An Introduction To Applied Calculus For Social And Life Sciences 1485117178, 9781485117179

Table of contents :
Front cover......Page 1
Title page......Page 2
Imprint page......Page 3
Contents......Page 4
About the authors......Page 8
1.1 Number systems......Page 10
1.2 Exponents and surds......Page 13
1.3 Logarithms......Page 17
2.1 Polynomials and rational expressions......Page 22
2.2 Solving equations......Page 26
2.3 Simultaneous equations......Page 30
3.1 Intervals......Page 35
3.2 Absolute values......Page 38
3.3 Quadratic inequalities......Page 41
4.1 Algebra of functions......Page 46
4.2 Composition of functions......Page 50
4.3 Quadratic functions......Page 51
4.6 Exponential functions......Page 54
4.7 Exponential growth and decay......Page 60
4.8 Trigonometric functions......Page 65
4.9 Solving trigonometric equations......Page 77
4.10 Piece-wise functions......Page 81
4.11 Graphing of basic functions......Page 84
4.12 Intersection of graphs......Page 91
4.13 Functional models......Page 95
5.1 The limit of a function......Page 103
5.2 Limits of various types of functions......Page 107
5.3 Limits involving infinity......Page 109
5.4 One-sided limits and continuity......Page 111
5.5 Continuity......Page 113
5.6 Continuity of polynomials and rational functions......Page 116
5.7 The Intermediate Value Theorem......Page 118
6.1 The derivative......Page 123
6.2 Differentiability and continuity......Page 127
6.3 Differentiation techniques......Page 129
6.4 Product and quotient rules......Page 131
6.5 Composition of functions......Page 133
6.6 The power rule......Page 135
6.7 Higher order derivatives......Page 136
6.8 Evaluating limits using L’Hôpital’s rule......Page 137
6.9 Implicit differentiation......Page 138
6.10 Related rates......Page 141
7.1 Derivatives of logarithmic functions......Page 149
7.2 Derivatives of exponential functions......Page 151
7.3 Logarithmic differentiation......Page 152
7.4 Derivatives of trigonometric functions......Page 154
7.5 Learning curves......Page 157
7.6 Logistic curves......Page 159
8.1 Increasing and decreasing functions......Page 164
8.2 Relative extrema......Page 168
8.3 Vertical and horizontal asymptotes......Page 179
8.4 Curve sketching......Page 181
8.5 Graphs involving exponential functions......Page 186
8.6 Graphs with oblique asymptotes......Page 188
8.7 Newton’s Method......Page 191
8.8 Optimisation......Page 194
8.9 Applied optimisation......Page 195
9.1 Anti-differentiation......Page 204
9.2 Rules for integration......Page 206
9.3 Integration by substitution......Page 208
9.4 Definite integral......Page 215
9.5 Integrating rational functions......Page 223
9.6 Area between curves......Page 224
9.7 The average value of a function......Page 228
9.8 Integration by parts......Page 229
9.9 Integration of trigonometric functions......Page 234
10.1 Mathematical modelling......Page 241
10.3 Differential equations......Page 242

Citation preview

13mm

y= 1 −3 x x

2 4 x + 1 y= 2 y= 2 x −1 x −1

AN INTRODUCTION TO APPLIED

y=f(x)=ax +bx+c 2

CALCULUS AN INTRODUCTION TO APPLIED

CALCULUS AN INTRODUCTION TO APPLIED

for Social and Life Sciences

for Social and Life Sciences

An Introduction to Applied Calculus for Social and Life Sciences is designed primarily for students majoring in Social Sciences and Life Sciences. It prepares students to deal with mathematical problems which arise from real-life problems encountered in other areas of study, such as Agriculture, Forestry, Biochemistry, Biology and the Biomedical Sciences. It is also of value to anyone intending to develop foundational undergraduate calculus for the Physical Sciences. The accent throughout the book is on computational skills, critical thinking and problemsolving with no emphasis on mathematical theory and proofs. The book is carefully crafted to motivate students while stimulating understanding and proficiency. Key features: • Step-by-step problem-solving techniques followed by meticulously chosen examples to explain concepts clearly. • Definitions and key concepts highlighted by shaded boxes provide easy referencing for the student. • Comprehensive exercises at the end of each chapter. • Straightforward and concise writing style.

Lesley Wessels is a Lecturer in the Department of Mathematical Sciences at Stellenbosch University. She has taught the first-year Mathematics (Bio) module since 2005, and has been the coordinator of the course since 2007. She is currently pursuing doctoral studies in Algebraic Graph Theory.

www.jutaacademic.co.za

F NYABADZA L WESSELS

About the Authors Farai Nyabadza is currently Associate Professor in the Department of Mathematical Sciences at Stellenbosch University. He has lectured undergraduate students in introductory mathematics for Life Sciences and Social Sciences. He has also lectured and developed undergraduate courses in the applications of mathematics in the Biomedical Sciences.

FA R A I N YA BA D ZA • LE S LE Y W E S S E LS

An Introduction to Applied Calculus for Social and Life Sciences Farai Nyabadza & Lesley Wessels

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An Introduction to Applied Calculus for Social and Life Sciences Juta and Company (Pty) PO Box 14373, Lansdowne 7779, Cape Town, South Africa ©2017 Juta and Company (Pty) Ltd ISBN 978 1 48511 717 9 (Print) ISBN 978 1 48512 488 7 (WebPDF)

All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage or retrieval system, without prior permission in writing from the publisher. Subject to any applicable licensing terms and conditions in the case of electronically supplied publications, a person may engage in fair dealing with a copy of this publication for his or her personal or private use, or his or her research or private study. See section 12(1)(a) of the Copyright Act 98 of 1978. Project manager: Debbie Pienaar Proofreader: Rod Prodgers Cover designer: Jacques Nel, DragnDrop Figures drawn by: Wouter Reinders, Purple Pocket Solutions Typesetter: LaTeX Typeset in Stone Serif The author and the publisher believe on the strength of due diligence exercised that this work does not contain any material that is the subject of copyright held by another person. In the alternative, they believe that any protected pre-existing material that may be comprised in it has been used with appropriate authority or has been used in circumstances that make such use permissible under the law. Acknowledgements The authors acknowledge the support of Stellenbosch University, the institution to which they are employed. We are also indebted to Dr Gareth Boxall whose editorial comments were invaluable.

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Contents About the authors

vii

1 Algebraic Reviews 1.1 Number systems . . . . . 1.1.1 Natural numbers . 1.1.2 Integers . . . . . . 1.1.3 Real numbers . . . 1.1.4 Rational numbers . 1.1.5 Irrational numbers 1.2 Exponents and surds . . . 1.2.1 Exponents . . . . . 1.2.2 Surds . . . . . . . 1.3 Logarithms . . . . . . . . 1.3.1 Logarithmic rules . Exercises . . . . . . . . . . . .

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2 Linear and Quadratic Equations 2.1 Polynomials and rational expressions . . 2.1.1 Polynomials . . . . . . . . . . . . 2.1.2 Rational expressions . . . . . . . 2.2 Solving equations . . . . . . . . . . . . . 2.2.1 Solving by factoring . . . . . . . 2.2.2 Solving by completing the square 2.3 Simultaneous equations . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . .

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3 Inequalities and Absolute Values 3.1 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Absolute values . . . . . . . . . . . . . . . . . . . . . 3.2.1 Properties . . . . . . . . . . . . . . . . . . . . 3.2.2 Solving inequalities involving absolute values 3.3 Quadratic inequalities . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .

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An Introduction to Applied Calculus for Social and Life Sciences 4 Functions 4.1 Algebra of functions . . . . . . . . . . . . . . . . . . . 4.2 Composition of functions . . . . . . . . . . . . . . . . 4.3 Quadratic functions . . . . . . . . . . . . . . . . . . . 4.4 Power functions . . . . . . . . . . . . . . . . . . . . . . 4.5 Rational functions . . . . . . . . . . . . . . . . . . . . 4.6 Exponential functions . . . . . . . . . . . . . . . . . . 4.6.1 Logarithmic functions . . . . . . . . . . . . . . 4.7 Exponential growth and decay . . . . . . . . . . . . . 4.7.1 Half-life . . . . . . . . . . . . . . . . . . . . . . 4.7.2 Doubling time . . . . . . . . . . . . . . . . . . 4.7.3 Carbon dating . . . . . . . . . . . . . . . . . . 4.8 Trigonometric functions . . . . . . . . . . . . . . . . . 4.8.1 Angles . . . . . . . . . . . . . . . . . . . . . . . 4.8.2 The sine and cosine functions . . . . . . . . . . 4.8.3 The graphs of y = sin θ and y = cos θ . . . . . . 4.8.4 Other trigonometric functions . . . . . . . . . . 4.8.5 Special triangles . . . . . . . . . . . . . . . . . 4.8.6 Trigonometric identities . . . . . . . . . . . . . 4.8.7 Sum and difference formulas . . . . . . . . . . . 4.9 Solving trigonometric equations . . . . . . . . . . . . . 4.10 Piece-wise functions . . . . . . . . . . . . . . . . . . . 4.11 Graphing of basic functions . . . . . . . . . . . . . . . 4.12 Intersection of graphs . . . . . . . . . . . . . . . . . . 4.12.1 Distance formula . . . . . . . . . . . . . . . . . 4.12.2 Intercepts of functions . . . . . . . . . . . . . . 4.13 Functional models . . . . . . . . . . . . . . . . . . . . 4.13.1 Algorithm for setting up a mathematical model 4.13.2 Proportionality . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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37 37 41 42 45 45 45 47 51 53 54 54 56 56 58 60 60 62 63 66 68 72 75 82 85 85 86 86 88 91

5 Limits 5.1 The limit of a function . . . . . . . . . 5.2 Limits of various types of functions . . 5.3 Limits involving infinity . . . . . . . . 5.4 One-sided limits and continuity . . . . 5.4.1 One-sided limits . . . . . . . . 5.4.2 Existence of a limit . . . . . . . 5.5 Continuity . . . . . . . . . . . . . . . . 5.6 Continuity of polynomials and rational 5.7 The Intermediate Value Theorem . . . Exercises . . . . . . . . . . . . . . . . . . .

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6 Differentiation 6.1 The derivative . . . . . . . . . . . . . . . . . 6.1.1 The rate of change and the slope . . 6.1.2 Differentiation using first principles . 6.2 Differentiability and continuity . . . . . . .

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Contents 6.3 6.4

Differentiation techniques . . . . . . . . . . . . . . . . Product and quotient rules . . . . . . . . . . . . . . . 6.4.1 The product rule . . . . . . . . . . . . . . . . . 6.4.2 The quotient rule . . . . . . . . . . . . . . . . . 6.5 Composition of functions . . . . . . . . . . . . . . . . 6.6 The power rule . . . . . . . . . . . . . . . . . . . . . . 6.7 Higher order derivatives . . . . . . . . . . . . . . . . . 6.8 Evaluating limits using L’Hôpital’s rule . . . . . . . . 6.9 Implicit differentiation . . . . . . . . . . . . . . . . . . 6.10 Related rates . . . . . . . . . . . . . . . . . . . . . . . 6.10.1 An algorithm for solving related rates problems 6.10.2 Rectilinear motion . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7 Derivatives of Exponential and Logarithmic Functions 7.1 Derivatives of logarithmic functions . . . . . . . . . . . . . 7.2 Derivatives of exponential functions . . . . . . . . . . . . 7.3 Logarithmic differentiation . . . . . . . . . . . . . . . . . . 7.4 Derivatives of trigonometric functions . . . . . . . . . . . 7.5 Learning curves . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Logistic curves . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8 Applications of Differentiation 8.1 Increasing and decreasing functions . . . . . 8.2 Relative extrema . . . . . . . . . . . . . . . 8.2.1 Critical numbers and critical points 8.2.2 Concavity and points of inflection . 8.2.3 Second derivative test . . . . . . . . 8.3 Vertical and horizontal asymptotes . . . . . 8.4 Curve sketching . . . . . . . . . . . . . . . . 8.5 Graphs involving exponential functions . . . 8.6 Graphs with oblique asymptotes . . . . . . 8.7 Newton’s Method . . . . . . . . . . . . . . . 8.8 Optimisation . . . . . . . . . . . . . . . . . 8.9 Applied optimisation . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . .

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9 Integration and its Applications 9.1 Anti-differentiation . . . . . . . . . . . . . . . . 9.2 Rules for integration . . . . . . . . . . . . . . . 9.3 Integration by substitution . . . . . . . . . . . 9.4 Definite integral . . . . . . . . . . . . . . . . . . 9.4.1 Area as a limit of a sum . . . . . . . . . 9.4.2 The Fundamental Theorem of Calculus 9.4.3 Substitution in a definite integral . . . . 9.5 Integrating rational functions . . . . . . . . . . 9.6 Area between curves . . . . . . . . . . . . . . . 9.7 The average value of a function . . . . . . . .

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An Introduction to Applied Calculus for Social and Life Sciences 9.8 Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . 220 9.9 Integration of trigonometric functions . . . . . . . . . . . . . . . . . 225 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 10 Modelling with Differential Equations 10.1 Mathematical modelling . . . . . . . . 10.2 Building models . . . . . . . . . . . . . 10.3 Differential equations . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . .

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About the authors Farai Nyabadza is currently Associate Professor in the Department of Mathematical Sciences at Stellenbosch University. He has lectured undergraduate students in introductory mathematics to Life Sciences and Social Sciences. He has also lectured and developed undergraduate courses in the applications of mathematics in the Biomedical Sciences. Lesley Wessels is a Lecturer in the Department of Mathematical Sciences at Stellenbosch University. She has taught the first-year Mathematics (Bio) module since 2005, and has been the coordinator of the course since 2007. She is currently pursuing doctoral studies in Algebraic Graph Theory.

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Chapter 1

Algebraic Reviews 1.1

Number systems

The concept of numbers is fundamental in science and mathematics. A number is a mathematical entity used for counting, labelling and measuring. One immediately thinks of numerals, i.e. the symbols that represent numbers. Different types of numbers are classified into groupings called number systems. We now discuss the common number systems.

1.1.1

Natural numbers

Natural numbers are counting numbers and are denoted by N. We can write this as a set so that N = {1, 2, 3, . . .}. One can also represent these numbers pictorially on a number line as follows:

These numbers preserve some order as you read them from left to right and right to left. The number on the left is smaller than the number on the right. There are two important properties of natural numbers. First, addition of two natural numbers will always give a natural number and, secondly, the multiplication of two natural numbers gives a natural number.

1.1.2

Integers

You may have noted that natural numbers do not contain negative numbers. By including negative whole numbers, a larger set of can be defined. The set of negative numbers combined with the set of natural numbers, including 0, give the set of integer numbers denoted by Z. In set notation, Z = {. . . , −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, . . .}.

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An Introduction to Applied Calculus for Social and Life Sciences The figure below depicts integers:

Note that the addition, subtraction and multiplication of integers will always give an integer.

1.1.3

Real numbers

Real numbers consist of two sets of numbers: rational numbers and irrational numbers. Real numbers are denoted by R. Rational numbers are denoted by Q. There is no set symbol for irrational numbers, we shall denote them by R \ Q. Note that R \ Q is simply the set of all real numbers ‘minus’ the set of rational numbers.

1.1.4

Rational numbers

A rational number is any number that can be written in the form pq , where q = 0 and p and q are integers. These are simply fractions, whose set notation form is   p | p, q ∈ Z and q = 0 . Q= q It is important to note that rational numbers can also be expressed in decimal form and the decimals can be terminating or non-terminating. The following examples are fractions with terminating decimals. example 1.1.1

(a)

5 = 0.625 8

(b)

1 = 0.025 40

(c)

52 = 6.5 8

The examples on the following page are fractions with non-terminating decimals.

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Chapter 1 Algebraic Reviews example 1.1.2

(a)

2 = 0.666666 . . . 3

(b)

17 7 = 1.54545454 . . . (c) = 0.31818181 . . . 11 22

For non-terminating decimal forms, the repeated digits can be written with a bar over them, for example, 2 = 0.6666666 . . . = 0.¯6. 3 Some terminating and non-terminating decimals can be written in the form pq . The following two examples provide a framework in which non-terminating decimals can be written in the form pq . example 1.1.3 Express the following examples in the form pq . (a) 0.63

(b) 1.35¯7

Solutions (a) We let x = 0.63.

(1.1)

Because two digits are repeated, we multiply both sides of the above equation by 100 so that (1.2) 100x = 63.63. The subtraction (1.1) from (1.2) yields 99x = 63 ⇒ x =

7 63 = . 99 11

(b) We let x = 1.35¯7. We approach this problem in a slightly different way. We multiply both sides by 100 so that the repeating part is just after the decimal point. This gives 100x = 135.¯7.

(1.3)

Because only one digit is repeated, we multiply both sides of equation (1.3) by 10 so that 1000x = 1357.¯7. (1.4) The subtraction (1.3) from (1.4) yields 900x = 1222 ⇒ x =

611 1222 = . 900 450

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An Introduction to Applied Calculus for Social and Life Sciences

1.1.5

Irrational numbers

A real number that cannot be expressed √ √as a quotient of two integers is said to be irrational. Numbers such as π, 3, 7. So far the set of real numbers is the largest of the discussed number systems. Using the symbol ⊂ which is ‘subset of’ or ‘contained in’ we have the following relation N ⊂ Z ⊂ Q ⊂ R.

1.2 1.2.1

Exponents and surds Exponents

Exponents present the short form for repeated multiplication of the same number by itself. For example, we write 4 × 4 × 4 × 4 × 4 as 45 . The integer 5 is called the exponent (also called the power or index) and the number 4 is called the base. In general for any positive integer n and any real number x we have · x · . . . · x . xn = x  · x · x  n times

The integer n is called the exponent (power or index) and x is the base. We read xn as the nth power of x or x to the power of n. example 1.2.1 A number expressed as a power of some base is said to be in exponential form. Note that  3 1 (a) 53 = 5 × 5 × 5 = 125 and 14 = 14 × 14 × 14 = 64  5 1 (b) (−4)5 = −1024 and − 13 = − 243

The following rules (called the laws of exponents) will help us to simplify and work through exponents. Laws of exponents Let x, y ∈ R and m, n positive integers. Then Law 1: xn · xm = xn+m n m

Law 2: (x )

n

nm

=x

Law 7: x−n =

n n

Law 3: (xy) = x y

n n Law 4: xy = xyn (y = 0) Law 5:

xn xm

(x = 0)

Law 6: x0 = 1

1

Law 8: x n =

1 xn

√ n

(x = 0)

x (x ≥ 0)

√ m Law 9: x n = ( n x)m

= xn−m (m = n)

(x ≥ 0)

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Chapter 1 Algebraic Reviews example 1.2.2 Simplify the following (a) (−3a3 b4 )(−4ab3 )

(b)

18a3 b3 (3a2 b4 )3

Solutions (a) We can quickly do this by combining like terms in the two expressions being multiplied. (−3a3 b4 )(−4ab3 ) = (−3)(−4) × (a3 )(a) × (b4 )(b3 ) = 12a(3+1) b(4+3) = 12a4 b7 (b)

18a3 b3 = (3a2 b4 )3



18 33



a3 (a2 )3



b3 (b4 )3

=

2 18 × (a(3−6) ) × (b(3−12) ) = a−3 b−9 27 3

example 1.2.3 Simplify the following (a)

(92n )(32n+1 ) (27n )−3

(c)

(ab2 c−3 − a−2 b−1 c3 ) a2 b−1 c4

(b)

Solutions

(c)

(x2 y −2 z)−2  2 −3 × 2y (xz −2 )3



36n+1 = 3(6n+1)−(−9n) = 315n+1 . 3−9n

1 x−4 y 4 z −2 x(−4−3) y (4−6) z (−2−(−6)) = × = 3 −6 6 x z 8y 8 −7 −2 4 4 z x y z = 7 2. = 8 8x y

(92n )(32n+1 ) (34n )(32n+1 ) (a) = = (27n )−3 (33n )−3 (b)

(x2 y −2 z)−2  2 −3 · 2y (xz −2 )3

(ab2 c−3 − a−2 b−1 c3 ) ab2 c−3 a−2 b−1 c3 = 2 −1 4 − 2 −1 4 2 −1 4 a b c a b c a b c = a(1−2) b(2+1) c(−3−4) − a(−2−2) b(0) c(3−4) = a−1 b3 c−7 − a(−4) c−1 = a−4 c−7 (a3 b3 − c6 ).

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An Introduction to Applied Calculus for Social and Life Sciences example 1.2.4 Simplify 1

4

(a) 36− 2

(b) 27− 3 

− 32 1 (c) 4

(d)

3

2x 2 2 y3

2 

−9

3x 2 2 2y 3



Solutions 1

1

1 . 6

4

4

1 . 81

(a) 36− 2 = (62 )− 2 = 6−1 = (b) 27− 3 = (33 )− 3 = 3−4 =

− 32 − 32 − 3  1 1 (c) = = 2−2 2 = 23 = 8 2 4 2    3 2  −9  

 −9   9 3 4x3 2x 2 3x 2 12x3− 2 6x− 2 3x 2 (d) = = = . 2 2 4 2 4 2 y2 y3 2y 3 y3 2y 3 2y 3 + 3 example 1.2.5 Solve the following equations 3

2

(a) a 5 −n = a− 5 (a ∈ N)

(b) 4x = 32

Solutions (a) The condition a ∈ N restricts a to the set of natural numbers so that a = {1, 2, 3, . . .}. 3

2

a 5 −n = a− 5

2 3 −n=− 5 5 3 2 ⇒ n = + = 1. 5 5

⇒

(b) 4x = 32 ⇒ (22 )x = 32 ⇒ 22x = 25 ⇒ 2x = 5 5 ⇒x= . 2

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Chapter 1 Algebraic Reviews

1.2.2

Surds

Surds are irrational numbers that √ are expressed in terms of roots √ of rational √ √ numbers. For example 2, 1 + 3 and 2+2√5 are surds. Note that 36 and 3 1000 are not surds since they can be simplified to 6 and 10 respectively. Because surds appear in many mathematical solutions, it is important we learn how to simplify √ them. We begin by simplifying expressions with radicals ( ). example 1.2.6 Simplify (a)

√

2+

√

32

(b)

2+

√

50

(c)

√ √ √ √ (3 2 + 3)(2 3 − 2)

Solutions √ √ √ √ √ √ √ (a) 2 + 32 = 2 + 16 × 2 = 2 + 4 2 = 5 2. √ √ √ (b) 2 + 50 = 2 + 25 × 2 = 2 + 5 2. √ √ √ √ √ √ √ √ √ √ √ √ (c) (3 2 + 3)(2 3 − 2) = 6 2 3 − 3 2 2 + 2 3 3 − 3 2 √ √ √ = 6 6 − 6 + 6 − 6 = 5 6. At times, an expression can have a denominator that is a surd. The process of writing the same expression without a surd in the denominator is referred to as rationalisation. Let us consider the following examples to better our understanding of rationalisation. So, the idea of rationalising of the denominator is to multiply the expression both top and bottom by the ‘conjugate’. By ‘conjugate’ we mean the expression where we change the√sign in the middle of the two terms. For instance, the conjugate of √ a + b is a − b, where a and b are real numbers. example 1.2.7 Rationalize the denominator for the following expressions 5 (a) √ 3

(b)

1 √ 2+ 3

(c)

1 1 √ − √ 2+ 3 2− 3

Solutions

√ √ √ 5 5 3 5 3 3 5 . (a) √ = √ × √ = √ √ = 3 3 3 3 3 3 √ √ √ 2− 3 1 1 2− 3 √ = √ × √ (b) = 2 − 3. = 4−3 2+ 3 2+ 3 2− 3 √ √ √ √ 1 (2 − 3) − (2 + 3) −2 3 1 √ − √ = √ √ = = −2 3. (c) 4 − 3 2+ 3 2− 3 (2 + 3)(2 − 3) 7

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An Introduction to Applied Calculus for Social and Life Sciences

1.3

Logarithms

Logarithms and exponents are closely related. The exponential equation 32 = 9, read as 3 raised to the power of 2 equals 9. The question, to which power is 3 raised in order to get 9, has its answer as 2. We say then that 2 is the logarithm of 9 with base 3. It is more convenient to write this as 32 = 9 ⇔ log3 9 = 2. The left- and right-hand sides of the above expression represent the same relationship between 3, the base, 2 the exponent and 9.

Definition of a logarithm If x is a positive number, then the logarithm of x to the base b (b > 0 and b = 1) denoted logb x, is the number y such that by = x, that is, y = logb x ⇔ by = x for x > 0. example 1.3.1

Evaluate (a) log10 1000

(b) log5

1 125

(c) log2 32

Solutions (a) Let y = log10 1000 so that 10y = 1000 = 103

(b) If y = log5

⇒ y = 3. 1 125

then 5y =

1 1 = 3 = 5−3 125 5

⇒ y = −3. (c) We set y = log2 32 so that 2y = 32 = 25 ⇒ y = 5.

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Chapter 1 Algebraic Reviews example 1.3.2 Solve the following equations 1 (b) log64 16 = x (a) log4 x = 2

(c) logx 27 = 3

Solutions (a) log4 x =

1 1 ⇒ 42 = x 2 ⇒ x = 2.

log64 16 = x ⇒ 64x = 16.

(b)

So (26 )x = 24 ⇒ 6x = 4 2 ⇒ x= . 3 (c) logx 27 = 3 ⇒ x3 = 27 = 33 ⇒ x = 3.

1.3.1

Logarithmic rules

Just like the exponents, logarithms are governed by the following rules.

Rules of logarithms Let b be any logarithmic base (b > 0, and b = 1.) Then logb 1 = 0 and logb b = 1 and if a, c, x and y are positive numbers, then (i) logb x = logb y ⇔ x = y. (ii) logb (xy) = logb x + logb y. (iii) logb xp = p logb x for any p ∈ R. x (iv) logb = logb x − logb y. y (v) logb bx = x. (vi) logb a =

logc a . logc b

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An Introduction to Applied Calculus for Social and Life Sciences example 1.3.3 Use logarithmic rules to evaluate the expression: log7 (x3



1 − y2 )

Solution log7 (x3



1 − y2 )

=

1

log7 x3 [(1 − y)(1 + y)] 2 1

log7 x3 + log7 [(1 − y)(1 + y)] 2 1 = 3 log7 x + log7 [(1 − y)(1 + y)] 2 1 1 = 3 log7 x + log7 (1 − y) + log7 (1 + y). 2 2 =

example 1.3.4 Simplify using the rules of logarithms √ √ log x + 1 + log x − 1 − log(x2 − 1)2 Solution √ √ log x + 1 + log x − 1

− log(x2 − 1)2 1 1 log(x + 1) + log(x − 1) − 2 log(x2 − 1) = 2 2 1 = [log(x + 1) + log(x − 1)] − 2 log(x2 − 1) 2 1 = [log(x + 1)(x − 1)] − 2 log(x2 − 1) 2  1 = log(x2 − 1) − 2 log(x2 − 1) 2 3 = − log(x2 − 1). 2

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Chapter 1 Algebraic Reviews

Exercises

1.1 Use the terms integer, rational, irrational to classify the given numbers: 17 7 √ (c) − 9

(b) 0.125

(a)

(e)

√

(d)

2−3

(f)

(g) 0.119

(h)

√ 3 5 7 √ 3

8

−27

1.2 Given that every odd integer is of the form 2n + 1, prove that the sum of any two integers is even and that their product is odd. 1.3 Classify the following as either rational or irrational: (a) (c)

π 3 √

(b) 2.141589 11

(d)

14 99

1.4 Write the following decimals in the form pq , (a) 1.¯ 1

(b) 0.014

(c) 3.212

(d) 0.168

p, q ∈ Z:

1.5 Simplify the following: (a)

3x3 y2



3 ÷

x2 y − 3 4

2 (b)

x+4 x − x2 − 16 x2 − 4x

(c)

x−2 − y −2 x−1 + y −1

(d) 2(3x 2 + y 2 )(2x 2 − y 2 )

(e)

542x · 32x+1 36x

(f)

(g)

3(2n+1 ) − 8(2n−1 ) 2n+1 − 2n

(h)

1

1

1

1

x−2 −y −2 (xy)−2

32x (81)x 92x

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An Introduction to Applied Calculus for Social and Life Sciences 1.6 Prove that if a = bn and nn = an−1 then bn = a. 1.7 Rationalise the denominator: 3 √ x+3− x

(b) √

x+4 √ x+1+ x−1

(d)

(a) √ (c) √

x−5 √ x− 5

1 2 √ − √ 2− x 2+ x

1.8 Determine the rational numbers a and b such that:   √ √ √ √ √ √ (b) 3 + 2 2 = a + b (a) 7 + 2 10 = a + b √ √ x+ y x+y+ and find . 1.9 If m = √ √ , show that m = x−y x− y 1.10 Express √

√ √ 1 √ in the form p 5 + q 3 where p and q are rational numbers. 5+ 3

1.11 Simplify: (a) 2log2 5 + log2 3 − 2log2 15

√ (b) 4log6 36

1 (c) 3log2 x + log2 x − 2log2 (x + 1) 2 (d) log10 (x2 − 1) − log10 (x + 1) +

1 logx 10

1.12 Solve for x in the following equations: (a) logx 48 = 2

(b) 2log25 2x = log5 (1 − x)

(c) log(x + 1) = log x − log 2

2 1 (d) logx =− 3 3

1.13 Use logarithmic rules to simplify each expression: √ (b) log[ 3 x2 − x]

(a) log2 (x4 y 3 ) 

(1 + x)2 √ (c) log x





1 1 + (d) log x x3



1.14 Solve the given equation for x: (a) log2 x = 4

(b) 3x+1 = 5

(c) log3 (2x − 1) = 2

(d) log x = 13 (log 16 + 2 log 2)

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Chapter 2

Linear and Quadratic Equations 2.1

Polynomials and rational expressions

An algebraic expression is a mathematical expression built up from constants (numbers), variables, and the algebraic operations of addition, subtraction, multiplication and division. A variable is any symbol (usually letters of the alphabet) used to represent a number in an expression.

2.1.1

Polynomials

Let n be a non-negative integer and an , an−1 , . . . , a0 be real numbers with an = 0. The function f (x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 is called a polynomial function of the variable x of degree n. The real numbers an , an−1 , . . . , a0 are called the coefficients of the polynomial with an as the leading coefficient and a0 is the constant. Polynomials of specific degrees have special names. If n = 0 we have a constant function representing the polynomial. We summarise these special polynomials in the table below.

Polynomial f (x) = a0 f (x) = a1 x + a0 f (x) = a2 x2 + a1 x + a0 a0 f (x) = a3 x3 + a2 x2 + a1 x + a0 f (x) = a4 x4 + a3 x3 + a2 x2 + a1 x + a0 f (x) = a5 x5 + a4 x4 + a3 x3 + a2 x2 + a1 x + a0

Degree (n) 0 1 2 3 4 5

Name Constant function Linear function Quadratic function Cubic function Quartic function Quintic function

We first begin by looking at constant functions. 13

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An Introduction to Applied Calculus for Social and Life Sciences Constant functions (of form f (x) = b, where b ∈ R ) A polynomial of degree 0, for instance f (x) = 5, is a typical example of a constant function. The graph of a constant function is a horizontal line.

Linear functions A polynomial of degree 1, for instance f (x) = 2x − 4, represents a straight line.

The general form of the equation of a straight line is given by f (x) = mx + c, where m is the slope, and c the y-intercept of the graph. 14

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Chapter 2 Linear and Quadratic Equations The slope of a line passing through two points whose coordinates are (x0 , y0 ) and (x1 , y1 ) is given by y1 − y 0 . m= x1 − x0 Two lines (non-vertical) L1 and L2 with respective gradients m1 and m2 are said to be • parallel if and only if m1 = m2 • perpendicular if and only if m2 = − m11 . To determine the equation of a straight line one needs: • A given point and its gradient. • Two given points. • A point and a parallel or perpendicular line. example 2.1.1 Write an equation for the line with the given properties: (a) through (−1, 2) with slope

2 3

(b) through (2, 5) and (1, −2) (c) through (−2, 3) and parallel to the line x + 3y = 5

Solutions (a) Given a point and the gradient m of the line, the equation of the line is of the form y − y0 = m(x − x0 ), so that y−2=

2 2 8 (x + 1) ⇒ y = x + . 3 3 3

We can also use the general form of the equation of the line, y = mx + c, so that 2

=

⇒c = Therefore y =

2 (−1) + c 3 2 8 2+ = . 3 3

2 8 x+ . 3 3 15

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An Introduction to Applied Calculus for Social and Life Sciences (b) To determine slope m, we set m

=

y2 − y1 Δy −2 − 5 = = 7. = Δx x2 − x1 1−2

Since we now have the slope, we can use one of the points to determine the equation. The choice of the point does not really matter. Let us choose the point (1, −2). We thus have y − y0 = m(x − x0 ) ⇒ y + 2 = 7(x − 1) so that y = 7x − 9. (c) Let L1 be the line x + 3y = 5. We first need to write x + 3y = 5 in standard form so that 5 1 L1 : y = − x + . 3 3 So, if L2 is the line that we seek, then its slope must be the same as that of L1 since they are parallel. We thus have the gradient of the line L2 as m2 = − 13 . The equation of L2 is given by y − y0 = m2 (x − x0 ) ⇒ y − 3 so that y

1 = − (x + 2) 3 2 1 = − x− +3 3 3 1 7 = − x+ . 3 3

Quadratic equations An equation of the form ax2 + bx + c = 0 for a = 0,

(2.1)

is called a quadratic equation, where a, b and c are real numbers. The following examples show how polynomials can be factored and used in simplifying expressions. example 2.1.2 Factor the following polynomials: (a) y = x2 + 8x + 12

(b) y = 3x2 − x − 14

(c) y = x2 − 81 Solutions (a) y = x2 + 8x + 12 = (x + 6)(x + 2) (b) y = 3x2 − x − 14 = (3x − 7)(x + 2) (c) y = x2 − 81 = x2 − 92 = (x − 9)(x + 9) 16

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Chapter 2 Linear and Quadratic Equations

2.1.2

Rational expressions

The quotient of two polynomials is called a rational expression. Typical examples are x3 + 1 x3 − x + 1 1 4 , , and . 2 2 x x −1 x −x+1 4 In general, a rational function is of the form

p(x) , where q(x) = 0. q(x)

example 2.1.3 Write each of the following as a rational expression in lowest terms 3



x − 7x2 + 10x x x+3 −2 + (b) (a) 2 x −1 x−1 x2 + 6x + 9 x−5 Solutions x −2 x −2 + = + x2 − 1 x − 1 (x − 1)(x + 1) x − 1 −2 + x2 + x = (x − 1)(x + 1) (x − 1)(x + 2) = (x − 1)(x + 1) (x + 2) , x = 1, −1. = (x + 1) 3



x − 7x2 + 10x x+3 x(x2 − 7x + 10)(x + 3) (b) = x2 + 6x + 9 x−5 (x + 3)2 (x − 5) x(x − 5)(x − 2)(x + 3) = (x + 3)2 (x − 5) x(x − 2) , x = 5, −3. = (x + 3) (a)

2.2

Solving equations

2.2.1

Solving by factoring

Writing polynomials in terms of their factors (a process known as factorisation) is often useful in simplifying mathematical expressions. Equations involving polynomials can be solved by factorisation.

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An Introduction to Applied Calculus for Social and Life Sciences example 2.2.1 Solve the following equations (a) x2 − 3x = 10

(b)

3 11x + 10 x + − 2 =0 x + 1 2x + 3 2x + 5x + 3

Solutions x2 − 3x = 10 Therefore, x = 5 or x = −2.

(a)

x2 − 3x − 10 = 0 (x − 5)(x + 2) = 0 (b)

3 11x + 10 x + − 2 x + 1 2x + 3 2x + 5x + 3 x(2x + 3) + 3(x + 1) − (11x + 10) (2x + 3)(x + 1) 2 2x + 3x + 3x + 3 − 11x − 10 (2x + 3)(x + 1) 2x2 − 5x − 7 (2x + 3)(x + 1) (2x − 7)(x + 1) (2x + 3)(x + 1) 2x − 7 ⇒ 2x + 3 the final answer is x = 72 .

2.2.2

= 0 Therefore, x =

7 2,

x =

−3 2 ; −1.

So

=0 =0 =0 =0 = 0.

Solving by completing the square

The quadratic equation (2.1) can be solved by completing squares as follows: Consider the quadratic equation ax2 + bx + c = 0 with a = 0. • First, when completing the square, we always need to have the coefficient of x2 as 1. We then write the equation as c b x2 + x = − . a a • Second, we divide the coefficient of x by 2 and add the square of the result to both sides. This results in x2 +

b2 b b2 c x+ 2 = 2 − . 2a 4a 4a a

This makes the left-hand side a complete square, hence the term ‘completing the square’.

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Chapter 2 Linear and Quadratic Equations • Third, we write the polynomial with the square, so that x+

b 2a

2 =

b2 − 4ac 4a2

• Finally, we solve for x so that b x=− ± 2a



b2 − 4ac . 4a2

We can easily see that this gives the well-known quadratic formula for solving quadratic equations √ −b ± b2 − 4ac . x= 2a The expression under the square root sign is called the discriminant and is defined by Δ = b2 − 4ac. The discriminant is used to determine the nature of the roots or solutions of a quadratic equation, i.e. the values of x at which the graph of the quadratic function crosses the x-axis. By ‘nature’, we mean the types of the roots or the solutions, which can either be real, rational, irrational or imaginary. We have the following cases: Sign of discriminant

Nature

Possibilities If Δ is a perfect square, the roots are rational numbers If Δ is not a perfect square, the roots are irrational numbers

b2 − 4ac > 0

Two real roots

b2 − 4ac = 0

One real root

The root is repeated

b2 − 4ac < 0

Two imaginary roots

Roots are complex (This is beyond the scope of this book)

example 2.2.2 Solve the given quadratic equation by completing the square (a) 2x2 + 11x + 15 = 0

(b) x2 + 5x + 11 = 0

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An Introduction to Applied Calculus for Social and Life Sciences Solutions 15 11 x+ 2 2 2 11 11 x2 + x + 2 4

2 11 x+ 4

2 11 x+ 4

2 11 x+ 4

(a) 2x2 + 11x + 15 = 0 ⇒ x2 +

x+

=0 15 =− + 2



11 4

=

121 15 − 16 2

=

121 − 120 16

=

1 16 

2

11 =± 4

1 16 11 1 x=− ± . 4 4

5 Therefore, x = −3 or x = − 10 4 = −2. 2 2 5 5 (b) x2 + 5x + 11 = 0 ⇒ x2 + 5x + = −11 + 2 2

2 25 5 − 11 = x+ 2 4 25 − 44 = 4 19 =− . 4

Here we have no real solution because the left-hand side is always positive or zero and yet the right-hand side is negative. example 2.2.3 Factorise and simplify the following expressions: (a) 12(x + 3)5 (x − 1)3 − 8(x + 3)6 (x − 1)2

1

3

(b) x− 4 (3x + 5) + 4x 4

Solutions (a) 4(x + 3)5 (x − 1)2 [3(x − 1) − 2(x + 3)] = 4(x + 3)5 (x − 1)3 (x − 9) 1

1

(b) x− 4 [(3x + 5) + 4x] = x− 4 (7x + 5)

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Chapter 2 Linear and Quadratic Equations example 2.2.4 Determine the nature of the solutions or roots of the quadratic equations (a) −3x2 + 6x + 2 = 0 (c) 2x2 + x − 3 = 0

(b) x2 − 6x + 8 = 0 1 2 (d) x2 − x + = 0 3 9

Solutions (a) The discriminant Δ = 62 − 4(−3)(2) = 60 > 0. Since Δ > 0 and is not a perfect square, we have two distinct roots that are irrational. In fact √ 15 . x=1± 3 (b) The discriminant Δ = (−6)2 − 4(8) = 4 > 0. Since Δ > 0 and is a perfect square, we have two rational roots, x = 2 or x = 4. (c) The discriminant Δ = 1 − 4(3)(2) = −23 < 0. Since Δ < 0, we have complex roots.

2 1 2 −4 (d) The discriminant Δ = − = 0. Since Δ = 0, we have a repeated 3 9 root. In fact √ 15 . x=1± 3

2.3

Simultaneous equations

Two equations, say for instance 2x − 4y = 3 and xy = 7, can be solved for values of x and y that satisfy both equations simultaneously. In this section we shall learn how to find the solutions of various simultaneous equations. example 2.3.1 Solve the following system of equations 4x − 3y = 7 x + 3y = 3

Solution If we add the two equations, we can easily eliminate y so that 5x = 10, which gives the solution x = 2. We can then substitute x = 2 in any one of the equations. If we chose to use the second equation, we have 2 + 3y = 3 ⇒ y =

1 . 3

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An Introduction to Applied Calculus for Social and Life Sciences

We will thus have the solution x = 2 and y =

1 . 3

example 2.3.2 Solve the following system of equations x+y xy

= =

9 8

(2.2) (2.3)

Solution From (2.2) we have x = 9 − y. We can then substitute the expression for x into (2.3) so that (9 − y)y = 8 ⇒ y 2 − 9y + 8 = 0. Factorising the quadratic equation gives (y − 8)(y − 1) = 0, so that y = 1 or y = 8. Using the expression x = 9 − y, we can easily see that x = 8 or x = 1. When y = 1, x = 8 and when y = 8, x = 1. The solutions to the simultaneous equations are x = 8, y = 1 x = 1, y = 8.

example 2.3.3 If the line 3x − 5y = 8 meets the curve the coordinates of the points.

3 1 − = 4 at two points, determine x y

Solution Rewriting the equation of the curve, we have 3y − x = 4xy, and from the line 8 + 5y x= . 3 Substituting the expression for x into the equation of the curve, we have



8 + 5y 8 + 5y 3y − = 4y ⇒ 9y − 8 − 5y = 4y(8 + 5y) 3 3 ⇒ 20y 2 + 28y + 8 = 0 ⇒ 5y 2 + 7y + 2 = 0 ⇒ (5y + 2)(y + 1) = 0. 2 So y = −1 or y = − and the corresponding x values are x = 1 or x = 2, 5 respectively.

2 The coordinates are thus (1, −1) and 2, − . 5 22

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Chapter 2 Linear and Quadratic Equations

Exercises

2.1 Find the slope and y-intercept of the graph of each of the following equations: (a) 4y − x =

8 3

(b) 5y − 2x + 2 = 0

(c) 2x − 3y = 18

(d) −4x + 2y = 9

2.2 Write down the equation of the line with gradient m and passing through the given point in the form y = mx + c: (a) m = −6,

(−2, 3)

(b) m = 32 , (−3, 2)

(c) m = −2, (2, −3)

2.3 Determine whether the following lines are parallel, perpendicular or neither: (a) y = 4x + 2 and 0.25y = x +

11 4

(b) x + 2y = 3 and 3x + 6y = 4 (c) y + 1 = 2x and 0.5(y + 3) = −x (d) y = 3x + 1 and 3y = −x + 3 2.4 Find the equation of the line that is (i) parallel, (ii) perpendicular to the given equation and passes through the given point: (a) y = 6x − 12 at (0, 6)

(b)

1 y = x + 1 at (3, 1) 2

(c) y = 2x at (2, 1) 2.5 Solve the following equations: (a) 4x+1 22x = 16 (c) 4x + 4−3−x =

(b) 3x−1 22x = 48 5 16

(d) 83−4x 4x+4 = 2

2.6 Solve the following simultaneous equations: 1 x + 2y = 0 2

(a) 2x − y = −4, x + y = −2

(b) 3x − 2y = −9,

(c) x − 5y = 25, 9x − 7y = 24

(d) x − 2y = 1, x2 + y 2 = 29

(e) 2y = x + 3, x2 + y 2 − 2x + 6y − 15 = 0 (f) xy + y 2 = 5, 2x + 3y = 7 23

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An Introduction to Applied Calculus for Social and Life Sciences 2.7 Factorise the following expressions: 3

5

9(x − 2)4 (−1 − x) − (x + 1)2 (x − 2)3 (x − 2)2 (x + 1)

(a) x− 4 (x + 3) + 3x 4

(b)

(c) x4 + 5x3 − 14x2

(d) x4 − 16

(e) x4 − 81

(f) x− 3 (x − 2) + 3x 3 (x − 2)2

2

2

2.8 Write down an equation for the line with the given properties: (a) through (0, 1) and parallel to x + 2y = 3 (b) through (−1, 2) and perpendicular to 2x − y = 1 2.9 Sketch the graphs and show all x- and y-intercepts: (a) f (x) = −x2 − 1

(b) f (x) = x2 + 2x + 1

(c) f (x) = (x + 2)2 − 1

(d) f (x) = x2 + 9

2.10 Find all the real numbers x that satisfy the given equations: (a) 3x 22x = 144

(b) 42x−1 = 16

(c) 4x ( 12 )3x = 8

(d) ( 19 )1−3x = 34x

2

2.11 Solve the following equations by factorisation: (a) x(3x + 10) = 77

(b) 4x2 + 12x + 9 = 0

(c) x2 − 2 = 0

(d) 1 +

(e) 2x(4x + 15) = 27

(f) 6x2 + 7x − 3 = 0

(g)

5 4 − 2 =0 x x

3 11x + 10 x + = 2 x + 1 2x + 3 2x + 5x + 3

(h) 3x2 − x − 2 = 0 2.12 Solve the following quadratic equations by completing the square: (a) x2 + 2x − 3 = 0

(b) 2x2 + 11x + 15 = 0

(c) 6x2 + 17x − 4 = 0

(d) −3x2 + 6x − 12 = 0

5 x 2 + = (f) 9x2 + 12x + 4 = 0 2 x 2 √ √ √ (g) 3 − x − 7 + x = 16 + 2x (h) −x2 + 7x − 4 = 0

(e)

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Chapter 2 Linear and Quadratic Equations 2.13 Solve the following equations simultaneously: (a) 2x − 3y = 4 3x − 5y = 2

(b) 2x2 − y 2 = −7 2x + y = 1

(c) 2y 2 − x2 = 1 x − 2y = 3 2.14 Determine the nature of the roots or solutions of the following quadratic equations: (a) 4x2 − 7x + 3 = 0

(b) x2 + bx + b2 = 0

(c) 9x2 + 6x + 1 = 0

(d) x2 − 3x + 3 = 0

2.15 Find the values of a ∈ R for which the equation 2x2 − ax + 8 = 0 has equal roots. 2.16 Given that x2 − 2kx + 4 = 0, where k is a real number, find the value(s) of k for which (a) the equation has equal roots. (b) the equation has no roots. 2.17 One side of a rectangular vineyard is 10 m shorter than the other side. Given that the area of the vineyard is 300 m2 , what are the dimensions of the field? 2.18 Mr Anele has 130 m of wire. He intends to fence a rectangular field. A stone wall runs along one side of the field so that no fencing is required there. If the area of the field to be fenced is 4200 m2 , what will be the dimensions of the field?

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Chapter 3

Inequalities and Absolute Values 3.1

Intervals

An interval is a set of real numbers between two given numbers. The set of all numbers satisfying 1 ≤ x ≤ 2 defines an interval which contains the numbers 1 and 2 as well as all the numbers between them. The numbers 1 and 2 are called the end points of the interval. Intervals can either be open, closed or neither. An open interval does not include the end points. It is often indicated by inequality signs or parenthesis. For example, 1 < x < 3 or (1, 3) or x ∈ (1, 3) represents numbers great than 1 and less than 3 or numbers lying between 1 and 3 but not including the end points 1 and 3. A closed interval on the other hand includes the end points and is denoted by non strict inequalities (≤, ≥) or square brackets. For example, a ≤ x ≤ b (here a and b are any real numbers) or [a, b] represents numbers between the end points a and b including the end points themselves. Intervals are best described using the set builder notation. Below we have the interval notation on the left and the corresponding set builder notation on the right.

Interval notation (a, b) [a, b) (a, b] [a, b]

Set builder notation {x ∈ R|a < x < b}, an open interval, read as x is a real number such that x lies in the open interval (a, b) {x ∈ R|a ≤ x < b}, a left closed, right open interval {x ∈ R|a < x ≤ b}, a left open, right closed interval {x ∈ R|a ≤ x ≤ b}, a closed interval

It is important to note that the intervals (a, a), [a, a) and (a, a] each represent a set with no elements, i.e. an empty set, represented by the symbol ∅, and [a, a] is a set with one element a, i.e. {a}. 26

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Chapter 3 Inequalities and Absolute Values The intervals described using the set builder notation can also be represented pictorially as follows:

The intervals we have just discussed have finite end points. An end point can be infinite, so that either a = ±∞ or b = ±∞. For example, the interval (a, +∞) is the set of all numbers to the right of a, but excluding the end point a itself. On the other hand (−∞, b] is the set of all numbers to the left of b including the end point b. Such an interval is called a right closed interval. The interval (−∞, +∞) = R is the set of all real numbers. The corresponding set builder notation is shown below. Interval notation (a, +∞) [a, +∞) (−∞, b] (−∞, b)

Set builder notation {x ∈ R|x > a}, a left open interval {x ∈ R|x ≥ a}, a left closed interval {x ∈ R|x ≤ b}, a right closed interval {x ∈ R|x < b}, a right open interval

example 3.1.1 Use inequalities to describe the following interval

Solution

−1 ≤ x ≤ 2.

example 3.1.2 Use inequalities to describe the following interval

Solution

x < 2.

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An Introduction to Applied Calculus for Social and Life Sciences example 3.1.3 Use inequalities to describe the following interval

Solution

x ≤ −1 or 0 < x ≤ 2.

example 3.1.4 Solve the following inequalities (a) 3x − 3 ≤ 9

(b) − 8 < 3x − 2 ≤ 1

(c)

1 − 3x > 9. 2

Solutions (a) Solving inequalities such as these is similar to solving ordinary equations with an equal sign instead of the inequality sign. 3x − 3 ≤ 9 ⇒ 3x ≤ 9 + 3 x ≤ 12 3 x ≤ 4. The graphical solution is

and in interval notation it is x ∈ (−∞, 4]. (b) There are two ways to solve this inequality. We can solve it directly or we can separate the inequality into two inequalities and solve each inequality as was done in (a). We shall put the two ways side by side. −8 < 3x − 2 and 3x − 2 ≤ 1 −8 + 2 < 3x and 3x ≤ 1 + 2 −6 3 3 < x and x ≤ 3 −2 < x and x ≤ 1.

−8 < 3x − 2 ≤ 1 −8 + 2 < 3x ≤ 1 + 2 −6 3 3 0. One can easily see that this is the set of real numbers whose distance from the origin is less than or equal to a. So the solutions should surely lie in the interval [−a, a] or −a ≤ x ≤ a. example 3.2.2 Solve the following inequality |x − 6| ≤ 2. Solution The problem here is to find x given that the distance of x − 6 from zero is less than or equal to 2. This is equivalent to −2 ≤ x − 6 ≤ 2

⇒ ⇒

−2 + 6 ≤ x ≤ 2 + 6 4 ≤ x ≤ 8.

Graphically, the inequality defines the set of numbers whose distance from 6 is less than or equal to 2.

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Chapter 3 Inequalities and Absolute Values In simple terms, the equation |x − 6| ≤ 2 is trying to answer the question: ‘what numbers on the number line are less than or equal to 2 units from 6?’ We can easily observe that 2 units to the right of 6 is 8 and 2 units to the left of 6 is 4. Let us now solve the inequality |x| ≥ a where a > 0. One can easily see that this is the set of real numbers whose distance from the origin is great than or equal to a. So each solution should lie in the interval (−∞, −a] or the interval [a, ∞) i.e. x ∈ (−∞, −a] ∪ [a, ∞). example 3.2.3 Solve the following inequality |x − 2| ≥ 5 Solution By definition |x − 2| =

=

⎧ ⎨

x−2

if x − 2 ≥ 0

⎩ −(x − 2) if x − 2 < 0 ⎧ ⎨ x − 2 if x ≥ 2 ⎩

2 − x if x < 2.

When solving such a problem we have to consider the two cases separately, i.e. we look at the cases x ≥ 2 and x < 2. Case 1: x ≥ 2

Case 2: x < 2

|x − 2| ≥ 5

|x − 2| ≥ 5

⇒ ⇒

x−2≥5 x≥7

⇒ 2−x≥5 ⇒ −x ≥ 5 − 2 ⇒ x ≤ −3.

So the solution lies in the interval (−∞, −3] or the interval [7, ∞) i.e. x ∈ (−∞, −3]∪ [7, ∞). Note the change of sign as a result of division or multiplication by –1. Graphically, the inequality defines the set of numbers whose distance from 2 is greater than or equal to 5. example 3.2.4 Solve the following inequality 1 < |x − 2| ≤ 2

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An Introduction to Applied Calculus for Social and Life Sciences Solution There are two pairs of simultaneous inequalities that can be drawn from this problem: 1 < |x − 2| and |x − 2| ≤ 2 ⇒ (x − 2 > 1 or x − 2 < −1) and ⇒ (x > 3 or x < 1) and

−2≤x−2≤2 0≤x≤4

⇒ x ∈ (−∞, 1) ∪ (3, ∞) and x ∈ [0, 4]. Note that, in sets or intervals, the words ‘or’ and ‘and’ are interpreted as corresponding to the union and intersection respectively. Our solution is thus  x ∈ (−∞, 1) ∪ (3, ∞) ∩ [0, 4].

The solution of the two inequalities is the set of all values of x ∈ [0, 1) ∪ (3, 4].

3.3

Quadratic inequalities

Let us now consider inequalities involving quadratic polynomials. Consider the inequality x2 − 2x − 3 < 0. Through factorisation, we have (x − 3)(x + 1) < 0. There are two approaches to solving this inequality. The first includes looking at intervals and the second looks at the graph of y = x2 − 2x − 3. Let us consider the interval method first. When one looks at the expression y = (x − 3)(x + 1), one notes that the points x = 3 and x = −1 are crucial as they divide the real line into three subintervals (−∞, −1), (−1, 3) and (3, ∞). We chose a value of x in each subinterval and substitute into the expression. If the sign of the expression in a subinterval is negative then that interval contains values of x such that y = x2 − 2x − 3 < 0. If the sign of the expression in a subinterval is positive then y = x2 − 2x − 3 > 0.

We are solving x2 − 2x − 3 < 0. The solution is given by the interval in which the function y = x2 − 2x − 3 is negative. Hence, the solution is x ∈ (−1, 3). Lets us now turn to the graphical method: The graph of the function y = x2 − 2x − 3 is shown in the figure on the next page. 32

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Chapter 3 Inequalities and Absolute Values

We are looking for the solutions or values of x where the graph is negative, i.e. less than zero or below the x-axis. We can see that for x ∈ (−1, 3) the graph is below the x-axis. example 3.3.1 Solve the following inequality −x2 + 4x + 12 ≤ 0. Solution We require the solutions for which the function y = −x2 + 4x + 12 is negative. So through factorisation, we have (−x − 2)(x − 6) ≤ 0. Considering the signs of the function, we have

The solution is given by the intervals in which the function y = −x2 + 4x + 12 is negative but with the other end points included. Hence, the solution is (−∞, −2] ∪ [6, ∞). The graph on the next page shows the same result.

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An Introduction to Applied Calculus for Social and Life Sciences

example 3.3.2 Solve the following inequality |3x + 5| ≤1 |1 − x| Solution We can rewrite the inequality as follows: |3x + 5| ≤ |1 − x| Since both sides of the inequality are positive, we can square both sides without changing the inequality sign so that (3x + 5)2 ≤ (1 − x)2 . So, 9x2 + 30x + 25 ≤ 1 − 2x + x2 ⇒ 8x2 + 32x + 24 ≤ 0 ⇒ 8(x2 + 4x + 3) ≤ 0 ⇒ 8(x + 1)(x + 3) ≤ 0. We can draw the graph of the quadratic equation, where we will easily see that the graph lies below the x-axis in the interval [−3, −1]. Our solution is thus x ∈ [−3, −1]. Alternatively we can determine the signs of the function f (x) = 8(x + 1)(x + 3) in the intervals (−∞, −3], [−3, −1] and [−1, ∞).

The function is negative in the interval x ∈ [−3, −1].

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Chapter 3 Inequalities and Absolute Values

Exercises

3.1 Express each of the shown intervals using inequalities and the interval notation:

(a)

(b)

(c) 3.2 Graph the following intervals and write as a single interval: (a) (3, 10] ∩ [8, ∞)

(b) (4, 8] ∪ [5, ∞)

(c) (−∞, −3) ∩ [−3, 8)

(d) (−∞, −5] ∪ [−8, 3]

3.3 Solve the inequality and express the solution in terms of intervals whenever possible: (a) |2x + 5| < 1

(b) |2 − x| ≤ 5

(c) |3x − 7| ≥ 5

(d) | − 11 − 7x| > 6

(e)|3x + 2| ≥ |5x − 8|

(f) |2x − 1| − 2|x + 1| > 0

3.4 Solve the inequality and express the solutions graphically: (a) |1 − 2x| ≤ 9 (c)

(b) |2x − 4| ≥ 5

|x + 3| |x| = |2 − x| x

(d) |x2 − 9| < 27

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An Introduction to Applied Calculus for Social and Life Sciences 3.5 Find the set of values of x for which: (a) 2x2 + 5x − 3 > 0

(b) x2 + 4x + 3 ≥ 0

(c) 3x2 − 3x ≤ 1

(d) 1 + x − 6x2 ≥ 0

(e) 2x2 + 5x − 3 > 0

(f)

(g) x < (i)

1 ≤2 2x

4x + 8 >3 x−1

(h) −1
0 ⇒ (3 − t)(3 + t) > 0. Therefore Df = (−3, 3). example 4.1.4 Find the range of the following functions. (a) f (x) = x2

(b) f (x) =

1 x−2

(c) f (x) = x2 + x − 2

Solutions (a) We first write the equation as y = x2 and then make x the subject of the √ expression. We will have x = ± y. We then find the values of y for which we √ have real values of x. Here y is defined for only the positive values and zero.

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Chapter 4 Functions So, Rf = R\(−∞, 0) or equivalently y ∈ [0, ∞). The graph shows that the y values are always positive or zero. 1 + 2y 1 and solve for x, so that x = . Again with x−2 y rational functions, we do not want division by zero. So,

(b) We first write y =

⇒ Rf

= {y ∈ R|y = 0} or y ∈ (−∞, 0) ∪ (0, ∞).

(c) From y = x2 + x − 2, we can rewrite the equation as x2 + x − 2 − y = 0. Using the quadratic formulae or by completing the square we have  −1 ± 1 + 4(2 + y) . x = 2 We can only have real values of x if 1 + 4(2 + y) ≥ 0, y ≥ − 94 .   So, Rf = y ∈ R|y ≥ − 94 or y ∈ [− 94 , ∞).

4.2

Composition of functions

Definition of composite functions Given two functions f (u) and g(x), the composite function f (g(x)) is the function of x formed by substituting u = g(x) for u in the formula for f (u). example 4.2.1 Find the composite function f (g(x)) given that (a) f (u) =

1 , g(x) = x − 1 u2

(b) f (u) = u2 , g(x) =

1 x−1

Solutions 1 1 = . (g(x))2 (x − 1)2

2 1 1 = . (b) f (g(x)) = (g(x))2 = x−1 (x − 1)2 (a) f (g(x)) =

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An Introduction to Applied Calculus for Social and Life Sciences example 4.2.2 Find the composite function (a) f (x − 2) where f (x) = 2x2 − 3x + 1 (b) f (x − 1) where f (x) = (x + 1)5 − 3x2

Solutions (a) f (x − 2) = 2(x − 2)2 − 3(x − 2) + 1 = 2x2 − 8x + 8 − 3x + 6 + 1 = 2x2 − 11x + 15. (b) f (x − 1) = [(x − 1) + 1]5 − 3(x − 1)2 = x5 − 3x2 + 6x − 3. example 4.2.3 Find functions h(x) and g(u) such that f (x) = g(h(x)), given that (a) f (x) = (x5 − 3x2 + 12)3

(b) f (x) =

√ 3

2−x+

4 2−x

Solutions (a) If f (x) = g(h(x)), and if we let h(x) ⇒ g(u)

= x5 − 3x2 + 12 = u3 .

(b) If f (x) = g(h(x)), and if we let

4.3

h(x)

=

⇒ g(u)

=

2−x √ 4 3 u+ . u

Quadratic functions

A quadratic function is a polynomial of degree 2, whose graph is a parabola. The function is of the form y = f (x) = ax2 + bx + c where a, b and c ∈ R and a = 0. The common properties of quadratic functions are:

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Chapter 4 Functions (a) The graph opens upwards if a > 0 and the general form of the graph is as shown below.

(b) The graph opens downwards if a < 0 and the general form is shown below.

(c) The equation ax2 + bx + c = 0 has no more than two roots (or x-intercepts).  −b (d) The turning point occurs at x = −b 2a , and corresponding y-value is y = f 2a . This can easily be established from the formulae for completing the square.

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An Introduction to Applied Calculus for Social and Life Sciences example 4.3.1 Sketch the graph of f (x) = x2 + 2x − 8 Solution The function f (x) = x2 + 2x − 8 has a = 1, b = 2 and c = −8 as coefficients of the quadratic expression. Factorising gives f (x) = (x + 4)(x − 2). The x-intercept(s) are obtained when f (x) = 0 ⇒ x = −4 or x = 2. The y-intercept is obtained when x = 0 so that f (0) = −8. The coordinates for the y-intercept are thus (0, −8). −b −2 The quadratic polynomial turns at x = = = −1. 2a 2(1) The corresponding y value is obtained from y = f (−1) = (−1)2 + 2(−1) − 8 = −9, and the coordinates of the turning point are (−1, −9). One can also use the formula for completing the square so that f (x) = (x2 + 2x + 1) − 8 − 1 = (x + 1)2 − 9. Clearly, f (−1) = −9, which also gives a turning point (−1, −9). Since a > 0, parabola opens upwards. We can show the details on the graph below.

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Chapter 4 Functions

4.4

Power functions

A function of the form f (x) = xn , n ∈ R is called a power √ function. This is because the exponents are referred to as powers. For instance, y = x, y = x12 or f (x) = x3 are examples of power functions.

4.5

Rational functions

A quotient p(x) q(x) of two polynomials p(x) and q(x) is called a rational function. Typical examples include y=

x2 + 1 1−x 4 or y = . , y = x3 x2 − 1 x2 − 1

example 4.5.1 Classify the following functions as a polynomial, a power function, or a rational function (or none). (a) f (x) = x14

(b) f (x) = −2x3 − 3x2 + 8

(c) f (x) = (3x − 5)(4 − x)2

(d) f (x) =

(e) f (x) =

2x + 9 x2 − 3

√

x + 3x

3

Solutions (a) power function and polynomial (c) polynomial (e) rational function

4.6

(b) polynomial (d) none of these

Exponential functions

Definition of an exponential function If b is a positive number other than 1 i.e. (b > 0, b = 1), then the unique exponential function with base b is defined by f (x) = bx , for every x ∈ R is called an exponential function.

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An Introduction to Applied Calculus for Social and Life Sciences It is important to note that, the graphs of functions y = bx and y = b−x (b > 0, b = 1) never intersect the x-axis; thus the exponential function is always positive. Below we give examples of sketches of exponential functions. example 4.6.1 Sketch the graphs of y = 2x and y =

 1 x 2

Solution

Basic properties of exponential functions For bases a > 0, b > 0 with a = 1 and b = 1, and x, y ∈ R, we have (i)

bx = by if and only if x = y)

(iv) bx · by = bx+y

(ii)

bx = bx−y by

(v)

(iii)

(ab)x = ax bx

(vi)

(bx )y = bxy

a x b

=

ax bx

example 4.6.2 Simplify 2 −3

(a) (x y

3

z)

(b)

x3 y −2 z4

16

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Chapter 4 Functions Solutions (a) (x2 y −3 z)3 = x6 y −9 z 3 . (b)

x3 y −2 z4

16 =

√ √ −1 1 x x x2 y 3  = = . 2 1 2 3 z3 y3z3 yz 2

example 4.6.3 Find all the real numbers x that satisfy the given equation (b) 62x−3 = 36(6)−x

(a) 3x .22x = 144

Solutions (a) 3x · 22x = 122 ⇒ ((3)(22 ))x = 122 We thus have 12x = 122 ⇒ x = 2. (b) 62x−3 = 62 6−x ⇒ 62x · 6−3 = 62 · 6−x ⇒

62x 62 = . 63 6x

So, cross-multiplication yields 62x · 6x = 62 · 63 ⇒ 63x = 65 . Here the exponents must be equal so that 3x = 5 ⇒ x =

4.6.1

5 . 3

Logarithmic functions

The natural exponential base e In algebra, it is common practice to use the base b = 10 for exponential functions or in some cases b = 2. In calculus it is more convenient to use the number e as a base. It is called Napier’s number or Napier’s constant named after John Napier, who worked on logarithms in the 16th century. Leonhard Euler introduced the letter e as the base of a natural logarithm, in the 18th century. The number e is an irrational number defined by e=

∞  1 = 2.71828 · · · n! n=0

It is an important mathematical constant that is the base of the natural logarithm. The natural logarithm is written as loge x = ln x. That is, y = ln x if and only if ey = x.

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An Introduction to Applied Calculus for Social and Life Sciences Properties of natural logarithm For p ∈ R and x, y ∈ R, with x > 0 and y > 0, we have (a) ln x = ln y if and only if x = y (b) ln(xy) = ln x + ln y (c) ln xp = p ln x x (d) ln = ln x − ln y y (e) ln 1 = 0 (f) ln e = 1 since ln e = loge e = 1. The graph of y = ln x is given below.

The inverse relationship between ex and ln x The following statements are true with respect to the functions ex and ln(x). eln x = x for x > 0 and ln ex = x for all x ∈ R. So, graphically, the graphs of y = ln(x) and y = ex are a reflection of each other on the line y = x. The relationship is shown in the figure on the next page.

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Chapter 4 Functions

example 4.6.4 Solve the given equation for x (a) ln x = 2(ln 3 − ln 5)

(b) 3x = e2

Solutions 3 (a) ln x = 2(ln 3 − ln 5) = 2 ln . 5

Therefore, ln x = ln

 3 2 5

⇒x=

 3 2 5

=

9 25 .

(b) 3x = e2 ⇒ ln 3x = ln e2 (By taking logarithms of both sides.) Using property (c) of natural logarithms we have x ln 3 = 2 ln e = 2(1) = 2. Therefore x =

2 . If we use the definition ln 3 3x = e2 ⇐⇒ log3 e2 = x,

we will have x = 2 log3 e = =

2 ln e ln 3

2 . ln 3

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An Introduction to Applied Calculus for Social and Life Sciences example 4.6.5 If log5 (2x) = 7, what is ln x? Solution loge 2x =7 loge 5 ln 2x = 7. ⇒ ln 5

log5 (2x) = 7 ⇒

We thus have in this case ln 2x = 7 ln 5. But ln 2x = ln 2 + ln x. This then gives ln 2 + ln x = ln 57 ⇒ ln x = ln 57 − ln 2 = ln

57 . 2

example 4.6.6 Write down the numerical values of the following expression (a) eln 5

(b) e2 ln 3

(c) e3 ln 2−2 ln 5

Solutions (a) eln 5 = 5. 2

(b) e2 ln 3 = eln 3 = 32 = 9. 3

2

(c) e3 ln 2−2 ln 5 = eln 2 · e− ln 5 ⇒ = eln 8 · e− ln 25 8

= eln( 25 ) 8 . = 25

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Chapter 4 Functions example 4.6.7 Solve for x 1 (a) ln x = (ln 16 + 2 ln 2) 3

(b)

5 =3 1 + 2e−x

Solutions (a)

1 1 (ln 16 + 2 ln 2) = (ln 16 + ln 4) 3 3 1 1 = ln (16 × 4) = ln 64 3 3 1 1 So ln x = ln 64 3 = ln(43 ) 3 = ln 4. ln x =

Therefore x = 4. (b)

5 5 = 3 ⇒ = 1 + 2e−x 1 + 2e−x 3 2 5 So 2e−x = − 1 = 3 3 1 −x ⇒ e = . 3 Taking logarithms on both sides, we obtain 1 −x ln e = ln ⇒ −x = − ln 3. 3 We therefore have x = ln 3.

4.7

Exponential growth and decay

Exponential growth and decay definitions A quantity Q(t) grows exponentially if Q(t) = Q0 ekt , k, Q0 > 0 and decays exponentially if Q(t) = Q0 e−kt , k, Q0 > 0. The figure on the next page shows exponential growth.

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An Introduction to Applied Calculus for Social and Life Sciences

The figure below shows exponential decay.

example 4.7.1 It is estimated that the population of South Africa grows exponentially. If the population in 1997 was 30 million and 40 million in 2007, what would the population be in 2022?

Solution Let Q(t) = Q0 ekt million be the size of the population after t years. Suppose that t = 0 in 1997, so that Q(0) = Q0 e0 ⇒ Q0 = 30. So, Q(t) = 30ekt . 52

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Chapter 4 Functions In 2007, t = 10, i.e.. Q(10) = 40 million. Therefore Q(10) ⇒ e10k ⇒ 10k ⇒k

30e10k = 40 4 = 3 4 = ln 3 4 1 ln . = 10 3 =

We are required to find Q(25) (in 2022 t = 25). We thus have Q(25) Instead of using k =

1 10

30e25k .

=

ln 43 , we use the fact that e10k = 43 , so that

Q(25)

=

10k 2.5

30(e

)

2.5 4 = 30 million. 3

example 4.7.2 It is projected that t years from now, the population of Worcester will be p(t) = 0.5e0.02t million (a) What is the current population? (b) What will the population be in 30 years from now? Solutions (a) p(0) = 0.5e0.02(0) = 0.5 million people (b) After 30 years, we set t = 30 so that p(30) = 0.5e0.02×30 = 0.5e0.6 million people.

4.7.1

Half-life

The half-life of a substance is the time required for the amount of that substance to fall to half its initial value. The term is commonly used in radioactive substances decay. The converse of half-life is doubling time, which we consider in the next section. A radioactive substance, for example, that decays exponentially and has initial size of Q0 , will have size Q(t) = Q0 e−kt after t years (where k > 0). Note that the negative sign on k indicates a decay process. 53

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An Introduction to Applied Calculus for Social and Life Sciences If the initial amount of the substance is Q0 then by the time it reaches its half-life it will be 12 Q0 . If we are to calculate when the substance will reach half its initial size, then we solve for t = th , where th is the time it takes for the substance to decay to half its original size. 1 1 Q0 ⇒ Q0 e−kth = Q0 2 2 1 1 −kth −kth So e = ⇒ ln e = ln 2 2 Simplifying, we have − kth = ln 1 − ln 2. ln 2 . ⇒ th = k Q(h) =

4.7.2

Doubling time

The time period required for a quantity to double in size or value is called the doubling time. It is often applied to the growth of populations, such as that of bacteria or even the size tumours in cancer growth. For a population that undergoes exponential growth, the doubling time is constant and can be evaluated from the growth rate. example 4.7.3 The amount of bacteria in a petri dish is given by Q(t) = Q0 ekt (k > 0) at any time t and the initial amount, given at time t = 0 is Q0 . If the bacteria doubles when t = td , show that td = lnk2 . Solution We know that Q(t) = Q0 ekt doubles when Q(t) = 2Q0 . Q0 ektd = 2Q0 ⇒ ektd = 2. Taking logarithms both sides we have ln ektd = ln 2 ⇒ ktd ln e = ln 2 ln 2 . ⇒ td = k

4.7.3

Carbon dating

Carbon dating (also referred to as carbon-14 dating) is a way of determining the age of a substance or matter (which was once living and presumed to be in equilibrium with the atmosphere) containing organic material by using the properties of radiocarbon, 14 C, a radioactive isotope of carbon. Below are some important facts about carbon dating. 54

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Chapter 4 Functions (a) Carbon dioxide in air contains two isotopes as shown below.

(b) Living organisms exchange carbon with the atmosphere in the form of carbon dioxide continuously in their life time. So the ratio of 14 C to 12 C in living organisms is the same as that in the atmosphere. When a plant or animal dies, the absorption of carbon dioxide ceases and the following happens: (i) The

12

C in the organism stays the same but the

14

C decays.

(ii) The ratio of 14 C to 12 C decays exponentially and the decay process is given by the function R(t) = R0 e−kt , with k as the decay constant. (iii) R0 is the ratio of 14 C to 12 C in the air when the organism dies. The decay process is assumed to begin at that particular time. (c) The half life of 14 C is known to be 5730 years. By comparing R(t) and R0 archaeologists can estimate the age of any former living thing. example 4.7.4 The amount of a certain radioactive substance remaining after t years is given by a function of the form Q(t) = Q0 e−0.003t Find the half-life of the substance. Solution Using the derived expression for the half-life of a decaying substance, we have th =

ln 2 1000 ln 2 ln 2 = = . k 0, 003 3

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An Introduction to Applied Calculus for Social and Life Sciences example 4.7.5 An archaeologist found a fossil at the Cradle of Humankind in South Africa in which the ratio of 14 C to 12 C was 13 the ratio found in the atmosphere. Approximately how old was the fossil? Solution Let R(t) be the ratio of by

14

C to

12

C in the fossil so that the decay process is given R(t) = R0 e−kt .

Given that R(t) = 13 R0 , then R0 e−kt =

1 1 R0 ⇒ e−kt = 3 3

Taking logarithms ln e−kt = ln

1 ⇒ −kt = − ln 3 3 ln 3 . so that t = k

We need to find the value of k. We know the half life, th = 5730 years. Then 5730 = Therefore, the fossil is t =

4.8 4.8.1

ln 3 ln 2 5730

=

ln 2 ln 2 ⇒ k= . k 5730

5730 ln 3 years old. ln 2

Trigonometric functions Angles

Angles are measured in degrees, whose symbol is ◦ . So ten degrees is written as 10◦ . One complete revolution (round a complete circle) has 360◦ . In calculus the most useful measure of angles is the circular or radian measure, with symbol rad. Angles are measured positively if the direction of movement is anticlockwise and negatively if it is clockwise. Consider a circle of radius r and an arc AB of length r on the same circle as shown in the figure on the next page.

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Chapter 4 Functions

ˆ is one radian. We can now define what a radian is. The angle AOB Definition of a radian A radian describes the angle made when an arc has a length that is equal to the radius of the circle. Generally, for any circle of radius r, if an arc of length s subtends an angle θ, then θ=

s radians. r

Since the circumference of a circle of radius r is 2πr, then the radians are θ=

2πr = 2π. r

So, one revolution gives 2π radians that are equal to 360◦ . Therefore 180◦ = π radians. We can convert angles in degrees to radians and vice versa. The following formulae are useful for such conversions.

Conversion formulae From degrees to radians:

θ◦ =

π  θ radians. 180

From radians to degrees:

Θ radians =

180 π

Θ degrees.

Note: We often write radians without specifying the units. We can write of π4 radians.

π 4

instead

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An Introduction to Applied Calculus for Social and Life Sciences example 4.8.1 Convert (a) 45◦ , 150◦ and 240◦ to radians (b)

π π 3π , and radians to degrees 6 2 8

Solutions

π  π 45 = radians. 180 4

π  5π ◦ 150 = radians. 150 = 180 6

π  4π radians. 240◦ = 240 = 180 3

π  180 π rad = (b) degrees = 30◦ . 6 6 π

3π 3π 180 rad = degrees = 270◦ . 2 2 π

π  180 π rad = degrees = 22.5◦ . 8 8 π

(a)

45◦ =

4.8.2

The sine and cosine functions

Suppose the line segment joining the points (0, 0) and (1, 0) on the x-axis is rotated through an angle of θ radians so that the free endpoint of the segment moves from (1, 0) to (x, y), as in the sketch below:

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Chapter 4 Functions Definitions of sine and cosine functions For any angle θ, cos θ = x and sin θ = y, where (x, y) is the point to which (1, 0) is carried by a rotation of θ radians about the origin. So, at θ θ θ θ

π π π ; x = cos = 0 and y = sin = 1, the coordinates are (0, 1). 2 2 2 = π; x = cos π = −1 and y = sin π = 0, the coordinates are (−1, 0). 3π 3π 3π ; x = cos = 0 and y = sin = −1, the coordinates are (0, −1). = 2 2 2 = 2π; x = cos 2π = 1 and y = sin 2π = 0, the coordinates are (1, 0). =

Basic properties of the sine and cosine functions For any point on a circle, a rotation of 2π radians sends that point to its original starting point. We will thus have the following important properties of the sine and cosine functions: sin(θ + 2π) = sin θ and cos(θ + 2π) = cos θ. So, the sine and cosine functions have a period 2π and are said to be 2π periodic. Also, sin(−θ) = − sin θ and cos(−θ) = cos θ. The two functions are often referred to as trigonometric co-functions of each other and are linked by the following relationships. 

π◦ 

π◦ − θ and cos θ = sin −θ . sin θ = cos 2 2 example 4.8.2 Find the exact values of (a) sin 3π (c) cos

(b) sin

−5π 2

7π 4



Solutions (a) sin 3π = sin(2π + π) = sin π = 0. √

π 2 π . = sin 2π − = sin − =− 4 4 2



−5π 5π 3π (c) cos = 0. = cos 4π − = cos 2 2 2

(b) sin

7π 4

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An Introduction to Applied Calculus for Social and Life Sciences

4.8.3

The graphs of y = sin θ and y = cos θ

4.8.4

Other trigonometric functions

The circular functions form the foundation of all other trigonometric functions.  Consider a right-angled triangle with sides x, y and a hypotenuse x2 + y 2 (from Pythagoras’ theorem). Recall that cos θ = x and sin θ = y. We can easily establish the following: sin θ 1 x cos θ y = , cot θ = = = , x cos θ tan θ y sin θ 1 1 , cosecθ = csc θ = , sec θ = cos θ sin θ

tan θ =

provided the denominators are not zero. example 4.8.3 Find the exact value(s) of: (a) tan θ if sec θ =

5 , for 0 ≤ θ ≤ π 4

(b) cos θ if tan θ =

3 π 3π , for ≤ θ ≤ 4 2 2

(c) sin θ if tan θ =

2 , for 0 ≤ θ ≤ 2π 3

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Chapter 4 Functions Solutions (a)

sec θ =

√ 4 5 ⇒ cos θ = so, y = 25 − 16 = 3, 4 5

Therefore tan θ =

3 . 4

(b)

tan θ =

√ 3 , and r = 9 + 16 = 5, 4

So cos θ = −

4 for the range of values of θ. 5

(c)

tan θ =

√ √ 2 , and r = 4 + 9 = 13, 3

2 2 Therefore sin θ = √ or sin θ = − √ 13 13

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An Introduction to Applied Calculus for Social and Life Sciences

4.8.5

Special triangles

There are two triangles that are special in trigonometry. These are the 45◦ −45◦ −90◦ and 30◦ − 60◦ − 90◦ triangle. The 45◦ − 45◦ − 90◦ triangle is an isosceles right triangle while the 30◦ − 60◦ − 90◦ is derived from an equilateral triangle. These triangles are ‘special’ because of their ability to give exact answers instead of decimal approximations when dealing with trigonometric functions. We first begin by looking at the 45◦ − 45◦ − 90◦ special right triangle. The triangle given by the figure below.

The 30◦ − 60◦ − 90◦ special right triangle is shown below.

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Chapter 4 Functions (45◦ − 45◦ − 90◦ ) special triangle

(30◦ − 60◦ − 90◦ ) special triangle

√ 1 2 π sin = √ = 4 2 2

4.8.6

cos

√ 1 2 π =√ = 4 2 2

tan

π =1 4

√ 3 π sin = 3 2 1 π sin = 6 2 1 π = 3 2√ 3 π cos = 6 2 √ 3 √ π = 3 tan = 3 1 √ 1 3 π tan = √ = 6 3 3 cos

Trigonometric identities

An identity is an equation that is true for all values of its variable or variables when both sides are defined. In this subsection we look at some of the common trigonometric identities. We make use of the following figure of a line of unit length being rotated an angle θ anticlockwise, so that we have:

We have already established the following identities that are true for all values of θ.

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An Introduction to Applied Calculus for Social and Life Sciences Basic identities tan θ =

sin θ cos θ

cot θ =

1 sin θ

cosecθ =

sec θ =

cos θ sin θ 1 cos θ

From the figure above, we have x2 + y 2 = 1 for all values of x and y. By Pythagoras’ theorem we have (cos θ)2 + (sin θ)2 = 1 written as cos2 θ + sin2 θ = 1.

(4.1)

Sine and cosine identities ≡

cos2 θ + sin2 θ = 1

cos2 θ = 1 − sin2 θ

≡

sin2 θ = 1 − cos2 θ

Dividing both sides of (4.1) by cos2 θ yields the following 1+

sin θ cos θ



2

⇒ 1 + tan2 θ

=

1 cos θ

=

sec2 θ.

2 for cos θ = 0

Tangent and secant identity 1 + tan2 θ = sec2 θ Also, if we divide both sides of (4.1) by sin2 θ we have



2

2

+1

=

1 sin θ

⇒ cot2 θ + 1

=

cosec2 θ.

cos θ sin θ

for sin θ = 0

Cotangent and cosecant identity 1 + cot2 θ = cosec2 θ

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Chapter 4 Functions example 4.8.4 Prove the following identities (a) (1 + tan2 θ) cos2 θ = 1 (c) sec2 x − tan2 x = 1

Solutions



(a) LHS (the left-hand side)=

1+

(b) cos2 θ − sin2 θ = 1 − 2 sin2 θ (d) sec θ − tan θ =

sin2 θ cos2 θ



cos2 θ =

1 sec θ + tan θ

cos2 θ + sin2 θ cos2 θ

cos2 θ

= cos2 θ + sin2 θ = 1 = RHS (the right-hand side). (b) LHS = (1 − sin2 θ) − sin2 θ = 1 − 2 sin2 θ = RHS. 1 sin2 x 1 − sin2 x cos2 x − = = = 1 = RHS. 2 2 2 cos x cos x cos x cos2 x



cos θ 1 − sin θ cos θ 1 = = (d) RHS = 1 sin θ 1 + sin θ 1 + sin θ 1 − sin θ cos θ + cos θ (c) LHS =

=

cos θ sin θ cos θ cos θ − sin θ cos θ cos θ − sin θ cos θ = − = cos2 θ cos2 θ cos2 θ 1 − sin2 θ

=

sin θ 1 − = sec θ − tan θ = LHS. cos θ cos θ

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An Introduction to Applied Calculus for Social and Life Sciences

4.8.7

Sum and difference formulas

For any angles A and B we have the following identities: Sum and difference identities

cos(A + B)

=

cos A cos B − sin A sin B

sin(A + B)

=

sin A cos B + cos A sin B

cos(A − B)

=

cos A cos B + sin A sin B

sin(A − B)

=

sin A cos B − cos A sin B

tan(A + B)

=

tan A + tan B 1 − tan A tan B

tan(A − B)

=

tan A − tan B 1 + tan A tan B

If we set A = B, we have a new set of identities given by:

Related identities ⎧ cos2 A − sin2 A ⎪ ⎪ ⎪ ⎪ ⎨ 1 − 2 sin2 A cos 2A = ⎪ ⎪ ⎪ ⎪ ⎩ 2 cos2 A − 1 sin 2A

=

2 sin A cos A

tan(2A)

=

2 tan A 1 − tan2 A

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Chapter 4 Functions example 4.8.5 Given that sin A = find:

4 5

and cos B =

−5 13 ,

where 0 ≤ A ≤

π 2

and π ≤ B ≤

3π 2 ,

(a) sin(A − B) (b) tan(A − B) Solutions Representing the given information on triangles on the cartesian plane, we have the following figure.

We use Pythagoras’ theorem to find the remaining sides of the two triangles. Using the information on the triangles, we have



4 −5 3 −12 − 5 13 5 13 16 −20 36 + = = 65 65 65

(a) sin(A − B) = sin A cos B − cos A sin B =

(b) tan(A − B) =

4 − 12 tan A − tan B = 3 4 512 1 + tan A tan B 1+ 3 · 5

=

−16 15 1 + 48 15

=−

16 . 63

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An Introduction to Applied Calculus for Social and Life Sciences

4.9

Solving trigonometric equations

We now consider solving trigonometric equations. It is important to have a clear understanding of the graphs of the basic trigonometric functions and quadrants. We present the graphs with marked quadrants in which they exist. The figure below illustrates trigonometric functions graphs for sin θ and cos θ showing the corresponding quadrants.

The figure below illustrates trigonometric functions graph for the tangent function and the corresponding quadrants.

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Chapter 4 Functions The signs for the cosine, sine and tangent functions in each quadrant are summarised in the figure below.

Let us begin by solving the equation cos θ = 0.5 where the solutions are assumed to lie in the interval 0 ≤ θ ≤ 2π. In other words we are looking for all the angles, θ, in the given interval for which the cosine is equal to 0.5. To get a clearer view of the solution we begin by sketching the graph of the function cos θ over the given interval.

The horizontal line on the graph indicates where cos θ = 0.5. The solutions of cos θ = 0.5 are obtained from the points where the horizontal line intersects the 69

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An Introduction to Applied Calculus for Social and Life Sciences graph of cos θ. It is also important to note the intersections take place in the first and fourth quadrants where indeed cos θ is positive (when cos θ = 0.5 then the cosine is positive.) The solution in the first quadrant is θ = π3 and the second one in the fourth quadrant is θ = 5π 3 . These are the only solutions in the given interval. So π 5π . θ= , 3 3 example 4.9.1 Solve the equation cos 2θ = −

√ 3 2

in the interval 0 ≤ θ ≤ 2π.

Solution It is important here to note that the cosine is a multiple angle, 2θ. It is easier to redefine the multiple angle as a single angle by setting α = 2θ. This is now equivalent to solving √ 3 for 0 ≤ α ≤ 4π. cos α = − 2 The graph of cos α over the new interval, with a horizontal dotted line at − shown below.

√ 3 2 ,

is

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Chapter 4 Functions We know that cos α is negative in the second and third quadrants. The solution in 7π the second quadrant is α = 5π 6 and in the third quadrant is α = 6 . Since cos α is 2π periodic, the second set of solutions are α=

17π 7π 19π 5π + 2π = and α = + 2π = . 6 6 6 6

So, α=

5π 7π 17π 19π , , , 6 6 6 6

2θ =

5π 7π 17π 19π , , , 6 6 6 6

θ=

5π 7π 17π 19π , , , . 12 12 12 12

Given that α = 2θ,

so that

example 4.9.2 Find all the values of θ in the specified interval that satisfy the given equation: (a) cos θ − sin 2θ = 0; 0 ≤ θ ≤ 2π (c) sin 2θ =

√

3 cos θ; 0 ≤ θ ≤ π

(b) 3 sin2 θ + cos 2θ = 2; 0 ≤ θ ≤ 2π (d) cos 2θ = cos θ; 0 ≤ θ ≤ π

Solutions (a)

cos θ − sin 2θ = 0 ⇒ cos θ − 2 sin θ cos θ = 0 ⇒ cos θ(1 − 2 sin θ) = 0. ⇒ cos θ = 0 or sin θ = π , 2 π Therefore θ = , 6 θ=

(b)

1 2

3π π 5π or θ = , 2 6 6 π 5π 3π , or . 2 6 2

3 sin2 θ + cos 2θ = 2 ⇒ 3 sin2 θ + cos2 θ − sin2 θ = 2 ⇒ 2 sin2 θ + (1 − sin2 θ) − 2 = 0 ⇒ sin2 θ − 1 = 0 ⇒ sin θ = ±1. 3π π . Therefore θ = or 2 2 Note that is is quicker to solve this problem by replacing cos 2θ by 1 − 2 sin2 θ. 71

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An Introduction to Applied Calculus for Social and Life Sciences (c)

sin 2θ =

√ √

3 cos θ

⇒ 2 sin θ cos θ = 3 cos θ √ ⇒ 2 sin θ cos θ − 3 cos θ = 0 √ ⇒ cos θ(2 sin θ − 3) = 0 ⇒ cos θ = 0 or sin θ = So, θ =

√ + 3 2

2π π π , or . 2 3 3

(d) cos 2θ = cos θ ⇒ (cos2 θ − sin2 θ) − cos θ = 0 ⇒ cos2 θ − cos θ − (1 − cos2 θ) = 0 ⇒ 2 cos2 θ − cos θ − 1 = 0 ⇒ (2 cos θ + 1)(cos θ − 1) = 0 −1 So cos θ = 1 or cos θ = 2 Therefore θ = 0,

4.10

2π . 3

Piece-wise functions

A function that is defined by more than one function, each for different values of x, is called a piece-wise function. In simple terms it is a function that is made up of pieces of functions defined on subintervals of the domain. Consider two functions defined on either side of x = 1, such that ⎧ 3 ⎪ if x < 1 ⎨x , f (x) = ⎪ ⎩ 3 − x, if x ≥ 1 The function f (x) is made up of two functions: f (x) = x3 on the left-hand side of x = 1 and f (x) = 3 − x on the right-hand side of x = 1. The inequality signs indicate that at x = 1 it is only the function on the right-hand side that is defined.

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Chapter 4 Functions We give a sketch of the function below.

example 4.10.1 Sketch the following function ⎧ ⎪ x2 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ f (x) = 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩x,

if 0 ≤ x < 1 if 1 ≤ x ≤ 2 if x > 2

Solution

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An Introduction to Applied Calculus for Social and Life Sciences

example 4.10.2 Sketch the following function

f (x) =

⎧ ⎪ ⎨x + 1,

if x = 1

⎪ ⎩ 3,

if x = 1

Solution

example 4.10.3 ⎧ ⎪ x2 + 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ f (x) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩x + 1,

if x < 0 if x = 0 if x > 0

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Chapter 4 Functions Solution

4.11

Graphing of basic functions

When one views a graph, it is important to always note that the graphs of many functions are simply transformations of the graphs of very basic functions such as √ 1 1 y = x2 , y = x3 , y = , y = 2 , y = ex , y = ln x, y = x etc. x x Graphs can be shifted vertically or horizontally. They can also be stretched. Let us begin by looking at the vertical shifting.

Vertical shifting • Let a be a positive real number. The graph of y = f (x) + a is the graph of y = f (x) shifted a units upwards. • Also, the graph of y = f (x) − a is the graph of y = f (x) shifted a units downwards. example 4.11.1 Given that f (x) = |x|, we can use the graph of f (x) to graph the functions f1 (x) = |x| + 2 and f2 (x) = |x| − 3.

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An Introduction to Applied Calculus for Social and Life Sciences Solution The first function is shifted 2 units upwards and the second function is shifted 3 units downwards. The figure below shows the plots of f1 and f2 .

We now consider the case when graphs are shifted horizontally.

Horizontal shifting • Let a be a positive real number. The graph of y = f (x − a) is the graph of y = f (x) shifted a units to the right. • Also, the graph of y = f (x + a) is the graph of y = f (x) shifted a units to the left.

example 4.11.2 Given a function f (x) = x2 , use the graph of f (x) to graph the functions g1 (x) = (x + 4)2 and g2 (x) = (x − 3)2 . Solution The first function is shifted 4 units to the left and the second function is shifted 3 units to the right.

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Chapter 4 Functions

There are cases when the graph of a function is a reflection of one of the graphs of the given basic function in the x-axis, y-axis or a line. We observed in the earlier sections that the graphs of y = ln x and y = ex are reflections of each other in the line y = x.

Reflection of graphs • The graph of the function y = f (−x) is the graph of y = f (x) reflected on the y-axis. • The graph of the function y = −f (x) is the graph of y = f (x) reflected on the x-axis.

Functions can either be stretched or shrunk through multiplication by a number. The stretching or shrinking can happen vertically or horizontally. We have the following.

Vertical stretching and shrinking of graphs • If a is a real number such that a > 1 then the graph of y = af (x) is the graph of y = f (x) stretched vertically by a factor of a. • In the case where 0 < a < 1, then the graph of y = af (x) is the graph of y = f (x) shrunk vertically by a factor of a.

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An Introduction to Applied Calculus for Social and Life Sciences example 4.11.3 Consider the function f (x) = cos x. Graph the following functions: y = 3 cos x and y = 0.5 cos x. Solution The function y = 3 cos x is vertically stretched by 3 units while y = 0.5 cos x is shrunk by half a unit vertically. This is shown graphically below.

Horizontal stretching and shrinking of graphs • If a is a real number such that a > 1 then the graph of y = f (ax) is the graph of y = f (x) that has been compressed horizontally by a factor of a. • In the case where 0 < a < 1, then the graph of y = f (ax) is the graph of y = f (x) stretched horizontally by a factor of a.

example 4.11.4 Use the graph of f (x) = x3 to graph the following functions: y = (2x)3 , y = (3x)3 and y = (4x)3 .

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Chapter 4 Functions Solution The function y = (2x)3 , is compressed horizontally by a factor of 8 units while y = (0.5x)3 is horizontally stretched by 8 units. This is shown graphically on the next page.

example 4.11.5 Sketch the graphs of the following functions by means of shifting basic function. (b) y = x2 + 4x + 5 (a) y = x2 − 2x

Solution (a) y = (x2 − 2x + 1) − 1 = (x − 1)2 − 1.

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An Introduction to Applied Calculus for Social and Life Sciences This is the graph of y = x2 shifted 1 unit to the right and then 1 unit down.

(b) y = x2 + 4x + 5 = (x2 + 4x + 4) + 1 = (x + 2)2 + 1. Again this is the graph of y = x2 shifted 2 units to the left and then 1 unit up.

The graphs of the following functions can be plotted by shifting and reflecting the basic function y = x1 . 1 is the graph of the basic function shifted 1 unit The graph of the function y = x−1 to the right. 80

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Chapter 4 Functions The graph of the function y = 2 − x-axis and then shifted 2 units up.

1 x−1

is the graph of y =

1 x−1

reflected on the

example 4.11.6 Sketch the graphs of the following functions x−4 x+3 (b) f (x) = (a) f (x) = x+2 x+2 Solutions (a) Using long division

x+2

1 x+ 3 , − (x + 2) 1

f (x) =

1 x+3 =1+ x+2 x+2

or f (x)

=

(x + 2) + 1 1 x+3 = =1+ . x+2 x+2 x+2

The graph of this function can be obtained from the basic function y = x1 . 1 is the graph of the basic function shifted 2 units to the The function y = x+2 1 1 left, so that the graph of the function y = 1 + x+2 , is the graph of y = x+2 shifted 1 unit up.

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An Introduction to Applied Calculus for Social and Life Sciences (b) Again using long division

x+2

1 x− 4 − (x + 2) −6

Therefore f (x) = 1 −

6 x+2

or f (x) =

x−4 (x + 2) − 6 6 = =1− . x+2 x+2 x+2

1 We again use the basic function y = x1 . The function y = x+2 is the graph of 6 the basic function shifted 2 units to the left and y = x+2 is a vertical stretch 6 by a factor of 6.Then y = − x+2 will be its reflection in the x-axis. So that 6 6 the graph of the function y = 1 − x+2 is the graph of y = − x+2 shifted 1 unit up.

4.12

Intersection of graphs

Suppose the graphs of f (x) and g(x) are sketched on the same coordinates system. Suppose that f (x) intersects g(x) at (a, f (a)) and (b, f (b)). 82

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Chapter 4 Functions

To find the two points, it is preferable to set the equations in the form y = f (x), then set the two equations for y equal to each other. You then solve for x in the resulting equation. The solution for x will be the x-coordinate for the point(s) of intersection. To find the y value, we then use the x-coordinate(s) and substitute into either of the original equations.

example 4.12.1 Find the point of intersection (if it exists) of the given curves and sketch their graphs and y = x − 1 y = x2 − x

Solution We begin by equating the two functions so that

x2 − x = x − 1 ⇒ x2 − 2x + 1 = 0 (x − 1)2 = 0 ⇒ x = 1.

Substitute x = 1 in either one of the functions, to find the corresponding y-value, so that y = (1) − 1 = 0. This means that we only have one point of intersection (1, 0). 83

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An Introduction to Applied Calculus for Social and Life Sciences

In fact the line y = x − 1 is a tangent line to the curve y = x2 − x.

example 4.12.2 Solve the following set of simultaneous equations 2x − 3y = −8

and

3x − 5y = −13

Solution Multiplying the first equation by 3 and the second equation by 2 so as to eliminate one of the variables, we have: 3(2x − 3y = −8) 2(3x − 5y = −13)

⇒ 6x − 9y = −24 ⇒ 6x − 10y = −26.

It is now possible to eliminate x by subtracting the second equation from the first. We get y = 2. Now substituting y = 2 in any equation will give us a value of x. Let us use the first equation so that 2x − 3(2) = −8 ⇒ 2x = −2. This gives x = −1. Therefore the point of intersection is (−1, 2). An alternative approach: Rewriting the equations in standard form gives

y

=

2 8 3 13 x + and y = x + . 3 3 5 5 84

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Chapter 4 Functions Equating the two equations gives 2 8 3 13 x+ = x+ 3 3 5 5

⇒

2 3 13 8 x− x = − 3 5 5 3 39 − 40 1 1 x = =− ⇒ 15 15 15 ⇒ x = −1.

Substituting x = −1 into the first equation gives y

3 13 10 (−1) + = = 2. 5 5 5

=

The result is the same.

4.12.1

Distance formula

The distance between the points (x1 , y1 ) and (x2 , y2 ) is given by

Distance formula D=

4.12.2

 (x1 − x2 )2 + (y1 − y2 )2 .

Intercepts of functions

To find the x-intercept(s) of a graph, set y = 0. The x-axis is the line y = 0. To find the y-intercept of a graph, set x = 0. The y-axis is the line x = 0. example 4.12.3 Sketch the graph of the following function. Show all intercepts. ⎧ ⎪ x 0 there exists a δ > 0 such that | f (x) − L |<  whenever 0 2.

Solution Here we check if conditions for continuity are satisfied.

At x = −2 (i) the function value f (−2) exists and is equal to 4. (ii) the limit does not exist, since 4=

lim f (x) = lim + f (x) = 3.

x→−2−

x→−2

Condition (b) is not satisfied, therefore the f (x) is not continuous at x = −2.

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An Introduction to Applied Calculus for Social and Life Sciences At x = 0 (i) f (0) = 0. (ii) the limit lim f (x) exists, since x→0

lim f (x) = 0 = lim+ f (x).

x→0−

x→0

(iii) Also, lim f (x) = 0 = f (0). x→0

Since all the 3 conditions are met, the function f (x) is continuous at x = 0. At x = 2 (i) the function value f (2) exists and is equal to 4. (ii) the limit lim f (x) exists, since x→2

lim f (x) = 2 = lim + f (x).

x→2−

x→−2

(iii) but, lim f (x) = 2 = 4 = f (2). x→2

Condition (iii) is not satisfied. Therefore the f (x) is not continuous at x = 2, but has what is called a removable discontinuity, since it is defined at x = 2. We will consider in depth such limits in the next section. example 5.5.2 Determine whether the function is continuous at the given point. ⎧ ⎪ ⎨x + 1 if x < 0, (a) f (x) = at x = 0. ⎪ ⎩ x − 1 if x ≥ 0, √ x−2 at x = 4 (b) f (x) = x−4 ⎧ 2 x −1 ⎪ if x < −1, ⎨ x2 −3 (c) f (x) = at x = −1. ⎪ ⎩ 2 x − 3 if x ≥ −1,

Solutions (a) For function f (x) =

⎧ ⎪ ⎨x + 1 ⎪ ⎩ x−1

if x < 0, at x = 0. if x ≥ 0.

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Chapter 5 Limits (i) f (0) = 0 − 1 = −1 (ii) lim (x + 1) = 1 and lim+ (x − 1) = −1 − x→0

x→0

Therefore, lim f (x) does not exist since lim (x + 1) = lim+ (x − 1) x→0−

x→0

x→0

Hence, f (x) is discontinuous at x = 0. √ x−2 (b) For f (x) = at x = 4 x−4 √ √ ( x − 2) ( x + 2) x−4 1 1 √ √ = lim = lim √ = . (i) lim x→4 (x − 4) ( x + 2) x→4 (x − 4)( x + 2) x→4 4 x+2 (ii) f (4) is not defined Therefore, f (x) is discontinuous at x = 4, since the function is not defined at x = 4 despite the fact that the limit exists. ⎧ 2 x −1 ⎪ if x < −1, ⎨ x2 −3 (c) f (x) = at x = −1. ⎪ ⎩ 2 x − 3 if x ≥ −1 (i) f (−1) = (−1)2 − 3 = −2 (ii)

x2 − 1 (x + 1)(x − 1) = lim = −2 − − x+1 x+1 x→−1 x→−1 and lim + (x2 − 3) = (−1)2 − 3 = −2 lim

x→−1

Therefore, lim f (x) = −2 x→−1

(iii) f (−1) = −2 = limx→−1 f (x) Therefore, f (x) is continuous at x = −1.

5.6

Continuity of polynomials and rational functions

If p(x) and q(x) are polynomials, then lim p(x) = p(c)

x→c

and lim

x→c

p(c) p(x) = q(x) q(c)

if q(c) = 0.

Therefore, a polynomial or a rational function is continuous whenever p(c) and q(c) are defined.

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An Introduction to Applied Calculus for Social and Life Sciences example 5.6.1 For what value of the constant A is the following function continuous for all real x? ⎧ ⎪ if x < 1 ⎨Ax + 5 f (x) = ⎪ ⎩ 2 x − 3x + 4 if x ≥ 1. Solution The expressions Ax + 5 and x2 − 3x + 4 are both polynomials. Therefore, both will be continuous everywhere, but we need to investigate the behaviour of f (x) at the point x = 1. So, to achieve continuity at x = 1, we must test the conditions for continuity at that point. (a) We know that f (1) is defined and in this case f (1) = 1 − 3 + 4 = 2. (b) The limit must exist and equal 2. The right-hand side limit is lim+ (x2 − 3x + 4) = 2. x→1

and the left-hand side limit exists and we require that lim Ax + 5 = 2.

x→1−

This implies that A + 5 = 2 ⇒ A = −3. (c) For A = −3, then lim f (x) = 2 = f (1). x→1

Therefore, f is continuous for all real values of x if A = −3. Removable discontinuity A function f (x) is said to have a removable discontinuity at c if and only if lim f (x) exists and is finite BUT, either f (x) is not defined at c or lim f (x) = x→c

x→c

f (c).

example 5.6.2 Show that f (x) =

x2 − 9 has a removable discontinuity at x = 3. x−3

Solution First we note that x2 − 9 (x − 3)(x + 3) = lim = lim x + 3 = 6. x→3 x − 3 x→3 x→3 x−3 This means that the limit of f (x) exists and is finite. The function f (x) is, however, not defined at x = 3. The function thus has a removable discontinuity at x = 3. lim

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Chapter 5 Limits

5.7

The Intermediate Value Theorem

Continuity on an interval A function f (x) is said to be continuous on an open interval a < x < b if it is continuous at each point x = c in that interval. Moreover, f is continuous on the closed interval a ≤ x ≤ b if it is continuous on the open interval a < x < b and lim f (x) = f (a) and

x→a+

lim f (x) = f (b).

x→b−

The Intermediate Value Theorem (IVT) Suppose f (x) is a continuous function on an interval a ≤ x ≤ b, and suppose L is a number between f (a) and f (b), then f (c) = L for some c between a and b. In other words, if f is continuous on some interval [a, b] and has the function values f (a) and f (b) at the end points of the interval, then it also has a value between f (a) and f (b) for some point in the interval.

Consider two points connected by a continuous curve as shown. If one draws a horizontal line then there will be at least one place where the line f (x) crosses the horizontal line. So as long as c ∈ (a, b) then f (c) ∈ (f (a), f (b)).

One of the immediate applications of the Intermediate Value Theorem is to show the existence of roots between intervals.

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An Introduction to Applied Calculus for Social and Life Sciences example 5.7.1 Show that the equation the interval 0 ≤ x ≤ 1.

√ 3

x = x2 + 2x − 1 must have at least one solution on

Solution √ We need to prove that the graph of y = 3 x intersects the graph of y = x2 + 2x − 1 on 0 ≤ x ≤ 1. Therefore, √ √ 3 x = x2 + 2x − 1 ⇒ 3 x − x2 − 2x + 1 = 0. √ Now let f (x) = 3 x − x2 − 2x + 1 (the left-hand side of the equation) so that f (x) = 0. We have to show that f (x) has a root/zero on 0 ≤ x ≤ 1. Using the Intermediate Value Theorem, if f (x) is continuous, for L = 0, c represents the solution we are looking for. √ 3 f (0) = 1 ≥ 0 and f (1) = 1 − 1 − 2(1) + 1 = −1 < 0. The change in the sign means that the f (x) must cut the x-axis between x = 0 and x = 1. Therefore, there must exist some number c such that f (c) = 0 where 0 ≤ c ≤ 1. Therefore, f (x) has at least one root on 0 ≤ x ≤ 1.

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Chapter 5 Limits

Exercises

5.1 Determine the limits of the following functions, if they exist: (a) lim (x2 + 7x − 1) x→3

(c) lim (1 − x − 2x4 ) x→−∞

x2 − 1 x→−1 x2 + 3x + 2

(e) lim

(g) lim− x→2

−5x x−2

4 + 6x − 5x2 x→−∞ 2 − 3x + 5x2

(i) lim

(k) lim+ x→1

x2 + 1 x−1

(b) limπ (3 sin θ cos θ) θ→ 2

3x2 − 3 x→1 x + 1 √ x−1 (f) lim x→1 x − 1 (d) lim

2x3 + 3x + 1 x→∞ 3x2 − 5x + 2

(h) lim

1 − 3x3 x→−∞ x4 − 4

(j) lim

(l)

lim

x→−1−

3x2 − 2 x+1

2x − 3 x+1

(m) lim1 (1 + 2x − x2 )

(n) lim

(o) lim (1 + x3 − 5x4 )

(p) lim

(q) limπ (3 sin θ cos θ)

(r) limπ (sin 2θ cos 2θ)

x→1

x→ 2

x→−∞

θ→ 4

1 − x − x3 x→−∞ 1 + x2 − 2x4

(s) lim

x2 + 1 x→−∞ x − 1

(u) lim

(w) lim+ x→1

x+1 x−1

x2 − x − 2 x→2 x2 − 4

θ→ 2

x4 − 1 x→∞ 2x − 3x4 √ x−2 (v) lim x→4 4 − x

(t) lim

(x) lim

x→1−

x+1 x−1

5.2 Find whether the limit of f (x) exists at x = −1 and x = 1: ⎧ 3 x + 3 if x ≤ −1 ⎪ ⎪ ⎪ ⎪ ⎨ −1 < x < 1 x2 + 1 if f (x) = ⎪ ⎪ ⎪ ⎪ ⎩ 3x − 1 if x≥1

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An Introduction to Applied Calculus for Social and Life Sciences 5.3 Determine whether the given function f is continuous at the given point: x−1 x=3 x+1 2x + 3 (b) f (x) = 2 2x − 4x − 30 (a) f (x) =

5.4 Given

x=5

⎧ |x| + 1 if x ≤ −1 ⎪ ⎪ ⎪ ⎪ ⎪ √ ⎪ ⎪ ⎪ ⎪ ⎨ x + 3 if −1 < x < 1 f (x) =

(a) Determine: (i) lim f (x) x→−1

4 ⎪ ⎪ ⎪ ⎪ ⎪ x+1 ⎪ ⎪ ⎪ ⎪ ⎩ 2 x −1

if 1 < x ≤ 3 if x > 3

(ii) lim f (x)

(iii) lim f (x)

x→1

x→3

(b) Is f (x) continuous at points x = −1; x = 1 and/or x = 3? 5.5 List the values of x at which the given function is not continuous ⎧ ⎪ ⎨2x + 3 if x < 1 (a) f (x) = ⎪ ⎩ 6x − 1 if x > 1 x (b) f (x) = (x + 5)(x − 1) ⎧ −x − 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1, ⎪ ⎪ ⎪ ⎪ ⎨ 0, f (x) = ⎪ ⎪ ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ x + 1, ⎪ ⎪ ⎪ ⎪ ⎩ −x + 3,

5.6 Given

x < −2 −2 < x < 0 x=0 02

(a) Draw the graph of f (x). (b) Determine (i) lim f (x) x→−2

(ii) lim f (x)

(iii) lim f (x)

x→0

x→2

(c) Is f (x) continuous at x = −2, x = 0 and x = 2? 5.7 For what values of c do the following functions have removable discontinuities? (a) f (x) =

x2 − 1 x−c

(b) f (x) =

x3 − 6x2 + 11x − 6 x−c

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Chapter 5 Limits 5.8 Show that the equation interval 0 ≤ x ≤ 8. 5.9 Show that the equation interval 0 ≤ x ≤ 9.

√ 3

2

x − 8 + 9x 3 = 29 has at least one solution in the

√

9 − x + 2x = 11 has at least one solution in the

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Chapter 6

Differentiation 6.1

The derivative

The foundations of calculus are built on the derivative. It gives the rate of change of a function with respect to its variable. The derivative is related to the slope of a function. Finding the derivative is equivalent to finding the slope of the tangent line to a function at any given point.

6.1.1

The rate of change and the slope

(a) It is known that A linear function of the form l(x) = mx + b : • changes at a constant rate m with respect to x • the rate of change is given by the slope of the function, m. (b) For a non-linear function, f (x) the rate of change will not be constant. The rate of change at x = c is given by the steepness of the graph of f (x) at the point (c, f (c)); this in turn is measured by the slope of the tangent line to the graph at (c, f (c)). If the slope of the tangent line at x = c is positive, the rate of change is said to be increasing at x = c. If the slope of the tangent line at x = c is negative, the rate of change is said to be decreasing at x = c.

6.1.2

Differentiation using first principles

Gradient function Consider the function f (x) = x2 . Suppose we desire to find the gradient of the curve at a general point A with coordinates (x, x2 ). If we move h units to the right, we will have a new point B whose coordinates are (x + h, (x + h)2 ).

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Chapter 6 Differentiation We can approximate the gradient of the tangent line at A by finding the gradient of the line AB. So, the gradient of the line AB =

BC AC

(x + h)2 − x2 h x2 + 2xh + h2 − x2 = h = 2x + h.

=

Now, as B moves closer and closer to A, h → 0. The limiting value of the gradient is thus given by the function 2x called the gradient function. In general, if the equation of the curve is y = f (x), the gradient function is denoted by f  (x) or simply the derivative of f (x). Definition of a derivative The derivative of f (x) with respect to x is the function f  (x) given by f  (x) = lim

h→0

f (x + h) − f (x) . h

The process of finding the derivative using the gradient function determined by the limit is called differentiation by first principles.

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An Introduction to Applied Calculus for Social and Life Sciences The slope of the tangent line to the curve y = f (x) at the point c, f (c) is m = f  (c). Also the rate of change of f (x) with respect to x when x = c is given by f  (c). example 6.1.1

√ Find the equation of the tangent line to the graph of the function f (x) = 2 x at the point x = 4.

Solution We first find the gradient function of f (x) and the determine its numerical value at x = 4. We will then be able to find the equation of a straight line if we have the gradient and a point. So, f (x + h) − f (x) h→0 h √ √ √ √ 2 x+h−2 x 2 x+h+2 x · √ = lim √ , h→0 h 2 x+h+2 x 4(x + h) − 4x √ = lim √ h→0 h(2 x + h + 2 x) 4h √ = lim √ h→0 h(2 x + h + 2 x) 4 = √ 4 x 1 =√ . x

f  (x) = lim

The slope of the tangent line to f (x) at x = 4 is given by 1 1 m = f  (4) = √ = . 2 4 √ The function value at x = 4 is f (4) = 2 4 = 4. So, the tangent has a slope passes through the point (4, 4).

1 2

and

The equation of the tangent line is y−4=

1 1 (x − 4) ⇒ y = x − 2 + 4, 2 2 1 = x + 2. 2

Why did we not choose x = − 12 ?

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Chapter 6 Differentiation The sign of f  (x) If the function is differentiable at x = c, then f increases at x = c if f  (c) > 0. See figure below.

If the function is differentiable at x = c, then f decreases at x = c if f  (c) < 0. See figure below.

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An Introduction to Applied Calculus for Social and Life Sciences Notation For the function y = f (x), the derivative is written as f  (x) or derivative at x = c (i.e. f  (c)) is written as

dy dx

or

df dx

and the

df dy |x=c or |x=c dx dx instead of ‘if y = x2 , then

dy dx

= 2x’, we can write dy d 2 = (x ) = 2x. dx dx

6.2

Differentiability and continuity

It is important now that we understand continuity to link it to differentiability of a function. A function is differentiable if the derivative exists. We say a function is differentiable at a point c if lim

h→0

f (x + h) − f (x) h

exists. Here is how the two are connected: Differentiability and continuity If a function f (x) is differentiable at x = c, then f is also continuous at x = c but, a function that is continuous at x = c is not necessarily differentiable at x = c. Differentiability implies continuity while continuity does not necessarily imply differentiability. We can quickly understand this by looking at the following example. Consider the graph of y = |x|. At the point x = 0, the function is continuous. However, the derivative does not exist. So the function is continuous but not differentiable at x = 0. It is easy to verify all the three conditions for continuity at x = 0; however, the derivative might not be that obvious. Given that  x if x ≥ 0, f (x) = −x if x < 0. We can determine the derivative by looking on either side of x = 0. For x > 0 we have x+h−x =1 f  (x) = lim h→0 h and for x < 0 we have f  (x) = lim

h→0

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Chapter 6 Differentiation We notice here that the derivatives of the function are not the same from either side of zero, hence the derivative does not exist at x = 0.

example 6.2.1 Find the slope of the tangent line to the graph of f (x) =

1 at x = 2. x2

Solution We first find the derivative using the first principles so that, f (x + h) − f (x) = lim h→0 h→0 h

f  (x) = lim

= lim

1 (x+h)2

−

1 x2

h x2 −(x+h)2 (x+h)2 x2

h −2xh − h2 1 × = lim h→0 (x + h)2 x2 h −2x − h = lim h→0 (x + h)2 x2 −2x = 2 2 (x )x −2 = 3. x x→0

Evaluating the derivative at x = 2 gives m = f  (2) =

−2 1 =− . 23 4

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An Introduction to Applied Calculus for Social and Life Sciences

6.3

Differentiation techniques

The following are the fundamental blocks of the techniques for differentiating functions.

Differentiation techniques d (c) = 0. dx This means that the slope of any horizontal line is zero.

• The derivative of a constant:

• The derivative of a power of x: d n [x ] = nxn−1 dx for any real number n = 0. • The derivative of a constant multiple of a power of x: d d (cf (x)) = c (f (x)) = c·nxn−1 . dx dx • The derivative of the sum/difference of functions is the sum/difference of the derivative of functions d d d [f (x) ± g(x)] = [f (x)] ± [g(x)] . dx dx dx example 6.3.1 Find the derivative of y = x6.7 Solution dy = 6.7x5.7 dx example 6.3.2 Find the derivative of the following (a) y =

4 3 πr 3

(c) y =

3 + e2 2t2

√ 4 (b) y = 2 x3 (d) f (x) =

√

1 x3 + √ x3

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Chapter 6 Differentiation Solutions (a)

dy dr

= 43 π(3)r2 = 4πr2 3

(b) If y = 2x 4 , then dy =2 dx

1 3 3 . x− 4 = √ 4 24x

(c) If y = 32 t−2 + e2 , then dy 3 3 = (−2)t−3 = −3t−3 = − 3 . dt 2 t 3

3

(d) If f (x) = x 2 + x− 2 , then

5 3 1 3 2 f (x) = x + − x− 2 2 2 3√ 3 = x− √ . 2 2 x5 

example 6.3.3 Find the derivative of y = line at (4, −7).

√

x3 − x2 + x162 and find the equation of the tangent

Solution We use differentiation techniques to find the derivative of the function. The derivative is given by 3 1 dy = x 2 − 2x − 32x−3 . dx 2 To determine the gradient of the tangent line to the graph at x = 4 we set m=

dy dx

= x=4

1 3 11 (4) 2 − 2(4) − 32(4)−3 = − . 2 2

Given that we now have the gradient of the line, the equation of the tangent line at (4, −7) is 11 (x − 4) 2 11 y = − x + 22 − 7 2 11 = − x + 15. 2

y − (−7) = −

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An Introduction to Applied Calculus for Social and Life Sciences

6.4 6.4.1

Product and quotient rules The product rule

If f (x) and g(x) are differentiable at x, then y = f (x) · g(x) has a derivative given by the product rule stated as follows:

Product rule d d d [f (x) · g(x)] = f (x) (g(x)) + g(x) (f (x)) dx dx dx or [f (x)g(x)] = f (x)g  (x) + f  (x)g(x)

6.4.2

The quotient rule

If f (x) and g(x) are differentiable functions, with g(x) = 0, then their quotient is differentiable and has a derivative given by the quotient rule that is stated as follows:

Quotient rule   d d (f (x)) − f (x) dx (g(x)) g(x) dx d f (x) if g(x) = 0 = dx g(x) (g(x))2 or 

f (x) g(x)

 =

g(x)f  (x) − f (x)g  (x) if g(x) = 0. g(x)2

example 6.4.1 Find the derivatives of the following functions (a) y = (2x − 1)(3x + 2) (c) y =

x+2 x−2

(b) y = (5 − x3 )(3x + 2) (d) y =

(2x − 1)(x + 3) (x + 1)

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Chapter 6 Differentiation Solutions (a) Let y = f (x) · g(x) where f (x) = 2x − 1 and g(x) = 3x + 2 dy = 2(3x + 2) + 3(2x − 1) dx = 6x + 4 + 6x − 3 = 12x + 1. dy If we had multiplied out, we would have had y = 6x2 + x − 2 so that dx = 12x + 1, giving us the same result. One may ask: Why do we have to use the product rule instead of multiplying out and then differentiating? The answer is that the functions may be too large or tedious to multiply out.

(b) y = (5 − x3 )(3x + 2)   dy = (−3x2 )(3x + 2) + 3(5 − x3 ) dx = −9x3 − 6x2 + 15 − 3x3 = −12x3 − 6x2 + 15. f (x) x+2 = with f (x) = x + 2 and g(x) = x − 2. x−2 g(x) Using the quotient rule we have

(c) y =

1(x − 2) − 1(x + 2) dy = dx (x − 2)2 −4 = . (x − 2)2 (d) y =

G(x) f (x) · g(x) (2x − 1)(x + 3) = = (x + 1) h(x) h(x)

where f (x) = 2x − 1, g(x) = x + 3 and h(x) = x + 1. The numerator is a product, so we use the product rule to differentiate it and the whole function is a quotient. dy G (x)h(x) − h (x)G(x) = dx h(x)2 [2(x + 3) + 1(2x − 1)](x + 1) − (2x − 1)(x + 3)(1) = (x + 1)2 (4x + 5)(x + 1) − (2x − 1)(x + 3) = (x + 1)3 2 4x + 9x + 5 − (2x2 + 5x − 3) = (x + 1)2 2 2x + 4x + 8 = (x + 1)2 2(x2 + 2x + 4) = . (x + 1)2 123

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An Introduction to Applied Calculus for Social and Life Sciences

6.5

Composition of functions

The flow of goods in an economy is usually from a manufacturer to a retailer and then to a consumer. Consider the sale of a new car by Toyota. Let x be the price of producing a new car. The manufacturer adds a markup of 20%. The function f (x) = 1.2x represents the price the retailer buys from the manufacturer. If the retailer sells to the consumer with a markup of 30% then the function g(x) = 1.3f (x) gives the price the consumer pays. We can represent the pricing chain as follows:

Note that the input of the function g(x) is the range of f (x). This is a typical example of a composition of functions where the output of one function is the input of another function. √ The function f (x) = 3 x3 − 7 can be written as a combined or composite function. It is a combination of two simple functions  g(x) = x3 − 7 and f (x) = 3 g(x). Here g(x) is called the core function. Note that f (x) is a function of g(x). So, f (x) is a function of a function. The derivatives of such functions are obtained through the chain rule.

Chain rule If y = f (u) is a differentiable function of u and u = g(x) is in turn a differentiable function of x, then the composite function y = f (g(x)) is a differentiable function of x whose derivative is given by the product dy du dy = · dx du dx or dy = f  (g(x)) · g  (x). dx

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Chapter 6 Differentiation example 6.5.1 Use the chain rule to find (a) y =

1 u2

dy dx .

u = x2 + 5

(b) y = u2

u=

1 x2 + 1

Solutions (a)

dy dy du = · dx du dx = −2u−3 · (2x) −4x = 3 u −4x = 2 . (x + 5)3

(b) Given that u =

1 x2 +1 ,

using the quotient rule we have du −2x = 2 . dx (x + 1)2

So, dy du dy = · dx du dx

−2x = 2u · (x2 + 1)2 1 −4x · = 2 (x + 1)2 x2 + 1 −4x = 2 . (x + 1)3 example 6.5.2 Find

dy dx

(a) y = (x2 − x + 1)7

(b) y =

2 (2x − 3)4

Solutions (a) Let u = x2 − x + 1 be the core function so that y = u7 . 7 If u = x2 − x + 1 then du dx = 2x − 1. Also if y = u then

dy du

= 7u6 .

Therefore dy dy du = · = 7(2x − 1)u6 = 7(2x − 1)(x2 − x + 1)6 . dx du dx 125

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An Introduction to Applied Calculus for Social and Life Sciences (b) Let u = 2x − 3 be the core function so that y = u24 = 2u−4 . dy −4 then du = −8u−5 . If u = 2x − 3 then du dx = 2 and if y = 2u Therefore dy du 16 dy = · = 2(−8u−5 ) = −16(2x − 3)−5 = − . dx du dx (2x − 3)5

6.6

The power rule

Power rule The power rule states that d n [x ] = nxn−1 . dx The general power rule states that d [u(x)]n = n[u(x)]n−1 · u (x) dx for any real number n and differentiable function u(x) for which [u(x)]n−1 is defined.

example 6.6.1 Find the derivative of the following functions  (b) f (x) = (a) f (x) = 5x6 − 6

4 (1 + x2 )3

Solutions 1

(a) Our function can be written as f (x) = (5x6 − 6) 2 . Using the general power rule, with u(x) = 5x6 − 6, we have f  (x) =

1 1 15x5 (5x6 − 6)− 2 · (30x5 ) = √ . 2 5x6 − 6

(b) The function can also be written as f (x) = 4(1 + x2 )−3 . The derivative is given by f  (x) = 4(−3)(1 + x2 )−4 (2x) −24x = . (1 + x2 )4

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Chapter 6 Differentiation example 6.6.2



Find the equation of the tangent line to the graph of f (x) = the point x = 3.

x+1 x−1

2 at

Solution  2 When x = 3 we can find the corresponding y-value so that f (x) = 42 = 22 = 4. So, we desire to find the equation of the tangent line at (3, 4). The derivative can be found using the general power rule and quotient rule.

1 x+1 1(x − 1) − 1(x + 1) x−1 (x − 1)2

x+1 −2 =2 x−1 (x − 1)2 −4(x + 1) . = (x − 1)3

f  (x) = 2

Alternatively,we can use the general power rule and product rule on f (x) = (x + 1)2 (x − 1)−2 . It is left as an exercise to show that we will indeed arrive at the same result. The = −2. slope at x = 3 is m = f  (3) = −4(4) 23 Therefore, the equation of the tangent line to f (x) at x = 3 is y − 4 = −2(x − 3) ⇒ y = −2x + 10.

6.7

Higher order derivatives

dy dy is If y is a function of x then dx is either a function of x or a constant. dx often referred to as the derivative. We can differentiate the derivative with respect d dy d2 y to x to get the second derivative . The derivative of the second = dx dx dx2 2 d d y d3 y derivative yields , which is the third derivative. This process can = dx dx2 dx3 dn y . The power of be continued for say n ∈ N times, so that the nth derivative is dxn the derivative defines the order of the derivative. f  (x) =

dy d2 y d3 y dn y , f  (x) = 2 , f  (x) = 3 , . . . , f ( n)(x) = n dx dx dx dx

respectively give the first, second, third and the nth derivatives of the function.

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An Introduction to Applied Calculus for Social and Life Sciences example 6.7.1 Find the second derivative of y = (x2 − x)(2x −

1 ) with respect to x. x

Solution We first find the first derivative. We thus have 1 dy 1 = (2x − 1)(2x − ) + (x2 − x)(2 + 2 ) dx x x 1 1 = 4x2 − 2x − 2 + + 2x2 − 2x + 1 − x x = 6x2 − 4x − 1. The second derivative is the derivative of the first derivative. In this case we will have d2 y d dy d (6x2 − 4x − 1) = 12x − 4. = = dx2 dx dx dx

6.8

Evaluating limits using L’Hôpital’s rule

L’Hôpital’s rule If lim f (x) = 0 and limx→c g(x) = 0, then x→c

f (x) f  (x) = lim  . x→c g(x) x→c g (x) lim

If limx→c f (x) = ∞ and limx→c g(x) = ∞, then f (x) f  (x) = lim  . x→c g(x) x→c g (x) lim

Note: (a) L’Hôpital’s rule dictates that we find the derivatives of the numerator and denominator separately; that is the quotient rule should not be applied. (b) L’Hôpital’s rule is only applicable for which the limit we are  if the quotient finding is of indeterminate form 00 or ∞ ∞ .

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Chapter 6 Differentiation example 6.8.1 Use L’Hôpital’s rule to find the given limits x3 − 3x2 x→0 3x2 + 2x

x2 (x − 1) x→0 3x2 + 2x − 5

(b) lim

(a) lim

Solutions x3 − 3x2 of form (a) lim 2 x→0 3x + 2x

0 . Using L’Hôpital’s rule, we have 0

0 x3 − 3x2 3x2 − 6x = lim = = 0. 2 x→0 3x + 2x x→0 6x + 2 2 lim

x2 (x − 1) 0 = = 0. x→0 3x3 + 2x − 5 −5

(b) lim

6.9

Implicit differentiation

Some functions can be explicitly expressed in terms of only one variable, for example y = x2 + 3x + 1, y =

 x3 + 1 , y = 1 − x2 2x − 3

or in general y = f (x). However, some functions can be described implicitly by a relation between variables, say x and y such as x2 + y 2 = −1, x3 + y 3 = 4xy and cos(xy) = xy 2 . Consider x2 y 3 − 5y 3 = x + 6

(6.1)

so that x+6 y = 2 x −5 3

 ⇒ y=

3

x+6 . x2 − 5

(6.2)

Finding the derivative of (6.2) is possible but tedious. There is a way of finding the derivative of (6.1) without writing y explicitly as a function of x. We use a technique based on the chain rule to find the derivative of (6.1). We shall illustrate the technique through the following examples.

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An Introduction to Applied Calculus for Social and Life Sciences example 6.9.1 Find the derivative

dy dx

by means of implicit differentiation (b) x2 + y 3 = 1

(a) xy = 7 (c) 3x +

1 = x2 y

(d) 5x − x2 y 3 = 2y

Solutions (a) Remember, y = f (x) is a function of x, so xy = 7 can be written as xf (x) = 7. If we are to find the derivative, we apply the product rule on the left of the ‘equal sign’ so that 1 · f (x) + x · f  (x) = 0 ⇒ So

xf  (x) = −f (x)

f (x) x y dy =− . ⇒ dx x f  (x) = −

Alternatively, for the function x · y = 7, we can find the derivative by applying the product rule. However, when we find the derivative of y with respect to x, we need to apply the chain rule, i.e. the derivative becomes 1·y+x·1·

dy dy =0 ⇒ x = −y dx dx dy ⇒ dx

y =− . x

Note: When we find the derivative of y, we multiply the derivative by

dy dx .

(b) Given x2 + y 3 = 1, then dy =0 dx dy = −2x ⇒ 3y 2 dx 2x dy = − 2. ⇒ dx 3y

2x + 3y 2

(c) Given x +

1 y

= x2 , then dy = 2x dx 1 dy = 2x − 1 − 2 y dx dy = −y 2 (2x − 1). dx

1 + (−1)y −2

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Chapter 6 Differentiation (d) Given 5x − x2 y 3 = 2y, then   dy dy 5 − 2xy 3 + x2 · 3y 2 =2 dx dx dy dy −2 =0 5 − 2xy 3 − 3x2 y 2 dx dx dy = 2xy 3 − 5 (−3x2 y 2 − 2) dx dy 2xy 3 − 5 =− 2 2 . dx 3x y + 2 Let us revisit (6.1), i.e. x2 y 3 − 5y 3 = x + 6. The derivative with respect to x is obtained as follows: dy dy ] − 5(3)y 2 =1 dx dx dy = 1 − 2xy 3 [3x2 y 2 − 15y 2 ] dx dy 1 − 2xy 3 = 2 2 . dx 3x y − 15y 2

[2xy 3 + x2 · 3y 2

example 6.9.2 Find (a)

dy dx

by means of implicit differentiation for (3xy 2 + 1)4 = 2x − 3y.

(b) the equation of the tangent line to the curve at the given point: (x2 + 2y)3 = 2xy 2 + 64

at

(0, 2).

Solutions (a) Implicit differentiation yields   dy dy 2 4(3xy + 1) 3y + 3x(2y) =2−3 dx dx dy dy +3 =2 4(3xy 2 + 1)3 3y 2 + 4(3xy 2 + 1)3 6xy dx dx 2

3

  dy 24xy(3xy 2 + 1)3 + 3 = 2 − 12y 2 (3xy 2 + 1)3 dx 2 − 12y 2 (3xy 3 + 1)3 dy = . dx 24xy(3xy 2 + 1)3 + 3

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An Introduction to Applied Calculus for Social and Life Sciences (b) Differentiating the function yields     dy dy 2 = 2y + 2x(2y) 3(x + 2y) 2x + 2 dx dx dy dy − 4xy = 2y 2 3(x2 + 2y)2 (2x) + 3(x2 + 2y)2 · 2 dx dx dy = 2y 2 − 6x(x2 + 2y)2 [6(x2 + 2y)2 − 4xy] dx 2

2

2y 2 − 6x(x2 + 2y)2 dy = . dx 6(x2 + 2y)2 − 4xy The slope is given as m=

dy 2(2)2 − 6(0)(0 + 2(2))2 |(x,y)=(0,2) = dx 6(02 + 2(2))2 − 4(0)(2) 1 . = 12

The tangent line is represented by 1 (x − 0) 12 1 y= x + 2. 12

y−2=

6.10

Related rates

6.10.1

An algorithm for solving related rates problems

(a) Draw a figure (if appropriate) and assign variables. (b) Find a formula (or formulas) relating the variables. (c) Use implicit differentiation to find how the rates are related. (d) Substitute any given numerical information into the equation in step (c) and find desired rate of change.

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Chapter 6 Differentiation example 6.10.1 A 10 m long ladder leans against a wall. The top of the ladder is sliding down the wall at a rate of 3 m/sec. How fast is the foot of the ladder moving away from the wall when the top is 6 m above the ground? Solution (a) The pictorial representation of the problem is given by the figure below.

(b) Find equations that relate x and y to each other. In this case, we use Pythagoras’ theorem so that x2 + y 2 = 102 .

(6.3)

(c) Use implicit differentiation to relate the rates of change. Remember, the word rate (of change) refers to finding the derivative with respect to time t. Therefore, we differentiate x as well as y implicitly with respect to time t. 2x dx dt dy dt

dy dx + 2y · =0 dt dt

(6.4)

gives the rate at which the length x increases at time t gives the rate at which the length y decreases at time t.

Therefore,

dy dt

will be negative and

dx dt

will be positive.

(d) The given numerical values give dy dt

= −3 m/sec and y = 6 m.

We wish to find

dx dt .

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An Introduction to Applied Calculus for Social and Life Sciences If y = 6 we find x by substituting for y in (6.3) so that x2 = 102 − 62 = 100 − 36 = 64. Therefore x = 8x Since x is distance, it is positive. Substituting the values of x and y into (6.4) we obtain 2(8)

dx dx + 2(6)(−3) = 0 ⇒ 16 = 36 dt dt 36 9 dx = = m/sec. ⇒ dt 16 4

example 6.10.2 It is estimated that t years from now, the population of suburb in Cape Town will be p(t) = 40 −

3 thousand t+1

(a) Derive a formula for the rate at which the population will be changing with respect to time t years from now. (b) At what rate will the population be growing 1 year from now? (c) By how much will the population actually increase during the second year? (d) At what rate will the population be growing 9 years from now? (e) What will happen to the rate of population growth in the long run? Solutions (a) The rate of change is given by the derivative of p(t) with respect to time. We thus have p (t) = −

0(t + 1) − 3(1) 3 = thousand people/year. 2 (t + 1) (t + 1)2

(b) After one year, the rate of growth is obtained by finding p (1). So, p (1) =

3 3 = thousand people/year 2 2 4 = 750 people/year.

(c) The population size after the second year is obtained from p(2). We have p(2) = (40 −

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Chapter 6 Differentiation The population size after one year is given by p(1) = (40 −

3 ) × 1 000 = (40 − 1.5) × 1 000 1+1 = 38 500 people.

The change in the second year is thus given by p(2) − p(1) = 39 000 − 38 500 = 500 people. (d) The growth rate after 9 years is given by p (9) = −

3 3 thousand people/year =− 2 (9 + 1) 100 = −30 people/year.

The population is increasing at a rate of 30 people/year. (e) The long run can simply be interpreted as after a very long time, i.e. as t → ∞. We thus have lim p (t) = lim

t→∞

t→∞

3 = 0. (t + 1)2

Therefore the rate of population growth will approach zero in the long run.

6.10.2

Rectilinear motion

Rectilinear motion If the position at time t of an object moving along a straight line is given by s(t), then the object has velocity v(t) = s (t) =

ds dt

and acceleration a(t) = v  (t) =

dv . dt

The object is advancing when v(t) > 0, retreating when v(t) < 0. It is stationary when v(t) = 0. It is accelerating when a(t) > 0 and decelerating when a(t) < 0.

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An Introduction to Applied Calculus for Social and Life Sciences example 6.10.3 The function s(t) that gives the position of a moving particle at any time t is given by s(t) = 2t3 − 18t2 + 30t + 50 for 0 ≤ t ≤ 6 . Find (a) the velocity of the particle (b) the acceleration of the particle (c) all times in the given interval when the particle is stationary (d) the acceleration at t = 4. Solutions (a) The velocity v is given by

ds dt

= s (t) so that v(t) = s (t) = 6t2 − 36t + 30.

(b) The acceleration of the particle is a(t) = v  (t) = 12t − 36. (c) Note that v(t) = 0 when 6t2 − 36t + 30 = 0. So 6t2 − 36t + 30 = 0 6(t2 − 6t + 5) = 0 6(t − 1)(t − 5) = 0. Therefore, t = 1 or t = 5. Here, both values fall within the interval 0 ≤ t ≤ 6. Therefore, the particle becomes stationary when t = 1 and t = 5. (d) The acceleration at t = 4 is given by a(4) = 12(4) − 36 = 12.

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Chapter 6 Differentiation

Exercises

6.1 Find the derivative of the function f (x) = √

1 x+1

by means of the first principles and find the equation of the line tangent to the curve at x = 4. 6.2 Find the tangent to the graph of f (x) at the given x-value: −2 at x = 1 x √ (d) f (x) = x3 at x = 1

(a) f (x) = x2 − 1 at x = −1 √ 3

(c) f (x) =

6.3 Calculate

(b) f (x) =

x2 at x = −1

dy dx :



√ (a) y = (3x2 + 1) x + 1

1−

(b) y =



1−

√

x

6.4 Determine the derivatives of the following functions: (a) f (x) =

1√ 1 3 + x− √ 2x 2 x

(b) y =

(c) y = (x3 − 7)2 (x − x2 ) (e) f (x) =

 3

1 x2 + 1

(d) f (x) = (x4 + 10x − 20)5

(1 − x2 )4

(f) y = e3x+2

(g) y = 3x5 − 4x3 + 9x − 6

(h) y =

√

5x3 +

5 1 − √ x2 53x

(i) y =

x2 + 1 2x − 3

(j) y = (x3 − 1)(2x3 + 3)

(k) y =

2x + 1 (x + 1)2

(l) y = x3

4

(m) f (x) = x 3 −



√ 1 1 3 + 2 x3 − + √ 3 3 3x x x4

(o) y = (x2 + 1)(2x3 − x) (q) f (x) =



(n) y =

2 +3 x3

x2 + 1 x3 − 1

(p) f (x) = (x4 + 10x2 − 1)5

(x2 + 1)3

(r) f (x) =

 4 (3x4 − 2x − 10)3

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An Introduction to Applied Calculus for Social and Life Sciences 6.5 Determine the fourth derivative f (4) (x) of f (x) = 6.6 Determine

d2 y if: dx2

(a) y = x2 − 4 (c) y =

1 . 1−x

(b) y = (x2 +

x+1 1 − 2x

(d) y =

6.7 Determine the derivative √



1 )(x2 − 1) x

1 − x2

dy at the value of x, by using the chain rule: dx

u, u = x2 − 2 for x = 2 1 , u = 2x2 + 1 for x = 0 (b) y = u+1 (a) y =

6.8 Find the derivative of the following functions: (a) f (x) = (4 + 5x2 − 2x3 )3

(b) f (x) =

(c) f (x) = (3 − x)3 (x2 + 1)5

(d) f (x) =

√

(e) f (x) =

5x + 1 x2 + 1 x+1

3

1 2(x3 + 1)3

6.9 Find an equation of the line that is tangent to the graph of f (x) for the given value of x  1 ; x = −1. f (x) = 3 x+2 6.10 Find the derivative,

dy , by implicit differentiation dx

(a) x2 − y 2 = 1

(b) x2 y + xy 2 = 3x

6.11 Find the equation of the tangent line to the curve at the given points: (a) x2 = y 3

at (8, 4)

(b) x y − 2xy = 6x + y + 1 2 3 2

3

2

(c) (x + 2y) = 2xy + 64 6.12 If y = x3 + 2x and

at (0, −1) at (0, 2)

dy dx = 6, find when x = 2. dt dt

6.13 Find the equation of the tangent line to the curve at the given point x2 + xy + y 2 = 3;

(1, 1).

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Chapter 6 Differentiation 6.14 An object is moving in a clockwise direction around the unit circle x2 +y 2 = 1. √ 1 3 As it passes through the point ( 2 , 2 ), its y-coordinate is decreasing at the rate of 3 units/second. At what rate is the x-coordinate changing at this point? 6.15 Use L’Hôpital’s rule, where applicable, to find the following limits: (a) lim

x→−2 x2

(c) lim

x→−∞

x2 − 1 x→−1 x + 1

x+2 + 3x + 2 x2 − 4x + 1 2 + x − 2x2

(b) lim

2x3 + 3x + 4 x→0 x − x3

(d) lim

6.16 Calculate the second derivative of the given function: (a) f (x) =

3 √ x − x4 − 2

(b) f (x) =

x3 − 5 1 + x2

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Chapter 7

Derivatives of Exponential and Logarithmic Functions 7.1

Derivatives of logarithmic functions

Derivatives of logarithmic functions The derivative of y = ln x is d 1 (ln x) = , dx x

for x > 0.

The derivative of y = ln[u(x)] is d 1 du 1 [ln(u(x))] = · = · u (x). dx u(x) dx u(x)

example 7.1.1 Differentiate the following logarithmic functions: (a) f (x) = (ln x)3

(b) f (x) = ln x3

(c) f (x) = x2 · ln x

(d) f (x) = ln



x+1 x−1

Solutions (a) For the function f (x) = (ln x)3 , we let u = ln x so that f (x) = u3 . We can thus use the chain rule to find the derivative 1 df (x) du = 3u2 , and = du dx x

⇒ f  (x) =

df (x) du df (x) 1 = · = 3(ln x)2 · dx du dx x

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Chapter 7 Derivatives of Exponential and Logarithmic Functions (b) Here we first apply one of the laws of logarithms so that f (x) = ln x3 = 3 ln x. The derivative is 1 3  f (x) = 3 = . x x (c) We use the product rule so that f  (x) = 2x · ln x + x2

1 x

= 2x ln x + x. (d) We can rewrite the logarithmic function as f (x) = ln[(x + 1)(x − 1)−1 ] then apply the product rule. f  (x) = = = = =

  1 (x − 1)−1 + (x + 1)(−1)(x − 1)−2 −1 (x + 1)(x − 1)  x−1 (x − 1)−2 ((x − 1) − (x + 1)) x+1 x−1 (x − 1)−2 (−2) x+1 −2(x − 1) (x + 1)(x − 1)2 −2 . (x + 1)(x − 1)

Alternatively, we can use the rules of logarithms so that

x+1 f (x) = ln x−1 = ln(x + 1) − ln(x − 1). The derivative is given by 1 1 − x+1 x−1 x − 1 − (x + 1) = (x + 1)(x − 1) −2 . = (x + 1)(x − 1)

f  (x) =

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An Introduction to Applied Calculus for Social and Life Sciences

7.2

Derivatives of exponential functions

Derivatives of exponential functions The derivative of y = ex is given by d x (e ) = ex . dx The derivative of y = eu(x) is d u(x) (e ) = eu(x) · u (x). dx

example 7.2.1 Differentiate the following exponential functions: 2

(a) f (x) = 3e4x

2

(b) f (x) = xe−x

+1

(c) f (x) = ln(e−x + x)

(d) h(x) =

e−x x2

Solutions 2

(a) If f (x) = 3e4x

+1

then

2  2 f  (x) = 3 e4x +1 · 8x = 24xe4x +1 .

2

(b) If f (x) = xe−x then by the product rule we have 2

2

2

f  (x) = (1)e−x + xe−x (−2x) = e−x (1 − 2x2 ). (c) If f (x) = ln(e−x + x) then  −x 1 −e + 1 +x 1 − e−x . = −x e +x

f  (x) =

(d) If h(x) =

e−x x2

e−x

then by the quotient rule we have −e−x (x2 ) − e−x (2x) x4 −x e (−x)(x + 2) = x4 −e−x (x + 2) = . x3

h (x) =

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Chapter 7 Derivatives of Exponential and Logarithmic Functions example 7.2.2 Use L’Hôpital’s rule to find the given limits 1

(a) lim x3 e−2x

(b) lim (1 + 2x) x

x→∞

x→0

Solutions

∞ x3 which is of the form , so we use L’Hôpital’s rule. 2x x→∞ x→∞ e ∞

∞ 2 3x which again is of the form = lim x→∞ 2e2x ∞

∞ 6x which again is of the form = lim x→∞ 4e2x ∞ 6 = lim x→∞ 8e2x = 0. So, as long as we still have an indeterminate form, we continue to apply the rule.

(a) lim x3 e−2x = lim

1

1

(b) To determine the limit lim (1 + 2x) x , we rewrite (1 + 2x) x by setting x→0

1

y = (1 + 2x) x . Taking logarithms of both sides we have 1

ln y = ln(1 + 2x) x =

ln(1 + 2x) 1 ln(1 + 2x) = . x x

Here we easily note that limx→0 ln(1+2x) has an indeterminate form, i.e. if we x substitute x = 0, we get ln01 = 00 . Applying L’Hôpital’s rule:

ln(1 + 2x) = lim x→0 x→0 x

lim ln y = lim

x→0

1 1+2x

 (2)

1 2 = lim x→0 1 + 2x = 2.

We need to find lim (y) instead of limx→0 (ln y) . Our solution gives x→0

ln y → 2 as x → 0 ⇒ eln y → e2 as x → 0. So,

1

y → e2 as x → 0 ⇒ (1 + 2x) x → e2 as x → 0.

7.3

Logarithmic differentiation

This process involves the use of logarithms to aid the differentiation process. The technique is used when it is easier to differentiate the logarithm of a function rather 143

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An Introduction to Applied Calculus for Social and Life Sciences than the function itself. When a function is composed of a product that can be turned into a sum through a logarithmic transformation it becomes easier to differentiate. At times when a function is raised to some power, logarithms can transform the function into one that can easily be differentiated. example 7.3.1 Find f  (x) using logarithmic differentiation 2 −x

(a) f (x) = x e

(c) f (x) =

3

(3x + 5)

(b) f (x) =

 4

2x + 1 1 − 3x

√ e−3x 2x − 5 (6 − 5x)4

Solutions   (a) Take the natural logarithm on both sides, i.e. ln f (x) = ln x2 e−x (3x + 5)3 Applying the rules of logarithms to the expression on the right-hand side gives:   ln f (x) = ln x2 e−x (3x + 5)3 = ln x2 + ln e−x + ln(3x + 5)3 = 2 ln x − x ln e + 3 ln(3x + 5). Take the derivative on both sides with respect to x for the function ln f (x) = 2 ln x − x ln e + 3 ln(3x + 5) to obtain

1 1 − ln e + 3 (3) x 3x + 5   2 9 −1+ , since ln e = 1 f  (x) = f (x) x 3x + 5   2 9 −1+ . = x2 e−x (3x + 5)3 x 3x + 5

1 · f  (x) = 2 f (x)

 (b) Given that f (x) =

4

2x + 1 , taking logarithms on either side gives 1 − 3x

1 2x + 1 4 ln f (x) = ln 1 − 3x

2x + 1 1 = ln . 4 1 − 3x This simplifies to ln f (x) =

1 4

[ln(2x + 1) − ln(1 − 3x)]. 144

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Chapter 7 Derivatives of Exponential and Logarithmic Functions Take the derivative with respect to x on both sides. 



 1 1 1 1 · f  (x) = (2) − (−3) f (x) 4 2x + 1 1 − 3x

=

1 4



2 3 + 2x + 1 1 − 3x

f  (x) = f (x) ·

1 = 4

 4

1 4



.

2 3 + 2x + 1 1 − 3x

2x + 1 1 − 3x



2 3 + 2x + 1 1 − 3x

.

(c) Expressing the function in logarithmic terms yields

−3x √ e 2x − 5 ln f (x) = ln (6 − 5x)4  −3x √ = ln e 2x − 5 − ln(6 − 5x)4 1

= ln e−3x + ln(2x − 5) 2 − 4 ln(6 − 5x) 1 = −3x ln e + ln (2x − 5) + 4 ln(6 − 5x) 2 1 = −3x + ln(2x − 5) − 4 ln(6 − 5x). 2 Take the derivative with respect to x on both sides to obtain



1 1 1 1 · f  (x) = −3 + (2) − 4 (−5) f (x) 2 2x − 5 6 − 5x   20 1 + f  (x) = f (x) −3 + 2x − 5 6 − 5x √   −3x 20 2x − 5 1 e + −3 + . = (6 − 5x)4 2x − 5 6 − 5x Note that we do not pay much attention to the simplification of some of the rational functions. This is left as an exercise to the reader.

7.4

Derivatives of trigonometric functions

If x is measured in radians, then (a)

d dx (sin x)

= cos x

(b)

d dx (cos x)

= − sin x

(c)

d dx (tan x)

= sec2 x 145

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An Introduction to Applied Calculus for Social and Life Sciences We give a table below that summarises the derivatives of the basic trigonometric functions dealt with so far. The derivative formulas for the common trigonometric functions are: Function sin(x) cos(x) tan(x) cot(x) sec(x) csc(x)

Derivative cos(x) − sin(x) sec2 (x) − csc2 (x) csc(x) is cosec(x) sec(x) tan(x) − csc(x) cot(x).

We show how to evaluate of derivative of sin(x) and that of tan(x). We begin by considering the derivative of sin(x). We are now ready to find the derivative of sin(x) from first principles. lim

h→0

f (x + h) − f (x) sin(x + h) − sin x = lim . h→0 h h

Using difference of 2 sines formula, we have



(x + h) − x (x + h) + x sin(x + h) − sin(x) = 2 sin cos 2 2

h h = 2 sin cos x + 2 2 We thus have

  2 sin h2 cos x + h2 f (x + h) − f (x) = lim lim h→0 h→0 h h 

sin h2 h = lim · lim cos x + . h h→0 h→0 2 2

sin(x) = 1 and the result follows. Hence x d sin (x) = cos (x). dx We now consider the derivative of tan(x).

d sin x d (tan x) = dx dx cos x cos x(cos x) − sin x(− sin x) , using the quotient rule = cos2 x cos2 x + sin2 x = cos2 x

2 1 = cos x We know that lim

x→0

= sec2 x. 146

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Chapter 7 Derivatives of Exponential and Logarithmic Functions We leave it as an exercise to the reader to show the remaining derivatives. When the argument of a trigonometric function is a function, then the chain rule is used to evaluate the derivative. For example, given that y = sin(2x2 + x + 1) the derivative is obtained by setting u = 2x2 + x + 1 so that y = sin(u). Here dy dy du = · = cos(u) · (4x + 1) = (4x + 1) cos(2x2 + x + 1). dx du dx In summary, this is the derivative of the main function (the sine, in this case) multiplied by the derivative of what is inside the argument.

Using the chain rule (a)

d (sin f (x)) = cos(f (x)) · f  (x) dx

(b)

d (cos f (x)) = − sin(f (x)) · f  (x) dx

(c)

d (tan f (x)) = sec2 (f (x)) · f  (x) dx

example 7.4.1 Differentiate the following functions (a) f (θ) = cos(θ3 + 2) u2 2

(c) f (u) = e−

(b) f (t) = cos2

cos (2πu)

(d) f (x) =

π 2

− 2t



sin 2x 1 + sin x 3

(e) f (x) = ln(tan2 x)

(f) f (θ) = [ln(tan θ)]

Solutions (a) f  (θ) = − sin(θ3 + 2)(3θ2 ) = −3θ2 sin(θ3 + 2)

π 1   

π − sin − 2t − 2t (−2) (b) f  (t) = 2 cos 2 

π 

π2 − 2t sin − 2t = 4 cos 2

2π  = 2 sin 2 − 2t . (using the identity sin 2A = 2 sin A cos A). 2 (c) f  (u) = cos (2πu)e− 2

− u2

= −e

u2 2

(−u) + e−

u2 2

(− sin (2πu)) (2π)

[u cos (2πu) + 2π sin (2πu)] . 147

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An Introduction to Applied Calculus for Social and Life Sciences 2 cos (2x)(1 + sin x) − sin 2x(cos x) (1 + sin x)2 2 cos(2x) + 2 cos(2x) sin x − sin 2x cos x = . (1 + sin x)2 You can play around with trigonometric identities to simplify the answer, especially the numerator.

(d) f  (x) =

(e) We can use properties of logarithms to simplify the expression before differentiating. We will thus have f (x) = ln(tan2 x) = ln(tan x)2 = 2 ln(tan x). Differentiating the function f (x) gives 2  2 sec x tan x 2 2 sec x . = tan x

f  (x) =

1 · sec2 θ tan θ sec2 θ = 3[ln(tan θ)]2 . tan θ

(f) f  (θ) = 3[ln(tan θ)]2 ·

7.5

Learning curves

A graph that depicts the increase in learning as experience in a task increases is called a learning curve. The experience is usually measured by time of exposure to a certain task. So the horizontal axis is often represented by time and the vertical axis by the acquired skill or proficiency.

Learning curve definition The graph of a function of the form Q(t) = K − Q0 e−kt , where K, Q0 and k are positive constants, is called a learning curve. To sketch the curve, we observe that the graph is always increasing since Q (t) = −Q0 (e−kt )(−k) = Q0 ke−k > 0. The y-intercept is given by Q(0) = K − Q0 and the maximum possible learning is lim (K − Q0 e−kt )

t→∞

= K.

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Chapter 7 Derivatives of Exponential and Logarithmic Functions Below is a figure depicting a typical learning curve.

example 7.5.1 Lecturers believe that when a student is asked to recall a set of formulas, the number of formulas recalled after t minutes is given by a function of the form Q(t)

= Q0 (1 − e−kt ), k positive constant

and Q0 is the total number of relevant formulas in the student memory. (a) What happens to the graph as t increases without bound? Explain the behaviour in practical term. (b) Sketch the graph of Q(t).

Solutions (a) A student can recall all the relevant formulas in his/her memory, if given enough time, so lim Q0 (1 − e−kt )

t→∞

= Q0 (1 − 0) = Q0 .

(b) Before we sketch the graph we need to note that at t = 0, Q = 0 and limt→∞ Q(t) = Q0 .

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An Introduction to Applied Calculus for Social and Life Sciences

7.6

Logistic curves

The graph of a function of the form Q(t)

=

L 1 + Q0 e−Lkt

where L, Q0 and k are positive constants, is called a logistic curve. To sketch the graph, find Q (t)

= L(−1)(1 + Q0 e−Lkt )−2 (Q0 e−Lkt )(−Lk) Q0 L2 ke−Lkt = > 0. (1 + Q0 e−Lkt )2

Since Q (t) > 0 for all t, the graph of Q(t) is always increasing. The y-intercept L . There is no vertical asymptote, but there is a horizontal given by Q(0) = 1+Q 0 asymptote y = lim

t→∞

L 1 + Q0 e−Lkt

=

L = L. 1+0

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Chapter 7 Derivatives of Exponential and Logarithmic Functions

Exercises

7.1 Use the chain rule to perform the given differentiation: (a)

d sin(ln(cos x)) dx

(b)

d ln(esin x + 2x) dx

7.2 Determine the derivative the following functions: (a) f (x) = ln x cos x

(b) f (x) = x2 esin x

(c) f (t) = tan2 (1 + t3 )

(d) f (x) = [ln(cos2 (1 − 3x2 )]2

7.3 Evaluate the derivative the following functions: 2

ln x 1 − cos x

(a) f (x) = e−2x sin x

(b) f (x) =

(c) f (t) = sin(1 − 2t3 )5

(d) f (x) = [ln(cos2 (3x2 + 1)]3

 (e) f (x) = tan( x2 + 1)

(f) f (x) =

(g) y = e3x+2

(h) y =

(i) y = ln(4 − x2 + x3 )

(j) y = x3 e−2x

(k) f (x) = e4x ln(3x2 + 1)

(l) y = ln(3x + 2)

e−x −3

(n) y = ln(x2 − 1)

(o) y = x−3 ln x

(p) f (x) = tan(sin

(m) y =

7.4 Find

x2

 3

(1 − x2 )4

e−x ln x



x2 + 1)

dy , by implicit differentiation: dx

(a) x2 y 2 + x tan y = 4

(b) 1 + x = sin(xy 2 )

(c) x2 cos y + sin (2y) = xy

(d)yex−x = x + y

(e) xe−y + ye−x = 3

(f) x ln y + x2 y = y + 1

2

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An Introduction to Applied Calculus for Social and Life Sciences 7.5 Use L’Hôpital’s rule, where applicable, to find the limits: cos x 1 + sin x

(b) lim

x2 x→0 1 − cos x

(d) lim

(a) limπ x→ 2

x→0−

3x3 x→0 2e3x − 2

(c) lim

(e) lim

x→∞

ex − 1 x2

ln x x

(f) lim xe−x x→∞

√

1

(g) lim (1 − 2x) x

(h) lim+

x→0

x→0

x sin x

7.6 Calculate the second derivative of the given function: (a) f (x) =

√

sin x + tan x

(b) f (x) =

cos x x3 + x

7.7 Differentiate the following: (a) y = tan x sec x (b) y = e(−x/2) sin(ax) where a is a constant (c) y = be

(−x2 /8)

where b a constant

2

(d) y = ln[sin(x )] 7.8 Use the technique of logarithmic differentiation to calculate the derivative of each of the following functions:

2x + 4 sin x(x4 + 2x) 4 (b) x (a) x2 (x + 7) x2 − x !

2

(x + 2)6 ex +1 (c) f (x) = x3 − 4x ! (e) f (x) =

3

(d) f (x) =

x2 e−3x2 tan(3x) 2

x3 ex3 tan(2x)

(g) f (x) = (3x + 1)4 (1 − x2 )5

4

(f) f (x) = 

(2x + 1)6 ex √ 3 1 − x2

+1

(x3 + 1)3

7.9 A petri dish has 10 000 bacteria present at 8:00 a.m and 15 000 present at 11:00 p.m. How many bacteria will there be at 12:00 p.m. if the population of bacteria grows exponentially?

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Chapter 7 Derivatives of Exponential and Logarithmic Functions 7.10 A hot chocolate drink is taken outside on a cold winter day when the air temperature is −5◦ C. According to a principle of physics called Newton’s law of cooling, the temperature T (in degrees Celsius) of the drink t minutes after being taken outside is given by a function of the form T (t) = −5 + Ae−kt where A and k are positive constants. Suppose the temperature of the drink is 80◦ C when it is taken outside and 20 minutes later is 25◦ C. (a) Determine A and k. (b) What happens to the temperature of the drink as t increases indefinitely? (c) What will the temperature be after 30 minutes? (d) When will the temperature reach 0◦ C? 7.11 The fish population in Klipdrift dam is growing according to the function y=

2 000 1 + 49e−0.3t

where y is the number of fish after t months. (a) How many fish were there in the beginning? (b) How many fish will there be after 10 months? (c) What is the maximum number of fish the lake can support (the carrying capacity)? (d) When will the fish population be growing at its fastest? (e) Sketch the graph of the function. 7.12 A cool object with temperature 10◦ C is placed in a room that is maintained at a constant temperature of 20◦ C. The rate at which the temperature of the object rises is proportional to the difference between the room temperature and the temperature of the object. Let y = f (t) be the temperature of the object at time t. (a) Write down a differential equation which describes the rate of change of f (t). (b) If the differential equation in (a) above is being solved, then the solution is y = f (t) = 20 − αe−kt , with α ≥ 0. Determine the value of α. (c) After 3 minutes the temperature of the object is 15◦ C Now determine the value of k and express it in terms of the ln-function. (d) After how many minutes is the temperature of the object 18◦ C? Express your answer in terms of the ln-function.

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An Introduction to Applied Calculus for Social and Life Sciences 7.13 The size of a certain insect population near Clanwilliam is given by P (t) = 100e0.02t , where t is measured in days. (a) How many insects were originally in the population? (b) When will the population double its size? (c) When will the population be 1 000? 7.14 It is estimated that t years from now, the population of a certain country will be 20 million. P (t) = 2 + 3e−0.06t (a) Sketch the graph of P (t) against time. (b) What is the current population? (c) What will be the population 50 years from now? (d) What will happen to the population in the long run?

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Chapter 8

Applications of Differentiation 8.1

Increasing and decreasing functions

Let f (x) be a function defined on the interval a ≤ x ≤ b, and let x1 and x2 be two numbers on the interval. Then f (x) is increasing on the interval if f (x2 ) > f (x1 ) whenever x2 > x1 . It is important to note that for increasing functions, the slope of any tangent line to the curve at a point on the interval (a, b) will be positive. The figure below shows an increasing function.

Also, let f (x) be a function defined on the interval a ≤ x ≤ b, and let x1 and x2 be two numbers on the interval. Then f (x) is decreasing on the interval if f (x2 ) < f (x1 ) whenever x2 > x1 . The slope of any tangent line to curve at a point on the interval (a, b) will be negative. 155

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An Introduction to Applied Calculus for Social and Life Sciences The figure below indicates a decreasing function.

The general procedure for solving problems involving increasing and decreasing functions over some intervals is simply to follow the following steps: (a) Find the derivative. (b) Set derivative equal to zero or determine points where the derivative is undefined. (c) Use sign table to find the intervals of increase or decrease. example 8.1.1 Find the intervals of increase and decrease for the given function. (a) f (x) = x3 − 3x − 4

(b) f (x) = 3x5 − 5x3

1 4 − t2  (e) h(u) = 9 − u2

(c) f (t) =

(d) f (x) = x +

9 x

Solutions (a) For f (x) = x3 − 3x − 4 the derivative is given by f  (x) = 3x2 − 3 = 3(x2 − 1) = 3(x − 1)(x + 1) Therefore f  (x) = 0 if x = ±1.

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Chapter 8 Applications of Differentiation Here x = ±1 are special points that divide the real line (number line) into three intervals. We make a sign table then use it to make conclusions depending on the sign of the derivative. The sign of the derivative determines whether Interval Sign of f  (x) Test value Value of derivative

(−∞, −1) + -2 f  (−2) = 9

(−1, 1) – 0 f  (0) = −3

(1, ∞) + 2 f  (2) = 9

a function is increasing or decreasing. The test value is substituted into the derivative to determine the sign of the derivative. For instance, take −2 and substitute into f  (x) = 3(x − 1)(x + 1). We observe that

f  (−2) = 3(−2 − 1)(−2 + 1) = +9.

Hence the sign of f  (x) is positive. We thus make the following conclusions: • f increases on (−∞, −1) and on (1, ∞), where the sign of the derivative is positive. • f decreases on (−1, 1) where the sign of the derivative is negative. (b)

f  (x) = 15x4 − 15x2 = 15x2 (x2 − 1) = 15x2 (x − 1)(x + 1) ⇒ f  (x) = 0 if x = 0 or x = ±1. Here we have three special points that divide the real line into four regions. Consider the sign table. Interval Sign of f  (x) Test value

(−∞, −1) + -2

(−1, 0) – -0.5

(0, 1) – 0.5

(1, ∞) + 2

• f (x) increases on (∞, −1) and on (1, ∞) • f (x) decreases on (−1, 0) and on (0, 1). (c) Note that f (t) =

1 = (4 − t2 )−1 . 4 − t2

2t 2t = . 2 2 2 (4 − t ) (2 − t) (2 + t)2 Therefore f  (x) = 0 when 2t = 0 ⇒ t = 0. f  (t) = −(4 − t2 )−2 (−2t) =

Also f  (t) is undefined when (4 − t2 )2 = 0 ⇒ t = ±2. So the special points are t = 0, t = ±2. Interval Sign of f  (x) Test value

(−∞, −2) – -3

(−2, 0) – -1

(0, 2) + 1

(2, ∞) + 3

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An Introduction to Applied Calculus for Social and Life Sciences Therefore • f (t) increases on (0, 2) and on (2, ∞); and • f (t) decreases on (−∞, −2) and on (−2, 0). (d) Given f (x) = x + 9x−1 , 9 x2 x2 − 9 (x − 3)(x + 3) = = . x2 x2

f  (x) = 1 − 9x−2 = 1 −

Note that f  (x) = 0 when (x − 3)(x + 3) = 0 ⇒ x = ±3; and f  (x) undefined when x = 0 Interval Sign of f  (x) Test value

(−∞, −3) + -4

(−3, 0) – -1

(0, 3) – 1

(3, ∞) + 4

Therefore • f (x) = 0 increases on (−∞, −3) and on (3, ∞) • f (x) decreases on (−3, 0) and on (0, 3). √ 1 (e) h(u) = 9 − u2 = (9 − u2 ) 2 Note that domain Dh = [−3, 3] and h (u) =

1 1 −u (9 − u2 )− 2 (−2u) = √ . 2 9 − u2

So, h (u) = 0 when u = 0; and h (u) undefined when  9 − u2 = 0 ⇒ 9 − u2 = 0 ⇒ u = ±3. So the special points are t = 0, t = ±3. Interval Sign of f  (x) Test value

(−∞, −3) undefined -4

(−3, 0) + -1

(0, 3) – 1

(3, ∞) undefined 4

Therefore • h(u) increases on (−3, 0) • h(u) decreases on (0, 3).

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Chapter 8 Applications of Differentiation

8.2

Relative extrema

Relative extrema The graph of f (x) has a relative maximum at x = c if f (c) ≥ f (x) for all x on an interval a < x < b containing c. The graph of f (x) has a relative minimum at x = c if f (c) ≤ f (x) for all x on an interval a < x < b containing c. The figure below illustrates respectively a relative maximum and a relative minimum.

The relative maximum and relative minimum points of a function are also called local extrema. We also have global extrema that define absolute minimums and maximums. We have the following definitions.

Definitions of local maxima and minima (a) The function f (x) has an absolute maximum at x = c if f (x) ≤ f (c) for all x in the entire domain under consideration. (b) The function f (x) has a local maximum at x = c if f (x) ≤ f (c) for all x on an open interval, a < x < b, containing c. (c) The function f (x) has an absolute minimum at x = c if f (x) ≥ f (c) for all x in the entire domain under consideration. (d) The function f (x) has a local minimum at x = c if f (x) ≥ f (c) for all x on an open interval, a < x < b, containing c.

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An Introduction to Applied Calculus for Social and Life Sciences

8.2.1

Critical numbers and critical points

Definitions of critical numbers and points A number c in the domain of f (x) is called a critical number if either f  (c) = 0 or f  (c) does not exist. The corresponding point (c, (f (c)) on graph of f (x) is called a critical point for f (x). Note the relative extrema only occur at critical points. Consider the figure below where extrema are described. We shall discuss global extrema in the coming sections.

example 8.2.1 Determine the critical numbers of the given function and classify each critical point as a relative maximum/minimum, or neither. (a) f (x) = 324x − 72x2 + 4x3

(b) g(x) = 4 −

3 2 + x x2

Solutions (a) We begin by finding the derivative of f (x). f  (x) = 324 − 144x + 12x2 = 12(27 − 12x + x2 ) = 12(9 − x)(3 − x).

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Chapter 8 Applications of Differentiation For the critical numbers we have (i) f  (x) = 0 ⇒ x = 3 or x = 9 and (ii) f  (x) is defined for all values of x. Interval Sign of f  (x) Test value

(−∞, 3) + 0

(3, 9) – 5

(9, ∞) + 10

Increasing/decreasing Note that if x = 3 then f (3) = 324(3) − 72(3)2 + 4(3)3 = 432, which is the corresponding y-value. Also, if x = 9, then f (9) = 0. If we look at the increasing/decreasing lines in the sign table, as we pass through the point (3, 432), it is easy to note that this is a relative maximum. On the other hand as we pass through (9, 0) the decreasing function turns at this point to become an increasing function indicating that we have a relative minimum at (9, 0). (b) Let g(x) = 4 − 2x−1 + 3x−2 , so that 2 6 − 3 x2 x 2x − 6 = x3 2(x − 3) = . x3

g  (x) = 2x−2 − 6x−3 =

The critical points are at g  (x) = 0 and where this derivative is not defined. We thus have g  (x) = 0 ⇒ 2(x − 3) = 0 ⇒ x = 3. Also, g  (x) is undefined at x3 = 0 which implies that x = 0 but x = 0 ∈ / Dg . So x = 0 is not a critical number. This means that the function is not defined at x = 0, hence in terms of intervals of increase and decrease we should consider it but the function cannot have a turning point at x = 0. In fact the function is asymptotic to the line x = 0. We shall deal with asymptotes in the coming sections. Interval Sign of g  (x) Test value

(−∞, 0) + -1

(0, 3) – 1

(3, ∞) + 4

Increasing/decreasing When x = 3, g(3) = 11 3 . On (−∞, 0), the function is always increasing and then it decreases on (0, 3) and increases again on (3, ∞). We have a relative  . minimum at 3, 11 3 161

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An Introduction to Applied Calculus for Social and Life Sciences

8.2.2

Concavity and points of inflection

Concavity If a function f (x) is differentiable on the interval a < x < b, then the graph of f is • concave upward on a < x < b if f  (x) is increasing on the entire interval; • concave downward on a < x < b if f  (x) is decreasing on the entire interval. So, f is concave upward on the interval if the derivative of f  (x) is positive (i.e. f  (x)>0) and f is concave downward on the interval if the derivative of f  (x) is negative (i.e. f  (x) < 0).

Inflection points An inflection point of the function f is a point (c, f (c)) on the graph of f where f is continuous and the concavity changes. The second derivative at that value f  (c) is either equal to zero or it doesn’t exist. example 8.2.2 1 Find the concavity of the function f (x) = x2 − as well as all the inflection x points. Solution For the function f (x) = x2 −

1 1 ⇒ f  (x) = 2x + 2 . x x

Therefore f  (x) = 2 −

2 2x3 − 2 = 3 x x3 2(x3 − 1) = x3 2(x − 1)(x2 + x + 1) = x3

The possible inflection points are at: (a) f  (x) = 0 ⇒ 2(x − 1)(x2 + x + 1) = 0 ⇒ x = 1. / Df (domain of f ), (b) f  (x) is undefined, so x3 = 0 ⇒ x = 0 but x = 0 ∈ therefore there is no inflection point at x = 0. 162

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Chapter 8 Applications of Differentiation Interval Sign of f  (x) Test value Concavity

(−∞, 0) + -1 Up

(0, 1) – 0.5 Down

(1, ∞) + 2 Up

(i) f (x) is concave upward on (−∞, 0) and on (1, ∞). (ii) f (x) is concave downward on (0, 1). Given that there is a change of concavity at (1, f (1)) = (1, 0), there is an inflection point at (1, 0). example 8.2.3 Find where f (x) is increasing/decreasing and where its graph is concave upward/downward. Sketch the graph. (a) f (x) = x3 + 3x2 + 1  (c) f (x) = x2 + 1

(b) f (x) =

x2 x2 + 3

Solutions (a) For the function f (x) = x3 + 3x2 + 1, we first find f  (x) to determine where the function is increasing or decreasing. (i) So

f  (x) = 3x2 + 6x = 3x(x + 2).

The critical points are obtained by setting f  (x) = 0 so that 3x(x + 2) = 0 ⇒ x = −2 or 0. The corresponding function values are f (−2) = 5 and f (0) = 1. So, we have our relative extrema at (−2, 5) and (0, 1). Interval Sign of f  (x) Test value

(−∞, −2) + -3

(−2, 0) – -1

(0, ∞) + 1

Increasing/decreasing The function f (x) is increasing on (−∞, −2) and on (0, ∞), and decreasing on (−2, 0). We can easily pick from the sign table that (−2, 5) is a relative maximum and (0, 1) is a relative minimum. (ii) The second derivative of f (x) is now used to determine the concavity of the function. So f  (x) = 6x + 6 = 6(x + 1). 163

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An Introduction to Applied Calculus for Social and Life Sciences Therefore

f  (x) = 0 if x = −1.

With respect to concavity our interval is only divided into two so that Interval Sign of f  (x) Test value Concavity

(−∞, −1) – -3 Down

(−1, ∞) + 0 Up

Therefore f concave up on (−1, ∞) and concave down on (−∞, −1) Given that f (−1) = 3, the inflection point is (−1, 3). (iii) To sketch the curve we use additional information such as the determination of the intercepts on the axes. First, the intercepts of the x-axis are found by setting f (x) = 0 this can not be easily solved since we cannot easily factorise the expression for f (x). We can, however, determine the y intercept by setting x = 0. We observe that (0, 1) is also a minimum point.

(b) For the function f (x) =

x2 +3

x2

(i) to find the intervals where f (x) increases/decreases we determine f  (x). f  (x) =

2x(x2 + 3) − x2 (2x) 6x = 2 2 2 (x + 3) (x + 3)2

The critical numbers are obtained by setting f  (x) = 0 considering where f  (x) is undefined. Therefore 6x = 0 ⇒ x = 0. For x = 0, f (0) = 0. So we have an extremum at the point (0, 0). Also note that f  (x) defined for all values of x. So, f increases on (0, ∞) and deceases on (−∞, 0). We thus have a relative minimum at (0, 0). 164

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Chapter 8 Applications of Differentiation (−∞, 0) – -1

Interval Sign of f  (x) Test value

(0, ∞) + 1

Increasing/decreasing (ii) To determine the concavity of graph of f we find f  (x). 6(x2 + 3)2 − 6x(2)(x2 + 3)(2x) (x2 + 3)4 [6(x2 + 3) − 24x2 ] = (x2 + 3)3 18 − 18x2 = 2 (x + 3)3 18(1 − x)(1 + x) = . (x2 + 3)3

f  (x) =

The function changes concavity when f  (x) = 0. This occurs at x = ±1 (also, f  (x) is defined for all values of x). Note that f (±1) = 14 . Interval Sign of f  (x) Test value Concavity

(−∞, −1) – -2 Down

(−1, 1) + 0 Up

(1, ∞) – 2 Down

So, f (x) is concave up on (−1, 1) and concave down on (−∞, −1) and on (1, ∞). The inflection points are thus (−1, 14 ) and (1, 14 ). A sketch of the function is as shown below.

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An Introduction to Applied Calculus for Social and Life Sciences (c) For the function f (x) =

√

x2 + 1

(i) f (x) increases/decreases in the intervals determined by the sign of f  (x). f  (x) =

1 1 2 x (x + 1)− 2 (2x) = √ . 2 2 x +1

The critical numbers are given by solving f  (x) = 0, so that x = 0 (noting that x2 + 1 is always positive for all values of x that are real.) If x = 0 then f (0) = 1. So (0, 1) is our extremum. The nature of this turning point is determined by the signs in the following sign table. (−∞, 0) – -1

Interval Sign of f  (x) Test value

(0, ∞) + 1

Increasing/decreasing So, f (x) is increasing on (0, ∞) and decreasing on (−∞, 0). So, (0, 1) is a relative minimum. (ii) We now look at the concavity of f (x) by considering the signs of f  (x) in some intervals.

√

√ 1 x2 + 1 − x 12 (x2 + 1)− 2 (2x) x2 + 1 √ f (x) = ×√ ( x2 + 1)2 x2 + 1 (x2 + 1) − x2 = 1 (x2 + 1)(x2 + 1) 2 1 = . 2 (x + 1)3 

There is no value of x for which f  (x) = 0. Also f  (x) is defined for all x and our function does not change concavity over the entire set of real numbers. Since f  (x) > 0 for all values of x, f (x) is concave up.

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Chapter 8 Applications of Differentiation (iii) A sketch of the function is shown below.

8.2.3

Second derivative test

We have been determining the nature of the extrema by looking at the sign of the derivative. We can also use the second derivative test to determine whether a point is a maximum or minimum for functions that can be differentiated twice. For a given function f (x), to the left of a maximum point f  (x) is positive and it becomes negative once it passes through the point. So f  (x) moves from being positive to negative. In fact f  (x) will be decreasing. This means that the gradient of the derivative will be negative at the maximum point i.e. f  (x) < 0. On the other hand f  (x) > 0 at a minimum point.

Definition of second derivative test Suppose f  (x) exists on an open interval that contains x = c and f  (c) = 0 • if f  (c) > 0, then f has a relative minimum at x = c. • if f  (c) < 0, then f has a relative maximum at x = c. In summary Minimum Point df = 0 and f  (x) > 0 dx

Maximum Point df = 0 and f  (x) < 0 dx

Note that if f  (c) = 0 or f  (c) does not exist, then the test is inconclusive.

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An Introduction to Applied Calculus for Social and Life Sciences example 8.2.4 Use the second derivative test to find the relative extrema of the following functions: (a) f (x) = 2x4 − 4x2 − 3

2 x (c) f (x) = x+4

(b) f (x) = (x2 − 4)2

Solutions (a) For the function f (x) = 2x4 − 4x2 − 3 f  (x) = 8x3 − 8x = 8x(x − 1)(x + 1). The turning points (extrema) are obtained from f  (x) = 0. This gives three solutions, x = 0 and x = ±1. The turning points are (0, −3), (±1, −5). Given that f  (x) = 24x2 − 8, • f  (0) = −8 < 0. So, (0, 3) is a relative maximum. • f  (−1) = 16 > 0. So (−1, −5) is a relative minimum. • f  (1) = 16 > 0. So (1, −5) is a relative minimum. A sketch of the function is shown below.

(b) Consider the function f (x) = (x2 − 4)2 . f  (x) = 2(x2 − 4)(2x) = 4x(x − 2)(x + 2). The second derivative is given by f  (x) = 4(x2 − 4) + 4x(2x) = 12x2 − 16. 168

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Chapter 8 Applications of Differentiation Therefore f  (x) = 0 if x = 0 or x = ±2. Here f (0) = 16 and f (±2) = 0. The function thus has relative extrema at (0, 16), (2, 0) and (−2, 0). Therefore • f  (0) = −16 < 0. So we have a relative maximum at (0, 16). • f  (−2) = 32 > 0. So we have a relative minimum at (−2, 0) and • f  (2) = 32 > 0 so we also have a relative minimum at (2, 0). A sketch of the function is given below.

(c) Consider the function f (x) =

x x+4

2 .

The derivative is given by

x 1(x + 4) − x(1) f (x) = 2 x+4 (x + 4)2

x 4 =2 x+4 (x + 4)2 8x = . (x + 4)3 

The second derivative is given by f  (x) =

8x + 32 − 24x 16(2 − x) 8(x + 4)3 − 8x(3)(x + 4)2 = = . (x + 4)6 (x + 4)4 (x + 4)4

So, if f  (x) = 0 then x = 0. Note that f (0) = 0 so that (0, 0) is a critical point. 32 Therefore f  (0) = = 0.125 > 0. The point (0, 0) is thus a relative 256 minimum. 169

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An Introduction to Applied Calculus for Social and Life Sciences A sketch of the graph of f (x) =

8.3

x x+4

2 is given below.

Vertical and horizontal asymptotes

The behavior of any function at infinity both in terms of the domain and range is of importance. We are interested in describing what happens to the function as x → ±∞ or when f (x) → ±∞. This is done through the consideration of asymptotes.

Vertical asymptotes The line x = c is a vertical asymptote of the graph of f (x) if lim f (x) = +∞ (or − ∞)

x→c−

or lim f (x) = +∞ (or − ∞).

x→c+

In general, the rational function R(x) = whenever q(c) = 0 and p(c) = 0.

p(x) q(x)

has a vertical asymptote x = c

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Chapter 8 Applications of Differentiation Horizontal asymptotes The horizontal line y = b is called a horizontal asymptote of the graph of y = f (x) if lim f (x) = b

x→−∞

or

lim f (x) = b

x→+∞

example 8.3.1 Find all the vertical and horizontal asymptotes of f (x) =

x2 − 9 x2 + 3x

if they exist. Solution Let us consider the vertical asymptotes first. A consideration of the denominator of f (x) gives x2 + 3x = x(x + 3) = 0 if x = −3 or x = 0. We need to find the one-sided limits at these points (if they exist). If the limits exist and are finite at the given point, then there will be no vertical asymptote. At x = 0 :

lim−

x→0

x2 − 9 (x − 3)(x + 3) = lim . x2 + 3x x→0− x(x + 3)

We can get rid of the factor ‘x + 3’ i.e. divide through by ‘x + 3’. So our answer x2 − 9 = +∞, since both the is infinite at x = 0. We can conclude that lim− 2 x→0 x + 3x numerator and denominator are always negative when we approach zero from the left. x2 − 9 = −∞, since the numerator is x→0 x2 + 3x negative for value very close to zero and the denominator is always positive when we approach zero from the right. For the right-hand side limit, we have lim+

In this case, we have a vertical asymptote at x = 0. At x = −3, lim

x→−3−

(x − 3)(x + 3) x(x + 3)

−3 − 3 =2 −3

=

and lim +

x→−3

(x − 3)(x + 3) x(x + 3)

=

−6 =2 −3

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An Introduction to Applied Calculus for Social and Life Sciences Therefore, the limit exists at x = −3. So x = −3 is not a vertical asymptote. To find the horizontal asymptote of f (x) =

x2 − 9 x2 + 3x

we consider x2 − 9 x→−∞ x2 + 3x lim

Therefore lim

x→−∞

=

lim

x→−∞

2x 2x + 3

(L’Hôpital’s rule).

2x 2 = lim = 1. x→−∞ 2x + 3 2

Similarly x2 − 9 =1 x→∞ x2 + 3x Using the method for limits at infinity, we divide by the highest power in x so that lim

x2 − 9 x→−∞ x2 + 3x lim

1 − 9/x2 = 1. x→−∞ 1 + 3/x

=

lim

Therefore y = 1 is a horizontal asymptote of y = f (x).

8.4

Curve sketching

An algorithm for sketching the graph of y = f (x) requires the following steps.

Guidelines for sketching a curve (a) Find the domain. (b) Find the x- and y-intercepts. (c) Find all the vertical and horizontal asymptotes. (d) Find f  (x) and find critical points. (e) Set up sign table to find intervals where f increases/decreases. (f) Classify critical points, i.e. the relative extreme points. (g) Find f  (x) and find points of inflection. (h) Set up sign table to find concavity of graph of f . (i) Sketch the curve.

A useful tip would be to set up the sign table for the function f (x) itself. This will help you place your graph on the coordinate system, i.e. where f (x) is above or below the x-axis. 172

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Chapter 8 Applications of Differentiation example 8.4.1 Sketch the following graphs x (a) f (x) = x−1

(b) f (x) =

x x2 − 1

Solutions (a) Consider the function f (x) =

x . x−1

(i) Df = {x ∈ R | x = 1} = (−∞, 1) ∪ (1, ∞). From the domain we can easily observe that the function is undefined when x = 1. This is the vertical asymptote. We will formally find the vertical asymptote later. (ii) To find x-intercept: Let f (x) = 0 ⇒ x = 0. So the y-intercept is (0, 0). (iii) For the horizontal asymptote lim

x→∞

and lim

x→−∞

x 1 = lim x − 1 x→∞ 1 −

=1

1 x

x 1 = lim x→−∞ x−1 1−

1 x

= 1.

Therefore the line y = 1 is the horizontal asymptote. (iv) For the vertical asymptote we consider the point x = 1 where the function is undefined. x = −∞ lim x→1− x − 1 and lim+

x→1

x x−1

=

+∞

So, x = 1 is a vertical asymptote. (v) To find intervals of increase/decrease we find the derivative of f (x). Therefore f  (x)

=

1(x − 1) − x(1) −1 = . 2 (x − 1) (x − 1)2

There are no values of x for which f  (x) = 0. It is, however, important to note that f  (x) is undefined at x = 1, which is not in the domain of f. From the table we can see that the function is decreasing on either side of the vertical asymptote x = 1. So f (x) is decreasing on (−∞, 1) and on (1, ∞). 173

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An Introduction to Applied Calculus for Social and Life Sciences (−∞, 1) – 0

Interval Sign of f  (x) Test value

(1, ∞) – 2

Increasing/decreasing (vi) The function f (x) has no extreme points. (vii) To determine concavity, we find the second derivative so that f  (x)

=

2 . (x − 1)3

Again there is no value of x for which f  (x) = 0. So, there are no points of inflection. However, f  (x) is not defined at x = 1. We thus determine concavity on either side of the vertical asymptote. Interval Sign of f  (x) Test value Concavity

(−∞, 1) – 0 Down

(1, ∞) + 2 Up

From the table we see that f (x) is concave upward on (1, ∞) and concave down on (−∞, 1). (viii) There are no inflection points. (ix) We can determine where the function f (x) lies, with a positive sign meaning it is above the x-axis and negative if it is below. The important points to consider are x = 0 (the x intercept) and x = 1 (the asymptote). Interval Sign of f (x) Test value Above/below x-axis

(−∞, 0) + -1 Above

(0, 1) – 0.5 Below

(1, ∞) + 2 Above

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Chapter 8 Applications of Differentiation (x) Sketch:

(b) Consider the function f (x)

=

x2

x −1

(i) The domain = {x ∈ R | x = ±1}

Df (ii) For the x-intercept, set f (x)

=

0 ⇒

x =0⇒x=0 x2 − 1

and for the y-intercept set x =

0 so that f (0) =

0 = 0. 0−1

So the only intercept is (0, 0). (iii) For the horizontal asymptote lim

x→∞ x2

1 x x = lim =0 − 1 x→∞ 1 − x12

The horizontal asymptote is y = 0 and the vertical asymptotes are x = 1 and x = −1. The vertical asymptotes can be determined by considering lim

x→1+

x 1 = +∞ and lim 2 = −∞. x2 − 1 x→1− x − 1 175

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An Introduction to Applied Calculus for Social and Life Sciences Similarly lim +

x→−1

x2

x = +∞ and −1

lim

x→−1−

x2

1 = −∞. −1

(iv) To determine intervals of increase and decrease we find the derivative of f (x). f  (x)

= −

x2 + 1 (x2 − 1)2

There are no points where f  (x) = 0. So, the function has no critical points. We now check on intervals of increase and decrease being guided by the vertical asymptotes. (−∞, −1) – -2

Interval Sign of f  (x) Test value

(−1, 1) – 0

(1, ∞) – 2

Increasing/decreasing The function is decreasing on (−∞, −1), (−1, 1) and on (1, ∞). (v) To determine the concavity of f (x) we determine the second derivative of the function so that

f  (x)

=

2x(x2 + 3) . (x2 − 1)3

2x(x2 + 3) = 0 ⇒ x = 0. We thus have (0, 0) as a point of (x2 − 1)3 inflection and it turns out that concavity changes at this point.

f  (x) = 0 ⇒

Interval Sign of f  (x) Test value Concavity

(−∞, −1) – -2 Down

(−1, 0) + -0.5 Up

(0, 1) – 0.5 Down

(1, ∞) + 2 Up

From the table, f (x) is concave up on (−1, 0) and on (1, ∞) and concave down (−∞, −1) and on (0, 1).

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Chapter 8 Applications of Differentiation (vi) To determine where the curve lies with respect to the x-axis we look for the sign of f (x) on the given intervals. Interval Sign of f (x) Test value Above/below x-axis

(−∞, −1) – -2 Below

(−1, 0) + -0.5 Above

(0, 1) – 0.5 Below

(1, ∞) + 2 Above

(vii) Sketch:

8.5

Graphs involving exponential functions

example 8.5.1 Sketch the graph of f (x) = (x2 − 1)e−x Solutions (a) The domain of f (x) is given by Df

= R.

(b) For the x-intercept, let f (x) = 0, so that (x2 − 1)e−x = 0 ⇒ x = ±1. For the y-intercept, we let x = 0. Therefore f (0) = −1e−0 = −1. The points (−1, 0) and (1, 0) are intercepts on the x-axis and the point (0, −1) is the y-intercept. 177

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An Introduction to Applied Calculus for Social and Life Sciences (c) For the horizontal asymptote(s) we have lim (x2 − 1)e−x

x→∞

=

x2 − 1 x→∞ ex 2 lim = 0 (Using L Hôpital s rule twice). x→∞ ex lim

The line y = 0 is a horizontal asymptote. Note that we do not have any vertical asymptotes for this function. (d) To determine intervals of increase/decrease we find f  (x) Therefore f  (x)

(−∞, 1 − – -1

Interval Sign of f  (x) Test value

√

= e−x (1 + 2x − x2 ). √ = 0 ⇒ x = 1 ± 2.

2)

(1 −

√

2, 1 + + 0

√

2)

(1 +

√

2, ∞)

– 3

Increasing/decreasing √ √ The √ function f (x)√is increasing on (1− 2, 1+ 2) and decreasing on (−∞, 1− 2) and on (1 + 2, ∞). √ (e) The function has √ a relative minimum when x = 1− 2 and a relative maximum when x = 1 + 2. Explicit y-value can be found through the substitution of these values into the function f (x). (f) For concavity, the second derivative of f (x) is f  (x) = e−x (1 − 4x + x2 ) ⇒ f  (x) = 0 if x = 2 ± Interval Sign of f  (x) Test value Concavity

(−∞, 2 − + 0 Up

√

3)

(2 −

√

3, 2 + – 2 Down

We note that f (x) is concave up on (−∞, 2 − √ √ concave down on (2 − 3, 2 + 3).

√

√

3)

√

(2 +

3.

√

3, ∞) + 5 Up

3) and on (2 +

√

3, ∞) and

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Chapter 8 Applications of Differentiation √ √ (g) There are two points of inflection at x = 2 − 3 and x = 2 + 3. The corresponding y values can be found by substituting the x values into the function f (x).

(h) The function is above the x- axis on (−∞, −1) ∪ (1, ∞) and below on (−1, 1) as depicted in the table below. Interval Sign of f (x) Test value Above/below x-axis

(−∞, −1) + -2 Above

(−1, 1) – 0 Below

(1, ∞) + 2 Above

(i) Using all the above facts we can now plot the graph of f (x).

8.6

Graphs with oblique asymptotes

We can redefine an asymptote to a curve as a line that the curve approaches as x tends to ±∞ or some fixed number. The line y = mx + c is an asymptote to the curve y = f (x) if  lim f (x) − (mx + c) = 0. x→±∞

If m = 0 then the line y = mx + c is referred to as an oblique asymptote to the curve y = f (x).

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An Introduction to Applied Calculus for Social and Life Sciences example 8.6.1 Sketch the graph of f (x) =

2x2 +4 x+1

Solutions (a) The domain Df

= {x ∈ R | x = −1}

(b) For the intercepts: If we set f (x)

0 ⇒ 2x2 + 4 = 0. There are no values of x for which 2x2 + 4 = 0.

=

For the y-intercept, set x = 0 so that f (0) = 4. So the only intercept is (0, 4). (c) For the asymptotes: We note that

6 2x2 + 4 = 2x − 2 + . x+1 x+1 We do not have a horizontal asymptote but rather the line y = 2x − 2 is the 2 +4 oblique asymptote to the curve f (x) = 2xx+1 . f (x) =

The vertical asymptote can be determined from the evaluation of the following limits: lim +

x→−1

2x2 + 4 = ∞ and x+1

lim

x→−1−

2x2 + 4 = −∞. x+1

So x = −1 is a vertical asymptote. (d) To determine intervals of increase and decrease we find the derivative of f (x). f  (x)

=

2(x2 + 2x − 2) (x + 1)2

The function has turning points where f  (x) = 0, i.e. x2 + 2x − 2 = 0. We √ note that x = −1 ± 3. We now check on intervals of increase and decrease, being guided by the vertical asymptotes. Interval Sign of f  (x) Test value

(−∞, −2.73) + -3

(−2.73, −1) – -2

(−1, 0.73) – 0

(0.73, ∞) + 1

Increasing/decreasing √ √ The function is decreasing √ and on (−1, −1 + 3) and in√ on (−1 − 3, −1) creasing on (−∞, −1 − 3) and on (−1 + 3, ∞). 180

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Chapter 8 Applications of Differentiation (e) To determine the concavity of f (x) we determine the second derivative of the function so that

f  (x) =

12 . (x + 1)3

Note that f  (0) = 0 so we do not have any points of inflection. We can, however, test for concavity since f  (x) is not defined at x = −1. Interval Sign of f  (x) Test value Concavity

(−∞, −1) – -2 Down

(−1, ∞) + 0 Up

From the table, we note that f (x) is concave up on (−1, ∞) and concave down (−∞, −1). (f) To determine where the curve lies with respect to the x-axis we look for the sign of f (x) on the given intervals. Interval Sign of f (x) Test value Above/below x-axis

(−∞, −1) – -2 Below

(−1, ∞) + 0 Above

(g) Sketch:

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An Introduction to Applied Calculus for Social and Life Sciences

8.7

Newton’s Method

Newton’s Method approximates roots to a polynomial numerically using the first derivative. Suppose that a differentiable function y = f (x) has a zero at x = c which is unknown. The procedure involves drawing a tangent line to y = f (x) at (x0 , f (x0 )), and letting x1 be the x-value of the point where the tangent line cuts the x-axis. The point x1 is a better approximation for the zero (root) than x0 . Using x1 as the new guess to the root, we draw another tangent at (x1 , f (x1 )), and let x2 be the x-value of the point where the tangent line cuts the x-axis. Surely x2 is a better approximation than x1 . If we continue the process, we continue to get closer and closer to the root c. It is important to note that it is not always possible to find solutions using this method. We now focus on the derivation of the formula.

We take a guess and say that the zero is at x = x0 . To find x1 given that we have a first guess x0 , we consider the slope of the tangent line at x0 is f  (x0 ). Therefore the equation of the tangent is y − f (x0 ) = f  (x0 )(x − x0 ).

(8.1)

The point (x1 , 0) lies on the tangent line. So substitute it into (8.1) so that 0 − f (x0 ) = f  (x0 )(x1 − x0 ) = x1 f  (x0 ) − x0 f  (x0 ). Therefore

x1 f  (x0 ) = x0 f  (x0 ) − f (x0 ).

So

f (x0 ) . f  (x0 ) Since x0 , f (x0 ) and f  (x0 ) are known, so we can find x1 . Now, we can use x1 in a similar manner to find x2 : x1 = x0 −

x2 = x1 −

f (x1 ) . f  (x1 )

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Chapter 8 Applications of Differentiation We then use x2 to find x3 , and so on, iteratively. We call this approximation of the zero Newton’s Method. The general formula for Newton’s Method: xn+1 = xn −

f (xn ) for n = 0, 1, 2, . . . f  (xn )

example 8.7.1 Find a positive real root of the following y = x4 + x − 3 correct to four decimal places. Solution Since we are not given the initial guess, we can use the Intermediate Value Theorem to choose x0 . We start at x = 0 and determine the function value. y(0) = −3 < 0 y(1) = 1 + 1 − 3 = −1 < 0 y(2) = 24 + 2 − 3 = 15 > 0. Therefore, a positive root lies between x = 1 and x = 2. It is important to note that the root could be closer to x = 1 than x = 2 due to the magnitude of the function values. Let us choose x0 = 1. To apply Newton’s Method, we require the derivative, dy = f  (x) = 4x3 + 1. So, dx x1 = x0 −

−1 1 6 f (x0 ) =1− =1+ =  3 f (x0 ) 4(1) + 1 5 5

x2 = x1 −

(1.2)4 + (1.2) − 3 f (x1 ) = 1.2 − = 1.16542 f  (x1 ) 4(1.2)3 + 1

x3 = x2 −

(1.16542)4 + 1.16542 − 3 f (x2 ) = 1.16542 − = 1.16404 f  (x2 ) 4(1.16542)3 + 1

x4 = 1.16404 −

(1.16404)4 + 1.16404 − 3 = 1.16404. 4(1.16404)3 + 1

Therefore y = x4 + x − 3 has a root at x = 1.1640 (correct to four decimal places).

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An Introduction to Applied Calculus for Social and Life Sciences example 8.7.2 Find the following correct to 3 decimal places. √ √ 4 (b) 2 (a) 7

Solutions (a) Set x =

√

7 so that x2 = 7 ⇒

x2 − 7 = 0.

We now set f (x) = x2 − 7 to find x such that f (x) = 0. We use the Intermediate Value Theorem to find an interval on which the positive root occurs. f (0) = −7 < 0 f (1) = 1 − 7 = −6 < 0 f (2) = 22 − 7 = −3 < 0 f (3) = 32 − 7 = 2 > 0 So, f (x) has a zero/root on 2 ≤ x ≤ 3. Choose x0 = 2.5 and use Newton’s Method given that f  (x) = 2x. −0.75 53 f (2.5) = 2.5 − = = 2.65  f (2.5) 5 20 f (2.65) x2 = 2.65 −  = 2.64575 f (2.65) f (2.64575) x3 = 2.64575 −  = 2.64575. f (2.64575) x1 = 2.5 −

√ Therefore 7 ≈ 2.646. √ (b) Set x = 4 2 so that x4 = 2 ⇒ x4 − 2 = 0.

Now, let f (x) = x4 − 2. By the Intermediate Value Theorem, we have that there is a zero on interval 1 ≤ x ≤ 2 since f (1) = −1 < 0 and f (2) = 16 − 2 = 14 > 0. Choose x0 = 1 and use the fact that f  (x) = 4x3 . (−1) f (1) =1− = 1.25  f (1) 4 f (1.25) x2 = 1.25 −  = 1.1935 f (1.25) f (1.1935) x3 = 1.1935 −  = 1.18923x4 f (1.1935) x1 = 1 −

So

√ 4

= 1.18921.

2 ≈ 1.189. 184

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Chapter 8 Applications of Differentiation

8.8

Optimisation

The aim of this section is to find the absolute maximum and/or the absolute minimum of a certain function on a set interval.

Extrema revisited Let f (x) be a continuous function defined on an interval , where  contains the number c. Then • f (c) is the absolute maximum of f on  if f (c) ≥ f (x) for all x on ; • f (c) is the absolute minimum of f on  if f (c) ≤ f (x) for all x on . Also, if f (x) is a continuous function on an interval a, b, for a, b ∈ R with a < b, then f (x) has both an absolute maximum and absolute minimum and these must occur at the end points or critical numbers (recalling that a critical number is a point in the domain where f  (x) = 0 or f  (x) is undefined).

example 8.8.1 Find the absolute maximum and absolute minimum (if any) of the given function on the specified interval. (a) f (x) = x3 − 3x2 + 3x − 1, −2 ≤ x ≤ 2 (c) f (u) = u +

1 , u

(b) g(x) =

x2

1 , 0≤x≤2 − 16

u>0

Solutions (a) Given that f (x) = x3 − 3x2 + 3x − 1, we have f  (x)

=

3x2 − 6x + 3 = 3(x − 1)2 .

At the turning points f  (x) = 0 ⇒ x = 1, which is in [−2, 2]. Therefore for the global (absolute) extrema we test the end points together with the turning point. So, f (−2) = −8 − 12 − 6 − 1 = −27 f (2) = 8 − 12 + 6 − 1 = 1 f (1) = 1 − 3 + 3 − 1 = 0 So the absolute minimum is −27 and occurs at (−2, −27) and the absolute maximum is 1 and occurs at (2, 1). The global extrema are at the end points of the interval. 185

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An Introduction to Applied Calculus for Social and Life Sciences 1 , we have x2 − 16

(b) For the function g(x) = g  (x)

= −1(x2 − 16)−2 (2x) =

−2x . (x2 − 16)2

At the turning point g  (x) = 0 ⇒ x = 0. In this case x = 0 lies on the interval 0 ≤ x ≤ 2. Therefore 1 (absolute maximum) 16 1 (absolute minimum) = − 12

= −

g(0) g(2)

(c) Given a function f (u) = u+ u1 , the derivative is given by f  (u) = 1− u12 = At the turning point f  (u)

u2 −1 u2 .

0 ⇒ u = ±1 but –1 is not in the interval, u > 0.

=

So f (1) = 2. How do we determine whether (1, 2) is an absolute maximum or absolute minimum? We use the second derivative test. Given that f  (u) Therefore f (1)

8.9

2 2 . We have f  (1) = 3 = 2 > 0. u3 1 = 2 is an absolute minimum of f (u) on u > 0.

=

Applied optimisation

In this section we will be applying the technique of modelling, and the optimisation technique of the previous section, Section 8.8, to maximise or minimise a specific variable. example 8.9.1 What positive number is exceeded by its cube root by the largest amount? Solution If x is the number, then P =

√ 3

x − x must be as large as possible.

So, we find the derivative and set it equal to zero. We need to find the turning point and check if it is indeed an absolute maximum. P  (x)

=

1 −2 x 3 − 1. 3

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Chapter 8 Applications of Differentiation 1 − 23 3x

Setting P  (x) = 0 ⇒

= 1. So x−2 = 27 resulting in 1 √ 27

x = To make sure the turning point at x = second derivative. P  (x) Substituting x =

√1 27

P 

√1 27

is an absolute maximum we find the

2 5 −2 = − x− 3 = √ . 3 9 9 x5

into P  , yields

1 √ 27

−2 −2 = 

5 = 9 · 3− 52 < 0. 3 √1 9 27

1 So x = √ is the absolute maximum. 27 example 8.9.2 Find two positive numbers x and y whose sum is 60 and which are such that xy 2 is as large as possible. Solution From the information we have, the following relations can be established x+y

=

60 and P = xy 2 .

We want to maximise P. So we find its derivative and set it equal to 0. But first, we need to write P in terms of a single variable. So, x+y So P (y)

60 ⇒ x = 60 − y = (60 − y)y 2 = 60y 2 − y 3 .

=

The derivative is given by P  (y) Therefore P  (y)

= −3y(y − 40) =

0 if y = 0 or y = 40

But y = 0, which means that y = 40. Using the second derivative test, we can establish that indeed y = 40 gives a maximum. P  (y) = 120 − 6y ⇒ P  (40) = 120 − 6(40) < 0. 187

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An Introduction to Applied Calculus for Social and Life Sciences This means that we have absolute maximum when y = 40. If y = 40, we have x = 60 − 40 = 20. Therefore the absolute maximum of P

=

(20)(40)2

=

32 000.

example 8.9.3 A university plans to build a rectangular playground having an area of 3 600 m2 and surround it by a fence. How can this be done using the least amount of fencing? Solution We are given that the area A = 3 600 m2 .

Let x and y be the width and length of playground as shown, so that A = xy = 3 600. The question is: What must be optimised? We desire to minimise the perimeter. Let P be the perimeter, so that P So, P (x)

=

2x + 2y

3 600 . = 2x + 2 x

(in terms of x and y) 3 600 ) (from y = x

The derivative of P (x) is given by

−3 600 2(x2 − 3 600) 7 200 P  (x) = 2 + 2 = . =2− 2 2 x x x2 So, for P  (x) = 0 we have x2 = 3 600 so that x = 60. 188

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Chapter 8 Applications of Differentiation The second derivative is P  (x) =

14 400 . x3

Substituting x = 60 into P  (x) yields P  (60)

=

14 400 > 0. (60)3

Therefore x = 60 m gives an absolute minimum value. The corresponding value of 600 = 60 m. The minimum amount of fencing will be 240 m. y is thus y = 3 60 example 8.9.4 An open box with square base needs to be constructed. The sides of the box will cost R3/m2 , and the base will cost R4/m2 . What are the dimensions of the box of greatest volume that can be constructed for R48?

Solution Consider the following figure. If it is open, then the top is not included in the calculation.

Given that the total cost is C = 48 then 48 = C

=

3 · 4(xy) + 4 · (x2 )

=

12xy + 4x2 .

So 4x2 + 12xy = 48 ⇒ y =

48 − 4x2 . 12x

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An Introduction to Applied Calculus for Social and Life Sciences We desire to maximise the volume. Let the volume be V so that = x2 y

48 − 4x2 1 2 = x = 4x − x3 . 12x 3

V V (x)

The derivative of V (x) is given by V  (x)

=

4 − x2 .

So, V  (x) = 0 ⇒ x2 = 4 ⇒ x = 2. Note that x = −2 does not make sense since x is the length of the base. Now, the second derivative is given by V  (x)

= −2x and V  (2) = −2(2) < 0

So at x = 2 we have an absolute maximum. The dimensions of the box with greatest volume are x =

2m and y =

32 4 48 − 4(2)2 = = m. 12(2) 24 3

example 8.9.5 A cylindrical can is to hold 8π cm3 of fruit juice. The cost per square centimeter to manufacture the metal lid and bottom is three times the cost per square centimeter of manufacturing the cardboard side. What are the dimensions of the cheapest can?

Solution

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Chapter 8 Applications of Differentiation Let h be the height of can and r the radius of can. We are also given that the volume V = 8π cm3 . From the formula for the volume of a cylinder we have V = πr2 h

= 8π

2

So r h

= 8 8 . = r2

⇒ h

Let C be the cost of manufacturing the can and let a be the cost per cm2 of the cardboard side. We thus have C

a × 2πrh +   

=

cardboard side

=

2aπrh + 6aπr

(3a) 

thrice the cost 2

×

2 2πr   lid and bottom

We want to minimise the cost of manufacturing the can. Therefore, we write C as a function of one variable. Let’s take r. So 8 8 C(r) = 2aπr given that h = 2 + 6aπr2 r2 r 16aπ + 6aπr2 . = r The derivative of the cost function is given by C  (r)

= −

16aπ 4aπ + 12aπr = 2 (3r3 − 4). r2 r

To find the critical point we set 

C (r)

0 ⇒

=

 To find out if indeed r = derivative test.

3

4 3

 3

4 3

 3

4 cm. 3

gives an absolute minimum value, we use the second

C  (r) When r =

4 giving us r = r = 3 3

=

32aπ + 12aπ. r3

we have   3 4 = C 3 

36aπ > 0.

 So when r = 3 43 we have an absolute minimum value for C(r) .  2 Also h = 8 34 3 cm.   2 So the cheapest can will have a radius of 3 43 cm and a height of 8 34 3 cm. 191

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An Introduction to Applied Calculus for Social and Life Sciences

Exercises

8.1 Find the intervals on which the following functions increase/decrease: m Determine the absolute minimum and absolute maximum (if such exist) of the following function on the specified interval:  (b) g(x) = 9 − x2 (a) f (x) = 3x2 − x3 (c) f (x) =

x x−4

8.2 Determine the absolute minimum and absolute maximum (if such exist) of the following function on the specified interval: (a) f (x) = (x2 − 4)5 , −3 ≤ x ≤ 2

(b) f (x) = x +

1 1 , ≤x≤3 x 2

(c) f (x) =

1 , x>0 x2

(d) f (x) =

(e) f (x) =

1 , x≥0 (x + 1)2

(f) f (x) = x4 − 2x2 , −2 ≤ x ≤ 2

2

(g) f (x) = x 3 (x − 5), −2 ≤ x ≤ 1 (i) f (t) =

t2 , −2 ≤ t ≤ 0 t−1

1 , x≥0 x+1

(h) f (x) = 3x5 − 5x3 , −1 ≤ x ≤ 0 (j) f (x) = x3 + 3x2 + 1, −3 ≤ x ≤ 2

x , indicating all local maxima and minima x2 + 3 together with concavity properties.

8.3 Sketch the graph of f (x) =

8.4 Sketch the graph of the function f (x) = and concavity.

x2 − 1 . Exhibit maxima, minima, x2 + 1

8.5 Consider the following functions: (a) f (x) = 3x2 − x3

(b) f (x) = x2 (x + 5)2

(c) f (x) =

3−x x+2

(d) f (x) =

(e) f (x) =

x+3 x−2

(f) f (x) = x2 (x + 5)2

(g) f (x) = x3 − x2 − 2x (i) f (x) =

(h) f (x) =

x x+1

(j) f (x) =

x2 x2 − 4



x2 − 4

x x2 − 1

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Chapter 8 Applications of Differentiation

(k) f (x) =

x2 − 4 x2 − 1

(l) f (x) = xe−x

(m) f (x) = (x − 1)e−x

(n) f (x) = (x + 3)ex

(o) f (x) =

x3 + 1 x2

(p) f (x) =

x2 − 4 x−1

(q) f (x) =

x3 − 4 x2 − 1

(r) f (x) =

x2 x+1

For each of the above functions, determine (i) the domain Df . (ii) the x-, y-intercepts (if they exist). (iii) the horizontal and vertical asymptotes (if they exist). (iv) f  (x) and the intervals where f (x) increases or decreases. (v) the relative extreme points. (vi) f  (x) and the intervals where f (x) is concave up or down. (vii) the coordinates of the inflection points (if any). Sketch the graph, showing all coordinates of important points. 8.6 Given the function f (x) = x3 − x − 1 (a) Use the Intermediate Value Theorem to find an interval on which a root of f (x) exists. (b) Use Newton’s Method to determine a root of f (x) to 3 decimal places. 8.7 Let f (x) = x3 − x2 . (a) Use the Intermediate Value Theorem to find an interval on which a root of f (x) exists. (b) Use Newton’s Method to determine a root of f (x) to 3 decimal places. 8.8 Approximate the following radicals to 3 decimal places: √ √ 4 (b) 2 (a) 3 (c)

√

8

(d)

√ 3

3

8.9 Find two positive numbers whose sum is 50 and whose product is as large as possible. 8.10 A Limpopo citrus grower estimates that if 60 orange trees are planted, the average yield per tree will be 400 oranges. The average yield will decrease by 4 oranges per tree for each additional tree planted on the same acreage. How many trees should the grower plant to maximise the total yield? 193

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An Introduction to Applied Calculus for Social and Life Sciences 8.11 What is the maximum possible volume of a cylindrical can with no top that can be made from 54π cm2 of metal? 8.12 Mr Thato wants to plant a rectangular garden with a fence around the sides. If only 10 m of fencing is available, find the dimensions of the largest such garden. 8.13 A closed box with square base and volume of 250 m3 . The material for the top and the bottom of the box costs R2/m2 , and the material for the sides costs R1/m2 . Can the box be constructed for less than R300? 8.14 Because of viral infection, the shape of a certain cone-shaped cell is changing. The height is increasing at the rate of 3 microns per minute. For metabolic reasons, the volume remains constantly equal to 20 cubic microns. At the moment that the radius is 5 microns, what is the rate of change of the radius of the cell? (The volume of a cone of radius r and height h is given by h V = πr2 .) 3 8.15 Mrs Wynand wants to plant a rectangular garden along a wall of her house, with a fence around the three other sides of the garden. If only 100 m of fencing is available, find the dimensions of the largest such garden. 8.16 A cylindrical container (with no top) is to be constructed to hold a volume of 24π cm3 , of liquid. The cost of the material used for the bottom is 3 cents/cm2 , and that for the curved side is 1 cent/cm2 . Derive a simple relationship between the radius and height of the least expensive container. 8.17 A closed empty TV box with a square base has a volume of 1 m3 . The material for the square top and bottom costs R2/m2 , and the material for the rectangular sides costs R6/m2 . Find the dimensions of the cheapest such box. 8.18 The radius plus the height of a cylinder cannot exceed 100 cm. Find the maximum volume of such a cylinder. 8.19 It is projected that x months from now the population of a Kayamandi will be 3 f (x) = 2x + 4x 2 + 5000. At what rate will the population be changing with respect to time 9 months from now?

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Chapter 9

Integration and its Applications 9.1

Anti-differentiation

If we differentiate the function f (x) = x3 + x2 + x + 1, we get df (x) = f  (x) = 3x2 + 2x + 1. dx The question we will attempt to answer in this chapter is: If we suppose then that we have f  (x) = 3x2 + 2x + 1, can we do a reverse operation to find f (x)? This reverse operation is called anti-differentiation.

The Indefinite integrals A function F (x) is called an anti-derivative of f (x) if F  (x) = f (x) for every x in the domain of f (x). The process of finding an anti-derivative is called anti-differentiation or integration. 1 For example, F (x) = x3 + 5x + 2 is the anti-derivative of f (x) = x2 + 5, since 3 1 F  (x) = (3)x2 + 5 = x2 + 5 = f (x). 3 A function can have more than one anti-derivative. For example, for f (x) = 3x2 any one of the following functions is an anti-derivative of f (x): F (x) = x3 − 10 or F (x) = x3 + 15 or F (x) = x3 + π.

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An Introduction to Applied Calculus for Social and Life Sciences Fundamental property of anti-derivatives Suppose F (x) is an anti-derivative of the continuous function f (x). Then any other anti-derivative of f (x) will have the form G(x) = F (x) + C, where C is a constant.

The process of anti-differentiation is called integration. We will now define what an integral is.

Indefinite integral The family of all anti-derivatives of f (x) is given by " f (x)dx = F (x) + C and is called the integral of f (x), where •

#

is the integral symbol

• f (x) is the integrand • dx indicates the variable of integration • F (x) is an anti-derivative • C is a constant. Is is important to note that if we differentiate an anti-derivative of f (x) we get f (x), that is "  d f (x)dx = f (x). dx

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Chapter 9 Integration and its Applications

9.2

Rules for integration

Integration rules " (a) The constant rule:

" kdx = k

dx = kx + C where k is a constant

" Also,

" kf (x)dx = k

f (x)dx where k is a constant

"

xn+1 + C, for all n = −1 n+1 " " 1 dx = x−1 dx = ln |x| + C (c) The logarithmic rule: x # (d) The exponential rule: ekx dx = k1 ekx + C, k constant, k = 0 " " " (e) The sum/difference rule: [f (x) ± g(x)]dx = f (x)dx ± g(x)dx (b) The power rule:

xn dx =

Please note: rules (b) and (c) can be combined ⎧ n+1 x ⎪ " ⎨ n+1 + C if n = −1 xn dx = ⎪ ⎩ ln |x| + C if n = −1

example 9.2.1 Determine the following the integrals " (a) −3dx

" "

" (c) " (e)

x

7e dx

(d)

2 √ dt t

(f)

"

" 5

2 4 − 2 + e3 dx x x

(Rule (a))

1 2 4 x dx = 4 x6 + C = x + C = x + C 6 6 3 5

4x dx = 4 "

(c)

x−0.3 dx

"

Solutions " (a) −3dx = −3x + C (b)

4x5 dx

(b)

(Rule (b))

" 7ex dx = 7

ex dx = 7ex + C

(Rule (d)) 197

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An Introduction to Applied Calculus for Social and Life Sciences "

x0.7 x−0.3+1 +C = + C (Rule (b)) −0.3 + 1 0.7  1   1 " " − 2 +1 √ 1 2 t t2 √ dt = 2t− 2 dt = 2 (e) +C =2 1 +C =4 t+C 1 −2 + 1 t 2

(d)

x−0.3 dx =

" (f)

(Rule (b))

" " " 2 4 1 − 2 + e3 dx = 2 dx − 4 x−2 dx + e3 1 · dx x x x 4 = 2 ln |x| + + e3 x + C (Multiple rules). x

example 9.2.2 Evaluate the following integrals

" 1 1 (a) − 3 dx x2 x (c)

" √

x3

"

(b)

√ 1 − √ + 2 dx 2 x

" (d)

3t2 −

√

 5t + 2 dt

1 2 (x + 1) dx x

Solutions

" " 1 1 − (a) dx = (x−2 − x−3 )dx x2 x3 x(−3+1) x(−2+1) − +C (−2 + 1) (−3 + 1) 1 1 = − + 2 + C. x 2x  3 3 "

 √ √ t t2 2 (b) 3t − 5t + 2 dt = 3 − 5 3 + 2t + C 3 2 √ 3 5 2 t 2 + 2t + C = t3 − 3 √ 2 5√ 3 t + 2t + C. = t3 − 3

" " √ √ 3 1 1 1 √ x3 − √ + 2 dx = x 2 − x− 2 + 2 dx (c) 2 2 x   1 ( 32 +1) √ 1 x(− 2 +1) x − = 3 + 2x + C 1 ( 2 + 1) 2 (− 2 + 1)  1 √ 2 5 1 x2 = x2 − + 2x + C 1 5 2 2 √ √ √ 2 x5 − x + 2x + C. = 5 =

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Chapter 9 Integration and its Applications " (d)

9.3

"

1 2 x + 2x + 1 dx x

" 1 = x+2+ dx x x2 + 2x + ln |x| + C. = 2

1 (x + 1)2 dx = x

Integration by substitution

The process of integration by substitution is the reverse of the chain rule for differentiation. For example, to find " x8 dx we use the power rule for integration so that " 1 x8 dx = x9 + C. 9 In some cases we need to find the integral of a composite function such as " (2x + 5)8 dx. We need to write the above integral in such a way that we can apply the power rule for integration. So we eliminate the ‘inside’ function by using a suitable substitution. For the integral " (2x + 5)8 dx we replace the inside function with another variable u. An important aspect to always remember is that if we replace the function of x by u, then we need to replace the ‘dx’ by ‘du’. This is why we find the derivative of u with respect to x. So we let du = 2 ⇒ du = 2dx dx 1 We thus have dx = du. 2 So, ‘dx’ is now in terms of ‘du’. The next step is to replace ‘2x + 5’ with ‘u’ and ‘dx’ with ‘ 12 du’, so that " " " 1 1 u8 du. (2x + 5)8 dx = u8 du = 2 2 u = 2x + 5 so that

Now, we can use the power rule for integration so that " 1 u9 1 8 u du = +C 2 2 9 1 9 u + C. = 18 199

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An Introduction to Applied Calculus for Social and Life Sciences We now replace ‘u’ by ‘2x + 5’, so that " 1 (2x + 5)8 dx = (2x + 5)9 + C. 18 We can quickly check if our answer is correct   d 1 1 9 (2x + 5) + C = (9)(2x + 5)8 (2) dx 18 18 = (2x + 5)8 . Substitution rule The substitution rule states that " "  f (g(x))g (x)dx = f (u)du, for u = g(x). There are five steps for integration by substitution: (i) Choose a new substitution function u = g(x) (ii) Determine the expression for dx by differentiating u (iii) Make the substitutions for the integrand and variable of integration (iv) Integrate resulting integral and (v) Substitute back to get the integral in terms of the original variable x. example 9.3.1 Find the integrals " (a) (3x + 6)3 dx

" (b) "

" t2 (t3 + 1)5 dt

(c) " (e)

√

(d)

√

4x − 1dx

y2 dy (y 3 + 5)2

2x3 − x dx x4 − x2 + 6

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Chapter 9 Integration and its Applications Solutions " du (a) For (3x + 6)3 dx set u = 3x + 6 ⇒ = 3. dx 1 Therefore dx = du. Replacing the inside function we have 3 " " 1 (3x + 6)3 dx = u3 du 3 " 1 u3 du = 3 1 u4 +C = 3 4 1 4 = u + C. 12 We thus have " (3x + 6)3 dx = " (b) The integral

√

1 (3x + 6)4 + C. 12

" 4x + 1dx =

1

(4x + 1) 2 dx.

Set u = 4x + 1 so that

1 du = 4 ⇒ dx = du. dx 4

Substituting the inner function, we have "

1 (4x + 1) dx = u du 4 " 1 1 u 2 du = 4 "

1 2

1 2

3

1 u2 +C = 4 32 1 3 = u 2 + C. 6 So "

1

(4x + 1) 2 dx = "

" (c) Note that

1 (4x + 1)3 + C. 6

t2 (t3 + 1)5 dt =

(t3 + 1)5 t2 dt.

Let u = t3 + 1 be the inside function so that du = 3t2 dt ⇒

1 du = t2 dt. 3

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An Introduction to Applied Calculus for Social and Life Sciences " 5 2

(t + 1) t dt = = = = We thus have

" (t3 + 1)5 t2 dt =

" (d) We have

1 u du 3 " 1 u5 du 3 1 u6 +C 3 6 1 6 u + C. 18

" 3

y2 dy = (y 3 + 5)2

"

5

1 3 (t + 1)6 + C. 18

(y 3 + 5)−2 y 2 dy.

We set u = y 3 + 5 so that du = 3y 2 dy ⇒ So

"

The solution is

"

1 du = y 2 dy. 3

"

1 u−2 du 3 " 1 = u−2 du 3

1 u−1 = +C 3 −1 −1 +C = 3u

(y 3 + 5)−2 · y 2 dy =

(y 3 + 5)−2 y 2 dy =

−1 + C. 3(y 3 + 5)

(e) Rewriting the integral, we have " " 1 2x3 − x √ dx = (x4 − x2 + 6)− 2 (2x3 − x)dx. 4 2 x −x +6 If we let u = x4 − x2 + 6 so that we have du = (4x3 − 2x)dx = 2(2x3 − x)dx we get

1 du = (2x3 − x)dx. 2

The integral reduces to " " 1 1 4 2 − 12 3 (x − x + 6) (2x − x)dx = u− 2 du 2 " 1 1 = u− 2 du 2  1 1 u2 +C = 1 2 2 1

= u 2 + C. 202

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Chapter 9 Integration and its Applications The final solution is "  1 (x4 − x2 + 6)− 2 (2x3 − x)dx = x4 − x2 + 6 + C. One interesting and important aspect of integration is the ability to recognise that expressions of the integrand may be related. For instance, for the integral # 2 2xex +3 dx, the derivative of x2 + 3 is part of the integrand. Integration by substitution also relies on one recognising such relations.

Integrating by recognition Given a function f (x) that is differentiable "  f (x) • dx = ln |f (x)| + C f (x) " • f  (x)ef (x) dx = ef (x) + C

example 9.3.2 Find the following integrals " 5y 4 dy (a) 5 y +3 " 1 dx (c) x ln x

" (b)

2

(−3x + 4)e−3x

+8x

dx

Solutions (a) The integral can be solved by recognising that the numerator is the derivative of the denominator. The answer then is simply "

5y 4 dy = ln |y 5 + 3| + C. +3

y5 (b) Notice that "

2

(−3x+4)e−3x

+8x

dx =

1 2

"

2 2 d 1 (−3x2 + 8x) e−3x +8x dx = e−3x +8x +C. dx 2

Alternatively, we let u = −3x2 + 8x. Then du = (−6x + 8)dx so that dx =

1 1 1 1 du = du ⇒ (−3x + 4)dx = du. (−6x + 8) 2 (−3x + 4) 2 203

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An Introduction to Applied Calculus for Social and Life Sciences We thus have

"

2

(−3x + 4)e(−3x

(c) First we note that

"

+8x)

" 1 eu du 2 1 = eu + C 2 2 1 = e(−3x +8x) + C. 2

dx =

1 dx = x ln x

"

1/x dx. ln x

Here the numerator is the derivative of the denominator so " 1 dx = ln(ln x) + C. x ln x We can, however, use substitution as follows " " 1 1 dx = (ln x)−1 dx x ln x x Let u = ln x so that du = Then

"

1 dx x

" 1 (ln x)−1 dx = u−1 du x = ln |u| + C = ln | ln x| + C.

example 9.3.3 Evaluate the following integrals " 4 (a) e 3 x dx " (c)

" (b)

3

3x2 ex

−1

dx

3

x2 e1−x dx

Solutions " 4 1 4 3 4 (a) e 3 x dx = 4 e 3 x + C = e 3 x + C. 4 3 (b) Notice that "

2 x3 −1

3x e

" dx =

3 3 d 3 (x − 1) ex −1 dx = ex −1 + C. dx 204

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Chapter 9 Integration and its Applications We can also have

"

2 x3 −1

3x e

"

3

ex

dx =

−1

3x2 dx.

Let u = x3 − 1 so that du = 3x2 dx. Then "

x3 −1

e

" 2

eu du

3x dx =

= eu + C 3

= ex " (c)

2 1−x3

x e

" dx =

−1

+C

3

e1−x x2 dx

Notice that

" " 3 3 3 d 1 1 (1 − x3 ) e1−x dx = − e1−x + C. x2 e1−x dx = − 3 dx 3 Alternatively, we can let u = 1 − x3 from which we have 1 du = −3x2 dx ⇒ − du = x2 dx. 3 So "

3



1 eu − du 3 " 1 1 eu du = − eu + C =− 3 3 1 1−x3 =− e + C. 3 "

e1−x x2 dx =

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An Introduction to Applied Calculus for Social and Life Sciences

9.4 9.4.1

Definite integral Area as a limit of a sum

Consider the shaded region under the curve y = f (x) over an interval a  x  b, where f (x)  0 and f (x) is continuous.

We determine the area of this region by dividing the region into equal partitions. Choose the least value of f (x) in each subinterval and construct a rectangle with the least value as the height. The sum of the areas of these rectangles is clearly lower than the area under the curve. This sum is called lower sum. This is shown in the figure below.

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Chapter 9 Integration and its Applications On the other hand we can choose the greatest value of f (x) in each subinterval and construct a rectangle with the chosen value as the height. The sums of these rectangles is clearly greater than the area under the curve as shown in the figure below. This sum is called upper sum.

The actual area lies between the lower and upper sums. If we increase the number of strips, we get new values of lower and upper sums. The difference between the upper and lower sums gets smaller with increasing partitions, meaning we are getting closer to the actual area. So, the difference between the lower and upper sums approaches zero as the number of subdivisions increases to infinity. The area under the curve is defined as the limit of either the lower or upper sum as the width of each division tends to zero.

Let us introduce some notations. Given a closed interval [a, b], we define a partition, Δ of the interval as the finite set Δ = {x0 , x1 , x2 , . . . , xn } of real numbers with the following property a = x0 < x1 < x2 < . . . < xn−1 < xn = b. 207

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An Introduction to Applied Calculus for Social and Life Sciences Suppose the distance between each width is Δx, and the largest and least values of f (x) in the ith are f (x∗i ) and f (xi ) respectively. We represent this in the figure below.

The lower sum is given by f (x1 )Δx + f (x2 )Δx + . . . + f (xn )Δx =

n 

f (xi )Δx.

i=1

The upper sum is given by f (x∗1 )Δx + f (x∗2 )Δx + . . . + f (x∗n )Δx =

n 

f (x∗i )Δx.

i=1

Letting A to be the area under the curve y = f (x), in the interval [a, b] we can write n 

f (xi )Δx ≤ A ≤

i=1

n 

f (x∗i )Δx.

i=1

The lower and upper sums are called the Lower and Upper Reimann sums respectively. The area is thus given by the limit of the lower or upper sum i.e. A = lim

n→∞

n 

f (xi )Δx = lim

n→∞

i=1

n 

f (x∗i )Δx.

i=1

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Chapter 9 Integration and its Applications Definite integral If f (x) is defined on a closed interval [a, b] and the limit lim

n→∞

n 

f (xi )Δx

i=1

exists, and xi is a point in the ith subinterval, then f (x) integrable on [a, b] and the limit is denoted by lim

n→∞

n 

" f (xi )Δx =

b

f (x)dx. a

i=1

The limit is called the definite integral of f (x) from a to b.

It is important to note that a sufficient condition for a function to be integrable on [a, b] is that it must be continuous on [a, b].

9.4.2

The Fundamental Theorem of Calculus

Fundamental Theorem of Calculus If a function f (x) is continuous on a  x  b, then "

b

f (x)dx = F (x) a

b a

= F (b) − F (a)

where F (x) is any anti-derivative of f (x) on a  x  b.

example 9.4.1 Find the integrals " 3 2 dx (a)

" (b)

−1

1

(x + 1)dx 0

Solutions "

3

(a) −1

3

2 dx = 2x

−1

= 2(3) − 2(−1) = 6 + 2 = 8.

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An Introduction to Applied Calculus for Social and Life Sciences

From the graph, clearly the required area is a rectangle of length 4 and width 3. So the area A = (4)(2) = 8, which agrees with the integral.



" 1 1 1 2 1 2 1 3 1 +1 − 0 +0 = . (b) (x + 1)dx = ( x2 + x) = 2 2 2 2 0 0

The area under the curve is a trapezium and the area of a trapezium is 1 2

(sum of parallel sides) times the height.

Here

3 1 [1 + 2] (1) = , 2 2 which again matches the value of the integral. A=

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Chapter 9 Integration and its Applications Rules for definite integration "

"

b

b

kf (x)dx = k

•

f (x)dx for constant k

a

"

a b

•

" [f (x) ± g(x)]dx =

a

"

b

" f (x)dx ±

a

b

g(x)dx a

a

f (x)dx = 0

• a

"

a

•

" f (x)dx = −

f (x)dx

b

"

a

"

b

f (x)dx =

•

b

a

"

c

b

f (x)dx + a

f (x)dx c

example 9.4.2 Find the definite integral

"

−1

−2

x2 + 1 dx x4

Solution

" −1 " −1 2 1 x +1 1 dx = + 4 dx x4 x2 x −2 −2 x−3 −1 1 −1 1 x−1 + =− − 3 −1 −3 −2 x 3x −2     1 1 1 1 − − = − − − (−1) 3(−1)3 −2 3(−2)3



1 1 1 = 1+ + − 3 2 24 19 4 13 = . = − 3 24 24 =

9.4.3

Substitution in a definite integral

Consider a simple definite integral in which we can integrate using substitution in the integration process. We approach the integration in two ways: • we integrate without the limits, then substitute the limits after the integration process. • we change the integration limits during the integration process, being guided by the substitution that we will have chosen.

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An Introduction to Applied Calculus for Social and Life Sciences example 9.4.3 Find the integral

"

2

(2x − 2)4 dx 1

Solutions (a) Method 1: This method demands that we first find the indefinite integral before substituting the limits. " • Find the indefinite integral (2x − 2)4 dx Let u = 2x − 2, du = 2dx ⇒ 12 du = dx. "

" (2x − 2) dx = 4

=

1 (u) du = 2 2 41



u5 5

+C

1 5 (2x − 2) + C 10

• Find the definite integral "

2  1 1  5 (2x − 2)5 = 2 − (0)5 10 10 1 32 16 1 (32) = = . = 10 10 5

2

(2x − 2)4 dx = 1

#2 (b) Method 2: For the integral 1 (2x − 2)4 dx, set u = 2x − 2 so that du = 1 2dx ⇒ du = dx. Now that we are doing a substitution from the variable x 2 to the variable u, the limits of integration must also be changed. Note that if x = 1 then u = 0 and if x = 2 then u = 2 So "

2

"

2

1 u4 du 2 0 5 2 1 u 1  5 (2) − (0)5 ) = = 2 5 0 10 32 16 = = . 10 5

(2x − 2)4 dx = 1

It is important to note that when we replace x with u, we need to change the upper and lower limits from x-values to u-values.

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Chapter 9 Integration and its Applications example 9.4.4 Find the following " 2 3x2 (a) dx 3 2 1 (x + 1) Solutions " (a) For

2

" (b)

1 2 1 3

e1/x dx 2x2

3x2 dx we let + 1)2

(x3

1

u = x3 + 1 so that du = 3x2 dx. We now change the limits. If x = 1 then u = 2 and if x = 2 then u = 23 + 1 = 9. So "

2

1

3x2 dx = 3 (x + 1)2

"

9

u−2 du

2

u−1 9 ) =( −1 2

1 1 = − + 9 2 7 = 18 "

1 2

e1/x dx we let u = x1 so that du = − x12 dx. 2 1 2x 3 We can change our limits here so that

(b) For

1 then u = 3, 3 1 and if x = then u = 2. 2 if x =

We thus have "

1 2 1 3

e1/x

" 1 2 u 1 dx = − e du 2x2 2 3 " 1 3 u e du Note the change in the order of integration. = 2 2 1 3 1 3 e − e2 . = eu = 2 2 2 213

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An Introduction to Applied Calculus for Social and Life Sciences

9.5

Integrating rational functions

(x) We consider polynomials of the form fg(x) where the degree of f (x) is greater than or equal to that of g(x) that can be integrated by first dividing the polynomials. The division of the polynomials expresses the rational polynomial in a simpler form that can easily be integrated sometimes. We consider the following examples.

example 9.5.1 Find the integrals " x dx (a) x+1

"

x2 + 1 dx x+1

(b)

Solutions (a) Note that the rational function So "

x dx = x+1

" 1−

1 x+1

" dx =

" Alternatively we can rewrite

x 1 =1− . x+1 x+1 " dx −

1 dx x+1

= x − ln |x + 1| + C. " x dx as x(x + 1)−1 dx. x+1

We let u = x + 1 ⇒ x = u − 1. Also note that du = dx. Therefore

" " " 1 1− x(x + 1)−1 dx = u−1 (u − 1)du = du = u − ln |u| + c u = (x + 1) − ln |x + 1| + c = x − ln |x + 1| + C, where C = c + 1. Please note that substitution is not, however, straightforward when the degree of the polynomial increases. It is always better to divide first. Let us consider the following example. " (b)

 "  2 (x − 1) + dx x+1 " " 2 dx = (x − 1)dx + x+1 1 = x2 − x + 2 ln |x + 1| + C. 2

x2 + 1 dx = x+1

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Chapter 9 Integration and its Applications

9.6

Area between curves

Area between curves If f (x) and g(x) are continuous with f (x)  g(x) on the interval a  x  b, then we define the area between y = f (x) and y = g(x) over the interval (see figure below) to be " A=

b

[f (x) − g(x)]dx.

a

The figure below shows the area between two functions.

example 9.6.1 Find the area bounded by (a) the lines y = x, y = −x and y = 1 (b) the x-axis and curve y = −x2 + 4x − 3 (c) by the curves y = x2 − 2x and y = −x2 + 4 (d) the curve y =

1 x2

and lines y = x, y =

x 8

in the first quadrant.

Solutions (a) We first sketch the area bounded by the given lines.

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An Introduction to Applied Calculus for Social and Life Sciences

The area bounded by the lines y = x, y = −x and y = 1 is given by "

1

(1 − x)dx

A=2 0

= 2(x −

x2 ) 2

1 0

= 2(0.5) = 1.

(b) The following figure shows the area bounded by the x-axis and curve y = −x2 + 4x − 3.

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Chapter 9 Integration and its Applications The shaded area is given by " 3 A= [(−x2 + 4x − 3) − 0]dx 1 3 1 = (− x3 + 2x2 − 3x) 3 1 1 3 1 2 = [− (3) + 2(3) − 3(3)] − [− + 2 − 3] 3 3 1 = −9 + 18 − 9 − (− − 1) 3 4 = . 3

(c) The area bounded by the curves y = x2 − 2x and y = −x2 + 4 is shown in the figure below.

We first find the points of intersection by setting x2 − 2x = −x2 + 4 2(x + 1)(x − 2) = 0 ⇒ x = −1 or x = 2.

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An Introduction to Applied Calculus for Social and Life Sciences The area is thus given by " A=

2

−1 " 2

[(−x2 + 4) − (x2 − 2x)]dx (−2x2 + 2x + 4)dx

= −1

2 2 = (− x3 + x2 + 4x) 3 −1 2 3 2 2 = [− (2) + 2 + 4(2)] − [− (−1)3 + (−1)2 + 4(−1)] 3 3 16 2 = − + 4 + 8 − ( + 1 − 4) 3 3 18 = − + 15 = −6 + 15 = 9. 3

(d) The two points of intersection are found by setting x =

1 1 x and = 2 . 2 x 8 x

x3 = 1 and x3 = 8 ⇒ x = 1 and x = 2.  The points of intersection are (1, 1) and 2, 14 . The area is shown in the figure below.

The area is determined by considering two regions, one in which y = x is above y = x8 and the other in which y = x12 is above y = x8 . 218

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Chapter 9 Integration and its Applications  " 2 1 x x − dx + dx 8 x2 8 0 1 2

−1 x2 2 x x2 1 − − + = 2 16 0 x 16 1





 1 1 1 1 1 − = + − − − −1 − 2 16 2 4 16 3 = . 4 "

1

So A =

9.7



x−

The average value of a function

One application of the definite integral is the average value of a function. Average function value Let f (x) be a function that is continuous on the interval a  x  b. The average value L of f (x) over a  x  b is given by the definite integral 1 L= b−a

"

b

f (x)dx. a

The integral is useful in determining say the average pollution level in a city over a certain time interval. We use the following example to illustrate the usefulness of the integral. example 9.7.1 The population of a community in the Vaal t years after the year 2000 is given by e0.1t million people. P (t) = 1 + e0.1t What was the average population of the community during the decade from 2000 to 2010? Solution The average population in the community is given by " 10 e0.1t 1 dt L= 10 − 0 0 1 + e0.1t " 10 1 (1 + e0.1t )−1 e0.1t dt = 10 0 " 1+e 1 1 du u−1 = 10 2 0.1

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An Introduction to Applied Calculus for Social and Life Sciences 1 du = e0.1t dt. Let u = 1 + e0.1t so that du = 0.1e0.1t dt ⇒ 0.1 The new limits are such that, if t = 0 then u = 2 and if t = 10 then u = 1 + e. So " 1+e 1 1 du u−1 10 2 0.1 " 1+e u−1 du =

L=

2

= ln |u|

1+e 2

= [ln(1 + e) − ln 2] = ln

9.8

(1 + e) 2

million people.

Integration by parts

Integration by parts is a technique for integration based on the product rule. It is a powerful method that can easily be remembered by simply writing it in terms of differentials. Let us put together the derivation of the differentials. Suppose u(x) and v(x) are differentiable functions of x, then the product rule gives dv du d [u(x)v(x)] = u(x) + v(x) dx dx dx dv Make u(x) dx the subject of the formula so that

u(x)

d du dv = [u(x)v(x)] − v(x) . dx dx dx

Integrate both sides with respect to x " " " d du dv [u(x)v(x)]dx − v(x) dx u(x) dx = dx dx dx " " u(x)dv = u(x)v(x) − v(x)du. We can write the above formula as " " udv = uv − vdu. The formula for integration by parts: For the indefinite integral "

" udv = uv −

vdu

(9.1)

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Chapter 9 Integration and its Applications example 9.8.1 Evaluate the following " (a) xex dx " (b)

x2 e−x dx

Solutions (a) The integral to be solved must be such that "

" xex dx =

" x ex dx ≡  u

udv

dv

i.e. set the integral equal to the left-hand side of the equation in (9.1) . We need to choose u and dv.

What to consider when choosing u and dv The following should be considered: (i) The derivative of u must be easy to determine. # (ii) One should be able to evaluate dv. # # (iii) vdu should be easier to evaluate than udv, except in those cases where you use integration by parts twice and then rearrange the equation.

We will always have two options when it comes to substitutions. Option 1: # For the integral xex dx we can set up a table to help us with the integration. Say we choose the following: u = ex dv = xdx

du = ex dx 1 v = x2 2

Note that whatever we choose for u needs to be differentiated, and what we choose for dv needs to be integrated.

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An Introduction to Applied Calculus for Social and Life Sciences " So,

" udv = uv −

vdu is

"

" 1 2 1 x (ex )dx xex dx = (ex )( x2 ) − 2 2 " 1 2 x 1 = x e + x2 ex dx. 2 2    more complicated

# How #do we know whether#we made the right choice? When we compare udv # 2 −x −x e dx is more complicated than xe dx. So, and vdu, we note that x # 2 −x x e dx is not simpler than the integral we started with. This means that our choice of u and dv is not the right one. Option 2: du = dx v = ex

u=x dv = ex dx

"

" xex dx = xex −

ex dx    (∗)

= xex − ex + C. Note that (*) is simpler than the original integral. (b) The solution involves integrating by parts twice. We begin by making the table so that du = 2xdx u = x2 dv = e−x dx v = −e−x "

x2 e−x dx = −x2 e−x −

"

= −x2 e−x +

−e−x (2x)dx " 2 xe−x dx   

.

Integrate again by parts

u=x dv = e−x dx "

du = dx v = −e−x

x2 e−x dx = −x2 e−x + 2 2 −x

= −x e

"

xe−x dx

  " −x −x + 2 −xe + e dx

= −x2 e−x − 2xe−x − 2e−x + C.

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Chapter 9 Integration and its Applications example 9.8.2 Find the following integrals using integration by parts. " " √ (b) −x x − 1dx (a) (x + 1)e2x dx " (c)

" ln xdx

e

ln x dx x2

(d) 1

"

1

"

2

x3 ex dx

(e)

(f)

0

2

√

0

x dx 4x + 1

Solutions " (a) For (x + 1)e2x dx the following table will be useful: u=x+1 2x

dv = e dx " (x + 1)e2x dx =

du = dx 1 v = e2x 2

1 1 (x + 1)e2x − 2 2

" e2x dx

1 1 (x + 1)e2x − e2x + C 2 4

1 1 2x x+ = e + C. 2 4

=

(b) For

#

√ −x x − 1dx we use the following table: u = −x 1 2

dv = (x − 1) dx "

du = −dx 2 v = (x − 1)3/2 3

√ 2 2 −x x − 1dx = (−x) (x − 1)3/2 − 3 3

" (x − 1)3/2 (−dx)

" 2 2 3/2 (x − 1)3/2 (dx) = − x(x − 1) + 3 3 2 (x − 1)5/2 2 +C = − x(x − 1)3/2 + 5 3 3 2 2 4 = − x(x − 1)3/2 + (x − 1)5/2 + C 3 15 2 = (x − 1)3/2 (−3x − 2) + C. 15 223

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An Introduction to Applied Calculus for Social and Life Sciences (c) For

#

ln xdx we use the following table:

"

u = ln x

du =

dv = dx

v=x

1 dx x

" 1 ln xdx = (x) ln x − (x) dx x " = x ln x − dx = x ln x − x + C.

(d) We use the following table: 1 dx x 1 v = −x−1 = − . x " e ln x 1 ln x e dx = − + dx 2 x2 x 1 x 1



ln e ln 1 1 − =− + − e 1 x 1 e 1 =− − e x 1

1 1 −1 =− − e e 2 =1− . e u = ln x 1 dv = 2 dx x

"

e

1

(e) We first rewrite

"

du =

"

2

x3 ex dx = since the integral

#

e 1

2

x2 xex dx

2

xex dx is easy to find. It is of the type " f  (x)ef (x) dx = ef (x) + C.

We use the following table: u = x2 x2

dv = xe dx

du = 2xdx 1 2 v = ex using substitution 2

For substitution to find v, set 1 s = x2 ⇒ ds = 2xdx, so that ds = xdx " 2 " 1 1 x2 es ds = es + C Therefore xe dx = 2 2 1 x2 = e + C. 2 224

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Chapter 9 Integration and its Applications So

"

(f) For

" 1 2 x2 1 x2 x e dx = x e − e (2x)dx 2 2 " 2 2 1 = x2 ex − xex dx 2 2 1 1 2 = x2 ex − ex + C. 2 2 3 x2

"

2

0

x √ dx = 4x + 1

"

2

(4x + 1)−1/2 xdx

0

we use the following table: u=x dv = (4x + 1)−1/2 dx "

2

(4x + 1)−1/2 xdx =

0

9.9

du = dx 1 (4x + 1)1/2 1 v= = (4x + 1)1/2 1/2 4 2 1 x(4x + 1)1/2 2

2

−

1 2

"

2

(4x + 1)1/2 dx 0 1 (4x + 1)3/2 1 2 1 1/2 = [2(9) − 0] − 2 2 3/2 4 0  1  3/2 =3− 9 −1 12 5 = . 6 0

Integration of trigonometric functions

Integrals for basic trigonometric functions " " cos x dx = sin x + C sin x dx = − cos x + C "

" csc2 x dx = − cot x + C

sec2 x dx = tan x + C "

" csc x cot xdx = − csc x + C

sec x tan x dx = sec x + C

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An Introduction to Applied Calculus for Social and Life Sciences example 9.9.1 Evaluate the following integrals involving trigonometric functions " " π2 cos2 xdx (a) 2 sin 2xdx (b) 0

" (c)

tan5 x sec2 xdx

Solutions " (a) For 2 sin 2xdx one can quickly notice that the derivative of the argument of sine is related to the factor multiplying sin 2x. So, if we let u = 2x, then du = 2dx. " " 2 sin 2xdx = sin udu = − cos u + C = − cos 2x + C. " (b) Given So

π 2

cos2 xdx, we consider the identity cos2 x =

0

" 0

π 2

1 + cos 2x . 2



1 + cos 2x cos xdx = dx 2 0 " π2 " π2 1 cos 2x dx + dx = 2 2 0 0 1 π2 sin 2x π2 = x + 2 0 4 0 π = . 4 "

2

π 2

" (c) For the integral So

tan5 x sec2 xdx, we set u = tan x so that du = sec2 xdx. "

" tan5 x sec2 xdx =

u5 du =

tan6 x u6 +C = + C. 6 6

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Chapter 9 Integration and its Applications example 9.9.2 Evaluate the following integrals " (a) x cos(1 − 3x2 )dx

" (b)

" (c)

x sin x dx

(d)

sin x dx (1 + cos x) " cos3 x sin x dx

Solutions (a) Using the method of substitution, we set u = 1 − 3x2 ⇒ du = −6xdx so that − 16 du = xdx. So

" " " 1 1 2 cos udu cos(1 − 3x )xdx = cos u − du = − 6 6 1 = − sin u + C 6 1 = − sin(1 − 3x2 ) + C. 6 (b) Rewriting the integral so that " " sin x dx = (1 + cos x)−1 sin xdx (1 + cos x) we set u = 1 + cos x ⇒ du = − sin xdx, so that − du = sin xdx. We thus have " " " 1 du (1 + cos x)−1 sin xdx = u−1 (−1)du = − u = − ln |u| + C = − ln |1 + cos x| + C. " (c) For

x sin xdx no substitution works. We try integrating by parts. We set

up that table of substitutions so that u=x dv = sin xdx

du = dx v = − cos x

"

" x sin xdx = −x cos x − = −x cos x +

(− cos x) dx " cos xdx

= −x cos x + sin x + C 227

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An Introduction to Applied Calculus for Social and Life Sciences (d) We can rewrite the integral so that " " cos3 x sin xdx = (cos x)3 sin xdx. We let u = cos x ⇒ du = − sin xdx giving − du = sin xdx. So " " (cos x)3 sin xdx = − u3 du u4 +C 4 1 = − (cos x)4 + C 4 1 = − cos4 x + C. 4 =−

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Chapter 9 Integration and its Applications

Exercises

9.1 Evaluate each of the following integrals: "

"

3

(x2 − 5x3 + 6)dx

(a)

(tan x − x2 cos x3 )dx

1

1

9.2 Determine: " (a) 3t4 dt

" (b)

"

√ 1 3 2 (d) x − √ + x dx 2x

" 1 1 (f) − 3 dx x2 x

(e − e )dt "

(e)

1 dx x2

"

1 2 x + − 2 +π x x 3

(g) " (i)

3

x− 2 dx

"

−t

t

(c)

2

(b)

" dx

(h)

√ 1 + x x dx 2

" (j)

√

x3 − 5x − 1 x3

dx

2

ln(e−x ) dx

9.3 Use the technique of substitution to find the following integrals: " " √ (a) 1 − 2x dx (b) e1−x dx " (c)

x2

xe "

 4

(e) " (g) " (i) " (k)

dx x2

(x3 + 8)3

" dx

(f)

(h) " (j) "

x dx 2x − 1

(l) "

x dx 2−x

(n)

2y 4 dy + 1)

(y 5 "

ln x2 dx x

√ 4

t(1 − t2 )3 dt

(d)

ln x dx x

" (m)

"

ln 3x dx x √

e x √ dx x √ x 1 − 2x dx −2 dx 3x + 5

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An Introduction to Applied Calculus for Social and Life Sciences "

2

xex

(o) " " (s) " (u)

" dx

x2 (x3 + 1)2/3 dx

(p) "

3t dt 2 t +8

(r)

ex + e−x dx ex − e−x

(t)

ln x2 dx 4x

(v)

√

(q)

−1

2y 4 (y 5 + 3)−3 dy

"

ln 5x dx 2x

"

2x ln(x2 + 1) dx x2 + 1

9.4 Determine: "

"

1

(a)

πdx

−2

"

(5 − 2t)dt 1

ln 2

(c)

"

(et − e−t )dt

x2 (x − 1)dx 1

2

(e) 1

"

4

(g)

"

x2 dx (x3 + 1)3

(f)

√ 2 xdx

(h)

"

1

1 dx x ln x e−x (4 − ex )dx

0

ln 2

(i)

"

(et − e−t )dt

3



(j)

1+

0

1

"

e+1

2

"

"

x dx x−1

(k)

dx

(t + 1)(t − 2)6 dt "

−1

1

1

(m)

1 1 + x x2

2

(l)

4

"

e2

e

1

"

6

(d)

0

"

4

(b)

3t dt

9

(n)

3

x− 2 dx

4

1/2

"

e1/x dx x2

(o) 1/3

(p) 0

1

6t dt t2 + 1

9.5 Find the integrals using any suitable method: " (a)

x "

(c)

−2/3

√ 3

1 √ 3 − + x dx x

5 2 s+ s

" (b)

" ds

(d)

(x4 − 5e−x/2 ) dx

(x + 2)(x2 + 4x + 2)−2 dx

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Chapter 9 Integration and its Applications "

"

x+2 dx x2 + 4x + 2

(e) "

"

2

5xe−x dx

(g)

" √ (i) "

4

(k) −1

v(v − 5)6 dv

(f)

(x + 2)(x2 + 4x + 2)5 dx

(h) "

ln x dx x

(j)

x dx x+5

(l)

sin t ecos t dt " x cos x dx "

" 4

cos x sin x dx

(m)

e−1



(n) 0

" √

(o) " (q) 0

1

"

w3 dw 1 + w2

(p) e

e2

x x+1

dx

1 dx x(ln x)2

x dx e2x

9.6 Use the substitution method to evaluate each of the following indefinite integrals: " " (a) (1 + sin x)2 2 sin x cos xdx (b) etan x sec2 xdx

9.7 Calculate the area under the given function and above the x-axis over the indicated interval: (a) f (x) = x2 + 2x + 7,

[2, 5]

(b) h(x) =

ln x , x

[1, e]

9.8 Calculate the area between the two given curves over the indicated interval: (a) f (x) = ex ,

g(x) = x 0 ≤ x ≤ 3

(b) f (x) = 3x − 8, 2

g(x) = −3x2 + 10

−1≤x≤1

9.9 Find the area of the region under the curve 2

y = xe−x over the interval 0 ≤ x ≤ 3.

9.10 Sketch the region bounded by the following curves and/or lines, and find the area: (a) y = x3 and y = x (b) y = ex , y = e−x and the lines x = −1, x = 1. 231

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Chapter 10

Modelling with Differential Equations 10.1

Mathematical modelling

A crude definition of mathematical modelling is that it is a process of describing some part or whole of a real–world phenomenon in mathematical terms. What we refer to as the real world may be a real physical system, a social system, an ecological system or any other system whose behaviour can be observed or tracked. Before the construction of a building, architects often produce a model for the building. Equally, mechanical engineers often produce some models before they produce a real machine. These models have one thing in common: they are objects used to represent a real thing and when one looks at them, they represent a real thing in a scaled down form that increases our understanding of what is being represented. However, they contain only the important features that define what they represent. On a building model, one will not be able to see the staircases in the building, for instance, or the engine of the car in the case of the car model. A mathematical description of a real-world situation is referred to as a mathematical model. Consider a community whose population size is N. If part of the population of the community is infected by a disease, say influenza, the number of people that are not infected, let’s say S, can be obtained from the relation S = N − I where I is the number of infected individuals. This is a mathematical model that helps us to find the number of individuals that are not infected in a community. The foundations of mathematical models are rooted in calculus, geometry, differential equations and all branches of mathematics. The biggest question is: why model? It is important to note that, first, modelling helps us to gain understanding through mathematical analysis of the model. Secondly, in the process of building the model we can determine which factors are most important and how entities of the system are related. Lastly, through simulations we are able to predict the future dynamics of the system being modelled. 232

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Chapter 10 Modelling with Differential Equations

10.2

Building models

The following steps are usually taken when building a model. We represent them through the following figure.

The process can simply be summarised as follows: • We start will a real-world problem that needs to be solved. • We draw up some assumptions to reduce the complexity of the model while considering the most important aspects of the real-world problem. • We then write down the mathematical equations. • We now do the mathematical manipulations of solving the equations and performing numerical simulations. • The mathematical analysis results will then be interpreted in the context of the real-world scenario being modelled. • The results are compared with reality. This may be comparisons with data. Once there is some agreement, we can deliver conclusions. Otherwise, we have to go back to the model and start the process again. Modelling is thus an iterative process. Let us now look at the simple models represented by differential equations.

10.3

Differential equations

A differential equation is an equation that contains derivatives, for example 2 dy dy dp dy 2 = 3x + 5 or = kp or + 2y = ex . +3 dx dt dx dx The differential equation dy = 3x2 + x + 5 dx 233

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An Introduction to Applied Calculus for Social and Life Sciences is solved by finding y = f (x) such that the derivative of y (i.e. the left-hand side) will yield the function on the right-hand side. Therefore, the integral with respect to x is applied to both sides: " " dy dx = (3x2 + x + 5)dx dx " " dy = (3x2 + x + 5)dx 1 y = x3 + x2 + 5x + C. 2

(10.1)

This is called the general solution. If we are given a point together with a differential equation, we call it an initial value problem. If the above differential equation is given with y = 2 when x = 0, then 10.1 gives C=2 The particular solution of

#

#

dy = (3x2 + x + 5)dx where y = 2 when x = 0 is 1 y = x3 + x2 + 5x + 2. 2

An initial value problem is therefore a differential equation coupled with a side condition that gives the state of the system at the starting point or at some point. dy = 3x2 + 1 with condition that y = 6 if For example, to find y = f (x) such that dx x = 0. We solve the initial value problem as follows dy = 3x2 + 1 ⇒ dy = (3x2 + 1)dx. dx Integrating both sides gives " y=

(3x2 + 1)dx = x3 + x + C.

We now solve for C using the point (x, y) = (0, 6). So y = f (0) = 6 ⇒ C = 6. Therefore y = f (x) = x3 + x + 6.

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Chapter 10 Modelling with Differential Equations example 10.3.1 Find the general solution of the given differential equation dy = x2 + 2x dx Solution

" "

dy = dx

" (x2 + 2x)dx "

dy =

(x2 + 2x)dx

y=

1 3 x + x2 + C. 3

It is important to note that for each value of C we have a different solution to the differential equation. We can make a plot of y against x and have a representation of a family of curves or a family of solutions that represent the general solution to the differential equation.

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An Introduction to Applied Calculus for Social and Life Sciences Separable differential equations A differential equation in the form h(x) dy = dx g(y) is called separable. The general solution of such an equation can be obtained by separating the variables and integrating both sides, i.e. "

g(y)dy = h(x)dx " g(y)dy = h(x)dx.

example 10.3.2 Find the general solution of the given differential equations (a)

dy = 2e−y dx

(b)

x2 + 6x + 3 dy = dx y2

(c)

y2 + 4 dy = dx 2xy

(d)

2 dy = 2xex −y dx

Solutions (a)

dy = 2e−y ⇒ dy = 2e−y dx dx " " y e dy = 2dx ey = 2x + C y = ln(2x + C).

(b)

x2 + 6x + 3 dy = dx y2

⇒ y 2 dy = (x2 + 6x + 3)dx " " Integrating gives y 2 dy = (x2 + 6x + 3)dx ⇒

1 3 1 y = x3 + 3x2 + 3x + C 3 3  3 y = x3 + 9x2 + 9x + 3C.

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Chapter 10 Modelling with Differential Equations The general solution for a few values of C is given in the following figure:



4 + y2 y2 + 4 1 dy = = (c) dx 2xy 2y x

4 + y2 1 ⇒ dy = dx 2y x " " 1 2y dy = dx y2 + 4 x ⇒ ln(y 2 + 4) = ln |x| + ln C (Note that ln C is a constant) y 2 + 4 = Cx √ ⇒ y = Cx − 4. (d) Note that

2 2 dy = 2xex −y = 2xex · e−y dx " " 2 y ⇒ e dy = 2xex dx 2

⇒ ey = ex + C.

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An Introduction to Applied Calculus for Social and Life Sciences example 10.3.3 Solve the initial value problem dy = (x3 + 1)y 2 ; dx

y = 1 whenever x = 2.

Solution dy = (x3 + 1)y 2 ⇒ dy = (x3 + 1)y 2 dx dx " " −2 ⇒ y dy = (x3 + 1)dx ⇒ −y −1 =

1 4 x + x + C. 4

Applying the initial condition we have y(2) = 1. −

1 1 = x4 + x + C y 4

1 4 2 +2+C 1=− 4

⇒ C = −7. So y = −

x4

4 . + 4x − 28

example 10.3.4 Solve the initial value problem dy sin(2t) = ; dt y2 Solution sin(2t) dy = ⇒ dt y2

"

" (y 2 )dy = ⇒

Substitute (x, y) = (0, 1) :

1 3

y(0) = 1

sin(2t)dt

1 1 3 y = − cos(2t) + C. 3 2

= − 12 cos(0) + C ⇒ C = 56 .

So 5 3 y 3 = − cos(2t) + 2 2 3 3 5 − cos(2t). y= 2 2

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Chapter 10 Modelling with Differential Equations

Exercises

10.1 Solve the given initial value problem for y = f (x): (a)

dy = e−x ; f (0) = 3 dx

(b)

x+1 dy = √ ; f (4) = 5 dx x

10.2 Find the general solution of the given differential equation: (a)

dy √ = t + e−t dt

(b)

dy = 3y dx

(c)

√ dy = ey x + 1 dx

(d)

√ dy = y ln t dt

(e)

ln t dx = dt ln x

10.3 Find the particular solution of the differential equation satisfying the indicated condition: (a)

y+1 dy = , y(1) = 2 dt t(y − 1)

(b)

dy sin(2t) = , y(0) = 1 dt y+1

(c)

dx = x2 cos t, x(π) = 1 dt

(d)

sin x − x π dy = , y( ) = 0 dx y 2

(e)

2 dy = xey−x , y(1) = 0 dx

(f)

√ dx = xt t + 1, x(0) = 1. dt

10.4 The slope f  (x) at each point (x, y) on a curve y = f (x) is given along with a point (a, b) on the curves. Determine f (x): 1

(a) f  (x) = 3x2 + 6x − 2; (0, 6)

(b) f  (x) = x− 2 + x; (1, 2)

(c) f  (x) = e−x + x2 ; (0, 4)

3 − 4; (1, 0) x √ (f) f  (x) = x + x; (1, 2)

(e) f  (x) = x2 + 2x − 2; (0, 4) (g) f  (x) = e−x +

1 ; (1, 2) x2

(i) f  (x) = −x(x + 1); (−1, 5)

(d) f  (x) =

(h) f  (x) =

3 + x; (1, 0) x

(j) f (x) = e−x + x2 ; (0, 4)

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An Introduction to Applied Calculus for Social and Life Sciences 10.5 Find the general solution of the given differential equation: (a)

dy = y2 dx

(b)

dy = ex+y dx

(c)

xt dx = dt 2t + 1

(d)

dy sin x = dx cos y

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