UCF EXCELApplications

ofCalculus II

spring 2010

Edited by:Scott C. Hagen, Ph.D., P.E.,Lillie E. Thomas, E.I.

Preface

One of the goals of the EXCEL program is to enhance understanding of mathematics (primarily calculus) for Science, Technology, Engineering and Mathematics (STEM) students. Mathematics is considered the cornerstone of the success of any student pursuing a degree in a STEM discipline. One way of achieving this goal at UCF is by introducing students to the Applications of Calculus courses. This book is the textbook material for the Applications of Calculus II course that you will be taking in the spring semester of your freshman studies at UCF. It is comprised of a series of chapters written by EXCEL faculty from various disciplines of the sciences, engineering and mathematics. These chapters and the presentations that the individual EXCEL faculty members will make in the Applications of Calculus II class will show you how material that you will be learning in your Calculus II class is actually used in the discipline of the faculty presenter. These presentations are correlated with the sections of your calculus book and will be presented just shortly after you cover those sections in your Calculus II class.

Each of the chapters of this book is organized in the same way. The chapters begin with a section titled Calculus Topic which includes at least one calculus problem taken directly from your calculus text along with a detailed solution. The purpose of including a problem from your calculus text is to 1) reinforce the techniques used to solve problems from that particular section and 2) to explain how a particular mathematical concept is related to a particular physical (real world) problem. The next section in each chapter is a Background/Motivation section. This section provides you with the basic understanding of the field of study that will follow and why the mathematics (in particular, a Calculus II topic) is necessary in that field of study. Following the Background/Motivation section you will find a series of Learning Objectives. Each Learning Objective is followed by text, figures and tables which lead the reader through the process of understanding that objective. Each of the Learning Objectives shown in a chapter will be a focus of the faculty presentation. The faculty presenters are outstanding educators and researchers in their respective fields. Each faculty member authored his/her chapter as well as the presentation that will be given in class.

We hope that you, the reader and student, find this book to be helpful in your journey through this course, that the course offers a useful enhancement to your understanding of Calculus and that the weekly presentations broaden your horizons and peek your interests in topics of mathematics, engineering and sciences that you may have never otherwise studied on your own.

Acknowledgements

First and foremost, we would like to acknowledge the support of the NationalScience Foundation grant DUE 05254209, who has entrusted us with the opportunity toeducate the STEM youth of tomorrow.

We also want to acknowledge the generous support from Provost Hickey, Vice–Provost Schell, Dean Morrison-Shetlar, the UCF Office of Research and Com-mercialization (in particular, the Vice–President of Research Dr. Soileau), the College ofEngineering and Computer Science Dean’s Office (in particular, Dean Simaan, AssociateDean Reilly and Ms. Falls), the College of Sciences Dean’s Office (in particular, DeanPanousis, Associate Dean Johnson and Ms. Kirkpatrick), the Burnett School ofBiomedical Sciences (in particular, Associate Director Roseann White), the Office ofOperational Excellence and Assessments (in particular, Dr. Lancey), the InstitutionalResearch Office (in particular, Ms. Ramsey), UCF Admissions Office (in particular, Dr.Chavis and Ms. Costello), Worthington, the UCF Orientation Office (in particular, Mr.Richie and Mr. Hicks), the UCF Housing Office (in particular, Mr. Paulick, Mr. Novakand Ms. Rutkowski) and Associate Provost Poisel, amongst many others.

This book would not have been possible without the hard work and dedication ofthe EXCEL faculty: Drs. Georgiopoulos and Young (EXCEL Directors), who have beeninvolved with all aspects of this project from planning, writing, coordinating, schedulingto the more thankless jobs of handling the finances of the project. Drs. Geiger, Hagen,and Islas (EXCEL Coordinators), all of which spent a tremendous amount of time inplanning and directing this project as well as their continuing projects. The EXCEL fac-ulty who have contributed their knowledge and expertise in order to make this book areality are Drs. Self (Chapter 1), Divo (Chapter 2), LaViola (Chapter 3), Clausen(Chapter 4), Dubey (Chapter 5) and Kassab (Chapter 6). The assessment and evaluationmaterials associated with this class were developed by the Faculty Center for Teachingand Learning director Dr. Crouse and Dr. Lancey from the Office of OperationalExcellence and Assessments.

Finally, the load of editing and integrating the information from seven differentEXCEL faculty into a cohesive volume of work fell upon Lillie Thomas, who did anoutstanding job under the guidance of the EXCEL coordinator, Dr. Hagen. In addition,Marc Ellie and Greg Territo provided valuable insight. We are very grateful for theireffort.

Table of Contents

Exponential and Logarithmic Functions in Biomendical Science 1–1

Integration by Parts - Applications in Engineering 2–1

Parametric Equations for Fun and Profit 3–1

Application of Polar Coordinates in Chemistry 4–1

Integrals in Physics 5–1

Application of Series in Heat Transfer - Transient Heat Conduction 6–1

Chapter 1

AApppplliiccaattiioonnss ooff CCaallccuulluuss IIII TTooppiicc

Exponential and Logarithmic Functions in BiomedicalScience

FFaaccuullttyy CCoonnttrriibbuuttoorr Dr. William Self Department of Molecular Biology and MicrobiologyBurnett School of Biomedical Science

WWeeeekkss:: January 25 and February 1 of Spring 2010

Exponential and Logarithmic Functions in Biomedical Science 1-2

In this section, we will discuss the use of exponential functions in the study ofmicrobiology and in the use of radioisotopes, both in research and in clinical practice.

This will be relevant for some future courses you make take including:

MCB 3020 – General Microbiology MCB 4414 – Microbial Metabolism

Background/Motivation

Infectious diseases are caused by both bacterial and viral pathogens, each causingdisease when present in significant numbers attacking the host. The ability of amicroorganism to multiply rapidly is critical to its ability to cause disease and to overtakethe defense mechanisms of the host (such as us humans!). The cultivation and study ofpure populations of microbes (those that exist in nature) require that the scientist becapable of determining the growth rate of the organism. Bacteria divide and multiply at alogarithmic rate and as such the equations used to determine growth, growth rate andpopulation densities are exponential functions.

Radioisotopes are used in both scientific research and clinical practice asdiagnostic tools and have proven to be invaluable in several areas of biomedical science.The ability to specifically label proteins and follow them in living cells systems has beenvital to our understanding of the proteome. The metabolic pathways of living organismshas also been built on the use of radioisotope-labeled metabolites (such as 14C-labeledglucose) and without the development of research radioisotopes our knowledge ofbiology would certainly be limited. Short-lived radioisotopes have also found their placein imaging technologies and have enabled the diagnosis and treatment of several diseases,most notably cancer and cardiovascular disease.

The mathematics behind these key tools in the biomedical scientist’s toolboxrequire good working knowledge of logarithmic and exponential functions. The materialpresented below will introduce you the use of these functions in determining bacterialgrowth, and the calculation of the half-life of radioisotopes. These are two commonlyused examples that relate directly with concepts being learned in Calculus II.

Calculus Concepts to be covered in this section – reinforcement

• Logarithmic and Exponential functions

Applied Concepts to be covered in this section – real world applications

• Growth of bacterial cells in culture • Calculation of bacterial populations in culture • Use of radioisotopes in research and clinical applications • Calculation of radioisotope half-life

Applied Learning Objective #1: Bacterial growth curve

Exponential and Logarithmic Functions in Biomedical Science 1-3

Under nutrient rich conditions (such as in a culture flask with ample carbon,nitrogen, sulfur, phosphorus, etc) bacteria grow and divide until either 1.) a nutrientbecomes limiting or 2.) a toxic metabolite accumulates and inhibits growth. Thisgrowth, since it follows division of one cell to two, (1, 2, 4, 8, 16, 32, 64, 128, etc)follows a logarithmic pattern. Shown below is a bacterial ‘growth curve’ showing thefour phases of growth:

There are four phases of growth:

Lag phase: Cells are adapting to their new environment (after inoculation into freshmedium). They are altering their proteome to best use the nutrients and increasing theirsize and protein content

Exponential phase: This is the phase at which cells divide rapidly, and their absolutenumbers increase in a logarithmic manner. Cells are large, and contain more protein andRNA than slower growth, nutrient limited cells.

Stationary phase: As nutrients become limiting (or toxic product inhibit growth), cellsreduce in size and produce less protein and RNA per cell. Cells also stop dividing andtheir growth is not logarithmic.

Death phase: Without removal of a toxic metabolite, or addition of the limiting nutrient,cells will eventually die. Alternatively some cells (such as the pathogen that causesAnthrax) will form a resistant spore that can survive extreme conditions.

Growth physiology:Growth physiology: Phases of Population GrowthPhases of Population Growth

time

ccceeellllll

mmma aas ss

sss

vvv iiiaaabbb

lll eee ccc

eeellllll sss

adapt fromstationary, cellsize increases

steady state,largest cells

substrate limited,product accumulation,

cell size decreases

energy (ATP & Δp)depleted, autolysis,

sporulation, cysts, etc.

log

Exponential and Logarithmic Functions in Biomedical Science 1-4

It is important to study bacterial populations in test tubes to be able to askexperimental questions about how they grow, how they respond to their environment, andhow best to inhibit their growth (i.e. use of an antibiotic). Most antibiotics that havebeen developed are actually derivatives of something called natural products. These arecompounds found in nature that are produced by eukaryotes and prokaryotes that areaimed at killing their ‘neighbors’ and allowing optimal growth of their ‘kind’. We mustfirst know the mechanism of action of a drug (such as a newly isolated natural product),and then we can determine whether this will be tolerated by a mammalian host(humans!). All of these experiments require that we study populations of bacterialcultures. The study of microorganisms in this way is generally termed bacteriology.

Pathogens – common: Although sometimes referred to as ‘bugs’, bacteria are defined by the lack of a

membrane-bound nucleus. Beyond that similarity there are a vast number of differenttypes of microorganisms that live in practically every niche on earth. As humans we areusually most interested in microbes that can cause disease in humans and animals. Belowis a description of some of the best studied, and most notorious microbes that arepathogens. These descriptions are taken from the “BAD BUG BOOK” published for freeaccess by the Food and Drug Administration.

Staphylococcus aureus S. aureus is a spherical bacterium (coccus) which on microscopic examinationappears in pairs, short chains, or bunched, grape-like clusters. These organismsare Gram-positive. Some strains are capable of producing a highly heat-stableprotein toxin that causes illness in humans.

Salmonella

Salmonella is a rod-shaped, motile bacterium -- nonmotile exceptions S.gallinarum and S. pullorum--, nonsporeforming and Gram-negative. There is awidespread occurrence in animals, especially in poultry and swine.Environmental sources of the organism include water, soil, insects, factorysurfaces, kitchen surfaces, animal feces, raw meats, raw poultry, and rawseafoods, to name a few.

Escherichia coli O157:H7 Normally E. coli serves a useful function in the body by suppressing the growth ofharmful bacterial species and by synthesizing appreciable amounts of vitamins. Aminority of E. coli strains are capable of causing human illness by several differentmechanisms. E. coli serotype O157:H7 is a rare variety of E. coli that produces largequantities of one or more related, potent toxins that cause severe damage to the lining ofthe intestine.

Infectious doses:

Exponential and Logarithmic Functions in Biomedical Science 1-5

Below are the infectious doses of some common bacterial and viral pathogens:

Figure 2. Summary of Infectious doses for several pathogens. From: Pathogenesis, Virulence, and Infective Dose Schmid-Hempel P, Frank SA PLoSPathogens Vol. 3, No. 10, e147 doi:10.1371/journal.ppat.0030147

The infectious dose is the number of bacteria (or virus) that is needed to cause disease inthe animal (human). It is clear from the figure above that different types of pathogenscan have very different infectious doses. This is due to their variable ability to grow onceintroduced into host, while attempting to overcome the host’s defense mechanisms.

Applied Learning Objective #2: Calculating the growth of a bacterial population

The growth rate of bacterial populations, assuming they are in the logarithmic phase ofgrowth, is characterized by the following equation:

x = x0 2y where:

x = anything that doubles each generationx0 = starting value y = number of generations

The generation time is defined as the time that is required for a population of cellsto double in number and can be defined by g. The number of generations can beestimated by dividing the time that has elapsed, t, by the generation time g as follows:

Y = t/g Given the above equations, we can do an example problem:

Exponential and Logarithmic Functions in Biomedical Science 1-6

Suppose you have a culture of cells that has a density of 1 x 108. You wish to subculture(reinoculate) fresh medium and at precisely 16 hours later have the same culture density (i.e. same number of cells). If g=2 hours, what should xo be (inoculum)?

Estimate the number of generations using the second equation. Y = 16/2 which giveseight generations.

Next substitute in Y to the first equation to give:

1 x 108 = xo 28 or xo = 108/28 or 3.9 x 105

What exactly does this mean?

We can predict, knowing the generation time of a bacterium, that if we inoculate 3.9 x 105 cells we will have 108 cells in 16 hours. This is very important in designingexperiments where a known number of cells are needed at a given time.

Applied Learning Objective #3: Use of radioisotopes in research and in the clinic

Radioisotopes in research.

The atom is the smallest unit of matter that is capable of entering into a chemicalreaction. Atoms are made of electrons, which circle the center of the atom, as well asprotons and neutrons within the center of the atom. The number of protons in the nucleusdefines the element (also called atomic number). The total number of neutrons andprotons is called the mass number. Isotopes are atoms of the same element (same numberof protons) but that carry different numbers of neutrons. A radioisotope is an atom thatcontains an unstable ratio of neutrons to protons. In order to gain stability a radioisotopecan release energy or particles from its nucleus, and this defines the type of radioisotope.

Radioactive decay is defined as the rate of decay of the unstable material to the morestable form. The unit used is called the Curie (named after Marie Curie), and a Curie isequal to 3.7 x 1010 disintegrations per second.

Radioisotopes have proven quite valuable in scientific research. Because a radioactiveisotope ‘behaves’ in a biological system just as the more stable isotope, it can be usedeffectively to trace the material. The conversion of the carbon from sugars into metabolicintermediates and cellular biomass was primarily followed using 14C-labeled glucose. This allowed for isolation of each of the enzymes that acts on carbon in the cell. Likewise the metabolism of nitrogen, phosphorus and trace metals can easily be followedby radiolabeling your compound of interest and introducing it into a living cell model.

Radioisotopes in Medicine.

Exponential and Logarithmic Functions in Biomedical Science 1-7

In addition to their valuable use in research, radioisotopes also play a critical role in theclinic. The half-life of isotopes varies depending on the rate of decay of the energy, sovery short-lived isotopes can be used safely to allow for imaging of internal organs.

Diagnostic Nuclear Medicine

Radioisotopes are used to diagnose many different types of disease. Gamma emittingisotopes (that emit gamma rays that can easily penetrate tissue) are used, and to minimizedamage to cells these isotopes have a very short half-life (3-4 hours).This type of imagingis better than x-rays since it can image both soft and hard tissues.

List of a few commonly used radioisotopes and their use (source – Uranium InformationCentre)

Molybdenum-99 (66 h): Used as the 'parent' in a generator to produce technetium-99m.

Technetium-99m (6 h): Used in to image the skeleton and heart muscle in particular, butalso for brain, thyroid, lungs (perfusion and ventilation), liver, spleen, kidney (structureand filtration rate), gall bladder, bone marrow, salivary and lacrimal glands, heart bloodpool, infection and numerous specialised medical studies.

Cobalt-57 (272 d): Used as a marker to estimate organ size and for in-vitro diagnostickits.

Gallium-67 (78 h): Used for tumour imaging and localisation of inflammatory lesions(infections).

Krypton-81m (13 sec) from Rubidium-81 (4.6 h): Kr-81m gas can yield functionalimages of pulmonary ventilation, e.g. in asthmatic patients, and for the early diagnosis oflung diseases and function.

Thallium-201 (73 h): Used for diagnosis of coronary artery disease other heartconditions such as heart muscle death and for location of low-grade lymphomas.

Radiation exposure from natural sources:

Although these radioisotopes are prepared and enriched from natural sources, there areradioisotopes all around us everyday. Below is a comparison of average annual exposurein different areas of the world (URI – source):

Exponential and Logarithmic Functions in Biomedical Science 1-8

Where does all that exposure come from? Well, besides the sun (constantly bombardingus with several types of radioactive particles), Here is a fascinating list comparing somecommon isotope sources and day to day isotopes.

Radioactivity of some natural and other materials1 adult human (100 Bq/kg) 7000 Bq 1 kg of coffee 1000 Bq 1 kg superphosphate fertiliser 5000 Bq The air in a 100 sq metre Australian home (radon) 3000 Bq The air in many 100 sq metre European homes (radon) 30 000 Bq 1 household smoke detector (with americium) 30 000 Bq Radioisotope for medical diagnosis 70 million Bq Radioisotope source for medical therapy 100 000 000 million Bq 1 kg 50-year old vitrified high-level nuclear waste 10 000 000 million Bq 1 luminous Exit sign (1970s) 1 000 000 million Bq 1 kg uranium 25 million Bq 1 kg uranium ore (Canadian, 15%) 25 million Bq 1 kg uranium ore (Australian, 0.3%) 500 000 Bq 1 kg low level radioactive waste 1 million Bq 1 kg of coal ash 2000 Bq 1 kg of granite 1000 Bq

Applied Concept #4: Calculation of half-life of radioisotopes

Radioisotope half life can be calculated using the following equation:

Exponential and Logarithmic Functions in Biomedical Science 1-9

A = A0e-λt Where: A = Current amount of radioactivity A0= Original amount of radioactivity e = base natural log (approximately 2.718)λ= the decay constant = 0.693/t1/2

(where t1/2= half-life) t = the amount of time elapsed from A0 to A

Here is an example: We have a test tube with Selenium-75 with a known activity of 10 μCi on October 31st.However some time has elapsed and you are now planning to do an experiment to labelClostridium botulinum to determine whether it makes selenoproteins. Your experimentwill be done on December 20th, 50 days later. The half-life of Selenium-75 is 144 days.

The A will be activity that will remain on Dec. 20. The Ao will represent the original activity (10 μCi )T will represent the elapsed time, 50 days.

First, calculate the decay constant λ.λ= 0.693 ÷144 days or 0.0048 days-1.

Then, we can calculate the remaining activity.A = A0e-λt A = 0.7866 (10) Activity remaining equal 7.86 μCi

Chapter 2

AApppplliiccaattiioonnss ooff CCaallccuulluuss IIII TTooppiicc

Integration by Parts

Applications in Engineering

FFaaccuullttyy CCoonnttrriibbuuttoorr Dr. Eduardo Divo Department of Engineering TechnologyCollege of Engineering and Computer Science

WWeeeekkss:: February 8 and February 15 of Spring 2010

Calculus Topic: Integration by Parts ( )Section 8.1

Background

Integration by parts is a technique employed to aid with the integration of the product

of two independent functions. In some cases, the integral of the product of two functions

can be computed directly without the need of integration by parts by employing other

techniques. For example:

( (c d c d? ÐBÑ ?ÐBÑ .B œ � G .B œ 68l?ÐBÑl � G

?ÐBÑ ? ÐBÑ

8 � " ?ÐBÑw 8

8�" w

or

are solved directly employing the general power formula and the general logarithmic

formula respectively. In both cases, the two functions appearing in each integrand are

clearly not independent from each other as one is the derivative of a power of the other.

The general formula for integration by parts can be easily derived from the formula

for differentiation of a product of two functions as:

.

.B?ÐBÑ@ÐBÑ œ ?ÐBÑ@ ÐBÑ � @ÐBÑ? ÐBÑc d w w

Integrating the left and right-hand sides of the formula above with respect to , we find:B

( ( (.

.B?ÐBÑ@ÐBÑ .B œ ?ÐBÑ@ ÐBÑ.B @ÐBÑ? ÐBÑ.Bc d w w�

Recall that the integral operation is an anti-derivative, therefore, the integral on the left-

hand side cancels out with the derivative to yield:

?ÐBÑ@ÐBÑ œ ?ÐBÑ@ ÐBÑ.B @ÐBÑ? ÐBÑ.B( (w w�

Rearranging the expression above we find:

( ( ( (?ÐBÑ@ ÐBÑ.B œ ?ÐBÑ@ÐBÑ � @ÐBÑ? ÐBÑ.B Ê ?.@ œ ?@ � @.?w w

The expression above is the general formula of integration by parts. Notice that the

integral on the left-hand side contains a product of two independent functions and?ÐBÑ@ ÐBÑw and the integration by parts formula simply ‘shifts’ the operation so that the integral

on the right-hand side is performed over the product of and with the ‘hope’@ÐBÑ ? ÐBÑw

that the resulting integral on the right-hand side is ‘simpler’ or directly solvable by any of

the general formulae.

When confronted with an integral that seems suited for integration by parts, it is

crucial to decide which of the two functions that appear as a product should be selected as

?ÐBÑ @ ÐBÑ and which should be selected as . For this purpose there are several rules andw

criteria that can be employed to achieve the goal of yielding a ‘simpler’ integral on the

right-hand side. A general guide for function selecting when integrating by parts is known

as the rule which can be used to decide which of the two functions in the productILATE

is the function by identifying the one that comes first in the following list:?ÐBÑ

Integration by Parts - Applications in Engineering 2-2

• : inverse trigonometric functionsI

• : logarithmic functionsL

• : algebraic functions (polynomials)A

• : trigonometric functionsT

• : exponential functionsE

It should be noted that integration by parts does not guarantee a solution to the

integral and the use of the rule just constitutes a general guide to achieve the goalILATE

of yielding a ‘simpler’ integral on the right-hand side. As a first example let us consider

the following integral ( :Section 8.1. Example 1)

( B =38B .B œ ?

Note that none of the general formulae (power, logarithm, etc.) can be directly

implemented to solve this integral that clearly shows the power of two independent

functions ( and ), therefore, integration by parts seems like the viable option. TheB =38Bfirst step is to decide which one of the two functions will be and which one will be?ÐBÑ@ ÐBÑ ÐBÑw . By using the rule it is clear that the algebraic (A) function appearsILATE

before the trigonometric (T) function in the list, therefore:Ð=38BÑ

?ÐBÑ œ B Ê ? ÐBÑ œ "w

The function is then selected as the other part of the product as:@ ÐBÑw

@ ÐBÑ œ =38B Ê @ÐBÑ œ =38B .B Ê @ÐBÑ œ � -9=Bw (With the explicit forms for and solved for, the integration by parts formula can? ÐBÑ @ÐBÑw

be implemented directly as:

( (B =38B .B œ � B -9=B � -9=B .B

Notice that the goal of yielding a ‘simpler’ integral on the right-hand side was achieved

by the proper implementation of the rule. Had the opposite choice been made inILATE

the function selection, the integral on the right-hand side would have ended up with a

higher level of complexity than the original one on the left-hand side.

The integral on the right-hand side can now be solved directly leading to:

( B =38B .B œ � B -9=B � =38B � G

There are cases for which it is necessary to apply the integration by parts formula

more than once to yield an integral on the right-hand side that can be solved. Let us

consider one of such examples ( ):Section 8.1. Example 3

( > / .> œ# > ?

Integration by Parts - Applications in Engineering 2-3

In this case the integrand is composed of a product of an algebraic function andÐ> Ñ#

an exponential function . By using the rule, the algebraic function is to beÐ/ Ñ> ILATE

selected as while the exponential function as , leading to:?Ð>Ñ @ Ð>Ñw

?Ð>Ñ œ > Ê ? Ð>Ñ œ #># w and @ Ð>Ñ œ / Ê @Ð>Ñ œ / .> Ê @Ð>Ñ œ /w > > >(Then, substitution of these expressions into the general formula of integration by parts

yields:

( (> / .> œ > / � # >/ .># > # > >

Notice that the integral on the right-hand side is ‘simpler’ than the original one, however,

it is not yet in a form that can be solved directly. Therefore, a second integration by parts

is necessary to simplify it even more. In this case:

?Ð>Ñ œ > Ê ? Ð>Ñ œ "w and @ Ð>Ñ œ / Ê @Ð>Ñ œ / .> Ê @Ð>Ñ œ /w > > >(Notice that the rule was employed once again to select the functions. If theILATE

opposite selection of functions on the second integration by parts had been made, the

right-hand side would have been restored as the original integral on the left-hand side.

Substitution of these new expressions on the right-hand side leads to:

( (Œ > / .> œ > / � # >/ � / .> œ# > # > > > ˆ ‰> � #> � # / � G# >

The integration by parts formula can also be implemented for definite integrals by

simply transferring the limits of integration to the right-hand side Let us consider theÞfollowing example:

("

/

68B .B œ ?

In this case it seems that the integrand is composed of a single function ,Ð68BÑhowever, it can be assumed that this function is multiplying a second function .Ð"ÑTherefore, by using the rule:ILATE

?ÐBÑ œ 68B Ê ? ÐBÑ œ"

Bw and @ ÐBÑ œ " Ê @ÐBÑ œ ".B Ê @ÐBÑ œ Bw (

Then, substitution of these expressions into the general formula of integration by parts

yields:

( (¹ Œ " "

/ /

"

/

68B .B œ B 68B � B .B œ"

B ¹B 68B � B œ / 68/ � / � " 68" � " œ "

"

/

Integration by Parts - Applications in Engineering 2-4

Applications

As a technique for integrating function products, integration by parts plays a major

role in all fields of engineering as its applications are commonly found in problems

ranging from electric circuits, heat transfer, vibrations, structures, fluid mechanics,

transport modeling, air pollution, electromagnetism, digital signal processing, and many

more. To illustrate this, three representative engineering applications are detailed next.

i. Heat Transfer Problem (EGN3358 and EML4142)

The time-dependent temperature of an object changes at a rate proportional toXÐ>Ñthe difference between the temperature of its surroundings and the temperature of the

object. This relation is expressed as the Newton's Law of cooling and is written as:

3- œ � XÐ>Ñ � X Ð>Ñ.X Ð>Ñ 2E

.> Z=

=c dHere, is the density of the object, is its specific heat, is the heat transfer3 - 2

coefficient between the object and its surroundings, is the surface area of the object, E Z=

is the volume of the object, and is the temperature of the surroundings. The solutionX Ð>Ñ=

of this equation is found through an integrating factor as:

XÐ>Ñ œ 7/ X Ð>Ñ/ .>�7> 7>=(

Where 7 œ 2E Î -Z= 3 . Assume that the object is a thermometer that is being used in an

experiment to read the temperature of a surrounding medium that is linearly changing in

time as , as seen in Figure 1. Here is the initial temperature of theX Ð>Ñ œ > � X X= 9 9"surroundings and is the rate at which the temperature of the surroundings is changing."Therefore, the expression for the thermometer temperature is given by:

XÐ>Ñ œ 7/ > � X / .>�7> 7>9( "

Using the rule for integration by parts the following function choices are made:ILATE

?Ð>Ñ œ > � X Ê ? Ð>Ñ œ " "9w and @ Ð>Ñ œ / Ê @Ð>Ñ œ / .> Ê @Ð>Ñ œ /

"

7w 7> 7> 7>(

Substitution of these terms in the integral expression leads to:

( ( " ""

> � X / .> œ > � X / � / .> œ"

7 79 9

7> 7> 7> œ > � X � / � G"

7 7Π"

"9

7>

Substitution of this result into the expression for the temperature leads to:

XÐ>Ñ œ > � X � � G7/7

Π""

9�7>

The constant of integration can be computed by imposing an initial conditionGÐXÐ!Ñ œ X Ñ3 . Assume, for example, that a thermometer with a spherical test section of

Integration by Parts - Applications in Engineering 2-5

radius is initially at a temperature ° , the< œ !Þ$-7 ÐZ œ < ß E œ % < Ñ X œ &! G%$

$ #= 31 1

density of the thermometer medium , its specific heat is3 œ "!!!51Î7$

- œ $!!!NÎ51 G 2 œ "&!![Î7 G° , the heat transfer coefficient ° , the initial#

temperature of the surroundings ° , and the rate of change of the surroundingsX œ "!! G9

temperature ° . The time-dependent temperature of the thermometer given by" œ & GÎ=the solution above is provided in the plot below for the first seconds along with the#!plot of the surrounding temperature where it can be clearly seen that the thermometer

reading follows and lags behind the linearly increasing surrounding temperature.

T(t)

V, c, ρ

As

h

Ts(t)=βt+To

0 5 10 15 2050

100

150

200

T t( )

Ts t( )

t

Figure 1: Thermometer and surroundings temperature evolution.

ii. Beam Deflection problem (EGN3331 and CES4100C)

The space-dependent deflection of a beam with Elastic Modulus and inertiaCÐBÑ IMÐBÑ 0ÐBÑ subjected to a distributed load is governed by the following equation with

solution through integration as:

. . CÐBÑ

.B .BIMÐBÑ œ 0ÐBÑ

# #

# #” •With solution for the deflection through successive integration as:CÐBÑ

CÐBÑ œ 0ÐBÑ.B.B .B.B"

IMÐBÑ( ( ( (Œ

Assuming that the distributed load is constant and downward, i.e. , the0ÐBÑ œ � 0two inner integrals are readily solved as:

CÐBÑ œ .B.B� 0 B

#I MÐBÑ( ( #

In addition, the inertia term is a geometric relation of the beam cross-section givenMÐBÑby . If the base of the beam is allowed to change exponentially asMÐBÑ œ ,ÐBÑ2 Î"#$

,ÐBÑ œ , / MÐBÑ œ M /9 9B B" ", as seen in Figure 2, then, the inertia can be expressed as ,

where and is the base at the free end of the beam. In this case, theM œ , 2 Î"# ,9 9 9$

deflection can be expressed as:CÐBÑ

Integration by Parts - Applications in Engineering 2-6

CÐBÑ œ B / .B.B� 0

#IM9

# � B( ( "

Concentrating on the inner integral:

( B / .B# � B"

And using the rule for integration by parts the following choices are made:ILATE

?ÐBÑ œ B Ê ? ÐBÑ œ #B @ ÐBÑ œ / Ê @ÐBÑ œ � /"# w w � B � B and " "

"

Yielding:

( (Œ B / .B œ � / � B/ .BB ## � B � B � B#

" " "

" "

Another integration by parts and substitution into the deflection solution leads to:

CÐBÑ œ � / � / � / � G .B� 0 B #B #

#IM9

#� B � B � B

# $ "( Π" " "

" " "

The expression above can be broken up into four separate integrals where the first two

will require the same integration by parts procedure employed in the inner integral. After

the implementation of these procedures and simplification, the general solution for the

deflection is:CÐBÑ

CÐBÑ œ G B � G � � � /0 B %B '

#IM9" #

#

# $ %� B” •Œ

" " ""

The two constants of integration and can be computed by imposing twoG G" #

boundary conditions at the fixed end of the beam. Namely, and .CÐPÑ œ ! C ÐPÑ œ !w

Assume that a beam with constant height , free-end base , fixed-2 œ !Þ!&7 , œ !Þ"79

end base , and length , is made of a material with Elastic modulus, œ !Þ#7 P œ %7P

I œ "!!KRÎ7 0 œ "5RÎ7# and a uniformly distributed load is applied downward

throughout the length of the beam. The space-dependent deflection can be seen inCÐBÑthe plot below revealing a maximum deflection at the free end of about .!Þ"(&7

bL

bo

hx

y

0 1 2 3 40.2−

0.15−

0.1−

0.05−

0

y x( )

x

Figure 2: Tapered cantilever beam and space-dependent deflection.

Integration by Parts - Applications in Engineering 2-7

iii. Electric Circuit Problem (EGN3373 and EEL3004)

The time-dependent current that passes through a circuit with applied voltage3Ð>ÑZ Ð>Ñ P V, constant inductance , and constant resistance , is described by the following

equation:

P � V 3Ð>Ñ œ Z Ð>Ñ.3Ð>Ñ

.>

The solution to the equation above for the time-dependent current is found3Ð>Ñthrough the use of an integrating factor as:

3Ð>Ñ œ Z Ð>Ñ/ .>/

P

� VÎP >VÎP >

(

Assume, that an alternating voltage is applied to the circuit with a form

Z Ð>Ñ œ Z =38Ð >Ñ Z7 7= =, where is the peak voltage and is the alternating frequency.

Then, the integral for the current takes the form:

3Ð>Ñ œ / =38Ð >Ñ/ .>Z

P7 � VÎP > VÎP > ( =

The integral above can be solved using integration by parts by selecting:

?Ð>Ñ œ =38Ð >Ñ Ê ? Ð>Ñ œ -9=Ð >Ñ= = =w and @ Ð>Ñ œ / Ê @Ð>Ñ œ /P

Vw VÎP > VÎP >

Substitution leads to:

( (=38Ð >Ñ/ .> œ =38Ð >Ñ/ � -9=Ð >Ñ/ .>P P

V V= = = = VÎP > VÎP > VÎP >

Note that the new integral on the right-hand side is not ‘simpler’ than the original one

therefore it seems that integration by parts did not achieve its goal. However, a second

integration by parts on the integral on the right-hand side will reveal an interesting

feature. Let:

?Ð>Ñ œ -9=Ð >Ñ Ê ? Ð>Ñ œ � =38Ð >Ñ= = =w and @ Ð>Ñ œ / Ê @Ð>Ñ œ /P

Vw VÎP > VÎP >

Substitution of these new terms and simplification yields:

( (” • Œ =38Ð >Ñ/ .> œ =38Ð >Ñ � -9=Ð >Ñ / � =38Ð >Ñ/ .>P P P

V V V= = = = = = VÎP > VÎP > # VÎP >

#

Note that the integral on the right-hand side is identical to the original one on the left-

hand side, therefore, the solution can be readily found as:

( ˆ ‰ ” •=38Ð >Ñ/ .> œ =38Ð >Ñ � -9=Ð >Ñ / � G" �

P

V= = = =

=

VÎP > V VÎP >P

# PV

#

Integration by Parts - Applications in Engineering 2-8

This recursive behavior of the integration by parts implementation is typical of cases

where the product of functions is composed of a trigonometric and an exponential

function. Substitution of this solution into the expression for the current leads to:3Ð>Ñ

3Ð>Ñ œ =38Ð >Ñ � -9=Ð >Ñ � G /" �

P Z

V P

ZV � VÎP >

# PV

#7

7

== = =ˆ ‰ ” •

The constant of integration can be computed by imposing an initial conditionGÐ3Ð!Ñ œ 3 Ñ9 . Assume, for example, that a given circuit has an applied voltage with peak

value Volts, frequency rad/s , inductance Z œ "#! 0 œ '!LD Ð œ # † '! Ñ P œ "!7 = 1Henry, resistance Ohms, and an initial current Amps. The time-V œ "!! 3 œ !9

dependent current from the expression above is illustrated in the plot below for the first

!Þ% seconds. It can be clearly seen how the oscillative current settles to a periodic

alternating state after just a few fractions of a second.

R L

V(t)

i(t)0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0.04−

0.02−

0

0.02

0.04

0.06

0.08

i t( )

t

Figure 3: Inductive alternating voltage circuit and time-dependent current evolution.

Integration by Parts - Applications in Engineering 2-9

1

Chapter 3

AApplpp icli atca iotionnss ooff CCaalculculuslus IIII TTopiop c ic

Parametric Equations for Fun and Profit

FFaaccuullttyy CConton ritr buibutortor Dr. Joseph J. LaViola, Jr. School of Electrical Engineering and ComputerScience

WWeeeks eks:: February 22 and March 1 of Spring 2010

3.1 Introduction

Parametric Equations for Fun and Profit 3-2

In the Cartesian coordinate system, a common way to describe a collection of(x, y) points is with functions of x or y. For example, we may describe a curvein the xy-plane by giving y as a function of x (y = f(x)) or x as a functionof y (x = f(y)). Unfortunately, there are many curves in mathematics thatwe cannot describe using this approach. For example, any curve that failsthe vertical line test cannot be written in the above form. To deal withthis problem, we can describe these types of curves by describing x and yas functions of a third variable, a parameter. This approach is referred to asdescribing curves with parametric equations or making use of parametric curves.

Parametric equation and applying Calculus to them are useful in a varietyof different applications. Interestingly enough, these mathematical tools arenot only used in science and engineering but also to create mathematicalartwork because they let us describe curves in both 2D and 3D that havebeautiful shapes. In this chapter, we will examine parametric equations andwill explore how they can be used in creating mathematical artwork. We willalso examine how parametric curves are used in gesture recognition.

3.1.1 Relevant Courses

The material covered in this chapter will be seen in a variety of different courses at UCF. These include math courses, such as

MAC 2313 Calculus III

MAP 2302 Diff erential Equations

physics courses, such as

PHY 2048 Physics for Engineering and Scientists I

PHY 2053 College Physics I

PHY 3220 Mechanics I

and computer science courses, such as

COT 4500 Numerical Calculus

CAP 4720 Computer Graphics

as well as several courses in engineering.

3.2 Parametric Equation Basics

Suppose x and y are both given as functions of a third variable t by theequations

x = f(t) (3.1)

and

y = g(t). (3.2)

We refer to these equations as parametric in nature (i.e., parametric equa-tions) because we use a parameter to describe the behavior of x and y. Eachvalue of t determines a point (x, y) which we can plot in the plane. As tvaries, (x, y) = (f(t), g(t)) varies and traces out a curve C, which is knownas a parametric curve.

For example, consider the parametric equations x(t) = t3 − t and y(t) =4t − 2. Varying t from -5 to 10 yields the graph shown in Figure 3.1.

Parametric Equations for Fun and Profit 3-3

Figure 3.1: The parametric curve generated from x(t) = t3 − t and y(t) =4t − 2.

In another example, consider the parametric equations x(t) = 4cos(t) andy(t) = 2sin(t). Varying t from 0 to 2π yields the graph shown in Figure 3.2.Note that we can move into the third dimension by adding another parametricequation to the ones found in Equations 3.1 and 3.2. Thus, we can create 3Dparametric curves by adding z = h(t). Each value of t would then determine

Figure 3.2: The parametric curve generated from x(t) = 4cos(t) and y(t) =2sin(t).

a point (x, y, z) in 3D space. For example, the parametric equations x(t) =3cos(3t), y(t) = 3sin(3t), z(t) = t, varying t from 0 to 3π, define a curvecalled a helix (see Figure 3.3).

In many cases, it is possible to simplify parametric equations by eliminat-ing the parameter. As example, consider the parametric equations presentedin Figure 3.1. Given y(t) = 4t − 2, we can solve for t to get

t =y + 2

4. (3.3)

We then substitute this equation for t in x(t) = t3 − t to get

x =(

y + 2

4

)3

− y + 2

4. (3.4)

After some Algebra,

x =y3 + 6y2 − 4y − 24

64(3.5)

This approach is relatively straightforward but does not always lead to a moresimple equation. For example, if we eliminate the parameter in the paramet-ric equations shown in Figure 3.2 by solving for t in the first equation we get

Parametric Equations for Fun and Profit 3-4

3.2. PARAMETRIC EQUATION BASICS

Figure 3.3: A helix defined using 3 parametric equations.

t = cos−1(

x

4

)

(3.6)

which leads to

y = 2 sin(

cos−1 x

4

)

. (3.7)

This equation is no simpler than the original parametric equations. Anotherapproach to take is to look for identities that can be used to make the theparametric equations simpler. In this example, we can use the Pythagoreanidentity, sin2 u+cos2 u = 1. If we square both equations, add them together,and use the Pythagorean identity we get

x2

16+

y2

4=

16 cos2 t

16+

4 sin2 t

4(3.8)

= cos2 t + sin2 t (3.9)

= 1 (3.10)

resulting in the equation for an ellipse, x2

Parametric Equations for Fun and Profit 1-53-5

3.3 The Lissajous Curve

Figure 3.4: An example of a closed Lissajous curve.

Up till now, we have seen some simple and rather academic parametricequation examples. In this section, we will explore a more interesting set ofparametric equations that are seen in Classical Dynamics, a subfield of Physics.Lissajous curves, first shown by Jules Lissajous in 1857, are used to describethe general case of two-dimensional oscillations. This generality stems fromthe fact that the angular frequencies for the motions in the x and y directionsneed not be equal. The parametric equations used to describe these curves are

x(t) = C cos(ωxt − a)(3.11)

y(t) = D cos(ωyt − b) (3.12)

Parametric Equations for Fun and Profit 333---666

where C and D are amplitude parameters, a and b are phase parameters, andωx and ωy are angular frequency parameters. Lissajous curves have aninteresting property of either being open or closed. Lissajous curves areclosed if the motion repeats itself at regular intervals of time and this onlyoccurs if the angular frequencies ωx and ωy are commensurable (i.e.,

aaannnddd PPPrrrooo

3.3. THE LISSAJOUS CURVE

ωx/ωy is a rational fraction). If the ratio of the angular frequencies is not a ra-tional fraction, the curve is open, meaning that the moving particle will nev-er pass through the same point twice with the same velocity. In these cases,after a sufficient amount of time, the curve will pass arbitrarily close to anypoint within the rectangle defined by 2C × 2D. Thus, the curve will”fill” the rectangle.

Figure 3.5: An example of an open Lissajous curve.

As an example, consider the Lissajous curve defined by the parametricequations shown in Figure 3.4. Since the angular frequency ratio is a rationalfraction, the Lissajous curve is closed, as shown by the motion trace of theparticle p in Figure 3.4. However, if we change ωx to be equal to 2.3, theangular frequency ratio is no longer a rational fraction and the Lissajous curveis open. Figure 3.5 showed this open curve. One can see the rectangle formingfrom the particle’s motion. If the simulation was run where t went from 0 toa large number, the entire rectangle would be filled from the particle trace.

Parametric Equations for Fun and Profit 3-7

3.4 Parametric Equations and Art

Parametric equations can yield some very interesting curves with uniquepatterns. Thus, we can use parametric equations to create mathematicalart. As a simple example, consider the butterfly curve discovered by TempleFay. The butterfly curve is defined as

x(t) = sin(t)(

ecos(t) − 2 cos(4t) − sin5(

t

12

))

(3.13)

y(t) = cos(t)(

ecos(t) − 2 cos(4t) − sin5(

t

12

))

. (3.14)

Figure 3.6 shows a visualization of the Butterfly curve. One can see thesymmetry in the curve and its complexity.

Figure 3.6: The butterfly curve.

The types of images one can create using parametric equations is endless.It simply depends on the imagination and on the mathematical equations

Parametric Equations for Fun and Profit 3-8

3.4. PARAMETRIC EQUATIONS AND ART

utilized. Using computers, we can easily try out diff erent parametricequations to see what images they create. Two examples of web applets thatlet users create parametric art are InterAct1 and the Parametric ArtGallery2. Figures 3.7 - 3.9 show some example artwork using these applets.

Figure 3.7: An image generated using InterAct. The parametric equationsused were x(t) = A sin(Bt) cos(t) and y(t) = A sin(Bt) sin(t).

1Go to http://www.math.uri.edu/ bkaskosz/flashmo/firstfin.html to use this applet.2Go to http://www.flashandmath.com/advanced/art/ to use this applet.

Parametric Equations for Fun and Profit 3-9

Figure 3.8: An image generated using InterAct. The parametric equationsused were x(t) = A sin(3t) and y(t) = 0.8B cos(7t).

Parametric Equations for Fun and Profit 3-10

3.4. PARAMETRIC EQUATIONS AND ART

Figure 3.9: An image generated using the Parametric Art Gallery.

Parametric Equations for Fun and Profit 3-11

3.5 Parametric Equations in Gesture

Recognition

Recognizing gestures, whether they be 2D using a stylus on a Tablet PC or3D using a Nintendo Wiimote, is an important part of many applicationsin pen computing and video games. There are many different approachesto performing gesture recognition that make use of a variety of differentalgorithms in computer science. One common method is to create templatesfor each gesture and then test any incoming gesture against the templates.If the incoming gesture closely matches a particular template, then it isclassified as that gesture. There are many different ways to create gesturetemplates. One approach is to create a collection of features, informationthat describes a particular gesture. These features are primarily based oncomputing statistical or geometric quantities from the data that representsthe gesture. To ensure accurate recognition, these features must be gooddiscriminators or provide quantities that can be used to tell the differencebetween one gesture from another.

As it turns out, the data we need to compute on for recognizing gesturescan be described using parametric equations. For example, a Nintendo Wiimoteprovides a series of acceleration values in the x, y, and z directions through time. Inanother example, a stylus on a Tablet PC provides the location of the styluspoint on the surface of the screen as points in x and y through time. Recallthat parametric equations can be used to describe a particle in motion in 2Dor 3D through time. Thus, we get a collection of points from these devicesthat we can use to compute the features we need for gesture recognition.

3.5.1 Arc Length

Of the many possible features that one could compute to create a gesturetemplate, one of the most common is arc length, the length of a curve. Wecan define arc length in terms of parametric equations. Thus, if we havea curve C defined by x = f(t) and y = g(t), then its arc length can becomputed as

L =∫ β

α

√

√√√

√

dx

dt

)2

+

(

dy

dt

)2

dt. (3.15)

Parametric Equations for Fun and Profit 3-12

3.5. PARAMETRIC EQUATIONS IN GESTURE RECOGNITION

For example, consider the parametric equations x = 3 sin(3t) and y =3 cos(3t), where 0 ≤ t ≤ 2π. To compute the arc length of this curve,we first find the derivatives of x and y,

dx

dt= 9 cos(3t) (3.16)

dy

dt= −9 sin(3t). (3.17)

Plugging in these derivatives into the arc length formula we obtain

L =∫ 2π

0

√

(9 cos(3t))2 + (−9 sin(3t))2dt (3.18)

=∫ 2π

0

√

81 cos2(3t) + 81 sin2(3t)dt (3.19)

=∫ 2π

0

√81dt (3.20)

=∫ 2π

09dt (3.21)

= 18π. (3.22)

Unfortunately, when we are dealing with gesture recognition, the inputdevices we use do not provide functions of time. Rather, they provide pointsthrough time. So, when calculating arc length in gesture recognition applica-tions, we need to compute an approximation. Suppose we are recognizing pengestures. A pen ink stroke is defined as a sequence of points s = p1p2...pn inthe xy-plane where pi = (xi, yi), 1 ≤ i ≤ n, p1 is the pen-down point, pn

is the pen-up point, and n is the number of points in the stroke. Thus, withthese xy points we have an approximation to the curve generated by the penstroke. The larger n is, the smaller the distance between the points and the betterthe approximation will be. To approximate arc length for any s, we compute

L̂ =n∑

i=2

‖pi − pi−1‖ (3.23)

where ‖pi−pi−1‖ =√

(xi − xi−1)2 + (yi − yi−1)2. As n goes to infinity, L̂ goesto L. Note that this formula works for 3D points as well, the z componentsimply needs to be added to the distance calculation.

Parametric Equations for Fun and Profit 3-13

3.5.2 Linear Classification

We can use arc length and other features, such as aspect ratio, point and anglehistograms, to create templates for each gesture in the gesture recognition process.Once we have the features calculated, we can train a machine to learn an algorithm,so it can tell us how incoming gestures should be classified. There are a widevariety of machine learning algorithms used in gesture recognition. In this chapter,we focus on one of the simplest, yet powerful approaches, the linear classifier.

Given a feature vector ~f, associated with a given gesture g in a gesturealphabet C, a linear evaluation function is derived over the features. Thelinear evaluation function is given as

gc = wc0 +F∑

i=1

wcifi (3.24)

where 0 ≤ c < C, F is the number of features, and wci are the weightsfor each feature associated with each gesture in C. The classification of thecorrect gesture g is the c that maximizes gc.

Training of this classifier is done by finding the weights wci from thegesture samples. First, a feature vector mean ~fc is calculated using

f̄ci =1

Ec

Ec−1∑

e=0

fcei (3.25)

where fcei is the ith feature of the eth example of gesture c and 0 ≤ e <Ec where Ec is the number of training samples for gesture c. The samplecovariance matrix for gesture c is

Σcij =Ec−1∑

e=0

(fcei − f̄ci)(fcej − f̄cj). (3.26)

The Σcij are averaged to create a common covariance matrix

Σij =

C−1∑

c=0Σcij

−C +C−1∑

c=0Ec

. (3.27)

Parametric Equations for Fun and Profit 333---111444

The inversion of Σij then lets us calculate the appropriate weights for thelinear evaluation functions,

aaannnddd PPPrrrooo

3.5. PARAMETRIC EQUATIONS IN GESTURE RECOGNITION

wcj =F∑

i=1

(Σ−1)ij f̄ci (3.28)

and

wc0 = −1

2

F∑

i=1

wcif̄ci (3.29)

where 1 ≤ j ≤ F .

3.5.3 Recognizing Gestures

Figure 3.10: The 25 gestures used in the 3D gesture database.

To test the gesture recognition algorithms, the Interactive Systems andUser Experience Lab at UCF created a 3D gesture database. The 3D gestureswere collected using a Nintendo Wii Remote. A total of 8500 gestures werecollected spanning 25 different gestures (see Figure 3.10). We tested the

Parametric Equations for Fun and Profit 3-15

linear classifier described above on this database, by examining how manytraining samples per gesture would aff ect the overall gesture recognitionaccuracy. A total of 5, 10 and 15 samples were used to create templates foreach gesture. The algorithm achieved 93.5%, 96.3%, and 98.5% accuracy,using 5, 10, and 15 training samples per gesture respectively.

3.6 Conclusion

In this chapter, we have examined parametric equations and how they canbe used to represent interesting curves in 2D and 3D space. We have ex-plored how these equations are used to create mathematical art and howwe can use Calculus on parametric equations to compute arc length. Arclength is an important feature that is used to describe gestures for gesturerecognition. These features are used in machine learning algorithms to cre-ate classifiers that take the feature data and create templates used in therecognition process. Finally, we showed a gesture recognition example forrecognizing Nintendo Wiimote gestures. These gestures can be used to cre-ate more naturally expressive interfaces in video games.

Parametric Equations for Fun and Profit 3-16

Chapter 4

AApppplliiccaattiioonnss ooff CCaallccuulluuss IIII TTooppiicc

Application of Polar Coordinates in Chemistry

FFaaccuullttyy CCoonnttrriibbuuttoorr Dr. Christian Clausen IIIDepartment of ChemistryCollege of Sciences

WWeeeekkss:: March 15 and March 22 of Spring 2010

Application of Polar Coordinates in Chemistry 4-2

CALCULUS TOPIC: POLAR COORDINATES

Section 11.3 (page 705 in your textbook)

In your previous math and science courses you have probably always usedCartesian coordinates to describe points in a plane. However, there are lots of othercoordinate systems in use as well, particularly in specialized applications. Onespecialized application in Chemistry that you will learn about in this section concerns theallowed energies and regions in space that an electron can occupy in a hydrogen-likeatom. The coordinate system that is used in performing the calculations about theelectron is known as Polar coordinates. This is the coordinate system that you studied inthis section. Let’s review what you have learned before moving on to the Chemistryapplication.

In the Cartesian system for a position of a point in a plane we fix an origin, apoint O and then take two lines which are perpendicular to each other passing through O. These two lines are the axes. The x-axis is horizontal, and the y-axis is vertical. Theposition of any point P in the plane is then described by noting its displacement from O inthe direction of the x-axis, the x coordinate, and also its displacement from O in thedirection of the y-axis, the y coordinate as shown in Figure 1.

Figure 1. Cartestian coordinate system

Notice that we talk about the displacement, and not the distance. This is because, in thissystem, we need to use the directions indicated by the arrows to help us to be able tospecify a point uniquely. Thus moving up from the origin is a positive displacement,whereas moving down from the origin is a negative displacement. Similarly, moving tothe right from the origin gives a positive displacement, and moving to the left gives anegative displacement.

In the Polar coordinate system we take an origin (or pole) O, and a fitted line OA. A point P is then described by specifying a distance O to P along the radius direction, andthe angle θ where we have had to turn from the initial line to be looking along the radiusdirection as shown in Figure 2. The point then has polar coordinates (r, θ).

Application of Polar Coordinates in Chemistry 4-3

Figure 2. Polar coordinate system

There are similar conventions about direction with polar coordinates as there arewith Cartesian coordinates. We use the convention that an angle is positive if measuredin the counter-clockwise direction from the polar axis and negative in the clockwisedirection. We can also extend the meaning of polar coordinates (r, θ) to the case in whichr is negative by agreeing that, as in Figure 3, the points (-r, θ) and (r, θ) lie on the sameline through O and at the same distance │r│ from O, but on opposite sides of O, thepoint (r, θ) lies in the same quadrant θ; if r < 0, it lies in the quadrant on the opposite sideof the pole. Notice that (-r, θ) represents the same point as (r, θ + π).

Figure 3. (-r, θ) and (r, θ)

Example: Problem 11.3-1(a). Plot the point whose polar coordinates are given. Then find two other pairs of polarcoordinates of this point, one with r > 0 and one with r < 0.

Solution:

Application of Polar Coordinates in Chemistry 4-4

Let’s consider the conversion of a point P with Cartesian coordinates (x, y) topolar coordinates (r, θ). We shall identify the pole with the origin of Cartesiancoordinates, and the initial line with the x-axis. From Figure 4, we see that there are two

Figure 4. Conversion between Polar and Cartesian coordinate systems

equations we can use to help us to move from one system to another:x = rcos(θ), y = rsin(θ)

to convert from polar coordinates to Cartesian coordinates and we can use the followingequation to convert from Cartesian coordinates to polar coordinates.

r2 = x2 + y2, tan(θ) = y/x Both of these sets of equations will be useful, but they must be used with care, as it iseasy to obtain an incorrect value of θ when the point does not lie in the first quadrant. Soit is always a good idea to plot the points first so that the point (r, θ) lies in the correctquadrant.

Example: Assuming a common origin and the x-axis as the initial line, find the Cartesiancoordinates of the point with the following polar coordinate: (2, - π/2)

Solution:

Application of Polar Coordinates in Chemistry 4-5

Curves in Polar Coordinates

The graph of a polar equation r = f(θ) consists of all points P that have at least onepolar representation (r, θ) whose coordinates satisfy the equation. For example let’s plotthe curve represented by the polar equation r = 4. The curve consists of all points (r, θ)with r = 4. Since r represents the distance from the point to the origin, the curve r = 4represents the circle with center O and radius 4. In general, the equation r = a representsa circle with center O and radius │a│ as shown in Figure 5.

Figure 5. Polar curve of radius a

When the polar equations become more complicated than the example we justdemonstrated then it sometimes helps to sketch out the equation in Cartesian coordinatesfirst to see how r changes with increasing values of θ. This is illustrated in the nextexample:

Example: Sketch the curve r = cos(2θ) first in Cartesian coordinates and then in polar coordinates.

Solution: First we sketch r = cos(2θ), 0 ≤ θ ≤ 2π, in Cartesian coordinate as shown in Figure 6. Asθ increase from 0 to π/4, Figure 6 shows that r decreases from 1 to 0 and so we draw thecorresponding portion of the polar curve in Figure 7 (this is indicated by [1]). As θincreases from π/4 to π/2, r goes from 0 to -1. This means that the distance from Oincreases from 0 to 1, but instead of being in the first quadrant this portion of the polarcurve (indicated by [2]) lies on the opposite side of the origin in the third quadrant. Theremainder of the curve is drawn in a smaller fashion with the final result being the four-leaf clover design shown in Figure 7.

Application of Polar Coordinates in Chemistry 4-6

Figure 6. r = cos(2θ) in Cartesian coordinates

Figure 7. Four-leaf clover r = cos(2θ)

In this section of your book you have only covered the use of polar coordinates intwo dimensions, later on in Calculus III you will add the third dimension and it then willbe called spherical polar coordinates. This is covered in section 13.7 of your text book. In Chemistry we will be using polar coordinates in three dimensions to calculate theallowed energy levels and locations of electrons in an atom. Even though we can use atwo dimensional model using r and θ a more complete picture of the atomic model can beattained by adding the 3rd dimension. Thus it is useful to discuss the addition of this 3rd

dimension at this time. The addition of this new dimension to the polar coordinatesystem involves the distance along the z-axis. This distance is related to the angle φ asshown in Figure 8.

Application of Polar Coordinates in Chemistry 4-7

Figure 8. Spherical polar coordinates

The relationship of the polar coordinates to the Cartesian coordinate system for x, y, and zis:

x = rcos(θ) y = rsin(θ) z = rcos(φ)

To convert from spherical to rectangular coordinates we use the equations: x = rsin(φ)cos(θ) y = rsin(φ)sin(θ)

z = rcos(φ) We have now reached the point where we can begin to discuss the application of

Polar coordinates in Chemistry. But first we will look at a brief historical discussionabout the evolution of the structure of the atom

Background/Motivation

Chemistry has a rich, colorful history, even some concepts and discoveries thatled temporarily along confusing paths have contributed to the heritage of chemistry. Thisbrief overview of early breakthroughs and false directions provides some insight into howmodern chemistry arose and how science progresses.

Chemistry has its origin in a pre-scientific past that incorporated threeoverlapping traditions, alchemy, medicine, and technology. Alchemy refers to the occultstudy of nature practiced in the 1st century AD by Greeks living in northern Egypt. Alchemists were influenced by the Greek idea that matter naturally strives towardperfection, and they searched for ways to change less valued substances into preciousones. Thus alchemy’s legacy to chemistry is mixed at best, but it did dominate thethinking about matter for some 1500 years.

Chemical investigation in the modern sense – inquiry into the causes of changesin matter – began in the early 18th century. A young French chemist Antoine Lavoisterput chemistry on a scientific basis by performing experiments using quantitative,reproducible measurements to prove his theories. Thus laws about matter began todevelop that still are true today, such as the law of mass conservation: the total mass ofsubstances does not change during a chemical reaction.

Application of Polar Coordinates in Chemistry 4-8

The mass laws led to John Dalton’s theory about the structure of matter in 1808which he proposed in a series of postulates. One of the postulates being that all matterconsists of atoms, tiny indivisible particles of an element that cannot be created ordestroyed. This model of the atom lasted until 1897, when J. J. Thomspson discoveredthe presence of electrons in the atom and thus disproved Dalton’s theory that the atomcould not be broken down into smaller particles. In 1910 Ernest Rutherford, through useof alpha particle scattering experiments, established that most of the mass of an atom waslocated in a tiny region of space within the atom, which he called the nucleus. Thus themodel of the atom now became one of a tiny, very dense, positive nucleus surrounded byelectrons.

One of several experimental observations about matter that the Rutherford modelof the atom could not explain is that when elements are heated to a high temperature theyemit electromagnetic radiation of certain characteristic frequencies. An example of this is the color of a neon sign. However, soon after Rutherford proposed his nuclear model,Niels Bohr, a young Danish physicist working in Rutherford’s laboratory, suggested amodel for the H atom that predicted the existence of line spectra. In his model, Bohrused Planck’s and Einstein’s ideas about quantized energy and proposed three postulates: 1) the H atom has only certain allowable energy levels, 2) the atom does not radiateenergy while in one of its stationary states, and 3) the atom changes to another stationarystate (the electron moves to another orbit) only by absorbing or emitting a photon whoseenergy equals the difference in energy between the two states. In Bohr’s model, thequantum number n (1, 2, 3,…) is associated with the radius of an electron’s orbit, whichis directly related to the electron’s energy: the lower the n value, the smaller the radius ofthe orbit, and the lower the energy level. Thus the Bohr model of the atom can beenvisioned as shown in Figure 9.

Figure 9. Bohr model of the atom

Despite its great success in accounting for the spectral lines of the H atom, theBohr model failed to predict the spectrum of any other atom, even that of helium, the nextsimplest element. Thus, the conclusion can be made that the Bohr model of the atom isincorrect. So what is the correct model? We will see that it took several amazing

Application of Polar Coordinates in Chemistry 4-9

discoveries in the early part of the 20th century before the correct model would beproposed.

Without going into any detail to describe the reasons, it has been established thatas a result of quantum theory and relativity theory, we can no longer view matter andenergy as distinct entities. This idea is embodied in Einstein’s famous equation E = mc2. Another amazing theory proposed by Louis deBroglie is that if energy is particle-like,perhaps matter is wave-like. The deBroglie wavelength is the idea that electrons (and allmatter) have wave-like motion. Thus, allowed atomic energy levels are related toallowed wavelengths of the electron’s motion. Electrons exhibit diffraction patterns, asdo waves of energy, and photons can exhibit transfer of momentum, as do particles ofmass. The wave-particle duality of matter and energy exists at all scales but is observableonly on the atomic scale. Now if an electron has the properties of both a particle and awave, what can we determine about its position in the atom. In 1927, the Germanphysicist Werner Heisenburg postulated the uncertainty principle, which states that it isimpossible to know the exact position and momentum of the electron simultaneously. Thus the electron cannot be in an orbit because we would know both its position andmomentum (i.e. energy).

Acceptance of the dual nature of matter and energy and of the uncertaintyprinciple culminated in the field of quantum mechanics, which examines the wave natureof objects on the atomic scale. In 1926, Erwin Schrödinger derived an equation that isthe basis for the quantum-mechanical model of the hydrogen atom. The model describesan atom that has certain allowed quantities of energy due to the allowed wave-likebehavior of an electron whose exact location is impossible to know. In order to accountfor these requirements Schrödinger introduced the wave function Ψ (psi), a mathematicalfunction of the position coordinates x, y, and z. The Schrödinger equation, of which thewavefunction is a solution, is:

In this equation Ψ is the amplitude of the wave function associated with the electron, E isthe total energy of the system, V is the potential energy of the system (equal to –e2/r forhydrogen), m is the mass of the electron, h is Planck’s constant, and x, y, and z are theusual Cartesian coordinates.

Since the system described by the Schrödinger equation is sphericallysymmetrical, it is most convenient to use spherical polar coordinates. When thesecoordinates are substituted for the Cartesian coordinates the Schrödinger equation takesthe form:

Although this equation at first appears more cumbersome than the Cartesiancoordinated equation it leads to a much simpler solution because it may be solved by thewell-known method of separation of variable, which leads to three ordinary differentialequations that can be exactly solved. Each of the three equations has only one of the

Application of Polar Coordinates in Chemistry 4-10

variables r, θ, and φ. To solve these we write the wave function Ψ, which is a function ofr, θ, and φ, as the product of three functions R(r), Θ(θ), and Φ(φ):

Ψ(r, θ, φ) = R(r)*Θ(θ)*Φ(φ) Substitution of this expression into the above equation and division by RΘΦ gives:

The partial derivatives have been replaced by ordinary derivatives since eachfunction now depends only on a single variable. Without going through the actualmanipulations, the Schrödinger equation as it is written above can be split into threeequations, one involving Φ and φ, one involving Θ and θ, and one involving R and r. These equations can be solved so as to eliminate the differentials.

In solving the differential equations, definite boundary conditions require thatcertain constants that enter into the solution of the wave equation take on only integralvalues. These constants are called quantum numbers and are designated by n, l, and ml. The letter n is called the principal quantum number and may take on the values 1, 2, 3, …; l may have the values of 0, 1, … up to n – 1 and ml can have values ranging from –lthrough 0 to +l. The wave functions, Ψ, which are solutions of the Schrödinger equation,are commonly called orbitals. Orbitals for which l = 0, 1, 2, 3, and 4 respectively, arecalled s, p, d, f, and g orbitals. Solutions of the Θ equation leads to the values for ahydrogen-like atom shown in Table 1 for five different values of the quantum number ml.

Table 1. Solutions of the Θ equation Value of ml Function

0 Θ0 = π2

1

1 Θ1 = θ1 π

cos

-1 Θ-1 = θ1 π

sin

2 Θ2 = πθ )2cos(

-1 Θ-2 = πθ )2sin(

Plots of these solutions in planar polar coordinates are shown in Figure 10. The reasonwhy plots of both θ and θ2 are shown is that θ2 is proportional to the region in spacewhere the probability of finding the electron is greatest. Solutions of the Φ equation are

Application of Polar Coordinates in Chemistry 4-11

Figure 10. Plots of θ and θ2

given in Table 2. These solutions are given for l values of 0, 1, and 2 and thecorresponding permitted ml values. Because the sine and cosine functions can havepositive and negative values, there are positive and negative regions of the wavefunctions. A plot of φ and φ2 for the functions listed in Table 2 are shown in Figure 11.

Table 2. Solutions of the Φ equation l ml Function

0 0 Φ = π1

1 0 Φ = ϕcos26

1 +1, -1 Φ = ϕsin23

2 0 Φ = ( )1cos3410 2 −ϕ

2 +1, -1 Φ = ϕϕ cossin215

2 +2, -2 Φ = ϕ2sin415

Application of Polar Coordinates in Chemistry 4-12

Figure 11. Plots of φ and φ2

The θ and φ wave functions are best considered together, since both are concernedwith the angular dependence of the orbitals. Plots of [Θ(θ)Φ(φ)]2 for various l and mlvalues are shown in Figure 12. It is to be emphasized that these plots are in no wayrelated to distance from the nucleus but only to the variation of the wave function withthe angles θ and φ. For given values of θ and φ, the length of the line joining the origin to the surface of the solid figure is the relative probability that the electron is to be foundin that direction. These Ψ(θ, φ) surfaces are often referred to as orbitals and indeed servevery satisfactorily for most arguments concerned with chemical bonding. It should berealized, however, that the variation of Ψ in space is the product of Ψ(R) and Ψ(θ, φ) atevery point in space.

Application of Polar Coordinates in Chemistry 4-13

Figure 12. Total angular dependence of the wave function Ψ(θ, φ) with all Cartesiancoordinates fixed

It is now time to look at the solution to the radial part of the wave equation R(r). The radial part of the wave function depends only on the n and l values and has an exponential term 0/ nare−

where a0 is 0.529Å, and a pre-exponential term involving apolynomial of the n -1 degree. The radial functions for n = 1, l = 0; n = 2, l = 0; and n =2, l = 1 are given in Table 3.

Table 3. Solutions of the R equationn l Function

1 0 R = 0/2/3

0

11 area

−⎟⎟⎟⎠

⎞⎜⎜⎜⎝

⎛

π

2 0 R = 02/

0

2/3

0

21241 are

ar

a−⎟⎟⎟

⎠

⎞⎜⎜⎜⎝

⎛−⎟⎟⎟

⎠

⎞⎜⎜⎜⎝

⎛

π

2 1 R = 02/

0

2/3

0

1641 are

ar

a−

⎟⎟⎟⎠

⎞⎜⎜⎜⎝

⎛⎟⎟⎟⎠

⎞⎜⎜⎜⎝

⎛

π

The probability of finding the electron in a volume element of fixed size located(essentially) at some given point at a distance r from the nucleus is given by R2dτ (wheredτ is the elemental volume unit). The probability that the electron is to be found at adistance r from the nucleus is given by (4πr2)R2dr. This is the probability that theelectron will be fond in a spherical shell of thickness dr at a distance from the nucleusranging from r to r + dr. Curves showing R2dτ and (4πr2)R2dr versus r are given inFigure 13. R2dτ (or more generally Ψdτ) is sometimes referred to as the electronprobability distribution density, or more simply as the electron density. The shape of thesurfaces for which Ψ (and consequently Ψ2) is constant is referred to as the shape of the

Application of Polar Coordinates in Chemistry 4-14

orbital. It should be noted that as n increases the maximum probability of finding theelectron lies further away from the nucleus.

Figure 13. Radial probability distribution functions of the hydrogen atom. Dashedcurves represent R2; solid curves represent r2R2

CONCLUSION

What we have learned is through the use of Polar coordinates in the solution ofthe Schrödinger equation we have determined the regions around the nucleus of ahydrogen-like atom where the electrons are most probably located. These regions arecalled orbitals and they have different shapes, sizes and energies. However, chemists arenot just interested in atoms, we are interested in how atoms come together to formmolecules. In class I will show you how we can take these atomic orbitals and combinethem together to form molecules.

Chapter 5

AApppplliiccaattiioonnss ooff CCaallccuulluuss IIII TTooppiicc

Integrals in Physics

FFaaccuullttyy CCoonnttrriibbuuttoorr Dr. Archana DubeyDepartment of PhysicsCollege of Sciences

WWeeeekkss:: March 29 and April 5 of Spring 2010

In this chapter, we will study the importance of Calculus in understanding Physics. For this purpose, we will have a quick review of velocity and acceleration, which was reviewed in the Applications of Calculus I in Physics. In addition to this, we will study the concept of work, electric force, electric potential and alternating current.

If x = f(t) is the position function of a particle that is moving in a straight line, then Δx/Δt represents the average velocity over a time period Δt, and v = dx/dt represents the instantaneous velocity (the rate of change of displacement with respect to time). The SI unit of instantaneous velocity is m/s.

dtdx

txv

tx =∆∆

=→∆ 0

lim

Figure 1: A graphical representation of a particle in motion.

The slope of velocity-time graph is the acceleration and has a SI unit of m/s2. We define instantaneous acceleration following the same procedure that we used to define instantaneous velocity. The instantaneous acceleration is the limit of the average acceleration as the time interval approaches zero. In the language of Calculus, instantaneous acceleration equals the instantaneous rate of change of velocity with time. Thus,

2

2

0lim

dtxd

dtdv

tv

a xx

tx ==∆∆

=→∆

.

When an object’s velocity and acceleration are in the same direction, the object is speeding up. When an object’s velocity and acceleration are in the opposite direction, the object is slowing down.

Integrals in Physics 5-2

Figure 2: The graphical representation of determining instantaneous acceleration.

It is possible to find the position of a particle if its velocity is known as a function of time. In Calculus, the procedure used to perform this task is referred to either as integration or as finding the antiderivative. Graphically, it is equivalent to finding the area under a curve as shown in Figure 3.

Figure 3: A following curve shows velocity versus time for a particle moving along the x axis.

The time interval tf - ti can be divided into many small intervals and each interval is of duration Δtn. The displacement of the particle during any small interval is given by

,, navgxnn tvx ∆=∆ where avgxnv , is the average velocity in that interval. Thus, we can say that

the displacement during this small interval is simply the area of the shaded rectangle in the figure. The total displacement, corresponding to the interval tf - ti is the sum of the areas of all the rectangles from ti to tf, and is given as follows:

Integrals in Physics 5-3

nn

avgxn tvx ∆=∆ ∑ ,

In the limit ,0or, →∆∞→ ntn the displacement is

nn

xnttvx

n∑→∆

=∆0

lim

The limit of the sum shown in this equation is called a definite integral and is written as

( )dttvtvf

in

t

txn

nxnt ∫∑ =

→∆ 0lim ,

where ( )tvx denotes the velocity at any time t.

We can derive kinematic equations with Calculus, which are demonstrated below. The

defining equation for acceleration dt

dva x

x = may be written as dtadv xx = , or in terms of an

integral,

dtavvt

xxixf ∫=−0

If the acceleration is constant, ax can be removed from the integral and we get

( ) tatadtavv xx

t

xxixf =−==− ∫ 00

---------------- (1)

Next, let us consider the defining equation for velocity: dtdxvx =

This equation can be written as dtvdx x= or in the integral form as dtvxxt

xif ∫=−0

.

Since tavvv xxixfx +== , this expression becomes

( ) ( )

−+−=+=+=− ∫∫∫ 0

20

2

000

tatvtdtadtvdttavxx xxi

t

x

t

xi

t

xxiif

2

21 tatvxx xxiif +=− ---------------- (2)

Equations (1) and (2) are two of the important constant acceleration equations.

Integrals in Physics 5-4

Now, based on our knowledge of displacement, velocity and acceleration, we will solve the following mechanics problems. To solve these problems, we will utilize integration by parts, Simpson’s rule, and trigonometric integrals.

• A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is m, the fuel is consumed at rate r, and the exhaust gases are ejected with constant velocity ve (relative to the rocket). A model for the velocity of the rocket at time t is given by the equation

( )m

rtmvgttv e−

−−= ln

where g is the acceleration due to gravity and t is not too large. If g = 9.8 m/s2, m = 30,000 kg, r = 160 kg/s, and ve = 3000 m/s, find the height of the rocket one minute after liftoff.

Solution: The rocket will have height ( )dttvH ∫=60

0

after 60 seconds.

( )

( ) ( )( ) ( )dtrtmvmvgH

mdtdtrtmvtgdtm

rtmvgtH

ee

ee

∫

∫ ∫∫

−−+−=

−−−

−=

−

−−=

60

0

60

0

60

0

60

0

60

0

ln60ln1800

lnln221ln

Let ( ) ( ) .,1,ln tvdtrrtm

dudtdvrtmu =−−

=⇒=−= Then

( ) ( )[ ] ( )

( ) ( )

( ) ( )

( ) ( ) .ln60ln6060ln60ln601800

ln60ln6060ln60

ln60ln60

160ln60lnln

60

0

60

0

60

0

600

mvrmrmv

rmvrmvmvgHSo

mrmrm

rmrm

rtmrmtrm

dtrtm

mrmdtrtm

rtrtmtdtrtm

eeeee −−++−−+−=

+−−−−=

−−−+−=

−+−+−=

−+−=− ∫∫∫

Substituting g = 9.8, m = 30,000, r =160, and ve = 3000 gives us H≈14,844 m.

• A particle that moves along a straight line has velocity ( ) tettv −= 2 meters per second after t second. How far will it travel during the first t seconds?

Integrals in Physics 5-5

Since v (t) > 0 for all t, the desired distance is ( ) ( ) .0

2

0

dwewdwwvts wtt

−∫∫ ==

First let

( ) [ ]

( ) [ ] [ ]( )( )

( )meterstteeteeteteet

eteetdweweetts

TheneVdwdUdwedVwUletNext

dwweewtsThenevwdwdudwedvwu

t

tttttt

twttt

wtwt

ww

twtwww

22222212

022

.,,

.2.,2,

2

22

02

00

2

00

22

++−=

+−−−=+−−+−=

−++−+−=

+−+−=

−==⇒==

+−=−==⇒==

−

−−−−−−

−−−−−−

−−

−−−−

∫

∫

• A radar gun was used to record the speed of a runner during the first 5 seconds of the race (see the table). Use Simpson’s Rule to estimate the distance the runner covered during those 5 seconds.

t(s) 0 .5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

v (m/s) 0 4.67 7.34 8.86 9.73 10.22 10.51 10.67 10.76 10.81 10.81

Solution: We use Simpson’s Rule with n= 10 and Δx = ½:

( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( ) ( )( )

( ) ( )

( ) m

fffffSdttv

735.4441.26861

]81.1081.10476.10

67.10451.10222.10473.9286.8434.7267.440[61

55.44125.04032

1distance 10

5

0

==

+++

+++++++=

+++++⋅

=≈= ∫

• A particle moves on a straight line with velocity function ( ) tttv ωω 2cossin= . Find its position function s = f (t) if f (0) = 0.

Integrals in Physics 5-6

Solution: ( ) duwuwutfst

∫==0

2cossin . Let .sincos duwuwdywuy −=⇒=

Then ( ).cos131

3111 3

cos

1

3cos

1

2 wtw

yw

dyyw

swtwt

−=

−=−= ∫

Work Done By a Constant Force

Suppose that you are trying to push a heavy object across a room floor. If you push on the object and it moves through a displacement, you have done work on the object.

Figure 4: A diagram of Force being applied to a heavy object.

If an object undergoes a displacement ,r∆ the work done by the constant force F

(SI unit of Newton, N) on the object is

(F cosθ)Δr. ------------- (3)

Work is a scalar quantity with the SI unit of Newton-meter (N·m); no direction is associated with it. The Newton.meter, when referred to work or energy, is called the Joule (J).

Integrals in Physics 5-7

Figure 5: A free body diagram.

WORK DONE BY A VARYING FORCE

Consider a particle being displaced along the x axis under the action of a force with an x component Fx that varies with position, as shown in Figure 6.

Figure 6: A graphical representation of Force varying in the x-direction.

The work done by a force is zero when the force is perpendicular to the displacement. That is, if θ = 90°, then cos 90°= 0 and W = 0. For example, consider the free body diagram for a block moving across a frictionless surface in Figure 5. The work done by the normal force and the gravitational force on the block during its horizontal displacement are both zero for the same reason: they are both perpendicular to the displacement.

The particle is displaced in the direction of increasing x from x = xi to x = xf. In such a situation, we can not use equation (3) to calculate the work done by the force because this relationship applies only when F

is constant in magnitude

and direction. Using Figure 6, suppose that the point of application of the force undergoes a small displacement in the x direction so that Δr = Δ x, the x component Fx of the force is approximately constant over this interval. Hence, we can approximate the work done by the force on the particle for the small displacement as

xFW x∆≈1

Integrals in Physics 5-8

Now suppose that the force versus position curve is divided into a large number of such intervals, hence the total work done is approximately equal to

xFWf

i

x

xx∆≈ ∑

If the displacements Δx are allowed to approach zero, the number of terms in the sum increases without limit, but the value of the sum approaches a definite value equal to the area under the curve bounded by Fx and the x axis as shown in Figure 7.

Figure 7: A graphical representation of determining work.

When more than one force acts on a system and we model the system as a particle, the total work done on the system is just the work done by the net force. The x component of the net force is∑ xF , the total work, done on the

particle as it moves from xi to xf is

( )dxFWWf

i

x

xxnet ∫ ∑∑ == .

• The table shows values of the force function f(x) where x is measured in meters and f(x) in Newtons. Use Simpson’s Rule to estimate the work done by the force in moving an object a distance of 18 m.

As you have learned in Calculus, this limit of the sum is represented by

∑ ∫=∆→∆

f

i

f

i

x

x

x

xxxxdxFxF

0lim .

The limits on the integral x = xi to x = xf define what is called a definite integral. This definite integral is numerically equal to the area under the curve of Fx versus x between xi and xf. Therefore, we can express the work done by Fx for the displacement from xi to xf as

dxFWf

i

x

xx∫=

Integrals in Physics 5-9

x 0 3 6 9 12 15 18

f(x) 9.8 9.1 8.5 8.0 7.7 7.5 7.4

Solution:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( ) ( )[ ] joules

fffffffSdxxfWork

1484.75.747.720.845.821.948.91

18154122946234036018

6

18

0

=++++++⋅=

++++++⋅−

=≈= ∫

ELECTRIC FORCE: COULOMB’S LAW

Electric forces between charged objects were measured quantitatively by Charles Coulomb using the torsion balance. He confirmed that the electric force between two small charged spheres is proportional to the inverse square of their separation distance r.

a. b.

Figure 8: (a) A torsion balance used by Coulomb; (b) Interaction of like and unlike electric charges.

Coulomb’s law describes the magnitude of the electrostatic force between two charged particles with charges q1 and q2 and separated by a distance r:

Integrals in Physics 5-10

221

rqq

kF ee =

where ( )229 /.1099.8 CmNke ×= is the Coulomb constant and the force is in Newtons, if the charges are in Coulombs and the separation distance is in meters. The constant ek is also written as

041πε

=ek

where the constant 0ε , known as the permittivity of free space, has the value 2212

0 /108542.8 mNC ⋅×= −ε . The equation for the electric force gives the magnitude of the Force.

Now, since we know how work and force are related to each other, let us solve the following problem to understand what electric potential is.

• The magnitude of the repulsive force between two point charges with the same sign, one of size 1 and the other of size q, is

𝐹𝐹 =𝑞𝑞

4𝜋𝜋𝜀𝜀0𝑟𝑟2

where r is the distance between the charges and ε0 is a constant. The potential V at a point P due to the charge q is defined to be the work expended in bringing a unit charge to P from infinity along the straight line that joins q and P. Find a formula for V.

Solution: If the distance between P and the point charge is d, then the potential V at P is

V = W = d

qtd

qr

qdrr

qFdrt

d

tt

dd

0002

0 411lim

41

4lim

4 πεπεπεπε−=

+−=

−==

∞→∞→∞∞∫∫

To understand more about electric potential, let us calculate electric potential due to continuous charge distribution. If the charge distribution is known, we consider the potential due to a small charge element dq, treating this element as a point charge.

Integrals in Physics 5-11

Figure 9: Electric potential at point P from a charge element, dq.

To calculate the total potential at point P, we integrate this equation to include contributions from all elements of the charge distributions. Since each element is at a different distance from point P, potential V can be expressed as

∫=r

dqkV e

In this equation, the electric potential is taken to be zero when point P is infinitely far from the charge distribution.

The SI unit of potential, 1J/C, is called 1 volt (1V), in honor of the Italian scientist Alessandro Volta.

Figure 10: An annulus.

The electric potential dV at some point P due to the charge element dq is

rdqkdV e=

where r is the distance from the charge element to point P. To calculate the total potential at point P, we integrate this equation to include contributions from all elements of the charge distributions.

Let us now calculate the electric potential at point P on the axis of the annulus shown in Figure 10, which has uniform charge density σ.

22 xr

dqkdV e

+=

In this equation rdrdAdq πσσ 2== .

Hence, 2222

2

xr

drrk

xr

dqkdV ee

+=

+=

πσ

[ 2222

2222 axbxk

xrdrrkV e

b

ae +−+=

+= ∫ πσπσ

Integrals in Physics 5-12

Alternating Current Most present-day household and industrial power-distribution systems operate with alternating current (AC). Any appliance that we plug into a wall outlet uses AC. Circuits in modern communications equipment, including radios and televisions, make extensive use of AC.

An AC circuit consists of circuit elements and a power source that provides an alternating voltage Δv. This time-varying voltage from the source is described by

tVv ωsinmax∆=∆

where ΔVmax is the maximum output voltage of the source, or the voltage amplitude. As shown in Figure 11, since the output voltage of an AC source varies sinusoidally with time, the voltage is positive during one half of the cycle and negative during the other half of the cycle.

Figure 11: An alternating voltage over time.

Similarly, the current in an AC circuit is an alternating current that varies sinusoidally with time. The angular frequency of the AC voltage is

Tf ππω 22 ==

where f is the frequency of the source and T is the period. The SI unit of frequency, f .is Hz or 1/s, and the SI unit of angular frequency, ω, is rad/s. The frequency of the current in an AC circuit is determined by the source connected to that circuit. Commercial electric power plants in the United States use a frequency of 60 Hz.

The average value of the current over one cycle is zero.

Integrals in Physics 5-13

In an AC circuit, it is an average value of current called rms current, is of importance. The notation rms stands for root-mean-square, which means the square root of the mean (average)

value of the square of the current: ( )avgrms iI 2= .

Figure 12: Current over time.

Since i2varies as tω2sin and the average value of i2 is 2max2

1 I , the rms current is

maxmax 707.02

II

I rms == .

Alternating voltage can also be described as rms voltage and can be represented as follows:

maxmax 707.02

VV

Vrms ∆=∆

=∆ .

After getting an insight into rms voltage and currents, let us solve the following problem, which utilizes trigonometric integrals.

• Household electricity is supplied in the form of alternating current that varies from 155 V to -155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation

( ) ( )ttE π120sin155=

where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of ( )[ ]2tE over one cycle.

(a) Calculate the RMS voltage of household current.

Integrals in Physics 5-14

(b) Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude, A, needed for the voltage ( ) ( )tAtE π120sin= .

Solution: (a) We want to calculate the square root of the average value of ( )[ ] ( )[ ] ( ).120sin155120sin155 2222 tttE ππ ==

First, we calculate the average value itself, by integrating ( )[ ]2tE over one cycle (between t = 0 and t = 1/60, since there are 60 cycles per second) and dividing by (1/60 – 0):

( )[ ] ( )[ ] ( )[ ]

( ) ( )2

155000601

21155.60240sin

2401

2115560

.240cos12115560120sin155

60/11

22

60/1

0

2

60/1

0

260/1

0

222

=

−−

−

=

−

⋅=

−⋅== ∫∫

tt

dttdtttE

ππ

ππ

The RMS value is just the square root of this quantity, which is .1102

155 V≈

( )[ ]

( )[ ] ( ) ( )[ ]

( ) ( )

.311222021220,Thus

21000

60130240sin

240130

240cos12160120sin

60/11220

220)b(

22

2260/1

0

2

60/1

0

2260/1

0

222

2

VAA

AAttA

dttAdttAtE

tE

ave

ave

≈=⇒=

=

−−

−=

−=

−===

⇒=

∫∫

ππ

ππ

For this write up, the following books have been used:

Physics for Scientists and Engineers, 7th edition, Serway and Jewett.

Calculus (Special edition for UCF), James Stewart.

College Physics, 8th edition, Young and Geller.

Integrals in Physics 5-15

Chapter 6

AApppplliiccaattiioonnss ooff CCaallccuulluuss IIII TTooppiicc

Application of Series in Heat Transfer

Transient Heat Conduction

FFaaccuullttyy CCoonnttrriibbuuttoorr Dr. Alain Kassab Mechanical, Materials and Aerospace EngineeringDepartment College of Engineering and Computer Science

WWeeeekkss:: April 12 and April 19 of Spring 2010

Calculus Topic: Infinite Sequences and series

Background on Series We will consider two series and their application in this chapter: (1) Taylor series andtheir particular form, the Maclaurin series, and (2) Fourier series.

1. Taylor and Maclaurin series ( )section 12.10 The Taylor series can be used to represent any function , that is infinitelyß 0ÐBÑdifferentiable about a point, , called the expansion point, and the series isB9

0ÐBÑ œ 0ÐB Ñ 0 ÐB ÑÐB B Ñ 0 ÐB Ñ ÞÞÞ œ Ð Ñ9 9 9 9w ww ÐBB Ñ 0 ÐB ÑÐBB Ñ

#x 8x8œ!

∞9 9 9

# 8 8( )1

where, denotes the first derivative, the second derivative, and so on and is the -th0 0 à 0 8w ww 8( )

derivative of The location where the series is evaluated can be taken anywhere within the0ÐBÑÞ Bradius of convergence of the series. In practice, when we compute using series, we can never addan infinite number of terms, so we truncate the series after -terms (stop after adding -terms)R Rand, from the Taylor remainder theorem, we have an approximation of the function as0ÐBÑ

0ÐBÑ œ 0 ÐB ÑÐB B Ñ 0

8x

0ÐBÑ

ðóóóóóóóóóñóóóóóóóóóò ðóóóóóóóóñóóóóóóóóò8œ!

R 8 8 89 9

( ) ( +1

terms we computeand add to approximate

the function

)Ð ÑÐB B Ñ

Ð8 "Ñx

0ÐBÑ

Ð Ñ0 9

8"

the term we truncateand drop from our

computation of is the truncation error (TE)

2

While we do not know exactly, we do know at least that its location is somewhere 0 0B BÞ9

More importantly, although we do not know the truncation error (TE) exactly, we know at leastthat it is proportional to . How many terms do we take in the series? We take enoughÐB B Ñ9

8"

terms such that adding more terms will not make a significant change in the value of the sum wecompute to approximate 0ÐBÑÞ The Maclaurin series is a Taylor series expanded about the particular expansion pointB œ ! „ B9 . For instance, the Maclaurin series for the exponential of is

/ œ " „ B „ ÞÞÞ œ Ð ÑB B B Ð "Ñ B

#x $x %x 8x„B

# $ % 8 8

8œ!

∞

3

which converges on the interval , and that means the series can be used to∞ B ∞represent the exponential for any value of . How do we utilize such series in practice? We willBconsider an example of the application of Taylor and Maclaurin series to the determination of thetemperature of a hot metal bar being quenched (rapidly cooled) in water bath.

2. Application of Maclaurin series to heat transfer: quenching of a metal bar at early times Quenching is used in metallurgy to harden steel; for instance, it is by repeated quenchingthat skilled blacksmiths crafted high quality swords throughout history, with the Japanese havingparticularly mastered this process to produce their famous swords that were the weaponKatana

Application of Series in Heat Transfer Transient Heat Conduction 6-2

of choice of the . We wish to study the temperature history in the metal bar that is beingSamuraiworked into what may eventually become such a sword. Suppose then that we take a long metalbar of width, , that is intially at anP œ "!-7elevated temperature we denote byX œ "!!! G3 ° , and we plunge this extremely hotmetal into a large bath of water. The water incontact with the metal surface will boil at aconstant temperature of ° , we denote as"!! GX œ "!! G9 ° . The temperature of the bar, wedenote as, , will decrease as a function of time,Xwe denote as, , and it will also vary as a>function of space, , as the heat leaves the metalB

x=0 x=LTo =100oC

Steelbar initiallyat 1000 Co

To =100oC

Water bathx

x=0 x=LTo =100oC

Steelbar initiallyat 1000 Co

To =100oC

Water bathx

bar; that is in general . The rate at which the heat leaves is a function of the thermalXÐBß >Ñdiffusivity, , that is particular to each metal. Using principles of conservation of energy andαanalytical methods you will be exposed to later in courses such as , the temperatureEML 4142distribution for the early part of the quenching process is given by

XÐBß >Ñ œ X ÐX X Ñ/<0 Ð Ño 3 9B

# >Š ‹Èα

4

where the error function, , is defined by the integral/<0ÐBÑ

/<0ÐBÑ œ / .B Ð Ñ#

!

BBÈ1

' # 5

This integral has no closed form solution, that is we cannot find a formula, a method, or acombination of tricks from calculus to integrate and find an explicit expression for the integral.How are we to then compute the temperature? In this case, we can utilize the Malaurin series.Let us inroduce the Maclaurin series for in the integral and integrate term by term (/B# section12.9),

/<0ÐBÑ œ / .B#

œ " B ÞÞÞ .B# B B B

#x $x %x

œ B ÞÞÞ# B B B B

$ #x% $x( %x*

œ# Ð "Ñ B

8xÐ#8 "Ñ

Ð Ñ

È (È ( ’ “È ’ “È

1

1

1

1

!

BB

!

B#

% ' )

$ % ( *

8œ!

∞ 8 #8

#

+1

6

so that the error function can be written as the series,

Application of Series in Heat Transfer Transient Heat Conduction 6-3

/<0ÐBÑ œ Ð Ñ#

8œ!

∞Ð"Ñ B8xÐ#8"ÑÈ1

8 #8 "+7

This series converges in [ , ] since that is the range of convergence for the Maclaurin∞ ∞series for . You will also encounter this function in statistics ( ) as it is also called/B STA 3032the probability function, and, in statistics, you will likely be told to use look-up tables to evaluatethis function. After this discussion, you may know another way than to use a table. We will now investigate how to compute a value for using this series, and then/<0ÐBÑapply this experience to evaluating the temperature in the metal bar as a function of space andtime. Let us take a typical value for the thermal diffusivity, , for steel which is aboutαα œ "! 7 Î= B œ " -7& #

9, and a location close to the surface, say , and let us ask what is thetemperature there second after the metal bar was plunged into the water bath. We compute the"argument of the error function for this combination,

B œ œ œ "Þ&)" Ð ÑB !Þ!"7

# > # "! ‚ "=

9

& 7=

È Éα #8

introducing this value into the equation for the series, Eq. (7), and summing we find that as weadd terms in the series, after ten terms the series converges (settles down) to a value of

/<0Ð"Þ&)"Ñ œ !Þ*(%'()& Ð Ñ9

to seven significant figures. A table of computed values for the error function as the number ofterms taken in the series is increased from to and a plot of these values is providedR " "!below,

1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

erf(1.581)

N – number of terms taken in the series

∑=

+

+−

=N

n

nn

nnxxerf

0

12

)12(!)1(2)(

π

1 2 3 4 5 6 7 8 9 100

0.5

1

1.5

erf(1.581)

N – number of terms taken in the series

∑=

+

+−

=N

n

nn

nnxxerf

0

12

)12(!)1(2)(

π

N123456789

10

0.2975891.41217710.74884831.07124320.93937750.98586040.97147540.97544120.97445570.9746785

erf(1.581)N123456789

10

0.2975891.41217710.74884831.07124320.93937750.98586040.97147540.97544120.97445570.9746785

erf(1.581)

which corresponds to a temperature

XÐ!Þ!!"7ß "=Ñ œ "!! G Ð"!!! G "!! GÑ ‚ !Þ*(%'()& Ð Ñ

œ *((Þ"* G

° ° ° 10°

If we check the temperature at that same location a little later, say seconds after the plunge, we&find that in this case

B œ œ œ !Þ(!( Ð ÑB !Þ!"7

# > # "! ‚ &=

9

& 7=

È Éα #11

Application of Series in Heat Transfer Transient Heat Conduction 6-4

and in this case examining the series,

1 2 3 4 5 6 7 8 9 100.660

0.665

0.67

0.675

0.680

0.685

erf (0.707)

∑=

+

+−

=N

n

nn

nnxxerf

0

12

)12(!)1(2)(

π

N – number of terms taken in the series

1 2 3 4 5 6 7 8 9 100.660

0.665

0.67

0.675

0.680

0.685

erf (0.707)

∑=

+

+−

=N

n

nn

nnxxerf

0

12

)12(!)1(2)(

π

N – number of terms taken in the series

N123456789

10

erf (0.707)

0.66484350.68477560.68240350.682634

0.68261520.68261650.68261640.68261640.68261640.6826164

N123456789

10

erf (0.707)

0.66484350.68477560.68240350.682634

0.68261520.68261650.68261640.68261640.68261640.6826164

we find that only the first few terms, say the first five terms are enough to obtain a convergedvalue for the series. In this case, for example, the first five terms sum to:

/<0Ð!Þ(!(Ñ œ !Þ(*((' !Þ"$#*# !Þ!"**$ #Þ$(#"& ‚ "! #Þ$!&&' ‚ "! Ð Ñ

œ !Þ')#'"&#

$ 4 12

and, with an additional two terms the sum settles to a value with seven significant figures,providing a value of This then provides for us a value of the/<0Ð!Þ(!(Ñ œ !Þ')#'"'%Þtemperature as

XÐ!Þ!!"7ß &=Ñ œ "!! G Ð"!!! G "!! GÑ ‚ !Þ')#'"'% Ð Ñ

œ ("%Þ%# G

° ° ° 13°

In such a manner, we can determine the temperature at various locations in the bar for varioustimes and a plot of these results is provided below

0 cm 1 cm 2 cm 3 cm 4 cm 5 cm 6 cm0

500oC

1000oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x ,( )

7 cmx

25s

x=0

To =100oC

x

100oC

steelbar

0 cm 1 cm 2 cm 3 cm 4 cm 5 cm 6 cm0

500oC

1000oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x ,( )

7 cmx

25s

x=0

To =100oC

x

100oC

steelbar

As you can see from this plot, and as you expect from logic, the effect of the cooling from thewater bath takes time to be felt at the interior of the bar. That is called the penetration depth andit is controlled by the parameter Now the solution we have been considering is valid untilÈα>Þthe penetration depth reaches the midpoint of the bar, , we then have to use another solutionPÎ#for the temperature and this is in terms of what is called a Fourier series, named after itsdiscoverer the famous French mathematician and physicist, Jean-Baptiste Fourier. This part ofthe solution will then comprise our second example of series applications in heat transfer.

Application of Series in Heat Transfer Transient Heat Conduction 6-5

3. Application of Fourier series to heat transfer: quenching of a metal bar at later times By later times," we mean after the middle of the bar has "felt the cooling effects of the water bath. In our example, themidpoint of a bar that is long will have felt the effect of the"!-7cooling bath after . The solution for the temperature at any#&=location in the bar and at any time after this is given by the infiniteseries,

XÐBß >Ñ œ X ÐX X Ñ G =38Ð BÑ/ Ð Ñ9 3 9 88œ"

∞8P

Ð Ñ >1 α 8P

#1 14

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

X=0 X=L

T=100oCT=100oC

x

steel bar

T(x,t)

where the coefficients are given in our case by the integralG8

G œ =38Ð BÑ.B Ð Ñ# 8

P P8

!

P( 115

You have learned in Calculus II how to carry out this integral ( ) , and you should besection 5.3able to show that,

G œ Ò" Ð "Ñ Ó Ð Ñ#

88

8

1 16

using the fact that when is an integer. All the-9=Ð8 Ñ œ Ð "Ñ 81 8

even coefficients are zero, and so we do not count even terms inthe series. The first few non-zero of these coefficients are given inthe table to right. This special infinite series is called a Fourier series afterJean-Baptiste Fourier (1768-1830) who lived in the days ofNapoleon whom he accompanied in his 1798 campaigns to Egyptand served as governor of Lower Egypt under French rule. Helater became Prefect of the Department of Isere in South-WesternÏ

n13579

cn

1.2730.4240.2550.1820.141

n13579

cn

1.2730.4240.2550.1820.141

France where he carried out the famous experimental and analytical research that laid thefoundations of heat transfer. It was a great accomplishment by Fourier to show that any function,with some reasonable degree to continuity, could be expressed in terms of summations oftrigonometric functions. Actually, some of the greatest mathematicians and physicists of histime were not convinced of this and made his life a bit miserable. He published these results andthe foundations of heat transfer in solids in his famous 1822 treatise Théorie analytique de lachaleur (the Analytical Theory of Heat) which got him elected as a member of the FrenchNational Academy of Science, of which he was later to be the Permanent Secretary. Let us investigate the calculation of the temperature utilizing this series. Again, we willnot take an infinite number of terms in the series, but rather a finite number. It turns out that theFourier series for the temperature converges slowly when the non-dimensional value of time thatappears in the exponential called the Fourier number, J9ß

J9 œ Ð Ñα>P# 17

Application of Series in Heat Transfer Transient Heat Conduction 6-6

is less than That is, for earlier times that correspond to smaller Fourier numbers, it takes!Þ#Þmany terms for the series to settle down (converge) to a value. Let us investigate what that limitcorresponds to in our case. Taking our bar to be , then a Fourier number of P œ "! -7 J9 œ !Þ#corresponds to a time of

> œ !Þ# ‚ œ !Þ# ‚ Ð ÑP Ð!Þ"7Ñ

"!

œ #!! =

# #

& 7=

α # 18

For times larger than , or in our case, the series converges rather quickly, andJ9 œ !Þ# #!! =you can obtain the solution by using only ! Let us see that this is the case. Suppose weone termwant to find the temperature at the midpoint at The one term solution is then,> œ #!! =Þ

X Ð!Þ!&7ß &!=Ñ œ "!! Ð Ñ ‚ "Þ#($ ‚ =38Ð !Þ!&7Ñ ‚ /!Þ"7

œ #&*Þ") G

°

° (19)

"!!! G "!! G° °1 "! Ð Ñ #!!=& #7#

= !Þ"71

For times that correspond to smaller Fourier numbers then more terms are required in the series.Let us look at the same point at time . which corresponds to . Aß B œ & -7 > œ #&= J9 œ !Þ!#&9

plot of the termperature at this point as a function of the number of terms in the series is givenbelow as well as a table for the temperature as we add terms in the series and the value of theindividual terms in the series is also tabulated:

0 5 10 15940oC

960oC

980oC

1000oC

n

T(5cm,25s)

0 5 10 15940oC

960oC

980oC

1000oC

n

T(5cm,25s)

n13579

111315

T 25s,( )995.354953.896954.376954.375954.375954.375954.375954.375

Terms0.995

-0.0465.333·10 -4

-1.021·10 -6

2.957·10 -10

-1.251·10 -14

00

5cm

Utilizing this series solution, we can reconstruct the temperature history in the metal bar, and theresult is shown for several representative times:

0 2 cm 4 cm 6 cm 8 cm 10 cm0

200oC

400oC

600oC

800oC

1000oC

1200oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x 25s,( )

T x 50s,( )

T x ,( )

x

100s

0 2 cm 4 cm 6 cm 8 cm 10 cm0

200oC

400oC

600oC

800oC

1000oC

1200oC

T x 1s,( )

T x 5s,( )

T x 15s,( )

T x 25s,( )

T x 50s,( )

T x ,( )

x

100s

Application of Series in Heat Transfer Transient Heat Conduction 6-7

Now, we always want to check that our solutions make sense. To this end, let us checkthat the solution reported to you in Eq. (14) reproduces the initial temperature at Since> œ !Þ/ œ " > œ !! , the series solution at is then,

XÐBß !Ñ œ X ÐX X Ñ G =38Ð BÑ Ð Ñ8

P

"B

9 3 9 8

8œ"

∞

ðóóóóóñóóóóóò1

this term must sum tofor any value of

20

with the coefficients were given in Eq. (16). For the the initial temperatureG œ Ò" Ð "Ñ Ó8#8

81

to be equal to , then the series must sum to one This actuallyX! regardless of the value of .Boccurs at all values of except at and , where since the and theB − Ò!ß PÓ B œ ! B œ P =38Ð!Ñ œ !=38Ð8 Ñ œ !1 , the series is zero. The series in Eq. (20) is actually called a half-range Fourier sineseries for . The series settles down to the value of (converges) very slowly, particularly close" "to and . A plot is provided below for the result of the sum obtained using , and! P #!ß &!ß %!!&!! B ! P œ !Þ"-7 terms at various values of from to .

0 cm 0.02 cm 0.04 cm 0.06 cm 0.08 cm 0.1 cm0

500oC

1000oC

1500oC

T (x,0)

T(x,0)

T(x,0)

T(x,0) N=400

x

N=200

N=50N=20

0 cm 0.02 cm 0.04 cm 0.06 cm 0.08 cm 0.1 cm0

500oC

1000oC

1500oC

T (x,0)

T(x,0)

T(x,0)

T(x,0) N=400

x

N=200

N=50N=20

Moreover, what is remarkable about this series representation is that since it utilizestrigonometric functions, which themselves are periodic, to represent a function (in this case aconstant) on the interval , the series representation is itself periodic! If we plot the series ofÒ!ß PÓEq. (20), say for the interval , we see that indeed, the series representation is that of aÒ !Þ'ß !Þ'Óperiodic (step) function that is constant with an alternating value of and over intervals of" "!Þ" -7.

Application of Series in Heat Transfer Transient Heat Conduction 6-8

N=1000

N=20

0.4 0.2 0 0.2 0.4

1

0

1

0.60.6

x

∑=

−−N

n

n xL

nn

0

)sin(])1(1[2 ππ

0.5 0.3 0.1 0.1 0.3 0.5

N=1000

N=20

0.4 0.2 0 0.2 0.4

1

0

1

0.60.6

x

∑=

−−N

n

n xL

nn

0

)sin(])1(1[2 ππ

0.5 0.3 0.1 0.1 0.3 0.5

This function can also be thougth of a waveform produced by an electronic device such as theclock in your computer. Indeed, Fourier series and some variations of such series findapplication in many other branches of engineering, for instance in electrical and electronicsengineering where they are routinely utilized in signal processing and control.

4. Finite Difference: a method to approximate derivatives and why we can't take limits tozero on the computer. The final application of Taylor series we will consider is their use in finite differencemethods (FDM). As the name implies, this is a technique utilized routinely in research andcommercial to approximate various derivatives that appear in the equationscomputer programsthat govern the behavior of the dependent variable of the problem of interest; in our case, thedependent variable is the temperature. Such equations are derived from conservation principles(balances) typically applied to mass, linear momentum, energy, and species, and they are calleddifferential equations. You will learn how to solve such differential equations in andMAP 3032 in most of your engineering courses, and you will encounter finite differences in andEML 3034 EML 4142. The reason that finite differences are used is that although we can solve certaindifferential equations analytically, we cannot do so in many cases especially when the geometryis complicated. You will see examples of such Finite Difference Methods applied to fluids andheat transfer in the lectures. Here, we simply focus on one of the basic devices utilized in FDM:approximation of the first derivative using one sided differences. We will develop and apply a finite difference method to estimate the heat flux (heat flowper unit area) through the walls of our quenched bar. For this purpose, we rely on a basicrelationship between the derivative of the temperature and the heat flux, , that was established;by Fourier, and that is mathematically stated as follows for one dimensional problems,

Application of Series in Heat Transfer Transient Heat Conduction 6-9

; œ 5 Ð Ñ`X

`B21

Here, has units of [ / ], and the proportionality constant is a property of each material; [ 7 5#

and it is called the thermal conductivity. For steel, a typical value is What this5 œ "&[Î7OÞequation tells us is that heat flows in proportion to the derivative of the temperature and,furthermore, since the derivative points in the direction of maximum increase of a function andheat flows from high to low temperature, the negative ensures the appropriate sign on the heatflux. A finite difference approximate of the derivative in Eq. (21) utilizes the Taylor series, seeEq. (2). We re-arrange the Taylor series solving for as0 ÐB Ñw

9

0 ÐB Ñ œ 0 ÐB Ñ 0 ÐB Ñ ÞÞÞ0ÐBÑ 0ÐB Ñ ÐB B Ñ ÐB B Ñ

ÐB B Ñ #x $xw ww www

9 9 99 9 9

9

#

Ð Ñ22

The location at which we take can be anywhere to the right or to the left of the point .B B9

Suppose that we wish to approximate the derivative of temperature at the left wall of the bar, thatis at the point . Since there is no solid to the left of , we then take withB œ ! B B œ B B9 9 9 ??B B denoting a distance (usually small) to the right of . Substituting into Eq. (22) we find that9

0 ÐB Ñ œ 0 ÐB Ñ 0 ÐB Ñ ÞÞÞ0Ð Ñ 0ÐB Ñ

#x $xw ww www

9 9 99ðóóóóóóóóóñóóóóóóóóóò ðóóóóóóóóóóóóóóóñóóóóóóóóóóóóóóóòB B B B

B9

#? ? ?

?term we compute terms we drop from the computation

(truncate) and these terms collectivelyconstitute the runcation rror (TE)T E

Ð Ñ23

This expression is called a for the first derivative, and since the truncationforward difference error (TE) is proportional to it is said to be a .?B first order accurate forward differenceTechnically, the symbol ( ) is used to express this fact; that is that the TE is of order . ByS B B? ?the way, the is due to a famous and brilliant Russian physicist and mathematician,big notationS Lev Landau. In summary, we have that

0 ÐB Ñ œ SÐ Ñw9

0Ð Ñ0ÐB ÑB BB

9 ??

9 ?B Ð Ñ24

We do not much care what the TE is exactly, rather, what we are interested to know is that theerror will reduce linearly with . We shall investigate what happens when we apply this to our?Bproblem and evaluate this approximation on the computer. Notice that if we take the limit that?Bp! in Eq. (29) we retrieve that exact definition of the derivative which you learned inCalculus I. The problem is that on the computer we cannot take that limit, as we will shortly findout, and we have to work with finite ; hence the name finite difference.?B Applying the forward difference to estimate the heat flux at the left wall, we find that

; œ 5 Ð Ñ`X

`B

œ 5 B

¹ ¹B œ! B œ!9 9

”XÐ Ñ XÐB ß >ÑB Bß >

B9 ?

?9

B œ!•

9

SÐ Ñ?

25

Application of Series in Heat Transfer Transient Heat Conduction 6-10

where, for large times corresponding to , we can use the one-term evaluation of theJ9 !Þ#temperature as we did in Eq. (19). Let us choose a time and estimate the heat flux at> œ #&!=9

the left wall utilizing the above first order forward finite difference, using various values for ?Branging from expecting the estimate for the heat flux to approach the?B œ !Þ!"ß !Þ!!" ÞÞÞ "!#"

exact value given by differentiating Eq. (14) with respect to and taking only one term,B

26; œ 5 Ð Ñ`X

`B

œ 5ÐX X ÑÐ ÑG -9=Ð BÑ/P P

œ %Þ&(*%( ‚ "! Ò[Î7 Ó

/B+->B œ! B œ!

3 9 " Ð Ñ >

% #

¹ ¹9 9

P#

91 1 α 1

Let us see what happens when we carry out this computation. The results are provided in thetable below and in the accompanying plot of the error vs. step size , where?B

/<<9<Ð BÑ œ %Þ&(*%( ‚ "! Ò[Î7 Ó 5? º Š% # ”XÐ Ñ XÐB ß >ÑB Bß >

B9 ?

?9

B œ!• º

9

‹ (27)

At first, the results behave as expected: the error reduces linearly with the reduction of the stepsize , until at some point close to ~ 10 the error begins to grow, and after ~? ? ?B B B'

10 the error is 100% as the computer has reached its limit of accuracy, called machine '1

precision, and does not recognize the difference between B B B9 9? and in the finite differenceequation, Eq. (25), and computes zero for the heat flux. Why did the error grow after ~?B10 ? This is because the computer makes very small mistakes every time it stores and'

computes with numbers due to the fact that it utilizes finite arithmetic; this error is called round-off error (RO) and there is no way to get around it. Round off will always exist due to the binarynature of representation of data on the computer (one and zeroes, called bits) on the computerand the limited amount of bits the computer uses to represent a number (in our case we used 64bits). Round off occurs then when in Eq. ( ) we compute 25 then use that value toB B9 ?

10-2

10-3

10-4

10-5

10-6

10-7

10-8

10-9

10-10

10-11

10-12

10-13

10-14

10-15

10-16

10-17

10-18

10-19

10-20

10-21

-4.5045101383·10 4

-4.5787152582·10 4

-4.5794609805·10 4

-4.5794684381·10 4

-4.5794685127·10 4

-4.5794685134·10 4

-4.5794685143·10 4

-4.5794685164·10 4

-4.5794685377·10 4

-4.5794685377·10 4

-4.5794621428·10 4

-4.5793768777·10 4

-4.5787373892·10 4

-4.5830006457·10 4

-4.4764192353·10 4

-4.2632564146·10 4

0

0

0

0

749.58375116027.5325520397

0.0753292002

7.5329423271·10 -4

7.5227799243·10 -6

2.7524947654·10 -7

8.2512706285·10 -6

2.9567549063·10 -5

2.4273036979·10 -4

2.4273036252·10 -4

0.0637061158

0.9163573988

7.3112420206

35.321322125

1.0304927815·10 3

3.1621209888·10 3

4.5794685134·10 4

4.5794685134·10 4

4.5794685134·10 4

4.5794685134·10 4

Forward differencecomputed heat fluxat the left wall xo=0:q(xo,to) [W/m2]

Step size From the left wall:Δx

Error in computedflux at xo=0

10-2

10-3

10-4

10-5

10-6

10-7

10-8

10-9

10-10

10-11

10-12

10-13

10-14

10-15

10-16

10-17

10-18

10-19

10-20

10-21

-4.5045101383·10 4

-4.5787152582·10 4

-4.5794609805·10 4

-4.5794684381·10 4

-4.5794685127·10 4

-4.5794685134·10 4

-4.5794685143·10 4

-4.5794685164·10 4

-4.5794685377·10 4

-4.5794685377·10 4

-4.5794621428·10 4

-4.5793768777·10 4

-4.5787373892·10 4

-4.5830006457·10 4

-4.4764192353·10 4

-4.2632564146·10 4

0

0

0

0

749.58375116027.5325520397

0.0753292002

7.5329423271·10 -4

7.5227799243·10 -6

2.7524947654·10 -7

8.2512706285·10 -6

2.9567549063·10 -5

2.4273036979·10 -4

2.4273036252·10 -4

0.0637061158

0.9163573988

7.3112420206

35.321322125

1.0304927815·10 3

3.1621209888·10 3

4.5794685134·10 4

4.5794685134·10 4

4.5794685134·10 4

4.5794685134·10 4

Forward differencecomputed heat fluxat the left wall xo=0:q(xo,to) [W/m2]

Step size From the left wall:Δx

Error in computedflux at xo=0

20 15 10 5 08

6

4

2

0

2

4

6

log(error(Δx))

log(Δx)

Round Off error

~ ε/Δx TruncationError ~ c Δx

beyond this pointthe computer doesnot recognizexo and xo+Δx. This iscalled machine precision ~ 10-16

when using 64 bit computing

x=0 x=L

T=100oC

x

steel bar

xo xo+Δx

Δx

20 15 10 5 08

6

4

2

0

2

4

6

log(error(Δx))

log(Δx)

Round Off error

~ ε/Δx TruncationError ~ c Δx

beyond this pointthe computer doesnot recognizexo and xo+Δx. This iscalled machine precision ~ 10-16

when using 64 bit computing

x=0 x=L

T=100oC

x

steel bar

xo xo+Δx

Δx

x=0 x=L

T=100oC

x

steel bar

xo xo+Δx

Δx

Application of Series in Heat Transfer Transient Heat Conduction 6-11

compute , to compute , and then again when we subtract these two values.XÐ Ñ X Ð ÑB Bß > B ß >9 9?Let us denote that little error by . Then, we see that the total error in computing the finite%difference is

/<<9<Ð BÑ œ - BB

? ?%

?í îRO

TE (28)

where - µ 0 ÐB ÑÎ#xww9 for the first order forward difference we are utilizing. This general

behavior of the total error is always true in any numerical computation; that is truncation errorreduces with decreasing step size ?B while round off error increases with decreasing step size,and, therefore, the two errors compete directly against each other. Moreover, at some point ofreduction of the step size in an effort to drive to zero, not only is the result getting worse but?Bwe reach a point where the computer no longer recognizes that we tried to compute B B9 ? ,and this is why we cannot take the limit that on the computer.?Bp ! If we wished to evaluate the heat flux at the right wall, , then we could not B œ P9 takeB œ B B9 ? in Eq. (22) as there is no solid to the right. Rather, we take a step back and chooseB œ B B9 ? in Eq. (22) we find that

0 ÐB Ñ œ 0 ÐB Ñ 0 ÐB Ñ ÞÞÞ0Ð Ñ 0ÐB Ñ

#x $xw ww www

9 9 99ðóóóóóóóóóñóóóóóóóóóò ðóóóóóóóóóóóóóóóñóóóóóóóóóóóóóóóòB B B B

B9

#? ? ?

?term we compute terms we drop from the computation

(truncate) and these terms collectivelyconstitute the runcation rror (TE)T E

Ð Ñ29

This expression is called a for the first derivative, and since the truncationbackward difference error (TE) is proportional to it is said to be a . The?B first order accurate backward differencecomputation follows the same procedure we used to evaluate the heat flux at the left wall. Thisshows that the finite difference representation for the first derivative is not unique, we just foundtwo ways to accomplish this task using Taylor series. There is yet another common way to approximate the first derivative by more complexmanipulation of the Taylor series and this is called a central difference whose advantage is thatthe truncation error is second order, that is TE~ ( ; or, for the central difference theS B Ñ? #

truncation error reduces with the square of the reduction in . In light of our discussions on the?Blimits of the computer, this is a big advantage. However, this discussion is beyond what we wishto discuss in this lecture, and you will have the opportunity to revisit this subject in EML 3034and .EML 4142

5. Conclusions This concludes our coverage of series applications in heat transfer. We have seen how tocompute certain integrals with Maclaurin series, how series converge to a value in severalpractical examples where series were utilized in the computation of temperature distributions ofquenched metals. We have also utilized Taylor series in finite difference approximations ofderivatives and taken a small glimpse into the nature of scientific and engineering computationswith finite differences.

Application of Series in Heat Transfer Transient Heat Conduction 6-12