dx

1

= [ (0 + 4.9) + 4(4.6 + 3 ) + 2(4 ) ] 3

= 14.4

6.




f (1) = 625, f (0) = 81,

f (-1) = 1, f (-2) = 1, f (-3) = 81

1

4

3(2 3)x dx

1

= [ 625 + 81 + 4(81 + 1) + 2(1) ]3

= 345.3

7.

f(1) = 4, f(1.5) = 81

16, f(2) = 4, f(2.5) =

25

16, f(3) = 0

𝜋 ∫ (3𝑥 − 𝑥2)2𝑑𝑥3

1

1 = (0.5)( ) [ (2 + 0) + 4( 2.25 + 1. 25) + 2( 2)]

3

= 𝜋(0.5) (1

3) [4 + 4 ×

81

16+ 2 × 4 + 4 ×

25

16+ 0]

= 77

12𝜋 ≅ 20.159

8.




Solve y = x – x2 and y = x2

5

and obtain x = 0 or x = -1.5.

0

5.1

2 )]()2

5[( dxxxx gives the required area.

0

5.1

2 ]5.1[ dxxx

f (0)= 0, f (-0.25) = -0.3125, f (-0.5) = -0.5, f (-0.75) = -

0.5625, f (-1) = -0.5, f (-1.25) = -0.3125, f (-1.5) = 0

0

5.1

2 ]5.1[ dxxx

= 3

25.0 [ 0+0+4(-0.3125 + -0.5625 + -0.3125) + 2( -0.5 +-0.5)]

1

(0.25) (-6.75) - 0.56253




Part E Problems involving Maxima and Minima

Examples Involving Minima and Maxima Problems Steps in Solving Applied Maximum and Minimum Problems

Refer to Student Manual (pp 103 – 104).

Task 3.3 Maxima and Minima Problems

1. Find the two numbers whose sum is 50 and whose product is

maximum.

2. A manufacturer knows that the total cost c of producing q units is

given by 800404.0 2 qqc . Find the value of q which

minimises the average cost.

3. Suppose the demand function for a certain product is

qp 4200 , where p is the price (in dollars) per unit (q) and the

average cost function (in dollars) is q

c200

2 . Find the value

of q which maximises the profit?

4. A hotel has 80 rooms that can be rented for $400 per month.

However, for each $20 per month increase, there will be two

vacancies with no possibilities of filling them. What rent per

room will maximise the monthly revenue?

5. Find the maximum area of a rectangular that can be inscribed in a

circle of radius 8.

6. A rectangle of cardboard is 16 cm by 8 cm. In order to form a box

without a cover, identical small squares are cut from each corner

and the remaining cardboard section is folded. Find the

dimensions of the box that maximises the volume and find that

volume.

7. A power station is on one side of a river, which is 2 km away

from the beach and a house is 4 km downstream on the other side

of the river. It costs $4 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 qp 4150 and the

average cost function is q

c100

8 , where q is number of units,

and both p and c are expressed in dollars per unit.

a. Determine the level of output at which profit is maximised.

b. Determine the price at which maximum profit occurs.

c. Determine the maximum profit.

d. If, as a regulatory device, the government imposes a tax of

$16 per unit on the monopolist, what is the new price for

profit maximisation?




Solutions Task 3.3

1. Let the numbers be x and y.

Then 50 50x y y x

Product (50 )xy x x

( ) 50 2p x x

25x , and 50 25 25y

The 2 numbers are 25 and 25.

2.

20.04 4 800c q

20.04 4 800c q qc

q q

Average cost 10.04 4 800q q

2( ) 0.04 800 0c q q

141.4q which minimises the average cost.

3. 200 4p q 200

2cq

2 200c c q q

Revenue 2200 4pq q q

Profit = revenue – cost

2200 4 (2 200)q q q

24 198 200q q

Max Profit when 8 198 0q

Maximum profit when 24.75q

4. p = price, q = apartment.

Hotel has 80 rooms, rent = $400 per month.

Revenue = pq = 80 400 $32000

$20 increase per month gives 2 vacancies.

Then, Revenue $(400 20) , rooms 80 2

If x represents the number of $20 increases

2

2

(400 20 )(80 2 )

32000 800 1600 40

32000 800 40

R x x

x x x

x x

( ) 800 80 0

10

R x x

x

Revenue is max when x=10; Revenue = $36000.

36000600

80 2 10

$600 per room will maximise profits.




5.

Triangle: Length = 2 28 x

Rectangle : Length = 2 22 8 x Width = 2x

Area of rectangle = l w

2 22 (2 8 )x x

2

2

8 256( ) 0

64

xA x

x

28 256x

32 5.66x

This is a square of size 11.32units

Maximum area = 11.32 x 11.32

= 128.1 2unit

Alternatively, use angle in a semicircle (consider diameter instead)

2 2 16 16x y y x

Area 16

A gives the same maximum area as above.

xy x x




6.

length = 16 – 2x, width = 8 – 2x , height = x

2 3(8 2 )(16 2 ) 128 48 4Vol x x x x x x

2( ) 128 96 12 0V x x x

x = 6.31 or 1.69(max)

height=1.69, width=4.62, length = 12.62

Max Volume = 98.533cm

7.

BX =

Total cost=C(x)

1

2 28( 8 20) 4x x x

1

2 2

2

2

1( ) 8( )( 8 20) (2 8) 4

2

4(2 8)4

8 20

3 24 44 0

C x x x x

x

x x

x x

x = 2.85 (or 5.15 unacceptable)

Hence power line on land is 2.85km and 2.31km under water. The

minimum cost is $29.88.

20848162)4( 2222 xxxxx




8. p = 150 – 4q

1008c

q

c cq

c = 8q + 100

revenue = pq

= (150 – 4q) (q)

= 150q – 42q

profit = revenue – cost

= (150q – 42q ) – (8q + 100)

= -42q + 142q – 100

(a) p' (q) = -8q + 142

q = 3

174

the profit is maximised

(b) price = 150 – 4q

= 150 – 71

= $79

Maximum profit occurs when price is $79 per item

(c) maximum profit= -42q + 142q – 100

= – 4

23

174

+ 142(3

174

) – 100

= $1160.25 is the maximum profit

(d) new price= $79 + $16

= $95 per unit




Part F Exponential Growth and Decay

Exponential

Growth and

Decay

Refer to Student Manual (pages 106 – 108).

Task 3.4 1) Show that if P=A+BeKt then

dPK P A

dt

2) If 0.03 400dP

Pdt

Write and equation for P in terms of t if t=0, P=500

Hence find P when t=15

3) Assume Newton’s Law of Cooling for the following

problem:

i.e. o

dTK T T

dt

An oven has been heated to 200oC.

An object with temperature of 15oC is placed in the oven

and after 20 minutes its temperature is 80oC.

i) What will be the temperature of the object

after 30 minutes?

ii) How long will it take the object to reach a

temperature of 180oC?

Solutions Task 3.4

1. ktBeAP

KeBdt

dP kt

ktKBe

But )( APBe kt

)( APkdt

dP

2. )400(03.0 Pdt

dP

t

t

eP

B

Be

BeP

03.0

0

03.0

100400

100

400500

400

15t , 1503.0100400 eP

= 463.76

Exponential Growth and Decay Definition:

Any quantity is said to have an exponential growth/decay model if

at each instant in time, its rate of increase/decrease is proportional

to the amount of the quantity present.




3. )( 0TTKdt

dT

ktBeTT 0

2000 T ; 0t oT 15 ; 20t

oT 80

ktBeT 200

0t , oT 15

kt

o

eT

B

Be

185200

185

20015 0

20t , oT 80

185

120ln

20

1

185

120ln20

185

120lnln

185

120

120185

18520080

20

20

20

20

k

k

e

e

e

e

k

k

k

k

0216.0 teT 0216.0185200

(i) 30t , 300216.185200 eT

23.103

(ii) teT 0216.0185200

0216.0185

20ln

185

20ln0216.0

185

20

20185

185200180

180

0216.0

0216.0

0216.0

t

t

e

e

e

T

t

t

t

103 minutes

1 hour 43 minutes




Part G Verhulst-Pearl Logistic Function – The Natural Law of Growth & Decay

Refer to the Student Manual (Pages 111 – 112).

Task 3.5 1. Suppose the number of members in a new RSL club was to be a

maximum of 2 000 persons. Last year, the initial number of

members was 200 persons and now there are 600. Assume the

enrolment follows a logistic function. How many members will

there be three years from now?

2. The health department in NSW is studying the spread of a disease

in a city of 80 000 people. At the beginning of the study, 1000

people were infected. One month later 8000 people are infected.

Find the number N of people that are infected 4 months after the

beginning of the study if N follows logistic growth function.

3. A factory performed a marketing study and found that there is a

potential of 200 000 customers for its products. When the study

started there were 60 000 customers. One year later there were

70 000. Assuming N follows logistic growth function, find the

number N of customers (in terms of t) t years after the beginning

of the study.

Solutions Task 3.5

1. M=2000

t=0, N = 200

now N=600, t=1

three years from now, t=4

N= ctbe

M1

200= )0(1

2000cbe

b=9

→600=ce 91

2000

e-c = 7/27

→ N = 4)27/7(91

2000

= 1921.86




2. M=80 000

t=0, N=1000

t=1, N=8000

t=4, N=?

1000=(0)

80000

1 cbe

b=79

8000=(1)

80000 9

1 79 79

c

ce

e

4

80000N = 78949.4

91+79

79

3. M=200 000

t=0, N=60 000

t=1, N=70 000

find N after t years

60 000=(0)

200 000

1 cbe

b=21

3

70 000=200 000

1 (7 / 3) ce

e-c = 39

49

N = 200 000

7 391

3 49

t

Assessment

Events 3 & 4

The Examination is to be conducted now.

Students are required to submit their Mathematical Terminology

Logbook now.


Appendix: Formulae


Appendix: Formulae

In the following table, A and b are constants and u, v, and y are functions of x.

Derivative Integral Other

0dA

dx 0dx A

1dx

dx dx x

dudx u

dx

d du

Au Adx dx

Au dx A u dx

d du dv

u vdx dx dx

u v dx u dx v dx

d dv du

uv u vdx dx dx

u dv uv v dx dy

udy u dxdx

1b bdx bx

dx

1

, 11

bb x

x dx bb

sin cosd

x xdx

cos sinx dx x

(cos ) sind

x xdx

sin cosx dx x

2(tan ) secd

x xdx

2sec tanx dx x

bx bxde be

dx

1bx bxe dx eb

1

lnd

xdx x

1

lndx xx

1

( )

'( )

nn n

n

f xx x

f x

2

du dvv u

d u dx dx

dx v v

bfhafhafafh

dxxfb

a ...222

2

n

abh

b

an afafafafafaf

hdxxf ...2...4

3)( 42310

b ah

n

duxufdxxufxu )()()(

1 ct

MN

be

level iii module 16: gac016 mathematics iii: calculus &...

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