Jacaranda Maths Quest Units 3&4 Mathematical Methods 12 for Queensland [1 ed.] 9780730368861, 9780730379959

Table of contents :
Half Title Page
Title Page
Copyright Page
Contents
About this resource
About eBookPLUS and studyON
Acknowledgements
TOPIC 1 The logarithmic function 2
1.1 Overview
1.2 Review of the index laws
1.3 Logarithmic laws and equations
1.4 Logarithmic scales
1.5 Indicial equations
1.6 Logarithmic graphs
1.7 Applications
1.8 Review: exam practice
Answers
REVISION UNIT 3 Further calculus
TOPIC 2 Calculus of exponential functions
2.1 Overview
2.2 Review of limits and differentiation
2.3 The exponential function
2.4 Differentiation of exponential functions
2.5 Applications of exponential functions
2.6 Review: exam practice
Answers
TOPIC 3 Calculus of logarithmic functions
3.1 Overview
3.2 The natural logarithm and the function y = loge(x)
3.3 The derivative of y = loge(x)
3.4 Applications of logarithmic functions
3.5 Review: exam practice
Answers
TOPIC 4 Calculus of trigonometric functions
4.1 Overview
4.2 Review of the unit circle, symmetry and exact values
4.3 Review of solving trigonometric equations with and without the use of technology
4.4 Review of graphs of trigonometric functions of the form y = A sin(B(x + C)) + D and y = A cos(B (x + C)) + D
4.5 Derivatives of the sine and cosine functions
4.6 Applications of trigonometric functions
4.7 Review: exam practice
Answers
TOPIC 5 Further differentiation and applications
5.1 Overview
5.2 The chain rule
5.3 The product rule
5.4 The quotient rule
5.5 Applications of differentiation
5.6 Review: exam practice
Answers
REVISION UNIT 3 Further calculus
TOPIC 6 Antidifferentiation
6.1 Overview
6.2 Antidifferentiation of rational functions
6.3 Antidifferentiation of exponential functions
6.4 Antidifferentiation of logarithmic functions
6.5 Antidifferentiation of sine and cosine functions
6.6 Further integration
6.7 Review: exam practice
Answers
TOPIC 7 Integration
7.1 Overview
7.2 Estimating the area under a curve
7.3 The fundamental theorem of calculus and definite integrals
7.4 Areas under curves
7.5 Areas between curves
7.6 Applications of integration
7.7 Review: exam practice
Answers
REVISION UNIT 3 Further calculus
PRACTICE ASSESSMENT 1
PRACTICE ASSESSMENT 2
TOPIC 8 The second derivative and applications of differentiation
8.1 Overview
8.2 Second derivatives
8.3 Concavity and points of inflection
8.4 Curve sketching
8.5 Applications of the second derivative
8.6 Review: exam practice
Answers
REVISION UNIT 4 Further functions and statistics
TOPIC 9 Cosine and sine rules
9.1 Overview
9.2 Review of trigonometric ratios and the unit circle
9.3 The sine rule
9.4 The cosine rule
9.5 Area of a triangle
9.6 Applications of the sine and cosine rules
9.7 Review: exam practice
Answers
REVISION UNIT 4 Further functions and statistics
TOPIC 10 Discrete random variables
10.1 Overview
10.2 Bernoulli distributions
10.3 Bernoulli random variables
10.4 Binomial distributions
10.5 The mean and variance of a binomial distribution
10.6 Applications
10.7 Review: exam practice
Answers
REVISION UNIT 4 Further functions and statistics
TOPIC 11 General continuous random variables
11.1 Overview
11.2 Continuous random variables and the probability density function
11.3 Cumulative distribution functions
11.4 Measures of centre and spread
11.5 Review: exam practice
Answers
TOPIC 12 The normal distribution
12.1 Overview
12.2 The normal distribution
12.3 Standardised normal variables
12.4 The inverse normal distribution
12.5 Applications of the normal distribution
12.6 Review: exam practice
Answers
REVISION UNIT 4 Further functions and statistics
TOPIC 13 Sampling and confidence intervals
13.1 Overview
13.2 Sample statistics
13.3 The distribution of p̂
13.4 Confidence intervals
13.5 Review: exam practice
Answers
REVISION UNIT 4 Further functions and statistics
PRACTICE ASSESSMENT 3
PRACTICE ASSESSMENT 4
Glossary
Index

Citation preview

JACARANDA MATHS QUEST

12 3&4

MATHEMATICAL METHODS

FOR QUEENSLAND

UNITS

JACARANDA MATHS QUEST

12 3&4

MATHEMATICAL METHODS

UNITS

FOR QUEENSLAND

KAHNI BURROWS | BEVERLY LANGSFORD WILLING | SUE MICHELL Raymond Rozen | Margaret Swale

CONTRIBUTING AUTHORS

First published 2020 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 11/14 pt TimesLTStd © John Wiley & Sons Australia, Ltd 2020 The moral rights of the authors have been asserted. ISBN: 978-0-7303-6886-1 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Trademarks Jacaranda, the JacPLUS logo, the learnON, assessON and studyON logos, Wiley and the Wiley logo, and any related trade dress are trademarks or registered trademarks of John Wiley & Sons Inc. and/or its affiliates in the United States, Australia and in other countries, and may not be used without written permission. All other trademarks are the property of their respective owners. Front cover image: © antishock/Shutterstock Back cover image: © MPFphotography/Shutterstock Illustrated by various artists, diacriTech and Wiley Composition Services Typeset in India by diacriTech Printed in Singapore by Markono Print Media Pte Ltd

10 9 8 7 6 5 4 3 2 1

CONTENTS About this resource ..................................................................................................................................................................................................

ix

About eBookPLUS and studyON

xii

.......................................................................................................................................................................

Acknowledgements ................................................................................................................................................................................................. xiii UNIT 3

FURTHER CALCULUS

1

TOPIC 1 THE LOGARITHMIC FUNCTION 2

1 The logarithmic function 2

1

1.1 Overview.............................................................................................................................................................................................. 1.2 Review of the index laws ................................................................................................................................................................ 1.3 Logarithmic laws and equations .................................................................................................................................................. 1.4 Logarithmic scales ........................................................................................................................................................................... 1.5 Indicial equations .............................................................................................................................................................................. 1.6 Logarithmic graphs .......................................................................................................................................................................... 1.7 Applications ........................................................................................................................................................................................ 1.8 Review: exam practice .................................................................................................................................................................... Answers ............................................................................................................................................................................................................

1 2 7 17 21 25 32 37 41

REVISION UNIT 3 TOPIC 1 Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 TOPIC 2 FURTHER DIFFERENTIATION AND APPLICATIONS 2

2 Calculus of exponential functions

48

2.1 Overview.............................................................................................................................................................................................. 2.2 Review of limits and differentiation ............................................................................................................................................. 2.3 The exponential function ................................................................................................................................................................ 2.4 Differentiation of exponential functions ..................................................................................................................................... 2.5 Applications of exponential functions ........................................................................................................................................ 2.6 Review: exam practice .................................................................................................................................................................... Answers ............................................................................................................................................................................................................

3 Calculus of logarithmic functions

48 49 54 61 64 74 78

83

3.1 Overview.............................................................................................................................................................................................. 83 3.2 The natural logarithm and the function y = loge (x)................................................................................................................. 84 3.3 The derivative of y = loge (x) .......................................................................................................................................................... 91 3.4 Applications of logarithmic functions ......................................................................................................................................... 94 3.5 Review: exam practice .................................................................................................................................................................... 97 Answers ............................................................................................................................................................................................................ 100

4 Calculus of trigonometric functions

107

4.1 4.2 4.3 4.4

Overview.............................................................................................................................................................................................. 107 Review of the unit circle, symmetry and exact values ........................................................................................................... 108 Review of solving trigonometric equations with and without the use of technology ................................................... 117 Review of graphs of trigonometric functions of the form y = A sin(B (x + C)) + D and y = A cos(B (x + C)) + D ................................................................................................................................................................. 122 4.5 Derivatives of the sine and cosine functions ............................................................................................................................ 130 4.6 Applications of trigonometric functions ..................................................................................................................................... 136 4.7 Review: exam practice .................................................................................................................................................................... 145 Answers ............................................................................................................................................................................................................ 148

CONTENTS v

5 Further differentiation and applications

156

5.1 Overview.............................................................................................................................................................................................. 156 5.2 The chain rule .................................................................................................................................................................................... 157 5.3 The product rule ................................................................................................................................................................................ 162 5.4 The quotient rule ............................................................................................................................................................................... 165 5.5 Applications of differentiation ....................................................................................................................................................... 170 5.6 Review: exam practice .................................................................................................................................................................... 188 Answers ............................................................................................................................................................................................................ 191 REVISION UNIT 3 TOPIC 2 Chapters 2, 3, 4 and 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 TOPIC 3 INTEGRALS

6 Antidifferentiation

197

6.1 Overview.............................................................................................................................................................................................. 197 6.2 Antidifferentiation of rational functions ...................................................................................................................................... 198 6.3 Antidifferentiation of exponential functions .............................................................................................................................. 202 6.4 Antidifferentiation of logarithmic functions ............................................................................................................................... 205 6.5 Antidifferentiation of sine and cosine functions ....................................................................................................................... 208 6.6 Further integration ............................................................................................................................................................................ 211 6.7 Review: exam practice .................................................................................................................................................................... 220 Answers ............................................................................................................................................................................................................ 224

7 Integration

228

7.1 Overview.............................................................................................................................................................................................. 228 7.2 Estimating the area under a curve ............................................................................................................................................... 229 7.3 The fundamental theorem of calculus and definite integrals ............................................................................................... 240 7.4 Areas under curves .......................................................................................................................................................................... 247 7.5 Areas between curves ..................................................................................................................................................................... 257 7.6 Applications of integration ............................................................................................................................................................. 265 7.7 Review: exam practice .................................................................................................................................................................... 275 Answers ............................................................................................................................................................................................................ 280 REVISION UNIT 3 TOPIC 3 Chapters 6 and 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 PRACTICE ASSESSMENT 1 Mathematical Methods: Problem solving and modelling task

................................................................................

286

PRACTICE ASSESSMENT 2 Mathematical Methods: Unit 3 examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 UNIT 4

FURTHER FUNCTIONS AND STATISTICS

302

TOPIC 1 FURTHER DIFFERENTIATION AND APPLICATIONS 3

8 The second derivative and applications of differentiation

302

8.1 Overview.............................................................................................................................................................................................. 302 8.2 Second derivatives ........................................................................................................................................................................... 303 8.3 Concavity and points of inflection ............................................................................................................................................... 306 8.4 Curve sketching ................................................................................................................................................................................ 311 8.5 Applications of the second derivative ........................................................................................................................................ 321 8.6 Review: exam practice .................................................................................................................................................................... 327 Answers ............................................................................................................................................................................................................ 332 REVISION UNIT 4 TOPIC 1 Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

vi CONTENTS

TOPIC 2 TRIGONOMETRIC FUNCTIONS 2

9 Cosine and sine rules

338

9.1 Overview.............................................................................................................................................................................................. 338 9.2 Review of trigonometric ratios and the unit circle .................................................................................................................. 339 9.3 The sine rule ....................................................................................................................................................................................... 347 9.4 The cosine rule .................................................................................................................................................................................. 355 9.5 Area of a triangle ............................................................................................................................................................................... 362 9.6 Applications of the sine and cosine rules .................................................................................................................................. 367 9.7 Review: exam practice .................................................................................................................................................................... 372 Answers ............................................................................................................................................................................................................ 375 REVISION UNIT 4 TOPIC 2 Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 TOPIC 3 DISCRETE RANDOM VARIABLES 2

10 Discrete random variables

378

10.1 Overview.............................................................................................................................................................................................. 378 10.2 Bernoulli distributions ...................................................................................................................................................................... 379 10.3 Bernoulli random variables ............................................................................................................................................................ 380 10.4 Binomial distributions ...................................................................................................................................................................... 385 10.5 The mean and variance of a binomial distribution .................................................................................................................. 393 10.6 Applications ........................................................................................................................................................................................ 398 10.7 Review: exam practice .................................................................................................................................................................... 401 Answers ............................................................................................................................................................................................................ 404 REVISION UNIT 4 TOPIC 3 Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406 TOPIC 4 CONTINUOUS RANDOM VARIABLES AND THE NORMAL DISTRIBUTION

11 General continuous random variables

407

11.1 Overview.............................................................................................................................................................................................. 407 11.2 Continuous random variables and the probability density function .................................................................................. 408 11.3 Cumulative distribution functions ................................................................................................................................................ 419 11.4 Measures of centre and spread .................................................................................................................................................... 429 11.5 Review: exam practice .................................................................................................................................................................... 441 Answers ............................................................................................................................................................................................................ 446

12 The normal distribution

449

12.1 Overview.............................................................................................................................................................................................. 449 12.2 The normal distribution ................................................................................................................................................................... 450 12.3 Standardised normal variables ..................................................................................................................................................... 457 12.4 The inverse normal distribution .................................................................................................................................................... 468 12.5 Applications of the normal distribution ...................................................................................................................................... 473 12.6 Review: exam practice .................................................................................................................................................................... 479 Answers ............................................................................................................................................................................................................ 482 REVISION UNIT 4 TOPIC 4 Chapters 11 and 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484

CONTENTS vii

TOPIC 5 INTERVAL ESTIMATES FOR PROPORTIONS

13 Sampling and confidence intervals

485

13.1 Overview.............................................................................................................................................................................................. 485 13.2 Sample statistics............................................................................................................................................................................... 486 13.3 The distribution of p̂ ......................................................................................................................................................................... 491 13.4 Confidence intervals ........................................................................................................................................................................ 499 13.5 Review: exam practice .................................................................................................................................................................... 504 Answers ............................................................................................................................................................................................................ 508 REVISION UNIT 4 TOPIC 5 Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 510 PRACTICE ASSESSMENT 3 Mathematical Methods: Unit 4 examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 511 PRACTICE ASSESSMENT 4 Mathematical Methods: Units 3 & 4 examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522

Glossary ....................................................................................................................................................................................................................... 532 Index

............................................................................................................................................................................................................................. 535

viii CONTENTS

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ACKNOWLEDGEMENTS xiii

1

The logarithmic function 2

1.1 Overview Let’s say that you were asked to multiply 167 893 by the square root of 283.983. Chances are that the first thing that you would do is reach for your calculator. But what if you did not have a calculator — could you still do it? And how long would it take you? Before small hand-held electronic calculators were developed in the early 1970s, calculations that could not be quickly done with pencil, paper and mental arithmetic were performed using a device called a slide rule. Invented in the early seventeenth century, a slide rule is essentially a ruler with a sliding central section. A normal ruler is marked with numbers that form a linear scale, with the marks for 1, 2 … 30 cm being equally spaced. In contrast, each of the three ruler sections of a slide rule is marked with numbers that form logarithmic scales. In this type of scale, there are equal distances between the marks for 1, 10, 100, 1000 and so on. Calculations were done by using a table of logarithms to identify the base 10 logarithms of the numbers you wanted to manipulate, sliding the central scale relative to one of the fixed outer scales until the appropriate numbers lined up with the cursor (a red line fitted in a sliding window), reading a number from a third scale and then using the logarithm table to determine your actual answer. Sounds like a lot more work than just pressing a few buttons on your calculator, doesn’t it? Yet, with practice, a slide rule (with a log table, a pencil and a piece of paper) can be used to perform calculations in nearly the same time.

LEARNING SEQUENCE 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Overview Review of the index laws Logarithmic laws and equations Logarithmic scales Indicial equations Logarithmic graphs Applications Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

CHAPTER 1 The logarithmic function 2 1

1.2 Review of the index laws As you may recall from your earlier studies, a number in index form has two parts: the base and the index (also called the logarithm, power or exponent). Such numbers are written as shown: Index, logarithm, power or exponent

ax Base

The ways in which combinations of numbers written in index form are treated are described by a set of index laws.

The index laws 1. When numbers with the same base are multiplied, the indices are added. ax × ay = ax+y

2. When numbers with the same base are divided, the indices are subtracted. ax ÷ ay = ax−y ax = ax−y ay

or

3. When numbers with an index or exponent are raised to another index or exponent, the indices are multiplied. (ax ) y = axy 4. When numbers have an index of 0, the answer is 1. a0 = 1

5. When a number has a negative index, it becomes a fraction with a positive index. a−x =

1 ax

and

1 = ax a−x

6. When a number has a fractional index, the denominator of the fraction becomes the root. ay = 1

√ y

a and

ay = x

√ y ax

or

WORKED EXAMPLE 1

( Simplify

)3

( )2 × 3 xy4

6x4 × 2xy4

2x2 y3

.

2 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

ay = x

(√ )x y a

(2x2 y3 )3 × 3(xy4 )2

THINK

WRITE

1.

Remove the brackets by multiplying the indices.

2.

Add the indices of x and add the indices of y. Simplify 23 to 8 and multiply the whole numbers.

3.

Subtract the indices of x and y. Divide 24 by 12.

6x4 × 2xy4

= =

23 x6 y9 × 3x2 y8 12x5 y4 24x8 y17 12x5 y4

= 2x3 y13

For negative indices and fractional or decimal indices, the same rules apply. WORKED EXAMPLE 2

Write the following in simplest form. a.

2

64 3

b.

Rewrite using the index law = (√ )x √ y y 2. Rewrite using ax = a .

THINK

x ay

a. 1.

32−0.4

ax .

a.

Simplify by taking the cube root of 64. 4. Square 4. 3.

b. 1.

2.

3.

Write as a fraction with a positive index. Change 0.4 to

64 3 =

WRITE

√ y

b.

2

√ 3 642 (√ )2 3 = 64

= 42 = 16

32−0.4 = =

4 . 10

=

Simplify the fractional index.

4.

Rewrite using the index law a y =

5.

Simplify by taking the 5th root of 32.

6.

Square 2.

x

√ y

ax .

1 320.4 1 4

32 10 1 2

32 5

1 = (√ )2 5 32

=

1 22 1 = 4

WORKED EXAMPLE 3

Simplify the following, leaving your answers with positive indices. −1 ⎛ 21 −1 ⎞ ( ) −1 a b 4 −4 a. a−2 b × a3 b b. ⎜ −1 2 ⎟ ÷ ⎜3 b ⎟ ⎝ ⎠

⎛ − 32 2 ⎞ ⎜ 3a b ⎟ ⎜ 3 1 ⎟ ⎝ a4 b2 ⎠

2

CHAPTER 1 The logarithmic function 2 3

THINK a. 1.

Remove the brackets by multiplying the indices.

2.

Add the indices of a and of b.

3.

Place a5 in the denominator with a positive index.

b. 1.

a−2 b4 × (a3 b−4 )−1 = a−2 b4 × a3 b4

WRITE a.

= a−5 b8

−1

⎛ 12 −1 ⎞ a b ⎟ b. ⎜ ⎜ 3−1 b2 ⎟ ⎠ ⎝

Remove the brackets by multiplying the indices.

−1

=

b8 a5

⎛ − 32 2 ⎞ 3a b ÷⎜ 3 1 ⎟ ⎜ ⎟ ⎝ a4 b2 ⎠

2

a 2 b 32 a−3 b4 = −2 ÷ 3 3b a2 b

2.

When dividing by a fraction, invert and multiply.

3.

Add the indices of a and of b in the numerator and add the indices of b in the denominator. Multiply the numbers.

4.

Subtract the indices of a and b.

−1

a 2b a2 b = −2 × −3 4 3b 9a b ab2 = 27a−3 b2 =

3

a4 27

WORKED EXAMPLE 4

Simplify

3n × 6n+1 × 12n−1 . 32n × 8n

THINK 1.

Write each number as the product of prime factors.

2.

Remove the brackets.

Add the indices of numbers with base 3 in the numerator and add indices of numbers with base 2 in the numerator. 4. Subtract the indices. 3.

5.

Write the term with a negative index in the denominator with a positive index.

6.

Simplify.

3n × 6n+1 × 12n−1 32n × 8n ( )n−1 n 3 × (3 × 2)n+1 × 22 × 3 = 32n × 23n WRITE

=

3n × 3n+1 × 2n+1 × 22n−2 × 3n−1 32n × 23n

=

33n × 23n−1 32n × 23n

=

3n 2

= 3n × 2−1 1 = 3n × 2

4 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Index laws can be applied in many situations, such as in exponential modelling. A simple example of an exponential model is presented in the following worked example. WORKED EXAMPLE 5

An antique chair worth $15 000 is increasing in value by 10% each year. a. Write an equation for the value of the chair, $v, in terms of the time, t, in years. b. Hence, find the value of the chair after 10 years. Give your answer correct to the nearest hundred dollars.

THINK a. 1.

Find by what percentage the chair appreciates each year.

2.

Write this as a decimal.

3.

Find the value after 1 year.

4.

Find the value after 2 years.

5.

Find the value after 3 year.

6.

Hence, find the formula. Note: A formula does not include the dollar sign. Substitute t = 10 in the equation. Evaluate 1.110 . Calculate v and express your answer correct to the nearest hundred dollars. Write your answer in a sentence.

b. 1. 2. 3. 4.

Units 3 & 4

The chair appreciates by (100 + 10) % or 110%. 110 110% = 100 = 1.1 After 1 year it is worth $ (15 000 × 1.1) = $16 500 After 2 years it is worth $ (15( 000 × 1.1) × (1.1) ) = $ 15 000 × 1.12 = $18 150 After ( 3 years it is) worth $ 15 000 × 1.13 = $19 965 v = 15 000 × 1.1t

WRITE

Area 1

Sequence 1

Concept 1

a.

b.

v = 15 000 × 1.110 = 15 000 × 2.593 742 5 = 38 906.138 ≈ 38 900 (to the nearest 100) The value of the chair after 10 years is about $38 900.

Review of the index laws Summary screen and practice questions

CHAPTER 1 The logarithmic function 2 5

Exercise 1.2 Review of the index laws Technology free 1.

Simplify the following. a. x × x4 b. x7 ÷ x2 ( 2 )3 x × x5 x4 × x5 f. e. ( 5 )2 x3 x ( 2 3 )2 ( )5 ( 2 )3 ( 4 )2 3 x y × 2 xy3 2xy × 5 x y WE1

3

WE2

a.

g.

( 2 )5 x

5x2 y4 × 4x5 y

h.

22 x3 y2

4x5 y3 × 3x2 y3 4x4 y2 × 3x5 y Simplify the following without using a calculator.

2

27 3

b.

3

16 4

c.

25 (

e.

d.

(

x−3

)2

3x3 y5 × 10xy4 5x2 y6

j.

i.

2.

c.

810.25

f.

361.5

g.

−3 2

9 49

d.

)1

(

2

( )− 3 )− 3 256 4 243 5 j. i. 32 81 WE3 3. Simplify the following, leaving your answers with positive indices. ( ( )−4 )−1 b. x4 y−1 × x−2 y3 a. 3x−3 y2 × x2 y ( )1 ( ) 3 1 2 −1 2 1 2 2 −1 3 2 3 3 3 3 2 2 4 2 c. 2x y × 9x y d. 5x y × 8 x y (

)− 3 ( )5 ( 1 2 −1 −1 2 −2 2 × 9x 5 y 2 e. x y ⎛ 23 −2 ⎞ a b c ⎟ g. ⎜ ⎜ − 1 −2 ⎟ ⎝ 3a 2 bc ⎠

−2

⎛ 32 3 ⎞ a b ÷ 3 ⎜ −1 2 ⎟ ⎜a c ⎟ ⎝ ⎠

)2

3

( )− 1 ( )1 2 2 1 2 2 −1 x5 y 4 × 4x 5 y 2 −2

⎛ − 23 34 ⎞ a b ⎟ h. ⎜ ⎜ ab2 ⎟ ⎝ ⎠

3

h.

27 64

÷

(

9a−3 b 4a2 b3

)1

2

3n × 9n−1 × 27n+1 c. 2n−1 × 3n × 6n+1 32 × 2−3 52 × 3−1 27 d. 2n × 3n+1 × 9n e. × 16 f. ÷ −2 3 5 125 × 9 92 5. Simplify the following, writing your answers as single fractions with positive indices. ( )2 1 a. x−1 + b. x−1 + x−2 −1 x ( )−1 1 1 c. + −1 d. 2x x2 − y2 − (x − y)−1 −1 x +1 x −1 6. If a = 23 , b = 2−3 , c = 62 and d = 3−1 , determine: 4.

Simplify the following. 2n × 4n+1 × 8n−1

1 f. 16 2

100 000

WE4

a.

a.

b.

1

c2 7. MC 3−x + 3x is equal to: A. 1

a 3 b−1 d b. c2 1

a2 b

B.

1 + 32x 3x

C. 3−x

2

6 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

D. 6

−3 5

Technology active 8.

WE5 A population of organisms is growing so that the number of organisms, N, after t days is given by the formula N = 500 × 20.1t .

Determine the number of organisms after 10 days. b. Determine the size of the population after 15 days. Give your answer to the nearest whole number. a.

9.

MC A car worth $10 000 is depreciating at 20% per annum, so that each year the car is worth 80% of its value the previous year. A model for the value of the car, $V, in terms of the time, t, in years is:

V = 10 000 × 20t V = 10 000 × (0.2)t V = 0.8 × 10 000t V = 10 000 (0.8)t 10. A ball is dropped from a window h m above the ground. When it lands on the ground it rebounds to 80% of its height. The equation showing the height of the ball, h metres, after r rebounds is h = 10 × (0.8)r . a. From how far above the ground was the ball dropped? b. How far above the ground does the ball reach on the fourth rebound? Give your answer to the nearest centimetre. c. What is the total vertical distance that the ball travelled when it hits the ground for the fourth time? A. B. C. D.

1.3 Logarithmic laws and equations 1.3.1 Writing numbers in logarithmic form The number 81 can be written as:

81 = 34

That is, given the base 3 and the exponent 4, we can find the number 81 by calculating 3 × 3 × 3 × 3. Note, however, that we need a calculator to compute 34.5 : 140.296 = 34.5

34 gives 140.296.

What do we do if we are given the number and the base, but need to find the power or exponent? 100 = 10x

CHAPTER 1 The logarithmic function 2 7

In this case, an easy calculation shows that 100 = 102 . However, how do we find x in an equation such as the following?

This is where logarithms are useful. If 200 = 10x

200 = 10x

then x = log10 200

= 2.301 (from calculator) In general, if N = ax then x = loga N. So, a number written in index form as ↓

Exponent, index or power

becomes

Base number → 102 = 100

log10 100 = 2 ← Exponent, index or power ↑

Base number

when written in logarithmic form. Logarithms can only be used when the base number and the exponent are positive numbers; that is, for loga N, a > 0 and N > 0. For any other values, the logarithm is not defined.

1.3.2 Natural logarithms Any number, provided that it is positive, can be used as the base number. A natural logarithm is one which uses Euler’s number (written as e) as its base number. Named for 18th-century Swiss mathematician Leonhard Euler (pronounced ‘oiler’), Euler’s number is the base number for many processes and formations in nature that can be described logarithmically, such as the decay of radioactive isotopes, the structure of a spiral galaxy or even the arrangement of leaves on a plant.

Euler’s number

1 Like 𝜋, e is an irrational number. It can be found by evaluating the expression 1 + n increasing values of n: ( )n ( )1 1 1 n=1 1+ = 1+ =2 n 1 ( )n ( )2 1 1 n=2 1+ = 1+ = 2.25 n 2 ( )n ( )3 1 1 n=3 1+ = 1+ = 2.370 37 n 3 ( )n ( )5 1 1 n=5 1+ = 1+ = 2.488 32 n 5

8 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

(

)n for

)n 1 n = 10 1 + n ( )n 1 n = 100 1 + n )n ( 1 n = 1000 1 + n ( )n 1 n = 10 000 1 + n (

1 As n increases, 1 + n (

)n

1 1+ 10

=

(

=

(

=

(

)10

1 1+ 100

= 2.593 74

)100

1 1+ 1000

= 2.704 81

)1000

1 1+ 10 000

= 2.716 92

)10 000

= 2.718 15

becomes closer and closer to 2.718 281 or e. ( )n 1 e = lim 1 + n→∞ n

Using a base e, if

then

=

(

N = ex

x = loge (N) (this can also be written as x = ln (N)).

1.3.3 The logarithmic laws

The indicial laws can be used to derive the logarithmic laws. 1. am × an = am+n ⇔ loga (m) + loga (n) = loga (mn) To prove this law: Let x = loga (m) and y = loga (n). So ax = m and ay = n. Now am × an = am+n . Thus, mn = ax + ay = ax+y . By applying the definition of a logarithm to this statement, we get loga (mn) = x + y

loga (mn) = loga (m) + loga (n). ( ) m m n m−n 2. a ÷ a = a ⇔ loga (m) − loga (n) = loga n To prove this law: Let x = loga (m) and y = loga (n). or

So ax = m and ay = n.

ax = ax−y . ay m ax Thus, = y = ax−y . n a Now

CHAPTER 1 The logarithmic function 2 9

By converting the equation into logarithm form, we get ( ) m loga =x−y n

( ) m loga = loga (m) − loga (n). n

or

Note: Before the first or second law can be applied, each logarithmic term must have a coefficient of 1. 3. (am )n = amn ⇔ loga (mn ) = n loga (m) To prove this law: Let x = loga (m). So ax = m. Now (ax )n = mn or anx = mn . By converting the equation into logarithm form, we have loga (mn ) = nx

or loga (mn ) = n loga (m). Applying these laws, we can also see the following: 4. As a0 = 1, then by the definition of a logarithm, loga (1) = 0. 5. As a1 = a, then by the definition of a logarithm, loga (a) = 1. 6. ax > 0; therefore, loga (0) is undefined, and loga (x) is defined only for x > 0 and a ∈ R+ \ {1}. Another important fact related to the definition of a logarithm is ( ) aloga m = m. This can be proved as follows: Let y = aloga (m) . Converting index form to logarithm form, we have loga (y) = loga (m). Consequently, aloga m = m. ( )

The logarithm laws 1. loga (m) + loga (n) = loga (mn) ( ) m 2. loga (m) – loga (n) = loga n 3. loga (mn ) = n loga (m) 4. loga (1) = 0 5. loga (a) = 1 6. loga (0) = undefined 7. loga (x) is defined for x > 0 and a ∈ R+ \ {1}. 8. aloga m = m ( )

WORKED EXAMPLE 6

Simplify the following without a calculator. a. log10 (5) + log10 (2) c.

log2 (16)

log4 (20) (√ −) log4 (5) 5 d. log5 x b.

10 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Rewrite using loga (mn) = loga (m) + loga (n).

log10 (5) + log10 (2) = log10 (5 × 2)

THINK

WRITE

a. 1.

a.

Simplify using loga (a) = 1. ( ) m = loga (m) − loga (n). b. 1. Rewrite using loga n 2. Simplify. 2.

3.

3. c. 1. 2. 3.

20 b. log4 (20) − log4 (5) = log4 5 = log4 (4)

Simplify using loga (a) = 1.

Rewrite 16 as a number with base 2. Rewrite using loga (mp ) = p loga (m).

c.

Simplify using loga (a) = 1.

(

)

=1

log2 (16) = log2 (24 )

= 4 log2 (2) =4×1

=4 ( ) (√ ) 1 5 d. log5 x = log5 x 5

1 √ x d. 1. Rewrite using a = a y .

2.

= log10 (10) =1

Simplify.

Rewrite using loga (mp ) = p loga (m).

=

1 log5 (x) 5

c.

log3 (9) log3 (27)

WORKED EXAMPLE 7

2 + log10 (3)

Simplify the following. a.

b.

3 log3 (6) − 3 log3 (18)

2 + log10 (3) = 2 log10 (10) + log10 (3)

THINK

WRITE

a. 1.

a.

Write 2 as 2 log10 (10) because log10 (10) = 1. 2. Rewrite using loga (mp ) = p loga (m).

= log10 (102 × 3)

Rewrite using loga (mn) = loga (m) + loga (n). 4. Write 102 as 100.

3.

5. b. 1. 2.

3.

4.

Rewrite using loga (mp ) = p loga (m).

Write 6 as 6 × 6 × 6 and 18 as 18 × 18 × 18. 3

3

Simplify.

Write the number with base 3. 6. Rewrite using loga (mp ) = p loga (m). 7. Simplify using loga (a) = 1. 5.

= log10 (100 × 3)

Multiply the numbers in the brackets.

Rewrite ( using ) m loga = loga (m) − loga (n). n

= log10 (102 ) + log10 (3)

b.

= log10 (300)

3 log3 (6) − 3 log3 (18) = log3 (63 ) − log3 (183 ) ( 3 ) 6 = log3 183 6×6×6 = log3 18 × 18 × 18 ( ) 1 = log3 33 −3 = log3 (3 ) = −3 log3 (3) = −3 × 1 = −3 (

)

CHAPTER 1 The logarithmic function 2 11

c. 1.

2. 3.

Write the numbers with the same base. It is not possible to cancel the 9 and the 27 because they cannot be separated from the log.

c.

log3 (9) log3 (27)

Rewrite using loga (mp ) = p loga (m).

= =

log3 (32 ) log3 (33 ) 2 log3 (3)

3 log3 (3) 2 = 3

Cancel the logs because they are the same.

1.3.4 Solving logarithmic equations

Solving logarithmic equations involves the use of the logarithm laws as well as converting to index form. As loga x is defined only for x > 0 and a ∈ R+ \ {1}, always check the validity of your solution. WORKED EXAMPLE 8

Solve the following for x, giving your answer correct to 3 decimal places where appropriate. a. loge (3) = loge (x) b. loge (x) + loge (3) = loge (6) THINK a.

Since the base is the same, equate the numbers.

b. 1.

loge (3) = loge (x) x=3 b. loge (x) + loge (3) = loge (6)

WRITE

Rewrite using loge (mn) = loge (m) + loge (n).

a.

Equate the number parts. 3. Solve for x.

2.

TI | THINK b. 1. On a Calculator page,

press MENU then select: 3: Algebra 1: Numerical Solve.

2. Complete the entry

line as: (ln(x) + ln(3) = ln(6), x) Then press the ENTER button.

3. The answer appears

on the screen.

WRITE

CASIO | THINK

loge (3x) = loge (6) 3x = 6 x=2

b. 1. On a Main Menu screen,

select Equation.

2. On an Equation screen,

select Solver by pressing F3.

3. Complete the entry line

as: ( ) ln(x) + ln(3) = ln(6), x When the equation has been entered, press the SOLVE button.

12 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WRITE

4. The answer appears on

the screen.

WORKED EXAMPLE 9

Solve the following equations for x. a. log4 (64) = x ( )2 c. log2 (x) = 3 − 2 log2 (x) THINK a. 1. 2. b. 1. 2.

Convert the equation into index form. Convert 64 to base 4 and evaluate.

Rewrite 3 in log form, given log2 (2) = 1. Apply the law loga (mn ) = n loga (m).

log2 (3x) + 3 = log2 (x − 2) d. log2 (2x) + log2 (x + 2) = log2 (6) b.

log4 (64) = x 4x = 64 4x = 43 ∴x = 3 b. log2 (3x) + 3 = log2 (x − 2)

WRITE a.

log2 (3x) + 3 log2 (2) = log2 (x − 2) log2 (3x) + log2 (23 ) = log2 (x − 2)

log2 (3x × 8) = log2 (x − 2)

Simplify the left-hand side by applying loga (mn) = loga (m) + loga (n). 4. Equate the logs and simplify.

3.

Identify the quadratic form of the log equation. Let a = log2 (x) and rewrite the equation in terms of a. 2. Solve the quadratic.

c. 1.

3.

d. 1. 2.

Substitute in a = log2 (x) and solve for x.

Simplify the left-hand side by applying loga (mn) = loga (m) + loga (n) . Equate the logs and solve for x.

24x = x − 2

c.

23x = −2 2 x=− 23

)2 log2 (x) = 3 − 2 log2 (x) Let a = log2 (x). a2 = 3 − 2a a2 + 2a − 3 = 0 (a − 1) (a + 3) = 0 a = 1, −3 log2 (x) = 1 log2 (x) = −3 (

x = 21 x = 2−3 1 ∴ x = 2, 8 d. log2 (2x) + log2 (x + 2) = log2 (6) log2 (2x(x + 2)) = log2 (6) 2x(x + 2) = 6 2x2 + 4x − 6 = 0 x2 + 2x − 3 = 0

(x − 1) (x + 3) = 0

3. 4.

Check the validity of both solutions. Write the answer.

x = 1, −3 x = −3 is not valid, as x > 0. ∴x=1 CHAPTER 1 The logarithmic function 2 13

1.3.5 Change of base rule Although any positive number can be used as the base for a logarithmic or indicial expression, calculators generally only allow you two options — the standard logarithm, log10 (x), or the natural logarithm, loge (x). In order to calculate values for bases other than 10 or e, the base must be changed to one of these options. Suppose y = loga (m). By definition, ay = m. Take the logarithm to the same base of both sides. logb (ay ) = logb (m)

y logb (a) = logb (m) logb (m) y= logb (a) Change the base logb (m) loga (m) = logb (a)

WORKED EXAMPLE 10 a. b.

Evaluate the following, correct to 4 decimal places. i. log7 (5) ii. log 1 (11) If p = log5 (x), find the following in terms of p. i. x ii. logx (81) 3

THINK

WRITE

a. i.

Apply the change of base rule and calculate.

ii.

Apply the change of base rule and calculate.

b. i.

Rewrite the logarithm in index form to find an expression for x.

ii. 1.

2.

3.

Rewrite logx (81) using loga (m ) = n loga (m). n

= 0.8271 log10 (7) log10 (11) ii. log 1 (11) = ( ) = −2.1827 log10 31 3 b. i. p = log5 (x)

a. i.

log7 (5) =

log10 (5)

x = 5p ( ) ii. logx (81) = logx 92

Apply the change-of-base rule so that x is no longer a base. Note: Although 9 has been chosen as the base in this working, a different value could be applied, giving a different final answer. Replace x with 5p and apply the law loga (mn ) = n loga (m).

14 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

= 2 logx (9) log9 (9) =2 log9 (x) 1 =2 log9 (x) =2

1 log9 (5p ) 2 = p log9 (5)

Units 3 & 4

Area 1

Sequence 1

Concept 2

Logarithmic laws and equations Summary screen and practice questions

Exercise 1.3 Logarithmic laws and equations Technology free

Simplify the following without using a calculator. log6 (3) + log6 (2) b. log10 (5) + log10 (2) d. log2 (10) − log2 (5) e. log2 (32) ( ) ( ) 1 1 h. log3 g. log5 5 27 2. Simplify the following. (√ ) (√ ) 3 a. log2 x b. log3 x (√ ) (√ ) x4 4 x e. log2 d. 4 log4 y2

1.

log3 (6) − log3 (2) f. log3 (81)

WE6

a.

3.

Simplify the following without using a calculator. a. 4 log2 (12) − 4 log2 (6) b. 3 log2 (3) − 3 log2 (6) c. 2 + log5 (10) − log5 (2) d. 2 + log5 (2) − log5 (10) e. 1 + log2 (5) f. 3 + log3 (2)

i.

5.

6.

7.

(√ ) 3 log3 3 x (√ ) x5 5 f. log3 y10

c.

WE7

g.

4.

c.

log2 (64)

log2 (8) (√ ) loga x loga (x)

Simplify without using a calculator. a. 5 log3 (x) + log3 (x2 ) − log3 (x7 ) c. 3 log4 (x) − 5 log4 (x) + 2 log4 (x) e. log10 (x2 ) + 3 log10 (x) − 2 log10 (x) 2 g. log5 (x + 1) + log5 (x + 1)

log5 (125)

h.

log5 (25) ( ) loga x2 j. ( ) loga x3

3 log2 (x) + log2 (x3 ) − log2 (x6 ) d. 4 log6 (x) − 5 log6 (x) + log6 (x) f. 4 log10 (x) − log10 (x) + log10 (x2 ) 3 h. log4 (x − 2) − 2 log4 (x − 2) b.

WE8 Solve the following for x. Give exact answers when appropriate; otherwise, give answers correct to 3 decimal places. a. loge (x) = loge (2) b. loge (x) = loge (5) c. loge (x) + loge (3) = loge (9) d. loge (x) + loge (2) = loge (8) e. loge (x) − loge (5) = loge (2) f. loge (x) − loge (4) = loge (3) g. 1 + loge (x) = loge (6) h. 1 − loge (x) = loge (7) i. loge (4) − loge (x) = loge (2) j. loge (5) − loge (x) = loge (25)

Solve the following for x. a. log5 (125) = x ( )2 c. 3 log2 (x) − 2 = 5 log2 (x) WE9

Solve the following for x. a. log3 (x) = 5

8. a.

WE10

b.

d. b.

log4 (x − 1) + 2 = log4 (x + 4)

log5 (4x) + log5 (x − 3) = log5 (7) log3 (x − 2) − log3 (5 − x) = 2

Evaluate the following, correct to 4 decimal places.

log7 (12)

If z = log3 (x), find the following in terms of z. i. 2x i.

b.

( ) 1 ii. log3 4

ii.

logx (27) CHAPTER 1 The logarithmic function 2 15

Technology active 9.

Rewrite the following in terms of base 10. a. log5 (9)

Express each of the following in logarithmic form. a. 63 = 216 b. 28 = 256 d. 10−4 = 0.0001 e. 5−3 = 0.008 11. Find the value of x.

b.

2

10.

log3 (81) = x

logx (121) = 2 12. Simplify the following. a. log2 (256 + log2 (64) − log2 (128) (√ ) 6 1 c. log4 64 a.

c.

e.

log5 (32)

3 log5 (16) Simplify the following. 2 a. log3 (x − 4) + log3 (x − 4) ( 2) ( 3) c. log5 x + log5 x − 5 log5 (x) 14. Evaluate the following, correct to 4 decimal places. 13.

log3 (7)

If n = log5 (x), find the following in terms of n. ( ) a. 5x b. log5 5x2 16. Solve the following for x. a.

log 1 (12) c. f.

) 1 =x b. log6 216 d. log2 (−x) = 7 (

5 log7 (49) − 5 log7 (343) ) ( 16 d. log4 256 (√ ) 6 log2 3 x f. ( ) log2 x5 b.

log7 (2x + 3)3 − 2 log7 (2x + 3) 3 2 d. log4 (5x + 1) + log4 (5x + 1) − log4 (5x + 1) b.

(

b.

log2

15.

a.

loge (2x − 1) = −3

log3 (4x − 1) = 3 e. 3 log10 (x) + 2 = 5 log10 (x) g. 2 log5 (x) − log5 (2x − 3) = log5 (x − 2) i. log3 (x) + 2 log3 (4) − log3 (2) = log3 (10) ( )2 k. log3 x = log3 (x) + 2 17. Express y in terms of x for the following equations. a. log10 (y) = 2 log10 (2) − 3 log10 (x) c.

18.

19. 20. 21.

34 = 81 71 = 7

1 121

)

c.

b. d. f. h. j. l.

logx (625)

( ) 1 =3 x log10 (x) − log10 (3) = log10 (5) ( ) log10 x2 − log10 (x + 2) = log10 (x + 3) log10 (2x) − log10 (x − 1) = 1 ( )( ( )) log10 (x) log10 x2 − 5 log10 (x) + 3 = 0 loge

log6 (x − 3) + log6 (x + 2) = 1

log4 (y) = −2 + 2 log4 (x) ( ) 2x c. log9 (3xy) = 1.5 d. log8 + 2 = log8 (2) y a. Find the value of x in terms of m for which 3 logm (x) = 3 + logm 27, where m > 0 and x > 0. ( ) 3y 100n2 b. If log10 (m) = x and log10 (n) = y, show that log10 =2+ − 5x. √ 2 m5 n Solve the equation 8 logx (4) = log2 (x) for x. Solve the following for x, correct to 3 decimal places. a. e2x − 3 = loge (2x + 1) b. x2 − 1 = loge (x) ( )( ) Find x, correct to 4 decimal places, if 3 log3 (x) 5 log3 (x) = 11 log3 (x) − 2. b.

16 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

1.4 Logarithmic scales Many scientific quantities are measured in terms of scales that are logarithmic rather than linear. For example, the pH scale used in chemistry to describe the acidity or alkalinity of a substance ranges from 1 for very strong acids through to 7 for neutral substances such as water, up to a pH of 14 for very strong bases or alkalis. This scale is based upon the concentration of hydrogen ions. A substance that has pH 5 is ten times more acidic than one that has a pH of 6, and has ten times the concentration of hydrogen ions. An acid with pH 3 has 10 000 times the acidity of water (pH 7). Other examples of logarithmic scales include: • the Richter scale, which describes the amount of energy released by the seismic waves of earthquakes in terms of magnitude • loudness of sound as a function of the sound’s intensity • the frequencies of musical notes • the intensity of the brightness of stars. WORKED EXAMPLE 11

The apparent brightness, B of a star can be found using the formula B = 6 − 2.5 log10 A, where A is the actual brightness of that star. Find the apparent brightness of a star with actual brightness of 3.16.

B = 6 − 2.5 log10 (A) When A = 3.16,

THINK

WRITE

Write the formula. 2. Substitute 3.16 for A.

1.

Evaluate log10 (3.16) using a graphics calculator. 4. Simplify. 5. Write your answer in a sentence. 3.

B = 6 − 2.5 log10 (3.16) = 6 − 2.5 × 0.5 = 4.75 The apparent brightness of the star is 4.75.

WORKED EXAMPLE 12

Loudness, L, in decibels (dB), is related to the intensity, I, of a sound by the equation L = 10 log10

(

I I0

)

( ) where I0 is equal to 10−12 watts per square metre W/m2 . (This value is the lowest intensity of sound that can be heard by human ears.) a. An ordinary conversation has a loudness of 60 dB. Calculate the intensity in W/m2 . b. If the intensity is doubled, what is the change in the loudness, correct to 2 decimal places?

CHAPTER 1 The logarithmic function 2 17

THINK a. 1.

Substitute L = 60 and simplify.

WRITE

( ) I a. L = 10 log10 I0 ) ( I 60 = 10 log10 10−12 60 = 10 log10 (1012 I)

2.

b. 1.

Convert the logarithm to index form and solve for I. Determine an equation for L1 .

6 = log10 (1012 I) 106 = 1012 I

b.

I = 10−6 W/m(2

L1 = 10 log10

I1

)

10−12 ) = 10 log10 1012 I1 ( ) = 10 log10 1012 + 10 log10 (I1 ) (

= 120 log10 (10) + 10 log10 (I1 )

2.

3.

The intensity has doubled; therefore, I2 = 2I1 . Determine an equation for L2 .

Replace 120 + 10 log10 (I1 ) with L1 . Answer the question.

Units 3 & 4

Area 1

Sequence 1

= 120 + 10(log10 (I) 1) 2I1 L2 = 10 log10 10−12 = 10 log10 (2 × 1012 I1 )

= 10 log10 (2) + 10 log10 (1012 ) + 10 log10 (I1 ) = 3.010 + 120 log10 (10) + 10 log10 (I1 )

= 3.01 + 120 + 10 log10 (I1 ) = 3.01 + L1

Doubling the intensity increases the loudness by 3.01 dB.

Concept 3

Logarithmic scales Summary screen and practice questions

Exercise 1.4 Logarithmic scales Technology active 1.

The loudness, L, of a jet taking off about 30 metres away ( ) I is known to be 130 dB. Using the formula L = 10 log10 , I0 where I is the intensity measured in W/m2 and I0 is equal to 10−12 W/m2 , calculate the intensity in W/m2 for this situation. WE12

18 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2.

3.

4. 5.

6.

The moment magnitude scale measures the magnitude, M, of an earthquake in terms of energy released, E, in joules, according to the formula ( ) E M = 0.67 log10 K where K is the minimum amount of energy used as a basis of comparison. An earthquake that measures 5.5 on the moment magnitude scale releases 1013 joules of energy. Find the value of K, correct to the nearest integer. Two earthquakes, about 10 kilometres apart, occurred in Iran on 11 August 2012. One measured 6.3 on the moment magnitude scale, and the other was 6.4 on the same scale. Use the formula from question 2 to compare the energy released, in joules, by the two earthquakes. An earthquake of magnitude 9.0 occurred in Japan in 2011, releasing about 1017 joules of energy. Use the formula from question 2 to find the value of K correct to 2 decimal places. To the human ear, how many decibels louder is a 500 W/m2 amplifier compared to a 20 W/m2 model? ( ) I Use the formula L = 10 log10 , where L is measured in dB, I is measured in W/m2 and I0 I0 = 10−12 W/m2 . Give your answer correct to 2 decimal places. Your eardrum can be ruptured if it is exposed to a noise that has an intensity of 104 W/m2 . Using the ( ) I , where I is the intensity measured in W/m2 and I0 is equal to 10−12 W/m2 , formula L = 10 log10 I0 calculate the loudness, L, in decibels that would cause your eardrum to be ruptured.

Questions 7–9 relate to the following information. Chemists define the acidity or alkalinity of a substance according to the formula [ ] pH = − log10 H+

[ ] where H+ is the hydrogen ion concentration measured in moles/litre. Solutions with a pH less than 7 are acidic, whereas solutions with a pH greater than 7 are basic. Solutions with a pH of 7, such as pure water, are neutral. 7. Lemon juice has a hydrogen ion concentration of 0.001 moles/litre. Find the pH and determine whether lemon juice is acidic or basic. 8. Find the hydrogen ion concentration for each of the following. a. Battery acid has a pH of 0. b. Tomato juice has a pH of 4. c. Sea water has a pH of 8. d. Soap has a pH of 12. 9. Hair conditioner works on hair in the following way. Hair is composed of the protein called keratin, which has a high percentage of amino acids. These acids are negatively charged. Shampoo is also negatively charged. When shampoo removes dirt, it removes natural oils and positive charges from the hair. Positively charged surfactants in hair conditioner are attracted to the negative charges in the hair, so the surfactants can replace the natural oils. a. A brand of hair conditioner has a hydrogen ion concentration of 0.000 015 8 moles/litre. Calculate the pH of the hair conditioner. b. A brand of shampoo has a hydrogen ion concentration of 0.000 002 75 moles/litre. Calculate the pH of the shampoo.

CHAPTER 1 The logarithmic function 2 19

“c01TheLogarithmicFunction2_print” — 2019/7/31 — 20:01 — page 20 — #20

10.

The number of atoms of a radioactive substance present after t years is given by N (t) = N0 e−mt .

The half-life is the time taken for the number of atoms to be reduced to 50% of the initial number of loge (2) atoms. Show that the half-life is given by . m b. Radioactive carbon-14 has a half-life of 5750 years. The percentage of carbon-14 present in the remains of plants and animals is used to determine how old the remains are. How old is a skeleton that has lost 70% of its carbon-14 atoms? Give your answer correct to the nearest year. 11. A basic observable quantity for a star is its brightness. The apparent magnitudes m1 and m2 for two stars are related to the corresponding brightnesses, b1 and b2 , by the equation a.

( m2 − m1 = 2.5 log10

b1 b2

) .

The star Sirius is the brightest star in the night sky. It has an apparent magnitude of −1.5 and a brightness of −30.3. The planet Venus has an apparent magnitude of −4.4. Calculate the brightness of Venus, correct to 2 decimal places. 12. Octaves in music can be measured in cents, n. The frequencies of two notes, f1 and f2 , are related by the equation ( ) f n = 1200 log10 2 . f1 Middle C on the piano has a frequency of 256 hertz; the C an octave higher has a frequency of 512 hertz. Calculate the C D E F G A B number of cents between these two Cs. 13. Prolonged exposure to sounds above 85 decibels can cause ( ) hearing damage or loss. A gunshot from a 0.22 rifle has an intensity of about 2.5 × 1013 I0 . Calculate the loudness, in decibels, of the gunshot sound and state if ear protection should be worn ( ) I , where I0 when a person goes to a rifle range for practice shooting. Use the formula L = 10 log10 I0 is equal to 10−12 W/m2 , and give your answer correct to 2 decimal places. 14. Early in the 20th century, San Francisco had an earthquake that measured 8.3 on the magnitude scale. In the same year, another earthquake was recorded in South America that was four times stronger than the one in San Francisco. Using the equation ( ) E M = 0.67 log10 , calculate the magnitude of the K earthquake in South America, correct to 1 decimal place.

20 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

1.5 Indicial equations

An indicial equation is an equation in which the power or index contains the unknown. When we solve an indicial equation such as 3x = 81, the technique is to convert both sides of the equation to the same base. For example, 3x = 34 ; therefore, x = 4. When we solve an equation such as x3 = 27, we write both sides of the equation with the same index. In this case, x3 = 33 ; therefore, x = 3. If an equation such as 52x = 2 is to be solved, then we must use logarithms, as the sides of the equation cannot be converted to the same base or index. To remove x from the power, we take the logarithm of both sides. ( ) log5 52x = log5 (2) 2x = log5 (2) 1 x = log5 (2) 2

Notes: 1. If ax = b, a solution for x exists only if b > 0.

2. log5 2 can be calculated using a graphics calculator or as

log10 2 log10 5

using a standard calculator.

WORKED EXAMPLE 13

Solve the following equations for x, giving your answers in exact form. 3x 3−x x−3 a. 4 ×16 = 256 b. 7 −3=0 2x x x c. (5 − 25)(5 + 1) = 0 d. 3 − 9 (3x ) + 14 = 0 THINK a. 1.

Convert the numbers to the same base.

2.

Simplify and add the indices on the left-hand side of the equation.

3.

As the bases are the same, equate the indices and solve the equation.

b. 1.

Rearrange the equation.

2.

Take the logarithm of both sides to base 7 and simplify.

3.

Solve the equation.

c. 1.

Apply the Null Factor Law to solve each bracket.

Convert 25 to base 5. 5x > 0, so there is no real solution for 5x = −1. d. 1. Let a = 3x and substitute into the equation to create a quadratic to solve. 2.

43x × 163−x = 256

WRITE

43x × (42 )3−x = 44 43x × 46−2x = 44

a.

4x+6 = 44 x+6=4

b.

c.

x = −2 −3=0

7x−3 = 3 ( ) log7 7x−3 = log7 (3) x−3

7

x − 3 = log7 (3) x = log7 (3) + 3

(5x − 25)(5x + 1) = 0

5x − 25 = 0

5x = 25 5x = 52 x=2 2x x d. 3 − 9 (3 ) + 14 = 0 Let a = 3x . a2 − 9a + 14 = 0

or

5x + 1 = 0

5x = −1

CHAPTER 1 The logarithmic function 2 21

Factorise the left-hand side. Apply the Null Factor Law to solve each bracket for a. 4. Substitute back in for a. 5. Take the logarithm of both sides to base 3 and simplify. 2. 3.

TI | THINK

WRITE

a.1. On a Calculator page,

press MENU then select: 3: Algebra 1: Numerical Solve.

2. Complete the entry

line as: nSolve (43x × 163−x = 256, x) Then press the ENTER button.

3. The answer appears

on the screen.

(a − 7) (a − 2) = 0 a − 7 = 0 or a − 2 = 0 a=7 a = 2 x x 3 = 7 or 3 = 2 x log3 (3 ) = log3 (7) log3 (3x ) = log3 (2) x = log3 (7)

CASIO | THINK

a.1. On an Equation screen,

select Solver by pressing F3.

2. Complete the entry line

as: 43x × 163−x = 256 When the equation has been entered, press the SOLVE button.

3. The answer appears on

the screen.

WORKED EXAMPLE 14

A tennis ball is dropped from a height of 100 cm, bounces and rebounds to 80% of its previous height. The height, h cm, of the ball after n bounces is given by the formula h = A × an , where A cm is the height from which the ball is dropped and a is the percentage of the height reached by the ball on the previous bounce. a. Find the values of A and a, and hence write the formula for h in terms of n. b. What height will the ball reach after 5 bounces? Give the answer to 1 decimal place. c. How many bounces will it take before the ball reaches less than 1 cm?

22 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x = log3 (2)

WRITE

THINK a. 1. 2. 3. b. 1. 2. 3. c. 1.

The ball is dropped from a height of A cm.

a.

The percentage of the height reached by the ball on the previous bounce is a. Substitute the values for A and a into the formula h = A × an . Substitute 5 for n. Evaluate using a calculator. Write your answer in a sentence.

Divide both sides by 100.

3.

Take thet log of both sides to base 10.

Divide both sides by log10 (0.8).

6.

Evaluate.

Bounces must be in whole numbers. 8. After 20 bounces the ball reaches more than 1 cm, but after 21 bounces the ball reaches less than 1 cm because it bounces to a smaller and smaller height. Write the answer in a sentence. 7.

Units 3 & 4

Area 1

Sequence 1

Concept 4

80 = 0.8 100

h = A × an h = 100 × (0.8)n

h = 100 × (0.8)5 = 32.768 The ball bounces to 32.8 cm after 5 bounces. c. 1 = 100 × (0.8)n b.

log10 (0.8)n = log10 (0.01)

Use loga (mp ) = p loga (m) to simplify.

5.

a = 80% =

(0.8)n = 0.01

Substitute 1 for h.

2.

4.

A = 100

WRITE

n log10 (0.8) = log10 (0.01) log10 (0.01) n= log10 (0.8) = 20.64

≈ 21 The ball reaches less than 1 cm after 21 bounces.

Indicial equations Summary screen and practice questions

Exercise 1.5 Indicial equations Technology free

Solve the following equations for x. 32x+1 × 272−x = 81 c. (4x − 16) (4x + 3) = 0 2. Solve the following equations for x. 1 a. 2x+3 − 64 = 0

1.

WE13

a.

3e2x − 5ex − 2 = 0 3. Solve the following equations for x. a. 72x−1 = 5 c. 25x − 5x − 6 = 0 4. Solve the following equations for x. a. 16 × 22x+3 = 8−2x 10 c. 2 (5x ) − 12 = − 5x c.

10(2x−1 − ) 5=0 x 2x d. 2 10 − 7 (10 ) + 3 = 0 b.

b. d.

22x − 9 = 0

e2x − 5ex = 0

x (3(x − 9) ) (3 − 1)x = 0 2x d. 6 9 − 19 (9 ) + 10 = 0

b.

2 × 3x+1 = 4 d. 4x+1 = 31−x b.

CHAPTER 1 The logarithmic function 2 23

Solve( the following equations for x. ) a. 2 2x−1 − 3 + 4 = 0 1 6. a. Simplify x−1 − . 1 1 − 1+x−1 5.

Solve 23−4x × 3−4x+3 × 6x = 1 for x. 7. Solve the following equations for x. b.

10. 11. 12.

13.

e4 + 1 = 3 2 d. ex + 2 = 4 b.

x

ex = 12 − 32e−x 5 d. ex − 12 = − ex b.

If y = m (10)nx , y = 20 when x = 2 and y = 200 when x = 4, find the values of the constants m and n. Solve the following for x, correct to 3 decimal places. x a. 2x < 0.3 b. (0.4) < 2 ( )2 Solve log3 (4m) = 25n2 for m. Solve the following for x. a. em−kx = 2n, where k ∈ R\ {0} and n ∈ R+ b. 8mx × 42n = 16, where m ∈ R\ {0} c. 2emx = 5 + 4e−mx , where m ∈ R\ {0} WE14 The diameter of a tree trunk increases according to the formula D = A × 100.04t , where D cm is the diameter of the trunk t years after it is first measured and A cm is the diameter of the trunk when it is first measured. a. Write an equation for D in terms of t if the trunk had a diameter of 20 cm when it was first measured. b. When will the diameter be 25 cm? c. After how many years will the diameter be greater than 30 cm?

Technology active 9.

( ) 2 51−2x − 3 = 7

2

ex−2 − 2 = 7 c. e2x = 3ex 8. Solve the following equations for x. a. e2x = ex + 12 c. e2x − 4 = 2ex a.

b.

If y = ae−kx , y = 3.033 when x = 2 and y = 1.1157 when x = 6, find the values of the constants a and k. Give your answers correct to 2 decimal places. 15. The compound interest formula A = Pert is an indicial equation. If a principal amount of money, P, is invested for 5 years, the interest earned is $12 840.25, but if this same amount is invested for 7 years, the interest earned is $14 190.66. Find the integer rate of interest and the principal amount of money invested, to the nearest dollar.

14.

24 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

1.6 Logarithmic graphs 1.6.1 The graph of y = loga (x)

The graph of the logarithmic function f : R+ → R, f (x) = loga (x), a > 1 has the following characteristics. The graph of y = loga (x) For f (x) = loga (x), a > 1: • the domain is (0, ∞) • the range is R • the graph is an increasing function • the graph cuts the x-axis at (1, 0) • as x → 0, y → −∞, so the line x = 0 is an asymptote • as a increases, the graph rises more steeply for x ∈ (0, 1) and is flatter for x ∈ (1, ∞).

y 4 3 y = log2(x) 2

y = log3(x)

1

y = log10(x) (1, 0)

0

–1

1

2

3

4

x

–1 –2 –3 –4

1.6.2 Dilations

Graphs of the form y = n loga (x) and y = loga (mx)

The graph of y = n loga (x) is the basic graph of y = loga x dilated by factor n parallel to the y-axis or 1 from the x-axis. The graph of y = loga (mx) is the basic graph of y = loga (x) dilated by factor parallel to m the x-axis or from the y-axis. The line x = 0 (the y-axis) remains the vertical asymptote and the domain remains (0, ∞). y

y

y = n log a (x) y = loga (mx) 1 –– m, 0

(

(1, 0) 0

x

x=0

0

) x

x=0

1.6.3 Reflections

Graphs of the form y = − loga (x) and y = loga (−x)

The graph of y = − loga (x) is the basic graph of y = loga (x) reflected in the x-axis. The line x = 0 (the y-axis) remains the vertical asymptote and the domain remains (0, ∞). CHAPTER 1 The logarithmic function 2 25

The graph of y = loga (−x) is the basic graph of y = loga (x) reflected in the y-axis. The line x = 0 (the y-axis) remains the vertical asymptote, but the domain changes to (−∞, 0). y

y y = –loga (x) y = loga (–x) (–1, 0)

(1, 0) x

0

x

0

x=0

x=0

1.6.4 Translations

Graphs of the form y = loga (x) + k and y = loga (x − h)

The graph of y = loga (x) + k is the basic graph of y = loga (x) translated k units parallel to the y-axis. Thus the line x = 0 (the y-axis) remains the vertical asymptote and the domain remains (0, ∞). The graph of y = loga (x − h) is the basic graph of y = loga x translated h units parallel to the x-axis. Thus, the line x = 0 (the y-axis) is no longer the vertical asymptote. The vertical asymptote is x = h and the domain is (h, ∞). y

y y = loga (x) + k

y = loga (x – h), h > 0

(a, 1 + k) 0

(a–k, 0)

x

0

x=0

(1 + h, 0)

x

x=h

WORKED EXAMPLE 15

Sketch the graphs of the following, showing all important characteristics. State the domain and range in each case. a. y = loge (x − 2) b. y = loge (x + 1) + 2 1 c. y = loge (2x) d. y = −loge (−x) 4 The basic graph of y = loge (x) has been translated 2 units to the right, so x = 2 is the vertical asymptote.

y = loge (x − 2) The domain is (2, ∞). The range is R.

THINK

WRITE

a. 1.

a.

26 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2.

x-intercept, y = 0: loge (x − 2) = 0

Find the x-intercept.

3.

Determine another point through which the graph passes.

4.

Sketch the graph.

e0 = x − 2 1=x−2 x=3 When x = 4, y = loge (2). ( ) The point is 4, loge (2) . y y = loge (x – 2) (3, 0)

(4, loge (2)) x

0

The basic graph of y = loge (x) has been translated 2 units up and 1 unit to the left, so x = −1 is the vertical asymptote. 2. Find the x-intercept.

b. 1.

y = loge (x + 1) + 2 The domain is (−1, ∞). The range is R. The graph cuts the x-axis where y = 0. loge (x + 1) + 2 = 0 x=2

b.

loge (x + 1) = −2

e−2 = x + 1

3.

Find the y-intercept.

4.

Sketch the graph.

x = e−2 − 1 The graph cuts the y-axis where x = 0. y = loge (1) + 2 =2 y

y = loge (x + 1) + 2

(0, 2)

(e–2 – 1, 0) 0

c. 1.

The basic graph of y = loge (x) has been dilated by 1 1 factor from the x-axis and by factor from the 4 2 y-axis. The vertical asymptote remains x = 0.

x

1 c. y = loge (2x) 4 The domain is (0, ∞). The range is R. x = –1

CHAPTER 1 The logarithmic function 2 27

2.

x-intercept, y = 0: 1 loge (2x) = 0 4 loge (2x) = 0

Find the x-intercept.

3.

Determine another point through which the graph passes.

4.

Sketch the graph.

e0 = 2x 1 = 2x 1 x= 2 When x = 1, y = loge (2). ( ) The point is 1, loge (2) . y 1 y = – loge (2x) 4

(1, loge(2)) 0

d. 1.

2.

The basic graph of y = loge (x) has been reflected in both axes. The vertical asymptote remains x = 0. Find the x-intercept.

3.

Determine another point through which the graph passes.

4.

Sketch the graph.

x

( ) 1– , 0 2

y = − loge (−x) The domain is (−∞, 0). The range is R. x-intercept, y = 0: − loge (−x) = 0 x=0

d.

loge (−x) = 0

e0 = −x x = −1 When x = −2, y = − loge (2). ) ( The point is −2, − loge (2) . y

y = –loge (–x) (–1, 0) 0

(–2, –loge (2)) x=0

28 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

TI | THINK a. 1. On a Calculator page,

press MENU then select: 2: Add Graphs.

in the f1(x) = tab as: ln(x − 2)

WRITE

CASIO | THINK

WRITE

a. 1. On a Main Menu screen,

select Graph.

2. Complete the entry line

2. Complete the function

3. Sketch the graph by

3. The asymptote can be

pressing the ENTER button.

asymptote, x = 2, select: Menu 3: Graph Entry/Edit 2: Relation.

4. To sketch the vertical

entry line in the Y1 tab as: ln(x − 2)

included in the sketch by changing the Y2 line to X2. To do this, select: TYPE X= Complete the entry line in X = as 2. 4. Sketch the graph by pressing either the DRAW button or the EXE button.

5. Complete the entry line

as: x=2 then press the ENTER button.

6. The sketch of the

graph and asymptote will appear on the screen as one.

The situation may arise where you are given the graph of a translated logarithmic function and you are required to find the rule. The information provided to you might include the equation of the asymptote, the intercepts and/or other points on the graph. As a rule, the number of pieces of information is equivalent to the number of unknowns in the equation.

CHAPTER 1 The logarithmic function 2 29

WORKED EXAMPLE 16

The rule for the function shown is of the form y = loge (x − a) + b. Find the values of the constants a and b. y (e2 – 3, 0) x

0

x = –3

The vertical asymptote corresponds to the value The vertical asymptote is x = −3. Therefore, a must be −3. So of a. y = loge (x + 3) + b. ( ) The graph cuts the x-axis at e2 − 3, 0 . 2. Substitute in the x-intercept to find b. ( 2 ) 0 = loge e − 3 + 3 + b ( ) −b = loge e2 −b = 2 b = −2 So y = loge (x + 3) − 2. 3. Write the answer. a = −3, b = −2 THINK

WRITE

1.

Resources Interactivity: Logarithmic graphs (int-6418)

Units 3 & 4

Area 1

Sequence 1

Concept 5

Logarithmic graphs Summary screen and practice questions

Exercise 1.6 Logarithmic graphs Technology free WE15 Sketch the graphs of the following functions, showing all important characteristics. State the domain and range for each graph. a. y = loge (x + 4) b. y = loge (x) + 2 c. y = 4 loge (x) d. y = − loge (x − 4) 2. Sketch the graphs of the following functions, showing all important characteristics. ( ) x a. y = log3 (x + 2) − 3 b. y = 3 log5 (2 − x) c. y = 2 log10 (x + 1) d. y = log2 − 2

1.

30 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Technology active

WE16 The rule for the function shown is y = loge (x − m) + n. Find the values of the constants m and n. 4. A logarithmic function with the rule of the form y = p loge (x − q) passes through the points (0, 0) and (1, −0.35). Find the values of the constants p and q. 5. Sketch the following graphs, clearly showing any axis intercepts and asymptotes. a. y = loge (x) + 3 b. y = loge (x) − 5 c. y = loge (x) + 0.5

3.

6.

7. 8.

9.

10.

11. 12.

y (e + 2, 3)

x

0

Sketch the following graphs, clearly showing any axis intercepts x=2 and asymptotes. a. y = loge (x − 4) b. y = loge (x + 2) c. y = loge (x + 0.5) Sketch the following graphs, clearly showing any axis intercepts and asymptotes. 1 a. y = 4 loge (x) b. y = 3 loge (x) c. y = 6 loge (x) Sketch the following graphs, clearly showing any axis intercepts and asymptotes. ( ) x a. y = loge (3x) b. y = loge c. y = loge (4x) 4 Sketch the following graphs, clearly showing any axis intercepts and asymptotes. ( ) 1 x a. y = 1 − 2 loge (x − 1) b. y = loge (2x + 4) c. y = loge +1 2 4 For each of the following functions, state the domain and range. Define the inverse function, f−1 , and state the domain and range in each case. (Hint: Recall your study of inverse functions from Units 1 and 2.) a. f (x) = 2 loge (3x + 3) b. f (x) = loge (2 (x − 1)) + 2 c. f (x) = 2 loge (1 − x) − 2 −1 For each the functions in question 10, sketch the graphs of f and f on the same set of axes. Give the coordinates of any points of intersection, correct to 2 decimal places. The equation y = a loge (bx) relates x to y. The table shows values for x and y. x

1

2

3

y

loge (2)

0

w

Find the integer values of the constants a and b. Find the value of w correct to 4 decimal places. 13. The graph of a logarithmic function of the form y = a loge (x − h) + k is shown. Find the values of a, h and k. 14. The graph of y = m log2 (nx) passes through the points ( ) 1 1 (−2, 3) and − , . Show that the values of m and n are 2 2 a.

b.

1.25 and −2 5 respectively. 15. Solve the following equations for x. Give your answers correct to 3 decimal places, a. x − 2 = loge (x) b. 1 − 2x = loge (x − 1)

y

0

7

16.

Solve the following equations for x. Give your answers correct to 3 decimal places, a. x2 − 2 < loge (x) b. x3 − 2 ≤ loge (x)

(1, 0)

x

(0, –2)

x = –1

CHAPTER 1 The logarithmic function 2 31

1.7 Applications Logarithmic functions can be used to model many real-life situations in which there is continuous growth or decay over time. Examples of this can be as diverse as the rate at which a hot cup of tea cools down, the increasing account balance of a long-term bank investment or the spread of bacteria through a population. As logarithmic functions are essentially the inverse of exponential functions, they can be used to solve exponential functions of the form A = A0 ekt , where A0 represents the initial value, t represents the time taken and k represents the rate constant. In order to determine the value of t in such a situation, the equation is first rearranged: A = ekt A0 Then we take the natural logarithm of both sides and rearrange for t: ) A loge = kt A0 ( ) A 1 t = loge k A0 (

WORKED EXAMPLE 17

If P dollars is invested into an account that earns interest at a rate of r for t years and the interest is compounded continuously, then A = Pert , where A is the accumulated dollars. A deposit of $6000 is invested at the Western Bank, and $9000 is invested at the Common Bank at the same time. Western offers compound interest continuously at a nominal rate of 6%, whereas the Common Bank offers compound interest continuously at a nominal rate of 5%. In how many years, correct to 1 decimal place, will the two investments be the same? THINK 1.

Write the compound interest equation for each of the two investments.

2.

Equate the two equations and solve for t. CAS could also be used to determine the answer.

A = Pert Western Bank: A = 6000e0.06t Common Bank: A = 9000e0.05t 6000e0.06t = 9000e0.05t WRITE

e0.06t 9000 = e0.05t 6000 3 e0.01t = 2 ( ) 3 0.01t = loge 2 0.01t = 0.4055 0.4055 t= 0.01 t = 40.5 years

32 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 18

A coroner uses a formula derived from Newton’s Law of Cooling to calculate the elapsed time since a person died. The formula is ( ) T−R t = −10 loge 37 − R where t is the time in hours since the death, T is the body’s temperature measured in °C and R is the constant room temperature in °C. An accountant stayed late at work one evening and was found dead in his office the text morning. At 10 am the coroner measured the body temperature to be 29.7 °C. A second reading at noon found the body temperature to be 28 °C. Assuming that the office temperature was constant at 21 °C, determine the accountant’s estimated time of death. THINK 1.

2.

Determine the time of death for the 10 am information. R = 21 °C and T = 29.7 °C. Substitute the values into the equation and evaluate.

Determine the time of death for the 12 pm information. R = 21 °C and T = 28 °C. Substitute the values into the equation and evaluate.

Determine the estimated time of death for each reading. 4. Write the answer. 3.

Units 3 & 4

) T−R t = −10 loge 37 − R ( ) 29.7 − 21 t = −10 loge 37 − 21 ( ) 8.7 = −10 loge 16 = −10 loge (0.54375) = 6.09 h ( ) T−R t = −10 loge 37 − R ) ( 28 − 21 t = −10 loge 37 − 21 ( ) 7 = −10 loge 16 = −10 loge (0.4376) = 8.27h 10 − 6.09 = 3.91 or 3.55 am 12 − 8.27 = 3.73 or 3.44 am The estimated time of death is between 3.44 and 3.55 am. WRITE

Area 1

Sequence 1

(

Concept 6

Applications of logarithms Summary screen and practice questions

CHAPTER 1 The logarithmic function 2 33

Exercise 1.7 Applications WE17 A deposit of $4200 is invested at the Western Bank, and $5500 is invested at the Common Bank at the same time. Western offers compound interest continuously at a nominal rate of 5%, whereas the Common bank offers compound interest continuously at a nominal rate of 4.5%. In how many years will the two investments be the same? Give your answer to the nearest year. a. An investment triples in 15 years. What is the interest rate that this investment earns if it is compounded continuously? Give your answer correct to 2 decimal places. b. An investment of $2000 earns 4.5% interest compounded continuously. How long will it take for the investment to grow to $9000? Give your answer to the nearest month. WE18 An elderly person was found deceased by a family member. The two had spoken on the telephone the previous evening around 7 pm. The coroner attended and found the body temperature to be 25 °C at 9 am. If the house temperature had been constant at 20 °C, calculate how long after the telephone call ( ) T−R the elderly person died. Use Newton’s Law of Cooling, t = −10 loge , where R is the room 37 − R temperature in °C and T is the body temperature, also in °C. The number of parts per million, ( )n, of a fungal bloom in a stream t hours after it was detected can be modelled by n (t) = loge t + e2 , t ≥ 0. a. How many parts per million were detected initially? b. How many parts of fungal bloom are in the stream after 12 hours? Give your answer to 2 decimal places. c. How long will it take before there are 4 parts per million of the fungal bloom? Give your answer correct to 1 decimal place. If $1000 is invested for 10 years at 5% interest compounded continuously, how much money will have accumulated after the 10 years? Let P(t) = 200kt + 1000 represent the number of bacteria present in a petri dish after t hours. Suppose the number of bacteria trebles every 8 hours. Find the value of the constant k correct to 4 decimal places. An epidemiologist studying the progression of a flu epidemic decides that the function

Technology active 1.

2.

3.

4.

5. 6.

7.

P(t) =

) 3( 1 − e−kt , k > 0 4

will be a good model for the proportion, P(t), of the earth’s population that will contract the flu after 1 t months. If after 3 months of the earth’s population has the flu, find the value of the constant k, 1500 correct to 4 decimal places. 8. Carbon-14 dating works by measuring the amount of carbon-14, a radioactive element, that is present in a fossil. All living things have a constant level of carbon-14 in them. Once an organism dies, the carbon-14 in its body starts to decay according to the rule Q = Q0 e−0.000124t

where t is the time in years since death, Q0 is the amount of carbon-14 in milligrams present at death and Q is the quantity of carbon-14 in milligrams present after t years.

34 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

If it is known that a particular fossil initially had 100 milligrams of carbon-14, how much carbon-14, in milligrams, will be present after 1000 years? Give your answer correct to 1 decimal place. b. How long will it take before the amount of carbon-14 in the fossil is halved? Give your answer correct to the nearest year. 9. Glottochronology is a method of dating a language at a particular stage, based on the theory that over a long period of time linguistic changes take place at a fairly constant rate. Suppose a particular language originally has W0 basic words and that at time t, measured in millennia, the number, W (t), of basic words in use is given by W (t) = W0 (0.805)t . a. Calculate the percentage of basic words lost after ten millennia. b. Calculate the length of time it would take for the number of basic words lost to be one-third of the original number of basic words. Give your answer correct to 2 decimal places. 10. The mass, M grams, of a radioactive element, is modelled by the rule a.

M = a − loge (t + b)

where t is the time in years. The initial mass is 7.8948 grams, and after 80 years the mass is 7.3070 grams. a. Find the equation of the mass remaining after t years. Give a correct to 1 decimal place and b as an integer. b. Find the mass remaining after 90 years. 11. The population, P, of trout at a trout farm is declining due to deaths of a large number of fish from fungal infections. The population is modelled by the function P = a loge (t) + c

where t represents the time in weeks since the infection started. The population of trout was 10 000 after 1 week and 6000 after 4 weeks. a. Find the values of the constants a and c. Give your answers correct to 1 decimal place where appropriate. b. Find the number of trout, correct to the nearest whole trout, after 8 weeks. c. If the infection remains untreated, how long will it take for the population of trout to be less than 1000? Give your answer correct to 1 decimal place. 12. In her chemistry class, Hei is preparing a special solution for an experiment that she has to complete. The concentration of the solution can be modelled by the rule C = A loge (kt)

where C is the concentration in moles per litre (M) and t represents the time of mixing in seconds. The concentration of the solution after 30 seconds of mixing is 4 M, and the concentration of the solution after 2 seconds of mixing was 0.1 M. a. Find the values of the constants A and k, giving your answers correct to 3 decimal places. b. Find the concentration of the solution after 15 seconds of mixing. c. How long does it take, in minutes and seconds, for the concentration of the solution to reach 10 M?

CHAPTER 1 The logarithmic function 2 35

13.

Andrew believes that his fitness level can be modelled by the function F (t) = 10 + 2 loge (t + 2)

where F (t) is his fitness level and t is the time in weeks since he started training. a. What was Andrew’s level of fitness before he started training? b. After 4 weeks of training, what was Andrew’s level of fitness? c. How long will it take for Andrew’s level of fitness to reach 15? 14. In 1947 a cave with beautiful prehistoric paintings was discovered in Lascaux, France. Some charcoal found in the cave contained 20% of the carbon-14 that would be expected in living trees. Determine the age of the paintings to the nearest whole number if Q = Q0 e−0.000124t

where Q0 is the amount of carbon-14 originally and t is the time in years since the death of the prehistoric material. Give your answer correct to the nearest year. 15. The sales revenue, R dollars, that a manufacturer receives for selling x units of a certain product can be modelled by the function ( ) x R (x) = 800 loge 2 + . 250 Furthermore, each unit costs the manufacturer 2 dollars to produce, and the initial cost of adjusting the machinery for production is $300, so the total cost in dollars, C, of production is C (x) = 300 + 2x.

Write the profit, P (x) dollars, obtained by the production and sale of x units. Find the number of units that need to be produced and sold to break even, that is, to reach P (x) = 0. Give your answer correct to the nearest integer. 16. The value of a certain number of shares, $V, can be modelled by the equation a.

b.

V = kemt

where t is the time in months. The original value of the shares was $10 000, and after one year the value of the shares was $13 500. a. Find the values of the constants k and m, giving answers correct to 3 decimal places where appropriate.

36 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Find the value of the shares to the nearest dollar after 18 months. After t months, the shares are sold for 1.375 times their value at the time. Find an equation relating the profit made, P, over the time the shares were owned. d. If the shares were kept for 2 years, calculate the profit made on selling the shares at that time. b.

c.

1.8 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. MC Simplifying 3 loge (5) + 2 loge (2) − loge (20) gives: ( ) 19 A. loge B. loge (109) C. loge (480) D. 2 loge (5) 20 ( ) 2. MC If 5 log10 (x) − log10 x2 = 1 + log10 (y), then x is equal to: √ 10 3 A. y B. 10y C. 10y D. y 3. MC The function h has the rule h (x) = a loge (x − m) + k, where m and k are positive constants and a is a negative constant. The maximal domain of h is: A. R+ B. R\ {m} C. R\ {n} D. (m, ∞) 4. MC If 7eax = 3, then x equals: ( ) ( ) loge 73 loge (3) 3 3 A. loge (a) B. a loge C. D. 7 7 a a loge (7) 2x+1 x − 4 × 3 + 1 = 0 is: 5. MC The exact solution of the equation 3 1 A. x = 0, x = −1 B. x = 0, x = 1 C. x = −1, x = 1 D. x = , x = 1 3 6. Solve the following equations for x. a. 2 loge (x) − loge (x − 1) = loge (x − 4) b. 2 loge (x + 2) − loge (x) = loge 3 (x − 1) ( 5) ( )2 c. 2 log4 (x) = 3 − log4 x 7. Express y in terms of x for the following equations, giving any restrictions for x. ( ) a. log2 (y) = 2 log2 (x) − 3 b. log3 (9x) − log3 x4 y = 2 8. Sketch the graphs of each of the following, showing any axis intercepts and the asymptote(s). State the domain and range in each case. a. y = loge (x − 1) + 3 b. y = loge (x + 3) − 1 c. y = 2 loge (−x) 9. The loudness of plant machinery at a manufacturing business is modelled by the equation ( ) I L = 10 log10 , where L is the loudness in decibels (dB), I is the intensity in W/m2 and I0 I0 = 10−12 W/m2 . a. If the loudness of the plant machinery at this business is known to be 90 dB, calculate the intensity for this situation. b. Calculate the loudness of the plant machinery if the intensity is 10−6 W/m2 . ( ) 5 10. If log2 5 = 2.321 and log2 9 = 3.17, find log2 . 9 11. Earthquake intensity is often reported on the Richter scale. The magnitude of R is given by ( ) a R = log10 + B, where a is the amplitude of the ground motion in microns at the receiving station, T T is the period of the seismic wave in seconds, and B is an empirical factor that allows for the weakening of the seismic wave with the increasing distance from the epicentre of the earthquake. Find the magnitude of the earthquake if the amplitude of the ground motion is 10 microns, the period is 1 second and the empirical factor is 6.8. CHAPTER 1 The logarithmic function 2 37

12.

Using previous knowledge of the transformation of the graphs of functions, match each equation (a–d) with a suitable graph (P–S). a. y = − log10 (x) b. y = 2 log10 (x) c. y = log10 (2x) d. y = log10 (x − 1) P

Q

y

0

(2, 0)

x

0

x=1 R

S

(1, 0)

x

(1, 0)

x

y

0

x

x=0

x=0

Complex familiar

(0.5, 0)

x=0

y

0

y

(

) 64q2 3y 13. If log4 (p) = x and log4 (q) = y, show that log4 = 3 − 3x + . √ 3 2 p q 14. The pH of a substance is a value that defines the acidity or alkalinity of that substance. It depends on the [ ] concentration of the hydrogen ion, H+ in moles/litre, and is calculated according to the formula [ ] pH = − log10 H+ .

Solutions with a pH less than 7 are acidic, solutions with a pH greater than 7 are basic, and solutions with a pH of 7 are neutral. a. For each of the following, f b. ind the pH and state whether the solution is acidic, basic or neutral. i. Vinegar has a hydrogen ion concentration of 0.01 moles/litre. ii. Ammonia has a hydrogen ion concentration of 10−11 moles/litre. c. Find the hydrogen ion concentration for each of the following. i. Apples have a pH of 3. ii. Sodium hydroxide has a pH of 14.

38 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

15.

The table gives values for x and y that relate to the equation y = a loge (bx). Find the exact values of a, b and m. x y

16.

−3 loge (2) 1

2

3

0

m

An object falls from a high tower. The distance it falls in a certain time is recorded in the table. t (s)

1

2

3

4

5

d (m)

4.7

18.8

42.3

75.2

117.5

If a relationship of the form d = Atn exists, find values for A and n. Use this relationship to predict the value of d after 7 seconds. Complex unfamiliar 17. The graph shown has the rule g (x) = loge (x − h) + k, where h and k are constants. a. b.

y

y = g(x) x

(0, 0)

a. b. c. 18.

State the value of h. Show that k = − loge (2).

x = –2

Hence, rewrite the rule in the form g (x) = loge

(

) x−h , where c is a constant. c

Carbon-14 dating measures the amount of radioactive carbon-14 in fossils. This can be modelled by the relationship Q = Q0 e−0.000 124t , where Q is the amount, in milligrams, of carbon-14 currently present in the fossil of an organism, t is the time in years since the organism’s death, and Q0 is the initial amount, in milligrams, of carbon-14 present.

A fossil shell initially has 150 milligrams of carbon-14 present. How much carbon-14 will be present after 2000 years? Give your answer correct to 3 decimal places. b. Find the number of years it will take for the carbon-14 in the shell to be halved. Give your answer correct to the nearest year. Q0 c. i. Suppose the amount of carbon-14 in the shell is . Find an equation relating n to t. n Q0 . Give ii. Hence, find how long it will be before the amount of carbon-14 in the fossil shell is 10 your answer to the nearest year. a.

CHAPTER 1 The logarithmic function 2 39

19.

The population of quokkas in a small corner of south-western Western Australia is currently described as vulnerable. The once-plentiful population of quokkas was drastically reduced after dingoes, foxes and wild pigs were introduced to Australia. Conservation efforts and dingo, fox and wild pig control programs have seen quokka populations recovering in some areas. In the Northern Jarrah forest, one of the areas where these conservation practices occur, there were known to be about 150 quokkas in 2008. Conservationists produced a model for the increase in population, P, which was given by P = a loge (t) + b

where t is the time in years since 2007 and a and b are constants. There were estimated to be about 6000 quokkas present in the forest in 2013. a. Determine the values of a and b. Give your answers correct to the nearest integer. b. Calculate the number of quokkas that is expected to be present in 2025. Give your answer correct to the nearest integer. c. Given that quokkas have a life expectancy of about 10 years, the model for the actual population is revised to PR = P − 0.25P

where PR is the revised population. i. Find the equation relating PR to t, the number of years since 2007. ii. Calculate the revised population prediction for 2025. Give your answer correct to the nearest integer. 20. The graph of the function f: (−5, ∞) → R, f (x) = loge (x + 5) + 1 is shown. a. Find the rule and domain of f−1 , the inverse function of f. b. On the same set of axes, sketch the graph of f−1 . Label the axis intercepts with their exact values. c. Find the coordinates of the point(s) of intersection correct to 3 decimal places. y y = loge(x + 5) + 1 (0, loge(5) + 1)

(e–1 – 5, 0) 0

x = –5

Units 3 & 4

Sit exam

40 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

Answers

4. a. 0 c. 0 e. 3 log10 (x)

The logarithmic function 2

g. log5 (x + 1) or

d.

7

b. x

1 4 3

g. j. 3. a.

d.

b. 8

1

e. 3

1000 3

h.

7 27 64 3

b.

x11 y2 20 x

e.

1 y4

h.

7. 8. 9. 10.

B

1+x

1 125

f. 216

9

i.

16

x6

c.

y4 5 243x 2

y2

3n+1

x 1

3

x+y 4 3

a. 1000

2n

f. 1

b.

b.

x4

c.

× 32n+1

1 − x2 2x

243

2. a.

1

2

log2 (x)

d. log4 (x) 3. a. 4 d. 1 g. 2 j.

2 3

1

log3 (x) ( 2) x e. log2 y b. −3 e. log2 (10) 3

b.

h.

3

2

z

10

log6 (216) = 3 log3 (81) = 4 log5 (0.008) = −3 4 11 7 1 c. − 2 5

16. a.

2 c. 7

c. 49.04 m

c. 1 f. 4 c. log3 (x)

x y2

c. 3 f. log3 (54) i.

1 2

+1

)

)

3 7 2 47 10

b. 2n + 1

b. log 1 (12) =

log (12) 10 ( ) 1 log 2

log2 (256) = 8 log10 (0.0001) = −4 log7 (7) = 1 −3 −128 −5 2

b. d. f. b. d. b.

10

d. −2 f.

2

b. log7 (2x + 3) d. 2 log4 (5x + 1) b. −6.9189

5

c.

b. e

−3

f. −

h.

5

4

n

5

6

5

4 3

j. 10 2 or 10

4

k. 9 or

17. a. y =

c. y =

1

l. 4

3 4

b. y =

x3 9

1

b. 2 +

20. a. −0.463, 0.675 21. 1.5518, 1.4422

16

x2 16

d. y = 64x

x

18. a. x = 3m 19. 16,

5 4

d. 15

e. 10

i.

f. log3

e

−3

g. 6

(

(9)

10. a. c. e. 11. a. c. 12. a.

1(

5 4 12 0.388 1

3

n+1

b. 1414

b. 1 e. 5 h. −3

b.

15. a. 5

Exercise 1.3 Logarithmic laws and equations 1. a. 1 d. 1 g. −1

d.

12 13. a. 3 log3 (x − 4) c. 0 14. a. 1.7712

1

b. 4.10 m

c.

e.

D a. 10 m

b.

1 − 2 3, 4

9. a. log5 (9) = log (5) 10

f. 8y 8

e.

3 (x + 1)2

6. a. 3

log

c. 2

2

j.

ii.

5 5 6x 4 y 3

3

2

i. 2

8. a. i. 1.2770 ii. −1.2619 z b. i. 2x = 2 × 3

27

b. 3

b. d. f. h.

7. a. 243

8

2a 2 b 2 6n+1

d. 2 × 3

6. a.

3

15 7

3

n

c.

5. a. c. e. g.

10x4 y2

2

a9 b3 6n−1 4. a. 2

d.

f. x i.

10

27x2 y14

g.

5. a.

6

h. 6x y

2. a. 9 d.

c. x

2 3

g. 5x y j.

5

e. x

x6

h. log4 (x − 2)

3 log5 (x + 1) 2 3 10 2.207 3

Exercise 1.2 Review of the index laws 1. a. x

b. 0 d. 0 f. 5 log10 (x)

3y 2

− 5x

b. 0.451, 1

CHAPTER 1 The logarithmic function 2 41

Exercise 1.4 Logarithmic scales

c.

1. 10 W/m2 2. K = 61 808 3. The 6.4 earthquake is 1.41 times bigger than the 6.3

earthquake.

amplifier.

6. 160 dB 7. pH[= 3] (acidic) + 8. a. H = 1 mole/litre

9. 10.

11. 12. 13. 14.

( loge

5+

√ ) 57

13. a. D = 20 × 10 b. 2 years and 5 months c. 5 years 14. a = 5, k = 0.25 15. r = 5%, P = $10 000

m

4

0.04 t

4. K = 3691.17 5. A 500-watt amplifier is 13.98 dB louder than a 20-watt

y

1.

[

y = loge (x + 4)

[

(–3, 0)

(0, loge (4)) x

0

a. Domain = (−4, ∞), range = R y b.

x = –4

y = loge (x) + 2

Exercise 1.5 Indicial equations 1. a. 3

2. a. −9

c. loge (2)

3. a.

1

2

log7 (5) +

1 2

c. log5 (3)

4. a. −

1 log10 (5) + 2 ( ) 2 1 , log10 (3) d. log10 2 1 b. log2 (9) 2 d. loge (5)

b.

c. 2

7 8

d. log9

( ) 2 3

b. 1, 3

c. loge

9. n =

√ ) 1+ 5

c.

2

y = 4 loge (x)

√

(2, 4 loge (2)) x

(1, 0)

Domain = (0, ∞), range = R x=0

b. 4 loge (2) d. ±

y

0

d.

loge (2)

y

b. 2 loge (2), 3 loge (2) ( ) d. loge 6 ±

√

,m=2 2 10. a. x < −1.737 b. x > −0.756 35n 1 11. m = ,m= 4 4 × 35n m − loge (2n) 12. a. , k ∈ R ∖ {0} , n ∈ R k 4 − 4n b. , m ∈ R ∖ {0} 3m 1

x=0

4

6. a.

8. a. 2 loge (2) (

( ) 5

x

(e–2, 0)

Domain = (0, ∞), range = R

( ) 3

5. a. 1

c. loge (3)

, log9

b. log3 (2) − 1

loge (12) b. 0

−x−1 x 7. a. 2 loge (3) + 2

0

b. 0, 2

d.

1

(1, 2)

1

loge c. 0, 1

, m ∈ R ∖ {0}

Exercise 1.6 Logarithmic graphs

] H+ = 0.0001 moles/litre ] −8 + c. H = 10 moles/litre [ +] −12 d. H = 10 moles/litre a. 4.8 (acidic) b. 5.56 (acidic) a. Sample responses can be found in the worked solutions in the online resources. b. 9 988 years old −437.97 n = 361 cents Ear protection should be worn as L = 133.98 dB. 8.7 b.

1

y = –loge (x – 4)

31

(5, 0) x (6, –loge (2))

0

Domain = (4, ∞), range = R

42 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x=4

2. a.

b.

y

y

y = log3 (x + 2) –3

y = loge (x) – 5

(25, 0) 0

x

0 (0, log3 (2) –3)

(1, –5)

x=0

x = –2

b.

x

(e5, 0)

y

c.

y

y = 3log5 (2 – x) y = loge (x) + 0.5 (0, 3log5 (2)) (1, 0) x

0

(1, 0.5)

0

x

(e , 0) –1 2

x=2 y

c.

y = 2 log10 (x + 1)

(1, 2 log10 (2)) x (0, 0)

6. a.

x=0 y

y = loge (x – 4) (10, loge (6))

x = –1

(5, 0)

y

d.

x

0

( )

x y = log2 – –2

(–2, 0)

x=4 x

0

(–1, –1)

y

b.

y = loge (x + 2) (0, loge (2)) 3. m = 2, n = 2 4. p = 5. a.

−7

20 loge (2) y

x=0

(–1, 0) 0

x

, q = −1

y = loge (x) + 3 (1, 3)

x = –2

0

(e–3, 0)

x

x=0

CHAPTER 1 The logarithmic function 2 43

c.

y

b.

y y = loge (x + 0.5)

x y = loge – 4 (4, 0)

()

(0.5, 0)

x

0 x

0

( c.

x=0 y y = loge (4x)

(1, loge(4))

x = – 0.5 7. a.

( ))

1 1, loge – 4

(0, loge (0.5))

y

(1, 0)

0

( (

x

(0.25, 0)

1 4, – y = –1 loge (x) 4 4

x

0

9. a.

x=0 y

x=0 b.

(2, 1 – 2 loge(1))

y y = 3 loge (x)

y = 1 – 2 loge (x – 1) (e0.5 + 1, 0)

(2, 3 loge (2))

x

0

(1, 0) x=1

x

0

b.

y

y = loge (2x + 4) (0, loge (4))

x=0 c.

y (2, 6 loge(2))

x = –2 c.

y = 6 loge (x)

y

()

x – loge – y= 1 4 +1 2

(1, 0) x

0

(4e–2, 0

x=0 8. a.

x

0 (–1.5, 0)

0) x

10. a. f (x) = 2 loge (3 (x + 1)), domain = (−1, ∞) and

y = loge (3x) (1, loge(3))

x=0

)

1 4, – loge (1) + 1 2

x=0

y

0

(

( –13 , 0)

x

range = R 1 x f−1 (x) = e 2 − 1, domain = R and range = (−1, ∞) 3 b. f (x) = loge (2 (x − 1)) + 2, domain = (1, ∞) and range = R 1 f−1 (x) = ex−2 + 1, domain = R and range = (1, ∞) 2

44 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

c. f (x) = 2 loge (1 − x) − 2, domain = (−∞, 1) and

range = R

11. a.

f−1 (x) = 1 −

domain = R and range = (−∞, 1)

1 e 2 (x+2) ,

y=x y = 2 loge(3x + 3)

12. a. a = −1, b = b. −0.4055

13. a. a =

2

loge (2)

1 1 − , 2 2

)

[1] − [2]:

(6.12, 6.12)

x

(– 2–3, 0)

(–0.77, –0.77) x = –1

y y=x

= m log2 −

y = loge (2(x – 1) + 2)

(0,

1 –2 –e 2

[1]

2

[2]

) ( n = m log2 (−2n) − m log2 − 2 2 ( ( )) 5 n = m log2 (−2n) − log2 2 2 ( ( )) −2n = m log2 n

5 4

into [1]: 3=

12 5

(1.23, 1.23)

n

)

1

Substitute m =

b.

(3.68, 3.68)

1

2

(

2

= m log2 22 = 2m 5 m= 4

y = –1

(2 loge (3), 0)

, h = −1, k = −2

= m log2 (4)

x

0

(0 , – 2–3)

⇒

3−

1 e–2 – 1 y=– 3

2

14. (−2, 3) ⇒ 3 = m log2 (−2n)

(

(0, 2 loge (3))

1

5 4

log2 (−2n)

= log2 (−2n)

2 5 = −2n 12

y = 1–2 e x – 2 + 1

n = 2 5 ÷ −2 12

)

+1

y=1

0

(

1– –2 e 2

x

)

+ 1, 0

x=1

= −2 5 7

15. a. 0.159 or 3.146 16. a. x ∈ (0.138, 1.564)

Exercise 1.7 Applications y

c.

1. 54 years 2. a. 7.32%

y=x

y = 2 loge(1 – x) – 2

b. 33 years 5 months

3. 8.46 pm, so the person died 1

(1 – e, 0) y=1

x

0 (–2, 0)

(0, 1 – e) (–0.81, –0.81)

(0, –2)

1 – (x + 2)

y = 1 – e2

x=1

b. 1.2315 [ ] b. x ∈ 0.136, 1.315

4. a. 2 parts per million b. 2.96 parts per million c. 47.2 hours 5. $1648.72 6. 0.1793 7. 0.0003 8. a. 88.3 mg 9. a. 88.57% lost 10. a. a = 12.5, b = 100 11. a. a = −2885.4, c = 10 000 b. 4000 c. 22.6 weeks 12. a. A = 1.440, k = 0.536 b. 3.00 M c. 32 minutes 14 seconds

3 4

hours after the phone call.

b. 5590 years b. 1.87 millennia b. 7.253 g

CHAPTER 1 The logarithmic function 2 45

13. a. 11.3863 14. 12 979 years

b. 13.5835

15. a. P(x) = 800 loge b. 16. a. b. c. d.

(

2+

x

)

250

750 k = 10 000, m = 0.025 $15 685.58 P = 13 750e0.025t − 10 000 $15 054.13

c. y = 2 loge (−x)

c. 10.18 weeks

− 300 − 2x

y

y = 2loge(–x)

(–1, 0)

x

0

1.8 Review: exam practice 1. 2. 3. 4. 5.

D C D C A

x=0

6. a. No solution

7. a. y =

b. y =

x2 8 1

b. 4

c.

where x > 0

1 64

Domain = (−∞, 0), range = R −3

9. a. 10 W/m2 10. −0.849 11. 7.8

or 2

12. a. S b. R c. Q d. P 13. Sample responses can be found in the worked solutions in

provided that x > 0 x3 8. a. y = loge (x − 1) + 3

the online resources. pH = 2 (acidic) pH = 11 (basic) 0.001 moles/litre 10−14 moles/litre ( ) 1 3 a = 3, b = and m = 3 loge 2 2 a. A = 4.7 and n = 2 b. 230.3 m a. h = −2 b. k = − loge (2) ( ) x+2 c. g (x) = loge 2 a. 117.054 mg b. 5590 years 0.000 124t c. i. n = e ii. 18 569 years a. a = 3265, b = 150 b. 9587 quokkas c. i. PR = 2448.75 loge (t) + 112.5 ii. 7 190 quokkas (x−1) −1 a. f (x) = e − 5, domain = R ) ( ) ( 1 −1 b. f loge (5) + 1, 0 , 0, − 5 e

14. a. i. ii. b. i. ii.

y

15. 16.

y = loge(x – 1) + 3 17.

(

b. 60 dB

1 – + 1, 0 e3

)

18.

x

0

19.

Domain = (1, ∞), range = R x=1

b. y = loge (x + 3) − 1

y

20.

y = loge (x + 3) – 1

y

(0, loge(5) + 1) y = f (x)

(0, loge (3) – 1) (e – 3, 0)

0

x

(–1e – 5, 0)

0

y=x (3.091, 3.091) (loge(5) + 1, 0) x y = f –1(x)

Domain = (−3, ∞), range = R

(–4.998, –4.998)

(0, –1e – 5)

x = –3

y = –5 c. (−4.998, −4.998) and (3.091, 3.091)

46 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x = –5

REVISION UNIT 3 Further calculus

TOPIC 1 The logarithmic function 2 • For revision of this entire topic, go to your studyON title in your bookshelf at www.jacplus.com.au. • Select Continue Studying to access hundreds of revision questions across your entire course.

• Select your course Mathematical Methods for Queensland Units 3&4 to see the entire course divided into syllabus topics. • Select the Area you are studying to navigate into the chapter level OR select Practice to answer all practice questions available for each area.

• Select Practice at the sequence level to access all questions in the sequence.

OR

• At sequence level, drill down to concept level.

• Summary screens provide revision and consolidation of key concepts. Select the next arrow to revise all concepts in the sequence and practise questions at the concept level.

REVISION UNIT 3 Further calculus 47

2

Calculus of exponential functions

2.1 Overview Rates of change and the concept of a limit are fundamental ideas in the study of calculus. Many processes in nature can be modelled by exponential functions. These include the growth in a population of bacteria and the decay of radioactive material. Other areas modelled by exponential functions include the value of an investment after a period of time and the temperature of a liquid after it has been cooling. In 1683, Jacob Bernoulli, a famous Swiss mathematician, was studying a problem relating to compound interest. What were the effects on the investment if smaller and smaller compounding intervals were used? In fact, he was attempting to find the ( )n 1 value of lim 1 + and discovered that the limit had to lie between 2 and 3. Other mathematicians in n→∞ n the 17th century were studying similar ideas and came close to finding the limit. Leonhard Euler, another famous Swiss mathematician, had studied under Bernoulli. While he was endeavouring to solve the problem of limits and investments proposed by Bernoulli, Euler discovered the constant e. He subsequently found many uses of the constant e and, in the mid 18th century published a paper showing all of his findings, including e to 18 decimal places. An approximation for e is 2.718 281 828 459 . . . Like 𝜋, Euler’s number, e, is irrational. Euler studied many areas of science, including mechanics, fluid dynamics, astronomy and physics. In mathematics, the influence of Euler is found in geometry, trigonometry, calculus and algebra. He also studied and wrote on the theory of music.

LEARNING SEQUENCE 2.1 2.2 2.3 2.4 2.5 2.6

Overview Review of limits and differentiation The exponential function Differentiation of exponential functions Applications of exponential functions Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

48 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2.2 Review of limits and differentiation 2.2.1 Limits The limit of a function, y = f (x), is the value that the function approaches as x approaches a given value. Consider the limit of f (x) = x + 1 as x approaches 1, using a spreadsheet. x

x+1

x

x+1

0.9

1.9

1.1

2.1

0.99

1.99

1.01

2.01

0.999

1.999

1.001

2.001

0.9999

1.9999

1.0001

2.0001

As x approaches 1 from the left-hand side, or below, the function approaches 2. As x approaches 1 from the right-hand side, or above, the function approaches 2. As both are equal, the limit exists and is written as: lim (x + 1) = 2 x→1

Since the function f(x) = x + 1 is continuous, the limit can be found by direct substitution. x2 − 1 y Consider a different function, g(x) = . 3 x−1 This function is undefined at x = 1. (x − 1) (x + 1) 2 However, g(x) can be simplified to g(x) = . (x − 1) 1

g(x) = (x + 1), x ≠ 1 –3

–2

–1

As shown above, since lim (x + 1) = 2, then x→1

0

1

2

3

x

–1

x2 − 1 = 2 or lim g(x) = 2. lim x→1 x − 1 x→1 The graph of y = g(x) is in fact a linear function with a point discontinuity at (1, 2), as shown.

–2 –3

WORKED EXAMPLE 1

Evaluate the following limits: a.

lim (3h − 5)

h→4

b.

x2 + 5x + 6 x→0 x+2 lim

THINK a. 1.

2.

Substitute h = 4 as the function is defined.

Answer the question.

c.

x2 + 5x + 6 x→−2 x+2 lim

WRITE a.

lim(3h − 5)

h→4

=3×4−5 =7 lim(3h − 5) = 7

h→4

CHAPTER 2 Calculus of exponential functions 49

b. 1.

2. c. 1.

Substitute x = 0 as the function exists for this value.

Answer the question. The function is undefined at x = −2. Factorise the numerator and simplify.

x2 + 5x + 6 x→0 x+2 6 = 2 =3 x2 + 5x + 6 lim =3 x→0 x+2 x2 + 5x + 6 c. lim x→−2 x+2 (x + 2) (x + 3) = lim x→−2 (x + 2) = lim (x + 3) b.

lim

x→−2

2.

Substitute x = −2.

3.

Answer the question.

= −2 + 3 =1 x2 + 5x + 6 lim =1 x→−2 x+2

2.2.2 The derivative as a limit In Year 11, you were introduced to differentiation, the process of finding the rate of change of a function at any point. Differentiation from first principles involves finding a limit as h approaches 0. For the function y = f (x): f (x + h) − f (x) dy = lim dx h→0 h

WORKED EXAMPLE 2

Calculate the derivative of f (x) = 3x2 − 4x + 7 from first principles. THINK 1.

State the function

2.

The derivative is equal to f (x + h) − f (x) lim . h→0 h

3.

Substitute for f (x).

4.

Expand and simplify.

WRITE

f (x) = 3x2 − 4x + 7 f(x + h) − f(x) f ′ (x) = lim h→0 h [3(x + h) 2 − 4(x + h) + 7] − [3x2 − 4x + 7] h→0 h 3x2 + 6xh + 3h2 − 4x − 4h + 7 − 3x2 + 4x − 7 = lim h→0 h 2 6xh + 3h − 4h = lim h→0 h

f ′ (x) = lim

50 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

5.

h(6x + 3h − 4) h = lim(6x + 3h − 4) = lim

Factorise and simplify.

h→0

h→0

Evaluate the limit as h approaches 0. 7. Answer the question. 6.

= 6x + 3 × 0 − 4 = 6x − 4 ′ f (x) = 6x − 4

2.2.3 Estimating a limit Technology, such as a spreadsheet, can be used to estimate the limit of a given expression. ah − 1 Consider the limit of as h → 0 for various values of a > 0. h For a = 2: (2h − 1) h

h 1

1

0.5

0.828 427 12

0.1

0.717 734 63

0.01

0.695 555 01

0.001

0.693 387 46

0.000 1

0.693 171 2

0.000 01

0.693 149 58

The limit is approaching 0.6931. For a = 3: (3h − 1) h

h 1

2

0.5

1.464 101 615

0.1

1.161 231 74

0.01

1.104 669 194

0.001

1.099 215 984

0.000 1

1.098 672 638

0.000 01

1.098 618 323

The limit is approaching 1.0986.

CHAPTER 2 Calculus of exponential functions 51

ah − 1 has a limiting value of 1 as h → 0? h The value of a would lie in the interval 2 < a < 3.

Can we find a value of a where the fraction

If

ah − 1 = 1, h ≠ 0, then ah − 1 = h ⇒ ah = 1 + h h 1

a = (1 + h) h Consider the value of a as h → 0. (1 +

h

( ) 1 h) h

1

2

0.5

2.25

0.1

2.593 742 46

0.01

2.704 813 829

0.001

2.716 923 932

0.000 1

2.718 145 927

0.000 01

2.718 268 237

0.000 001

2.718 280 469

0.000 000 1

2.718 281 694

0.000 000 01

2.718 281 786

0.000 000 001

2.718 282 052

1E − 10

2.718 282 053

The fraction is approaching 2.718 282 05 … as h → 0. Euler’s number, e ≈ 2.718 28, is the value of a that gives the limit of the fraction to be 1. Like 𝜋, e is an irrational number. Scientific and graphics calculators have an ex function that is treated in the same way as any other function. An answer given in terms of e is an exact answer.

Units 3 & 4

Area 2

Sequence 1

Concept 1

Review of limits Summary screen and practice questions

52 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Exercise 2.2 Review of limits and differentiation Technology free 1.

Evaluate the following limits. lim(5h + 4) b. lim (4 − 6h)

WE1a

a.

h→3

c.

h→−2

Evaluate the following limits. 2x2 + 7x + 3 x2 + 4x a. lim b. lim x→0 x→2 x + 2 x−1 3. WE1c Evaluate the following limits. h2 − h − 12 h2 + 4h a. lim b. lim h→−3 h→0 h+3 h 4. Evaluate the following limits. 2.

WE1b

6. 7.

8.

c.

x2 + 4x x→−3 x + 1

c.

lim

lim

x2 − x − 6 x→3 3−x

3x2 h + 4h2 h→0 h→0 h WE2 For the function f (x) = x2 − 6x, calculate the derivative from first principles. Use first principles to differentiate the function f (x) = 5 + 3x − 2x2 . ah − 1 Estimate to 5 decimal places, using technology, the limit of as h → 0, where: h b. a = 2.6 c. a = 2.7 a. a = 2.5 d. a = 2.8 e. a = 2.9 f. a = 2.718 28 Evaluate the following, giving your answers to 4 decimal places. a.

5.

lim(6h2 − 3h + 2)

h→0

lim(4x2 + 5xh − h2 )

b.

lim

e2 b. e3 9. Evaluate the following, giving your answers to 3 decimal places. √ 3 a. 2e−1 b. e a.

1

c.

e2

c.

4+e e

Use first principles to differentiate the function y = 8x − x2 . b. Calculate the gradient of the tangent to the curve y = 8x − x2 at the point where x = 2. c. Hence, determine the equation of the tangent to the curve at x = 2. Use differentiation by first principles to determine the gradient of the curve y = x3 − 3x2 at any point and hence the equation of the tangent at the point where the curve crosses the positive x-axis. Consider the function f (x) = x3 − 4x. a. Determine the x-intercepts of this function. b. Use first principles to differentiate the function. c. Calculate the gradient of the tangents at the points where the function crosses the x-axis. d. Deduce that two of the tangents are parallel. 1 1 a. Simplify the fraction − . x+h x 1 b. Hence, use first principles to determine the derivative of the function f (x) = , x ≠ 0. x 1 −h 1 a. Show that − = . (x + h − 2) (x − 2) (x − 2)(x + h − 2) −1 b. Evaluate lim . h→0 (x − 2)(x + h − 2) 1 c. Hence, use first principles to determine the gradient function for y = , x ≠ 2. (x − 2) d. Determine the x-value(s) on the curve where the tangent is parallel to 9x + y − 7 = 0.

10. a.

11. 12.

13.

14.

CHAPTER 2 Calculus of exponential functions 53

2.3 The exponential function 2.3.1 Review of exponential functions, f(x) = ax where a ∈ R+ \ {1} The graph of f (x) = ax , a > 1 has the following features: • The y-intercept is (0, 1). ( ) 1 • The key points are (1, a) and −1, . a • The maximal domain is x ∈ R. • The range is y ∈ R+ or y > 0. • As x → ∞, f (x) → ∞. • As x → −∞, f (x) → 0. • The horizontal asymptote is y = 0 (the x-axis). • It is a one-to-one function. The graph of f (x) = a−x , a > 1 is a reflection of f (x) = ax for a > 1 over the y-axis. The graph of f (x) = a−x , a > 1 has the following features: • The y-intercept is (0, 1). ) ( 1 • The key points are (−1, a) and 1, . a The maximal domain is x ∈ R. • • The range is y ∈ R+ or y > 0. • As x → ∞, f (x) → 0. • As x → −∞, f (x) → ∞. • The horizontal asymptote is y = 0 (the x-axis). • It is a one-to-one function.

y

(1, a)

(–1, 1a )

2.

3.

To sketch the graph of f (x) = 2x , first determine the y-intercept, which occurs when x = 0. To help determine the shape of the curve, it is useful to know any two other points on the graph. For example, determine the coordinates of the points at x = ±1.

y = a–x, a > 1

State the equation of the horizontal asymptote.

f (x) = 2x f (0) = 20 f (0) = 1 y-intercept (0, 1) f (1) = 21 and f (−1) = 2−1 1 f (−1) = 2 ( ) 1 (1, 2) and −1, 2 y=0 f (1) = 2

54 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

(–1, a)

WRITE a.

y=0

y

Sketch the following exponential functions, showing all important features. a. f (x) = 2x b. f (x) = 2−x c. f (x) = −2x

a. 1.

(0, 1)

0

WORKED EXAMPLE 3

THINK

y = ax, a > 1

(0, 1)

0

(1, 1a ) y=0 x

4.

Sketch the graph.

y f(x) = 2x (1, 2)

(–1, 12 )

(0, 1) y=0

0 b. 1.

2.

To sketch the graph of f (x) = 2−x , first determine the y-intercept, which occurs when x = 0.

b.

To help determine the shape of the curve, it is useful to know any two other points on the graph. For example, determine the coordinates of the points at x = ±1.

f (x) = 2−x f (0) = 2−0 f (0) = 1 y-intercept (0, 1) f (1) = 2−1 and f (−1) = 2−1 1 f (−1) = 2 f (1) = 2 ( ) 1 1, and (−1, 2) 2 y=0

State the equation of the horizontal asymptote. 4. Sketch the graph. Note: This is a reflection of the curve in part a over the y-axis. 3.

y f(x) =

2–x (–1, 2) (0, 1)

(1, 12) y=0

0 c. 1.

2.

To sketch the graph of f (x) = −2x , first determine the y-intercept, which occurs when x = 0. To help determine the shape of the curve, it is useful to know any two other points on the graph. For example, determine the coordinates of the points at x = ±1.

State the equation of the horizontal asymptote. 4. Sketch the graph. Note: This is a reflection of the curve in part a over the x-axis. 3.

x

c.

x

f(x) = −2x f (0) = −20 f (0) = −1 y-intercept (0, −1) f (1) = −21 and f (−1) = −2−1 1 f (−1) = − 2 ( ) 1 (1, −2) and −1, − 2 y=0 f (1) = −2

y 0

(

–1, – 1 2

)

(0, –1)

y=0 x

(1, –2) f(x) = –2x

CHAPTER 2 Calculus of exponential functions 55

2.3.2 The exponential function, f(x) = ex It has been established that e ≈ 2.718 28, giving 2 < e < 3. Therefore, the graph of the exponential function, y = ex , lies between y = 2x and y = 3x . y The graph of f (x) = ex has the following features: 4 y = 3x • The y-intercept is (0, 1). ( ) 1 3 (1, 3) • The key points are (1, e) and −1, . e (1, 2.718) 2 • The maximal domain is x ∈ R. (1, 2) + x • The range is y > 0 or y ∈ R . y=e y = 2x 1 (0, 1) • As x → ∞, f (x) → ∞. y=0 • As x → −∞, f (x) → 0. x • The horizontal asymptote is y = 0. 0 –1 –2 1 2 • It is a one-to-one function. Three exponential functions, y = 2x , y = ex and y = 3x , are shown. Graphs of f (x) = aenx + b where a, n, b ∈ R can be sketched using your knowledge of transformations.

Sketching y = ex and its transformations The graph of y = kex is a dilation of y = ex by a factor of k from the x-axis (or perpendicular to the y-axis). 1 The graph of y = enx is a dilation of y = ex by a factor of from the y-axis (or parallel to the x-axis). n The graph of y = −ex is a reflection of y = ex in the x-axis. The graph of y = e−x is a reflection of y = ex in the y-axis. The graph of y = ex + k is a translation of y = ex by k units vertically (or parallel to the y-axis). The graph of y = e(x−h) is a translation of y = ex by h units horizontally (or parallel to the x-axis), giving the vertical asymptote x = h. WORKED EXAMPLE 4

Sketch the function f (x) = 2ex , showing all important features. b. State the transformation required to map f (x) = ex onto f (x) = 2ex . a.

THINK a. 1.

2.

3.

WRITE x

To sketch the graph of f (x) = 2e , first determine the y-intercept, which occurs when x = 0.

To help determine the shape of the curve, it is useful to know any two other points on the graph. For example, determine the coordinates of the points at x = ±1.

State the equation of the horizontal asymptote.

a.

f (x) = 2ex f (0) = 2e0 f (0) = 2 × 1 f (0) = 2 y-intercept (0, 2) f (1) = 2e1 and f (−1) = 2e−1 f (1) = 2e

f (−1) =

(1, 2e) ≈ (1, 5.44) ( ) 2 −1, ≈ (−1, 0.74) e y=0

56 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2 e

4.

y

Sketch the function.

(1, 2e) f(x) = 2 ex (0, 2)

(–1, 2e ) 0 b.

State the transformation.

b.

y=0

x

The transformation required to map f (x) = ex onto f (x) = 2ex is a dilation by a factor of 2 from the x-axis.

WORKED EXAMPLE 5

Sketch the function f (x) = 2 + e−x , showing all important features. b. State the transformations required to map f (x) = ex onto f (x) = 2 + e−x . a.

THINK a. 1.

WRITE −x

To sketch the graph of f (x) = 2 + e , first determine the y-intercept, which occurs when x = 0.

a.

To help determine the shape of the curve, it is useful to know any two other points on the graph. For example, determine the coordinates of the points at x = ±1. 3. State the equation of the horizontal asymptote. 4. Sketch the function. 2.

f (x) = 2 + e−x f (0) = 2 + e−0 f (0) = 2 + 1 f (0) = 3 y-intercept (0, 3) f (1) = 2 + e−1 and f (−1) = 2 + e−1 f (−1) = 2 + e (1, 2 + e−1 ) ≈ (1, 2.37) (−1, 2 + e) ≈ (−1, 4.72) y=2 y (–1, 2 + e) f(x) = 2 + e–x (0, 3)

(1, 2 + e–1)

y=2 0 b.

State the transformations.

b.

x

The transformations required to map f (x) = ex onto f (x) = 2 + e−x are a reflection in the y-axis and a vertical translation upwards by 2 units.

CHAPTER 2 Calculus of exponential functions 57

2.3.3 Indicial equations with e Determining the x-intercepts of exponential functions may involve solving equations with e. • The laws of indices apply in the same way if e is the base. • Equations involving e are solved using the same methods as any equation involving indices. • Solving may require the use of logarithms with base e. • The laws of logarithms apply, using the notation loge (x) or ln (x) as found on your calculator. • Since ex > 0, not all equations have real solutions. For example, ex = −1 has no real solution. WORKED EXAMPLE 6

Consider the function f (x) = 2 − e−x . a. Determine the coordinates of any axis intercepts. b. Sketch the function f (x) = 2 − e−x , showing all important features. c. State the transformations required to map f (x) = ex onto f (x) = 2 − e−x . THINK

For the y-intercept, x = 0. • Substitute x = 0. • Evaluate. 2. For the x-intercept, y = 0. • Substitute y = 0. • Rearrange the equation. • Take the log (base e) of both sides. • Use log laws to simplify.

a. 1.

3.

b. 1.

2.

3. 4.

WRITE a.

f (x) = 2 − e−x f (0) = 2 − e0 f (0) = 2 − 1 = 1 2 − e−x = 0 2 = e−x loge 2 = loge e−x −x loge e = loge 2 x = − loge 2

Axis intercepts: (− loge 2, 0) and (0, 1) Approximately: (−0.693, 0) and (0, 1) −x To sketch the graph of f (x) = 2 − e , use b. Axis intercepts: the x- and y-intercepts found in part a. (− loge 2, 0) and (0, 1) Approximately: (−0.693, 0) and (0, 1) To help determine the shape of the curve, it is f (1) = 2 − e−1 and f (−1) = 2 − e−1 useful to know any two other points on the f (−1) = 2 − e −1 graph. For example, determine the (1, 2 − e ) ≈ (1, 1.632) coordinates of the points at x = ±1. (−1, 2 − e) ≈ (−1, −0.718) State the equation of the horizontal y=2 asymptote. Sketch the function. y State the coordinates of the axis intercepts. (Hint: An approximation may be useful.)

y=2 f(x) = 2–ex

(1, 2 – e) (0, 1)

(–0, 0.693) (1, 2 – e–1)

58 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

0

x

c.

State the transformations.

c.

The transformations required to map f (x) = ex onto f (x) = 2 − e−x are reflections in both the x- and y-axes and a vertical translation upwards by 2 units.

WORKED EXAMPLE 7

Solve 6ex = 15 + ex for x, giving your answer: a. as an exact number b. correct to 3 decimal places. THINK a. 1. 2. 3. 4. 5. b. 1. 2.

WRITE

Write the equation. a. 6ex = 15 + ex Collect like terms. 5ex = 15 Find ex . ex = 3 Take the log of both sides. (Note: loge (x) = ln(x)) ln(ex ) = ln(3) State the solution as an exact number. x = ln(3) Use your calculator to determine the approximation. b. x = 1.098 61 State the solution correct to 3 decimal places. x = 1.099

WORKED EXAMPLE 8

Solve ex − 3e−x = 2 for x, giving your answer(s) correct to 2 decimal places. THINK

WRITE

1.

Write the equation.

2.

Rewrite without negative indices.

3.

Multiply each term by ex . Recognise a quadratic equation in ex . Let a = ex . Factorise the quadratic expression. Solve for a. Substitute ex for a. Solve for x.

4. 5. 6. 7. 8. 9. 10.

State the solution correct to 2 decimal places.

ex − 3e−x = 2 3 ex − x = 2 e (ex ) 2 − 3 = 2ex (ex ) 2 − 2ex − 3 = 0 a2 − 2a − 3 = 0 (a − 3)(a + 1) = 0 a = 3 or a = −1 ex = 3 or ex = −1 ln(ex ) = ln(3), ex ≠ −1 x = ln(3) x = 1.10 (to 2 decimal places)

Resources Interactivity: Exponential functions (int-5959)

CHAPTER 2 Calculus of exponential functions 59

Units 3 & 4

Area 2

Sequence 1

Concept 2

The exponential function Summary screen and practice questions

Exercise 2.3 The exponential function Technology free 1. 2. 3. 4. 5. 6. 7.

8.

Sketch the following exponential functions, showing all important features. a. f (x) = 4x b. f (x) = 4−x c. f (x) = −4x x −x On the same set of axes, sketch the graphs of y = 10 and y = 10 . WE4 a. Sketch the function f (x) = 4ex , showing all important features. b. State the transformations required to map f (x) = ex onto f (x) = 4ex . a. Sketch the function f (x) = −5ex , showing all important features. b. State the transformations required to map f (x) = ex onto f (x) = −5ex . WE5 a. Sketch the function f (x) = e−x + 3, showing all important features. b. State the transformations required to map f (x) = ex onto f (x) = e−x + 3. a. Sketch the function f (x) = e2x + 3, showing all important features. b. State the transformations required to map f (x) = ex onto f (x) = e2x + 3. WE6 Consider the function f (x) = e2x − 3. a. Determine the coordinates of any axis intercepts for this function. b. Sketch the function f (x) = e2x − 3, showing all important features. c. State the transformations required to map f(x) = ex onto f (x) = e2x − 3. Consider the function f (x) = 4 − 2e−x . a. Determine the coordinates of any axis intercepts for this function. b. Sketch the function f (x) = 4 − 2e−x , showing all important features. c. State the transformations required to map f (x) = ex onto f (x) = 4 − 2e−x . WE3

x

9. a.

Sketch the function f (x) = 4e 2 , showing all important features. x

b.

State the transformations required to map f (x) = ex onto f (x) = 4e 2 .

Technology active x − 10. a. Determine the coordinates of any axis intercepts for the function y = 3e 2 − 6. b. Sketch the function, showing all important features. x

State the transformation required to map y = ex onto y = 3e− 2 − 6. WE7 Solve 3ex + 8 = 5ex for x, giving your answer: a. as an exact number b. correct to 3 decimal places. Solve for x, giving your answer correct to 3 decimal places. 1 a. ex = 5 b. ex = c. ex = 2.6 2 d. e−x = 6 e. 3 = 2ex f. 3e−x − 10 = 0 Solve for x in each of the following, giving your answer in exact form. a. (ex − 1)(ex − 2) = 0 b. (ex − 1)(ex + 3) = 0 −x 2x c. (e − 1)(e − 4) = 0 d. (3e−x − 2)(2ex − 1) = 0 e. (2ex + 1)(ex − 4) = 0 f. (3ex − 2)(ex + 4) = 0 x −x WE8 Solve e − 15e = 2 for x, giving your answer(s) correct to 2 decimal places. Solve for x in each of the following, giving your answers in exact form. a. 5ex − 12e−x − 11 = 0 b. 3ex + 6e−x = 11 x −x c. 2e = 9 + 5e d. ex = 25e−x c.

11.

12.

13.

14. 15.

60 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Solve for x in each of the following, giving your answers in exact form. a. ex > 1 b. e−x < e c. e2x ≥ 4 17. a. Sketch the curve y = 2 e−x + 1. b. For what values of x is y < 3? c. Discuss why 2 e−x + 1 < 0 has no real solutions. 18. a. Sketch the curve f(x) = 4 − ex , stating all axis intercepts in exact form. b. For what values of x is y > 0? c. Discuss the range of the function if the domain is x ≥ 0. 16.

d.

e1−x ≤ 6

2.4 Differentiation of exponential functions The derivative of the exponential function can be found using first principles. If f (x) = ex , then f(x + h) − f(x) h→0 h x+h e − ex = lim h→0 h x h e e − ex = lim h→0 h ex (eh − 1) = lim h→0 h h e −1 = ex lim h→0 h

f ′ (x) = lim

eh − 1 ah − 1 = 1 for a = e; that is, lim = 1. h→0 h→0 h h eh − 1 Substituting into f ′ (x) = ex lim : h→0 h

In subtopic 2.2, it was shown that lim

f ′ (x) = ex The derivative of y = ef (x) can be found using the chain rule, studied in Year 11. If y = ef (x) , let u = f (x) . Then y = eu . dy du = eu and = f ′ (x) du dx The chain rule states: Substitute: Replace u as f (x):

dy dy du = × dx du dx dy = eu × f ′ (x) dx dy = f ′ (x)e f(x) dx

CHAPTER 2 Calculus of exponential functions 61

The differential of an exponential d(ex ) = ex dx And with the chain rule: d(e f(x) ) = f ′ (x) e f(x) dx or u d(e ) du = eu × where u = f(x) dx dx

WORKED EXAMPLE 9

Using the chain rule, differentiate y = e−5x with respect to x. THINK

WRITE

Write the equation. 2. Substitute u = −5x. dy du and . 3. Determine du dx

y = e−5x y = eu and u = −5x dy du = eu and = −5 du dx dy dy du = × dx du dx dy = eu × (−5) dx dy = −5e−5x dx y = e−5x dy = −5ex dx

1.

dy . dx

4.

Use the chain rule to find

5.

State the derivative in terms of x. Alternatively, recognise and apply the formula d(e f (x) ) = f ′ (x)e f(x) where f (x) = −5x. dx

WORKED EXAMPLE 10

Determine the derivative of y = e2x+1 . THINK 1.

Write the equation.

2.

Recognise and apply the formula d(e f (x) ) = f ′ (x)e f (x) where f (x) = 2x + 1. dx

WRITE

y = e2x+1 dy = 2e2x+1 dx

62 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 11

Differentiate the following. a.

f (x) = ex (ex − 3)

b.

THINK

e3x − 2e−x ex

WRITE

f (x) = ex (ex − 3) f (x) = e2x − 3ex f ′ (x) = 2e2x − 3ex e3x − 2e−x b. f (x) = ex e3x 2e−x f (x) = x − x e e 2x −2x f (x) = e − 2e f ′ (x) = 2e2x + 4x−2x 4 f ′ (x) = 2e2x + 2x e

Write the equation. 2. Write the equation in expanded form. 3. Differentiate each term.

a. 1.

b. 1.

f (x) =

a.

Write the equation.

2.

Separate the terms and simplify.

3.

Differentiate.

WORKED EXAMPLE 12 (

Determine the derivative of the function y = e

3

)

x −x

.

THINK

WRITE

Write the equation. 2. Substitute u = x3 − x. dy du 3. Determine and . du dx

y = e(x −x) y = eu and u = x3 − x dy du = eu and = 3x2 − 1 du dx dy dy du = × dx du dx ( ) dy = eu × 3x2 − 1 dx 3 dy = (3x2 − 1) e(x −x) dx

3

1.

dy . dx

4.

Use the chain rule to find

5.

State the derivative in terms of x.

Alternatively: 3

1.

Write the equation

2.

Recognise and apply the formula d(e f (x) ) = f ′ (x)e f (x) where f (x) = x3 − x. dx

Units 3 & 1

Area 2

Sequence 1

y = e(x −x) 3 dy = (3x2 − 1) e(x −x) dx

Concept 3

Differentiation of exponential functions Summary screen and practice questions

CHAPTER 2 Calculus of exponential functions 63

Exercise 2.4 Differentiation of exponential functions Technology free 1.

WE9

Differentiate the following.

y = e10x d. y = e−x 2. WE10 Differentiate the following. a. y = e6x−2 d. y = 4e7−2x 3. a. y = 10e6−9x a.

1

x

b.

y = e3x e. y = 2e3x

c.

y = e8−6x y = −3e8x+1 b. y = −5e3x+4

y = 2e5x+3 y = −2e6−5x c. y = 6e−7x

y = e4 f. y = 4e−5x

b. e.

c. f.

x

x

x

2− y = 2e 2 + 1 e. y = 3e 3 3x+2 MC The derivative of y = e is equal to: 3x+2 A. 3e B. (3x + 2)e3x+2 WE11 Differentiate the following. a. f (x) = 2(ex + 1) c. f (x) = 5(e−4x + 2x) WE11 Differentiate the following. 3e3x + e−6x a. f (x) = ex WE12 Determine the derivatives of the following.

d.

4. 5.

6.

7.

a.

2

y = ex +3x

b.

f.

y = −4e 4 +5

C. 3e3x

D. 3xe3x+2

f (x) = 3e2x (ex + 1) d. f (x) = (ex + 2)(e−x + 3) b.

2

y = ex −3x+1

4e7x − 2e−x e−2x

b.

f (x) =

c.

y = ex −2x

2

d.

f(x) = e2−5x

d.

y = −5e1−2x−3x

f (x)

8.

9.

d(e ) = f ′ (x) e f (x) to differentiate the following functions. dx 2 3 2 a. f (x) = e6−3x+x b. g(x) = ex +3x−2 c. h(x) = 3e4x −7x

Use the formula

2

3

MC

The derivative of 6ex −5x is equal to: 3

B. (3x2 − 5)ex −5x

3

3

D. 6(3x2 − 5)e3x −5

A. 6(3x2 − 5)ex −5x

2

C. 6(x3 − 5x)ex −5x 10.

If f (x) = 5e9−4x , determine the exact value of f ′ (2).

11.

If g(x) = 2ex −3x+2 , determine the exact value of g′ (0).

12.

Calculate the exact value of h′ (−1) if h(x) = −5ex +3x .

2

2

2

Determine the equation of the tangent to the curve y = ex +3x−4 at the point where x = 1. 14. Determine the equations of the tangent and the line perpendicular to the curve y = e−3x − 2 at the point where x = 0. 15. Determine the derivative of the function f (x) = e−2x+3 − 4e and hence find: a. f ′ (−2) in exact form b. {x : f ′ (x) = −2} e3x + 2 and hence find: 16. Determine the derivative of the function f (x) = ex a. f ′ (1) in exact form b. {x : f ′ (x) = 0}

13.

2.5 Applications of exponential functions Functions involving the exponential function y = ex can be used to model many real-life situations. These include: • population growth and decay, for example bacteria • radioactive decay • growth of investments • cooling of heated objects. 64 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

In modelling that involves exponential growth and decay, the rate of change of the function often implies dy the change with respect to time, . dt It may be necessary to restrict the domain of the exponential function to suit the context of the problem. For example, time (t) cannot be negative. WORKED EXAMPLE 13

The number of bacteria on a culture plate, N, can be defined by the rule N(t) = 2000e0.3t , t ≥ 0 where t is the time, in seconds, the culture has been growing. a. How many bacteria were initially present? b. How many bacteria, to the nearest whole number, are present after 10 seconds? c. At what rate is the bacteria population increasing after 10 seconds? Give your answer correct to the nearest whole number. THINK a. 1.

2. b. 1.

2.

c. 1.

WRITE

The initial time is when t = 0. Substitute a. N(t) = 2000e0.3t t = 0 and evaluate. N(0) = 2000e0.3(0) = 2000 Write the answer. Initially there were 2000 bacteria present. After 10 seconds, t = 10. Substitute b. N(t) = 2000e0.3t t = 10. N(10) = 2000e0.3(10) = 2000e3 = 40 171.074 Write the answer. After 10 seconds there were 40 171 bacteria present. ( ) d(e f(x) ) Differentiate = f ′ (x) e f(x) to c. N(t) = 2000e0.3t dx find rate of change with respect to time. dN = 2000 × 0.3e0.3t dt dN = 600e0.3t dt

CHAPTER 2 Calculus of exponential functions 65

2.

After 10 seconds, t = 10. Substitute t = 10 and evaluate.

3.

Write the answer with correct units.

TI | THINK a.1. On a Calculator page,

press MENU, then select: 2: Add Graphs. Complete the entry line in the f1(x) = tab as: 2000e0.3x 2. Sketch the graph by

pressing the ENTER button.

3. To calculate the initial

value, select: Menu 8: Geometry 1: Points & Lines 2: Point On.

4. Move the cursor and

select the curve representing f1 (x). Press the ESC (escape) button. This allows you to move the text box indicating the coordinates ( ) of the point P x, y . Complete the entry line in the textbox as 0 for the x-value. 5. The answer appears on the screen. In this case, when x = 0, y = 2000.

WRITE

dN = 600e0.3(10) dt = 600e3 = 12 051.322 After 10 seconds the bacteria is increasing at a rate of 12 051 bacteria/second. CASIO | THINK a.1. On a Main Menu screen,

select: Graph. Complete the entry line in the Y1 tab as: 2000e0.3x Note: The independent variable t has been replaced with x. 2. Sketch the graph by pressing either the DRAW or EXE button.

3. To calculate the initial

value, select: G-Solv (SHIFT F5) Y-CAL. Complete the entry line in X: as 0.

4. The answer appears on

the screen. In this case, when x = 0, y = 2000.

5. To calculate the number

of bacteria after 10 seconds, select: G-Solv (SHIFT F5) Y-CAL. Complete the entry line in X: as 10.

66 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WRITE

b.1. Complete the entry

line in the textbox as 10 for the x-value. The answer appears on the screen. In this case, when x = 10, y = 40 171.074. Note: You can also calculate an x-value by entering a desired y-value. c.1. On a Calculator page, press MENU, then select: 4: Calculus 1: Numerical Derivative at a Point.

2. Complete the value

entry line as: 10 Press the OK button.

3. Complete the entry

line as: ) d ( 2000e0.3x dx Press the ENTER button to complete the calculation.

b.1. The answer appears on

the screen. In this case, when x = 10, y = 40 171.074.

c.1. On a Run Matrix screen,

select: Math

d

dx

.

2. Complete the entry line

as: ) d ( 2000e0.3x dx Enter 10 in the x-value box. Press the EXE button to complete the calculation. 3. The answer appears on

the screen. In this case, when x = 10, dy = 12 051.322. dx

The answer appears on the screen. In this case, when x = 10, dy = 12 051.322. dx

CHAPTER 2 Calculus of exponential functions 67

WORKED EXAMPLE 14

The mass, M grams, of a radioactive substance is initially 20 grams. The substance has been stored for 30 years already. The mass in any year is given by M(t) = M0 e−0.00152t where M0 is a constant and t is the time in years. a. Determine the value of M 0 . b. Calculate the amount of the substance remaining after a further 30 years. Give your answer correct to 2 decimal places. c. Determine the rate of decay at this time. Give your answer correct to 2 decimal places. THINK a. 1.

The mass, M grams, of a radioactive substance is initially 20 grams. Substitute t = 0, M(0) = 20.

Write the answer. b. 1. Rewrite the equation with M0 = 20. 2. The mass, M grams, of a radioactive substance is initially 20 grams. After 30 years (t = 30), a further 30 years is when t = 60. Substitute t = 60 and evaluate. 3. Write the answer. 2.

c. 1.

Differentiate to find the rate of change with respect to time.

Evaluate the rate of change at t = 60 by substitution. 3. Answer the question with correct units. 2.

WRITE a.

M (t) = M0 e−0.00152t M (0) = M0 e−0.00152(0) = 20 M0 e0 = 20

⇒ M0 = 20 M0 = 20 b. M(t) = 20e−0.00152t M(60) = 20e−0.00152(60) = 18.2567

After a further 30 years, the mass is 18.26 grams. c. M(t) = 20e−0.001 52t M′ (t) = 20 × (−0.00152) e−0.00152t M′ (t) = −0.0304e−0.00152t M′ (60) = −0.0304e−0.00152(60) = −0.027 75 The rate of decay after 60 years is 0.03 grams/year. Note: The question asked for the rate of decay, so the negative sign is not included in the final answers. The negative indicates a rate that is decreasing.

68 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 15

The population of foxes on the outskirts of a city is starting to increase. Data collected suggest that a model for the number of foxes is given by N(t) = 480 − 320e−0.3t , t ≥ 0 where N is the number of foxes t years after the observations began. a. How many foxes were present initially at the start of the observations? b. By how many had the population of foxes grown at the end of the first year of observations? c. After how many months does the model predict the number of foxes would double its initial population? d. Sketch the graph of N versus t. e. Explain why this model does not predict that the population of foxes will grow to 600. THINK a. 1.

Calculate the initial number, that is when t = 0.

Answer the question. b. 1. Calculate the number of foxes, N, after 1 year, t = 1. 2.

Express the change over the first year in context. c. 1. Calculate the required value of t. Note: An algebraic method requiring logarithms has been used here. 2.

WRITE a.

N(t) = 480 − 320e−0.3t When t = 0, N(0) = 480 − 320e0

= 480 − 320 = 160 There were 160 foxes present initially. b. When t = 1, N(1) = 480 − 320e−0.3 ≈ 242.94 After the first year 243 foxes were present. Over the first year the population grew from 160 to 243, an increase of 83 foxes. c. Let N = 2 × 160 = 320. 320 = 480 − 320e−0.3t 320e−0.3t = 160 e−0.3t =

2.

Answer the question.

1 2

( ) 1 −0.3t = loge 2 ( ) 1 1 t=− loge 0.3 2 t ≈ 2.31 0.31 × 12 ≈ 4 The population doubles after 2 years and 4 months.

CHAPTER 2 Calculus of exponential functions 69

d.

Sketch the graph.

d.

N(t) = 480 − 320e−0.3t The horizontal asymptote is N = 480. The y-intercept is (0, 160). N 480

(0, 160)

N = 480

N = 480 – 320e–0.3t

0 e.

Give an explanation for the claim that this model will not predict the population of foxes to grow to 600.

Units 3 & 4

Area 2

Sequence 1

e.

t

The presence of an asymptote on the graph shows that as t → ∞, N → 480. Hence, N can never reach 600. The population will never exceed 480 according to this model.

Concept 4

Applications of exponential functions Summary screen and practice questions

Exercise 2.5 Applications of exponential functions Technology active 1.

WE13

The number of bacteria on a culture plate, N, can be defined by the rule N(t) = 500e0.46t , t ≥ 0

where t is the time, in hours, that the culture has been growing. a. How many bacteria were initially present? b. How many bacteria, to the nearest whole number, are present after 5 hours? c. At what rate is the bacteria population increasing after 5 hours? Give your answer correct to the nearest whole number. 2. The bilby is an endangered species that can be found in the Kimberley in Western Australia as well as some parts of South Australia, the Northern Territory and Queensland. The gestation time for a bilby is 2–3 weeks and when they are born, they are only about 11 mm in length. The growth of a typical bilby can be modelled by the rule L = L0 e0.599t where L0 is its length in millimetres at birth and L is the length of the bilby in millimetres t months after its birth. a. Determine the value of L0 . b. What is the rate of change of length of the bilby at time t months? c. At what rate is the bilby growing when it is 3 months old? Give your answer correct to 3 decimal places.

70 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

3.

WE14 The mass, M grams, of a particular radioactive substance can be modelled by the exponential function

M(t) = M0 e−0.005t

4.

5.

6.

7.

where M0 is a constant and t is the time in days, Initially the substance weighed 50 grams. a. Determine the value of M0 . b. Calculate the amount of the substance remaining after 10 days. Give your answer correct to 2 decimal places. c. Determine the rate of decay at this time. Give your answer correct to 2 decimal places. Changing 𝛿-gluconolactone into gluconic acid can be modelled by the equation y = y0 e−0.6t where y is the number of grams of 𝛿-gluconolactone present t hours after the process has begun. Suppose 200 grams of 𝛿-gluconolactone is to be changed into gluconic acid. a. Determine the value of y0 . b. How many grams of 𝛿-gluconolactone will be present after 1 hour? Give your answer correct to the nearest gram. c. How long will it take to reduce the amount of 𝛿-gluconolactone to 50 grams? Give your answer correct to the nearest quarter of an hour. d. Determine the rate of change in the 𝛿-gluconolactone after 2 hours. Give your answer to 1 decimal place. The decay of radon-222 gas is given by the equation y = y0 e−0.18t , where y is the amount of radon remaining after t days. If initially there was 10 grams of radon-222 gas, determine: a. the value of y0 b. the amount of gas, to the nearest integer, remaining after 2 days c. how many days it will take for the mass to reach half its original mass d. the rate of decay after 5 days, correct to 2 decimal places. An amount of $1000 is invested in a building society where the 5% p.a. interest is compounded continuously. The amount in the account after t years can be modelled by the equation A = A0 × ert , where A0 is the initial value of the investment and r is the continuous interest rate expressed as a decimal. a. State the value of A0 and r. b. Calculate the amount in the account, correct to the nearest cent, after: i. 1 year ii. 5 years iii. 10 years. c. At what rate was the investment increasing, correct to the nearest cent per year, after: i. 1 year? ii. 5 years? iii. 10 years? d. Estimate how long, to the nearest year, it would take for the investment to double in value. A body that is at a higher temperature than its surroundings cools according to Newton’s Law of Cooling, which states that T = T0 e−zt where T0 is the original excess of temperature, T is the excess of temperature in degrees Celsius after t minutes, and z is a constant. a. The original temperature of the body was 95 °C and the temperature of the surroundings was 20 °C. Find the value of T0 . b. At what rate is the temperature decreasing after a quarter of an hour if it is known that z = 0.034? Give your answer correct to 3 decimal places.

CHAPTER 2 Calculus of exponential functions 71

The number of people living in Boomerville at any time, t years, after the first settlers arrived can be modelled by the equation P = P0 ekt . Initially, on 1 January 1850, Boomerville had a population of 500 people. By 1 January 1860, the population had grown to 675. a. Determine the value of P0 . b. Calculate the value of k correct to 2 decimal places. c. Using this value of k, determine the population on 1 January 1900. Give your answer to the nearest 10 people. d. At what rate was the population increasing on 1 January 1900? Give your answer to the nearest whole number. 9. The mass, m kg, of a radioactive isotope remaining in a sample t hours after observations began is given by the rule m(t) = ae−kt . Initially there is 2 kg of the isotope. After 3 hours the mass of the isotope has decreased to 1.1 kg. a. Determine the values of a and k. Give your answers correct to 1 decimal place where necessary. b. Using these values, find the rate of change of the isotope as a function of t. c. Calculate the rate of decay of the isotope after 6 hours. Give your answer to 2 decimal places. d. Determine the half-life of this isotope, that is, the time it takes for the isotope to reduce to half its original mass. Give your answer correct to 1 decimal place. 10. An unstable gas decomposes in such a way that the amount present, A units, at time t minutes is given by the equation 8.

A = A0 e−kt where k and A0 are constants. It was known that initially there were 120 units of unstable gas. a. Find the value of A0 . b. After 2 minutes there were 90 units of the gas left. Find the value of k. c. At what rate is the gas decomposing when t = 5? Give your answer correct to 3 decimal places. d. Will there ever be no gas left? Explain your answer. 11. The population of Australia since 1950 can be modelled by the rule P = P0 e0.016t where P0 is the population in millions at the beginning of 1950 and P is the population in millions t years after 1950. It is known that there were 8.2 million people in Australia at the beginning of 1950. a. Calculate the population in millions at the beginning of 2015, correct to 1 decimal place. b. During which year and month did the population reach 20 million? c. Determine the rate of change of population at the turn of the century, namely the year 2000, correct to 2 decimal places. d. In which year was the rate of increase of the population predicted to exceed 400 000 people per year? 12. The pressure of the atmosphere, P cm of mercury, decreases with the height, h km above sea level, according to the law P = P0 e−kh where P0 is the pressure of the atmosphere at sea level and k is a constant. At 500 m above sea level, the pressure is 66.7 cm of mercury, and at 1500 m above sea level, the pressure is 52.3 cm of mercury. a. What are the values of P0 and k, correct to 2 decimal places? b. Find the rate at which the pressure is falling when the height above sea level is 5 km. Give your answer correct to 2 decimal places.

72 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

13.

The cane toad, originally from South America, is an invasive species in Australia. Cane toads were introduced to Australia from Hawaii in 1935 in an attempt to control cane beetles, though this proved to be ineffective. In a controlled experiment at a particular waterhole, it was observed that at the beginning of the experiment there were an estimated 30 000 tadpoles (future cane toads) in the water. The number of tadpoles increased by about 60 000 a day over the period of a week. This growth pattern can be defined by the equation T = T0 ekt

where T0 is the initial number of cane toad tadpoles (in thousands) at the waterhole during the time of the experiment, T is the number of cane toad tadpoles (in thousands) at the waterhole t days into the experiment, and k is a constant. a. Calculate the value of T0 . b. How many cane toad tadpoles are in the waterhole after 7 days if it is known that k = 0.387? Give your answer to the nearest thousand. c. Determine the rate at which the cane toad tadpole numbers are increasing after 3 days. 14. WE 15 The population of possums in an inner city suburb is starting to increase. Observations of the numbers present suggest a model for the number of possums in the suburb given by P(t) = 83 − 65e−0.2t , t ≥ 0, where P is the number of possums observed and t is the time in months since observations began. a. How many possums were present at the start of the observations? b. By how many had the population of possums grown at the end of the first month of observations? Give your answer to the nearest whole number. c. When does the model predict the number of possums would be twice the initial population? d. Sketch the graph of P versus t. e. Explain why this model does not predict the population of possums will grow to 100. 15. Manoj pours himself a mug of coffee but gets distracted by a phone call before he can drink the coffee. The temperature of the cooling mug of coffee is given by T = 20 + 75e−0.062t , where T is the temperature of the coffee t minutes after it was initially poured into the mug. a. What was the initial temperature of the coffee when it was first poured? b. Sketch the graph of temperature, T °C against time, t minutes. c. To what temperature will the coffee cool if left unattended? d. How long does it take for the coffee to reach a temperature of 65 °C? Give your answer correct to 2 decimal places. e. Determine the rate of change in the temperature of the coffee after 10 minutes, correct to 1 decimal place. Explain why the rate of change is negative. 16. Newton’s Rule of Cooling states that the rate of change of the temperature of a particle is proportional to the difference between the temperature of the particle and the constant temperature of the surrounding medium. The temperature, T °C, of a particle when placed in a medium with a constant temperature of A °C can be modelled by the equation T = T0 e−kt + A where t is time and T0 is a constant. CHAPTER 2 Calculus of exponential functions 73

A metal ball has been heated to a temperature of 200 °C and is placed into a room that is maintained at a constant temperature of 30 °C. After 5 minutes, the temperature of the ball has dropped to 150 °C. a. State the value of A and hence determine the value of T0 . b. Calculate the value of k correct to 4 decimal places. c. Using the values found above, state the equation for this model and sketch its graph. d. Determine the temperature of the rod, correct to 1 decimal place, after a further 10 minutes. e. Calculate the rate of change in the temperature, correct to 1 decimal place, at this time. f. Determine how long it will take for the temperature of the ball to reach 40 °C. Give your answer to 2 decimal places. g. Verify, with reasons, that the metal ball would never reach 10 °C if left in the room.

2.6 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. Evaluate the following limits.

2.

a.

lim(6x − 1)

c.

lim

2x2 − 6x x→3 x − 3 3x − 5 d. lim x→0 2x − 1

b.

x→3

2x2 + 3x − 5 x→1 x2 − 1

Using first principles, calculate

lim

dy for the following functions. dx

y = 4 − x2 b. y = x2 + 4x c. y = x(x + 1) Given f (x) = (x + 5)2 : a. find f ′ (x) using first principles b. calculate f ′ (−5) and explain its geometric meaning c. calculate the gradient of the tangent to the curve y = f (x) at its y-intercept d. calculate the instantaneous rate of change of the function y = f (x) at (−2, 9). Solve the following for x, giving your answers to 3 decimal places. a. ex+1 = 6 b. 2e4−x − 5 = 0 −2x c. e =8 d. 4 − ex−2 = 0 Solve the following for x, giving your answers in exact form. a. e2x − 2ex = 0 b. (ex + 1)(ex − 3) = 0 2x x c. e + 2e = 8 d. 2e2x − 9ex + 4 = 0 Consider the function f(x) = −5x . a. Evaluate f(2). b. On the same set of axes sketch the graphs of y = 5x , y = −5x and y = 5−x . c. Express y = 5−x in an equivalent form. Sketch the following graphs and state the domain and range of each graph. a.

3.

4.

5.

6.

7.

a.

y = 2ex + 1

x

b.

y = 3 − 3e− 2

c.

1 y = − ex+1 4

Sketch the following graphs and state the domain and range of each graph. a. y = −2ex − 3 b. y = 4e−3x − 4 c. y = 5ex−2 9. a. Sketch the graph of y = 2e1−3x − 4, labelling any intercepts with the coordinate axes with their exact coordinates. b. Sketch the graph of y = 3 × 2x − 24 and state its domain and range.

8.

74 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

A possible equation for the graph shown is: A. y = 2 − ex B. y = 2 − e−x C. y = 2 + e−x D. y = e−x − 2 11. Determine the derivative of each of the following functions. 10.

y

MC

a.

y=e

y=2 (0, 1)

−1x 3

0

x

2

y = 3x4 − e−2x 4ex − e−x + 2 c. y = 3e3x 2x d. y = (e − 3)2 b.

12.

1 Consider the function defined by the rule f (x) = e3x + e−x . 2 Determine the gradient of the curve when x = 0.

Complex familiar 13. a. The graph shown is of the function f(x) = aex + b. Determine the values of a and b and write the function as a mapping.

y

(0, 0)

b.

y = 11

x

The graph shown has an equation of the form y = Aenx + k. Determine its equation. y (–1,

4 + e2) f(x) (0, 5) y=4 0

x

CHAPTER 2 Calculus of exponential functions 75

14.

2

The graph of y = Ae−x , where A is a constant, is shown. Answer the following questions correct to 2 decimal places where appropriate. a. Determine the value of A. dy . b. Find dx c. Determine the gradient of the tangent to the curve at the point where: i. x = −0.5 ii. x = 1 d. Show that the equation of the tangent at the point where x = 1 is given by 10x + ey − 15 = 0.

y (0, 5)

0

x

The mass, m g, of a radioactive isotope remaining in a sample t hours after observations began is given by the rule m(t) = ae−kt . Initially there are 4 grams of the isotope. After 6 hours, the mass of the isotope has decreased to 2.8 g. a. Evaluate the values of a and k. Give your answers correct to 3 decimal places where necessary. b. Calculate the rate of decay of the isotope as a function of t. c. Calculate the rate of decay after 6 hours. Give your answer correct to 2 decimal places. 16. The population of a certain town was 250 000 at the beginning of the year 2000 and 400 000 at the beginning of the year 2010. It was found that the relationship between the population, P thousands, and time, t years, could be modelled by the relationship P(t) = A ekt where A and k are constants. a. Determine the values of A and k, giving your answers correct to 3 decimal places where necessary. b. How many people lived in the coastal town at the beginning of the year 2015? Give your answer to the nearest thousand. c. The local council has determined that the population should not exceed 750 000. In what year would this likely occur?

15.

Complex unfamiliar 17. Microbiologists have been working with a certain type of bacteria that continues to thrive providing it has a favourable growth medium. For a particular experiment, they started with 500 bacteria and observed that the population doubles every 8 hours. The relationship between the number of bacteria, P, and the time, t hours since the bacteria started multiplying, is given by P(t) = P0 ekt , where P0 and k are constants. 1 a. State the value of P0 and show that k = loge 2. 8 b. The growth phase lasts for 40 hours. How many bacteria are present in the colony at this time? c. Show that the rate of increase in the colony size after 8 hours is 125 loge 2 bacteria/hour. d. Determine when the rate of increase in the colony size would be double the rate found after 8 hours.

76 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

The mass, m mg, of a radioactive substance remaining in a sample t hours after observations began is given by the rule m(t) = ae−kt . Initially, 30 mg of the substance is present. After 2 hours the scientist observes that 20% had disintegrated. a. State the value of a. b. Determine the mass of the substance after 2 hours. c. Calculate the value of k, correct to 4 decimal places. d. Determine the amount of the substance remaining after a further 3 hours. Give your answers correct to 2 decimal places. e. Determine when the rate of decay of the substance is 1 mg/hour. Give your answer correct to 1 decimal place. 19. When money is invested in a bank at a constant rate of r% with continuously compounding interest, the accumulated amount $A at a time t years after the start of the investment is modelled by the equation 18.

A = A0 ert where A0 is the amount to be invested. a. Determine the amount to which $10 000 will grow in 6 years if invested at 4.5%. b. If the rate did not change, estimate the time it would take for the investment to triple in value. Give your answer to the nearest month. 20. Newton’s Rule of Cooling states that the rate of change of the temperature of a particle is proportional to the difference between the temperature of the particle and the constant temperature of the surrounding medium. The temperature, T °C, of a particle when placed in a medium with a constant temperature of A °C can then be modelled by the equation T = T0 e−kt + A where t is time and T0 is a constant. A saucepan is filled with water and heated until the temperature of the water reaches 100 °C. The saucepan is then placed on a bench in a room that is kept at a constant temperature of 28 °C. After 3 minutes, the temperature of the water in the saucepan is 76 °C. a. State the value of A and determine the value of T0 . ( ) 1 3 b. Show that k = ln . 3 2 c. State the equation for this model and sketch a graph to represent it. d. Determine the temperature of the water after a further 3 minutes. e. Verify, with reasons, that the water would never reach 25 °C if left in the room.

Units 3 & 4

Sit exam

CHAPTER 2 Calculus of exponential functions 77

Answers

b.

(–1, 4)

2 Calculus of exponential functions

y = 4–x

Exercise 2.2 Review of limits and differentiation 1. a. 19

b. 16

2. a. −3 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

13. 14.

a. −7 2 a. 4x

y

c. 2

3

b. 3

c.

b. 4 2 b. 3x

c. −5

2

f ′ (x) = 2x − 6 f ′ (x) = 3 − 4x a. 0.916 29 b. 0.955 51 c. 0.993 25 d. 1.029 62 e. 1.064 71 f. 1.000 00 a. 7.3891 b. 20.0855 c. 1.6487 a. 0.736 b. 1.396 c. 2.472 dy a. = 8 − 2x b. 4 c. y = 4x + 4 dx dy = 3x2 − 6x; y = 9x − 27 dx a. (−2, 0), (0, 0), (2, 0) dy = 3x2 − 4 b. dx c. At x = −2, gradient = 8. At x = 0, gradient = −4. At x = 2, gradient = 8. d. The gradient of the tangent at x = −2 and x = 2 is m = 8. Therefore, since the gradients are equal, the tangents are parallel. dy −h −1 a. b. = 2 (x + h)x dx x a. Sample responses can be found in the worked solutions in the online resources. −1

b.

x

0 c.

y y=0 x

0 (–1, –0.25)

(0, –1) y = –4x

(1, –4) 2.

y (1, 10)

(–1, 10)

y = 10–x

y = 10x

(0, 1) 0 y = 0 (1, 0.1)

(–1, 0.1) 3. a.

(x − 2)2 dy −1 c. = dx (x − 2)2 5 7 d. x = , 3 3

(0, 1) (1, 0.25)

y=0

x

y (1, 10.873)

y = 4ex

Exercise 2.3 The exponential function 1. a.

(0, 4)

y (–1, 1.472) (1, 4) 0 y = 4x

x

x

b. The function f(x) = e has been dilated by a factor of 4

from the x-axis to give f(x) = 4ex .

(0, 1) (–1, 0.25) y =0 0

y=0

x

78 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

y

4. a.

b.

y

(0.549, 0) x

0

y=0 x

0 (–1, –1.84)

y = e2x – 3

(0, –5)

(0, –2) y = –3

y = –5e x (1, –13.59)

x

c. The function f(x) = e has been dilated by a factor of

2 from the y-axis and translated vertically down by 3 units to give f(x) = e2x − 3. 8. a. Intercepts: (− ln(2), 0) and (0, 2)

x

b. The function f (x) = e has been reflected in the x-axis

and dilated by a factor of 5 from the x-axis to give f (x) = −5ex . 5. a.

1

b.

y

y=4

y

(0, 2)

(–1, 5.718)

(–0.693, 0) x

0

y = 3 + e–x

y = 4 – 2e–x

(0, 4) (1, 3.368) x

c. The function f (x) = e has been dilated by a factor of 2

y=3

from the x-axis, reflected in the x-axis, reflected in the y-axis and translated vertically up by 4 units to give f (x) = 4 − 2e−x . 9. a.

y x

x

0

y = 4e 2

x

b. The function f (x) = e has been reflected in the y-axis

(1, 6.595)

and translated vertically up by 3 units to give f (x) = e−x + 3. 6. a.

(0, 4) (–1, 2.426)

y (1, 10.389)

y=0 x

0 x

y = 3 + e2x

b. The function f (x) = e has been dilated by a factor of 2

from the y-axis and dilated by a factor of 4 from the x

(0, 4) (–1, 3.135)

x-axis to give f (x) = 4e 2 . 10. a. Intercepts: (−2 ln(2), 0) and (0, −3)

y=3

b.

y

x

0

x

b. The function f (x) = e has been dilated by a factor of

1

2 from the y-axis and translated vertically up by 3 units to give f (x) = e2x + 3. ( ) 1 7. a. Intercepts: ln(3), 0 and (0, −2) 2

–x 2

y = 3e

x

0

(–1.386, 0) –6

(0, –3)

y = –6 x

c. The function y = e has been dilated by a factor of 2

from the y-axis and dilated by a factor of 3 from the x-axis, reflected in the y-axis and translated vertically down by 6 units to give y = 3e

−

x 2

− 6.

CHAPTER 2 Calculus of exponential functions 79

11. a. x 12. a. x c. x e. x 13. a. x

= ln (4) ≈ 1.609 ≈ 0.956 ≈ 0.405 = 0, ln(2)

x = 1.386 x ≈ −0.693 x ≈ −1.792 x ≈ −1.204 x=0 ( ) 3 d. x = ln , − ln(2) 2 ( ) 2 f. x = ln 3

b. b. d. f. b.

c. x = 0, ln(2) e. x = ln(4) 14. x = 1.61 15. a. x = ln(3)

b. x = ln(3), ln

c. x = ln(5)

( ) 2 3

d. x = ln(5)

16. a. x > 0 c. x ≥ ln(2) 17. a.

b. x > −1 d. x ≥ 1 − ln(6)

y

(0, 3) (1, 1.736) y=1 x

0

b. x > 0 −x c. For 2 e + 1 < 0, the curve would be below the x-axis.

But y > 1 for all x values, so the curve is never below the x-axis. Hence, 2 e−x + 1 < 0 has no real solutions. y

y=4

9. 10. 11. 12. 13.

d. −5e 2

6−3x+x 2

4x −7x

2

x −3x+1

2−5x

2

b. 3(x + 1)e

3

x +3x−2

d. 10(1 + 3x)e

2

1−2x−3x

A −20e −6e2 −5e−2 y = 5x − 4

14. Tangent: y = −3x − 1; perpendicular: y = 7

2

16. a. 2e −

b.

2 e

1 3

x−1

3 2

b. 0

1. a. 500 bacteria c. 2294 bacteria/hour

b. 4987 bacteria

2. a. 11

b.

c. 3. a. c. 4. a. c. 5. a. c. 6. a. b.

c.

(loge4, 0) x

0 b. x < ln(4) c. For x = 0, y = 3.

Observing the graph: For x ≥ 0, y ≤ 3. If the domain ] is restricted to x ≥ 0, the range is y ≤ 3 or y ∈ (−∞, 3 .

Exercise 2.4 Differentiation of exponential functions 10x

−e−x 6e6x−2 −8e7−2x −90e6−9x x e 2 +1

2

x −2x

c. 3(8x − 7)e

(0, 3)

y = 4 – ex

d. 2. a. d. 3. a.

8. a. (2x − 3)e

b. (2x − 3)e

Exercise 2.5 Applications of exponential functions

y = 2e–x + 1

1. a. 10e

2

x +3x

c. 2(x − 1)e

15. a. −2e

(–1, 6.437)

18. a.

7. a. (2x + 3)e

d. 4. A x 5. a. 2e −4x c. −10(2e − 1) 2x −7x 6. a. 6e − 7e

b. e. b. e. b.

1

x e3

c.

3 6e3x −6e8−6x −24e8x+1 −15e3x+4

e. −e

2−

f. c. f. c.

x 3

f. 2x

x

1

x e4

4 −20e−5x 10e5x+3 10e6−5x −42e−7x x −e 4 +5

b. 3e (3e + 2) x −x d. 3e − 2e 9x x b. 36e − 2e

d. 7. a. 8. a. c.

39.742 mm/month 50 0.24 grams/day 200 1 2 hours 4 10 4 days A0 = 1000; r = 0.05 i. $1051.27 ii. $1284.03 iii. $1648.72 i. $52.56/year ii. $64.20/year iii. $82.44/year 14 years 75 500 2240

9. a. a = 2; k = 0.2 c. 0.12 kg/h 10. a. 120 ( ) b.

1

2

ln

dL dt

= 6.589e0.599t

b. 47.56 grams b. 110 grams d. −36.1 grams/hour b. 7 grams d. 0.73 grams/day

b. 1.531 °C/ min b. 0.03 d. 67 people/year

dm

= −0.4e−0.2t dt d. 3.5 h

b.

4

3

c. 8.411 units/min d. As t → ∞, A → 0. Technically the graph approaches the

11. a. c. 12. a. b. 13. a. b. c.

line A = 0 (asymptotic behaviour). However, the value of A would be so small that in effect, after a long period of time, there is no gas left. 23.2 million b. August 2005 0.29 million/year d. 2019 75.32 cm; k = 0.24 5.45 cm of mercury/km 30 450 000 tadpoles 37 072.2 tadpoles/day

80 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

14. a. b. c. d.

18 possums 12 possums 1.62 months

4. a. 0.792 5. a. ln(2) c. ln(2) 6. a. −25 b.

P(t) (possums) P = 83

c. −1.040 d. 3.386 b. ln(3) d. − ln(2), ln(4)

b. 3.084

y = 5–x

y

(–1, 5)

y = 5x (1, 5)

P = 83 – 65e–0.2t (1.6, 36) (1, 30) (0, 18)

(0, 1) y=0 x

0 (0, –1)

0

t (months)

e. The presence of the asymptote at P = 83 shows that as

(1, –5)

t → ∞, P → 83. The population can never exceed 83, so the population cannot grow to 100 possums. 15. a. 95 °C

y = –5x

( )x 1

or y = (0.2)x 5 7. a. Domain: x ∈ R, range: y ∈ (1, ∞)

b. T (°C)

c. y =

(0, 95)

y

y = 75e–0.062x + 20

y = 2ex + 1 y = 20 (0, 3)

0

t (minutes)

y=1

c. Approximately 20 °C d. 8.24 min e. After 10 minutes, the coffee is cooling at a rate of

2.5 °C/minute. The temperature is decreasing, so the rate of change will be negative. 16. a. A = 30; T0 = 170 b. k = 0.0697 −0.0697t c. T = 170e + 30

x

0 b. Domain: x ∈ R, range: y ∈ (−∞, 3)

y

T (°C)

y=3 y = 3 – 3e

(0, 200) T = 170e–0.0697t + 30

x –– 2

x

(0, 0)

y = 30 0 d. e. f. g.

t (minutes)

89.8 °C −4.2 °C/ min 40.65 min From the graph, the temperature of the metal ball is always greater than 30 °C, the temperature of the room, so if left in the room it will never reach a temperature of 10 °C.

c. Domain: x ∈ R, range: y ∈ (−∞, 0)

y

y=0

2.6 Review: exam practice 1. a. 17

b. 6

c.

7

2 c. 2x + 1

(–1, –– –14) d. 5

2. a. −2x b. 2x + 4 ′ 3. a. f (x) = 2x + 10 b. 0; The function has a stationary point at x = −5. c. 10 d. 6

x

0

(0, – –4e)

1 ex + 1 y = –– 4

CHAPTER 2 Calculus of exponential functions 81

8. a. Domain: x ∈ R, range: y ∈ (−∞, −3)

10. A

y

1

1 − x

3 8

−2x

3

11. a. − e 3

x

0

c. − e

3

y = –3 12.

(0, –5)

b. 12x + 4xe

4 + e−4x − 2e−3x 3

4x

d. 4e − 12e

2

−2x

2x

1 2

x

13. a. a = −11, b = 11; f : R → R, f(x) = −11e + 11 −2x b. y = e +4 14. a. 5

y = –2ex – 3

b.

y

15. a. a = 4; k = 0.059

y = 4e–3x – 4

c. 16. a. c. 17. a. b. c. d. 18. a. c. e. 19. a. b. 20. a. b.

(0, 0) x

y = –4 c. Domain: x ∈ R, range: y ∈ (0, ∞)

y y = 5ex – 2 (2, 5)

y=0

b.

dm dt

= −0.236e−0.059t

0.17 g/h A = 250; k = 0.047 b. 506 000 2023 P0 = 500 16 000 bacteria. At t = 8, P′ (8) = 125 ln(2) bacteria/hour 16 hours 30 b. 24 mg k = 0.1116 d. 17.17 mg After 10.8 hours $13 099.65 24 years and 5 months A = 28; T0 = 72 Sample responses can be found in the worked solutions in the online resources. ( )

c. T = 72e

x

0

dx

2

= −10xe−x

c. i. 3.89 ii. −3.68

b. Domain: x ∈ R, range: y ∈ (−4, ∞)

(0, 5e–2)

dy

−

1 3 ln t 3 2

+ 28

T (°C) 9. a.

y

(0, 100)

(0, 2e – 4) 0

(–13 (1 – log (2)), 0)

x

y = 72e–

e

1 ln 3 x 3 2

+ 28

y = 28

y = 2e1 – 3x – 4 y = –4 0

b. Domain R, range (–24, ∞)

t (minutes)

d. 60 °C e. Since the room is kept at a constant temperature of

y

28 °C, the water will never cool below this level. y = 3 × 2x – 24 (3, 0) 0

(0, –21)

x

y = –24

82 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland



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B

 DPT (Y ϴ ۧ) ϴ    ϴۧ ф Y ф ۧ

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C

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྾ М  ྾ ྾ ྾ ྾ Y Ҳ И ྾ Ҭ И ྾ Ҭ И ྾ Ҭ     ྾ ྾ ྾ ྾ И И Ҳ И     ྾ ྾ ྾ ྾ YҲ И И И    

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C

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4PMWF UIF GPMMPXJOH FRVBUJPOT B  DPT(ྶ) Ҭ ‫ ٹ‬Ҳ  GPS  ‫ ܓ ྶ ܓ‬྾ C UBO(Y) Ҭ ‫ ٹ‬Ҳ  GPS ž ‫ ܓ‬Y ‫ž ܓ‬ D  DPT(ྶ) Ҳ  GPS ҭ྾ ‫ ܓ ྶ ܓ‬྾  B 4PMWF UIF FRVBUJPO  TJO(ྶ) Ҭ  Ҳ  ž ‫ž ܓ ྶ ܓ‬ C 4PMWF TJO(Y) Ҳ  ҭ྾ ‫ ܓ‬Y ‫ ܓ‬྾  8& 4PMWF UIF GPMMPXJOH FRVBUJPOT B  DPT(ྶ) ҭ ‫ ٹ‬Ҳ  GPS  ‫ ܓ ྶ ܓ‬྾ 

8&

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$)"15&3  $BMDVMVT PG USJHPOPNFUSJD GVODUJPOT 

5FDIOPMPHZ BDUJWF 

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–2π

– 3π — 2

–π

– –π 2

0 –1

y = sin(x)

π – 2

π

3π — 2

y 1 2π

x 3π –2π – — 2

–π

π –– 2

0 –1

y = cos(x)

π – 2

π

3π — 2

2π

x

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 (SBQIT PG " TJO(#Y) Ҭ % BOE " DPT(#Y) Ҭ % 'PS HSBQIT JO UIF GPSN Z Ҳ " TJO(#Y) Ҭ % PS Z Ҳ " DPT(#Y) Ҭ % UIF WBMVF PG " BGGFDUT UIF BNQMJUVEF BOE EJSFDUJPO PG UIF TJOF BOE DPTJOF GVODUJPOT ю *G " Ҵ  UIF BNQMJUVEF JT " ю *G " ҳ  UIF BNQMJUVEF JT " BOE UIF HSBQI JT BMTP SFáFDUFE JO UIF YBYJT Z Ҳ  TJO(Y) "NQMJUVEF Ҳ  /PUF 5IF HSBQI TIPXT B EJMBUJPO PG  GSPN UIF YBYJT

2 1

–, 2 ( ( ) 2 ) π, 1 – 2

y = sin (x)

0 (0, 0) –1 –2

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(2π, 0)

π – 2

π

 GSPN 

0 –1 –2

(0, – –12)

(π, –12) y = –0.5 cos (x) π (π, –1)

π – 2

2 1 0 –1

x

2π

3π — 2

(3π—2, –1) (3π—2, –2)

2 (0, 1) y = cos (x) 1

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y = 2 sin (x)

π – 2

(

3π — 4

π

5π — 4

)

1 2π, – – 2

y = cos (2x)

(0, 1) y = cos (x)

π – 4

3π — 2

(2π, 1) x

(2π, 1)

3π — 2

7π — 4

2π

x

–2

 GSPN UIF ZBYJT 

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y 2 y = sin (x)

1

0

π – 2

x y = sin – 2

()

π

3π — 2

2π

–1

7π — 2

3π

5π — 2

x

4π

–2

5IF WBMVF PG % BGGFDUT UIF FRVJMJCSJVN PS NFBO QPTJUJPO BCPVU XIJDI UIF TJOF BOE DPTJOF GVODUJPOT PTDJMMBUF ю 5IF HSBQI PTDJMMBUFT BCPVU UIF MJOF Z Ҳ % ю 5IF SBOHF PG UIF GVODUJPO DIBOHFT

y 2 y = sin (x)

1

Equilibrium y = 0 π – 4 –1 y = 0

π – 2

0

Z Ҳ TJO(Y) ҭ  -JOF PG PTDJMMBUJPO Z Ҳ ҭ /PUF 5IJT DBO CF EFTDSJCFE BT B WFSUJDBM USBOTMBUJPO EPXO CZ  VOJUT PS B USBOTMBUJPO PG  JO UIF OFHBUJWF EJSFDUJPO QBSBMMFM UP UIF ZBYJT

π

3π — 4

5π — 4

3π — 2

7π — 4

2π

Equilibrium y = –2

–2 y = sin (x) − 2 –3

803,&% &9".1-& 

4LFUDI UIF HSBQI PG Z   TJO (Y) Ҭ   ф Y ф ۧ 5)*/, 

4UBUF UIF QFSJPE BOE BNQMJUVEF PG UIF HSBQI



4UBUF UIF NFBO QPTJUJPO BOE UIF SBOHF



$POTUSVDU BQQSPQSJBUF TDBMFT PO UIF BYFT BOE TLFUDI UIF HSBQI

83*5&

Z Ҳ  TJO(Y) Ҭ И  ‫ ܓ‬Y ‫ ܓ‬྾ ྾ Ҳ ྾ 5IF QFSJPE JT  5IF BNQMJUVEF JT  5IF NFBO QPTJUJPO JT Z Ҳ  5IF SBOHF PG UIF HSBQI JT Ъ ҭ И  Ҭ Ы Ҳ ЪИ Ы  y 7

(–π4 , 7)

5π , 7) (–– 4

4 (0, 4)

(2π, 4)

(π, 4) 7π , 1) (–– 4

3π , 1) (–– 4

1 0

y = 3 sin(2x) + 4

π – 4

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π – 2

3π — 4

π

5π — 4

3π — 2

7π — 4

2π

x

x

803,&% &9".1-& 

5IF EJBHSBN TIPXT UIF HSBQI PG B DPTJOF GVODUJPO 4UBUF JUT NFBO QPTJUJPO BNQMJUVEF BOE QFSJPE BOE HJWF B QPTTJCMF FRVBUJPO GPS UIF GVODUJPO

y

(–π3 , 4) 0

π – 3

x

(0, –8) 5)*/, 

%FEVDF UIF NFBO QPTJUJPO



4UBUF UIF BNQMJUVEF



4UBUF UIF QFSJPE



%FUFSNJOF B QPTTJCMF FRVBUJPO GPS UIF HJWFO HSBQI

83*5&

5IF NJOJNVN WBMVF JT ҭ BOE UIF NBYJNVN WBMVF JT  TP UIF NFBO ҭ Ҭ  Ҳ ҭ QPTJUJPO JT Z Ҳ  5IF BNQMJUVEF JT UIF EJTUBODF GSPN UIF NFBO QPTJUJPO UP FJUIFS JUT NBYJNVN PS NJOJNVN 5IF BNQMJUVEF JT  ҭ ҭ "Ҳ Ҳ   ྾ "U Y Ҳ UIF HSBQI JT IBMGXBZ  ྾  UISPVHI JUT DZDMF TP JUT QFSJPE JT  -FU UIF FRVBUJPO CF Z Ҳ " DPT(#Y) Ҭ % 5IF HSBQI JT BO JOWFSUFE DPTJOF TIBQF TP " Ҳ ҭ ྾ 5IF QFSJPE JT  # ྾ ྾ Ҳ #  #Ҳ 5IF NFBO QPTJUJPO JT Z Ҳ ҭИ TP % Ҳ ҭ 5IF FRVBUJPO JT Z Ҳ ҭ DPT(Y) ҭ 

( ( )) ( ( ))  (SBQIT PG Z Ҳ " TJO # Y Ҭ $ Ҭ % BOE Z Ҳ " DPT # Y Ҭ $ Ҭ % #FMPX JT B TVNNBSZ PG UIF USBOTGPSNBUJPOT PG Z Ҳ TJO(Y) UP Z Ҳ " TJO(# (Y Ҭ $)) Ҭ % ю " XIFSF " Ҵ  EJMBUJPO CZ B GBDUPS PG " GSPN UIF YBYJT SFáFDUJPO JO UIF YBYJT ю " ҳ   ю # XIFSF # Ҵ  EJMBUJPO CZ B GBDUPS PG GSPN UIF ZBYJT # SFáFDUJPO JO UIF ZBYJT ю # ҳ  USBOTMBUJPO IPSJ[POUBMMZ PS QBSBMMFM UP UIF YBYJT PG $ UP UIF MFGU JG $ Ҵ  PS $ UP UIF ю $ SJHIU JG $ ҳ  USBOTMBUJPO WFSUJDBMMZ PS QBSBMMFM UP UIF ZBYJT PG % ю %

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5SBOTGPSNBUJPOT PG Z Ҳ DPT(Y) UP Z Ҳ " DPT(# (Y Ҭ $)) Ҭ % GPMMPX UIF TBNF QBUUFSOT BT UIPTF GPS Z Ҳ TJO(Y) 1SPQFSUJFT PG UIFTF USJHPOPNFUSJD GVODUJPOT BSF TVNNBSJTFE CFMPX 1SPQFSUJFT PG HSBQIT JO UIF GPSN Z Ⴝ " TJO(# (Y Ⴛ $)) Ⴛ % PS Z Ⴝ " DPT(# (Y Ⴛ $)) Ⴛ % ྾ ю 1FSJPE # ю "NQMJUVEF "И " Ҵ  ю -JOF PG PTDJMMBUJPO NFBO QPTJUJPO  Z Ҳ % ю %PNBJO Y ٝ 3 ю 3BOHF Z ٝ Ъ% ҭ "И % Ҭ "Ы

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4UBUF UIF QFSJPE BNQMJUVEF NFBO QPTJUJPO BOE IPSJ[POUBM USBOTMBUJPO CZ SFXSJUJOH UIF GVODUJPO JO UIF GPSN Z Ҳ " DPT(#(Y Ҭ $)) Ҭ % /PUF *U JT B DPNNPO FSSPS OPU UP GBDUPSJTF UP àOE #

4LFUDI UIF HSBQI XJUIPVU UIF IPSJ[POUBM USBOTMBUJPO Z Ҳ  DPT(Y)

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y = 4 cos(2x)

2 0 –2

π – 4

–4

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π – 2

3π — 4

π

5π — 4

(3π—2 , –4)

x 3π — 2



$BMDVMBUF UIF DPPSEJOBUFT PG UIF FOEQPJOUT PG UIF EPNBJO PG UIF HJWFO GVODUJPO



$BMDVMBUF PS EFEVDF UIF QPTJUJPOT PG UIF YJOUFSDFQUT



"QQMZ UIF IPSJ[POUBM USBOTMBUJPO UP LFZ QPJOUT PO UIF HSBQI BMSFBEZ TLFUDIFE BOE IFODF TLFUDI UIF GVODUJPO PWFS JUT HJWFO EPNBJO

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π f(x) = 4 cos 2x + – 3

(

2 (0, 2) π – 12 0 π – 4 –2

7π – 12 π – 2

3π — 4

13π – 12 π 5π — 4

)

x 3π — 2 3π — , –2 2

(

)

–4 C 

4UBUF UIF QFSJPE BNQMJUVEF MJOF PG PTDJMMBUJPO BOE SBOHF /PUF H (Y) JT SFMBUFE UP G (Y) GSPN QBSU B 5IF DVSWF G (Y) IBT CFFO SFáFDUFE JO UIF YBYJT BOE USBOTMBUFE WFSUJDBMMZ VQXBSET CZ 

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$)"15&3  $BMDVMVT PG USJHPOPNFUSJD GVODUJPOT 



4UBUF UIF USBOTMBUJPOT



$BMDVMBUF UIF FOEQPJOUT GPS UIF HJWFO EPNBJO /PUF 4JODF H (Y) JT B SFáFDUJPO BOE WFSUJDBM USBOTMBUJPO PG G (Y) UIF FOEQPJOUT DPVME CF PCUBJOFE FBTJMZ VTJOH UIF FOEQPJOUT PG G (Y)

྾ UP UIF MFGU  7FSUJDBM USBOTMBUJPO ( VQXBSET ) CZ  VOJUT ྾ H () Ҳ ҭ DPT Ҭ   H () Ҳ ҭ Ұ Ҭ   H () Ҳ  ( ) ( ) ྾ ྾ ྾ Ҳ ҭ DPT  Ұ Ҭ Ҭ H    ( ) ( ) ྾ ྾ Ҳ ҭ DPT Ҭ H   ( ) ҭ ྾ Ҳ ҭ Ұ Ҭ H   ( ) ྾ Ҳ H  ) ( ྾ И  5IF FOEQPJOUT BSF (И ) BOE  5IFSF YJOUFSDFQUT TJODF ( BSF OP )  ྾ ‫ ܍‬ DPT Y Ҭ   )PSJ[POUBM USBOTMBUJPO

6TJOH UIF USBOTGPSNBUJPOT PG G (Y) UIF GVODUJPO H (Y) XJMM OPU IBWF BOZ YJOUFSDFQUT  4LFUDI UIF GVODUJPO Z Ҳ H (Y) VTJOH UIF HSBQI PG G (Y) BMPOH XJUI UIF FYUSB JOGPSNBUJPO BOE USBOTGPSNBUJPOT 

y 10

5

π — , 10 3

(

( )

)

4π —, 10 3

( ) 3π —,8 2

π g(x) = 6 – 4 cos 2x + — 3

(

(0, 4)

( ) 5π —,2 6

0

π – 4

Resources *OUFSBDUJWJUJFT 4JO BOE DPTJOF HSBQIT JOU

5IF VOJU DJSDMF TJOF BOE DPTJOF HSBQIT JOU

0TDJMMBUJPO JOU

$PNQMFNFOUBSZ QSPQFSUJFT PG TJO BOE DPT JOU

6OJUT   

"SFB 

4FRVFODF 

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π – 2

3π — 4

π

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3π — 2

x

)

&YFSDJTF (( 3FWJFX )) PG HSBQIT PG USJHPOPNFUSJD GVODUJPOT PG UIF GPSN Z Ҳ " TJO # Y Ҭ $ Ҭ % BOE Z Ҳ " DPT(# (Y Ҭ $)) Ҭ % 5FDIOPMPHZ GSFF  

    



4LFUDI UIF GPMMPXJOH HSBQIT GPS  ‫ ܓ‬Y ‫ ܓ‬྾ B Z Ҳ TJO(Y) C Z Ҳ TJO(Y) D Z Ҳ  TJO(Y) E Z Ҳ TJO(Y) Ҭ  F Z Ҳ  ҭ TJO(Y) 4LFUDI UIF GPMMPXJOH HSBQIT GPS ҭ྾ ‫ ܓ‬Y ‫ ܓ‬྾ ( ) Y B Z Ҳ DPT(Y) C Z Ҳ DPT D Z Ҳ  DPT(Y) E Z Ҳ  ҭ DPT(Y) F Z Ҳ DPT(Y) ҭ   8& 4LFUDI UIF HSBQI PG Z Ҳ  DPT(Y) ҭ И  ‫ ܓ‬Y ‫ ܓ‬྾ 4LFUDI UIF HSBQI PG Z Ҳ  ҭ  TJO(Y)И  ‫ ܓ‬Y ‫ ܓ‬྾ 4LFUDI UIF HSBQI PG Z Ҳ ҭ DPT(Y) GPS  ‫ ܓ‬Y ‫ ܓ‬྾ TUBUJOH BOZ BYJT JOUFSDFQUT  'PS ҭ྾ ‫ ܓ‬Y ‫ ܓ‬྾ TLFUDI UIF GVODUJPO Z Ҳ DPT(Y) Ҭ   ( ) Y

MPDBUJOH BOZ JOUFSDFQUT XJUI UIF DPPSEJOBUF 4LFUDI UIF HSBQI PG G ЪИ ྾Ы ‫ צ‬3И G (Y) Ҳ  ҭ  TJO  BYFT 8& 5IF EJBHSBN TIPXT UIF HSBQI PG B TJOF GVODUJPO 4UBUF JUT NFBO QPTJUJPO y (1.5, 13) BNQMJUVEF BOE QFSJPE BOE HJWF B QPTTJCMF FRVBUJPO GPS UIF GVODUJPO

(0, 5) (2, 5) x

0 (0.5, –3) 

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y 4 3 2 1 –π

π – – 2

0 –1

π – 2

π

3π x — 4

–2

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MENU, then select: 2: Add Graphs. Complete the entry line in the f1 (x) = tab as: ( ) 𝜋x −3 cos + 18 9 Note: The independent variable t has been replaced with x. 2. Sketch the graph by pressing the ENTER button.

for when y = 20, select: Menu 8: Geometry 1: Points & Lines 2: Point On.

3. To calculate the x-value(s)

4. Move the cursor and select

the curve representing ( ) 𝜋x + 18 f1 (x)= −3 cos 9 Press the ESC (escape) button. This allows you to move the textbox indicating the coordinates of the point P (x, y). Complete the entry line in the textbox as 20 for the y-value. Press the ENTER button to perform the calculation. 5. Determine the next solution by moving the cursor point on the line to a position that is close to the desired solution. Complete the entry line in the textbox as 20 for the y-value. Press the ENTER button to perform the calculation. The second solution exists at x = 11.4.

WRITE

CASIO | THINK

WRITE

b. 1. On a Main Menu

screen, select Graph. Complete the entry line in the ( Y1)tab as: 𝜋x −3 cos + 18 9 Note: The independent variable t has been replaced with x. 2. Sketch the graph by

pressing either the DRAW or EXE button.

3. To calculate the

x-value(s) for when y = 20, select: G-Solv (SHIFT F5) X-CAL. Complete the entry line in Y: as 20. Press the EXE button to perform the calculation. 4. The answer appears on the screen. The first solution exists at x = 6.6. Note: The family of solutions can be calculated by pressing the directional cursor button either left or right.

5. Determine the next

solution by pressing the directional cursor button to the right. The second solution exists at x = 11.4.

CHAPTER 4 Calculus of trigonometric functions 139

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5

Further differentiation and applications 5.1 Overview Differential calculus is a branch of mathematical analysis concerned with determining how a change in one variable will affect another related variable. Calculus is the study of change, the slopes of curves, and the rate of change between two variables. It is generally thought that Sir Isaac Newton, in England, and Gottfried Leibniz, in Germany, independently discovered calculus in the mid 17th century. Both of them were building on earlier studies of motion and areas: Newton was investigating the laws of motion and gravity as well as geometry, whereas Leibniz was focused on understanding tangents to curves. Although Leibniz was the first to publish his results, controversy remained between the two as to who invented the notation which is still used today. Although early study of differential calculus involved ratios and geometry, during the 18th century it became more algebraic in nature. Today, calculus is used in many different areas. In economics and commerce, examples of rates of change include marginal costs, the increase or decrease in production costs if another unit is produced, and predictions on the stock market. In science, the rate of growth of bacteria or the rate of decay of a substance can be expressed as a differential equation. In engineering, optimisation — determining the value of one variable that would either maximise or minimise a related variable — is used extensively along with graphing curves. Sir Isaac Newton (1643–1727)

Gottfried Wilhelm Leibniz (1646–1716)

LEARNING SEQUENCE 5.1 5.2 5.3 5.4 5.5 5.6

Overview The chain rule The product rule The quotient rule Applications of differentiation Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

156 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

5.2 The chain rule 5.2.1 Composite functions A composite function, also known as a function of a function, consists of two or more functions nested within each other. Consider g(x) = x4 and h(x) = 2x + 1. If f (x) = g(h(x)), then: f (x) = g(2x + 1) = (2x + 1)4

The function f (x) can be differentiated if it is expanded.

f (x) = 16x4 + 32x3 + 24x2 + 8x + 1

f ′ (x) = 64x3 + 96x2 + 48x + 8

= 8(8x3 + 12x2 + 6x + 1) = 8(2x + 1)3

The chain rule allows us to reach this same outcome without having to expand. The chain rule also allows us to differentiate composite functions that we cannot expand. In complex functions, the chain rule may need to be applied more than once. For an example of this, see Worked example 3.

5.2.2 The chain rule The chain rule If y = f (g(x)),

dy = f ′ (g(x)) × g′ (x) dx

Alternatively, if y = f (u) and u = g(x), dy dy du = × dx du dx

5.2.3 Proof of the chain rule

The proof of the chain rule is as follows. If f (x) = m(n(x)), then f (x + h) = m(n(x + h)). f (x + h) − f (x) m(n(x + h)) − m(n(x)) Therefore, = . h h Multiply the numerator and the denominator by n(x + h) − n(x), as it is expected that at some stage n′ (x) will appear somewhere in the rule.

CHAPTER 5 Further differentiation and applications 157

f (x + h) − f (x) n(x + h) − n(x) m(n(x + h)) − m(n(x)) = × h h n(x + h) − n(x) [ ] m(n(x + h)) − m(n(x)) n(x + h) − n(x) ′ f (x) = lim × h→0 n(x + h) − n(x) h [ ] [ ] m(n(x + h)) − m(n(x)) n(x + h) − n(x) = lim × lim h→0 h→0 n(x + h) − n(x) h

n(x + h) − n(x) . Also, if we let n(x) = A and n(x + h) = A + B, then h→0 h n(x + h) − n(x) = A + B − A, so that By definition, n′ (x) = lim

m(n(x + h)) − m(n(x)) m(A + B) − n(A) = . n(x + h) − n(x) B

Also, as h → 0, B → 0. m(A + B) − m(A) = m′ (A). Consequently, lim B→0 B [ ] m(n(x + h)) − m(n(x)) Therefore, lim = m′ (n(x)). h→0 n(x + h) − n(x) Bringing this all together, we can see that

The derivative of f(x) = m(n(x)) If f (x) = m (n (x)),

f ′ (x) = m′ (n (x)) × n′ (x).

Using Leibnitz notation, this becomes dy dy du = × , where y = f (u) and u is a function of x. dx du dx Consider again y = f (x) = (2x + 1)4 . The chain rule can be used to find the derivative of this function. du Let u = 2x + 1. ∴ =2 dx dy Also let y = u4 . ∴ = 4u3 . du dy dy du = × dx du dx By the chain rule, = 4u3 × 2 Since u = 2x + 1,

= 8u3

dy = 8(2x + 1)3 . dx

158 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 1

Use the chain rule to determine the derivative of y = (x2 + 3x + 5)7 . THINK

y = (x2 + 3x + 5)7 y = u7 and u = x2 + 3x + 5 dy du = 7u6 and = 2x + 3 du dx dy dy du = × dx du dx dy = 7u6 × (2x + 3) dx dy = 7(2x + 3)(x2 + 3x + 5)6 dx

WRITE

Write the function to be derived. 2. Let u equal the inner function and rewrite. dy du 3. Differentiate to determine and . du dx 1.

4.

Apply the chain rule.

5.

Substitute for u and simplify.

WORKED EXAMPLE 2

Determine

2 dy for the function y = (4x − 7) 3 . dx

THINK 1.

Write the function to be derived.

Let u equal the inner function and rewrite. dy du 3. Differentiate to determine and . du dx

2.

4.

Apply the chain rule.

5.

Substitute for u and simplify.

y = (4x − 7) 3

WRITE

2

y = u 3 and u = 4x − 7 dy du 2 −1 = u 3 and =4 du 3 dx dy dy du = × dx du dx dy 2 −1 = u3 ×4 dx 3 dy 8 = √ 3 dx 3 4x − 7 2

WORKED EXAMPLE 3

Determine the derivative of y = cos2 (e2x ). b. Evaluate the derivative when x = 0, giving your answer correct to 4 decimal places. a.

THINK a. 1. 2.

Write the function to be derived. Let u equal the inner function.

y = cos2 (e2x ) WRITE

y = [cos(e2x )] 2 y = u2 and u = cos(e2x )

CHAPTER 5 Further differentiation and applications 159

3.

Use the chain rule to differentiate this inner function.

4.

Consider the outer function.

5.

Differentiate.

6.

Apply the chain rule.

7.

Substitute for u and simplify

b. 1.

Substitute x = 0 into

2.

Answer the question.

Units 3 & 4

Area 2

du = − sin(e2x ) × 2e2x dx du = −2e2x sin(e2x ) dx y = u2 dy = 2u du dy dy du = × dx du dx dy = 2u(−2e2x sin(e2x )) dx dy = −4e2x cos(e2x ) sin(e2x ) dx dy = −4e0 cos(e0 ) sin(e0 ) dx dy = −4 cos(1) sin(1) dx dy = −1.81859485 dx dy = −1.8186 to 4 decimal places. dx

dy . dx

Sequence 4

Concept 1

The chain rule Summary screen and practice questions

Exercise 5.2 The chain rule Technology free 1.

2.

Differentiate each of the following functions. √ a. y = (5x − 4)3 b. y = 3x + 1

c.

d.

f.

WE1

y=

WE2

a.

1 7 − 4x

Determine

y = (3x + 2)2 y=

e.

y = (5x + 3)−6

dy for each of the following functions. dx y = (7 − x)3 √ e. y = 5x + 2

b.

1 (4 − 2x)4 3. Determine the derivatives of the following functions. √ a. f (x) = (4 − 3x)5 b. y = 3x2 − 4 ( )6 1 3 −2 d. g(x) = (2x + x) e. g(x) = x − x d.

160 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

y=

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y = (4 − 3x) 3

4

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c.

f (x) = (x2 − 4x) 3

1 2x − 5 3 f. y = √ 3x − 2

f.

y = (x2 − 3x)−1

1

Use the chain rule to determine the derivatives of the following. y = sin2 (x) b. y = ecos(3x) dy 𝜋 3 5. If y = sin (x), determine the exact value of when x = . 3 dx 6. Determine the derivatives of the following functions.

4.

WE3

a.

√ g(x) = 3(x2 + 1)−1 b. g(x) = ecos(x) c. g(x) = (x + 1)2 + 2 √ 1 d. g(x) = e. f (x) = x2 − 4x + 5 sin2 (x) Simplify each of the following functions and use the chain rule to determine g′ (x). √ ( 2 )3 x +2 6x − 5 a. g (x) = b. g (x) = √ 6x − 5 x2 + 2 For each of the following functions, use the chain rule to determine f ′ (x). ( )−2 2 2 2 3x −1 3 a. f (x) = 3 cos(x − 1) b. f (x) = 5e c. f (x) = x − x2 √ 2−x d. f (x) = e. f (x) = cos3 (2x + 1) 2−x ( ) 2 𝜋 sin (x) ′ If f (x) = e , determine f . 4 Differentiate the following functions and hence determine the gradients at the given x-values. ( ) 2 −2 ′ 1 . b. f (x) = e2x ; determine f ′ (−1). a. f (x) = (2 − x) ; determine f 2 ( ) √ 𝜋 3 ′ 5 ′ 2 4 d. f (x) = (cos(3x) − 1) ; determine f c. f (x) = (3x − 2) ; determine f (1). . 2 If f (x) = sin2 (2x), determine the points where f ′ (x) = 0 for x ∈ [0, 𝜋]. dy 3 cos(5x) MC If y = e , then is: dx A. 15 sin(5x)e3 cos(5x) B. −15 sin(5x)e3 cos(5x) C. e−15 sin(5x) D. −15 cos(5x)e−3 sin(5x) dy 2 MC If y = sin (5x), then is: dx A. 2 sin(5x) B. −2 sin(5x) cos(5x) C. −10 sin(5x) cos(5x) D. 10 sin(5x) cos(5x) 4x MC Let f : R → R be a differentiable function. For all real values of x, the derivative of f (e ) with respect to x will be: A. 4e4x f ′ (x) B. e4x f ′ (x) 4x ′ 4x C. 4e f (e ) D. 4f ′ (e4x ) √ dy For y = 7 − 2f(x), is equal to: dx 2f ′ (x) −1 A. √ B. √ 7 − 2f(x) 2 7 − 2f(x) −f ′ (x) 1√ C. 7 − 2f ′ (x) D. √ 2 7 − 2f ′ (x) √ 2 − 1 and the function g has the rule g(x) = x + 3, calculate the a. If the function f has a rule f (x) = x√ integers m and n such that f(g(x)) = (x + m)(x + n). b. If h(x) = f (g(x)), determine h′ (x). a.

7.

8.

9. 10.

11. 12.

13.

14.

15.

16.

CHAPTER 5 Further differentiation and applications 161

5.3 The product rule 5.3.1 Differentiation using the product rule

Many functions are the product of two or more functions, such as f (x) = x sin(x) or y = e2x (3x + 1). See Worked example 4. To differentiate such functions, the product rule is applied.

The product rule If y = f (x) × g (x)

dy = f (x) × g′ (x) + g (x) × f ′ (x) dx

or If y = uv

dy dv du = u× +v× dx dx dx

5.3.2 Proof of the product rule Let f (x) = u(x)v(x) so f (x + h) = u(x + h)v(x + h).

f (x + h) − f (x) u(x + h)v(x + h) − u(x)v(x) = h h

Add and subtract u(x)v(x + h), as it is expected that at some stage v′ (x) will appear somewhere in the rule. f (x + h) − f(x) u(x + h)v(x + h) − u(x)v(x + h) + u(x)v(x + h) − u(x)v(x) = h h [u(x + h) − u(x)]v(x + h) + u(x)[v(x + h) − v(x)] = h f(x + h) − f(x) f ′ (x) = lim h→0 h [u(x + h) − u(x)]v(x + h) + u(x)[v(x + h) − v(x)] = lim h→0 h [ ] [ ] u(x + h) − u(x) v(x + h) − v(x) = lim × v(x + h) + lim × u(x) h→0 h→0 h h u(x + h) − u(x) v(x + h) − v(x) = lim × lim v(x + h) + lim × lim u(x) h→0 h→0 h→0 h→0 h h ′ ′ = u (x)v(x) + v (x)u(x) = u(x)v′ (x) + v(x)u′ (x)

Using Leibnitz notation, if y = uv,

dy dv du =u +v . dx dx dx

162 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 4

Differentiate the following functions. f (x) = x sin (x) b. y = e2x (3x + 1) a.

THINK a. 1.

Define u and v as functions of x.

2.

Differentiate with respect to x.

3.

Apply the product rule and simplify.

b. 1.

Define u and v as functions of x.

2.

Differentiate with respect to x.

3.

Apply the product rule and simplify.

f (x) = x sin(x) u = x and v = sin(x) du dv = 1 and = cos(x) dx dx dy dv du =u× +v× dx dx dx dy = x × cos(x) + sin(x) × 1 dx dy = x cos(x) + sin(x) dx y = e2x (3x + 1) u = e2x and v = (3x + 1) dv du = 2e2x and =3 dx dx dy dv du =u× +v× dx dx dx dy = e2x × 3 + (3x + 1) × 2e2x dx dy = 3e2x + 6xe2x + 2e2x dx dy = e2x (6x + 5) dx WRITE

WORKED EXAMPLE 5

The graph of f : R → R, f (x) = x2 ex is shown. Using calculus, determine the coordinates where f ′(x) = 0.

y

y = x2ex

0 THINK 1.

Define u and v as functions of x.

2.

Differentiate u and v with respect to x.

x

f (x) = x2 ex Let u(x) = x2 and v(x) = ex . u′(x) = 2x

WRITE

v′(x) = ex

CHAPTER 5 Further differentiation and applications 163

3.

4.

5.

6.

Apply the product rule to determine f ′ (x). Solve f ′(x) = 0.

= x2 ex + 2xex x2 ex + 2xex = 0

ex > 0 for all values of x. Either x = 0 or x + 2 = 0.

∴ x = 0, –2 When x = −2,

y = (−2)2 e−2

= 4e−2 When x = 0,

y = (0)2 e0 =0 The coordinates where the gradient is zero are (0, 0) and (−2, 4e−2 ).

Write the answer.

Area 2

= x2 × ex + ex × 2x

ex x(x + 2) = 0

Substitute the x-values to determine the corresponding y-values.

Units 3 & 4

f ′(x) = u(x)v′(x) + v(x)u′(x)

Sequence 4

Concept 2

The product rule Summary screen and practice questions

Exercise 5.3 The product rule Technology free 1.

For each of the following functions, determine the derivative function. f (x) = sin(3x) cos(3x) b. f (x) = x2 e3x c. f (x) = (x2 + 3x − 5)e5x dy Determine for the following functions. dx 5 4 3 5 a. y = x2 (x + 1) b. y = x3 (2x − 1) c. y = (4x + 1) (3x − 2) Determine the derived functions for the following. √ √ √ 5 a. f (x) = (x + 1) x b. f (x) = x x + 1 c. f (x) = e4x x Differentiate the following. √ a. x2 e5x b. x−2 (2x + 1)3 c. x cos(x) d. 2 x(4 − x) Differentiate the following. √ 2 3 a. 3x−2 ex b. e2x 4x2 − 1 c. x2 sin (2x) d. (x − 1)4 (3 − x)−2 ( ) 𝜋 If f (x) = 2x4 cos(2x), determine f ′ . 2 WE5 Given the function f (x) = (x + 1) sin(x), determine f ′ (x) and hence determine the gradient of the function when x = 0. Given f (x) = 2x2 (1 − x)3 , use calculus to determine the coordinates where f ′ (x) = 0. WE4

a.

2.

3. 4. 5.

6. 7. 8.

164 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

9.

10.

11.

12.

13.

14. 15.

y The graph of f: R+ → R, f (x) = e− 2 sin(x) is shown. a. Calculate the values of x when f (x) = 0 for x ∈ [0, 3𝜋]. b. Use calculus to determine the values of x when f ′ (x) = 0 for x ∈ [0, 3𝜋]. Give your answers correct to 2 decimal places. 0 2 4 6 Determine the derivative of the following functions and hence determine the gradients at the given points. a. f (x) = xex ; determine f ′ (−1). b. f (x) = x(x2 + x)4 ; determine f ′ (1). √ √ 2 c. f (x) = x sin (2x2 ); determine f ′ ( 𝜋). 3 MC If f (x) = (x − a) g(x), the derivative of f (x) is equal to: A. 3(x − a)g′ (x) B. 3(x − a)2 g′ (x) ′ C. 3g (x) D. 3(x − a)2 g(x) + (x − a)3 g′ (x) 8 MC The derivative of 12p(1 − p) with respect to p is equal to: A. 96p(1 − p)8 B. −96p(1 − p)8 7 C. 12(1 − p) (1 − 9p) D. 12(1 − p)8 (1 − 9p) dy 𝜋 3 MC Let y = 2x sin(x). The derivative when x = is: 2 dx 3𝜋 2 3𝜋 𝜋2 A. B. 4𝜋 2 C. D. 2 2 2 2 ′ ′ Evaluate f (2a) if f (x) = (x − a) g (x), given g (2a) = 6 and g (2a) = 3. ( ) ′ 𝜋 If f (x) = g(x) sin(2x) and f = −3𝜋, calculate the constant a if g(x) = ax2 . 2 x

8

10

12

x

5.4 The quotient rule 5.4.1 Differentiation using the quotient rule

ex x or y = , we have the cos(x) x2 + 1 quotient of the two functions. To differentiate such functions, the quotient rule is applied. See Worked example 6. Before applying the quotient rule, always check if the function can be first simplified. For example, 5x2 − 2x y= can be simplified to y = 5x−1 − 2x−2 and differentiated by the basic differentiation rule. 3 x When one function is divided by a second function, such as f (x) =

The quotient rule If y =

f(x) where g(x) ≠ 0, g(x)

g(x) × f ′ (x) − f(x) × g′ (x) dy = . dx [g(x)]2

or u If y = where v(x) ≠ 0, v

v× dy = dx

du dx

−u× v2

dv dx

.

CHAPTER 5 Further differentiation and applications 165

5.4.2 Proof of the quotient rule This rule can be proven as follows by using the product rule. u(x) If f(x) = , then f(x) = u(x) × [v(x)]−1 . v(x) f ′ (x) = u(x) × −1 × [v(x)]−2 × v′ (x) + [v(x)]−1 × u′ (x) =−

= =

u(x)v′ (x) u′ (x) + [v(x)] [v(x)]2

u′ (x)v(x) u(x)v′ (x) − [v(x)]2 [v(x)]2 v(x)u′ (x) − u(x)v′ (x) [v(x)]2

− u dv v du dy u dx dx In Leibnitz notation, the quotient rule states that if y = , then . = 2 v dx v WORKED EXAMPLE 6

Determine the derivatives of the following functions. x ex a. f (x) = b. y = x2 + 1 cos (x) THINK a. 1.

Define u and v as functions of x.

2.

Differentiate u and v with respect to x.

3.

Apply the quotient rule and simplify.

b. 1.

Define u and v as functions of x.

2.

Differentiate u and v with respect to x.

3.

Apply the quotient rule and simplify.

4.

Factorise the numerator.

f(x) =

WRITE

x x +1 2

u = x and v = x2 + 1

du dv = 1 and = 2x dx dx

v × du − u × dv dy dx dx = dx v2 dy (x2 + 1) × 1 − x × 2x = dx (x2 + 1)2 dy 1 − x2 = 2 dx (x + 1)2 ex y= cos(x) u = ex and v = cos(x)

du dv = ex and = −sin(x) dx dx dv du dy v × dx − u × dx = dx v2 dy cos(x) × ex − ex × (− sin(x)) = dx (cos(x))2 dy ex (cos(x) + sin(x)) = dx cos2 (x)

166 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 7

Determine the derivative of y =

sin (2t) t2

with respect to t. y=

THINK

WRITE

1.

Define u and v as functions of t.

2.

Differentiate u and v with respect to t.

3.

Apply the quotient rule to determine simplify.

dy and dt

sin(2t) t2 Let u = sin(2t) and v = t2 . du = 2 cos(2t) dt dv = 2t dt dv du dy v dt − u dt = dt v2 t2 2 cos(2t) − sin(2t) × 2t = (t2 )2 2t(t cos(2t) − sin(2t)) = t4 2(t cos(2t) − sin(2t)) = t3

WORKED EXAMPLE 8

Determine the derivative of f (x) = x = 0.

cos (3x) and hence determine the gradient at the point where 2ex − x f(x) =

THINK

WRITE

1.

Define u and v as functions of x.

2.

Differentiate u and v with respect to x.

3.

Apply the quotient rule to determine and simplify.

4.

Evaluate f ′(0).

dy dx

cos(3x) 2ex − x Let u(x) = cos(3x) and v(x) = 2ex − x. u′(x) = −3 sin(3x)

v′(x) = 2ex − 1 v(x)u′(x) − u(x)v′(x) f ′(x) = v2 x (2e − x) × −3 sin(3x) − cos(3x) × (2ex − 1) = (2ex − x)2 x −3(2e − x) sin(3x) − (2ex − 1) cos(3x) = (2ex − x)2 −3(2e0 − 0) sin(0) − (2e0 − 1) cos(0) f ′(0) = (2e0 − 0)2 0−1 = 4 1 =− 4

CHAPTER 5 Further differentiation and applications 167

5.4.3 The derivative of tan(x)

To determine a rule for the derivative of the tangent function, we can write tan(x) =

quotient rule. Let u = sin(x) and v = cos(x). du dv = cos(x) and = −sin(x). dx dx By the quotient rule,

sin(x) and apply the cos(x)

( ) d d sin(x) (tan(x)) = dx dx cos(x) v du − u dv dx dx = 2 v cos(x) × cos(x) − sin(x) × − sin(x) = (cos(x))2 cos2 (x) + sin2 (x) = cos2 (x) 1 = (by the Pythagorean identity) cos2 (x)

WORKED EXAMPLE 9

Find the derivative of y = tan (3x). THINK 1

Write the equation.

du 2 Express u as a function of x and find . dx dy 3 Express y as a function of u and find . du dy 4 Find using the chain rule. dx

Units 3 & 4

y = tan(3x) WRITE

Area 2

Sequence 4

Concept 3

Let u = 3x so

du = 3. dx dy 1 y = tan(u) so = . du cos2 (u) dy 3 = dx cos2 (u) 3 = 2 cos (3x)

The quotient rule Summary screen and practice questions

168 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Exercise 5.4 The quotient rule Technology free x+3 u 1. WE6 If y = is expressed as y = , determine: x+7 v du dv a. u and v b. and dx dx x2 + 2x u 2. If f(x) = is expressed as f(x) = , determine: 5−x v du dv a. u and v b. and dx dx 3. Determine the derivative of each of the following. 2x x2 + 7x + 6 a. y = b. y = 3x + 2 x2 − 4 8 − 3x2 4. MC If h(x) = , then h′ (x) equals: x −3x2 + 8 −3x2 − 8 8 − 9x2 A.

5.

6. 7. 8. 9.

10.

11.

12.

13.

14.

B.

C.

c.

dy dx

c.

f ′ (x)

c.

f(x) =

4x − 7 10 − 3x

D.

−3x2 + 8 x

x2 x2 x2 WE7 Use the quotient rule to determine the derivatives of: cos(3t) e2x a. b. x e +1 t3 x+1 Determine the derivative of 2 . x −1 sin(x) WE8 If y = , determine the gradient of the function at the point where x = 0. e2x 5x If y = 2 , calculate the gradient of the function at the point where x = 1. x +4 Differentiate the following. sin2 (x2 ) 3x − 1 e−x ex b. y = d. a. y = c. x cos(2x + 1) x−1 2x2 − 3 Differentiate the following. √ 3 x sin(x) (5 − x)2 x − 4x2 a. √ b. f(x) = √ c. f(x) = d. y = √ x+2 2 x x 5−x WE9 Differentiate each of the following. ( ) ( ) x −3x a. y = tan(2x) b. y = tan(−4x) c. y = tan d. y = tan 5 4 Calculate the gradient at the stated point for each of the following functions. sin(2x + 𝜋) 2x 𝜋 ,x= a. y = ,x=1 b. y = 2 2 cos(2x + 𝜋) x +1 2 x+1 5−x c. y = √ ,x=5 d. y = ,x=0 ex 3x + 1 2x Calculate the gradient of the tangent to the curve with equation y = at the point where x = 1. 3 (3x + 1) 2 ( ) −2 sin(x) d 1 + cos(x) Show that =− . dx 1 − cos(x) (cos(x) − 1)2

CHAPTER 5 Further differentiation and applications 169

d 15. a. Show that dx

(

sin(x) cos(x)

)

=

1 . cos2 (x)

𝜋 Hence, determine the gradient of the curve y = tan(x) at the point where x = . 4 √ 2x − 1 2 16. Given that f(x) = √ , calculate m such that f ′ (m) = √ . 2x + 1 5 15 b.

5.5 Applications of differentiation

Differentiation can be applied to many different situations. These include: • finding tangents to curves at specific points • curve sketching • optimisation — finding where maximum or minimum values occur within given constraints • kinematics, the study of motion • rates of change — investigating how a change in one variable affects another related variable. This section introduces some of these concepts. Chapter 8 covers these situations in more detail. The various rules of differentiation may have to be used first before an application problem can be solved.

5.5.1 Tangents and curve sketching The derivative of a function gives the gradient of the tangent to the curve at any point. The derivative also shows whether the function has a stationary point, or, as x increases, if the function is increasing or decreasing. dy • If = 0, the function has a stationary point. dx dy • If > 0, the function is increasing. dx dy • If < 0, the function is decreasing. dx Identifying stationary points provides information that assists curve sketching. Stationary points are classified as: • local minimum turning points • local maximum turning points • stationary points of inflection (or horizontal points of inflection). The nature of a stationary point is determined by examining the slope of the tangent to the curve immediately before and after the stationary point. The word ‘local’ means that the point is a minimum or maximum in a particular locality or neighbourhood. Beyond this section of the graph, there could be other points on the graph that are lower than the local minimum or higher than the local maximum. The nature of a stationary point is summarised in the following table. Minimum turning point

Maximum turning point

Stationary point of inflection

Stationary point or Slope of tangent or

170 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 10

Given that y = e2x (x + 1)2 , evaluate

curve at the point (0, 1).

dy and hence determine the equation of the tangent to the dx y = e2x (x + 1)2 Let u = e2x and v = (x + 1)2 . du = 2e2x dx dv = 2(x + 1) dx dy dv du =u +v dx dx dx dy = e2x × 2(x + 1) + (x + 1)2 × 2e2x dx = 2e2x (x + 1) + 2e2x (x + 1)2

THINK

WRITE

1.

Define u and v as functions of x.

2.

Differentiate u and v with respect to x.

3.

Apply the product rule to determine simplify.

dy and dx

= 2e2x (x + 1)(1 + x + 1)

= 2e2x (x + 1)(x + 2)

dy when x = 0. dx

4.

Evaluate

5.

Determine the equation of the tangent.

When x = 0, then dy = 2e0 (0 + 1)(0 + 2) dx =4 If m = 4 and (x1 , y1 ) = (0, 1), y − y1 = m(x + x1 ) y − 1 = 4(x − 0) y − 1 = 4x y = 4x + 1

WORKED EXAMPLE 11

Consider the function f (x) = ex (x − 2)3 . ′ a. Calculate f (x) and hence determine the coordinates of the stationary points. ′ b. By investigating the sign of f (x), state the nature of these stationary points. c. Investigate the values of f (x) as x → ± ∞. State the equations of any asymptotes. d. Calculate any axis intercepts. e. Sketch the curve of y = f (x), showing all important features. f. State the domain and range of the function. THINK a. 1. 2.

Define u and v as functions of x. Differentiate with respect to x.

f (x) = ex (x − 2)3 u = ex and v = (x − 2)3 du dv = ex and = 3(x − 2)2 dx dx WRITE

CHAPTER 5 Further differentiation and applications 171

3.

Apply the product rule and simplify by factorising.

4.

Stationary points exist when

5.

Determine y-values for x = 2, x = −1.

b. 1.

dy = 0. dx

Construct a table of values for f ′ (x) for suitable values of x.

dy dv du =u× +v× dx dx dx dy = ex × 3(x − 2)2 + (x − 2)3 × ex dx dy = ex (x − 2)2 [3 + (x − 2)] dx f ′ (x) = ex (x − 2)2 (x + 1)

ex (x − 2)2 (x + 1) = 0 ex = 0 is undefined, so x = 2 or x = −1 ) ( −27 The stationary points are −1, and (2, 0). e f ′ (x) = ex (x − 2)2 (x + 1) x

f ′ (x)

−2

−1

0

2

3

0

4

0

4e3

\

—

/

—

/

−16e−2

) −27 −1, is a minimum stationary point; (2, 0) e is a horizontal point of inflexion (or stationary point of inflexion) f (x) = ex (x − 2)3 ex → ∞; (x − 2)3 → ∞ ∴ as x → ∞, f (x) → ∞ ex → 0; (x − 2)3 → −∞ ∴ as x → −∞, f (x) → 0 from the negative side. y = 0 is an asymptote. (

2.

c. 1.

State the nature of the stationary points by considering the direction of the tangents. Consider the behaviour of f (x) as x → ∞ (or as x becomes very large).

Consider the behaviour of f(x) as x → −∞ (or as x becomes very small). 3. State equations of asymptotes Note: y = 0 is only an asymptote for small values of x. d. 1. For x-intercepts, y = 0. 2.

For y-intercepts, x = 0. 3. State the axis intercepts. 2.

e. 1.

Draw axes and plot the axis intercepts and stationary points, noting their nature.

ex (x − 2)3 = 0 ∴ x = 2 or (2, 0) f (0) = e0 (−2)3 ∴ y = −8 or (0, −8) The axis intercepts are (2, 0) and (0, −8). y

10 y = ex(x – 2)3

5 (2, 0)

y=0 –6

–4

0

–2

2

–5 (0, –8)

(

–10 27 –1, – e

172 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

)

4

6

x

Remember the x-axis is an asymptote on the left, as x → −∞. f. 1. State the domain. 2.

2.

State the range.

TI | THINK a.1. The graphing function

can be used to calculate the location of any stationary point(s). On a Calculator page, press MENU, then select: 2: Add Graphs. 2. Complete the entry line

in the f 1(x) = tab as: ex (x − 2)3 Press ENTER to sketch the graph.

3. On a Calculator page,

press MENU, then select: 6: Analyze Graph 2: Minimum.

4. Inspect the graph and set

lower and upper bounds to either side of the minimum value.

5. The answer appears on

the screen.

WRITE

f (x) = ex (x − 2)3 f (x) is defined for all values of x. The domain is x ∈ R. −27 The minimum y-value is e [ ) −27 −27 The range is y ≥ or y ∈ ,∞ . e e CASIO | THINK

WRITE

a.1. On a Main Menu

screen, select: Graph.

2. Complete the function

entry line in the Y1 tab as: ln(x − 2)

3. Press the DRAW

button to sketch the graph.

4. Determine the

minimum value by selecting: SHIFT F5 (G-Solv) MIN.

5. The answer appears

on the screen.

CHAPTER 5 Further differentiation and applications 173

The stationary point of inflection can be determined by locating the x intercept as described in part d. The stationary point of inflection is (2, 0). d.1. On a Calculator page, press MENU, then select: 6: Analyze Graph 2: Minimum. 6.

2. Inspect the graph and set

lower and upper bounds to either side of the x-intercept.

3. The answer appears on

the screen.

4. On a Calculator page,

select: Menu 8: Geometry 1: Points & Lines 2: Point On.

d.1. Determine the

x-intercept by selecting: SHIFT F5 (G-Solv) ROOT.

2. The answer appears

on the screen.

3. Determine the

y-intercept by selecting: SHIFT F5 (G-Solv) ROOT.

4. The answer appears

on the screen.

5. Move the cursor and

select the curve representing f 1(x)ex (x − 2)3 . Press the ESC (escape) button. Complete the entry line in the textbox as 0 for the x-value. Press ENTER to perform the calculation. The answer appears on the screen.

5.5.2 Maximum and minimum problems To solve optimisation problems, apply the following steps. 1. Draw diagrams when necessary. 2. Determine the variables and any connection between them. 3. State or determine the function to be optimised. 174 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

4. Differentiate the function and determine the stationary points and their nature. 5. Reject any unrealistic solution. 6. Answer the question. WORKED EXAMPLE 12

The profit, $P, per item that a store makes by selling n items of a certain type each day is √ P = 40 n + 25 − 200 − 2n. a. Determine the number of items that need to be sold to maximise the profit on each item. b. Calculate: i. the maximum profit per item ii. the total profit per day made by selling this number of items. THINK a. 1.

WRITE

Rewrite with powers.

√ P = 40 n + 25 − 200 − 2n 1

P = 40 (n + 25) 2 − 200 − 2n 2.

Differentiate with respect to n.

3.

Simplify.

4.

Solve

5.

Draw a sign diagram to justify your answer.

6.

dP = 0 for n. dn

State the answer.

b. i. 1.

Calculate P(75).

State the answer. ii. 1. Calculate the total profit for selling 75 items per day. 2. State the answer. 2.

−1 1 dP = 40 × × (n + 25) 2 × 1 − 2 dn 2 dP 20 −2 =√ dn n + 25 20 −2=0 √ n + 25 20 =2 √ n + 25 √ 20 = 2 n + 25 √ n + 25 = 10 n + 25 = 100 n = 75

n

70

75

80

dp dn

≈ 0.052

0

≈ −0.048

slope

/

—

\

The maximum profit per item is obtained when 75 items are sold each day. √ P = 40 n + 25 − 200 − 2n √ P = 40 75 + 25 − 200 − 2 × 75 P = 40 × 10 − 200 − 150 P = 50 The maximum profit per item is $50. Total profit = $50 × 75 The maximum total profit per day for selling 75 items is $3750.

CHAPTER 5 Further differentiation and applications 175

WORKED EXAMPLE 13

Find the minimum distance from the curve y = 2x2 to the point (4, 0), correct to 2 decimal places. You do not need to justify your answer.

y y = 2x2

(4, 0) 0

x

Let P be the point on the curve such that the P = (x, y) distance from P to the point (4, 0) is a minimum. √ 2. Write the formula for the distance between d(x) = (x2 − x1 )2 + (y2 − y1 )2 √ the two points. = (x − 4)2 + (y − 0)2 √ = (x − 4)2 + y2 y = 2x2 3. Express the distance between the two points √ as a function of x only. ∴ d(x) = (x − 4)2 + (2x2 )2 THINK

WRITE

1.

= (x2 − 8x + 16 + 4x4 ) 2

1

4.

5.

Differentiate d(x).

Solve d′ (x) = 0 using technology.

6.

Evaluate d(0.741).

7.

Write the answer.

d′ (x) =

1 −1 × (4x4 + x2 − 8x + 16) 2 × (16x3 + 2x − 8) 2

16x3 + 2x − 8 = √ 2 4x4 + x2 − 8x + 16

8x3 + x − 4 =√ 4x4 + x2 − 8x + 16 8x3 + x − 4 0= √ 4x4 + x2 − 8x + 16 0 = 8x3 + x − 4

x = 0.741 √ d(0.741) = 4(0.741)4 + (0.741)2 − 8(0.741) + 16 = 3.439 The minimum distance is 3.44 units (to 2 decimal places).

WORKED EXAMPLE 14

A new window is to be made to allow more light into a room. The window will have the shape of a rectangle surmounted by a semicircle. The frame of the window will be made from aluminium measuring 336 cm.

176 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

1 Show that the area, A cm2 , of the window is A = 336x − (4 + 𝜋) x2 , where x is the radius of 2 the semicircle in cm. b. Hence, determine the width of the window for which the area is greatest. Give your answer to the nearest cm. c. Structural limitations mean that the width of the window should not exceed 84 cm. What should the dimensions of the window of maximum area now be? Give your answer to the nearest cm. a.

THINK a. 1.

WRITE

Draw a diagram to illustrate the window where radius of the semicircle is x cm and the height of the rectangle is h cm.

h

P = 𝜋x + 2x + 2h 336 = 𝜋x + 2x + 2h 2x

Use the perimeter to form an expression connecting the two variables, x and h. 3. Express h in terms of x. 2.

4.

Express the area as a function of x by substituting for h.

5.

Express the area in the required form.

b. 1. 2.

Differentiate. Determine the stationary point and its nature.

2h = 336 − 𝜋x − 2x 1 h = (336 − 𝜋x − 2x) 2 1 A = 2xh + 𝜋x2 2 1 1 A = 2x × (336 − 𝜋x − 2x) + 𝜋x2 2 2 1 A = 336x − 𝜋x2 − 2x2 + 𝜋x2 2 1 A = 336x − 2x2 − 𝜋x2 2 1 A = 336x − (4 + 𝜋) x2 2 dA = 336 − (4 + 𝜋)x dx 336 − (4 + 𝜋)x = 0 336 x= 4+𝜋 x ≈ 47.05

CHAPTER 5 Further differentiation and applications 177

x

40

50

dA dx

336 (4 + 𝜋)

positive

0

negative

slope

/

—

\

The maximum area occurs when x =

c. 1.

State the restrictions for the window.

State the restricted domain for A(x). 3. Sketch the graph of A(x).

2.

336 cm or 4+𝜋

x ≈ 47 cm (to the nearest cm). The width of the window is not to exceed 84 cm. 2x ≤ 84 x ≤ 42 Restricted domain of area: x ∈ (0, 42] A 8000 6000 4000 2000

The maximum area occurs when x = 42 cm. 0

4.

Determine, for the restricted domain, the value of x for the greatest area.

5.

Calculate the height, h.

6.

Calculate the dimensions.

7.

State the answer.

42 47

94

x

h=

1 (336 − 𝜋 × 42 − 2 × 42) 2 h = 60.0266 Width = 2x = 84 Total height = h + x = 60 + 42 = 102 With the restrictions, the area of the window will be greatest if the width is 84 cm and the total height is 102 cm.

5.5.3 Rates of change and kinematics

The instantaneous rate of change, or simply the rate of change, of the function y = f(x) is given by the dy derivative, or f ′ (x). dx dV The derivative could be the rate of change of volume with respect to time. dt Kinematics, the study of the motion of a particle moving in a straight line, involves determining rates of change of displacement and velocity with respect to time. Displacement, x, gives the position of a particle by specifying both its distance and direction from a fixed point, the origin. 178 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Commonly used conventions for motion in a horizontal straight line are as follows: • If x > 0, the particle is to the right of the origin. • If x < 0, the particle is to the left of the origin. • If x = 0, the particle is at the origin. Common units for displacement are cm, m and km. dx Velocity, v, measures the rate of change of displacement with respect to time, so v = . dt For a particle moving in a horizontal straight line, the sign of the velocity indicates direction: • If v > 0, the particle is moving to the right. • If v < 0, the particle is moving to the left. • If v = 0, the particle is stationary, or instantaneously at rest. Common units for velocity are cm/s, m/s and km/h. dv Acceleration, a, measures the rate of change of velocity with respect to time, so a = . Common units dt for acceleration include m/s2 . The term ‘initially’ means at the start, or when t = 0. WORKED EXAMPLE 15

The number of mosquitoes, N, around a dam on a certain night can be modelled by the equation N=

400 + 100t + 1000 2t + 1

where t equals hours after sunset. Find: the initial number of mosquitoes b. the rate of change at any time, t c. the rate of change when t = 4 hours. a.

THINK a. 1. 2.

Write the rule.

Calculate N when t = 0.

Answer the question. b. 1. Differentiate N with respect to t. 3.

2.

Simplify.

c. 1.

Calculate

N=

400 + 100t + 1000 2t + 1 400 N= + 0 + 1000 1 N = 1400 Initially, there were 1400 mosquitoes. −1 b. N = 400 (2t + 1) + 100t + 1000

WRITE

dN when t = 4. dt

a.

dN = 400 × −1 (2t + 1)−2 × 2 + 100 dt dN −800 = + 100 dt (2t + 1)2 dN −800 c. = + 100 dt (2 × 4 + 1)2 −800 = + 100 (9)2 7300 = 81 ≈ 90.1

CHAPTER 5 Further differentiation and applications 179

2.

Answer the question.

After 4 hours, the rate of change is approximately 90.1 mosquitoes per hour.

WORKED EXAMPLE 16

The displacement, x metres, of a particle after t seconds is given by the equation x = 4sin (2t) + 3. a. Derive an expression for the velocity, v m/s, of the particle. b. Determine the time at which the particle is first at rest and its position at this time. c. Derive an expression for acceleration, a m/s2 , of the particle and its initial acceleration. THINK a. 1. 2.

State the displacement function. Differentiate with respect to t.

State the expression for velocity. b. 1. At rest, v = 0. 2. Solve for t. 3.

The first time the particle is at rest is the lowest value of t. ( ) 𝜋 4. Substitute to determine x . 4 3.

5.

c. 1.

Answer the question.

State the velocity function.

2.

Differentiate with respect to t.

3.

For initial acceleration, substitute t = 0.

4.

Answer the question.

x = 4 sin(2t) + 3 dx v= dt v = 4 cos(2t) × 2 v = 8 cos(2t) b. 8 cos(2t) = 0 cos(2t) = 0 𝜋 3𝜋 ,… 2t = , 2 2 𝜋 3𝜋 t= , ,… 4 4 𝜋 t = seconds 4 ( ) 𝜋 x = 4 sin 2 × +3 4 ( ) 𝜋 +3 = 4 sin 2 =4+3 =7 𝜋 The particle is first at rest after 4 seconds and it is 7 metres to the right of the origin. c. v = 8 cos(2t) dv a= dt = 8 × (− sin(2t)) × 2 = −16 sin(2t) a = −16 sin(2 × 0) = −16 sin(0) =0 The acceleration of the particle is given by the equation a = −16 sin(2t) and the initial acceleration is 0 m/s2 .

WRITE a.

180 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 17

A particle moves in a straight line such that its displacement, x metres, from a fixed origin at time t seconds is modelled by x = t2 − 4t − 12, t ≥ 0. a. Identify its initial position. b. Determine its velocity function and hence state its initial velocity and describe its initial motion. c. At what time and position is the particle momentarily at rest? d. Show the particle is at the origin when t = 6, and calculate the distance it has travelled to reach the origin. Calculate the value of x when t = 0.

x = t2 − 4t − 12, t ≥ 0 When t = 0, x = −12. Initially the particle is 12 metres to the left of the origin. dx Calculate the rate of change required. b. v = dt v = 2t − 4 Calculate the value of v at the given When t = 0, v = −4. instant. The initial velocity is −4 m/s. Describe the initial motion. Since the initial velocity is negative, the particle starts to move to the left with an initial speed of 4 m/s. Calculate when the particle is momentarily c. The particle is momentarily at rest when its at rest. velocity is zero. Note: This usually represents a change of When v = 0, direction of motion. 2t − 4 = 0

THINK

WRITE/DRAW

a.

a.

b. 1.

2. 3.

c. 1.

2.

d. 1.

2.

Calculate where the particle is momentarily at rest.

Calculate the position to show the particle is at the origin at the given time.

Track the motion on a horizontal displacement line and calculate the required distance.

t=2 The particle is at rest after 2 seconds. The position of the particle when t = 2 is x = (2)2 − 4(2) − 12 = −16 Therefore, the particle is momentarily at rest after 2 seconds at the position 16 metres to the left of the origin. d. When t = 6, x = 36 − 24 − 12 =0 The particle is at the origin when t = 6. The motion of the particle for the first 6 seconds is shown. t=2

t=6 t=0

–16 –12

0

x

Distances travelled are 4 metres to the left, then 16 metres to the right.

CHAPTER 5 Further differentiation and applications 181

The total distance travelled is the sum of the distances in each direction. The particle has travelled a total distance of 20 metres.

Resources Interactivities: Stationary points (int-5963) Rates of change (int-5960) Kinematics (int-5964)

Units 3 & 4

Area 2

Sequence 4

Concepts 4, 5 & 6

Equations of tangents Summary screen and practice questions Curve sketching Summary screen and practice questions Maximum and minimum problems Summary screen and practice questions

Exercise 5.5 Applications of differentiation Technology free 1.

WE10

3.

4. 5.

6.

√

3x2 + 2x, determine:

dy dx b. the equation of the tangent to the curve at x = 2. a.

2.

For the function y =

Use the chain rule to determine the derivative of y =

1 and hence determine the equation of the (2x − 1)2

tangent to the curve at the point where x = 1. 3 a. Evaluate f ′ (−1) if f (x) = √ . 5 − 4x b. Hence, determine the equation of the tangent to the curve y = f (x) at the point where x = −1. For the function with the rule y = xex , determine the equations of the tangent and the line perpendicular to the curve at the point where x = 1. √ The function h has a rule h(x) = x2 − 16 and the function √ g has the rule g(x) = x − 3. a. Determine the integers m and n such that h(g(x)) = (x + m)(x + n). b. State the maximal domain of h(g(x)). c. Determine the derivative of h(g(x)). d. Determine the gradient of the function h(g(x)) at the point when x = −2. x WE11 Consider the function f (x) = . 2 x +1 a. Calculate f ′ (x) and hence determine the coordinates of the stationary points. b. By investigating the sign of f ′ (x), state the nature of these stationary points. c. Investigate the values of f (x) as x → ± ∞. State the equations of any asymptotes. d. Calculate any axis intercepts. e. Sketch the curve of y = f (x), showing all important features. f. State the domain and range of the function.

182 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Consider the function f (x) = ln(x2 + 1). a. Calculate f ′ (x) and hence determine the coordinates of the stationary point. b. By investigating the sign of f ′ (x), state the nature of the stationary point. c. Investigate the values of f (x) as x = ±1, ±2, ±3. Explain why, in this logarithmic function, x-values can be negative. d. Sketch the curve of y = f (x), showing all important features. e. State the domain and range of the function. 8. Consider the function y = (x − 2)2 (x + 3)2 . a. Differentiate the function with respect to x. b. Determine the coordinates of any stationary points and their nature. c. Sketch the function, clearly showing all important features. d. State the domain and range of the function.

7.

The graph of y = e−x (1 − x) is shown. a. Determine the coordinates of the points where the graph cuts the x- and y-axes. b. Determine the coordinates of the points where the gradient is 0, giving your answers correct to 3 decimal places. c. Determine the equation of the tangent to the curve at the point where the curve intersects the x-axis. d. Determine the equation of the line perpendicular to the curve where the curve crosses the y-axis. e. Where do the tangent and the perpendicular line from parts c and d intersect? Give your answer correct to 2 decimal places. 10. The graph of the function f : R → R, f(x) = 3x3 e−2x is shown. The derivative may be written as f ′ (x) = ae−2x (bx2 + cx3 ) where a, b and c are constants. a. Calculate the exact values of a, b and c. b. Calculate the exact coordinates where f ′ (x) = 0. c. Determine the equation of the tangent to the curve at x = 1. 9.

2

y

x

0

y

0

x

Technology active 11. The length of a snake, L cm, at any time t weeks after it is born is modelled as:

L = 12 + 6t + 2 sin

𝜋t , 0 ≤ t ≤ 20. 4

Calculate: a. the length at: i. birth ii. 20 weeks b. R, the rate of growth, at any time, t c. the maximum and minimum growth rate.

CHAPTER 5 Further differentiation and applications 183

12.

The graph of the function f : R+ → R, f (x) = y

1 + loge (x) is shown. 10x 1 + log (x) y= — e 10x

x

0

x=0

Use calculus to determine the coordinates of the minimum turning point. 13. The profit, $P, per item that a store makes by selling n items of a certain type each day is given √ by P = 80 n + 8 − 15 − 5n. a. Determine the number of items that need to be sold to maximise the profit on each item. b. Calculate: i. the maximum profit per item ii. the total profit per day made by selling this number of items. 14. The population of cheetahs, P, in a national park in Africa since 1 January 2010 can be modelled as WE12

N = 100te

−

t 12

+ 500 where t is the number of years.

When does this model predict that the maximum population will be reached? What is the maximum population of cheetahs that will be reached? c. How many cheetahs will there be on 1 January in: i. 2034? ii. 2094? 15. The amount of money in a savings account t years after the account was opened on 1 January 2009 is given by the equation a.

b.

A(t) = 1000 − 12te

3

4−t 8

for t ∈ [0, 6].

How much money was in the account when the account was first opened? b. What was the least amount of money in the account? c. When did the account contain its lowest amount? Give the year and month. d. How much money was in the account at the end of the six years? a.

184 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

16.

WE13

Determine the minimum distance from the point (1, 1) to the straight line y = x − 4. y y=x–4 1

P (1, 1)

0

1

4 Q (x, y)

x

–4 17.

√ Find the minimum distance from the line y = 2 x to the point (5, 0). y

0 (0, 0)

(5, 0) x

√ A rectangle with its base on the x-axis is inscribed in the semicircle y = 4 − x2 . √ a. Show that the area, A, of the rectangle is A = 2x 4 − x2 . b. Hence, determine the dimensions of the largest rectangle that can be inscribed in this semicircle. c. State the maximum area of the rectangle. 19. The owner of an apartment wants to create a stained glass feature in the shape of a rectangle surmounted by an isosceles triangle of height equal to half its base. This will be adjacent to a door opening on to a balcony. The owner has 150 cm of plastic edging to place around the perimeter of the figure, and wants to determine the dimensions of the figure with the greatest area. ) ( √ a. Show that the area, A in cm2 , of the stained glass feature is A = 150x − 2 2 + 1 x2 . 18.

WE14

y

Hence, determine the width and the height of the figure for which the area is greatest. Give your answers correct to 1 decimal place. 2x c. Due to structural limitations, the width of the feature should not exceed 30 cm. What should the dimensions of the stained glass feature of maximum area now be? Give your answer correct to 1 decimal place. 20. A metal gutter is to be formed from a sheet of metal 30 cm wide and 5 m long. The three sides of the gutter are to be equal in length, forming a trapezoidal cross section. The sides are folded so the angle θ θ between the vertical and the side is 𝜃, as shown in the diagram. a. Show that the area, A, of the cross section is given by A = 100 cos(𝜃)(1 + sin(𝜃)). 10 b. State the restrictions on the value of 𝜃 for this metal gutter. c. Determine the value of 𝜃 that gives a maximum area of the cross section. d. Hence, calculate the maximum volume of the gutter that can be formed from this sheet of metal. 21. WE15 A colony of viruses can be modelled by the rule b.

N(t) =

2t + 0.5 (t + 0.5)2

where N hundred thousand is the number of viruses on a nutrient plate t hours after they started multiplying.

CHAPTER 5 Further differentiation and applications 185

How many viruses were present initially? Find N′ (t). c. What is the maximum number of viruses, and when will this maximum occur? d. At what rate were the virus numbers changing after 10 hours? 22. A population of butterflies in an enclosure at a zoo is modelled by a.

b.

N = 220 −

23.

24.

25.

26.

27.

150 ,t≥0 t+1

where N is the number of butterflies t years after observations of the butterflies commenced. a. How long did it take for the butterfly population to reach 190 butterflies, and at what rate was the population growing at that time? b. At what time was the growth rate 12 butterflies per year? Give your answer correct to 2 decimal places. c. Determine the growth rate after 10 years. Give your answer correct to 2 decimal places. d. Sketch the graphs of population versus time and rate of growth versus time, and explain what happens to each as t → ∞. WE16 The displacement, x metres, of a particle after t seconds is given by the equation x = 2 cos(4t) − 5. a. Derive an expression for the velocity, v m/s, of the particle. b. The particle is initially at rest. Determine the next time the particle is at rest and its position at this time. c. Derive an expression for acceleration, a m/s2 , of the particle and its initial acceleration. The displacement, x metres from the origin, of a particle moving in a straight line after t hours is given ( ) 𝜋 by the equation x (t) = 6 − 4 sin t for 0 ≤ t ≤ 24. 6 a. State the period and amplitude for the function. b. Determine the initial position of the particle. c. Derive an expression for velocity, v m/h. d. Determine the position of the particle when it is first at rest. e. Sketch the function, x (t). What observations can you make from the graph? WE17 A particle moves in a straight line such that its displacement, x metres, from a fixed origin at time t seconds is given by x = 2t2 − 8t, t ≥ 0. a. Identify its initial position. b. Derive an expression for its velocity and hence state its initial velocity and describe its initial motion. c. At what time and position is the particle momentarily at rest? d. Show that the particle is at the origin when t = 4 and calculate the distance it has travelled to reach the origin. 1 The position, in metres, of a particle after t seconds is given by x(t) = − t3 + t2 + 8t + 1, t ≥ 0. 3 a. Find its initial position and initial velocity. b. Calculate the distance travelled before it changes its direction of motion. c. What is its acceleration at the instant it changes direction? The position, x m, relative to a fixed origin of a particle moving in a straight line at time t seconds is 2 x = t3 − 4t2 , t ≥ 0. 3 a. Show the particle starts at the origin from rest. b. At what time and at what position is the particle next at rest? c. When does the particle return to the origin? d. What are the particle’s speed and acceleration when it returns to the origin?

186 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

A ball is thrown vertically upwards into the air so that after t seconds its height h metres above the ground is h = 50t − 4t2 . a. At what rate is its height changing after 3 seconds? b. Calculate its velocity when t = 5. c. At what time is its velocity −12 m/s and in what direction is the ball then travelling? d. When is its velocity zero? e. What is the greatest height the ball reaches? f. At what time and with what speed does the ball strike the ground? 29. The profile of water waves produced by a wave machine in a scientific laboratory is modelled by the trigonometric function f (x) = a sin(x) + b cos(x). 28.

y

(0, 7)

y = f (x)

( ) π, 3 – 2

0

π – 2

π

3π – 2

2π

x

( a.

Given that the graph of the wave profile passes through the points (0, 7) and

) 𝜋 , 3 , calculate the 2

constants a and b. Determine the maximum and minimum swells for the wave profile, correct to 1 decimal place. Hence, state the range of the function. c. Determine {x : f(x) = 0, 0 ≤ x ≤ 2𝜋}, giving your answers correct to 4 decimal places. d. Evaluate, correct to 3 decimal places, the gradient at the x-values found in part c. Comment on your results. 30. A country town has decided to construct a new road. The x-axis is also the position of the railway line that connects Sydney with Brisbane. The road can be approximated by the equation y = (4x2 − 5x)ex . b.

y

y = (4x2 – 5x)ex 0

T

x

B

The post office for the town is positioned at (−2, 3.5). They want the new road to be adjacent to the post office. Have they made a sensible decision regarding the placement of the road? b. Determine the coordinates of the point T where the road crosses the railway line. c. Use calculus to determine the coordinates of the point B. Give your answer correct to 3 decimal places. a.

CHAPTER 5 Further differentiation and applications 187

5.6 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar dy 1. Determine for each of the following functions. dx a. y = 3(2x2 + 5x)5 b. y = (4x − 3x2 )−2 )6 ( 1 c. y = x + d. y = 4(5 − 6x)−4 x 2. Use the product rule to differentiate each of the following. a. y = x2 sin(x) b. y = 3x sin(x) c. y = x5 cos(3x + 1) f. y = 5 cos(2x) sin(x) d. y = sin(x) cos(x) e. y = 8 sin(5x) loge (5x) 3. Differentiate each of the following, expressing your answer in simplest form. ( ) 4x a. y = sin cos(x) b. y = 2x−3 sin(2x + 3) c. y = 4e−5x sin(2 − x) 3 1 e. y = sin x loge (x) f. y = 𝜋x cos(2𝜋x) d. y = √ cos(6x) x 4. Determine the derivative of each of the following. sin(x) sin(4x) cos(x) a. y = b. y = c. y = x cos(2x) x (√ ) sin x cos(x) 2 cos(3 − 2x) d. y = e. y = f. y = x e x x2 2 5. MC The derivative of f (x) = x sin(2x) is: A. f ′ (x) = 4x cos(2x) B. f ′ (x) = 2x sin(2x) + x2 cos(2x) C. f ′ (x) = 2x sin(2x) + 2x2 cos(2x) D. f ′ (x) = 2x sin(x) + 2x2 cos(x) sin(4x) 6. MC The derivative of f (x) = is: 4x + 1 4(4x + 1) cos(x) − 4 sin(4x) (4x − 1) cos(4x) − 4 sin(4x) A. f ′ (x) = B. f ′ (x) = 2 (4x + 1) (4x + 1)2 4 sin(4x) − 4(4x − 1) cos(4x) 4(4x − 1) cos(4x) − 4 sin(4x) D. f ′ (x) = C. f ′ (x) = 2 (4x + 1) (4x + 1)2 2 3x 7. Given that y = (x + 1)e , determine the equation of the tangent to the curve at x = 0. 8. A curve is represented by the equation y = ax cos(3x) where a is a constant. dy = −5 when x = 𝜋, what is the value of a? a. If dx 𝜋 b. Determine the equation of the line perpendicular to the curve at x = . 3 9. The function f : R → R, f (x) = 6 loge (x2 − 4x + 8) has one stationary point. a. Use calculus to determine the coordinates of this stationary point. b. Determine the nature of this stationary point. c. Use technology to sketch the function, f. 10. Sketch the following functions by determining their stationary points and any axis intercepts. State the range of each function. 4 a. y = x4 − 4x3 b. y = 2 x +1 11. A particle moves in a straight line so that at time t seconds its displacement, x metres, from a fixed origin O is given by x(t) = t3 − 6t2 + 9t, t ≥ t ≥ 0. 188 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

How far is the particle from O after 2 seconds? What is the velocity of the particle after 2 seconds? c. After how many seconds does the particle reach the origin again, and what is its velocity at that time? d. What is the particle’s acceleration when it reaches the origin again? √ 12. A particle moves in a straight line so that its displacement a point, O, at any time, t, is x = 3t2 + 4. Determine: a. the velocity as a function of time b. the acceleration as a function of time c. the velocity and acceleration when t = 2. Complex familiar 13. Metal box guttering has to be formed on a common wall b cm between two adjacent town houses. The cross section of the box guttering is shown. For the most efficient elimination of rain water, this box guttering needs to have a maximum cross-sectional area h cm 20 cm 20 cm within the given dimensions. a. Determine an expression for h, the height of the trapezium, in terms of the angle x in radians, as shown. x x b. Determine an expression for b, the base length of the trapezium, 10 cm in terms of x. 2 c. Show that the cross-sectional area of the box guttering, A cm , is given by A = 200 sin(x)(2 cos(x) + 1). d. Determine, correct to 3 decimal places, the value of x that gives maximum cross-sectional area, and find this maximum area correct to the nearest cm2 . 14. A bushwalker can walk at a rate of 5 km/h through clear land and B 3 km/h through bushland. She has to get from point A to point B following a route indicated ( ) on the diagram. Clear distance Bush Note: Time = 3 km speed a.

b.

x

Determine the distance walked in terms of x: i. through the bush ii. through clear land. A 2 km b. If the total time taken is T hours, express T as a function of x. dT c. Derive . dx d. Hence, determine the value of x so that the route is covered in a minimum time. e. Calculate the minimum time to complete this route. Give your answer in hours and minutes. 1 1 15. The line perpendicular to the graph y = g( f(x)), where f (x) = and g(x) = x − 2 , is given by x x y = −x + a, where a is a real constant. Calculate the possible value(s) of a. 16. Consider the functions f(x) = 2 sin(x) and h(x) = ex . a. State the rule for: i. m(x) = f (h(x)) ii. n(x) = h( f (x)) b. Determine when m′ (x) = n′ (x) over the interval x ∈ [0, 3], correct to 3 decimal places. Complex unfamiliar ( −3x ) e−x (a + be−2x ) d e 17. If = , calculate the exact values of a and b. dx e2x + 1 (e2x + 1)2 18. a. Consider the function f : R → R f(x) = x4 e−3x . The derivative f ′ (x) may be written in the form f ′ (x) = e−3x (mx4 + nx3 ), where m and n are real constants. Calculate the exact values of m and n. a.

CHAPTER 5 Further differentiation and applications 189

b.

The graph of f is shown. Determine the coordinates of the stationary points. y

y = f (x)

x

0

19.

Let f: [−2, 2] → R, f (x) =

sin(2x − 3) . The graph of this function is shown. ex (a, b)

–2

y

0 (c, d)

2

x

The stationary points occur at (a, b) and (c, d). Calculate the values of a, b, c and d, giving your answers correct to 3 decimal places. b. Determine the gradient of the tangent to the curve at the point where x = 1, correct to 3 decimal places. 20. The population of rabbits on a particular island t weeks after a virus is introduced is modelled by P = 1200e−0.1t , where P is the number of rabbits. Determine: a. the time taken for the population to halve (to the nearest week) b. the rate of decrease of the population after: i. 2 weeks ii. 10 weeks. a.

After 15 weeks the virus has become ineffective and the population of rabbits starts to increase again according to the model p = p0 + 10 (t − 15) loge (2t − 29) where t is the number of weeks since the virus was first introduced. Calculate: c. the value of P0 d. the population after 30 weeks e. the rate of change of the population after i. 20 weeks ii. 30 weeks f. how many weeks the population takes to get back to its original number.

Units 3 & 4

Sit exam

190 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

16. a. m = 2, n = 4

Answers

b. h (x) = √ ′

5 Further differentiation and applications Exercise 5.2 The chain rule 1. a. c. e. 2. a.

dy dx dy dx dy dx dy dx dy

= 15(5x − 4)2 = =

b.

−8

(2x + 3)5 −30

d.

(5x + 3)7

f.

= 6(3x + 2)

b.

−2

=

(2x − 5)2 5 e. = √ dx 2 5x + 2 c.

3. a. b. c.

dx dy

dy

dx dy dx dy dx dy

d. f.

= −15(4 − 3x)4 = √ =

2

3

dx dy dx dy dx dy dx dy dx dy dx

Exercise 5.3 The product rule

3 = √ 2 3x + 1 4 = (7 − 4x)2 √ 3 = −4 4 − 3x

dy

= −3(7 − x)2 = =

(4 − 2x)5 −9

(x − 2)(x2 − 4x)

−

8

2(3x − 2) 2 3

2 3

= −2(6x2 + 1)(2x3 + x)−3 ( )( ) 1 1 5 =6 1+ 2 x− e. dx x x dy 2 −2 = −(2x − 3)(x − 3x) f. dx dy dy 4. a. = 2 cos(x) sin(x) b. = −3 sin(3x)ecos(3x) dx dx 9 d.

5.

dx dy

8

6. a. − d.

+

−2 cos(x)

1)2

sin3 (x)

7. a. g (x) = ′

b. − sin(x)e

6x (x2

−3

(6x −

3 5) 2

8. a. −6x sin(x − 1)

10. 11. 12. 13. 14. 15.

x−2

x2 − 4x + 5

b. g (x) = 5x(x + 2) 2 ′

2

3x −1 2

6x5 + 8 1 d. ( )3 3 2 2(2 − x) 2 x3 x3 − 2 x e. −6 sin(2x + 1) cos2 (2x + 1) √ e 16 2 a. b. −4e c. 8 27 ( )( )( ) 𝜋 3𝜋 𝜋 (0, 0), ,1 , ,0 , , 1 , (𝜋, 0) 4 2 4 B D C D c. −

9.

e. √

c. √

2

b. 30xe

′

2. a.

b. c.

5. a.

6. 7.

3

8. 9. 10. 11. 12. 13. 14.

d. 15

dy

dx dy dx dy

2

2

= x(x + 1)4 (7x + 2)

= x2 (2x − 1)3 (14x − 3)

= 3(4x + 1)2 (3x − 2)4 (32x − 3)

(x + 1)4 (11x + 1) √ dx 2 x (3x + 2) dy = √ b. dx 2 x+1 dy e4x (1 + 8x) c. = √ dx 2 x dy 4. a. = 5x2 e5x + 2xe5x dx dy 2(x − 1)(2x + 1)2 = b. dx x3 dy c. = −x sin(x) + cos(x) dx dy 4 − 3x d. = √ dx x

x+1

x2 + 2x + 3

cos(x)

1. a. f (x) = 3 cos (3x) − 3 sin (3x) ′ 2 3x 3x b. f (x) = 3x e + 2xe ′ 2 5x c. f (x) = (5x + 17x − 22)e

3. a.

3x

3x2 − 4

(x + 3)

(x + 2)(x + 4)

15.

dx dy

dy

=

=

6ex (x2 − 1) 2

x3 2e (4x2 + 2x − 1) b. = √ dx 4x2 − 1 dy = 2x sin2 (2x)[3x cos(2x) + sin(2x)] c. dx dy 2(x − 5)(x − 1)3 d. = dx (x − 3)3 −𝜋 3 f ′ (x) = sin(x) + (x + 1) cos(x); gradient = 1 ( ) 2 216 (0, 0), (1, 0), , 5 3125 a. x = 0, 𝜋, 2𝜋, 3𝜋 b. x = 1.11, 4.25, 7.39 a. 0 b. 112 c. 0 D C A f ′ (2a) = 3a (a + 4) 6 a= 𝜋 dx dy

2x

Exercise 5.4 The quotient rule 1. a. u = x + 3; v = x + 7 c.

dy dx

=

(x + 7)2 4

b.

du dx

= 1;

dv dx

=1

CHAPTER 5 Further differentiation and applications 191

2. a. u = x + 2x; v = 5 − x 2

b.

du dx

= 2x + 2;

3. a. b.

dy

dx dy dx

= =

4. C 5. a. b.

dy dx dy

dt −1

=

=

(x − 1)2 7. 1 3

6.

8.

5

9. a. b. c. d. 10. a. b. c. d. 11. a. c.

dy dx dy dx dy dx dy

(x2 −

= =

−1

3. a.

3x + 4x − 4 (3x + 2)2 19

c. √

(ex + 1)2 −3(t sin(3t) + cos(3t))

4x2 sin(x2 ) cos(x2 ) − sin2 (x2 ) x2 −6x + 4x − 9

(2x2 − 3)2 e cos(2x + 1) + 2ex sin(2x + 1) 2

=−

cos2 (2x + 1)

or 2x − 9y + 11 = 0

b. 2

(3x + 1) = √ dx 3x2 + 2x

1

x+

(

x−3

d.

2e

)

2e

7} −5 3

y 1

x y = –––– x2 + 1

(1, 0.5)

y=0 –4

b.

1 + 2e2

b. {x : x ≤ −1} ∪ {x : x ≥

1)2

dy

=

−4

cos2 (−4x) −3 d. = ( −3x ) dx 4 cos2 4 7 c. d. −5 64 dx dy

Exercise 5.5 Applications of differentiation dy

9

e.

32 14. Sample responses can be found in the worked solutions in the online resources. 15. a. Sample responses can be found in the worked solutions in the online resources. b. 2 16. m = 2

1. a.

11

( ) ( ) 1 1 Stationary points: −1, − and 1, 2 2 ) ( 1 b. Local minimum stationary point at −1, − 2 ( ) 1 Local maximum stationary point at 1, 2 x c. As x → ∞, → 0 (positive side). x2 + 1 x → 0 (negative side). Equation of As x → −∞, 2 x +1 asymptote: y = 0 d. Intercept at (0, 0)

t4

xe−x

9

x+

x2 − 6x − 7 1 − x2 ′ 6. a. f (x) = (x2 + 1)2

2

x

x

2

5. a. m = −7, n = 1

e (2 + e ) 2x

7

4. Tangent: y = 2e x − e; normal: y = −

2

(10 − 3x)

9

b. y =

4)2

(x − 2x cos(x) − sin(x) = √ dx 2x x dy −3 √ = 5−x dx 2 √ 1 f ′ ( x) = √ − 3 x 4 x dy 6 − 3x = √ dx 2 x(x + 2)2 dy 2 = dx cos2 (2x) dy 1 (x) = dx 5 cos2 5 dx dy

12. a. 0 13.

=

dv

(5 − x)2 −2(x2 + 4)

c. f (x) = ′

1 x + or 7x − 4y + 2 = 0 4 2 2. y = −4x + 5 2

= −1 dx 10 + 10x − x2

c. f (x) = ′

b. y =

–3

(0, 0) 0 –2 –1 (–1, –0.5) –1

f. Domain: x ∈ R; range −

7. a. f (x) = ′

1 2

≤y≤

1

2

3

4

1 2

; stationary point: (0, 0) x2 + 1 b. Local minimum turning point 2 2 c. For all x-values, x ≥ 0 and x + 1 ≥ 1 . Hence, this logarithmic function is defined and is greater than or equal to zero, even when x-values are negative. 2x

d.

y 5 y = ln(x2 + 1) –15

–10

–5

0 (0, 0) 5

10

15

x

–5 e. Domain: x ∈ R; range: y ≥ 0

= 2 (x − 2) (x + 3) (2x + 1) dx b. Local minimum stationary points at (−3, 0) and (2, 0) ) ( −1 625 Local maximum stationary point at , 2 16

8. a.

dy

192 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

b.

40 (–0.5, 39.063) (0, 36) 30 y = (x – 2)2(x + 3)2

20 10 (–3, 0) –6

0.5 − 2t2

21. a. 50 000 viruses

y

c.

c. d. 22. a. b. c. d.

(t + 0.5)4 150 000 viruses after half an hour −1641 viruses/hour 4 years; 6 butterflies/year 2.54 years 1.24 butterflies/year N

(2, 0)

–4

–2

0

2

4

6

x

n = 220

220

–10

( )

150 N = 220 – – t+1

d. Domain: x ∈ R; range: y ≥ 0 9. a. (0, 1) and (1, 0) b. (−0.366, 1.195) and (1.366, −0.057) c. y = −

1

x+

1

or x + ey − 1 = 0

e e y=x+1 (−0.46, 0.54) a = 3, b = 3, c = −2 Stationary point of inflection at (0, 0); local maximum ( ) 3 81 turning point at , 3 2 8e 3 c. y = x or 3x − e2 y = 0 e2 11. a. i. 12 cm ii. 132 cm ( ) 𝜋 𝜋t dL =6+ cos b. dt 2 4 𝜋 c. Maximum rate of growth = 6 + ≈ 7.571 cm/week; 2 𝜋 ≈ 4.429 cm/week minimum rate of growth = 6 − 2 ( ) 12. 0.1, 1 − loge (10) d. e. 10. a. b.

56 items i. $345 ii. $19 320 12 years, 1 January 2022 941 cheetahs i. 824 cheetahs ii. 507 cheetahs $1000 b. $980.34 May 2010 d. $1000 √ 2 2 units 4 units a. Sample responses can be found in the worked solutions in the online √ resources. √ b. Base: 2 2 units; height: 2 units c. 4 square units a. Sample responses can be found in the worked solutions in the online resources. b. Width: 39.2 cm; height: 47.3 cm c. Width: 30 cm; height: 53.8 cm a. Sample responses can be found in the worked solutions in the online resources. 𝜋 b. 0 ≤ 𝜃 < 2 𝜋 c. 𝜃 = 6√ 3 d. 37 500 3 cm

13. a. b. 14. a. b. c. 15. a. c.

16. 17. 18.

19.

20.

(0, 70) dN 150 – =– dt ( t + 1)2

As t → ∞, N → 220 and 0

23. a. v = −8 sin(4t) b.

c. 24. a. b. c.

𝜋

dN dt

s; −7 m 4 a = −32 cos(4t); −32 m/s2 12 h; 4 6 m to the right of the origin ( ) 2𝜋 𝜋 v=− cos t 3 6 2 m to the right of the origin

t

→ 0.

d. e. x (m)

(9, 10)

10 5

(0, 6)

25. a. b. c. d. 26. a. 27. a. b. c. d. 28. a. b. c. d. e. f.

5

(24, 6) y = 6 – 4 sin π x 6 (15, 2) 15 20 25 t (h)

( )

(3, 2) 0

(21, 10)

10

The particle is at rest at the turning points of the curve where the displacement is 2 m and 10 m. The particle oscillates between these two positions. At the origin −8 m/s; travelling to the left of the origin at 8 m/s 2 s; −8 m (8 m to the left of the origin) 16 m 2 1 m; 8 m/s b. 27 m c. −6 m/s2 3 Sample responses can be found in the worked solutions in the online resources. 1 4 s; 21 m to the left of the origin 3 6s 24 m/s; 16 m/s2 26 m/s 10 m/s 7.75 s travelling downwards 6.25 s 156.25 m 12.5 s; 50 m/s

CHAPTER 5 Further differentiation and applications 193

29. a. a = 3, b = 7 b. Maximum swell = 7.6 units; c. d.

30. a. b. c.

minimum swell = −7.6 units; range = [−7.6, 7.6] x = 1.9757 or 5.1173 f ′ (1.9757) = −7.616; f ′ (15.1173) = 7.616 The gradients are equal in magnitude (size), just differing in direction. One is when the swell is going down, and the other is when the swell is rising. This is due to the symmetry of the curve representing the swell. Yes ( ) 5 T= ,0 4 B = (0.804, −3.205)

c.

y

25

y = 6 ln(x2 – 4x + 8)

20 15 10 (2, 8.318)

5

–10

0

–5

5

10

x

y

10. a.

5.6 Review: exam practice 1. a. 15(4x + 5)(2x + 5x) 2 −3 b. −4(2 − 3x)(4x − 3x ) 2

4

10 y = x4 – 4x3

) ( ) 1 5 1 1− 2 6 x+ x x 96(5 − 6x)−5 x2 cos(x) + 2x sin(x) 3x cos(x) + 3 sin(x) 5x4 cos(3x + 1) − 3x5 sin(3x + 1) cos2 (x) − sin2 (x) 8 sin(5x) + 40 cos(5x) loge (5x) x 5 cos(x) cos(2x) − 10 sin(x) sin(2x) ( ) ( ) 4 4x 4x cos(x) cos − sin(x) sin 3 3 3 4x−3 cos(2x + 3) − 6x−4 sin(2x + 3) −4e−5x cos(2 − x) − 20e−5x sin(2 − x) −6 sin(6x) cos(6x) − √ √ x 2 x3 1 sin(x) + cos(x) loge (x) x 𝜋 cos(2𝜋x) − 2𝜋 2 x sin(2𝜋x) x cos(x) − sin(x) (

c.

d. 2. a. b. c. d. e. f. 3. a. b. c. d. e. f. 4. a. b. c. d.

f.

2

(3, –27)

b.

Stationary points: (0, 0) and (3, −27) Axis intercepts: (0, 0) and (4, 0) Range: y ≥ −27 y 5 (0, 4)

y = 24 x +1 y=0 –2

x3

2, 12 loge (2)

)

b. Local minimum

b. y =

0

–1

Stationary point: (0, 4) Axis intercept: (0, 4) Range: 0 < y ≤ 4 11. a. 2 m c. 3 s; at rest (v = 0) 3t 12. a. v = √ 3t2 + 4

2x2 4x sin(3 − 2x) − 4 cos(3 − 2x)

(

x

–30

ex ) (√ (√ ) x − 2 sin x

8. a. 5

4

–20

x2 ( ) − sin(x) + cos(x) x cos

(4, 0)

–10

cos2 (2x) −x sin(x) − cos(x)

5. C 6. A 7. y = 3x + 1 9. a.

2

x2 4 cos(2x) cos(4x) + 2 sin(2x) sin(4x)

√ e.

(0, 0) 0

1 5

x−

26𝜋 15

c. v = 1.5; a =

1

2

b. −3 m/s d. 6 m/s2

b. a = (√

3

x

12

3t2 + 4

)3

13. a. h = 20 sin(x) b. b = 10 + 40 cos(x) c. Sample responses can be found in the worked solutions

16

in the online resources.

c

d. For a maximum, the angle x is 0.936 and the maximum 2

area is 352 cm .

194 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

4 + x2 km ii. (3 − x) km 3 x 1√ b. T = − + 4 + x2 5 5 3 x 1 dT − = √ c. dx 5 3 4 + x2 d. 1.5 km e. 1 h 8 min 15. a = −3 2 sin(x) x 16. a. i. m(x) = 2 sin(e ) ii. n(x) = e b. x = 1.555, 2.105, 2.372, 2.844 17. a = −5, b = −3 14. a. i.

√

18. a. m = −3, n = 4

19. a. b. 20. a. b. c. d. e. f.

) 4 256 , 3 81e4 a = −1.088, b = 2.655, c = 0.483, d = −0.552 0.707 7 weeks i. 98.25 rabbits/week ii. 44.15 rabbits/week 267 782 rabbits i. 33 rabbits/week ii. 44 rabbits/week 39 weeks (

b. (0, 0),

CHAPTER 5 Further differentiation and applications 195

REVISION UNIT 3 Further calculus

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196 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland



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Answers

3

12. a. 2x +

6 Antidifferentiation x2 8 3

3. 4.

5.

6.

7.

8. 9.

10.

11.

5

2

b. 2x + 2x + c

x2 − 8x + c

x4

+ x3 − 3x2 − 9x + c 2 2 2 a. x + 5x + c 3 2 b. x + 2x − 10x + c 3 5 4 c. 2x + x + 2x + c 2 −2 6 1 4 x + x − 2x3 + x2 + c d. 3 4 1 4 1 e. x + 12x − x3 + c 4 3 √ √ 1 8 √ +c b. 4x x + 6 x + 8x + c a. x x + 3 x 1 5 1 4 1 a. +c x +c b. x + c c. − 25 8 9x3 2 3 3 5 16 7 d. x 2 + c e. x 3 + c f. x4 + c 3 5 7 7 4 5 9 a. x 7 + c b. − +c c. − + c 2 4 2x x √ 2 12 d. f. √ + c +c e. 16 x + c x5 x 1 3 a. x − 2x2 − 21x + c 3 5 3 x + 5x2 − 5x + c b. 3 1 4 7 3 c. x − x + 2x2 − 28x + c 4 3 1 4 d. x + x3 − 2x2 + c 4 √ 1 4 1 2 2 5 4 3 a. x + x + c b. x 2 + x 2 − 2 x + c 4 2 5 3 −2 −1 2 c. −5x + x + x + c 1 1 f ( x) = x3 + + c 3 x 1 4 7 3 1 a. x + c b. x + +c 4 3 5x2 16 2 √ 7 3 4 2 d. x x+c c. x − x + x − x + c 3 5 3 2 4 3 1 4 a. f (x) = x − x + x + c 4 3 2 √ 1 4 b. 6 x − x − +c 5x2 1 4 1 3 15 2 x − x − x +c c. 2 3 2 √ √ 3 3 1 d. x x− x x+c 7 3 √ 3 1 a. 4 x − + 2 +c x 4x 1 4 1 3 1 b. x − x + x2 + 4x + c 2 3 2 c. x +

2.

4

+c

x2 − 6x + c

2

d.

1

x2 + x −

1

+c 2 x 1 7 4 +c d. x − 4x − 7 5x5

b.

√ 4 √ x x− 8 x+c 3 √ 1 4 13. y = x − 2x x + c 4 1 2 3 14. y = x + 3x + +c 2 x √ 2 √ 15. y = x x + 2 x + c 3 c.

Exercise 6.2 Antidifferentiation of rational functions 1. a.

5

Exercise 6.3 Antidifferentiation of exponential functions 1. a.

x8 8

d. −

+c 1

b.

e−3x + c

3

1 4

e. e

e4x + c

5x

c. −e

+c

x

x

e. e − e

−x

6. a. 7. a.

1 5 1 2

1 6

2

x

d. −9 e 3 + c

+c

f.

e2x − 2x −

4. y = x + e 5. y =

e4x + c

−x

x

2

+c

b. 0.4e 4 + c = 5 e 4 + c

c. 6 e 2 + c

1

4

x

2. a. 3 e 3 + c

3. y =

7

f.

−x

2x

+

1 4

1 2

1 2

ex +

1 2

e−x + c

e−2x + c

e4x + c

e6x + 4 e3x + 36x + c

x5 + ex −

1 4 1 4

e−4x + c e−6x + c

x

− x 8. e + 4e 2 −

1

e2x +

4

−

x 2

+c 3 2 1 b. e2x− e3x + e4x + c 2 3 4 b.

4 1

e

1

e−2x + c 2 2x 9. f (x) = 2e + 8x + c 1 4x 10. y = e − x + c 4

12. 13. 14. 15.

3x

x

+ 3x3 − 4e 2 + c 1 6x 3 1 e − 3 ex − e−4x + e−9x + c 6 4 9 −2x a. −4 b. f (x) = −2e − 4x + c a = −9, b = 3 m = 10, n = 2, p = 4, q = −3

11. y = 2e

Exercise 6.4 Antidifferentiation of logarithmic functions 1. a. 3 ln(x) + c d. 2. a. c. e. 3. a. c.

7

ln(x) + c

b. 8 ln(x) + c e.

4

3 7 ln(x + 3) + c −2 ln(x + 4) + c 4 ln(3x + 2) + c 3 8 ln(5x + 6) + c 5 5 − ln(3 + 2x) + c 2

224 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

c.

6

5

ln(x) + c

ln(x) + c b. 3 ln(x + 3) + c d. −6 ln(x + 5) + c

b.

3 2

d. −

ln(2x − 5) + c 2 7

ln(6 + 7x) + c

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y

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(0, 1)

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y = x3 – 4

C

x y = x3

(0, –2)

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y = x3 – 2

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(0, –4) C G (Y) Ⴝ Y Ⴛ  y  B

F

(0, 4)

y = –sin(2x) + 4

(0, 2)

y = –sin(2x) + 2

(0, 0) π

π – 2

3π – 2

y = –sin(2x) x 2π y = –sin(2x) – 2

H  B C D

(0, –5)

y = –sin(2x) – 5

(0, –7)

y = –sin(2x) – 7

E F

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C

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(0, 0) y = x3 + 1

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(0, 3) y = x3 + 3

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y = –e –3x + 2 y = –e –3x + 1

(0, 1)

y = –e –3x

x

(0, –1) y = –e –3x – 2 (0, –3) y = –e –3x – 4 (0, –5)

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7

Integration

7.1 Overview In the previous two chapters, you studied various aspects of calculus. Calculus examines how a change in one variable will affect another related variable. Historically, integration began with mathematicians attempting to find the area under curves. Various methods of estimating the areas under curves are investigated in this chapter. This leads to the fundamental theorem of calculus and its applications to the definite integral. Integration has many real-life applications. In economics and commerce, if marginal costs are known, predictions can be made for total costs incurred. In science, given the rate of change of a substance, such as an oil spill, the total area affected can be ascertained. Motion in a straight line, or kinematics, is yet another application of both differential and integral calculus. This chapter continues on from the antidifferentiation chapter, studying in particular the definite integral together with its applications in real-life situations.

LEARNING SEQUENCE 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Overview Estimating the area under a curve The fundamental theorem of calculus and definite integrals Areas under curves Areas between curves Applications of integration Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

228 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

7.2 Estimating the area under a curve There are several different ways to estimate or approximate the area between a curve and the x-axis. This section considers three methods: • the left end-point rectangle method • the right end-point rectangle method • the trapezoidal method.

7.2.1 The left end-point rectangle method

Consider the curve defined by the function f (x) = x2 + 2 and the area between this curve and the x-axis from x = 0 to x = 3. • Construct rectangles of width 1 unit. • The height of the rectangle is given by the function on the left side, so the height of the first rectangle is f (0), the height of the second rectangle is f (1) and so on. • Approximate area:

• • •

A = 1 × f (0) + 1 × f (1) + 1 × f (2) =2+3+6 = 11 square units

y y = f(x)

(2, f(2)) (1, f(1)) (0, f(0)) 0

1

2

3

y

Construct rectangles of width 0.5 units. The height is again given by the function on the left side of the rectangle. Approximate area: A = 0.5 × f (0) + 0.5 × f (0.5) + 0.5 × f (1) + 0.5 × f (1.5) + 0.5 × f (2) + 0.5 × f(2.5) = 0.5 × ( f (0) + f (0.5) + f (1) + f (1.5) + f (2) + f (2.5)) = 0.5 × (2 + 2.25 + 3 + 4.25 + 6 + 8.25) = 12.875 square units

Observe from these examples that: • the narrower the rectangles, the closer the approximation is to the actual area under the curve • the approximate area is less than the actual area • if the function was a decreasing function, for example f (x) = 10 − x2 , then the area using the left end-points would be greater than the actual area under the curve.

x

4

y = f(x)

(2, f(2)) (1, f(1)) (0, f(0)) 0

1

2

3

4

x

y

y = f (x)

0

1

2

3

4

x

CHAPTER 7 Integration 229

7.2.2 The right end-point rectangle method

Again, consider the curve defined by the function f (x) = x2 + 2 and the area between this curve and the x-axis from x = 0 to x = 3. • Construct rectangles of width 1 unit. • The height of the rectangle is given by the function on the right side, so the height of the first rectangle is f (1), the height of the second rectangle is f (2) and so on. • Approximate area: A = 1 × f (1) + 1 × f (2) + 1 × f (3) = 3 + 6 + 11 = 20 square units

• •

•

Construct rectangles of width 0.5 units. The height is again given by the function on the right side of the rectangle. Approximate area:

A = 0.5 × f (0.5) + 0.5 × f (1) + 0.5 × f (1.5) + 0.5 × f (2) + 0.5 × f (2.5) + 0.5 × f(3) = 0.5 × ( f (0.5) + f (1) + f (1.5) + f (2) + f (2.5) + f (3)) = 0.5 × (2.25 + 3 + 4.25 + 6 + 8.25 + 11) = 17.375 square units

y (3, f(3)) y = f(x)

(2, f(2))

(1, f(1))

0

230 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

3

4

x

(3, f(3)) y = f(x)

(2, f(2))

(1, f(1))

1

2

3

4

x

4

x

y

y = f(x)

0

The rectangle methods can also be described as the lower rectangle method (if the tops of the rectangles are below the curve) and the upper rectangle method (if the tops of the rectangles are above the curve). • If the function is an increasing function, then the left end-point rectangles are lower rectangles and the right end-point rectangles are upper rectangles. • If the function is a decreasing function, then the left end-point rectangles are upper rectangles and the right end-point rectangles are lower rectangles.

2

y

0

Observe from these examples that: • the narrower the rectangles, the closer the approximation is to the actual area under the curve • the approximate area is greater than the actual area • if the function was a decreasing function, for example f (x) = 10 − x2 then the area using the right end-points would be less than the actual area under the curve.

1

1

2

3

7.2.3 The trapezoidal method

1 (a + b) × h, where a and b are the lengths of the 2 parallel sides and h is the width of the trapezium. Recall that for a trapezium, area A =

b a h

Once again, consider the curve defined by the function f (x) = x2 + 2 and the area between this curve and the x-axis from x = 0 to x = 3. • Construct trapeziums of width 1 unit. • The lengths of the parallel sides are defined by f (x) . 1 • Area of T1 = ( f (0) + f (1)) × 1 2 y 1 = (2 + 3) 2 5 = 2 1 • Area of T2 = ( f (1) + f (2)) × 1 2 1 = (3 + 6) 2 9 = 2 1 • Area of T3 = ( f (2) + f (3)) × 1 2 1 0, f(0)) = (6 + 11) 2 17 = 2 0 5 9 17 Total area = + + • 2 2 2 31 = 2 = 15.5 square units

y = f(x)

(2, f(2))

(1, f(1))

T1

T2

T3 x

1

2

3

4

Note: When the widths of the rectangles and the trapezia are the same, the average of the areas found using left end-point rectangles and right end-point rectangles gives the same result as the trapezoidal rule. For example, using a width of 1 unit: Area using left end-points = 11 Area using right end-points = 20 Average of areas = (11 + 20)/2 = 15.5 square units, the area of the trapezium.

CHAPTER 7 Integration 231

WORKED EXAMPLE 1

Points on a function, y = f (x), are shown in the diagram. An approximation of the area between the function y = f (x) and the x-axis from x = 1 to x = 7 is to be found using the given information. Determine the approximate area using: a. the left end-point rectangle method b. the right end-point rectangle method c. the trapezoidal method.

y 20 18 16 14 12 10 8 6 4 2 0

THINK a. 1.

Draw the left end-point rectangles on the graph. State the width and heights of the rectangles.

(1, 20) (3, 18) y = f (x) (5, 12)

(7, 2) 1

3

5

7

WRITE a.

y

(1, 20) (3, 18) y = f (x) (5, 12)

0

2.

b. 1.

3

5

x

7

The width of each rectangle is 2 units. The heights of the rectangles are 20, 18 and 12 units. A = 2 × 20 + 2 × 18 + 2 × 12 = 100 The area is approximately 100 square units.

Calculate the approximate area by adding the areas of the rectangles. Draw the right end-point rectangles on the graph. State the widths and heights of the rectangles.

1

b.

y (3, 18) y = f (x) (5, 12)

(7, 2) 0

1

3

5

7

x

The width of each rectangle is 2 units. The heights of the rectangles are 18, 12 and 2 units.

232 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

2.

c. 1.

A = 2 × 18 + 2 × 12 + 2 × 2 = 64 The area is approximately 64 square units.

Calculate the approximate area by adding the areas of the rectangles. Draw trapeziums on the graph. State the widths of the trapeziums.

c.

y

(1, 20) (3, 18) y = f (x) (5, 12)

(7, 2) 0

2.

Use the formula to calculate the area of each trapezium.

3.

Calculate the approximate area by adding the areas of the trapeziums.

1

3

5

7

x

The width of each trapezium is 2 units. 1 T1 = (20 + 18) × 2 = 38 2 1 T2 = (18 + 12) × 2 = 30 2 1 T3 = (12 + 2) × 2 = 14 2 Area = 38 + 30 + 14 = 82 The area is approximately 82 square units.

WORKED EXAMPLE 2

The graph of the function defined by the rule f (x) = ex is shown. Give your answers to the following correct to 2 decimal places. a. Use the left end-point rectangle method with rectangles of width 0.5 units to estimate the area between the curve and the x-axis from x = 0 to x = 2.5. b. Use the right end-point rectangle method with rectangles of width 0.5 units to estimate the area between the curve and the x-axis from x = 0 to x = 2.5.

y

y = ex

(0, 1) –1 –0.5 0

0.5

1

1.5

2

2.5

3

x

y=0

CHAPTER 7 Integration 233

THINK a. 1.

Draw the left-end-point rectangles on the graph. State the widths and heights of the rectangles.

WRITE

y

a.

y = ex

(0, 1) –1

2.

b. 1.

0

0.5

1

1.5

2

2.5

3

x

y=0

The width of each rectangle is 0.5 units.The heights of the rectangles are f (0), f (0.5), f (1), f (1.5) and f (2). A = 0.5f (0) + 0.5f (0) + 0.5f (1) + 0.5f (1.5) + 0.5f (2) = 0.5 [f (0) + f (0.5) + f (1) + f (1.5) + f (2)]

Determine the approximate area by adding the areas of all the rectangles.

Draw the right end-point rectangles on the graph. State the widths and heights of the rectangles.

–0.5

= 0.5 [e0 + e0.5 + e1 + e1.5 + e2 ]

= 0.5 × 17.2377 ≈ 8.62 The area is approximately 8.62 square units. y

b.

y = ex

(0, 1) –1

2.

Determine the approximate area by adding the areas of all the rectangles.

–0.5

0

0.5

1

1.5

2

2.5

3

x

y=0

The width of each rectangle is 0.5 units. The heights of the rectangles are f (0.5), f (1), f (1.5), f (2) and f (2.5). A = 0.5f (0.5) + 0.5f (1) + 0.5f (1.5) + 0.5f (2) + 0.5f (2.5) [ ] = 0.5 f (0.5) + f (1) + f (1.5) + f (2) + f (2.5) [ ] = 0.5 e0.5 + e1 + e1.5 + e2 + e2.5

= 0.5 × 28.4202 ≈ 14.21 The area is approximately 14.21 square units.

234 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 3

Determine an approximation for the area enclosed by the graph of f (x) = 0.2 x2 + 3, the x-axis and the lines x = 1 and x = 3 using the trapezoidal rule with interval widths of 0.5 units.

THINK a. 1.

WRITE

Sketch the graph of f (x). y

f(x) = 0.2x2 + 3

3

Draw trapeziums of width 0.5 units from x = 1 to x = 3. 3. Evaluate the height of each vertical side of the trapeziums by substituting the appropriate x-value into f (x). 2.

4.

Calculate the area of each trapezium.

5.

Calculate the approximate area by adding the areas of the trapeziums.

–2

0

1

2

3

4

x

f (1) = 0.2(1)2 + 3 = 3.2

f (1.5) = 0.2(1.5)2 + 3 = 3.45 f (2) = 0.2(2)2 + 3 = 3.8

f (2.5) = 0.2(2.5)2 + 3 = 4.25

f (3) = 0.2(3)2 + 3 = 4.8 1 T1 = (3.2 + 3.45) × 0.5 = 1.6625 2 1 T2 = (3.45 + 3.8) × 0.5 = 1.8125 2 1 T3 = (3.8 + 4.25) × 0.5 = 2.0125 2 1 T4 = (4.25 + 4.8) × 0.5 = 2.2625 2 Total area of trapeziums = 1.6625 + 1.8125 + 2.0125 + 2.2625 = 7.75 Therefore, the area under the curve is approximately 7.75 square units.

Resources Interactivity: Estimation of area under a curve (int-6422)

Units 3 & 4

Area 3

Sequence 2

Concepts 1 & 2

Estimating areas under curves — rectangle method Summary screen and practice questions Estimating areas under curves — trapezoidal method Summary screen and practice questions

CHAPTER 7 Integration 235

Exercise 7.2 Estimating the area under a curve Technology free 1.

y

For the curve f (x) shown, determine an approximation for the area between the curve and the x-axis from x = 1 to x = 5. Use lower rectangles with a width of 2 units.

f(x)

WE1

(3, 3)

3 (1, 2) 2

0 2.

1

5

3

For each of the following curves, determine an approximation for the area between the curve and the x-axis from x = 1 to x = 5 by calculating the areas of the shaded rectangles. a.

b.

y

y

f(x)

(5, 4) 4

(4, 19)

19

(1, 2)

(3, 12)

12

2 f(x)

(2, 7)

7 (1, 4)

4 3 0 3.

1

5

x

0

2

1

4

3

5 x

The left end-point rectangle method and the right end-point rectangle method are shown for the 1 calculation of the approximate area between the curve f (x) = , x > 0, and the x-axis from x = 0.5 to x x = 2.5. WE2

y

y

1 y= – x

0 x=0

0.5

1 y= – x

1

1.5

2

2.5

3

x

y=0 0 x=0

0.5

236 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

1

1.5

2

2.5

3

x

y=0

x

Calculate the approximate area under the curve: a. using the left end-point rectangle rule b. using the right end-point rectangle rule. 4. The graph of f: [0, 4] → R, f(x) = −(x − 1)2 + 9 is shown. y

y

y = –(x – 1)2 + 9

y = –(x – 1)2 + 9 (0, 8) (0, 8)

(4, 0) 0

1

2

3

4

x

0

1

2

3

4

(4, 0) x

Use the left end-point rule with rectangles 1 unit wide to estimate the area between the curve and the x-axis from x = 0 to x = 4. b. Use the right end-point rule with rectangles 1 unit wide to estimate the area between the curve and the x-axis from x = 0 to x = 4. 5. WE3 Calculate an approximation for the area enclosed by the y y = x2 graph of f (x) = x2 , the x-axis and the lines x = 1 and x = 3 with interval widths of 1 unit. Use the trapezoidal method. a.

0 6.

2

1

3

x

For each of the following curves, calculate an approximate area between the curve and the x-axis over the interval indicated by calculating the areas of the shaded trapeziums. a.

y

b.

f(x)

(2, 3)

3

(6, 5)

(4, 3)

3 2

0

f(x)

5

(4, 5)

5

y

2

4

x

0

(2, 2)

2

4

6

x

CHAPTER 7 Integration 237

7.

Using width intervals of 1 unit, calculate an approximation for the area between the graph of f (x) = x2 + 4 and the x-axis from x = 1 to x = 4 using: a. lower rectangles b. upper rectangles c. averaging of the lower and upper rectangle areas.

y y = x2 + 4

0 8.

2

x

4

3

For each of the following figures, determine the approximate area between the curve and the x-axis over the interval indicated by calculating the area of the shaded rectangles. Give your answers in exact form. a.

y

b. y

y = ex

0 1 –1 x = –1 to x = 2 9.

1

y = loge(x)

x 2

0

x 1 2 3 4 5 x = 1 to x = 5

For each of the following figures, determine the approximate area between the curve and the x-axis over the interval indicated by calculating the area of the shaded rectangles. Give your answers in exact form. a. f(x) = –x2 – 4x

y

b.

x = 2 to x = 6

y 1

2

3

4

x

x = –3 to x = –1

238 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

6 x

0

–3 –2.5 –2 –1.5 –1 0

5

y = x3 – 6x2

10.

For each of the following figures, determine the approximate area between the curve and the x-axis over the interval indicated by calculating the shaded area using the trapezoidal rule. Give your answers in exact form. a.

b.

y

y = ex

y

y = 10 – x2

–1 – 12–

0

–1 2

1

x 0

2

1

3

x

Calculate approximations for the area between the graph of f(x) = (x − 1)3 and the x-axis between x = 1 and x = 4 using the trapezoidal rule and interval widths of: a. 1 unit b. 0.5 units. Give your answers correct to 1 decimal place. 1 12. Calculate approximations for the area under the graph of y = between x = 0.5 and x = 2.5 using the x trapezoidal rule and interval widths of: a. 1 unit b. 0.5 units. Give your answers correct to 2 decimal places. 13. Consider the function defined by the rule f : R → R, y f (x) = −0.01x3 (x − 5)(x + 5), x ≥ 0. The graph of the y = –0.01x3(x – 5)(x + 5) function is shown. Use the left end-point rule with rectangles 1 unit wide to approximate the area bound by the curve and the x-axis.

Technology active 11.

(5, 0) (0, 0)

14.

1 The graph of the function f (x) = 4 − x2 is shown. Estimate 4 the area bound by the curve and the x-axis using the right end-point method with rectangles of width 1 unit.

1

2

(0, 4)

3

y

4

5

6

x

1

f(x) = 4 – –4 x2

(4, 0)

(–4, 0) –5 –4 –3 –2 –1 0

1 2 3 4 5

x

CHAPTER 7 Integration 239

15.

The graph of f (x) =

√

x(4 − x) for x ∈ [0, a] is shown. y

f(x) = x(4 – x)

(a, 0) x

0

a. b.

The graph intersects the x-axis at the point (a, 0) as shown. Calculate the value of the constant a. Use both the left end-point and the right end-point rules to determine the approximate area between the curve and the x-axis from x = 0 to x = a. Use a rectangle width of 1 and give your answers correct to 2 decimal places.

7.3 The fundamental theorem of calculus and definite integrals

7.3.1 The fundamental theorem of calculus

Consider the function y = f (x), where f (x) ≥ 0, that is continuous for x ∈ [a, b]. Let A(x) represent the area between the curve and the x-axis from x = a to x. Similarly, let A(x + 𝛿x) represent the area between the curve and the x-axis from x = a to x + 𝛿x. y

y

y = f(x)

y = f(x)

A(x + δx)

A(x) 0

b

a

x

b

x

0

a

x x + δx

t

The difference between these areas is A (x + 𝛿x) − A(x). From the previous section, we know that this area lies between the areas of the left end-point rectangle and the right end-point rectangle. y

y y = f(x)

y y = f(x)

y = f(x) (x, f(x + δx))

(x, f(x)) A(x + δx) – A(x)

0

x x + δx

x

Left end-point rectangle method

0

x x + δx

Actual area under the curve

240 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

0

x x + δx

x

Right end-point rectangle method

Therefore, f (x) × 𝛿x ≤ A (x + 𝛿x) − A (x) ≤ f (x, +, 𝛿, x) × 𝛿x

A (x + 𝛿x) − A (x) ≤ f(x + 𝛿x) 𝛿x

f (x) ≤

As the width of each rectangular strip becomes smaller, the areas become closer together. This means as 𝛿x → 0, f (x + 𝛿x) → f (x) A (x + 𝛿x) − A (x) = f (x). 𝛿x→0 𝛿x

or lim

A (x + 𝛿x) − A (x) d (A(x)). = 𝛿x→0 𝛿x dx

By definition of differentiation from first principles, lim d (A(x)) = f (x). dx Integrating both sides with respect to x gives:

Thus,

d (A(x)) dx = f (x) dx ∫ dx ∫ Or simply: Let the antiderivative of f (x) be F(x). Then: Or: When x = a: But:

This gives: Let x = b:

A(x) =

∫

f (x) dx

A(x) = F (x) + c ∫ ∫ ∫

f (x) dx = F (x) + c

f (x) dx = F (a) + c

f (x) dx = 0, as the area defined is zero at x = a.

∴ c = −F(a)

∫ ∫

f (x) dx = F (x) − F(a)

f (x) dx = F (b) − F(a)

That is, the area between the graph of y = f (x), the x-axis and x = a and x = b is given by F (b) − F (a), where F (x) is the antiderivative of f (x).

This is the fundamental theorem of integral calculus. It allows areas under graphs to be calculated exactly. The fundamental theorem can be stated as: Area =

b

f (x) dx

= [F(x)]ba ∫a

= F (b) − F(a)

CHAPTER 7 Integration 241

The fundamental theorem of integral calculus b

∫a

f(x) dx = [F(x)]ba

= F(b) − F(a)

where F(x) is the antiderivative of f (x).

7.3.2 The definite integral and its properties •

• • •

• •

b

∫a

f (x) dx is called the definite integral as it evaluates to a real number and not a function.

The indefinite integral,

f (x) dx, involves finding only an antiderivative of the function. ∫ There is no need to add + c as the two c constants in F (a) and F (b) would cancel each other out. The variables a and b are called the terminals of the definite integral and indicate the range of the values of x over which the integral is taken. The function to be integrated, f (x), is called the integrand. Definite integrals have the following properties, assuming f and g are continuous functions for x ∈ [a, b] or a ≤ x ≤ b and k is a real constant.

Properties of definite integrals a

∫a b

∫a b

∫a b

∫a b

∫a

f(x) dx = 0

f(x) dx = −

a

f(x) dx

∫b

k f(x) dx = k

b

∫a

f(x) dx

( f(x) ± g(x)) dx = f(x) dx =

c

∫a

b

∫a

f(x) dx ±

f(x) dx +

b

∫c

b

∫a

g(x) dx

f(x) dx provided a < c < b

WORKED EXAMPLE 4

Evaluate the following definite integrals. 3

a.

∫0

4 dx ∫1 (2x + 1)3 2

b.

(3x2 + 4x − 1) dx

242 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

THINK

WRITE

( 2 ) 3x + 4x − 1 dx ∫0 [ ]3 = x3 + 2x2 − x 0 [ 3 ] [ ] 2 = 3 + 2 (3) − 3 − 03 + 2 (0)2 − 0 3

a. 1.

Antidifferentiate each term of the integrand and write in the form [F (x)]ba .

Substitute the values of a and b into F (b) − F (a). 3. Evaluate the integral. 2.

b. 1.

Express the integrand in simplest index form.

a.

= 42 − 0 = 42 2 2 4 b. dx = 4 (2x + 1)−3 dx ∫1 (2x + 1)3 ∫1 ]2 4 (2x + 1)−2 = 2 × −2 1 [ ]2 = − (2x + 1)−2 [ ]2 1 −1 = (2x + 1)2 1 [ ] [ ] 1 −1 = − 2 − 5 32 [

2.

Antidifferentiate by rule.

3.

Express the integral with a positive index number.

4. 5.

Substitute the values of a and b into F (b) − F (a).

TI | THINK a. 1. On a Calculator page,

press MENU, then select: 4: Calculus 2: Numerical Integral.

2. Complete the entry

line as: 3

=−

1 1 + 25 9 16 = 225

Evaluate the integral.

3x + 4x − 1 dx 2

∫0 then press the ENTER button.

3. The answer appears

on the screen. Note: This calculation can also be performed using the Scratchpad features, as shown in Worked examples 9 and 10.

WRITE

CASIO | THINK

WRITE

a. 1. On a Run-Matrix

page, select: SHIFT F5 ∫

dx.

2. Complete the entry

line as:

3x2 + 4x − 1 dx ∫0 then press the EXE button. 3

3. The answer appears

on the screen. Note: This calculation can also be performed using the GRAPH function, as shown in Worked examples 9 and 10.

CHAPTER 7 Integration 243

WORKED EXAMPLE 5 𝜋 2

Evaluate: a.

∫0

cos (x) dx

2

b.

THINK a. 1.

2.

∫0

(e−x + 2) dx 𝜋 2

WRITE

Antidifferentiate the given function and specify the end points for the calculation using square brackets.

a.

Substitute the upper and lower end points into the antiderivative and calculate the difference between the two values.

∫0

cos(x)dx 𝜋

= [sin (x)]02 ( ) 𝜋 = sin − sin (0) 2 =1−0 =1

(e−x + 2)dx ∫0 = [−e−x + 2x]20 2

b. 1.

2.

Antidifferentiate the given function and specify the end points for the calculation using square brackets. Substitute the upper and lower end points into the antiderivative and calculate the difference between the two values.

b.

= (−e−2 + 2(2)) − (−e0 + 2(0)) 1 =− 2 +4+1 e 1 =5− 2 e

Even if the function is unknown, we can use the properties of definite integrals to evaluate the values of related integrals. This is demonstrated in the following worked example. WORKED EXAMPLE 6 3

a.

Given that 3

i. b.

∫1

∫1

f (x)dx = 8, determine: 3

2f (x)dx

ii. k

Calculate k if

∫1

∫1

(x + 2)dx = 0.

(f (x) + 1)dx

THINK

1

iii.

∫3

3

f (x)dx

WRITE b

a. i.

Apply the definite integral property b

∫a

kf (x)dx = k

b

∫a

a. i.

f(x)dx.

244 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

∫a

iv.

2f (x)dx = 2

∫1

(f (x) − x)dx

3

f (x)dx ∫1 =2×8 = 16

3

ii. 1.

Apply the definite integral property b

( f (x) ± g(x))dx =

b

f (x)dx ±

ii.

b

g(x)dx. ∫a ∫a ∫a 2. Integrate the second function and evaluate.

∫1

= 8 + [x]31 = 8 + (3 − 1) = 10 1

Apply the definite integral property

iii.

b

∫a iv. 1.

∫b

f (x)dx.

b

∫3 3

Apply the definite integral property ∫a

b.

f (x)dx = −

iii.

a

( f (x) ± g(x)) dx =

b

∫a

f (x)dx ±

iv.

b

∫a

g(x)dx.

2.

Integrate the second function and evaluate.

1.

Antidifferentiate and substitute the values of 1 and k.

2.

Simplify and solve for k.

3.

Write the answer.

( f (x) + 1)dx =

∫1

f (x)dx = −

3

∫1

f (x)dx +

3

∫1

1dx

3

∫1 = −8

f (x)dx

( f (x) − x)dx =

3

∫1

f (x)dx −

3

∫1

xdx

]3 1 2 =8− x 2 1 ) ( 1 2 1 2 (3) − (1) =8− 2 2 ( ) 9 1 =8− − 2 2 =8−4 =4 [

b.

0=

(x + 2)dx ∫1 [ ]k 1 2 0 = x + 2x 2 )1 ( ) ( 1 2 1 2 (1) + 2 (1) 0= k + 2k − 2 2 5 1 0 = k2 + 2k − 2 2 2 0 = k + 4k − 5 0 = (k + 5) (k − 1) k = −5 or k = 1 k = 1, −5 k

Resources Interactivity: The fundamental theorem of calculus (int-6423)

Units 3 & 4

Area 3

Sequence 2

Concepts 3 & 4

The fundamental theorem of calculus Summary screen and practice questions Properties of definite integrals Summary screen and practice questions

CHAPTER 7 Integration 245

Exercise 7.3 The fundamental theorem of calculus and definite integrals Technology free 1.

Evaluate the following definite integrals.

WE4

1

a.

3

x2 dx

b.

1 dx ∫2 x 2

e.

∫0

∫0 2

6

d. 2.

4

x3 dx (

∫0

c.

) x3 + 3x2 − 2x dx

Evaluate the following definite integrals. 3 5 2x3 + 5x2 3 a. dx b. dx ∫1 ∫1 5x x 7

∫3

(x2 − 2x) dx

−4 dx ∫0 (3x − 4)5 1

c.

0

1 6 dx dx e. √ √ ∫3 2x − 5 ∫−2 8 − 3x 3. WE5 Evaluate the following definite integrals. d.

a.

𝜋 2

∫0

sin(x) dx

4.

∫𝜋

WE5

∫𝜋

( ) x 2 sin dx 3

∫0 2

d. 5.

∫1

∫−𝜋

e.

a.

∫0

6.

(3e + 5x) dx 6x

(3x − 2x + 3) dx 2

7.

∫2𝜋

∫−2 4(

e.

( ) x sin dx 3

∫1

2

∫−3

WE6

(x + 1)3 dx

Given that

5

a.

∫2

2

c.

∫5

x 5 + e2 x

c.

)

2x3 + 3x2 b. dx ∫1 x

c.

∫2

5

∫2

( ) x dx 4

8 cos(4x) dx

−4e−2x dx

∫−1

(

) e2x − e−2x dx

[ ( )] x f. cos (2x) − sin dx ∫ −𝜋 2 3

1

m(x) dx = 7 and

∫−1

𝜋 2

2

b. 5

∫ −𝜋

1

dx ∫−3 √1 − 3x

e.

𝜋 2

5 sin

dx

2

−1

∫0

1

x

e 3 dx

b.

Evaluate: a.

f.

𝜋

2

0

e4x dx

4𝜋

d.

cos(2x) dx

Determine the exact value of the following definite integrals.

Evaluate the following. 3

c.

0

2

a.

3 sin(4x) dx

2

2𝜋

d.

𝜋

b.

∫0

(ex + e−x )2 dx

n(x) dx = 3, calculate: 5

3m(x) dx

b.

(m(x) + 3) dx

d.

∫2

5

∫2

(2m(x) − 1) dx

(2m(x) + n(x) − 3) dx

246 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

5

5

8.

Given that

∫0

f(x) dx = 7.5 and

∫0

g(x) dx = 12.5, determine:

5

a.

∫0

0

−2f (x) dx

b.

(g(x) + f(x)) dx

e.

5

d.

∫0

∫5

5

g (x) dx

c.

(8g(x) − 10f(x)) dx

f.

5

∫0

(3 f (x) + 2) dx

∫0 3

∫0

5

g(x) dx +

∫3

g(x) dx

k

9.

If

10.

If

∫0

3x2 dx = 8, determine the value of k.

k

2 dx = loge (9), determine the value of k. ∫1 x a

x

e 2 dx = 4. ∫0 ) ( a 1 1 −2x 12. Determine a if e dx = 1− 8 . ∫0 2 e 11.

Determine the value of a if

Technology active 13.

The graph of the function f : R → R, f(x) = x3 − 8x2 + 21x − 14 is shown. a. The graph cuts the x-axis at the point (a, 0). Calculate the value of the constant a.

y

5

b. 14. a.

b. 15. a. b.

Evaluate

∫a

If y = x sin (x), determine

dy . dx

Hence, determine the value of 3

f(x) = x3 – 8x2 + 21x – 14

(x3 − 8x2 + 21x − 14) dx.

2

If y = ex −3x , determine

dy . dx

Hence, determine the value of

(a, 0) 𝜋 2

∫−𝜋

0

2x cos (x)dx.

1

∫0

x

3

2

(x2 − 2x)ex −3x dx.

7.4 Areas under curves 7.4.1 Definite integrals and areas The area between a continuous function, y = f (x), and the x-axis between x = a and x = b is shown in the diagram. As we have already seen, this area can be approximated by dividing it into a series of thin vertical strips or rectangles of width 𝛿x. The approximate value of the area is the sum of the areas of all the rectangles, whether they are left end-point or right end-point rectangles. As the number of strips increases, 𝛿x → 0.l From the fundamental theorem of calculus, the shaded area, A, can be expressed as the limiting sum of the rectangles or the definite integral.

CHAPTER 7 Integration 247

A = lim ∑ f(x) 𝛿x = 𝛿x→0

x=b

b

x=a

∫a

f(x)dx

y

y

y = f(x)

y = f(x)

0

x=a

x=b

x

0

x=a

x

x=b

δx

WORKED EXAMPLE 7

Determine the area bound by the curve defined by the rule y = e−x + 3 and the x-axis from x = 0 to x = 3. THINK 1.

WRITE

y

Sketch the graph of the given function and shade the required area.

(0, 4) y = e–x + 3

x=0

0

2.

Write the integral needed to determine the area.

3.

Antidifferentiate the function and evaluate.

A=

Write the answer.

3

(e−x + 3) dx ∫0 A = [−e−x + 3x]30 3

= (−e−3 + 3(3)) − (−e0 + 3(0))

= −e−3 + 9 + 1

4.

x=3

= −e−3 + 10 The area is 10 − e−3 square units.

248 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

7.4.2 Signed areas When we calculate the area between the graph of a function, y = f (x), and the x-axis from x = a to x = b b

using the definite integral

f (x) dx, the area can either be positive or negative. ∫a y The definite integral will be positive if f (x) > 0, since the curve is above the x-axis and the height of the rectangular strips is positive. The definite integral will be negative if f (x) < 0, since the curve is below the x-axis and the height of the rectangular strips is negative. Since area cannot be negative, if f (x) < 0, the answer needs to be made positive by either subtraction (that is, × (−1)), reversing the terminals, or taking its absolute value. For a region above the x-axis: Since f (x) > 0,

y = f (x)

b

f(x) dx > 0.

∫a

0

b

Thus, the shaded area

=

f (x)dx. ∫a For a region below the x-axis: a Since f (x) < 0, f (x) dx < 0. ∫b =−

∫a

x

b

y y = f (x) a

b

Thus, the shaded area

a

b x

0

f (x)dx.

b

Alternatively,

∫a

f (x) dx (reversing terminals)

| | b | f (x) dx|. | |∫a

or simply

Combining regions y

For regions that are combinations of areas above and below the x-axis, calculate separate integrals, one for each area above and one for the area below. Note that, on the diagram, A2 would have a negative value. Shaded area

y = f (x)

A1

= A1 + (−A2 ) c

=

∫b

c

or =

∫b

( f (x) dx + − a

a

∫c

) f (x) dx

a

0

A2

c

b

x

| | f (x) dx + | f (x) dx| ∫ | c |

A sketch of the function, including x-intercepts, is imperative to determine if the function is above or below the x-axis or a combination of both. This will ensure the correct use of the definite integrals required. The procedure is illustrated in the following example.

CHAPTER 7 Integration 249

[ 4 ]2 ( 3 ) x 4x2 x − 4x dx = − The definite integral ∫−2 4 2 −2 [ 4 ]2 x 2 = − 2x 4 −2 ( 4 ) ( ) (−2)4 2 2 2 = −2×2 − − 2 × (−2) 4 4 2

= (4 − 8) − (4 − 8)

= (−4) − (−4)

=0 However, when you draw the function y = x3 − 4x and shade the region defined by the definite integral 2

(

) x3 − 4x dx, you can see the area is not 0 square units.

∫−2 The function has x-intercepts at x = −2, 0 and 2. The area on the left is above the x-axis, so it is positive, and the area on the right is below the x-axis, so it is negative. The total area equals: ( ) 0 2 ( 3 ) ( 3 ) Area = x − 4x dx + − x − 4x dx ∫−2 ∫0 ]0 ]2 [ 4 [ 4 x x 2 2 − 2x − 2x = − 4 4 −2 0 = (0 − (4 − 8)) − ((4 − 8) − 0) = (4) − (−4) = 8 square units. Alternatively, using symmetry of this function, the area 0 ( 3 ) would be 2 x − 4x dx, giving the same result. ∫−2

y

y = x3 – 4x

(0, 0)

(–2, 0) 0

–2

(2, 0) x

2

WORKED EXAMPLE 8 y

Calculate the shaded area.

y = x2 – 4x

x

0 1

THINK 1.

Express the area in definite integral notation, showing a negative sign in front of the integral as the region is below the x-axis.

WRITE

Area = −

250 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

3

∫1

) x2 − 4x dx

(

3

4

]3 1 3 2 = − x − 2x 3 1 {[ ] [ ]} 1 3 1 3 2 2 (3) − 2 (3) − (1) − 2 (1) =− 3 3 ]} { [ 1 −2 = − [9 − 18] − 3 [ ( )] 2 = − −9 − −1 3 [ ] 2 = − −9 + 1 3 ) ( 1 = − −7 3 1 =7 3 1 The area is 7 square units. 3 [

2.

Antidifferentiate the integrand.

3.

Evaluate.

4.

State the solution.

WORKED EXAMPLE 9

Calculate the area bound by the curve y = (x2 − 1)(x2 − 4) and the x-axis from x = −2 to x = 2. THINK 1.

Make a careful sketch of the given function. Shade the required region.

The graph cuts the y-axis where x = 0. ∴ the y-intercept is (0, 4). ∴ the graph cuts the x-axis where y = 0: (x2 − 1)(x2 − 4) = 0 (x − 1)(x + 1)(x − 2)(x + 2) = 0 x = ±1, x = ±2

WRITE

y

(0, 4)

(–2, 0)

0 (1, 0) (–1, 0)

y = (x2 – 1)(x2 – 4)

(2, 0)

x

CHAPTER 7 Integration 251

2

1

Express the area using definite integrals. Account for the negative regions by subtracting these from the positive areas. Note that the region from x = −2 to x = −1 is the same as the region from x = 1 to x = 2 due to the symmetry of the graph.

2.

3.

Antidifferentiate and evaluate.

4.

Write the answer.

A=

∫−1

(x2 − 1)(x2 − 4) dx − 2

1

=

∫−1

∫1

(x2 − 1)(x2 − 4) dx

2

(x4 − 5x2 + 5) dx − 2

∫1

(x4 − 5x2 + 5) dx

[

]1 ]2 [ 1 5 5 3 1 3 5 3 = x − x + 5x − 2 x − x + 5x 5 3 5 3 −1 1 ( ) ) ( 1 5 5 3 1 5 5 3 (1) − (1) + 5 (1) − (−1) − (−1) + 5 (−1) = 5 3 5 3 [( ) ( )] 1 5 5 3 1 2 5 3 (2) − (2) + 5 (2) − (1) − (1) + 5 (1) −2 5 3 5 3 ) ( ) ( 1 5 1 5 − +5 − − + −5 = 5 3 5 3 ) ( )] [( 1 5 32 40 −2 − + 10 − − +5 5 3 5 3 ( ) 1 5 1 5 32 40 1 5 = − + − +5−2 − + 10 − + − 5 5 3 5 3 5 3 5 3 2 10 64 80 2 10 = − + 10 − + − 20 + − + 10 5 3 5 3 5 3 = −12 + 20 =8 The area is 8 square units.

Resources Interactivity: Area under a curve (int-6424)

Units 3 & 4

Area 3

Sequence 2

Concept 5

Areas under curves Summary screen and practice questions

Exercise 7.4 Areas under curves Technology free 1.

For each of the following graphs: express the shaded area as a definite integral hence, calculate the shaded area, giving your answer in exact form.

WE7

i. ii.

252 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

a.

y

b.

c.

y

y y = 3x2

4

y = x2

0

x

4

0

1

y=4–x d.

x

2

e.

y

x

0 –3

–1

y

y = x3 – 9x2 + 20x

0

x

–2 0

x 1

2.

y = –x3 – 4x2 – 4x

3

For each of the following graphs: i. express the shaded area as a definite integral ii. hence, calculate the shaded area, giving your answer in exact form. a.

y

b.

y = ex

y

y = e–2x

1

–1 c.

0

1

x

0 d.

y

1

4

x

y

y = 2 sin 2x 2 y = cos –x 3

1

0

π – 2

x

0

3π — 2

x

CHAPTER 7 Integration 253

3.

WE8 Express the following shaded areas as definite integrals. Hence, calculate the shaded areas, giving your answers in exact form.

a.

b.

y

y = x2 – 4

y = – 4 – 2x

–2

–1

c.

y

y

x

0

–2 0

–1

0

x

x

2

y = 1– x2 d.

e.

y

y

y = x3

x

0 –2

–1

x

0

1

y = x3 + 2x2 – x –2

4.

Express the following shaded areas as definite integrals. Hence, calculate the shaded areas, giving your answer in exact form. a.

b.

y

y 1 – 2

x

0 –1

1

x

y = –ex

c.

1

0

y = –e–2x d.

y

y

y = sin x

x

0

π

5.

3π — 2

–2π

–π

0

2π

x y = 2 cos – 2

( )( ) Consider the function y = x2 − 1 x2 − 9 . a. Sketch the graph of the function, stating all axis intercepts. b. Determine the area enclosed by the function, the lines x = −3 and x = 3, and the x-axis. WE9

254 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

Technology active 6.

7.

8. 9. 10.

11. 12. 13. 14.

y For the curve shown, calculate the area between the curve and the x-axis from: a. x = −2 to x = 1 b. x = 1 to x = 3 c. x = −2 to x = 3. For each of the following functions: i. sketch the function, showing clearly all axis intercepts ii. calculate the area bounded by the graph of the function x 0 1 2 3 –2 –1 and the x-axis. a. g(x) = 8 − x2 b. g(x) = x3 − 4x2 c. f(x) = x(x − 2)(x − 3) y = x3 – 2x2 – 5x + 6 3 2 d. f(x) = x − 4x − 4x + 16 e. g(x) = x3 + 3x2 − x − 3 f. h(x) = (x − 1)(x + 2)(x + 5) 1 1 Determine the exact area between the curve y = , the x-axis and the lines x = and x = 2. Express x 2 your answer in simplest form. Calculate the exact area of the region enclosed by the x-axis, y = e3x and the lines x = 1 and x = 2 𝜋 Calculate the exact area of the region enclosed by the x-axis, y = −cos(x) and the lines x = and 3 5𝜋 x= . (Use a sketch graph to assist your calculation.) 6 √ Determine the area bound by the curve defined by the rule y = 2 x, x ≥ 0 and the x-axis from x = 0 to x = 25. Calculate the area bounded by the curve y = 2 sin(2x) + 3, the x-axis and the lines x = 0 and x = 𝜋. Sketch the graph of y = 1 − e−x and hence calculate the exact area between the curve and the x-axis from x = −1 to x = 1. 1 The graph of y = 2 , x < 0 is shown. Calculate the area of the shaded region (that is for x −2.5 ≤ x ≤ −0.5).

y

1 y = —2 x

x –2.5

–2

–1.5

–1

–0.5

0

CHAPTER 7 Integration 255

15.

The graph of the function y = 2 sin(x) is shown. Using calculus, calculate the area between the curve and the x-axis from x = 0 to x = 4𝜋. y 2

y = 2 sin(x)

(0, 0) 0

π – 2

π

3π – 2

2π

5π – 2

(4π, 0) x 3π 7π – 4π 2

–2

Differentiate x loge (x) for x > 0. b. Hence, determine an antiderivative of loge (x). c. Determine the area bounded by the graph of loge (x), the x-axis, x = 1 and x = 4, giving your answer in exact form. y 17. The graph of y = 2xe2x is shown. d ( x2) a. Determine e . dx 2 b. Hence, calculate the exact area between the curve y = 2xex y = 2xe2x and the x-axis from x = −1 to x = 1. 16. a.

–1.5

18. a.

–1

256 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

0

0.5

1

1.5

y=

x x2 + 2

x

y

Differentiate loge (x2 + 2).

x b. Hence, determine an antiderivative of . 2 x +2 x c. The graph of y = is shown. Calculate the area 2 x +2 between the graph, the x-axis, x = −1 and x = 1. Give your answer in exact form.

–0.5

–5

–1 0

1

5

x

19.

The graph of f: R → R, f(x) = 3x3 is shown.

y f(x) = 3x3 3

0

1

x

Calculate the area bounded by the curve and the x-axis from x = 0 to x = 1. Hence, or otherwise, calculate the area of the shaded region. 3𝜋 3𝜋 20. The graph of y = 2 sin (x), − ≤x≤ is shown. 2 2 𝜋 a. b.

a.

Calculate

2

y

∫a b. Hence, or otherwise, calculate the area of the shaded region.

– – 3π 2

–π

y = 2 sin(x)

2

2 sin(x) dx.

0

– –π2

π – 2

3π – 2

π

x

–2

7.5 Areas between curves

When we find areas between two continuous functions, f (x) and g(x), for an interval a ≤ x ≤ b, our approach depends on whether the curves intersect or not in this interval.

7.5.1 When f(x) and g(x) do not intersect in the interval a ≤ x ≤ b y

Example 1: If the region is above the x-axis, the lower function is subtracted from the higher function to ensure a positive answer. b [ ] Blue shaded area = f (x) − g (x) dx ∫a

g(x)

0

Example 2: If the region is below the x-axis, the lower function is subtracted from the higher function to again ensure a positive answer. b [ ] Blue shaded area = f (x) − g (x) dx ∫a

f(x)

a

x

b

y

a

0

x

b

f(x) g(x)

CHAPTER 7 Integration 257

y

Example 3: If the region crosses the x-axis, the lower function is again subtracted from the higher function to ensure a positive answer. b [ ] Shaded area = f (x) − g (x) dx ∫a In the case where the region crosses the x-axis, there is no need to consider the x-intercepts because, if the two functions were both translated vertically by the same factor, k units, to ensure they are above the x-axis, the constants would cancel each other. That is, b b [ ] [ ] ( f (x) + k) − (g (x) + k) dx = f (x) − g (x) dx. ∫a ∫a

f(x) g(x)

0

b

a

WORKED EXAMPLE 10

Consider the functions y = x and y = x2 − 2. a. Determine the values of x where the functions intersect. b. Sketch the graphs of the functions on the same axes, stating the coordinates of the points of intersection. c. Hence, calculate the area bounded by the two functions. THINK

State the two functions. 2. Determine where the curves intersect. 3. Solve for x.

a. i. 1.

ii.

Determine the key points of each function and sketch.

y = x and y = x2 − 2 For points of intersection: x = x2 − 2 x2 − x − 2 = 0 (x − 2)(x + 1) = 0 x = 2 or x = −1 ii. For y = x, when x = 0, y = 0; when x = 2, y = 2; when x = −1, y = −1. The line passes through (0, 0), (2, 2) and (−1, −1). For y = x2 − 2, when x = 0, y = −2. Hence, the y-intercept is −2. The parabola also passes through (2, 2) and (−1, −1).

WRITE a. i.

y

y = x2 – 2

(2, 2) y=x

0 (–1, –1)

258 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

x

Define f(x) and g(x) . 2. Write the area as a definite integral between the values of x at the points of intersection.

iii. 1.

3.

Antidifferentiate.

4.

Evaluate the integral.

5.

State the area.

f (x) = x and g(x) = x2 − 2. Area 2 [ ] = f (x) − g(x) dx ∫−1 2 [ ( )] = x − x2 − 2 dx ∫−1 2 ( ) = x − x2 + 2 dx ∫−1 ]2 [ 1 2 1 3 = x − x + 2x 2 3 −1 ] [ [ ] 1 2 1 3 1 1 2 3 = (2) − (2) + 2(2) − (−1) − (−1) + 2(−1) 2 3 2 3 ( ) ( ) 8 1 1 = 2− +4 − + −2 3 2 3 ( ) ( ) 1 1 = 3 − −1 3 6 1 1 =3 +1 3 6 1 =4 2 1 The area is 4 square units. 2

iii. Let

7.5.2 When f(x) and g(x) intersect in the interval a ≤ x ≤ b

Let the points of intersection of f (x) and g(x) in the interval a ≤ x ≤ b be x = c and x = d, where c < d. The area is found by considering the separate intervals: a ≤ x ≤ c, c ≤ x ≤ d and d ≤ x ≤ b For each section, the lower function is subtracted from the higher function to ensure a positive answer. c d [ ] [ ] Shaded area = g (x) − f (x) dx + f (x) − g (x) dx ∫a ∫c b [ ] + g (x) − f (x) dx ∫d

y

g(x)

a c

0

d

x

b

f(x)

WORKED EXAMPLE 11

4 and g(x) = x. x a. Determine the values of x where the functions intersect. b. Sketch the graphs of the functions on the same axes. Shade the region between the two functions and x = 1 and x = 3. Consider the functions f (x) =

CHAPTER 7 Integration 259

c.

Hence, calculate the area between f (x) and g(x) from x = 1 to x = 3.

THINK a. 1.

State the two functions.

a.

Let f(x)= g(x) to calculate the values of x where the graphs intersect. 3. Solve for x.

2.

Sketch f(x) and g(x) on the same axis and shade the region between the two curves from x = 1 to x = 3.

b.

4 f(x) = , g(x) = x x 4 For points of intersection, x = . x

WRITE

x2 = 4 x2 − 4 = 0 (x + 2) (x − 2) = 0

b.

x = −2 and x = 2 y

– f(x) = 4 x g(x) = x

0

c. 1.

State the area as the sum of two integrals for the two subintervals.

2.

Antidifferentiate.

3.

Evaluate the two integrals.

4.

Simplify.

5.

State the area.

c.

1

2

3

x

) ) 2( 3( 4 4 Area = − x dx + x− dx ∫1 x ∫2 x [ ]2 [ ]3 1 2 1 2 = 4 loge (x) − x − x − 4 loge (x) 2 1 2 2 ] ] [ [ 1 2 1 2 = 4 loge (2) − (2) − 4 loge (1) − (1) 2 2 {[ ] [ ]} 1 2 1 2 (3) − 4 loge (3) − (2) − 4 loge (2) + 2 2 ] [ ] [ 1 = 4 loge (2) − 2 − 4 loge (1) − 2 {[ ] } [ ] 9 + − 4 loge (3) − 2 − 4 loge (2) 2 1 9 = 4 loge (2) − 2 − 0 + + 2 2 (3) − 2 + 4 loge (2) −4 log( e ) 4 = 4 loge +1 3 ( ) 4 The area is 4 loge + 1 or approximately 3 2.151 square units.

260 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 12

The graphs of f (x) = 3 sin (2x) and g(x) = 3 cos (2x) are shown for x∈ [0, 𝜋]. a. Determine the coordinates of the point(s) of [ ] 𝜋 intersection of f and g for the interval 0, . 2 b. Using calculus, determine the area enclosed [ ] 𝜋 between the curves on the interval 0, . 2

y (π, 3)

3 y = g(x)

x

0

π – 2

π

y = f(x) –3

THINK a. 1.

Use simultaneous equations to determine where the graphs intersect, and equate the two equations.

] 𝜋 . 2. Solve for 2x for x ∈ 0, 2 [

3.

Calculate the corresponding y-value.

4.

Write the solution.

b. 1.

2.

3 sin(2x) = 3 cos(2x)

WRITE

Determine when f > g and f < g. Express each area individually in definite integral notation.

3 sin(2x) 𝜋 = 1, 0 ≤ x ≤ . 3 cos(2x) 2 𝜋 tan(2x) = 1, 0 ≤ x ≤ 2 𝜋 2x = 4 𝜋 ∴x = ( )8 ( ) 𝜋 𝜋 f = 3 sin 8 4 1 =3× √ 2 √ 3 2 = ( √ ) 2 𝜋 3 2 The coordinates are , . 8 2 𝜋 b. When 0 < x < , g > f. 8 𝜋 𝜋 When < x < , f > g. 8 2 The area is equal to: A 𝜋= 𝜋 a.

8

∫0

(3 cos(2x) − 3 sin(2x))dx +

2

∫𝜋 8

(3 sin(2x) − 3 cos(2x))dx

CHAPTER 7 Integration 261

] 𝜋2 ] 𝜋8 [ 3 3 3 3 = sin(2x) + cos(2x) + cos(2x) − sin(2x) 𝜋 2 2 2 2 0 ( ) ( ) ( )8 3 𝜋 𝜋 3 3 3 = − sin + cos − sin(0) + cos(0) 2 4 2 4 2 2 ( ( ) ( )) 𝜋 𝜋 3 3 3 3 + sin + cos(𝜋) − sin(𝜋) − − sin 2 2 2 4 2 4 √ √ √ √ 2 3 2 2 3 2 3 3 3 3 = × + × −0− + −0+ × + × 2 2 2 2 2 2 2 2 2 2 √ √ √ √ 3 2 3 2 3 2 3 2 + + + = 4 4 4 4 √ =3 2 √ The area is 3 2 square units. [

3.

Use calculus to antidifferentiate and evaluate.

4.

Write the answer.

Units 3 & 4

Area 3

Sequence 2

Concept 6

Areas between curves Summary screen and practice questions

Exercise 7.5 Areas between curves Technology free 1.

WE10

a.

Calculate the shaded area in each of the following diagrams. y

y

b.

y = x2 y = 3x

0

0

x

y = 8 – x2

y = 4 – x2 c.

d.

y

x

2

–2

1

y

y = x3

y = ex y = 3x

0 x

0 3

262 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

–1

x 1 y = 9 – x2

e.

y

f.

y=x

y

y = x2 – 5 0

1

x

2

–1

x

0

1 y = –4

y = – ex 2.

MC Which one of the following does not equal the shaded area shown?

5

A.

∫1

5

B.

∫1

5

C.

∫1

5

D. 3.

g (x) dx +

f (x) dx −

f(x)

5

f (x) dx

∫1

g(x)

1

f (x) dx

∫5 5

∫1

0

g (x) dx

B. C. D. MC

[ ] g(x) − f(x) dx

−3

∫−1

−1

∫−3

−1

∫−3

−1

A.

∫0

3

B.

∫0

4

C.

∫0

3

D.

[

[

∫0

[

f(x) + g(x) dx

f(x)

]

f(x) − g(x) dx

0

–4

] f(x) − g(x) dx

–3

x

–1

y

g(x)

]

[ ] g(x) − f(x) dx +

] [ g(x) − f(x) dx [

g(x)

f(x) − g(x) dx

[ ] g(x) − f(x) dx [

y

]

∫−3 The shaded area shown is equal to: 4

x

5

1

MC The area bounded by the curves f (x), g (x) and the lines x = −3 and x = −1 is equal to:

A.

4.

∫1

g (x) dx −

y

] f(x) − g(x) dx +

4

∫3

[

f(x) − g(x) dx

f(x)

]

0 4

∫3

3

4

x

[ ] g(x) − f(x)

CHAPTER 7 Integration 263

Consider the functions f (x) = x3 and g (x) = x. a. Determine the values of x where the functions intersect. b. Sketch the graphs of the functions on the same axes. c. Hence, calculate the area between f (x) and g(x), giving an exact answer. 6. Consider the functions f (x) = x3 + 2x and g (x) = 3x2 . a. Determine the values of x where the functions intersect. b. Sketch the graphs of the functions on the same axes. c. Hence, calculate the√ area between f (x) and g (x), giving an exact answer. x and the line f(x) = 4 are shown. 7. The graphs of g(x) = y Determine the coordinates of the point of intersection between f and g, and hence calculate the area of the shaded region 5.

WE11

f(x) = 4

Calculate the area enclosed between the curve f(x) = (x − 3)2 g(x) = x and the line g(x) = 9 − x. 9. WE12 Consider the functions f (x) = sin(x) and g (x) = −cos(x) for x ∈ [0, 𝜋]. a. Determine the coordinates of the points of intersection of f and g for the given domain. 0 b. Using calculus, calculate the area enclosed between the curves for the given interval. √ 𝜋 10. Calculate the area between the curves y = 3 − sin 2x and y = sin 2x from x = 0 to x = . 4 11. Calculate the exact area bounded by the curves y = ex and y = 3 − 2e−x . 12. The graphs of f(x) = 3x3− x4 and g(x) = −x + 3 are shown. y a. The graphs intersect at the points (a, b) and (c, 0). f(x) = 3x3 – x4 Calculate the constants a, b and c. b. Calculate the area enclosed between the curves from x = a to x = c. 8.

x

(a, b)

(c, 0) x

0

g(x) = –x + 3

Technology active 13.

The graph shows the cross-section of a bricked archway. (All measurements are in metres.) a. Determine the x-intercepts of f(x). b. Determine the x-intercepts of g(x). c. Determine the cross-sectional area of the brickwork.

y

1 x2 f(x) = 4 – – 4

x

0 1 x2 g(x) = 3 – – 3

264 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

The edge of a garden bed can be modelled by the rule ( ) x y = 0.5 sin + 2. The bed has edges defined by 2 y = 0, x = 0 and x = 4𝜋. All measurements are in ( ) metres. x a. Sketch the graph of y = 0.5 sin + 2 along with 2 y = 0, x = 0 and x = 4𝜋 as edges to show the shape of the garden bed. b. Calculate the area of the garden bed, correct to the nearest square metre. c. Topsoil is going to be used on the garden bed in preparation for new planting for spring. The topsoil is to be spread so that it is uniformly 50 cm thick. Determine the amount of soil, in cubic metres, that will be needed for the garden bed. 15. A stone footbridge over a creek is shown along with the mathematical profile of the bridge. The arch of the footbridge can be modelled by a quadratic function for x ∈ [−5, 5], with all measurements in metres. 14.

y 7 5

–6 –5

0

5 6

x

Determine the equation for the arch of the bridge Determine the area between the curve and the x-axis from x = −5 to x = 5. c. Determine the area of the side of the bridge represented by the shaded area. d. The width of the footbridge is 3 metres. Determine the volume of stones used in the construction of the footbridge.

a. b.

7.6 Applications of integration 7.6.1 Total change as the integral of instantaneous change If we are given the equation for the instantaneous rate of change of a certain item, such as revenue or area, the amount by which the item has changed over a particular time period would be found by integrating the rate of change equation using the starting and finishing times as the terminals. dV For example, if water is flowing into a holding tank at litres per minute, the amount of liquid that has dt ) 30 ( dV flowed into the tank in the first 30 minutes would be dt litres. ∫0 dt In economics, the instantaneous rate of change with respect to the number of items is also referred to as the marginal rate of change. For example, if the revenue function for selling x units is given as R(x) dollars, then the marginal revenue dR is given by dollars per unit. The marginal revenue is the extra revenue received for selling one more unit dx after a particular number of units have been sold.

CHAPTER 7 Integration 265

WORKED EXAMPLE 13

It is a common practice to include heating dV –– dt in concrete slabs when new residential πt πt dV homes or units are being constructed, –– = 2 cos – + sin – + 3 3 9 dt because it is more economical than installing heating later. A typical reinforced (0, 8) concrete slab, 10–15 centimetres thick, has tubing installed on top of the reinforcement, then concrete is poured on top. When the system is complete, hot water runs through the tubing. The concrete slab absorbs the heat from the water and releases it into the area above. The number of litres / minute of water t 0 12 16 20 24 28 32 4 8 flowing through the tubing over t minutes can be modelled by the rule ( ) ] [ ( ) dV 𝜋t 𝜋t + sin +3 . = 2 cos dt 3 9 The graph of this function is shown. a. What is the rate of flow of water, correct to 2 decimal places, at: i. 4 minutes? ii. 8 minutes? b. State the period of the given function. c. Determine the volume of water that flows through the tubing during the time period for one whole cycle.

[ ( )

THINK a. i.

Substitute t = 4 into the given equation and evaluate.

ii.

Substitute t = 8 into the given equation and evaluate.

b.

Determine the cycle for the function by analysing the shape of the graph, or the period of the functions

( ) ]

[ ( ) ( ) ] 𝜋t 𝜋t dV = 2 cos + sin +3 a. i. dt 3 9 When t = 4, [ ( ) ( ) ] dV 4𝜋 4𝜋 = 2 cos + sin +3 dt 3 9 = 6.97 The rate at 4 minutes is 6.97 litres/minute. ii. When t = [ 8, ( ) ( ) ] dV 8𝜋 8𝜋 = 2 cos + sin +3 dt 3 9 = 5.68 The rate at 8(minutes ) is 5.68 litres/minute. 𝜋t 2𝜋 b. Period of cos = ( 𝜋 ) = 6 hours 3 3 ( ) 𝜋t 2𝜋 Period of sin = ( 𝜋 ) = 18 hours 9 9 Period of combined function = 18 hours WRITE

266 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

) ( ) ) 𝜋t 𝜋t c. A = 2 cos + sin + 3 dt ∫0 3 9 ( ) ( ) ) 18 ( 𝜋t 𝜋t =2 cos + sin + 3 dt ∫0 3 9 [ ( ) ( ) ]18 pt pt 3 9 =2 sin − cos + 3t p 3 p 9 )0 {( 3 9 sin(6p) − cos(2p) + 54 =2 p p ( )} 3 9 − sin(0) − cos(0) p p ) ( )} {( 9 9 =2 0 − + 54 − 0 − p p ( ) 9 9 = 2 − + 54 + p p = 2 × 54 = 108 The volume of water that passes through the tubing during one cycle is 108 litres. 18

c. 1

The area under the curve of the equation of the rate of flow gives the total volume that has flowed through the tubing. To calculate the area, antidifferentiate and evaluate.

2.

Write the answer.

(

(

WORKED EXAMPLE 14

√ A manufacturer of a game knows that the revenue, $R, for selling x games is R = 500 x. The √ costs, $C, to produce x games is C = 2000 + x x. a. Calculate the profit made when 25 games were sold. b. Determine the average profit per game when 25 games were sold. c. For 100 games, calculate and interpret: i. the marginal revenue ii. the marginal cost iii. the marginal profit. THINK

Substitute x = 25 and evaluate. 3. State the cost equation.

a. 1.

State the revenue equation.

2.

Substitute x = 25 and evaluate. 5. Profit = revenue − cost

4.

b.

Average profit per game =

√ R = 500 x √ R = 500 25 = 2500 √ C = 2000 + x x √ C = 2000 + 25 25 = 2125 Profit = 2500 − 2125 Profit = $375

WRITE

profit number

Profit per game 375 = 25 = $15

CHAPTER 7 Integration 267

c. i. 1. 2. 3.

ii. 1. 2. 3.

State the revenue equation. Differentiate with respect to x.

Substitute x = 100 and evaluate.

State the cost equation. Differentiate with respect to x.

Substitute x = 100 and evaluate.

Write an equation for profit. 2. Differentiate with respect to x.

iii. 1.

3.

Substitute x = 100 and evaluate.

Note: Observe that marginal profit = marginal revenue − marginal cost.

√ R = 500 x

R = 500x 2 1

dR 1 −1 = 500 × x 2 dx 2 dR 250 = √ dx x dR 250 =√ dx 100 = 25

The marginal revenue at x = 100 is $25, so the approximate revenue from selling the 101st game is $25. √ C = 2000 + x x C = 2000 + x 2 3

dC 3 1 = x2 dx 2 √ dC 3 x = dx 2 √ dC 3 100 = dx 2 dC = 15 dx The marginal cost at x = 100 is $15, so the approximate cost for making the 101st game is $15. Profit = revenue − costs √ √ = 500 (x − (2000 + x ) x) = 500 x 2 − 2000 − x 2 1

3

dP 1 −1 3 1 = 500 × x 2 − x 2 dx 2 2 √ dP 250 3 x = √ − dx 2 x √ 3 100 dP 250 =√ − dx 2 100 dP = 25 − 15 dx = 10 The marginal profit at x = 100 is $10, so the approximate profit for selling the 101st game is $10.

268 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WORKED EXAMPLE 15

On any day, the cost per item for a machine producing x items is given by

where x ∈ [0, 200] and C is the cost in dollars Use the rate to calculate the marginal cost of producing the 100th item. b. What is the total cost of producing the first 100 items? c. Determine the average cost of production for the first 100 items.

dC = 50 − 4e0.02x , dx

a.

THINK a. 1.

State the marginal rate of change.

2.

Substitute x = 100 and evaluate.

3.

Answer the question.

b. 1.

2.

3.

Area under the rate of change equation of cost gives the total cost for a given interval. Antidifferentiate and evaluate.

Answer the question.

Average cost of production = total cost / number produced. 2. Answer the question.

c. 1.

dC = 50 − 4e0.02x dx dC = 50 − 4e0.02×100 dx = 50 − 4e2 = 20.4438 The marginal cost of producing the 100th item is approximately $20.44. 100 ( ) Total cost = 50 − 4e0.02x dx ∫0 WRITE

[ ]100 1 0.02x = 50x − 4 × e .02 [ ]100 0 = 50x − 200e0.02x 0 ) ( ( ) = 50 × 100 − 200e2 − 0 − 200e0

= 5000 − 200e2 + 200 = 3722.19 The total cost of producing the first 100 items is $3722.19. Average cost of production = $3722.19 / 100 = $37.22 The average cost of production for the first 100 items is approximately $37.22 each.

7.6.2 Displacement, velocity and acceleration The relationships between displacement, velocity and acceleration have been discussed in Chapters 5 and 6. Your knowledge about the definite integral and the area under curves gives you additional skills for the calculation of facts relating to kinematics. From equations or graphs of velocity as a function of time, the following may be obtained: dv • acceleration, the gradient of the velocity function or dt • displacement, the signed area or definite integral • distance travelled, the area under the curve.

CHAPTER 7 Integration 269

WORKED EXAMPLE 16

A particle starting from rest accelerates according to the rule a = 3t(2 − t). Determine a relationship between the velocity of the particle, v metres/second, and the time, t seconds. b. Determine the displacement of the particle after 4 seconds. c. Sketch the graph of velocity versus time for the first 4 seconds of the motion. d. Calculate the distance travelled by the particle in the first 4 seconds. a.

THINK a. 1.

Antidifferentiate the acceleration equation to determine the velocity equation.

v=

WRITE a.

∫

a(t)dt

= (3t(2 − t)dt ∫

= (6t − 3t2 )dt ∫

2.

b. 1.

2. c.

Apply the initial conditions to determine v in terms of t.

Integrate v between t = 0 and t = 4. As we are finding displacement, there is no need to sketch the graph.

Write the answer. Sketch a graph of v versus t.

= 3t2 − t3 + c When t = 0, v = 0, so c = 0. ∴ v = 3t2 − t3

(3t2 − t3 )dt ∫0 [ ]4 14 3 = t − t 4 0 ( ) ( ) 1 1 = 43 − (4)4 − 03 − (0)4 4 4 =0 After 4 seconds the displacement is 0. c. y-intercept: (0, 0) t-intercepts: 0 = 3t2 − t3 b.

x=

4

= t2 (3 − t) t = 0, 3 When t = 4, v = 3 × 42 − 43 = −16. v

(0, 0)

(3, 0) t

2 v = 3t2 – t3

d. 1.

The area under the curve of a velocity–time graph gives the distance covered. Set up the integrals and subtract the negative region.

d.

D=

(4, –16)

3

∫0

(3t − t )dt − 2

3

270 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

4

∫3

(3t2 − t3 )dt

2.

Antidifferentiate and evaluate.

3.

Write the answer.

1 1 = [t3 − t4 ]30 − [t3 − t4 ]43 4 4 (( ) ( )) 4 3 04 3 3 = 3 − − 0 − 4 4 ) ( )) (( 4 34 4 − 33 − − 43 − 4 4 81 81 = 27 − − 0 − 64 + 64 + 27 − 4 4 162 = 54 − 4 = 13.5 The distance travelled by the particle in 4 seconds is 13.5 metres.

Alternative working for the distance travelled in the first 4 seconds: 1. Determine where the particle is at rest.

2.

3.

Determine the displacement of the particle initially and at t = 3 and 4. Note: We were not told where the particle was initially, so we cannot find the constant, c.

Draw a motion diagram to represent the displacement of the particle during the first 4 seconds.

At rest: v = 0 3t2 − t3 = 0

t2 (3 − t) = 0 t = 0 or 3 The particle is at rest initially and after 3 seconds, so it changes direction after 3 seconds. 1 x = t3 − t4 + c 4 At t = 0: x=c 81 At t = 3: x = 27 − +c 4 27 x= +c 4 At t = 4: x = 64 − 64 + c x=c t=4

t=3

t=0 c

Distance travelled: =

27 –+c 4

x

27 27 + 4 4 27 = = 13.5 metres 2

Units 3 & 4

Area 3

Sequence 2

Concept 7

Applications of integration Summary screen and practice questions

CHAPTER 7 Integration 271

Exercise 7.6 Applications of integration Technology active 1.

Heat escapes from a storage tank at a rate of kilojoules per day. This rate can be modelled by ( ) dH 𝜋t 𝜋2 =1+ sin , 0 ≤ t ≤ 100 9 dt 45 WE13

where H(t) is the total accumulated heat loss in kilojoules, t days after 1 June. What is the rate of escape of the heat, correct to 2 decimal places, at: i. 15 days? ii. 60 days? b. State the period of the function. c. Calculate the total accumulated heat loss after 45 days. Give your answer in exact form. 2. The average rate of increase, in cm/month, in the length of a baby boy from birth until age 36 months is given by the rule a.

dL 4 =√ dt t where t is the time in months since birth and L is the length in centimetres. Determine the average total increase in length of a baby boy from 6 months of age until 36 months of age. Give your answer correct to 1 decimal place.

3.

A number of apprentice bricklayers are competing in a competition in which they are required to build a fence. The competitors must produce a fence that is straight, neatly constructed and level. The winner will also be judged on how many bricks they have laid during a 30-minute period. The winner laid bricks at a rate defined by the rule dN = 0.8t + 2 dt

where N is the number of bricks laid after t minutes. a. Sketch the graph of the given function for 0 ≤ t ≤ 30. b. Shade the region defined by 10 ≤ t ≤ 20. c. How many bricks in total did the winner lay in the 10-minute period defined by 10 ≤ t ≤ 20? 4. The rate of growth of mobile phone subscribers with a particular company in the UK can be modelled by the rule dN = 0.853e0.1333t dt

where N million is the number of subscribers with the company since 1998 and t is the number of years since 1998, the year the company was established. Determine how many millions of mobile phone subscribers have joined the company between 1998 and 2015, correct to 1 decimal place.

272 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

5.

6.

7.

8.

9.

10.

A manufacturer of a game knows that the revenue, $R, for selling x games is √ √ R = 100( x + 4 − 2) The costs, $C, to produce x games is C = 50 + x x. a. Calculate the profit made when 10 games were sold. b. Determine the average profit per game when 10 games were sold. c. For 20 games, calculate and interpret: i. the marginal revenue ii. the marginal cost iii. the marginal profit. √ The weekly profit of a factory, P (in hundreds of dollars), is given by P = 8n − n n, where n is the number of employees. a. Calculate the weekly profit of a factory with 16 employees. b. Determine the average weekly profit per employee when there are 16 employees. c. Calculate the marginal weekly profit, in dollars per employee, when the number of employees is: i. 10 employees ii. 25 employees. WE15 A manufacturer has found that the cost per item to produce x items is given by dC = 20 + x + e−0.05x , where x ∈ [0, 50] and C is the cost in dollars. dx a. Use the rate to calculate the marginal cost of producing the 10th item. b. What is the total cost of producing the first 10 items? c. Determine the average cost of production for the first 10 items. dC On any day the cost per item for a machine producing n items is given by = 40 − 2e0.01n, where dn n ∈ [0.200] and C is the cost in dollars. a. Use the rate to determine the cost of producing the 100th item. b. Express C as a function of n. c. What is the total cost of producing the first 100 items? d. Determine the average cost of production for the first 100 items. WE16 A particle moves in a line so that its velocity, v metres/second, from a fixed point, O, is defined √ by v = 1 + 3 t + 1, where t is the time in seconds. a. Determine the initial velocity of the particle. b. What is the acceleration of the particle when: i. t = 0? ii. t = 8? c. Sketch the graph of v versus t for the first 10 seconds. d. Determine the distance covered by the particle in the first 8 seconds. An object travels in a line so that its velocity, v metres/second, at time t seconds is given by ( ) t 𝜋 v = 3 cos − , t ≥ 0. 4 2 √ Initially the object is −3 2 metres from the origin. a. Determine the relationship between the displacement of the object, x metres, and time, t seconds. b. What is the displacement of the object when time is equal to 3𝜋 seconds? c. Sketch the graph of v versus t for 0 ≤ t ≤ 4𝜋. d. Determine the distance travelled by the object after 3𝜋 seconds. Give your answer in metres, correct to 2 decimal places. e. Determine a relationship between the acceleration of the object, a metres/second2 , and time, t seconds. f. What is the acceleration of the object when t = 3𝜋 seconds? WE14

CHAPTER 7 Integration 273

The rate of deflection from a horizontal position y(m) of a 3-metre diving board when an 80-kilogram x(m) person is x metres from its fixed end is given by 0 Deflection dy Board = −0.03(x + 1)2 + 0.03, where y is the dx deflection in metres. a. What is the deflection when x = 0? b. Determine the equation which measures the deflection. c. Hence, determine the maximum deflection. 12. The rate of change of position (velocity) of a racing car travelling down a straight stretch of dx road is given by = t(16 − t), where x is dt measured in metres and t in seconds. a. Determine the velocity when: i. t = 0 ii. t = 4. b. Determine: i. when the maximum velocity occurs ii. the maximum velocity. dx against x for 0 ≤ t ≤ 16. c. Sketch the graph of dt d. Determine the area under the graph between t = 0 and t = 10. e. What does this area represent? 13. The rate of flow of water into a hot water system during a 12-hour period on a certain day is thought to ( ) dV 𝜋t be = 10 + cos , where V is in litres and t is the number of hours after 8 am. dt 2 dv a. Sketch the graph of against t. dt b. Determine the length of time for which the rate is above 10.5 L/h. c. Determine the volume of water that has followed into the system between: i. 8 am and 2 pm ii. 3 pm and 8 pm. v 14. A particle moves in a straight line. At time t seconds its velocity, v metres per second, is defined by the rule v = e−0.5t − 0.5, t ≥ 0. The graph of the motion is shown. (0, 0.5) a. Determine the acceleration of the particle, a m/s2 , in (2 loge (2), 0) terms of t. t 0 b. Determine the displacement of the particle, x m, if x = 0 when t = 0. v = e–0.5t – 0.5 c. Determine the displacement of the particle after 4 seconds. d. Calculate the distance covered by the particle in the fourth second. Give your answer correct to 4 decimal places. 11.

274 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

15.

The maintenance costs for a car increase as the car gets older. It has been suggested that the increase in maintenance costs of dollars per year could be modelled by dC = 15t2 + 250 dt

where t is the age of the car in years and C is the total accumulated cost of maintenance for t years. a. Sketch the graph of the given function for 0 ≤ t ≤ 10. b. Determine the total accumulated cost of maintenance for t = 5 to t = 10 years.

7.7 Review: exam practice

A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. Using the figure shown, calculate the approximate area under the curve from x = 2 to x = 4, using the left-hand, or lower, rectangles. y

(3.5, 9) (3, 6) (2.5, 4) (2, 3)

0

x 2

2.

4

3

Using the figure shown, calculate the approximate area under the curve from x = −5 to x = −1, using the upper rectangles. y (–5, 8) (–4, 6) (–3, 5) (–2, 4)

0 –5

x

–1

CHAPTER 7 Integration 275

3.

A student is using the trapezoidal rule to approximate the area under the curve shown from x = 0 to x = 4. What would their answer be? y (4, 10) (3, 7) (2, 5) (1, 4)

x 0

1

2

3

4

Apply the trapezoidal method to determine the area between y = e2x−1 and the x-axis from x = 0 to x = 4, using intervals that are 1 unit wide. (Give your answer correct to 2 decimal places.) 5. Consider the area under the curve y = loge (x) from x = 2 to x = 4. With interval widths of 0.5 units, determine an approximation to the area using: a. left end-point rectangles b. right end-point rectangles c. the average of the rectangles. Give your answers in exact form. 6. Determine: 1 2( ) √ 2 a. 3x + 6 x + 1 dx b. (ex + 1)(ex − 1) dx ∫0 ∫0 4.

0

c.

∫−1

9 dx (2x + 3)4

7.

Given that

8.

Given that

k

∫0

5

5

a.

∫1

5

b.

∫1

5

c.

∫1

∫1

d.

2𝜋 3

∫𝜋

cos 2x dx

(4x − 5) dx = −2, determine two possible values for k. f (x) dx = 4 and

(4f(x) + 1) dx

3

5

∫1

g (x) dx = 3, determine:

(2f(x) − g(x)) dx

(3f(x) + 2g(x) − 5) dx

Sketch the graph of the function f(x) =

1 . x−2 b. Calculate the exact area between the graph of f(x), the x-axis and the lines x = 3 and x = 6.

9. a.

276 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

10.

Determine the area bounded by the curve y = x(x + 2)(x − 3) and the x-axis.

y

y = x(x + 2)(x – 3)

0 m

11.

Determine m if

∫1

2

12.

x

6 (2x − 1)2 dx = 1.

A particle starts at the origin and travels in a straight line with a velocity, v m/s, modelled by v = t2 − t − 2, where t is the time in seconds. a. State the equation for the acceleration of the particle. b. Determine when the particle is at rest. c. What is the displacement of the particle after 3 seconds? d. Determine the distance covered by the particle in the first 3 seconds. e. Hence, determine the average speed of the particle for the first 3 seconds.

Complex familiar 13. a. Determine any point(s) of intersection between the two curves f(x) = x3 − 3x + 2 and g(x) = x + 2. b. Sketch f(x) and g(x) on the same set of axes. Label the point(s) of intersection and any x- and y-intercepts. c. Evaluate the area between the two curves. 14. A manufacturer of a new gadget knows that the revenue, $R,

for selling x gadgets is R =√ e 20 − 1. The costs, $C, to produce x gadgets is C = 40 + x − x. a. Calculate the profit or loss made when: i. 50 gadgets are sold ii. 100 gadgets are sold. b. Determine the average profit per gadget when 100 gadgets are sold. c. For 120 gadgets, calculate and interpret: i. the marginal revenue ii. the marginal cost iii. the marginal profit. 15. The velocity of a particle is given by v = 2 sin(2t) + 3, where x is the displacement in metres and t is the time in seconds. Initially the particle is at the origin. a. Show that the displacement is given by x = −cos(2t) + 3t + 1. 𝜋 b. Determine the displacement when t = seconds. 2 𝜋 c. What distance has the particle travelled in the first seconds? 2 16. The graphs of y = 2 sin(x) + k and y = k cos(x) are shown. The shaded region is equal to (3𝜋 + 4) square units. Determine the value of the constant k. x

CHAPTER 7 Integration 277

y

y = 2 sin(x) + k

( ) 3π –, 0 2

x

0 y = k cos(x)

Complex familiar √ 17. The graphs with equations y = x and y = 2 − x are shown. y

y= x y=2–x 0 (0, 0)

x

Determine the point of intersection of the graphs. Calculate the area of the blue shaded region. c. Hence, or otherwise, determine the area of the pink shaded region. 18. An arched window has width 6 metres and height 3 metres, as shown in the diagram. The arch can be approximated to a parabola. a. b.

y

3m

x 3m

3m

278 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

b.

Determine the equation of parabola. Calculate the area of the window.

c.

Show that the area of the window is

a.

2 the base of the arch times the height. 3 d. Hence, calculate the area of a similar arched window with width 8 metres and height 4.5 metres. dA 1 19. A ground-cover plant can cover the ground at a rate modelled by = 2t + 6t2 − t3 , where A is the dt 4 area in square centimetres and t is the time in weeks. a. If the plant initially covers 10 cm2 , how long, to the nearest week, will it take to cover 0.6 m2 ? b. What is the maximum area covered by the plant? 20. The cross-section of a platform is shown. y (All measurements are in metres.) a. Determine the derivative of x loge (x). 1 b. Hence, determine an antiderivative of loge (x). c. Calculate the height of the platform. e–2 d. Calculate the cross-sectional area of the platform. 0 1 e x e. Calculate the volume of concrete required to build this platform if it is 20 metres long. Give your answer to the f(x) = loge (x) nearest cubic metre.

Units 3 & 4

Sit exam

CHAPTER 7 Integration 279

Answers

13. a. 1

7 Integration

14. a.

Exercise 7.2 Estimating the area under a curve 1. 10 sq. units 2. a. 8 sq. units 3. a. b.

4. 5. 6. 7.

8. 9. 10.

11. 12. 13. 14. 15.

25 12 77

b. 42 sq. units

sq. units or approx. 2.08 sq. units

dy dx dy

b. 21

= x cos(x) + sin(x)

( ) = 3 x2 − 2x e dx ) 1 ( −2 b. e −1 3

15. a.

1. a. i.

∫0 2

b. i.

∫1

(4 − x) dx

ii. 8 sq. units

( 2) x dx

ii. 2

∫−3 3

d. i.

∫1

1

(ex ) dx

ii.

( −2x ) e dx

ii.

∫−1 4

∫1

c. i.

1. a.

b. 20

3

e. 3. a. c. e. 4. a. c.

4

b.

3 85

256 3.65 (to 2 decimal places) 1 √ 20 − 10 2 ≈ 5.86 0 ) 1( 8 e −1 4 ( ) 2 e−2 − e2 2

e. 5 ln(4) + 2e − 5. a. 27

e.

1

5

3𝜋 2

e. 8

3

b. 0 d. −6 f. 0 (

−2

1( 2

1( 2

) e−2 − e−8 sq. units

x cos dx 3

ii. 3 sq. units b. 5

7

1 3

sq. units

d. 4 sq. units

) e−1 − e−2 sq. units 2 c. 1 sq. unit d. 4 sq. units 5. a. x-intercepts: x = 1, −1, 3, −3; y-intercept: y = 9

)

1− e 3

b. 3

∫0

sq. units 24 2 e. 2 sq. units 3 ( ) −1 4. a. e − e sq. units c.

d. 2

d.

d. i.

( ) e − e−1 sq. units

ii. 2 sq. units

3. a. 1 sq. unit

ln(5) ≈ 0.966

e12 + 15 − e6

b.

1(

y

)

10 (0, 9)

1

6 4 (√ 3

= 16 4 4 7. a. 21 b. 11 8. a. −15 b. −12.5 c. 32.5 9. 2 10. 3 11. 2 loge (3) 12. 4 6. a.

3

( ) 2 sin(2x) dx

1 2e 2

b. 9

d. 0

65

d.

3

1

2. a. 37 c.

c. 5

∫0

sq. units

ii. 22 sq. units

∫−2 1 ii. 1 sq. units 3

b. i.

3

( 3 ) −x − 4x2 − 4x dx

0

2. a. i.

1

ii. 26 sq. units

( 3 ) x − 9x2 + 20x dx

e. i.

Exercise 7.3 The fundamental theorem of calculus and definite integrals 1

( 2) 3x dx

−1

c. i.

𝜋 2

1

b. 𝜋 + 2

) ( 3 2 x −3x

4

60 a. 30 sq. units b. 22 sq. units 9 sq. units a. 8 sq. units b. 13 sq. units a. 26 sq. units b. 41 sq. units 1 c. 33 sq. units 2 −1 a. (1 + e + e ) sq. units b. loge (24) sq. units a. 7.25 sq. units b. 100 sq. units 1 a. 19 sq. units 4 ) 1( 1 + 2e + 2e2 + e3 sq. units b. 2 a. 22.5 sq. units b. 20.8 sq. units a. 1.87 sq. units b. 1.68 sq. units 12 sq. units 21 sq. units a. 4 b. 7.56 sq. units

1

3

Exercise 7.4 Areas under curves

sq. units or approx. 1.28 sq. units

1

1

c. 0

)

√

10 − 2 b. 2 + c. −16 d. 20

f.

1 2

2

2−

1

e − e 2

√ 3 3 4

(–1, 0) (–3, 0)

–2

−2

d. 8 e. 25

(1, 0) 0

–10 f. 12.5

b. 52

4 15

sq. units

280 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2

(3, 0) x

6. a. 15 c. 21

3 4 1

sq. units

sq. units 12 2 7. a. i. y = 8 − x

b. 5

1 3

d. i. y = x − 4x − 4x + 16 3

sq. units

2

y

15 10

y 10

5 (2, 0)

(–2, 0) –2 (2.828, 0) x 5

(–2.828, 0) 0

–5

1

3

4

x

x-intercepts: x = −2, 2, 4 1 ii. 49 sq. units 3 3 2 e. i. y = x + 3x − x − 3

x-intercepts: x = ± 8 or ±2 2 √ 64 2 ii. sq. units 3 3 2 b. i. y = x − 4x

y 5

√

(–1, 0) (1, 0)

(–3, 0) –4

–3

–2

0

–1

y 10

x

1

–5

x-intercepts: x = −3, −1, 1

(0, 0)

ii. Area = 8 sq. units f. i. y = (x − 1) (x + 2)(x + 5)

(4, 0)

0

–5

2

(4, 0)

–5

–10

√

–1

0

5

x

y 10

5 (–5, 0)

–10

x-intercepts: x = 0, 4 1 ii. 21 sq. units 3 c. i. y = x(x − 2)(x − 3)

–6

0

(3, 0) 2

x

9. 10.

–2

x-intercepts: x = 0, 2, 3 1 ii. 3 sq. units 12

–3

–2

–1

0

1

x

–10

8.

(2, 0)

–4

–5

y 2

(0, 0)

–5

(1, 0)

(–2, 0)

11. 12. 13. 14. 15. 16.

x-intercepts: x = −5, −2, 1 1 ii. 40 sq. units 2 2 ln(2) sq. units ) 1( 6 e − e3 sq. units 3 √ 3− 3 sq. units 2 2 166 sq. units 3 3𝜋 units ) ( sq. −1 e + e − 2 sq. units 1.6 sq. units 16 sq. units dy a. = 1 + ln(x) dx

ln(x) dx = x ln(x) − x + c ∫ c. (4 ln(4) − 3) sq. units 2 dy 17. a. = 2xex b. 2(e − 1) sq. units dx b.

CHAPTER 7 Integration 281

18. a. b. c. 19. a. 20. a.

=

(x2 + 2) dx ) x 1 ( dx = ln x2 + 2 + c ∫ (x2 + 2) 2 ( ) ( ) 3 sq. units ln(3) − ln(2) sq. units, or ln 2 3 1 sq. units b. 2 sq. units 4 4 2 b. 2 (𝜋 − 2) sq. units dy

2x

Exercise 7.5 Areas between curves 1. a. 2

1 6

b. 21 c. d. e. f. 2. 3. 4. 5.

C D D

8. 20 9. a.

5 6 (

sq. units 1 ,√ 4 2

3𝜋

√ 𝜋 3

)

√

b. 2

2 sq. units

sq. units (≈ 0.45 sq. units) 12 11. (3 ln(2) − 2 ) sq. units 12. a. a = 1, b = 2, c = 3 b. 9.6 sq. units

10.

13. a. (−4, 0), (4, 0) 14. a.

b. (−3, 0), (3, 0)

c. 9

1 3

m2

v

sq. units 1

x y = 0.5 sin 2 + 2

()

sq. units

3 1 2 sq. units 4 52 1 + − e ≈ 14.98 sq. units 3 e 3 2 e − e + ≈ 6.17 sq. units 2 1 1 sq. units 3

(4π, 2) (0, 2)

y=2

0

2π

t

4π

2

b. 25 m 3 c. 12.5 m

15. a. y = 5 −

a. x = 0, 1 or −1 b. y

c. 50

2

2 3

1 5

x2

b. 33

m2

1 3

m2

d. 152 m

3

g(x) = x

Exercise 7.6 Applications of integration

(1, 1) (0, 0) 0

2

–2 f(x) = x3 (–1, –1)

1. a. i. 1.95 kJ/day ii. 0.05 kJ/day b. 90 days c. Accumulated heat loss after 45 days is (45 + 10𝜋) kJ, or

x

approx. 76.42 kJ.

2. 28.4 cm dN 3. a, b. —

–2

1

dt

sq. units

6. a. x = 0, 1 or 2 y b. c.

2

(30, 26) 26

12

dN — = 0.8t + 2 dt

(2, 12)

10 8 (0, 2)

6

0

f(x) = x3 + 2x

4

(1, 3) 2 g(x) = 3x2

(0, 0) –0.5 c.

1 2

0

0.5

sq. units

7. (16, 4); 21

1 3

sq. units

1

1.5

2

x

10

20

30

t

c. 140 bricks 4. 55.3 million 5. a. $92.54 b. $9.25 c. i. Marginal revenue at x = 20 is $10.21, so the

approximate revenue from selling the next game is $10.21. ii. Marginal cost at x = 20 is $6.71, so the approximate cost of manufacturing the next game is $6.71.

282 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

iii. Marginal profit at x = 20 is $3.50, so the

6. a. b. c. 7. a. 8. a. b. c. d. 9. a. b.

ii. 64 m/s

approximate profit from selling the next game is $3.50. $6 400 $400 i. $32.57/employee ii. $2/employee $30.61 b. $257.87 c. $25.79/item $34.56 C = 40n − 200 e0.01n + 200 $3 656.34 $36.56/item 4 m/s i. 1.5 m/s2 ii. 0.5 m/s2

c.

c.

dx – = t(16 – t) dt

64

0

8

t (s)

16

2

m 3 e. The area represents the distance travelled in the first 10 s.

d. 466

v (10, 10.95)

dx (m/s) dt

13. a. dV L

dt y = 1 + 3√x + 1

H

11 10 9

( )

dV = 10 + cos πt – – 2 dt

(0, 4)

0 0 d. 60 m

10. a. x = 6 sin b. −3 c.

(

√

(

t 2

2m

−

𝜋

10

)

0, 3 2 2

t π v = 3 cos 2 – 4

(

(

t (h)

12

)

(4π, 3 2 2 (

= −0.5 e−0.5t dt −0.5t b. x = 2 − 2 e − 0.5t c. −0.2707 m d. 0.3244 m

14. a. a =

15. a.

0

8

b. 4 h c. i. 60 L ii. 50.6 L

4

v 3

4

t

(7π–2 , 0 (

(3π–2 , 0 (

dv

dC – dt 1750

(10, 1750)

4π t dC – = 15t2 + 250 dt

–3 d. 20.49 m e. a =

√

f. 11. a. b. c. 12. a. b.

dv

3 2

dt

=−

3 2

( sin

t 2

−

𝜋

(0, 250)

)

0

4

m/s2 4 0m y = −0.01 (x + 1)3 + 0.03x + 0.01 54 cm downwards i. 0 m/s ii. 48 m/s i. 8 s

10

t

b. $5625

7.7 Review: exam practice 11 sq. units. 23 sq. units. 21 sq. units. 719.72 sq. units ( ) 1 105 5. a. ln sq. units 2 2 1. 2. 3. 4.

CHAPTER 7 Integration 283

1

b.

2 1

c. 6. a. b. c. d. 7.

1 2

14. a. i. A loss of $71.75 ii. A profit of $17.41 b. $0.17/gadget c. i. The marginal revenue at x = 120 is $20.17, so the

ln(105) sq. units

(2 ln(105) − ln(2)) sq. units 4 √ 8+8 2 1 e−1 2 4 1 9√ 3 − 2

,2

8. a. 20 9. a. y

c. −2

b. 5

15.

4 y=

1 x–2

2

16. 17. 1

0

2

4

6

x

–2 –4 18.

b. loge (4) sq. units 10. 21 11. 1

1

12

sq. units

12. a. a =

= 2t − 1 dt b. The particle is at rest at t = 2 s. dv

c. After 3 s, the displacement is −3

of the origin.

19.

1

1 m, or 3 m to the left 3 3

d. The distance travelled in first 3 s is 5

1

m.

6 13 e. The average speed for the first 3 s is 1 m/s. 18 13. a. (−2, 0), (0, 2), (2, 4) b.

20.

approximate revenue from selling the next gadget is $20.17. ii. The marginal cost at x = 120 is $0.95, so the approximate cost of manufacturing the next gadget is $0.95. iii. The marginal profit at x = 120 is $19.22, so the approximate profit from selling the next gadget is $19.22. a. Sample responses can be found in the worked solutions in ) resources. ( the online 3𝜋 m b. 2 + 2 ( ) 3𝜋 c. 2 + m 2 2 a. (1, 1) 1 b. The blue shaded area is 1 sq. units. 6 5 c. The pink shaded area is sq. units. 6 ) 1( 2 a. y = − x −9 3 2 b. 12 m c. Sample responses can be found in the worked solutions in the online resources. 2 d. 24 m a. 19 weeks 2 b. 0.75 m dy a. = 1 + loge (x) dx loge (x) dx = x loge (x) − x + c ∫ c. 3 ( m −2 ) 2 d. e − e m or approx. 2.58 m2 3 e. 52 m b.

y 10 f(x) = x3 – 3x + 2

(–2, 0) –3

–2

–1

5 g(x) = x + 2 (2, 4) (0, 2) 0

1

2

x

c. The area between the curves is 8 sq. units.

284 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

REVISION UNIT 3 Further calculus

TOPIC 3 Integrals • For revision of this entire topic, go to your studyON title in your bookshelf at www.jacplus.com.au. • Select Continue Studying to access hundreds of revision questions across your entire course.

• Select your course Mathematical Methods for Queensland Units 3&4 to see the entire course divided into syllabus topics. • Select the Area you are studying to navigate into the chapter level OR select Practice to answer all practice questions available for each area.

• Select Practice at the sequence level to access all questions in the sequence.

OR

• At sequence level, drill down to concept level.

• Summary screens provide revision and consolidation of key concepts. Select the next arrow to revise all concepts in the sequence and practise questions at the concept level.

REVISION UNIT 3 Further calculus 285

PRACTICE ASSESSMENT 1 Mathematical Methods: Problem solving and modelling task Unit Unit 3: Further calculus

Topics Topic 2: Further differentiation and applications 2 Topic 3: Integrals

Conditions Duration

Mode

Individual/group

4 weeks

Written report, up to 10 pages (maximum 2000 words) excluding appendix

Individual

Resources permitted The use of technology is required, for example: • graphing calculator • computer • internet • graphing packages. Milestones Week 4 Week 5 Week 6 Week 7 (assessment submission) Criterion*

Marks allocated

Formulate *Assessment objectives 1, 2, 5

4

Solve *Assessment objectives 1, 6

7

Evaluate and verify *Assessment objectives 4, 5

5

Communicate *Assessment objectives 3

4

Total

20

Result

Scaffolding Please refer to the fourth page of this assessment for an appropriate approach to problem solving and modelling. *© State of Queensland (Queensland Curriculum & Assessment Authority), Mathematical Methods General Senior Syllabus 2019 v1.2, Brisbane. For the most up to date assessment information, please see www.qcaa.qld.edu.au/senior.

286 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Context Three engineers are discussing various cross-sections for a proposed new irrigation canal. The canal needs to be 14 metres across at the surface, with a maximum depth of 6 metres. As the canal should be able to carry as much water as possible, the engineers are analysing their proposals to find which gives the greatest area of the crosssection and so the greatest volume. When making calculations, they decided to express all lengths in metres, to 6 decimal places where necessary, and all angles in degrees, correct to 1 decimal place.

The first engineer proposes a canal with straight sloping sides of length 5 metres with a curved section at the base. He has sketched his proposal, not drawn to scale, and shown it to his colleagues. He maintains that the curved section would be quadratic, cubic, quartic or exponential in shape.

14 m

6m

Ground level

5m

60º

The second engineer proposes a simple trapezoidal canal, as he estimates it would be the most cost effective. He maintains it could easily be constructed with the same area as any other proposal, and with the same length sloping sides of 5 metres.

14 m

Ground level

5m

θ

PRACTICE ASSESSMENT 1 287

The third engineer proposes a canal which is the lower half of a sine function. She has stated that the function would be of the form y = a sin(𝜋nx) + k and has presented her sketch to her colleagues.

14 m Ground level

6m

Task You are helping the engineers with their calculations to determine the shape of the irrigation canal that gives the greatest volume. For the first engineer, determine the equations of the curves in his diagram and the areas of the cross-sections. Determine the angle the second engineer would find for 𝜃 if his diagram is to have the same area as the largest area of the first engineer’s proposal. Discuss the method used to find this angle. The third engineer maintains her canal would be greater in area than both of the others. Determine whether she is, in fact, correct. In making a final decision on the cross-section for the new canal, the three engineers will check all regulations for irrigation canals. One regulation requires, for safety reasons, that the angle made between the side and the surface of the canal needs to be as small as possible. Discuss, with reasons, which of the cross-sections would give the best result for the proposed new irrigation canal when this regulation is considered. As an engineering student, you are asked to alter the first engineer’s proposal to meet the added requirement that the angle between the surface and the side be as small as possible. Prepare your altered design to present to the engineers with justifications for why it is a reasonable model.

288 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Approach to problem-solving and modelling Formulate In this task, you will develop equations of various cross-sections from the given information. Using these equations and your knowledge of calculus, you will solve the given problem. Appropriate mathematical procedures and use of technology will be used to develop your response. Your report should include: • the various types of functions • algebraic solutions • solutions derived using technology. Remember to state all assumptions and variables along with how you made use of technology. Solve Select and apply your knowledge of mathematical functions to determine the various equations needed to solve the problem. You must demonstrate how you obtained the functions. You must use both technological and algebraic solutions efficiently and show detailed calculations demonstrating the procedures used to formulate your solutions. In your solutions, include all diagrams, clearly labelled.

Is it solved? Evaluate and verify Justify and explain all the procedures you have used to come to the decision you have made on the most appropriate cross-section. You must interpret the restrictions placed on the irrigation canal to ensure your solution is valid.

Is the solution verified? Communicate Once you have completed all necessary calculations, you should consider how you have communicated the aspects of your report. Your response should have: • used appropriate mathematical language • included any graphs or diagrams • shown all calculations or stated how they were obtained • drawn conclusions, noting particularly how key concepts were addressed.

PRACTICE ASSESSMENT 1 289

PRACTICE ASSESSMENT 2 Mathematical Methods: Unit 3 examination Unit Unit 3: Further calculus

Topics Topic 1: The logarithmic function 2 Topic 2: Further differentiation and applications 2 Topic 3: Integrals

Conditions Technique

Response Type

Duration

Reading

Paper 1: Technology free Paper 2: Technology active

Short response

Paper 1: 60 minutes Paper 2: 60 minutes

5 minutes

Resources • • • •

QCAA formula sheet Notes not permitted Paper 1: Calculator not permitted Paper 2: Non-CAS graphics calculator permitted

Criterion Foundational knowledge and problem solving *Assessment objectives 1, 2, 3, 4, 5 and 6

Instructions • Show all working. • Write responses using a black or blue pen. • Unless otherwise instructed, give answers to two decimal places.

Marks allocated Paper 1

Marks allocated Paper 2

47

53

Result

*© State of Queensland (Queensland Curriculum & Assessment Authority), Mathematical Methods General Senior Syllabus 2019 v1.2, Brisbane. For the most up to date assessment information, please see www.qcaa.qld.edu.au/senior.

290 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Paper 1 — Technology free — total marks: 47 Part A: Simple familiar — total marks: 27 Question 1 (1 + 3 = 4 marks) ( )5 dy a. If y = 4x2 − 3x , determine . dx b. If f (x) = x sin (2x), evaluate f ′ (𝜋).

Question 2 (3 marks) ( ) Solve 2 cos 2𝜃 = −√3 for 0 ≤ 𝜃 ≤ 2𝜋.

Question 3 (1 + 3 = 4 marks) a. Determine ∫ (3 + cos (2x − 1))dx. 5

b.

2 dx = ln(k). Determine the value of k. ∫3 (x − 2)

PRACTICE ASSESSMENT 2 291

Question 4 (2 + 3 = 5 marks) Solve the following for x. a. loge (4x − 3) = 0 b. log5 (3x) − log5 (x − 2) = 1

Question 5 (4 marks) If f (x) =

2x2 − 3 x , determine the coordinates of the point(s) at which the gradient is 5.

Question 6 (2 + 2 + 3 = 7 marks) 1 a. On one set of axes, sketch the graphs of f : [0, ∞) → R, f(x) = x2 and g: (0, ∞) → R , g(x) = x . b. Use algebra to find the coordinates of the point of intersection of the two graphs. c. Determine the exact area between the two curves from their point of intersection to x = 2.

292 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Part B: Complex familiar — total marks: 10 Question 7 (3 marks) Solve the following for the value of x. 5 2x − 4 = 3 (5x )

Question 8 (2 + 2 + 1 + 2 = 7 marks) A particle is attached to a spring that moves up and down in a straight line. The velocity of the particle, v cm/s, after t seconds is given by ( v = 2𝜋 sin

𝜋t 4

) , t ≥ 0.

a. Determine the rule relating the position of the particle, x cm, to t if the particle starts from rest at the origin. b. What is the maximum displacement of the particle? c. Find an expression for the acceleration of the particle at any time t. d. Determine the particle’s displacement and acceleration after 9 seconds. Express your answers in the form a + b√c.

PRACTICE ASSESSMENT 2 293

Part C: Complex unfamiliar — total marks: 10 Question 9 (2 + 1 + 2 + 2+ 3 = 10 marks) A section of a rollercoaster track at a local fun park is shown. h

0

25

x

The track can be described by the rule ( ) 𝜋x + 5, 0 ≤ x ≤ 25 h (x) = 4 cos 25 where h is the height in metres above ground level and x is the horizontal distance in metres from the top of the descent. Note that the x-axis represents the ground. a. How high is a rollercoaster car from the ground: i. at the beginning of its descent? ii. at the end of the ride? b. Find h′ (x). c. Sketch the function y = h′ (x) for 0 < x < 25. d. Determine where the rollercoaster track is steepest, and determine the gradient at this point. e. The section below the rollercoaster track is to be enclosed on either side for safety reasons. Determine the volume of the enclosure if the track is 2.5 metres wide.

294 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Paper 2 — Technology active — total marks: 53 Part A: Simple familiar — total marks: 33 Question 1 (4 + 2 = 6 marks) ( ) a. Differentiate loge 2x3 + 4 and hence find the value of

1

∫−1

(

x2 x3 + 2

) dx, expressing your answer in the form

k loge p. b. Given that ∫ aebx dx = − 41 e4x + c, find the exact values of a and b.

Question 2 (3 + 2 = 5 marks) Solve the following for x. a. log6 (x − 3) + log6 (x + 2) = 1 b. 22x+1 × 82−x = 16

PRACTICE ASSESSMENT 2 295

Question 3 (1 + 1 + 1 = 3 marks) The temperature in °C on a particular winter’s day in Toowoomba can be modelled by the function ( ) 𝜋t + 12, 0 ≤ t ≤ 24 T = 2 sin 9 where t is the time in hours after 8:00 am. a. Calculate the temperature, correct to the nearest degree, at 12 noon. dT b. Find . dt c. What is the rate of change of temperature at midnight? Give your answer correct to 3 decimal places.

Question 4 (1 + 1 + 1 + 2 + 1 = 6 marks) The population of possums in a reserve is starting to increase. The numbers observed currently can be defined by the rule P(t) = 100 − 60e−0.1t where P is the population of possums, t is the time in months since observations began, and t ≥ 0. Give all answers to the nearest integer. a. How many possums were present initially? b. How many possums were present after: i. 1 month? ii. 1 year? c. How long does it take for the possum number to double? d. Sketch the graph of the function representing the possum population for the first year.

296 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Question 5 (1 + 1 + 1 + 2 + 1 = 6 marks) A square piece of cardboard of side length 5 cm is to be formed into an open-top box by cutting out the corners of length x cm, as shown. 5 cm

x

a. Show that the volume, V cm3 , is given by V = 4x3 − 20x2 + 25x. b. For which values of x would the function of V be defined? dV c. State . dx d. Determine the value of x that would give a maximum volume. e. What is the maximum volume that a box formed from this cardboard could hold? Give your answer correct to 2 decimal places.

PRACTICE ASSESSMENT 2 297

Question 6 (3 + 4 = 7 marks) 1 The graphs of y = −3x and y = − x3 are shown. 3

y

y = –3x

(0, 0) 0

x 1 y = – – x3 3

a. Determine the coordinates of the points of intersection of the two graphs. b. Calculate the area of the shaded region shown.

298 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Part B: Complex familiar — total marks: 10 Question 7 (3 + 2 + 1 = 6 marks) Children are playing on a seesaw. Initially the seesaw is in a horizontal position. The children begin playing, going up and down on the seesaw. The height that a child is above the ground can be modelled by the rule h(t) = a sin(n𝜋t) + k where h is the height in metres a child is above the ground, and t is the time in seconds since the seesaw motion began. The greatest height above the ground that a child can be is 1.5 metres, and the least height is 0.5 metres. It takes 3 seconds for the child to seesaw between the two heights. a. Find the values of the constants a, n and k. b. Draw the graph of the function for 0 ≤ t ≤ 12. c. For what length of time during the first 12 seconds is the child 1 or more metres above the ground?

Question 8 (2 + 2 = 4 marks) The loudness of a sound, S, in decibels (dB), is related to the intensity, I, of the sound by the equation ( S = 10 log10

I

)

10−12

where I is measured in W/m2. a. Show that I = 10

S−120 10 .

b. Being in an environment where the intensity of noise is above 0.003 W/m2 is not recommended for long periods of time, as hearing loss may occur. If the loudness of a typical rock concert is measured to be 115 dB, is it safe to attend many rock concerts due to the noise level?

PRACTICE ASSESSMENT 2 299

Part C: Complex unfamiliar — total marks: 10 Question 9 (1 + 3 + 2 + 1 + 3 = 10 marks) Consider the graph defined by the function f (x) = x2 e a. Differentiate to find an expression for f ′ (x).

(1−x)

.

b. Hence, determine the coordinates of the stationary point(s). c. Sketch the function y = f (x), showing clearly all important features, and state the equations of any asymptotes. ( ) d. The point P k, f (k) lies on the curve. Determine the gradient of the line joining the point P to the origin, O. e. Hence, determine the coordinates of any non-stationary point(s) on the curve where the tangent passes through the origin, and determine the equation of this tangent.

300 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Examination marks summary Question number

Simple familiar

Complex familiar

Complex unfamiliar

Topics

Paper 1 — Technology free 1

4

2

2

3

2

3

4

1, 3

4

5

1

5

4

2

6

7

3

7

3

1

8

7

2, 3

9

10

2, 3

Paper 2 — Technology active 1

6

1, 2, 3

2

5

1

3

3

2

4

6

1

5

6

2

6

7

3

7

6

2

8

4

1

9

10

Totals

60

20

20

Percentage

60%

20%

20%

1, 2

PRACTICE ASSESSMENT 2 301

8

The second derivative and applications of differentiation

8.1 Overview This is the final chapter in the area of differential calculus, the study of how a change in one variable will affect a related variable. You have learned about differentiation and its applications to curve sketching and determining equations of tangents to curves, as well as practical applications. You have also studied antidifferentiation, or integration, along with applications such as determining areas under curves. In this chapter we will explore the second derivative — the rate of change of the derivative. You will discover that using the second derivative can help in discussing the behaviour of functions. For example, it allows us to answer questions such as: what is the concavity of the function? Is the critical point the largest value of the function? The second derivative also has many practical applications. In problems of optimisation, the second derivative assists in determining the nature of the critical points. In kinematics, the second derivative of the displacement of a particle with respect to time is the acceleration of the function. Scientists, economists and engineers are all concerned with determining critical values.

LEARNING SEQUENCE 8.1 8.2 8.3 8.4 8.5 8.6

Overview Second derivatives Concavity and points of inflection Curve sketching Applications of the second derivative Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

302 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

8.2 Second derivatives From previous chapters, you are familiar with the relationships between displacement, velocity and acceleration as functions of time. If x = f(t) is the displacement of a particle from the origin at time t, then: dx • velocity, the rate of change of displacement with respect to time, is v = = f ′ (t) dt ( ) dv d dx . • acceleration, the rate of change of velocity with respect to time, is a = = dt dt dt This is, in fact, the ‘derivative of the derivative’ with respect to time. If y = f(x) is the equation of the curve, then: dy • the first derivative, = f ′ (x), is the rate of change of y with respect to x, or in other words, the dx gradient of the curve • the rate of change of the derivative with respect to x is the second derivative. Notation for the second derivative ( ) d2 y d dy = 2 = f ″ (x) dx dx dx

When we differentiate to determine either the first of second derivatives, we may need to use the rules for differentiation, such as the product or quotient rules.

WORKED EXAMPLE 1

Determine

d2 y if y = x4 + 2x3 − 4x2 + 5. 2 dx

THINK

WRITE

1.

Write the equation.

2.

Differentiate to determine the first derivative and simplify.

3.

Differentiate to determine the second derivative and simplify.

y = x4 + 2x3 − 4x2 + 5 dy = 4 × x3 + 2 × 3x2 − 4 × 2x dx dy = 4x3 + 6x2 − 8x dx d2 y = 4 × 3x2 + 6 × 2x − 8 dx2 d2 y = 12x2 + 12x − 8 2 dx

WORKED EXAMPLE 2

If f (x) =

4√x5 , calculate f ″ (9). 3x2

CHAPTER 8 The second derivative and applications of differentiation 303

THINK 1.

WRITE

Express the function in simplified form using index laws.

2.

Determine the first derivative, using the basic laws for differentiation, and simplify.

3.

Determine the second derivative, using the basic laws for differentiation, by differentiating the first derivative again; then simplify.

4.

Substitute in the value for x.

5.

State the final result.

f(x) =

4√x5 3x2 5

4x 2 = 2 3x 4 5 −2 = x2 3 4 1 = x2 3 4 1 −1 f ′(x) = × x 2 3 2 2 −1 = x 2 3 2 −1 − 3 f ″ (x) = × x 2 3 2 −1 = 3√x3 −1 f ″ (9) = 3√93 −1 = 3 × 27 1 f ″ (9) = − 81

WORKED EXAMPLE 3

Determine the second derivative,

d2 y , of y = x2 ln (3x + 5). dx2

THINK 1.

Write the equation.

2.

Use the product rule and simplify. If y = uv, then y′ = uv′ + vu′ .

3.

Differentiate with respect to x again, using the product rule for the first term and the quotient rule dy for the second term of . dx

4.

Simplify.

WRITE

y = x2 ln(3x + 5) dy 1 = ln(3x + 5) × 2x + x2 × ×3 dx (3x + 5) 3x2 = 2x ln(3x + 5) + (3x + 5) 2 dy 1 = {ln(3x + 5) × 2 + 2x × × 3} 2 (3x + 5) dx (3x + 5) × 6x − 3x2 (3) + (3x + 5)2 ( 2 ) 9x + 30x d2 y 6x = 2 ln(3x + 5) + + (3x + 5) dx2 (3x + 5)2

Note: Sometimes you may be able to combine terms with common denominators, but this is not necessary unless you are asked to express your answer in a particular form.

304 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Resources Interactivity: Stationary points (int-5963)

Units 3 & 4

Area 4

Sequence 1

Concept 1

Second derivatives Summary screen and practice questions

Exercise 8.2 Second derivatives Technology free WE1

a.

Determine

d2 y

for each of the following functions. dx2 a. y = x4 − 5x3 + x2 − 9 b. y = x3 − 4x2 4 e. y = (2x − 1) d. y = x2 (8 − x) 2. Determine the second derivatives of the following. 1 a. x√x b. x2 ( ) 2x d. cos e. 3 sin(4x − 𝜋) 5 3. Determine f ″ (x) if f (x) is given by: 1.

x ln(x)

b.

2

e3x

c.

y = 4 − x2

c.

4 e2x+3

c.

ln(x + 1)

8√x3 , calculate f ″ (4). 3x ( ) ( ) x 𝜋 ″ 5. If f (x) = 8 cos , calculate f . 2 3 4x2 6. a. If f (x) = , calculate f ″ (4). 3√x 2 , calculate f ″ (1). b. If f (x) = 3x − 5 7. a. If f (x) = 4 loge (2x − 3), calculate f ″ (3).

4.

WE2

9. 10.

11.

2

If f (x) = ex , calculate f ″ (1). ( 2 ) d2 y 3 WE3 Determine for y = x log 2x + 5 . e dx2 d2 y x4 Determine 2 for y = 3x . dx e d2 y Determine 2 if: dx ) ( a. y = loge x2 + 4x + 13 d2 y Determine 2 if: dx a. y = x3 e−2x b.

8.

If f (x) =

b.

e3x cos(4x)

b.

x2 cos(3x)

Technology active 12.

Consider the function f (x) = esin(x) . a. Show that the gradient of the function at x = 𝜋 is −1. b. Determine f ″ (𝜋). CHAPTER 8 The second derivative and applications of differentiation 305

Determine the values of a and b if it is known that, for all x, the second derivative of the function f (x) = 2 sin(3x) + 4 cos(2x) is given by f ″ (x) = a sin(3x) + b cos(2x). 14. Consider the function y = ex sin(x). 3𝜋 a. Show that the function has a stationary point at x = . 4 d2 y 3𝜋 when x = . Give your answer correct to 2 decimal places. b. Evaluate 2 4 dx 15. The displacement, x metres, of a particle at any time, t seconds, is given by the equation ) ( 𝜋 (2t − 1) . x = 6 sin 4 a. Where is the particle initially? b. When is the particle first at rest? c. Calculate the acceleration of the particle at 3.5 seconds. 13.

8.3 Concavity and points of inflection 8.3.1 Concavity The shape of continuous functions is often described in terms of its concavity. They are said to be either concave up (sometimes referred to as convex) or concave down. A function is concave up when the gradient of the function is increasing, so the rate of change of the gradient is positive.

–2 2 –1

1 0

A function is concave down when the gradient of the function is decreasing, so the rate of change of the gradient is negative. 0 –1

1

2

–2

Remember, the rate of change of the gradient with respect to x is the second derivative. d dx

(

dy dx

) =

d2 y 2

dx

= f ″ (x)

306 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Concavity may be summarised as follows: Shape Concave up (or convex)

Concave down

First derivative

Second derivative

dy is increasing dx

d2 y

dy is decreasing dx

d2 y

dx2

dx2

>0

0 24 > 6x x4 The function is concave down for x > 4. At x = 4: f(4) = 12 × 16 − 64 = 128 The point of inflection is (4, 128).

The function is concave up when f ″ (x) > 0. Solve for x.

The function is concave down when f ″ (x) < 0. Solve for x.

b. The

point of inflection is where concavity changes. State the point at x = 4.

WORKED EXAMPLE 6

By investigating the concavity of the curve y = x4 , explain why the curve does not have a point of inflection. A sketch of the curve may be useful. THINK 1.

2.

For concavity, determine the second derivative.

The point of inflection exists when d2 y = 0 and changes sign. dx2

WRITE

y = x4 dy = 4x3 dx d2 y = 12x2 dx2 12x2 = 0 x=0 x

0−

0

0+

d2 y dx2

>0

=0

>0

CHAPTER 8 The second derivative and applications of differentiation 309

3.

State a reason for your decision.

4.

Sketch the curve.

The second derivative does not change sign either side of x = 0. Therefore, x = 0 is not a point of d2 y inflection. 2 ≥ 0 for all x, so the curve is dx always concave up. A sketch of y = x4 is shown, demonstrating that it is always concave up. y 4 3 y = x4

2 1

–2

Units 3 & 4

Area 4

Sequence 1

–1

0

1

2

x

Concept 2

Concavity and points of inflection Summary screen and practice questions

Exercise 8.3 Concavity and points of inflection Technology free 1.

2.

3.

4.

5.

6. 7.

Describe the shape of the curve y = x3 − 9x2 + 8 at the point where: i. x = 4 ii. x = −4 b. Determine the coordinates of the point of inflection. a. Describe the shape of the curve y = x3 + 6x2 at the point where: i. x = −3 ii. x = 3 b. Determine the coordinates of the point of inflection. a. Describe the shape of the curve y = 4x2 − x3 at the point where: i. x = 0 ii. x = 1 b. Determine the coordinates of the point of inflection. WE5 Consider the function f (x) = x3 + 9x2 . a. Determine where the function is: i. concave up ii. concave down. b. Hence, state the coordinates of the point of inflection. a. For the function y = x3 + 2x2 − 3x + 1, determine where the function is: i. concave up ii. concave down. b. Hence, state the coordinates of the point of inflection. WE6 By investigating the concavity of the curve y = 6 − x4 , explain why the curve does not have a point of inflection. A sketch of the curve may be useful. By investigating the concavity of the curve y = 2x6 − 4, explain why the curve does not have a point of inflection. A sketch of the curve may be useful. WE4

a.

310 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Determine the coordinates of the point(s) of inflection for the following functions. a. y = x3 − 3x2 − 9x + 5 b. y = −x3 + 9x2 − 15x − 20 9. Consider the function f (x) = x4 + 4x3 − 16x + 3. a. Determine where the function is: i. concave up ii. concave down. b. State, with reasons, the coordinates of the point(s) of inflection of the function. 1 10. Determine the point(s) of inflection of the function f (x) = x2 − 3x4 . Hence, state where the function is 2 concave down. Technology active 3 11. a. Consider the curve y = (2x − 3) + 4. d2 y i. Determine . dx2 d2 y = 0. ii. Solve dx2 iii. Determine the coordinates of any point of inflection. 4 b. Consider the curve y = (2x − 3) + 4. d2 y . i. Determine dx2 d2 y ii. Solve = 0. dx2 iii. Determine the coordinates of any point of inflection. c. Discuss the similarities and differences between the curves in parts a and b in relation to any point of inflection. A sketch, using technology, of the curves may be of assistance. 12. Determine the value of k if the function f(x) = 2x3 − kx2 + 3x has a point of inflection when x = 3. 13. Consider the function f (x) = x4 + kx3 . a. Determine the value for k if the function f has a point of inflection at x = 1. b. Hence, determine the interval where the function is concave up. 14. Show that the function f (x) = x loge (x), x > 0 is always concave up. 15. a. Sketch the graph of y = 2 sin(x) + 3, x ∈ [0, 2𝜋]. b. Determine where the function is: i. concave up ii. concave down. c. Hence, state the coordinates of the point(s) of inflection. 8.

8.4 Curve sketching 8.4.1 Reviewing polynomial shapes You will be familiar with the general shapes of polynomials. These are summarised in the following table. Remember, these graphs may also be inverted. Quadratics (degree 2)

Cubics (degree 3)

Quartics (degree 4)

CHAPTER 8 The second derivative and applications of differentiation 311

Knowledge about the first and second derivatives of a function will allow us to make important observations about the function and help us sketch the graph more accurately.

8.4.2 Stationary points, points of inflection and derivatives A stationary point on a curve is defined as a point where the gradient is 0; that is, where

dy = f ′ (x) = 0. dx

You will be familiar with three types of stationary points: • relative maximum turning points • relative minimum turning points • horizontal (or stationary) points of inflection. The word ‘relative’ means that the point is a maximum or a minimum in a particular locality or neighbourhood. Beyond this section of the graph, there could be other points on the graph that are higher than the relative maximum or lower than the relative minimum. Previously, you have determined the nature of the stationary points by calculating the slope of the tangent, dy or f ′ (x), on either side of the point. This method is still useful. However, the nature of the turning points dx can also be determined by considering concavity: • At a maximum turning point, the curve is concave down. • At a minimum turning point, the curve is concave up. Care needs to be taken to determine the nature of a horizontal (or stationary) point of inflection. At a stationary point of inflection: • the first derivative equals 0 • the second derivative equals 0, and the sign of the second derivative changes either side of the point, since there is a change in concavity. Non-stationary points of inflection occur when the first derivative is not 0, and the second derivative equals 0 and changes sign. Determining points of inflection is discussed in the previous section of this chapter. Turning points and points of inflection are summarised in the following table.

Shape

First derivative

Second derivative

Maximum turning point

dy =0 dx

d2 y

Minimum turning point

dy =0 dx

d2 y

312 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

dx2

dx2

0

Shape

First derivative

Stationary point of inflection (tangent is horizontal)

dy = 0 dx

Second derivative d2 y dx2

= 0 and changes sign either side –

Non-stationary point of inflection (non horizontal tangent)

dy ≠ 0 dx

d2 y dx2

+

= 0 and changes sign either side +

–

8.4.3 Curve sketching Using the first and second derivatives of a function allows curves to be sketched with greater accuracy. The first derivative of a function gives the gradient at any point: dy • If > 0, the function is increasing. dx dy • If < 0, the function is decreasing. dx dy • If = 0, the function has a stationary point. dx First derivative Increasing function

Decreasing function

Stationary point

The second derivative of a function gives its concavity: d2 y • If 2 > 0, the function is concave up. dx d2 y • If 2 < 0, the function is concave down. dx d2 y • If 2 = 0 and changes sign, there is a point of inflection. dx

CHAPTER 8 The second derivative and applications of differentiation 313

Second derivative Concave up

Concave down

Point of inflection

To illustrate the relationships between a function and its derivatives, the following functions are shown. y = x4 − 4x3 + 10

d2 y

dy = 4x3 − 12x2 dx

dx2

= 12x2 − 24x

y 40 Stationary point of inflection 4

Point of inflection 20 (0, 10)

3

y = x – 4x + 10 –2

0

2 (2, –6)

–20

6

4 (3, –17)

x

Local minimum turning point

y 40 dy = 4x3 – 12x2 dx

20 (0, 0)

–2

0 –20

(3, 0) 2

4

6

2 (2, 0) 4

6

x

(2, –18)

y 40 d 2y = 12x2 – 24x dx2

20 (0, 0)

–2

0 –20

314 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

When sketching the graph of a function, f (x), you may need to consider the following: • Determine the y-intercept by evaluating f(0). • Determine the x-intercept(s) by solving f (x) = 0, if possible. • Determine the coordinates of the stationary point(s) and their nature. • Determine the coordinates of any point(s) of inflection. • Consider restrictions on the domain. • Calculate the coordinates of the end points of the domain, where appropriate. • Identify vertical and horizontal asymptotes, where appropriate. • Consider the direction of f (x) as x → ±∞. WORKED EXAMPLE 7

Sketch the graph of the function f : R → R, f (x) = x3 + 6x2 + 9x by determining the coordinates of all axis intercepts as well as any stationary points and their nature. Include on your sketch the coordinates of any point(s) of inflection. THINK 1.

2.

State the function and differentiate to determine the first and second derivatives. For x-axis intercepts: • factorise the function • solve for f (x) = 0.

3.

For stationary points, f ′ (x) = 0.

4.

Determine the nature of the stationary points using the second derivative and determine the corresponding values of f (x).

5.

For points of inflection, f ″ (x) = 0 and changes sign either side of that value.

WRITE

f (x) = x3 + 6x2 + 9x f ′ (x) = 3x2 + 12x + 9 f ″ (x) = 6x + 12 f (x) = x3 + 6x2 + 9x = x (x2 + 6x + 9) x (x + 3) (x + 3) = 0 x (x + 3)2 = 0 The x-intercepts are (0, 0) and (−3, 0). f ′ (x) = 3x2 + 12x + 9 ( ) = 3 x2 + 4x + 3 3 (x + 3) (x + 1) = 0 x = −3 or x = −1 When x = −3: f ″ (−3) = −18 + 12 = −6 < 0, so concave down f (−3) = 0 The point (−3, 0) is a maximum turning point. When x = −1: f ″ (−1) = −6 + 12 = 6 > 0, so concave up f (−1) = −1 + 6 − 9 = −4 The point (−1, −4) is a minimum turning point. f ″ (x) = 6x + 12 6x + 12 = 0 x = −2 Check for change of sign either side of x = −2. x

−2−

−2

−2+

d2 y dx2

0

CHAPTER 8 The second derivative and applications of differentiation 315

6.

Determine f (x) for the point and make a statement.

7.

Sketch the curve, showing all important features.

f (−2) = −8 + 24 − 18 = −2 The second derivative has changed sign, so there is a point of inflection at (−2, −2). y 3 2 f(x) = x3+ 6x2 + 9x (–3, 0) –5

–4

–3

–2

1 (0, 0) –1 0 –1

1

2

3

x

–2 (–2, –2) –3 –4 –5 (–1, –4)

WORKED EXAMPLE 8

Sketch the graph of y = x4 − 4x3 , stating the coordinates of all axis intercepts, any stationary points and their nature, and any point(s) of inflection. b. State the values of x where the function is: i. decreasing ii. concave up. a.

THINK a. 1.

2.

3.

WRITE

State the function and differentiate to determine the first and second derivatives.

For x-axis intercepts: • factorise the function • solve for y = 0. For stationary points, solve for

a.

y = x4 − 4x3 dy = 4x3 − 12x2 dx d2 y = 12x2 − 24x dx2 y = x4 − 4x3 x3 (x − 4) = 0 The x-intercepts are (0, 0) and (4, 0).

dy =0 dx

dy = 4x3 − 12x2 dx 4x2 (x − 3) = 0 x = 0, 3

316 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

4.

5.

Determine their nature using the second derivative and determine the corresponding y-value. Remember: If the second derivative is 0, you need to check either side by taking a value of x close to the point. (For example, either side of x = 0, take x = ±0.1)

For points of inflection: d2 y = 0 and changes sign. dx2

6.

Determine the y-value for the point

7.

Sketch the curve, showing all of the critical points.

When x = 0: d2 y = 12 × 0 − 24 × 0 = 0 dx2 Possible point of inflection; check for change of sign either side of x = 0. x

0−

0

0+

d2 y dx2

>0

0

0, so concave up dx2 y = 34 − 4 × 33 = −27 The point (3, −27) is a minimum turning point. d2 y = 12x2 − 24x dx2 12x (x − 2) = 0 x = 0, 2 x

2−

2

2+

d2 y dx2

0

y = 24 − 4 × 23 = −16 The concavity has changed, so there is a point of inflection at (2, −16). y

y = x4 – 4x3

10 (4, 0) x

(0, 0) –3 –2 –1

0

1

2

3

4

–10 (2, –16) –20

–30 b i.

dy < 0. dx Read from the graph. Remember the gradient cannot equal 0.

For a decreasing function,

b. i.

(3, –27)

The function is decreasing when x < 0 or 0 < x < 3.

CHAPTER 8 The second derivative and applications of differentiation 317

ii.

d2 y

> 0. dx2 Read from the graph, noting the points of inflection already found. For concave up,

ii.

There are points of inflection at (0, 0) and (2, −16). The function is concave up when x < 0 or x > 2.

WORKED EXAMPLE 9

Sketch the graph of f (x) = 2x ex , x ≤ 1 showing the important features, including stationary points and points of inflection. Give your answers correct to 3 decimal places where necessary. THINK 1.

State the function and differentiate using the appropriate rules to determine the first and second derivatives.

WRITE

f (x) = 2xex f ′ (x) = 2x × ex × 1 + ex × 2 = 2xex + 2ex f ″ (x) = (2x × ex × 1 + ex × 2) + 2ex × 1 = 2xex + 2ex + 2ex

2.

For x-axis intercepts, solve for f (x) = 0.

= 2xex + 4ex f (x) = 2x ex

2x ex = 0 x=0 The x-intercept is (0, 0). 3. For stationary points, solve for f ′ (x) = 0. f ′ (x) = 2xex + 2ex

4.

Determine the nature of the stationary points using the second derivative and determine the corresponding value of f (x).

5.

For points of inflection, f ″ (x) = 0 and changes sign on either side.

6.

Determine f (x) for the point and make a statement.

2ex (x + 1) = 0 x = −1 When x = −1: f ″ (−1) = −2e−1 + 4e−1 = 2e−1 > 0, so concave up 2 f(−1) = −2 e−1 = − ( )e 2 The point −1, − ≈ (−1, −0.736) is a e minimum turning point f ″ (x) = 2xex + 4ex 2ex (x + 2) = 0 x = −2 x

−2−

−2

−2+

d2 y dx2

0

4 e2 The concavity has changed, so the point ( ) 4 −2, − 2 ≈ (−2, − 0.541) is a point of inflection. e f (−2) = −4 e−2 = −

318 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

7.

Determine the coordinates of the end point of the restricted domain.

8.

Consider the behaviour of the graph to the left, as x becomes very small.

9.

Sketch the curve, showing all important features. Include a closed circle at the end point.

f (1) = 2 e The end point is (1, 2e) ≈ (1, 5.437) correct to 3 decimal places As x → −∞, ex → 0 ∴ as x → −∞, x ex → 0 The function approaches the x-axis, which will be a horizontal asymptote on the left side of the graph. y 6

4

(1, 5.437)

f (x) = 2xex

2 (–4, –0.147)

–6

–4 (–2, –0.541)

–2

0 –2

2

x

(–1, –0.736)

Resources Interactivities: Graph of a derivative function (int-5961) Equations of tangents (int-5962)

Units 3 & 4

Area 4

Sequence 1

Concept 3

Curve sketching Summary screen and practice questions

Exercise 8.4 Curve sketching Technology free

Sketch the graph of the function f : R → R, f (x) = x3 − 4x2 + 4x by determining the coordinates of all axis intercepts as well as any stationary points and their nature. Include on your sketch the coordinates of any point(s) of inflection. 2. Sketch the graphs of each of the following by determining the coordinates of all axis intercepts and any stationary points, and establishing their nature. Also determine the coordinates of the points of inflection. a. y = x3 − 27x b. y = 9x − x3 3. Sketch the graphs of each of the following by determining the coordinates of all axis intercepts and any stationary points, and establishing their nature. Also determine the coordinates of the points of inflection. State where the functions are: ii. concave up. i. increasing 3 2 a. y = x + 12x + 36x b. y = −x3 + 10x2 − 25x 1.

WE7

CHAPTER 8 The second derivative and applications of differentiation 319

Sketch the graphs of each of the following by determining the coordinates of all axis intercepts and any stationary points, and establishing their nature. Also determine the coordinates of the points of inflection. a. y = x3 − 3x2 − 9x − 5 b. y = −x3 + 9x2 − 15x − 25 Technology active 5. WE8 a. Sketch the graph of y = 8x3 − x4 , stating the coordinates of the axis intercepts, any stationary points and their nature, and any point(s) of inflection. b. State the values of x where the function is: i. decreasing ii. concave up. 6. a. Sketch the graph of the function f (x) = x4 − 8x2 − 9, stating the coordinates of the axis intercepts, any stationary points and their nature, and any point(s) of inflection. b. State the values of x where f (x) is both increasing and concave up. 3 7. a. Sketch the graph of the function f (x) = (x − 1) + 8, showing all important features. b. State the values of x where f (x) is both increasing and concave down. 8. Sketch the graphs of each of the following by determining the coordinates of all axis intercepts and any stationary points, and establishing their nature. Also determine the coordinates of the points of inflection. The functions have been given in both factorised and expanded form for ease of calculations. 3 a. y = x4 + 4x3 − 16x − 16 = (x − 2) (x + 2) 3 b. y = x4 − 6x2 + 8x − 3 = (x − 1) (x + 3) 9. WE9 Sketch the graph of f (x) = 3x e−x , showing the important features, including stationary points and points of inflection. Give your answers correct to 3 decimal places where necessary. 10. Sketch the graph of f (x) = 4 − 10xex , −4 ≤ x ≤ 0, showing the important features, including stationary points and points of inflection. Give your answers correct to 3 decimal places where necessary. 11. The function f (x) = x3 + bx2 + cx + d has a stationary point of inflection at (1, −2). Determine the values of b, c and d. 12. A cubic polynomial, y = x3 + bx2 + cx + d, crosses the y-axis at y = 5 and has a point of inflection at (1, −21). Determine the equation of the tangent at the point of inflection. 13. The function f (x) = x3 + bx2 + cx + d crosses the x-axis at x = 3 and has a point of inflection at (2, −4). Calculate the values of b, c and d. 1 14. Consider the function f (x) = loge (x2 + 1). 2 a. State the domain of f (x). b. Determine the coordinates and nature of the stationary point. c. By considering the second derivative, show that the function has two points of inflection. State the coordinates of these points. d. Sketch the graph of f (x) for a suitable domain. 10 loge (x) 15. Consider the function f (x) = . x a. State the domain of f (x). b. Determine the coordinates and nature of the stationary point. c. By considering the second derivative, show that the function has one point of inflection. State the coordinates of this point. d. Determine the coordinates of the x-intercept. e. By considering various large values of x, discuss the behaviour of the function as x increases. f. Sketch the graph of f (x) for a suitable domain. 4.

320 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

8.5 Applications of the second derivative 8.5.1 Acceleration The relationships between displacement, velocity and acceleration have been discussed in Chapters 5, 6 and 7. dv Acceleration is the rate of change of velocity with respect to time, . dt Velocity is the rate of change of displacement with respect to time,

dx . dt

Acceleration is, in fact, the second derivative of displacement with to time,

Acceleration d dv = a= dt dt

(

dx dt

) =

d2 x . dt2

d2 x dt2

WORKED EXAMPLE 10

An object travels in a straight line so that its displacement from the origin, x m/s, at time t seconds is given by x (t) = t e−t , t ∈ [0, 5] . a. Determine an expression for: i. velocity as a function of time ii. acceleration as a function of time. b. Calculate the initial speed of the object. c. When is the object at rest? Determine the position of the object at this time. d. Determine when the acceleration is positive. THINK a i. 1.

2.

ii. 1.

2.

WRITE

Differentiate displacement with respect to time to determine velocity. Use the product rule.

Simplify. Differentiate velocity with respect to time to determine acceleration. Use the product rule. Simplify.

b. Substitute c. 1.

t = 0 into velocity equation.

At rest, v = 0.

x (t) = t e−t dx dt = e−t × 1 + t × (e−t × −1) = e−t − t e−t

v (t) =

v (t) = e −t (1 − t) dv d2 x a (t) = = 2 dt dt = (1 − t) × (e−t × −1) + e−t × (−1) = −e−t + t e−t − e−t a (t) = e−t (t − 2) v (0) = e0 (1 − 0) v = 1 m/s (t) v = e−t (1 − t) e−t (1 − t) = 0

CHAPTER 8 The second derivative and applications of differentiation 321

t = 1s x (t) = t e−t

Solve for t. 3. For position, calculate displacement when t = 1. 2.

4.

1 ≈ 0.368 e The object is at rest after 1 second and its 1 position is metres or approximately 0.368 e metres to the right of the origin. a (t) = e−t (t − 2) e−t (t − 2) > 0 t >2 Acceleration is positive when t ∈ (2, 5] seconds. x (1) = e−1 =

Answer the question.

State the expression for acceleration. 2. Solve for a(t) > 0.

d. 1.

3.

Answer the question and note the restricted domain for time.

8.5.2 Optimisation In many practical situations, it is necessary to determine the maximum or minimum value of the function that describes it. For example, if you were running your own business, you would always want to minimise the production costs while maximising the profits. Optimisation was discussed in Chapter 5. Checking the nature of the stationary point(s) using the second derivative may now be easier and quicker instead of using a sign diagram of the first derivative. When solving optimisation problems, the following steps may be useful: • Draw a diagram if possible and label it with as few variables as possible. • Determine a connection between the variables from the information given. These may include: • Pythagoras’ theorem • measurement formulas • trigonometry • similar triangles. • Determine an expression for the quantity to be maximised or minimised in terms of the one variable. • Differentiate the expression to determine the stationary point(s). • Check the nature of the stationary point(s) by either substituting into the second derivative or using a sign diagram of the first derivative. • Check whether the answer is the absolute maximum or minimum by evaluating the end points of the domain. • Sketch the graph the function to check for realistic values. • Answer the actual question. WORKED EXAMPLE 11

The sum of two positive numbers is 10. Determine the numbers if the sum of their squares is a minimum. THINK

WRITE

Define the two positive numbers. 2. Write an expression for the sum, S, of the squares of the two numbers.

Let the two positive numbers be x and y. S = x2 + y2 [1]

1.

322 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

State the relationship between the variables. 4. Express y in terms of x using [2]. 5. Express S in terms of one variable, x. 6. Simplify the expression. 3.

x + y = 10 [2] y = 10 − x S = x2 + (10 − x)2 S = x2 + 100 − 20x + x2 = 2x2 − 20x + 100 dS = 4x − 20 dx 4x − 20 = 0 x=5

dS . dx 8. For the maximum or minimum, solve dS = 0. dx 7.

Differentiate to obtain

9.

Determine the second derivative to verify the minimum.

State the value of y. 11. Answer the question. 10

d2 S =4>0 dx2 S (x) is concave up for all x, so S is minimum at x = 5. When x = 5, y = 10 − 5 = 5. The two numbers that add to 10 and have the minimum sum of their squares are 5 and 5.

WORKED EXAMPLE 12

A cuboid container with a base length twice its width is to be made with 48 m2 of metal. 8 2x a. Show that the height, h m, is h = − , where x m is the width of the base. x 3 b. Express the volume, V m3 , in terms of x. c. Determine the dimensions of the container with maximum volume. d. Hence, calculate the maximum volume of the container. THINK a. 1.

Draw a diagram of the cuboid.

WRITE a.

h

2x x

Let x = width of base and hence express length in terms of x. 3. Calculate the total surface area (TSA) of the cuboid in terms of x and h.

2.

Let x = width and h = height, so length = 2x. TSA = 2 [2x (x) + 2x (h) + x (h)] ( ) = 2 2x2 + 3xh = 4x2 + 6xh

CHAPTER 8 The second derivative and applications of differentiation 323

4.

Express h in terms of x.

As TSA = 48 m2 4x2 + 6xh = 48 6xh = 48 − 4x2 48 − 4x2 6x 48 4x2 − = 6x 6x 8 2x h= − x 3 b. V = x (2x) h h=

b. 1.

2.

c. 1.

2.

Determine the volume, V, in terms of x and h. Express the volume in terms of x by substituting for h.

Differentiate to obtain

dV . dx

For maximum/minimum values, dV solve = 0. dx

3.

Determine the second derivative to verify the maximum.

4.

Calculate the dimensions by substituting x = 2 into expression for h.

5.

State the dimensions of the cuboid.

) 8 2x − V (x) = 2x x 3 4x3 = 16x − 3 dV 4 c. = 16 − × 3x2 dx 3 dV = 16 − 4x2 dx 16 − 4x2 = 0 (

2

x2 = 4 x = ±2 Reject x = −2, as width cannot be negative. d2 V = −8x dx2 When x = 2: d2 V = −16 < 0 dx2 V (x) is concave down, so there is a maximum turning point at x = 2. 8 2x h= − x 3 8 4 = − 2 3 8 = 3 8 x = 2, 2x = 4, h = 3 The dimensions of the cuboid with 8 maximum volume are 2 m by 4 m by m. 3

324 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

d. 1.

2.

Calculate the volume, V = lbh. Alternatively, substitute into the expression for volume.

State the answer.

d.

8 64 3 = m or 3 3 4 V (x) = 16x − x3 3 4 V (2) = 16 × 2 − × 23 3 32 = 32 − 3 64 = 3 V=2×4×

The maximum volume is

64 3 m. 3

Resources Interactivity: Kinematics (int-5964)

Units 3 & 4

Area 4

Sequence 1

Concept 4

Applications of the second derivative Summary screen and practice questions

Exercise 8.5 Applications of the second derivative Technology free 1.

WE10

An object travels in a straight line so that its displacement from the origin, x m/s, at time −

t

t seconds is given by x (t) = 8te 2 , t ∈ [0, 6]. a. Determine an expression for: i. velocity as a function of time ii. acceleration as a function of time. b. Calculate the initial speed of the object. c. When was the object at rest? Determine the position of the object at this time. d. Determine when the acceleration is positive. 2. An object travelling in a straight line has its displacement (in metres) after t seconds given by x(t) = 2 cos(3t − 1) + 3. a. Determine the maximum and minimum displacement. b. Determine when the velocity is first equal to 0. c. How long after the object is first at rest is it next at rest? d. Determine an expression for the acceleration. 3. The velocity of an object which is initially 3 m left of O is given by v(t) = 3t2 − 2t − 5 m/s. Determine: a. the displacement from O at any time t b. the acceleration at any time t c. when the object is at rest d. the distance travelled in the first second e. the acceleration when the velocity is 0.

CHAPTER 8 The second derivative and applications of differentiation 325

4.

5.

6. 7. 8.

9.

10.

A particle is travelling in a straight line with its displacement, x metres, at any time, t seconds, given as 16 x (t) = , t ≥ 0. Determine the acceleration of the particle after 2 seconds. (t + 2) An object initially starts at the origin travelling in a straight line at 15 m/s and speeds up with acceleration, a m/s2 , at any time, t s, given as a(t) = 12t2 − 4t + 4, t ≥ 0. a. Determine the equation for the velocity of the object with respect to time. b. Determine the equation for the position of the object with respect to time. c. Calculate the distance travelled in the first 2 seconds. WE11 The sum of two positive numbers is 32. Determine the numbers if their product is a maximum. The sum of two positive numbers is 8. Determine the numbers if the sum of the cube of one and the square of the other is a minimum. WE12 The total surface area of a closed cylinder is 200 cm2 . The base radius is r cm and the height is h cm. a. Express h in terms of r. b. Show that the volume, V cm3 , is V = 100r − 𝜋r3 . c. Hence, show that for maximum volume the height must equal the diameter of the base. d. Calculate, to the nearest integer, the minimum volume if 2 ≤ r ≤ 4. An open rectangular storage bin is to have a volume of 12 m3 . The cost of the materials for its sides is $10 per square metre and the material for the reinforced base costs $25 per square metre. If the dimensions of the base are x and y metres and the bin has a height of 1.5 m, determine, with justification, the cost, to the nearest dollar, of the cheapest bin that can be formed under these conditions. A section of a rose garden is enclosed by edging to form the shape of a sector ABC of radius r metres and arc length l metres. The perimeter of this section of the garden is 8 metres. A

l

θ C

r

B

If 𝜃 is the angle in radian measure subtended by the arc at C, express 𝜃 in terms of r. 1 b. The formula for the area of a sector is Asector = r2 𝜃. Express the area of a sector in terms of r. 2 c. Calculate the value of 𝜃 when the area is greatest. 11. A rectangular box with an open top is to be constructed from a rectangular sheet of cardboard measuring 16 cm by 10 cm. The box will be made by cutting equal squares of side length x cm out of the four corners and folding the flaps up. a. Express the volume as a function of x. b. Determine the dimensions of the box with greatest volume and give this maximum volume. a.

Technology active 12.

A cylinder has a surface area of 220𝜋 cm2 . Determine the height and radius of each end of the cylinder so that the volume of the cylinder is maximised, and determine the maximum volume for the cylinder. Give answers correct to 2 decimal places.

326 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

13.

A colony of blue wrens, also known as superb fairy wrens, survives in a national park because the wooded areas have rich undergrowth and a plentiful supply of insects, the wrens’ main food source. Breeding begins in spring and continues until late summer. The population of the colony any time t months after 1 September can be modelled by the function P(t) = 200te

−

t 4

+ 400, 0 ≤ t ≤ 12

where P is the number of birds in the colony. Determine: a. the initial population of the birds b. when the largest number of birds is reached c. the maximum number of birds, to the nearest bird. 14. A rower is in a boat 4 km from the nearest point, O, on a straight beach. His destination is 8 km along the beach from O. If he is able to row at 5 km/h and walk at 8 km/h, what point on the beach should he row to in order to reach his destination in the least possible time? Give your answer correct to 1 decimal place.

A

4 km

B O

x km

C

8 km 15.

A cone is 10 cm high and has a base radius of 8 cm. Determine the radius and height of a cylinder which is inscribed in the cone such that the volume of the cylinder is a maximum. Determine the maximum volume of the cylinder, correct to the nearest cubic centimetre.

8.6 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. MC The graph of y = (x + 2)3 has: A. 1 turning point B. 2 turning points C. 1 point of inflection D. 0 stationary points 3 2 2. MC The graph of x + 2x + x − 2 has A. 2 points of inflection B. 1 turning point and 1 point of inflection C. 3 turning points D. 2 turning points

CHAPTER 8 The second derivative and applications of differentiation 327

3.

The graph of g(x) has the following properties: ′ g (x) = 0 if x = −3, 1 and 4 • ′ • g (x) < 0 if x < −3 and 1 < x < 4 • g′ (x) > 0 for all other x. Which of these diagrams shows the graph of g(x)? A. B. y MC

y

y = g(x) 0

–3 C.

1

4

0

x

−3 D.

y

1

0

4.

MC

A. B. C. D.

5.

1

4

x

y

y = g(x)

−3

4 y = g(x)

y = g(x) x

0 −3

1

The graph of f ′ (x) shown indicates that the graph of f(x) has: a turning point at x = 2 and x = −4 a turning point at x = 2 and a point of inflection at x = −4 a turning point at x = −4 and a point of inflection at x = 2 2 points of inflection at x = −4 and x = 2

4

x

y y = f (x)

0 −4

y

x

MC

MC

x

The value of y in terms of x is:

A. 40 − 2x C. 40 − x 7.

x

A particle moves in a straight line so that at time t seconds its displacement, x metres, from a fixed origin O is given by x(t) = t3 − 6t2 + 9t, t ≥ 0. a. How far is the particle from O after 2 seconds? b. What is the velocity of the particle after 2 seconds? c. After how many seconds does the particle reach the origin again, and what is its velocity at that time? d. What is the particle’s acceleration when it reaches the origin again?

Questions 6 to 9 relate to the isosceles triangle shown, which has a perimeter of 40 cm.

6.

2

B. 20 − x D. 20 − 2x

The height of the triangle in terms of x is:

A. √400 − 40x

B. 20 − √40x

C. √400 − 40x + 2x2

D. √400 − 40x − x2

328 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

8.

9.

10.

MC

The area in terms of x is:

A. x√400 − 40x + x2

B. 2x√400 − 40x + x2

C. x√400 − 40x

D. 2x√400 − 40x

The maximum area of the triangle is obtained if x equals: 2 A. 6 cm B. 10 cm 3 2 C. 5 cm D. 10 cm 5 MC

A playground is being constructed by the local council. The shape of the playground is shown. All measurements are in metres. The perimeter of the playground is known to be 96 metres. b a

a

2.5b

Determine the values of a and b that give a maximum area for the playground. Determine the maximum area. dv = 4et − 6t + 1 m/s2 , where v is the 11. The acceleration of a particle moving in a straight line is given by dt velocity at any time. If the particle starts at the origin with a velocity of −1 m/s, determine: a. the velocity at any time t b. the displacement, x, from the origin at any time t c. the displacement from the origin after 1 second. 12. The position of a particle, x metres, from the origin at time t seconds is given by 1 x (t) = e2t − 4t2 − 3t + 10. 4 a. Determine the initial position of the particle. b. Calculate, correct to 2 decimal places, the velocity of the particle after 2 seconds. c. Derive an expression for the acceleration at any time. d. Determine when the acceleration of the particle is negative. Complex familiar 13. The displacement of an object, x metres, from a fixed point at any time, t hours, is given by ( ) 𝜋t x (t) = 2 cos + 10. 12 a. Determine the initial position of the particle. b. Calculate the velocity of the particle after 3 hours. c. Show that the particle was initially at rest and determine when the particle is again at rest. d. Calculate the position and acceleration of the particle at this time. 14. Sketch the graphs of each of the following by determining the coordinates of all axis intercepts and any stationary points, and establishing their nature. Also determine the coordinates of the point(s) of inflection. a. y = x3 − x2 − 16x + 16 b. y = −x3 − 5x2 + 8x + 12 c. y = x4 + 6x3 + 9x2 a. b.

CHAPTER 8 The second derivative and applications of differentiation 329

A manufacturing company is required to produce cylindrical cans (for tuna) of volume 50 cm3 . The tin used to produce the cans costs 40 cents per 100 cm2 . a. Determine the area of tin required, A, in terms of the radius, r. b. Calculate the radius of the can (to the nearest tenth) for minimum area. c. Hence, calculate the minimum area (to the nearest tenth). d. What is the cost of tin to produce 10 000 such cans? Give your answer to the nearest $20. 16. Water is being poured into a vase. The volume, V mL, of water in the 2 vase after t seconds is given by V = t2 (15 − t), 0 ≤ t ≤ 10. 3 a. What is the volume after 10 seconds? b. At what rate is the water flowing into the vase at t seconds? c. What is the rate of flow after 3 seconds? d. When is the rate of flow the greatest, and what is the rate of flow at this time? 17. Complex familiar The amount of antibiotic drugs, A (t) milligrams, in the body of a patient after 15.

−

t

time t hours, where t ≥ 0, is given by A (t) = 500te 4 . a. Determine the expression for the rate of change of the amount of the drug in the patient’s body at time t. b. Determine the rate of change of the amount of the drug in the patient’s body after 2 hours. c. Determine the time in hours when the amount of the drug in a patient’s body is a maximum. d. Determine the maximum amount of the drug in the patient’s body. Give your answer in milligrams, correct to 2 decimal places. e. Sketch the graph of A (t) versus time. f. For how many hours is the amount of the drug in the body in excess of 350 milligrams? Give your answer to the nearest minute. 18. A cylinder of cheese is to be removed from a spherical piece of cheese of radius 8 cm. What is the maximum volume of the cylinder of cheese? Express your answer to the nearest unit. 18 8 cm 19. Consider the function y = . 2 x −9 a. State the domain of the function and the coordinates of any axis intercepts. dy −36x =( b. Show that )2 . dx x2 − 9 Show that the function does not have any points of inflection. d. Determine the coordinates and nature of any stationary points. e. By considering extreme values of x, discuss the behaviour of the function as x → ±∞. f. Sketch the graph to represent the function, showing clearly all important features, including the equation(s) of any asymptotes. c.

330 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Sketch the following functions, showing any stationary points or points of inflection. 2 3 i. y = (x − 2) + 1 ii. y = (x − 2) + 1 5 4 iv. y = (x − 2) + 1 iii. y = (x − 2) + 1 b. Discuss the similarities and differences of the functions in part a in terms of their stationary points. n c. Consider the function y = (x − h) + k, n ≥ 2 and n ∈ Z. By discussing various values of n, what conclusions can you draw about the shape of functions that can be expressed in this form?

20. a.

Units 3 & 4

Sit exam

CHAPTER 8 The second derivative and applications of differentiation 331

Answers

5. a. i. x > −

8 The second derivative and applications of differentiation

ii. x < −

(

2

1. a. 12x − 30x + 2 c. −2 2 e. 48 (2x − 1)

b. 6x − 8 d. 16 − 6x

3

2. a.

b.

4√x c. 16 e

2x+3

6 x4

d. −

4 25

e. −48 sin(4x − 𝜋) 3. a.

1 x

c. 4.

6.

2

b. 6e

3x

( cos

2x

)

5

( 2 ) 6x + 1

12 5. −√3 1 7. a.

2 −16

9( ) 20x3 2x2 + 7

10.

b.

−9 11.

2

10. a.

11. 12. 13. 14.

( x2

( iii.

15. a. Initially, the particle has a position of −3√2 m, or

3√2 m to the left of the origin.

d2 y

c. Similarities: Both curves had

dx2

= 0 at x =

3𝜋 2

m/s2 .

ii. Concave up ii. Concave up

3 2

.

Differences: Consider the second derivatives: For part a, the second derivative was a linear function, so 3 it changed sign either side of x = . 2 For part b, the second derivative was a perfect square, so it did not change sign. 12. 18 13. a. −2 b. x < 0 or x > 1 14. Sample responses can be found in the worked solutions in

6 y = 2 sin(x) + 3

4

ii. Concave down

ii. x < −3

2

iii. None

15. a. y

2

Exercise 8.3 Concavity and points of inflection (3, −46) i. Concave down (−2, 16) i. Concave up ( ) 4 20 b. ,4 3 27 4. a. i. x > −3 b. (−3, 54)

2

3

the online resources.

b. The particle is first at rest after 1.5 s.

i. Concave up

2

2 ) ,4

ii. x =

2

c. The acceleration of the particle at 3.5 s is

3

3

i. 48(2x − 3)

b.

+ 4x + 13) ( ) 3x b. −e 7 cos(4x) + 24 sin(4x) ( 2 ) −2x a. 2xe 2x − 6x + 3 ( ) 2 b. 2 − 9x cos(3x) − 12x sin(3x) a. See worked solutions b. 1 a = −18, b = −16 a. Sample responses can be found in the worked solutions in the online resources. b. −14.92 (correct to 2 decimal places)

1. a. b. 2. a. b. 3. a.

3 ) 16

2

ii. x =

b. 6e

+ 6x ln(2x2 + 5) (2x2 + 5)2 ( 2 ) −3x 2 9. 3x 3x − 8x + 4 e ( ) −2 x2 + 4x − 5

8.

8. 9.

−1

(x + 1)2 −1

6. a.

7.

3 2

− ,3 3 27 The second derivative does not change sign either side of d2 y x = 0. In fact, 2 ≤ 0 for all x, so the curve is always dx concave down. The second derivative does not change sign either side of d2 y x = 0. In fact, 2 ≥ 0 for all x, so the curve is always dx concave up. a. (1, −6) b. (3, −11) a. i. x < −2 or x > 0 ii. −2 < x < 0 b. (−2, 19), (0, 3) ( ) ( ) 1 5 1 5 a. − , , , 6 432 6 432 1 1 b. x < − or x > 6 6 a. i. 24(2x − 3)

b.

Exercise 8.2 Second derivatives

2

(π, 3)

(0, 3)

(2π, 3)

2

0

2

b. i. x ∈ (0, 𝜋) ii. x ∈ (𝜋, 2𝜋) c. (𝜋, 3)

332 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

4

6

8 x

Exercise 8.4 Curve sketching 1.

i.

f(x) = x3 – 4x2 + 4x

y

5 3

−2 ii. x > −4 b.

x

0

f(x) = x4 – 8x2 – 9

(1.155, –17.889)

(–1.155, –17.889)

y y = –x3 + 10x3 – 25x (0, 0)

(–2, –25)

(5, 0) x

0

(

7 10 –, –9 – 3 27

(

14 5 – , –18 – 3 27

)

b. −2 < x < −

2√3 3

(2, –25)

or x > 2

) CHAPTER 8 The second derivative and applications of differentiation 333

7. a.

y

10.

f(x) = (x –

1)3

(–1, 7.679)

+8

y

(–2, 6.707) f (x) = 4 – 10xex (–4, 4.733)

(1, 8)

(0, 4)

(0, 7) (–1, 0) x

0

b. x < 1 8. a.

11. 12. 13. 14.

y

(2, 0) x

0

(–2, 0)

x

0

b = −3, c = 3, d = −3 y = −27x + 6 b = −6, c = 15, d = −18 a. x ∈ R b. (0, 0); relative minimum ( ) ( ) 1 1 c. −1, ln(2) , 1, ln(2) or (−1, 0.347), (1, 0.347) 2 2 y

d.

y = (x – 2)(x + 2)3

f (x) =

1 loge(x2+1) 2

(0, –16)

(1, –27) (–1, 0.347)

(1, 0.347) 0 (0, 0)

b.

x

y y = (x – 1)3(x + 3)

15. a. x > 0

(

( 0

x

(1, 0)

c.

e

d. (1, 0) e. As the values of x increase, the values of f (x) approach 0,

(0, –3)

the x-axis. Since the curve only crosses the x-axis at x = 1, the curve will be approaching the axis. The x-axis will be a horizontal asymptote.

(–1, –16) f.

(–2, –27)

9.

)

≈ (2.718, 3.679); relative maximum ) 3 −3 e 2 , 15e 2 ≈ (4.482, 3.347) e,

b.

(–3, 0)

10

y

(2.718, 3.679)

y (1, 0) 0

(1, 1.104) (2, 0.812) (0, 0) 0

x

x=0

f(x) = 3xe–x

334 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

(4.482, 3.347)

f(x) =

10 loge(x) x x

Exercise 8.5 Applications of the second derivative −

t

1. a. i. v (t) = 4e 2 (2 − t) b. 8 m/s

b. c. d. 3. a. c. e. 4.

1 2

(

t

ii. a (t) = 2e 2 (t − 4)

16

m ≈ 5.886 m from the origin e t ∈ (4, 6] Maximum: 5 m; minimum: 1 m 1 s 3 𝜋 s 3 a (t) = −18 cos(3t − 1) x (t) = t3 − t2 − 5t − 3 b. a (t) = 6t − 2 2 1 s d. 5 m 3 2 8 m/s

c. 2 s; d. 2. a.

−

t

4

b. x (t) = t − c. 48

2

3 6. 16, 16 7. 2 and 6 8. a. h =

12. 13. 14. 15.

4e −

d. 0 ≤ t < ln √8 or t ∈ [0, ln 2√2 13. a. 12 m from the fixed point

(

−

𝜋 √2

)

m/h ≈ −0.370 m/h 12 c. After 12 hours ( 2) 𝜋 d. 8 m from the fixed point; m/h2 ≈ 0.137 m/h2 72 b.

y

14. a.

(–2, 36)

(0, 16)

2

2 3

(–13 , 10 –2722 )

(–4, 0)

(4, 0)

t3 + 2t2 + 15t

0

(

m

x 8, 22 – –14 – 3 27

)

y = x3 – x2 – 16x + 16 b.

100 − 𝜋r2

y

𝜋r

(–23 , 14 –2722 ) (0, 12)

b, c. Sample responses can be found in the worked

11.

− 5t − 4

m/s2 3

10.

19

2

)

m ≈ 1.373 m from the origin 2 12. a. 10.25 m from the origin b. 8.30 m/s 2t c. a(t) = e − ( )) (8) c.

5. a. v (t) = 4t − 2t + 4t + 15

9.

t2

3

b. x (t) = 4e − t +

solutions in the online resources. 3 d. 175 cm $370 8 − 2r 2 c b. A = 4r − r c. 2 a. 𝜃 = r 3 2 a. V = 4x − 52x + 160x b. Length: 12 cm; width: 6 cm; height: 2 cm; volume 144 cm3 Radius: 6.06 cm; height: 12.11 cm; volume: 1395.04 cm3 a. 400 birds b. At the end of December c. 694 birds 3.2 km to the right of point O 1 1 Radius: 5 cm; height: 3 cm; volume: 298 cm3 3 3

(–6, 0)

(–1, 0)

(– –35 , –10 –2716)

(2, 0) x

0

(–4, –36) y = x3 – 5x2 + 8x + 12 c.

y = x4 + 6x3 + 9x2 y

(–1.5, 5.063)

8.6 Review: exam practice 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

C D D C a. 2 m c. 3 s; 0 m/s

B A C A

a. a = 12 m, b = 9.6 m 2 b. 403.2 m t 2 ( ) 11. a. v t = 4e − 3t + t − 5

(–0.63, 2.25) (–2.37, 2.25) (0, 0)

(–3, 0)

x

0

b. −3 m/s d. 6 m/s2 2

15. a. A = 2𝜋r + c. 75.1 cm 16. a. 333

1

2

mL 3 c. 42 mL/s

100 r

b. 2.0 cm d. $3000

dV

= 20t − 2t2 dt d. 5 s; 50 mL/s

b.

CHAPTER 8 The second derivative and applications of differentiation 335

−

t

3 ii. y = (x − 2) + 1

′ 17. a. A (t) = 125e 4 (4 − t)

b.

250 √e

y = (x – 2)3 + 1

y

mg/h (or 151.63 mg/h to 2 decimal places)

c. After 4 hours d. 735.76 mg e. t (h)

(2, 1) 800

(4, 735.759) x

0 600 400

t

A(t) = 500te – –4 200 (0, 0) 0

(0, –7) 2 4 6 8 10 12 A (mg)

f. 10 hours 10 minutes 3 18. 1238 cm 19. a. x ∈ R\ ± 3; axis intercept (0, −2) b, c. Sample responses can be found in the worked

4 iii. y = (x − 2) + 1

y y = (x – 2)4 + 1

solutions in the online resources. d. Maximum turning point at (0, −2) e. For both large and small values of x, the curve is approaching the x-axis. f.

x = –3

(0, 17)

x=3 y 6 4 2

(2, 1) x

0 –8

–6

–4

–2

0

2 (0, –2)

4

6

–2 –4 –6 20. a.

8

x 5 iv. y = (x − 2) + 1

y 18 y=– x2 – 9

(2, 1)

2

i. y = (x − 2) + 1

y = (x – 2)5 + 1 x

0

y (0, 5)

(0, –31) y = (x – 2)2 + 1 (2, 1) x 0 (0, –2)

b. Similarities: All have a stationary point at (2, 1), so for

dy

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336 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

REVISION UNIT 4 Further functions and statistics

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REVISION UNIT 4 Further functions and statistics 337



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a = 12

c=8

C = 35°

A

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a = 12 c=8 C = 35° A′

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Side c

a = 15 C = 25°

c=8

c=8 A′

A

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B

C

A x

D

L

R 52°

63°

16 cm

t

1.9 km 45°

50°

N

89 mm

T

S

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x H 74° 74°

B

C

D

N 80°

B 19.4 km



84°

59°

q

M

m 62° P

y

85°

27°

C

A

35.3 cm M   







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B

C

A 100° B

P

32 cm

29.5 m

θ

46 cm

D

L

ϕ

79 mm

153 mm

C R Q 60° 18.9 m

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M

117°

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REVISION UNIT 4 Further functions and statistics

TOPIC 2 Trigonometric functions 2 • For revision of this entire topic, go to your studyON title in your bookshelf at www.jacplus.com.au. • Select Continue Studying to access hundreds of revision questions across your entire course.

• Select your course Mathematical Methods for Queensland Units 3&4 to see the entire course divided into syllabus topics. • Select the Area you are studying to navigate into the chapter level OR select Practice to answer all practice questions available for each area.

• Select Practice at the sequence level to access all questions in the sequence.

OR

• At sequence level, drill down to concept level.

• Summary screens provide revision and consolidation of key concepts. Select the next arrow to revise all concepts in the sequence and practise questions at the concept level.

REVISION UNIT 4 Further functions and statistics 377

10

Discrete random variables

10.1 Overview Binomial probability involves events made up of trials that have only two outcomes — success or failure. Tossing a coin and rolling a die are two of the most commonly used examples of this type of trial, often with the caveat that they are ‘fair’ or ‘unweighted’. A die is said to be ‘loaded’ or ‘weighted’ if it has been engineerped to produce a particular result more often than a fair die would. The most common method of doing this is to drill down into the pips of the opposite face from the one desired, fill the cavities with a heavy material such as lead, and then repaint the pips in the original colour. Loading dice in this way is much easier if they are made of an opaque material. The dice used in casinos are traditionally made from transparent materials so that it is easier to tell if they have been rigged. Although coins can be made to have one side heavier than the other — usually by etching back the opposite face to remove metal from it — this will only make a difference to the outcome if the coin is spun on its edge rather than tossed. However, this doesn’t mean that tossing the coin will remove any bias; scientific studies have found that, when tossed from a flat hand up into the air, nearly 51% of the time the coin will land on the ground with the same face up as when it was originally thrown. This is due to a combination of effects including the rotation rate of the coin as it rises and falls, the average speed at which people toss the coins up into the air, and air resistance.

LEARNING SEQUENCE 10.1 10.2 10.3 10.4 10.5 10.6 10.7

Overview Bernoulli distributions Bernoulli random variables Binomial distributions The mean and variance of a binomial distribution Applications Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

378 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

10.2 Bernoulli distributions In probability theory, the Bernoulli distribution is a discrete probability distribution of the simplest kind. It is named after the Swiss mathematician Jacob Bernoulli (1654–1705). The term ‘Bernoulli trial’ refers to a single event that has only 2 possible outcomes, a success or a failure, with each outcome having a fixed probability. The following are examples of Bernoulli trials. • Will a coin land Heads up? • Will a newborn child be a male or a female? • Are a random person’s eyes blue or not? • Will a person vote for a particular candidate at the next local council elections or not? • Will you pass or fail an examination?

WORKED EXAMPLE 1

Determine which of the following can be defined as a Bernoulli trial. a. Interviewing a random person to see if they have had a flu vaccination this year b. Rolling a die in an attempt to obtain an even number c. Choosing a ball from a bag which contains 3 red balls, 5 blue balls and 4 yellow balls

THINK

WRITE

a.

Check for the characteristics of a Bernoulli trial.

b.

Check for the characteristics of a Bernoulli trial.

c.

Check for the characteristics of a Bernoulli trial.

Units 3 & 4

Area 6

Sequence 1

Concept 1

Yes, this is a Bernoulli trial, as there are 2 possible outcomes. A person either has or has not had a flu vaccination this year. b. Yes, this is a Bernoulli trial, as there are 2 possible outcomes. The die will show either an odd number or an even number. c. No, this is not a Bernoulli trial, as success has not been defined.

a.

Bernoulli distributions Summary screen and practice questions

CHAPTER 10 Discrete random variables 379

Exercise 10.2 Bernoulli distributions Technology active 1. WE 1 Determine which of the following can be defined as a Bernoulli trial. a. Spinning a spinner with 3 coloured sections b. A golfer is at the tee of the first hole of a golf course. As she is an experienced golfer, the chance of her getting a hole in one is 0.15. Will she get a hole in one at this first hole? c. A card is drawn from a standard pack of 52 cards. What is the chance of drawing an ace? 2. Determine which of the following can be defined as a Bernoulli trial. a. A new drug for arthritis is said to have a success rate of 63%. Jing Jing has just been prescribed the drug to treat her arthritis, and her doctor is interested in whether her symptoms improve or not. b. Juanita has just given birth to a baby, and we are interested in the gender of the baby, in particular whether the baby is a girl. c. You are asked what your favourite colour is. d. A telemarketer rings random telephone numbers in an attempt to sell a magazine subscription and has a success rate of 58%. Will the next person he rings subscribe to the magazine? 3.

State clearly why the following are not Bernoulli trials. a. A bag contains 12 balls, 5 of which are black, 3 of which are white and 4 of which are red. Paul has just drawn a ball from the bag without returning it. Now it is Alice’s turn to draw a ball from the bag. Does she get a red one? b. A die is tossed and the outcome is recorded. c. A fairy penguin colony at Phillip Island in Victoria is being studied by an ecologist. Will the habitat be able to sustain the colony in the future?

10.3 Bernoulli random variables Bernoulli distributions are controlled by the probability of success, p. Given that there are only two possible outcomes for a single Bernoulli trial and that the sum of probabilities for that trial is 1, it can be seen that the probability of failure, q, is equal to 1 − p. This means that the probability distribution table for a single Bernoulli trial looks like this: r

0

1

P (X = r)

1−p

p

where r represents the number of successes.

380 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

10.3.1 Mean of a Bernoulli distribution As you will recall from your earlier studies of probability, the mean of a discrete random variable distribution is referred to as the expected value, represented by E(X) or 𝜇. The expected value of a discrete random variable is the sum of each value of X in the distribution multiplied by its probability: E (X) = ∑ r P (X = r) In the case of the probability distribution for a Bernoulli trial E (X) = ∑ r P (X = r) = 0 (1 − p) + 1 × p =p

10.3.2 Variance and standard deviation of a Bernoulli distribution The variance (written as Var (X) or 𝜍 2 ) and standard deviation (written SD (X) or 𝜍) of any distribution are measures of spread used to indicate the range over which an outcome deviates from the expected value. Substituting the values from the Bernoulli probability distribution table into the equation for variance: ( ) Var (X) = E X2 − [E (X)]2 = [02 (1 − p) + 12 p] − p2 = p − p2 = p (1 − p) As the standard deviation is calculated by taking the square root of the variance, we can determine an expression for the standard deviation of a Bernoulli distribution: SD (X) = √Var (X) = √p (1 − p) Note: When calculating Var(X) and SD(X), it is conventional to round to 4 decimal places unless otherwise indicated. WORKED EXAMPLE 2

A new cream has been developed for the treatment of dermatitis. In laboratory trials the cream was found to be effective in 72% of the cases. Hang’s doctor has prescribed the cream for her. Let X be the effectiveness of the cream. a. Construct a probability distribution table for X. b. Determine E(X). c. Determine the variance and the standard deviation of X, correct to 4 decimal places. THINK a.

b. 1.

Construct a probability distribution table and clearly state the value of p.

State the rule for the expected value.

WRITE a.

b.

p = success with cream = 0.72 r

0

1

P (X = r)

0.28

0.72

E(X) = ∑ rP(X = r) all r

CHAPTER 10 Discrete random variables 381

Substitute the appropriate values and evaluate. c. 1. Determine E(X2 ). 2.

2.

3.

Calculate the variance.

Calculate the standard deviation.

TI | THINK a. 1. On a New Document

WRITE

E(X) = 0 × 0.28 + 1 × 0.72 = 0.72 2 c. E(X ) = 02 × 0.28 + 12 × 0.72 = 0.72 Var(X) = E(X2 ) − [E(X)]2 Var(X) = 0.72 − (0.72)2 = 0.2016 SD(X) = √0.2016 = 0.4490 CASIO | THINK a. 1. On a Main Menu

page, select: 4: Add Lists & Spreadsheets.

screen, select: Statistics.

2. Define the list names of

2. Complete the entry

each column by completing the entry lines: list 1 list 2

line in List 1 as: 0 1 Complete the entry line in List 2 as: 0.28 0.72

3. Complete the entries in

3. Select List 2 as a

list 1 as: 0 1 Complete the entries in list 2 as: 0.28 0.72 The probability table has been constructed. b.1. To complete the statistical calculations, select: MENU 4: Statistics 1: Stat Calculations 1: One-Variable Statistics. 2. Set the number of lists

to 1 and press the OK button.

WRITE

frequency list by pressing the SET button. Press the LIST button. Complete the entry line as: 2 Press the EXE button twice. 4. The probability table has been constructed.

Continued b.

To complete the statistical calculations, press the 1-VAR button. The expected value, E(X), can be read from the screen.

382 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

TI | THINK

WRITE

CASIO | THINK

3. Complete the entry

WRITE

c. The standard deviation

lines as: X1list: ′list1 Frequency List: ′list2 1st Result Column: c[] Press the OK button.

can be read from the screen. Note: The variance can be calculated using SD(X) = √Var (X).

4. The expected value

E(X) can be read from the screen.

c.

The variance can be read by scrolling down the screen.

Resources Interactivity: Bernoulli distribution (int-6430)

Units 3 & 4

Sequence 1

Area 6

Concept 2

Bernoulli random variables Summary screen and practice questions

Exercise 10.3 Bernoulli random variables Technology active 1. WE2 Caitlin is playing basketball for her local club. The chance that Caitlin scores a goal is 0.42. The ball has just been passed to her and she shoots for a goal. Let X be the random variable that defines Caitlin getting a goal. (Assume X obeys the Bernoulli distribution.) a. Set up a probability distribution for this discrete random variable. b. Determine E(X). c. Determine: i. Var(X) ii. SD(X) 2. A discrete random variable, Z, has a Bernoulli distribution as follows.

z

0

1

P (Z = z)

0.37

0.63

a.

Determine E(Z).

b.

Determine Var(Z).

c.

Determine SD(Z).

CHAPTER 10 Discrete random variables 383

3.

4.

5.

6.

7.

8.

9.

10.

Eli and Jacinta are about to play a game of chess. As Eli is a much more experienced chess player, the chance that he wins is 0.68. Let Y be the discrete random variable that defines the fact the Eli wins. a. Construct a probability distribution table for Y. b. Evaluate: i. E(Y) ii. Var(Y) iii. SD(Y) c. Calculate P(𝜇 − 2𝜍 ≤ Y ≤ 𝜇 + 2𝜍). During the wet season, the probability that it rains on any given day in Cairns in northern Queensland is 0.89. I am going to Cairns tomorrow and it is the wet season. Let X be the chance that it rains on any given day during the wet season. a. Construct a probability distribution table for X. b. Evaluate: i. E(X) ii. Var(X) iii. SD(X) c. Determine P(𝜇 − 2𝜍 ≤ X ≤ 𝜇 + 2𝜍). X is a discrete random variable that has a Bernoulli distribution. It is known that the variance for this distribution is 0.21. a. Determine the probability of success, p, where p > 1 − p. b. Determine E(X). Y is a discrete random variable that has a Bernoulli distribution. It is known that the standard deviation for this distribution is 0.4936. a. Determine the variance of Y correct to 4 decimal places. b. Determine the probability of success, p, if p > 1 − p. c. Determine E(Y). It has been found that when breast ultrasound is combined with a common mammogram, the rate in which breast cancer is detected in a group of women is 7.2 per 1000. Louise is due for her two-yearly mammography testing, which will involve an ultrasound combined with a mammogram. Let Z be the discrete random variable that breast cancer is detected. a. What is the probability that Louise has breast cancer detected at this next test? b. Construct a probability distribution table for Z. c. Determine P(𝜇 − 2𝜍 ≤ Z ≤ 𝜇 + 2𝜍). A manufacturer of sweets reassures their customers that when they buy a box of their ‘All Sorts’ chocolates there is a 33% chance that the box will contain one or more toffees. Kasper bought a box of ‘All Sorts’ and selected one. Let Y be the discrete random variable that Kasper chose a toffee. a. Construct a probability distribution table for Y. b. Determine E(Y). c. Determine P(𝜇 − 2𝜍 ≤ Y ≤ 𝜇 + 2𝜍). Z is a discrete random variable that has a Bernoulli distribution. It is known that the variance of Z is 0.1075. a. Determine the probability of success, correct to 4 decimal places, if P(success) > P(failure). b. Construct a probability distribution table for Z. c. Evaluate the expected value of Z. Y is a discrete random variable that has a Bernoulli distribution. It is known that the standard deviation of Y is 0.3316. a. Calculate the variance correct to 2 decimal places. b. Calculate the probability of success correct to 4 decimal places if P(success) > P(failure).

384 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

10.4 Binomial distributions A binomial distribution results when a Bernoulli trial is carried out a number of times. A binomial distribution has the following characteristics: 1. The trials must be independent. 2. Only two possible outcomes must exist for each trial — success and failure. 3. The probability for success, p, is the same for each trial. If a discrete random variable, X, has a binomial distribution, we can say that X ~ Bi (n, p), where p is the probability of success in a single trial and n is the number of trials. Consider the case of Cassandra, a Science student who has 5 multiple choice questions left to answer on her exam. These questions are on a topic that she has not studied. Each question has 5 different choices for the correct answer, and she plans to randomly guess the answer for every question. It can be seen that this situation represents a binomial distribution. The five questions can be considered as five independent Bernoulli trials, each having only two possible outcomes: either Cassandra guesses the correct answer to the 1 4 question (a success, with p = ), or she guesses incorrectly (a failure, with 1 − p = ). 5 5 There are six possible results in this case: Cassandra can get all 5 questions incorrect; 1 correct and 4 incorrect; 2 correct and 3 incorrect; 3 correct and 2 incorrect; 4 correct and 1 incorrect; or all 5 correct. Recall the concept of combinations, as the order of the correct answers out of the 5 questions is not an important consideration in this case. As you will recall from previous study of the binomial theorem, the probability of getting r successes out of n trials can be calculated using the equation ( ) n P (X = r) = pr (1 − p) n−r r ( ) n! n . where = nC r = r (n − r)! r! If X represents the number of questions answered correctly, then we may calculate the probabilities for 4 1 the distribution as shown in the following table given that n = 5, p = and 1 − p = . 5 5 C represents a correctly guessed answer and I represents an incorrect answer.

0

IIIII

1

CIIII, ICIII, IICII, IIICI, IIIIC

( ) n Probability P (X = r) = pr (1 − p) n −r r ( ) ( )0 ( )5 1 4 1024 5 P (X = 0) = = = 0.3277 0 5 5 3125 ( ) ( )1 ( )4 1 4 1280 5 P (X = 1) = = = 0.4096 1 5 5 3125

2

CCIII, ICCII, IICCI, IIICC, CICII, ICICI, IICIC, CIICI, ICIIC, CIIIC

( ) ( )2 ( )3 1 4 640 5 P (X = 2) = = = 0.2048 2 5 5 3125

Number of correct answers, r

Possible outcomes

(Continued) CHAPTER 10 Discrete random variables 385

(Continued) Number of correct answers, r

( ) n Probability P (X = r) = pr (1 − p) n −r r

Possible outcomes

3

IICCC, ICICC, CIICC, ICCIC, CICIC, CCIIC, ICCCI, CICCI, CCICI, CCCII

4

ICCCC, CICCC, CCICC, CCCIC, CCCCI

5

CCCCC

( ) ( )3 ( )2 4 1 160 5 P (X = 3) = = = 0.0512 3 5 5 3125 ( ) ( )4 ( )1 1 4 20 5 P (X = 4) = = = 0.0064 4 5 5 3125 ( ) ( )5 ( )0 1 4 1 5 P (X = 5) = = = 0.0003 5 5 5 3125

The results can then be written in the form of a probability distribution table: r

0

1

2

3

4

5

P (X = r)

0.3277

0.4096

0.2048

0.0512

0.0064

0.0003

From this table, we can see that it is most likely that Cassandra will correctly guess the answer to 1 out of the 5 questions and least likely that she will guess all 5 answers correctly. It should be noted that, if the order is specified for a particular scenario, then the binomial probability distribution rule cannot be used. Although we can calculate the chances of Cassandra getting 3 correct answers, we cannot determine the chances of her specifically getting the first, fourth and fifth questions correct using the binomial distribution. WORKED EXAMPLE 3

A new drug for hay fever is known to be successful in 40% of cases. Ten hay-fever sufferers take part in the testing of the drug. Determine the probability that: a. 4 people are cured b. no people are cured c. at least 2 people are cured. THINK a. 1.

Check that all the characteristics have been satisfied for a binomial distribution.

Write down the rule for the binomial probability distribution. 3. Define and assign values to variables. 2.

WRITE a.

This is a binomial distribution with n independent trials and two outcomes, p and (1 − p). ( ) n r P (X = r) = p (1 − p)n−r r n = 10 Let X = number of people cured. Therefore, r = 4. p = 0.4 (1 − p) = 0.6

386 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

(

4.

Substitute the values into the rule.

5.

Evaluate.

Round the answer off to 4 decimal places. 7. Answer the question. 6.

b. 1.

Define and assign values to variables.

2.

Substitute the values into the rule.

3.

Evaluate.

Round off the answer to 4 decimal places. 5. Answer the question. 4.

c. 1.

The condition that at least 2 people are cured means that the probability will be the sum of probabilities for which 2 ≤ r ≤ 10, that is ( ) 10 r 10 P (X ≥ 2) = ∑r=2 p (1 − p)10−r . r

This expression would require the summation of 9 terms. It would be easier in this case to determine the probability of the complementary event and subtract it from 1. 2. Define and assign values to variables.

3.

Substitute the values into the expression.

4.

Evaluate.

Round off the answer to 4 decimal places. 6. Answer the question. 5.

) 10 (0.4)4 (0.6)6 P (X = 4) = 4 = 210 × 0.0256 × 0.046 656 = 0.250 822 656 = 0.2508 The probability that 4 people are cured is 0.2508. b. n = 10 Let X = number of people cured. Therefore, r = 0. p = 0.4 (1 − p) = 0.6( ) 10 (0.4)0 (0.6)10 P (X = 0) = 0 = 1 × 1 × 0.006 046 617 6 = 0.006 046 6176 = 0.0060 The probability that no people are cured is 0.0060. c. P (X ≥ 2) = 1 − P(X > 2) P (X ≥ 2) = 1 − [P (X = 0) + P (X = 1)]

From b, P (X = 0) = 0.0060. For P (X = 1): n = 10 Let X = number of people cured. Therefore, r = 1. p = 0.4 (1 − p) = 0.6 ( ) 10 (0.4)1 (0.6)9 ] P (X ≥ 2) = 1 − [(0.0060) + 1 = 1 − [(0.0060) + (10 × 0.4 × 0.010 077 696)] = 1 − [0.0060 + 0.040 310 784] = 0.953 689 216 = 0.9537 The probability that at least 2 people are cured is 0.9537.

CHAPTER 10 Discrete random variables 387

TI | THINK a. 1. On a Calculator page,

select: MENU 6: Statistics 5: Distributions A: Binomial Pdf …

2. Complete the entry

lines as: n: 10 p: 0.4 X Value: 4 then press OK.

3. The answer appears

on the screen.

WRITE

CASIO | THINK a.1. On a Statistics screen,

select: DIST BINOMIAL Bpd.

2. Select Variable by

pressing the Var button.

3. Complete the entry

lines as: x: 4 Numtrial: 10 p: 0.4 then press EXE.

The answer appears on the screen.

c. 1. On a Calculator page,

select: MENU 6: Statistics 5: Distributions B: Binomial Cdf …

2. Complete the entry

lines as: n: 10 p: 0.4 Lower Bound: 2 Upper Bound: 10 then press the OK button. 3. The answer appears

on the screen.

c.1. On a Statistics screen,

select: DIST BINOMIAL Bcd.

2. Complete the entry

lines as: Lower: 2 Upper: 10 Numtrial:10 p: 0.4 then press the EXE button. 3. The answer appears

on the screen.

388 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

WRITE

WORKED EXAMPLE 4

It is known that 52% of the population participates in sport on a regular basis. Five random individuals are interviewed and asked whether they participate in sport on a regular basis. Let X be the number of people who regularly participate in sport. a. Construct a probability distribution table for X. b. Determine the probability that 3 people or fewer play sport. c. Determine the probability that at least one person plays sport, given that no more than 3 people play sport. d. Determine the probability that the first person interviewed plays sport but the next 2 do not. THINK

WRITE

Write the rule for the probabilities of the binomial distribution. 2. Substitute r = 0 into the rule and simplify.

a. 1.

3.

Substitute r = 1 into the rule and simplify.

4.

Substitute r = 2 into the rule and simplify.

5.

Substitute r = 3 into the rule and simplify.

6.

Substitute r = 4 into the rule and simplify.

7.

Substitute r = 5 into the rule and simplify.

8.

Construct a probability distribution table and check that ∑ P(X = r) = 1.

a.

all r

X ~ Bi(5, 0.52) P(X = r) = n Cr (1 − p)n−r pr P(X = 0) = 5 C0 (0.48)5 = 0.025 48 P(X = 1) = 5 C1 (0.48)4 (0.52) = 0.138 02 P(X = 2) = 5 C2 (0.48)3 (0.52)2 = 0.299 04 P(X = 3) = 5 C3 (0.48)2 (0.52)3 = 0.323 96 P(X = 4) = 5 C4 (0.48) (0.52)4 = 0.175 48 P(X = 5) = 5 C5 (0.52)5 = 0.038 02

x

P(X = r)

0

0.025 48

1

0.138 02

2

0.299 04

3

0.323 96

4

0.175 48

5

0.038 02

∑ P(X = r) = 1 all r

Interpret the question and write the probability to be found. 2. State the probabilities included in P(X ≤ 3).

b. 1.

b.

P(X ≤ 3) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) P(X ≤ 3) = 1 − (P(X = 4) + P(X = 5))

CHAPTER 10 Discrete random variables 389

3.

Substitute the appropriate probabilities and evaluate.

State the rule for conditional probability. 2. Evaluate P(X ≥ 1 ∩ X ≤ 3).

c. 1.

3.

Substitute the appropriate values into the rule.

Simplify. d. 1. Order has been specified for this question. Therefore, the binomial probability distribution rule cannot be used. The probabilities must be multiplied together in order. 2. Substitute the appropriate values and evaluate. 4.

P(X ≤ 3) = 1 − (0.175 48 + 0.038 02) P(X ≤ 3) = 0.7865 P(X ≥ 1 ∩ X ≤ 3) c. P(X ≥ 1|X ≤ 3) = P(X ≤ 3) P(X ≥ 1 ∩ X ≤ 3) = P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) = 0.138 02 + 0.299 04 + 0.323 96 = 0.761 02 P(X ≥ 1 ∩ X ≤ 3) P(X ≥ 1|X ≤ 3) = P(X ≤ 3) 0.761 02 = 0.7865 P(X ≥ 1|X ≤ 3) = 0.9676 d. S = plays sport, N = doesn’t play sport P(SNN) = P(S) × P(N) × P(N)

P(SNN) = 0.52 × 0.48 × 0.48 = 0.1198

WORKED EXAMPLE 5

The probability of an Olympic archer hitting the centre of the target is 0.7. What is the smallest number of arrows he must shoot to ensure that the probability he hits the centre at least once is more than 0.9? THINK

WRITE

Write the rule for the probabilities of the binomial distribution. 2. The upper limit of successes is unknown, because n is unknown. Therefore, P(X ≥ 1) cannot be found by adding up the probabilities. However, the required probability can be found by subtracting from 1 the only probability not included in P(X ≥ 1). 3. Substitute in the appropriate values and simplify.

X ~ Bi(n, 0.7) P(X ≥ 1) > 0.9 P(X ≥ 1) = 1 − P(X = 0)

1.

P(X ≥ 1) = 1 − P(X = 0) 1 − P(X = 0) > 0.9 1 −n Cx (1 − p)n−r pr > 0.9 1 −n C0 (0.3)n (0.7)0 > 0.9 1 − 1 × (0.3)n × 1 > 0.9 1 − (0.3)n > 0.9

390 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

4.

5.

1 − 0.9 > (0.3)n

Rearrange and take the log of both sides to determine the value of n.

log10 (0.1) > log10 (0.3)n log10 (0.1) > n log10 (0.3) log10 (0.1) n > log10 (0.3) n > 1.91249 n = 2 (as n must be an integer) The smallest number of arrows the archer needs to shoot in order to guarantee a probability of 0.9 of hitting the centre is 2.

Interpret the result and answer the question.

Units 3 & 4

Area 6

Sequence 1

Concept 3

Binomial distributions Summary screen and practice questions

Exercise 10.4 Binomial distributions Technology active 1. Which of the following constitutes a binomial probability distribution? a. Rolling a die 10 times and recording the number that comes up b. Rolling a die 10 times and recording the number of 3s that come up c. Spinning a spinner numbered 1 to 10 and recording the number that is obtained d. Tossing a coin 15 times and recording the number of Tails obtained e. Drawing a card from a fair deck, without replacement, and recording the number of picture cards f. Drawing a card from a fair deck, with replacement, and recording the number of black cards g. Selecting 3 marbles from a jar containing 3 yellow marbles and 2 black marbles, without replacement 2. Seven per cent of items made by a certain machine are defective. The items are packed and sold in boxes of 50. What is the probability of 5 items being defective in a box? 3. Nadia has a 40% chance of getting a red light on the way to work. What is the probability of Nadia getting a red light on 4 out of 5 days? 4. Peter has 4 chances to knock an empty can off a stand by throwing a ball. On each throw, the 1 probability of success is . Determine the probability that he will knock the empty can off the stand: 5 a. once b. twice c. at least once. 5. WE 3 Fifty-five per cent of the local municipality support the local council. If 8 people are selected at random, determine the probability that: a. half support the council b. all 8 support the council c. 5 support the council d. 3 oppose the council. 6. The probability of Colin beating Maria at golf is 0.4. If they play once a week throughout the entire year and the outcome of each game is independent of any other, determine the probability that they will have won the same number of matches. 7. It is known that 5 out of every 8 people eat Superflakes for breakfast. Determine the probability that half of a random sample of 20 people surveyed eat Superflakes.

CHAPTER 10 Discrete random variables 391

8.

9.

10.

11.

12.

13. 14.

On a certain evening, during a ratings period, two television stations put their best shows on against each other. The ratings showed that 39% of people watched Channel 6, while only 30% of people watched Channel 8. The rest watched other channels. A random sample of 10 people were surveyed the next day. Determine the probability that exactly: a. 6 watched Channel 6 b. 4 watched Channel 8. WE 4 Jack is an enthusiastic darts player and on average is capable of achieving a bullseye 3 out of 7 times. Jack will compete in a five-round tournament. Let Y be the discrete random variable that defines the number of bullseyes Jack achieves. a. Construct a probability distribution table for Y, giving your answers correct to 4 decimal places. b. Determine the probability that Jack will score at most 3 bullseyes. c. Determine the probability that Jack will score more than 1 bullseye, given that he scored at most 3 bullseyes. d. Determine the probability that his first shot missed, his second shot was a bullseye and then his next 2 shots missed. At a poultry farm, eggs are collected daily and classified as large or medium. Then they are packed into cartons containing 12 eggs of the same classification. From experience, the director of the poultry farm knows that 42% of all eggs produced at the farm are considered to be large. Ten eggs are randomly chosen from a conveyor belt on which the eggs are to be classified. Let Z be the discrete random variable that gives the number of large eggs. a. Determine P(Z = 0), P(Z = 1) … P(Z = 9), P(Z = 10) for this binomial distribution. b. Construct a probability distribution table for Z. c. Determine P(Z ≥ 5|Z ≤ 8). WE 5 The probability of winning a prize in a particular competition is 0.2. How many tickets would someone need to buy in order to guarantee them a probability of at least 0.85 of winning at least one prize? Lizzie and Matt enjoy playing card games. The probability that Lizzie will beat Matt is 0.67. How many games do they need to play so that the probability of Matt winning at least one game is more than 0.9? If X has a binomial distribution so that n = 15 and p = 0.62, calculate: a. P(X = 10) b. P(X ≥ 10) c. P(X < 4|X ≤ 8) A particular medication used by asthma sufferers has been found to be beneficial if used 3 times a day. In a trial of the medication it was found to be successful in 63% of the cases. Eight random asthma sufferers have had the medication prescribed for them. a. Construct a probability distribution table for the number of sufferers who have benefits from the medication, X. b. Determine the probability that no more than 7 people will benefit from the medication. c. Determine the probability that at least 3 people will benefit from the medication, given that no more than 7 will. d. Determine the probability that the first person won’t benefit from the medication, but the next 5 will.

392 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Lilly knows that the chance of her scoring a goal during a basketball game is 0.75. What is the least number of shots that Lilly must attempt to ensure that the probability of her scoring at least 1 goal in a match is more than 0.95? 16. The tram that stops outside Maia’s house is late 20% of the time. If there are 12 times during the day that the tram stops outside Maia’s house, calculate, correct to 4 decimal places: a. the probability that the tram is late 3 times b. the probability that the tram is late 3 times for at least 6 out of the next 14 days.

15.

10.5 The mean and variance of a binomial distribution When we are working with a binomial probability distribution, it is very useful to know the mean (or expected value), the variance and the standard deviation. These values are calculated the same way as for other probability distributions that you would have encountered in Year 11. For a binomial distribution X ~ Bi (n, p), the mean is calculated simply from the following equation: 𝜇 = E (X) = np This can be proven as follows.

( ) n pr (1 − p)n−r , where r = 0, 1, 2, … n. If we define a value q We have already seen that P (X = r) = r as being the probability of failure, then q = 1 − p and we can rewrite our expression as ( ) ( ) n! n n r n−r = . P (X = r) = p q where r r (n − r) ! r! For any probability distribution, n

𝜇 = E (X) = ∑ rP (X = r) r=0 ( ) n n r n−r 𝜇 = ∑r pq r r=0 n! n = ∑r=0 r pr qn−r (n − r) ! r! As the first term of this sum will equal 0, r=1

n

∑r r=0

n! n! pr qn−r = ∑ r pr qn−r (n − r)! r! (n − r)! r! n

Therefore, n

𝜇 = ∑r r=1 n

n! pr qn−r (n − r) ! r!

n! pr qn−r (n (r − r) ! − 1) ! r=1

=∑

Taking the factor np outside the summation gives n

(n − 1)! pr−1 qn−r . (n − r) ! (r − 1) ! r=1

𝜇 = np ∑

CHAPTER 10 Discrete random variables 393

n

(n − 1)! pr−1 qn−r is the binomial expansion of (p + q)n−1 , then (n (r − r) ! − 1) ! r=1

As ∑

𝜇 = np (p + q)n−1 = np (1)n−1 = np Hence, for a binomial distribution, 𝜇 = E(X) = np. The variance of the binomial distribution X ~ Bi (n, p) is calculated using the equation 𝜍 2 = Var (X) = np (1 − p) . Let’s look at why. First, we know that, for any probability distribution ( ) 𝜍 2 = Var (X) = E X2 − [E (X)]2 Mathematically, it is clear that x2 = x (x − 1) + x, so we can write ( ) E X2 = E [X (X − 1) + X] = E [X (X − 1)] + E (X) As we know that E (X) = np, ( ) E X2 = E [X (X − 1)] + np n

= np + ∑ r (r − 1) P (X = r) r=0 n

= np + ∑ r (r − 1) r=0 n

n! pr qn−r r! (n − r) !

n! pr qn−r (r (n − 2) ! − r) ! r=0

= np + ∑

n

(n − 2) ! pr qn−r (r (n − 2) ! − r) ! r=2

= np + n (n − 1) ∑

n

(n − 2) ! pr−2 qn−r (r (n − 2) ! − r) ! r=2 = np + n (n − 1) p2 (p + q)n−2 = np + n (n − 1) p2 ∑

= np + n (n − 1) p2 (1)n−2

Thus, ( ) E X2 = np + n (n − 1) p2 .

394 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

We may now substitute this relation into our earlier expression for variance: ( ) 𝜍 2 = E X2 − [E (X)]2 = [np + n (n − 1) p2 ] − (np)2 = np + n2 p2 − np2 − n2 p2 = np − np2 = np (1 − p) Therefore, 𝜍 2 = Var(X) = np (1 − p). Standard deviation As the standard deviation 𝜍 = SD (X) = √Var (X), 𝜍 = SD (X) = √np(1 − p).

WORKED EXAMPLE 6

A test consists of 20 multiple choice questions, each with 5 alternatives for the answer. A student has not studied for the test, so she chooses the answers at random. Let X be the discrete random variable that describes the number of correct answers. i. Determine the expected number of correct questions answered. ii. Determine the standard deviation of the correct number of questions answered, correct to 4 decimal places. b. A binomial random variable, Z, has a mean of 8.4 and a variance of 3.696. i. Determine the probability of success, p. ii. Determine the number of trials, n. a.

THINK a. i. 1. 2.

WRITE

Write the rule for the expected value. Substitute the appropriate values and simplify.

a. i.

E(X) = np n = 20, p = E(X) = np

1 5

= 20 ×

3. ii. 1.

Write the answer. Write the rule for the variance.

2.

Substitute the appropriate values and evaluate.

3.

Write the rule for the standard deviation.

1 5

=4 The expected number of questions correct is 4. ii. Var(X) = np(1 − p) 1 4 Var(X) = 20 × × 5 5 16 = 5 = 3.2 SD(X) = √Var(X)

CHAPTER 10 Discrete random variables 395

4.

Substitute the variance and evaluate.

b. i. 1.

Write the rules for the variance and expected value.

2.

Substitute the known information and label the two equations.

To cancel out the n, divide equation [2] by equation [1]. 4. Simplify. 3.

Write the answer. ii. 1. Substitute p = 0.56 into E(Z) = np and solve for n. 5.

2.

Write the answer.

16 √5 = 1.7889 b. i. E(Z) = np Var(Z) = np(1 − p) 8.4 = np [1] Var(Z) = np(1 − p) [2] 3.696 = np(1 − p) np(1 − p) 3.696 [2] ÷ [1]: = np 8.4 𝜎=

1 − p = 0.44 p = 0.56 The probability of success is 0.56. ii. E(Z) = np 8.4 = n × 0.56 n = 15 There are 15 trials.

Resources Interactivity: Effects of n and p on the binomial distribution (int-6432)

Units 3 & 4

Area 6

Sequence 1

Concept 4

The mean and variance of a binomial distribution Summary screen and practice questions

Exercise 10.5 The mean and variance of a binomial distribution Technology active 1. WE6a A fair die is tossed 25 times. Let X be the discrete random variable that represents the number of ones achieved. Determine, correct to 4 decimal places: a. the expected number of ones achieved b. the standard deviation of the number of ones achieved. 2. WE6b A binomial random variable, Z, has a mean of 32.535 and a variance of 9.021 95. a. Determine the probability of success, p. b. Determine the number of trials, n. 3. A fair coin is tossed 10 times. Determine: a. the expected number of Heads b. the variance for the number of Heads c. the standard deviation for the number of Heads. 4. A card is selected at random from a standard playing pack of 52 and then replaced. This procedure is completed 20 times. Determine: a. the expected number of picture cards b. the variance for the number of picture cards c. the standard deviation for the number of picture cards. 396 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

5.

6.

7.

8. 9. 10.

11. 12.

13.

14.

15.

Six out of every 10 cars manufactured are white. Twenty cars are randomly selected. Calculate: a. the expected number of white cars b. the variance for the number of white cars c. the standard deviation for the number of white cars. A fair die is rolled 10 times. Determine: a. the expected number of 2s rolled b. the probability of obtaining more than the expected number of 2s. Eighty per cent of rabbits that contract a certain disease will die. If a group of 120 rabbits contract the disease, how many would you expect to: a. die? b. live? A binomial random variable has a mean of 10 and a variance of 5. Determine: a. the probability of success, p b. the number of trials, n. A binomial random variable has a mean of 12 and a variance of 3. Determine: a. the probability of success, p b. the number of trials, n. For each of the following binomial random variables, calculate: i. the expected value ii. the variance. a. X ~ Bi ( 45, 0.72) ) ( 1 b. Y ~ Bi 100, 5 ) ( 2 c. Z ~ Bi 72, 9 Four per cent of pens made at a certain factory do not work. If pens are sold in boxes of 25, determine the probability that a box contains more than the expected number of faulty pens. A statistician estimates the probability that a spectator at a Brisbane Lions versus Collingwood AFL 1 match barracks for Brisbane is . At an AFL grand final between these two teams there are 100 000 2 spectators. Determine: a. the expected number of Brisbane supporters b. the variance of the number of Brisbane supporters c. the standard deviation of the number of Brisbane supporters. Thirty children are given 5 different yoghurts to try. The yoghurts are marked A to E, and each child has to select his or her preferred yoghurt. Each child is equally likely to select any brand. The company running the tests manufactures yoghurt B. a. How many children would the company expect to pick yoghurt B? b. The tests showed that half of the children selected yoghurt B as their favourite. What does this tell the company manufacturing this product? A binomial experiment is completed 16 times and has an expected value of 10.16. a. Determine the probability of success, p. b. Determine the variance and the standard deviation. A large distributor of white goods has found that 1 in 7 people who buy goods from them do so by using their lay-by purchasing system. On one busy Saturday morning, 10 customers bought white goods. Let X be the number of people who use the lay-by purchasing system to buy their goods. Determine E(X) and Var(X).

CHAPTER 10 Discrete random variables 397

10.6 Applications The binomial distribution has important applications in medical research, quality control, simulation and genetics. In this section we will explore some of these areas. WORKED EXAMPLE 7

It has been found that 9% of the population have diabetes. A sample of 15 people were tested for diabetes. Let X be the random variable that gives the number of people who have diabetes. a. Determine P(X ≤ 5). b. Determine E(X) and SD(X). THINK a. 1.

WRITE

Define and assign values to variables.

2.

Substitute the values into the rule.

3.

Evaluate.

Round the answer off to 4 decimal places. 5. Answer the question. 4.

State the rule for the expected value. 2. Substitute the appropriate values and simplify. 3. Determine the variance.

b. 1.

4.

Determine the standard deviation.

Units 3 & 4

Area 6

Sequence 1

a.

n = 15 Let X = number of people who have diabetes. Therefore, 0 ≤ r ≤ 5. p = 0.09 (1 − p) = 0.91 ( ) 15 5 (0.09)r (0.91)15−r P (X ≤ 5) = ∑r=0 r ( ) ( ) 15 15 (0.09)1 (0.91)14 (0.91)15 + = 0 1 ( ) ( ) 15 15 (0.09)2 (0.91)13 + (0.09)3 (0.91)12 + 2 3 ( ) ( ) 15 15 4 11 (0.09) (0.91) + (0.09)5 (0.91)10 + 4 5 = 0.243 008 + 0.360 507 + 0.249 582 + 0.106 964 + 0.031 736 + 0.006 905 = 0.998 702 = 0.9987

The probability that 5 or fewer people of the 15 selected have diabetes is 0.9987. b. E(X) = np E(X) = 15 × 0.09 = 1.35 Var(X) = np(1 − 9) = 15 × 0.09 × 0.91 = 1.2285 SD(X) = √Var(X) = √1.2285 = 1.1084

Concept 5

Applications Summary screen and practice questions

398 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Exercise 10.6 Applications Technology active 1. WE7 It is thought that about 30% of teenagers receive their spending money from part-time jobs. Ten random teenagers were interviewed about their spending money and how they obtained it. Let Y be the random variable that defines the number of teenagers who obtain their spending money by having a part-time job. a. Determine P(Y ≥ 7). b. Determine E(Y) and SD(Y).

In Australia, it is estimated that 30% of the population over the age of 25 have hypertension. A statistician wishes to investigate this, so he arranges for 15 random adults over the age of 25 to be tested to see if they have high blood pressure. Let X be the random variable that defines the number of adults over the age of 25 with hypertension. a. Determine P(X ≤ 5). b. Determine E(X) c. Determine SD(X). 3. The proportion of defective fuses made by a certain company is 0.02. A sample of 30 fuses is taken for quality control inspection. a. Determine the probability that there are no defective fuses in the sample. b. Determine the probability that there is only 1 defective fuse in the sample. c. How many defective fuses would you expect in the sample? d. The hardware chain that sells the fuses will accept the latest batch for sale only if, upon inspection, there is at most 1 defective fuse in the sample of 30. What is the probability that they accept the batch? e. Ten quality control inspections are conducted monthly for the hardware chain. Determine the probability that all of these inspections will result in acceptable batches. 4. Suppose that 85% of adults with allergies report systematic relief with a new medication that has just been released. The medication has just been given to 12 patients who suffer from allergies. Let Z be the discrete random variable that defines the number of patients who get systematic relief from allergies with the new medication. a. Determine the probability that no more than 8 people get relief from allergies. b. Given that no more than 8 people get relief from allergies after taking the medication, determine the probability that at least 5 people do. c. Calculate: i. E(Z) ii. SD(Z) 5. Consider a woman with the genotype XX and a man with the genotype XY. = XX XY Their offspring have an equal chance of inheriting one of these genotypes. What is the probability that 6 of their 7 offspring have the genotype XY? 2.

XX

XY

CHAPTER 10 Discrete random variables 399

6.

Silicon chips are tested at the completion of the fabrication process. Chips either pass or fail the inspection, and if they fail they are destroyed. The probability that a chip fails an inspection is 0.02. What is the probability that in a manufacturing run of 250 chips, only 7 will fail the inspection?

7.

A manufacturer of electric kettles has a process of randomly testing the kettles as they leave the assembly line to see if they are defective. For every 50 kettles produced, 3 are selected and tested for any defects. Let X be the binomial random variable that is the number of kettles that are defective, so that X ~ Bi(3, p). a. Construct a probability distribution table for X, giving your probabilities in terms of p. b. Assuming P(X = 0) = P(X = 1), determine the value of p where 0 < p < 1. c. Determine: i. 𝜇 ii. 𝜍 The probability of a person in Australia suffering anaemia is 1.3%. A group of 100 different Australians of differing ages were tested for anaemia. a. Determine the probability that at least 5 of the 100 Australians suffer from anaemia. Give your answer correct to 4 decimal places. b. Determine the probability that 4 of the 100 Australians suffer from anaemia, given that less than 10 do. Give your answer correct to 4 decimal places. Edie is completing a multiple choice test of 20 questions. Each question has 5 possible answers. a. If Edie randomly guesses every question, what is the probability, correct to 4 decimal places, that she correctly answers 10 or more questions? b. If Edie knows the answers to the first 4 questions but must randomly guess the answers to the other questions, determine the probability that she correctly answers a total of 10 or more questions. Give your answer correct to 4 decimal places. Six footballers are chosen at random and asked to kick a football. The probability of a footballer being able to kick at least 50 m is 0.7. a. Determine the probability, correct to 4 decimal places, that: i. only the first three footballers chosen kick the ball at least 50 m ii. exactly three of the footballers chosen kick the ball at least 50 m iii. at least three of the footballers chosen kick the ball at least 50 m, given that the first footballer chosen kicks it at least 50 m. b. What is the minimum number of footballers required to ensure that the probability that at least one of them can kick the ball 50 m is at least 0.95? Lori is a goal shooter for her netball team. The probability of her scoring a goal is 0.85. In one particular game, Lori had 12 shots at goal. Determine the probability, correct to 4 decimal places, that: a. she scored at least 9 goals b. only her last 9 shots were goals c. she scored exactly10 goals, given that her last 9 shots were goals. The chance of winning a prize in the local raffle is 0.08. What is the least number of tickets Siena needs to purchase so that the chance of both her and her sister each winning at least one prize is more than 0.8?

8.

9.

10.

11.

12.

400 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

13.

A regional community is trying to ensure that their local water supply has fluoride added to it, as a medical officer found that a large number of children aged between 8 and 12 have at least one filling in their teeth. In order to push their cause, the community representatives have asked a local dentist to check the teeth of ten 8– 12-year-old children from the community. Let X be the binomial random variable that defines the number of 8– 12-year-old children who have at least one filling in their teeth: X ~ Bi(10, p). Determine the value of p, correct to 4 decimal places, if P(X ≤ 8) = 0.9.

10.7 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. MC Which of the following does not represent a binomial distribution? A. Rolling a die four times and recording the number of 2s B. Tossing a coin 10 times and recording the number of Heads C. Rolling two dice simultaneously 20 times and recording the outcomes D. Drawing a card with replacement and recording the number of aces obtained 2. MC A Bernoulli random variable, X, has a probability of failure of 0.35. The expected value and variance of X are respectively: A. 0.35 and 0.2275 B. 0.35 and 0.65 C. 0.65 and 0.2275 D. 0.65 and 0.35 3. MC Suppose that X is a binomial random variable with a mean of 12 and a standard deviation of 3. The probability of success, p, in any trial is: A. 0.25 B. 0.35 C. 0.50 D. 0.75 4. MC The probability that the 7:35 am bus arrives on time is 0.45. What is the probability that the bus is on time at least once in the next 5 days? 5 A. 1 − (0.55) 5 B. (0.55) 5 C. (0.45) 5 D. 1 − (0.45)

5.

1 If X is a binomial random variable with n = 15 and p = , the mean and variance of X are 5 closest to: A. 𝜇 = 3, 𝜍 2 = 12 B. 𝜇 = 12, 𝜍 2 = 2.4 2 C. 𝜇 = 2.4, 𝜍 = 12 D. 𝜇 = 3, 𝜍 2 = 2.4 MC

CHAPTER 10 Discrete random variables 401

6.

One-quarter of all customers at a particular bookstore buy non-fiction books. If 5 customers purchase a book on a particular day, what is the probability that 3 of them purchased a non-fiction book?

One in every 100 new cars is returned with faulty steering. A survey is taken of 300 buyers. Determine the probability that: a. none have cars with faulty steering b. one has a car with faulty steering. 8. A binomial random variable has a mean of 10 and variance of 8. Determine: a. the probability of success, p b. the number of trials, n. 9. Six out of every 10 cars manufactured are white. If 20 cars are selected at random, how many would you expect to be white? 7.

A random variable X follows a Bernoulli sequence with a probability of success of 0.86. What is the variance of X correct to 2 decimal places? 11. A coin is biased such that the probability of a tail is 0.7. What is the probability that at most 1 Tail will be observed when the coin is tossed 5 times? 12. A binomial experiment is completed 16 times and has an expected value of 10.16. a. Determine the probability of success, p. b. Determine the variance and the standard deviation. Complex familiar 13. Five per cent of watches made at a certain factory are defective. Watches are sold to retailers in boxes of 20. Determine: a. the expected number of defective watches in each box b. the probability that a box contains more than the expected number of defective watches per box c. the probability of a ‘bad batch’, if a ‘bad batch’ entails more than a quarter of the box being defective. 14. Ten per cent of all Olympic athletes are tested for drugs at the conclusion of their event. One per cent of all athletes use performance-enhancing drugs. Of the 1000 Olympic wrestlers competing from all over the world, Australia sends 10. Determine: a. the expected number of Australian wrestlers who are tested for drugs b. the probability that half the Australian wrestlers are tested for drugs c. the probability that at least 2 Australian wrestlers are tested for drugs d. the expected number of drug users among all Olympic wrestlers. 10.

402 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

One-fifth of Australia’s population has a British background. Fifty Australians are randomly selected and questioned about their ancestry. Determine (correct to 4 decimal places) the probability that at least 96% of the selected people have a non-British background. 16. Speedy Saverio’s Pizza House claims to cook and deliver 90% of pizzas within 15 minutes of the order being placed. If your pizza is not delivered within this time, it is free. On one busy Saturday night, Saverio has to make 150 deliveries. a. How many deliveries are expected to be made within 15 minutes of placing the order? b. What is the probability of receiving a free pizza on this night? c. If Saverio loses an average of $4 for every late delivery, how much would he expect to lose on late deliveries this night? Complex unfamiliar 17. An experiment consists of 3 independent trials. Each trial results in a success or failure. The probability of success in a trial is p. Determine in terms of p the probability of exactly 1 success given at least 1 success. 18. Keepers at a zoo are concerned that their herd of 10 giraffes are low in iron. In order to investigate this, they ask the zoo vet to take blood samples from all the giraffes to check the iron levels. Let X be the binomial random variable that defines the number of giraffes that have low iron levels. For this distribution, X ~ Bi (10, p). Determine the value of p, correct to 4 decimal places, if P (X ≤ 8) = 0.9. 15.

While on holiday at the Gold Coast, Jordan and Bronte play a total of n games of mini-golf. The probability that Jordan wins any game is 0.15. How many games of mini-golf must they play if the probability of Jordan winning exactly 2 games is 0.2759? 20. A barrel contains 100 balls, some of which have a stripe painted on them. Five balls are randomly selected from the barrel with replacement after each ball has been withdrawn. Let p be the proportion of striped balls in the barrel such that 0 < p < 1. Using technology, determine the value of p for which the probability that exactly 1 of the 5 balls chosen has a stripe will be greatest. 19.

Units 3 & 4

Sit exam

CHAPTER 10 Discrete random variables 403

Answers

4. a. 0.4096 b. 0.1536 c. 0.5904 5. a. 0.2627 b. 0.0084 c. 0.2568 d. 0.2568 6. 0.0381 7. 0.0924 8. a. 0.1023 b. 0.2001 9. a. *See the table at the bottom of the page. b. 0.8891 c. 0.9315 d. 0.0800 10. a, b. *See the table at the bottom of the page. c. 0.4164 11. 9 tickets 12. 6 games 13. a. 0.1997 b. 0.4665 c. 0.0034 14. a. *See the table at the bottom of the page. b. 0.9752 c. 0.9655 d. 0.0367 15. 3 shots 16. a. 0.2362 b. 0.0890

10 Discrete random variables Exercise 10.2 Bernoulli distributions 1. b and c are Bernoulli trials. 2. a, b and d are Bernoulli trials. 3. a. The probability of success is not able to be calculated (as

Paul has not replaced the ball he drew). b. There are more than two outcomes. c. The probability of success is not able to be calculated.

Exercise 10.3 Bernoulli random variables 1. a.

x P(X = x) b. 0.42 c. i. 0.2436 2. a. 0.63 3. a. y P(Y = y) b. i. 0.68 c. 1 4. a. x P(X = x) b. i. 0.89 c. 0.89 5. a. 0.7 6. a. 0.2436 7. a. 0.0072 b. z P(Z = z) c. 0.9928 8. a. y P(Y = y) b. 0.33 c. 1 9. a. 0.8775 b. z P(Z = z) 10. a. 0.11

0 0.58

1 0.42 ii. 0.4936 c. 0.4828

b. 0.2331

0 0.32

1 0.68 ii. 0.2176

0 0.11

iii. 0.4665

Exercise 10.5 The mean and variance of a binomial distribution

1 0.89 ii. 0.0979 b. 0.7 b. 0.58

0 0.9928

c. 0.58

1 0.0072

0 0.67

8.

1 0.33

9. 10.

0 0.1225

1 0.8775

11. 12. 13.

b. 0.8742

Exercise 10.4 Binomial distributions

14.

1. b, d and f are binomial; a, c, e and g are not binomial. 2. 0.1359 3. 0.0768

*9. a.

15.

y

0

1

2

3

4

5

P(Y = y)

0.0609

0.2285

0.3427

0.2570

0.0964

0.0145

*10. a, b.

*14. a.

4.1667 Var (X) ≈ 3.472; SD (X) = 1.8634 0.7227 b. 45 5 b. 2.5 c. 1.58 4.62 b. 3.55 c. 1.88 12 b. 4.8 c. 2.19 1.67 b. 0.2248 96 b. 24 1 a. b. 20 2 3 b. 16 a. 4 a. i. 32.4 ii. 9.072 b. i. 20 ii. 16 c. i. 16 ii. 12.4 0.2642 a. 50 000 b. 25 000 c. 158.11 a. 6 children b. Yoghurt B is far more popular than expected. a. 0.635 b. Var (X) = 3.7084; SD (X) = 1.9257 E (X) = 1.4286; Var (X) = 1.2245

1. a. b. 2. a. 3. a. 4. a. 5. a. 6. a. 7. a.

iii. 0.3129

z

0

1

2

3

4

5

6

7

8

9

10

P(Z = z)

0.0043

0.0312

0.1017

0.1963

0.2488

0.2162

0.1304

0.0540

0.0147

0.0024

0.0002

x

0

1

2

3

4

5

6

7

8

P(X = x)

0.0004

0.0048

0.0285

0.0971

0.2067

0.2815

0.2397

0.1166

0.0248

404 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Exercise 10.6 Applications

11. a. 0.9078 12. 37 tickets 13. 0.6632

0.0106 E (Y) = 3; SD (Y) = 1.4491 0.7216 b. 4.5 0.5455 b. 0.3340 0.8795 e. 0.2769 0.0922 0.9992 i. 10.2 ii. 1.2369 5. 0.0547 6. 0.1051

1. a. b. 2. a. 3. a. d. 4. a. b. c.

c. 1.7748 c. 0.6

x

0

1

2

3

P(X = x)

(1 − p)3

3(1 − p)2 p

3(1 − p) p2

p3

1 4

c. i. ii.

3 4 3

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

4

8. a. 0.0101 9. a. 0.0026 10. a. i. 0.0093 ii. 0.1852 iii. 0.1320 b. 3 footballers

b. 0.0319 b. 0.0817

c. 0.0574

10.7 Review: exam practice

7. a.

b.

b. 0.0008

17. 18. 19. 20.

C C A A D 0.0879 a. 0.0490 a. 0.2 12 0.12 0.030 78 a. 0.635 a. 1 b. 0.2642 a. 1 b. 0.0015 0.0013 a. 135 b. 0.1 c. Expected loss of $60 ( )2 3p 1 − p ( )3 1− 1−p 0.6632 10 p = 0.2

b. 0.1486 b. 50

b. 1.9257 c. 0.2639

c. 0.0003 d. 10

CHAPTER 10 Discrete random variables 405

REVISION UNIT 4 Further functions and statistics

TOPIC 3 Discrete random variables 2 • For revision of this entire topic, go to your studyON title in your bookshelf at www.jacplus.com.au. • Select Continue Studying to access hundreds of revision questions across your entire course.

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406 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

11

General continuous random variables 11.1 Overview Continuous random variables such as height have values that depend upon the precision of the devices used to measure them. Modern technology has allowed standardised measurements of length to a reasonable level of precision; however, the first ‘standardised’ units of length used over five thousand years ago were a bit unreliable. Length was described in terms of cubits, where one cubit was equal to the distance between the elbow and the tips of the finger. Obviously, with so much variation between human beings, an ‘official cubit’ had to be decided upon, but even then there were problems. Cubit rods found by archaeologists have varied from 47.2 to 52.5 cm in length, with each cubit broken up into 5 to 7 ‘palms’ and a palm broken up into anywhere between 4 and 7 ‘fingers’. So, depending on the cubit rod used, the same person could be described as being 3 cubits, 2 palms and 1 finger tall; 3 cubits, 5 palms and 5 fingers tall; or somewhere in between. Today, when people measure their height they can be confident that they will get the same value each time when rounded to the nearest whole centimetre. The size of this centimetre is standardised as one-hundredth of a metre. The metre itself is 1 defined as being the distance travelled by light in a vacuum in of a second. 299 792 458

LEARNING SEQUENCE 11.1 11.2 11.3 11.4 11.5

Overview Continuous random variables and the probability density function Cumulative distribution functions Measures of centre and spread Review: exam practice

Fully worked solutions for this chapter are available in the Resources section of your eBookPLUS at www.jacplus.com.au.

CHAPTER 11 General continuous random variables 407

11.2 Continuous random variables and the probability density function

In Chapter 10, we dealt with discrete random variables, that is, data which is finite or countable. The number of white cars in a car park, the number of students in a class and the number of lollies in a jar are all examples of discrete random variables. A continuous random variable, however, can assume any value within a given range.

11.2.1 Discrete and continuous variables For example, imagine that you are measuring the length of a piece of string. Using a metre ruler, you find that its length is somewhere between 9 and 10 centimetres. But is there any way of determining an exact value for its length? This is the fundamental difference between discrete and continuous variables: although you can count the number of white cars in a car park and get an exact number, there is no such thing as an exact value for a continuous variable. You can use more and more precise instruments to measure that string length. For example, you might get 9.5 centimetres using a school ruler, 9.492 centimetres using engineering callipers, or even 9.492 000 000 862 centimetres using a high precision laser. But this is only narrowing the range of values within which a theoretical ‘exact’ value would lie — determining if it lies between 9 and 10, between 9.4 and 9.5, between 9.49 and 9.50 centimetres and so on. As a result, a continuous random variable is described in terms of the interval in which its value lies. Examples of continuous random variables include time, height and weight — all quantities that must be measured rather than counted. Consider an Australian health study that was conducted. The study targeted young people aged 5 to 17 years old. They were asked to estimate the average number of hours of physical activity they participated in each week. The results of this study are shown in the following histogram.

Frequency

Physical activity y 400 350 300 250 200 150 100 50 0

364

347

156 54 0

1

2

3 4 Hours

32

10 5

7 6

7

x

The frequencies of individual activity times cannot be determined due to the fact that the times have been grouped into class intervals. This limits the information we are able to extract from the histogram. For example, to determine the number of young people who did less than 2 hours of activity, we simply add the frequencies of the 0 < x < 1 and 1 < x < 2 class intervals; that is, 364 + 347 = 711 people. However, we would not be able to determine the number of people who did less than 1.5 hours of physical activity as this value lies within a class interval rather than being an end point. Since the value that a continuous random variable can assume is always measured in some way, exact values are unable to be obtained. As there are an infinite number of values that the variable can have within that interval, the probability of a continuous random variable assuming an exact value is zero. 408 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

11.2.2 Probability and relative frequency Because histograms are drawn such that the intervals for each class are of uniform width, the probability of a variable lying within a particular interval is also equal to the fraction of the histogram area that the column for that interval occupies. In other words, the probability that x lies within a particular interval on a histogram is approximately equal to the relative frequency of the interval: P (a < x < b) =

f (a < x < b) ∑f

So, the probability that a randomly selected person in the study did between 1 and 2 hours of activity can be determined: P (1 < x < 2) =

=

f (1 < x < 2) ∑f

347 364 + 347 + 156 + 54 + 32 + 10 + 7 347 = 970 = 0.358

Similarly, we could find the probability that a person did more than 4 hours of activity per week by adding the interval frequencies for all intervals higher than 4: f (4 < x < 5) + f (5 < x < 6) + f(6 < x < 7) ∑f 32 + 10 + 7 = 970 49 = 970 = 0.051

P (x > 4) =

WORKED EXAMPLE 1

Nose length (mm) 27.5 < X ≤ 32.5 32.5 < X ≤ 37.5 37.5 < X ≤ 42.5 42.5 < X ≤ 47.5 47.5 < X ≤ 52.5 52.5 < X ≤ 57.5 57.5 < X ≤ 62.5 62.5 < X ≤ 67.5 67.5 < X ≤ 72.5

Frequency 2 5 17 21 11 7 6 5 1

Frequency

In a study, the nose lengths, X millimetres, of 75 adults were measured. The results of the study are shown in the table and histogram.

y 35 30 25 20 15 10 5 0

Nose length

27.5 32.5 37.5 42.5 47.5 52.5 57.5 62.5 67.5 72.5 Length in mm

x

Determine the probability of someone in the study having a nose length that is: a. between 42.5 mm and 47.5 mm b. less than or equal to 47.5 mm.

CHAPTER 11 General continuous random variables 409

THINK a. 1.

Use the table to determine the frequency of the interval and the sum of the frequencies for the study.

2.

Substitute values into the equation and evaluate.

3.

Answer the question.

b. 1.

Determine the total of the frequencies for all of the intervals such that x ≤ 47.5.

2.

Substitute values into the equation and evaluate.

3.

Answer the question.

f (42.5 < x < 47.5) = 21 ∑ f = 2 + 5 + 17 + 21 + 11 + 7 + 6 + 5 + 1 = 75

WRITE a.

f (42.5 < x < 47.5) ∑f 21 = 75 = 0.28 The probability of a person having a nose length between 42.5 and 47.5 mm is 0.28. b. f (x ≤ 47.5) = 2 + 5 + 17 + 21 = 45 P (42.5 < x < 47.5) =

f (x ≤ 47.5) ∑f 45 = 75 = 0.6 The probability of a person having a nose length less than or equal to 47.5 mm is 0.6.

P (x ≤ 47.5) =

11.2.3 Modelling continuous random variables The histogram shown displays the masses (in kg) of members joining the Toowong branch of a popular weight loss program.

Masses of people joining a weight loss program 12 Frequency

10 8 6 4 2 0 60 70 80 90 100 110 120 130 Mass (kg)

A frequency polygon is drawn by joining the midpoints at the top of each bar representing an interval.

12 Frequency

10 8 6 4 2 0 60 70 80 90 100 110 120 130 Mass (kg) 410 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

This is a relatively small set of data, as the study only involved 40 people. However, if we increased the size of the study so that the members of more branches of the weight loss program were included and the class intervals were made smaller, the frequency polygon would take on the appearance of a smooth curve. As the shape of a curve may be modelled by a mathematical formula, it should therefore be possible to develop a function, f (x), that will allow us to approximate relative frequency values — and therefore probability values — for intervals regardless of the original interval boundaries that generated the function curve. f (x)

f 3000 2500 2000 1500 1000

0.38

500 0

0

60 70 80 90 100 110 120 130 x Mass (kg)

60 70 80 90 100 110 120 130 x Mass (kg)

For our weight loss program curve, 38% of the area under the curve is bounded by the x-axis and the line x = 84. This represents the relative frequency of members having a mass of 84 kg or less. This also means that the probability of a randomly chosen person in the program having a mass of 84 kg or less is approximately 0.38; that is, P (x < 84) = 0.38. The absolute area bounded by the curve and the x-axis can have any value greater than 0. However, if the curve is scaled down in such a way that the total area under the curve is equal to 1, it is referred to as the probability density curve (or pdf) for the continuous random variable.

11.2.4 The probability density curve As you will recall from Chapter 7, the area under a curve f (x) for any interval may be determined by evaluating the integral of the function over that interval. Thus, the probability that a continuous random variable X lies between the values a and b is given by: P (a < X < b) =

f (x)

b

∫a

f (x) dx 0

a

b

x

Using this relationship, we can clarify why probabilities are assessed for continuous random variables lying within intervals rather than at exact values. Consider a continuous variable X such that X = x where x ∈ R. P (X = x) =

x

∫x

f (x) dx

= [F (x)]xx

= F (x) − F (x) P (X = x) = 0

CHAPTER 11 General continuous random variables 411

Thus, P (X = x) = 0 when x ∈ [a, b]. There are two critical conditions that constrain a probability density function. First, as probability cannot be a negative number, the probability density function must be greater than or equal to zero over the interval being considered; that is, f (x) ≥ 0 for all x ∈ [a, b].

Secondly, the sum of probabilities for all possible values of x must be equal to 1: ∞

∫−∞

f (x) dx = 1

(This is analogous to the condition for the discrete probability distribution that ∑ P (X = x) = 1.) all x

b

In the event that an interval is bounded by a and b such that everywhere else outside this interval.

∫a

f (x) dx = 1, then the function must be 0

WORKED EXAMPLE 2

Sketch the graph of each of the following functions and state whether each function is a probability density function. { { 2(x − 1), 1 ≤ x ≤ 2 0.5, 2 ≤ x ≤ 4 a. f (x) = b. f (x) = 0, elsewhere 0, elsewhere { −x 2e 0≤x≤2 c. f (x) = 0, elsewhere Sketch the graph of f (x) = 2(x − 1) over the domain 1 ≤ x ≤ 2, giving an x-intercept of 1 and an end point of (2, 2). Make sure to include the horizontal lines for y = 0 either side of this graph. Note: This function is known as a triangular probability function because of its shape.

THINK

WRITE

a. 1.

a.

f (x) (2, 2)

2 f(x) = 2(x – 1)

0

Inspect the graph to determine if the function is always positive or zero, that is, f (x) ≥ 0 for all x ∈ [a, b]. 3. Calculate the area of the shaded region to 2.

2

determine if

∫1

2(x − 1) dx = 1.

(1, 0)

(2, 0)

x

Yes, f (x) ≥ 0 for all x-values. Method 1: Using the area of triangles 1 Area of shaded region = × base × height 2 1 = ×1×2 2 =1

412 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Method 2: Using calculus Area of shaded region =

=

1

∫2 1

∫2

2(x − 1) dx

(2x − 2)dx

= [x2 − 2x]21 = (22 − 2(2)) − (12 − 2(1))

4.

b. 1.

=0−1+2 =1 f (x) ≥ 0 for all values, and the area under the curve equals 1. Therefore, this is a probability density function.

Interpret the results. Sketch the graph of f (x) = 0.5 for 2 ≤ x ≤ 4. This gives a horizontal line with end points of (2, 0.5) and (4, 0.5). Make sure to include the horizontal lines for y = 0 on either side of this graph. Note: This function is known as a uniform or rectangular probability density function because of its rectangular shape.

b.

f(x)

0.5

(2, 0.5)

(2, 0)

f(x) = 0.5

(4, 0.5)

(4, 0) x

0

Inspect the graph to determine if the function is always positive or zero, that is, f (x) ≥ 0 for all x ∈ [a, b]. 3. Calculate the area of the shaded region to

2.

4

determine if

∫2

0.5 dx = 1.

Yes, f(x) ≥ 0 for all x-values.

Again, it is not necessary to use calculus to determine the area. Method 1: Area of shaded region = length × width = 2 × 0.5 =1 Method 2: Area of shaded region =

2

∫4

0.5 dx

= [0.5x]42

4.

Interpret the results.

= 0.5(4) − 0.5(2) =2−1 =1 f (x) ≥ 0 for all values, and the area under the curve equals 1. Therefore, this is a probability density function.

CHAPTER 11 General continuous random variables 413

c. 1.

Sketch the graph of f (x) = 2e−x for 0 ≤ (x ≤ 2. The ) end points will be (0, 2) and 2, e−2 . Make sure to include the horizontal lines for y = 0 on either side of this graph.

c.

f (x) (0, 2) f(x) = 2e–x

(2, –e2) 2

(0, 0) 2.

3.

Inspect the graph to determine if the function is always positive of zero, that is, f (x) ≥ 0 for all x ∈ [a, b]. Calculate the area of the shaded region to 2

determine if

4.

∫0

2e−x dx = 1.

Interpret the results.

(2, 0)

x

Yes, f (x) ≥ 0 for all x-values. 2

∫0

2e dx = 2 −x

2

∫0

e−x dx

= 2 [−e−x ]20 ( ) = 2 −e−2 + e0 ( ) = 2 −e−2 + 1

= 1.7293 f (x) ≥ 0 for all values. However, the area under the curve does not equal 1. Therefore, this is not a probability density function.

WORKED EXAMPLE 3

Given that the following functions are probability density functions, determine the value of a in each function. { a(x − 1)2 , 0 ≤ x ≤ 4 a. f (x) = 0, elsewhere { −4x ae , x > 0 b. f (x) = 0, elsewhere THINK

WRITE 4

a. 1.

As the function has already been defined as a probability density function, this means that the area under the graph is definitely 1.

a. 4

∫0

∫0

a(x − 1)2 dx = 1 4

2.

Remove a from the integral, as it is a constant.

a

3.

Antidifferentiate and substitute in the terminals.

a

∫0

4

414 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

f(x)dx = 1

∫0

(x − 1)2 dx = 1

(x − 1)2 dx = 1

[ a

b. 1.

Solve for a.

As the function has already been defined as a probability density function, this means that the area under the graph is definitely 1.

2.

Remove a from the integral, as it is a constant.

3.

To evaluate an integral containing infinity as one of the terminals, we determine the appropriate limit.

4.

Antidifferentiate and substitute in the terminals.

b.

∞

∫0

∞

∫0

a

f (x)dx = 1

ae−4x dx = 1 ∞

∫0

e−4x dx = 1

a × lim

Solve for a. Remember that a number divided by an extremely ( )large number is effectively 0, so 1 lim = 0. k→∞ e4k

k

k→∞ ∫

a × lim

0

e−4x dx = 1

0

k→∞ ∫

e−4x dx = 1

]k 1 −4x a × lim − e =1 k→∞ 4 0 ) ( −4k 1 e + =1 a × lim − k→∞ 4 4 ( −4k ) e 1 a × lim − + =1 k→∞ 4 4 ( ) 1 1 a × lim − 4k + =1 k→∞ 4 4e ( ) 1 a 0+ =1 4 a =1 4 a=4 [

5.

=1

]4

]0 33 (−1)3 a − =1 3 3 ) ( 1 =1 a 9+ 3 28 a× =1 3 3 a= 28 [

4.

(x − 1)3 3

k

Resources Interactivity: Continuous random variables (int-6433)

CHAPTER 11 General continuous random variables 415

Exercise 11.2 Continuous random variables and the probability density function Technology free

Which of the following are continuous random variables? a. The population of your town or city b. The types of motorbike in a parking lot c. The heights of people in an identification line-up d. The masses of babies in a group e. The languages spoken at home by students in your class f. The time spent watching TV g. The number of children in the families in your suburb h. The air pressure in your car’s tyres i. The number of puppies in a litter j. The types of radio program listened to by teenagers k. The times for swimming 50 metres l. The quantity of fish caught in a net m. The number of CDs you own n. The types of shops in a shopping centre o. The football competition ladder at the end of each round p. The lifetimes of torch batteries q. The number of people attending a rock concert r. Exam grades s. The types of magazine sold at a news agency t. Hotel accommodation ratings 2. WE1 The frequency histogram shows the distribution of masses (in kilograms) of 60 students in Year 7 at Northwood State High School. Determine the probability that a random student has a mass: a. between 40 and 60 kilograms b. less than 45 kilograms c. greater than 55 kilograms. 3. A small car-hire firm keeps note of the age and kilometres covered by each of the cars in their fleet. Generally, cars are no longer used once they have either covered 350 000 kilometres or are more than 5 years old. The following information describes the ages of the cars in their current fleet. Frequency

1.

3 −1) P(−1 < Y < 2) = P(Y > −1) P(−1 < Y < 2) = P(−1 < Y < 0) + P(0 ≤ Y < 2) P(Y < 2|Y > −1) =

1 1 = − ydy + ydy ∫−1 9 ∫0 9 0

3.

4.

To calculate the probabilities we need to determine the areas under the curve.

Antidifferentiate and evaluate after substituting the terminals.

2

1 1 ydy + ydy =− ∫0 9 ∫−1 9 [ ]0 [ ]2 1 2 1 2 + =− y y 18 1 18 0 ( ) 1 1 1 1 2 2 =− (0) − (−1) + (2)2 − (0)2 18 18 18 18 2

0

4 1 + 18 18 5 = 18

=

5.

Determine P(Y > −1). As the interval is across two functions, the interval needs to be split.

P(Y > −1) = P(−1 < Y < 0) + P(0 ≤ Y ≤ 3) 1 1 = − y dy + y dy ∫−1 9 ∫0 9 0

6.

7.

To determine the probabilities we need to determine the areas under the curve. As P(0 ≤ Y ≤ 3) covers exactly half the area under the curve, 1 P(0 ≤ Y ≤ 3) = . (The entire 2 area under the curve is always 1 for a probability density function.) Antidifferentiate and evaluate after substituting the terminals.

=−

3

0

∫−1

1 1 y dy + 9 2

]0 1 2 1 =− y + 18 −1 2 ( ) 1 1 1 2 2 =− (0) − (−1) + 18 18 2 1 9 = + 18 18 10 = 18 5 = 9 [

CHAPTER 11 General continuous random variables 421

8.

P(−1 < Y < 2) Now substitute into the formula P(Y < 2|Y > −1) = P(Y > −1) to determine 5 5 P(−1 < Y < 2) = ÷ P(Y < 2|Y > −1) = . 18 9 P(Y > −1) 5 9 = × 18 5 1 = 2

11.3.2 The cumulative distribution function The cumulative distribution function F (x) describes the probability that a continuous random variable X has a value less than or equal to x; that is, F (x) = P (x ≤ c) =

c

∫−∞

f (x) dx.

The cumulative distribution function (or cdf) of a continuous random variable has three fundamental properties: • F (x) must be a non-decreasing function; that is, F (a) ≤ F (b) when a < b. • As x approaches −∞, F (x) approaches or equals 0; that is, lim F (x) = 0.

•

As x approaches ∞, F (x) approaches or equals 1; that is, lim F (x) = 1. x→−∞

x→∞

The probability of a variable falling within a particular interval is easily determined by using the cumulative distribution function. Given that P (X ≤ b) = P (X ≤ a) + P (a < X ≤ b), P (a < X ≤ b) = P (X ≤ b) − P (X ≤ a). Thus, P (a < X ≤ b) = F (b) − F (a). Consider the probability distribution function f (x) such that f (x) = {

0.5x 0 ≤ x ≤ 2 . 0, elsewhere

The probabilities for consecutive intervals across the distribution can be tabulated as shown. a≤x≤b

P(a ≤ x ≤ b)

0.25 ≤ x < 0.5

0.015 625

1 ≤ x < 1.25

0.109 375

1.75 ≤ x ≤ 2

0.203 125

0 ≤ x < 0.25

0.5 ≤ x < 0.75

0.046 875

1.25 ≤ x < 1.5

0.140 625

0.75 ≤ x < 1

0.078 125

1.5 ≤ x < 1.75

0.171 875 0.234 375

422 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

As P (a ≤ X ≤ b) = P (a ≤ X ≤ c) + P (c < X ≤ b), a < c < b,

P (X ≤ 0.5) = P (0 ≤ X ≤ 0.25) + P (0.25 < X ≤ 0.5) P (X ≤ 0.75) = P (X ≤ 0.5) + P (0.5 < X ≤ 0.75) P (X ≤ 1) = P (X ≤ 0.75) + P (0.75 < X ≤ 1)

and so on. We can tabulate the cumulative probability P (x ≤ c) across the interval 0 ≤ c ≤ 2: x≤c

P (x ≤ c)

x ≤ 0.5

0.015 625

x ≤ 1.25

0.25

x≤2

0.765 625

x ≤ 0.25

c

Note that as

∫−∞

f (x) dx =

c

∫0

x ≤ 0.75

0.062 5

x ≤ 1.5

0.390 625

x≤1

0.140 625

x ≤ 1.75

0.562 5 1

f (x) dx for this pdf, P (x ≤ c) = P (0 ≤ x ≤ c) = P (−∞ ≤ x ≤ c).

When F (x) is graphed against x, the curve formed is continuous across the range of the probability function. F(x) 1 0.8 0.6 0.4 0.2

0

As F (x) =

x 0.25

0.5

0.75

1

1.25

1.5

1.75

2

2.25

2.5

( ) x dx for 0 ≤ x ≤ 2, ∫0 2 [ 2 ]x x F (x) = . 4 0 x

Thus, F (x) =

x2 for 0 ≤ x ≤ 2. 4

CHAPTER 11 General continuous random variables 423

The cumulative distribution function for the continuous random variable X will, in this case, be such that: 0, x≤0 ⎧ ⎪ x2 F (x) = , 0 < x ≤ 2. ⎨ 4 ⎪ x>2 ⎩ 1,

It is important to remember that the area under the probability density curve for an interval will be equivalent to a point on the cumulative distribution curve. Probability density function f(x)

1

0.5 P(X < 1.25) x

0

0.25

0.5

0.75

1

1.25

1.5

1.75

2

Cumulative distribution function F(x)

1

0.5 P(X < 1.25)

0

x 0.25

0.5

0.75

1

1.25

1.5

1.75

2

WORKED EXAMPLE 5

A continuous random variable Y has a probability density function such that ⎧ y, 0≤y≤1 ⎪ f (y) = ⎨ 2 − y, 1 < y ≤ 2. ⎪ 0, elsewhere ⎩

Determine the cumulative distribution function, F(y), of Y. b. Determine: i. P(Y ≤ 0.8) ii. P(0.6 ≤ Y < 1.2) a.

424 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

THINK a. 1.

2.

WRITE

Integrate f (y) to determine F (y) over the domains 0 ≤ y ≤ 1 and 1 < y ≤ 2.

a.

For 0 ≤ y ≤ 1, F (y) =

Identify the appropriate equation and substitute the appropriate values.

F (y) =

b. i.

2.

3.

1

∫0

y dy +

y

∫1

(2 − y) dy

P (Y ≤ 0.8) = F (Y ≤ 0.8) =

(0.8)2 2 = 0.32 The probability of Y being less than 0.8 is 0.32

Evaluate. 3. Answer the question. Identify the appropriate equation.

y2 . 2

[ ]y y2 1 = + 2y − 2 2 1 [( ) ( )]y y2 1 12 = + 2y − − 2×1− 2 2 2 1 2 y = 2y − −1 2 ⎧ 0, y≤0 ⎪ ⎪ y2 ⎪ , 02 ⎩

2. ii. 1.

y dy =

∫0 For 1 < y ≤ 2, F(y) = P (Y ≤ y) = P (Y ≤ 1) + P(1 < Y ≤ y)

Write F (y) as a hybrid function for all y ∈ R.

b. i. 1.

y

ii.

Evaluate.

P (a ≤ Y < b) = F (b) − F (a)

P (0.6 ≤ Y < 1.2) = F (1.2) − F (0.6) ] [ ] [ (0.6)2 (1.2)2 −1 − = 2 (1.2) − 2 2 = 0.68 − 0.18

= 0.5 The probability of Y being between 0.6 and 1.2 is 0.5.

Answer the question.

Resources Interactivity: Cumulative density functions (int-6434)

Units 3 & 4

Area 7

Sequence 1

Concept 1

Probability density functions Summary screen and practice questions

CHAPTER 11 General continuous random variables 425

Exercise 11.3 Cumulative distribution functions Technology free 1.

WE4

The continuous random variable Z has a probability density function given by −z + 1, 0 ≤ z < 1 f (z) = { z − 1, 1 ≤ z ≤ 2 . 0, elsewhere

Sketch the graph of f . b. Determine P(Z < 0.75). c. Determine P(Z > 0.5). 2. The continuous random variable X has a probability density function given by a.

f (x) = {

4x3 , 0 ≤ x ≤ a 0,

elsewhere

where a is a constant. a. Determine the value of the constant a. b. Sketch the graph of f . c. Determine P(0.5 ≤ X ≤ 1). 3. WE5 The continuous random variable X has a uniform rectangular probability density function defined by 1 , 1≤x≤6 f (x) = { 5 . 0, elsewhere Determine the cumulative distribution function, F (x), for X. Determine: i. P (x ≤ 4) ii. P(2.2 < x ≤ 4.5) 4. Let X be a continuous random variable with a probability density function defined by a.

b.

1 sin(x), 0 ≤ x ≤ 𝜋 f (x) = { 2 . 0, elsewhere Determine the cumulative distribution function F (x) for X. ( ) 𝜋 b. Determine P X ≤ . 2 ) ( 3𝜋 𝜋 c. Determine P

|X < . 4 4 a.

426 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

5.

A probability density function is defined by the rule

k(2 + x), −2 ≤ x < 0 f (x) = { k(2 − x), 0 ≤ x ≤ 2 0, elsewhere

Frequency

where X is a continuous random variable and k is a constant. a. Sketch the graph of f . b. Determine the value of k. c. Determine the cumulative distribution function F (x). d. Determine P (−1 ≤ X ≤ 1). ( ) e. Determine P X ≥ −1|X ≤ 1 . 6. The amount of petrol sold daily by a busy service station is a uniformly distributed probability density function. A minimum of 18 000 litres and (18, k) (30, k) a maximum of 30 000 litres are sold on any given k day. The graph of the function is shown. a. Determine the value of the constant k. b. Determine the formula (including cases) for the probability density function f (x). c. Determine the formula (including cases) for the cumulative distribution function F (x). d. Determine the probability that between 20 000 and 25 000 litres of petrol are sold on a given day. 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 e. Determine the probability that as much as 26 000 litres Petrol sold (thousands of litres) of petrol were sold on a particular day, given that it was known that at least 22 000 litres were sold. Technology active 7. The continuous random variable X has a probability density function defined by 3 2 x , 0≤x≤2 f (x) = { 8 . 0, elsewhere Determine the cumulative distribution function for f (x). b. Determine P(X > 1.2). c. Determine the value of n such that P(X ≤ n) = 0.75. 8. The continuous random variable Z has a probability density function defined by a.

1 , 1 ≤ z ≤ e2 f (z) = { 2z . 0, elsewhere e

a.

Sketch the graph of f and shade the area that represents e

b.

Determine

∫1

∫1

2

f (z) dz.

2

f (z) dz. Explain your result.

CHAPTER 11 General continuous random variables 427

The continuous random variable U has a probability function defined by f (u) = {

e4u , u ≥ 0 0,

elsewhere

.

a

c.

Sketch the graph of f and shade the area that represents e

d. 9.

10.

Determine the exact value of the constant a if

∫0

f (u) du, where a is a constant.

2

∫1

a

f (z) dz is equal to

∫0

f (u) du.

The continuous random variable Z has a probability density function defined by 1 𝜋 𝜋 cos(z), − ≤ z ≤ 2 2 . f (z) = { 2 0, elsewhere

) ( 𝜋 𝜋 , correct to 3 decimal places. Determine P − ≤ Z ≤ 6 4

The continuous random variable U has a probability density function defined by 1 1 − (2u − 3u2 ), 0 ≤ u ≤ a 4 f (u) = { 0, elsewhere

where a is a constant. Determine: a. the value of the constant a b. P(U < 0.75) c. P(0.1 < U < 0.5) d. P(U = 0.8). 11. The continuous random variable Z has a probability density function defined by f (z) = { where a is a constant. Determine: a.

a

the value of the constant a such that

e

−

z 3,

0,

0≤z≤a

elsewhere

f (z) dz = 1

P(0 < Z < 0.7) P(Z < 0.7|Z > 0.2), correct to 4 decimal places d. the value of 𝛼, correct to 2 decimal places, such that P(Z ≤ 𝛼) = 0.54. 12. The continuous random variable X has a probability density function given as b. c.

∫0

f (x) = {

3e−3x , x ≥ 0 0,

Determine P(0 ≤ X ≤ 1), correct to 4 decimal places b. Determine P(X > 2), correct to 4 decimal places.

elsewhere

a.

428 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

.

13.

The continuous random variable X has a probability density function defined by f (x) = {

loge (x2 ), x ≥ 1 0,

elsewhere

.

Determine, correct to 4 decimal places: a

a.

P(1.25 ≤ X ≤ 2).

∫1

f (x) dx = 1

The graph of the probability function f (z) = b.

14.

the value of the constant a if

1 is 𝜋(z + 1)

f (z)

2

shown. a. Determine, correct to 4 decimal places, P(−0.25 < Z < 0.25). b. Suppose another probability density function is defined as

(0, 1–π ) 1 f (z) = — π(z2 + 1)

1 , −a ≤ x ≤ a f (x) = { x + 1 . 0, elsewhere 2

–3

Determine the value of the constant a.

–2

–1

0

1

2

3 z

11.4 Measures of centre and spread The commonly used measures of central tendency and spread in statistics are the mean, median, variance, standard deviation and range. These same measures are appropriate for continuous probability functions.

11.4.1 Measures of central tendency: the mean Remember that for a discrete random variable, E(X ) = 𝜇 = ∑ xn P(X = xn ). x=n x=1

∞

This definition can also be applied to a continuous random variable. We define E(X ) = 𝜇 =

∫−∞

xf (x) dx.

The expected value of a continuous random variable If f(x) = 0 everywhere except for x ∈ [a, b], where the function is defined, then E(X) = 𝜇 =

b

∫a

xf(x) dx.

Consider the continuous random variable, X, which has a probability density function defined by f (x) = {

x2 , 0 ≤ x ≤ 1 0,

elsewhere

.

CHAPTER 11 General continuous random variables 429

For this function, E(X) = 𝜇 =

1

∫0

=

xf (x) dx

1

∫0

=

x(x2 ) dx

1

x3 dx ∫0 [ 4 ]1 x = 4 0 4 1 −0 = 4 1 = 4 Similarly, if the continuous random variable X has a probability density function of f (x) = {

7e−7x , x ≥ 0 0,

then E(X ) = 𝜇 =

elsewhere

∞

∫0

x f (x)dx

= lim

k→∞ ∫

k

= 0.1429 0

7xe−7x dx

where technology is required to determine the integral. The mean of a function of X is similarly found.

The mean of a continuous random variable The function of X, g(x), has a mean defined by E(g(x)) = 𝜇 =

So if we again consider f (x) = {

∞

∫−∞

g(x) f(x) dx.

x2 , 0 ≤ x ≤ 1 0,

elsewhere

430 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

then E(X2 ) =

=

1

∫0

x2 f (x) dx

1

x4 dx ∫0 [ 5 ]1 x = 5 0 15 = −0 5 1 = 5 This definition will be important when we investigate the variance of a continuous random variable.

11.4.2 Median and percentiles The median is also known as the 50th percentile, Q2 , the halfway mark or the middle value of the distribution. The median of a continuous random variable For a continuous random variable, X, defined by the probability function f, the median, m, can be m

found by solving

∫−∞

f(x) dx = 0.5.

Other percentiles that are frequently calculated are the 25th percentile or lower quartile, Q1 , and the 75th percentile or upper quartile, Q3 . The interquartile range The interquartile range is calculated as:

IQR = Q3 − Q1

Consider a continuous random variable, X, that has a probability density function of f (x) = {

0.21e2x−x , −3 ≤ x ≤ 5 2

0,

elsewhere

.

To determine the median, m, we solve for m as follows: m

∫−3

f(x)

0.21e2x−x dx = 0.5

2

f (x) = 0.21e2x – x Area = 0.5

2

–3

0

x=1

5

x

The area under the curve is equated to 0.5, giving half of the total area and hence the 50th percentile. Solving using technology, the result is that m = 0.9897 ≃ 1. This can be seen on a graph as shown. CHAPTER 11 General continuous random variables 431

Consider the continuous random variable X, which has a probability density function of x3 , 0≤x≤2 f (x) = { 4 . 0, elsewhere

The median is given by P(0 ≤ x ≤ m) = 0.5:

x3 dx = 0.5 ∫0 4 [ 4 ]m x 1 = 16 0 2 m

1 m4 −0= 16 2 m4 = 8

√ 4 m=± 8

m = 1.6818 (0 ≤ m ≤ 2)

To determine the lower quartile, we make the area under the curve equal to 0.25. Thus the lower quartile is given by P(0 ≤ x ≤ a) = 0.25: x3 dx = 0.25 ∫0 4 [ 4 ]a x 1 = 16 0 4 a

a4 1 −0= 16 4 a4 = 4

√ 4 a=± 4

a = Q1 = 1.4142 (0 ≤ a ≤ m)

Similarly, to determine the upper quartile, we make the area under the curve equal to 0.75. Thus, the upper quartile is given by P(0 ≤ x ≤ n) = 0.75: x3 dx = 0.75 ∫0 4 [ 4 ]n 3 x = 16 0 4 n

n4 3 −0= 16 4 n4 = 12 √ 4 n = ± 12

n = Q3 = 1.8612 (m ≤ x ≤ 2)

432 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

So, the interquartile range is given by

Q3 − Q1 = 1.8612 − 1.4142 = 0.4470

These values are shown on the following graph.

f (x) 2

3 f (x) = x 4 Upper quartile x = 1.8612

(2, 2)

Median x = 1.6818

Lower quartile x = 1.4142 (0, 0)

x

(2, 0)

WORKED EXAMPLE 6

A continuous random variable, Y, has a probability density function, f , defined by f (y) =

{

ky, 0 ≤ y ≤ 1 0, elsewhere

where k is a constant. Sketch the graph of f . b. Determine the value of the constant k. c. Determine: i. the mean of Y ii. the median of Y. d. Determine the interquartile range of Y. a.

graph f (y) = ky is a straight line with end points at (0, 0) and (1, k). Remember to include the lines f (y) = 0 for y > 1 and y < 0.

THINK

WRITE

a. The

a.

f (y) k

(0, 0)

(1, k)

(1, 0)

y

CHAPTER 11 General continuous random variables 433

1

b. Solve

∫0

ky dy = 1 to determine the value of k.

1

b.

∫0

ky dy = 1

y dy = 1 ∫0 [ 2 ]1 y k =1 2 0 k(1)2 −0=1 2 k =1 2 k=2 Using the area of a triangle also enables you to find the value of k. 1 ×1×k=1 2 k =1 2 k=2 1

k

c. i. 1.

State the rule for the mean.

𝜇=

c. i.

=

2.

1

∫0

y(2y) dy

1

2y2 dy ∫0 ]1 [ 2 3 = y 3 0

Antidifferentiate and simplify.

=

2(1)3 −0 3 2 = 3

m

ii 1.

State the rule for the median.

ii

∫0

f (y) dy = 0.5

m

∫0 2.

3. d. 1.

Antidifferentiate and solve for m. Note that m must be a value within the domain of the function, so within 0 ≤ y ≤ 1.

[ 2 ]m y = 0.5

m2 − 0 = 0.5 √ m = ± 0.5 Median = 0.7071 (as 0 < m < 1) 0

Write the answer.

State the rule for the lower quartile, Q1 .

2y dy = 0.5

a

d.

∫0

f (y) dy = 0.25 a

∫0

2y dy = 0.25

434 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2.

[ 2 ]a y = 0.25

a2 − 0 = 0.25 √ a = ± 0.25

Antidifferentiate and solve for Q1 .

0

n

3.

4.

State the rule for the upper quartile, Q3 .

∫0

a = Q1 = 0.5 (as 0 < Q1 < 0.7071)

f (y) dy = 0.75

2y dy = 0.75 ∫0 [ 2 ]n y = 0.75 n

n2 − 0 = 0.75 √ n = ± 0.75

Antidifferentiate and solve for Q3 .

0

n = Q3 = 0.8660 (as 0.7071 < Q3 < 1) IQR = Q3 − Q1 = 0.8660 − 0.5

State the rule for the interquartile range. 6. Substitute the appropriate values and simplify. 5.

= 0.3660

11.4.3 Measures of spread: variance, standard deviation and range The variance and standard deviation are important measures of spread in statistics. From previous calculations for discrete probability functions, we know the following. Variance and standard deviation for √ discrete probability functions 2 2 Var(X) = E(X ) − [E(X)] and SD(X) = Var(X).

For continuous probability functions, Var(X ) = =

=

∞

∫−∞ ∞

∫−∞ ∞

∫−∞

(x − 𝜇)2 f (x) dx

(x2 − 2x𝜇 + 𝜇 2 )f (x) dx x2 f (x) dx −

= E(X2 ) − 2𝜇

∞

∫−∞

∞

∫−∞

2xf (x)𝜇 dx +

xf (x) dx + 𝜇 2

= E(X2 ) − 2𝜇 × E(X) + 𝜇 2 = E(X2 ) − 2𝜇 2 + 𝜇 2

∞

∞

∫−∞

∫−∞

𝜇 2 f (x) dx

1f (x) dx

= E(X2 ) − 𝜇 2

= E(X2 ) − [E(X)]2

CHAPTER 11 General continuous random variables 435

∞

∞

Two important facts were used in this proof: f (x) dx = 1 and xf (x) dx = 𝜇 = E(X). ∫−∞ ∫−∞ √ Substituting this result into SD(X) = Var(X) gives us SD(X) =

√

E(X2 ) − [E(X)]2 .

The range is calculated as the highest value minus the lowest value. So, for the probability density 1 , 1≤x≤6 function given by f (x) = { 5 , the highest possible x-value is 6 and the lowest is 1. 0, elsewhere Therefore, the range for this function = 6 − 1 = 5. WORKED EXAMPLE 7

For a continuous random variable, X, with a probability density function, f , defined by ⎧ 1 ⎪ x + 2, −4 ≤ x ≤ −2 f (x) = ⎨ 2 ⎪ 0, elsewhere ⎩ determine: a. the mean b. the median c. the variance d. the standard deviation, correct to 4 decimal places. THINK a. 1.

2.

State the rule for the mean and simplify.

−2

WRITE a.

𝜇=

∫−4

−2

xf (x) dx

) 1 = x x + 2 dx ∫−4 2 ) −2 ( 1 2 = x + 2x dx ∫−4 2 ]−2 [ 1 = x3 + x2 6 −4 ( ) ( ) 1 1 3 2 3 2 = (−2) + (−2) − (−4) + (−4) 6 6 4 32 =– +4+ − 16 3 3 8 2 = – = −2 3 3

Antidifferentiate and evaluate.

(

m

b. 1.

State the rule for the median.

b. m

∫−4

(

∫−4

f (x) dx = 0.5

) 1 x + 2 dx = 0.5 2

436 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

[ 2.

Antidifferentiate and solve for m. The quadratic formula is needed as the quadratic equation formed cannot be factorised. Alternatively, use technology to solve for m.

(

1 2 x + 2x 4

]m

= 0.5

−4 ) ( ) (−4)2 1 2 m + 2m − + 2(−4) = 0.5 4 4 1 2 m + 2m + 4 = 0.5 4 m2 + 8m + 16 = 2

So,

m=

m=

−8 ±

−8 ± 2

m2 + 8m + 14 = 0

√ (8)2 − 4(1)(14) 2(1)

√

8

= −4 ±

Write the answer. c. 1. Write the rule for variance. 3.

2.

3.

( ) Determine E X2 first.

( ) Substitute E(X ) and E X2 into the rule for variance.

√ 2 √ ∴ m = −4 + 2 as m ∈ [−4, 2]. √ The median is −4 + 2. c. Var(X)2 = E(X2 ) − [E(X)]2 ( ) E X2 =

b

x2 f (x) dx ) ( −2 1 2 x + 2 dx = x ∫−4 2 ) −2 ( 1 3 2 = x + 2x dx ∫−4 2 1 2 = [ x4 + x3 ]−2 18 3 −4 ( ) ( ) 1 2 1 2 4 3 4 3 (−2) + (−2) − (−4) + (−4) = 8 3 8 3 16 128 =2− − 32 + 3 3 112 = −30 + 3 22 = 3 Var(X) = E(X2 ) − [E(X )]2 ( )2 22 8 = − − 3 3 22 64 = − 3 9 66 64 = − 9 9 2 = 9 ∫a

CHAPTER 11 General continuous random variables 437

d. 1.

2.

Write the rule for standard deviation.

d.

SD(X ) = =

√

Var(X )

√

2 9 = 0.4714

Substitute the variance into the rule and evaluate.

Resources Interactivities: Mean (int-6435) Median and percentiles (int-6436) Variance, standard deviation and range (int-6437)

Units 3 & 4

Area 7

Sequence 1

Concepts 4 & 5

Mean and median Summary screen and practice questions Variance and standard deviation Summary screen and practice questions

Exercise 11.4 Measures of centre and spread Technology free 1.

WE6

The continuous random variable Z has a probability density function of f (z) = {

4, 1 ≤ z ≤ a 0, elsewhere

where a is a constant. a. Determine the value of the constant a. b. Determine: i. the mean of Z ii. the median of Z. 2. The continuous random variable, Y, has a probability density function of f (y) = {

2y, 0 ≤ y ≤ a 0, elsewhere

where a is a constant. Determine: a. the value of the constant a b. E(Y) c. the median value of Y.

438 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

3.

WE7

For the continuous random variable Z, the probability density function is f (z) = {

4.

2z – 4, 2 ≤ z ≤ 3 0,

elsewhere

.

Determine the mean, median, variance and standard deviation. The function f (x) = {

3e−3x , x ≥ 0 0,

elsewhere

defines the probability density function for the continuous random variable X. Determine the mean, median, variance and standard deviation of X. 5. Let X be a continuous random variable with a probability density function of f (x) =

⎧ ⎨ ⎩

1 √ , 0≤x≤1 2 x . 0,

elsewhere

Prove that f is a probability density function. b. Determine E(X). c. Determine the median value of f . 6. The time in minutes that an individual must wait in line to be served at the local bank branch is defined by a.

f (t) = 2e−2t , t ≥ 0

where T is a continuous random variable. a. What is the mean waiting time for a customer in the queue, correct to 1 decimal place? b. Calculate the standard deviation for the waiting time in the queue, correct to 1 decimal place. c. Determine the median waiting time in the queue, correct to 2 decimal places. √ 7. a. Determine the derivative of 4 − x2 . b. Hence, determine the mean value of the probability density function defined by f (x) = 8.

⎧

3

𝜋 4 − x2 ⎨ 0, ⎩ √

, 0≤x≤

√

3

.

elsewhere

Consider the continuous random variable X with a probability density function of h(2 − x), 0 ≤ x ≤ 2 f (x) = { h(x − 2), 2 < x ≤ 4 0, elsewhere

where h is a positive constant. a. Determine the value of the constant h. b. Determine E(X). c. Determine Var(X).

CHAPTER 11 General continuous random variables 439

9.

Consider the continuous random variable X with a probability density function of f (x) = {

k, a ≤ x ≤ b 0, elsewhere

where a, b and k are positive constants. a. Sketch the graph of the function f . 1 b. Show that k = . b−a c. Determine E(X) in terms of a and b. d. Determine Var(X) in terms of a and b. Technology active 10.

The continuous random variable Y has a probability density function defined by √ y2 3 , 0≤y≤ 9 f (y) = { 3 . 0, elsewhere

Determine, correct to 4 decimal places: a. the expected value of Y b. the median value of Y c. the lower and upper quartiles of Y d. the interquartile range of Y. 11. The continuous random variable Z has a probability density function defined by a , 1≤z≤8 f (y) = { z 0, elsewhere where a is a constant. a. Determine the value, correct to 4 decimal places, of the constant a. b. Determine E(Z) correct to 4 decimal places. c. Determine Var(Z) and SD(Z). d. Determine the interquartile range for Z. e. Determine the range for Z. 12. X is a continuous random variable. The graph of the probability 1 density function f (x) = (sin(2x) + 1) for 0 ≤ x ≤ 𝜋 is shown. 𝜋 Show that f (x) is a probability density function. Calculate E(X) correct to 4 decimal places. c. Calculate, correct to 4 decimal places: i. Var(X) ii. SD(X) d. Determine the median value of f correct to 4 decimal places. 13. The continuous random variable X has a probability density function defined by

f (x) 2 – π

1 f (x) = – π (sin(2x) + 1)

a.

b.

f (x) = {

ax − bx2 , 0 ≤ x ≤ 2 . 0, elsewhere

Determine the values of the constants a and b if E(X) = 1.

440 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

(0, –π1 ) 0

(π, –π1)

0.25 0.5 0.75

1

x

14.

The continuous random variable Z has a probability density function of 3 , 1≤z≤a f (z) = { z2 0, elsewhere

where a is a constant. 3 a. Show that the value of a is . 2 b. Determine the mean value and variance of f correct to 4 decimal places. c. Determine the median and interquartile range of f . 15. The continuous random variable Y has a probability density function ( ) y 0.2 loge , 2 ≤ y ≤ 7.9344 f (y) = { . 2 0, elsewhere Verify that f is a probability density function. Determine E(Y ) correct to 4 decimal places. c. Determine Var(Y ) and SD(Y ) correct to 4 decimal places. d. Determine the median value of Y correct to 4 decimal places. e. State the range. 16. The continuous random variable Z has a probability density function a.

b.

f (z) = {

√

z − 1, 1 ≤ z ≤ a 0, elsewhere

where a is a constant. a. Determine the value of the constant a correct to 4 decimal places. b. Determine, correct to 4 decimal places: i. E(Z ) ii. E(Z2 ) iii. Var(Z )

iv.

SD(Z )

11.5 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. Which of the following represent continuous random variables? a. The number of goals scored at a football match b. The heights of students in a Maths B class c. Shoe sizes d. The number of girls in a five-child family e. The time taken to run a distance of 10 kilometres in minutes 2. X is a continuous random variable with a probability function defined by f (x) = {

2 sin(2x), a ≤ x ≤ b . 0, elsewhere

Given that a = 0 and 0 ≤ b ≤ 𝜋, what is the value of b?

CHAPTER 11 General continuous random variables 441

3.

A survey was taken to determine the amount of time, X hours, that teenagers spend interacting with digital devices during a 24-hour period. The table of findings and histogram are shown. Teenagers’ interaction with a digital device

0≤x≤1

Frequency

3 30

0

4.

0≤x≤2

5.

x>2

6.

1 f (z) = – 2z

7.

( (1, 0)

8

1 e2, – 2e2

(e2, 0)

) z

9 8

b. E (Y ) =

2 3

√

2 1 =2 ;m=2+ ; Var (Z ) = ; 3 2 18 1 SD (Z ) = √ 3 2 1 1 1 1 E (X ) = ; m = – loge (0.5); Var (X ) = ; SD (X ) = 3 3 9 3 a. Sample responses can be found in the worked solutions in the online resources. 1 b. E (X ) = 3 c. m = 0.25 a. E (T ) = 0.5 min b. SD (T ) = 0.5 min 1 c. m = − loge (0.5) 2 dy x 3 a. = −√ b. E (X) = 𝜋 2 dx 4−x

3. E (Z ) =

8. a. f (z)

0.5

ii. m =

9

1 c. m = √ 2

x 174) 0.32 = 2 = 0.16 Thus, P (X > 174) ≈ 0.16.

= 0.32

CHAPTER 12 The normal distribution 453

c. 1.

2.

Determine how many standard deviations 147 cm is from the mean.

c.

Using symmetry, calculate P(X < 147).

𝜇 − 𝜎 = 165 − 9 = 156 𝜇 − 2𝜎 = 165 − 2 × 9 = 147 147 cm is 2 standard deviations from the mean. The-corresponding upper value is 183(165 + 2 × 9). P(147 ≤ X ≤ 183) ≈ 0.95

0.95

147

165

183

x

Thus, P(X < 147) ∪ P(X < 183) ≈ 1 − 0.95 = 0.05

P(X < 147) = P(X > 183) ≈

and by symmetry,

0.05 2 = 0.025 Thus, approximately 2.5%, of the population of women in this particular town are shorter than 147 cm.

Resources Interactivities: The normal distribution (int-6438) The 68-95-99 rule (int-6439)

Units 3 & 4

Area 7

Sequence 2

Concept 1

The normal distribution Summary screen and practice questions

Exercise 12.2 The normal distribution Technology free 1.

WE1

The probability density function of a normal distribution is given by 1 −1 f (x) = √ e 2 3 2𝜋

(

x−2 3

)2

.

State the mean and the standard deviation of the distribution. b. Sketch the graph of the probability function. a.

454 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2.

A normal distribution has a probability density function of 2 1 − 1 (x−3) f (x) = √ e 2 . 2𝜋

State 𝜇 and 𝜍. Sketch the graph of the probability function. 3. WE2 The results of a Mathematical Methods test are normally distributed with a mean of 72 and a standard deviation of 8. a. What is the approximate probability that a student who sat the test has a score which is greater than 88? b. What approximate proportion of the students who sat the test had a score which was less than 48? c. What approximate percentage of the students who sat the test scored less than 80? 4. The length of pregnancy for a human is normally distributed with a mean of 275 days and a standard deviation of 14 days. A mother gave birth in less than 233 days. What is the approximate probability of this happening for the general population? 5. Consider the normal probability density function a.

b.

1 −1 f (x) = √ e 2 4 2𝜋

6.

(

x+2 4

)2

Identify 𝜇. A normal probability density function is defined by

10 − 1 f (x) = √ e 2 3 2𝜋

(

10(x−1) 3

, x ∈ R.

)2

, x ∈ R.

Determine the values of 𝜇 and 𝜍. State what effect the mean and standard deviation have on the graph of the normal distribution. c. Sketch the graph of the function, f. 7. A normal probability density function is given by a. b.

f (x) =

1 √

10 2𝜋

−1 e 2

(

x+4 10

)2

, x ∈ R.

Determine the values of 𝜇 and 𝜍. b. State what effect the mean and standard deviation have on the graph of the normal distribution. c. Determine: ( ) i. Var (X) ii. E X2 d. Verify that this is a probability density function. a.

5(x−2) 5 −1 8. f (x) = √ e 2 2 , x ∈ R defines a normal probability density function. 2 2𝜋 a. Determine the values of 𝜇 and 𝜍. ( ) b. Calculate E X2 . c. Determine: ( ) i. E (5X) ii. E 5X2 (

)2

CHAPTER 12 The normal distribution 455

Technology active 9.

10.

11.

12.

13.

14.

15.

Scores on a commonly used IQ test are known to be normally distributed with a mean of 120 and a standard deviation of 20. a. Determine: i. 𝜇 ± 𝜍 ii. 𝜇 ± 2𝜍 iii. 𝜇 ± 3𝜍 b. Determine: i. P(X < 80) ii. P(X > 180) The results of a Year 12 Biology examination are known to be normally distributed with a mean of 70 and a standard deviation of 6. What approximate percentage of students sitting for this examination can be expected to achieve a score that is greater than 88? A continuous random variable, X, is known to be normally distributed with a mean of 15 and a standard deviation of 5. Determine the range between which approximately: a. 68% of the values lie b. 95% of the values lie c. 99.7% of the values lie. A normal probability density function, X, has a mean of 24 and a standard deviation of 7. Determine the approximate values for: a. P(X < 31) b. P(10 < X < 31) c. P(X > 10|X < 31). The number of pears harvested from each tree in a large orchard is normally distributed with a mean of 230 and a standard deviation of 25. Determine the approximate probability that the number of pears harvested from a randomly selected tree is: a. less than 280 b. between 180 and 280 c. is greater than 180, given that less than 280 pears were harvested. The annual rainfall in a particular area of Australia, X mm, is known to be normally distributed with a mean of 305 mm and a standard deviation of 50 mm. a. Calculate the approximate value of P(205 < X < 355). b. Determine k such that P(X < k) ≈ 0.025. c. Determine h such that P(X < h) ≈ 0.0015. A normally distributed probability density function is given by 2 5 − 1 (5(x−1)) f (x) = √ e 2 , x ∈ R. 2𝜋

Calculate Var(X ( 2 )), giving your answer correct to 2 decimal places. b. Calculate E X , giving your answer correct to 2 decimal places. c. Determine: i. E(2X + 3) ii. E ((X + 1)(2X − 3)) 16. A continuous random variable, X, is normally distributed with a mean of 72.5 and a standard deviation of 8.4. Determine the approximate values for: a. P(64.1 < X < 89.3) b. P(X < 55.7) c. P(X > 47.3 | X < 55.7) d. m such that P(X > m) ≈ 0.16. a.

456 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

12.3 Standardised normal variables 12.3.1 The standard normal distribution Suppose we are comparing the results of two students who took similar Maths tests. Michelle obtained 92 on one test, for which the results were known to be normally distributed with a mean of 80 and a standard deviation of 6. Samara obtained 88 on her test, for which the results were known to be normally distributed with a mean of 78 and a standard deviation of 10. Which student was more successful? This question is very difficult to answer unless we have some common ground for a comparison. This can be achieved by using a transformed or standardised form of the normal distribution called the standard normal distribution. The variable in a standard normal distribution is always denoted by Z, so that it is immediately understood that we are dealing with the standard normal distribution. The standard normal distribution always has a mean of 0 and a standard deviation of 1, so z in the following formula indicates how many standard deviations the corresponding X-value is from the mean. To calculate the value of z, we determine the difference between the x-value and the mean, x − 𝜇. To find how many standard deviations this equals, we divide by the standard deviation, 𝜍. The result is known as the z-value or z-score. The z-value formula

Therefore, if z =

z=

x−𝜇 𝜍

x−𝜇 , 𝜇 = 0 and 𝜍 = 1, the probability density function is given by 𝜍 1 − 1 z2 f (z) = √ e 2 , z ∈ R. 2𝜋

Remember that 𝜇 ± 3𝜍 encompasses approximately 99.7% of the data, so for the standard normal curve, these figures are 0 ± 3 (× 1 = 0) ± 3. Therefore, approximately 99.7% of the data lies between −3 and 3. For Michelle: X ~ N 80, 62 f (z) x−𝜇 𝜍 92 − 80 = 6 12 = 6 =2

z=

( ) For Samara: X ~ N 78, 102

–3

–2

–1

0

1

2

3

z

x−𝜇 𝜍 88 − 78 = 10 10 = 10 =1

z=

CHAPTER 12 The normal distribution 457

Michelle’s mark lies within 2 standard deviations of the mean, so it lies in the top 2.5%, whereas Samara’s mark is 1 standard deviation from the mean, so it is in the top 16%. Hence, Michelle performed better than Samara. For the standard normal distribution, we say Z ~ N (0, 1). Let us return to the comparison between Michelle and Samara. Obviously, not all data values will lie exactly 1, 2 or 3 standard deviations from the mean. In these cases, once the z-value is obtained, the corresponding probability may be found using the cumulative normal distribution (CND) function on your graphics calculator. WORKED EXAMPLE 3

Calculate the values of the following probabilities, correct to 4 decimal places. i. P(Z < 2.5) ii. P(−1.25 ≤ Z ≤ 1.25) ( ) 2 b. X is a normally distributed random variable such that X ~ N 25, 3 . i. Calculate P(X > 27) correct to 4 decimal places. ii. Convert X to a standard normal variable, Z. a.

THINK a. i. 1.

WRITE

Sketch a graph to help understand the problem.

a. i.

f (z)

P(z < 2.5) = 0.9938 –3

Use your graphics calculator to calculate the probability. The upper limit is 2.5 and the lower limit is −∞. The mean is 0 and the standard deviation is 1. ii. 1. Sketch a graph to help understand the problem. It is important to remember that P (−1.25 ≤ Z ≤ 1.25) = P (−1.25 < Z < 1.25) since P (Z = z) = 0 for all values of z. This is because we cannot measure a continuous random variable exactly. 2.

ii.

–1

0

1

2

3

P(−1.25 < Z < 1.25) = 0.7887

2.

z

f (z)

–3

Use your graphics calculator to calculate the probability. The upper limit is 1.25 and the lower limit is −1.25. b. i. 1. Sketch a graph to help understand the problem.

–2

–2

–1

0

1

2

3

z

b. i.

25 27

458 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

x

2.

ii. 1.

2.

P(X > 27) = 0.2525

Use your graphics calculator to calculate the probability. The upper limit is ∞ and the lower limit is 27. The mean is 25 and the standard deviation is 3. Write the rule to standardise X.

ii.

a. 1. On a Calculator page,

select: MENU 6: Statistics 5: Distributions 2: Normal Cdf …

2. Complete the entry

lines as: Lower bound: −9e999 Upper bound: 2.5 𝜇: 0 𝜍: 1 Press the OK button. 3. The answer appears

on the screen.

WRITE

x −𝜇 𝜎

27 − 25 3 2 = 3

z=

Substitute the mean and standard deviation.

TI | THINK

z=

CASIO | THINK

WRITE

a. 1. On a Statistic screen,

select: DIST NORM Ncd

2. Complete the entry lines

as: Variable Lower: −1 × 1099 Upper: 2.5 𝜍: 1 𝜇: 0 Press the EXE button. 3. The answer appears on the

screen.

12.3.2 Calculating probabilities using cumulative normal distribution (CND) tables If we are working without the aid of a graphics calculator, we can also find the probability corresponding to a particular z-value by using a cumulative normal distribution (CND) table. An example of such a table is shown below. This table represents the probability of observing a result (or area) from −∞ to a particular positive value of the variable z, that is, P (Z < z). If this were to be represented graphically it would correspond to the shaded region shown in the diagram.

0

z

z

CHAPTER 12 The normal distribution 459

460 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Cumulative normal distribution table (CND table) z 0 1 2 3 4 5 0 3 4 0.0 0.5 0.504 0.508 0.512 0.516 0.5199 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.2 0.5793 0.5832 0.5871 0.591 0.5948 0.5987 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.4 0.6554 0.6591 0.6628 0.6664 0.67 0.6736 0.5 0.6915 0.695 0.6985 0.7019 0.7054 0.7088 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7 0.758 0.7611 0.7642 0.7673 0.7703 0.7734 0.8 0.7881 0.791 0.7939 0.7967 0.7995 0.8023 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 1.5 0.9332 0.9345 0.9357 0.937 0.9382 0.9394 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 6 5 0.5239 0.5636 0.6026 0.6406 0.6772 0.7123 0.7454 0.7764 0.8051 0.8315 0.8554 0.877 0.8962 0.9131 0.9279 0.9406 0.9515 0.9608 0.9686 0.975

7 6 0.5279 0.5675 0.6064 0.6443 0.6808 0.7157 0.7486 0.7793 0.8078 0.834 0.8577 0.879 0.898 0.9147 0.9292 0.9418 0.9525 0.9616 0.9693 0.9756 0.5319 0.5714 0.6103 0.648 0.6844 0.719 0.7517 0.7823 0.8106 0.8365 0.8599 0.881 0.8997 0.9162 0.9306 0.9429 0.9535 0.9625 0.9699 0.9761

8

9 8 0.5359 0.5753 0.6141 0.6517 0.6879 0.7224 0.7549 0.7852 0.8133 0.8389 0.8621 0.833 0.9015 0.9177 0.9319 0.9441 0.9545 0.9633 0.9706 0.9767

Mean differences 8 9 4 8 12 16 20 4 8 12 16 20 4 8 12 15 19 4 8 11 15 19 4 7 11 14 18 3 7 10 14 17 3 6 10 13 16 3 6 9 12 15 3 6 8 11 14 3 5 8 10 13 2 5 7 9 12 2 4 6 8 10 2 4 6 7 9 2 3 5 6 8 1 3 4 6 7 1 2 4 5 6 1 2 3 4 5 1 2 3 3 4 1 1 2 3 4 1 1 2 2 3 24 24 23 23 22 21 19 18 17 15 14 12 11 10 8 7 6 5 4 4

28 28 27 26 25 24 23 21 19 18 16 14 13 11 10 8 7 6 5 4

32 32 31 30 29 27 26 24 22 20 18 16 15 13 11 10 8 7 6 5

36 35 35 34 32 31 29 27 25 23 21 19 16 14 13 11 9 8 6 5

CHAPTER 12 The normal distribution 461

2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.2 3.4 3.5 3.6 3.7 3.8 3.9

0.9772 0.9821 0.9861 0.9893 0.9918 0.9938 0.9953 0.9965 0.9974 0.9981 0.9987 0.999 0.9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 1

0.9778 0.9826 0.9864 0.9896 0.992 0.994 0.9955 0.9966 0.9975 0.9982 0.9987 0.9991 0.9993 0.9995 0.9997 0.9998 0.9998 0.9999 0.9999 1

0.9783 0.983 0.9868 0.9898 0.9922 0.9941 0.9956 0.9967 0.9976 0.9982 0.9987 0.9991 0.9994 0.9995 0.9997 0.9998 0.9999 0.9999 0.9999 1

0.9788 0.9834 0.9871 0.9901 0.9925 0.9943 0.9957 0.9968 0.9977 0.9983 0.9988 0.9991 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1

0.9793 0.9838 0.9875 0.9904 0.9927 0.9945 0.9959 0.9969 0.9977 0.9984 0.9988 0.9992 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1

0.9798 0.9842 0.9878 0.9906 0.9929 0.9946 0.996 0.997 0.9978 0.9984 0.9989 0.9992 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1

0.9803 0.9846 0.9881 0.9909 0.9931 0.9948 0.9961 0.9971 0.9979 0.9985 0.9989 0.9992 0.9994 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1

0.9808 0.985 0.9884 0.9911 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 0.9992 0.9995 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1

0.9812 0.9854 0.9887 0.9913 0.9934 0.9951 0.9963 0.9973 0.998 0.9986 0.999 0.9993 0.9995 0.9996 0.9997 0.9998 0.9999 0.9999 0.9999 1 0.9998 0.9999 0.9999 0.9999 1

0.9817 0.9857 0.989 0.9916 0.9936 0.9952 0.9964 0.9974 0.9981 0.9986 0.999 0.9993 0.9995 0.9997

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0

2 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0

3 2 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0

3 3 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0

4 3 3 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0

4 4 3 2 2 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0

WORKED EXAMPLE 4

Using the cumulative normal distribution (CND) table, calculate the values of the following. a. P(Z < 2) b. P(Z < 0.74) c. P(Z ≤ 1.327) d. P(Z ≤ 0.5369) THINK a. 1.

Draw a diagram and shade the region required.

WRITE a.

0

Using the CND table, go down to the row containing 2.0 and move across to the column headed 0. 3. Write down the required value. b. 1. Draw a diagram and shade the region required.

2

z

2.

b.

P(Z < 2) = 0.9772

0 0.74

Using the CND table, go down to the row containing 0.7 and move across to the column headed 4. 3. Write down the required value. c. 1. Draw a diagram and shade the region required.

z

2.

c.

P(Z < 0.74) = 0.7703

0

Using the CND table, go down to the row containing 1.3 and move across to the column headed 2. Write down the corresponding value. 3. Move across again to the mean difference column headed 7. Add the value from this final column into the last two digits of the previous answer. Note: The value of 11 really represents 0.0011. 4. Write down the required value.

1.327

z

2.

Round off the z-value to 3 decimal places since this is the limit of the CND table. 2. Draw a diagram and shade the region required.

d. 1.

P(Z ≤ 1.327) = 0.9066 + 0.0011 = 0.9077

d.

P(Z ≤ 1.327) = 0.9077

P (Z ≤ 0.5369) = P (Z ≤ 0.537)

0 0.537

462 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

z

Using the CND table, go down to the row containing 0.5 and move across to the column headed 3. Write down the corresponding value. 4. Move across again to the mean difference column headed 7. Add the value from this final column onto the last two digits of the previous answer. Note: The value of 24 really represents 0.0024. 5. Write down the required value. 3.

P(Z ≤ 0.5369) = P(Z ≤ 0.537) P(Z ≤ 0.537) = 0.7019 + 0.0024 = 0.7043 P(Z ≤ 0.537) = 0.7043

As mentioned previously, the CND table represents the probability of observing a result (or area) from −∞ to a particular positive value of the variable z, that is, P(Z < z). However, it does not directly allow us to observe a result from a particular positive value of the variable z to +∞, that is, P(Z > z). This problem can easily be solved by drawing a diagram of the situation, such as the one shown, and using the fact that the area between the graph and the horizontal axis is 1. From the graph, it can be seen that P(Z > z) + P(Z < z) = 1. Transposing the equation, we obtain P(Z > z) = 1 − P(Z < z).

0

z

z

WORKED EXAMPLE 5

Using the cumulative normal distribution (CND) table, calculate the values of the following. a. P(Z < 3.2) b. P(Z ≥ 2.3741) THINK a. 1.

Draw a diagram and shade the region required.

Write down the rule for obtaining P(Z > z). 3. Use the CND table to obtain the value required. 4. Evaluate. b. 1. Draw a diagram and shade the region required.

WRITE a.

P(Z > 3.2) = 1 − P(Z < 3.2) = 1 − 0.9993 = 0.0007 0

2.

b.

Round off the z-value to 3 decimal places since this is the limit of the CND table. Write down the rule for obtaining P(Z ≥ z). 3. Use the CND table to obtain the value required. 4. Evaluate.

z

P(Z ≥ 2.3741) = P(Z ≥ 2.374) = 1 − P(Z < 2.374) = 1 − (0.9911 + 0.0001) = 1 − 0.9912 = −0.0088 0

2.

3.2

2.374 z

CHAPTER 12 The normal distribution 463

So far we have observed probabilities relating to the positive values of z, that is, values of z greater than the mean. Now we will investigate probabilities associated with negative values of z, that is, values of z less than the mean.

–z

0

z

0

z

z

From the figure above, and using the fact that the curve is symmetrical, it can be seen that P(Z < −z) = P(Z > z) = 1 − P(Z < z) (from previous calculations)

and the figure below clearly shows that P(Z > −z) = P(Z < z).

–z

0

z

0

z

z

WORKED EXAMPLE 6

Using the cumulative normal distribution (CND) table, calculate the values of the following. a. P(Z < −1.23) b. P(Z ≥ −0.728) THINK a. 1.

2.

Draw a diagram and shade the region required.

WRITE a.

Write down the rule for obtaining P(Z < −z).

Use the CND table to obtain the value required. 4. Evaluate. b. 1. Draw a diagram and shade the region required.

P(Z < −1.23) = P(Z > 1.23) = 1 − P(Z < 1.23) = 1 − 0.8907 = 0.1093 –1.23 0

3.

b.

P(Z ≥ −0.728) = P(Z < 0.728) = 0.7642 + 0.0024 = 0.7666 –0.728 0

Write down the rule. 3. Use the CND table to obtain the value required. 4. Evaluate. 2.

z

z

Note: The inequality signs > and ≥ can be interchanged, as they give the same probability. 464 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

It is also important to be able to determine the probability of z falling between two values, say a and b. As with all of these types of problems, a diagram is essential as it allows us to see the situation clearly and hence solve the problem. Consider the equation P(a < Z < b) = P(z < b) − P(z < a). The following figure clearly demonstrates this situation. y

y

y

–

= 0 b

a

x

a

0 b

x

a

0

x

y

y Common region is cancelled out

a

0 b

x

a

0 b

x

WORKED EXAMPLE 7

Using the cumulative normal distribution (CND) table, calculate the values of the following. a. P(1.5 < Z < 2.32) b. P(−2.02 ≤ Z ≤ 1.59) c. P(−0.235 < Z < −0.108) THINK a. 1.

2.

Draw a diagram and shade the region required.

WRITE a.

P(1.5 < Z < 2.32) = P(Z < 2.32) − P(Z < 1.5) = 0.9898 − 0.9332 = 0.0566

Write down the rule required.

Use the CND table to obtain the value required and evaluate. b. 1. Draw a diagram and shade the region required. 3.

0 1.5 2.32

z

0 1.59

z

b.

–2.02

CHAPTER 12 The normal distribution 465

2.

3.

Write down the rule for obtaining P(a ≤ Z ≤ b).

Using the symmetry of the graph, obtain P(z < −2.02).

Use the CND table to obtain the value required and evaluate. c. 1. Draw a diagram and shade the region required. 4.

P(−2.02 ≤ Z ≤ 1.59) = P(Z < 1.59) − P(Z < −2.02) = P(Z < 1.59) − P(Z > 2.02) = P(Z < 1.59) − [1 − P(Z < 2.02)] = 0.9441 − [1 − 0.9783] = 0.9441 − 0.0217 = 0.9224

P(−0.235 ≤ Z ≤ −0.108) = P(Z < −0.108) − P(Z < −0.235) = P(Z > 0.108) − P(Z > 0.235) = [1 − P(Z < 0.108)] − [1 − P(Z < 0.235)] = [1 − (0.5398 + 0.0032)] − [1 − (0.5910 + 0.0019)] = [1 − 0.5430] − [1 − 0.5929] = 0.4570 − 0.4071 = 0.0499 –0.235

2.

Write down the rule.

Using the symmetry of the graph, obtain P(z < −0.108) and P(Z < −0.235). 4. Use the CND table to obtain the values required and evaluate. 3.

0 –0.108

z

Resources Interactivities: Calculation of probabilities (int-6440) The standard normal distribution (int-6441)

Units 3 & 4

Area 7

Sequence 2

Concepts 3 & 4

The 65–95–99.7% rule Summary screen and practice questions Standardised normal variables Summary screen and practice questions

Exercise 12.3 Standardised normal variables Technology free

Calculate the values of the following probabilities correct to 4 decimal places. i. P(Z < 1.2) ii. P(−2.1 < Z < 0.8) ( ) b. X is a normally distributed random variable such that X ~ N 45, 62 . i. Calculate P(X > 37) correct to 4 decimal places. ii. Convert X to a standard normal variable, Z. 2. If Z ~ N (0, 1), determine: a. P(Z ≤ 2) b. P(Z ≤ −2) c. P(−2 < Z ≤ 2) d. P(Z > 1.95) ∪ P(Z < −1.95) 1.

WE3

a.

466 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

3.

Use the CND table in section 12.3.2 to determine the values of the following. P(Z ≤ 1) b. P(Z < 2.3) c. P(Z ≤ 1.52) d. P(Z ≤ 0.74) e. P(Z < 1.234) f. P(Z < 2.681) WE5 Use the CND table in section 12.3.2 to determine the values of the following. a. P(Z > 2) b. P(Z ≥ 1.5) c. P(Z ≥ 1.22) d. P(Z > 0.16) e. P(Z > 1.111) f. P(Z ≥ 2.632) WE6 Use the CND table in section 12.3.2 to determine the values of the following. a. P(Z ≤ −2) b. P(Z < −1.3) d. P(Z > −2.71) c. P(Z < −1.75) e. P(Z ≥ −1.139) f. P(Z > −0.642) WE7 Use the CND table in section 12.3.2 to determine the values of the following. a. P(−1.6 ≤ Z ≤ 1.4) b. P(−2.21 < Z < 0.34) c. P(−0.645 ≤ Z ≤ 0.645) d. P(−0.72 ≤ Z ≤ −0.41) X is a continuous random variable and is known to be normally distributed. a. If P(X < a) = 0.35 and P(X < b) = 0.62, determine: i. P(X > a) ii. P(a < X < b) b. If P(X < c) = 0.27 and P(X < d) = 0.56, determine: i. P(c < X < d ) ii. P(X > c|X < d ) c. A random variable, X, is normally distributed with a mean of 20 and a standard deviation of 5. i. Determine k if P(X > 32) = P(Z > k). ii. Determine n if P(X < 12) = P(Z > n). If X ~ N (20, 25), determine: a. P(X > 27) b. P(X ≥ 18) c. P(X ≤ 8) d. P(7 ≤ X ≤ 12) e. P(X < 17|X ≤ 25) f. P(X < 17|X < 𝜇) Light bulbs have a mean life of 125 hours and a standard deviation of 11 hours. Determine the probability that a randomly selected light bulb lasts: a. longer than 140 hours b. less than 100 hours c. between 100 and 140 hours. The heights jumped by Year 9 high-jump contestants follow a normal distribution with a mean jump height of 152 cm and a variance of 49 cm. Determine the probability that a competitor jumps: a. at least 159 cm b. less than 150 cm c. between 145 cm and 159 cm d. between 140 cm and 160 cm e. between 145 cm and 150 cm, given that she jumped over 140 cm. WE4

a.

4.

5.

6.

7.

8.

9.

10.

Technology active

For a particular type of laptop computer, of time, X hours, between charges of the battery is ( the2length ) normally distributed such that X ~ N 50, 15 . Calculate P(50 < X < 70). 12. Convert the variable in each of the following expressions to a standard normal variable, Z, and use it to write an equivalent expression. Use your calculator to evaluate each probability. a. P(X < 61), X ~ N (65, 9) b. P(X ≥ 110), X ~ N (98, 225) c. P(−2 < X ≤ 5), X ~ N (2, 9) 11.

CHAPTER 12 The normal distribution 467

The volume of milk in a 1-litre carton is normally distributed with a mean of 1.000 litres and a standard deviation of 0.006 litres. A randomly selected carton is known to have more than 1.004 litres. Determine the probability that it has less than 1.011 litres. 14. A radar gun is used to measure the speeds of cars on a freeway. The speeds are normally distributed with a mean of 98 km/h and a standard deviation of 6 km/h. What is the probability that a car picked at random is travelling at: a. more than 110 km/h b. less than 90 km/h c. a speed between 90 km/h and 110 km/h?

13.

Teresa has taken her pulse each day for a month after going for a brisk walk. Her pulse rate in beats per minute is known to be normally distributed with a mean of 80 beats per minute and a standard deviation of 5 beats per minute. After her most recent walk she took her pulse rate. What is the probability that her pulse rate was: a. in excess of 85 beats per minute b. equal to or less than 75 beats per minute c. between 78 and 82 beats per minute, given that it was higher than 75 beats per minute? 16. The labels on packets of sugar say the bags have a weight of 1 kg. The actual mean weight of the bags is 1.025 kg in order to minimise the number of bags that are underweight. If the weight of the bags is normally distributed with a standard deviation of 10 g, determine the percentage of bags that would be expected to weigh: a. more than 1.04 kg b. less than 996 g, the legal meaning of underweight. 15.

12.4 The inverse normal distribution 12.4.1 Using technology to determine probability Technology provides an easy way to determine a Z or X value, given a probability for a normal distribution. Suppose X is normally distributed with a mean of 32 and a standard deviation of 5. We wish to determine P(X ≤ a) = 0.72. The key information to enter into your calculator is the known probability, that is, the area under the curve. It is essential to input the correct area so that your calculator knows if you are inputting the ‘less than’ area or the ‘greater than’ area.

468 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

0.72

32 a

x

WORKED EXAMPLE 8

If X is a normally distributed random variable, determine: ( ) m given that P(X ≤ m) = 0.85, X ~ N 15.2, 1.52 ( ) 2 b. n given that P (X > n) = 0.37, X ~ N 22, 2.75 ( ) c. p given that P(37.6 − p ≤ X ≤ 37.6 + p) = 0.65, X ~ N 37.6, 122 . a.

P(X ≤ m) = 0.85, 𝜇 = 15.2, 𝜎 = 1.5 m = 16.7547 b. P(X > n) = 0.37, 𝜇 = 22, 𝜎 = 2.75 n = 22.9126

THINK

WRITE

Use the probability menus on your graphics calculator to determine the required X value. b. Use the probability menus on the graphics calculator to determine the required X value. Note: It may be a requirement to input the ‘less than’ area, so P(X < n) = 1 − 0.37 = 0.63 c. 1. Sketch a graph to visualise the problem. Due to symmetry, the probabilities either side of the upper and lower limits can be calculated.

a.

a.

c.

0.65 0.175

0.175

1 − 0.65 = 0.35 P(X < 37.6 − p) = P(X > 37.6 + p) 0.35 = 2 = 0.175 P(X < 37.6 − p) = 0.175 37.6 − p = 26.38 = 37.6 − 26.38 = 11.22 37.6 – p 37.6

2.

Determine p by determining X given that P(X < 37.6 − p) = 0.175. Note: p could also be found by using the upper limit.

TI | THINK a. 1. On a Calculator page,

select: MENU 6: Statistics 5: Distributions 3: Inverse Normal … 2. Complete the entry

lines as: Area: 0.85 𝜇:15.2 𝜍:1.5 Press the OK button.

WRITE

CASIO | THINK

37.6 + p

x

WRITE

a. 1. On a Statistic screen,

select: DIST NORM InvN.

2. Complete the entry

lines as: Data: Variable Tail: Left Area: 0.85 𝜍:1.5 𝜇:15.2 Press the EXE button.

CHAPTER 12 The normal distribution 469

3. The answer appears

on the screen.

3. The answer appears

on the screen.

12.4.2 Using a CND table to determine probabilities WORKED EXAMPLE 9

Determine the value of c in each of the following. a. P(Z < c) = 0.57 b. P(Z ≤ c) = 0.25 THINK a. 1.

Draw a diagram of the situation.

P(Z ≥ c) = 0.91

c. WRITE a.

57%

Using the CND table in section 12.3.2, look up the z-value corresponding to the given probability. Obtain the closest probability value to 0.57 (0.5675 when c = 0.17) and then go to the mean differences column headed 6 (which gives 0.0024). b. 1. Draw a diagram of the situation.

P(z < c) = 0.57 c = 0.17 + 0.006 = 0.176

2.

0 c

z

0

z

b.

25%

c

This value is not contained in the table, so we must use the symmetry of the standard normal distribution. Redraw the graph displaying the value of c on the opposite side of the mean. Call this c1 and shade the equivalent section. Note: c = −c1 3. Determine the unshaded section of the curve. 2.

4.

Use the CND table to determine the z-value of this probability.

5.

Using the symmetry of the standard normal distribution, determine the value of z. As the required z-value is to the left of the mean, it will be negative.

25%

0

c1

z

25% is shaded; therefore, 75% is unshaded. P(Z < c1 ) = 0.75 c1 = 0.67 + 0.004 c1 = 0.674 For P (Z ≤ c) = 0.25, c = −0.674.

470 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

c. 1.

Draw a diagram of the situation.

c.

91%

c 2.

Redraw the graph displaying the value of c on the opposite side of the mean. Call this c1 and shade the equivalent section. Note: c = −c1 .

z

0

91%

P(Z < c1 ) = 0.91 c1 = 1.34 0

Using the CND table, it is possible to determine the value from −∞ to the positive z-value that corresponds to a probability of 0.91. 4. Using the symmetry of the standard normal distribution, determine the value of z. As the required z-value is to the left of the mean, it will be negative.

3.

c1

z

For P(Z ≥ c) = 0.91, c = −1.34.

12.4.3 Quantiles and percentiles Quantiles and percentiles are terms that enable us to convey information about a distribution. Quantiles refer to the value below which there is a specified probability that a randomly selected value will fall. For example, to determine the 0.7 quantile of a standard normal distribution, we determine a such that P(Z < a) = 0.7. Percentiles are very similar to quantiles. For the example of P(Z < a) = 0.7, we could also be asked to determine the 70th percentile for the standard normal distribution. WORKED EXAMPLE 10

For the normally distributed variable X, the 0.15 quantile is 1.9227 and the mean is 2.7. Determine the standard deviation of the distribution. b. X is normally distributed so that the 63rd percentile is 15.896 and the standard deviation is 2.7. Determine the mean of X. a.

THINK a. 1.

Write the probability statement.

2.

Determine the corresponding standardised value, Z, by using your calculator.

3.

Write the standardised formula connecting z and x.

4.

b. 1.

Substitute the appropriate values and solve for 𝜎. Write the probability statement.

WRITE

The 0.15 quantile is 1.9227. P(X < 1.9227) = 0.15 P(Z < z) = 0.15 z = −1.0364 x−𝜇 z= 𝜎 1.9227 − 2.7 −1.0364 = 𝜎 −1.0364 = −0.7773 𝜎 = 0.75 b. The 63rd percentile is 15.896. P(X < 15.896) = 0.63 a.

CHAPTER 12 The normal distribution 471

2.

Calculate the corresponding standardised value, Z, by using your calculator.

3.

Write the standardised formula connecting z and x.

4.

Substitute in the appropriate values and solve for 𝜇.

Units 3 & 4

Area 7

Sequence 2

P(Z < z) = 0.63 z = 0.3319 x−𝜇 z= 𝜎 15.896 − 𝜇 0.3319 = 2.7 0.8960 = 15.896 − 𝜇 𝜇 = 15

Concept 5

The inverse normal distribution Summary screen and practice questions

Exercise 12.4 The inverse normal distribution Technology free 1.

2. 3.

4.

5.

Use the CND table in section 12.3.2 to determine the value of c in each of the following. a. P(Z < c) = 0.9 b. P(Z < c) = 0.6 c. P(Z ≤ c) = 0.3 d. P(Z < c) = 0.2 e. P(Z ≥ c) = 0.8 f. P(Z > c) = 0.54 Use the CND table to determine the value of Z in the following. a. P(Z < z) = 0.39 b. P(Z ≥ z) = 0.15 c. P(−z < Z < z) = 0.28 Let X ~ N (22, 25). Calculate k if: a. P(22 − k ≤ X ≤ 22 + k) = 0.7 b. P(22 − k < X < 22 + k) = 0.24 c. P(X < k|X < 23) = 0.32 ( ) If X ~ N 37.5, 8.622 , calculate a correct to 2 decimal places such that: a. P(X < a) = 0.72 b. P(X > a) = 0.32 c. P(37.5 − a < X < 37.5 + a) = 0.88 For a standard normal distribution, calculate: a. the 0.57 quantile b. the 63rd percentile. WE9

Technology active 6.

Calculate the value of(a, correct ) to 2 decimal places, if X is normally distributed and: P(X ≤ a) = 0.16, X ~ N 41, 6.72 ( ) 2 b. P(X > a) = 0.21, X ~ N 12.5, 2.7 ( ) c. P(15 − a ≤ X ≤ 15 + a) = 0.32, X ~ N 15, 42 Calculate the values of m and n if X is normally distributed and P(m ≤ X ≤ n) = 0.92 when 𝜇 = 27.3 and 𝜍 = 8.2. The specified interval is symmetrical about the mean. WE10 X is distributed normally with a mean of 112, and the 42nd percentile is 108.87. Calculate the standard deviation of the distribution, correct to 1 decimal place. ( ) X is a normally distributed random variable such that X ~ N 𝜇, 4.452 . If the 0.11 quantile is 32.142, calculate the value of 𝜇, correct to 1 decimal place. If X is distributed normally with 𝜇 = 43.5 and 𝜍 = 9.7, calculate: a. the 0.73 quantile b. the 24th percentile. WE8

a.

7. 8. 9. 10.

472 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

11. 12. 13. 14. 15. 16. 17. 18.

X is distributed normally with a standard deviation of 5.67, and P(X > 20.952) = 0.09. Calculate the mean of X, giving your answer correct to 2 decimal places. X is distributed normally with a standard deviation of 3.5, and P(X < 23.96) = 0.28. Calculate the mean (for X, rounded to the nearest whole number. ) X ~ N 115, 𝜍 2 and the 76th percentile is 122.42. Calculate the value of 𝜍, giving your answer correct to 1 decimal place. X is distributed normally with 𝜇 = 41 and P(X > 55.9636) = 0.11. Calculate 𝜍, giving your answer correct to 1 decimal place. X is distributed normally such that P(X < 33.711) = 0.36 and P(X < 34.10) = 0.42. Calculate the mean and the standard deviation of X, giving your answers correct to 1 decimal place. X is distributed normally such that P(X > 18.376) = 0.31 and the 45th percentile is 15.15. Calculate 𝜇 and 𝜍 for X, giving your answers correct to 1 decimal place. X is normally distributed with a mean of 𝜇 and a standard deviation of 3. If 35% of X-values are at least 27, calculate the mean. The time taken for Grade 4 students to complete a small jigsaw puzzle follows a normal distribution with a standard deviation of 30 seconds. If 70% of Grade 4 students complete the puzzle in 4 minutes or less, calculate the mean completion time for Grade 4 students.

12.5 Applications of the normal distribution Application problems involving the normal distribution cover a wide range of topics. Such questions will not only incorporate theory associated with the normal distribution but may also include other areas of probability you have previously studied. WORKED EXAMPLE 11

The amount of instant porridge oats in packets packed by a particular machine is normally distributed with a mean of 𝜇 grams and a standard deviation of 6 grams. The advertised weight of a packet is 500 grams. a. Calculate the proportion of packets that will be underweight (less than 500 grams) when 𝜇 = 505 grams. b. Calculate the value of 𝜇 required to ensure that only 1% of packets are underweight. c. As a check on the setting of the machine, a random sample of 5 boxes is chosen and the setting is changed if more than one of them is underweight. Calculate the probability that the setting on the machine is changed when 𝜇 = 505 grams.

CHAPTER 12 The normal distribution 473

THINK

WRITE

Rewrite the information in the question using appropriate notation. 2. Use your graphics calculator to calculate P(X < 500). b. 1. State the known probability. 2. Calculate the corresponding standardised value, Z, by using a graphics calculator or CND table.

a.

a. 1.

3. 4.

Write the standardised formula connecting z and x.

Substitute the appropriate values and solve for 𝜇.

The wording of the question (sample of 5 boxes) indicates that this is now a binomial distribution. Rewrite the information in the question using appropriate notation. 2. Using a graphics calculator or CND table, calculate the probability.

c. 1.

Units 3 & 4

Area 7

Sequence 2

X is the amount of instant porridge ( ) oats in a packet and X ~ N 505, 62 . P(X < 500) = 0.2023

P(X < 500) = 0.01 P(Z < z) = 0.01 z = −2.3263 x−𝜇 z= 𝜎 500 − 𝜇 −2.3263 = 6 −13.9581 = 500 − 𝜇 𝜇 = 513.96 g c. Let Y = the number of underweight packets. Y ~ Bi(5, 0.2023) b.

P(Y > 1) = 1 − Pr(Y ≤ 1) = 1 − 0.7325 = 0.2674

Concept 6

Applications of the normal distribution Summary screen and practice questions

Exercise 12.5 Applications of the normal distribution Technology free 1.

WE11 Packages of butter with a stated weight of 500 g have an actual weight of W g, which is normally distributed with a mean of 508 g. a. If the standard deviation of W is 3.0 g, calculate: i. the proportion of packages that weigh less than 500 g ii. the weight that is exceeded by 99% of the packages. b. If the probability that a package weighs less than 500 g is not to exceed 0.01, calculate the maximum allowable standard deviation of W.

474 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

2.

Chocolate Surprise is a toy that is packed inside an egg-shaped chocolate. A certain manufacturer provides four different types of Chocolate Surprise toy — a car, an aeroplane, a ring and a doll — in the proportions given in the table. Toy

Proportion

Car

6k2 + 2k

Aeroplane Ring Doll

3k2 + 2k k2 + 2k 3k

Show that k must be a solution to the equation 10k2 + 9k − 1 = 0. Calculate the value of k. In response to customer demand, the settings on the machine that produce Chocolate Surprise have been changed so that 25% of all Chocolate Surprises produced contain rings. A sample of 8 Chocolate Surprises is randomly selected from a very large number produced by the machine. c. What is the expected number of Chocolate Surprises in the sample that contain rings? Give your answer correct to the nearest whole number. d. What is the probability, correct to 4 decimal places, that this sample has exactly 2 Chocolate Surprises that contain rings? 3. A particular brand of car speedometer was tested for accuracy. The error measured is known to be normally distributed with a mean of 0 km/h and a standard deviation of 0.76 km/h. Speedometers are considered unacceptable if the error is more than 1.5 km/h. Calculate the proportion of speedometers that are unacceptable. a. b.

4.

The heights of adult males in Perth can be taken as normally distributed with a mean of 174 cm and a standard deviation of 8 cm. Suppose the Western Australia Police Force accepts recruits only if they are at least 180 cm tall. a. What percentage of Perth adult males satisfy the height requirement for the Western Australia Police Force? b. What minimum height, to the nearest centimetre, would the Western Australia Police Force have to accept if it wanted a quarter of the Perth adult male population to satisfy the height requirement?

CHAPTER 12 The normal distribution 475

Technology active

Farmer David grows avocados on a farm on Mount Tamborine, Queensland. The average weight of his avocados is known to be normally distributed with a mean weight of 410 g and a standard deviation of 20 g. i. Calculate the probability that an avocado chosen at random weighs less than 360 g. ii. Calculate the probability that an avocado that weighs less than 360 g weighs more than 340 g. b. Farmer Jane grows avocados on a farm next to farmer David’s. If Y represents the average weight of Jane’s avocados, the weights of which are also normally distributed where P(Y < 400) = 0.4207 and P(Y > 415) = 0.3446, calculate the mean and standard deviation of the weights of Jane’s avocados. Give answers correct to the nearest integer. 6. A manufacturer produces metal rods whose lengths are normally distributed with a mean of 145.0 cm and a standard deviation of 1.4 cm. a. Calculate the probability, correct to 4 decimal places, that a randomly selected metal rod is longer than 146.5 cm. b. A metal rod has a size fault if its length is not within d cm either side of the mean. The probability of a metal rod having a size fault is 0.15. Calculate the value of d, giving your answer correct to 1 decimal place. c. A random sample of 12 metal rods is taken from a crate containing a very large number of metal rods. Calculate the probability that there are exactly 2 metal rods with a size fault, giving your answer correct to 4 decimal places. d. The sales manager is considering what price, $x, to sell each of the metal rods for, whether they are good or have some kind of fault. The materials cost is $5 per rod. The metal rods are sorted into three bins. The staff know that 15% of the manufactured rods have a size fault and another 17% have some other fault. The profit, $Y, is a random variable whose probability distribution is shown in the following table.

5. a.

Bin

Description

A

Good metal rods that are sold for x dollars each

B

Metal rods with a size fault — these are not sold but recycled.

C

Metal rods with other faults — these are sold at a discount of $3 each.

Profit ($y) x−5 0

x−8

P(Y = y) a 0.15 0.17

Calculate the value of a, correct to 2 decimal places. Calculate the mean of Y in terms of x. iii. Hence or otherwise, calculate, correct to the nearest cent, the selling price of good rods so that the mean profit is 0. iv. The metal rods are stored in the bins until a large number is ready to be sold. What proportion of the rods ready to be sold are good rods? 7. A company sells two different products, X and Y, for $5.00 and $6.50 respectively. Regular markets exist for both products, with sales being normally distributed and averaging 2500 units (standard deviation 700) and 3000 units (standard deviation 550) respectively each week. It is company policy that if in any one week the sales for a particular product fall below half the average, that product is advertised as a ‘special’ for the following week. i. ii.

476 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Calculate the probability, correct to 4 decimal places, that product X will be advertised as a ‘special’ next week. b. Calculate the probability, correct to 4 decimal places, that product Y will be advertised as a ‘special’ next week. c. Calculate the probability, correct to 4 decimal places, that both products will be advertised as a ‘special’ next week. d. If 40% of the company’s product is product X and 60% is product Y, calculate the probability that: i. one product is a ‘special’ ii. if one product is advertised as ‘special’, then it is product X. 8. The height of plants sold at a garden nursery supplier are normally distributed with a mean of 18 cm and a standard deviation of 5 cm. a. Complete the following table by calculating the proportions for each of the three plant sizes, correct to 4 decimal places. a.

Description of plant

Plant size (cm)

Cost in $

Small

Less than 10 cm

2.00

Medium

10–25 cm

3.50

Large

Greater than 25 cm

5.00

Proportion of plants

Calculate the expected cost, to the nearest dollar, for 150 plants chosen at random from the garden nursery. 9. A fruit grower produces peaches whose weights are normally distributed with a mean of 185 g and a standard deviation of 20 g. Peaches whose weights exceed 205 g are sold to the cannery, yielding a profit of 60c per peach. Peaches whose weights are between 165 g and 205 g are sold to wholesale markets at a profit of 45 cents per peach. Peaches whose weights are less than 165 g are sold for jam at a profit of 30c per peach. a. Calculate the percentage of peaches sold to the canneries. b. Calculate the percentage of peaches sold to the wholesale markets. c. Calculate the mean profit per peach. 10. The Lewin Tennis Ball Company makes tennis balls whose diameters are distributed normally with a mean of 70 mm and a standard deviation of 1.5 mm. The tennis balls are packed and sold in cylindrical tins that each hold five tennis balls. A tennis ball fits in the tin if its diameter is less than 71.5 mm. a. What is the probability, correct to 4 decimal places, that a randomly chosen tennis ball produced by the Lewin company fits into the tin? b.

CHAPTER 12 The normal distribution 477

The Lewin management would like each ball produced to have a diameter between 68.6 mm and 71.4 mm. What is the probability, correct to 4 decimal places, that a randomly chosen tennis ball produced by the Lewin company is in this range? c. A tin of five balls is selected at random. What is the probability, correct to 4 decimal places, that at least one ball has a diameter outside the range of 68.6 mm to 71.4 mm? d. Lewin management wants engineers to change the manufacturing process so that 99.5% of all balls produced have a diameter between 68.6 mm and 71.4 mm. The mean is to stay at 70 mm but the standard deviation is to be changed. What should the new standard deviation be, correct to 4 decimal places? 11. The Apache Orchard grows a very juicy apple called the Fuji apple. Fuji apples are picked and then sorted by diameter in three categories: • small — diameter less than 60 mm • jumbo — the largest 15% of the apples • standard — all other apples. Diameters of Fuji apples are found to be normally distributed with a mean of 71 mm and a standard deviation of 12 mm. a. A particular apple is the largest possible whose diameter lies within 2 standard deviations of the mean. What is the diameter? Give your answer correct to the nearest millimetre. b. Calculate, correct to 4 decimal places, the probability that a Fuji apple, selected at random, has a diameter less than 85 mm. c. What percentage of apples (to the nearest 1 per cent) is sorted into the small category? d. Calculate, correct to the nearest millimetre, the minimum diameter of a jumbo Fuji. e. An apple is selected at random from a bin of jumbo apples. What is the probability, correct to 4 decimal places, that it has a diameter greater than 100 mm? f. The Apache Orchard receives the following prices for Fuji apples: • small — 12 cents each • standard — 15 cents each • jumbo — 25 cents each. What is the expected income, correct to the nearest dollar, for a container of 2500 unsorted apples? g. Some apples are selected before sorting and are packed into bags of six to be sold at the front gate of the orchard. Calculate the probability, correct to 4 decimal places, that one of these bags contains at least two jumbo apples. 12. A brand of disinfectant is sold in two sizes: standard and large. For each size, the contents, in litres, of a randomly chosen bottle is normally distributed with a mean and standard deviation as shown in the following table. b.

Bottle size

Mean

Standard deviation

Standard

0.765 L

0.007 L

Large

1.015 L

0.009 L

Calculate the probability, correct to 4 decimal places, that a randomly chosen standard bottle contains less than 0.75 L. b. Calculate the probability that a box of 12 randomly chosen large bottles contains at least 4 bottles whose contents are each less than 1 L. 13. Amalie is gathering data on two particular species of yellow butterflies: the lemon emigrant and the yellow emigrant. These two species can be very difficult to tell apart. Both species are equally likely to be caught in a particular area of Australia. One technique for telling them apart is by measuring the lengths of their antennae. For the lemon emigrant, the antennae are distributed normally with a mean of 22 mm and a standard deviation of 1.5 mm. a.

478 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Calculate the probability, correct to 4 decimal places, that a randomly chosen lemon emigrant butterfly will have antennae which are shorter than 18 mm. b. Amalie knows that 8% of the yellow emigrants have antennae that are shorter than 15.5 mm, and 8% of yellow emigrant butterflies have antennae that are longer than 22.5 mm. Assuming that the antenna lengths are normally distributed, calculate the mean and standard deviation of the antenna length of yellow emigrant butterflies, giving your answers correct to the nearest 0.1 mm. c. In the region where Amalie is hunting for yellow butterflies, 45% of the yellow butterflies are lemon emigrants and 55% are lemon emigrants. Calculate the probability, correct to 4 decimal places, that a random sample of 12 butterflies from the region will contain 5 yellow emigrant butterflies. 14. The daily error (in seconds) of a particular brand of clock is known to be normally distributed. Only those clocks with an error of less than 3 seconds are acceptable. a. Calculate the mean and standard deviation of the distribution of error if 2.5% of the clocks are rejected for losing time and 2.5% of the clocks are rejected for gaining time. b. Determine the probability that fewer than 2 clocks are rejected in a batch of 12 such clocks. a.

12.6 Review: exam practice A summary of this chapter is available in the Resources section of your eBookPLUS at www.jacplus.com.au. Simple familiar 1. The diagram shows two normal distributions with means 𝜇1 and 𝜇2 and variances 𝜍12 and 𝜍22 . Blue: X1 ~ N (𝜇1 , 𝜍12 )

2. 3.

4. 5. 6.

Pink: X2 ~ N (𝜇2 , 𝜍22 ) Which of the two distributions has: a. the higher mean x b. the smaller standard deviation? If X ~ N (20, 25), calculate: a. P(X > 27) b. P(X ≥ 18) c. P(7 ≤ X ≤ 12) d. P(X < 17|X ≤ 25) The variable X is normally distributed with mean 𝜇 = 9 and standard deviation 𝜍 = 3. Standardise the following X-values. a. X = 10 b. X = 7.5 c. X = 12.4 X is normally distributed with a mean of 10 and a standard deviation of 2. Calculate x1 if: a. P (X ≤ x1 ) = 0.72 b. P (X < x1 ) = 0.4 Calculate the value of c given that P (−c ≤ Z ≤ c) = 0.38. Tennis balls are dropped from a height of 2 m. The rebound height of the balls is normally distributed with a mean of 1.4 m and a standard deviation of 0.1 m. What is the probability that a ball rebounds more than 1.25 m?

CHAPTER 12 The normal distribution 479

7.

The results by Justine in Chemistry, Mathematical Methods and Physics are shown in the table. The marks, X, the mean, 𝜇, and standard deviation, 𝜍, for each examination are given. Mark, X

Mean, 𝜇

Standard deviation, 𝜎

Chemistry

72

68

5

Mathematical Methods

75

69

7

Physics

68

61

8

Subject

Standardised mark, Z

Complete the table by calculating Justine’s standardised mark for each subject and use this to determine in which subject she did best when compared to her peers. 8. If Z ~ N (0, 1), identify the 0.35 quantile. 9. The results of a state-wide Science exam are normally distributed. If 68% of entrants scored between 8 and 12 points on the exam, what are the approximate values of the mean and the standard deviation? 10. The heights of Year 9 students are known to be normally distributed with a mean of 160 cm and a standard deviation of 8 cm. a. How tall is Theo if he is taller than 95% of Year 9 students? b. How tall is Luisa if she is shorter than 80% of Year 9 students? ( ) 11. If X ~ N 20, 𝜍 2 and P (X ≥ 19) = 0.7, calculate the standard deviation, 𝜍. 12. Describe the relationship between the mean, mode and median of a standard normal distribution. Complex familiar 13. Peter has a strawberry farm in Stanthorpe. The average length of a strawberry is normally distributed with a mean of 3.5 cm and a standard deviation of 0.8 cm. Strawberries that are longer than 4.5 cm are sold to a restaurant supplier for $6.50 per kilogram. Strawberries that are between 2.5 cm and 4.5 cm long are sold to a supermarket supplier for $4.50 per kilogram, and strawberries that are less than 2.5 cm long are sold to a jam manufacturer for $1.75 per kilogram. a. Calculate the percentage of strawberries that are sold to the supermarket supplier. b. Calculate the percentage of strawberries that are sold to the jam manufacturer. c. Calculate the mean profit for a kilogram of strawberries. 14. Jing Jing scored 85 on the mathematics section of a scholarship examination, the results of which were known to be normally distributed with a mean of 72 and a standard deviation of 9. Rani scored 18 on the mathematics section of a similar examination, the results of which were normally distributed with a mean of 15 and a standard deviation of 4. Assuming that both tests measure the same kind of ability, which student has the higher score? 15. X is a normally distributed variable for which P(X < 47) = 0.3694 and P(X > 56) = 0.3385. Calculate the mean and standard deviation of X. 16. The lengths of fish caught in a certain river follow a normal distribution, with mean 32 cm and standard deviation 4 cm. Fish that are less than 27 cm long are considered to be undersized and must be returned to the river. Calculate the expected number of fish that a fisherman could take home if he catches 20 fish in one afternoon and follows the rules for undersized fish.

480 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Complex unfamiliar 17. The lengths of certain sunflower stems follow a normal distribution with a mean of 75 cm and a standard deviation of 8 cm. Stems are measured and awarded grades depending on their lengths. The top 10% receive an A grade, the next 10% a B grade and the third 10% a C grade. Give a range of sunflower stem lengths to 2 decimal places for which: a. an A grade is awarded b. a B grade is awarded c. a C grade is awarded. 18. Under certain circumstances, a random variable X that follows a binomial distribution with n trials and a probability of success p can be approximated to a normal distribution. The mean of this normal distribution approximation is defined by 𝜇 = np, the variance 𝜍 2 = npq, and standard deviation √ 𝜍 = npq. Use a normal approximation to determine the probability that an archery contestant will score at least 240 bullseyes out of 500 shots given that she hits the bullseye 49% of the time during practice. 19. P(a < X < b) = 0.52 and the specified interval is symmetrical about the mean. If X is normally distributed with a mean of 42.5 and a standard deviation of 10.3, calculate P(X > a|X < b). 20. X is distributed normally such that P(X < 39.9161) = 0.579 and P(X > 38.2491) = 0.4798. Calculate the mean and the standard deviation of X, giving your answers correct to 2 decimal places.

Units 3 & 4

Sit exam

CHAPTER 12 The normal distribution 481

Answers 12 The normal distribution Exercise 12.2 The normal distribution 1. a. 𝜇 = 2; 𝜍 = 3 b.

8.

f (x) 1 (2, — 3 2π )

1 — 3 2π

9.

2. a. 𝜇 = 3, 𝜍 = 1 b.

–7

–4

–1 0

2

5

8

11

10. 11. 12. 13. 14. 15.

x

(3, )

16.

−∞

x+4

dx = 0.9999 ≈ 1 10 10 2𝜋 f (x) ≥ 0 for all values of x, and the area under the curve is 1. Therefore, this function is a probability density function. 2 a. 𝜇 = 2; 𝜍 = = 0.4 5 104 b. = 4.16 25 c. i. 10 ii. 20.8 a. i. 100 and 140 ii. 80 and 160 iii. 60 and 180 b. i. 0.025 ii. 0.0015 0.15% a. 10 and 20 b. 5 and 25 c. 0 and 30 a. 0.84 b. 0.815 c. 0.9702 a. 0.975 b. 0.95 c. 0.9744 a. 0.815 b. 205 c. 155 a. 0.04 b. 1.04 c. i. 5 ii. −1.92 a. 0.815 b. 0.025 c. 0.94 d. 80.9 d.

∫∞

1 √

e

−

1 2

(

)2

Exercise 12.3 Standardised normal variables 3. a. 0.025 4. 0.0015 5. 𝜇 = −2

b. 0.0015

6. a. 𝜇 = 1; 𝜍 = 0.3 or

1. a. i. 0.8849

c. 0.84

b. i. 0.9088

3

10 10 b. Dilation of factor parallel to the y-axis, dilation of 3 3 parallel to the x-axis, translation of 1 unit in factor 10 the positive x-direction c.

f (x) 10 (1, — 3 2π)

10 — 3 2π

0.005 0

0.1

7. a. 𝜇 = −4; 𝜍 = 10 b. Dilation of factor

0.4

0.7

1

1.3

ii. 0.7703

1.6

1.9

x

2. a. 0.9772 3. a. 0.8413 e. 0.8914 4. a. 0.0228 e. 0.1333 5. a. 0.0228 e. 0.8727 6. a. 0.8644 7. a. i. 0.65 b. i. 0.29 c. i. 2.4 8. a. 0.0808 e. 0.3260 9. a. 0.0863 10. a. 0.1587 e. 0.2391 11. 0.4088 12. a. 0.0912 13. 0.8676 14. a. 0.0228 15. a. 0.1587 16. a. 6.68%

0.0228 0.9893 0.9963 0.0668 0.0043 0.0968 0.7395 0.6195 ii. 0.27 ii. 0.5179 ii. 1.6 b. 0.6554 f. 0.5486 b. 0.0115 b. 0.3874

b. b. f. b. f. b. f. b.

ii. −

4

3 c. 0.9545 c. 0.9357

d. 0.0512 d. 0.7703

c. 0.1112

d. 0.4364

c. 0.0401

d. 0.9966

c. 0.4810

d. 0.1051

c. 0.0082

d. 0.0501

c. 0.9022 c. 0.6826

d. 0.8302

b. 0.2119

c. 0.7501

b. 0.0912 b. 0.1587 b. 0.19%

c. 0.8860 c. 0.3695

Exercise 12.4 The inverse normal distribution 1

from the x-axis, dilation of factor

10 10 from the y-axis, translation of 4 units in the negative x-direction c. i. 100 ii. 116

1. a. e. 2. a. 3. a. 4. a. 5. a. 6. a.

1.282 −0.842 −0.2793 5.18 42.52 0.1764 34.34

482 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

b. f. b. b. b. b. b.

0.253 −0.100 1.0364 1.525 41.53 0.3319 14.68

c. −0.524 c. 0.3585 c. 17.525 c. 13.40 c. 1.65

d. −0.842

7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

m = 12.9444; n = 41.6556 15.5 37.6 a. 49.4443 b. 36.6489 13.35 26 10.5 12.2 𝜇 = 34.6; 𝜍 = 2.5 𝜇 = 15.8; 𝜍 = 5.2 25.844 3 min 44 s

9. a. 15.87% b. 68.27% 10. a. 0.8413 b. 0.6494 11. a. 95 mm b. 0.8783 e. 0.052 f. $399 12. a. 0.0161 b. 0.0019 13. a. 0.0038 b. 𝜇 = 19.0 mm; 𝜍 = 2.5 mm c. 0.2225 14. a. 𝜇 = 0; 𝜍 = 1.5306 1. a. Pink 2. a. 0.0808

3. 4. 5. 6.

7.

8.

4. 5. 6. 7.

10k2 + 9k − 1 = 0

8. 9. 10.

10 c. 2 d. 0.3115 0.0484 or 4.84% a. 22.66% b. 179 cm a. i. 0.0062 ii. 0.9625 b. 𝜇 = 405; 𝜍 = 25 a. 0.1420 b. 2.0 c. 0.2924 d. i. 0.68 ii. 0.85x − 4.76 iii. $ 5.60 iv. 80% a. 0.0371 b. 0.0032 c. 0.0001 d. i. 0.0167 ii. 0.8856 a. *See the table at the bottom of the page. b. $531

*8. a.

d. 0.4987 d. 83 mm

b. 0.8816

11. 12. 13. 14. 15. 16. 17.

18. 19. 20.

= 0.3̇

b. Pink b. 0.6554

c. 1.13̇

c. 0.0501

d. 0.3260

b. −0.5 3 a. 11.166 b. 9.494 0.496 0.9332 Chemistry: 0.8; Mathematical Methods: 0.857; Physics: 0.875. Justine did best in Physics. −0.385 𝜇 = 10; 𝜍 = 2 a. Theo is 173.16 cm tall. b. Luisa is 153.26 cm tall. 1.908 Mean = median = mode a. 10.56% b. 10.56% c. $4.42/kg Jing Jing did better. 𝜇 = 51; 𝜍 = 12 17 a. A grade for stems longer than 85.25 cm b. B grade for stem lengths between 81.73 cm and 85.25 cm c. C grade for stem lengths between 79.19 cm and 81.73 cm 0.6725 0.6842 𝜇 = 37.68; 𝜍 = 11.21

3. a.

1. a. i. 0.0038 ii. 501.0210 g b. 3.4389 g 2 2 2 2. a. 3k + 2k + 6k + 2k + k + 2k + 3k = 1

1

45c 0.8845 18% 0.2236

12.6 Review: exam practice

Exercise 12.5 Applications of the normal distribution

b.

c. c. c. g.

1

Description of plant

Plant size (cm)

Cost in $

Proportion of plants

Small

Less than 10 cm

2.00

0.0548

Medium

10–25 cm

3.50

0.8644

Large

Greater than 25 cm

5.00

0.0808

CHAPTER 12 The normal distribution 483

REVISION UNIT 4 Further functions and statistics

TOPIC 4 Continuous random variables and the normal distribution • For revision of this entire topic, go to your studyON title in your bookshelf at www.jacplus.com.au. • Select Continue Studying to access hundreds of revision questions across your entire course.

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484 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland



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Relative frequency

0.14 0.12 0.1 0.08 0.06 0.04 0.02 0

0.26 0.30 0.34 0.38 0.42 0.46 0.50 0.54 0.58 0.62 0.66 0.70 0.74

1

pˆ

"T UIFTF HSBQIT TIPX UIF WBMVF PG Q EPFTOU NBUUFS 5IF EJTUSJCVUJPO PG Q˜ JT TZNNFUSJDBM BCPVU Q

803,&% &9".1-& 

$POTJEFS B QPQVMBUJPO TJ[F PG  BOE B TBNQMF TJ[F PG  *G Q   XPVME UIJT TUJMM CF B MBSHF TBNQMF *G OPU IPX CJH XPVME UIF TBNQMF OFFE UP CF 5)*/, 

*T O ‫ ܓ‬/



*T OQ ‫ ܔ‬



$BMDVMBUF B WBMVF GPS O UP NBLF B MBSHF TBNQMF CZ TPMWJOH OQ Ҳ 



$IFDL UIF PUIFS DPOEJUJPOT

83*5&

O Ҳ  BOE / Ҳ  O Ҳ  5IFSFGPSF O ‫ ܓ‬/ Q Ҳ М OQ Ҳ М Ұ  Ҳ ‫ܔ‬   5IF TBNQMF JT OPU MBSHF OQ Ҳ  МO Ҳ  O Ҳ  O Ҳ  Ұ  Ҳ  Ҳ/ OR Ҳ  Ұ М Ҳ  OR ‫ ܔ‬ " TBNQMF TJ[F PG  XPVME CF OFFEFE GPS B MBSHF TBNQMF

803,&% &9".1-& 

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83*5& B

ྺQ˜ Ҳ Q Ҳ М

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8& $POTJEFS B QPQVMBUJPO TJ[F PG  BOE B TBNQMF TJ[F PG  *G Q Ⴝ ᆡ JT UIJT B MBSHF TBNQMF *G OPU IPX CJH EPFT UIF TBNQMF OFFE UP CF  *G UIF QPQVMBUJPO TJ[F XBT   BOE Q Ⴝ ᆡ XIBU XPVME CF B MBSHF TBNQMF TJ[F  8& *G / Ⴝ  O Ⴝ  BOE Q Ⴝ ᆡ B DBMDVMBUF UIF NFBO PG UIF EJTUSJCVUJPO C DBMDVMBUF UIF TUBOEBSE EFWJBUJPO PG UIF EJTUSJCVUJPO DPSSFDU UP  EFDJNBM QMBDFT  *G / Ⴝ  O Ⴝ  BOE Q Ⴝ ᆡ B DBMDVMBUF UIF NFBO PG UIF EJTUSJCVUJPO C DBMDVMBUF UIF TUBOEBSE EFWJBUJPO PG UIF EJTUSJCVUJPO DPSSFDU UP  EFDJNBM QMBDFT



$)"15&3  4BNQMJOH BOE DPOýEFODF JOUFSWBMT 



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0.95

0.025

–1.96σpˆ

0.025

p

1.96σpˆ

803,&% &9".1-& 

5IFSF BSF  QFPQMF JO ZPVS DMBTT BOE  BSF QMBOOJOH PO CVZJOH UIFJS MVODI &TUJNBUF UIF QSPQPSUJPO PG UIF TDIPPM QPQVMBUJPO UIBU XJMM QVSDIBTF UIFJS MVODI UPEBZ %FUFSNJOF B  DPOàEFODF JOUFSWBM GPS ZPVS FTUJNBUF HJWFO [  

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83*5&

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83*5&

5P EFUFSNJOF PUIFS DPOàEFODF JOUFSWBMT XF DBO UBML JO HFOFSBM BCPVU B  Ⴜ ᆀ DPOàEFODF JOUFSWBM *O UIJT DBTF UIF UBJMT DPNCJOFE ᆀ XJMM IBWF BO BSFB PG ᆀ BT FBDI UBJM IBT BO BSFB PG  *O UIJT DBTF  ᆀ UIF [WBMVF UIBU IBT B UBJM BSFB PG JT VTFE 

1–α α – 2

α – 2 z–α σpˆ 2

p

z–α σpˆ 2

803,&% &9".1-& 

1BVM TBNQMFT  QFPQMF BOE EFUFSNJOFT UIBU  PG UIFN MJLF ESJOLJOH DPDPOVU NJML &TUJNBUF UIF QSPQPSUJPO PG UIF QPQVMBUJPO UIBU MJLFT ESJOLJOH DPDPOVU NJML %FUFSNJOF B  DPOàEFODF JOUFSWBM GPS ZPVS FTUJNBUF DPSSFDU UP  EFDJNBM QMBDF 5)*/, 

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83*5&

O Ҳ   Q˜ Ҳ  Ҳ М 'PS UIF ҩ DPOàEFODF JOUFSWBM ҩ XJMM CF JO UIF UBJMT TP Мҩ JO FBDI UBJM 5IFSFGPSF UIF BSFB VOEFS UIF OPSNBM EJTUSJCVUJPO DVSWF UP UIF MFGU PG [ JT М [ Ҳ М Q( ˜ ҭ Q)˜ [࿁Q˜ Ҳ [ ‫ٻ‬ O М Ұ М Ҳ М ‫ٻ‬  Ҳ М Q˜ ҭ [࿁Q˜ Ҳ М ҭ М Ҳ М Q˜ Ҭ [࿁Q˜ Ҳ М Ҭ М Ҳ М 8F DBO CF ҩ DPOàEFOU UIBU CFUXFFO Мҩ BOE Мҩ PG UIF QPQVMBUJPO MJLF ESJOLJOH DPDPOVU NJML

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REVISION UNIT 4 Further functions and statistics

TOPIC 5 Interval estimates for proportions • For revision of this entire topic, go to your studyON title in your bookshelf at www.jacplus.com.au. • Select Continue Studying to access hundreds of revision questions across your entire course.

• Select your course Mathematical Methods for Queensland Units 3&4 to see the entire course divided into syllabus topics. • Select the Area you are studying to navigate into the chapter level OR select Practice to answer all practice questions available for each area.

• Select Practice at the sequence level to access all questions in the sequence.

OR

• At sequence level, drill down to concept level.

• Summary screens provide revision and consolidation of key concepts. Select the next arrow to revise all concepts in the sequence and practise questions at the concept level

510 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

PRACTICE ASSESSMENT 3 Mathematical Methods: Unit 4 examination Unit Unit 4: Further functions and statistics

Topics Topic 1: Further differentiation and applications 3 Topic 2: Trigonometric functions 2 Topic 3: Discrete random variables 2 Topic 4: Continuous random variables and the normal distribution Topic 5: Interval estimates for proportions

Conditions Technique

Response type

Duration

Reading

Paper 1: Technology free Paper 2: Technology active

Short response

Paper 1: 60 minutes Paper 2: 60 minutes

5 minutes

Resources

Instructions

• • • •

• Show all working. • Write responses using a black or blue pen. • Unless otherwise instructed, give answers to two decimal places.

QCAA formula sheet Notes not permitted Paper 1: Calculator not permitted Paper 2: Non-CAS graphics calculator permitted

Criterion Foundational knowledge and problem solving *Assessment objectives 1, 2, 3, 4, 5 and 6

Marks allocated Paper 1

Marks allocated Paper 2

50

50

Result

*© State of Queensland (Queensland Curriculum & Assessment Authority), Mathematical Methods General Senior Syllabus 2019 v1.2, Brisbane. For the most up to date assessment information, please see www.qcaa.qld.edu.au/senior.

PRACTICE ASSESSMENT 3 511

Paper 1 — Technology free — total marks: 50 Part A: Simple familiar — total marks: 30 Question 1 (1 + 2 + 2 = 5 marks) A random variable, X, has the following probability distribution. x P(X = x)

1

2

3

4

5

0.1

0.2

0.2

a

0.3

a. Find the value of a. b. Calculate the expected value of X. ( ) c. Find P X = 3 | X > 2 .

Question 2 (2 marks) An A-frame house has an angle at the roof apex of 30° and two sides each 7 m long as shown. Find the exact area of the front of the house.

30º 7m

Question 3 (3 marks) In the triangle shown below, find the exact length of side x.

x 30º

12 cm 45º

512 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Question 4 (3 marks) The probability that the 7:35 am bus arrives on time is 0.8. What is the probability that the bus is on time at least once in the next 3 days?

Question 5 (3 marks) ( If f (x) = x sin(x), calculate f ″

) 3𝜋 . 2

Question 6 (2 + 2 + 2 = 6 marks) The continuous random variable X has a probability density function defined by f ( x) = {

kx2 , 0 ≤ x ≤ 2 0, elsewhere

where k is a constant. a. Find the value of k. b. Determine: i. the mean ii. the median.

PRACTICE ASSESSMENT 3 513

Question 7 (1 + 3 + 1 + 3 = 8 marks) Each year the Year 7 class from Green High School visits a theme park. One hundred of the students decided to go on the monster rollercoaster, and 10 of them complained of feeling dizzy afterwards. a. What is the value of the sample proportion? b. Write an expression for the 95% confidence interval for the likelihood of feeling dizzy. c. Write an expression for the margin of error, M, for the 95% confidence interval. d. If only 50 people had decided to go on the rollercoaster, what would be the effect on the margin of error?

514 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Part B: Complex familiar — total marks: 10 Question 8 (2 + 1 + 2 = 5 marks) X has a normal distribution with a mean of 4 and a standard deviation of 2. Find: a. P (4 ≤ X ≤ 8) b. P (X > 0) c. P (X > 0 | X < 4)

Question 9 (1 + 1 + 1 + 1 + 1 = 5 marks) An object moves in a straight line with a displacement of x cm from a fixed point, O, at any time, t seconds, where t ∈ [0, 6]. The object is decelerating at a constant rate of 2 cm/s2 . Initially, the object was 12 cm to the right of O, moving away from the point with a speed of 4 cm/s. a. Determine an expression for: i. velocity as a function of time ii. displacement as a function of time. b. When did the object come to rest? Calculate its position at this time. c. Calculate the distance travelled by the object. d. Determine the object’s average speed during the given time period.

PRACTICE ASSESSMENT 3 515

Part C: Complex unfamiliar — total marks: 10 Question 10 (10 marks) The function y =

a

, where a and b are non-zero constants, has a point of inflection at (−1, 2). Determine the x +b values of a and b, and hence find the coordinates and the nature of any stationary point(s). 2

Paper 2 — Technology active — total marks: 50 Part A: Simple familiar — total marks: 30 Question 1 (3 marks) Show that

d2 dx2

[loge (ax + b)n ] = −

a2 n (ax + b)2

.

516 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Question 2 (1 + 2 + 2 = 5 marks) A triathlon requires the competitors to run 230 m along a shore line from a point marked A directly north to a point B on the shore. From point B, they swim to point C on the edge of shore. They then cycle 230 m from point C directly to the starting point A. The bearing of C from A has been recorded as 074°T. B

Sea

Land

C

N Land A a. State the value of angle A.

b. Find the distance the competitors had to swim, correct to the nearest metre. c. Find the bearing of C from B, correct to the nearest degree.

Question 3 (2 + 1 + 1 = 4 marks) It is thought that only 40% of puppies adopted come from rescue shelters. Ten random families were surveyed about the places from which they recently adopted their puppies. Let X be the random variable that defines the number of families who adopt their puppy from a rescue shelter. Find, correct to 4 decimal places: a. the probability that at least 8 families adopt their puppy from a shelter b. the expected number of families that will adopt from a shelter c. the standard deviation of X.

Question 4 (3 marks) Wholistic Enterprises believe that 75% of people prefer their services to those of other companies. If they want their margin of error to be less than 5%, how many people do they need to survey to be 95% confident about their claim?

PRACTICE ASSESSMENT 3 517

Question 5 (3 marks) Krypton Industries are 99% certain that between 67% and 83% of people prefer their product. How many people were sampled for this level of confidence?

Question 6 (2 + 3 + 2 = 7 marks) A four-wheel-drive vehicle leaves a camp site and travels across a flat sandy plain in a direction of S65°E for a distance of 8.2 km. It then heads due south for 6.7 km to reach a waterhole. a. i. How far is the waterhole from the camp site? Give your answer to 1 decimal place. ii. What is the bearing of the waterhole from the camp site? Give your answer to the nearest degree. b. A search plane sets off to find the vehicle. It is on a course that takes it over points A and B, two locations on level ground. At a certain time, the angle of elevation from point A to the plane is 72°. From point B, the angle of elevation is 47°. If A and B are 3500 m apart, find the height of the plane off the ground, to the nearest metre.

Question 7 (2 + 3 = 5 marks) Consider the function f (x) = x4 − 8x3 + 18x2 − 8x + 2. a. Show that the point (2, 10) is a maximum turning point. b. Determine the coordinates of the point(s) of inflection, and hence determine the values of x where the function is concave up.

518 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Part B: Complex familiar — total marks: 10 Question 8 (4 marks) Farmer Jane grows avocados and has found that the average weight of her avocados is normally distributed. Jane has determined that the probability of an avocado weighing less than 400 grams is 0.4207, whereas the probability of an avocado weighing more than 420 grams is 0.2445. Determine the mean and standard deviation of the weights of Jane’s avocados. Give your answers correct to the nearest integer.

Question 9 (2 + 1 + 1 + 2 = 6 marks) The times taken for a class of Year 12 swimmers to swim 50 metres freestyle are normally distributed with a mean of 23.6 seconds and a variance of 0.36 seconds. a. Determine the approximate percentage of swimmers who can swim the 50 metres freestyle in a time between 23 seconds and 24.2 seconds. Give your answer to 1 decimal place. b. If a swimmer is chosen at random, what is the probability, correct to 4 decimal places, that they can swim the 50 metres freestyle in less than 23.7 seconds? c. Find the percentage of swimmers, to the nearest percent, who cannot swim the 50 metres freestyle in less than 23 seconds. d. What is the probability, correct to 4 decimal places, that one swimmer can swim the 50 metres freestyle in less than 23.6 seconds, given that the swimmer is randomly chosen from the group who can swim the 50 metres freestyle in less than 23.7 seconds?

PRACTICE ASSESSMENT 3 519

Part C: Complex unfamiliar — total marks: 10 Question 10 (3 + 2 + 1 + 1 + 2 + 1 = 10 marks) In a factory manufacturing sweets, candy-covered peanuts are produced from one of two machines: machine A and machine B. The time, X seconds, that machine A takes to produce a candy-covered peanut is distributed normally with a mean of 1.5 seconds and a standard deviation of 0.5 seconds. The time, Y seconds, that machine B takes to produce a candy-covered peanut has the following probability density function: 0, ⎧ y ⎪ , ⎪ f(y) = 6 ⎨1 ⎪ −0.5(y−1) , ⎪ e ⎩2

y1

a. Find, correct to 4 decimal places: i. P(1 ≤ X ≤ 3) ii. P(1 ≤ Y ≤ 3) b. Find E(Y) correct to 4 decimal places. c. Determine the probability of a candy-covered peanut taking 1 or less seconds to produce if it is made on machine A. d. Show that the probability of a candy-covered peanut produced on machine B taking 1 or less seconds to 1 . produce is 12 e. The candy-covered peanuts from machine B went into a large bin. A random sample of 20 candy-covered peanuts was chosen from this bin. Determine the probability that at most 2 of these peanuts took 1 seconds or less to produce. Give your answer correct to 4 decimal places. f. The manufacturer knows that their profits rely on producing the candy-covered peanuts in as little time as possible without compromising the product. Both machines produce excellent results. If only one machine could be used, which machine should the manufacturer use? Give a reason for your answer.

520 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Examination marks summary Question number

Simple familiar

Complex familiar

Complex unfamiliar

Topics

Paper 1 — Technology free 1

5

3

2

2

2

3

3

2

4

3

3

5

3

1

6

6

4

7

8

5

8

5

4

9

5

1

10

10

1

Paper 2 — Technology active

1

3

1

2

5

2

3

4

3

4

3

5

5

3

5

6

7

2

7

5

1

8

4

4

9

6

4

10

10

Totals

60

20

20

Percentage

60%

20%

20%

3, 4

PRACTICE ASSESSMENT 3 521

PRACTICE ASSESSMENT 4 Mathematical Methods: Units 3 & 4 examination Unit Unit 3: Further calculus

Topics Topic 1: The logarithmic function 2 Topic 2: Further differentiation and applications 2 Topic 3: Integrals

Unit Unit 4: Further functions and statistics

Topics Topic 1: Further differentiation and applications 3 Topic 2: Trigonometric functions 2 Topic 3: Discrete random variables 2 Topic 4: Continuous random variables and the normal distribution Topic 5: Interval estimates for proportions

Conditions Technique

Response Type

Duration

Reading

Paper 1: Technology free Paper 2: Technology active

Short response

Paper 1: 60 minutes Paper 2: 60 minutes

5 minutes

Resources

Instructions

• • • •

• Show all working. • Write responses using a black or blue pen. • Unless otherwise instructed, give answers to two decimal places.

QCAA formula sheet Notes not permitted Paper 1: Calculator not permitted Paper 2: Non-CAS graphics calculator permitted

Criterion Foundational knowledge and problem solving *Assessment objectives 1, 2, 3, 4, 5 and 6

Marks allocated Paper 1

Marks allocated Paper 2

50

50

Result

*© State of Queensland (Queensland Curriculum & Assessment Authority), Mathematical Methods General Senior Syllabus 2019 v1.2, Brisbane. For the most up to date assessment information, please see www.qcaa.qld.edu.au/senior.

522 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Paper 1 — Technology free — total marks: 50 Part A: Simple familiar — total marks: 30 Question 1 (1 + 2 + 1 = 4 marks) a. Determine

dy dx

if y = (4 − 3x)6 .

b. Consider the function f (x) =

ln(x) . x

i. Find f ′ (x). ii. Evaluate f ′ (e).

Question 2 (2 + 2 = 4 marks) 9

a. Evaluate

1 dx. ∫4 √x

b. If f ′ (x) = 1 − cos(2x), determine f (x) if f (𝜋) = 3.

Question 3 (1 + 2 = 3 marks) Solve the following for x. a. loge (2 − 3x) = 0 b. log10 (2x) − log10 (x − 1) = 1

Question 4 (3 + 2 = 5 marks) Consider the function f (x) = x4 − 4x3 . a. Determine the coordinates and nature of the stationary point(s). b. Sketch the function, showing all important features.

PRACTICE ASSESSMENT 4 523

Question 5 (3 marks) Two rowers set out from the same point. One rows on a bearing of N30°E for 2 km and the other rows on a bearing of S30°E for 3 km. How far apart are the two rowers?

Question 6 (1 + 2 + 3 = 6 marks) A discrete random variable, X, takes the values shown in the table, along with its probabilities. x

1

2

3

4

5

P(X = x)

k

2k

3k

4k

5k

Determine: a. the constant, k b. E(X), the expected value of X c. Var(X), the variance of X.

Question 7 (3 + 2 = 5 marks) Let X be a continuous random variable with a probability function defined by f (x) = 24x2 , 0 ≤ x ≤ a where a is a constant. Determine: a. the value of a b. P(0.2 < X < 0.4)

524 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Part B: Complex familiar — total marks: 10 Question 8 (2 + 2 = 4 marks) The depth of water, d metres, in a canal changes with the tides according to the rule ( ) 𝜋t d = 6 + 2 cos 6 where t is the time in hours after high tide. On a particular day, high tide was at 10:00 am. a. Calculate the depth of the water at low tide and when it next occurs. b. Determine the rate of change of the depth of water with respect to time at 2:00 pm.

Question 9 (2 + 1 + 3 = 6 marks) X has a normal distribution with a mean of 12 and a standard deviation of 3. Determine: a. P(12 ≤ X ≤ 21) b. P(X > 3) c. P(X > 3 |X < 12)

PRACTICE ASSESSMENT 4 525

Part C: Complex unfamiliar — total marks: 10 Question 10 (2 + 1 + 3 + 2 + 2 = 10 marks) The graph of the function f (x) =

10x 2

x +4

is shown.

y f (x) =

10x x2 + 4

0

x

a. Determine the coordinates of the stationary points. ( ) 20x x2 − 12 ″ b. Show that f (x) = ( )3 . x2 + 4 c. Determine the coordinates of any points of inflection. ( ) d. Differentiate g (x) = ln x2 + 4 and hence find the antiderivative of f (x). e. Calculate the area enclosed between the curve, the x-axis and the vertical lines through the points of inflection.

526 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Paper 2 — Technology active — total marks: 50 Part A: Simple familiar — total marks: 30 Question 1 (1 + 3 = 4 marks) Solve the following equations for x. a. 3(x+3) −

1 =0 81

b. 2 ln(x + 2) − ln(x) = ln(3) + ln(x − 1)

Question 2 (2 + 2 = 4 marks) A discrete random variable, X, has the following probability distribution. x P(x)

0

1

2

3

0.12

0.36

0.38

0.14

a. Calculate the expected value of X. b. Determine P(X = 3|X > 1).

Question 3 (3 marks) A cross-country runner starts at checkpoint A and runs for 8 km on a bearing of 050°T to reach checkpoint B. The runner then heads directly east for 10 km to checkpoint C. How far is checkpoint C from the starting point, A? Give your answer correct to 2 decimal places.

Question 4 (4 marks) For the function y = e2x sin(3x),

d2 y

can be expressed in the form e2x (a sin(3x) + b cos(3x)) where a and b are dx2 constants. Determine the values of a and b.

PRACTICE ASSESSMENT 4 527

Question 5 (4 marks) Smart Industries surveyed 150 people and believe that between 80% and 95% of their customers are satisfied with their level of service. How confident are they about that claim? Give your answer to the nearest percentage.

Question 6 (4 marks) An object, starting from rest, accelerates according to the equation a = 4 − t, where a represents acceleration in m/s2 and t is time in seconds. Calculate the distance the object has travelled in the first 10 seconds.

Question 7 (1 + 2 + 4 = 7 marks) For a large population of people, it is fair to assume that dates of birth are evenly distributed throughout the year, 1 so that the probability of a person having their birthday in a particular month is . 12 a. Find the probability that if 8 people are randomly chosen, exactly 3 will have their birthday in August. Give your answer correct to 4 decimal places. b. Find the probability that of the 8 randomly chosen people, at least 1 will have their birthday in November. Give your answer correct to 4 decimal places. c. If n people are chosen at random, find the smallest value of n so that the probability that at least 1 will have their birthday in March is greater than 0.9.

528 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Part B: Complex familiar — total marks: 10 Question 8 (1 + 3 = 4 marks) a. Show that the function f (x) = 2e−2x , x ≥ 0 is a probability density function. b. The time, T (in minutes), that an individual must wait in line to be served at the local coffee shop is defined by f(t) = {

2e−2t , t ≥ 0 0, elsewhere

where T is a continuous random variable. Determine the median waiting time in the queue, correct to 2 decimal places.

Question 9 (2 + 4 = 6 marks) ( ) dy = 8x ln(2x + 1) + 4x − 2. a. Show that if y = 4x2 − 1 ln(2x + 1), then dx 1 b. The graph of f (x) = 8x ln(2x + 1), x > − is shown. Calculate the area of the region bounded by the curve, the 2 x-axis, x = 0 and x = 1. y 10

8 6 4 2

–1.5 –0.5

0

f (x) = 8xln(2x + 1), x > –1 2 0.5

1.5

x

PRACTICE ASSESSMENT 4 529

Part C: Complex unfamiliar — total marks: 10 Question 10 (1 + 1 + 1 + 2 + 3 + 2 = 10 marks) The Apache Orchard grows a very juicy apple called the Fuji apple. Fuji apples are picked and then sorted by diameter into three categories: • small — diameter less than 60 mm • jumbo — the largest 15% of the apples • standard — all other apples. The diameters of Fuji apples are found to be normally distributed with a mean of 71 mm and a standard deviation of 12 mm. a. Calculate, correct to 4 decimal places, the probability that a Fuji apple selected at random has a diameter less than 85 mm. b. Determine the percentage of apples, to the nearest per cent, that are sorted into the small category. c. What is the minimum diameter, correct to the nearest millimetre, of a jumbo Fuji apple? d. An apple is selected at random from a bin of jumbo Fuji apples. Determine the probability, correct to 4 decimal places, that it has a diameter greater than 100 mm. e. The Apache Orchard receives the following prices for Fuji apples: • small — 12 cents each • standard — 15 cents each • jumbo — 25 cents each. Calculate the expected income, correct to the nearest dollar, for a container of 2500 unsorted apples. f. Some apples are selected before sorting and are packed into bags of 6 to be sold at the front gate of the orchard. Determine the probability, correct to 4 decimal places, that one of these bags contains at least 2 jumbo apples.

530 Jacaranda Maths Quest 12 Mathematical Methods Units 3 & 4 for Queensland

Examination marks summary Question number

Simple familiar

Complex familiar

Complex unfamiliar

Topics

Paper 1 — Technology free 1

4

3.2

2

4

3.3

3

3

3.1

4

5

3.2, 4.1

5

3

4.2

6

6

4.3

7

5

4.4

8

4

3.2, 4.1

9

6

4.4

10

10

3.2, 3.3, 4.1

Paper 2 — Technology active 1

4

3.1

2

4

4.3

3

3

4.2

4

4

3.2, 4.1

5

4

4.5

6

4

3.2, 3.3, 4.1

7

7

4.3

8

4

4.4

9

6

3.3, 4.1

10

10

Totals

60

20

20

Percentage

60%

20%

20%

4.3, 4.4

Note: 3.2 = Unit 3, Topic 2

PRACTICE ASSESSMENT 4 531

GLOSSARY antiderivative Where the function f ′ (x) is a derivative of f (x), f (x) is an antiderivative of f ′ (x). antidifferentiation the reverse process to differentiation; allows for f (x) to be found when f ′ (x) is given Bernoulli distribution a discrete probability distribution involving one Bernoulli trial Bernoulli trial a single event that has only two possible outcomes: success or failure. Each outcome has a fixed probability. binomial distribution a discrete probability distribution involving n Bernoulli trials. Each trial is an independent trial with the same probability, p. cdf see cumulative distribution function chain rule Use the chain rule to differentiate a function that is a composite function of two simpler dy dy du functions; = × . dx du dx composite function a function that can be expressed as a composition of two simpler functions concave down describes parabolas that open downwards and have a decreasing gradient concave up describes parabolas that open upwards and have an increasing gradient, also referred to as convex concavity a description of the shape of a continuous function; see concave up, concave down confidence intervals the level of confidence that a population parameter will lie in a)particular interval. The ( √ √ p(1 ̂ − p)̂ p(1 ̂ − p)̂ 95% confidence interval is given by p̂ − 1.96 , p̂ + 1.96 . n n continuous random variable a random variable that assumes quantities that can be measured, such as weights, time, height, in a given range cosine rule the extended version of Pythagoras’ theorem. The cosine rule for a triangle ABC can be expressed as a2 = b2 + c2 − 2bc cos(A). cumulative distribution function a function that describes the probability that a continuous random variable X has a value less than or equal to x; F(x) = P(x ≤ c) = b

definite integral evaluates to a real number not a function;

∫a

c

∫−∞

f (x)dx

f (x)dx = F(b) − F(a) where F′ (x) = f(x), a is

the lower terminal and b is the upper terminal dilation a linear transformation that enlarges or reduces the size of a figure by a scale factor k parallel to either axis or both axes empirical rule In a normal distribution: • approximately 68% of the population will fall within 1 standard deviation of the mean • approximately 95% of the population will fall within 2 standard deviations of the mean • approximately 99.7% of the population will fall within 3 standard deviations of the mean. This is also known as the 68–95–99.7% rule. expected value a measure of the central tendency of the probability distribution of a random variable. For a discrete random variable, X, E(X) = 𝜇 = ∑ xP(x). exponent an index or power; for the number n = ap , the base is a and the power, or index, or exponent is p frequency polygon created by joining the midpoints at the top of each bar representing an interval in a histogram fundamental theorem of integral calculus allows areas under graphs to be calculated exactly; b

∫a

f(x)dx = F(b) − F(a)

indefinite integral

532 GLOSSARY

∫

f(x)dx = F(x) + c where F′ (x) = f(x). F(x) is an antiderivative of f(x).

index form the form ax , where a is the base and x is the index (also called the logarithm, power or exponent) index laws the mathematical laws that apply to combinations of numbers written in index form indicial equation an equation in which the unknown variable is an exponent integrand the function that is to be integrated, f (x) integration (applied to calculate areas under curves) the process of forming the limiting sum of the area of a large number of strips of very small width in order to obtain the total area under the curve interquartile range the middle 50% of the data; IQR = Q3 − Q1 (the value of the upper quartile – the value of the lower quartile) interval a set of numbers described by two numbers where any number that lies between those two numbers is also included in the set kinematics the study of the motion of a particle in a straight line logarithm an index or power. If n = ax , then x = loga (n) is an equivalent statement. margin of error the distance between the endpoints of the confidence interval and the sample estimate. For √ p(1 ̂ − p)̂ a 95% confidence interval, the margin of error, M, is given by M = 1.96 . n normal distribution curve a continuous probability distribution characterised by a symmetrical bell-shape

x−𝜇 1 −1 curve. It is given by the function f(x) = √ e 2 𝜍 . 𝜎 2𝜋 parameter a characteristic of a population that stays constant during an analysis of that particular population pdf see probability density curve percentile the value below which there is a specified probability that a randomly selected value will fall. The 70th percentile means that 70% of the data lies below this value. point of inflection a point on a continuous curve where the concavity changes population size the number of members included in the total population, symbol N population the set or group from which information is required probability density curve or probability density function (pdf); a function that represents the probabilities of a given scenario for the interval x ∈ [a, b]. Two conditions must be met for a function to be classified (

)2

f(x) dx = 1 and f(x) ≥ 0 for all x ∈ [a, b]. ∫a quartiles values that divide an ordered set into four (approximately) equal parts. There are three quartiles: the first (or lower) quartile, Q1 ; the second quartile (or median), Q2 ; and the third (or upper) quartile, Q3 . random sample a sample where every member of a population has the same probability of being selected rate of change the derivative of a function evaluated at an instant reflection a transformation of a figure defined by the line of reflection where the image point is a mirror image of the pre-image point number of times an event has occurred relative frequency refers to actual data obtained; relative frequency = number of trials sample proportion the proportion of a sample size that has is a successful outcome; number in the sample with the favourable attribute p̂ = sample size sample size the number of members included in a sample, symbol n sample the group or subset within the population that information is collected from sampling distribution the distribution of the sample proportion, represented in a probability distribution table or a graph second derivative For a function f (x), the second derivative is the rate of change of the derivative with d2 y respect to x, written as 2 or f ′′(x). dx b

as a pdf:

GLOSSARY 533

self-selected sample a sample where the participants choose to participate; this is rarely representative of the entire population sine rule the rule that, in a triangle, the ratios of the side lengths to the sines of the opposite angles are a b c equal; expressed as = = sin(A) sin(B) sin(C) √ standard deviation a measure of the spread of a data set using the equation SD(X) = 𝜎 = Var(X). The larger the standard deviation, the more spread out the data. If the standard deviation is small, the data is clumped about the mean. standard normal distribution the standardised form of the normal distribution where 𝜇 = 0 and 𝜎 = 1, commonly written as Z ~ N (0, 1) statistic a characteristic of a sample. Different samples of the same population can give different results. stratified random sample a sample where subgroups within a population are represented in a similar proportion. Each member within a subgroup has the same chance of being selected. systematic sample every kth member of the population is sampled b

terminals values that indicate the range over which a definite integral is taken; in

∫a

f (x)dx, the variable a is

the lower terminal and b is the upper terminal translation a transformation of a figure where each point in the plane is moved a given distance in a horizontal or vertical direction unit circle in trigonometry, the circle x2 + y2 = 1 with the centre at the origin and a radius of 1 unit variance a measure of the spread of a data set using the equation Var(X) = 𝜎 2 = E(X2 ) − [E(X)]2

534 GLOSSARY

INDEX A

acceleration 269–71, 303, 321–2 ambiguous case of sine rule 352 antiderivatives defined 198 indefinite integral 198 notation for 198 xn , n ≠ −1 198–9 antidifferentiation defined 198 of cosine functions 208–9 of exponential functions 202–4 of logarithmic functions 205–6 of rational functions 198–201 of sine functions 208–9 area 247–8 between curves 257–62 estimating, under a curve 229–35 of a triangle 362–5 under curves 247–52 B

bell curve see normal distribution Bernoulli distributions 379 expected value 381 mean 381 random variables 380–2 standard deviation 381–2 variance 381–2 Bernoulli random variables 380–2 Bernoulli trial 379 Bernoulli, Jacob 48, 379 binomial distributions 385–91 applications of 398 mean 393–6 revision of 493 variance 393–6 binomial theorem 385 boundary angles 342 C

calculus fundamental theorem of 240–2 Cavalieri, Bonaventura 197 cdf (cumulative distribution function) 422 central measurements mean 429–31 median 431–5 percentiles 431–5 central tendency 429–31 chain rule 61, 157

composite functions 157 defined 157 proof of 157–60 change of base rule, logarithm 14 circle see unit circle composite functions 157 concave down 306 concave up 306 concavity 306 confidence intervals defined 499 margin of error 502 normal approximation to distribution of p̂ 499–502 continuous random variables modelling 410–11 probability and relative frequency 409–10 probability density function 411–15 continuous variables 408 vs. discrete variables 408 cos(𝜃) 339, 342 cosine function 339 antidifferentiation of 208–9 defined 342 differentiation of 133–4 domain and range 343 cosine ratio 339 and unit circle 109–10 cosine rule 355–6 applications of 367–9 calculate angle sizes using 357–60 calculate side lengths using 356–7 cumulative distribution functions (cdf) defined 422 cumulative normal distribution (CND) tables calculating probabilities using 459–66 determine probabilities using 470–1 curve sketching 170–4, 311–19 curves area between 257–62 D

definite integral 242, 247–52 and areas 247–52 properties of 242

degrees 341 derivatives cosine functions 130–4 as limit 50–1 sine functions 130–4 tangent functions 168 y = loge (x) 90–1 y = loge ( f (x)) 91–3 differential calculus 156 differentiation applications of 170–82 curve sketching 170–4 of exponential functions 61–3 kinematics 178–82 maximum and minimum problems 178–82 product rule 162 quotient rule 165–6 rate of change 178–82 tangents 170–4 dilations exponential graphs 56 logarithmic graphs 25 sine and cosine graphs 123–7 discrete random variables Bernoulli distributions 379 Bernoulli random variables 380–3 binomial distributions 385–91 expected value 381 variance 381 discrete variables 408 displacement 269–71 distribution of p̂ 491–7 binomial distributions 493 normal approximation 499–502 sample proportion 491–2 sampling distribution of p̂ 493–7 E

empirical rule 451 estimating limits value 51–2 Euler, Leonhard 48, 107 Euler’s number 8 exact values 113–14 expected value 381 exponent 2 exponent as logarithm 84 exponential functions antidifferentiation 202–4 applications of 64–70 differentiation of 61–3

INDEX 535

f (x) = ex 56–7 indicial equations with e limits 49–50 exponential graphs dilations 56 reflections 56 translations 56

58–9

F

first derivative 303 frequency polygon 410 fundamental theorem of integral calculus 241–2 G

Gaussian curve see normal distribution H

hybrid probability density functions 419–22 I

indefinite integral 198 index form 2 index laws 2–5 index as logarithm 84 indicial equations 21–3 with e 58–9 initial conditions 212–14 instantaneous rate of change 265 integral calculus fundamental theorem of 241–2 integrand 242 integration 228 applications of 265–71 by recognition 214–15 defined 198 form f (ax + b) 211–12 general formula 198 indefinite 199 initial conditions 212–14 integration functions see antidifferentiation linear motion 215–17 properties of 199–201 interquartile range 431 intersections f (x) and g(x) intersecting in interval 259–62 f (x) and g(x) not intersecting in interval 257–9 interval 408 inverse normal distribution 468–72

536 INDEX

inverse relationship, y = ex and y = loge (x) 85–6

K

kinematics

178–82

L

large sample values 494–5 left end-point rectangle method 229 Leibniz, Gottfried 156, 197 limits 49–50 derivative as limit 50–1 estimation 51–2 linear motion 215–17 logarithm 2, 84 logarithmic form, writing numbers in 7–8 logarithmic functions antidifferentiation of 205–6 applications of 32–4, 94–5 change of base rule 14 equations 12–13 index laws 2–5 laws 9–12 logarithmic form 7–8 natural logarithm 8–9, 84–5 logarithmic graphs dilations 25 reflections 25–6 translations 26–30 y = −loga (x) and y = loga (−x) 25–6 y = loga (x) 25 y = loga (x) − k and y = loga (x − h) 26–30 y = loga (x) and y = loga (mx) 25 logarithmic scales 17–18 M

margin of error 502 maximum and minimum problems 174–8 mean Bernoulli distribution 381 binomial distribution 393–6 central tendency 429–31 median 431–5 N

Napierian logarithm 84 Napier’s bones 83 natural logarithms 8–9 defined 84–5 function y = loge (x) 84–8 inverse relationship of y = ex and y = loge (x) 85–6

sketching y = loge (x) and transformations 86–8 negative angles 111 Newton, Isaac 107, 156, 197 non-stationary points 312 normal approximation to distribution of p̂ 499–502 normal distribution 450 applications of 473–4 intervals and probabilities 451–4 inverse 468–72 properties of 450 standard 457–9 normal distribution curve 450 normal ruler 1 number of degrees, in one radian 341–2 O

optimisation

322–5

p̂ distribution 491–7 binomial distributions 493 sample proportion 491–2 sampling distribution of p̂ 493–7 parameters 487–8 pdf (probability density curve) 411–15 percentiles 431–5, 471–2 periodicity of trigonometric functions 343–4 points of inflection 307, 312–13 polynomial functions 198 polynomial shapes 311–12 populations size 486 vs. samples 487–8 power 2 power as logarithm 84 probabilities using cumulative normal distribution tables 459–66, 470–1 using technology to determine 468–70 probability and relative frequency 409–10 probability density curve (pdf) 411–15 probability theory 379 product rule differentiation using 162 proof of 162 Pythagoras’ theorem 339 Pythagorean triple 340 P

Q

quantiles 471–2 quartiles 431 quotient rule differentiation using proof of 166

165

R

radians 108–9, 340–1 and unit circle 108–109, 340–1 number of degrees in one radian 341–2 random samples 488–9 range 435–8 rate of change 178–82 rational functions antiderivatives 198–9 antidifferentiation 198 integration properties 199–201 reflections exponential graphs 56 logarithmic graphs 25–6 sine and cosine graphs 123–7 regions, combining 249–52 relative frequency 409–10 Richter scale 83 right end-point rectangle method 230–1 right-angled triangle 339–40 S

sample defined 486 parameters 487–8 populations 486–8 proportion 491–2 random samples 488–9 size 486 statistics 487–8 using technology 489 sampling distribution of p̂ 493–4 standard deviation 495–7 expected value of large samples 494–5 variance 495–7 second derivative 302, 303

applications of 321–5 notation for 303 self-selected sample 488 signed areas 249–52 sin(𝜃) 339, 342 sine functions antidifferentiation of 208–9 defined 342 differentiation of 133–4 domain and range 343 first principles 132 investigating differential of y = sin(x) using technology 130–1 sine ratio 339 and unit circle 109–10 sine rule 347, 350 ambiguous case of 352 applications of 367–9 calculate side lengths using 348 derivation of 347–51 determine angle sizes using 351 slide rule 1 special angles 342 special values and unit circle 113 spread measurement 435–8 standard deviation 435–8, 495–7 Bernoulli distribution 381–3, 395 standard normal distribution 457–9 standardised normal variables 457–66 stationary points 312–13 statistics and parameters 487–8 sample 486–9 stratified random sample 488 symmetry and unit circle 110–12 systematic sample 488

tangents 169–73 terminals 242 translations exponential graphs 56 logarithmic graphs 26–30 sine and cosine graphs 123–7 trapezoidal method 231–5 triangle, area of 362–5 trigonometric functions 107 applications of 135 derivatives of 130–4 domains and ranges of 343–5 graphs of 122–8 graphs of A sin(Bx) + D and A cos(Bx) + D 123–5 graphs of y = sin(x) and y = cos(x) 122–3 graphs of y = A sin(B(x + C) + D and y = A cos(B(x + C) + D 125–8 periodicity of 343 unit circle 108–14 with technology 117–21 without technology 117–21 trigonometric ratios 339 sin, cos and tan 339 turning points 312

T

V

tan(𝜃) 339, 342 tangent functions differentiation of 168 defined 342 domain and range 343 tangent ratio 339 and unit circle 109–10

variance 435–8, 495–7 Bernoulli distribution 381–3, 395 binomial distribution 393–6 velocity 269–71, 303, 321

U

unit circle 108–109, 340 cosine ratios 109–10 exact values 113–14 and radians 108–9 sine ratios 109–10 special values 113 symmetry 110–12 tangent ratios 109–10

Z

z-value 457

INDEX 537