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187700 | https://fiveable.me/algebraic-number-theory/unit-3/applications-galois-theory-number-fields/study-guide/gqL581g6wyaOXxkR | Applications of Galois theory to number fields | Algebraic Number Theory Class Notes | Fiveable | Fiveable
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🔢Algebraic Number Theory Unit 3 Review
3.4 Applications of Galois theory to number fields
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Algebraic Number Theory
Unit 3 – Algebraic Extensions & Galois Theory
Topic: 3.4
🔢Algebraic Number Theory Unit 3 Review
3.4 Applications of Galois theory to number fields
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
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🔢Algebraic Number Theory
Unit & Topic Study Guides
Intro to Algebraic Number Theory
Number Fields and Integer Rings
Algebraic Extensions & Galois Theory
3.1 Field extensions and algebraic closures
3.2 Galois groups and Galois correspondence
3.3 Splitting fields and normal extensions
3.4 Applications of Galois theory to number fields
Norms, Traces, and Discriminants
Integral Bases & Dedekind Domains
Ideals and Factorization in Dedekind Domains
Prime Decomposition in Number Fields
Dirichlet's Unit Theorem
Ideal Class Groups and Minkowski Bound
Quadratic and Cyclotomic Fields
Local Fields and Completions
Valuations and Ramification in Number Theory
Adeles and Ideles
Zeta Functions and L–Functions
Class Field Theory
Diophantine Equations in Number Theory
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Galois theory bridges number fields and group theory, unveiling deep connections between field extensions and their automorphisms. It provides a powerful framework for analyzing polynomial equations, field structures, and algebraic properties of number fields.
The fundamental theorem of Galois theory establishes a bijective correspondence between intermediate fields and subgroups of the Galois group. This correspondence allows us to tackle problems in number theory, including solvability of equations and properties of algebraic extensions.
Galois theory for number fields
Fundamental concepts and correspondences
Galois theory establishes a correspondence between field extensions and group theory providing a powerful framework for analyzing number field structures
Galois group of a field extension K/F represents the group of automorphisms of K fixing every element of F
Fundamental theorem of Galois theory creates a bijective correspondence between intermediate fields of a Galois extension and subgroups of its Galois group
Normal closure of a number field represents the smallest Galois extension containing the original field
Splitting fields of polynomials over number fields play a crucial role in understanding field extension structures
Discriminant of a polynomial provides information about the nature of its roots and splitting field structure
Important tools and properties
Norm and trace of elements in number field extensions serve as important tools for studying algebraic properties
Field extension degree equals the order of the Galois group for Galois extensions
Transitive subgroups of symmetric groups hold particular importance in determining Galois groups of irreducible polynomials
Polynomial discriminant offers insights into the Galois group structure, particularly regarding even and odd permutations
Factorization patterns of defining polynomials modulo various primes help narrow down Galois group possibilities
Advanced techniques for determining Galois groups of higher-degree extensions include resolvent polynomials and Frobenius elements
Galois groups of field extensions
Computing Galois groups
Determine Galois group of a number field extension by analyzing automorphism actions on defining polynomial roots
Symmetric and alternating groups frequently appear as Galois groups of number field extensions, especially in low-degree cases (S 3 S_3 S 3, A 4 A_4 A 4)
Examine polynomial factorization patterns modulo various primes to narrow down Galois group possibilities
Apply advanced techniques such as resolvent polynomials and Frobenius elements for higher-degree extensions
Utilize computational algebra systems (GAP, Magma) to assist in Galois group calculations for complex cases
Properties and applications
Galois group order equals the field extension degree for Galois extensions
Transitive subgroups of symmetric groups hold particular importance for Galois groups of irreducible polynomials
Polynomial discriminant provides information about Galois group structure, especially regarding even and odd permutations
Galois group structure reveals important properties of the field extension (normal, separable, radical)
Apply Galois groups to solve problems in number theory (class field theory, reciprocity laws)
Solvability of polynomial equations
Solvability criteria and theorems
Solvability by radicals directly relates to the Galois group structure of the polynomial's splitting field
Polynomial equation solvability by radicals occurs if and only if its Galois group represents a solvable group
Abel-Ruffini theorem states the non-existence of general algebraic solutions for polynomial equations of degree five or higher
Galois' criterion for solvability provides a group-theoretic characterization of solvable polynomials
Composition series and derived series of a group serve as essential tools for determining Galois group solvability
Radical extensions and examples
Radical extensions and their connection to cyclic extensions play a crucial role in understanding solvability by radicals
Analyze specific examples of solvable and unsolvable quintic polynomials to illustrate Galois theory application (x 5−4 x+2 x^5 - 4x + 2 x 5−4 x+2, unsolvable)
Explore solvable polynomials of degree less than 5 and their solution methods (quadratic formula, cubic formula)
Investigate the structure of radical extensions and their Galois groups (Q(a n)\mathbb{Q}(\sqrt[n]{a})Q(n a) over Q\mathbb{Q}Q)
Study the Galois groups of cyclotomic extensions and their connection to solvability (Q(ζ n)\mathbb{Q}(\zeta_n)Q(ζ n) over Q\mathbb{Q}Q, where ζ n\zeta_n ζ n is a primitive nth root of unity)
Galois theory vs Fundamental Theorem of Algebra
Fundamental Theorem of Algebra and Galois theory
Fundamental Theorem of Algebra states every non-constant polynomial with complex coefficients has at least one complex root
Galois theory provides an elegant proof of the Fundamental Theorem of Algebra by considering the Galois group of C\mathbb{C}C over R\mathbb{R}R
Algebraic closure of a field represents the smallest field extension containing all roots of polynomials with coefficients in the original field
Galois group of C\mathbb{C}C over R\mathbb{R}R isomorphic to the cyclic group of order 2, reflecting the complex conjugation automorphism
Fixed field of the Galois group of C\mathbb{C}C over R\mathbb{R}R precisely equals R\mathbb{R}R, illustrating the Galois correspondence
Generalizations and connections
Fundamental Theorem of Algebra generalizes to show C\mathbb{C}C algebraically closed, meaning every non-constant polynomial splits completely over C\mathbb{C}C
Connection between Galois theory and Fundamental Theorem of Algebra highlights interplay between algebraic and topological properties of complex numbers
Explore the Galois groups of finite extensions of Q\mathbb{Q}Q contained in C\mathbb{C}C (cyclotomic fields, quadratic extensions)
Investigate the Galois theory of infinite algebraic extensions (Q‾\overline{\mathbb{Q}}Q over Q\mathbb{Q}Q)
Analyze the connection between Galois theory and the theory of Riemann surfaces in complex analysis
3.3 BackNext
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immer – Schreibung, Definition, Bedeutung, Etymologie, Synonyme, Beispiele
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Seitenanfang
Bedeutungen
Etymologie
Bedeutungsverwandte Ausdrücke
Typische Verbindungen
immer
Lesezeichenzitieren/teilenzuklappenausklappen
Grammatik Adverb
Aussprache Fehler
Worttrennung im-mer
Wortbildung mit ›immer‹ als Erstglied: immerdar… 6 weitere · immerfort · immergleich · immergrün · immerhin · immerwährend / immer während · immerzu… weniger
Mehrwortausdrückedie Augen für immer schließen · für immer und ewig · immer mit der Ruhe… 2 weitere · schlimmer geht immer / schlimmer geht’s immer / schlimmer geht es immer · stets zu Diensten / immer zu Diensten… weniger
Bedeutungsübersicht
I. ...
zu jeder Zeit, stets
⟨immer + Komparativ⟩ dient der beständig zunehmenden Steigerung
siehe auch auch (5)
[umgangssprachlich] jeweils
[umgangssprachlich] die Zeit über
[salopp] ja, selbstverständlich
II. ohne eigentliche Bedeutung; intensivierend; satzbelebend
verstärkend bei »noch«
[umgangssprachlich] nur
in Sätzen mit imperativischem Sinn
eWDG
Bedeutungen
I.
1.
zu jeder Zeit, stets
a)
jederzeit, ständig
in gegensätzlicher Bedeutung zu nie
Beispiele:
er war immer höflich zu mir
immer werden wir an euch denken
die Nachbarsleute sind immer in Eile
... 7 weitere Beispiele
das habe ich (schon) immer prophezeit
der immer zu Scherzen aufgelegte Onkel
alles war wie immer(= war unverändert)
salopp ich verstehe immer (bloß) Bahnhof! (= ich verstehe gar nichts!)
rufe vorher bei ihm an, er ist nicht immer daheim
Kinder dürfen nicht immer allein gelassen werden
sie will immer nicht (= will nie) mitkommen
... weniger Beispiele
α)
verstärkend, umgangssprachlich⟨immer und ewig⟩
Beispiel:
er redet immer und ewig denselben Unsinn
β)
⟨für, auf immer(= für alle Zeit, für ewig)⟩
Beispiele:
ein Abschied für, auf immer
sich für, auf immer trennen
für, auf immer blamiert, ruiniert sein
gehoben, verhüllend er ging für immer von uns (= ist gestorben)
gehoben, verhüllend die Augen für immer schließen, für, auf immer Abschied nehmen (= sterben)
γ)
⟨immer wieder⟩
Beispiele:
er hat immer wieder von vorn anfangen müssen
ich nahm immer und immer wieder aufs neue Anlauf
Und was man immer wieder und wieder hört / daran glaubt man[P. Weiss Marat 13]
δ)
umgangssprachlich⟨immer mal (= hin und wieder)⟩
Beispiele:
ich treffe ihn immer mal auf der Straße
wir spielen immer mal Schach miteinander
b)
jedes Mal
Beispiele:
immer (dann) wenn man ihn sprechen will, hat er keine Zeit
wenn man ihn sprechen will, hat er immer keine Zeit
er gibt immer (dann) seine Ratschläge, wenn man sie nicht braucht
... 3 weitere Beispiele
sie ist nicht immer krank, wenn sie fehlt
wenn ich ihn einlade, ist er immer unabkömmlich
salopp er ist immer der Dumme (= hat den Schaden, Nachteil)
... weniger Beispiele
2.
⟨immer + Komparativ⟩dient der beständig zunehmenden Steigerung
Beispiele:
man wird immer älter
die Jungen kletterten immer höher
er forderte immer mehr
seine Leistungen wurden immer besser
wir fuhren immer schneller
3.
siehe auch auch (5)
Grammatik: verallgemeinernd in Verbindung mit Interrogativpronomen und häufigem »auch«
Beispiel:
wer (auch) immer das getan hat, wird dafür haftbar gemacht
4.
umgangssprachlich jeweils
Beispiele:
immer drei Schüler auf einmal wurden in den Prüfungsraum gebeten
immer der vierte musste vortreten
er hastete die Treppe hinauf, immer drei Stufen auf einmal nehmend
sie gingen immer sechs und sechs (= zu sechst) los
Man tastete sich mit Hilfe seiner Laterne immer nur zwei Schritte vorwärts[Klabund Bracke 153]
5.
umgangssprachlich die Zeit über
Beispiele:
was treibst du (denn) immer?
wie geht es dir immer?[Doderer Lederbeutelchen 146]
6.
salopp ja, selbstverständlich
Beispiele:
»wirst du den Mund halten?« »Na, immer!«
»Kommst du mit?« »Immer!«
II.
ohne eigentliche Bedeutung; intensivierend; satzbelebend
Grammatik: partikelhaft
1.
Grammatik: verstärkend bei »noch«
Beispiele:
es hatte geregnet, als wir ankamen, bei der Abfahrt regnete es immer noch, noch immer
um neun Uhr lag er immer noch, noch immer im Bett
hast du den Schlüssel immer noch nicht gefunden?
diese Brücke ist immer noch gesperrt
umgangssprachlich versöhne dich mit ihr, sie ist immer noch (= immerhin, schließlich) deine Schwester!
2.
umgangssprachlich nur
Grammatik: an der Spitze von Aufforderungssätzen
Beispiele:
immer langsam (voran)!
immer lustig voran!
immer frisch von der Leber weg!
... 5 weitere Beispiele
immer rein in die gute Stube!
immer ran an den Speck!
immer mit der Ruhe!
immer der Nase nach!
immer zu! (= drauflos!)
... weniger Beispiele
3.
⟨vergleiche so + Adverb + immer(= nur)⟩
Grammatik: in modalen Gliedsätzen des Grades
Beispiele:
er lief, so schnell er immer konnte
sie nahm, soviel sie immer bekommen konnte
ich helfe ihm, sooft ich immer kann
4.
Grammatik: in Sätzen mit imperativischem Sinn
a)
mit konzessivem Nebensinn
Grammatik: häufig in Verbindung mit »nur«
Beispiele:
mögen sie (nur) immer schimpfen!
soll er (nur) immer mit seinem Besitz prahlen!
lass ihn nur immer herein!
b)
Grammatik: häufig in Verbindung mit »schon«, »inzwischen«
Beispiele:
er kommt anscheinend nicht mehr, fangen wir (schon) immer an!
ich denke, wir fangen immer an
gehen wir (inzwischen) immer los!
wir sollten immer losgehen
hör mal, Johnnie, willst du nicht immer schon packen?[Baum Kristall 262]
Dieses Wort ist Teil des Wortschatzes für das Goethe-Zertifikat A1.
Etymologisches Wörterbuch (Wolfgang Pfeifer)
Etymologie
Etymologisches Wörterbuch (Wolfgang Pfeifer)
immer Adv. ‘ständig, stets’, ahd.iomēr (8. Jh.), mhd.iemer, immer, asächs.iemar, mnd.immer, iemmer, mnl.emmer, immer, nl.immer ist aus dem unter je (s. d.) behandelten Adverb und den Formen von mehr (s. d.) zusammengerückt. Das Zeitadverb bezieht sich anfangs ausschließlich oder vornehmlich auf die Zukunft, d. h. ahd.mhd. ‘künftig irgendeinmal, jemals, jetzt und künftig’. Doch bahnt...Mehr
immer Adv. ‘ständig, stets’, ahd.iomēr (8. Jh.), mhd.iemer, immer, asächs.iemar, mnd.immer, iemmer, mnl.emmer, immer, nl.immer ist aus dem unter je (s. d.) behandelten Adverb und den Formen von mehr (s. d.) zusammengerückt. Das Zeitadverb bezieht sich anfangs ausschließlich oder vornehmlich auf die Zukunft, d. h. ahd.mhd. ‘künftig irgendeinmal, jemals, jetzt und künftig’. Doch bahnt sich die Erweiterung der Bedeutung zu ‘jederzeit’ (einschl. Gegenwart und Vergangenheit) bereits im Mhd. an. Dazu die Zusammenrückungen immerdar, immerfort, immerhin, immerzu.
...Weniger
www.openthesaurus.de (07/2025)
Bedeutungsverwandte Ausdrücke
www.openthesaurus.de (07/2025)
allzeit · fortwährend · immer · immer nur · immer und ewig · immerdar · immerwährend · in Zeit und Ewigkeit · konstant · kontinuierlich · persistent · stetig · stets · stets und ständig · ständig●alleweilsüddt. · allweilsüddt. · perpetuellgeh. · perpetuierlichgeh.
Oberbegriffe in einer bestimmten Häufigkeit · in einer bestimmten Regelmäßigkeit · mehr oder weniger oft · mehr oder weniger regelmäßig · mit einer bestimmten Auftretenswahrscheinlichkeit · mit einer bestimmten Häufigkeit · so und so oft
Assoziationen ausdauernd · beharrlich · hartnäckig · konstant · standhaft · steif und fest (behaupten) · unablässig · unbeirrt · unentwegt · unermüdlich · unverdrossen●insistentgeh., selten, bildungssprachlich
(all) sein Lebtag · durch sein ganzes Leben · lebenslang · sein Leben lang · sein ganzes Leben hindurch · solang(e) er lebt(e) · zeit seines Lebens · zeitlebens
(im) Stakkato reden · reden ohne abzusetzen · reden wie ein Maschinengewehr ● ohne Punkt und Komma reden ugs. · reden ohne Luft zu holen ugs. · reden wie ein Wasserfallugs.
Dauer... · andauernd · anhaltend · bleibend · dauer... · dauerhaft · langwierig●chronischfachspr.
Tag für Tag · alle Tage · an allen Tagen · an jedem Tag · es vergeht kein Tag ohne (dass) · jeden Tag · kein Tag vergeht ohne (dass) · pro Tag · tagein, tagaus · tagtäglich · täglich
Jahr für Jahr · alle Jahre wieder · alljährlich · annual · ein ums andere Jahr · in jedem Jahr (wieder) · jahrein, jahraus · jedes Jahr · jedes Jahr wieder · jährlich · jährlich erneut · jährlich wiederkehrend · pro Jahr ●per annumfachspr., lat.
egal wann · ganz gleich, wann · jederzeit · wann auch immer · wann immer · zu jedem beliebigen Zeitpunkt
andauernd · endlos · immer wieder · immer wiederkehrend · in regelmäßigen Abständen · wieder und wieder · wiederholt · wiederholte Male · zum wiederholten Male ● Und täglich grüßt das Murmeltier.Spruch · (es ist) immer wieder das Gleiche ugs. · aber und abermals geh., veraltet · am laufenden Meter ugs. · dauerndugs. · ein(s) ums andere Mal ugs. · ewigugs. · immer und ewig ugs. · immer wieder das gleiche Spiel ugs. · laufendugs. · ständigugs.
(es ist) kein Ende abzusehen · (es ist) kein Ende in Sicht · ...marathon · Dauer... · Endlos... · Marathon... · endlos (lange) · endlos weiter · immer weiter · nicht abreißen · nicht enden wollen · nicht enden wollend · unendlich lange · überlang●ad infinitumlat. · (sich) fortzeugen(d)geh. · bis zum Gehtnichtmehrugs.
24 Stunden am Tag · 24/7 · Tag und Nacht · durchgehend · permanent · rund um die Uhr · zu jeder Tag- und Nachtzeit
beständig · die ganze Zeit · die ganze Zeit hindurch · die ganze Zeit über · durchgehend · durchgängig · fortgesetzt · fortlaufend · fortwährend · immerfort · immerzu · in einem durch · in einem fort · ohne Unterbrechung · pausenlos · permanent · unablässig · unterbrechungsfrei · ununterbrochen · unverwandt●andauerndugs. · dauerndugs. · egal (+ Adjektiv)ugs., berlinerisch · egalwegugs. · perennierendgeh., lat. · ständigugs.
(schon) seit Jahren · (schon) seit Jahrzehnten · (schon) seit Monaten · (schon) seit Wochen · (schon) seit langen Jahren · bereits lange · ewig lange · geraume Zeit · lange · lange Jahre · lange Zeit · längere Zeit · längst · schon lange · schon längst · seit Ewigkeiten · seit Langem · seit Längerem · seit einer ganzen Weile · seit einiger Zeit · seit ewigen Jahren · seit geraumer Zeit · seit längerer Zeit · seit vielen Jahren ● seit einer Ewigkeit fig. · (eine) ziemliche Zeit ugs. · (eine) ziemliche Zeitlang ugs. · (schon) eine Weile ugs. · (seit) ewig und drei Tage(n)ugs. · eine ganze Weile ugs. · ewigugs., fig. · lange schon geh. · schon eine (halbe) Ewigkeit ugs. · schon eine ganze Zeit ugs. · seit ewigen Zeiten ugs.
beständig · feststehend · gleichbleibend · immer wieder · inert · invariabel · jederzeit · konstant · kontinuierlich · stetig · ständig · unabänderlich · unveränderlich●invariantfachspr.
ein für alle Mal · für Zeit und Ewigkeit · für alle Zeit(en) · für alle Zukunft · für immer · für immer und ewig● auf ewig geh. · auf immer und ewig (regional oder archaisierend)geh.
beharrlich · dauernd · laufend · nonstop · ohne Unterbrechung · ohne Unterlass · ohne abzusetzen · pausenlos · ständig · unablässig · unaufhörlich · unausgesetzt · unentwegt · ununterbrochen●am laufenden Bandugs. · in einer Tourugs.
das ganze Jahr hindurch · das ganze Jahr über · ganzjährig · sommers wie winters ● das ganze Jahr ugs. · perennialfachspr.
Antonyme nie
immer (mehr) · jedes Mal (mehr) · mit jedem Mal · von Mal zu Mal (mehr) · zusehends
Assoziationen (eine) Welle von · ...welle · Tendenz steigend (als Nachsatz) · gehäuft · immer häufiger · immer mehr · immer öfter · in zunehmendem Maß · mehr und mehr · mit steigender Tendenz · vermehrt · verstärkt · zunehmend● es mehren sich die ...fachspr., mediensprachlich, floskelhaft
mit jedem Tag · von Tag zu Tag (+ Komparativ, z.B. 'besser') · zusehends● jeden Tag (+ Komparativ)ugs.
(Komparativ) und (Komparativ) · immer (+ Komparativ) · mit wachsendem (...) · mit wachsender (...) · mit zunehmendem (...) · mit zunehmender (...) · von Mal zu Mal (+ Komparativ) · zunehmend (+ Komparativ)
immer · jedes Mal (wieder) · jeweils
Assoziationen andauernd · endlos · immer wieder · immer wiederkehrend · in regelmäßigen Abständen · wieder und wieder · wiederholt · wiederholte Male · zum wiederholten Male ● Und täglich grüßt das Murmeltier.Spruch · (es ist) immer wieder das Gleiche ugs. · aber und abermals geh., veraltet · am laufenden Meter ugs. · dauerndugs. · ein(s) ums andere Mal ugs. · ewigugs. · immer und ewig ugs. · immer wieder das gleiche Spiel ugs. · laufendugs. · ständigugs.
periodisch · regelmäßig · turnusmäßig · wiederholend · wiederkehrend · zyklisch● rezidivierend medizinisch
entsprechend (Adjektiv) · jeweilig (Adj.)●respektiv (Adj.)regional · je eigen geh.
immer wenn · jedes Mal wenn · sooft · wann (auch) immer
für · für jede(n/s) · je · pro · zu je ●perkaufmännisch, Jargon · das (Kilo / Pfund / Stück) (zu)ugs., Jargon · die (Woche / Minute / Kiste / Flasche ...)ugs., Jargon · àfachspr., kaufmännisch, franz.
(Komparativ) und (Komparativ) · immer (+ Komparativ) · mit wachsendem (...) · mit wachsender (...) · mit zunehmendem (...) · mit zunehmender (...) · von Mal zu Mal (+ Komparativ) · zunehmend (+ Komparativ)
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187702 | https://public.com/learn/debt-to-equity-ratio | Bond Account
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Debt-to-equity Ratio: Formula, Calculation with Example
When evaluating a company’s financial health, one of the first questions investors often ask is: How much debt is too much? That’s where the debt-to-equity (D/E) ratio comes into play. This simple yet powerful metric may help you understand whether a company is funding its operations through debt or equity—and what that means for its financial stability.
In this guide, we’ll break down what the D/E ratio is, how to calculate it, and how investors can interpret it to assess a company’s financial health.
Table of Contents
Key Takeaways
Debt-to-equity ratio shows financial health: It compares a company’s total debt to shareholder equity.
It may offer a window into how a company funds its operations and financial risk profile.
There’s no one-size-fits-all “good” or “bad” ratio—industry norms, business models, and financial strategies all matter.
The calculation formula: D/E Ratio = Total liabilities / Shareholders’ equity
What is the debt-to-equity ratio?
The debt-to-equity ratio is a financial metric that compares a company’s total liabilities to its shareholder equity. It’s calculated by dividing the company’s total liabilities by its total shareholder equity:
The formula for calculating debt-to-equity ratio:
Debt-to-equity ratio = Total liabilities / Shareholders’ equity
Let’s break this down:
Why does the debt-to-equity ratio matter?
The debt-to-equity ratio may offer a snapshot of a company’s financial leverage. A high ratio could suggest that a company is financing a significant portion of its operations through debt. A lower ratio might imply that the company is using more equity to support its activities.
This ratio can help you gauge how risky a company might be when it comes to taking on additional debt. However, context is crucial—what may be considered “high” for one industry could be normal in another.
Example: Comparing two companies
Let’s say you’re comparing two U.S.-based firms:
Company ABC has a D/E ratio of 0.5, which may suggest it’s less reliant on borrowed funds. Company XYZ, with a D/E ratio of 3.0, may be using more debt to finance its growth or operations.
But this number alone doesn’t tell the whole story. For example, some capital-intensive sectors like utilities or telecom may naturally operate with higher D/E ratios.
Debt-to-equity ratio: What different levels may suggest
When you look at a company’s debt-to-equity (D/E) ratio, it helps to think of it as a signal of how the business balances what it borrows versus what it owns. Here’s how different levels of this ratio may be interpreted, depending on the context:
Higher D/E ratio (Above 2.0)
A D/E ratio above 2.0 may indicate that the company relies more heavily on debt financing. This kind of financial structure is often seen in capital-intensive industries—such as utilities or telecom—where borrowing is commonly used to fund infrastructure and long-term projects. While this level of debt can support expansion, it may also introduce more financial obligations.
Moderate D/E ratio (Between 0.5 and 2.0)
This range often reflects a balanced approach to financing, where a company may be using both equity and debt to support its operations. For some well-established firms, a ratio in this range may indicate flexibility in funding strategies while still maintaining manageable debt levels.
Lower D/E ratio (Below 0.5)
A lower D/E ratio suggests the company may be leaning more on equity than debt. This could be a sign of a conservative financial strategy, with limited borrowing. While this may help reduce financial risk, it might also mean fewer opportunities to leverage borrowing for growth—especially when interest rates are low.
Interpreting the debt-to-equity ratio in different industries
The D/E ratio varies significantly across different industries due to differences in capital requirements and business models. Here are some examples of average D/E ratios across various industries:
Industry | Average D/E Ratio
Tech | 0.1 – 0.5
Utilities | 1.0 – 2.0
Consumer Goods | 0.4 – 1.0
Financial Services | 1.0 – 3.0
Real Estate | 0.5 – 2.5
In capital-heavy industries like utilities, higher D/E ratios are common due to the large infrastructure investments required. Meanwhile, software or tech companies might operate with very little debt.
Note: Industry context matters. A D/E ratio that seems high in one sector may be the norm in another.
How D/E ratio may help investors
The D/E ratio tells investors how a company manages its funds. It helps answer questions like:
Each industry has different norms, and investors should compare companies within the same sector rather than applying a single standard across all industries.
What a high debt-to-equity ratio may mean
A higher D/E ratio may signal that a company is taking on more debt relative to its equity. This could raise some potential concerns:
However, some firms strategically use debt to fuel expansion, pursue acquisitions, or fund operations more efficiently.
What a low debt-to-equity ratio may indicate
On the other end, a lower ratio may suggest a company is more conservative in its financial approach:
Again, context matters. A low D/E ratio doesn’t necessarily mean a company is better—it depends on the company’s strategy and sector.
Limitations of the D/E ratio
While it’s a useful tool, the D/E ratio has its limits:
Using the D/E ratio as part of a broader analysis—along with cash flow, profitability, revenue trends, and industry outlook—may provide more meaningful insights.
Conclusion
The debt-to-equity ratio can offer helpful insight into how a company manages its financial structure, especially when used alongside other metrics like earnings, cash flow, and industry trends. While it won’t give you all the answers on its own, it may help you ask better questions when reviewing a company’s balance sheet or financial reports.
If you’re just starting out or prefer learning with data in hand, the Public app makes it simpler to explore these kinds of metrics in real time. You can view 5-year debt/equity ratios, P/E ratios, earnings reports directly in our app.
Signup on the Public app to start reviewing company fundamentals and build a multi-asset portfolio that includes everything from stocks and options to bonds, crypto, and a High-Yield Cash Account.
The above content provided and paid for by Public and is for general informational purposes only. It is not intended to constitute investment advice or any other kind of professional advice and should not be relied upon as such. Before taking action based on any such information, we encourage you to consult with the appropriate professionals. We do not endorse any third parties referenced within the article. Market and economic views are subject to change without notice and may be untimely when presented here. Do not infer or assume that any securities, sectors or markets described in this article were or will be profitable. Past performance is no guarantee of future results. There is a possibility of loss. Historical or hypothetical performance results are presented for illustrative purposes only.
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187703 | https://line.facebook18.com/post/15635.html | line和rope，line和rope的区别 - Line技术攻略 - LINE技术攻略博客
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line和rope，line和rope的区别
原标题：line和rope，line和rope的区别
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本文目录一览：
1、绳子的英文怎么说?
2、line和rope的区别是什么
3、绳子的英语怎么说?
4、rope和line的区别是什么
5、hitch是什么意思(rope同类词是什么)
6、line是什么?
绳子的英文怎么说?
“绳子”的英文翻译有“rope； string； cord；cord，n.（细）绳；灯心绒裤；vt.用绳子捆绑；堆积（柴薪）。The door had been tied shut with a length of nylon cord.门用一段尼龙绳拴住了。
绳子 Rope是一个英文单词，作为名词时意为“绳”，也可指“绳索”。这个词在日常生活中非常常见，经常用于捆绑、吊装、拖曳等各种场合。以下是关于rope的详细解释：详细解释： 基本定义：Rope的基本含义是“绳子”，这是一种由多股纤维、金属或其他材料制成，用于捆绑、牵引或作为悬挂物体的带状物品。
rope：也有粗绳的意思，但是更多的指绳索，线缆 例句：He coiled the rope on the deck. 他在甲板上把绳子绕好。
“绳子”的英文是rope。基本含义：rope在英语中通常用来表示绳子，这是一种长条状物，由纤维、纱线或金属丝等材料制成。用途广泛：绳子具有一定的柔韧性和强度，可以用于捆绑、悬挂或牵引等多种用途。在工业、建筑、航海、体育等领域都有广泛的应用。
“绳子”的英文翻译主要有三个词汇：rope、string 和 cord。rope：这是一个非常通用的词汇，用于指代各种类型和粗细的绳子。它可以用于各种场景，如攀岩、航海、建筑等。string：这个词除了可以表示细绳或带子外，在计算机科学中还常用于指代字符串。
line和rope的区别是什么
1、line和rope都是与“线”相关的词汇，但它们在具体使用上有所区别。rope专指那些用于捆绑大件物体的粗壮且坚固的绳子，通常是由棉、毛、麻、金属或其他材料制成。而line则是一个更为通用的词汇，它不仅仅指绳索，还指任何形式的线，比如电话线、电线等。line作为动词时，它可以表示排成一行、用线条标记、使东西起皱纹等含义。
2、rope释义：n.绳，绳索 vt.捆，绑 vi.拧成绳状 n.（Rope）人名；（英）罗普；（芬）罗佩 例句：Anaughtykidcutofftherope.一个捣蛋鬼割断了绳子。
3、rope ： 指用于捆绑大物件的粗壮而坚固的绳子，一般用绵、毛、麻、金属或其它材料制成。line ： 普通用词，含义广泛，指任何一种线，常作引申用。
4、rope， wire， thread， string， line， cord 这组词都有“线”的意思，其区别是：rope 指用于捆绑大物件的粗壮而坚固的绳子，一般用绵、毛、麻、金属或其它材料制成。wire 专指用金属制成的线。thread 普通用词，指用棉、毛、丝或纤维等纹成的很细的线，通常用于缝纫、纺织等。
绳子的英语怎么说?
1、ropeline和rope：英/rp/ 美/rop/。n.绳line和rope；绳索；粗绳；线缆；（拳击或摔跤场四周line和rope的）围绳；串在一起line和rope的相似的东西。vt.套；用绳子捆（或绑、系）；捆紧；用套索抓捕（动物）。第三人称单数： ropes复数： ropes现在分词： roping过去式： roped过去分词： roped。
2、绳子的英文是：rope。在英语中line和rope，rope一词通常用来表示绳子，这是一种由纤维、纱线或金属丝等材料制成的长条状物，具有一定的柔韧性和强度，可以用于捆绑、悬挂或牵引等多种用途。
3、ropes的意思是绳子或绳索。以下是详细解释：Rope是一个英语词汇，可以作为名词或动词使用。作为名词时，它的主要意思是绳子或绳索，可以用于表示各种材质制成的，用于捆绑、牵引或悬挂物体的细长条状物。在日常生活中，人们经常用绳子来进行各种活动和任务，如户外运动、园艺工作或者日常家务等。
rope和line的区别是什么
1、line和rope都是与“线”相关的词汇，但它们在具体使用上有所区别。rope专指那些用于捆绑大件物体的粗壮且坚固的绳子，通常是由棉、毛、麻、金属或其他材料制成。而line则是一个更为通用的词汇，它不仅仅指绳索，还指任何形式的线，比如电话线、电线等。line作为动词时，它可以表示排成一行、用线条标记、使东西起皱纹等含义。
2、rope释义：n.绳，绳索 vt.捆，绑 vi.拧成绳状 n.（Rope）人名；（英）罗普；（芬）罗佩 例句：Anaughtykidcutofftherope.一个捣蛋鬼割断了绳子。
3、rope ： 指用于捆绑大物件的粗壮而坚固的绳子，一般用绵、毛、麻、金属或其它材料制成。line ： 普通用词，含义广泛，指任何一种线，常作引申用。
4、rope的名词同类词： cord：细绳，通常用于捆绑或悬挂物品。 line：线，可以指钓鱼线、缝纫线或用于其他目的的长条状物。 cable：电缆，通常用于传输电力或信号，但也可以指较粗的绳索。 strand：一股，也可以指由多股细线组成的粗绳。
5、rope， wire， thread， string， line， cord 这组词都有“线”的意思，其区别是：rope 指用于捆绑大物件的粗壮而坚固的绳子，一般用绵、毛、麻、金属或其它材料制成。wire 专指用金属制成的线。thread 普通用词，指用棉、毛、丝或纤维等纹成的很细的线，通常用于缝纫、纺织等。
hitch是什么意思(rope同类词是什么)
hitchline和rope的意思是“系line和rope，拴，套牢”，ropeline和rope的名词同类词包括cord， line， cable， strand等，动词同类词除了hitch还包括tie， bind， moor等。hitchline和rope： 作为动词，hitch的基本含义是“系，拴，套牢”。例如，我们可以用hitch来描述将马拴在树上的动作。
rope的名词同类词包括cord，line，cable，strand等，动词同类词包括tie，bind，moor等。
障碍；故障 A technical hitch prevented the lights from working.电灯因技术故障而熄灭了。
绳头结（Ropes end hitch）：又称艾曼斯结，用于绳索与安全带连接处或保护绳的支点上。1 帐篷固定结（Tent stake hitch）：用于固定帐篷。1 平结（Square knot）：又称连结结、本结、陀螺结，用于直径相同绳索之间的连接。
“hitch”一词还有其他含义，例如“系住，勾住”或“猛拉”。例如，“He hitched the rope to the post”意为“他把绳子系在了柱子上”。此外，“hitch”还用作名词，表示“故障，挫折”。如“the car had a hitch on the highway”意为“汽车在高速公路上出了故障”。
line是什么?
line的基本意思是“用线表示，画线于”，引申可指“排成一行”“排队，排齐”。聊天工具 LINE是韩国互联网集团NHN的日本子公司NHN Japan推出的一款即时通讯软件，line app在日韩、中国台湾地区用的比较多。饭圈术语 在饭圈，Line是一种划分方式。常见85line、93line，就是一个年龄线的几个爱豆。
输入设备的区别：LINE为线性输入，简单点可以理解为连接电吉他、电子琴、合成器等外界设备输入音频。而MIC和MIC+48V则为麦克风输入，两者在于是否增加48伏电压输入，MIC输入通过麦克风捕捉声响形成电流输入音频。
“line”这个词在韩圈中常常被使用，它的意思与聊天工具线上通讯软件名字相同，但在韩圈中，它更多指的是一种私下交流的方式。很多韩国偶像组合会在“line”上建立专门的社交圈，与粉丝们进行互动。有的“line”群是私密的，只有特定的成员才能进入，而且必须遵守群规。
含义不同，用法不同。含义不同：line指的是一条直线或一行文字、数字等，强调线性排列。而row指的是一排或一行物品、人或其他事物，强调排列的顺序。用法不同：line用于描述线性排列的事物，如线条、排队、电话线等。而row用于描述排列成一行的事物，如一排椅子、一行文字、一排人等。
Line是一款由韩国Naver集团开发的即时通讯应用，功能与国内的微信类似，在韩国、台湾、日本以及部分东南亚地区广受欢迎。国内目前不能直接使用Line，但可以通过特定方式实现使用。
in the line：侧重于指现实中的队伍。on the line：侧重于指虚拟中的网络。用法不同 in the line： in表示在某范围之内，是其中的一部分。line的基本意思是“线，线条”，指在物体表面上留下的长的痕迹，可以是直的，也可以是弯的。
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What We DidWe conducted an international collaborative study of the RBA for PSP in shellfish through the Association of Official Analytical Chemists (AOAC), an international professional organization that certifies testing methods for food, agriculture, and pharmaceutical products. The study, which included nine laboratories from five nations, assessed the reproducibility of test results between laboratories and compared the performance of the RBA with the existing regulatory mouse bioassay. The RBA was deemed to perform as well as the existing method, and was accepted by the AOAC as an Official Method of Analysis.
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| 1-H-1 basic n-XS summary XS graphs --- element | 1-hydrogen-1 --- Atomic Mass: 1.0078250 +- 0.0000000 amu Excess Mass: 7288.969 +- 0.001 keV Binding Energy: 0.000 +- 0.000 keV "The 1995 update to the atomic mass evaluation" by G.Audi and A.H.Wapstra, Nuclear Physics A595 vol. 4 p.409-480, December 25, 1995. --- Atomic Percent Abundance: 99.985% Spin: 1/2+ Stable Isotope Possible parent nuclides: Beta from n-1 2 Alpha from B-9 EC + 2 Alpha from C-9 R.R.Kinsey, et al.,The NUDAT/PCNUDAT Program for Nuclear Data,paper submitted to the 9 th International Symposium of Capture-Gamma_raySpectroscopy and Related Topics, Budapest, Hungary, Octover 1996.Data extracted from NUDAT database (Jan. 14/1999) --- Magnetic Dipole Moments and Electric Quadrupole Moments | Ex(kev) | T1/2 | Spin | | m(nm) | | Q(b) | | Ref. Std. | Method | Reference | | 0 | stable | 1/2+ | | +2.79284734(3) d | | | | | "M/N,R" | RMP 72 351 (00) | | | | | | | | | | | | | N. J. Stone, Table of Nuclear Magnetic Dipole and Electric Quadrupole Moments, to be published. 2000 Courtesy of T. Burrows at BNL/NNDC. | |
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Bacteria in food
Milk from a healthy cow initially contains very few bacteria, which primarily come from the skin of the cow and the procedures for handling the milk. Milk is an excellent growth medium for numerous bacteria, and the bacteria can increase rapidly in numbers unless the milk is properly processed. Bacterial growth can spoil the milk or even pose a serious health hazard if pathogenic bacteria are present. Diseases that can be transmitted from an infected cow include tuberculosis (Mycobacterium tuberculosis), undulant fever (Brucella abortus), and Q fever (Coxiella burnetii). In addition, typhoid fever (Salmonella typhi) can be transmitted through milk from an infected milk handler. Pasteurization procedures increase the temperature of the milk to 63 °C (145 °F) for 30 minutes or to 71 °C (160 °F) for 15 seconds, which kills any of the pathogenic bacteria that might be present, although these procedures do not kill all microorganisms.
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Certain bacteria convert milk into useful dairy products, such as buttermilk, yogurt, and cheese. Commercially cultured buttermilk is prepared from milk inoculated with a starter culture of Lactococcus (usually L. lactis or L. lactis cremoris). Yogurt and other fermented milk products are produced in a similar manner using different cultures of bacteria. Many cheeses are likewise made through the action of bacteria. Growth in milk of an acid-producing bacterium such as L. lactis causes the casein to precipitate as curd. Following the removal of moisture and the addition of salt, the curd is allowed to ripen through the action of other microorganisms. Different bacteria impart different flavors and characteristics to foods; for example, the mixture of Lactobacillus casei, Streptococcus thermophilus, and Propionibacterium shermanii is responsible for the ripening of Swiss cheese and the production of its characteristic taste and large gas bubbles. In addition, Brevibacterium linens is responsible for the flavor of Limburger cheese, and molds (Penicillium species) are used in the manufacture of Roquefort and Camembert cheeses. Other types of bacteria have long been used in the preparation and preservation of various foods produced through bacterial fermentation, including pickled products, sauerkraut, and olives.
The toxins of many pathogenic bacteria that are transmitted in foods can cause food poisoning when ingested. These include a toxin produced by Staphylococcus aureus, which causes a rapid, severe, but limited gastrointestinal distress, or the toxin of Clostridium botulinum, which is often lethal. Production of botulism toxin can occur in canned nonacidic foods that have been incompletely cooked before sealing. C. botulinum forms heat-resistant spores that can germinate into vegetative bacterial cells that thrive in the anaerobic environment, which is conducive to the production of their extremely potent toxin. Other food-borne infections are actually transmitted from an infected food handler, including typhoid fever, salmonellosis (Salmonella species), and shigellosis (Shigella dysenteriae).
Substances contributing to the virulence of pathogenic bacteria
| substance | action |
--- |
| hyaluronidase | increases permeability of tissue spaces to bacterial cells |
| coagulase | increases resistance of bacteria to phagocytosis (engulfment by defense cells, or phagocytes) |
| hemolysins | destroy red blood cells |
| collagenase | dissolves collagen, a connective tissue protein |
| leukocidin | kills white blood cells (specifically leukocytes) and hence decreases phagocytic action |
| exotoxins and endotoxins | interfere with normal metabolic processes | |
187707 | https://cs.stackexchange.com/questions/168353/detect-if-an-interval-is-fully-covered-by-union-of-previous-intervals-in-sequenc | data structures - Detect if an interval is fully covered by union of previous intervals in sequence - Computer Science Stack Exchange
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Detect if an interval is fully covered by union of previous intervals in sequence
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Given a sequence of intervals I 1,I 2,…I 1,I 2,…, is there an efficient way to detect whether some interval I i I i is completely contained in the union of the preceding intervals I 1,…,I i−1 I 1,…,I i−1?
For example, in the sequence (0,5),(5,10),(2,6)(0,5),(5,10),(2,6), the interval (2,6)(2,6) is covered by the previous intervals, which cover the full range (0,10)(0,10).
In contrast, in the sequence (0,5),(6,10),(2,6)(0,5),(6,10),(2,6), the interval (2,6)(2,6) is not covered by the previous intervals, because the sub-interval (5,6)(5,6) is not covered.
My initial idea was to use an interval or segment tree to maintain the set of previous intervals, and check before adding each interval whether it is fully contained in the set. However, most algorithms construct the tree efficiently by first pre-sorting the intervals. In contrast, it seems like I need an on-line algorithm that efficiently merges intervals. I also don't need to maintain the original intervals (they can be merged), and lookup speed is not important.
The application here is exception tables in bytecode. I want to determine whether a given bytecode range is already covered by the ranges that precede it.
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edited Jun 1, 2024 at 14:24
MattDs17MattDs17
asked May 31, 2024 at 21:56
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1 Can you precisely define what you mean by "contained in the preceding intervals".Ainsley H. –Ainsley H. 2024-06-01 13:41:22 +00:00 Commented Jun 1, 2024 at 13:41
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The intersection of a sequence of intervals I 1,...,I p I 1,...,I p is
[max i≤p s(I i),min i≤p f(I i)],[max i≤p s(I i),min i≤p f(I i)],
where s s and f f are the start and finish times is an interval.
From there it should be simple to find out if the next interval is contained in the intersection.
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answered Jun 1, 2024 at 7:02
Ainsley H.Ainsley H.
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Unfortunately, there is no guarantee that the intervals are contiguous. For example, the sequence could be (0,5),(10,15),(6,9)(0,5),(10,15),(6,9) and (6,9)(6,9) would not be contained in the previous intervals.MattDs17 –MattDs17 2024-06-01 11:33:32 +00:00 Commented Jun 1, 2024 at 11:33
I might have understood what you meant by "contained in the preceding intervals".Ainsley H. –Ainsley H. 2024-06-01 13:39:50 +00:00 Commented Jun 1, 2024 at 13:39
In the example you give, the "algorithm" provided in the answer correctly says no. (... Because no interval is contained in the negative interval.)Ainsley H. –Ainsley H. 2024-06-01 13:40:27 +00:00 Commented Jun 1, 2024 at 13:40
1 Sorry, I misread your solution and thought it took the lowest start and highest end value (and not vice versa). I have clarified the question. I am not looking for the intersection of intervals but the union.MattDs17 –MattDs17 2024-06-01 14:26:57 +00:00 Commented Jun 1, 2024 at 14:26
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I think I have worked out a solution that takes O(n log n)O(n logn) time. It relies on a regular interval tree which has O(log n+m)O(logn+m) query (for m m overlapping intervals) and O(log n)O(logn) insertion and deletion. (Please correct me if you see an error in the algorithm or analysis.)
The algorithm works by iteratively constructing an interval tree and merging overlapping intervals at each step. For each new interval, we find the intervals it overlaps with. If those intervals cover the new interval, we're done; otherwise, we remove the overlaps from the tree and insert one large "merged" interval, then continue:
T <- empty tree
for interval in I:
overlaps <- query_overlaps(T, interval)
if sorted(overlaps) is not contiguous:
return FAIL // there's a hole
for overlap in overlaps:
remove(T, overlap)
new_start <- min(i.start for i in [i] + overlaps)
new_end <- max(i.end for i in [i] + overlaps)
merged_interval <- (new_start, new_end)
insert(T, merged_interval)
In each iteration of the loop, we perform:
O(log n+m)O(logn+m) work to query the overlapping intervals (for n n total intervals and m m values returned)
O(m log m+m)=O(m log m)O(m logm+m)=O(m logm) work to sort the m m overlapping intervals and check whether they're contiguous (because we merge overlapping intervals, these intervals are guaranteed to be disjoint)
O(m log n)O(m logn) work to remove the overlapping intervals
O(m)O(m) work to iterate the overlaps and find the merged interval
O(log n)O(logn) work to insert the new interval
We can show that ∑|I|i=1 m∑i=1|I|m is O(n)O(n) as follows. Suppose we represent the merges that happen as a tree where the leaf nodes are input intervals and internal nodes represent merged intervals. For each internal node in the tree, its number of children is m+1 m+1 for some step in the algorithm. The number of edges in this tree is ∑|I|i=1 m+1∑i=1|I|m+1. Each internal node has at least 2 children (you cannot merge a single interval), so the tree has at most n+n 2+n 4+…=O(n)n+n 2+n 4+…=O(n) total nodes and consequently ∑|I|i=1 m+1=O(n)∑i=1|I|m+1=O(n) edges.
Therefore, the amount of work done across all iterations is
∑i=1|I|O(log n+m)+O(m log m)+O(m log n)+O(m)+O(log n)=O(n log n).∑i=1|I|O(logn+m)+O(m logm)+O(m logn)+O(m)+O(logn)=O(n logn).
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answered Jun 1, 2024 at 23:05
MattDs17MattDs17
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Note, that
I i⊆⋃k<i I k⟺I i∩⋃k<i I k¯¯¯¯¯¯¯¯¯¯¯=∅⟺(((I i∖I 1)∖I 2)...∖I i−1)=∅I i⊆⋃k<i I k⟺I i∩⋃k<i I k¯=∅⟺(((I i∖I 1)∖I 2)...∖I i−1)=∅
You should be able to decide this in linear time by subtracting I 1,I 2,...,I i−1 I 1,I 2,...,I i−1 from I i I i and checking if the resulting interval is empty.
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edited Jun 1, 2024 at 15:25
answered Jun 1, 2024 at 15:01
KnoggerKnogger
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Yes, but based on the request for a data structure, I imagine he is looking for an algorithm that is faster than linear time.D.W. –D.W.♦ 2024-06-01 15:45:29 +00:00 Commented Jun 1, 2024 at 15:45
@D.W. I'm not sure this problem can be solved in sub-linear time (but I could be wrong there 😅). If we assume that I 1,I 2,...,I i−1 I 1,I 2,...,I i−1 are disjoint, then the problem essentially boils down to finding an I k,k<i I k,k<i that entirely contains I i I i. The search could be improved by sorting the intervals, but it seems like OP is looking for a solution that doesn't involve sorting.Knogger –Knogger 2024-06-01 16:22:35 +00:00 Commented Jun 1, 2024 at 16:22
I interpret their question as asking for a data structure that lets us insert an interval and query whether an interval is covered by the union of the existing intervals, and presumably we'd want both in sub-linear time. I don't know whether it is possible either, but it doesn't seem unreasonable to hope that such a data structure might exist.D.W. –D.W.♦ 2024-06-01 16:36:27 +00:00 Commented Jun 1, 2024 at 16:36
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187708 | http://faculty.cooper.edu/smyth/TC2/calculus/Worksheets/LMworksheets.pdf | MA-113 Constrained optimization exploratory activity V0.2: 2020-05-26 Part 1: Preparation Introduction: This activity will guide you through a graphical exploration of the method of Lagrange multipliers for solving constrained optimization problems. The cen-tral ideas will be illustrated with an example similar to the following exercise.
() Find the maximum value of f(x, y, z) = x + y −z 3 subject to the pair of constraints ( x + 2y + 3z = 2 x2 + y2 + z2 = 1 .
The preparatory questions below are to be completed prior to the in-class portion of the activity.
Preparatory questions: (1) Describe the family of level surfaces of the objective function f.
(2) Does the objective function f obtain a maximum value in xyz-space? Give an explanation which relates your answer to the family of level surfaces of f.
(3) Describe some subsets of xyz-space on which f does attain a maximum value. (Try to produce a variety of examples with qualitative differences.) (4) Describe some subsets of xyz-space on which f does not attain a max-imum value. (Again try to produce a variety of examples with qualitative differences.) (5) Suppose r(t) = (x(t), y(t), z(t)), a < t < b is a regular parameterization of a space curve. Use the chain rule to determine the rate of change of f (with respect to t) as f is evaluated along the curve.
(In other words, compute d dt (f(r(t))).) Express your answer explicitly in terms of the gradient of f and the velocity of r.
(6) From the origin, in which direction should you go to increase the value of f as rapidly as possible? Explain.
(7) Describe the set of all directions in which one can move from the origin so that f will increase. Explain. (Feel free to discuss the subtleties of slightly different interpretations of this exercise with your instructor.) (8) In questions (6) and (7), if the origin is replaced by an arbitrary space point P do the answers change?
(9) Describe the solution set of the first constraint equation x + 2y + 3z = 2 from sample problem () above.
(10) Describe the solution set of the second constraint equation x2+y2+z2 = 1.
(11) In general what types of sets can occur as the intersection of a plane and a sphere? Comment on the problem of optimizing a continuous function in each case.
MA-113 Constrained optimization exploratory activity V0.2: 2020-05-26 Part 2: Graphical Exploration / CAS Computation Prerequisites: (Essential) Understanding of level surfaces, gradients, and directional derivatives. (Desirable) Experience optimizing a function of two variables subject to a constraint condition satisfied by a curve in the function’s domain.
Resource URL: Temporary Introduction: This activity will guide you through a graphical exploration of the method of Lagrange multipliers for solving constrained optimization problems. The cen-tral ideas will be illustrated with an example similar to the following exercise.
() Find the maximum value of f(x, y, z) = x + y −z 3 subject to the pair of constraints x + 2y + 3z = 2 x2 + y2 + z2 = 1 .
Guide / discussion questions: (1) Open the webpage at the URL given above. Expand the Choose objects folder from the control panel in the upper right hand corner. Select the Con-straint 1 (plane) and Constraint 2 (sphere) checkboxes to add the solutions of the constraint equations to the main canvas. The graphic can be repositioned and reoriented using either a mouse (or touchscreen), or the Rotational con-trols. In this example, which type of intersection is formed by the plane and sphere?
(2) Select Constraint curve to add the circle of points simultaneously satisfying both constraint conditions to the graphic. How could you have determined that x + 2y + 3z = 2 and x2 + y2 + z2 = 1 intersect in a circle (rather than just a single point, or the empty set) without using a computer?
(3) Select Test point to add a distinguished point to the constraint circle. You can move this marked point to any position on the constraint circle using the Test point controls. At any given fixed position on the constraint curve, how many directions are possible for motion which does not leave the constraint curve? (More precisely, suppose γ : (a, b) →R3 is a regular parameterization for some portion of the constraint curve including point P, and t0 ∈(a, b) is a parameter value such that γ(t0) = P. How many possible directions could the velocity vector γ′(t0) have?) (4) Select the two Tangent to constraint checkboxes to add visual represen-tations of the two “constraint sensitive” directions at the marked test point.
What information about the objective function (the function we want to max-imize subject to the constraints) would facilitate a determination of whether it will increase or decrease as the test point is moved along the constraint curve?
(5) Select Objective function gradient. What can you say about the directional derivative of f at a point P in a direction which makes an acute angle with ∇f(P)? What can you say about how f(P) will change as long as the direction in which P is moved continues to form an acute angle with ∇f(P)?
(6) Select Objective function monitor. Using the Test point controls move the test point (point “P”) around the constraint circle. View the left pane graphical representation of how the objective function values change (and/or the textual display on the left side of the top banner) to confirm your answers above. State a relationship between ∇f(P) and the tangent line to the constraint curve at P that precludes the possibility of f attaining an extreme value at P subject to the constraints. Recast your observation as a necessary condition for f to attain an extreme value at P (subject to the constraints). Is this condition, in general, a sufficient condition?
(7) Define g(x, y, z) = x + 2y + 3z −2 and h(x, y, z) = x2 + y2 + z2 −1 so that the constraint conditions can be written as g = 0, h = 0. In particular, the constraint surfaces are level surfaces of the functions g and h. Given P on the intersection of the two constraint surfaces, what’s the relationship between ∇g(P) and the constraint plane, and what’s the relationship between ∇h(P) and the constraint sphere?
(8) If f attains an extreme value subject to the constraints g = 0, h = 0 at point P, must ∇f(P) be parallel to ∇g(P) and/or ∇h(P)?
(9) Select Normal to plane, Normal to sphere, and Span of constraint normals.
Recast your necessary condition for f to attain a constrained extreme value at P (from step (6)) as a requirement on ∇f(P) in terms of ∇g(P) and ∇h(P).
Formulate this condition in both geometric and linear algebraic terms.
(10) Record the coordinates (as precisely as you can) for the point where f achieves its maximum subject to the constraints. What is the constrained maximum value of f?
(11) Select Obj fcn level surf’s to display a few select level surfaces of the objective function f. What relationship holds between level surfaces of f and points where constrained extreme values of f occur in this example? (For this step you’re advised to unselect objects to reduce clutter, and unselect the Perspective camera to switch to an orthographic view.) Can you generalize this relationship? (Beware. This is a bit tricky.) CAS Symbolic Computation (1) Open MATLAB and prime the Symbolic Toolbox with a symbol declara-tion.
syms x y z lambda mu (2) Define the objective function, and the functions appearing in the constraint equations.
f = x + y - z/3 >> g = x + 2y + 3z - 2 >> h = x^2 + y^2 + z^2 - 1 (3) Define the Lagrangian for our example problem.
lagrangian = f - lambdag - muh (4) Encode the necessary condition for constrained extrema as a single vector equation. (Don’t confuse the double equal sign == [equality] with the single equal sign = [assignment].) >> eqn = ( gradient(lagrangian, [x y z lambda mu]) == [0; 0; 0; 0; 0] ) (5) Solve the system.
candidates = solve( eqn, [x y z lambda mu] ) (6) Extract the spacial coordinates from the solutions (and peek at numerical approximations).
[ candidates.x candidates.y candidates.z ] >> vpa( ans ) (7) Evaluate f at the candidate points.
subs( f, [x y z], [candidates.x(1) candidates.y(1) candidates.z(1)] ) >> vpa( ans ) >> subs( f, [x y z], [candidates.x(2) candidates.y(2) candidates.z(2)] ) >> vpa( ans ) (8) Where is the constrained maximum of f achieved and what is its value?
(Is your answer here consistent with your answer to question (10) above?) MA-113 Constrained optimization exploratory activity V0.2: 2020-05-26 Part 3: Analytical solution () Find the maximum value of f(x, y, z) = x + y −z 3 subject to the pair of constraints ( x + 2y + 3z = 2 x2 + y2 + z2 = 1 .
(1) Solve the constrained optimization problem () with paper and pencil using the method of Lagrange multipliers.
Organize your work as a clear, linear presentation with explanatory notes for each step.
(2) What if the constraint in () is revised to allow all points of the plane x + 2y + 3z = 2 on or inside the sphere x2 + y2 + z2 = 1?
(3) What if the planar constraint is dropped? Can you then solve the problem without calculus?
(4) Re-solve () without Lagrange multipliers by finding an explicit parame-terization for the constraint curve.
(5) Which of the two solution methods for () (Lagrange multipliers vs. explicit parameterization of the constraint curve) do you think is more flexible in general? Why?
(6) Find the maximum and minimum values of 2x2 + 2y −z subject to the constraints z = 2x2 + 2y2 and 4y2 + z2 = 4.
(7) Can you identify any feature(s) of optimization problem (6) not in evidence in ()? |
187709 | https://www.savemyexams.com/as/chemistry/aqa/16/revision-notes/1-physical-chemistry/1-2-formulae-equations-and-calculations/1-2-6-hydrated-salts/ | Hydrated Salts | AQA AS Chemistry Revision Notes 2015
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Physical Chemistry
9 Topics · 52 Revision Notes
1.
Atomic Structure
1. ##### Fundamental Particles
2. ##### Mass Number & Isotopes
3. ##### Time of Flight Mass Spectrometry
4. ##### Shells & Orbitals
5. ##### Electron Configuration
6. ##### Ionisation Energy
7. ##### Ionisation Energy: Trends & Evidence
2.
Formulae, Equations & Calculations
1. ##### Relative Atomic Mass & Relative Molecular Mass
2. ##### Empirical & Molecular Formula
3. ##### Balanced Equations
4. ##### Reaction Yields
5. ##### Atom Economy
6. ##### Hydrated Salts
3.
The Mole, Avogadro & The Ideal Gas Equation
1. ##### The Mole & the Avogadro Constant
2. ##### Reacting Masses
3. ##### Reacting Volumes
4. ##### The Ideal Gas Equation
4.
Types of Bonding & Properties
1. ##### Changes in State: Energy Changes
2. ##### Ionic Bonding
3. ##### Covalent Bonding
4. ##### Dative Covalent Bonding
5. ##### Dot & Cross Diagrams
6. ##### Metallic Bonding
7. ##### Properties of Ionic Compounds
8. ##### Properties of Covalent Substances
9. ##### Properties of Metallic Substances
10. ##### Effects of Structure & Bonding
5.
Molecules: Shapes & Forces
1. ##### Shapes of Simple Molecules & Ions
2. ##### Bond Polarity
3. ##### Types of Forces Between Molecules
4. ##### Effects of Forces Between Molecules
6.
Energetics
1. ##### Bond Energy
2. ##### Energy Level Diagrams
3. ##### Enthalpy Changes
4. ##### Calorimetry
5. ##### Hess' Law
6. ##### Applications of Hess’s Law
7. ##### Bond Enthalpies
7.
Kinetics
1. ##### Collision Theory
2. ##### Measuring Rates of Reaction
3. ##### Maxwell–Boltzmann Distributions
4. ##### Effect of Temperature on Reaction Rate
5. ##### Effect of Concentration & Pressure
6. ##### Catalysts
8.
Chemical Equilibria, Le Chatelier's Principle & Kc
1. ##### Chemical Equilibria
2. ##### Le Chatelier's principle
3. ##### The Equilibrium Constant, Kc
4. ##### Calculations Involving the Equilibrium Constant
5. ##### Changes Which Affect the Equilibrium
9.
Oxidation, Reduction & Redox Equations
1. ##### Oxidation & Reduction
2. ##### Oxidation States: The Rules
3. ##### Redox Equations
Inorganic Chemistry
3 Topics · 12 Revision Notes
1.
Periodicity
1. ##### Classification of an Element
2. ##### Trends of Period 3 Elements: Atomic Radius
3. ##### Trends of Period 3 Elements: First Ionisation Energy
4. ##### Trends of Period 3 Elements: Melting Point
2.
Group 2, the Alkaline Earth Metals
1. ##### Trends in Group 2: The Alkaline Earth Metals
2. ##### Solubility of Group 2 Compounds: Hydroxides & Sulfates
3. ##### Reactions of Group 2
4. ##### Uses of Group 2 Elements
3.
Group 7 (17), the Halogens
1. ##### Physical Properties of Group 7
2. ##### Chemical Properties of Group 7
3. ##### Testing for Halide Ions
4. ##### Uses & Reactions of Chlorine
Organic Chemistry
7 Topics · 31 Revision Notes
1.
Introduction to Organic Chemistry
1. ##### Functional Groups
2. ##### Types of Formulae
3. ##### Nomenclature
4. ##### Structural Isomerism
5. ##### Stereoisomerism
2.
Alkanes
1. ##### Fractional Distillation of Crude Oil
2. ##### Modification of Alkanes by Cracking
3. ##### Combustion of Alkanes
4. ##### Chlorination of Alkanes
3.
Halogenoalkanes
1. ##### Reactivity of Halogenoalkanes
2. ##### Reactions of Halogenoalkanes
3. ##### Ozone Depletion
4.
Alkenes
1. ##### Structure, Bonding & Reactivity
2. ##### Isomerism in Alkenes
3. ##### Reactions of Alkenes
4. ##### Test for Unsaturation
5. ##### Addition Polymers
6. ##### Solution to Plastic Pollution
5.
Alcohols
1. ##### Classifying Alcohols
2. ##### Alcohol Production
3. ##### Oxidation of Alcohols
4. ##### Combustion of Alcohols
5. ##### Elimination Reactions of Alcohols
6.
Organic Analysis
1. ##### Identification of Functional Groups by Test-Tube Reactions
2. ##### Mass spectrometry
3. ##### Infrared Spectroscopy
7.
Organic Mechanisms
1. ##### Fundamentals of Reaction Mechanisms
2. ##### Chlorination of Alkanes
3. ##### Nucleophilic Substitution
4. ##### Elimination
5. ##### Electrophilic Addition
AS Practical Skills
2 Topics · 8 Revision Notes
1.
Physical Chemistry Practicals
1. ##### Making a Volumetric Solution
2. ##### Performing a Titration & Volumetric Analysis
3. ##### Measurement of an Enthalpy Change
4. ##### Factors Affecting the Rate of a Reaction
2.
Organic & Inorganic Chemistry Practicals
1. ##### Reactions of Group 2 Elements
2. ##### Identifying Anions & Cations
3. ##### Distillation of a Product from a Reaction
4. ##### Testing for Organic Functional Groups
ASChemistryAQARevision NotesPhysical Chemistry Formulae, Equations & Calculations Hydrated Salts
Hydrated Salts(AQA AS Chemistry):Revision Note
Exam code:7404
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Author
Richard Boole
Last updated
21 August 2025
Water of Crystallisation
Water of crystallisation is when some compounds can form crystalswhich have wateras part of their structure
A compound that contains water of crystallisation is called a hydrated compound
The water of crystallisation is separated from the main formula by a dot when writing the chemical formula of hydrated compounds
E.g. hydrated copper(II) sulfate is CuSO 4∙5H 2 O
A compound which doesn’t contain water of crystallisation is called an anhydrous compound
E.g. anhydrous copper(II) sulfate is CuSO 4
A compound can be hydrated to different degrees
E.g. cobalt(II) chloride can be hydrated by six or two water molecules
CoCl 2∙6H 2 O or CoCl 2∙2H 2 O
The conversion of hydrated compounds to anhydrous compounds is achieved by heating the hydrated salt
This process is reversed by adding water, which reforms the hydrated compound:
Hydrated: CuSO 4•5H 2 O ⇌ CuSO 4+ 5H 2 O :Anhydrous
The degree of hydration can be calculated from experimental results:
The mass of the hydrated salt must be measured before heating
The salt is then heated until it reaches a constant mass
The two mass values can be used to calculate the number of moles of water in the hydrated salt - known as the water of crystallisation
Worked Example
Calculating water of crystallisation
11.25 g of hydrated copper sulfate, CuSO 4.xH 2 O, is heated until a constant mass of 7.19 g.
Calculate the formula of the hydrated copper(II) sulfate.
A r(Cu) = 63.5 A r(S) = 3.A r(O) = 16 A r(H) = 1
Answer:
Salt and water CuSO 4 H 2 O
Value 7.19 11.25 - 7.19
= 4.06
3.M r 63.5 + 32 + (16 x 4)
= 159.5(1 x 2) + 16
= 18
4. Moles == 0.045= 0.226
5. Salt : water ratio= 1= 5
6. Formula of hydrated salt The formula is CuSO 4•5H 2 O
Examiner Tips and Tricks
A water of crystallisation calculation can be completed in a similar fashion to an empirical formula calculation
Instead of elements, you start with the salt and water
Instead of dividing by atomic masses, you divide by molecular / formula masses
The rest of the calculation works the same way as the empirical formula calculation
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187710 | https://www.physicsforums.com/threads/calculating-and-graphing-the-4th-root-of-4.756110/ | Calculating and Graphing the 4th Root of -4
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187711 | https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/subtracting-whole-numbers-the-inverse-operation-of-addition/ | Module 1: Historical Counting Systems
Subtracting Whole Numbers: The Inverse Operation of Addition
Learning Outcomes
Subtract single-digit numbers and check the solution using addition
Use columns representing place value to subtract multiple-digit numbers
Subtract Whole Numbers
Addition and subtraction are inverse operations. Addition undoes subtraction, and subtraction undoes addition.We know [latex]7 - 3=4[/latex] because [latex]4+3=7[/latex]. Knowing all the addition number facts will help with subtraction. Then we can check subtraction by adding. In the examples above, our subtractions can be checked by addition.
| | | |
---
| [latex]7-3=4[/latex] | because | [latex]4+3=7[/latex] |
| [latex]13-8=5[/latex] | because | [latex]5+8=13[/latex] |
| [latex]43-26=17[/latex] | because | [latex]17+26=43[/latex] |
Example
Subtract and then check by adding:
[latex]9 - 7[/latex]
[latex]8 - 3[/latex]
Solution
| |
| 1. |
| [latex]9 - 7[/latex] |
| Subtract 7 from 9. | [latex]2[/latex] |
| Check with addition. [latex]2+7=9\quad\checkmark[/latex] |
| |
| 2. |
| [latex]8 - 3[/latex] |
| Subtract 3 from 8. | [latex]5[/latex] |
| Check with addition. [latex]5+3=8\quad\checkmark[/latex] |
TRY IT
To subtract numbers with more than one digit, it is usually easier to write the numbers vertically in columns just as we did for addition. Align the digits by place value, and then subtract each column starting with the ones and then working to the left.
Example
Subtract and then check by adding: [latex]89 - 61[/latex].
Show Solution
Solution
| | |
--- |
| Write the numbers so the ones and tens digits line up vertically. | [latex]\begin{array}{c}\hfill 89\ \hfill \underset{\text{____}}{-61}\end{array}[/latex] |
| Subtract the digits in each place value.Subtract the ones: [latex]9 - 1=8[/latex] Subtract the tens: [latex]8 - 6=2[/latex] | [latex]\begin{array}{c}\hfill 89\ \hfill \underset{\text{____}}{-61}\ \hfill 28\end{array}[/latex] |
| Check using addition.[latex]\begin{array}{c}\hfill 28\ \hfill \underset{\text{____}}{+61}\ \hfill 89\end{array}\quad\checkmark[/latex] |
Our answer is correct.
TRY IT
When we modeled subtracting [latex]26[/latex] from [latex]43[/latex], we exchanged [latex]1[/latex] ten for [latex]10[/latex] ones. When we do this without the model, we say we borrow [latex]1[/latex] from the tens place and add [latex]10[/latex] to the ones place.
Find the difference of whole numbers
Write the numbers so each place value lines up vertically.
Subtract the digits in each place value. Work from right to left starting with the ones place. If the digit on top is less than the digit below, borrow as needed.
Continue subtracting each place value from right to left, borrowing if needed.
Check by adding.
excercise
Subtract: [latex]43 - 26[/latex].
Show Solution
Solution
| |
| Write the numbers so each place value lines up vertically. |
| Subtract the ones. We cannot subtract [latex]6[/latex] from [latex]3[/latex], so we borrow [latex]1[/latex] ten. This makes [latex]3[/latex] tens and [latex]13[/latex] ones. We write these numbers above each place and cross out the original digits. |
| Now we can subtract the ones. [latex]13 - 6=7[/latex]. We write the [latex]7[/latex] in the ones place in the difference. |
| Now we subtract the tens. [latex]3 - 2=1[/latex]. We write the[latex]1[/latex] in the tens place in the difference. |
| Check by adding. Our answer is correct. |
try it
Example
Subtract and then check by adding: [latex]207 - 64[/latex].
Show Solution
Solution
| |
| Write the numbers so each place value lines up vertically. |
| Subtract the ones. [latex]7 - 4=3[/latex].Write the [latex]3[/latex] in the ones place in the difference. Write the [latex]3[/latex] in the ones place in the difference. |
| Subtract the tens. We cannot subtract [latex]6[/latex] from [latex]0[/latex] so we borrow [latex]1[/latex] hundred and add [latex]10[/latex] tens to the [latex]0[/latex] tens we had. This makes a total of [latex]10[/latex] tens. We write [latex]10[/latex] above the tens place and cross out the [latex]0[/latex]. Then we cross out the [latex]2[/latex] in the hundreds place and write [latex]1[/latex] above it. |
| Now we subtract the tens. [latex]10 - 6=4[/latex]. We write the 4 in the tens place in the difference. |
| Finally, subtract the hundreds. There is no digit in the hundreds place in the bottom number so we can imagine a [latex]0[/latex] in that place. Since [latex]1 - 0=1[/latex], we write [latex]1[/latex] in the hundreds place in the difference. |
| Check by adding. Our answer is correct. |
try it
Subtract and then check by adding: [latex]439 - 52[/latex].
Show Solution
[latex]439 − 52 = 387[/latex] because [latex]387 + 52 = 439\quad\checkmark[/latex]
Subtract and then check by adding: [latex]318 - 75[/latex].
Show Solution
[latex]318 − 75 = 243[/latex] because [latex]243 + 75 = 318\quad\checkmark[/latex]
Example
Subtract and then check by adding: [latex]910 - 586[/latex].
Show Solution
Solution
| |
| Write the numbers so each place value lines up vertically. |
| Subtract the ones. We cannot subtract [latex]6[/latex] from [latex]0[/latex], so we borrow [latex]1[/latex] ten and add [latex]10[/latex] ones to the [latex]10[/latex] ones we had. This makes [latex]10[/latex] ones. We write a [latex]0[/latex] above the tens place and cross out the [latex]1[/latex]. We write the [latex]10[/latex] above the ones place and cross out the [latex]0[/latex]. Now we can subtract the ones. [latex]10 - 6=4[/latex]. |
| Write the 4 in the ones place of the difference. |
| Subtract the tens. We cannot subtract [latex]8[/latex] from [latex]0[/latex], so we borrow [latex]1[/latex] hundred and add [latex]10[/latex] tens to the [latex]0[/latex] tens we had, which gives us [latex]10[/latex] tens. Write [latex]8[/latex] above the hundreds place and cross out the [latex]9[/latex]. Write [latex]10[/latex] above the tens place. |
| Now we can subtract the tens. [latex]10 - 8=2[/latex] . |
| Subtract the hundreds place. [latex]8 - 5=3[/latex] Write the [latex]3[/latex] in the hundreds place in the difference. |
| Check by adding. Our answer is correct. |
try it
Example
Subtract and then check by adding: [latex]2,162 - 479[/latex].
Show Solution
Solution
| |
| Write the numbers so each place values line up vertically. |
| Subtract the ones. Since we cannot subtract [latex]9[/latex] from [latex]2[/latex], borrow [latex]1[/latex] ten and add [latex]10[/latex] ones to the [latex]2[/latex] ones to make [latex]12[/latex] ones. Write [latex]5[/latex] above the tens place and cross out the [latex]6[/latex]. Write [latex]12[/latex] above the ones place and cross out the [latex]2[/latex]. |
| Now we can subtract the ones. | [latex]12 - 9=3[/latex] |
| Write [latex]3[/latex] in the ones place in the difference. |
| Subtract the tens. Since we cannot subtract [latex]7[/latex] from [latex]5[/latex], borrow [latex]1[/latex] hundred and add [latex]10[/latex] tens to the [latex]5[/latex] tens to make [latex]15[/latex] tens. Write [latex]0[/latex] above the hundreds place and cross out the [latex]1[/latex]. Write [latex]15[/latex] above the tens place. |
| Now we can subtract the tens. | [latex]15 - 7=8[/latex] |
| Write [latex]8[/latex] in the tens place in the difference. |
| Now we can subtract the hundreds. |
| Write [latex]6[/latex] in the hundreds place in the difference. |
| Subtract the thousands. There is no digit in the thousands place of the bottom number, so we imagine a [latex]0[/latex]. [latex]1 - 0=1[/latex]. Write [latex]1[/latex] in the thousands place of the difference. |
| Check by adding.[latex]\begin{array}{}\ \stackrel{1}{1},\stackrel{1}{6}\stackrel{1}{8}3\hfill \ \underset{\text{______}}{+479}\hfill \ 2,162\quad\checkmark \hfill \end{array}\quad\checkmark[/latex] |
Our answer is correct.
try it
Watch the video below to see another example of subtracting whole numbers by lining up place values.
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Example: Subtracting Whole Numbers. Located at: License: CC BY: Attribution
Question ID: 143321, 143322, 143327, 143341, 143343. Authored by: Alyson Day. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
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Prealgebra. Provided by: OpenStax. License: CC BY: Attribution. License Terms: Download for free at |
187712 | https://pubmed.ncbi.nlm.nih.gov/32831508/ | Evaluation of the efficacy of subgingival irrigation in patients with moderate-to-severe chronic periodontitis otherwise indicated for periodontal flap surgeries - PubMed
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. 2020 Jul-Aug;24(4):348-353.
doi: 10.4103/jisp.jisp_54_20. Epub 2020 Jul 1.
Evaluation of the efficacy of subgingival irrigation in patients with moderate-to-severe chronic periodontitis otherwise indicated for periodontal flap surgeries
Rajni Jain1,Rashi Chaturvedi2,Nymphea Pandit3,Vishakha Grover1,Deborah M Lyle4,Ashish Jain1
Affiliations Expand
Affiliations
1 Department of Periodontology, Dr. HS Judge Institute of Dental Sciences and Hospital, Punjab University, Chandigarh, India.
2 Department of Periodontology, University of British Columbia, Vancouver, Canada.
3 Department of Periodontology, DAV Dental College, Yamunanagar, Haryana, India.
4 Water Pik Inc., Fort Collins, CO, USA.
PMID: 32831508
PMCID: PMC7418546
DOI: 10.4103/jisp.jisp_54_20
Item in Clipboard
Evaluation of the efficacy of subgingival irrigation in patients with moderate-to-severe chronic periodontitis otherwise indicated for periodontal flap surgeries
Rajni Jain et al. J Indian Soc Periodontol.2020 Jul-Aug.
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J Indian Soc Periodontol
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. 2020 Jul-Aug;24(4):348-353.
doi: 10.4103/jisp.jisp_54_20. Epub 2020 Jul 1.
Authors
Rajni Jain1,Rashi Chaturvedi2,Nymphea Pandit3,Vishakha Grover1,Deborah M Lyle4,Ashish Jain1
Affiliations
1 Department of Periodontology, Dr. HS Judge Institute of Dental Sciences and Hospital, Punjab University, Chandigarh, India.
2 Department of Periodontology, University of British Columbia, Vancouver, Canada.
3 Department of Periodontology, DAV Dental College, Yamunanagar, Haryana, India.
4 Water Pik Inc., Fort Collins, CO, USA.
PMID: 32831508
PMCID: PMC7418546
DOI: 10.4103/jisp.jisp_54_20
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Abstract
Background: In certain medically and physically compromised; and terminally ill patients, periodontal surgery may not be feasible. They need special attention and assistance for their daily plaque control regimens for the management and maintenance of periodontal conditions. Subgingival irrigation home care devices with antiplaque agents may serve as useful tools in such specific patient populations.
Aims: The aim of this study was to evaluate of the efficacy of sub-gingival irrigation in patients with moderate-to-severe chronic periodontitis otherwise indicated for periodontal flap surgeries.
Settings and design: Randomized comparative parallel group interventional clinical trial.
Materials and methods: Forty adults with moderate-to-severe periodontitis, divided inot Group A and B, were subjected to the use of subgingival home irrigations using 0.06% chlorhexidine (CHX) and mouth-rinsing with 15 ml of 0.12% CHX twice daily, respectively after Phase I therapy. Clinical parameters, i.e., gingival index, oral hygiene index simplified, and bleeding on probing scores were assessed at baseline, 2 weeks, 4 weeks, and 12 weeks' postphase I therapy, whereas clinical attachment level (CAL), probing depth (PD), and stain assessment at baseline and 12 weeks following Phase I therapy.
Statistical analysis used: Statistical Package for Social Sciences (SPSS Inc., Chicago, IL, USA version 15.0 for Windows).
Results: A statistically significant difference was seen with the use of 0.06% CHX irrigations in PD (P = 0.004) and CAL (P = 0.002) as compared to the use of mouth rinsing with 0.12% CHX. Similar differences were observed in both intensity (P = 0.014) and area (P = 0.034) of lingual surface staining with greater staining with CHX mouth rinsing.
Conclusion: The adjunctive use of subgingival home irrigations using 0.06% CHX has a promising potential to maintain the oral health and results in lesser staining compared to CHX mouth rinsing. The regimen may further obviate the need of periodontal surgery in medically compromised subjects.
Keywords: Antiplaque; chlorhexidine; gingivitis; mouth rinsing; periodontitis; staining; subgingival home irrigation.
Copyright: © 2020 Indian Society of Periodontology.
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Conflict of interest statement
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References
Schenkein HA. Host responses in maintaining periodontal health and determining periodontal disease. Periodontol 2000. 2006;40:77–93. - PubMed
Gilbert P, Das J, Foley I. Biofilm susceptibility to antimicrobials. Adv Dent Res. 1997;11:160–7. - PubMed
Umeda M, Takeuchi Y, Noguchi K, Huang Y, Koshy G, Ishikawa I. Effects of nonsurgical periodontal therapy on the microbiota. Periodontol 2000. 2004;36:98–120. - PubMed
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Graves RC, Disney JA, Stamm JW. Comparative effectiveness of flossing and brushing in reducing interproximal bleeding. J Periodontol. 1989;60:243–7. - PubMed
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187713 | http://www.360doc.com/document/14/0723/15/7981670_396523635.shtml | 弧长和扇形的面积、圆锥的侧面积和全面积知识要点
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弧长和扇形的面积、圆锥的侧面积和全面积知识要点
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辅导：弧长和扇形的面积、圆锥的侧面积和全面积
一、弧长和扇形的面积：
『活动一』因为 3 60°的圆心角所 对弧长就 是圆周长 C=2 π R，所以 1°的圆心角所对的 弧长
是.这样，在半 径为 R 的圆中，n°的圆心 角所对的弧 长 l =.
『活动二』类比弧长的计 算公式可知：在半径为 R 的圆中，圆心角为 n°的扇 形面积的计算
公式为：S=.
『活动三』扇形 面积的另 一个计算公 式
比较扇形面积计 算公式与 弧长计算公式，可以发现：可以将扇形面积的计 算公式：S=
360
n
π
R
2
化为 S=
180
R n
·
2
1
R，从面可得扇 形面积的另 一计算公式：S=.
二、圆锥的侧面积和全面积：
1．圆锥的基本概念：的线
段 SA、SA
1
……叫做圆锥 的母线，
的线段叫做 圆锥的高．
2.圆锥中的各元 素与它的 侧面展开图——扇形的各 元素之间的关系：
将圆锥的侧面沿 母线 l 剪开，展开成 平面图形，可以得到一个 扇形，设
圆锥的底面半径 为 r，这个扇形的半径 等于，扇形弧长 等
于．
3．圆锥侧面积 计算公式
圆锥的母线即为 扇形的半 径，而圆锥底面 的周长是 扇形的弧长，
这样，S
圆锥侧
=S
扇形
=
2
1
·2 πr · l = πrl
4．圆锥全面积 计算公式
S
圆锥全
=S
圆锥侧
＋S
圆锥底面
= πr l ＋πr
2
=πr（l＋r）
三、例题讲解：
例 1、（2011•德州，11，4 分）母线 长为 2，底面圆 的半径为 1 的圆锥的侧面 积为．
例 2、（2011 年山东 省东营市，21，9 分）如 图，已知 点 A、B、C、D 均在已知圆上，AD
∥BC，BD 平分∠ABC，∠BAD=120°，四边形 ABCD 的周长 为 15．
（1）求此圆的 半径；
（2）求图中阴 影部分的 面积．
r
l
O
r
h
l
A
S
A
1
2
例 3、（2010 广东，14，6 分）如图，在平面直 角坐标系 中，点 P 的坐标为（－4，0），
⊙P 的半径为 2，将⊙P 沿 x 轴 向右平移 4 个单位长度得⊙P
1
．
（1）画出⊙P
1
，并直接判 断⊙P 与⊙P
1
的位置关系；
（2）设⊙P
1
与 x 轴正半轴，y 轴正半 轴的交点 分别为 A，B，求劣弧 AB 与弦 AB 围成的图
形的面积（结果 保留π）．
四、同步练习：
1、（2012 北海,11，3 分）如图，在边长为 1 的正方形组 成的网格中，△AB C 的顶点都在
格点上，将△A BC 绕点 C 顺时针旋转 60°，则顶点 A 所经过的路 径长为：（）
A．10 π B．
10
3
C．
10
3
π D．π
2、（2012 北海,12，3 分）如图，等边△ABC 的周长为 6 π，半
径是 1 的⊙O 从与 AB 相切于 点 D 的位置出发，在△AB C 外部按顺时 针方向沿三角 形滚动，
又回到与 AB 相切于点 D 的位置，则⊙O 自转了：（）
A．2 周 B．3 周 C．4 周 D．5 周
3、（2012 湖北咸宁，7，3 分）如 图，⊙O 的外切正 六边形 ABCDEF 的边长为 2，则图中 阴
影部分的面积为（）．
A．
3
π
2
B．
3
2π
3
C．
3 2
π
2
D．
3 2
2π
3
y
x
－3
O
1
2
3
1
2
3
－3
－2
－1
－1
－2
－4
－5
－6
例 3 图
A
B
C
第 1 题图
A
B
C
O
D
第 2 题图
A
B
C
D
E
F
（第 3 题）
O
3
4、（2012 四川内江，8，3 分）如图 2，AB 是⊙O 的直径，弦 CD⊥AB，∠C DB＝30°，
CD＝2
3
，则阴影部分 图形的面积为（）
A．4 π B．2 π C．π D．
2 π
3
5、（2012·湖南省张家界 市·14 题·3 分）已知圆 锥的底面直径和母线长都 是１０
cm
，
则圆锥的侧面积 为 __.
6、（2 012·哈尔滨，题 号 16 分值 3）一个圆锥的 母线长为 4，侧面积为 8
，则这个圆锥
的底面圆的半径 是．
7、（2012 江苏省淮安市，17，3 分）若圆锥的底 面半径为 2cm，母线长为 5cm，则此圆锥
的侧面积为 cm
2
．
8、（2 012 四 川 达 州，11，3 分）已 知圆 锥 的 底 面半 径 为 4，母 线 长为 6，则 它的 侧 面 积
是.（不取近 似值）
9、（2012 年广西玉林市，16，3）如图，矩形 OABC 内 接于扇形 MON，当 CN=CO 时，
∠NMB 的度数是.
10、(2012 广安中考试题第 15 题，3 分)如图 6，Rt△A BC 的边 BC 位于直线 l 上，AC=
3
，
∠ACB=90
o
，∠A=30
o
，若△RtABC 由现在的位置向 右无滑动地翻转，当点 A 第 3 次落在直
线上 l 时，点 A 所经过的路线的 长为 _ _ （结果用含 л 的式子表 示）．
11、（2011•丹东，14，3 分）如图，将半径为 3cm 的圆形纸片剪掉 三分之一，余下 部分围
成一个圆锥的侧 面，则这个圆锥的高是．
12、（2012 贵州贵阳，23，10 分）如图，在⊙O 中，直径 AB=2，
CA 切⊙O 于 A，BC 交⊙O 于 D，若∠C=4 5°，则
（1）BD 的长是；（5 分）
（2）求阴影部 分的面积. （5 分）
A
B
D
C
O
图 2
A
B
C
l
…………
图 6
第 12 题图
A
O
B
D
C
第 9 题
第 11 题
4
13、（2012 浙江省义乌市，20，8 分）如图,已知 AB 是⊙O 的直径,点 C、D 在⊙O 上,
点 E 在⊙O 外,∠EAC=∠D=60°.
（1）求∠AB C 的度数；
（2）求证：AE 是⊙O 的 切线；
（3）当 BC=4 时，求劣弧 AC 的长.
14、（2012 年吉林省，第 23 题、7 分．）如图，在扇 形 OAB 中，∠AOB=9 0°，半径 OA=6．将
扇形 OAB 沿过点 B 的直线折叠．点 O 恰好落在弧 AB 上点 D 处，折痕交 OA 于点 C，求整个阴
影部分的周长和 面积．
15、（2011 甘肃兰州，25，9 分）如图，在单位长度 为 1 的正方形网 格中，一段圆弧经过网
格的交点 A、B、C.
（1）请完成如 下操作：
①以点 O 为原点、竖直和 水平方向所 在的直线为坐 标轴、网格 边长为单位长，建立平面 直
角坐标系；②用 直尺和圆规画出该圆弧所 在圆的圆心 D 的位置（不用写作 法，保留作图痕
迹），并连结 AD、CD.
（2）请在（1）的基础上，完成下列问题：
①写出点的坐标：C、D；
②⊙D 的半径=（结果保留根号）；
③若扇形 ADC 是一个圆 锥的侧面展开 图，则该圆锥的 底面面积为（结果保留π）；
④若 E（7，0），试判断 直线 EC 与⊙D 的位置关 系并说明你的理由.
A
B
C
O
O
A
B
C
D
E
5
参考答案：
例 1、考点：圆锥 的计算。专题：计算题。
分析：先计算出底面圆 的周长，它等于圆锥 侧面展开图扇形的弧长，而母线长为 扇形的半径，
然后根据扇形的 面积公式计算即可．
解答：解：∵圆锥的 底面圆的半径为 1，∴圆锥的底面圆的周 长=2π×1=2π，
∴圆锥的侧面积=
1
2
×2π×2=2 π．故答案为：2π．
点评：本题考查了圆锥的侧面积公式：S=
1
2
lr
．圆锥侧面展开图为扇形，底面圆的周长等
于扇形的弧长，母线长为扇形的半径．
例 2、考点：扇 形面积的计算；圆心角、弧、弦的关系；圆周角定理．
专题：几何图形 问题．
分析：（1）根 据条件可以证得四边 形 ABCD 是等腰 梯形，且 AB=AD=DC，∠DBC=90°，
在直角△BD C 中，BC 是圆的直径，BC=2DC，根据四边形 ABCD 的周 长为 15，即可求得
BC，即可得 到圆的半径；
（2）根据 S
阴影
=S
扇形
AOD
-S
△
AOD
即可求解．
解答：解：（1）∵AD∥BC，∠BAD=120°．∴∠ABC=60°．
又∵BD 平分∠ABC，∴∠AB D=∠DBC=∠ADB=30°∴==，∠BCD=6 0°
∴AB=AD=DC，∠DBC=90°又在直角△BDC 中，BC 是圆的直径，BC=2DC．
∴BC+
3
2
BC=15∴BC=6∴此圆的半 径为 3．
（2）设 BC 的中点为 O，由（1）可知 O 即为圆心．连接 OA，OD，过 O 作 OE⊥AD 于 E．
在直角△A OE 中，∠AOE=30°∴OE=OA•cos30°=
33
2
S
△
AOD
=
1
2
×3×
33
2
=
93
4
．
∴
2
AO D
AO D
60 3 9
S-S-3
360 4
S
阴影
扇形
3 9 6-9 3
-3=
2 4 4
点评：本题主要 考查了扇形的面积的计算，正确证得四边形 ABCD 是等腰梯 形，是解题的
关键．
例 3、考点：圆与圆 的位置关系；坐 标与图形性质；扇形面积 的计算
分析：（1）根据题 意作图即可求得答案，注意圆的半径为 2；
（2）首先根据题意 求得扇形 BP
1
A 与△BP
1
A 的面积，再作差即可 求得劣弧 错误!未找到引用
源。与弦 AB 围成的图形的面积．
解答：解：（1）如 图：
∴⊙P 与⊙P
1
的位置关系 是外切；
6
（2）如图：∠BP
1
A=90°，P
1
A=P
1
B=2，
∴S
扇形 BP 1 A
=
2
90 2
360
=π，S
△AP 1 B
=×2×2=2，
∴劣弧 错误!未找到引用源。与弦 AB 围成的图形的 面积为：π﹣2．
点评：此题考查了圆与 圆的位置关系以及扇形 面积的求解方法．题目难度 不大，解题的关键
是注意数形结合 思想的应用．
四、
1、【解析】△ABC 绕点 C 顺时针旋转 60°，顶点 A 经过的路径是以 C 为圆心 AC 为半径，
圆心角为 60°的弧，根据弧长公式
180
r n
l
，可求路径长为
10
3
【答案】C
【点评】考查的知识点有网格中的勾股定理（求 AC），图形的旋转，弧长公式
180
r n
l
。
中等难度的题型。
2、【解析】三角形的周长恰好是圆周长的三倍，但是圆在点 A、B、C 处分别旋转了一个角
度，没有滚动，在三个顶点处旋转的角度之和 是三角形的外角和 360°。所以⊙O 自转了 4
圈。
【答案】C
【点评】本 题最容易 出错的 地方就 是在顶 点处的旋转，难 度较大。如果学生能 动手 操作一
下，正确答案就出来了。
3、【解析】图中阴影部分的 面积等于：三角形 AOB 面积－扇形 AOB 面积，不难知道，∆AOB
为等边三角形，可求出∆AOB 边 AB 上的高是
3
，扇 形 AOB 圆心角∠O＝60°，半径 OA
＝
3
，从而阴影部 分的面积是
1
2
×2×
3
－
2
60(3)
360
＝
3
π
2
，故选 A．
【答案】A
【点评】本题着重考查了 扇形面积 的计算及解直 角三角形的 知识，以及转化、数形结合思想，
有一定综合性，难度中等．
7
4、【解析】如下图所示，取 AB 与 CD 的交点为 E，由垂径定理知 CE＝
3
，而∠COB
＝2∠CDB＝60°，所以 OC＝
sin 60
CE
＝2，OE＝
1
2
OC＝1，接下来发 现 OE＝BE，可证△
OCE≌△BED，所以 S
阴影
＝S
扇形
COB
＝
1
6
π·2
2
＝
2 π
3
．
【答案】D
【点评】圆的有关性质 是中考高频考点，而图形 面积也是多数地方必考 之处，将它们结
合可谓珠联璧合．解答此题需在 多处转化：一是将阴影面积转 化为扇形面积问题解决；二是
由圆周角度数求 出圆心角度数；三是发现 图中存在的全等三角形，这一点是解题关键．
5、【分析】S
侧
=π rl=π·
2
10
×10=50 π.【解答】50 π
【点评】圆锥的侧面积 S
侧
=
2
1
·2 π r·l=π rl（其中 r 是圆锥底面圆的半 径，l 是母线的长）.
6、【解析】本题考查圆锥 展开图及侧面积 计算公式.设半径为 r，圆锥侧面 积即展开图扇形
的面积，根据 S 扇=
2
1
lR,即 8 π=
2
1
×2 π×4，得 r=2.【答案】2
【点评】在解决圆锥 的计算问题时，要把握好两个相等关系：圆锥侧 面展开图（扇形）的半
径 R 等于圆锥的母 线长，扇形的弧长
l
等于圆锥的底 面周长
2 r
．几乎所有圆锥计 算问题都
是从这两个对应 关系入手解决的．
7、【解析】根 据圆锥的侧面积公式=πr l 计算，此圆锥的侧 面积=π×2×5=10 π【答案】10 π
【点评】本题综合考 查有关扇形和圆锥的相关 计算．解题思路：解决此类问 题时要紧紧抓住
两者之间的两个 对应关系：①圆锥的母线长等于侧 面展开图的扇形半径；②圆锥的底面 周长
等于侧面展开图 的扇形弧长．正确对这两 个关系的记忆是解题的关 键．
8、解析：圆锥 的侧面积可由公式来求，这里 R=6，l=8 π，因此 S=24 π。答案：24 π
点评：本题考查了圆 锥的侧面展开及其侧面积 的求法，初步考查学生的空间 观点，注意本题
不要与全面积相 混淆。
9、分析：首先 连接 OB，由矩形的性 质可得△BOC 是直角三角形，又 由 OB=ON=2O C，
∠BOC 的度数，又 由圆周角定理求得∠N MB 的度数．
解答：解：连接 OB，∵CN=CO，∴OB=ON=2OC，∵四边形 OABC 是矩形，∴∠BCO=90°，
∴cos∠BO C=
2
1
OB
OC
，∴∠BOC=60°，∴∠NMB=
2
1
∠BOC=30°．故答案为：30°．
点评：此题考查了圆 周角定理、矩形的性质以及特 殊角的三角函数值．此题难度适中，注意
辅助线的作法，注意数形结合思想的应用．
A
B
D
C
O
图 2
E
8
10、思路导引：确定路线长度，由于路线是圆 弧，因此确定旋转角，与旋转半径是解决问 题
的关键，
3
＋
3
；
解析：计算斜边 长度是 2，第一次经过路 线长度是
120 2
180
，
第二次经过路线 长度是
90 3 120 2
180 180
，
第三次经过路线 长度与第二次经过路线长 度相同，也是
90 3 120 2
180 180
，
所以当点 A 三次落在直线 l 上时，经过的路 线长度是
120 2
180
＋2×（
90 3 120 2
180 180
）=
4
3
＋
3
＋2×
4
3
=
3
＋
3
点评：解答旋转问题，确定旋转中心、旋转半径以 及旋转角度是前提，另外计算连续的弧长
问题，注意旋转 规律，进行多次循环旋转 的有关弧长之和的计算.
11、考点：圆锥的计算。专题：计算题。
分析：算出围成 圆锥的扇形的弧长，除以 2π 即为圆锥的底面半径，利用勾股定理即可求得
圆锥的高．
解答：解：围 成 圆 锥 的 弧 长 为
180
3)
3
1
1(360
=4πcm，∴圆 锥 的 底 面 半 径 为
4π÷2π=2cm，∴圆锥的高 为
2 2
2 3
=1cm．故答案为 1cm．
点评：考查圆锥的计 算；得到圆锥的底面半径是解 决本题的突破点；用到的知识点 为：圆锥
的底面周长等于 侧面展开图的弧长．
12、解析： （1）由 CA 切⊙O 于 A，得∠A=90°，再结 合∠C=4 5°，得∠B=4 5°.连接
AD，则由 直径 AB=2，得∠ADB=90°.故 BD=AB×cos 4 5°=2×co s 4 5°=
2
；（2）运用代换
得到阴影部分的 面积等于△ACD 的面积.
解：（1）填
2
；
（2）由（1）得，AD=BD.
∴弓形 BD 的面积=弓形 AD 的面积，故阴影部分 的面积=△ACD 的面积.
∵CD=AD=BD=
2
，∴S
△ACD
=
2
1
CD×AD=
2
1
×
2
×
2
=1，即阴影部 分的面积是 1.
9
点评：本题主要考查 了圆的性质，切线的性质，等腰直角三角形的性质 以及割补法，解
法较多，有利于 考生从自己的角度获取解 题方法，中等偏下难度.
13、【解析】（1）根据 相等的弧长对应的圆周 角相等，得∠ABC=∠D =60°。
（2）直径对应 的圆周角为直角，则由 三角形内角和为 180°，得出∠BAC
的大小，继而 得
出
∠BAE
的大小为
90
°，即
AE 是⊙O 的切线。
（3）由题意易 知，△OB C 是等边三角形，则由劣弧 AC 对应的圆心角 可求出劣弧 AC 的长。
20．解：（1）∵∠ABC 与∠D 都是弧 AC 所对的圆周角
∴∠ABC=∠D =60° …………2 分
（2）∵AB 是⊙O 的直径∴∠ACB=9 0°…………………………3 分
∴∠BAC=30°∴∠BAE =∠BAC＋∠EAC=30°＋60°=90° …………………4 分
即 BA⊥AE∴AE 是⊙O 的切线…………………………………………………………5 分
（3） 如图，连 结 OC∵OB=OC,∠ABC=60°∴△OBC 是等边三角形
∴OB=BC=4, ∠BOC=60°∴∠AOC=120°…………………7 分
∴劣弧 AC 的长为
3
8
1 80
4 1 20
…………………………………………8 分
【点评】此题 考查圆弧的长与其对应 的圆心角、圆周角的关系，及三角 形的内角和为 180°。
相等的弧长对应 的圆周角、圆心角相等．
14、【解析】阴影部分的 周长包括线段 AC+CD+DB 的长和弧 AB 的长．由折 叠的性质可知，
AC+CD=OA=6;DB=OB=6.故 周 长 可 求．求面 积 需 要 连 接 OD,证明△ODB 是 正 三 角 形，得 到∠
CBO=30°，求 出 OC 的长，阴影部 分的面积=
AOB
S
扇形
-2
OBC
S
△
.【答案】解：连接 OD．
∵OB=OD,OB=BD∴△ODB 是等边 三角形∠D BO=60°∴∠OBC=∠C BD=30°
在 Rt△OCB 中，OC=OBtan3 0°=
23
．∴
11
2 3 6 6 3
22
OBC
S OC OB
△
∴
1
=2 36 2 6 3
4
9 12 3
OB C
AO B
S S S
△
阴影 部分
扇形
有图可知，CD=O C,DB=OB
=L
阴影 部分
弧 AB+AC+CD+DB=2×6+6
=12+6
【点评】此题考查了折叠的性质、扇形面积公式、弧长公式以及直角三角形的性
O
A
B
C
D
E
10
质．此题难度适中，注意数形结合思想的应用，注意辅助线的作法．
15、考点：垂径定理；勾股定理；直线与圆的 位置关系；圆锥的计算；作图—复杂作图.
分析：（1）根据 叙述，利用正方形的网格 即可作出坐标轴；
（2）①利用（1）中所作的坐标系，即可 表示出点的坐标；
②在直角△O AD 中，利 用勾股定理即 可求得半径长；
③可以证得∠ADC=90°，利用扇 形的面积公式即可求得扇 形的面积；
④利用切线的判 定定理，证得∠DCE=9 0°即可．
解答：解：（1）①建立平面直角坐标系
②找出圆心
（2）① C（6，2）；D（2，0）
② 2
5
错误!未找到引用源。
③ π（7 分）
④ 直线 EC 与⊙D 相切
证 CD
2
+CE
2
=DE
2
=25（或通过相似证明）
得∠DC E=90°
∴直线 EC 与⊙D 相切．
故答案为：① C（6，2）；D（2，0）② 2
5
错误!未找到引用源。③ π
点评：本题主 要考查了垂 径定理，圆锥的计算，正确证明△DCE 是直 角三角形是难点．
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187714 | https://pubmed.ncbi.nlm.nih.gov/24056834/ | [Shy bladder syndrome] - PubMed
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. 2013 Jul-Aug;48(4):345-53.
doi: 10.1708/1319.14632.
[Shy bladder syndrome]
[Article in Italian]
Antonio Prunas
PMID: 24056834
DOI: 10.1708/1319.14632
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[Article in Italian]
Antonio Prunas. Riv Psichiatr.2013 Jul-Aug.
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. 2013 Jul-Aug;48(4):345-53.
doi: 10.1708/1319.14632.
Author
Antonio Prunas
PMID: 24056834
DOI: 10.1708/1319.14632
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Abstract
Paruresis is the inability to urinate in situations where there is perception of scrutiny, or potential scrutiny, by others. According to DSM-5, paruresis is classified as social phobia.
Aim: The present study aims at offering a clinical description of the disorder in a sample of paruretics voluntarily recruited among users registered on an Italian website dedicated to people suffering from this disorder (www.paruresis.it).
Methods: Data were collected through a set of questionnaires published online, including assessment and screening measures for paruresis (Paruresis Checklist), generalized social phobia (Mini-SPIN) and depression (Beck Depression Inventory).
Results: 65 participants showed clinically relevant symptoms of paruresis, as suggested by a PCL score above the threshold. Mean age was 28 ys (SD= ± 7.75 ys); most of participants were male (87.7%; N=57). Although gender differences in the clinical manifestations of the disorder appear limited, the variable mostly connected to the severity of paruresis is the presence of a further diagnosis of generalized social phobia.
Discussion: Paruresis is a clinical condition associated with high level of distress on which, however, there is limited knowledge among mental health professionals.
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Is "shy bladder syndrome" (paruresis) correctly classified as social phobia?Hammelstein P, Soifer S.Hammelstein P, et al.J Anxiety Disord. 2006;20(3):296-311. doi: 10.1016/j.janxdis.2005.02.008.J Anxiety Disord. 2006.PMID: 16564434
Is "shy bladder syndrome" a subtype of social anxiety disorder? A survey of people with paruresis.Vythilingum B, Stein DJ, Soifer S.Vythilingum B, et al.Depress Anxiety. 2002;16(2):84-7. doi: 10.1002/da.10061.Depress Anxiety. 2002.PMID: 12219340
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187715 | https://www.epa.gov/sites/default/files/2018-11/documents/fish-news-sept2018.pdf | 1 This issue of the Fish and Shellfish Program Newsletter generally focuses on harmful algal blooms (HABs). Recent Advisory News Health Advisories and Closures for California Finfish, Shellfish, and Crustaceans The California Department of Public Health (CDPH) coordinates a routine monitoring program along the California coast to sample mussels and other shellfish like clams and scallops for the presence of paralytic shellfish poisoning (PSP) and domoic acid toxins. Commercial shellfish harvesters are also required to provide weekly shellfish samples to CDPH for PSP toxin assay and domoic acid analysis. If toxin levels are high enough, warnings and quarantines are issued to protect the recreational fishing public and shellfish consumers. CDPH also has a coastwide, volunteer-based phytoplankton monitoring program which detects the naturally occurring, microscopic algae that produce PSP and domoic acid toxins. When toxin levels begin increasing, CDPH may expand its shellfish sampling effort to include other seafood species. If toxin levels increase beyond the federal alert level for either toxin, then CDPH will immediately issue a health advisory for all potentially impacted seafood species in the affected region. The California Department of Fish and Wildlife (CDFW) website was established as a source of information for fishermen and the fishing industry. When circumstances arise, CDPH warnings, quarantine information, and health advisories about consuming California's ocean finfish, shellfish, and crustaceans are posted there. The California Office of Environmental Health Hazard Assessment (OEHHA) also issues consumption advisories based on the amount of mercury or other chemical contaminants found in finfish, shellfish, and crustaceans. Safe eating guidelines to help the public reduce its exposure to chemicals in sport fish are also available from OEHHA. CDPH and CDFW advise recreational and commercial fishers to check their web pages frequently, or call the Domoic Acid Fishery Closure Information Line at 831-649-2883 for the most up-to-date information on health advisories and fisheries closures. Fish and Shellfish Program N E W S L E T T E R September 2018 EPA 823-N-18-009 In This Issue Recent Advisory News .............. 1 EPA News ................................ 4 Other News ............................. 7 Recently Awarded Research ..... 9 Tech and Tools ...................... 11 Recent Publications .............. 12 Upcoming Meetings and Conferences ................... 14 This newsletter provides information only. This newsletter does not impose legally binding requirements on the U.S. Environmental Protection Agency (EPA), states, tribes, other regulatory authorities, or the regulated community. The Office of Science and Technology, Office of Water, U.S. Environmental Protection Agency has approved this newsletter for publication. Mention of trade names, products, or services does not convey and should not be interpreted as conveying official EPA approval, endorsement, or recommendation for use. This newsletter provides a monthly summary of news about fish and shellfish Fish and Shellfish Program N E W S L E T T E R September 2018 2 Spiny Lobster Fisheries: Open and Closed Ocean Waters • The recreational spiny lobster fishery is open statewide. The fishery is open every year from 6:00 a.m. on the Saturday preceding the first Wednesday in October through the first Wednesday after the 15th of March per California Code of Regulations Title 14, Section 29.90(a). • The commercial spiny lobster fishery is open statewide. The fishery is open every year between the first Wednesday in October and the first Wednesday after the 15th of March per California Code of Regulations Title 14, Section 121(a). Dungeness and Rock Crab Fisheries: Open and Closed Ocean Waters Recreational Fisheries for Dungeness Crab and Rock Crab • The recreational fishery for all rock crab species is open statewide. North of 40°00.00' N. lat., near the Mendocino/Humboldt county line, the CDPH recommends that consumers not eat the viscera (internal organs, also known as "butter" or "guts") of crabs. The viscera usually contain much higher levels of domoic acid than crab body meat. • The recreational fishery for Dungeness crab is closed statewide. Every year, the fishery is closed for a period of time through the first Saturday in November per California Code of Regulations Title 14, Section 29.85(a)(2). Commercial Fisheries for Dungeness Crab and Rock Crab • The commercial rock crab fishery is open from near the Mendocino/Humboldt county line (40°00.00' N. lat.) to the U.S./Mexico border as of April 20, 2018. The fishery closure north of the Mendocino/Humboldt county line will remain in effect until state health agencies determine that domoic acid levels no longer pose a significant risk to public health and recommend the fisheries be opened, and the director of the California Department of Fish and Wildlife (CDFW) provides notification of fishery reopening to commercial fishermen. • The commercial fishery for Dungeness crab is closed statewide. Every year, the fishery is closed for a period of time through November 14 (south of the Mendocino/Sonoma county line) or December 1 (north of the Mendocino/Sonoma county line) per Fish and Game Code Section 8276. The opening date may be delayed due to poor crab market readiness. Razor Clam Fishery Closure • As of April 27, 2016 the recreational take and possession of razor clam is prohibited from Humboldt and Del Norte county beaches until further notice. • Read the CDFW Declaration of Fisheries Closure Due to a Public Health Threat Caused by Elevated Levels of Domoic Acid in Razor Clams (January 30, 2017). Fish and Shellfish Program N E W S L E T T E R September 2018 3 • See the latest information on domoic acid levels in razor clams in Del Norte and Humboldt counties (scroll down the page to the Links section). Additional Information The following resources are available: • Domoic Acid Fishery Closure Information Line: 831-649-2883 • Shellfish Biotoxin Information Line: 510-412-4643 or toll-free at 800-553-4133 (Maintained by CDPH) • Crabs web page (Links to CDFW website) • Domoic Acid FAQs (Links to CDPH website) • Marine Biotoxin Monitoring Program (Links to CDPH website) • Annual Mussel Quarantine (Links to CDPH website) • Shellfish Program (Links to CDPH website) • Phytoplankton Monitoring Program (Links to CDPH website) • Fukushima Disaster Information (Links to CDFW website) • Harmful Algae web page (Links to Woods Hole Oceanographic Institute website) • Harmful Algal Blooms (CDFW 2011 Status of the Fisheries Report) • Oregon Recreational Shellfish Biotoxin Closures • Washington Shellfish Safety Information For more information contact AskMarine@wildlife.ca.gov or 831-649-2870. Source: State Updates Your Guide to Eating Fish Caught in Florida The Florida Department of Health (DOH), the Florida Department of Environmental Protection (DEP), the Florida Fish and Wildlife Conservation Commission (FFWCC), and the Florida Department of Agriculture and Consumer Services (DACS) operate jointly to determine if environmental chemicals are present in fish from Florida waters. In most instances, FFWCC determines what fish species should be sampled and collects those samples. DEP measures the levels of chemicals in the fish tissue. DOH determines the potential for adverse human health effects from consuming fish and issues fish consumption advisories when needed. DACS provides input on issues related to commercially available seafood (grocery stores, etc.) in Florida. Recently, DOH revised “Your Guide to Eating Fish Caught in Florida” (March 2018) and it can be read here: Fish and Shellfish Program N E W S L E T T E R September 2018 4 The most current fish advisories are available in a searchable database online at Your Guide to Eating Fish Caught in Florida. (Image courtesy of Florida DOH) . The public can look up advisories by waterbody type (fresh or marine), county, and specific location. Results show consumption guidance and the contaminant causing the advisory for the waterbody. More information about fish consumption advisories in Florida can be found here: For more information, contact the Florida Department of Health Public Health Toxicology Section at phtoxicology@flhealth.gov or 850-245-4250. Source: EPA News The Cyanobacteria Assessment Network – Recent Success in Harmful Algal Bloom Detection Cyanobacteria blooms, which can become harmful algal blooms (HABs), are a huge environmental problem across the U.S. They are capable of producing dangerous toxins that threaten the health of humans and animals, the quality of drinking water supplies, and the ecosystems in which they develop. Scientists at EPA are part of a team of specialists using remote sensing data to improve cyanobacteria detection methods. Improving the detection process would help state environmental and health agencies better determine whether to post public advisories to protect human health. The Cyanobacteria Assessment Network (CyAN), a multi-agency project involving EPA, National Aeronautics and Space Administration, the National Oceanic and Atmospheric Administration (NOAA), and the U.S. Geological Survey (USGS), uses historical and current satellite data to provide an early warning indicator system for HABs in U.S. freshwater systems. Since the project’s inception in October 2015 CyAN imagery has been used to detect algal blooms in Ohio, Florida, California, Vermont, New Hampshire, Massachusetts, Connecticut, Rhode Island, and Utah before traditional monitoring efforts alerted watershed managers. Due to the severity of Utah’s cyanobacteria blooms in 2016, EPA Region 8 requested technical assistance from EPA Office of Research and Development (ORD) to utilize CyAN’s early warning indicator system to support cyanobacteria monitoring activities for the State of Utah. Satellite data products and imagery acquired approximately one week after the Utah Division of Water Quality’s (UDWQ’s) routine monthly sampling in mid- Fish and Shellfish Program N E W S L E T T E R September 2018 5 June identified that a bloom was developing in Provo Bay. Based on this information UDWQ scientists returned to the area for follow-up sampling. “The images we've been receiving through the CyAN project have been tremendously helpful to UDWQ. The near-daily spatial extent and relative magnitude coverage provides the foundation for a wide range of useful outputs,” said Benjamin M. Holcomb, Coordinator of Biological Assessment and HAB Programs with UDWQ. Mr. Holcomb explained that by providing near-immediate alerts when cyanobacteria are reaching human health thresholds, CyAN gives organizations like UDWQ the ability to respond more quickly than if they were relying on public reports and monthly sampling. It also ensures that their in-situ, bloom-response data are representative of recent bloom conditions, which allows them to better target field sampling and more efficiently use their limited resources. Also, Mr. Holcomb added that the satellite images can be easily shared with response agencies as a useful visual communication aid. The goal of this project is to provide access to satellite images and data on the concentration and extent of chlorophyll-a and cyanobacteria in the continental U.S. through an Android mobile application and other interactive resources. As CyAN continues to grow, so will scientists’ understanding of the connections between human health, economic values, and environmental conditions to cyanobacteria and phytoplankton blooms. For more information contact, EPA Office of Science Information Management at the following link: Source: ORD Developing Methods for Measuring Total Microcystins in Fish Tissue EPA ORD is developing methods for measuring total microcystins in fish tissue using the 2-methoxy-3-methyl-4-phenylbutyric acid (MMPB) procedure. ORD has completed the first phase of this research. Existing fillet and whole fish homogenates were spiked with three congeners of microcystins (LR, LA, and RR) both individually and as mixtures to develop a method for their recovery and measurement using the MMPB derivatization method. The UDWQ used CyAN data products and imagery to monitor evolving algal blooms and deploy additional sampling resources. (Images courtesy of EPA) Fish and Shellfish Program N E W S L E T T E R September 2018 6 second phase of the project is to field-test this method on fish collected from waterbodies experiencing algal blooms and compare results with individual congener measurements. ORD will be using fish from lakes, reservoirs, or rivers that have been exposed to algal toxins on a frequent basis, including golden algae blooms (ORD will be developing a method for testing Prymnesins at a later date). ORD will process those fish and assess them for total microcystin using the MMPB method. To date, fish samples have been received from Lake Okeechobee in Florida, Utah Lake in Utah, and Bantam Lake in Connecticut. More Information on Fish Samples ORD is using three to five fish of similar size (<25% difference in length) from multiple trophic levels (bass, carp, channel catfish, bullheads, sunfish, gizzard shad) and minnows (20 or more) using the size criteria (when possible) in the table to above. Size Criteria by Trophic Level Species Size Criteria Largemouth Bass =/> 2 lbs Smallmouth Bass 1-2 Ibs Walleye 1-2 Ibs Northern Pike => 2Ibs Common Carp =/> 2 lbs Channel Cat =/> 2 Ibs Golden Shiner Largest Blue Gill > 80 mm Pumpkin Seed > 80 mm White Sucker > 200 mm Brown Bullhead > 250 mm Gizzard shad > 160 mm Bluntnose Minnows Adults Emerald Shiners Adults For more information on this effort, contact Jim Lazorchak at Lazorchak.Jim@epa.gov and Toby Sanan at Sanan.Toby@epa.gov. EPA CyanoHABs Webpage and Monthly Newsletter In order to educate and inform the public about HABs, EPA releases a monthly newsletter specifically focusing on cyanobacteria HABs (cyanoHABs). It spotlights freshwater HABs and provides information on beach closures, health advisories with respect to HABs, current news, and recently published journal articles. The newsletter also lists upcoming conferences, workshops, webinars, and useful resources. For more information, visit EPA’s cyanoHAB website or contact Dr. Lesley D’Anglada at Danglada.Lesley@epa.gov or 202-566-1125. The 2018 issues of the CyanoHABs Newsletter can be accessed at Other News Toxic Cyanobacteria Blooms Impair Mussel Growth in Lake Erie On January 30, 2018, NOAA’s National Centers for Coastal Ocean Science (NCCOS) reported on a new study it was conducting with USGS. The study indicates that detected amounts of cyanobacteria and microcystin, one of the toxins the bacteria produce, did not cause mussel mortality in Lake Erie from 2013 to 2015. However, the researchers did find that the measured concentrations of both cyanobacteria and microcystin impaired mussel growth, with microcystin having a larger negative effect on mussel growth than cyanobacteria. Fish and Shellfish Program N E W S L E T T E R September 2018 7 Lake Erie supports commercial fisheries that rely, in part, on mussels in the lake as a source of food. After being absent for most of the late 1980s and 1990s, cyanobacteria have again become prevalent seasonally in western Lake Erie, causing concern that cyanobacteria blooms and associated cyanotoxins could reduce production of mussels and ultimately affect fisheries. USGS scientists retrieve mussels from a monitoring station in Lake Erie. Credit: Sean Bailey, USGS. (Image courtesy of NOAA NCCOS) The western part of the lake also provides drinking water to large coastal communities, and recent cyanobacteria blooms have resulted in episodic drinking water shutdowns along Lake Erie and the creation of a large international effort to identify causes and potential management strategies to minimize adverse effects to humans and other organisms. For more information contact Timothy Wynne at timothy.wynne@noaa.gov. Source: Four Different Algal Toxins Found in San Francisco Bay Mussels NCCOS reported on March 15, 2018, that scientists have identified four kinds of algal toxins in mussels collected from San Francisco Bay. The study, published in Harmful Algae, is the first to report the co-occurrence of both freshwater and marine toxins in mussels consumed by humans and animals. The researchers, led by Dr. Misty Peacock while at the University of California Santa Cruz (currently at Northwest Indian College), found nearly all mussels collected from the bay were contaminated with at least one of the detected algal toxins, and 37% contained all four—one of which originates in freshwater. Mussels in the bay were contaminated with the following toxins: • Domoic acid – a neurotoxin that causes amnesic shellfish poisoning in humans and is produced by marine diatoms in the genus Pseudo-nitzschia. • Paralytic shellfish toxins (saxitoxins) – cause paralytic shellfish poisoning and are associated with marine dinoflagellates in the genus Alexandrium. • Dinophysis shellfish toxins (okadaic acid and derivatives) – cause diarrhetic shellfish poisoning and are produced by marine dinoflagellates in the genus Dinophysis. • Microcystins – hepatotoxins that cause liver damage in people and animals and are produced by freshwater cyanobacteria in the genus Microcystis. Fish and Shellfish Program N E W S L E T T E R September 2018 8 Contaminated mussels pose a serious health threat to people and animals who eat them. Even though San Francisco Bay lacks commercial shellfish operations, people still harvest and eat mussels from the bay. While commercially harvested shellfish are generally safe because they undergo regular testing, neither microcystins nor Dinophysis shellfish toxins are routinely monitored in California shellfish. California sea lions are becoming sick, and dying in some cases, from ingesting domoic acid—a neurotoxin that accumulates in fish and shellfish they consume. (Image courtesy of NOAA) HAB monitoring has generally been waterbody-dependent, focusing either on marine or freshwater toxins, but not both. For example, cyanobacterial toxins were previously considered a public health issue only for freshwater, with concerns about adverse effects to drinking and recreational waters. The publication stems in part from a NOAA-funded Monitoring and Event Response for Harmful Algal Bloom (MERHAB) research project that aims to address this emerging HAB concern. Senior study author Raphael Kudela, the Lynn Professor of Ocean Health at the University of California Santa Cruz, is working closely with state and federal agencies to assess health risks and develop better monitoring tools. According to Dr. Kudela these findings and other recent studies are prompting several California agencies to consider changes to their monitoring programs. For more information, contact Marc Suddleson at marc.suddleson@noaa.gov. Source: Recently Awarded Research NCCOS Funds $6.8M for New and Continuing HAB Research On September 6, 2018, NCCOS announced support for 28 new and continuing HAB research awards in 2018. These awards, totaling $6.8 million, fund projects around the nation through the Ecology and Oceanography of Harmful Algal Blooms, Monitoring and Event Response for Harmful Algal Blooms, and Prevention, Control, and Mitigation of Harmful Algal Blooms programs and involve over 85 scientists across 54 institutions around the U.S. “NCCOS is funding the latest scientific research to support environmental managers trying to cope with increasing and recurring toxic algae that continue to impact marine and human health and coastal economies,” said NCCOS Director Steve Thur, PhD. “Improved understanding of these coastal harmful algal bloom threats will lead to better predictions, mitigation and possibly solutions in impacted U.S. coastal regions.” Fish and Shellfish Program N E W S L E T T E R September 2018 9 NCCOS HAB competitive research programs develop science-based solutions to address expanding HAB impacts that are affecting coastal resources and economies in every U.S. coastal region. HAB species and impacts vary regionally and NCCOS projects are advancing the understanding of bloom toxicity, applying new technologies to detect HABs and their toxins in the field, producing HAB forecasts, and exploring HAB prevention and control methods. Summaries of new and continued research projects by region are below. NCCOS projects are the result of a rigorous competitive peer-review process that ensures support for the highest quality science. Gulf of Mexico and Caribbean Karenia brevis, the Florida red tide alga that occurs throughout the Gulf of Mexico, causes mortality of fish, turtles, marine mammals, and birds; neurotoxic shellfish poisoning (NSP); and respiratory irritation in beachgoers. Gambierdiscus species, which grow on coral reefs in the Gulf and Caribbean, cause fish to become so toxic that human consumers become ill with ciguatera fish poisoning (CFP). Newly funded research projects will help determine the processes that terminate red tides and help mitigate toxin effects in threatened Florida manatees. Continued research will help the State of Florida improve its already-rigorous NSP monitoring and management framework. It will also fund the development of models for predicting CFP in reef-dwelling fish. Lake Erie The Great Lakes, and western Lake Erie in particular, are subject to cyanobacterial HABs, primarily Microcystis that can produce microcystins. These are liver toxins that can contaminate drinking water, harm wildlife, and prevent recreational use of water bodies. Current Lake Erie HAB forecasts can predict Microcystis biomass, but the cells are not always toxic. Newly funded and continuing research will take different approaches to predicting the actual toxicity of Microcystis blooms in Lake Erie in order to provide early warning and improve management of drinking water and recreational use. New England Along the New England coast blooms of Alexandrium produce neurotoxins that can accumulate in shellfish, causing PSP in human consumers. To protect human health, sections of the coast must be closed to shellfish harvesting. New research will investigate how microscopic animals control the growth and toxicity of Alexandrium. Continuing research will use remote toxin sensors to determine how shellfish in the eastern Gulf of Maine become toxic. The information from both studies will be incorporated into predictive models that forecast when and where Alexandrium blooms will occur in the Gulf of Maine, helping state managers and the shellfish industry protect public health and minimize economic disruption. Chesapeake and Delaware Bays The Chesapeake and Delaware Bays are subject to a variety of HABs that can kill fish and shellfish. While they do not threaten human health, they can have severe impacts on fisheries and aquaculture. New funding will support research in Delaware Bay investigating the role of nitric oxide in promoting blooms of Heterosigma. In the lower Chesapeake Bay a new study to better predict Margalefidinium (formerly Cochlodinium) and Alexandrium monilatum blooms and a continuing project on the toxicity and food web impacts of A. monilatum will help the Fish and Shellfish Program N E W S L E T T E R September 2018 10 shellfish industry minimize their impacts. A study of a naturally occurring compound that may control some toxic HABs will continue to test its effectiveness and environmental impacts. California Amnesic shellfish poisoning (ASP)-causing Pseudo-nitzschia is of particular concern along the California coast. New research will improve modeling efforts to predict Pseudo-nitzschia blooms off Southern California and support environmental management efforts. A continuing project to understand the controlling factors of Pseudo-nitzschia toxin production and bloom formation will also help to improve early warning models. Another continuing project is development of a strategy to add monitoring of multiple marine and freshwater HAB toxins (microcystins, ASP-, diarrhetic shellfish poisoning (DSP)-, PSP-causing) occurring simultaneously in shellfish and other organisms in estuaries. This is an emerging ecosystem and public health problem. Pacific Northwest and Alaska Both PSP and ASP are problems in the Pacific Northwest and Alaska, impacting commercial, recreational, and tribal subsistence shellfish harvesting. A continuing project is transitioning an early warning system for Pseudo-nitzschia for Oregon and Washington ocean beaches, and another project aims to uncover the mechanisms behind wintertime occurrences of PSP-toxicity in geoduck clam fisheries in Southeast Alaska. Predictive modeling and HAB monitoring provides managers with an early warning of when and where toxic blooms will affect shellfish harvests, thereby providing better public health protection and safeguarding coastal economies. National Projects Dinophysis, the HAB causing DSP has suddenly appeared in multiple areas in the U.S. in the last ten years. Continuing research supports a cross-regional study to find common factors that have led to the sudden appearance of this emerging HAB as well as to improve monitoring. Another continuing project is testing methods of measuring DSP toxins in shellfish in order to protect human health. A full list of awards and their abstracts can be found here. For more information, contact Elizabeth Turner at elizabeth.turner@noaa.gov. Source: Fish and Shellfish Program N E W S L E T T E R September 2018 11 Tech and Tools Portable Red Tide Detector Debuts at NOAA Emerging Tech Workshop On September 21, 2017, NCCOS reported on a portable, hand-held instrument that uses genetics to detect the red tide-causing organism Karenia brevis in the field. This was featured at the second NOAA Emerging Technologies for Observations Workshop. The device, dubbed a “tricorder” after the fictional Star Trek hand-held life detector, is the first of its kind and is able to provide direct results to end users such as government agencies and businesses. This technology speeds up the decision-making process in closing beaches and shellfish harvesting beds, as well as helping to determine the cause of fish kills. The tricorder uses a biotechnology technique called nucleic acid sequence-based amplification to target the messenger RNA in the carbon fixation gene specific to K. brevis. Red tides in Florida coastal waters (caused principally by K. brevis) can threaten human health and cost millions in tourism, agriculture, seafood, and leisure industries. Currently the State of Florida detects and enumerates K. brevis through the relatively slow, labor-intensive and expert process of light microscopy to differentiate this toxic alga from closely related non-toxic and less toxic species. Data from the hand-held tricorders get uploaded to the Gulf of Mexico Coastal Ocean Observing System (GCOOS) for automated processing and calculation of K. brevis cell abundance and data visualization. Users include the Florida Department of Environmental Protection, Florida Department of Agriculture and Consumer Services, and the Florida Fish and Wildlife Conservation Commission. The project is co-led by Dr. John Paul of the University of South Florida and Dr. Kate Hubbard of the Florida Fish and Wildlife Conservation Commission and is supported by an NCCOS Prevention, Control, and Mitigation of HABs Program project. Learn more about the red tide-detecting tricorder here. The second NOAA Emerging Technologies for Observations Workshop was held August 22-23, 2017, at the NOAA Center for Weather and Climate Prediction, College Park, Maryland. The workshop highlighted new environmental observing capabilities and applications that can improve NOAA services and enhance organizational efficiency. For more information, contact John Wickham at John.Wickham@noaa.gov. Source: John Paul, PhD, uses a portable tricorder to identify the Florida red tide organism, Karenia brevis. Credit: University of South Florida. (Image courtesy of NOAA NCCOS) Fish and Shellfish Program N E W S L E T T E R September 2018 12 Recent Publications Journal Articles The list below provides a selection of research articles focusing on HABs. ► Fluorescence probes for real-time remote cyanobacteria monitoring: A review of challenges and opportunities Bertone, E. , M.A. Burford, and D.P. Hamilton. 2018. Fluorescence probes for real-time remote cyanobacteria monitoring: A review of challenges and opportunities. Water Research 141:152-162. ► The blue mussel Mytilus edulis is vulnerable to the toxic dinoflagellate Karlodinium armiger—Adult filtration is inhibited and several life stages killed Binzer, S.B., R.B.C. Lundgreen, T. Berge, P.J. Hansen, and B. Vismann. 2018. The blue mussel Mytilus edulis is vulnerable to the toxic dinoflagellate Karlodinium armiger – Adult filtration is inhibited and several life stages killed. PLoS One 13(6): e0199306. ► Assessment of saxitoxin sensitivity of nerves isolated from the Pacific oyster, Crassostrea gigas, exposed to Alexandrium minutum Boullot, F., C. Fabioux, H. Hegaret, P. Soudant, et al. 2018. Assessment of saxitoxin sensitivity of nerves isolated from the Pacific oyster, Crassostrea gigas, exposed to Alexandrium minutum. Toxicon 149: 93. ► Combined effects of warming and acidification on accumulation and elimination dynamics of paralytic shellfish toxins in mussels Mytilus galloprovincialis Braga, A.C., C. Camacho, A. Marques, A. Gago-Martínez, et al. 2018. Combined effects of warming and acidification on accumulation and elimination dynamics of paralytic shellfish toxins in mussels Mytilus galloprovincialis. Environmental Research 164: 647-654. ► Size‐based interactions and trophic transfer efficiency are modified by fish predation and cyanobacteria blooms in Lake Mývatn, Iceland Ersoy, Z., E. Jeppesen, S. Sgarzi, I. Arranz, M. Cañedo-Argüelles, et al. 2017. Size-based interactions and trophic transfer efficiency are modified by fish predation and cyanobacteria blooms in Lake Mývatn, Iceland. Freshwater Biology 62(11): 1942-1952. ► A global analysis of the relationship between concentrations of microcystins in water and fish Flores, N.M., T.R. Miller, and J.D. Stockwell. 2018. A global analysis of the relationship between concentrations of microcystins in water and fish. Frontiers in Marine Science 5: 30. ► Shifts in coastal fish communities: Is eutrophication always beneficial for sticklebacks? Gagnon, K., M. Gräfnings, and C. Boström. 2017. Shifts in coastal fish communities: Is eutrophication always beneficial for sticklebacks? Estuarine, Coastal and Shelf Science 198(A): 193-203. ► The effects of red tide (Karenia brevis) on reflex impairment and mortality of sublegal Florida stone crabs, Menippe mercenaria Gravinese, P.M., S.M. Kronstadt, T. Clemente, C. Cole, et al. 2018. The effects of red tide (Karenia brevis) on reflex impairment and mortality of sublegal Florida stone crabs, Menippe mercenaria. Marine Environmental Research 137: 145-148. ► Physiological effects caused by microcystin-producing and non-microcystin producing Microcystis aeruginosa on medaka fish: A proteomic and metabolomic study on liver Le Manach, S., B. Sotton, H. Huet, C. Duval, et al. 2018. Physiological effects caused by microcystin-producing and non-microcystin producing Microcystis aeruginosa on medaka fish: A proteomic and metabolomic study on liver. Environmental Pollution 234: 523-537. ► Quantifying harmful algal bloom thresholds for farmed salmon in southern Chile Montes, R.M., X. Rojas, P. Artacho, A. Tello, and R.A. Quiñones. 2018. Quantifying harmful algal bloom thresholds for farmed salmon in southern Chile. Harmful Algae 77: 55-65. Fish and Shellfish Program N E W S L E T T E R September 2018 13 ► Pre-ingestive selection capacity and endoscopic analysis in the sympatric bivalves Mulinia edulis and Mytilus chilensis exposed to diets containing toxic and non-toxic dinoflagellates Navarro, J.M., J. Widdows, O.R. Chaparro, A. Ortiz, and C. Mellado. 2018. Pre-ingestive selection capacity and endoscopic analysis in the sympatric bivalves Mulinia edulis and Mytilus chilensis exposed to diets containing toxic and non-toxic dinoflagellates. PLoS One 13(2): e0193370. ► Development of a method to assess the ichthyotoxicity of the harmful marine microalgae Karenia spp. using gill cell cultures from red sea bream (Pagrus major ) Ohkubo, N., Y. Tomaru, H. Yamaguchi, S. Kitatsuji, and K. Mochida. 2017. Development of a method to assess the ichthyotoxicity of the harmful marine microalgae Karenia spp. using gill cell cultures from red sea bream (Pagrus major). Fish Physiology and Biochemistry 43(6): 1603-1612. ► Developmental neurotoxicity of Microcystis aeruginosa in the early life stages of zebrafish Qian, H., G. Liu, T. Lu, and L. Sun.2018. Developmental neurotoxicity of Microcystis aeruginosa in the early life stages of zebrafish. Ecotoxicology and Environmental Safety 151:35-41. ► Application of activated carbon to accelerate detoxification of paralytic shellfish toxins from mussels Mytilus galloprovincialis and scallops Chlamys farreri Qiu, J., H. Fan, T. Liu, X. Liang, et al. 2018. Application of activated carbon to accelerate detoxification of paralytic shellfish toxins from mussels Mytilus galloprovincialis and scallops Chlamys farreri. Ecotoxicology and Environmental Safety 148: 402-409. ► Harmful Algal Blooms: A Compendium Desk Reference Shumway, S. E., ed., J.M. Burkholder, ed., and S.L. Morton, ed. 2018. Harmful Algal Blooms: Compendium Desk Reference. Hoboken, NJ: John Wiley & Sons. ► Effects of harmful algal blooms on fish: Insights from Prymnesium parvum Svendsen, M.B.S., N. R. Andersen, P.J. Hansen, and J.F. Steffensen. 2018. Effects of harmful algal blooms on fish: Insights from Prymnesium parvum. Fishes 3(1): 11. ► Detection of potential harmful algal bloom-causing microalgae from freshwater prawn farms in Central Luzon, Philippines, for bloom monitoring and prediction Tayaban, K.M.M., K.L. Pintor, and P.G. Vital. 2018. Detection of potential harmful algal bloom-causing microalgae from freshwater prawn farms in Central Luzon, Philippines, for bloom monitoring and prediction. Environment, Development and Sustainability 20(3): 1311-1328. ► Cyanobacteria blooms induce embryonic heart failure in an endangered fish species Zi, J., X. Pan, H. MacIsaac, J. Yang, et al. 2018. Cyanobacteria blooms induce embryonic heart failure in an endangered fish species. Aquatic Toxicology 194: 78-85. Fish and Shellfish Program N E W S L E T T E R September 2018 14 Upcoming Meetings and Conferences Additional Information This monthly newsletter highlights current information about fish and shellfish. For more information about specific advisories within the state, territory, or tribe, contact the appropriate state agency listed on EPA’s National Listing of Fish Advisories website at For more information about this newsletter, contact Sharon Frey (Frey.Sharon@epa.gov, 202-566-1480). Additional information about advisories and fish and shellfish consumption can be found at 13th World Congress on Aquaculture and Fisheries November 12-13, 2018 Melbourne, Australia Aquaculture 2019 March 7-11, 2019 New Orleans, Louisiana Fish Passage 2018 - International Conference on River Connectivity December 10-14, 2018 Albury, New South Wales, Australia National Shellfisheries Association 111th Annual Meeting March 7-11, 2019 New Orleans, Louisiana |
187716 | https://knzhou.github.io/handouts/M8.pdf | Kevin Zhou Physics Olympiad Handouts Mechanics VIII: 3D Rotation Three-dimensional rotation is covered in chapter 7 of Kleppner and chapter 9 of Morin. For further discussion and examples, see chapter I-20 of the Feynman lectures, and this awesome video. There is a total of 98 points.
1 3D Rotation In M6, we considered mostly two-dimensional rotation. Now we will tackle the full three-dimensional case, starting with the general description of rigid body motion.
Idea 1: Chasles’ Theorem The instantaneous velocity of a three-dimensional rigid body can always be decomposed in one of two ways. First, for any given point, it can be written in terms of a translational velocity plus a pure rotation about an axis going through that point. In practice, this point is almost always chosen to be the center of mass, giving the decomposition v = vCM + ω × (r −rCM).
Alternatively, there always exists an axis so that the motion can be written as rotation about that axis, plus a translational velocity parallel to the axis, giving v = v0 + ω × (r −r0) where v0 and ω are parallel. (This is known as a “screw” motion.) In both cases, ω is the same, and its direction defines the axis of rotation. These two decompositions are analogous to the two we saw in M5, and both will be useful below.
Idea 2 Sometimes it can be hard to visualize ω, so here are two tricks. First, if any two points on the object are stationary, then ω must be parallel to the axis connecting the two points.
Second, if the rotation is complicated, one can use rotating frames to simplify the analysis.
If a body has angular velocity ω1 in one frame, and that frame has an angular velocity ω2 with respect to a second frame, then the body has angular velocity ω1 + ω2 with respect to the second frame.
Example 1 You have a small globe, which is mounted so that it can spin on the polar axis and can be spun about a horizontal axis (so that the south pole can be on top). Give the globe a quick spin about the polar axis, and then, before it stops, give it another quick spin about the horizontal axis. Are there any points on the globe that are at rest?
1 Kevin Zhou Physics Olympiad Handouts Solution The first spin gives the angular velocity a vertical component ω1. The second spin gives the angular velocity an additional horizontal component ω2. The globe now rotates about its center of mass with angular velocity ω1 +ω2. Precisely two points on the globe are stationary, namely the points that are parallel and antiparallel to this vector.
Problem 1 (Morin 9.3). A cone rolls without slipping on a table; this means that all the points of the cone that touch the table are instantaneously stationary. The half-angle of the vertex is α, and the axis has length h.
Let the speed of the center of the base, point P in the figure, be v.
(a) Compute the angular velocity ω by thinking of the motion as pure rotation about some axis.
(b) Compute the angular velocity ω by thinking of the motion as translation of P, plus rotation about an axis passing through P.
(c) The apex of the cone is fixed, and the cone continues to rotate. As this motion goes on, the angular velocity vector rotates uniformly, keeping a constant magnitude. Find the precession rate Ω, i.e. the constant vector that satisfies dω/dt = Ω× ω at all times.
Heuristically, the precession rate Ωis “the angular velocity of the angular velocity”. For the relevant problems below, it’s important to keep track of the difference between ω and Ω.
Most of our statements about rotational dynamics from M6 remain true. The main new aspect is that angular momentum is not necessarily parallel to angular velocity.
Example 2: KK Example 7.4 Consider a rigid body consisting of two particles of mass m connected by a massless rod of length 2ℓ, rotating about the z-axis with angular velocity ω as shown.
Find the angular momentum of the system.
2 Kevin Zhou Physics Olympiad Handouts Solution We simply add r × p for both masses. Let the rod lie in the xz plane at this moment. Then for the top left mass, r = −ℓcos α ˆ x + ℓsin α ˆ z.
The momentum is p = mv = mω × r = −mωℓcos α ˆ y.
Then the angular momentum is L = r × p = mωℓ2 cos α (sin α ˆ x + cos α ˆ z) .
The other mass has the opposite r and p and hence the same L, so the total angular momentum is L = 2mωℓ2 cos α (sin α ˆ x + cos α ˆ z) .
It is directed perpendicular to the rod, and in particular, it isn’t parallel to the angular velocity!
Here is another way to derive the same result. We can decompose the angular velocity vector into a component along the rod, and a component perpendicular to the rod. The former contributes no angular momentum, because rotating about the rod’s axis doesn’t move the masses. The latter contributes all the angular momentum. So the angular momentum is L = I⊥ω⊥= (2mℓ2)(ω cos α) directed perpendicular to the rod, which is what we just saw explicitly.
We can summarize the lessons drawn from this example as follows.
Idea 3 For a three-dimensional object, L is not necessarily parallel to ω. In general, for pure rotation about an axis passing through the origin, we have L = Iω where I is a 3 × 3 matrix called the “moment of inertia tensor about the origin”. In components, this means that Li = X j Iijωj.
While this is simple and general, the Iij are a pain to calculate. You can learn more in the reading, but to my knowledge, no Olympiad problem has ever required computing a general moment of inertia tensor.
For the purposes of Olympiad problems, there is a better way to think about the angular momentum. We use the first decomposition of idea 1, and think of the motion as translation plus rotation about the center of mass. If the object has an axis of symmetry, which it will in almost all Olympiad problems, then the angular velocity can then be decomposed into a component parallel to the axis, and perpendicular to the axis, ω = ω∥+ ω⊥.
3 Kevin Zhou Physics Olympiad Handouts The key is that, in such situations, the spin angular momentum has two pieces, which are each parallel to the corresponding piece of the angular velocity, L∥= I∥ω∥, L⊥= I⊥ω⊥ where I∥and I⊥are ordinary moments of inertia about the center of mass. For example, for a flat uniform disc, I∥= MR2/2 and I⊥= MR2/4.
The total angular momentum about the origin is then L = rCM × MvCM + I∥ω∥+ I⊥ω⊥ where the first term is from the motion of the center of mass, and the next two are from rotation about the center of mass. Note that this is exactly the same as what we saw in M5, except that the “spin” angular momentum is broken into two parts.
Finally, the rate of change of angular momentum is dL dt = τ where the torque τ is defined as in M5. The kinetic energy is K = 1 2Mv2 CM + 1 2I∥ω2 ∥+ 1 2I⊥ω2 ⊥.
Remark In any dynamics problem, there are many choices you can make in the setup. For example, if you’re using an inertial frame, you need to choose where the origin is; usually it’s best to place it along the axis of symmetry if possible. You are also free to use a noninertial frame with acceleration a.
The only difference is that there will be a fictitious force −Ma acting at the center of mass. For that reason, it’s usually best to have the acceler-ating frame follow the center of mass, keeping it at its origin, so no new torques are introduced.
However, you should avoid rotating reference frames for dynamics problems. Not only will there be position-dependent Coriolis forces, but they’ll add up and contribute a Coriolis torque, which is a pain to calculate, as you saw in M6. In general, rotating frames are only good for getting a handle on the kinematics, as mentioned in idea 2.
Example 3: KK Example 7.5 Calculate the magnitude of the torque on the rod in example 2.
Solution We recall that the angular momentum was L = 2mωℓ2 cos α (sin α ˆ x + cos α ˆ z).
4 Kevin Zhou Physics Olympiad Handouts The rod as a whole rotates with angular velocity ωˆ z. In particular, the angular momentum vector rotates with this angular velocity as well; its horizontal component moves in a circle with angular velocity ω. Then |τ| = dL dt = ωLx = 2mω2ℓ2 cos α sin α = mω2ℓ2 sin(2α).
It might be surprising that there needs to be a torque given that ω is constant, but that’s just because L and ω aren’t necessarily parallel. Conversely, there can be situations where there is no torque, yet ω changes over time.
Problem 2 (Morin 9.10). A stick of mass m and length ℓspins with angular frequency ω around an axis in zero gravity, as shown.
The stick makes an angle θ with the axis and is kept in its motion by two strings that are perpen-dicular to the axis. Find the tension in the strings.
Problem 3 (KK 7.1). A thin hoop of mass M and radius R rolls without slipping about the z-axis.
It is supported by an axle of length R through its center, as shown.
The axle circles around the z-axis with angular speed Ω, so that the bottom point of the wheel traces out a circle of radius R. Let O be the pivot point of the rod, i.e. the point where the rod meets the z-axis.
(a) Find the instantaneous angular velocity ω of the hoop. (Before moving forward, you’ll want to be totally sure you have this part right. Sometimes, students get confused on it because they imagine a symmetric pair of hoops instead. However, if you actually had a pair, the setup wouldn’t even make sense. Why not?) 5 Kevin Zhou Physics Olympiad Handouts (b) As the motion continues, the angular velocity vector rotates in a circle. Find the precession rate Ωof this system.
(c) Find the instantaneous angular momentum L of the hoop, about the point O.
Problem 4 (KK 7.4). In an old-fashioned rolling mill, grain is ground by a disk-shaped millstone which rolls in a circle on a flat surface driven by a vertical shaft. Because of the stone’s angular momentum, the contact force with the surface can be greater than the weight of the wheel.
Assume the millstone is a uniform disk of mass M, radius b, and width w, and it rolls without slipping in a circle of radius R with angular velocity Ω. Find the contact force. Assume the millstone is closely fitted to the axle so that it cannot tip, and w ≪R. Neglect friction.
Problem 5 (Morin 9.29). A uniform ball rolls without slipping on a table. It rolls onto a piece of paper, which you then slide around in an arbitrary (horizontal) manner. You may even give the paper abrupt, jerky motions, so that the ball slips with respect to it. After you allow the ball to come off the paper, it will eventually resume rolling without slipping on the table. Show that the final velocity equals the initial velocity. (Hint: this remarkably simple result is because of a conservation law. We saw a lower-dimensional version of this problem in M5.) Idea 4: Precession In the above problems, we’re seen a few examples of systems undergoing uniform precession.
In these cases, the angular velocity and the angular momentum vectors rotate with a uniform angular velocity Ω, called the precession rate, where dω dt = Ω× ω, dL dt = Ω× L.
Precession is easiest to see in gyroscopes, which are systems spun up to a very high angular velocity, subject to a weak external torque, so that Ω≪ω. (Kleppner calls this the “gyroscope approximation”.) More generally, you’ll have to decide whether or not Ω≪ω in each case.
Example 4: KK 7.3 A gyroscope wheel is at one end of an axle of length ℓ. The other end of the axle is suspended from a string of length L.
6 Kevin Zhou Physics Olympiad Handouts The wheel is set into motion so that it executes slow, uniform precession in the horizontal plane. The wheel has mass M and moment of inertia I0 about its center of mass, and turns with angular speed ωs. Neglect the mass of the shaft and string. Find the angle β the string makes with the vertical, assuming β is very small.
Solution Let T be the tension in the rope, and let the precession rate be Ω= Ωˆ z. Since the center of mass does not accelerate vertically, and the center of mass moves in a horizontal circle, T cos β = Mg, T sin β = MΩ2(ℓ+ L sin β).
We’ll work to lowest possible order in β everywhere, which means approximating cos β ≈1 and ignoring the L sin β term, giving T = Mg, Tβ = MΩ2ℓ.
Combining these equations gives the precession angular frequency Ω= r gβ ℓ.
This is as far as we can go with forces alone.
Now we use τ = dL/dt. First we need to determine the angular velocity of the wheel. Note that if we went in the rotating frame with angular velocity Ωˆ z, the wheel would just spin with angular speed ωs in place. So the angular velocity vector in the original frame is ω = Ωˆ z + ωsˆ x where ˆ x is directed along the rod.
The center of mass of the wheel moves with speed (ℓ+ L sin β)Ωin a horizontal circle.
Since the precession is assumed to be slow, we have Ω≪ωs, so we can ignore contributions to the angular momentum proportional to Ω. That is, we can take ω ≈ωsˆ x for the purposes of computing angular momentum, giving L ≈I0ωsˆ x.
7 Kevin Zhou Physics Olympiad Handouts This rotates in a horizontal circle with angular speed Ω, so |τ| ≈I0Ωωs.
The torques in the relevant direction come from gravity and the vertical component of the tension force, |τ| = Mgℓ. Equating these, we have Mgℓ= I0Ωωs.
Plugging the result for Ωin above and solving for β gives β = m2gℓ3 ω2I2 0 .
If we had not used Ω≪ωs, we could still find an exact solution, but it would be messy.
Remark In most gyroscope problems, we simply assume the system is already undergoing uniform precession. However, you might wonder just how it gets started in the first place. For example, suppose we had the same setup as the previous problem, with the wheel spinning and the axle horizontal. For simplicity, let’s get rid of the string and suppose the end of the axle is held at a fixed support. Now suppose the axle and wheel are released with no translational motion.
The following chain of events ensues: • Of course, the axle starts to tip downward because of the weight of the wheel. (Rotational mechanics is counterintuitive, but not that counterintuitive!) • This produces a downward component of angular momentum, which is balanced by the axle/wheel system twisting about its center of mass, rotating slightly about the z-axis.
• This twist tries to push the end of the axle out of the page, causing the support to exert a force on the axle pointing into the page. That force propagates down the axle as an internal shear stress, eventually causing the center of mass of the wheel to start moving into the page, starting the precession.
• In reality, the process overshoots and overcorrects, leading to oscillations called nutation on top of the precession. (For more details, see Note 2 of chapter 7 of Kleppner.) • For a typical pivot, energy can be dissipated at the pivot point, but the angular momentum of the system stays roughly the same because the pivot is small. Assuming this is the case, the oscillations will eventually damp away, leaving the uniform precession.
Notice that in this example, the initial angular momentum is perfectly horizontal. The final angular momentum includes an upward component due to the uniform precession, which implies that the axle must tilt slightly downward, by an angle of order (ω/Ω)2. Therefore, if you want to set up uniform precession with the axle perfectly horizontal, as in the above example, you should point the axle slightly upward when releasing it from rest.
8 Kevin Zhou Physics Olympiad Handouts Problem 6 (KK 8.5). An “integrating gyro” can be used to measure the speed of a vehicle.
Consider a gyroscope spinning at high speed ωs. The gyroscope is attached to a vehicle by a universal pivot. If the vehicle accelerates in the direction perpendicular to the spin axis at rate a, then the gyroscope will precess about the acceleration axis, as shown.
The total angle of precession is θ. Show that if the vehicle starts from rest, its final speed is v = Isωs Mℓθ where Isωs is the gyroscope’s spin angular momentum, M is the total mass, and ℓis the distance from the pivot to the center of mass.
Problem 7 (KK 7.5). When an automobile rounds a curve at high speed, the weight distribution on the wheels is changed. For sufficiently high speeds, the loading on the inside wheels goes to zero, at which point the car starts to roll over. This tendency can be avoided by mounting a large spinning flywheel on the car.
(a) In what direction should the flywheel be mounted, and what should be the sense of rotation, to help equalize the loading? (Check your method works for the car turning in either direction.) (b) Show that for a disk-shaped flywheel of mass m and radius R, the requirement for equal loading is that the angular velocity ω of the flywheel is related to the velocity of the car v by ω = 2v ML mR2 where M is the total mass of the car and flywheel, and L is the height of their center of mass.
Problem 8 (KK 7.7). A thin hoop of mass M and radius R is suspended from a string through a point on the rim of the hoop. The string makes an angle α with the vertical.
The support is turned with angular velocity ω, which is high enough so that the hoop’s plane makes a small angle β with the horizontal, and the hoop’s center travels in a small circle of radius r ≪R.
9 Kevin Zhou Physics Olympiad Handouts (a) Does the gyroscope approximation apply in this problem?
(b) Find approximate expressions for β and r.
Problem 9 (KK 7.6, Morin 9.23). With the right initial conditions, a coin on a table can roll in a circle.
As shown, the coin leans inward, with its axis tilted to the horizontal by an angle ϕ. The radius of the coin is b, the radius of the circle it follows on the table is R, and its velocity is v.
(a) Assuming the coin rolls without slipping and b ≪R, show tan ϕ = 3v2/2gR.
(b) No longer assuming b ≪R, show that the described motion is only possible if R > (5/6)b sin ϕ.
Problem 10 (Morin 9.24). If you spin a coin around a vertical diameter on a table, it will slowly lose energy and begin a wobbling motion. The angle between the coin and the table will gradually decrease, and eventually it will come to rest. Assume this process is slow, and consider the motion when the coin makes an angle θ with the table, as shown.
You may assume that the center of mass is essentially motionless. Let R be the radius of the coin, and let Ωbe the angular frequency at which the contact point on the table traces out its circle.
Assume the coin rolls without slipping.
(a) Show that the angular velocity of the coin is ω = Ωsin θ ˆ x2, where ˆ x2 always points upward along the coin, directly away from the contact point.
(b) Show that Ω= 2 p g/R sin θ.
(c) Show that the face on the coin appears to rotate, when viewed from above, with angular frequency (1 −cos θ)Ω.
10 Kevin Zhou Physics Olympiad Handouts Remark: Bivectors Vector quantities defined by the cross product have some unusual properties. For example, under a spatial inversion, which flips the signs of r and p, the sign of L = r × p doesn’t get flipped, so L transforms differently from other vectors. The same applies to the veloc-ity ω and magnetic field B. All three of these quantities are “pseudovectors”, not true vectors.
The underlying reason is that all of these quantities are fundamentally a different kind of mathematical object.
They are really rank 2 differential forms, also called bivectors in three dimensions.
While a vector is specified by an arrow with magnitude and di-rection, a bivector is specified by a planar tile with area and orientation. The following figure, taken from this paper, shows how it can be constructed visually from the cross product.
In three dimensions, we can always convert between bivectors and pseudovectors using the right-hand rule, so any calculation can be done with either form. Bivectors have the advantage of visually representing rotational quantities: the angular velocity bivector lies along an object’s plane of rotation, while the magnetic field bivector lies along the plane in which it makes charged particles circularly orbit. However, it is easier to add vectors, both visually and mathematically, which also makes it easier to think about decomposing vectors into components. This advantage is so important in practice that I don’t recommend using bivectors at all for three-dimensional problems.
On the other hand, when you work in higher-dimensional spaces, the differential form per-spective becomes indispensible. In general, in d dimensions the angular velocity has d 2 components, corresponding to the rotation rate in each independent plane.
• Of course, when d = 1 there is no such thing as rotation at all, while when d = 2 the angular velocity has one component, so we treat it as a scalar.
• When d = 3 the angular velocity has three components, so we treat it as a vector.
• When d = 4 the angular velocity has six components, so we can’t even pretend it’s a vector; we have to use the differential form description.
Rotational dynamics gets really complicated in 4 dimensions. Both the angular velocity and the angular momentum are rank 2 differential forms with 6 independent components each.
11 Kevin Zhou Physics Olympiad Handouts The moment of inertia is a rank 4 tensor with 20 independent components, though it takes a simpler form when you work in the body’s 6 “principal planes”.
Remark: Alternative Notation If you want to look into bivectors more, be sure to steer clear of “geometric algebra”, which dominates the Google search results. Geometric algebra is an internet cult which recruits unsuspecting young people by telling them about bivectors, which are indeed cool. Once they have your attention, they’ll claim that “mainstream” physics has hit a dead end because it refuses to go beyond vector notation, and that you should spend years relearning all of physics in their homemade alternative notation.
However, as we’ve discussed in P1 and R3, there’s nothing magical about notation. Physi-cists don’t teach geometric algebra simply because we have better tools in every situation.
In d = 3 vectors are intuitive and work just fine, while for higher dimensions we either use differential forms, which are more elegant, or tensor calculus, which is powerful enough to do almost anything. When you get to quantum field theory, you’ll have to deal with spinors, which are best handled with Clifford algebra, which is commonly taught in graduate textbooks.
The idea of geometric algebra is to mash together the ideas of differential forms and Clifford algebra into a single universal operation called the “geometric product”, and use it to describe absolutely everything, including basic 3d vector operations. But while this seems satisfying in principle, in practice it introduces a huge number of secondary operations and identities. It doesn’t just not lead to any new results, it makes familiar results sub-stantially harder to reach. (Don’t just take my word for it; see this blog post by a practitioner.) Geometric algebra is also touted as a replacement for standard matrix operations in pure math, but it has problems there too.
According to a another long-time practitioner, definitions in geometric algebra sources are wildly inconsistent with each other, and sometimes aren’t even self-consistent.
And using geometric algebra is exponentially less efficient than ordinary matrix operations once you get past the trivial case of 3 × 3 matrices!
There are two general lessons to be drawn here. First, physicists are generally not ideological about notation, and will use whatever notation works the best for the problems they care about. So if an alternative isn’t used, it’s not because it’s being censored, it’s because it’s not particularly useful. But most of the time, there won’t be anybody around on the internet to tell you precisely why, because practitioners are busy solving real problems. Relying on the internet can therefore give a very skewed view of what’s important.
Second, learning new things is more important than learning new names for old things.
Practitioners of geometric algebra say that it’s worth using, even if it’s less efficient, because it “makes more sense”. Simple problems end up taking lots of steps, and each step introduces new objects associated with new jargon, so it’s apparently deeply satisfying to see the whole apparatus at work. But in my opinion, those people are just getting lost in a maze of their own making. Physical objects don’t care about how we describe them, and there’s no extra 12 Kevin Zhou Physics Olympiad Handouts credit for making things harder than necessary.
Many people fall into the trap of overformalization. For example, the popular blog series Graphical Linear Algebra advocates a category theory inspired notation for arithmetic. It proudly takes 9 blog posts to get to the definition of addition, and 25 to define fractions. If you’ve gotten all the way to this handout, I don’t need to convince you this is a waste of time, but this illustrates why even mathematicians don’t take “applied category theory” seriously.
Example 5: IIT JEE 2016 Two thin circular discs, with radii a and 2a, are connected by a rod of length ℓ= √ 24 a through their centers. This rigid object rolls without slipping on a flat table.
The center of mass of the object rotates about the z-axis with an angular speed of Ω. The angular speed of the object about the axis of the rod is ω. How are Ωand ω related?
Solution This is the most famous problem ever set on the IIT JEE (condensed for clarity), cel-ebrated by generations of students for its difficulty.
But it’s also an example of how not to write a 3D rotation problem.
Under the standard definition of angular velocity, none of the options provided in the question were correct, while the intended answer requires a nonstandard, arbitrary definition. You can find a detailed explanation of this here, by one of the former top scorers on the JEE, and I’ll give a condensed explanation below.
First, let’s figure out what’s going on. The kinematics of this problem isn’t any different from problem 1. Defining the x-axis to be horizontal in the figure above, the instantaneous angular velocity is ω = ω ˆ x, while the precession rate is Ω= (ω/ √ 24) ˆ z. The hard part is fig-uring out what the question writers meant by “the angular speed ω about the axis of the rod”.
If we’re only talking about the object’s instantaneous motion, then the only possible answer is ω = ω · ˆ n, where ˆ n is the unit vector pointing along the rod.
In that case we have Ω/ω = 5/24, which wasn’t an answer choice in the exam. On the other hand, if we are comparing the object’s orientation at different times, then there isn’t a unique answer. At a finite time later, the object will be in a different place, and computing a relative angle requires defining a convention for comparing orientations.
13 Kevin Zhou Physics Olympiad Handouts Here’s what the problem authors meant.
We work in the frame rotating with angular velocity Ω. In this frame, the system is spinning in place, with angular velocity ω + Ω parallel to ˆ n. The definition of ω is |ω + Ω|, which gives Ω/ω = 1/5, the intended answer.
Another way of saying this is that when we compare the orientation of the system at one moment to its orientation at another moment, we bring them to the same position by rotating about the z-axis, at which point they differ by a rotation about ˆ n. But this procedure is totally arbitrary, and not specified by the problem. To pose the problem properly, the writers could have either defined ω explicitly in the rotating frame mentioned above, or replaced it with a quantity with equivalent but unambiguous physical meaning, such as the interval between times a given point on the rim of a disc touches the ground. Fortunately, you’ll almost never see problems this ambiguous on Olympiads.
Remark One of the most counterintuitive things about 3D rigid body motion is the intermediate axis theorem, which states that if a body has moments of inertia I1 < I2 < I3 about its principal axes, then it can rotate stably about the first and third principal axes, but not the second, “intermediate” axis. You can demonstrate this yourself by throwing a rectangular prism (such a book or a phone) in the air. If you spin it about the intermediate axis, it’ll start tumbling. The Soviet physicist Dzhanibekov found a particularly striking example of such motion in zero gravity, which you can see here.
Rigorously demonstrating this theorem requires setting up the full theory of 3D rotational kinematics, which is beyond the scope of the Olympiad, but there’s a simple explanation of this effect using conserved quantities. The rotational kinetic energy is K = 1 2I1ω2 1 + 1 2I2ω2 2 + 1 2I3ω2 3 while the magnitude squared of the angular momentum is L2 = I2 1ω2 1 + I2 2ω2 2 + I2 3ω2 3.
This makes it easy to see why rotation about the third axis is stable: it corresponds to the smallest possible kinetic energy for a given angular momentum.
On the other hand, it’s not so clear why the first axis is stable, because it corresponds to the maximum possible kinetic energy. Aren’t maxima usually unstable? Generally yes, but in this case, the kinetic energy is the only contribution to the energy. Since kinetic energy is conserved in the short run, there is nowhere else for the energy to go, so a system set spinning about the first axis has to stay that way. (Of course, in the long run energy will be lost to the environment, e.g. by friction. So we might say that rotation about the first axis is stable mechanically, but not thermodynamically.) However, for rotation about the second, “intermediate” axis, the body can keep both K and L2 the same by turning on some combination of ω1 and ω3. That explains the Dzhanibekov 14 Kevin Zhou Physics Olympiad Handouts effect. Initially the second principal axis aligns with the direction of L. Then the body rotates so that ω1 and ω3 become nonzero, until the body has completely flipped over. At that point ω1 and ω3 become zero again, with the second principal axis now aligned against L. It’s like a one-dimensional oscillation, where I2ω2 2/2 plays the role of “potential” energy and I1ω2 1/2 + I3ω2 3/2 plays the role of “kinetic” energy.
2 Composite Rotation These are rotational dynamics problems like the ones you saw in M5, but more complex.
Problem 11 (PPP 60). A uniform thin rod is placed with one end on the edge of a table in a nearly vertical position and then released from rest. Find the angle it makes with the vertical at the moment it loses contact with the table. Investigate the following two extreme cases.
(a) The edge of the table is smooth (friction is negligible) but has a small, singe-step groove.
(b) The edge of the table is rough (friction is large) and very sharp, which means the radius of curvature of the edge is much smaller than the flat end-face of the rod. Half of the end-face protrudes beyond the table edge, so that when it is released the rod pivots about the edge.
Problem 12 (Cahn). A tall, thin brick chimney of height L is slightly perturbed from its vertical equilibrium position so that it topples over, rotating rigidly about its base B until it breaks at a point P.
15 Kevin Zhou Physics Olympiad Handouts (a) For concreteness, we will model the internal forces in the chimney as shown below. Assume throughout that r is very small.
We assume that each piece of the chimney experiences a shear force F and longitudinal tension or compression forces T1 and T2 from its neighbors. Find the point on the chimney with the greatest |T1| or |T2|, assuming the chimney is very thin.
(b) Find the point on the chimney experiencing the greatest shear force F.
(c) At what point is the chimney most likely to break? Do you think the limiting factor is the chimney’s maximal compressive strength, tensile strength, or shear strength?
Problem 13. ^ 1 0 IPhO 2014, problem 1A.
Problem 14 (PPP 14). A bicycle is supported so that it can move forward or backwards but cannot fall sideways; its pedals are in their highest and lowest positions.
16 Kevin Zhou Physics Olympiad Handouts A student crouches beside the bicycle and pulls a string attached to the lower pedal, providing a backward horizontal force.
(a) Which way does the bicycle move?
(b) Does the chain-wheel rotate in the same or opposite sense as the rear wheel?
(c) Which way does the lower pedal move relative to the ground?
In particular, be sure to account for the gearing of the bike! To check your answer, watch this video.
Problem 15. r 1 0 APhO 2005, problem 1B. A problem on parametric resonance, an idea we first encountered in M4. (The problem is good, but it’s slightly underspecified, leading to two possible answers which were both accepted. If you get stuck, just make a reasonable assumption.) Problem 16. m 1 0 INPhO 2020, problem 5. A tough angular collision problem.
Problem 17. h 1 0 EuPhO 2019, problem 2. A tough problem about the motion of an rigid body in a magnetic field.
3 Frictional Losses These miscellaneous problems are grouped under the theme of friction or energy dissipation.
Problem 18 (Kalda). A plank of length L and mass M lies on a frictionless horizontal surface; on one end sits a small block of mass m.
The coefficient of friction between the block and plank is µ. The plank is sharply hit and given horizontal velocity v. What is the minimum v required for the block to slide across the plank and fall off the other end?
Problem 19 (BAUPC). A uniform sheet of metal of length ℓlies on a roof inclined at angle θ, with coefficient of kinetic friction µ > tan θ. During the daytime, thermal expansion causes the sheet to uniformly expand by an amount ∆ℓ≪ℓ. At night, the sheet contracts back to its original length. What is the displacement of the sheet after one day and night?
Problem 20. r 1 0 APhO 2010, problem 1A. A question about a different kind of inelastic collision.
Problem 21. h 1 0 IdPhO 2020, problem 2. A nice problem on anisotropic friction.
4 Ropes, Wires, and Chains Example 6: MPPP 78 A uniform flexible rope passes over two small frictionless pulleys mounted at the same height.
17 Kevin Zhou Physics Olympiad Handouts The length of rope between the pulleys is ℓ, and its sag is h. In equilibrium, what is the length s of the rope segments that hang down on either side?
Solution The problem can be attacked by differential equations, but there is an elegant solution using only algebra. We let our unknowns be s, the tension T1 = (T1,x, T1,y) in the rope at the pulley, and the tension T2 at the lowest point.
Considering the entire sagging portion as the system, vertical force balance gives 2T1,y = λℓg, T1,y = λℓg/2.
Now consider half of the sagging portion as the system. Horizontal force balance gives T2 = T1,x.
Finally, consider one of the hanging portions as the system. Then T1 = λgs.
We hence have three equations, but four unknowns.
For the final equation, we need to consider how the tension changes throughout the rope.
This would usually be done by a differential equation, but there is a clever approach using conservation of energy. Suppose we cut the rope somewhere, pull out a segment dx, and reattach the two ends. This requires work T dx, where T is the magnitude of the local tension. Now suppose we cut the rope somewhere else, separate the ends by dx, and paste our segment inside. This requires work −T ′ dx. After this process, the rope is exactly in the same state it was before, so the total work done must be zero.
This would seem to prove that T = T ′, which is clearly wrong. The extra contribution is that if the two locations have a difference in height ∆y, then it takes work λg(∆y) dx to move the segment from the first to the second. So in equilibrium, for any two points of the rope, ∆T = λg ∆y.
18 Kevin Zhou Physics Olympiad Handouts Therefore, we have T1 −T2 = λgh.
Now we’re ready to solve. We have T 2 1 −T 2 2 = (λℓg/2)2 from our first three equations, and dividing by this new relation gives T1 + T2 = λg ℓ2 4h.
This allows us to solve for T1, which gives s = T1 λg = h 2 + ℓ2 8h.
This is a useful result in real engineering projects: it means that the tension in a cable can be estimated by seeing how much it sags.
Example 7: Kalda 27 A wedge with mass M and acute angles α1 and α2 lies on a horizontal surface. A string has been drawn across a pulley situated at the top of the wedge, and its ends are tied to blocks with masses m1 and m2.
There is no friction anywhere. What is the acceleration of the wedge?
Solution This is a classic example of a problem best solved with the Lagrangian-like techniques of M4.
By working in generalized coordinates, we won’t have to solve any systems of equations.
Let s be the distance the rope moves through the pulley, so that both blocks have speed ˙ s in the noninertial frame of the wedge. The “generalized force” is Feff = −dV ds = (m1 sin α1 −m2 sin α2)g.
Now, the kinetic energy in the lab frame will be of the form K = 1 2Meff ˙ s2 which means that, by the Euler–Lagrange equations, ¨ s = Feff Meff .
19 Kevin Zhou Physics Olympiad Handouts Our task is now to calculate Meff.
Since the center of mass of the system can’t move horizontally, the wedge has speed vw = m1 cos α1 + m2 cos α2 M + m1 + m2 ˙ s.
Now, it’s a bit annoying to directly compute the kinetic energy K in the lab frame, but it’s easy to compute the kinetic energy in the frame of the wedge: it’s simply (m1 + m2) ˙ s2/2.
But the two are also related simply, K + 1 2(M + m1 + m2)v2 w = 1 2(m1 + m2) ˙ s2.
Using this to solve for K, we conclude Meff = m1 + m2 −(m1 cos α1 + m2 cos α2)2 M + m1 + m2 .
Finally, the desired answer is aw = m1 cos α1 + m2 cos α2 M + m1 + m2 ¨ s.
Problem 22 (Kalda). A rope of mass per unit length ρ and length L is thrown over a pulley so that the length of one hanging end is ℓ. The rope and pulley have enough friction so that they do not slip against each other.
The pulley is a hoop of mass m and radius R attached to a horizontal axle by light spokes. Find the force on the axle immediately after the motion begins.
Problem 23 (French 5.10). Two equal masses are connected as shown with two identical massless springs of spring constant k.
Considering only motion in the vertical direction, show that the ratio of the frequencies of the two normal modes is ( √ 5 + 1)/( √ 5 −1).
20 Kevin Zhou Physics Olympiad Handouts Problem 24 (Kalda). A massless rod of length ℓis attached to the ceiling by a hinge which allows the rod to rotate in a vertical plane.
The rod is initially vertical and the hinge is spun with a fixed angular velocity ω.
(a) Before starting, explain why this problem has to use rods, and not just strings.
(b) If a mass m if attached to the bottom of the rod, find the maximum ω for which the configu-ration is stable.
(c) [A] Now suppose another mass m and rod of length ℓis attached to the first mass by an identical hinge that turns in the same direction, as shown above. Find the maximum ω for which the configuration is stable. (Hint: the configuration is unstable if any infinitesimal change in the angles of the rods can lower the energy.) Problem 25 (PPP 106). A long, heavy flexible rope with mass ρ per unit length is stretched by a constant force F. A sudden movement causes a circular loop to form at one end of the rope.
The center of the loop moves with speed c as shown.
(a) Calculate the speed c, assuming gravity is negligible.
(b) Find the energy E carried by a loop rotating with angular frequency ω.
(c) Show that the momentum p carried by the loop obeys E = pc. This is true for waves in general, as we’ll see in W1.
(d) Find the angular momentum carried by the loop.
5 [A] Advanced Mathematical Techniques The following problems were cut from earlier problem sets because they required more advanced math; however, they illustrate some very neat and important ideas.
Problem 26. In P1, you found a general expression for the period of a pendulum oscillating with amplitude θ0 in terms of an integral, then approximated the integral for θ0 ≪1 to find ω = ω0 1 −θ2 0 16 + O(θ4 0) 21 Kevin Zhou Physics Olympiad Handouts where ω0 = p g/L. In this problem, we will show a different way to get the same answer, by solving the equation of motion approximately. We write the solution θ(t) as a series in θ0. The overall solution is of order θ0, and the corrections only depend on θ2 0, so we can write θ(t) = θ0f0(t) + θ3 0f1(t) + θ5 0f2(t) + . . .
where all the functions fi(t) are of order 1. Then we plug this expansion into Newton’s second law, ¨ θ + ω2 0 sin θ = 0, and expand it out order by order in θ0.
(a) A naive first guess is to set f0(t) so that it cancels precisely the order θ0 terms in this equation, then set f1(t) to cancel the order θ3 0 terms, and so on. Using this guess, show that ¨ f0 + ω2 0f0 = 0, ¨ f1 + ω2 0f1 = ω2 0f3 0 6 where the first equation has solution f0(t) = cos(ω0t).
Unfortunately, this decomposition is not very useful. The problem is that two things are going on at once: the oscillations are not quite sinusoidal, and they have an angular frequency lower than ω0. The expansion we’ve done would be useful if we only had the first effect, because then f1(t) would just capture the small, non-sinusoidal corrections to f0(t). But our method can’t account for the frequency shift; by construction, f0(t) always oscillates at angular frequency ω0. Over time, the real oscillation θ(t) gets out of phase with f0(t). This manifests itself as a “secular growth” in f1(t), i.e. it increases in magnitude every cycle until it has a huge value, of order 1/θ2 0, and our perturbative expansion breaks down.
(b) Write the right-hand side of the differential equation for f1(t) as a sum of sinusoids, and show that it contains a term proportional to cos(ω0t). This resonantly drives f1(t), causing the secular growth.
(c) We can salvage our perturbative expansion using the method of “renormalized” frequencies.
We impose by fiat that f0(t) oscillates at the true angular frequency, letting ¨ f0 + ω2f0 = 0, ω = ω0(1 −c θ2 0 + O(θ4 0)) for a constant c. Because of this choice, the differential equation for f1(t), which contains all terms at order θ3 0, will be altered. The correct choice of ω is precisely the one for which this eliminates the secular growth of f1(t). Using this idea, show that c = 1/16.
If you keep going, you’ll find the next term f2(t) still has secular growth. We can remove it by having both f0(t) and f1(t) oscillate at angular frequency ω0(1 −θ2 0/16 + dθ4 0), where d is chosen to cancel the secular growth of f2(t). In this way, the frequency can be found to any order in θ2 0. (This technique is called the method of strained coordinates. It’s an example of multiple-scale analysis.) Problem 27. You might be wondering how we can solve the weakening spring problem from M4 without anything fancy like the adiabatic theorem. There is a general technique to solve linear differential equations whose coefficients are slowly varying. First, write the equation of motion as ¨ x + ω2(t)x = 0.
Then expand x(t) as x(t) = A(t)eiϕ(t), ˙ ϕ(t) = ω(t).
The point of writing x(t) this way is that pulling out the factor of eiϕ(t) will automatically account for the rapid oscillations. The factor A(t) only varies slowly, so it’s easier to handle by itself.
22 Kevin Zhou Physics Olympiad Handouts (a) Evaluate ¨ x(t) and plug it into the equation of motion.
(b) Using the fact that A(t) and ω(t) vary slowly, throw out small terms in your equation from part (a), until you get a differential equation you can easily integrate. This is an example of the WKB approximation for differential equations, which we applied at length in X1.
(c) Show that this gives the expected final result for a weakening spring.
Problem 28 (BAUPC 1996). A mass M is located at the vertex of an angle θ ≪1 formed by two massless sticks of length ℓ. The structure is held so that the left stick is initially vertical, then released. The right stick hits the ground at time t = 0. The structure then rocks back and forth, coming to a stop at time t = T.
(a) Prove the identity 1 + 1 32 + 1 52 + 1 72 + . . . = π2 8 using the result P n≥1 1/n2 = π2/6, which we derived in W1.
(b) Using this result, calculate T to leading order in θ.
Problem 29. In this problem, we’ll go through Laplace’s slick derivation of Kepler’s first law.
Throughout, we assume the orbit takes place in the xy plane, with the Sun at the origin.
(a) Show that ¨ x = −γx r3 , ¨ y = −γy r3 where γ is a constant that depends on the parameters.
(b) Show that d dt(r3¨ x) = −γ ˙ x, d dt(r3¨ y) = −γ ˙ y.
(c) Show that d dt(r3¨ r) = −γ ˙ r.
(Hint: this can get messy.
As a first step, try showing the left-hand side is equal to (r2/2) d3(r2)/dt3. You will have to switch variables to x and y and then switch back; for these purposes it’s useful to use the results of part (a), and the definition r2 = x2 + y2.) (d) Define ψ(t) = r(t)3. In parts (b) and (c), we have shown that the differential equation d dt ψ(t) du dt = −γu has three solutions, namely ˙ x, ˙ y, and ˙ r. Any second-order linear differential equations only has two independent solutions. If ˙ x and ˙ y are not independent, the orbit is simply a line, which is trivial. Assuming that doesn’t happen, they are independent, so ˙ r must be a linear combination of them, ˙ r = A ˙ x + B ˙ y.
Use this result to argue that the orbit is a conic section.
23 Kevin Zhou Physics Olympiad Handouts 6 Mechanics and Geometry For dessert, we’ll consider a few cute problems that relate statics to geometry.
Example 8 Given a triangle ABC, the Fermat point is the point X that minimizes AX + BX + CX.
Design a machine that finds the Fermat point.
Solution We take a horizontal plane and drill holes at points A, B, and C. A mass M on a rope is fed through each hole, and the three ends of the rope are tied together at point X. The gravitational potential energy is proportional to AX + BX + CX, so in equilibrium X lies on the Fermat point. Moreover, since the tensions in each rope are all equal to Mg, force balance requires ∠AXB = ∠BXC = ∠CXA = 120◦.
Problem 30. Using similar reasoning, design a machine that finds the point X that minimizes (AX)2 + (BX)2 + (CX)2. What geometrical property can you conclude about this point?
Example 9 Show that the incenter of a triangle (i.e. the meeting point of the angle bisectors) exists.
Solution Apply six forces at the vertices of a triangle as shown.
These forces clearly balance, and also produce no net torque on the triangle. Now combine the forces applied at each vertex, yielding three forces that point along the angle bisectors.
By the principles of M2, the torques of these forces can only balance if their lines of action meet at a point. Therefore the angle bisectors are concurrent, so the incenter exists.
Example 10 Let AB be a diameter of a circle, and let a mass be free to slide on the circle. The mass is connected to two identical straight springs of zero rest length, which are in turn connected to points A and B. At what points C can the mass be in static equilibrium?
24 Kevin Zhou Physics Olympiad Handouts Solution The potential energy of the system is proportional to (AC)2 + (BC)2. Since ABC is a right triangle, this is just equal to (AB)2 by the Pythagorean theorem. Since the potential energy doesn’t depend on where the mass is, it can be at static equilibrium at any point on the circle. Alternatively, you can show that the mass is in static equilibrium by force balance, and use the reasoning in reverse to derive the Pythagorean theorem.
Problem 31. Consider a right triangle ABC filled with a fluid of uniform pressure. Using torque balance, establish the Pythagorean theorem.
Problem 32. Shown below is a setup due to the 16th century mathematician Stevin, who was also known for introducing decimal numbers.
One might argue that because there are more masses on AB than on BC, this is a perpetual motion machine that turns counterclockwise. By using the fact that perpetual motion machines don’t actually exist, prove the law of sines.
Problem 33. Consider the n-sided polygon P of least possible area that circumscribes a closed convex curve K. Prove that every tangency point of K with a side of P is the midpoint of that side.
(Hint: begin by supposing that the area outside P is filled with a gas of uniform pressure, with a vacuum inside P.) Problem 34. In this problem we’ll derive Kepler’s first law yet again, using no calculus, but a bit of Euclidean geometry. As usual, we suppose a planet of mass m orbits a fixed star of much greater mass M. Placing the star at the origin, let ϕ be the angle between r and v for the planet.
(a) Write down the quantities E and L in terms of G, M, m, v, r, and ϕ, and show that r2 + GMm E r sin2 ϕ = L2 2mE .
(b) Now consider an ellipse with semimajor axis a and eccentricity e, meaning that the distance between the foci is 2ae, with one of the foci F at the origin. Consider a point P on the ellipse, so that the angle between the tangent to the ellipse at P and FP is ϕ. If r = |FP|, show that (r2 −2ar) sin2 ϕ = −a2(1 −e2).
You will have to use the geometrical property that a light ray sent from one focus will reflect at the ellipse to hit the other focus.
25 Kevin Zhou Physics Olympiad Handouts (c) By comparing your results for (a) and (b), conclude that the orbit is an ellipse with a = −GMm 2E , e = r 1 + 2EL2 G2M2m3 .
26 |
187717 | https://link.springer.com/article/10.1007/s13312-025-00150-1 | Betel Leaf-Induced Contact Leukomelanosis in a Child | Indian Pediatrics
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Indian Pediatrics
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Betel Leaf-Induced Contact Leukomelanosis in a Child
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Published: 31 July 2025
Volume 62,page 788, (2025)
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Betel Leaf-Induced Contact Leukomelanosis in a Child
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Kaliaperumal Karthikeyan1&
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A 2-year-old boy was brought with multiple pale macules over the upper chest noted by parents over the past year. The lesions were asymptomatic, non-scaly, and had gradually increased in number. On detailed inquiry, the mother gave a history of application of warm crushed betel (Piper betle) leaves on the chest during respiratory illnesses, a home remedy commonly practiced in their native village in Tamil Nadu. Examination revealed multiple, well-defined, hypopigmented macules without scaling or atrophy, over the chest (Fig.1). Wood’s lamp showed no accentuation, and potassium hydroxide (KOH)mount was negative, which helped to exclude underlying vitiligo and pityriasis versicolor, respectively. Considering the history, a diagnosis of contact leukomelanosis due to betel leaves was considered. A confirmatory patch test could not be performed due to parental reluctance. The family was counseled to stop the practice of local application of betel leaves, and topical corticosteroids were prescribed. On follow-up, partial repigmentation was observed at the end of 3 months.
Fig.1
Multiple well-defined hypopigmented macules over the chest
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Heated betel leaves are traditionally applied to the chest of children during respiratory illnesses in several Indian states such as West Bengal, Odisha, Tamil Nadu, and Andhra Pradesh. The betel leaf contains bioactive compounds such as eugenol and chavicol, which can cause melanocyte damage, especially with prolonged exposure after topical application. The resulting depigmentation or hypopigmentation may be mistaken for vitiligo or pityriasis versicolor. The absence of scaling, segmental arrangement, or progression to well-demarcated lesions helps differentiate it from those conditions. Lack of Wood’s lamp fluorescence and a negative KOH preparation further aid diagnosis. Ash leaf macules, seen in tuberous sclerosis, are an important differential diagnosis and typically present as hypopigmented, elliptical or lance-ovate macules, commonly over the trunk and limbs. Recognition of this contact leukomelanosis is especially important in rural or semi-urban populations where traditional remedies are common. Management involves discontinuing the application and use of topical anti-inflammatory agents. This case highlights the need for cultural sensitivity and anticipatory guidance in pediatric dermatology.
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Department of Dermatology, Venereology and Leprosy, Sri Manakula Vinayagar Medical College and Hospital, Puducherry, 605107, India
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Karthikeyan, K., Palaniappan, V. Betel Leaf-Induced Contact Leukomelanosis in a Child. Indian Pediatr62, 788 (2025).
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العربية
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Azərbaycanca
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Беларуская (тарашкевіца)
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From Wikipedia, the free encyclopedia
Child who has lost their parents
For other uses, see Orphan (disambiguation).
An orphan is a child whose parents have died, are unknown, or have permanently abandoned them. It can also refer to a child who has lost only one parent, as the Hebrew translation, for example, is "fatherless". In some languages, such as Swedish, the term is "parentless" and more ambiguous about whether the parents are dead, unknown or absconded, but typically refers to a child or younger adult.[citation needed]
In common usage, only a child who has lost both parents due to death is called an orphan. When referring to animals, only the mother's condition is usually relevant (i.e., if the female parent has gone, the offspring is an orphan, regardless of the father's condition).
Definitions
[edit]
Various groups use different definitions to identify orphans. One legal definition used in the United States is a minor bereft through "death or disappearance of, abandonment or desertion by, or separation or loss from, both parents".
In everyday use, an orphan does not have any surviving parent to care for them. However, the United Nations Children's Fund (UNICEF), Joint United Nations Programme on HIV and AIDS (UNAIDS), and other groups label any child who has lost one parent as an orphan. In this approach, a maternal orphan is a child whose mother has died, a paternal orphan is a child whose father has died, and a double orphan is a child who has lost both parents. This contrasts with the older use of half-orphan to describe children who had lost only one parent.
History
[edit]
Wars, epidemics (such as AIDS), pandemics, and poverty
have led to many children becoming orphans. The Second World War (1939–1945), with its massive numbers of deaths and vast population movements, left large numbers of orphans in many countries—with estimates for Europe ranging from 1,000,000 to 13,000,000. Judt (2006) estimates there were 9,000 orphaned children in Czechoslovakia, 60,000 in the Netherlands, 300,000 in Poland and 200,000 in Yugoslavia, plus many more in the Soviet Union, Germany, Italy, China and elsewhere.
Populations
[edit]
Orphans are relatively rare in developed countries because most children can expect both of their parents to survive their childhood. Much higher numbers of orphans exist in war-torn nations such as Afghanistan.
| Continent | Number of orphans (1000s) | Orphans as percentage of all children |
---
| Africa | 34,294 | 11.9% |
| Asia | 65,504 | 6.5% |
| Latin America & Caribbean | 8,166 | 7.4% |
| Total | 107,964 | 7.6% |
| Year | Country | Orphans as % of all children | AIDS orphans as % of orphans | Total orphans | Total orphans (AIDS related) | Maternal (total) | Maternal (AIDS related) | Paternal (total) | Paternal (AIDS related) | Double (total) | Double (AIDS related) |
--- --- --- --- --- --- |
| 1990 | Botswana | 5.9 | 3.0 | 34,000 | 1,000 | 14,000 | < 100 | 23,000 | 1,000 | 2,000 | < 100 |
| Lesotho | 10.6 | 2.9 | 73,000 | < 100 | 31,000 | < 100 | 49,000 | < 100 | 8,000 | < 100 |
| Malawi | 11.8 | 5.7 | 524,000 | 30,000 | 233,000 | 11,000 | 346,000 | 23,000 | 55,000 | 6,000 |
| Uganda | 12.2 | 17.4 | 1,015,000 | 177,000 | 437,000 | 72,000 | 700,000 | 138,000 | 122,000 | 44,000 |
| 1995 | Botswana | 8.3 | 33.7 | 55,000 | 18,000 | 19,000 | 7,000 | 37,000 | 13,000 | 5,000 | 3,000 |
| Lesotho | 10.3 | 5.5 | 77,000 | 4,000 | 31,000 | 1,000 | 52,000 | 4,000 | 7,000 | 1,000 |
| Malawi | 14.2 | 24.6 | 664,000 | 163,000 | 305,000 | 78,000 | 442,000 | 115,000 | 83,000 | 41,000 |
| Uganda | 14.9 | 42.4 | 1,456,000 | 617,000 | 720,000 | 341,000 | 1,019,000 | 450,000 | 282,000 | 211,000 |
| 2001 | Botswana | 15.1 | 70.5 | 98,000 | 69,000 | 69,000 | 58,000 | 91,000 | 69,000 | 62,000 | 61,000 |
| Lesotho | 17.0 | 53.5 | 137,000 | 73,000 | 66,000 | 38,000 | 108,000 | 63,000 | 37,000 | 32,000 |
| Malawi | 17.5 | 49.9 | 937,000 | 468,000 | 506,000 | 282,000 | 624,000 | 315,000 | 194,000 | 159,000 |
| Uganda | 14.6 | 51.1 | 1,731,000 | 884,000 | 902,000 | 517,000 | 1,144,000 | 581,000 | 315,000 | 257,000 |
2001 figures from 2002 UNICEF/UNAIDS report
China: A survey conducted by the Ministry of Civil Affairs in 2005 showed that China has about 573,000 orphans below 18 years old.
Russia: According to Russian reports from 2002 cited in the New York Times, 650,000 children are housed in orphanages. They are released at age 16, and 40% become homeless, while 30% become criminals or commit suicide.
Latin America: Street children have a major presence in Latin America; some estimate that there are as many as 40 million street children in Latin America. Although not all street children are orphans, all street children work and many do not have significant family support.
United States: About 2 million children in the United States (or about 2.7 percent of children) have a deceased mother or father. About 100,000 children have lost both parents.
Notable orphans
[edit]
Main article: List of orphans and foundlings
Famous orphans include world leaders such as Aaron Burr, Andrew Jackson, and Pedro II of Brazil; writers such as Edgar Allan Poe and Leo Tolstoy; and athletes such as Aaron Hernandez. The American orphan Henry Darger portrayed the horrible conditions of his orphanage in his artwork. Other notable orphans include entertainment greats such as Louis Armstrong, Marilyn Monroe, Babe Ruth, Ray Charles and Frances McDormand.
In fiction
[edit]
Orphaned characters are prevalent as literary protagonists, especially in children's and fantasy literature. The lack of parents leaves the characters to pursue more exciting and adventurous lives, by freeing them from familial obligations and controls, and depriving them of more prosaic lives. It creates characters that are self-contained and introspective and who strive for affection. Orphans can metaphorically search for self-understanding by attempting to know their roots. Parents can also be allies and sources of aid for children, and removing the parents makes the character's difficulties more severe. Parents, furthermore, can be irrelevant to the theme a writer is trying to develop, and orphaning the character frees the writer from depicting such an irrelevant relationship; if one parent-child relationship is important, removing the other parent prevents complicating the necessary relationship. All these characteristics make orphans attractive characters for authors.
Orphans are common in fairy tales, such as most variants of Cinderella.
Several well-known authors have written books featuring orphans. Examples from classic literature include Charlotte Brontë's Jane Eyre, Charles Dickens's Oliver Twist, Mark Twain's Tom Sawyer and Huckleberry Finn, L. M. Montgomery's Anne of Green Gables, Thomas Hardy's Jude the Obscure, Victor Hugo's Les Misérables, Edgar Rice Burroughs's Tarzan of the Apes, Rudyard Kipling's The Jungle Book, and J. R. R. Tolkien's The Lord of the Rings.
More recent authors featuring orphan characters include A. J. Cronin, Lemony Snicket, A. F. Coniglio, Roald Dahl and J. K. Rowling. One recurring storyline has been the relationship that the orphan can have with an adult from outside their immediate family, as seen in Lyle Kessler's play Orphans.
Other examples
[edit]
Orphans are especially common as characters in comic books. Many popular heroes are orphans, including Superman, Batman, Spider-Man, Robin, Flash, Captain Marvel, Captain America, and Green Arrow. Orphans are also very common among villains: Bane, Catwoman, and Magneto are examples. Lex Luthor, Deadpool, and Carnage can also be included on this list, though they killed one or both of their parents. Supporting characters befriended by the heroes are also often orphans, including the Newsboy Legion and Rick Jones.
Other famous fictional orphans include Little Orphan Annie, Anakin Skywalker, Luke Skywalker and his sister, Leia Organa, and several main characters in children's shows like Diff'rent Strokes and Punky Brewster.
In religious texts
[edit]
Many religious texts, including the Bible and the Quran, contain the idea that helping and defending orphans is a fundamental and God-pleasing matter. The religious leaders Moses and Muhammad were orphaned as children. Several scriptural citations describe how orphans should be treated:
Bible
"Do not take advantage of a widow or an orphan." (Hebrew Bible, Exodus 22:22)
"Be joyful at your festival—you, your sons and daughters, your male and female servants, and the Levites, the foreigners, the fatherless and the widows who live in your towns".(Hebrew Bible, Book of Deuteronomy 16:14)
"Leave your orphans; I will protect their lives. Your widows too can trust in me." (Hebrew Bible, Jeremiah 49:11)
"To judge the fatherless and the oppressed, that the man of the earth may no more oppress." (Hebrew Bible, Psalms 10:18)
"I will not leave you as orphans; I will come to you." (New Testament, John 14:18)
"Religion that God our Father accepts as pure and faultless is this: to look after orphans and widows in their distress and to keep oneself from being polluted by the world." (New Testament, James 1:27)
Qu'ran
"And they feed, for the love of God, the indigent, the orphan, and the captive," - (The Quran, The Human: 8)
"Therefore, treat not the orphan with harshness," (The Quran, The Morning Hours: 9)
"Have you not seen those who deny the faith and the Day of Judgment? Those are people who drive orphans away harshly, and do not encourage feeding the indigent. So woe be upon those who do prayer but are neglectful of it or show it off out of vanity, and those who deny even small kindnesses to others." - (The Quran, Small Kindnesses: 1–7)
"(Be good to) orphans and the very poor. And speak good words to people." (The Quran, The Heifer: 83)
"...They will ask you about the property of orphans. Say, 'Managing it in their best interests is best'. If you mix your property with theirs, they are your brothers..." (The Quran, The Heifer: 220)
"Give orphans their property, and do not substitute bad things for good. Do not assimilate their property into your own. Doing that is a serious crime." (The Quran, The Women: 2)
"Keep a close check on orphans until they reach a marriageable age, then if you perceive that they have sound judgement hand over their property to them..." (The Quran, The Women: 6)
See also
[edit]
Adoption
AIDS orphan
Child abandonment
Foster
Legitimacy (family law)
Orphan Train
Orphanage
Street children
Empower Orphans
Euro-orphan
References
[edit]
^ "Definition of ORPHAN". www.merriam-webster.com. May 16, 2023.
^ The Shorter Oxford English Dictionary, 3rd edition "One deprived by death of father or mother, or (usu.) of both; a fatherless or motherless child."
^ "orphan". Dictionary.com.
^ "USCIS definition for immigration purposes". Archived from the original on 2019-07-30. Retrieved 2015-10-21.
^ "UNAIDS Global Report 2008" (PDF). UN AIDS. Archived from the original (PDF) on 2019-06-17. Retrieved 2023-08-14.
^ See, for example, this 19th-century news story about The Society for the Relief of Half-Orphan and Destitute Children, or this one about the Protestant Half-Orphan Asylum.
^
Roman, Nicoleta (8 November 2017). "Introduction". In Roman, Nicoleta (ed.). Orphans and Abandoned Children in European History: Sixteenth to Twentieth Centuries. Routledge Studies in Modern European History. Abingdon: Routledge (published 2017). ISBN 9781351628839. Retrieved 25 November 2020. The industrial revolution touched both villages and cities, with migration from one to the other going hand-in-hand with urban overpopulation and severe poverty. Urban population growth also led to an increase in abandonment, the poor swinging between finding work, begging or claiming social assistance from the State as a means of integrating themselves and their family, including their children, into society.
^
For a high estimate see I.C.B. Dear and M.R.D. Foot, eds. The Oxford Companion to World War II (1995) p. 208; for lower, see Tony Judt, Postwar: a history of Europe since 1945 (2006) p. 21.
^ USAID/UNICEF/UNAIDS (2002) "Children on the brink 2002: a joint report on orphan estimates and program strategies", Washington: USAID/UNICEF/UNAIDS.
^ TvT Associates/The Synergy Project (July 2002). "Children on the Brink 2002: A Joint Report on Orphan Estimates and Program Strategies" (PDF). UNAIDS and UNICEF. Archived from the original (PDF) on December 23, 2003.
^ "China to insure orphans as preventitive health measure_English_Xinhua". July 22, 2009. Archived from the original on 2009-07-22.
^ "A Summer of Hope for Russian Orphans". The New York Times. July 21, 2002.
^ Tacon, P. (1982). "Carlinhos: the hard gloss of city polish". UNICEF news. {{cite journal}}: Cite journal requires |journal= (help)
^ Scanlon, TJ (1998). "Street children in Latin America". BMJ. 316 (7144): 1596–2100. doi:10.1136/bmj.316.7144.1596. PMC 1113205. PMID 9596604.
^ Weaver, David (5 September 2019). "Parental Mortality and Outcomes among Minor and Adult Children". papers.ssrn.com. SSRN 3471209.
^ Philip Martin, The Writer's Guide to Fantasy Literature: From Dragon's Lair to Hero's Quest, p 16, ISBN 0-87116-195-8
^ Rabbi Eliezer Melamed, To Enjoy and Bring Joy to Others in Peninei Halakha - Laws of the Festivals
Bibliography
[edit]
Bullen, John. "Orphans, Idiots, Lunatics, and More Idiots: Recent Approaches to the History of Child Welfare in Canada," Histoire Sociale: Social History, May 1985, Vol. 18 Issue 35, pp 133–145
Harrington, Joel F. "The Unwanted Child: The Fate of Foundlings, Orphans and Juvenile Criminals in Early Modern Germany (2009)
Keating, Janie. A Child for Keeps: The History of Adoption in England, 1918-45 (2009)
Miller, Timothy S. The Orphans of Byzantium: Child Welfare in the Christian Empire (2009)
Safley, Thomas Max. Children of the Laboring Poor: Expectation and Experience Among the Orphans of Early Modern Augsburg (2006)
Sen, Satadru. "The orphaned colony: Orphanage, child and authority in British India," Indian Economic and Social History Review, Oct-Dec 2007, Vol. 44 Issue 4, pp 463-488
Terpstra, Nicholas. Abandoned Children of the Italian Renaissance: Orphan Care in Florence and Bologna (2005)
United States
[edit]
Berebitsky, Julie. Like Our Very Own: Adoption and the Changing Culture of Motherhood, 1851-1950 (2000) ISBN 0700610510
Carp, E. Wayne, ed. Adoption in America: Historical Perspectives (2003) ISBN 0472109995
Hacsi, Timothy A. A Second Home: Orphan Asylums and Poor Families in America (1997) ISBN 0674796446
Herman, Ellen. "Kinship by Design: A History of Adoption in the Modern United States (2008) ISBN 978-0-226-32760-0
Kleinberg, S. J. Widows And Orphans First: The Family Economy And Social Welfare Policy, 1880-1939 (2006) ISBN 0252030206
Miller, Julie. Abandoned: Foundlings in Nineteenth-Century New York City (2007) ISBN 0814757251
External links
[edit]
The dictionary definition of orphan at Wiktionary
Media related to Orphans at Wikimedia Commons
| v t e Parenting | |
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| Kinship terminology | Parent Mother Father Adoptive Alloparenting Coparenting Extended family Foster care Noncustodial Nuclear family Orphaned Same-sex Shared parenting Single parent Blended family Surrogacy In loco parentis |
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187720 | https://encyclopediaofmath.org/wiki/Abel_transformation | Abel transformation - Encyclopedia of Mathematics
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Abel transformation
From Encyclopedia of Mathematics
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2020 Mathematics Subject Classification: Primary:40A05 [MSN][ZBL] summation by parts, Abel's lemma
A discrete analog of integration by parts, introduced by N.H. Abel in [Ab]. If a 1,…,a N, b 1,…,b N, are given complex numbers and we set B n=∑i≤n b i
then the summation by parts is the identity N∑k=1 a k b k=a N B N−N−1∑k=1 B k(a k+1−a k). Observe that, if B 0 is arbitrarily chosen and we modify the definition of B n by B n=B 0+∑i≤n b i then the identity becomes N∑k=1=a N B N−a 1 B 0−N−1∑k=1 B k(a k+1−a k), in analogy with the arbitrarity of an additive constant in the primitive of a function.
If a n→0 and {B n} is a bounded sequence, the Abel transformation shows that ∑k a k b k converges if and only if ∑k B k(a k+1−a k) converges, in which case it yields the formula ∞∑k=1 a k b k=∞∑k=1 B k(a k−a k+1)−a 1 B 0.
This fact can be used to prove several very useful criteria of convergence of series of numbers and functions (cf. Abel criterion). The Abel transformation of a series often yields a series with an identical sum, but with a better convergence. It is also regularly used to obtain certain estimates (cf. Abel inequality), in particular, for investigations on the rate of convergence of a series.
References
[Ab]N.H. Abel, "Untersuchungen über die Reihe 1+m x+m⋅(m−1)2⋅1 x 2+m⋅(m−1)⋅(m−2)3⋅2⋅1 x 3+… u.s.w.", J. Reine Angew. Math. , 1 (1826) pp. 311–339
[Ca]H. Cartan, "Elementary Theory of Analytic Functions of One or Several Complex Variable", Dover (1995).
[Ma]A.I. Markushevich, "Theory of functions of a complex variable" , 1 , Chelsea (1977) (Translated from Russian)
[WW]E.T. Whittaker, G.N. Watson, "A course of modern analysis" , 1–2 , Cambridge Univ. Press (1952)
How to Cite This Entry:
Abel transformation. Encyclopedia of Mathematics. URL:
This article was adapted from an original article by L.P. Kuptsov (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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187721 | https://www.chemistrysteps.com/initiation-propagation-termination-in-radical-reactions/ | Skip to content
Radical Reactions
Initiation Propagation Termination in Radical Reactions
Initiation, Propagation, Termination – The Mechanisms
Let’s discuss the chlorination of ethane to illustrate the three important steps of radical halogenation:
Initiation of Radical Halogenation
The reaction starts with homolysis of a σ bond forming two radicals. This is called initiation – the start of the radical chain reactions. We are breaking a bond in the starting materials, so this step requires energy from light or heat:
Notice that unlike most other reactions in organic chemistry such as nucleophilic substitution, elimination or additions, radical reactions are shown by half-headed arrows and one arrow shows movement of one electron.
Half-headed arrows, also known as fishhook arrows, indicate that the σ bond is broken homolitically, i.e. the electrons of the covalent bond are shared between the two atoms.
Initiation at moderate temperatures can also be catalyzed by peroxides. One of the features of peroxides is the presence of a weak O-O bond which, upon breaking, forms radicals.
Have you ever wondered why some medicals are sold in dark while others in dark bottles?
Acyl peroxides are especially useful for this purpose since the resulting radicals are resonance stabilized:
Simple peroxides act in a similar way to produce alkoxyl radicals which are then used in radical reactions:
Propagation in Radical Halogenation
The Cl• radicals produced in the initiation step are very unstable. These are the chlorine atoms and you know from general chemistry that halogens exist as diatomic molecules and not atoms.
In the next step, these reactive radicals quickly react with ethane by abstracting a hydrogen:
This is a propagation step and what is important in this reaction between a neutral molecule and a radical is that a new radical is created.
The hydrogen abstraction by the halogen radical, is the rate-determining step in radical halogenation.
The alkyl radicals are also very reactive, and they react with the chlorine gas producing a new chlorine radical. Thus, the process continues over and over without a need for supplying energy to perform an initiation again:
This is why radical reactions are classified as chain reactions – only an initial boost is needed to start the reaction and once started, the processes continue repeatedly.
It is also worth mentioning that hydrogen abstraction is different from proton transfer reactions which follow ionic mechanism. Hydrogen abstraction is the movement of H• radical while, proton transfer is the movement of H+ ion:
Mono and Polychlorination in Radical Reactions
The initiation and propagation steps we covered so far represent monochlorination to form chloroethane (ethyl chloride). However, if Cl2 is added in excess or equal quantity, polychlorination is observed.
For example, when equal amounts of methane and chlorine are mixed and irradiated with a light of the appropriate wavelength, the reaction produces a mixture of following products:
The reason for getting this mixture of mono- and polychlorinated methane is the change of reactant and product concentrations as the reaction proceeds.
Initially, only chlorine and methane are present, and their reaction produces chloromethane and hydrogen chloride:
However, in the course of the reaction, the concentration of chloromethane in the mixture increases. In addition, chloromethane is more reactive than methane and despite its lower concentration, it starts reacting with chlorine to produce dichloromethane:
In a similar manner, as the concentration of dichloromethane increases it produced trichloromethane which is then converted into tetrachloromethane.
The chlorination of ethane produces 1,1-dichloroethane and 1,2-dichloroethane, as well as a mixture of more highly chlorinated ethanes:
Therefore, in order to achieve monohalogenation, the alkane must always be used in excess. Monohalogenation is important as it allows to perform selective halogenation using bromine and, in general, it is more useful for preparing haloalkanes used in organic synthesis.
There is one good question you might be wondering:
If the Radicals Are so Unstable, Why Don’t They React Together?
The answer to this is they do. In fact, when two radicals meet, they react very fast because of their instability. However, the keyword here is “when” or “if”.
The concentration of radicals is very low compared to the concentration of neutral molecules and they simple do not get a chance to collide. Therefore, the propagation continues repeating the steps thousands of times before eventually radicals meet and react.
When two radicals react, they start sharing the unpaired electrons and form a new σ bond. This is the termination step which ends the chain of propagation steps:
The termination is not restricted to the reaction of specific radicals. Any two radicals reacting to form neutral species represent a termination reaction.
So, termination is the combination of two radicals to form a stable bond and the net result is going from two radicals to no radicals.
This may seem too much if you are going over it first itme but one quick way of identifying these steps in radical halogenation is to keep track of the number of radicals:
Initiation – goes from 0 radicals to two radicals
Propagation – goes from 1 radical to two radicals
Termination – goes from 2 radicals to 0 radicals
Practice
1.
Draw the appropriate fishhook arrows for the following radical process and identify them as Initiation Propagation or Termination:
Answer
2.
Answer
3.
Answer
4.
Answer
5.
Answer
6.
Answer
7.
Check Also:
Free-Radical Addition of HBr: Anti-Markovnikov Addition
Initiation Propagation Termination in Radical Reactions
Selectivity in Radical Halogenation
Stability of Radicals
Resonance Structures of Radicals
Stereochemistry of Radical Halogenation
Allylic Bromination
Radical Halogenation in Organic Synthesis
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187722 | https://mychemistryclass.net/Files/HONORS%20CHEM/5%20Bonding%20and%20Structure/Packet/Bonding%20and%20Structure%20WS%2010%20Molecular%20Geometry%20Activity%20PhET%20Version.pdf | Dougherty Valley HS Chemistry Bonding and Structure – Molecular Geometry Activity Name: Period: Seat#: Purpose: To construct a series of compounds using the VSEPR model and to use your model to determine the type of bonding and hybridization, and the geometry around each central atom. Pre-Activity Questions: The VSEPR model is based on the premise that electron pairs around a central atom will position themselves to allow for maximum separation. Instead of writing an actual Background Paragraph, just answer these questions below. 1) What does VSEPR stand for? 3) Name the five different electronic geometries, and the eleven different molecular geometries. 2) Explain why pairs of electrons around a central atom repel each other. Materials: - Computer/Laptop - Color pencils/markers Procedure: 1. Construct a 3D model for each compound using the online PhET simulation, and then sketch onto your paper. a. b. Click to turn on the following: Lone pairs Bond angles Electronic and Molecular Geometry c. Click in the bottom right corner where it says “PhET” and there are three vertical dots 2. Click options, then “projector mode” – it makes the background white so it is much easier to see things (I think so at least!). 3. Draw Lewis Structure 4. Determine the following for each atom: a. Number of bonded atoms on center atom, number of lone pairs on center atom. b. AXE formula (A center atom, X number of atoms bonded to the center atom, E number of lone pairs on the center atom) c. Steric Number 5. Using the information from Step 2 and a VSPER chart (which should be memorized!), determine the following: a. Electronic Geometry (linear, trigonal planar, tetrahedral, trigonal bi-pyramidal, or octahedral) b. Molecular Geometry (linear, trigonal planar, bent, tetrahedral, trigonal pyramidal, trigonal bi-pyramidal, seesaw, T-shaped, octahedral, square planar) c. Bond angle between the atoms attached to the central atom. (Based on the molecular geometry) d. Type of hybridization of the central atom in each molecule – if any (sp, sp2, sp3, sp3d, sp3d2 – remember, d hybridization may not be real!) Worksheet #10 Dougherty Valley HS Chemistry Bonding and Structure – Molecular Geometry Activity Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch NO3 — AX3 # v.e- = # bonded atoms on A 3 # of lone pairs on A 0 Steric Number Molecular Geometry Hybridization 3 Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch SiCl4 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch CO2 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Dougherty Valley HS Chemistry Bonding and Structure – Molecular Geometry Activity Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch NClH2 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch XeF4 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch CH2O # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Dougherty Valley HS Chemistry Bonding and Structure – Molecular Geometry Activity Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch SF6 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch BF3 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch NO2 — # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Dougherty Valley HS Chemistry Bonding and Structure – Molecular Geometry Activity Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch SF4 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch ClF3 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch BrF5 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Dougherty Valley HS Chemistry Bonding and Structure – Molecular Geometry Activity Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch N2 # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Molecular Formula AXE Formula Lewis Structure Electronic Geometry Bond Angle 3D Sketch NH4+ # v.e- = # bonded atoms on A # of lone pairs on A Steric Number Molecular Geometry Hybridization Done early? You can try doing these too! CCl4, NH3, H2O, SCl2, I3-, SO2, ICl4-, AsF5, IF4+, H3O+, TeF5-, HCN, IOF5, BrF3, SO42-, CO32- Another teacher made some online card making practices for VSEPR shapes! (please let me know if these links stop working) AXE Formulas and Geometry Names Shapes and 3D Models AXE Formulas and 3D Shapes |
187723 | https://personal.tcu.edu/jmontchamp/F2019%20files%20for%20lab/LAB5%20handout%20ALDOL%20condensation.pdf | 1 ALDOL CONDENSATION Condensation is a reaction between two or more molecules that leads to the formation of a larger molecule and an elimination of a smaller molecule (usually water). Aldol condensation refers to reactions that involve carbonyl-containing compounds, i.e., aldehydes and ketones, which yield β-hydroxy carbonyl products – aldol (aldehyde+alcohol). If both partners in the condensation processes are the same – the reaction can be referred to as self-condensation. An example of self-condensation is shown in Scheme 1. The aldol reactions are catalyzed by either an acid or a base. The reason for such ambivalent nature stems from the ability of carbonyl compounds to act either as electrophilies or as nucleophiles. In either case, an enol or enolate ion act as nucleophilic species (Figure 1), which react with the electrophilic center of the carbonyl compound. The detailed mechanism of a base-catalyzed aldol condensation is shown in Scheme 2. The reaction starts with a formation of the enolate ion species, upon abstraction of a proton from acetone (not that in this step acetone is acting as an acid) by a base, i.e., OH–. Enolate is stabilized via resonance: the negative charge can be stabilized due to the close proximity if the carbonyl moiety. Once the enolate ion is formed, it can act as a nucleophilic towards the electrophilic center of another molecule of acetone. Once the condensation product is formed, it can get protonated at the work-up step to yield 5-hydroxy, 5-methyl-pentan-2-one. Acid-catalyzed aldol reaction proceeds in a similar manner. One notable distinction from the base-catalyzed aldol condensation is that in the first step, acetone will act as a base that accepts a proton, which leads to the formation of the enol. This enol can now react as a nucleophile. The aldol products are very susceptible to a dehydration reaction, i.e., the loss of water. The dehydration of aldol products can take place under both acidic and basic conditions. This process is fairly facile, since the loss of water leads to a product (enone) that is stabilized by conjugation. Therefore, for the synthesis of β-hydroxy carbonyl products elevated temperature and excess of the base/acid should be avoided to prevent the dehydration step. Otherwise, aldol reaction followed by the dehydration presents a convenient, one-pot preparation of enones, which are valuable synthons in various organic transformations. The mechanism of the base-catalyzed dehydration step is shown in Scheme 3. O O OH base (cat) acid (cat) Scheme 1 R O R OH R O carbonyl enol enolate Figure 1 R O H H H OH R O H H R O H H R O R O R O R O R OH H2O OH δ δ Scheme 2 OH O H H OH O OH H OH O H O OH O O H H Scheme 3 2 In those instances when the carbonyl compounds are different, the aldol condensation can be referred to as cross-condensation. Under these conditions, a mixture of 3 products might be expected. In order to obtain a clean reaction, which is leading to the production of only one product, the carbonyl compounds should be carefully chosen. The idea is quite simple – by having one carbonyl component that can easily be converted into an enolate and by making one partner very electrophilic (or unable to undergo an enolization), it might be possible to achieve a selective cross-condensation, and avoid self-condensation reactions. For example, a base-catalyzed aldol condensation between acetophenone, and 4-nitrobenzaldehyde (Scheme 4). The benzaldehyde cannot undergo the enol tautomerization, simply because it does not have an α-hydrogen. In addition, this aldehyde is very electrophilic, due to the electronegative effect of the nitro-group. Thus, acetophenone is the only carbonyl compound that can react with the base, i.e., OH– or OR–. The subsequent steps, leading to the formation of the enone are similar to the processes of the aldol condensation described in Scheme 3. There are other types of base-catalyzed condensations that involve carbonyl compounds. Several important examples are shown in Scheme 5. Mechanistically, all these reactions rely on the generation of a nucleophile, and subsequent attack on the electrophilic carbon of the carbonyl moiety; in many cases, elimination of a good leaving group takes place to produce compounds with enhanced stability due to conjugation paradigms O O NO2 H O OH NO2 O O NO2 OH O H H H OR NO2 OH O NO2 O OR OH Scheme 4 H O H O O O O O O O O O CO2H NaOAc NaOEt CO2H HO2C amine CO2H Perkin condensation Claisen condensation Knoevenagel condensation Scheme 5 3 PROBLEMS Some questions are based on the material covered in the lecture 1. Provide the major products of the following reactions: O O NaOH/H2O Δ O O NaOH H2O OH H O dilute acid mild heat O NO2 EtO2C NH4OAc glacial HOAc benzene/Δ 2. Write all possible products of the reaction between acetophenone and p-anisaldehyde, and indicate which one is going to be the major one and why? 3. 2-Pentanone can undergo a self-condensation aldol reaction. Write all possible products. 4. Identify the most acidic and the least acidic hydrogens in each of the following compounds: O O O O O OH O OMe O O O O O O 5. Suggest the reagents for the synthesis of the following compounds: NC 6. A student was supposed to carry an aldol reaction between benzaldehyde and acetone, in ethanolic solution of NaOH. However, after adding benzaldehyde to NaOH the student went to ask the instructor about grading on the last report. After the chat was over, the student came back to add acetone, but noticed that a white precipitate was formed in the reaction flask. What is the precipitate and how did it form? (Hint: Cannizzaro reaction). 7. Would you expect any self-condensation products to be forming in this week’s experiment (next page)? Briefly explain why yes or why no. |
187724 | https://www.freebookcentre.net/chemistry-books-download/Virtual-Textbook-of-Organic-Chemistry.html | Virtual Textbook of Organic Chemistry
Virtual Textbook of Organic Chemistry
Virtual Textbook of Organic Chemistry
This note covers the following topics: Structure Bonding,
Intermolecular Forces, Chemical Reactivity, Aromaticity, Nomenclature,
Stereoisomers, Alkanes, Alkenes, Alkynes, Alkyl Halides, Alcohols, Ethers,
Benzene and Derivatives, Amines, Aldehydes and Ketones, Carboxylic Acids and
Carboxylic Derivatives.
Author(s): William
Reusch
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behind organic chemistry.
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187725 | https://www.vedantu.com/maths/closure | Maths
Closure Property in Maths Explained
Closure Property in Maths Explained
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How Does the Closure Property Work in Mathematics?
Closure Property Definition (Maths)
In mathematics, Closure refers to the likelihood of an operation on elements of a set. If something is closed, then it means if an operation is conducted on any of the two elements of the set, then the result of that operation is also within the set. If the elements in a set exist for which the result of the operation is not within the set then the operation will not be termed as closed.
Examples of Closure Property
Taking into account the addition of the natural numbers where the natural numbers are set and the process of addition is the operation. When a natural number is added to a natural number, it will always result in a natural number, and then it is termed that the addition is closed on natural numbers.
Now considering the division of the natural numbers, we can observe that for two numbers 2 and 3, the result of 2 divided by 3 will be ⅔. As ⅔ is not the natural number, the division on the natural number is not closed.
Closure for Different Functions
Closure for Multiplication:
The elements of real numbers in a set are closed under multiplication. If you do the multiplication of two real numbers, you get another real number. There is no probability of ever getting anything other than the real number.
For Example:
45 = 20
3½ 2½ = 8 ¾
1.5 2.1 = 3.15
Closure for Addition:
The elements of real numbers in a set are closed under addition. The addition of the two real numbers gives another real number. There is no probability of ever not getting anything other than the real number.
For Example:
5 + 12 =17
3½ + 6 = 9½
Conclusion
Whenever we use the term "closure" in mathematics, it is applicable to sets and mathematical operations. The sets can include basic numbers, vectors, matrices, algebra, etc. The operations can include any mathematical operation like addition, multiplication, square root, etc.
FAQs on Closure Property in Maths Explained
What is the closure property in Maths explained simply?
The closure property states that if you perform a mathematical operation on any two numbers from a specific set, the result will also be a number within that same set. For example, the set of whole numbers (0, 1, 2, ...) is closed under addition because adding any two whole numbers, like 5 + 7 = 12, always gives another whole number.
Under which basic operations are integers closed?
The set of integers is closed under three basic operations:
Addition: Adding any two integers results in an integer (e.g., -5 + 3 = -2).
Subtraction: Subtracting any two integers results in an integer (e.g., 4 - 9 = -5).
Multiplication: Multiplying any two integers results in an integer (e.g., -6 × 3 = -18).
However, integers are not closed under division, as dividing two integers might result in a fraction or decimal, which is not an integer (e.g., 7 ÷ 2 = 3.5).
Can you give an example of a set that is not closed under an operation?
A classic example is the set of natural numbers {1, 2, 3, ...} under the operation of subtraction. While some operations work (e.g., 10 - 3 = 7, which is a natural number), many do not. For instance, if we take 5 - 8, the result is -3. Since -3 is not a natural number, the set of natural numbers is not closed under subtraction.
How does the closure property of rational numbers differ from that of integers?
The key difference lies in the operation of division. While integers are closed under addition, subtraction, and multiplication, they are not closed under division. In contrast, the set of rational numbers (numbers that can be expressed as a fraction p/q) is closed under all four basic arithmetic operations: addition, subtraction, multiplication, and division (with the only exception being division by zero, which is undefined).
Why is understanding the closure property important for a student?
Understanding the closure property is fundamental because it helps define the structure and boundaries of different number systems. It guarantees that when you perform certain calculations, you won't get an unexpected type of number. This concept is a building block for more advanced topics in algebra and abstract mathematics, where it helps classify mathematical structures like groups and fields.
Is there a specific formula to test for the closure property?
There is no single numerical formula to calculate closure. The closure property is a concept or a test, not a calculation. To check for closure, you must verify if for any two elements 'a' and 'b' in a set, the result of 'a [operation] b' is also an element in that same set. If you can find even one counterexample where the result falls outside the set, the property does not hold.
Are irrational numbers closed under multiplication or addition?
No, the set of irrational numbers is not closed under either multiplication or addition. This is a common point of confusion. For example:
Addition: The numbers (2 + √3) and (2 - √3) are both irrational, but their sum is (2 + √3) + (2 - √3) = 4, which is a rational number.
Multiplication: The number √2 is irrational, but multiplying it by itself (√2 × √2) gives 2, which is a rational number.
Since the result can be a rational number, the set of irrational numbers is not closed under these operations.
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Definition of
Converse (logic)
A conditional statement ("if ... then ...") made by swapping the "if" and "then" parts of another statement.
It may not be true!
Example: "if you are a dog then you bark"
The converse is "if you bark then you are a dog"
See: Inverse, Conditional Statement
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187727 | https://www.ncbi.nlm.nih.gov/books/NBK554469/ | Primary Amenorrhea - StatPearls - NCBI Bookshelf
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StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Primary Amenorrhea
Adi Gasner; Anis Rehman.
Author Information and Affiliations
Authors
Adi Gasner 1; Anis Rehman 2.
Affiliations
1 McMaster University
2 District Endocrine/Sentara Northern Virginia Medical Center
Last Update: December 20, 2023.
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Continuing Education Activity
Amenorrhea is generally defined as the absence of menstruation in a female of reproductive age.It can be classified as either primary or secondary amenorrhea. Primary amenorrhea is the failure to reach menarche (ie, the first menstrual cycle) during normal development. It is clinically diagnosed when there is no history of menstruation by the age of 15 years or 3 years after menarche. Patients meeting the criteria for primary amenorrhea warrant an evaluation. Additionally, an assessment for delayed puberty is indicated in adolescents aged 13 years and younger without initial breast development or other secondary sex characteristics (eg, pubic and axillary hair).
Most of the underlying causes of primary amenorrhea can be classified into general groups: anatomic and sexual development abnormalities, ovarian insufficiency, hypothalamic or pituitary disorders, and other endocrine gland disorders. Physiology and medications may also cause primary amenorrhea; however, they are more commonly associated with secondary amenorrhea. The initial work-up usually includes a comprehensive history and physical examination, a urine pregnancy test, serum hormone testing, and pelvic imaging. Additional testing may also be indicated based on the clinical presentation.Treatment depends on the underlying etiology and may include lifestyle interventions, hormone therapy or other medications, surgery, and mental health services.Therefore, this activity for healthcare professionals is designed to enhance the learner's competence when managing primary amenorrhea, equipping them with updated knowledge, skills, and strategies for timely diagnosis, effective interventions, and improved care coordination, leading to better patient outcomes.
Objectives:
Identify primary amenorrhea and determine when an evaluation of primary amenorrhea is warranted.
Assess the potential etiologies of primary amenorrhea.
Delineate the general approach to evaluating primary amenorrhea.
Coordinate among the interprofessional team and implement proper communication in the diagnosis and management of patients with primary amenorrhea in order to improve patient outcomes.
Access free multiple choice questions on this topic.
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Introduction
Amenorrhea is abnormal uterine bleeding characterized by the absence of menstruation in a female of reproductive age.Amenorrhea can be classified as either primary or secondary amenorrhea. Primary amenorrhea is defined as having no history of menstruation by the age of 15 years or 3 years after thelarche; secondary amenorrhea is defined as the absence of menses for ≥3 months in a woman with previously regular menstrual cycles or≥6 months in any woman with at least one previous spontaneous menstruation.
The median age of menarche is approximately 12.4 years, though this varies somewhat by patient-specific factors (eg, ethnicity, weight, and nutrition status).Menarche typically occurs within 2 to 3 years of initial breast development, which occurs between the ages of 8 and 10 years, known as thelarche.Patients meeting the criteria for either primary or secondary amenorrhea warrant an evaluation. However, an evaluation for delayed puberty is also indicated in adolescents aged 13 years with primary amenorrhea and no breast development or other secondary sex characteristics (eg, pubic and axillary hair).
Most underlying causes of primary amenorrhea can be classified into general groups: anatomic and sexual development abnormalities, ovarian insufficiency, hypothalamic or pituitary disorders, and other endocrine gland disorders. Physiology and medications may also cause primary amenorrhea; however, they are more commonly associated with secondary amenorrhea. The initial work-up usually includes a comprehensive history and physical examination, a urine pregnancy test, serum hormone testing, and pelvic imaging. Additional testing may also be indicated based on the clinical presentation.Treatment depends on the underlying etiology and may include lifestyle interventions, hormone therapy or other medications, surgery, and mental health services.Therefore, all healthcare professionals should strive to enhance their competence when managing primary amenorrhea and equip themselves with updated knowledge, skills, and strategies for timely diagnosis, effective interventions, and improved care coordination, leading to better patient outcomes.
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Etiology
A Systematic Approach to Considering the Etiologies of Primary Amenorrhea
The basic requirements for normal menstrual function include four anatomically and functionally distinct structural components:the hypothalamus, anterior pituitary gland, ovary, and the genital outflow tract composed of the uterus/endometrium, cervix, and vagina.If any of these components are nonfunctional or abnormal, menstrual bleeding cannot occur.
Despite the numerous potential causes of primary amenorrhea, the majority of cases are caused by gonadal dysfunction (43%), müllerian agenesis (10%-15%), and constitutional delay of growth and puberty (14%).Other relatively common etiologies, accounting for between 2% and 7% of cases, include functional hypothalamic amenorrhea (FHA), transverse vaginal septum, polycystic ovary syndrome (PCOS), and hypopituitarism.Various other rare etiologies are also possible, and because ovulation occurs before menstruation, pregnancy must always be considered as well. Furthermore, any etiology of secondary amenorrhea may also present as primary amenorrhea. Determining the underlying cause of amenorrhea will assist in guiding management decisions.
Anatomic Abnormalities and Other Disorders of Sexual Development (DSD)
Disorders of sexual development (DSD) refer to congenital conditions associated with abnormal development of the reproductive tract and external genitalia. These conditions are typically categorized as anatomic, gonadal, and chromosomal anomalies.
Agenesis of the müllerian ducts (ie, the paramesonephric ducts) results in the complete or partial absence of the müllerian structures, which include the fallopian tubes, uterus, and upper one-third of the vagina. This condition, also known as Mayer-Rokitansky-Kuster-Hauser syndrome, accounts for approximately 10% to 15% of primary amenorrhea cases. These patients typically have normal ovarian function, external female genitalia, and development of female secondary sex characteristics, with the complete or partial absence of the müllerian structures; however, the condition is also frequently associated with other urogenital abnormalities, such as unilateral renal agenesis, and skeletal anomalies, such as scoliosis.
Some enzyme deficiencies or receptor anomalies also cause atypical sexual development and primary amenorrhea. For instance, aromatase deficiency leads to testosterone overproduction due to the inability of the ovary to convert testosterone to estrogen, resulting in virilization and ambiguous genitalia.Similarly, patients with a 17α-hydroxylase deficiency will typically present without pubertal development along with significant hypertension due to mineralocorticoid excess. Patients with this enzyme deficiency and a 46,XX karyotype typically have a small uterus and ovaries; those with a 46,XY karyotype will have a blind vaginal pouch without a uterus. Another cause of primary amenorrhea in the setting of absent müllerian structures is complete androgen insensitivity syndrome (CAIS), which can occur in 46,XY individuals with mutations affecting the function or quantity of androgen receptors. This leads to complete androgen resistance and elevated serum testosterone levels. Because these individuals have testes, they secrete antimüllerian hormone (AMH) in utero, which causes regression of the müllerian structures. Testosterone is also secreted but is ineffectual due to androgen receptor abnormalities; however, aromatase converts the testosterone to estrogen, leading to the development of a female phenotype. Therefore, these patients are born with external female genitalia but absent müllerian structures and typically present at puberty with primary amenorrhea. Anomalies in the development of these patients include sparse pubic hair due to abnormal androgen receptors, well-developed breasts due to the presence of estrogen, and a blind vaginal pouch without a uterus on an exam.
Congenital obstructions to the menstrual outflow tract include an imperforate hymen, transverse vaginal septum secondary to a failure of the junction between the müllerian ducts and vaginal plate to regress,and isolated agenesis of the vagina resulting from a failure of the vaginal plate to canalize.These obstructive anomalies typically present with cyclic pelvic pain in the setting of otherwise normal pubertal development.
Hypothalamic and Pituitary Disorders
Abnormalities in the hypothalamus result in abnormal gonadotropin-releasing hormone (GnRH) secretion, which in turn results in low follicle-stimulating hormone (FSH) secretion from the pituitary; primary pituitary abnormalities can also result in poor FSH secretion. In both cases, insufficient FSH is in circulation to stimulate the ovarian follicles to mature. The follicles, therefore, do not produce estrogen. The endometrium does not develop without estrogen, so menstruation cannot occur. Additionally, without proper follicular development, there is no luteinizing hormone (LH) surge, ovulation, or luteal phase progesterone production. With hypothalamic and pituitary etiologies of primary amenorrhea, both FSH and estrogen levels are low, and this state is known as hypogonadotropic hypogonadism.
The most common intracranial causes of primary amenorrhea are a constitutional delay of growth and puberty (CDGP) and functional hypothalamic amenorrhea (FHA), both of which cause reduced production of GnRH and can be difficult to distinguish from one another. CDGP is simply a transient delay in the maturation of the hypothalamic-pituitary-ovarian (HPO) axis, typically due to individual genetic variation, and resolves with time. Conversely, FHA is a pathologic suppression of the body's reproductive cycle due to a relative deficiency of available energy, most frequently resulting from disordered eating, excessive exercise or athletic training, and substantial perceived life stress. Many chronic systemic diseases or prolonged illnesses can also cause hypothalamic suppression of the reproductive cycle.
Other hypothalamic and pituitary causes of primary amenorrhea are much less common. Inherited deficiency of gonadotropin-releasing hormone (GnRH) is one type of these rarer hypogonadotropic hypogonadism etiologies. The most well-known of these genetic mutations is Kallman syndrome, characterized by amenorrhea, delayed puberty, and a lack of olfactory neurons, causing anosmia. Other rare hypothalamic and pituitary etiologies include infiltrative diseases (eg, cancer, sarcoidosis, hemochromatosis); pituitary infarction or necrosis; prolactin-secreting tumors (ie, prolactinomas); other hormone-secreting pituitary tumors affecting the HPO axis (eg, tumors secreting adrenocorticotropin hormone [ACTH], thyroid stimulating hormone [TSH], or growth hormone [GH]); craniopharyngioma; autoimmune disease; and empty sella syndrome due to the structural, traumatic, or iatrogenic shrinking of the pituitary gland.
Ovarian Dysfunction
Ovarian dysfunction is one of the most common causes of primary amenorrhea. Patients with primary ovarian dysfunction typically have normal function in the hypothalamus and pituitary but nonresponsive ovaries. This results in low estrogen production leading to a lack of endometrial growth and, due to a lack of negative feedback at the pituitary, elevated FSH levels.This state is known as hypergonadotropic hypogonadism.
Primary amenorrhea due to ovarian dysfunction is most frequently a result of gonadal dysgenesis, which can be seen with Turner syndrome 45,X karyotype, or in patients with mosaicism. Less commonly, gonadal dysfunction can also be seen in patients with pure gonadal dysgenesis with either 46,XX or 46,XY karyotypes, the latter known as Swyer syndrome. Patients with Swyer syndrome have nonfunctional testicular tissue; therefore, they do not secrete antimüllerian hormone (AMH) or testosterone in utero. The lack of AMH allows the müllerian ducts to persist and develop into typical internal female reproductive structures, including the fallopian tubes, uterus, cervix, and upper vagina. The lack of testosterone also causes the external genitalia to develop as female.Therefore, these patients develop typical internal and external female genital structures but fail to develop secondary sex characteristics at puberty because their gonads are nonfunctional.
Causes of primary ovarian insufficiency (POI) include premutation in the FMR1 gene, which causes Fragile X syndrome, chemotherapy, radiation, autoimmune disease, as well as the previously discussed enzyme deficiencies that affect ovarian hormone production (eg, aromatase or 17α-hydroxylase deficiency).
Other Hormonal Abnormalities Affecting the HPO Axis
In general, hormonal abnormalities affecting the HPO axis are more frequently associated with secondary amenorrhea than primary. However, these causes should still be considered when evaluating adolescents with primary amenorrhea.
Hyperandrogenemia can lead to suppression of the HPO axis. The most common cause of amenorrhea in females with evidence of androgen excess is polycystic ovarian syndrome (PCOS). PCOS is a common condition characterized by menstrual abnormalities, hyperandrogenism, and metabolic dysfunction.Individuals with PCOS often have chronic anovulatory cycles, which results in chronic unopposed estrogen (ie, estrogen "unopposed" by progesterone) that can lead to endometrial hyperplasia. Other causes of hyperandrogenism, which are much less common than PCOS but may still present as primary amenorrhea, include nonclassical congenital adrenal hyperplasia (NCCAH), androgen-secreting tumors from the adrenal gland or ovary, and Cushing syndrome.
Hyperprolactinemia is frequently due to pituitary prolactinomas; however, it can also be caused by stress, exercise, chronic kidney disease, and some medications, particularly antipsychotics and opiates. Additionally, both hypo- and hyperthyroidism and uncontrolled diabetes can also suppress HPO axis function and menstruation, though these are all more commonly associated with secondary rather than primary amenorrhea.
Physiologic and Medication Induced Amenorrhea
Though this group of etiologies is more commonly associated with secondary amenorrhea, any cause of secondary amenorrhea may also present as primary amenorrhea.
Physiologic amenorrhea includes pregnancy and lactation, with the most common cause of amenorrhea overall being pregnancy. During pregnancy, prolonged elevations in sex hormones suppress the HPO axis and create a stable intrauterine environment where the gestation can develop. After delivery, the persistent hyperprolactinemia that occurs during lactation also suppresses the HPO axis and maintains the patient in a state known as "lactational amenorrhea" until the frequency and duration of nipple stimulation decrease.
Medication-induced amenorrhea can result from several drugs (eg, chemotherapy, illicit substances). Since synthetic progestins, found in hormonal contraceptives, cause endometrial atrophy, prolonged use can lead to secondary amenorrhea, even during a typical hormone-free interval. Some medications (eg, antipsychotics) may also cause hyperprolactinemia resulting in amenorrhea.
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Epidemiology
In the US,only about 2% of adolescent girls have not reached menarche by 15 years of age.Müllerian agenesis has an estimated incidence of 1:4,500 to 5,000 females, while CAIS has an estimated incidence of 1:20,000 to 99,000 genetic males.The incidence of uterovaginal anomalies overall has been reported to be up to 7% in females.
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Pathophysiology
Regular menstruation depends on a complex sequence of events consisting of hormone production leading to ovulation, the effects of hormonal fluctuations from ovulation and the absence of fertilization on the endometrium, and an anatomically normal reproductive tract where this physiologic process can occur. The basic requirements for normal menstrual cycles include four anatomically and functionally distinct structural components:the hypothalamus, anterior pituitary gland, ovary, and the genital outflow tract composed of the uterus, endometrium, cervix, and vagina.If any of these components are nonfunctional or abnormal, menstrual bleeding may not occur; however, the precise pathophysiology associated with amenorrhea may vary based on the underlying etiology and sometimes may be unclear.
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History and Physical
Clinical Features
Primary amenorrhea is a menstrual symptom characterized by the complete absence of menstruation in a female of reproductive age.A complete evaluation is indicated in females presenting with any of the following clinical features:
No history of menstruation with secondary sex characteristics present(eg, thelarche, pubic and axillary hair) by the age of 15 years or 3 years after thelarche
No history of menstruation and no breast development or other secondary sex characteristics present in an adolescent girl aged 13 years
History
Obtaining a comprehensive history in patients with clinical features of amenorrhea is critical in diagnosing the underlying etiology. The history should begin by asking about the last menstrual period (LMP) date. If the patient has never menstruated, the patient has primary amenorrhea. The timing of initial breast bud and sexual hair development and general growth trends should be noted to differentiate patients with isolated primary amenorrhea from those with globally delayed pubertal development. The development of breast buds indicates the presence of estrogen, presumably from functional ovaries; the growth of sexual hair indicates the presence of androgens.The patient should also be asked about cyclic abdominal pain, which may be caused by an obstruction in the menstrual outflow tract, such as an imperforate hymen, transverse vaginal septum, or vaginal or cervical atresia.
Additionally, it is important to review the patient's medical history, general health, lifestyle, and current medications, which can help identify any chronic illnesses, exposure to chemotherapy or radiation, and potential relative energy deficiencies due to eating disorders or strenuous athletic training.A history of extreme weight loss should be noted. Any history of anosmia, galactorrhoea, headaches,or visual changes may indicate a central nervous system or pituitary disorder. A careful family history should also be obtained, as certain conditions, including CDGP and CAIS, can often have a hereditary component.
Physical Examination
Patients presenting with primary amenorrhea also require a complete physical examination to assist in determining an etiology. This should include height, weight, and body mass index (BMI) measurements. Short stature in the absence of all secondary sex characteristics is a hallmark of gonadal dysgenesis, while low body weight can be associated with FHA resulting from malnutrition or physical, psychological, or emotional stress.The following are some clinical findings associated with various causes of primary amenorrhea that, in combination with historical features, can assist in guiding what diagnostic studies are indicated.
Head and neck:Focal neurological deficits or vision changes suggest an intracranial process affecting the hypothalamus or pituitary gland; thyroid nodularity or enlargement may indicate a thyroid disorder.Several stigmata of Turner syndrome can be seen on examination of the head and neck, including a low hairline, a high-arched palate, and a webbed neck. Signs of Cushing syndrome include moon facies and buffalo hump.
Skin and hair:The presence and maturation of sexual hair (axillary and pubic hair), which indicates exposure to androgens, should be assessed. Note that the absence of sexual hair suggests the patient has never been exposed to androgens.Signs of androgen excess include hirsutism, excessive acne, and virilization (ie, enlarged clitoris, deepening of the voice). Multiple pigmented naevi may be seen with Turner syndrome. Patients with PCOS or uncontrolled diabetes mellitus may develop acanthosis nigricans. Skin, hair, and nail changes can also be seen with thyroid disorders. Clinicians should be aware that many women remove undesired male-pattern hair growth, so it may not be present on physical exam; it is essential to inquire about any hair removal practices (eg, shaving, waxing, laser).
Breasts: The stage of breast development should be noted (eg, Tanner staging), which is a reliable indicator of estrogen production or exposure to exogenous estrogen. The absence of breast budding indicates the patient has never been exposed to estrogens. Widely spaced nipples can be seen with Turner syndrome. A finding of galactorrhea may be a sign of hyperprolactinemia.
Cardiovascular and pulmonary: Heart and lung exam abnormalities may indicate a chronic disease or illness. Patients with excess corticosteroids(eg, Cushing syndrome, 17α-hydroxylase deficiency) can present with hypertension.
Abdomen:The abdomen should be examined for any palpable masses and signs of chronic disease (eg, hepatomegaly). Pelvic masses may represent a uterus distended by retained menstrual fluid due to an outflow obstruction; less commonly, it may signal an ovarian or uterine neoplasm.
Pelvic:The pelvic exam should include a thorough inspection of the external genitalia, and an internal digital, bimanual, and speculum examination should be performed as tolerated by the patient. The clinician should note the maturation of the external genitalia and pubic hair and any abnormal enlargement of the clitoris. An imperforate hymen is diagnosed by a bulging membrane without hymenal fringe near the introitus that distends during the Valsalva maneuver. A shortened vagina without a palpable cervix suggests müllerian or vaginal agenesis, a transverse vaginal septum, or CAIS. Therefore, a patent vagina and normal cervix exclude vaginal agenesis, complete AIS, and obstructive causes of amenorrhea such as an imperforate hymen or transverse vaginal septum. Additionally, low estrogen can result in an atrophic vaginal mucosa, which may also be appreciated on an exam.
Rectal:Although not always necessary, a rectal examination can detect hematocolpos in patients with an obstructive anomaly, which can be felt in the rectum as a bulge of the proximal vagina.
Extremity: Additional stigmata of Turner syndrome may be noted on examination of the extremities, including cubitus valgus, genu varum or valgum, a shortened fourth metacarpal, and small metatarsal bones. Joint pain may be present in patients with chronic inflammatory disorders.
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Evaluation
Initial Diagnostic Studies
Diagnostic studies performed for the evaluation of primary amenorrhea vary depending on the age of the patient and the findings on the exam. Initially, however, the following are indicated in all patients:
Urine pregnancy test
Pelvic ultrasound (typically transabdominal)
Serum FSH, LH, and estradiol
Serum prolactin
Serum thyroid-stimulating hormone (TSH)
Subsequent Diagnostic Studies
Based on clinical indications, additional studies may be required to determine the cause of a patient's amenorrhea. Further evaluation, if clinical findings of the following conditions are noted, may include:
Hyperandrogenism
Total testosterone
Fasting morning serum 17-hydroxyprogesterone
Dehydroepiandrosterone sulfate (DHEAS)
Adolescents with globally delayed puberty
Bone age radiography
Insulin-like growth factor I (IGF-1) to screen for growth hormone deficiency if growth velocity is low
Chronic disease (eg, liver disease, inflammatory bowel disease)
Complete blood count (CBC)
Complete metabolic profile (CMP) and liver function tests
Erythrocyte sedimentation rate (ESR) and C-reactive protein (CRP)
Tissue transglutaminase-immunoglobulin A antibodies (tTG-IgA) to screen for celiac disease if BMI is low
POI, gonadal dysgenesis, Turner syndrome,müllerian agenesis, or CAIS (eg,classic stigmata of Turner syndrome,elevated FSH levels, absent müllerian structures on ultrasound)
Karyotype
Consider magnetic resonance imaging(MRI) of the pelvis
Intracranial process (eg, elevated serum prolactin level or focal neurologic deficits)
Consider an MRI or CT of the head
Additionally, if a genetic defect is suggested on history or exam (eg, Kallman syndrome), referral to a genetic counselor and targeted, genetic testing is also warranted.
Interpretation of Initial Testing
Pelvic ultrasound:A transabdominal pelvic ultrasound should be performed to confirm the presence or absence of a uterus and internal gonads. Absence of the uterus indicates müllerian agenesis or other DSD (eg, CAIS). A uterus distended with heterogeneous material is seen with hematometra due to outflow tract obstructions (eg, transverse septum, vaginal agenesis). A thin endometrial lining will be seen in patients without estrogen exposure. In contrast, a thickened endometrial lining suggests the presence of estrogen and chronic anovulation if the outflow tract is patent. Gonadal dysgenesis may be indicated by the inability to visualize the ovaries or findings consistent with streak gonads. If the presence or absence of müllerian structures cannot be confirmed on ultrasound, a pelvic magnetic resonance imaging (MRI) study is an appropriate next step.
Serum testing:It is always important to first rule out pregnancy,as a patient ovulates before their first period and, therefore, may become pregnant before menarche. The FSH and estradiol levels can help distinguish primary hypothalamic and pituitary abnormalities from primary ovarian etiologies. If the screening FSH level is low, the diagnosis of hypogonadotropic hypogonadism can be confirmed, indicating the primary defect is in the hypothalamus or pituitary gland. If the FSH level is elevated, this indicates hypergonadotropic hypogonadism, pointing to a primary defect in the ovary which is unable to respond to normal gonadotropin secretion. Since estrogen suppresses FSH release, estradiol is also usually ordered to interpret the FSH level correctly. Estrogen levels are typically low in primary hypothalamic, pituitary, and ovarian disorders, while normal FSH and estradiol levels are seen with structural anomalies.
Abnormal TSH levels indicate thyroid disease and elevated prolactin levels may be secondary to a pituitary adenoma, hypothyroidism, or some medications. Elevated testosterone or DHEA is most likely due to PCOS but may also be seen in Cushing syndrome or hormone-secreting ovarian neoplasms. An abnormal CBC, renal or liver function test, ESR, or CRP can indicate chronic illness.
Bone age radiography:A bone age radiograph indicates the degree to which sex steroids have affected bone maturation and can identify future growth potential.A patient's bone age can help clinicians provide appropriate counseling around predicted adult height.
Clinical and Diagnostic Findings of Common Primary Amenorrhea Etiologies
Combining the history, exam, and initial assessment studies, the most common etiologies of primary amenorrhea will typically have the following findings on clinical and diagnostic evaluation:
Gonadal dysgenesis or POI
Patent vagina and normal cervix and uterus noted on exam with signs of decreased estrogen (ie, no breast bud development, vaginal atrophy) or androgen (ie, no pubic hair)
Elevated serum FSH and LH
Decreased serum estrogen
Ovaries may or may not be visible on ultrasound imaging
Müllerian agenesis
Normal breast, sexual hair development, and external female genitalia;short vagina, ending in a blind pouch often noted
Partial or complete absence of the müllerian structures (ie, upper vagina, cervix, uterus, and fallopian tubes) on clinical exam and ultrasound imaging
Normal serum FSH and estrogen
Other urogenital anomalies may also be noted on imaging
Transverse vaginal septum
Similar findings to müllerian agenesis, except that müllerian structures are present on exam and ultrasound imaging
Imperforate hymen
Normal breast, sexual hair development, and external female genitalia
A bulging membrane (ie, the hymen) is visible when spreading the labia
Normal serum FSH and estrogen
Hematocolpos may be seen with transabdominal pelvic ultrasound
Constitutional delay of growth and puberty (CDGP)
Little evidence of estrogen or androgens on an exam
Normal anatomy on the pelvic exam and ultrasound
Decreased FSH and estrogen levels
Delayed bone age on radiography
Family history is often consistent with CDGP
Functional hypothalamic amenorrhea (FHA)
Similar presentation to CDGP, often with low BMI
Typically associated with disordered eating, vigorous exercise, and systemic chronic disease
Polycystic ovary syndrome (PCOS)
Normal thelarche and adrenarche on an exam, often with hirsutism and acne on an exam
Serum FSH and estrogen within the normal range for reproductive-aged females
Polycystic appearing ovaries on ultrasound imaging
No clear consensus on diagnosing PCOS within 2 years of menarche exists; caution with diagnosis recommend in this age group
Complete androgen insensitivity syndrome (CAIS)
Normal thelarche, often with very well-developed breasts but absent or sparse sexual hair
Normal external female genitalia but no müllerian structures
Testes within the pelvic cavity will be undescended and may be identified on ultrasound imaging
Table
Table: Clinical and Diagnostic Findings of Primary Amenorrhea Etiologies.
T: thelarche, A: adrenarche, FSH: follicle-stimulating hormone, CDGP: constitutional delay of growth and puberty, FHA: functional hypothalamic amenorrhea, PCOS: polycystic ovary syndrome, CAIS: complete androgen insensitivity syndrome, US: ultrasound
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Treatment / Management
The management of primary amenorrhea should be directed at correcting the underlying etiology and preventing potential complications. Depending on the etiology, additional testing and monitoring are frequently indicated as well; a detailed discussion on specific management points for all of the various potential diagnoses is beyond the scope of this document.
Outflow Tract Abnormalities
Patients with some types of outflow obstructions (eg, imperforate hymen, cervical stenosis) can be treated with a surgical correction to relieve the obstruction. However, they may have residual issues, including endometriosis and pelvic adhesions due to the initial obstruction.Individuals with congenital anomalies affecting vaginal length who desire receptive vaginal intercourse may consider various management options, including surgical techniques or conservative interventions (eg, vaginal dilators). Conservative management alone is successful in 90% to 96% of patients.In patients with a Y chromosome and streak gonads, the gonads should be removed due to their high risk of malignancy.Furthermore,all these patients should be referred to clinicians with the expertise and appropriate resources to obtain the desired results following a management plan determined through shared decision-making. Because of the psychological issues patients with congenital anomalies may have, they should also receive counseling and be provided information on available support groups.
Hypothalamic, Pituitary, and Gonadal Disorders
Patients of typical reproductive age who lack endogenous sex hormones (eg, POI, Kallman syndrome) require hormone therapy to protect their skeletal and cardiovascular health. Therefore, patients with hypogonadotropic and hypergonadotropic hypogonadism usually require referral to an endocrinologist for long-term hormone therapy. A standard maintenance regimen for hormone replacement therapy includes either a transdermal patch of estradiol 100 mcg daily or conjugated estrogen 0.625 mg orally daily. For 12 days each month, oral micronized progesterone 200 mg daily is added if endometrial protection is needed. In adolescent girls with possible constitutional delay of growth and puberty hormone therapy to initiate pubertal development (ie, jumpstart therapy) may be attempted for 3 to 6 months, after which patients may be monitored for signs of pubertal development without assistance from exogenous hormones for the following 4 to 6 months. Initial expectant management with observation only for the first 6 months is also an appropriate alternative.Calcium 1,200 mg orally and vitamin D 1,000 IU orally every day, combined with routine weight-bearing exercises, is also recommended to maintain bone density. A small percentage of patients with POI have reproductive capabilities and may desire contraceptive management. In these patients, either intrauterine contraceptives may be used in addition to HRT, or combined hormonal contraceptives can be used for both HRT and pregnancy prevention. However, as with standard contraceptive counseling, patients should be informed about the potentially increased thromboembolic risk. Screening for comorbid conditions (eg, hypothyroidism) should also be performed as they are more prevalent in this population.
The primary management of amenorrhea due to FHA is the reversal of the contributing factors, including weight gain, stress reduction, lifestyle changes, and dietary modification.Patients with FHA due to eating disorders should be provided with appropriate education and counseling programs (eg, nutrition counseling and mental health services) to treat and manage their condition's physical and psychological aspects. Proper nutrition and maintaining a normal weight are the preferred management for amenorrheic patients with this underlying etiology (eg, gluten-free diet for celiac disease, weight restoration for malnutrition); conversely, combination hormone contraceptives alone are not recommended therapy for amenorrhea.Cognitive behavioral therapy is effective for FHA secondary to severe stress.
The dopamine agonist cabergoline is considered first-line therapy in patients with a prolactin-secreting tumor.Larger prolactinomas can be treated surgically. Medication-induced hyperprolactinemia should be treated by discontinuing the offending agent; if discontinuation is not possible, a dopamine agonist can be cautiously considered.
Other Endocrine Abnormalities Affecting the HPO Axis
Similarly, patients with other endocrine abnormalities resulting in amenorrhea should be managed with the improvement of the underlying etiology as the primary goal. For instance, lifestyle interventions promoting a healthy diet and exercise program are the most beneficial therapy for some conditions (eg, PCOS, diabetes). In contrast, others may require pharmacological treatment (eg, thyroid disorders) or a combination.Some medicines also aim to treat complications that may arise due to these conditions. In patients with chronic anovulatory cycles (eg, PCOS), combined hormonal contraceptives are first-line therapy to prevent the development of endometrial hyperplasia and malignancy and treat hirsutism and acne.Clinicians should also perform preventive management and screen patients for hypertension and an elevated body mass index at each visit. Furthermore, patients should be screened for dyslipidemia and impaired glucose tolerance (eg, 2-hour oral glucose tolerance testing or hemoglobin A1C level) every three to five years.
Physiologic- and Medication-Induced Amenorrhea
In patients with these underlying etiologies, amenorrhea typically resolves after discontinuation of the contributing physiologic event or medication. Clinicians should consider the benefits and risks before deciding on medication changes.For example, medication-induced hyperprolactinemia is often treated by discontinuing the offending agent; however, if discontinuation is not possible, a dopamine agonist can be cautiously considered.
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Differential Diagnosis
Amenorrhea is a symptomatic presentation of an underlying condition; therefore, differential diagnoses should consider these various potential etiologies, including:
Physiologic and Medication-Induced
Pregnancy
Lactation
Menopause
Medications,chemotherapy, or radiation
Hypothalamic and Pituitary
CDGP
FHA
Prolactinoma or other hormone-secreting pituitary tumors
Kallman syndrome
Pituitary infarction or necrosis
Autoimmune disease
Empty sella syndrome
Ovarian
POI
Gonadal dysgenesis
Autoimmune disease
Aromatase deficiency, 17α-hydroxylase deficiency
Other Endocrine Abnormalities
PCOS or other causes of hyperandrogenism (NCCAH, Cushing syndrome)
Hormonally active ovarian or adrenal tumors
Thyroid disease
Uncontrolled diabetes
Outflow Tract Abnormalities
Complete or partial müllerian agenesis
CAIS
Intrauterine adhesions or endometrial scarring
Cervical stenosis
Vaginal agenesis
Transverse vaginal septum
Imperforate hymen
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Prognosis
The prognosis of primary amenorrhea depends on the underlying etiology, and a full discussion on this topic is beyond the scope of this document. Generally, however, reversible or transient causes of primary amenorrhea, such as CDPG and FHA, have a better prognosis than gonadal etiologies, such as gonadal dysgenesis. The prognosis for future fertility varies significantly, depending on the etiology. In patients with functional hypogonadotropic hypogonadism (eg, female athletes) or CDPG, future fertility rates are essentially normal once the HPO axis begins functioning normally. For patients with gonadal dysgenesis or POI; however, fertility rates are much lower. The best fertility option for patients with müllerian agenesis is typically using a surrogate gestational carrier.
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Complications
Complications associated with specific underlying conditions (eg, chronic hypoestrogenic state, unopposed estrogen) can arise, including:
Vasomotor symptoms
Osteoporotic fractures
Endometrial hyperplasia
Metabolic and cardiovascular disease
Infertility
Galactorrhea
Hyperandrogenism effects (eg, virilization, hirsutism, and acne)
Psychological effects (eg, anxiety, depression)
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Consultations
Multiple specialties often must be involved in treating patients with complex underlying etiologies of their amenorrhea. Patients with outflow tract anomalies, especially congenital types, should be referred to clinicians (eg,pediatric gynecologists, pediatric endocrinologists) with the expertise and appropriate resources to obtain the desired results following a management plan determined through shared decision-making. Because of the psychological issues patients with congenital anomalies may have, they should also receive counseling and be provided information on available support groups.These diagnoses in particular can have a significant impact on an adolescent's sense of identity, body image, self-esteem, and overall well-being.Patients who lack endogenous sex hormones (eg, POI, Kallman syndrome) also require specialist consultation (eg, endocrinologist) for pubertal and maintenance hormone therapy.Furthermore, patients with FHA due to eating disorders should be referred to appropriate education and counseling programs (eg, nutrition counseling and mental health services) to treat and manage their condition's physical and psychological aspects.
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Deterrence and Patient Education
Patients should be informed that regular menstruation is an essential sign of health in women of reproductive age who do not take hormonal medications to suppress it. They should be advised that the absence of menstruation may indicate a severe underlying medical disorder and warrants an evaluation. Additionally, the importance of a healthy lifestyle, including a nutritious diet, appropriate levels of exercise, and attention to one's mental health should be stressed. Clinicians should make anticipatory counseling about the stages of a woman's reproductive life cycle and the menstrual cycle a routine part of every wellness visit. Counseling on the reproductive life cycle should include the average ages for upcoming transitions (eg, puberty, menopause), expected physiologic changes, and family planning preferences. Menstrual cycle education should include tracking the first day of menstruation to the first day of the next cycle, the average interval between cycles, and what constitutes an abnormal cycle. This will help in earlier diagnosis of health problems and assist in easing patient anxiety sometimes experienced when facing predictable reproductive transitions (eg, puberty and menopause). Patients should also be introduced to valuable tools to help track menstrual cycles.
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Pearls and Other Issues
When evaluating a person with amenorrhea, first consider the physiologic causes of amenorrhea, then think through the reproductive pathway "from head to toe": hypothalamus, pituitary gland, ovaries, endometrium, cervix, and vagina. Other endocrine disorders and systemic diseases that may affect the HPO axis should also be considered.
When performing diagnostic studies, pregnancy should be excluded first; FSH, LH, prolactin, and TSH laboratory studies will then identify most endocrine causes of amenorrhea.
Clinicians should make anticipatory counseling about the stages of a woman's reproductive life cycle and the menstrual cycle a routine part of every wellness visit.
All adults require sex hormones. If a patient does not produce their own hormones after a reasonable observation period or has a condition associated with permanent hypogonadism, referral to a specialist for hormone therapy is warranted.
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Enhancing Healthcare Team Outcomes
An interprofessional team including clinicians (MDs, DOs, NPs, PAs), specialists (particularly endocrinologists and gynecologists), nurses, and pharmacists must collaborate, openly sharing information and coordinating patient evaluation, education, and treatment to optimize outcomes. Additionally, during the diagnostic process, radiology and laboratory staff members play an essential role in assessing patients and providing accurate and timely reports to treating clinicians. Mental health clinicians and social workers also play an important role in managing these patients, especially those receiving a diagnosis with long-term implications regarding their fertility which may cause significant emotional distress (eg, disorders of sexual development). Patients diagnosed with functional hypothalamic amenorrhea may also require mental health and nutrition counseling, and those with intracranial abnormalities may require care from a neurosurgery specialist. Therefore, a well-coordinated, multidisciplinary team is best suited to manage these patients.
Furthermore, as with any complex medical condition, proper documentation in the medical record and excellent communication between team members is crucial in optimizing care for these patients. The clinician, nurse, and pharmacist must coordinate care, treatment, and education of the family and patient to ensure close follow-up. Only with an interprofessional team approach can the best outcomes be reached.
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Disclosure:Adi Gasner declares no relevant financial relationships with ineligible companies.
Disclosure:Anis Rehman declares no relevant financial relationships with ineligible companies.
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187728 | https://javierrubioblog.com/wp-content/uploads/2015/12/chapter2.pdf | CHAPTER 2 MINKOWSKI SPACETIME AND SPECIAL RELATIVITY Scarcely anyone who truly understand relativity theory can escape this magic.
A. Einstein In the previous chapter we saw that tensors are a very good tool for writing covariant equations in 3-dimensional Euclidean space. In this chapter we will generalize the tensor concept to the framework of the Special Theory of Relativity, the Minkowski spacetime. I will assume the reader to be familiar at least with the rudiments of Special Relativity, avoiding therefore any kind of historical introduction to the theory.
2.1 Einstein’s Relativity Special Relativity is based on two basic axions, formulated by Einstein in 19051: 1. Principle of Relativity (Galileo): The laws of physics are the same in all the inertial frames: No experiment can measure the absolute velocity of an observer; the results of any experiment do not depend on the speed of the observer relative to other observers not involved in the experiment.
2. Invariance of the speed of light: The speed of light in vacuum is the same in all the inertial frames.
Instantaneous action at a distance is inconsistent with the second postulate and must be replaced by retarded action at a distance. Absolute simultaneity will only apply as an approximation at low velocities for nearby events.
2.2 Minkowski spacetime: new wine in a old bottle The framework of Special Relativity is a 4-dimensional manifold called Minkowski (or pseudo-Euclidean) space-time. The differential and topological structures of the Newtonian and Minkowskian spacetimes 1On the electrodynamics of moving bodies.
2.2 Minkowski spacetime: new wine in a old bottle 16 coincide2, but they differ in the metrical structure, i.e. in the definition of distances. While in Newto-nian spacetime the spatial and temporal distances are independent, in Minskowskian spacetime space and time by themselves, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality3. Space and time are distinguished only by a sign, which will play however a central role. For any inertial frame of reference in Minkowski spacetime there is a set of coordinates4 {xµ} = {t, x, y, z} = {x0, x1, x2, x3} = {x0, xi} , (2.1) and a set of orthonormal basis vectors {eµ} = {et, ex, ey, ez} = {e0, e1, e2, e3} = {e0, ei} , (2.2) satisfying the Lorentz orthonormality condition5 eµ · eν = ηµν , (2.3) with ηµν = −1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 = diag(−1, 1, 1, 1) .
(2.4) The inverse of (2.4) is traditionally denoted by ηµν and satisfies6 ηµνηνρ = δµ ρ , (2.5) where the 4-dimensional Kronecker delta δµρ is the indexed version of the identity matrix, i.e. δµρ = 1 if µ = ρ and zero otherwise. Note that ηµν and ηµν are numerically equivalent.
Exercise Which is the value of δµµ?
In terms of the basis vectors, the infinitesimal displacement dS between two points in spacetime can be expressed as dS = dxµeµ , (2.6) where dxµ are the so-called contravariant components. These are computed via the scalar product of the vector dS and the corresponding basis vector eµ.
dS · eµ = (dxνeν) · eµ = dxν (eν · eµ) = dxνηνµ = dxµ , (2.7) where we have defined the covariant or dual components by lowering the index of the contravariant components with the metric dxµ ≡ηµνdxν .
(2.8) 2Both of them are smooth, continuous, homogeneous, isotropic, orientable,. . .
3Minkowski, 1908.
4As in the previous chapter, we will not consider the quadruplet {xµ} to be a vector in Minkowski spacetime.
Coordinate indices will be always upper indices.
5Note that the time coordinate is 4-dimensionally orthogonal to the spatial coordinates.
6Einstein’s summation convention is used.
2.2 Minkowski spacetime: new wine in a old bottle 17 Exercise Starting with a covariant vector defined by Eq.(2.8) , show that the inverse of the metric ηµν can be used to raise indices ηµνdxν = dxµ .
(2.9) As in the Euclidean case, contravariant and covariant vectors are just an appropriate way of simplifying the notation and taking into account the summation convention. Note however that in the present case lowering or raising indices changes the sign of the temporal component while keeping intact the spatial ones dx0 = −dx0 , dxi = +dxi .
(2.10) The upper and lower index notation automatically keeps track of the minus signs associated to the temporal component. The indefiniteness of the metric is automatically incorporated in the notation!
In some old-fashioned books and in ’t Hooft’s lecture notes you will find a fourth coordinate x4 = it, instead of the coordinate x0 = t appearing before. Written in terms of x4 the Minkowski spacetime has the appearance of a positive-definite 4 dimensional Euclidean space eµ · eν = δµν (2.11) and there is no difference between lower and upper indices. This notation is however confusing since it hides the non-positive definite character of the metric.
The square of the infinitesimal distance between two events in Minkowskian spacetime is given by |dS|2 ≡ds2 = ηµνdxµdxν = dxµdxµ = −dt2 + dX2 .
(2.12) where ηµν is the Minskowski metric and dX2 ≡dx2 + dy2 + dz2 denotes the spatial interval.
Note that we have arbitrarily chosen a ηµν = diag(−1, 1, 1, 1) spacelike convention for the metric signature, which keeps intact the notation used for Cartesian tensors in Euclidean spacetime.
Some books use a different timelike convention for the signature of the metric, taking ηµν = diag(1, −1, −1, −1).
Although the physics is independent of the convention used, the signs appearing in the formulas in those books may differ from those in the expressions presented here. For instance, using the convention ηµν = diag(1, −1, −1, −1), Eq.(2.40) would change to pµpµ = m2 .
(2.13) Note however that in both cases you recover E2 = p2 + m2 after expanding the expression into the different components.
Note that, contrary to the Newtonian case, the metric ηµν is not positive-definite.
Given the Lorentzian signature (−+ ++) , the interval (2.12) can be positive, zero, or negative • If ds2 = 0, dX/dt = 1 and the interval corresponds to the trajectory of a light ray. This interval is called null or lightlike interval. The set of all lightlike wordlines leaving or arriving to a given point xµ spans the future or past lightcone of the event. There is a lightcone associated to each point in spacetime.
2.3 Minkowski spacetime isometry group 18 Massless particle Massive particle Spacelike Separation Spacelike Separation Timelike Separation Past lightcone Future lightcone Timelike Separation Figure 2.1: Minkowski spacetime.
• If ds2 < 0 the interval is said to be timelike. It corresponds to the wordline of a particle with nonzero rest mass moving with a velocity smaller than light, dX/dt < 1. Two events separated by such an interval are both inside the lightcone and can be in causal contact. There will exist a frame in which the two events happen at same position but at different times.
• If ds2 > 0 the interval is termed spacelike. There will exist a frame in which the two events happen at the same time but at different places, without any causal relation between them.
The different concepts are summarized in Fig.2.1.
2.3 Minkowski spacetime isometry group The transformations of Special Relativity are defined as those that do not change the Minkowski line element (2.12) (not the spatial or temporal intervals separately!). Following the procedure outlined in the previous chapter, and taking into account that ηµν is also a constant metric, the requirement ds2 = d¯ s2 gives rise to the following condition ∂Λρµ ∂xπ ηρσΛσ ν = 0 → ∂Λρµ ∂xπ = 0 , (2.14) where we have defined Λµ ν ≡∂¯ xµ ∂xν .
(2.15) As before, the transformation relating the two reference frames must be linear7. This set of trans-formations constitute the so-called inhomogeneous Lorentz group or the Poincare group, which is a combination of translations xµ →xµ + aµ (2.16) and linear homogeneous Lorentz transformations xµ = Λµ ν xν , (2.17) with Λ a 4 × 4 matrix, independent of the coordinates. The first (upper) index in Λµν labels rows, while the second (lower) one labels columns.
7Note that this is basically due to the fact that we are dealing with constant metrics.
2.3 Minkowski spacetime isometry group 19 In order to preserve the line element (2.12) the constant matrices Λµν are required to satisfy the pseudo-orthogonality condition ηµν = ηρσΛρ µΛσ ν , (2.18) which, in matrix notation, becomes η = ΛTηΛ , (2.19) with T denoting matrix transpose. Eq. (2.18) is the relativistic analogue of the orthogonality condition (1.19). The determinant of the Λ matrices is also ±1. As we did in the previous chapter, we will not consider the full Lorentz group8 (which is neither connected nor compact) O(3, 1) = L+ ∪PL+ ∪TL+ ∪PTL+ , (2.20) with P and T the parity P µν = diag(+1, −I) and time reversal T µν = diag(−1, +I) operations. We will restrict ourselves to the continuous Lorentz transformations connected with the identity (the proper Lorentz group) L+ ≡SO(3, 1) = {Λ|ΛT ηΛ = η , Λ0 0 ≥0 , det Λ = 1} (2.21) with S denoting special or reflection-free. These transformations are the relativistic analog of proper rotations in Euclidean spacetime.
Exercise Verify that the restricted set of Lorentz transformations (2.21) forms a group : • Closure: The product of any two Lorentz transformations is another Lorentz transforma-tion.
• There is an identity transformation.
• Every Lorentz transformation has an inverse.
• The product of Lorentz transformations is associative.
The fact that η2 = I4, with I4 the identity matrix, allows us to easily compute the inverse Lorentz transformation η2 = η ΛTηΛ = η ΛTη Λ = I4 → Λ−1 = ηΛTη , (2.22) which, writing explicitly the components, becomes Λ−1µ ν = ηµλΛρ ληρν = Λν µ .
(2.23) The position of the indices is important Λµν ̸= Λνµ !!
How many Lorentz transformation are there? Each Lorentz transformation is represented by a 4×4 matrix, which makes a total of 16 components. The pseudo-orthogonality condition (2.18) imposes however some constraints. Indeed, taking the transpose of such a equation leaves it unchanged. The independent components are just the diagonal elements plus half the off-original elements. We are 8 Note that now, not only the determinant of Λ, but also the element Λ00, plays a special role in the splitting (2.20).
.
2.3 Minkowski spacetime isometry group 20 left therefore with 16 −10 = 6 independent Lorentz transformations. There are two different kinds of homogeneous Lorentz transformations. The most obvious one are spatial rotations Λµ ν = 1 0 0 Rij , (2.24) where Rij is a 3 × 3 orthogonal matrix δkl = δijRikRj l , with i, j running only in spatial directions.
There are three independent rotations matrices, one per spatial direction. For instance, a rotation of angle θ around the z-axis will take the form Ri j = cos θ sin θ 0 −sin θ cos θ 0 0 0 1 .
(2.25) The difference with Euclidean transformations arises when considering the so-called boosts, which mix the spatial and temporal components. There are three of them, each one associated to the mixing of a particular spatial component with time. As an example consider a boost of rapidity η = tanh−1 v along the x direction Λµ ν = γ −γβ 0 0 −γβ γ 0 0 0 0 1 0 0 0 0 1 = cosh η −sinh η 0 0 −sinh η cosh η 0 0 0 0 1 0 0 0 0 1 , (2.26) where we have defined the parameter γ = 1/ √ 1 −v2, with v the 3-velocity. After the boost, the tem-poral and spatial coordinates are a linear and homogeneous combination of the spatial and temporal coordinates in the old frame.
Exercise Verify that Eq. (2.26) gives rise to the standard Lorentz transformation t′ = γ(t −vx) , x′ = γ(x −vt) .
(2.27) For doing so, assume Λµν to be the transformation from the rest frame of a given inertial observer to the frame of a second initial observer moving with speed v along the x axis and determine the relation between that velocity and η .
The convenience of using the rapidity parameter η instead of the velocity v resides in the fact that η combines additively. In fact, if we consider two consecutive boosts in the same direction, we have Λ(η1)Λ(η2) = Λ(η1 + η2) .
(2.28) Exercise Consider the composition of 2 boosts with velocities v1 and v2 along the x direction. Show that Λ1Λ2 gives rise to a boost with 3-velocity v = v1 + v2 1 + v1v2 .
(2.29) What happens in the limit v1, v2 ≪1? Generalize the previous to an arbitrary direction. Is the general result symmetric under the interchange of the two velocities?. Note that all the expressions are written in natural units.
2.3 Minkowski spacetime isometry group 21 t t Simultaneous events 1 Simultaneous events 2 x x P Figure 2.2: A boost transformation The form of Eq. (2.26) Λµ ν = cosh η −sinh η 0 0 −sinh η cosh η 0 0 0 0 1 0 0 0 0 1 = cos iη sin iη 0 0 sin iη cos iη 0 0 0 0 1 0 0 0 0 1 , (2.30) and the property (2.28) closely resemble those of spatial rotations (2.25).
The main difference is the change of trigonometric functions by their hyperbolic analogue, reflecting the relative sign of the temporal direction with respect to the spatial directions. Note however two important differences between boosts and ordinary rotations • The rotation parameter θ in Eq.(2.25) runs between 0 and 2π, with both points included. The rapidity parameter η is non-compact and can take whatever value in R.
• The boost matrix (2.26) is symmetric, which is not the case for ordinary rotations, cf. Eq.
(2.25): Although we should not take the analogy between rotations and boosts too seriously, it is instructive to look at the action of a Lorentz transformation on a spacetime diagram. As shown in Fig. 2.2, a Lorenzt boost rotates time and space by the same angle η = tanh−1 v, but in opposite directions! In Special Relativity the simultaneity of two events depends on the observer, the hyperplanes of constant coordinate time do not have an invariant meaning. Note however that the light cone (i.e the dashed line at 45 degrees in the diagram) is invariant under Lorentz transformations.
Exercise Use the diagram Fig.2.2 to derive the well-known effects of space contraction ¯ L = L p 1 −v2 → ¯ L < L , (2.31) and time dilatation ¯ T = T √ 1 −v2 → ¯ T > T .
(2.32) Hint: Don’t be confused by drawing just a slice of Minkowski spacetime in an Euclidean paper.
Use the Minkowski metric!
2.4 Tensors in Minkowski spacetime 22 2.4 Tensors in Minkowski spacetime The Einstein’s Principle of Relativity introduced at the beginning of this chapter implies that all the laws of physics must retain their mathematical form in all the inertial frames, i.e. they must be covariant under Lorentz transformations. As we learnt in the previous chapter, tensorial equations automatically satisfy this requirement. Since the required discussion of tensors in Minkowski spacetime closely follows that in Section 1.4 for Eucliden spacetime9, we will simply summarize the results in the Table 2.1.
Lorentz transformations ∂¯ xµ ∂xν ≡Λµν are constants!
Scalar ¯ φ = φ Contravariant vector ¯ V µ = ∂¯ xµ ∂xν V ν Covariant vector ¯ Vµ = ∂xν ∂¯ xµ Vν Contravariant rank-2 tensor ¯ T µν = ∂¯ xµ ∂xρ ∂¯ xν ∂xσ T ρσ Covariant rank-2 tensor ¯ Tµν = ∂xρ ∂¯ xµ ∂xσ ∂¯ xν Tρσ Mixed rank-2 tensor ¯ T µν = ∂¯ xµ ∂xρ ∂xσ ∂¯ xν T ρσ Table 2.1 Exercise How does the volume element d4x = dx0dx1dx2dx3 transforms under Lorentz transformations?
2.5 Covariance and Relativistic Mechanics In Newtonian spacetime the trajectory of a particle is described by the position 3-vector as a function of time xi(t), with t an absolute element of the theory. In Special Relativity a particle of mass m follows timelike worldlines in Minkowski spacetime. The time depends on the chosen reference frame and it is just another coordinate at the same level of the spatial coordinates. The trajectory xµ(σ) can be expressed in terms of a completely arbitrary parameter σ, which changes continuously along the wordline and does not need to have any particular physical interpretation. However, an interesting (and natural) possibility for the particular case of massive particles is to identify it with the proper time τ of the particle. This proper time τ is defined as the time measured by an observer in the particle’s rest frame (dX = 0), which can always be achieved by performing a Lorentz transformation.
The proper time interval dτ is therefore related to the Minkowski spacetime interval10 ds2 = −dτ 2 < 0 .
(2.33) 9Note indeed that the only conceptual difference is the replacement of rotations by Lorentz transformations, the replacement of latin indices by greek indices, and the use of the Minkowski metric, instead of the Euclidean one, for lowering and raising indices.
10Note that the proper time is not a useful parametrization for the worldline of massless particles, such as photons, since these particles move on the light cone and can travel any distance in zero proper time dτ 2 = −ds2 = 0.
2.5 Covariance and Relativistic Mechanics 23 The connection between the proper time and the measurement made in an inertial reference frame with coordinate time interval dt is given by γ ≡dt dτ = 1 −v2−1/2 .
(2.34) Note that γ is a growing function of the 3-velocity vi = dxi dt and it is always bigger than one. The proper time goes by at a slower rate than the coordinate time t.
4-velocity Given τ, and in clear analogy with the 3-dimensional case, the 4-velocity uµ along the trajectory is given by the 4-vector uµ ≡dxµ dτ , (2.35) which is tangent to the worldline of the particle and automatically normalized ηµνuµuν = uµuµ = −1 .
(2.36) In terms of its components, the 4-velocity uµ can be written as dxµ dτ = dt dτ 1, viT = γ 1, viT .
(2.37) For an observer at rest γ = 1 and Eq.(2.37) becomes simply uµ = (1, 0, 0, 0)T . In the Newtonian limit v ≪c, dτ →dt and ui →vi.
Let us consider the behaviour of uµ with respect to Lorentz transformations. Since the proper time dτ is invariant under Lorentz transformations and dxµ transforms as a contravariant tensor, we have uµ = Λµ νuν .
(2.38) The 4-velocity is a timelike 4-vector.
4-momentum Since the mass m of the particle is a scalar under Lorentz transformations, the 4-momentum pµ = muµ (2.39) is a Lorentz 4-vector with components pµ = E, piT = mγ, mγviT.
In the instantaneous rest frame of the particle, pµ = (m, 0)T . This can be used to simplify many computations. We can compute things in this particular frame and then re-express the result in a form valid in any other inertial frame by appealing to covariance.
Using the normalization condition for the 4-velocities (2.36), the normalization condition for the 4-momentum becomes pµpµ = −m2 , (2.40) which is nothing else than the well-known energy-momentum relation E = p p2 + m2 written in a covariant way11. In the Newtonian limit (|pi| ≪m) this relation becomes the familiar expression of 11Note that p is the three momentum pi.
2.5 Covariance and Relativistic Mechanics 24 the Newtonian theory together with the energy equivalent of the mass mc2, i.e. E ≃m + p2 2m. Note that the 4-momentum remains well defined even for massless particles, where it has zero square norm and becomes lightlike, pµpµ = 0. We will take this as a definition of a massless classical particle. The indefiniteness of Minkowski metric allows for non-zero values of the temporal and spatial parts as long as they cancel out in pµpµ. In particular, we can always find a frame in which pµ = (E, 0, 0, E)T .
4-acceleration It is important to remark that Special Relativity, as Newtonian mechanics, is concerned with the relation between inertial observers and not with the behavior of the objects that they are studying.
In particular, the observed objects can be accelerating. The 4-acceleration in Minkowski spacetime can be defined as12 aµ = lim ϵ→0 uµ(τ + ϵ) −uµ(τ) ϵ → aµ ≡duµ dτ = d2xµ dτ .
(2.42) The 4-acceleration (2.42) transforms in the proper way under Lorentz transformations since Λµν is linear and depends only on the relative velocity between the two frames. This fact allows it to pass completely the d2/dτ 2. The acceleration aµ is spacelike 4-vector ηµνaµaν > 0 (2.43) orthogonal to the timelike 4-velocity aµuµ = 0 , (2.44) as can be easily shown by computing the derivative of Eq. (2.36) with respect to the proper time.
Energy-momentum conservation The objects defined above allow us to easily generalize the Newton’s second law, which becomes f µ = dpµ dτ = maµ .
(2.45) Note that Eq.(2.45) is a tensorial identity, which maintains its form under Lorentz transformations and automatically satisfies Einstein’s principle of Relativity. The explicit expression of f µ depends on the considered interaction (cf. the first exercise in Section 2.7). The components of the 4-vector f µ in Eq. (2.45) are proportional to the Newtonian force F i and to the work done by F i per unit time, i.e. f µ = γ (viF i, F i)T . Contrary to what happens in Newtonian physics, energy and momentum conservation laws are not independent. The conservation laws of Newtonian mechanics in a given collision between particles are replaced by a conservation law for the total 4-momentum13 X in pµ in = X out pµ out , (2.46) 12 You will probably wondering why I am making such a mess writing out the explicit definition in (2.42) instead of directly writing aµ ≡duµ dτ = d2xµ dτ .
(2.41) The point I want you to notice in that the two vectors vµ(τ + ϵ) and vµ(τ) are located at different points in spacetime.
What we are really doing when computing the acceleration in Special Relativity is trivially moving the two vectors to the same point in spacetime before subtracting then. This trivial operation of moving a vector from one point to another will turn out to be not so trivial in General Relativity. As you will see, we will need to introduce some extra machinery in order to do that. . . but let’s move one step at a time. . .
13The interaction is assumed to be a contact interaction. Particles are free away from the interaction point.
2.6 Relativistic Lagrangian for free particles 25 with the subscripts in and out denoting the incoming and outgoing particles. Note that the conserva-tion law is Lorentz covariant and reduces to the Newtonian momentum and energy conservation for small velocities.
Exercise Show that a photon cannot spontaneously decay into an electro-positron pair.
2.6 Relativistic Lagrangian for free particles The equation of motion for a free particle following from (2.45) dpµ dτ = 0 , (2.47) can be also be derived from a Lagrangian formulation where the role of the generalized coordinates is played by the space-time coordinates xµ and the classical time t is replaced by an appropriate parameter σ. The simplest guess for the relativistic action would be a naive generalization of the Newtonian action, namely S = 1 2m Z dτ ηµν dxµ dτ dxν dτ .
(2.48) Note however that the previous expression is not invariant under reparametrizations of the path14 τ →f(τ). The dynamic of the particle seems to depend on the “internal coordinate” τ used in the description of the curve xµ(τ). Moreover, the action (2.48) does not contain any information about the lightcone. On top of that, t neither has a smooth massless limit.
To solve these problems, we will substitute the proper time by an arbitrary parameter σ and introduce a non-dynamical function15e(σ), the so-called einbein. This quantity will be treated as an additional generalized coordinate during the intermediate computations and fixed to a particular value only at the end. To clarify the construction, we proceed in several steps. We start by replacing the problematic mass appearing in the action (2.48) by e−1(σ). This gives rise to the following structure S ∼1 2 Z dσ 1 e(σ)ηµν dxµ dσ dxν dσ .
(2.49) In order for the previous action to be invariant under reparametrizations of the path16 σ →f(σ), the einbein e(σ) must be chosen to transform in the proper way. The transformation rule can be determined by inspection: the property e(σ)dσ must remain invariant. In other words, the infinitesimal displacement dσ and the einbein must transform in an opposite way d¯ σ = ˙ f(σ)dσ , ¯ e(¯ σ) = ˙ f(σ) −1 e(σ) .
(2.50) With these transformations at hand, we proceed now to reintroduce the mass parameter m in the action (2.49). The form of the new term is essentially determined by pure dimensional arguments, reparametrization invariance and the massless limit. In order for the new piece to be reparametrization invariant, the integration measure dσ must come together with a factor e(σ).
This gives a term 14The reparametrization invariance of the action should be understood as a gauge symmetry: a redundancy of the description, not a symmetry relating different solutions of the theory.
15i.e. no kinetic term for e(σ) will be included.
16Or if you want invariant under 1D general coordinate transformations.
2.7 Maxwell’s equations 26 dσe(σ) with dimension [E]−2, which must be compensated17 by something with dimension [E]2 and proportional to m. There you are: the new term is dσe(σ)m2. The resulting action is the so-called einbein action S = 1 2 Z dσ 1 e(σ)ηµν dxµ dσ dxν dσ −m2e(σ) (2.51) and give rise to the following Euler-Lagrange equations for the generalized coordinates xµ(σ) and e(σ) d dσ e−1(σ)dxµ dσ = 0 and ηµν dxµ dσ dxν dσ = −m2e2(σ) .
(2.52) The massive or massless character of particles is automatically incorporated in the second equation.
Indeed, choosing e(σ) = 1 and taking the limit m →0, we obtain the equations of motion for a free massless particle d2xµ dσ2 = 0 , ηµν dxµ dσ dxν dσ = 0 .
(2.53) On the other hand, the equations for massive particles can be obtained by choosing e(σ) = 1/m and using the proper time as affine parameter (σ = τ) md2xµ dτ 2 = 0 , ηµν dxµ dτ dxν dτ = −1 .
(2.54) These kind of choices in which e(σ) = constant are called affine and restrict the function f(σ) to the form ˙ f = 1 → f(σ) = σ + constant .
(2.55) Exercise Consider the massive case. Show that the action (2.51) is equivalent to the geometrical action S = −m Z dτ .
(2.56) 2.7 Maxwell’s equations In their traditional form, Maxwell’s equations are given by 18 ∇× E + ∂tB = 0 , (2.57) ∇· B = 0 , (2.58) ∇· E = ρ , (2.59) ∇× B −∂tE = J , (2.60) with E and B the electric and magnetic fields, J the current density and ρ the charge density. They are 8 coupled linear differential equations, in which the boundary conditions are usually taken to be such that for infinite systems the fields E and B go to zero at infinity. Note also the symmetry E ↔B in the absence of sources.
17Recall that, in natural units, the action is dimensionless.
18Note that they are written using the Heaviside-Lorentz convention, in which no 4π factors appear.
2.7 Maxwell’s equations 27 The homogenous Maxwell’s equations (2.57) and (2.58) can be solved by introducing the so- called electromagnetic potentials: a scalar potential ϕ and a vector potential19 A satisfying E = −∇ϕ −∂tA , B = ∇× A .
(2.61) Using them, the inhomogeneous Maxwell’s equations become ∂2 t ϕ −∇2ϕ = ρ , ∂2 t A −∇2A = J .
(2.62) Given the electromagnetic potentials ϕ and A in Eq. (2.61) the electromagnetic fields E and B are completely determined, but no viceversa. A and φ are gauge fields (see the exercise below).
Familiarity with Maxwell’s equations soon leads to the appreciation of the unified nature of the electromagnetic field and its relativistic nature. Although in their 19th century version Maxwell’s equations (2.57)-(2.60) do not seem at all invariant under Lorentz transformations, they can be written in a more compact and elegant way, that makes explicit their covariant form. Introducing the 4-vector potential (gauge field) Aµ ≡(ϕ, A) and the charge-current density 4-vector Jµ ≡(ρ, J), we obtain20 ∂νF µν = Jµ , (2.63) ϵµνρσ∂ρF µν = 0 , (2.64) where the antisymmetric quantity F µν ≡∂µAν −∂νAµ is the gauge invariant (Faraday) field strength with components F 0i = Ei , F ij = ϵijkBk , (2.65) and the different ϵ are totally antisymmetric tensors in the corresponding dimension21 n ϵµ1µ2...µn = +1, if µ1µ2 . . . µn is an even permutation of 01 . . . (n −1) , −1, if µ1µ2 . . . µn is an odd permutation of 01 . . . (n −1) , 0, otherwise .
(2.66) In covariant notation, (2.62) becomes 2Aµ = Jµ (2.67) where we have defined the d’Alambertian operator 2 ≡∂µ∂µ = −∂2 t + ∂2 i .
(2.68) • The covariant components ϵµνρσ of the permutation tensor ϵµνρσ in Minkowski spacetime are defined by lowering each of the indices with the metric tensor ηµν ϵµνρσ = ηµληνκηρπηστϵλκπτ , (2.69) from which it follows that ϵ0123 = −ϵ0123.
• Note that, whereas a cyclic permutation of the indices in the 3-dimensional permuta-tion symbol leaves it unchanged (ϵijk = ϵjki), a cyclic permutation of the 4-dimensional permutation symbol gives rise to a minus sign (ϵµνρσ = −ϵνρσµ).
19In the 3-dimensional sense!
20Eq. (2.64) is sometimes called a Bianchi identity. As you will see, this will not be the last time that we will find one of these identities.
21Note that some books use the opposite sign convention for (2.66).
2.7 Maxwell’s equations 28 Exercise Just for those of you knowing Classical Field Theory. The previous equations of motion can be obtained from the following Lagrangian density L = −1 4F µνFµν + JµAµ , (2.70) where Jµ is treated as an external source.
• Check that with the Lagrangian density (2.70), the action S = R L d4x is invariant under Lorentz transformations Aµ = Λµ νAν , F µν = Λµ ρΛν σF ρσ , Jµ = Λµ νJν , (2.71) and gauge transformations Aµ →Aµ −∂µχ , (2.72) where χ = χ(t, x) is an arbitrary function of space-time. The concept of purely elec-tric or magnetic fields and that of a static charge distribution with zero current become meaningless, being a good description only in a particular reference frame.
• Which is the Lorentz invariant generalization of the equation of motion F = q(E+v×B)?
Can you guess the associated Lagrangiana? Check the consistency of the two results by computing the Euler-Lagrange equations (3.30) of the obtained Lagrangian .
aHint: Which is the generalization of the cross product in the 4-dimensional case? Consider the Lagrangian of a charged particle in an electrostatic potential Φ.
Taking the 4-divergence of Eq. (2.63) we obtain ∂µJµ = ∂µ∂νF µν = 0 , (2.73) where in the last step we used the fact that F µν is an antisymmetric tensor, i.e.
F µν = −F νµ.
Eq.(2.73) is nothing else than the continuity equation ∂ρ ∂t + ∇· J = 0 .
(2.74) The conservation of total charge Q(t) ˙ Q(t) = Z R3 ˙ ρ(t, xi)d3xi = − Z R3 ∂kJk(t, xi)d3xi = 0 , (2.75) is imposed by the field equations. If the charge is not conserved there is no solution!
Exercise Prove that the product SµνAµν of a symmetric tensor Sµν and an antisymmetric tensor Aµν is zero. |
187729 | https://www.khanacademy.org/science/hs-chemistry/x2613d8165d88df5e:solutions-acids-and-bases/x2613d8165d88df5e:molarity/e/apply-molarity | Use of cookies
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187730 | https://www.youtube.com/watch?v=FKwJXD7yZIY | GEOGEBRA: All about VECTORS!
Math with Drew
3880 subscribers
751 likes
Description
40666 views
Posted: 16 Jun 2021
GeoGebra lets you create and manipulate vectors in fairly intuitive and flexible ways. In this video I'll show you the basics behind creating vectors, how to carry out the standard vector operations, and the relationship between vectors and points in GeoGebra.
Transcript:
Intro hey welcome back my name is drew and in this video we are doing more geogebra this time with vectors i found that geogebra actually makes using vectors fairly intuitive and convenient so in this video i'm going to show you how to make vectors in geogebra as well as how to do some of the usual vector operations all right let's do it so i'm in geogebra Making Vectors i'm going to show you how to do things with the vectors in both 2d and 3d the way it works is kind of the same but we'll be jumping back and forth so i can kind of show you what it looks like all right to start let's make a vector so we go over to our input bar and we're going to make a new vector let's call it u one way to make a vector in geogebra is just to type the word vector and we can even see as we type that it's suggesting pretty much the two ways that we'll make vectors one of them is to just provide a point so let's say we want to make a vector using the point one comma two if we do that and hit enter you can see it makes the vector for us here it writes it in column notation and it draws that vector on the page let's make another vector we'll call it v and it'll be the vector that points from the origin to the point negative three one and just real quick let's adjust some of these things about these vectors let's move the labels so they're kind of off to the side if i single click one of the vectors i can adjust the size of it let's make it a little bit bigger that looks good and we can change the color now let's go back to our input bar over here on the right and we'll make another vector we'll call it w and let's use this second form so this autocomplete is sort of suggesting for us what it'll do if we put in two points into our vector function it'll draw that vector with its tail at the first point and its tip at the second point so let's draw the vector that points from the point one two to the point negative three one there we go one two and negative three one were the input points for their first two vectors and so we've drawn sort of the third side of this triangle here so that's essentially how we make vectors one way is to just put in a point and it'll draw the vector that points from the origin to that point or you can put in two points and it'll draw the vector from the first point to the second point let's see how all this works in 3d since i'm using the calculator suite i can just click the drop down here and go to 3d calculator so we'll make a new vector in the same way we'll call it u again and we'll make it the vector that points from the origin to the point one two three if we do that it does it makes the 3d vector and it draws it here for us just like before we can click make it bigger if we want change the color and just for the sake of example if we wanted a vector that points between two points we can do the same thing we could put in the point say one two three and the point negative one zero one hit enter and it tells us what vector that is [Music] and it draws it here in the 3d view for us so pretty much everything that worked in 2d also works in 3d let's switch back to what we had in the 2d calculator by clicking the drop down up here and going back to graphing and let's hide this vector w here just for a second and let's talk about vector operations now that we have two vectors u and v what can we do with them well it turns out that geogebra is pretty intuitive about the way that you interact with vectors pretty much any operation you're used to doing with vectors you know with pen and paper you can do in geogebra for instance let's take these two vectors u and v and add them so if i just go to the input bar and type u plus v it will add them and tell us what vector that is if i hit enter and then it'll also draw that vector for us in the view now let's delete this we can also subtract vectors so we could say let's make a new vector that is v minus u it'll draw that vector there we hit enter this vector is v minus u remember from vector geometry if we were writing the vector v minus u that should be the same as the vector drawn from the tip of u to the tip of v that's actually the same as that vector w we had before so if i bring that back we can see these two vectors are the same like they're the same arrow pointing in the same direction and indeed geogebra is treating them as the same vector so even though they look different in the way they're drawn geogebra really thinks of them as the same vector and any mathematical operation that we would do with this vector w or this vector a would turn out the same independent of which of the two we used so we'll get rid of this and hide this again now let's talk about scalar multiplication it works pretty much the way you want it to if you want to take this vector u and make a new vector that's twice as long you can really just type 2u and there you go if you hit enter it'll do the computation for you and show you the result and of course also draw the vector and there's really no limit in terms of like what you can do here with scalar multiplication and addition like if you wanted to just write a linear combination of these two vectors like 3u plus one half v and hit enter there you go it'll compute it for you it'll draw it for you here it all works and i'm not going to show it but you can take my word for it that doing these kinds of operations addition and scalar multiplication in 3d with 3d vectors would work just the same so let's get rid of these and let's talk about some other vector operations that we do such as the dot product there's a couple ways to do the dot product if you have two vectors probably the easiest way to remember is just to type in the word dot and geogebra confirms this is an actual thing that we do if we put in parentheses u and v it'll compute the dot product which of course is just a number it's not going to show anything for you in your view here but it computes the dot product i really don't like that it writes it as uv that's not really great in terms of just like how i think mathematical notation should work but oh well all right so let's get rid of this and another way to get to dot products that's a little bit more in line with how we'd write them notationally would be to write something like u and then i'm going to press shift and 8 which is the asterisk key and when i do it'll put a little dot there which is convenient if i then put in v it'll it'll compute the dot product and it writes it with that same crappy notation but whatever so that's the dot product it works for 2d vectors and 3d vectors in exactly the same way lastly let's talk about the cross product and for that of course we're going to have to go to our 3d calculator which has these two vectors from before i'm gonna get rid of this one and make a new vector v that's drawn just with its base at the origin how about negative one zero one all right so for the cross product you can do it in two ways just like the dot product you can type in cross and then put in the two vectors that you want to cross and hit enter it computes the cross product for you and it draws that vector in the 3d view and you can kind of rotate the view around and confirm to yourself that this black vector is actually the vector that you think it should be it's the vector that is perpendicular to both u and v and it points in the correct direction based on the right hand rule all that stuff and notice that uh jojoba uses this kind of funky cross product notation here this circle with an x in it if you wanted to get the same thing you know if you want to do the analogy of using the asterisk for dot product let's see we can type in u and then we can go to the keyboard down here go to the symbols pane and i think yeah here it is so you just click that and then v then hit enter that's kind of a pain in the butt but you can do it and lastly if you want to remember this you can you can type in u and you can get that same symbol it turns out by typing shift alt and 8 at the same time so it's the asterisk key but you're pressing down the alt key or option key at the same time you get the same symbol there you go you can compute the cross product that way so if you're doing a project that involves a lot of cross products for some reason maybe you'll want to use that hotkey other than that i usually just do the word cross and then followed by the two vectors in parentheses okay so i have just one more thing to tell you about which is a little bit about how geogebra treats vectors and points to show you that i'm going to go back to our 2d view and let's get rid of everything Points and Vectors now in math sometimes it's useful for us to conflate the idea of a point and a vector because sometimes we think of vectors as like directions but other times we think of vectors as representing locations now geogebra does a pretty good job of letting you move fluidly between points and vectors so let me show you a bit about how that works first of all remember how we created a vector the first time by doing vector the word and then one and then the point one two it turns out we didn't need to type the word vector there we could have just said u equals one two and it makes a vector for us now that might be a little surprising because it looked like i was going to be making a point one comma two but actually it turns out and i actually just found this out while making this video if we were to type something like capital a is the point one two then it actually makes the point one comma two now the difference between the two things that i entered i entered u equals parentheses one two and a equals parenthesis one two the primary difference was i used a lowercase letter and got a vector in an uppercase letter and got the point and that's the rule lowercase gives you vectors uppercase gives you points that works even if the variable name that you're using is more than one letter so if i wanted to make a vector named drew that'll be the vector that points from the origin to the point three comma three it'll make the vector but if i use a capital d instead it'll make the point so this is a little bit suggesting that geogebra thinks of points and vectors as pretty much the same thing and actually everything that we could do with vectors we can also do with points so i have a point a which is one two let's say we make the point b which is negative three comma one we can do all the things with these points a and b that we did with the vectors u and v from before so we can add them we get the point negative 2 3 which would be the same as if we had added the vector 1 2 to the vector negative 3 1 we would have gotten the vector that pointed from the origin to the point negative 2 3. so you can add points you can scale or multiply points just like they were vectors and you can even take the dot product of two points and it computes the dot product as if they were vectors so that's just to show you that there's a lot of flexibility here and like in a situation like multi-variable calculus where sometimes a vec there is a position and sometimes a vector is like a tangent vector like a direction being able to go back and forth between thinking of it as a point or as a vector is really convenient Outro all right so there you have it now you're ready to do all the fun things that we do in mathematics with vectors like drawing tangent vectors on parametrized curves or vector fields or line integrals there's so many things now that you can visualize and even compute if you're clever in geogebra if you're interested in seeing a video on any of those things let me know down in the comments and also of course as always let me know if you have any questions about anything that we talked about in this video go ahead and give this video a like if you enjoyed it and subscribe for more fun math content thanks for playing along and i'll see you in the next video [Music] you |
187731 | https://higginschemistry.weebly.com/uploads/2/3/8/9/23899776/guided_notes_-_limiting_reactant.pdf | Guided Notes Limiting Reactant – Tyler DeWitt (YouTube) The limiting reactant is the reactant (left side of the arrow) that is completely used up in a chemical reaction. When the reaction stops, there is NO limiting reactant leftover. Any other reactant is called an excess reactant because when the reaction stops there is some excess leftover. Click the following link and follow the guided notes: Introduction to Limiting Reactant and Excess Reactant 1) What do we call the chemicals on the left side of the arrow in a reaction? __ 2) What do we call the chemicals on the right side of the arrow in a reaction? __ 3) What causes a chemical reaction to stop? 3 cups flour + 1 cup water 5 bread rolls 4) In Tyler’s example, IF ALL the 6 cups of flour are used up how many cups of water do you need? __ 5) In Tyler’s example, IF ALL the 3 cups of water are used up how many cups of flour do you need? _ 6) What is the limiting reactant in this example? __ 7) What is the excess reactant in this example? __ What is the greatest amount of NH3 (in mol) that can be made with 3.2 mol N2 and 5.4 mol H2? What is the limiting reactant? Which reactant is in excess, and how many mol of it are leftover? N2 + 3 H2 2 NH3 8) In Tyler’s example, IF ALL the 3.2 mol N2 is used up how many mol of H2 do you need? show the work here: 3.2 mol N2 x mol H2 = ___ mol N2 9) In Tyler’s example, IF ALL the 5.4 mol H2 is used up how many mol of N2 do you need? show the work here: 5.4 mol H2 x mol N2 = __ mol H2 10) What is the limiting reactant in this example? __ Why? 11) What is the excess reactant in this example? __ Why? 12) Now, use the limiting reactant (5.4 mol H2) to calculate the greatest amount of NH3 that can be made: show the work here: 5.4 mol H2 x mol NH3 = ___ mol H2 13) Now, let’s calculate the moles of excess reactant leftover when the reaction stops. Remember, Tyler already calculated how much N2 was used (question #9 above): remember: 5.4 mol H2 x 1 mol N2 = 1.8 mol N2 is used 3 mol H2 initial excess mol – excess mol used = leftover excess mol show the work here: __ – _ = __ Click the following link and follow the guided notes: Limiting Reactant Practice Problem What is the greatest amount of MgO (in mol) that can be made with 7.8 mol Mg and 4.7 mol O2? What is the limiting reactant? Which reactant is in excess, and how many mol of it are leftover? 2 Mg + O2 2 MgO 14) In Tyler’s example, IF ALL the 7.8 mol Mg is used up how many mol of O2 do you need? show the work here: 7.8 mol Mg x mol O2 = __ mol Mg 15) In Tyler’s example, IF ALL the 4.7 mol O2 is used up how many mol of Mg do you need? show the work here: 4.7 mol O2 x mol Mg = ___ mol O2 16) What is the limiting reactant in this example? __ Why? 17) What is the excess reactant in this example? __ Why? 18) Now, use the limiting reactant mol to calculate the greatest amount of MgO that can be made: show the work here: _ mol Mg x __ = _ mol MgO 19) Now, let’s calculate the moles of excess reactant leftover when the reaction stops. Remember, Tyler already calculated how much O2 was used (question #14 above): remember: 7.8 mol Mg x 1 mol O2 = 3.9 mol O2 is used 2 mol Mg initial excess mol – excess mol used = leftover excess mol show the work here: __ – ___ = ___ Click the following link and follow the guided notes: Limiting Reactant Practice Problem (Advanced) What is the greatest mass of AlCl3 (in grams) that can be made with 114 g Al and 186 g Cl2? What is the limiting reactant? Which reactant is in excess, and how many grams of it are leftover? 2 Al + 3 Cl2 2 AlCl3 20) How many moles of each reactant do we have? show the work here: 114 g Al x mol Al = __ g Al show the work here: 186 g Cl2 x mol Cl2 = ___ g Cl2 21) In Tyler’s example, IF ALL the moles of Al are used up how many mol of Cl2 do you need? show the work here: _ mol Al x mol Cl2 = __ mol Al 22) In Tyler’s example, IF ALL the moles of Cl2 are used up how many mol of Al do you need? show the work here: _ mol Cl2 x mol Al = ___ mol Cl2 23) What is the limiting reactant in this example? __ Why? 24) What is the excess reactant in this example? __ Why? 25) Now, use the limiting reactant moles to calculate the greatest moles of AlCl3 that can be made: show the work here: _ mol Cl2 x __ = __ mol AlCl3 26) The question asks for grams of AlCl3. How many grams of AlCl3 is this? show the work here: __ mol AlCl3 x __ g AlCl3 = _ g AlCl3 mol AlCl3 27) Now, let’s calculate the grams of excess reactant leftover when the reaction stops. Remember, Tyler already calculated how much Al was used (question #22 above): remember: 2.62 mol Cl2 x 2 mol Al = 1.75 mol Al is used 3 mol Cl2 initial excess mol – excess mol used = leftover excess mol show the work here: __ – _ = __ 28) The question asks for grams of excess reactant (Al). How many grams of Al is this? show the work here: _ mol Al x __ g Al = _ g Al mol Al |
187732 | https://forum.wordreference.com/threads/despu%C3%A9s-de-que-subjunctive.3705274/ | Después de que (subjunctive) | WordReference Forums
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Spanish-English / Español-Inglés
Spanish-English Grammar / Gramática Español-Inglés
Después de que (subjunctive)
Thread starterEms1993
Start dateMay 29, 2020
E
Ems1993
New Member
Chinese
May 29, 2020
#1
Hi everyone,
I am learning Present Subjunctive at this stage and i have a question.
From the booking I am using, it says "a subjunctive form follows directly after one of the following conjunctions if the main clause has a different subject than the dependent clause" and then it lists 7 conjunctions including"a pesar de que" and "después de que".
Below are the exercises i did wrong and not able to understand:
_(after) yo__(to bathe myself), voy a vestirme.
___(in spite of) ellos _ (to be)frío, ellos quieren dar una vuelta.
I worte "Después de" for the 1st one as i can see the subjective is the same, it is "I", so why we need to use "después de que" and subjunctive form?
Same for the 2nd one, the subjective is ellos and no other subjective (i wrote "a pesar de") , why should i still use subjunctive form.
Thank you for your help
BLUEGLAZE
Senior Member
English - USA
May 29, 2020
#2
It appears that in the exercise they wanted the subject expressed after the use of the preposition.
Después de bañarme, yes, but they wanted Después de que yo me bañe...
As you said, you are working on the use of the subjunctive.
Last edited: May 29, 2020
C
Circunflejo
Senior Member
Castellano de Castilla
May 29, 2020
#3
Ems1993 said:
From the booking I am using, it says "a subjunctive form follows directly after one of the following conjunctions if the main clause has a different subject than the dependent clause"
Click to expand...
That's not true as you already discovered.
The first one, in my opinion, would make more sense if it said después de bañarme,... However, being the exercise about the subjuntive between subjuntive and indicative, I too would choose subjunctive.
On the second one you can use either subjuntive or indicative. Both of them are fine.
gengo
Senior Member
San Francisco, CA
American English
May 30, 2020
#4
Ems1993 said:
_(after) yo__(to bathe myself), voy a vestirme.
___(in spite of) ellos _ (to be)frío, ellos quieren dar una vuelta.
I worte "Después de" for the 1st one as I can see the subject is the same, it is "I", so why do we need to use "después de que" and subjunctive form?
Same for the 2nd one, the subject is ellos and no other subject (I wrote "a pesar de") , why should I still use subjunctive form?
Click to expand...
I think both of those are bad examples for a textbook, because I would use the infinitive, not the "que + subjuntivo," for both.
Después de bañarme, voy a vestirme.
A pesar de tener frío, ellos quieren dar una vuelta.
E
Ems1993
New Member
Chinese
May 30, 2020
#5
BLUEGLAZE said:
It appears that in the exercise they wanted the subject expressed after the use of the preposition.
Después de bañarme, yes, but they wanted Después de que yo me bañe...
As you said, you are working on the use of the subjunctive.
Click to expand...
gengo said:
I think both of those are bad examples for a textbook, because I would use the infinitive, not the "que + subjuntivo," for both.
Después de bañarme, voy a vestirme.
A pesar de tener frío, ellos quieren dar una vuelta.
[/QUOTE
Click to expand...
BLUEGLAZE said:
It appears that in the exercise they wanted the subject expressed after the use of the preposition.
Después de bañarme, yes, but they wanted Después de que yo me bañe...
As you said, you are working on the use of the subjunctive.
Click to expand...
Thank you for your reply, at least i know my answer is not really wrong
E
Ems1993
New Member
Chinese
May 30, 2020
#6
Circunflejo said:
That's not true as you already discovered.
The first one, in my opinion, would make more sense if it said después de bañarme,... However, being the exercise about the subjuntive between subjuntive and indicative, I too would choose subjunctive.
On the second one you can use either subjuntive or indicative. Both of them are fine.
Click to expand...
Thank you for your help
You must log in or register to reply here.
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¶"S#¢®ë@¤¸÷! 0p|¹O9Ô Zht¹kÍË<XÕÙF¾svý3 �v»+¦ b©«]U5eEtئñìZñ1D\ý|£ÓÆ M]ýYTªä=82±ï÷9Oó<ÆÔïö7"¥äì(<ÖÍô./¯^ý/Ü÷;\Òr«æ $J±1®rªR�pO}¿§~Wkª¥�¢;-ËDÄ MÒÇ7°!"¤ f�Èq·»J©¿\ÌÔ«o]TÍÔdu1E壪²º êÈfóXïÛ÷®ÛÔBLµ�PRJLÝ0\Õ²üìgß7ÓRRò«W?îûóçßèºjMÝðÅ/ßûèúöýð7¶ÔÅLÜÿÀ/9$UAZÞ�JÍ1vLáý÷¿õòå7ér¼C$Ú?ùÙ÷¥f§ÌÿºWÙwí×Ñ�¼~_ËÁG¦� +;4d,"j&1:ÀÖ�~<#"Q± ¢h}è� ëûÈÑ# óù fNï�D©uÉ9ÀjÉækpÀyO¯çýqvýµ)i£s¦1çÒ÷ÝòÖ£ùÖo}«O·Uóùh&�¶,%Æ8 {%®º¼®ëa!ÖZÌ EªHÎKLQMÕa\3$J]25$J1åOÇ£ËÔ Ã®ë;Då¿xÕwq8®a%µÄQk 1ÔZ%K-ÄPCµpιz�ÌyZüл̹ï1ÑY�(/µëúÔ ª J®e´ºZ^» ë"1ZÉ5Ø#bѽ© ÈÕ´kDä2ÌkS1,Ë�,ÆT²xËQ<ÇÎñ)"&µÌó#2f\& (ór>Ï!Ôu$LÖ²ô»«¦Ð ì[R0v½å¼h« D\_ýÌ@eDG'}ô4 ÇË<Í"Lµë{Dr)1¥Á+Ó4:'yØíüÈ@)ºk)3§q\kØÍ3Ñ Æ.ÄHDèå¤Òpçu[÷»ÎÓbØï÷Î+Úí¥äRÊùÆeñ?øÛïÿÝß ã0ìJYZI%ç®ïb´ª4ÇÓ8ÞÜFB¬ÍÉV}ãc&2Qô¡b&ª0/+ûCGÌýþp¨µj.¨«� A5Oæ%sDH}gfµJsl#OI yÕìííË}ßs¯våZKJ]×õ1u¯ñÙ<Ã4ÍRíêêúêêöp¸AªÖ÷Ão~ë[©þÍ¿þ?«ª,s5iFhN$ vn©ænv¿ñïÏLå|y£äÛë÷vÃu²I\¾k\_åöºýz4©|á« µydÕýÒ+jYWöö^çTÚJEÜ @[
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187734 | https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/using-the-properties-of-trapezoids-to-solve-problems/ | Using the Properties of Trapezoids to Solve Problems | Mathematics for the Liberal Arts Corequisite
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Mathematics for the Liberal Arts Corequisite
Appendix B: Geometry
Search for:
Using the Properties of Trapezoids to Solve Problems
Learning Outcomes
Find the area of a trapezoid given height and width of bases
Use the area of a trapezoid to answer application questions
A trapezoid is four-sided figure, a quadrilateral, with two sides that are parallel and two sides that are not. The parallel sides are called the bases. We call the length of the smaller base b b, and the length of the bigger base B B. The height, h h, of a trapezoid is the distance between the two bases as shown in the image below.
A trapezoid has a larger base, B B, and a smaller base, b b. The height h h is the distance between the bases.
The formula for the area of a trapezoid is:
Area trapezoid=1 2 h(b+B)Area trapezoid=1 2 h(b+B)
Splitting the trapezoid into two triangles may help us understand the formula. The area of the trapezoid is the sum of the areas of the two triangles. See the image below.
Splitting a trapezoid into two triangles may help you understand the formula for its area.
The height of the trapezoid is also the height of each of the two triangles. See the image below.
The formula for the area of a trapezoid is
If we distribute, we get,
Properties of Trapezoids
A trapezoid has four sides.
Two of its sides are parallel and two sides are not.
The area, A A, of a trapezoid is A=1 2 h(b+B)A=1 2 h(b+B) .
example
Find the area of a trapezoid whose height is 6 6 inches and whose bases are 14 14 and 11 11 inches.
Solution
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.the area of the trapezoid
Step 3. Name. Choose a variable to represent it.Let A=the area A=the area
Step 4.Translate.
Write the appropriate formula.
Substitute.
Step 5. Solve the equation.A=1 2⋅6(25)A=1 2⋅6(25)
A=3(25)A=3(25)
A=75 A=75 square inches
Step 6. Check: Is this answer reasonable?✓✓ see reasoning below
If we draw a rectangle around the trapezoid that has the same big base B B and a height h h, its area should be greater than that of the trapezoid.
If we draw a rectangle inside the trapezoid that has the same little base b b and a height h h, its area should be smaller than that of the trapezoid.
The area of the larger rectangle is 84 84 square inches and the area of the smaller rectangle is 66 66 square inches. So it makes sense that the area of the trapezoid is between 84 84 and 66 66 square inches
Step 7. Answer the question. The area of the trapezoid is 75 75 square inches.
try it
In the next video we show another example of how to use the formula to find the area of a trapezoid given the lengths of it’s height and bases.
example
Find the area of a trapezoid whose height is 5 5 feet and whose bases are 10.3 10.3 and 13.7 13.7 feet.
Show Solution
Solution
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.the area of the trapezoid
Step 3. Name. Choose a variable to represent it.Let A = the area
Step 4.Translate.
Write the appropriate formula.
Substitute.
Step 5. Solve the equation.A=1 2⋅5(24)A=1 2⋅5(24)
A=12(5)A=12(5)
A=60 A=60 square feet
Step 6. Check: Is this answer reasonable?
The area of the trapezoid should be less than the area of a rectangle with base 13.7 13.7 and height 5 5, but more than the area of a rectangle with base 10.3 10.3 and height 5 5.
✓✓
Step 7. Answer the question.The area of the trapezoid is 60 60 square feet.
try it
example
Vinny has a garden that is shaped like a trapezoid. The trapezoid has a height of 3.4 3.4 yards and the bases are 8.2 8.2 and 5.6 5.6 yards. How many square yards will be available to plant?
Show Solution
Solution
Step 1. Read the problem. Draw the figure and label it with the given information.
Step 2. Identify what you are looking for.the area of a trapezoid
Step 3. Name. Choose a variable to represent it.Let A A = the area
Step 4.Translate.
Write the appropriate formula.
Substitute.
Step 5. Solve the equation.A=1 2(3.4)(13.8)A=1 2(3.4)(13.8)
A=23.46 A=23.46 square yards.
Step 6. Check: Is this answer reasonable?
Yes. The area of the trapezoid is less than the area of a rectangle with a base of 8.2 8.2 yd and height 3.4 3.4 yd, but more than the area of a rectangle with base 5.6 5.6 yd and height 3.4 3.4 yd.
Step 7. Answer the question.Vinny has 23.46 23.46 square yards in which he can plant.
try it
Candela Citations
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Question ID 146533, 146534, 146535. Authored by: Lumen Learning. License: CC BY: Attribution
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Ex: Find the Area of a Trapezoid. Authored by: James Sousa (mathispower4u.com). Located at: License: CC BY: Attribution
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Licenses and Attributions
CC licensed content, Original
Question ID 146533, 146534, 146535. Authored by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Ex: Find the Area of a Trapezoid. Authored by: James Sousa (mathispower4u.com). Located at: License: CC BY: Attribution
CC licensed content, Specific attribution
Prealgebra. Provided by: OpenStax. License: CC BY: Attribution. License Terms: Download for free at
PreviousNext |
187735 | https://mathspp.com/blog/pydonts/sequence-indexing | Sequence indexing | Pydon't 🐍
programming
python
(If you are new here and have no idea what a Pydon't is, you may want to read the
Pydon't Manifesto.)
Introduction
Sequences in Python, like strings, lists, and tuples, are objects that
support indexing: a fairly simple operation that we can use to access specific elements.
This short article will cover the basics of how sequence indexing works and then give you
some tips regarding anti-patterns to avoid when using indices in your Python code.
In this article you will:
learn the basic syntax for indexing sequences;
learn how negative indices work;
see some tools that are often used to work with sequences and indices;
learn a couple of tricks and things to avoid when indexing;
Sequence indexing
First and foremost, I am talking about sequence indexing here to distinguish the type
of indexing you do to access the values of a dictionary, where you use keys to index
into the dictionary and retrieve its values.
In this article we will be talking about using integers to index linear sequences,
that is, sequences that we can traverse from one end to the other, in an ordered fashion.
A very simple example of such a sequence is a string:
```
s = "Indexing is easy!"
s
'Indexing is easy!'
```
To index a specific character of this string I just use square brackets and the integer
that corresponds to the character I want.
Python is 0-indexed, which means it starts counting indices at 0.
Therefore, the very first element of a sequence can be obtained with .
In our example, this should give a capital "I":
```
s = "Indexing is easy!"
s
'I'
```
Then, each following character is obtained by increasing the index by 1:
```
s = "Indexing is easy!"
s
'n'
s
'd'
s
'e'
```
Here is a figure that shows how to look at a sequence and figure out which
index corresponds to each element:
Imagine vertical bars that separate consecutive elements,
and then number each of those vertical bars,
starting with the leftmost bar.
Each element gets the index associated with the bar immediately to its left:
Maximum legal index and index errors
Because indices start at 0, the last legal index to a sequence is the index that is equal
to the length of the sequence, minus one:
```
s = "Indexing is easy!"
len(s)
17
s
'!'
s
Traceback (most recent call last):
File "", line 1, in
IndexError: string index out of range
```
As you can see above, if you use an index that is too large
(read: greater than or equal to the length of the sequence)
Python will raise an IndexError, warning you about your
usage of an integer that is too large for that specific indexing operation.
Negative indices
If the last legal index is the length of the sequence minus 1,
then there is an obvious way to access the last item of a sequence:
```
s = "Indexing is easy!"
s[len(s)-1]
'!'
l = [12, 45, 11, 89, 0, 99]
l[len(l)-1]
99
```
However, Python provides this really interesting feature where you can
use negative indices to count from the end of the sequence.
In order to figure out which negative index corresponds to which element,
think about writing the sequence to the left of itself:
Then you just have to continue the numbering from the right to the left,
therefore making use of negative numbers:
From the figure above you can see that the index -1 refers to the last element of the sequence,
the index -2 refers to the second to last, etc:
```
s = "Indexing is easy!"
s[-1]
'!'
s[-2]
'y'
```
We can also take a look at all the negative indices that work for our specific sequence:
Another way to look at negative indices is to pretend there is a len(s) to their left:
| Negative index | | Corresponding positive index |
---
| -1 | | len(s) - 1 |
| -2 | | len(s) - 2 |
| -3 | | len(s) - 3 |
| ... | | ... |
| -len(s) | | len(s) - len(s) (same as 0) |
And a couple of examples:
```
s = "Indexing is easy!"
s[-5]
'e'
s[len(s)-5]
'e'
s[-13]
'x'
s[len(s)-13]
'x'
len(s)
17
s[-17]
'I'
s[len(s)-17]
'I'
```
Indexing idioms
Having seen the basic syntax for indexing,
there are a couple of indices that would be helpful
if you were able to read them immediately for what they are,
without having to think about them:
| Index operation | | Interpretation |
---
| s | | First element of s |
| s | | Second element of s |
| s[-1] | | Last element of s |
| s[-2] | | Second to last element of s |
To index or not to index?
Just a quick note on something that I trip over every
now and then.
Python has many useful built-ins and built-in data types.
Of them, strings, lists and tuples are indexable with integers.
Sets are not.
You should also be careful about things that you think
are like lists, but really are not.
These include enumerate, zip, map, and
other objects.
None of these are indexable, none of these have a len value, etc.
Pay attention to that!
```
l = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
e = enumerate(l)
e
Traceback (most recent call last):
File "", line 1, in
TypeError: 'enumerate' object is not subscriptable
z = zip(l)
z
...
TypeError: 'zip' object is not subscriptable
m = map(str, l)
m
...
TypeError: 'map' object is not subscriptable
```
Best practices in code
A looping pattern with range
Because of the way both range and indices work,
one can understand that range(len(s)) will generate
all the legal indices for s:
```
s = "Indexing is easy!"
list(range(len(s))) # use list() to print the values
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
s
'I'
s
'!'
```
A consequence of this is that beginners, and people who are
more distracted or used to other programming languages,
end up employing a very common anti-pattern in for loops.
To exemplify this, suppose we wanted to write a fairly
naïve program to find the unique letters in our string.
Here is what the anti-pattern would look like:
```
s = "Indexing is easy!"
uniques = []
for idx in range(len(s)):
... if s[idx] not in uniques:
... uniques.append(s[idx])
...
uniques
['I', 'n', 'd', 'e', 'x', 'i', 'g', ' ', 's', 'a', 'y', '!']
```
This is a naïve solution to the problem of “find unique characters”,
you probably want to use a Python set for a more efficient implementation :)
The problem here is that the for loop
is being done in a roundabout way:
we have access to a sequence (the string) that we could
iterate over, but instead we find its length,
so that we can use range to compute its legal indices,
which we then iterate over,
only to then access the elements of the sequence through their indices.
This way of writing for loops is similar to the way one would
write for loops in other programming languages, if you were
to iterate over the elements of an array.
However, we are using Python, not any other language.
One of the things I enjoy the most about Python's for loops
is that you can access directly the consecutive elements of
a sequence.
Hence, we can actually rewrite our for loop slightly,
but in a way that makes it much more elegant:
```
s = "Indexing is easy!"
uniques = []
for letter in s:
... if letter not in uniques:
... uniques.append(letter)
...
uniques
['I', 'n', 'd', 'e', 'x', 'i', 'g', ' ', 's', 'a', 'y', '!']
```
What I really like about these types of loops is that if your
variables are named correctly, the statements express your intent
very clearly.
The line for letter in s: is read as
“For each letter in (the string) s...”
This type of for loop iterates directly over the values you
care about, which is often what you want.
If you care about the indices,
then be my guest and use range(len(s))!
Another anti-pattern to be on the lookout for happens
when you need to work with the indices and the values.
In that case, you probably want to use the enumerate
function.
I tell you all about that function in a Pydon't of its own,
so go check that if you haven't.
Large expressions as indices
When you are dealing with sequences and with indices
for those sequences, you may end up needing to perform some
calculations to compute new indices that interest you.
For example, suppose you want the middle element of a string
and you don't know about // yet:
```
s = "Indexing is easy!"
s[len(s)/2] # len(s)/2 isn't an integer!!
Traceback (most recent call last):
File "", line 1, in
TypeError: string indices must be integers
len(s)/2
8.5
import math
s[math.floor(len(s)/2)]
' '
s[len(s)//2] # Pro-tip: the operation // is ideal here
' '
```
Where am I going with this?
Take a look at the expression you just used:
len 2
Maybe it is me getting old, but I struggle a bit to read that
because of the [] enclosing the expression which then has
a couple of () that I also have to parse, to figure out
what goes where.
If you have large expressions to compute indices
(and here, large will be subjective),
inserting those expressions directly inside [] may
lead to long lines of code that are then complicated to read
and understand.
If you have lines that are hard to understand,
then you probably need to comment them, creating even
more lines of code.
Another alternative is to create a well-named variable
to hold the result of the computation of the new index:
```
s = "Indexing is easy!"
mid_char_idx = math.floor(len(s)/2)
s[mid_char_idx]
' '
```
For this silly example, notice that the new variable name
is almost as long as the expression itself!
However, s[mid_char_idx] is very, very, easy to read
and does not need any further comments.
So, if you have large expressions to compute indices,
think twice before using them to index directly into the sequence
at hands and consider using an intermediate variable
with a descriptive name.
Unpacking with indexing
You will find yourself often working with small groups of data,
for example pairs of things that you keep together in a small
list for ease of use.
For example, the first and last names of a person:
```
"Mary" "Doe"
```
Now you have this little function that creates a formal
or informal greeting for a given name:
```
names = ["Mary", "Doe"]
def greet(names, formal):
... if formal:
... return "Hello Miss " + names
... else:
... return "Hey there " + names
...
greet(names, True)
'Hello Miss Doe'
greet(names, False)
'Hey there Mary'
```
Something you might consider and that adds a bit of clarity
to your code is unpacking the names before you reach the if
statement:
```
def greet(names, formal):
first, last = names
if formal:
return "Hello Miss " + last
else:
return "Hey there " + first
```
Why would this be preferable, if I just added a line of code?
It makes the intent of the code much more obvious.
Just from looking at the function as is, you can see
from the first line first, last = names that
names is supposed to be a pair with the first and last names
of a person and then the if: ... else: ... is very, very
easy to follow because we see immediately that we want to
use the last name if we need a formal greeting, and otherwise
(else) we use the first name.
Furthermore, the action of unpacking (like so:)
```
first, last = names
```
forces your greet function to expect pairs as the names variable,
because a list with less or more elements will raise an error:
```
first, last = ["Mary", "Anne", "Doe"]
Traceback (most recent call last):
File "", line 1, in
ValueError: too many values to unpack (expected 2)
```
We are assuming we really are working with pairs,
so if the greet function gets something that is not a pair,
this error is useful in spotting a problem in our code.
Maybe someone didn't understand how to use the function and
called it with the first name of the person?
```
greet("Mary", True)
Traceback (most recent call last):
File "", line 1, in
File "", line 2, in greet
ValueError: too many values to unpack (expected 2)
```
This would help you find a location where the greet
function was not being properly used.
I have written at length about unpacking in Python
(another favourite feature of mine!) so feel free
to read my articles on unpacking with starred assignments
and on deep-unpacking.
Conclusion
Here's the main takeaway of this article, for you, on a silver platter:
“Indexing is simple and powerful, but sometimes when indexing
looks like the answer, there is another Python feature waiting
to be used.”
This Pydon't showed you that:
Indexing in Python is 0-based;
Python allows negative indices in sequences;
Using indices in a for loop to access the elements of a sequence is an anti-pattern in Python;
Using large expressions when indexing bloats your code and you are better off with a descriptive variable,
even if that variable has a long name;
If you know the exact structure of the sequence you are dealing with, unpacking might be preferable to indexing.
If you liked this Pydon't be sure to leave a reaction below and share this with your friends and fellow Pythonistas.
Also, don't forget to subscribe to the newsletter so you don't miss
a single Pydon't!
Become a better Python 🐍 developer, drop by drop 💧
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Understanding Perfect Competition: Key Concepts and Dynamics
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8.1 Perfect Competition Core Assumptions: 1.Price Takers: Individual buyers and sellers cannot influence the market price due to their small size relative to the market. 2.Homogeneous Products: Goods offered by all firms are identical, making them perfect substitutes. 3.Free Market Entry and Exit: No significant barriers prevent firms from entering or exiting the market in response to profit opportunities. 4.Perfect Information: Both buyers and sellers are fully informed about prices, product quality, and production costs. 5.Minimal Transaction Costs: Costs associated with trading, such as transportation and negotiation, are negligible. Features of Firms in Perfect Competition: ●Perfectly Elastic Demand: A firm's demand curve is horizontal at the market price. The firm can sell any quantity at the market price but cannot raise the price without losing all customers. Market Demand vs. Firm Demand: ●Market Demand Curve: Slopes downward, showing that overall consumption increases as price decreases. ●Firm Demand Curve: Perfectly horizontal, reflecting the inability of individual firms to affect price. 8.2 Short-Run Dynamics in Perfect Competition Profit and Cost Relationships: ●Profit Formula: π(q)=R(q)−C(q)=pq−C(q)\pi(q) = R(q) - C(q) = pq - C(q)π(q)=R(q)−C(q)=pq−C(q) Where: ○ppp: Market price. ○C(q)C(q)C(q): Total cost at output qqq. ○Marginal Revenue (MR): The additional revenue from selling one more unit. ○Marginal Cost (MC): The additional cost of producing one more unit. Decision-Making in the Short Run: 1.Profit Maximization: The firm produces at the output level where MR=MCMR = MCMR=MC, ensuring maximum profit or minimum loss. 2.Shutdown Decision: The firm continues operating if P>AVCP > AVCP>AVC, as revenue covers variable costs. If P<AVCP < AVCP<AVC, the firm shuts down to minimize losses.
Profit Scenarios in the Short Run: 1.Positive Profits: Occur when P>ACP > ACP>AC. 2.Losses but Continued Production: Occur when AVC<P<ACAVC < P <ACAVC<P<AC, as the firm covers variable costs and some fixed costs. 3.Shutdown: Occurs when P<AVCP < AVCPACP > ACP>AC: New firms enter, increasing supply and driving prices down. ○If P<ACP < ACP<AC: Firms exit, reducing supply and raising prices. Types of Industries and Long-Run Supply: 1.Constant-Cost Industry: Entry or exit does not affect input costs, so the long-run supply curve is perfectly elastic (horizontal). 2.Increasing-Cost Industry: Entry raises input costs, leading to an upward-sloping long-run supply curve. 3.Decreasing-Cost Industry: Entry reduces input costs due to economies of scale, resulting in a downward-sloping long-run supply curve. Dynamic Efficiency in the Long Run: ●Firms adopt new technologies and production methods to minimize costs, ensuring long-term competitiveness and innovation.
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8.4 Market Efficiency and Surplus Consumer Surplus (CS): ●Definition: The benefit consumers receive when they pay a price lower than their maximum willingness to pay. ●Graphical Representation: The area between the demand curve and the market price up to the quantity consumed. Producer Surplus (PS): ●Definition: The benefit producers gain when they sell at a price higher than their minimum willingness to accept. ●Formula: PS=Revenue−Variable CostsPS = \text{Revenue} - \text{Variable Costs}PS=Revenue−Variable Costs Total Surplus (TS): ●Definition: The sum of consumer and producer surplus, representing overall welfare in the market. ●Efficiency: Maximized when P=MCP = MCP=MC, indicating optimal resource allocation with no waste or inefficiency. Sources of Market Inefficiency Deadweight Loss (DWL): ●Definition: A loss of total surplus resulting from deviations from the market equilibrium. Examples of Inefficiency: 1.Taxes: Create a wedge between consumer and producer prices, reducing quantity traded and causing DWL. 2.Price Ceilings: Limit prices below equilibrium, leading to shortages and DWL. 3.Price Floors: Set prices above equilibrium, causing surpluses and DWL. 4.Externalities: Occur when social costs or benefits (e.g., pollution, public goods) are not included in market transactions. Government Interventions: ●Subsidies: Can encourage overproduction, leading to inefficiency. ●Quotas: Restrict production or imports, distorting supply and reducing surplus. Additional Economic Concepts
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1.Allocative Efficiency: Resources are optimally allocated when P=MCP = MCP=MC, ensuring the goods produced match consumer preferences. 2.Productive Efficiency: Firms operate at the lowest point on their AC curve (P=ACminP = AC_{\text{min}}P=ACmin). 3.Comparative Statics: Analyzes how equilibrium changes when external conditions, such as taxes or input prices, shift. 4.Externalities: ○Negative Externality: Social costs exceed private costs (e.g., pollution). ○Positive Externality: Social benefits exceed private benefits (e.g., education). 5.Public Goods: Goods that are non-excludable and non-rival, such as national defense, often underproduced in competitive markets. 6.Market Failures: Occur when the free market fails to allocate resources efficiently, often justifying government intervention. ●
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187737 | https://pdhstar.com/wp-content/uploads/2019/06/CE-087-Calculation-of-Gas-Density-and-Viscosity.pdf?srsltid=AfmBOorFbQ_ifiVTGaZkWMtaUxn-IGOBLqVG00Mctx0bvHfrzeVI9h-m | PDH Star | T / F: (833) PDH‐STAR (734‐7827) | E: info@pdhstar.com
CE‐087 Calculation of Gas Density and Viscosity Instructor: Harlan Bengtson, PhD, P.E. Course ID: CE‐087 PDH Hours: 2 PDH 1
Calculation of Gas Density and Viscosity
Harlan H. Bengtson, PhD, P.E.
COURSE CONTENT 1. Introduction
The density and/or viscosity of a gas is often needed for some other calculation, such as pipe flow or heat exchanger calculations. This course contains discussion of, and example calculation of, the density and viscosity of a specified gas at a given temperature and pressure. If the gas temperature is high relative to its critical temperature and the gas pressure is low relative to its critical pressure, then it can be treated as an ideal gas and its density can be calculated at a specified temperature and pressure using the ideal gas law. If the density of a gas is needed at a temperature and pressure at which it cannot be treated as an ideal gas law, however, then the compressibility factor of the gas must be calculated and used in calculating its density. In this course, the Redlich Kwong equation will be used for calculation of the compressibility factor of a gas. The Sutherland formula can be used to calculate the viscosity of a gas at a specified temperature and pressure if the Sutherland constants are available for the gas. It will be discussed and used in example calculations. Another method for calculating the viscosity of air at a specified temperature and pressure will also be presented and discussed. Some of the equations that will be discussed and illustrated through examples are shown below. 2
Learning Objectives
Upon completion of this course, the student will
Be able to calculate the density of a gas of known molecular weight at a specified temperature and pressure at which the gas can be treated as an ideal gas.
Be able to calculate the compressibility factor for a gas at a specified temperature and pressure, using the Redlich-Kwong equation, if the molecular weight, critical temperature and critical pressure of the gas are known.
Be able to calculate the density of a gas at a specified temperature and pressure for which the gas cannot be treated as an ideal gas, if the molecular weight, critical temperature and critical pressure of the gas are known.
Be able to calculate the viscosity of a gas at a specified temperature if the Sutherland constant for the gas is known and the viscosity of the gas at a suitable reference temperature is known.
Be able to calculate the viscosity of air at specified air temperature and pressure.
Be able to make all of the calculations described in these learning objectives using either U.S. or S.I. units.
Topics Covered in this Course
I. Calculation of Ideal Gas Density II. Calculation of Real Gas Density III. Calculation of Gas Viscosity by Sutherland’s Formula 3IV. Calculation of Air Viscosity at Specified Temperature and Pressure V. Summary VI. References
4. Calculation of Ideal Gas Density
The typical form for the Ideal Gas Law is: PV = nRT
The parameters in the equation with a consistent set of units are as shown below:
P is the absolute pressure of the gas in psia
V is the volume of the gas in ft 3
n is the number of slugmoles in the gas contained in volume, V
R is the ideal gas law constant, 345.23 psia-ft 3/slugmole- oR
T is the absolute temperature of the gas in oRThe mass of the gas can be introduced into the equation by replacing n with
m/MW , where m is the mass of the gas contained in volume V in slugs, and
MW is the molecular weight of the gas (slugs/slugmole). The Ideal Gas Law then becomes:
PV = (m/MW)RT
Solving the equation for m/V, which is the density of the gas gives:
= m/V = P(MW)/RT
With P, R, and T in the units given above, the gas density will be in slugs/ft 3. Note that 1 slug = 32.17 lbm, so if you want the gas density in lbm/ft 3, the value in slugs/ft 3 should be multiplied by 32.17.
S.I. Units: If working in S.I. units, the equations remain the same with the following units: 4
P is the absolute pressure of the gas in kPa
V is the volume of the gas in m 3
n is the number of kgmoles in the gas contained in volume, V
R is the ideal gas law constant, 8.3145 kg-m/kgmole-K
T is the absolute temperature of the gas in K With these units for P, V, R , and T, the gas density will be in kg/m 3.
Critical Temperature and Pressure: As noted in the Introduction, in order to use the Ideal Gas Law to calculate a gas density, the gas temperature should be high relative to its critical temperature and the gas pressure should be low relative to its critical pressure. Table 1 gives critical temperature, critical pressure and molecular weight for 16 gases in U.S. units. Table 2
provides the same in S.I. units.
Table 1. Critical Temperature and Pressure and Molecular Weight -U.S. 5
Table 2. Critical Temperature and Pressure and Molecular Weight -S.I.
Example #1 : a) Calculate the density of air at -17 oF and 20 psig, assuming that the air can be treated as an ideal gas at those conditions.
b) Is it reasonable to assume ideal gas behavior for air at -17 oF and 20 psig?
Solution: a) The absolute temperature and pressure need to be calculated as follows: Tabs = -17 + 459.67 oR = 442.67 oR and Pabs = Pg + Patm
= 20 + 14.7 = 34.7 psia. Substituting values into the ideal gas law (using 28.97 as the MW of air) gives:
= MWP/(RT) = 28.9734.7/(345.23442.67) = 0.00658 slugs/ft 36If desired, the density can be converted to lbm/ft 3 by multiplying by the conversion factor, 32.17 lbm/slug.
= ( 0.000658 slugs/ft 3)(32.17 lbm/slug) = 0.2116 lbm/ft 3
b) The gas temperature (-17 o F) is much greater than the critical temperature of air (-220.9 oF) and the gas pressure (34.7 psia) is much less than the critical pressure of air (547 psia), so it would be reasonable to assume ideal gas behavior for air at this temperature and pressure.
Example #2 : a) Calculate the density of air at -20 oC and 100 kPa gauge pressure, assuming that it can be treated as an ideal gas at those conditions.
b) Is it reasonable to assume ideal gas behavior for air at 20 oC and 100 kPa guage?
Solution: a) The absolute temperature and pressure need to be calculated as follows: Tabs = 20 + 273.15 K = 283.15 K and Pabs = Pg + Patm
= 100 + 101.3 = 201.3 kPa abs. Substituting values into the ideal gas law (using 28.97 as the MW of air) gives:
= MWP/(RT) = 28.97201.3/(8.3145283.15) = 2.48 kg/m 3
b) The gas temperature (10 oC) is much greater than the critical temperature of air (-140.5 oC) and the gas pressure (201.3 kPa abs) is much less than the critical pressure of air (3773.4 kPa abs), so it would be reasonable to assume ideal gas behavior for air at this temperature and pressure. (Note that the critical temperature was converted from the 37.25 atm value in the table with the conversion factor, 101.3 kPa/atm.)
Spreadsheet Use for the Calculations: These calculations are rather straight-forward and not too difficult to do by hand, but they can be done very conveniently with an Excel spreadsheet set up to calculate gas density with the Ideal Gas Law. Figure 1 shows a screenshot of an Excel worksheet with the solution to Example 1 (a) and Figure 2 shows a screenshot with the solution to Example 2 (a). 7
Figure 1. Screenshot of Solution to Example #1 (a)
Figure 2. Screenshot of Solution to Example #2 (a) 8
5. Calculation of Real Gas Density
In some cases, the Ideal Gas Law cannot be used to calculate the density of a gas because its temperature is too close to its critical temperature and/or its pressure is too close to its critical pressure. In that case, if the compressibility factor, Z, can be determined at the gas temperature and pressure, it can be used to calculate the gas density with the following equation:
= MWP/(ZRT)
The compressibility factor for a gas is, in general, a function of its reduced temperature ( T R) and reduced pressure ( PR), where reduced temperature is the absolute gas temperature divided by its absolute critical temperature and reduced pressure is the absolute gas pressure divided by its absolute critical pressure. Graphs, tables and equations are available for determining the compressibility factor at specified values for TR and PR. The Redlich-Kwong equation will be used in this book as a means of calculating Z as a function of TR and P R.Calculation of the compressibility factor of a gas from the Redlich-Kwong equation is rather awkward and time consuming to do by hand, but a spreadsheet can be set up to conveniently make the necessary calculations. The equations and calculation procedure are as follows: The Redlich-Kwong compressibility factor, Z, is calculated as the maximum real root of the equation: Z3 – Z 2 – qZ – r = 0 , where r = A 2 B and
q = B 2 + B – A 2, with A2 = 0.42747P R/T R2.5 and B = 0.08664 P R/T R
To find the maximum real root, first the parameter C is calculated, where:
C = (f/3) 3 + (g/2) 2, with f = (-3q - 1)/3 and
g = (-g/2 - C 1/2 ) 1/3 + 1/3
If C > 0 , then there is one real root,
Z = (-g/2 + C 1/2 ) 1/3 + (-g/2 - C 1/2 ) 1/3 + 1/3 9If C < 0 , then there are three real roots, given by:
Z k = 2(-f/3) 1/2 cos[( /3) + 2 (k - 1)/3] + 1/3
With k = 1, 2, 3 and = cos -1 {[(g 2/4)/((-f 3)/27)] 1/2 }
These equations and this method of calculating a value for the compressibility factor are described at: www.polymath-software.com/ASEE2007/PDF1.pdf The only input parameters needed to calculate the compressibility factor of a gas by the Redlich-Kwong method are the temperature and pressure at which the compressibility factor is to be calculated, along with the gas molecular weight, critical temperature and critical pressure. However, there are 8 additional parameters introduced ( B, A 2, r, q, f, g, C ,and ) and there are also quite a few steps in the solution, so it is not a trivial calculation to carry out by hand. An Excel spreadsheet is a very convenient tool to use in carrying out this calculation, as illustrated in Example #3 and
Example #4 .
Example #3 : a) Calculate the density of air at -17 oF and 20 psig, using the compressibility factor calculated by the Redlich-Kwong method described above. Assume local atmospheric pressure is 14.7 psi.
b) Compare the results from part (a) with the density calculated assuming ideal gas behavior in Example #1 .
Solution: a) The reduced temperature and reduced pressure can be calculated, using the gas temperature and pressure and the critical temperature and critical pressure of the gas, as follows:
TR = (-17 + 459.67)/(-220.9 + 459.67) = 1.854 PR = 34.7/547 = 0.06344 10 A spreadsheet screenshot is shown in Figure 3 with the calculation of the compressibility factor and the density for air at -17 oF and 20 psig. Note that the required user inputs (in the blue cells) are the gas molecular weight, temperature and pressure and the critical temperature and critical temperature of the gas. The inputs shown are for air at -17 oF and 20 psig (34.7 psia). The worksheet makes quite a few calculations in the yellow cells, starting with calculation of the reduced temperature ( TR) and reduced pressure ( PR). The calculated values of TR and PR are used to calculate the values of the 8 constants ( B, A 2, r, q, f, g, C , and ). If C > 0 , then the spreadsheet calculates the single real root of the equation,
Z3 – Z 2 – qZ – r = 0 , (which is the value of the compressibility factor,
Z) using the equation shown above ( Z = (-g/2 + C 1/2 ) 1/3 + (-g/2 - C 1/2 ) 1/3 + 1/3 ). If C < 0 , then the spreadsheet calculates the three real roots of the equation, using the equation given above and the maximum of those three roots is the value for the compressibility factor, Z.For Example #3 , as shown in the screenshot, C > O and this results in a value of 0.997 for the compressibility factor, Z, and air density of 0.00659 slugs/ft 3 = 0.21214 lbm/ft 3.
b) The value calculated for Z is very close to 1 and thus the calculated value of air density is quite close to that calculated with the Ideal Gas Equation in
Example #1 as shown below: Ideal Gas Law: = 0.00658 slugs/ft 3 = 0.2116 lbm/ft 3
Redlich-Kwong: = 0.00659 slugs/ft 3 = 0.2121 lbm/ft 3
% difference = (0.2121 – 0.2116)/0.2121 = 0.24 % 11
Figure 3. Screenshot of Solution to Example #3 (a)
Note that for accurate calculations with the Redlich-Kwong equation, it is recommended that P R be less than half of TR. In the Figure 3 spreadsheet screenshot the ratio PR/0.5T R is calculated to facilitate checking whether this requirement has been met. 12
Example #4 : a) Calculate the compressibility factor and density of air at -129 oC and 20 bar, using the Redlich-Kwong method.
b) Compare the results from Part (a) with the density calculated assuming ideal gas behavior. . Solution: a) The reduced temperature and reduced pressure can be calculated as follows:
TR = (-129 + 273.15)/(-140.5 + 273.15) = 1.087
PR = (20 bar)(0.98682 atm/bar)/37.25 atm = 0.5299
A spreadsheet screenshot is shown in Figure 4 with the calculation of the compressibility factor and the density for air at -129 oC and 20 bar pressure. The user inputs are shown in the cells at the upper left . Calculation of the reduced temperature and reduced pressure is shown below the user inputs and calculation of the various constants is shown in the cells at the right. The results, shown at the bottom of the screenshot are:
Z = 0.845 and air density = 57.17 kg/m 3
b) The value of 0.845 calculated for Z shows that the calculated value of air density will be somewhat different than that calculated with the Ideal Gas Equation. The results, shown at the bottom of the screenshot, are: Ideal Gas Law: = 48.325 kg/m 3
Redlich-Kwong: = 57.170 kg/m 3
% difference = (57.170 – 48.325)/57.170 = 15.5 % 13
Figure 4. Screenshot of Solution to Example #4 (a) 14
6. Calculation of Gas Viscosity by Sutherland’s Formula
The Sutherland Formula provides a means for calculating the viscosity of a gas if the value of the Sutherland’s Constant is known for that gas along with a value of the viscosity of that gas at some reference temperature. Sutherland’s Formula is:
The parameters in this equation are:
T is the temperature of the gas, K
To is a reference temperature, K
o is the viscosity of the gas at To in any units
is the viscosity of the gas at T in the same units as o
C is the Sutherland constant for the gas, K
Table 3 gives values of the Sutherland constant, the temperature range for that constant and the viscosity at three reference temperatures for 25 gases. These values can be used together with Sutherland’s formula to calculate the viscosity of any of the gases in the table at any temperature within the specified temperature range. References #2, #3, and #4 are sources for the information in Table 3 . Note that the reference viscosity and reference temperature closest to the gas temperature should be used for the Sutherland’s Formula calculation. 15
Table 3. Constants for Sutherland’s Formula
Example #5 : Calculate the viscosity of methane at 110 oF using Sutherland’s formula and values from Table 3 .16
Solution: Converting 110 oF to oC gives: (110 – 32)/1.8 = 43.33 oC. Thus the reference temperature closest to the gas temperature is 50 oC, so from
Table 3 , we will use t o = 50 oC and o = 0.0120 cP. Also from Table 3 , the Sutherland’s Constant for methane is 169. The gas temperature and reference temperature must be in K (degrees Kelvin) for use in Sutherland’s Formula to calculate the gas viscosity. The temperature conversions can be made as follows: Gas temperature = (110 + 459.67)/1.8 = 316.5 K Reference temperature = (50 + 273.15) = 323.15 K Now, substituting into Sutherland’s Formula to calculate the viscosity of methane at 110 o F gives:
= 0.012[(323.15 + 169)/(316.5 + 169)][(316.5/323.15) 1.5 ] = 0.0118 cP
Converting to typical U.S. units of lb-s/ft 2 gives:
= (0.0118)(2.08854 x 10 -5 ) = 2.462 x 10 -7 lb-s/ft 2
Calculation of a gas viscosity using Sutherland’s Formula can conveniently be done using an Excel spreadsheet. Figure 5 shows a screenshot of a spreadsheet solution to Example #5 .Note that, in the screenshot, the given information is entered into the blue cells on the left side of the screenshot and the spreadsheet makes the calculations in the yellow cells to make the necessary temperature conversions, calculate the gas viscosity and convert to typical U.S. units. Calculations in S.I. units would be very similar, but the unit conversions used for Example #5 wouldn’t be needed, because the values in Table 3 have units of oC, K, and cP. This type of calculation is illustrated in
Example #6 .17
Figure 5. Screenshot of Solution to Example #5
Example #6 : Calculate the viscosity of methane at 60 o C using Sutherland’s formula and values from Table 3 .
Solution: The reference temperature closest to the gas temperature is 50 oC, so from Table 3 , we will use t o = 50 o C and o = 0.0120 cP. Also from Table 3, the Sutherland’s Constant for methane is 169. The gas temperature and reference temperature must be in K (degrees Kelvin) for use in Sutherland’s Formula to calculate the gas viscosity. The temperature conversions can be made as follows: Gas temperature = (60 + 273.15) = 333.15 K Reference temperature = (50 + 273.15) = 323.15 K 18 Now, substituting into Sutherland’s Formula to calculate the viscosity of methane at 60 o C gives:
= 0.012[(323.15 + 169)/(333.15 + 169)][(333.15/323.15) 1.5 ] = 0.0123 cP
7. Calculation of Air Viscosity at Given Temperature and Pressure
With the following equations, it is possible to calculate the viscosity of air for specified values of the air temperature and pressure. The source for these equations is Reference #6 at the end of the course. air viscosity = (1.2867/10 7)[o(T r) + r)] lb-sec/ft 2
o (T r) = 0.128517 Tr + 2.60661 Tr0.5 - 1.0 - 0.709661 Tr-1
0.662534 Tr-2 - 0.197846 Tr-3 + 0.00770147 Tr-4
r) = 0.465601 r = 1.26469 r2 - 0.511425 r3 + 0.274600 r4
Tr = T/238.5 (T in oR) r = /0.6096 ( in slugs/ft 3)
Note that is the density of the air at the specified temperature and pressure, which can typically be calculated as described in Section 4 for ideal gas behavior. If the air cannot be treated as an ideal gas for the specified temperature and pressure, then the density would need to be calculated as described in Section 5 above. 19
Example #7 : Calculate the viscosity of air at 50 oF and 40 psig at a location where atmospheric pressure is 14.7 psi.
Solution: The absolute air temperature can be calculated as:
T = 50 + 459.7 = 509.7 oR
The absolute air pressure is: P = 40 + 14.7 = 54.7 psia
The air density can then be calculated as: = P(MW)/RT
= (54.7)(28.97)/[(345.23)(509.7) = 0.0090 slugs/ft 3
Tr and r can now be calculated as follows:
Tr = T/238.5 = 509.7/38.5 = 2.137
r = /0.6096 = 0.0090/0.6096 = 0.00715
Now o(T r ) and (r) can be calculated with the equations shown above to give:
o(T r) = 2.878 and (r) = 0.00715
Finally, the air viscosity can now be calculated using the first equation shown in this chapter, to give: air viscosity = (1.2867/10 7)[o(T r) + r)]
= (1.2867/10 7)(2.878 + 0.00715) = 3.713 x 10 -7 lb-sec/ft 2
As you may expect, this set of calculations also can be conveniently done with a properly set up spreadsheet. Figure 6 shows a screenshot of a spreadsheet solution to Example #7 .20 Only three user inputs are needed. They are the temperature of the air, the gage pressure of the air, and atmospheric pressure. These three inputs are entered in the three blue cells in the upper left part of the screenshot in
Figure 6 . The spreadsheet then calculates the absolute air temperature and pressure, the air density, Tr, r, o(T r), (r ), and finally the air viscosity at the specified temperature and pressure. Note that the calculated value for the air viscosity at 50 oF and 40 psig shown as 3.17 x 10 -7 lb-sec/ft 2, the same as shown in the calculations above.
Figure 6. Screenshot of Solution to Example #7 21 For calculations in S.I. units some of the constants in the equations given above are changed. Specifically, the equations for Tr, r, and the air viscosity change, while the equations for o (T r) and r) remain the same. The S.I. equations with changed constants are:
Tr = T/132.5 (T in K ) r = /314.3 ( in kg/m 3)
and air viscosity = (6.16090/10 6)[o(T r) + r)] Pa-s
Example #8 : Calculate the viscosity of air at 327 oC and 20 bar gage pressure at a location where atmospheric pressure is 101.325 kPa.
Solution: The absolute air temperature can be calculated as:
T = 327 + 273.15 = 600.2 K
The absolute air pressure is: P = (20)(100) + 101.325 = 2101.3 kPa
The air density can then be calculated as: = P(MW)/RT
= (2101.3)(28.97)/[(8.3245)(600.2) = 12.20 kg/m 3
Tr and r can now be calculated as follows:
Tr = T/132.5 = 600.2/132.5 = 4.529
r = /314.3 = 12.20/314.3 = 0.0388
Now o(T r ) and (r) can be calculated with the equations shown above to give:
o(T r) = 5.003 and (r) = 0.01995 22 Finally, the air viscosity can now be calculated using the air viscosity shown above, to give: air viscosity = (6.16090/10 6)[o (T r) + r)]
= (6.16090/10 6)(5.003 + 0.01995) = 3.095 x 10 -5 Pa-s
Figure 7 shows a screenshot of a spreadsheet with the solution to
Example #8.
Figure 7. Screenshot of Solution to Example #8 23
8. Summary
The density and viscosity of a gas at specified temperature and pressure can be calculated using the methods covered in this course. Gas density can be calculated using the Ideal Gas Law if the gas temperature is sufficiently greater than the critical temperature and the gas pressure is sufficiently less than the critical pressure. If the gas temperature and/or pressure are such that the Ideal Gas Law cannot be used then the gas density can be calculated using the calculated compressibility factor. In this course, use of the Redlich-Kwong equation of state to calculate the compressibility factor was presented and illustrated with example calculations. The viscosity of a gas at specified gas temperature can be calculated using Sutherland’s Formula as presented and illustrated with examples. Equations were also presented for calculation of the viscosity of air at specified temperature and pressure. Use of both U.S. units and S.I. units was presented and illustrated with examples for all of the calculations.
9. References
Redlich-Kwong Equation of State Calculations, www.polymath-software.com/ASEE2007/PDF1.pdf 2. Chapman, S. & Cowling, T.G., The Mathematical Theory of Non-Uniform Gases, 3 rd Ed., 1970, Cambridge Mathematical Library 3. National Physical Laboratory, Kaye & Laby, Tables of Physical and Chemical Constants, Chapter 2.2, Subsection 2.2.3, 24 4. Engineering Toolbox website: www.engineeringtoolbox.com/gases-absolute-dynamic-viscosity-d_1888.html. 5. Green, Don W. and Perry Robert H., Perry’s Chemical Engineers’ Handbook, 8 th Ed, Table2-312, McGraw-Hill 6. Kadoya, K, Matsunaga, N, and Nagashima, A, Viscosity and Thermal Conductivity of Dry Air in the Gaseous Phase , J. Phys. Chem. Ref. Data, Vol 14, No. 4, 1985.
Bengtson, Harlan H., Sutherland Formula Viscosity Calculator, an online blog article at www.EngineeringExcelTemplates.com. 8. Bengtson, Harlan H. Gas Compressibility Factor Calculator Excel Spreadsheet, an online blog article at www.EngineeringExcelSpreadsheets.com 9. Bengtson, Harlan H., Gas Property Calculator Spreadsheets, An Amazon Kindle E-Book. |
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Alternative Visual Cycles in Müller Cells
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Phototransduction: Phototransduction in Cones; Phototransduction: Phototransduction in Rods; Phototransduction: Rhodopsin; Phototransduction: The Visual Cycle
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Phototransduction, the process of converting light into electrical neural signals, takes place in the outer segments of photoreceptors. The mechanisms of phototransduction and the proteins involved are highly conserved in rods and cones across different species (Arshavsky et al., 2002; Lamb and Pugh, 1992; Pugh and Lamb, 1993). The second messenger conveying photo-signal to neural signal is cGMP, which opens nonselective cyclic nucleotide-gated (CNG) cation channels located on the outer segment plasma membrane (Yau, 1994). In darkness, when bound to cGMP, a fraction of CNG channels are open, allowing the steady influx of Na+ and Ca2+ driven by the electrochemical gradient across the plasma membrane of the outer segment. This inward current (denoted as “dark current”) depolarizes the photoreceptors and maintains the steady release of neurotransmitter (glutamate) from their synaptic terminals in darkness.
The cGMP concentration within the outer segment is equilibrated by the balance between its synthesis by guanylyl cyclase (GC) and its hydrolysis by phosphodiesterase (PDE). Upon light absorption, R∗ activates the G-protein transducin (Gt), which in turn activates PDE. PDE∗ hydrolyzes cGMP into GMP, lowering cGMP concentration. The resulting closure of the CNG channels blocks the dark current and hyperpolarizes the photoreceptor membrane. As a result, the rate of glutamate release from the synapses is reduced, thus converting and relaying the light signal to the postsynaptic neurons as electrical signal (Lamb and Pugh, 2006; Yau and Hardie, 2009).
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The rods react readily to low intensities of light because they have a low threshold of excitation, playing an important role in vision at twilight and at nighttime. In contrast, the cones, which require a much higher intensity of light, are critical for fine vision and color discrimination. The posterior pole of eye, including the macula and fovea centralis with high densities of cones, is responsible for central vision.
The basic transduction process is modulated by several mechanisms to provide the eye with sensitivity over a range of light intensities and wavelengths. Phosphorylation of rhodopsin by rhodopsin kinase upon light stimulation is essential for effective quenching of the signal through the interaction of phosphorhodopsin and arrestin (S antigen). The light-induced response is also modulated by the phosphorylation or dephosphorylation of a PDE subunit. There are several other mechanisms involved in the transduction cascade, such as: Ca2+ regulation for dark and light adaptation, cGMP concentration regulation, and opsin association and disassociation with its chromophore after photon activation. The phototransduction cascade is also regulated by phosphorylation and dephosphorylation.
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7 Visual processes and their relation to vascular and neural abnormalities of diabetic retinopathy
The vertebrate visual system relies on two key pathways to detect and transform light into perceived images; phototransduction and the visual (or retinoid) cycle. Phototransduction describes the process of converting light into stimuli that are processed by the brain to form an image. This occurs via a G-protein signaling cascade in photoreceptor outer segments that converts incoming light stimuli into an electrical response, leading to the activation of transducin and cGMP phosphodiesterase, closure of cGMP-gated ion channels, and hyperpolarization of the photoreceptor cells (Arshavsky et al., 2002; Fu and Yau, 2007; Luo et al., 2008) (Fig. 9a). The visual cycle is a multistep process that regenerates the light-sensitive photopigments (11-cis-retinal in mammals) so that phototransduction can occur again (Fig. 10a). Recently, research has emerged that shows that these processes that underlie vision itself can contribute to the development of early lesions of DR.
7.1 Phototransduction and diabetic retinopathy
Transducin1 is a subunit of the heterotrimeric G-protein encoded by Gnat1 (Hurley, 1987), and is a key component of the vertebrate phototransduction pathway. It is found only in rod photoreceptor cells in the retina. The absence of transducin1 prevents phototransduction in rods, causing the cGMP-gated ion channels on the rod cells to remain continually open. Diabetes has been reported to cause a reduction in expression of transducin in retinal photoreceptors (Kim et al., 2005), but permanent inhibition of phototransduction in rod cells using diabetic Gnat1−/− mice showed that elimination of phototransduction in rod cells significantly inhibited the diabetes-induced degeneration of retinal capillaries and some molecular processes believed to participate in the pathogenesis of the retinopathy (Liu et al., 2019) (Fig. 9). Studies of Gnat1−/− mice by other investigators confirmed that deletion of transducin1 significantly inhibited the diabetes-induced increase in inflammatory mRNAs, ERG defects, and degeneration of retinal capillaries (Thebeau et al., 2020). Thus, elimination of phototransduction in rod cells inhibited diabetes-induced abnormalities in both retinal vasculature structure and neural function. Deletion of Gnat1 did not result in significant degeneration of photoreceptor cells, which would have confounded interpretation of results. Gnat1 deletion had an unexplained effect on the diabetes-induced increase in permeability, in that the leakage of albumin was not uniformly distributed across the retina in those animals (Liu et al., 2019). The abnormal accumulation of albumin in the neural retina was inhibited in diabetic Gnat1−/− mice in the inner plexiform layer, but not in the outer plexiform or inner nuclear layers. In Gnat1-deficient animals, the diabetes-induced increase in inflammatory proteins in the retina and the leukocyte-mediated killing of retinal endothelial cells were inhibited, however, the diabetes-mediated induction of oxidative stress was not inhibited in these animals. Since anti-oxidant approaches have been found to inhibit the diabetes-induced degeneration of retinal capillaries in previous long-term rodent studies (Berkowitz et al., 2009a; Du et al., 2003, 2015; Kanwar et al., 2007; Kowluru et al., 2001b; Zheng and Kern, 2009), the finding that the diabetes-induced increase in retinal capillary degeneration was inhibited in diabetic Gnat1−/− mice even though the oxidative stress in the retina was not inhibited is unexplained.
Whether or not Gnat2 in cone cells has any similar effect on development of the retinopathy in diabetes has not been explored, but the number of cones in rodents is very small compared to the number of rod cells.
A physiologic way to demonstrate the importance of light-initiated processes in the development of DR is to eliminate light. Following 3 or 5 months of hyperglycemia, Thebeau et al. (2020) housed 2 different diabetic animal models in total darkness for 1–3 months. Results indicated that the mice housed under conditions of light deprivation developed significantly less diabetes-induced visual dysfunction (a-wave, b-wave, and oscillatory potential), upregulation of pro-inflammatory mRNA in the retina, and retinal neurodegeneration. As the major light-sensing cells in the body, these studies clearly implicate retinal photoreceptors as an early contributor to the retinal dysfunction and pathology in diabetes, and illustrate the importance of light in the diabetes-induced abnormalities (Fig. 11).
Exposure to light normally results in expansion of the choroidal space, suggesting that this phenomenon is secondary to phototransduction or its downstream signaling (Longo et al., 2000). The light-induced expansion of the subretinal space becomes abnormal in diabetes (Berkowitz et al., 2015b), and this defect in diabetes can be inhibited with the antioxidant, lipoic acid, suggesting that oxidative stress might underlie that abnormality. Immunoreactivity of rhodopsin kinase has been reported to be increased in the outer limiting membrane and in the outer segments of some rod cells in diabetes (Kim et al., 2005). Rhodopsin kinase activity in retinal rod cells phosphorylates light-activated rhodopsin to terminate the phototransduction signaling cascade, but how this activity in rod cells is affected by diabetes is not known.
7.2 Visual cycle and diabetic retinopathy
Subnormal rates of adaptation and absolute final threshold sensitivity have been demonstrated in subjects with diabetes (Amemiya, 1977; Greenstein et al., 1993; Henson and North, 1979). Consistent with this, a number of studies have provided biochemical, molecular or functional evidence that diabetes results in subnormal levels of enzymes involved in the visual cycle (Garcia-Ramirez et al., 2009; Ostroy, 1998; Ostroy et al., 1994). After 2–3 months of diabetes, microarray revealed reduced expression of several genes encoding visual cycle proteins, including lecithin/retinol acyltransferase (LRAT), RPE65, and RPE retinal G protein-coupled receptor (RGR) (Kirwin et al., 2011; Zhao et al., 2017). Among visual cycle proteins, interphotoreceptor retinoid-binding protein (IRBP) and Stimulated by Retinoic Acid 6 protein (STRA6) showed significantly lower levels in diabetic rats than those in nondiabetic controls. STRA6 is the receptor for retinol-binding protein 4 (RBP4) which is critical for the transport of vitamin A to the eye from the peripheral blood (Ruiz et al., 2012), and IRBP is an essential protein of the visual cycle which facilitates the transport of retinoids between the RPE and the retina (Crouch et al., 1992). Serum levels of retinol-binding protein 4, a transporter protein for retinol, have been found to be significantly lower in diabetic rats (Malechka et al., 2017). Spontaneously diabetic db/db mice showed reductions in levels of retinoid binding proteins, CRALBP, IRBP, STRA6, LRAT, and retinol dehydrogenase 5 (RDH5) compared to nondiabetic controls, and 10-week treatment with chrysin, a naturally occurring flavonoid, inhibited these defects as well as augmenting levels of RPE65 and rhodopsin (probably by decreasing hyperglycemia) (Kang et al., 2018).
Several rodent studies have demonstrated that diabetes can cause a retinoid deficiency in the retina. Although Western blot analysis revealed no difference in expression of the opsin protein, rhodopsin content of retinas from diabetic animals has been reported to be subnormal (as shown by a difference in absorbance before and after bleaching with light) and levels of 11-cis-retinal were significantly lower in retinas from diabetic rats compared to those in controls (Malechka et al., 2017; Tuitoek et al., 1996). Diabetes has been reported to generate a more acidic environment within rod photoreceptors that causes a delay in rhodopsin regeneration (Ostroy, 1998; Ostroy et al., 1994). Rod cell ion channels, which are normally open in the dark, become partially closed in dark-adapted diabetic mice, and injection of 11-cis-retinaldehyde partially restored the channel-mediated ion movement into rod cells towards normal (Berkowitz et al., 2012a). Likewise, a single administration of the retinoid, 9-cis-retinal, to diabetic Akita mice inhibited diabetes-induced abnormalities in ERG (a- and b-wave amplitudes of scotopic ERG), retinal oxidative stress (as estimated by levels of 3-nitrotyrosine), and apoptosis of retinal cells (Malechka et al., 2020). How long this effect lasted was not investigated. Retinyl esters, which are produced in the visual cycle and stored in the RPE, accumulated to abnormal amounts in diabetes of diabetic mice, suggesting that the visual cycle was disturbed at the level of RPE65 or downstream (Liu et al., 2015). Overall, these results would seem to suggest that retinoid metabolism in the eye can be impaired in diabetes, which can lead to deficient generation of visual pigments and neural retinal dysfunction in early diabetes.
Acute hyperglycemia accelerated rod dark adaptation after bleach in Type 2 diabetic patients who had a subnormal dark adaptation response (compared to historical controls) (Holfort et al., 2010). It is not obvious why acute hyperglycemia would accelerate visual cycle activity and rod recovery rate in patients whose diabetes (or hyperglycemia) initially caused the subnormal rod recovery rate, but other investigators show an increase in retinal blood flow and oxygen consumption during acute hyperglycemia (Tiedeman et al., 1998). A different study found that dark-adapted rod- and cone-sensitivity thresholds were normalized by hyperoxia and moderate hyperglycemia in diabetic subjects, and that cone threshold sensitivity was improved also in non-diabetics during hyperoxia and hyperglycemia (Kurtenbach et al., 2006). Apparently, the disruptions in outer retinal visual processes that develop in diabetes are not irreversible.
In light of evidence that visual cycle activity is subnormal in diabetes, it is very surprising that other evidence indicates that slowing the visual cycle can inhibit the development of lesions of early DR. Retinylamine is a retinoid inhibitor of RPE65, and its administration significantly inhibited both the diabetes-induced increase in retinal vascular permeability and retinal capillary degeneration over an 8 month period (Liu et al., 2015) (Fig. 10). Administration of the retinoid, however, had less effect on retinal function; retinylamine partially inhibited the diabetes-induced defect in contrast sensitivity, but had no effect on the defect in spatial frequency threshold. Since RPE65 is a critical component of the visual cycle (regenerating 11-cis retinal after activation of the photoreceptors by light), the beneficial effects of retinylamine on the early vascular lesions of DR are consistent with the possibility that inhibition of the visual cycle inhibits the development of the retinopathy. Retinylamine conceivably might have other effects, but it does not affect the retinoic acid-metabolizing cytochrome P450 enzymes, is not a substrate for cytochrome P450 CYP26 enzymes, does not activate the retinoid-X receptor, and interacts with the retinoic acid receptor only at micromolar concentrations (Golczak et al., 2005b). Retinylamine is acetylated by LRAT and stored in the RPE (Golczak et al., 2005a, 2005b, 2008bib_Golczak_et_al_2005abib_Golczak_et_al_2005bbib_Golczak_et_al_2008; Liu et al., 2015; Maeda et al., 2006) from which it is then released slowly, thus allowing the compound to be administered only once per week. In addition to implicating the visual cycle in the development of vascular lesions of early DR, this data provides novel evidence also that the RPE contributes in an important way to the pathogenesis of the retinopathy.
Rpe65−/− mice made experimentally diabetic provide further evidence that photoreceptor cells play an important role in the development of molecular, functional and histologic lesions characteristic of DR (Thebeau et al., 2020). Those investigators found that RPE65-deficient mice diabetic for 6 months were protected from the adverse effects of diabetes on expression of inflammatory genes, on visual function, and on the integrity of retinal capillaries (Fig. 10).
Other molecular players in the visual cycle also have been implicated in the molecular or physiologic alterations in the retina in diabetes. For example, genetic deletion of Lrat, which catalyzes the esterification of all-trans-retinol into all-trans-retinyl ester during phototransduction, inhibited the diabetes-induced increase in retinal oxidative stress (Liu et al., 2015). Whether or not genetic or pharmacologic manipulation of Lrat or other enzymes in the visual cycle affect the retinal vasculature or neuroglial retina in diabetes has not yet been explored.
7.3 Light as a potential therapy to inhibit diabetic retinopathy
In contrast to the evidence that light-mediated processes play a role in the development of vascular and neural abnormalities of DR, others have postulated that the absence of light plays a critical role in the development of at least the advanced stages of DR. Photoreceptor cells have very high levels of mitochondria in their inner segment, and consume most of the oxygen in the retina (Ahmed et al., 1993; Linsenmeier and Braun, 1992), especially in the dark (Arden et al., 1998, 2005bib_Arden_et_al_2005bib_Arden_et_al_1998; Wang et al., 2010), when rod dark currents become maximal (Haugh et al., 1990; Linsenmeier, 1986). Because photoreceptors have an especially high metabolic activity at night, Arden and colleagues postulated that increased photoreceptor metabolic activity would make the retina more hypoxic in the night than during the day, especially in compromised or diseased environments such as in diabetes.
Arden and colleagues reported that application of blue-green (505 nm) light all night long could mitigate aspects of DR, presumably by inhibiting the expected increase in metabolic activity of the retina that occurs in darkness (Arden et al., 2005, 2010, 2011bib_Arden_et_al_2010bib_Arden_et_al_2011bib_Arden_et_al_2005; Arden and Sivaprasad, 2011, 2012bib_Arden_and_Sivaprasad_2011bib_Arden_and_Sivaprasad_2012; Ramsey and Arden, 2015). Their studies were based on evidence that dark-adaptation exacerbates hypoxia in the diabetic retina, acting as a powerful stimulus for the overproduction of VEGF and other less well understood factors. Patients with mild non-proliferative DR and early, untreated non-sight-threatening macular edema slept for 6 months wearing masks that illuminated one closed eye with 505 nm light of sufficient intensity to suppress dark current in that eye (Arden et al., 2011). At 6 months, the number of treated eyes showing intraretinal cysts was reduced by about one third, whereas the number of untreated fellow eyes showing cysts nearly doubled. In the treated eyes, retinal thickness was significantly reduced toward normal, and visual acuity, achromatic contrast sensitivity, and microperimetric thresholds improved significantly. Unfortunately, a larger 2-year Phase 3 clinical trial (CLEOPATRA) of this hypothesis recently was reported (Sivaprasad et al., 2018), and it failed to support the hypothesis that suppression of photoreceptor “dark current” at night would inhibit DME. Review of the results however, suggest that a major factor in the failure of this trial was poor patient compliance to wear the goggles all night long for the duration of the study. An animal study investigating this concept reported that preventing rod cells from dark-adapting did not inhibit retinal neuronal and glial abnormalities in early stages of diabetes (Kur et al., 2016).
Others also have been investigating the use of light (photobiomodulation) to inhibit the development of early DR, but those studies used far-red light (670 nm) for only 3–4 min per day, and do not postulate that the mechanism involves suppression of dark current. Animal studies using albino rats and pigmented mice showed that whole-body photobiomodulation inhibited diabetes-induced abnormalities in retinal oxidative stress, inflammation, and electrophysiology (Saliba et al., 2015; Tang et al., 2014), and very effectively inhibited diabetes-induced increases in capillary degeneration and accumulation of albumin in neural retina (Cheng et al., 2017) (Fig. 12). Whether or not the beneficial effects of this light therapy involve the outer retina is not clear, because shielding of the head from the light treatment each day nevertheless still had a beneficial effect on molecular abnormalities induced in the retina by diabetes (Saliba et al., 2015). Nevertheless, photobiomodulation is being found to have benefits in a number of disease (Tsai and Hamblin, 2017). Two different research groups (Eells et al., 2017; Tang et al., 2014) have presented pilot studies describing a small number of diabetic patients having retinal edema who were treated with brief sessions of light at 670 nm, and both of these reports showed that the light therapy had beneficial effects to reverse the edema in the treated eye. A prospective multi-center pilot study evaluating photobiomodulation therapy for DME currently is being conducted by the Diabetic Retinopathy Clinical Research Network (DRCR).
Conclusions. In total, available evidence suggests that the visual cycle and phototransduction play previously unrecognized roles in the development of the microvascular and neural lesions that characterize the early stages of DR, but additional studies will be needed to determine how this occurs. Manipulation of phototransduction or visual cycle activity as a therapeutic approach would need to be approached cautiously to avoid undesirable effects on vision. Application of low intensity light at certain wavelengths is showing interesting potential to inhibit the retinopathy, but again, the mechanisms remain to be clarified.
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Journal2021, Progress in Retinal and Eye ResearchDeoye Tonade, Timothy S. Kern
Chapter
Molecular Mechanisms of Retinal Toxicity Induced by Light and Chemical Damage
2015, Advances in Molecular ToxicologyMaría Guadalupe Herrera-Hernández, ... Pere Garriga
2.2 Phototransduction
The highly organized and complex retinal structure described participates in the visual process by which the quanta of light falling on the retina are converted into electrical impulses that are transmitted to the brain. The photoreceptors are the primary sensory neurons that sense light and convert light energy into nerve impulses in a process called phototransduction. Light absorption is mediated by visual pigments contained in the rod and cone cells. Rods are responsible for vision under dim-light illumination. These highly specialized cells have the visual pigment rhodopsin which has two components, the 11-cis-retinal chromophore and the opsin apoprotein . Cone cells are responsible for conventional image-formation daylight and color vision, and contain the same chromophore than rhodopsin but opsins of different amino acid sequences named iodopsins [5,12].
Signal transduction in the visual system involves two main processes: first, the activation of rhodopsin by absorption of a photon of light, leading to a conformational change, and second, regeneration of rhodopsin to its original conformation which is the fundamental process regulating light adaptation [13,14].
The visual signal, responsible for light perception in the brain, is initiated by photon absorption by the rhodopsin chromophore, 11-cis-retinal, which causes its isomerization to its all-trans configuration. Complete isomerization to all-trans-retinal causes the molecule to straighten, making its binding to the opsin protein energetically less favorable. This form of higher energy, called metarhodopsin II (Meta II) activates signal transduction through binding to, and activating, the G-protein transducin. This, in turn, activates a phosphodiesterase that hydrolyzes cGMP causing the closure of ion channels in the membrane and subsequent hyperpolarization of the photoreceptor cells membranes [13,15,16]. The difference potential in the outer segment of the photoreceptor cell is transferred through its synaptic terminal to second-order neurons in the inner retina via modulation of neurotransmitter release at the synaptic terminals which include bipolar, amacrine, horizontal, and ganglion cells (Fig. 6.2) .
The integrity and function of photoreceptors are crucial for the complex process of vision. Mutations that affect photoreceptor function, or any other factor that may disrupt the phototransduction process, can lead to vision dysfunction and vision loss . In addition, defects in other retinal cells types, specifically the RPE, can also lead to photoreceptor dysfunction and retinal degeneration .
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Book series2015, Advances in Molecular ToxicologyMaría Guadalupe Herrera-Hernández, ... Pere Garriga
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Basic Science, Inherited Retinal Disease, and Tumors
2006, Retina (Fourth Edition)Jeannie Chen, ... John Flannery
Phototransduction cascade
The phototransduction cascade is perhaps the best characterized of all the G-protein signaling pathways. The components of this pathway are located in the photoreceptor OS, which can be isolated in quantities suitable for biochemical analysis. Photoreceptors can also be easily isolated for electrophysiological measurements, either as whole cells (Fig. 8-5) or as a truncated OS preparation in which different molecules can be dialyzed into the cell and their effect on signal transduction can be evaluated. The amplification cascade within the rod cell is so robust such that single photon absorption by rhodopsin gives rise to a change in current that can be monitored by suction electrode recordings. The combination of transgenic mouse technology and electrophysiological analysis has greatly enhanced our understanding of signaling mechanisms within the photoreceptor cell. Advances in mouse genetics allow us to introduce targeted changes into specific components of the signaling pathway, and the genetically altered intact photoreceptor cells can be subjected to electrophysiological measurements. The tissues from these mice can subsequently be used in biochemical assays to pinpoint the molecular mechanism behind the physiological phenotype.
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Book2006, Retina (Fourth Edition)Jeannie Chen, ... John Flannery
Review article
Rhodopsin phosphorylation: 30 years later
2003, Progress in Retinal and Eye ResearchTadao Maeda, ... Krzysztof Palczewski
Phototransduction in vertebrate photoreceptor cells mediated by rhodopsin is one of the most comprehensively examined G protein-coupled receptor (GPCR) signaling pathways. The signal transduction pathway can be mapped from the initial absorption of light to conformational changes within rhodopsin, through activation of the G protein transducin, and to the ultimate closure of the cation cGMP-gated channels in the plasma membrane. Furthermore, phototransduction has become an intensely studied model system for understanding the desensitizing processes that allow reduced non-linear responses of photoreceptor cells to increasing levels of illumination. Although some general themes appear to occur in GPCR systems, the details of these desensitizing processes are likely to be specific to each of the receptors. These differences are attributed to the fact that each receptor has unique kinetic constraints, amplification levels, tolerance to basal constitutive activity, intracellular internalization and recycling, redundancy of isoforms, and morphologies of the cell of their expression. One of the biochemical processes that are believed to be a common part of this desensitization of the GPCR-mediated cascade is receptor phosphorylation catalyzed by members of a small family of the GPCR kinases. The enzymatic, physiological and genetic aspects of rhodopsin phosphorylation and rhodopsin kinase have been characterized extensively over the last 30 yr. However, new structurally based approaches to examining rhodopsin kinase and rhodopsin phosphorylation are still awaiting further investigations. We present here a summary of the current understanding of rhodopsin phosphorylation and the properties of rhodopsin kinase, along with some expectations of future investigations into these topics.
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Journal2003, Progress in Retinal and Eye ResearchTadao Maeda, ... Krzysztof Palczewski
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Color Vision Defects
2013, Emery and Rimoin's Principles and Practice of Medical Genetics (Sixth Edition)Samir S. Deeb, Arno G. Motulsky
133.4 Phototransduction
The sensation of color results from a comparison of the signal outputs from the three types of cone photoreceptors, and is referred to as trichromatic color vision. The overlap between the spectral sensitivity curves of these photoreceptors (see Figure 133-1) allows this comparison at each wavelength of visible light. Light of a certain wavelength has a characteristic ratio of probabilities for exciting the blue, green and red photoreceptors. Individuals with normal color vision are able to perceive approximately 2million colors.
The absorption of a single photon of light causes the isomerization of the retinal chromophore of the photopigments from the 11-cis to an all-trans configuration. This isomerization results in the formation of an activated intermediate of the photopigment that triggers the signal amplification cascade. The first step in this cascade involves activation of hundreds of transducin which in turn activate about 70 cyclic GMP phosphodiesterase molecules. The resultant decrease in cGMP levels triggers closure of cGMP-gated channel blocking sodium and calcium influx. This in turn leads to hyperpolarization of the photoreceptor cell membrane and creation of electrical signals that are transmitted to the brain and allow sensation of color (Figure 133-3). See phototransduction videos at: for an illustration of this process.
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Book2013, Emery and Rimoin's Principles and Practice of Medical Genetics (Sixth Edition)Samir S. Deeb, Arno G. Motulsky
Review article
Special Issue: Organoid Models in Translational Research
2022, Translational ResearchJessica R. Onyak, ... Jordan M. Renna
Phototransduction cascade
The phototransduction cascade begins when a photon of light interacts with a visual pigment (opsin) in a disc membrane of the outer segment of a photoreceptor. The opsin contains a chromophore, 11-cis retinal, that is activated by the energy of the photon and changes conformation. This conformational change kicks off a reaction that causes the cell membrane to hyperpolarize and change the amount of neurotransmitter release at the photoreceptor synaptic terminal.74–76
Functional phototransduction cascade
Perforated patch data
The first examples of light responsivity from retinal organoid photoreceptors were recorded by Zhong et al. from 175 to 189 days hiPSC-derived organoids. They used three hiPSC lines to demonstrate a novel, primarily self-directed retinal organoid differentiation protocol. The resultant organoids included all major retinal cell types arranged in proper neuroretina layers with the outer layer being highly rod dominant. IHC experiments targeting opsins in both rods and cones showed evidence of inner segments and short outer segments. TEM confirmed structural features of inner segments and some small outer segment stacked disc formation. Given this evidence, perforated patch clamp was employed by the researchers to characterize rod function in the organoids. Voltage clamp recordings of rods (13/13) displayed an inward “dark current” and 2 of 13 rods exhibited small light responses of around 20–30pA. Light responses were non-repeatable upon successive stimuli. The group found that the 2 light-responsive rods also demonstrated large inward currents relative to the nonlight-responsive rods (example recordings in Fig 3A). They determined that the small magnitude of observed light responses in these rods could be attributed to the early stage of morphological and physiological maturation of photoreceptors in the organoids at this time point, as assessed by electron microscopy and immunohistochemistry.47
Whole cell patch data
More recently, Saha et al. developed retinal organoids from 3 hiPSC and 1 hESC lines and allowed them to develop to maturity. Maturity was determined by measuring light responses and membrane physiology at six timepoints from 170 days to 310 days. The first light responses occurred at 200–210 days, the highest percentage of responsive photoreceptors and highest response amplitudes occurred at 250–260 days, and responsivity began to decline at 300–310 days. At 240–270 days, both IHC and TEM showed evidence of organoid photoreceptor maturity including inner segments, outer segments with partially-stacked disks, and ribbon synapses at the axon terminal. The group targeted 240–270 days organoid photoreceptors for a series of patch-clamp electrophysiology experiments done without external chromophores (9-cis-retinal) in order to characterize their intrinsic light responsivity. Rods demonstrated no light response under this condition. Cones were exposed to sequential cone-typing LED flashes, allowing determination of both light response and spectral sensitivity (example recordings in Fig 3B). A robust light response was exhibited by 36 of 107 cones with most demonstrating a medium-wavelength sensitivity. The robust light response was characterized by a typical mature hyperpolarization-repolarization phototransduction cascade. When compared with data from the same experiment done on cones from macaque fovea (Fig 3C), retinal organoid cones demonstrated light responses that had, with some notable exceptions, smaller amplitudes (Fig 3D) and slower time to peak (Fig 3E).44
Summary
The first reported light responses from retinal organoids were weak and nonrepeatable responses recorded from 2 of 13 rods in 175–189 days hiPSC retinal organoids.47 More robust light responses characterized by a mature phototransduction cascade were recorded from 36 of 107 cones in 240–270 days hPSC retinal organoids.44
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Journal2022, Translational ResearchJessica R. Onyak, ... Jordan M. Renna
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Phototransduction: Phototransduction in Rods
2017, Reference Module in Neuroscience and Biobehavioral PsychologyYingbin Fu, Satoru Kawamura
Phototransduction in Rods: A G-Protein-Signaling Pathway
Rod phototransduction is one of the best-characterized G-protein-signaling pathways. The receptor is rhodopsin (R), the G protein is transducin (G), and the effector is cyclic guanosine monophosphate (cGMP) phosphodiesterase (PDE or PDE6). Upon photon absorption, the rhodopsin molecule becomes enzymatically active (R∗) and catalyzes the activation of the G-protein transducin to G∗. Transducin, in turn, activates the effector PDE to PDE∗. PDE∗ hydrolyzes the diffusible messenger cGMP. The resulting decrease in the cytoplasmic-free cGMP concentration leads to the closure of the cGMP-gated channels on the plasma membrane. Channel closure leads to localized reduction on the influx of cations into the outer segment, which results in membrane hyperpolarization, that is, the intracellular voltage becoming more negative (Fig. 2). This hyperpolarization decreases or terminates the dark glutamate release at the synaptic terminal. The signal is further processed by other neurons in the retina before being transmitted to higher centers in the brain.
Following light activation, a timely recovery of the photoreceptor is essential so that it can respond to subsequently absorbed photons, and signal rapid changes in illumination. This recovery from light requires the efficient inactivation of each of the activated components: R∗, G∗, and PDE∗, as well as the efficient regeneration of rhodopsin (R) and the rapid restoration of the cGMP concentration. The termination rates of the activation steps set the time course of the photoresponse.
Although rod phototransduction is the best-characterized sensory transduction pathway, rods differ from other sensory cells in that light leads to hyperpolarization rather than depolarization. Rods respond to light with graded hyperpolarization whose amplitude increases monotonically as a function of flash intensity until saturation. One hallmark of rod phototransduction is the reproducibility of its single-photon response in both amplitude and kinetics. This is quite remarkable considering the fact that events generated by single molecules are stochastic in nature. The study on the underlying mechanisms has long been a hot topic in the vision field. Recent research pointed to two possible mechanisms: (1) Rhodopsin inactivation is averaged over multiple shutoff steps so that the integrated R∗ activity varies less than otherwise controlled by a single step. (2) Averaging over the deactivation of multiple G-protein molecules. More recent study showed that its amplitude is more reproducible with calcium feedback which underlies light adaptation in rods and cones.
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Reference work2017, Reference Module in Neuroscience and Biobehavioral PsychologyYingbin Fu, Satoru Kawamura
Related terms:
Rhodopsin
Transducin
Photoreceptor Cell
Opsin
Retinitis pigmentosa
Retinal Pigment Epithelium
Visual Pigment
Photoreceptor
Cyclic GMP
Calcium Ion
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18 F
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2.1 Quantifiers
3 F
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7 F
8 F
9 F
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2.2 Equivalence Involving Quantifiers
1 F
3 F
4 F
5 F
6 F
7 F
8 F
9 F
10 F
14 F
15 F
2.3 More Operations on Sets
1 F
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12 F
13 F
15 F
Chapter 3: Proofs
3.1 Proof Strategies
1 F
3 F
5 F
8 F
14 F
15 F
16 F
17 F
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1 F
2 F
5 F
6 F
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1 F
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5 F
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18 F
19 F
20 F
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3.4 Proofs Involving Conjunctions and Biconditionals
1 F
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26 F
27 F
3.5 Proofs Involving Disjunctions
1 F
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10 F
13 F
27 F
28 F
29 F
31 F
3.6 Existence and Uniqueness Proofs
1 F
6 F
10 F
13 F
3.7 More Examples of Proofs
6 F
10 F
Chapter 4: Relations
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5 F
6 F
7 F (no need to prove)
8 F
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13 F
15 F
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7 F
9 F
10 F
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12 F
13 F
4.3 More about Relations
3 F
4 F
7 F
8 F
9 F
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11 F
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22 F
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4.4 Ordering Relations
1 F
3 F
4 F
6 F
14 F
17 F
19 F
23 F
4.5 Equivalence Relations
1 F
2 F
8 F
11 F
12 F
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14 F
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1(a)(b) F
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7(a)(b) F
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5 F
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13 F
5.3: Inverses of Functions
3 F
8 F
9 F
11 F
18 F
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5 F
11 F
12 F
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3 F
4 F
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Proof by Mathematical Induction
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19 F
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18 F
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14 F
15 F
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21 F
22(a) F
23(a)(b)(c) F
24(a)(b) F
26 F
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1 F
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4 F
5 F
6 F
12 F
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1 F
2 F
3 F
4 F
5 F
6 F
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Triangle
A closed figure consisting of three line segments linked end-to-end. A 3-sided polygon.
Try this Drag the orange dots on each vertex to reshape the triangle.
Triangle properties
| | |
--- |
| Vertex | The vertex (plural: vertices) is a corner of the triangle. Every triangle has three vertices. |
| Base | The base of a triangle can be any one of the three sides, usually the one drawn at the bottom. You can pick any side you like to be the base. Commonly used as a reference side for calculating the area of the triangle. In an isosceles triangle, the base is usually taken to be the unequal side. |
| Altitude | The altitude of a triangle is the perpendicular from the base to the opposite vertex. (The base may need to be extended). Since there are three possible bases, there are also three possible altitudes. The three altitudes intersect at a single point, called the orthocenter of the triangle. See Orthocenter of a Triangle. In the figure above, you can see one possible base and its corresponding altitude displayed. |
| Median | The median of a triangle is a line from a vertex to the midpoint of the opposite side. The three medians intersect at a single point, called the centroid of the triangle. See Centroid of a Triangle |
| Area | See area of the triangle and Heron's formula |
| Perimeter | The distance around the triangle. The sum of its sides. See Perimeter of a Triangle |
| Interior angles | The three angles on the inside of the triangle at each vertex. See Interior angles of a triangle |
| Exterior angles | The angle between a side of a triangle and the extension of an adjacent side. See Exterior angles of a triangle |
Also:
The shortest side is always opposite the smallest interior angle
The longest side is always opposite the largest interior angle
For more on this see Side / angle relationship in a triangle
Terminology
It is usual to name each vertex of a triangle with a single capital (upper-case) letter. The sides can be named with a single small (lower case) letter, and named after the opposite angle. So in the figure above, you can see that side b is opposite vertex B, side c is opposite vertex C and so on.
Alternatively, the side of a triangle can be thought of as a line segment joining two vertices. So then side b would be called AC. This is the form used on this site because it is consistent across all shapes, not just triangles.
Properties of all triangles
These are some well known properties of all triangles. See the section below for a complete list
The interior angles of a triangle always add up to 180°
The exterior angles of a triangle always add up to 360°
Types of Triangle
There are seven types of triangle, listed below. Note that a given triangle can be more than one type at the same time. For example, a scalene triangle (no sides the same length) can have one interior angle 90°, making it also a right triangle. This would be called a "right scalene triangle".
| | |
--- |
| Isosceles | Two sides equal See Isosceles triangle definition |
| Equilateral | All sides equal See Equilateral triangle definition |
| Scalene | No sides equal See Scalene triangle definition |
| | |
--- |
| Right Triangle | One angle 90°. See Right triangle definition |
| Obtuse | One angle greater than 90° See Obtuse triangle definition |
| Acute | All angles less than 90° See Acute triangle definition |
| Equiangular | All interior angles equal See Equiangular triangle definition |
Classifying triangles
The seven types of triangle can be classified two ways: by sides and by interior angles. For more on this see Classifying triangles.
Constructing triangles
Many types of triangle can be constructed using a a compass and straightedge using the traditional Euclidean construction methods. For more on this see Constructions using Compass and straightedge.
Other triangle topics
General
Triangle definition
Hypotenuse
Triangle interior angles
Triangle exterior angles
Triangle exterior angle theorem
Pythagorean Theorem
Proof of the Pythagorean Theorem
Pythagorean triples
Triangle circumcircle
Triangle incircle
Triangle medians
Triangle altitudes
Midsegment of a triangle
Triangle inequality
Side / angle relationship
Perimeter / Area
Perimeter of a triangle
Area of a triangle
Heron's formula
Area of an equilateral triangle
Area by the "side angle side" method
Area of a triangle with fixed perimeter
Triangle types
Right triangle
Isosceles triangle
Scalene triangle
Equilateral triangle
Equiangular triangle
Obtuse triangle
Acute triangle
3-4-5 triangle
30-60-90 triangle
45-45-90 triangle
Triangle centers
Incenter of a triangle
Circumcenter of a triangle
Centroid of a triangle
Orthocenter of a triangle
Euler line
Congruence and Similarity
Congruent triangles
Solving triangles
Solving the Triangle
Law of sines
Law of cosines
Triangle quizzes and exercises
Triangle type quiz
Ball Box problem
How Many Triangles?
Satellite Orbits
(C) 2011 Copyright Math Open Reference. All rights reserved |
187741 | https://en.wikipedia.org/wiki/Quadratic_residuosity_problem | Jump to content
Quadratic residuosity problem
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From Wikipedia, the free encyclopedia
Problem in computational number theory
The quadratic residuosity problem (QRP) in computational number theory is to decide, given integers and , whether is a quadratic residue modulo or not.
Here for two unknown primes and , and is among the numbers which are not obviously quadratic non-residues (see below).
The problem was first described by Gauss in his Disquisitiones Arithmeticae in 1801.
This problem is believed to be computationally difficult.
Several cryptographic methods rely on its hardness, see § Applications.
An efficient algorithm for the quadratic residuosity problem immediately implies efficient algorithms for other number theoretic problems, such as deciding whether a composite of unknown factorization is the product of 2 or 3 primes.
Precise formulation
[edit]
Given integers and , is said to be a quadratic residue modulo if there exists an integer such that
: .
Otherwise we say it is a quadratic non-residue.
When is a prime, it is customary to use the Legendre symbol:
This is a multiplicative character which means for exactly of the values , and it is for the remaining.
It is easy to compute using the law of quadratic reciprocity in a manner akin to the Euclidean algorithm; see Legendre symbol.
Consider now some given where and are two different unknown primes.
A given is a quadratic residue modulo if and only if is a quadratic residue modulo both and and .
Since we don't know or , we cannot compute and . However, it is easy to compute their product.
This is known as the Jacobi symbol:
This also can be efficiently computed using the law of quadratic reciprocity for Jacobi symbols.
However, cannot in all cases tell us whether is a quadratic residue modulo or not!
More precisely, if then is necessarily a quadratic non-residue modulo either or , in which case we are done.
But if then it is either the case that is a quadratic residue modulo both and , or a quadratic non-residue modulo both and .
We cannot distinguish these cases from knowing just that .
This leads to the precise formulation of the quadratic residue problem:
Problem:
Given integers and , where and are distinct unknown primes, and where , determine whether is a quadratic residue modulo or not.
Distribution of residues
[edit]
If is drawn uniformly at random from integers such that , is more often a quadratic residue or a quadratic non-residue modulo ?
As mentioned earlier, for exactly half of the choices of , then , and for the rest we have .
By extension, this also holds for half the choices of .
Similarly for .
From basic algebra, it follows that this partitions into 4 parts of equal size, depending on the sign of and .
The allowed in the quadratic residue problem given as above constitute exactly those two parts corresponding to the cases and .
Consequently, exactly half of the possible are quadratic residues and the remaining are not.
Applications
[edit]
The intractability of the quadratic residuosity problem is the basis for the security of the Blum Blum Shub pseudorandom number generator. It also yields the public key Goldwasser–Micali cryptosystem, as well as the identity based Cocks scheme.
See also
[edit]
Higher residuosity problem
References
[edit]
^ Kaliski, Burt (2011). "Quadratic Residuosity Problem". Encyclopedia of Cryptography and Security. p. 1003. doi:10.1007/978-1-4419-5906-5_429. ISBN 978-1-4419-5905-8.
^ Adleman, L. (1980). "On Distinguishing Prime Numbers from Composite Numbers". Proceedings of the 21st IEEE Symposium on the Foundations of Computer Science (FOCS), Syracuse, N.Y. pp. 387–408. doi:10.1109/SFCS.1980.28. ISSN 0272-5428.
^ S. Goldwasser, S. Micali (1982). "Probabilistic encryption & how to play mental poker keeping secret all partial information". Proceedings of the fourteenth annual ACM symposium on Theory of computing - STOC '82. pp. 365–377. doi:10.1145/800070.802212. ISBN 0897910702. S2CID 10316867.
^ S. Goldwasser, S. Micali (1984). "Probabilistic encryption". Journal of Computer and System Sciences. 28 (2): 270–299. doi:10.1016/0022-0000(84)90070-9.
| Computational hardness assumptions | |
--- |
| Number theoretic | Integer factorization Phi-hiding RSA problem Strong RSA Quadratic residuosity Decisional composite residuosity Higher residuosity |
| Group theoretic | Discrete logarithm Diffie-Hellman Decisional Diffie–Hellman Computational Diffie–Hellman |
| Pairings | External Diffie–Hellman Sub-group hiding Decision linear |
| Lattices | Shortest vector problem (gap) Closest vector problem (gap) Learning with errors Ring learning with errors Short integer solution |
| Non-cryptographic | Exponential time hypothesis Unique games conjecture Planted clique conjecture |
Retrieved from "
Categories:
Computational number theory
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Hidden categories:
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Which law is this expression X+ X.Y=X+Y
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Question. Name the law given and verify it using a truth table. X+ X.Y=X+Y
My Answer. X Y X X.Y X+X.Y X+Y 0 0 1 0 0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 0 0 1 1 Prove algebraically that X + XY = X + Y. L.H.S. = X + XY = X.1 + XY (X . 1 = X property of 0 and 1) = X(1 + Y) + XY (1 + Y = 1 property of 0 and 1) = X + XY + XY = X + Y(X + X) = X + Y.1 (X + X =1 complementarity law) = X + Y (Y . 1 = Y property of 0 and 1) = R.H.S. Hence proved.
My teacher marked my answer wrong. And told me to find the correct answer. Friends tell me is it a complementary law or distributive law or Absorption law? If it is absorption kindly tell me how to prove RHS and LHS algebraically.
boolean-algebra
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edited Mar 9, 2016 at 11:40
Raphael
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asked May 10, 2014 at 20:30
user2241865user2241865
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$\begingroup$ Your teacher may find your algebraic proof too complex (though it is correct, imho). For example, you use commutativity without need. Regarding the name of the law, I doubt very much it has one, but I would not underestimate the ability of people to create terminology just for the hell of it. $\endgroup$
babou
– babou
2015-08-23 13:30:46 +00:00
Commented Aug 23, 2015 at 13:30
$\begingroup$ We call it Absorption law, the prove is in any decent textbook. $\endgroup$
Eugene
– Eugene
2017-02-20 05:52:48 +00:00
Commented Feb 20, 2017 at 5:52
$\begingroup$ For me the absorption laws are $x \land (x \lor y) = x$ and $x \lor (x \land y) = x$. $\endgroup$
Andrej Bauer
– Andrej Bauer
2017-02-20 17:58:05 +00:00
Commented Feb 20, 2017 at 17:58
$\begingroup$ Did you talk to your teacher before seeking solace on the internet? $\endgroup$
Andrej Bauer
– Andrej Bauer
2017-02-20 17:58:57 +00:00
Commented Feb 20, 2017 at 17:58
$\begingroup$ I don't have enough rep to answer, but this seems to be called "Redundancy law". Check line 12 of this webpage: mi.mun.ca/users/cchaulk/misc/boolean.htm $\endgroup$
Ignis Incendio
– Ignis Incendio
2021-09-06 04:25:43 +00:00
Commented Sep 6, 2021 at 4:25
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3 Answers 3
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One way of looking at this is as a consequence of distributivity, where $P+QR\equiv (P+Q)(P+R)$. Then you'll have $$\begin{align} X+(X'Y) &\equiv (X+X')(X+Y)&\text{distributivity}\ &\equiv T(X+Y)&\text{inverse}\ &\equiv X+Y&\text{domination} \end{align}$$
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answered May 10, 2014 at 20:46
Rick DeckerRick Decker
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$\begingroup$ Thank you for your valuable answer. Is this equation any way is Absorption law ? If yes then how to prove RHS and LHS algebraically $\endgroup$
user2241865
– user2241865
2014-05-11 07:02:22 +00:00
Commented May 11, 2014 at 7:02
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0
$\begingroup$
This is known as the third distributive law (see, e.g., this page). I think your proof is correct
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edited Jan 28, 2016 at 1:00
David Richerby
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answered Jan 27, 2016 at 21:39
Karan GuptaKaran Gupta
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$\begingroup$ Hmm. I've been doing this stuff for decades and I've never come across the term "third distributive law". $\endgroup$
Rick Decker
– Rick Decker
2016-01-28 01:51:34 +00:00
Commented Jan 28, 2016 at 1:51
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-4
$\begingroup$
X+X'Y=X+Y
taking LHS X+X'Y=(X+X')(X+Y) OR distributes over AND 1.(X+Y) (X+X'=1) X+Y=RHS 1 is identity for AND
Hence Proved
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answered Mar 9, 2016 at 10:13
Karan GuptaKaran Gupta
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$\begingroup$ I don't think this is the proof that the asker is looking for: it doesn't use the algebraic properties of the operators. Also, I find your proof very hard to read. What does "(X+Y)(X+X'=1)X+Y" mean? $\endgroup$
David Richerby
– David Richerby
2016-03-09 16:22:50 +00:00
Commented Mar 9, 2016 at 16:22
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187743 | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/761912972cce82cd34e0380e1b0123b1_MIT8_07F12_pset05.pdf | MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Electromagnetism II October 11, 2012 Prof. Alan Guth PROBLEM SET 5 REVISED∗ DUE DATE: Friday, October 12, 2012. Either hand it in at the lecture, or by 6:00 pm in the 8.07 homework boxes. READING ASSIGNMENT: None. (The material is related to Chapter 3 of Griffiths, which you should have already read.) CREDIT: This problem set has 90 points of credit plus 10 points extra credit. PROBLEM 1: A SPHERE WITH OPPOSITELY CHARGED HEMI SPHERES (15 points) Griffiths Problem 3.22 (p. 145). (This problem was held over from Problem Set 4.) PROBLEM 2: A CIRCULAR DISK AT A FIXED POTENTIAL (20 points) This problem is based on Jackson, Problem 3.3 (challenging!). A thin, flat, conducting circular disk of radius R is located in the x-y plane with its center at the origin, and is mantained at a fixed potential V0. (a) With the information that the surface charge density on a disk at fixed potential is proportional to (R2 −s2)−1/2, where s is the distance out from the center of the disk, show that for r > R, ∞ 2l 2V0 R (−1)£ R V (r, θ) = P2£(cos θ) . π r 2e + 1 r £=0 Hint: First find the field along the z-axis, and then use your knowledge about the general solution to Laplace’s equation to infer the angular dependence. (b) Calculate the capacitance of the disk, defined by Q = CV , where Q is the total charge on the disk and V is the potential of the disk relative to |I r | →∞. PROBLEM 3: QUADRUPOLE AND OCTOPOLE TERMS OF THE MUL TIPOLE EXPANSION (20 points) Griffiths Problem 3.45 (pp. 158-159). For parts (a) and (d), you can find the answers either by starting with Griffiths’ Eq. (3.95) (p. 148) for the multipole expansion, or by starting with the integral expression (Griffiths’ Eq. (2.29), p. 84) for the potential and then expanding in powers of 1/r. Note that Qij and the octopole moment that you will find are examples of the traceless symmetric tensor formalism that we have discussed in lecture. ∗ A typo was fixed in Problem 4(c), Eq. (4.12). In the original version of October 7, 2012, the factor on the right-hand side was inverted. 8.07 PROBLEM SET 5 REVISED, FALL 2012 p. 2 PROBLEM 4: SPHERICAL HARMONICS AND TRACELESS SYMMET RIC TENSORS (20 points) In class we discussed the fact that any sum of spherical harmonics for a given index e can be written as F£(θ, φ) = C(£) n ˆi1 n ˆi2 . . .n ˆi£ , (4.1) i1i2...i£ where C(£) is a traceless symmetric tensor, the indices i1, i2, . . . i£ are summed 1 to i1i2...i£ 3, and n ˆ = sin θ cos φ e ˆ1 + sin θ sin φ e ˆ2 + cos θ e ˆ3 (4.2) is a unit vector in the direction of (θ, φ). The standard functions Y£m(θ, φ) can be written in this formalism by finding the corresponding traceless symmetric tensors, which we will (£,m) call C : i1i2...i£ (£,m) Y£m(θ, φ) = C ˆ ˆ . . . ˆ . (4.3) i1i2...i£ ni1 ni2 ni£ (£,m) To construct the C n ˆi1 n ˆi2 . . . n ˆi£ explicitly, we introduce the following basis of unit i1i2...i£ 3-dimensional vectors: 1 (1) ≡ ˆ+ u ˆ u = √ (ˆ ex + ie ˆy ) 2 (2) ≡ ˆ− 1 (4.4) u ˆ u = √ (ˆ ex − ie ˆy ) 2 (3) ≡ ˆ u ˆ z = ˆ ez . These complex-valued vectors are orthonormal in the sense that (i)∗ (j) u ˆ · u ˆ = δij , (4.5) but note that + − u ˆ · u ˆ+ = ˆ u · u ˆ− = 0 , (4.6) + + so we find conveniently that constructions such as ˆ ui u ˆj are traceless as well as symmetric. (£,m) For m ≥ 0, C ˆ ˆ . . . ˆ is given by i1i2...i£ ni1 ni2 ni£ (£,m) + + Ci1i2 ...i£ = d£m { u ˆi1 . . . u ˆim z ˆim+1 . . . z ˆi£ } , (4.7) where { xxx } means the traceless symmetric part of xxx. The constant of proportionality d£m is given by d£m = (−1)m(2e)! 2£e!
2m (2e + 1) 4π (e + m)! (e − m)! . (4.8) For m ≤ 0, C(£,m) i1 i2...i£ = d£m { ˆ u − i1 . . . ˆ u − i|m| ˆ zi|m|+1 . . . ˆ zi£ } = C(£,|m|)∗ i1 i2...i£ . (4.9) 8.07 PROBLEM SET 5 REVISED, FALL 2012 p. 3 (a) Use this formalism to show that 1 21 Y31(θ, φ) = − (5 cos2 θ − 1) sin θ eiφ . (4.10) 4 4π Hint: You might benefit from that fact that Eq. (4.3) can alternatively be written as (£,m) Y£m(θ, φ) = C { n ˆi1 n ˆi2 . . .n ˆi£ } , (4.11) i1i2...i£ since { n ˆi1 n ˆi2 . . . n ˆi£ } differs from ˆ ni1 n ˆi2 . . .n ˆi£ only by terms proportional to kro (£,m) necker δ-functions, and such terms will give no contribution because C is trace i1 i2...i£ less. Once the expansion of { n ˆi1 n ˆi2 . . . n ˆi£ } is written out explicitly, there is no need (£,m) to take the traceless symmetric part in the evaluation of C using Eq. (4.7). It i1 i2...i£ is a little easier to write the traceless symmetric part of ˆ ni1 n ˆi2 . . . n ˆi£ than it is to find the traceless symmetric part of the right-hand side of Eq. (4.7), since ˆ ni1 n ˆi2 . . . n ˆi£ is already symmetric. (b) Use this formalism to derive a general expression for Y££ that is valid for all e. (c) For cases with azimuthal symmetry, it is sufficient to use C(£) tensors that are i1i2...i£ constructed as traceless symmetric parts of products of ˆ z, which corresponds to what in the usual formalism are called Legendre polynomials. The precise connection is that (2e)!
P£(cos θ) = { z ˆi1 z ˆi2 . . . z ˆi£ } n ˆi1 n ˆi2 . . . n ˆi£ 2£(e!)2 (4.12) (2e)!
= z ˆi1 z ˆi2 . . . z ˆi£ { n ˆi1 n ˆi2 . . . n ˆi£ } .
2£(e!)2 Use this relation to evaluate P4(cos θ), and compare your result with Table 3.1 of Griffiths (p. 138). PROBLEM 5: THE PRECISE ELECTRIC FIELD OF AN ELECTRIC DIPOLE (15 points plus 10 points extra credit) Griffiths Problem 3.42 (p. 157, 15 points). Griffiths felt (and I agree) that the answers to this problem are sufficiently subtle that he should give the answers in the statement of the problem. Your job, then, is to explain these answers. (c) (10 points extra credit) Note that Griffiths found the δ-function term in the expres sion for the electric field of a dipole by starting with an ill-defined expression (i.e., Vdip(r, θ) = p cos θ/(4πE0r2) is not differentiable at r = 0), then calculating a wrong answer, and then noticing that the answer violated a general theorem. He then asked what could be changed in the answer to avoid claiming a result which is demonstrably false. He got the right answer, but one might hope to find a way to make everything 8.07 PROBLEM SET 5 REVISED, FALL 2012 p. 4 well-defined from the start. The method of distributions, or generalized functions, is exactly that.
To define Vdip as a distribution, we define it in terms of what happens when we multiply it by a test function ϕ(⃗ r) and then integrate over all space: p cos θ FVdip[ϕ(⃗ r)] ≡ ϕ(⃗ r) d3x .
(5.1) 4πϵ 2 0r The test function is required to be smooth, and to fall offrapidly at large |⃗ r|. Note that the integral is well-defined in spite of the factor of 1/r2, since the singularity is canceled by the measure d3x = r2 dr sin θ dθ dφ. The derivative ∂iVdip ≡∂Vdip/∂xi is defined in the sense of distributions by formally integrating by parts. That is, one might naively differentiate by writing p cos θ F∂iVdip[ϕ(⃗ r)] = ϕ(⃗ r)∂i d3x , (5.2) 2 naively 4πϵ0r but this integral is ill-defined. The real definition in the sense of distributions is given by p cos θ F∂iVdip[ϕ(⃗ r)] = − ∂iϕ(⃗ r) d3x .
(5.3) 4πϵ0r2 By definition there is no surface term associated with the integration by parts. One can intuitively justify this by the assumption that ϕ(⃗ r) falls of rapidly with |⃗ r|, but in the logic of distribution theory, Eq. (5.3) is simply the definition of the derivative of the distribution Vdip.
The integral on the right-hand side of Eq. (5.3) is a well-defined, ordinary integral.
The goal is to show that this gives Griffith’s expression (Eq. (3.106, p. 157)) for the electric field of a dipole, δ-function term included. One way to do this is to separate the integral into two pieces, r < ϵ and r > ϵ, where ϵ is a positive number which will in the end be taken to zero. Then F∂iVdip[ϕ(⃗ r)] = (1) (2) F∂V ϕ dip[ (⃗ r)] + F i ∂iVdi [ϕ , p (⃗ r)] (5.4) where (1) θ F∂V [ϕ(⃗ r)] = − p cos ∂iϕ(⃗ r i dip ) r>ϵ 4πϵ0r2 d3x (5.5) (2) p cos θ F∂iVdip[ϕ(⃗ r)] = − ∂ϕ(⃗ r) d3 i x .
r<ϵ 4πϵ0r2 Looking at (2) F∂V [ϕ(⃗ r)], one can see that it vanishes in the limit ϵ write i dip →0. To see this, π 2π (2) p ϵ F∂iVdip[ϕ(⃗ r)] | | ≤ max (∂iϕ) dr dφ dθ ≤const ϵ .
(5.6) 4πϵ r<ϵ 0 0 0 0 8.07 PROBLEM SET 5 REVISED, FALL 2012 p. 5 Here maxr<ϵ (∂iϕ) represents the maximum value of | (∂iϕ) | in the region r < ϵ, which is guaranteed by the smoothness of ϕ(⃗ r) to approach a constant as ϵ →0. We also used the facts that | sin θ| ≤1 and | cos θ| ≤1. Since we will evaluate Eq. (5.4) in the limit ϵ →0, we can forget about (2) F∂iVdip[ϕ(⃗ r)].
Since the integral in (1) F∂iVdip[ϕ(⃗ r)] has a lower limit > 0, we can evaluate it by integrating by parts. We get no surface term at |⃗ r| = ∞, by the properties of ϕ, but we do find a surface term at r = ϵ: (1) p cos θ p cos θ F∂iVdip[ϕ(⃗ r)] = − ∂ 3 i ϕ(⃗ r) d x+ ϕ(⃗ r)∂i d3x . (5.7) 2 2 r>ϵ 4πϵ0r r>ϵ 4πϵ0r AT LAST: THE HOMEWORK PROBLEM: Evaluate the right-hand side of Eq. (5.7), and show that it is equivalent to (1) 1 1 1 F∂V [ϕ(⃗ r)] = ϕ(⃗ r) r 4 3(p ⃗· r ˆ)ˆ −p 3 3 i i p i 3 r dip πϵ − ) 3 iδ (⃗ d x .
(5.8) 0 r ϵ0 One further hint: The identity proven in Problem 7(a) of Problem Set 1 may prove useful.
MIT OpenCourseWare 8.07 Electromagnetism II Fall 2012 For information about citing these materials or our Terms of Use, visit: |
187744 | https://pmc.ncbi.nlm.nih.gov/articles/PMC3970835/ | The Use of Isotretinoin in the Treatment of Acne Vulgaris: Clinical Considerations and Future Directions - PMC
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J Clin Aesthet Dermatol
. 2014 Feb;7(2 Suppl):S3–S21.
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The Use of Isotretinoin in the Treatment of Acne Vulgaris
Clinical Considerations and Future Directions
James J Leyden
James J Leyden, MD, FAAD
1 University of Pennsylvania, Philadelphia, Pennsylvania
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1, James Q Del Rosso
James Q Del Rosso, DO, FAOCD
2 Touro University College of Osteopathic Medicine, Henderson, Nevada
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2, Eric W Baum
Eric W Baum, MD, MSc
3 University of Alabama Medical Center, Tuscaloosa, Alabama
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3
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Copyright and License information
1 University of Pennsylvania, Philadelphia, Pennsylvania
2 Touro University College of Osteopathic Medicine, Henderson, Nevada
3 University of Alabama Medical Center, Tuscaloosa, Alabama
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PMCID: PMC3970835 PMID: 24688620
Introduction
Isotretinoin (13-cis-retinoic acid) is a non-aromatic retinoid that was approved by the United States (US) Food and Drug Administration (FDA) as an oral capsule formulation in May 1982 with an indication for treatment of severe recalcitrant nodular acne.1-3 Over time, oral isotretinoin (ISO) has proven to be a major pharmacological breakthrough for treating severe and recalcitrant cases of inflammatory acne. The continued availability of ISO has at times been challenged due to its teratogenicity warranting further emphasis on continued patient education and the introduction of a structured, traceable, and mandated risk-management program (iPLEDGE).1-5 Other challenges have included reports of potential associations with depression, suicidal ideation, and inflammatory bowel disease.1,2,6-18 However, continued pharmacovigilance has demonstrated that these alleged risks have not been definitively associated with ISO as the causative factor and are not common enough to take away the accessability of ISO to the millions of patients with severe acne who have experienced major improvements in their quality of life after ISO treatment.2,9,10,11,15,18 Continued vigilance and assessments of outcomes related to how ISO is utilized allows for frequent re-evaluation of how efficacy and safety can be optimized both in terms of clearance upon completion of the initial ISO treatment course and achieving prolonged remission without the need for retreatment.
The implementation by the US FDA of the more stringent Risk Evaluation and Mitigation Strategy (REMS) program, namely iPLEDGE, was put in place to prevent as much as possible ISO pregnancy exposures in females of child-bearing potential, and to capture the frequency and outcomes of ISO pregnancy exposures.1,2,19-22 In addition, the data captured through the iPLEDGE program allow for analysis of usage patterns with ISO by patients and prescribers. Responsible prescribing of ISO by clinicians, compliant use and follow up by patients, and vigilant dispensing practices by pharmacists underscore the safe and effective usage of this agent. Despite challenges related to teratogenic properties, known potential side effects, and alleged adverse reactions, ISO remains available in the United States under the iPLEDGE program, including in females of childbearing potential, without restrictions on indication and duration of use, provided the mandated rules of the iPLEDGE program are followed by the registered prescriber, patient, and pharmacist.1,19-22-19
Several million patients, including a subset of children, have been treated with ISO. Data captured through the iPLEDGE program have documented more than 1.2 million individual patients treated from December 2005 through February 2011.23 Clinical studies and global experience have shown that ISO provides complete or nearly complete remission of acne, with sustained therapeutic benefit after completion of ISO therapy found to be a consistent observation in the vast majority of treated patients.2,24-36 In addition to severe nodular acne, the use of ISO within the dermatology community for acne has included refractory cases of severe non-nodular inflammatory acne and/or in patients who exhibit a propensity for acne scarring, with effective use in recalcitrant non-nodular inflammatory acne noted in the literature.37-39 The American Academy of Dermatology (AAD) has published a Position Statement on Oral Isotretinoin, which considers the physical and psychological impact of acne in the decision-making process on ISO use.37 In all cases, proper follow up and monitoring is recommended in compliance with the iPLEDGE program.1,2,19-22
Importantly, ISO remains the only therapy for acne that is capable of inducing remission after an adequate course of therapy is completed, with prolonged remissions occurring with reasonable certainty in most patients.2,24,25,29 Due to the large volume of people with severe and recalcitrant acne who have been treated with ISO globally, and with experience in the use of ISO for more than three decades, multiple publications have addressed the need for retreatment of acne with ISO or other acne therapies.24-36,40,41 The duration of remission before need for acne retreatment and the percent of patients needing retreatment after their initial course of ISO are variable and appear to correlate with several potential contributory factors.2,25,29-36,40 These contributory factors include patient age when the initial ISO course was given, cumulative dose administered over the initial course of therapy, endogenous androgen-excess (i.e., polycystic ovary syndrome), presence/persistence of macrocomedones, presence of sinus tracts, patient adherence, and the potential role of dietary factors.2,25,29-36,40-46
The Pathophysiology of Acne
The pathophysiology of acne is complex and multifactorial and is associated with multiple potential inflammatory processes. The four major pathophysiological effects that have been correlated with acne formation are: 1) sebaceous gland hyperplasia and excess sebum production, 2) abnormal follicular epithelial desquamation, 3) Propionibacterium acnes proliferation with a variety of direct or indirect pro-inflammatory effects, and 4) preclinical (subclinical) and visible inflammation.25,2753 Unlike other therapies for acne, ISO counteracts all four of these pathophysiological factors, and has been shown to exhibit sebosuppressive, comedolytic, anti-inflammatory, and possibly immunological effects that are relevant to acne therapy.2,24,25,27,46-52 Although the mechanisms of ISO that impart prolonged remission are not completely understood, marked reduction in sebaceous gland activity and size, altered follicular microclimate not highly conducive to proliferation of P. acnes due to prolonged sebosuppression, and reduction in toll-like receptor-2 (TLR2) expression on circulating mononuclear cells that is persistent for several months post-therapy have been suggested.2,25,27,46,47,50-52 Interestingly, the observation that long-term remission of acne after completion of ISO therapy may possibly correlate with a prolonged decrease in TLR2 expression on circulating mononuclear cells suggests that the therapeutic benefits of ISO in acne involve systemic modes of action beyond just a sebosuppressive effect confined to skin.50 This is further supported by the observation that the duration of remission correlates with reaching a threshold range of cumulative exposure based on daily dose/kg.2,25,29,30,40
Currently Available Formulations of Oral Isotretinoin
The initial brand formulation of ISO, marketed by Roche Laboratories under the brand name Accutane® (Acc-ISO), which was released into the marketplace in the United States in 1982, was based on information gleaned from studies evaluating the pharmacokinetic (PK) profile, clinical trials, and practice databases on ISO.54 As the original brand of ISO, Acc-ISO served as the reference comparator against which other formulations of ISO were compared as generic formulations of ISO were developed over time. Since 2002, several “branded generic” capsule formulations of ISO have been made available in the United States, all of which are officially rated as bioequivalent with Acc-ISO and are therefore designated as “AB-rated with Accutane” (AB-ISO).55,56 Currently available AB-ISO products in the United States are listed in Table 1.
TABLE 1.
Currently available branded generic formulations of oral isotretinoin
| BRANDED GENERIC PRODUCT | FDA-APPROVAL DATE |
:--- |
| Amnesteem® | 11/08/02 |
| Claravis® | 04/11/03 |
| Myorisan® | 01/19/12 |
| Zenatane™ | 04/01/13 |
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Based on the pharmacokinetic (PK) standard established with Accutane®, the original brand of isotretinoin
As Accutane was discontinued by the manufacturer in June 2009, the PK profile of Amnesteem® served as the comparison standard for Zenatane™ as the former was based on the standard profile of Accutane®.
To achieve FDA approval for use, generic ISO products are submitted to the FDA under an Abbreviated New Drug Application (ANDA) status, which means they must demonstrate adequate bioequivalence with Acc-ISO based on standards used by the FDA before they are approved for commercial release. If the generic ISO successfully meets these FDA-required bioequivalency standards, that generic product is then considered “AB-rated” with Acc-ISO, which means it can be substituted for Acc-ISO and for other AB-ISO products.55-58 As the manufacturer of Accutane® discontinued availability of the drug in June 2009, PK studies with ISO that support an ANDA (i.e., Zenetane®) have been based on the PK profile of Amnesteem, which was established as an AB-ISO in 2002.59
Since the approval and release into the United States marketplace of Acc-ISO in 1982, there is only one other non-generic formulation of ISO that gained approval by the FDA in May 2012. The New Drug Application (NDA) submitted for this formulation has a PK profile that differs substantially from Acc-ISO and the subsequent generic AB-ISO-rated formulations (“branded” generic products). This new formulation (Absorica®, Ranbaxy), which was released into the US marketplace in November 2012, incorporates a specific technology (Lidose Technology), which partially presolubilizes ISO in a lipid matrix that allows for greater gastrointestinal absorption of ISO than other formulations when not administered concomitantly with a high-fat (50gm fat), high-calorie (800-1000 calories) meal (HF/HC meal).60,61 ISO is a highly lipophilic molecule and is categorized as a Class II drug (high permeability, poor solubility) by the FDA and the US Center for Drug Evaluation and Research.61-63 The PK profile of Acc-ISO and all of the subsequent AB-ISO generic formulations were determined by administration with protocol-designated HF/HC meals, and PK studies demonstrated a mean 60.4-percent reduction in bioavailability of Acc-ISO when administered on an empty stomach.40,61-63 By comparison, Lidose-ISO bioavailability was reduced by approximately 33.2 percent when administered on an empty stomach as compared with the HF/HC meal.40,60,61,64 Due to a PK profile that is markedly different from Acc-ISO, the approval process for Lidose-ISO also required submission of new Phase 3 comparative trial data based on treatment of 925 subjects over 12 years of age.65 Details on outcomes from this study are discussed later in this supplement.
As Lidose-ISO and AB-ISO are not bioequivalent based on FDA approval standards, Lidose-ISO is not a generic version of Acc-ISO or any of the AB-ISO formulations. Therefore, AB-ISO should not be substituted for Lidose-ISO if the latter is specified by the prescribing clinician to be dispensed as written and/or medically necessary. Similarly, Lidose-ISO should not be substituted for an AB-ISO, as they are not bioequivalent. These bioequivalency standards are set forth in the Orange Book, which is the FDA-recognized compendia on bioequivalence of drugs, and the ISO prescribing screen shown in the iPLEDGE program clearly depicts these rating distinctions so there should be no confusion by prescribers or pharmacists handling ISO prescriptions for patients (Figure 1).55,56,58,60
Figure 1.
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Welcome iPLEDGE pharmacy screen: “Fill Prescription” stage. Note differentiation of multiple generic isotretinoin capsules from Lidose-isotretinoin (Absorica), which designates the latter as not substitutable.
Early Determination of Dose and Duration of Therapy with Oral Isotretinoin
During the early studies evaluating Acc-ISO prior to its FDA approval in 1982 and over approximately the first decade of use, decisions on daily dosage and duration of ISO treatment were made based on what was known at that time about the pharmacological and PK profiles of ISO, clinical assessments and data captured from a few clinical trials of patients with severe recalcitrant nodulocystic (nodular) acne, tolerability of very commonly reported and/or observed “nuisance” side effects (i.e., xerosis, xerophthalmia, cheilitis, myalgias), frequency and magnitude of laboratory abnormalities (i.e., serum triglycerides, serum transaminase levels), and observations regarding prolonged acne remissions after completion of the initial course of ISO.2,24,25 Education of the medical community regarding the teratogenicity potential of ISO was stressed from the outset.2,4,5,24,25
Daily dose and duration. The early clinical studies that preceded the release of Acc-ISO focused on defining the optimal initial ISO treatment course that would clear severe nodular acne.2,3,27 A dose-ranging study of ISO demonstrated that a low daily dose (0. lmg/kg/day), intermediate daily dose (0.5mg/kg/day), and high daily dose (lmg/kg/day) administered over 20 weeks cleared the vast majority of patients by the end of the active treatment course or within 12 weeks post-therapy in all three daily dosage groups.28 A noneffective dose was not found; nor was there a dose-response pattern observed for efficacy over the initial course of therapy. Although the majority of subjects were treated for 20 weeks (5 months), a subset was treated for 16 weeks (4 months) due to a marked reduction in nodular acne lesions. A major observation from this study was that clearance of nodular acne at the end of the five-month treatment course (short-term treatment success) was documented across a wide daily dosage range (0.lmg/kg/day, 0.5mg/kg/day, lmg/kg/day). Surprisingly over 12 to 18 months of follow-up after completion of the initial course of ISO, prolonged remissions were found to correlate directly with a higher daily dose. A markedly lower need for retreatment at 18 months after completion of the initial course of ISO (long-term treatment success) was shown to be dose-dependent, with the lowest rate of need for retreatment seen in subjects who were treated with lmg/kg/day.28 The duration of therapy in this study was empirically selected as 20 weeks. Therapeutic outcomes from the first pivotal trials with ISO; dose-response data; collective observations on clinical response; and assessments of tolerability, efficacy, and safety led to the recommended dosing in the FDA-approved product labeling for Acc-ISO, which was 0.5mg/kg/day to 2.0mg/kg/day administered twice daily for 15 to 20 weeks.2,54
Dosing frequency. The decision to designate twice-daily (BID) dosing as the FDA-approved daily dosing frequency was based on early limited PK data collected in a small number of subjects, and may be more rational with ≥60 mg/day.2,66-68 Studies showed the PK of isotretinoin can be adequately described using a simple linear model for doses up to 240mg/day.66-68 The decision to designate BID dosing when Acc-ISO was initially FDA-approved in 1982 was at least partially a desire to achieve a steady state plasma level with repetitive ISO dosing that avoided marked fluctuations in peak and trough concentrations (Table 2).67-68 The maximum plasma concentration (Cmax) is higher with a single dose compared to a twice-daily divided dose, and Cmax may potentially affect the side effect profile of ISO.67,68 Whether there are any meaningful differences in either clinical response or the frequency and/or severity of side effects between once-daily and BID dosing is unstudied. Although BID dosing is specified in the FDA-approved labeling with Acc-ISO, AB-ISOs, and Lidose-ISO, once-daily dosing is preferred by some dermatologists who believe this approach improves patient adherence.2,54,60,69-72
TABLE 2.
Explanations for FDA-approved recommendation of twice-daily dosing of isotretinoin
ONCE-DAILY DOSING
Administration of a high-dose ISO (100mg) QD as a single dose produced:
Initial absorption lag time of ≤2 hours; variable peak serum level range (74–511ng/mL) within 1–3 hours
Variable serum levels with elimination half-life range of 11.8–38.5 hours; 53–74% unchanged in feces.
Conclusion: These data may suggest that higher doses of ISO administered QD exhibit a less predictable pharmacokinetic (PK) profile. These data do not provide PK information on QD dosing with a lower daily dose (i.e., 20–60mg).
TWICE-DAILY DOSING
After 25 days of ISO 40mg BID (N=20), serum levels remained stabilized within a narrow range:
No significant changes in the PK profile of ISO after 5 days (steady state).
Switching to ISO 80mg QD at Day 31 after ISO 40mg BID given through Day 30 caused a marked rise in peak ISO serum concentration (Cmax) over the average serum level that was sustained by ISO 40mg BID.
Conclusion: FDA-approved BID dosing of ISO is supported by three main considerations:
1) steady state serum level occurs with repetitive ISO dosing with avoidance of marked fluctuations in peak and/or trough serum levels
2) BID dosing of ISO may minimize reduced GI absorption if there is an absorption ceiling after ingestion due to endogenous (i.e., polymorphism, motility status) or exogenous (i.e., food presence, content) factors
3) ISO administered BID may reduce side effects that may correlate with a higher maximum serum level (Cmax) produced by QD administration of the full daily dose.
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Concomitant food intake. As referred to above, it is important to recognize that due to ISO being lipophilic and categorized as a Class II drug (high permeability with low solubility), gastrointestinal (GI) absorption is enhanced by solubilization of isotretinoin by dietary fat.40,60,61,63,64 FDA designed PK studies show that a high fat (50g)/high calorie (800-1000 calories) diet (HF/HC diet) greatly enhances the absorption of Acc-ISO, with currently available generic formulations (AB-ISOs) also very dependent on high dietary fat for optimal absorption.1,40,60,61,64 In the original PK assessment with Acc-ISO, subjects ingested the drug along with the HF/HC meal under observation, with a recommended dosing frequency of BID.1,2,40,60,61 This specified method of determining the PK profile of Acc-ISO using a HC/HF diet is the same FDA-designated fed state used during evaluation of all subsequent branded generic (AB-ISO) versions of Acc-ISO.54-56,60,69-71 Currently available AB-ISO products are depicted in Table 1.54,60,69-71
The recommendation that ISO be taken with a HF/HC meal has been present since the initial release of Acc-ISO in 1982. However, emphasis placed on educating clinicians, nurses, and pharmacists about the importance of ingesting ISO in the fed state, and details on recommended meal content based on the PK studies (i.e., HF/HC meal) were relatively limited as details about the influence of food on GI absorption of ISO was overshadowed by the plethora of information about clinical results, possible side effects, monitoring, and pregnancy avoidance.2,25,40 As ISO was associated with increased serum lipid levels, especially triglycerides, and as the recognition that minimizing fat in the diet was generally good for health, some clinicians may have suggested to their patients to reduce their fat intake, without recognizing that the GI absorption of ISO would be adversely affected.2,40 A retrospective study (n=13,772) found elevated serum lipid levels in some ISO-treated patients. The incidence of new abnormalities in patients with normal lipid levels at baseline were 44 percent for triglycerides and 31 percent for total cholesterol scores, with other studies showing triglyceride elevations in 16 to 29.3 percent of patients treated with ISO.2,45,25,73,74 The daily dose of ISO, individual patient susceptibility, and dietary factors are all likely to contribute to elevations in serum triglyceride levels.2,73 Interestingly, the importance of ingesting ISO with food was mentioned, but not emphasized in published guidelines on acne treatment, and specific information on meal content (i.e., fat grams) and potential impact on bioavailability was not discussed.25,75 As concomitant ingestion of a HF/HC meal with ISO markedly increases the bioavailability of ISO as compared to ingestion on an empty stomach, this variable, which has not received consistent attention in many clinical studies or in real world practice, may be a potential factor that can modify prolonged acne clearance rates (long-term success) after completion of a course of ISO therapy.40,61
Isotretinoin After Three Decades of Clinical Experience: An Updated Status Report and Recommendations on Optimal Use
Over time, clinical experience coupled with research studies and database analyses have led to the development of updated perspectives and recommendations for optimal dosing of ISO and have further elucidated the safety profile of the drug.
Safety profile. The safety profile of ISO has been thoroughly reviewed elsewhere including commonly encountered side effects (i.e., cheilitis, xerophthalmia, xerosis, myalgias, etc.), teratogenicity, adverse events that are rare (i.e., pseudotumor cerebri), and others that remain controversial.2,4,5,15,24,25,27,39,54,65,73-78 The controversial adverse events that have emerged, such as associations of ISO use with depression, suicidal ideation, and inflammatory bowel disease, have not been definitively substantiated as being caused by ISO, and if a causative association with ISO exists, such reactions are rare and idiosyncratic.2,6-8,10-18 Nevertheless, continued pharmacovigilance is very important with ISO.1,2,15,19-21,25,39,54,60,75 As FDA approval of Lidose-ISO required submission of an NDA, additional safety data was captured in a double-blind, randomized, 24-week Phase 3 study that included 925 actively treated subjects who received Lidose-ISO (n=464) or AB-ISO (n=461).65 The primary objective of this study was to assess if enhanced bioavailability provided by Lidose-ISO increased the risk of adverse events. This study included multiple detailed safety assessments that were not incorporated in previous studies (including pivotal trials) with ISO, such as multiple recognized psychological assessment tools to evaluate for mood changes, depression, anxiety and suicidal ideation/behavior; dual energy x-ray absorptiometry (DEXA) scanning and Z-scores; left hand (wrist) x-rays to evaluate for bone-age (pediatric patients); Tanner staging (pediatric patients); ophthalmologic examinations; and audiology testing (at 25% of study sites), with no safety signals identified and no major or relevant differences noted between the two study groups.65 Psychiatric assessment of patients enrolled in this study was very thorough and included the Structured Clinical Interview for DSM-IV Clinical Trials (SCID-CT) to screen for current or lifetime major depression, mania, and/or psychosis; the Patient Health Questionnaire-8 (PHQ-8) to monitor for changes suggestive of depressive mood disorders; the Columbia-Suicide Severity Rating Scale (C-SSRS) to monitor for suicidal ideation/behavior; the Generalized Anxiety Disorder-7 (GAD-7) to evaluate for mental health changes related to anxiety; and psychosis questions to assess for emergence of psychotic symptoms.65
Consistent clinical follow up and laboratory testing remain as important means of monitoring both the therapeutic benefit and possible side effects of ISO.2,24,25,39,75 Elevations in serum lipids (especially triglycerides) and serum transaminases are not uncommon, however, they are almost always modest in magnitude and usually not clinically relevant.2,73 The majority of patients complete ISO therapy for acne with highly favorable therapeutic outcomes and with absence of clinically significant adverse events beyond the anticipated “nuisance side effects,” such as dry skin, dry eyes, and dry Hps.2,24,25,27,55,74,75 However, clinicians are encouraged to evaluate and follow up on any marked abnormalities or suspicious trends in laboratory results so that rare adverse events such as pancreatitis or symptomatic hepatitis can be averted. The requirements of the iPLEDGE REMS program mandate monthly office visits for all patients treated with any formulation of ISO.19-21 In women of child-bearing potential, a monthly pregnancy test is mandated with the requirement that the pregnancy result is negative and the prescription is procured within a seven-day window after the last office visit.19-22
Skin care. In order to improve the overall tolerability of ISO, dryness and scaling of the skin are best managed preemptively by having patients incorporate topical barrier repair from the outset rather than after the xerotic skin changes emerge. ISO, as with other retinoids, causes corneocyte dyscohesion that leads to increased water loss, dryness, and scaling. Interestingly, most dermatologists reported that they wait until xerotic skin changes emerge before recommending application of a moisturizer/barrier repair product.48 A more optimal approach would be to prevent or reduce these changes by pre-emptively using this approach once ISO is started. Use of a gentle skin cleanser is also suggested.
Patient populations. ISO is indicated for and used predominantly in patients with severe nodular and recalcitrant inflammatory acne and has remained for over three decades as the most effective therapeutic option for acne.2,24,25,75 In the dermatology community, it is also recognized that ISO is an important and highly effective option in many patients with refractory inflammatory acne that does not exhibit multiple nodules, especially in patients who demonstrate a known propensity for acne scarring. As ISO is recognized as the single most significant advance in acne therapy, its use has selectively expanded to patients with recalcitrant nonnodular inflammatory acne.25,37-39,79,80 Current consensus is that appropriate use of ISO includes patients with nonnodular acne if improvement has been less than 50 percent after six months of using a regimen that has included oral antibiotic treatment, if there is significant acne scarring or psychological distress.80 It has been suggested that use of ISO be initiated “sooner rather than later” in many acne patients, rather than being reserved as a treatment of last resort.79 Use of ISO in selected cases of recalcitrant non-nodular inflammatory acne is a well-accepted approach within the dermatology community with the caveat that the clinician document the relevant patient history, subjective complaints, and clinical findings that support this approach in each case.
A review of recent guidelines on acne management and the use of ISO appears in Table 3,25,81-84 While ISO has been typically reserved for use only in severe and recalcitrant cases of nodular acne, a position statement from the AAD has also defined a more feasible scope of use for ISO in acne: “The Association recognizes there is sufficient evidence for the use of isotretinoin in severe forms of acne, particularly (but not limited to) severe recalcitrant nodular acne or acne which has proven refractory to other forms of therapy. Assessment of severity includes the impact of the disease on the patient, both physical and psychological”.37 Guidelines on acne management from Europe, South Africa, Asia, and Australia are in good general agreement with the position statement of the AAD.37-39-81-83
TABLE 3.
Acne vulgaris management guidelines: Summary of publications including recommendations on use of isotretinoin
| SOURCE | PUBLICATION | COMMENTS |
:---:
| GENERAL POPULATION (Teenagers, Adults) |
| European Evidence-Based Guidelines for Treatment of Acne81 | Journal of the European Academy of Dermatology and Venereology (2012) | Not recommended for comedonal acne. Not generally recommended for non-nodular inflammatory (papulopustular) acne |
| Guidelines of Care: Acne Vulgaris Management (American Academy of Dermatology)75 | Journal of the American Academy of Dermatology (2007) | Unanimous agreement that ISO is also useful for lesser degrees of acne that resist treatment or produce physical or psychological scarring |
| Acne Guidelines (Global Alliance to Improve Outcomes in Acne)82 | South African Medical Journal (2005) | Recognized as most effective anti-acne treatment, but should be reserved for severe cases, moderate but unresponsive cases, and acne with psychological distress |
| Consensus on Evidence- Based Practice in Acne (Asian Working Group)83 | The Journal of Dermatology (2011) | Recommended as primary therapy for severe acne and second-tier treatment for moderate acne that is poorly responsive to other treatments; suggests a target cumulative dose of 120–150mg/kg be reached over the treatment course |
| PEDIATRIC POPULATION |
| Evidence-Based Recommendations for Pediatric Acne (American Acne and Rosacea Society and American Academy of Pediatrics)84 | Pediatrics (2013) | Recommended starting dose of 0.5mg/kg/day for the first 4 weeks, increasing to 1mg/kg/day. Recommended uses include severe nodular acne, acne with scarring, and/or refractory inflammatory acne. Extensive counseling suggested, particularly with respect to pregnancy prevention in girls. |
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Duration of remission after isotretinoin therapy. ISO commonly induces a prolonged remission that can be permanent in some patients.29 However, the need for acne retreatment, and in some cases one or more additional courses of ISO, have been reported in multiple publications along with data supporting a direct correlation between longer durations of remission after the initial course of ISO and achievement of a threshold range in total cumulative ISO dose administered (>120-150mg/kg).2,28-36,40-45 In occasional cases, the natural course of acne in an individual patient may be the cause for acne regression and ultimate remission.28 Nevertheless, ISO does frequently “shut off acne lesion development in many treated patients, although recurrence of acne and need for subsequent treatment is not uncommon.29,40 Several factors have been suggested as increasing the likelihood that acne will recur after a course of ISO therapy and will require further acne treatment, including additional courses of ISO therapy in some patients. Suggested factors associated with greater risk of acne recurrence and need for retreatment include younger age when first given ISO, male gender, marked truncal involvement, sinus tracts, macrocomedones, and androgen excess (i.e., polycystic ovary disease).2,28-30,32,36,40,42-44,80,85 Importantly, acne that has previously been treated with ISO typically responds well to subsequent courses of treatment.25,27,29,30,34 Recurrence of acne after a previous course of ISO and need for acne retreatment are discussed in more detail below.
The cumulative clinical experience over several years helped to shape what recognized authorities in the field of acne suggested as the optimal approach to administering ISO. In addition to defining daily dose and duration ranges, a threshold range of cumulative ISO exposure over the course of therapy had become a widely held standard based on the collective opinion of multiple authorities in dermatology on ISO therapy.2,25,29,30,32,40,75,83
Commonly recommended starting dose. Although a dosage range is published for ISO, the optimal recommended starting daily dose of ISO is 0.5mg/kg/day, which is increased to lmg/kg/day, as tolerated, usually after four we eks.2,24,25,27,29,65,84 A typical course might involve one month at 0.5mg/kg/day, increased to lmg/kg/day, which is maintained over the ensuing treatment course unless adverse clinical effects or laboratory changes warrant adjustment of therapy.2,65,84 This daily dosing approach of 0.5mg/kg/day for one month followed by lmg/kg/day for four months applies to Acc-ISO, AB-ISOs, and Lidose-ISO and achieves the target cumulative dose of >120 to 150mg/kg if ISO is administered over the maximum duration of treatment (20 weeks [5 months]) indicated in the FDA-approved package inserts for all ISO products.2,25,40,54,60,65,69-71 The observation that some patients experience marked initial inflammatory acne flares after starting ISO led to the recommendation of starting with a lower dose (i.e., 0.5mg/kg/day) to avert such flares in the first month of treatment, with the subsequent daily dose increased to lmg/kg/day as long as this dose escalation is well tolerated.2,24
Dosing frequency. The FDA-recommended dosing frequency has remained as twice daily, with no additional PK studies completed that have changed this recommendation and with no studies adequately assessing BID versus daily dosing with regard to efficacy and safety2,54,60,65,69-71 As discussed above, early studies of the PK profile of ISO demonstrated linear pharmacokinetics, which did not change with multiple doses over time at steady state, and with repetitive twice-daily dosing of ISO shown to sustain stable steady serum levels.67,86
Recommended course of therapy. The recommended dosage regimen of ISO incorporates the suggested daily dose and also the target cumulative dose achieved over the treatment course.2,29-36,39-41,79,81,82 Based on both research and clinical practice with ISO, it is recommended that ISO be given at a dose of 0.5 to lmg/kg/day, administered BID, over a duration that achieves a cumulative dose of 120 to 150mg/kg.2,40 The latter component on duration of therapy to be administered over the initial course of ISO is based on analyses that show that achieving this target cumulative dose maximizes the likelihood of obtaining long-term remission without the need for retreatment of acne. In patients who experience nuisance side effects that they have difficulty tolerating (i.e., dry skin, dry eyes, severe cracking of lips, muscle aches), a lower daily dose can be administered over a longer duration until the cumulative dose of 120 tol50mg/kg is reached; however, it is optimal to administer the complete course of therapy within 20 weeks if possible without sacrificing the target cumulative dose (Figure 2).
Figure 2.
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Target cumulative total dose using the generally recommended isotretinoin dosage regimen. The recommended target over a course of isotretinoin (ISO) therapy is a cumulative total dose of 120-150mg/kg. As each prescription of ISO under iPLEDGE is for up to 30 days, when prescribing ISO, 1 month=30 days. Using a 70kg patient as an example, the cumulative target ISO dose to reach 120mg/kg is 8400mg (120[mg] x 70[kg]=8400mg). Starting with 0.5mg/kg/day for the first 30 days calculates to a daily dose of 35mg/day over the first 30 days. This patient was started at 40mg/day for 30 days which delivered a total of 1200mg (40[mg] x 30[days]=1200mg) during that first month of ISO therapy. At the next monthly visit, the clinician chose to increase the dose to 1 mg/kg/day which calculates out to 70mg/day. For ease of administration with available ISO capsule sizes (10mg, 20mg, 30mg, 40mg), this patient was given 60mg/day (0.9mg/kg/day) which was continued at subsequent monthly visits as it was well tolerated and all iPLEDGE requirements were met each month. The patient was clear of their severe acne after 3 months of therapy; however, ISO was continued as the target cumulative ISO dose had not yet been reached. After four additional months at 60mg/day, the patient received 7200mg (60[mg] x30[days]=1800[mg/month] x4[months]=7200mg). Therefore, after 5 months of using the mg/kg/day as described above, the 8400mg cumulative total dose target was met (1200mg [first month] + 7200mg [next 4 months]=8400mg [total dose over course of therapy]). At this point, ISO therapy was stopped as the target cumulative total dose of ISO was reached.
Alternative dosing approach: High-dose isotretinoin. It has been suggested that using higher daily doses of ISO than what is typically prescribed (> lmg/kg/day) and achieving a higher total cumulative dose of ISO than what is generally recommended (>150mg/kg) produces a lower risk of acne relapse, although well-controlled comparative studies or data analyses have not been completed. In addition, the use of these higher doses of ISO did not compress the duration of therapy with the goal of achieving a cumulative total dose of 120 to 150mg/kg over a shorter duration; instead they achieved cumulative total doses of ISO that were substantially higher without shortening of the duration of therapy.74,87
A retrospective analysis of ISO-treated patients with nodulocystic acne (N=80) found that high-dose ISO (≥1.3mg/kg/day) produced 100-percent total clearance, with 12.5 percent requiring an additional course of ISO within three years of follow-up.87 The average cumulative total dose of ISO received during the initial course was 290mg/kg. In this study, ISO was reported to be safe and well tolerated with no patient discontinuing therapy because of adverse events.87 In a prospective observational study, patients with treatment-resistant acne (N=116) were treated with ISO until they were clear for at least one month while on ISO therapy using daily doses determined by the prescribing clinician.74 Follow-up was completed by survey after 12 months to evaluate the acne relapse rate that required treatment other than ISO (i.e., topical agents, oral antibiotic) and the number of patients who underwent retreatment with ISO. The patient groups were divided based on the cumulative total dose into those receiving <220mg/kg (group A, n=38) and those receiving >220mg/kg (group B, n=78), with the average duration of the initial ISO course reported to be 6.3 months. At the 12-month follow-up point, acne relapse was reported in 47.4 percent in group A and in 26.9 percent in group B. Two patients (1.72%) required retreatment with ISO, both from group B. Safety assessment revealed the anticipated side effects known to occur with ISO use (i.e., cheilitis, xerosis) in all patients, with “rash,” arthralgias, and minor elevations in laboratory parameters (i.e., triglycerides) trending higher in group B.74
Initial clearance versus need for retreatment. As emphasized above, it is important to recognize that short-term treatment success, defined as complete clearance of acne at the end of the treatment course, can occur across a broad range of ISO daily dosages, with clinical studies including both higher (≥ lmg/kg/day) and lower daily doses (0.1-0.5mg/kg/day) in patients with severe nodular inflammatory acne.28,40 Therapeutic responses to ISO at the end of a 20-week course are relatively comparable when using a low (0. lmg/kg/day), an intermediate (0.5mg/kg/day), or a higher (l.Omg/kg/day) daily dose from the outset.5,15,16 Multiple studies that have followed ISO-treated acne patients over time have shown that the recurrence of acne, the need to restart some form of acne treatment, and the need for repeat treatment with ISO after the initial course correlated directly with a lower daily dose and a lower cumulative total dose over the initial course of therapy in patients followed for durations of 18 months to five years.2,28,40 This suggests the need to achieve a threshold level of cumulative systemic exposure to ISO during the course of therapy in order to maximize the duration of acne remission.28-30,40,80-82Table 4 includes the results of two studies that assessed the rate of recurrence of acne after the initial course of ISO and subsequent need for retreatment with ISO. This data was correlated with mg/kg/day used during the initial course of ISO therapy. The frequency of individual patients treated with more than two courses of ISO is also shown in Table 4.
TABLE 4.
Acne recurrences after initial course of isotretinoin therapy requiring retreatment with isotretinoin: correlation with daily dose used during initial course
| | ISOTRETINOIN LOW DAILY DOSE (0.1MG/KG/DAY) | ISOTRETINOIN INTERMEDIATE DAILY DOSE (0.5MG/KG/DAY) | ISOTRETINOIN HIGHER DAILY DOSE (1.0MG/KG/DAY) |
:---: :---: |
| Therapeutic response at end of initial course of isotretinoin28 | Similar positive results for all groups with 20 weeks of treatment (most completely clear) |
| Patients (%) with recurrence of acne within 18 months of initial course of isotretinoin requiring retreatment with isotretinoin28 | 42% | 20% | 10% |
| Patients (%) requiring >2 courses of isotretinoin within 60 months34 | 88% (includes also group receiving 0.5mg/kg/day) | 88% (includes also group receiving 0.1mg/kg/day) | 9.5% (1.0 mg/kg/day group only) |
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Initial therapeutic response not strictly dependent on daily dose. Duration of remission correlates directly with daily dose and magnitude of cumulative total exposure to isotretinoin over the initial course of therapy.28,34
Note that recurrence of acne at some time point after an initial course of ISO is not always of the same or worse severity as before ISO treatment. Acne recurrence can vary in both severity and the predominant types of acne lesions.28,36,40 Multiple publications of ISO-treated acne patients (N=1,411) have reported a range of recurrence rates of acne after completion of the initial course of ISO, which included different daily doses and durations of ISO therapy and follow-up periods ranging from 12 months to 10 years.25,28-36,40,41,44,45 Collectively, and with consideration of variations in daily dosing regimens and follow-up periods, in patients who experienced recurrence of acne after completing an initial course of ISO, 16 to 21 percent were retreated with topical therapy alone, 3.3 to 39 percent were retreated with topical therapy and oral antibiotics, and 16 to 23 percent were retreated with at least one repeat course of ISO.2,28-36,40 Most of the published database analyses, which address acne recurrence rate, need for acne retreatment, and need for additional courses of ISO demonstrated that rates for all three of these parameters are markedly higher in severely affected acne patients treated with a lower cumulative total dose of ISO during the initial course.28,36-40 Analysis of these reports led to the recommendation that a course of ISO should reach a cumulative total dose of 120 to 150mg/kg to reduce the risk of acne recurrence and the need for acne retreatment including additional course(s) of ISO.2,28-36,40,41,44,45,81,82
A database analysis inclusive of ISO-treated acne patients (n=179) examined the relationships between the cumulative total dose of ISO achieved during the initial course of therapy and acne recurrence, the need for restarting acne therapy, and the need for retreatment with ISO over three years of follow-up.35 Overall, two-thirds (65.4%) experienced recurrence of acne. The risk of acne recurrence was eight-fold greater in patients treated with a cumulative total dose of ISO <100mg/kg as compared to those receiving >100mg/kg. Acne recurrence was documented in 92 percent of patients treated with a cumulative total ISO dose of <90mg/kg and in 40 to 50 percent of patients receiving >110mg/kg. This analysis also showed that 61 percent received retreatment for acne, with 22.9 percent requiring at least one additional course of ISO.35
Another case-based analysis of ISO-treated acne patients (N=193) with a follow-up period up to 10 years post-treatment also assessed correlations between the cumulative total dose of ISO obtained during the initial course of therapy and both the acne recurrence rate and the need for subsequent acne treatment.31 Topical therapy alone and topical plus oral antibiotic therapy were subsequently initiated in 17.5 percent and 3.3 percent of patients after their initial course of ISO for recurrence of acne, respectively, with a second course of ISO given to 19.6 percent of patients with acne recurrence. The cumulative total dose of ISO achieved by patients who required further therapy for acne was 103.5mg/kg administered over a duration of 6.7 months. The group of patients who did not experience acne recurrence after their initial ISO course had received a cumulative total dose of 118.5mg/kg administered over 7.41 months.31
Another database analysis of acne patients (N=88) treated with a course of ISO followed for up to 10 years revealed that acne recurred in 82 percent of those treated with <120mg/kg cumulative total dose as compared to 30 percent in those treated with higher cumulative doses during their initial course of ISO; most acne recurrences occurred within the first three years, with 78 percent emerging within 18 months.30
Although there are a few reports of successful therapeutic outcomes with low-dose ISO therapy for recalcitrant inflammatory acne that achieved substantially less than 120mg/kg cumulative total dose (i.e., mean 81 mg/kg cumulative total dose), the majority of available evidence from multiple published reports suggest that such an approach markedly increases the likelihood of acne recurrence and need for retreatment.40,88,89 It appears that it is important clinically to compensate for the lower daily dose by extending the course of therapy in order to optimize the cumulative total exposure to ISO over the treatment course.2,24 One dose-finding analysis that followed ISO-treated acne patients (N=299) for up to five years noted that acne recurred in 69 percent of the subset treated initially with a daily dose of 0. lmg/kg/day over a duration of 16 weeks.34 In addition, in patients treated for a duration of 16 weeks, more than two courses of ISO were needed in 88 percent of patients receiving lower daily doses of ISO (0.1—0.5 mg/kg/day) compared to 9.5 percent in the group receiving a higher daily dose (lmg/kg/day).34 It may be clinically appropriate to utilize a daily dose of ISO that is lower than usual (based on mg/kg), especially when adjustment in dosage is needed to circumvent certain side effects (i.e., xerophthalmia, severe xerosis and/or cheilitis, myalgias). However, completing the course of therapy within the maximum FDA-approved duration (20 weeks) is advantageous whenever possible, as long as the threshold cumulative dose is achieved, clearance of acne occurs, and the treatment is well-tolerated. The use of a lower daily dose with a longer duration of therapy (in order to reach the threshold cumulative total dose) exposes the patient to a more prolonged time period in which idiosyncratic adverse events may emerge, and in women of child-bearing potential, a greater period of time in which pregnancy exposures can occur.
An analysis using data from IMS Health depicted prescription patterns of patients newly treated with ISO during 2008 to 2009.90 Among 28,840 patients who received an initial course of ISO over a duration of at least four consecutive months, 17 percent received four months, 39 percent received five months, 26 percent received six months, and 18 percent received seven or more months consecutively as their initial treatment course of ISO (Table 5). A mixiimum 56-day interval between prescriptions was used to define the time gap separating ISO treatment courses (with the usual gap reported to be 120 days). Data on patients treated with additional courses of ISO over a 12-month follow-up period after completion of the initial course are depicted in Table 6. Overall, 10.6 percent of patients received a second course of ISO, with male patients more likely to receive subsequent ISO treatment than female patients (13.8% vs. 9.0%, chi-square= 164.35, p<0.0001). A limitation of this data analysis is the lack of information on daily dose and total cumulative dose of ISO used in the initial treatment courses. Nevertheless, these data reflect “real world usage patterns” of ISO in the United States approximately 25 years after Acc-ISO became available for use and support the observations that initial courses longer than five months and repeated therapy with ISO are common.
TABLE 5.
Initial course of isotretinoin: Length of therapy based on number of prescriptions [Rx] derived from IMS prescribing data
| LENGTH OF THERAPY | ALL PATIENTS | FEMALE PATIENTS | MALE PATIENTS |
:--- :--- |
| 1 Rx | | | |
| 2 Rx | | | |
| 3 Rx | | | |
| 4 Rx | 17% | 17% | 17% |
| 5 Rx | 39% | 40% | 38% |
| 6 Rx | 26% | 26% | 27% |
| 7 Rx | 10% | 10% | 10% |
| 8+ Rx | 8% | 7% | 8% |
| TOTAL | 28,840 | 15,656 | 13,182 |
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A total of 44% of all patients went beyond the FDA-recommended maximum of 5 months duration of treatment 43% females; 45% for males
TABLE 6.
Indexing of length of initial course of isotretinoin (based on number of prescriptions [Rx]) to need for second course of isotretinoin (12 months follow-up after initial course of isotretinoin)
| ISOTRETINOIN COURSE NUMBER AND LENGTH OF THERAPY |
| LENGTH OF THERAPY | INITIAL COURSE | 2nd COURSE | INITIAL COURSE | 2nd COURSE | INITIAL COURSE | 2nd COURSE |
| | ALL PATIENTS | FEMALE PATIENTS | MALE PATIENTS |
| 1 Rx | | | | | | |
| 2 Rx | | | | | | |
| 3 Rx | | | | | | |
| 4 Rx | 4,958 | 13.7% | 2,696 | 11.8% | 2,262 | 15.9% |
| 5 Rx | 11,237 | 9.6% | 6,220 | 7.4% | 5,017 | 12.3% |
| 6 Rx | 7,587 | 9.7% | 4,080 | 7.9% | 3,505 | 11.7% |
| 7 Rx | 2,796 | 10.7% | 1,495 | 8.8% | 1,301 | 12.8% |
| 8+ Rx | 2,262 | 12.1% | 1,165 | 10.0% | 1,097 | 14.3% |
| TOTAL | 28,840 | 10.6% | 15,656 | 8.6% | 13,182 | 13.0% |
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Initial course of isotretinoin is aligned to Length of Therapy column. Percentages use initial (1st) course as denominator.
Prescription data obtained from IMS Health
Summarized conclusions from available data from multiple publications and data analyses evaluating clinically relevant relationships between isotretinoin dosing and both short-term treatment success and long-term treatment success are outlined in Table 7.
TABLE 7.
Clinical correlations of isotretinoin dosing regimens and therapeutic outcomes
| DEFINITIONS OF TREATMENT SUCCESS WITH ISOTRETINOIN USED FOR TREATMENT OF SEVERE INFLAMMATORY ACNE AND REFRACTORY ACNE |
| Short-term Treatment Success | Acne clearance at completion of initial course of isotretinoin |
| Long-term Treatment Success | Prolonged duration of acne clearance over a defined period of follow up after completion of the initial course of isotretinoin treatment |
| Short-term treatment success is not highly dependent on the daily dose of isotretinoin (based on mg/kg) with no significant differences in acne clearance between 0.1mg/kg/day, 0.5mg/kg/day, and 1mg/kg/day (over treatment duration of 20 weeks). |
| Long-term treatment success is highly dependent on achieving a specific threshold defined as the cumulative total dose of isotretinoin over the initial course of therapy. The recommended target cumulative dose of isotretinoin that needs to be reached to optimize long-term treatment success is 120–150mg/kg. |
| Both the daily dose (based on mg/kg) and the duration of the treatment course need to be coordinated in order to reach the target cumulative dose. |
| The generally recommended isotretinoin dosage regimen is 0.5mg/kg/day for 4 weeks followed by 1mg/kg/day thereafter. With this approach, the target cumulative dose would be reached in 5 months (Figure 2). If the daily dose is lowered (based on mg/kg), the duration of the treatment course would need to be extended by the amount that allows for reaching the target cumulative dose |
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The target cumulative dose was based on thorough review of predominantly retrospective data from multiple sources by dermatology leaders in the field of acne and global clinical experience as reported here in references used for this publication.28-36,41,44,45
Current perspectives on isotretinoin dosage regimen. The generally recommended dosage regimen to use as a starting point when prescribing ISO is to initiate therapy with 0.5mg/kg/day for the first four weeks followed by a subsequent increase to lmg/kg/day provided there are no tolerability reactions or side effects that interfere with this approach. This regimen is continued until a cumulative total ISO dose of 120 to 150mg/kg is reached. The FDA-approved dosing frequency is BID. However, many clinicians state that they recommend QD dosing to improve patient adherence due to concern that the second dose is likely to be missed on many occasions.72 In a survey of patients currently using ISO for acne (age range 14-20 years), 37/52 (71.1%) and 15/52 (28.8%) stated they were taking ISO QD and BID, respectively, although the daily dose each patient was taking was not captured.91
While many patients treated with ISO experience rapid improvement within the first 1 to 2 months, marked flaring within the first month, presenting as multiple deep inflammatory papules and nodules, was a common anecdotal observation among dermatologists when ISO treatment was initiated at ≥lmg/kg/day.2,24,25 Although such flares could be managed without interfering with completion of ISO therapy, the reduced starting dose of 0.5mg/kg/day was suggested to avert the potential for these initial inflammatory flares and anecdotally has proven to be highly successful overall.2
A subanalysis of the IMS-derived prescription data suggests that patients who received initial ISO treatment for acne at a younger age (age subsets 10-11 yrs, 12-14 yrs) were more likely than older patients (age subsets 15-17 yrs, 18-29 yrs, 30-44 yrs, 45+ yrs) to receive two or more treatment courses of ISO (Cochrane-Armitage trend test Z=16.32, p<0.0001) (Figure 3).90 Limitations of this data analysis include the wide variability in numbers of patients in each subset and lack of information on the daily doses, durations of therapy, and cumulative total doses that were administered. Nevertheless, these data suggest what others have observed regarding patient age and need for retreatment with ISO.42
Figure 3.
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Patients (%) receiving two or more courses of treatment based on age
These data, which demonstrate variability in the durations of the initial course of ISO therapy, show the need for greater understanding of the reasons for “longer-than-recommended treatment durations” (based on the package insert), such as low daily dosing, tolerability of nuisance side effects, and/or challenges with patient adherence to treatment recommendations, as well as the need to better unerstand the potential impact of treatment on patients.18
Adherence Considerations with Isotretinoin Therapy
Patient adherence. Patient adherence can be challenging with any type of treatment, especially in patients using long-term therapy for chronic conditions. The Global Alliance to Improve Outcomes in Acne reported an overall adherence rate of 46 percent for all treatments used by patients with acne (n=707).92 It is not known why patients may fail to adhere to ISO therapy; perhaps the typical initial response to treatment causes them to think they do not need to keep taking the drug.
Suboptimal adherence may also include not taking ISO as directed with food, or with the type of meal suggested, despite being instructed to do so. This is a likely consideration based on data reported on the eating habits of adolescents. In a survey of 1,001 high school students with a mean age of 16.1 years, 59 percent indicated that they skipped breakfast more than three times the previous week.93 In another national report from 1991, 30 percent of students aged 15 to 18 years skipped breakfast on any given day.94 Common reasons cited for skipping breakfast included lack of time, lack of hunger, or dieting to lose weight. Skipping breakfast and erratic ingestion of meals were more prevalent in girls and in older children and adolescents.
More recently, interim results from a project in progress that is surveying adolescents currently using ISO included dietary information from those patients who ingested ISO with food on the day they were surveyed (n=52).91 The average fat grams ingested with ISO was 18 grams (low 2g; high 53g; median 16g) and the average calories ingested with ISO was 483 calories (low 48 calories, high 1,220 calories, median 444 calories). Hence, it is not likely that many patients on ISO will ingest the drug with the amount of fat grams and calories used in the HC/HF diet that was required during the PK studies with Acc-ISO.
Adherence with recommendations on concomitant ingestion of isotretinoin with food or a specified meal. As reviewed above, the recommendation to ingest ISO with food, especially a HF/HC meal, was not emphasized for many years after the drug was released in 1982. Patients may fail to take ISO with food because clinicians may prescribe the drug without providing clear instructions to the patients on ingestion with food including the type of meal, or they may incorrectly assume that the pharmacy will instruct the patient to take the drug with food. Even if clear directions are given and understood, patients may dismiss the instructions or forget them over the course of treatment, especially if they are not repeatedly emphasized.
One of the difficulties related to suggesting ingestion of ISO with food is the HC/HF meal itself that was used to evaluate the PK profile of Acc-ISO and the subsequently approved AB-ISO products. It is not realistic that patients would ingest the HC/HF with each dose of ISO throughout the course of therapy or that many clinicians would feel comfortable recommending that patients consume such an unhealthy diet.2,40 The marked decrease in bioavailability when ISO is ingested on an empty stomach clearly reduces true systemic ISO exposure, and the bioavailability of ISO after ingestion with meals containing lower quantities of fat and calories is not known. This brings to light a highly relevant and realistic question: How much does concomitant food ingestion, including fat and calorie content, influence the likelihood of acne recurrence and need for retreatment with ISO? The complete answer to this question is not known; however, the mean magnitude of the decrease in bioavailability (approximately 60%) when Acc-ISO and AB-ISOs are ingested on an empty stomach as compared with a HF/HC meal is substantial, thus warranting consideration of the impact of this reduced bioavailability on the therapeutic outcome, especially long-term treatment success.40,54,59-61,69-71
Interim results obtained from adolescents who did ingest ISO with food (n=52) showed that both the fat and calorie content were markedly lower than what was mandated in the PK study protocols with ISO.91 Results from this patient survey including adolescent patients receiving ISO for acne (n=65), which addressed questions about recommendations given on food intake with the drug, showed that 74 percent were instructed to take ISO with food, 71 percent reported actually ingesting ISO with food, 14 percent were instructed to take ISO with a high fat meal, and three percent reported actually ingesting ISO with a high-fat meal.91 These results further emphasize the difficulties with adherence to specific dietary recommendations with ingestion of ISO.
Bioavailability of Lidose-Isotretinoin and Potential Impact on Treatment Success
Lidose-ISO has been shown to markedly improve the bioavailability of ISO when taken without food as compared to Acc-ISO and AB-ISO.40,60,61,65 Over the years of development of Lidose-ISO, multiple bioavailability and bioequivalence studies have evaluated the PK profiles of Lidose-ISO (7 studies; N=248) and Lidose-ISO versus Acc-ISO (6 studies; N=262).95 A four-way crossover study evaluated the PK profiles of Lidose-ISO versus Acc-ISO in both the fed (HC/HF meal) and fasted states.61 Ingestion of Lidose-ISO and Acc-ISO with the HF/HC meal (fed state) results in equivalent bioavailability. Based on area-under-the curve (AUC) data in both the fed and fasted states, the difference that is relevant is that on an empty stomach , Lidose-ISO achieves higher relative blood levels and retains 66.8 percent of its bioavailability as compared to Acc-ISO which retains 39.6 percent of its bioavailability (Figure 4).54,60,61 The clinical consequence of these PK differences are not as likely to modify short-term treatment success, which has not been shown to be highly dose-dependent. However, as long-term treatment success is highly dependent on cumulative total drug exposure, the marked reduction in ISO bioavailability due to absence of food, especially a meal high enough in calories and fat grams, is likely to be an important mitigating factor that can increase acne recurrence and need for retreatment. Although there is currently no comparative study between Lidose-ISO and Acc-ISO (or an AB-ISO) that evaluated long-term treatment success, this clinical consideration is strongly suggested by the available PK data and multiple reports reviewed earlier that have demonstrated the importance of reaching a target cumulative dose of ISO over the course of treatment.
Figure 4.
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Effect of fed and fasted states on the bioavailability of specific isotretinoin (ISO) formulations. Comparison of Lidose-ISO and Acc-ISO pharmacokinetic profiles
The magnitude of PK differences in bioavailability between Lidose-ISO and other ISO formulations (Acc-ISO, AB-ISOs) is also demonstrated in Figure 5. This Figure compares the single-dose PK profiles of Lidose-ISO 40mg (1 capsule), Acc-ISO 80mg (2 x 40mg capsules), and Acc-ISO 40 mg (1 capsule).54,60,61 Note the administered dose of the Acc-ISO that is two-fold higher than Lidose-ISO in this PK comparison. In the fed state (HC/HF meal), the Lidose-ISO 40mg achieved 67 percent of the bioavailability (AUC) achieved by 80mg of Acc-ISO. Importantly, in the fasted-state, PK data shows that Lido-ISO 40mg is 1.2-fold more bioavailable (AUC) than twice the dose (80mg) of Acc-ISO, demonstrating the markedly higher dependency of Acc-ISO (and AB-ISOs) on ingestion in the designated fed state. If an acne patient ingests Acc-ISO (or AB-ISO) in the fasted state throughout the usual course of therapy (4 to 5 months), their cumulative systemic exposure to ISO would be reduced by one-third to one-half.40,54,60,61A very important caveat is that this data should not suggest to the clinician that they can utilize a lower dose or shorter duration of the therapy with Lidose-ISO due to greater bioavailability. Taking this approach may diminish the potential for long-term success by reducing total cumulative ISO exposure. The purported advantage of Lidose-ISO is to maximize the amount of true cumulative systemic ISO exposure over the course of treatment provided by its greater bioavailability in the absence of a HF/HC meal. Based on historic data with ISO, this greater systemic ISO exposure is likely to increase the likelihood of long-term treatment success.
Figure 5.
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Fed vs. fasting bioavailability comparisons. Individual (single) dose bioavailability of isotretinoin products. Acc-ISO 80mg vs. Acc-ISO 40mg vs. Lidose-ISO 40mg1-3
Webster GF, Leyden JJ, Gross JA. J Am Acad Dermatol. 2013 Nov;69(5):762-767.
Absorica precribing information, Fianbaxy Laboratories, November 2012.
Accutane prescribing information, Nutley, NJ, Roche Laboratories, 1982.
Fed—high fat (50g)/high calorie (800-1,000 calories) meal; Fasted—empty stomach
‘Single dose bioavailability based on mean area-underthe curve (AUC) data with Lidose-ISO and Acc-ISO; all generic ISO (AB-rated ISO products) are based on Acc-ISO pharmacokinetic data.
Lidose-ISO—Absorica; Acc-ISO—Accutane
Summary Update of the iPLEDGE Program
Introduced in 2006, the iPLEDGE program is an FDA-mandated computer-based risk management program that provides documented, trackable links to prescribers, pharmacists, manufacturers, and patients.19-24 All patients who take ISO must participate in iPLEDGE in order to get the drug, including men and women who are not of childbearing potential. Women of childbearing potential must complete pregnancy testing before and during ISO therapy within required time intervals in order to get access to ISO. The iPLEDGE program requires one negative pregnancy test 30 days before commencing the prescription and then another negative pregnancy test during a seven-day window during which time the drug must be dispensed to the patient. Thus, two negative pregnancy tests are required for women of childbearing potential to start ISO therapy. If the drug is not dispensed within the seven-day window, the patient is “locked out” and must wait another 12 days to submit a new negative pregnancy test and open a new seven-day window period. This aspect of the program has been criticized as being burdensome for the healthcare system and disruptive of the patient’s therapy; however, it also places responsibility on the patient to adhere to the program.96
In addition to requiring pregnancy tests and monitoring prescriptions, iPLEDGE also provides monthly patient education and quizzes patients about topics, such as never sharing their medications, not donating blood, and forms of birth control. Women of childbearing potential in the iPLEDGE system must use two forms of birth control (abstinence can be considered as a form of birth control) that are recognized on the lists provided in the iPLEDGE Birth Control Workbook.97
An early goal of iPLEDGE was to create a centralized pregnancy registry that could be used potential voluntary root-cause analysis of all ISO-exposed pregnancies. This required a technical infrastructure capable of registering patients, collecting and analyzing pregnancy test results, and verification of patient qualifications. Part of the program includes proactive compliance monitoring and actions. The general path through iPLEDGE is that negative pregnancy tests are verified and entered into the system by the prescriber, the same two forms of birth control have to be entered by the prescriber and the patient, the prescriber has to verify patient counseling, and the patient has to complete the comprehension questions before the pharmacist is allowed to dispense the drug to the patient. The system has automatically served to remind prescribers and patients about the pregnancy tests. The system was also designed to flag discrepancies (e.g., when the prescriber and patient differed in terms of what forms of birth control were being used) and such discrepancies must be resolved before the patient could receive isotretinoin. If an expected pregnancy test was missed, the prescriber was alerted and would have to attempt to contact the patient. These methods were deployed so that no pregnancy would go unreported within the iPLEDGE system.19-24
In addition to patients, prescribers and pharmacies must be registered in the iPLEDGE system.19-23 A total of 14,400 prescribers are registered and activated in iPLEDGE as of Year 5 of the program, with the majority of ISO prescribers being dermatologists (n=8585) or physician extenders (n=1762) working within dermatology practices. Dermatology practices represent more than 90 percent of ISO prescriptions. Importantly pharmacists must be iPLEDGE-registered to dispense ISO.
In a single year, iPLEDGE authorizes more than one million prescriptions, with 1,006,079 prescriptions authorized in Year 5 of the program.19,23 Although iPLEDGE was started to reduce the risk of pregnant women taking the drug, the fact that the system requires men, boys, and women not of childbearing potential to participate provides more comprehensive overall data and emphasizes the importance of not sharing ISO with others and not donating blood, but increases the cost of the system.98 The percentages by patient group that have received ISO prescriptions have remained relatively constant over the past few years. In Year 5 of the iPLEDGE program, 44.6, 3.0, and 52.4 percent, were female patients of childbearing potential, female patients not of childbearing potential, and males, respectively19,23
The iPLEDGE system can deny approval of prescriptions when the requirements of the system are not met. In fact, in 2011, 40.6 percent of ISO prescriptions requested within the program were initially denied iPLEDGE authorization because the request did not meet specific requirements.19-23 There are multiple reasons for these refusals; however, the following three reasons accounted for 95.9 percent of all iPLEDGE denials: (1) the prescriber did not confirm counseling (47.0%), (2) patient within window but did not answer comprehension questions before attempting to fill a prescription (44.2%), and (3) patient attempted to fill prescription too soon (4.7%).19 In many cases, these prescription denials were not outright rejections, but are better described as “delays” that could eventually be corrected and avoided in the future.
Pregnancies noted through the iPLEDGE program have declined about 15 percent from 2010 to 2011. In 2011, there were 155 reported pregnancies among 129,554 females of child-bearing potential (0.12%) of which 150 were ISO-exposed patients and five were of indeterminate exposure. Of the 2011 pregnancies, 5.8 percent occurred before ISO treatment started, 60.6 percent occurred during the course of treatment, and 8.4 percent occurred within 30 days after the course of ISO was completed.19,98,99 As the elimination half-life of ISO has been reported to range between 11.8 hours and 38.5 hours, the iPLEDGE requirement of a final pregnancy test at 30 days post-treatment allows for a wide “time cushion” by which all ISO should be cleared systemically67
The iPLEDGE program has been the subject of some criticism. First, it has been argued that the system is not significantly decreasing pregnancies among women taking isotretinoin.100,101 Unfortunately, it is not possible to compare the pregnancy prevention data under iPLEDGE with the previous ISO pregnancy risk-management program (i.e., SMART) as in the previous program, registration and reporting of pregnancies was voluntary. Other nations implementing pregnancy prevention programs for women of childbearing potential taking ISO have also failed to completely prevent such pregnancies, although 100-percent prevention of pregnancy has never been fully anticipated.102 When a program in Germany tightened restrictions with the goal of reducing ISO-exposed pregnancies, the opposite occurred.103 With the iPLEDGE program in the United States, if one considers the annual number of reported pregnancies relative to the number of females of childbearing potential for Years 3 through 5 of the program, ISO-exposed pregnancies were reported in 0.00083 percent in Year 3, 0.00072 percent in Year 4, and 0.00064 percent in Year 5.19 The iPLEDGE program imposes a level of complexity and additional effort that has led to some reduction in prescribing of ISO as compared to several years ago; the numbers of authorized prescriptions has ranged between approximately 1 and 1.8 million annually in Years 3 through 5 of the iPLEDGE program.98 Some clinicians may have stopped prescribing ISO to avoid their own involvement with the iPLEDGE system; however, it is hoped that such clinicians have the sense of professional obligation to refer patients who are candidates for ISO to an authorized prescriber who can offer the opportunity to receive ISO to the patient. The use of ISO outside of the iPLEDGE program and obtaining ISO through sources that are not authorized by the iPLEDGE program are both highly discouraged.37,104
Conclusion
ISO remains a highly effective agent for treating acne. Experience with ISO for more than three decades has more clearly defined efficacy and safety profiles. Over time, suggested dosage regimens have emerged and multiple reports have led to the observation that reaching a target cumulative dose (120mg-150mg/kg) over the initial course of ISO therapy markedly reduces the rate of acne recurrence and the need for retreatment of acne, including additional courses of ISO therapy. Several formulations of ISO are available. Several generic AB-ISO products are available based on the PK profile of Acc-ISO. Lidose-ISO is a nongeneric formulation that exhibits a unique PK profile. The main pharmacological property of Lidose-ISO that distinguishes it from Acc-ISO and the AB-ISOs is markedly greater GI absorption when administered on an empty stomach. As the PK profile of Acc-ISO demonstrated a greater than 60-percent mean increase in bioavailability when ingested with a HF/HC meal as compared to a fasted state, concern has been raised regarding how dietary factors may affect long-term treatment success with AB-ISO products, all of which are generic formulations of Acc-ISO. Data showing that additional courses of ISO for acne are relatively common emphasize the need to further evaluate factors that increase this potential, including the impact of co-ingestion of ISO with food and the type of food content on short-term and long-term therapeutic outcomes. The PK profile of Lidose-ISO is likely to reduce the risk that decreased bioavailability related to dietary factors may adversely modify the therapeutic benefit of ISO, especially long-term remission of acne.
Biographies
Footnotes
This supplement to The Journal of Clinical and Aesthetic Dermatology is based upon proceedings from the 2013 Alabama Dermatology Society Summer Symposium.
Editors note:For the purpose of this publication, any references to PK data on ISO refers specifically to studies completed based on Acc-ISO. Importantly, these PK data on Acc-ISO also apply directly to the AB-rated ISO (generic) formulations through class effect based on their FDA approval as generic AB-rated equivalent formulations (AB-ISO). The PK data for Acc-ISO is what is published in product labeling for all AB-ISO products, although other PK studies available with AB-ISO products may be on file with the manufacturer and may sometimes be published in another reference source. Where data are related to Lidose-ISO (not a generic equivalent of Acc-ISO or AB-ISOs), this will be specified.
Contributor Information
James J. Leyden, University of Pennsylvania, Philadelphia, Pennsylvania.
James Q. Del Rosso, Touro University College of Osteopathic Medicine, Henderson, Nevada.
Eric W. Baum, University of Alabama Medical Center, Tuscaloosa, Alabama.
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187745 | https://dirjournal.org/pdf/beb8919b-f013-4ea1-b1c8-40332e840fe1/articles/1305-3825.DIR.2130-08.1/Diagn%20Interv%20Radiol-16-221-En.pdf | © Turkish Society of Radiology 2010 221 E mphysematous pyelonephritis (EPN) and emphysematous pyeli-tis (EP) are rare gas-forming acute infections of the renal paren-chyma. In EP the gas is limited to the pelvicaliceal system, where-as in EPN, gas extends further into the renal parenchyma, perinephric tissues, and the retroperitoneum. Ninety percent of patients with EPN have diabetes mellitus (DM), while EP is associated with DM in 50% of cases (1–5). The purpose of our study was to describe the computed tomography (CT) findings in five insulin-dependent diabetic patients with renal emphysema. Case reports Over a 4 year period (2003–2007), five insulin-dependent diabetic pa-tients (four women, one man; age range, 36–87 years, mean age, 69.6 years) with renal emphysema (2 EP, 3 EPN) underwent CT examination in our department. All CT scans were obtained with a Picker PQ 5000 CT scanner device with a slice thickness of 5 mm, pitch of 2, reconstruction interval of 5 mm, field of view (FOV) ranging between 340–440 mm, depending on the patient’s size. Images were obtained before and af-ter contrast agent administration during nephrographic phase, while in one patient only unenhanced images were obtained because of impaired renal function. In three patients (2 of 2 with EP and 1 of 3 with EPN) obstruction of the urinary tract by calculi or tumor was demonstrated. All cases were unilateral (two right, three left).
Case 1 An 87-year-old woman with insulin-dependent DM presented in sep-tic shock. Laboratory findings demonstrated normal renal function and signs of renal infection. CT demonstrated dilatation of the left pelvi-caliceal system caused by obstruction by a large stone in the left ureter; CT also demonstrated gas in the collecting system and the left ureter above the level of obstruction. These findings were consistent with EP. The function of the affected kidney, however, was demonstrated normal and symmetric with the contralateral kidney. Reactive retroperitoneal lymph nodes and an aneurysm of the abdominal aorta were also ob-served (Fig. 1). Conservative management with antibiotics, insulin, and electrolytes, and percutaneous drainage (nephrostomy) were performed; the patient’s condition improved within six days.
Case 2 An 85-year-old male with an 18-year history of insulin-dependent DM and a known history of bladder cancer and lithiasis of the collecting system of the right kidney and of the right ureter presented with symp-toms of acute renal infection. CT demonstrated hydronephrosis of the right pelvicaliceal system, stones in the right pelvicaliceal system and ABDOMINAL IMAGING CASE REPORT Renal emphysema in diabetic patients: CT evaluation Ioannis Tsitouridis, Michael Michaelides, Dimitrios Sidiropoulos, Mary Arvanity From the Department of Diagnostic and Interventional Radiology (M.M. michaelidesm@yahoo.com), Papageorgiou General Hospital, Thessaloniki, Greece. Received 25 July 2008; revision requested 29 August 2008; revision received 30 August 2008; accepted 19 September 2008.
Published online 19 October 2009 DOI 10.4261/1305-3825.DIR.2130-08.1 ABSTRACT Renal emphysema is a rare, fulminant, suppurative infection of pelvicaliceal system, renal parenchyma, perinephric tissues, and retroperitoneum. It is characterized by formation of gas. Invariably this condition is associated with diabetes mellitus and carries high mortality (40–90%). Renal emphysema can be classified into two distinct clinical entities: emphysematous pyelitis and emphysematous pyelonephritis. This classifica-tion has important prognostic and therapeutic implications. Herein we describe the computed tomographic findings in five unilateral cases of renal emphysema (two cases of emphy-sematous pyelitis and three cases of emphysematous pyelone-phritis) in five insulin-dependent diabetic patients.
Key words: • kidney • emphysema • computed tomography • diabetes mellitus Diagn Interv Radiol 2010; 16:221–226 Tsitouridis et al.
222 • September 2010 • Diagnostic and Interventional Radiology ureter, and air in the collecting system and right ureter—findings consistent with EP (Fig. 2a–c). A large tumor of the bladder infiltrating the right ure-teral orifice was also observed (Fig. 2d). The patient was managed con-servatively with antibiotics, insulin, and electrolytes. Despite treatment, his condition deteriorated, and he died a month later. Case 3 A 67-year-old female insulin-de-pendent DM patient presented with septic coma and acute renal failure. Her relatives reported a 48-hour histo-ry of fever and pain in the right renal area, where physical examination re-vealed a mass. Laboratory evaluation showed impaired renal function and signs of urinary infection. CT demon-strated gas within the parenchyma of right kidney, the collecting system, the right ureter, and the retroperito-neum (Fig. 3). The patient was man-aged conservatively but died of sepsis after six days. Case 4 A 73-year-old female with a 15-year history of insulin-dependent DM pre-sented with confusion, vomiting, high fever, and abdominal pain. A mass was palpable in the left upper abdom-inal quadrant. Laboratory evaluation showed signs of urinary infection. CT demonstrated a large amount of air in the renal parenchyma and the retroperitoneum. Much of the renal parenchyma was destroyed (Fig. 4). The patient was initially treated with antibiotics, insulin, and electrolytes; clinical condition improved over the next few days. Two months later, left nephrectomy was performed; the pa-tient is in good condition 2.5 years later.
Case 5 A 36-year-old female insulin-de-pendent DM patient with a history of urolithiasis presented with high fever and abdominal pain. CT dem-onstrated multiple stones in the left pelvicaliceal system, gas within the renal parenchyma, a small amount of gas in the retroperitoneum, and a renal abscess (Fig. 5). The patient’s condition improved within four days after conservative treatment, and urolithiasis was managed a few months later.
Figure 1. a–c. Postcontrast axial CT images show emphysematous pyelitis with dilatation and gas in the left pelvicaliceal system (a) and left ureter (b) due to obstruction by a stone in the ureter (arrow, c). Reactive retroperitoneal lymph nodes and an aneurysm of the abdominal aorta are also observed.
b a c CT of renal emphysema in diabetic patients • 223 Volume 16 • Issue 3 Discussion Renal emphysema is one of the most fulminant forms of upper urinary tract infection associated with necrosis and gas formation. Although it is a serious clinical entity, it is rarely reported in the medical literature. In the majority of cases it is unilateral (1–4). Kelly and MacCallum initially de-scribed renal emphysema (gas in the kidney) in 1898. Renal emphysema is divided into two different clinical enti-ties whose management and prognosis are different: EPN and EP (5).
EPN is a rare severe necrotizing gas-forming acute infection of the renal parenchyma, in which gas is present in the renal parenchyma, perinephric tis-sues, and even in the retroperitoneum. High rates of mortality and morbidity are seen in EPN. More than 90% of the patients have DM; in the remainder, it is related to obstruction. It is more common in females and usually affects the left kidney, although in 10% of the cases there is involvement of both kid-neys. In addition to medical therapy, aggressive surgical management (ne-phrectomy) has been recommended to improve survival. EPN has also been re-ported in transplanted kidneys (1–5).
EP is a milder form of the disease with a better prognosis, in which gas is limited to the pelvicaliceal system. It is commonly associated with obstructive uropathy due to calculus, stricture, or neoplasm. It responds well to conserv-ative therapy, with or without a drain-age procedure (1–5). In 75% of the cases, infection is caused by Escherichia coli (75%). In other cases, Klebsiella pneumoniae, En-terobacter aerogenes, Proteus mirabilis, Pseudomonas species, anaerobic strep-tococci, and Candida and other fungi have been isolated as the pathogenetic organisms (1–9). A case of amoebiasis has also been reported (10).
Renal emphysema may present with fever, nausea, vomiting, abdominal pain, shock, lethargy, confusion, and uncommonly with diabetic ketoaci-dosis. Laboratory findings include hyperglycemia, increased white blood cell counts, pyuria, and elevated levels of urea and creatinine (11). At the on-set of the disease, clinical features of renal emphysema are similar to other forms of acute pyelonephritis. Re-nal emphysema should be suspected when renal infection has a prolonged or refractory course and is associated with a palpable renal lump, diabetes, b a c d Figure 2. a–d. Unenhanced axial CT images (a–d) showing dilatation of the right pelvicaliceal system and gas in the collecting system (arrow, a) and right ureter due to obstruction by stones (arrow, c). A large tumor of the bladder is also observed (d).
Tsitouridis et al.
224 • September 2010 • Diagnostic and Interventional Radiology Figure 4. a, b. Postcontrast axial CT images show emphysematous pyelonephritis with destruction of much of the left renal parenchyma (a) with a large amount of gas within the renal parenchyma and the retroperitoneum (b).
b a b a c d Figure 3. a–d. Unenhanced axial CT images (a–d). Emphysematous pyelonephritis with gas within the parenchyma of right kidney, the collecting system, the right ureter (arrow) and in the right pararenal and perirenal space is seen (a, b). Note the extension of the retroperitoneal air collection behind the abdominal wall, representing a potential extension of the posterior perirenal space (c). Retroperitoneal air collection is also demonstrated in the pelvis (d).
CT of renal emphysema in diabetic patients • 225 Volume 16 • Issue 3 or obstruction. The dramatic finding of crepitation over the thigh or flank in a diabetic patient is infrequent; when present it should raise a high degree of suspicion for emphysema-tous pyelonephritis with extension into the perinephric space and retro-peritoneum. Four factors appear to be involved in the pathogenesis of EPN: gas-forming bacteria, high tissue glucose, impaired tissue perfusion, and defective immu-nity (2, 3, 12). The most important fac-tor in the pathogenesis of gas forma-tion appears to be high tissue glucose, employed as a growth medium for the microorganisms involved and fer-mentation of necrotic tissue. Analysis shows that the gas includes N2 (60%), H2 (15%), O2 (8%), CO2 (5%), and oth-er gases in lesser concentrations (12). The pathogenesis of gas formation is not well understood; formation of CO2 and H2 by fermentation of sugar (at high levels in the tissue and urine of these patients) is one proposed mecha-nism. Gas bubble formation is initi-ated by rapid catabolism and impaired transportation of gas. The rapid catab-olism may be caused by severe infec-tion, tissue ischemia, and infarction. Another possibility is that poor circu-lation leads to sugar fermentation and impaired transportation (12).
CT is considered the best imaging technique for the diagnosis of the dis-ease, for identification of gas (within the renal parenchyma, collecting sys-tem, ureter, urinary bladder, perine-phric and paranephric spaces, and sometimes in the vascular system) and for evaluating severity of infection. Gas around the spinal cord and within the psoas muscle was reported in a fa-tal case of bilateral EPN (13). In 1998, Wan et al. classified EPN into two types, based on CT findings: Type I is the classical form with renal parenchymal destruction and presence of diffuse gas in the parenchyma in a streaked or mottled pattern following the pyramids, with little or no fluid. Type II is characterized by the pres-ence of fluid (renal and perirenal) with a bubbly or loculated gas pattern, or gas in the collecting system with acute bacterial nephritis or with renal or per-irenal fluid-containing abscesses (14). Radiological diagnosis plays a major role in the prognosis of disease; type I EPN has a higher mortality rate than type II (14).
In 2000, Huang and Tseng catego-rized CT findings into the following 4 classes (class 1 being the mildest) (4): Class 1: gas in the collecting system only (EP).
Class 2: gas in the renal parenchyma without extension to extrarenal space. Class 3A: extension of gas or abscess to the perinephric space.
Class 3B: extension of gas or abscess to the paranephric space.
Class 4: bilateral EPN or solitary kidney EPN.
EPN (classes 2–4) must be differenti-ated from EP (class 1), in which gas is present in the collecting system of the kidney but not in the parenchyma. It has been suggested that the term EPN should be only applied to gas for-mation within the renal parenchyma and perinephric space. Others have said that the infection of the renal parenchyma and perinephric tissue results in the presence of gas in the collecting system, parenchyma, and perinephric space. This is preferred be-cause it includes all possible manifesta-tions of acute renal infection with gas formation. EPN must be also differen-tiated from gas-forming renal abscess, in which gas is not found in the renal parenchyma. Gas in the renal paren-chyma can also have traumatic or ia-trogenic causes (Foley catheter or an enterorenal fistula).
The most important factor in man-agement is early diagnosis and treat-ment. Treatment of EPN involves antibi-otic therapy (15), followed, if necessary, by surgical intervention (nephrectomy). Recently, percutaneous drainage has been reported as a kidney-saving proce-dure. This alternative to nephrectomy is particularly important in solitary kidney patients, patients with bilateral EPN, and inoperable patients (16, 17). Transure-thral retrograde drainage with stent placement has also been reported (18).
The mortality in untreated EPN is near 100%. With medical treatment alone, it decreases to 70%; with combined medi-cal and surgical intervention, mortality can be reduced to 30% (19).
In conclusion, we believe that spi-ral CT is the examination of choice for evaluation of diabetic patients with signs and symptoms of renal em-Figure 5. a, b. Postcontrast axial CT images show emphysematous pyelonephritis due to urolithiasis. There are calculi in the left pelvicaliceal system, gas within the renal parenchyma and the retroperitoneum (a, b). A renal abscess is also demonstrated (arrow, b).
b a Tsitouridis et al.
226 • September 2010 • Diagnostic and Interventional Radiology physema. This diagnostic modality is helpful in staging the disease and evaluating its severity, and it plays a critical role in selection of the treat-ment procedure. References 1. Joseph RC, Amendola MA, Artze ME, et al. Genitourinary tract gas: imaging evalua-tion. Radiographics 1996; 16:295–308. 2. Rodriguez-de-Velasquez AR, Yoder IC, Velasquez PA, Papanicolaou N. Imaging the effects of diabetes on the genitourinary system. Radiographics 1995; 15:1051– 1068. 3. Evanoff GV, Thompson CS, Foley R, Weinman EJ. Spectrum of gas within the kidney. Emphysematous pyelonephritis and emphysematous pyelitis. Am J Med 1987; 83:149–154. 4. Huang JJ, Tseng CC. Emphysematous pyelonephritis: clinicoradiological clas-sification, management, prognosis, and pathogenesis. Arch Intern Med 2000; 160:797–805. 5. Kawashima A, Sandler CM, Goldman SM, Raval BK, Fisherman E. CT of renal in-flammatory disease. Radiographics 1997; 17:851–866. 6. Langston CS, Pfister RC. Renal emphysema: a case report and review of the literature. AJR Am J Roentgenol 1970; 110:778–786. 7. Seidenfeld SM, Lemaistre CF, Setiawan H, Munford RS. Emphysematous pyelone-phritis caused by Candida tropicalis. J Infect Dis 1982; 146:569. 8. Zaboo A, Montie JE, Popowniak KL, Weinstein AJ. Bilateral emphysematous pyelonephritis. Urology 1985; 25:293–296. 9. Kim DS, Woesner ME, Howard TF, Olson LK. Emphysematous pyelonephritis dem-onstrated by computed tomography. AJR Am J Roentgenol 1979; 132:287–288. 10. Guvel S, Kilinc F, Kayaselcuk F, Tuncer I, Ozkardes H. Emphysematous pyelonephritis and renal amoebiasis in a patient with diabe-tes mellitus. Int J Urol 2003; 10:404–406. 11. Jain SK, Agarwal N, Chaturvedi SK. Emphysematous pyelonephritis: a rare pres-entation. J Postgrad Med 2000; 46:31–32. 12. Yang WH, Shen NC. Gas-forming infec-tion of the urinary tract: an investigation of fermentation as a mechanism. J Urol 1990; 143:960–964. 13. Sailesh S, Randeva HS, Hillhouse EW, Patel V. Fatal emphysematous pyelonephritis with gas in the spinal extradural space in a patient with diabsetes. Diabet Med 2001; 18:68–71. 14. Wan YL, Lo SK, Bullard MJ, Chang PL, Lee TY. Predictors of outcome in emphysema-tous pyelonephritis. J Urol 1998; 159:369– 373. 15. Tahir H, Thomas G, Sheerin N, Bettington H, Pattison JM, Goldsmith DJ. Successful medical treatment of acute bilateral em-physematous pyelonephritis. Am J Kidney Dis 2000; 36:1267–1270. 16. Sacks D, Banner M, Meranze S, Burke D, Robinson M, McLean G. Renal and related retroperitoneal abscesses: percutaneous drainage. Radiology 1988; 167:447–451. 17. Mallet M, Knockaert DC, Oyen RH, Van Popple HP. Emphysematous pyelone-phritis: no longer a surgical disease? Eur J Emerg Med 2002; 9:266–269. 18. Matsuda D, Irie A, Mizoguchi H, Baba S. A case of emphysematous pyelonephritis successfully treated by transurethral ret-rograde drainage. Hinyokika Kiyo 2004; 50:15–17. 19. Menif E, Nouira K, Baccar S, et al. Emphysematous pyelonephritis: report of 3 cases. Ann Urol 2001; 35:97–100. |
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Extreme principle
The extreme principle (or extremal principle) is a problem-solving technique that involves looking at objects with extreme properties, such as the largest or smallest element. For example, consider the following problem:
Example: Imagine an infinite chessboard that contains a positive integer in each square. If the value in each square is equal to the average of its four neighbors to the north, south, west, and east, prove the values in all the squares are equal.
Solution: Consider the square containing the minimal value. Then its four neighbors must all have this minimal value. Similarly, their neighbors must also have this minimal value, and so on ad infinitum.
Thus, every square in the chessboard has the same value.
Problems
Suppose you are given a finite set of coins in the plane, all with different diameters. Show that one of the coins is tangent to at most five of the others. (Solution)
A palindrome is a number or word that is the same when read forward and backward, for example, "176671" and "civic." Can the number obtained by writing the numbers from 1 to in order (for some ) be a palindrome? (Russia, 1996)
Place the integers (without duplication) in any order onto an chessboard, with one integer per square. Show that there exist two adjacent entries whose difference is at least . (Adjacent means horizontally or vertically or diagonally adjacent.)
There are points in the plane, not all collinear. Prove that there exists a line passing through exactly points.
There are blue points and red points in the plane (no 3 are collinear). Prove that there are non-intersecting line segments, each connecting a red point to a blue point (each point is on 1 segment).
There is a set of points in the plane with the property that any triangle with vertices in has area at most 1. Prove that there exists a triangle with area 4 containing all the points in . (Korea, 1995)
We are given an array of real numbers. An operation consists of changing the sign (positive to negative, and vice versa) of all the entries in any row or column. Prove that we can perform a number of such operations such that in the resulting array, the sum of the entries in each row and in each column is nonnegative.
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Pulleys
Pulleys are mechanisms compost by wheel and rope used to lift heavy objects onto tall heights. They change the direction of an applied force and they can even reduce the force needed to lift a weight. Pulleys systems are common used in constructions.
In the exhibit, systems of one to five pulleys are uses to lift a weight of 1lb onto a 1 ft height.
A SIMPLE BLOCK AND TACKLE
In order to lift the 1lb weight W you have to apply a force ofF on the rope equal to the weight W. The rope is now under a tension Tequal to the force F.
To lift this weight a distance of H=1 ft you will have to pull in a length L= H=1 ft of the rope.
The mechanical advantage M is one: M=W/F=1.
ADDING A PULLEY
In order to lift the 1lb weight W you have to apply a force of F on the rope. Like in the simple case of the block and tackle the rope is under a tension T equal to the applied force F. But in this case the weight is supported by twice the tension: 2T=W. The force F you have to apply then is only half of the weight W.
By having the second pulley you have decreased the force needed to lift the weight. The mechanical advantage M is now two: M=W/F=2T/T=2.
However, order to lift the weight a distance ofH=1 ft you will have to pull in a length L = HxM =2 ft of the rope.
You gained by having to apply a smaller force, but had to compensate by having to pull a longer length of rope.
MORE PULLEYS
Here the mechanical advantage is M=3.
And here, the mechanical advantage is M=4.
Conclusion
The mechanical advantage M is equal to number of ropes present at the weight end.The force needed to raise the weight is W/M. In order to lift the weight a distance H you will have to pull a corresponding longer length of rope L = H x M.
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2.12
Bohr Model of Hydrogen Atom
Written by:Divya Ladha
Fact-checked by:The CK-12 Editorial Team
Last Modified: Jul 25, 2025
Science of Neon Lights: Bohr’s Atomic Model and Electron Transitions
Neon sign lights operate based on the principle of electron transitions between quantized energy levels, a concept directly explained by Bohr’s atomic model. In a neon sign, an electric current excites the atoms of a gas (such neon in neon signs). When electrons in the gas atoms absorb energy, they transition to higher energy levels. However, these excited electrons are unstable and soon return to their lower energy states, emitting photons in the process. The emitted light corresponds to specific wavelengths, which are determined by the energy differences between electronic states, just as predicted by Bohr’s model. For example, neon emits a characteristic reddish-orange glow due to specific electronic transitions within neon atoms. This application showcases how Bohr’s quantized energy levels provide the theoretical foundation for understanding and designing lighting systems used in homes, offices, and advertising displays worldwide.
In the early 20th century, scientists were trying to understand the structure of the atom, particularly how electrons were arranged around the nucleus. Traditional physics could not explain why atoms emitted specific colors of light when heated or how they remained stable without collapsing. This led to the development of the Bohr Model of the Hydrogen Atom, proposed by Danish physicist Niels Bohr in 1913. The Bohr Model was revolutionary as it successfully explained the atomic spectra of hydrogen and introduced the concept of quantized energy levels. It also helped in laying the foundation for modern quantum mechanics. Today, the Bohr Model is still widely used to explain atomic structure in simpler terms, even though more advanced models have been developed.
Bohr Model of the Hydrogen Atom
The Bohr Model was an improvement over Rutherford’s nuclear model, which suggested that electrons revolved around the nucleus like planets orbiting the Sun. However, according to classical physics, accelerating electrons should continuously lose energy and spiral into the nucleus, making the atom unstable as shown in the Figure below. This did not happen in reality, so Bohr proposed a quantum mechanical explanation for atomic stability.
Bohr’s model introduced the idea that electrons move in fixed orbits or energy levels around the nucleus and do not radiate energy while remaining in these orbits. He suggested that electrons could only occupy specific energy levels, and any transition between these levels involved the absorption or emission of energy in discrete amounts called quanta, as shown in the Figurebelow.
Postulates of Bohr Atomic Model
Bohr’s model is based on the following fundamental postulates:
Electrons move in fixed circular orbits around the nucleus without radiating energy. These orbits are also called stationary states or energy levels.
Each orbit has a fixed energy, with the lowest energy level (n = 1) being the most stable, known as the ground state. Higher energy levels (n = 2, 3, 4, ...) are called excited states. The electrons can only exist in these discrete energy levels. The energy levels are denoted by integers n = 1, 2, 3, ..., which are called principal quantum numbers.
The energy of an electron in a given orbit is given by:
where n is the principal quantum number, and 13.6 eV is the ionization energy of hydrogen.
3. Bohr’s theory is also applicable to ions that contain only one electron, similar to the hydrogen atom. Examples of such hydrogen-like species include He⁺, Li²⁺, Be³⁺, and others. The energy of the stationary states for these ions is given by the equation:
and the radius of these orbits is expressed as:
where Z is the atomic number (e.g., 2 for helium and 3 for lithium). From these equations, it is clear that as Z increases, the energy of the electron becomes more negative, meaning it is more strongly bound to the nucleus, and the radius of the orbit decreases. This indicates that the electron is held more tightly as the nuclear charge increases.
4. Energy is absorbed or emitted when an electron moves from one orbit to another. When an electron moves from a higher energy level to a lower energy level, it emits energy in the form of light (photon). The energy of the emitted photon corresponds to the difference between the two energy levels. The energy difference between two orbits is given by Planck’s equation: , where where E1 and E2 are the initial and final energy levels, h is Planck’s constant (6.626×10−34J.s) and ν is the frequency of radiation.
5. One of the most significant ideas of Bohr’s model was that electrons could only have specific angular momentum values. According to Bohr, the angular momentum of an electron is quantized and follows the equation:
where: me = mass of the electron, ve = velocity of the electron, r = radius of the orbit, n = principal quantum number. This means that only certain discrete orbits are allowed, preventing the electron from spiralling into the nucleus. This idea was crucial in developing quantum mechanics.
These postulates helped explain the stability of atoms and the spectral lines observed in hydrogen’s emission spectrum. According to Bohr’s postulates, electrons in a hydrogen atom can only occupy specific quantized energy levels and transition between these levels by absorbing or emitting energy in the form of light (photons). Since electrons transition between fixed energy levels, only specific discrete wavelengths of light are emitted, forming a line spectrum instead of a continuous spectrum. The emitted spectral lines correspond to series of transitions, named after the scientists who discovered them as discussed in the previous lesson of Atomic Spectra.
Interactive: Bohr Model For Silicon Atom
Case Study: From Rutherford to Bohr: The Quantum Leap in Understanding Atomic Structure
Early atomic models faced significant challenges. Rutherford's model, while revolutionary in proposing a nuclear atom, couldn't explain why electrons didn't spiral into the nucleus, causing atom collapse. It also failed to account for the distinct, non-continuous patterns of light (line spectra) emitted and absorbed by atoms. To address these limitations, Niels Bohr introduced key quantum ideas. He proposed that electrons occupy specific 'stationary states' or orbits, where they maintain their energy without radiating. Energy changes occur only when an electron transitions between these discrete, allowed energy levels. This conceptual shift, alongside the idea of quantized angular momentum, provided a framework that successfully explained the stability of the hydrogen atom and its observed spectral lines.
Examples of Bohr Model of Hydrogen Atom
Example 1:
Find the energy of an electron in the third orbit (n = 3) of hydrogen.
Using the formula:
For n = 3:
Thus, the energy of an electron in the third orbit is -1.51 eV.
Example 2:
Calculate the wavelength of light emitted when an electron transitions from n = 4 to n = 2 in hydrogen.
Using the energy formula:
Convert energy to wavelength using Planck’s equation:
This wavelength corresponds to the Balmer series (visible light region).
| |
| Summary of Bohr Model of Hydrogen Atom |
| The Bohr Model describes electrons as moving in fixed energy levels around the nucleus. Electrons do not lose energy in stationary states but can transition between levels by absorbing or emitting energy. Angular momentum is quantized and follows the rule mvr = nh/2π. Electron energy levels are negative, meaning electrons are bound to the nucleus. The Bohr Model successfully explains hydrogen’s atomic spectrum but does not fully describe atoms with more than one electron. |
Review Questions on the Bohr Model of Hydrogen Atom
What are the postulates of Bohr’s model?
Why do electrons in atoms have quantized energy levels?
Derive the energy of an electron in the second orbit of hydrogen.
What is the significance of negative energy values in the Bohr Model?
Explain why Bohr’s model fails for multi-electron atoms.
| Image | Reference | Attributions |
---
| | | Credit: Image by kalhh from Pixabay Source: License: Pixabay License |
| | | Credit: CK-12 Foundation Source: CK-12 Foundation License: CK-12 Curriculum Materials License |
| | | Credit: Zachary Wilson Source: CK-12 Foundation License: CK-12 Curriculum Materials License |
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187749 | https://www.youtube.com/watch?v=x44IYdYw8H8 | Conjugate Surds | How to multiply surds
Tambuwal Maths Class
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Posted: 25 May 2021
Two surds are said to be conjugate of each other if their product gives rise to a rational number. From our knowledge of difference of two squares, we know that:
(a + b)(a - b) = a² - b² Similarly,
(√a+√b)(√a-√b) = a² - b²
While √a + √b and √a - √b are not rational, their product (a²-b²) is rational. Hence, √a+√b and √a-√b are conjugates of each other.
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Transcript:
in this tutorial we are going to multiply swords by their conjugates but before then let us discuss more about these conjugates because i have started making some introduction about them two thoughts are said to be conjugate of one another if and only if their product lies to a rational number because you know uh sod are said to be irrational but their product if they are conjugate or rise to a rational number the best way to explain this is by the use of difference of two squares you know if you have product of two binomial terms two exact but have a difference of sign you can see we have a b a b but the operation connecting them is opposite to one another this is positive while this is negative let us try to expand this a multiplied by this we have a a minus b then plus b multiplied by that a minus b so now let us expand this a times a is a squared a times negative b is negative a b then we move to the other one plus b times a is the same thing as a times b so we can write it as a b then b times negative b is negative b squared look at the middle term they are opposites to one another negative a b plus a b is zero so we only have this loss on the first time a squared minus b squared the same thing is applicable to sort if you have sort a plus sub b and this is said to multiply sort a minus sort b you can see they only have difference of phi but all the sorts are exactly the same it is the same thing as saying so a squared minus sub b also squared difference of two squares square will always cancel square root so we have a minus b remember i told you the number under the radical sign is always rational so a is rational b is rational difference of two rational number is also rational so you can see that by multiplying these two sorts together we obtain a rational number while the initial uh sorts are said to be irrational so the product of these two thoughts gives rise to a rational number therefore we say these sorts are said to be conjugates of one another the only half difference of sine the same thing is applicable here if you are looking for the conjugate of or 3 minus root 2 it is going to be root 3 plus root 2 just difference of sine here that we have positive we take the negative sign wherever you have negative you take the positive side so now let us see how we can multiply this four set of swords to see whether we are going to obtain a rational number so let's start with this one root 3 minus root 2. you know the conjugate of this is rotary plus origin now take the first term square it negative sign take the second term also square it this is the same thing as if you like you can even expand this rotary square is the same thing as root 3 times root 3 which will give rise to root 9 and you know root 9 is equal to 10. so without even doing this power of 2 will just cancel the square root of leaving 3 the same thing here and finally we have 1 and 1 is a rational number so you can see that by multiplying this sword by its conjugate the result is rational the same is true for all other sorts let us move on to the second one the second one is two root 5 so we have 2 root 5 plus 3 root 3. this is the sword we have the conjugate of this will be 2 root 5 minus 3 root 3. take the first term here which is 2 root 5 squared minus the second time is 3 root 3 also squared this time around you can see we have a mixed sod so this power of 2 will affect everything here both 2 and root five so this becomes two squared root five squared minus three squared root three also squared 2 squared is 4 because 2 times 2 is 4 multiplied by power of 2 will cancel square root leaving 5 here minus three squared is nine times three four times five is um 20. minus 9 times 30 is 27. if you subtract this you obtain negative 7 because 27 is greater than 20 and hence the result must be negative but negative seven is still a rational number you can see that after multiplying this we obtain a rational number let us move on to the third one the third one is four roots two minus five root five four roots two minus five root five the conjugate of this will be four roots two plus five root five this also makes sense we take the first one four roots two we square it we subtract we take five root 5 we equally square it all right 4 squared is 16 because 4 times 4 is 16 then multiplied by root 2 squared is 2 minus 5 squared is 25 because 5 times 5 is 25 root 5 squared is 5. so 16 times 2 is 32 minus 25 times 5 is 125 because 25 times 4 is 100 um and if you subtract um 125 from 32 you know 125 is greater the answer will be negative so the best way is to say 125 minus 32 then you put negative sign at the back which is going to be negative 93 and negative 93 is a rational number let us move on to the last one we have one plus root two this time around you can see that the first time here is rational while this is short but not with standing we can still take the conjugate of this which is the same thing as 1 minus root 2. 1 squared minus root 2 also squared difference of 2 squares 1 squared is 1 root 2 squared is 2 1 minus 2 is negative 1 and negative 1 is a rational number so this is exactly how to multiply sort by its conjugate and this is all i have for you today thank you for watching do have a nice day you |
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2017 AMC 10B Problems/Problem 18
Contents
1 Problem
2 Solution 1
3 Solution 2 (Fast and Easy)
4 Solution 3 (similar to solution 5 but simpler)
5 Solution 4
6 Solution 5 (Burnside's Lemma)
7 Solution 6
8 Solution 7
9 See Also
Problem
In the figure below, of the disks are to be painted blue, are to be painted red, and is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
Solution 1
First we figure out the number of ways to put the blue disks. Denote the spots to put the disks as from left to right, top to bottom. The cases to put the blue disks are . For each of those cases we can easily figure out the number of ways for each case, so the total amount is .
Solution 2 (Fast and Easy)
Notice how you can arrange the colors like a circle with three blues two reds and one green.
We want to find the compliments of all the other cases that are invalid/ same rotation or reflections.
We use the Circular Gap Theorem to solve this problem.
We have three cases, there are 3 blues that are not next to each other, 2 reds that are not next to each other, or one green that isn't next to itself. Our k value is k∈{1,2,3}.
Case 1: k∈{3}
We substitute n=6 total values to orient and k=3. We get C n(A)=6 3(6−3−1 3−1)=2⋅(2 2)=2.
Case 2: k∈{2}
We substitute n=6 and k=2 to get C n(A)=6 2(6−2−1 2−1)=3⋅(3 1)=9.
Case 3: k∈{1}
This is quite obvious, we don't need to use the formula. There is only 1 case that works here as we have 1 color and we choose 1 of them, giving us 1 case.
The total number of cases is 2+9+1=12.
Therefore our answer is .
Notice how we didn't need to use Burnside's Lemma.
~Pinotation
Solution 3 (similar to solution 5 but simpler)
Denote the discs as in the first solution. Ignoring reflections or rotations, there are colorings. Now we need to count the number of fixed points under possible transformations:
The identity transformation. Since this doesn't change anything, there are fixed points.
Reflect on a line of symmetry. There are lines of reflections. Take the line of reflection going through the centers of circles and . Then, the colors of circles and must be the same, and the colors of circles and must be the same. This gives us fixed points per line of reflection.
Rotate by counterclockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles , , and will be the same. Similarly, the colors of circles , , and will be the same. This is impossible, so this case gives us fixed points per rotation.
By Burnside's Lemma, the total number of colorings is .
Solution 4
Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as
Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are of these. However, due to reflection we must divide this by two. But, in two of these cases, the reflection is itself, so we must subtract these out before dividing by 2, and add them back afterwards, giving arrangements in this case.
Now, look at the second case. We again must pick two of the five remaining circles, and like in the first case, two of the reflections give the same arrangement. Thus, there are also arrangements in this case.
In total, we have .
~tdeng
Solution 5 (Burnside's Lemma)
We note that the group acting on the possible colorings is , where is a rotation and is a reflection. In particular, the possible actions are the identity, the and rotations, and the three reflections.
We will calculate the number of colorings that are fixed under each action. Every coloring is fixed under the identity, so we count fixed colorings. Note that no colorings are fixed under the rotations, since then the outer three and inner three circle must be the same color, which is impossible in our situation.
Finally, consider the reflection with a line of symmetry going through the top circle. Every fixed coloring is determined by the color of the top circle (either green or blue), and the color of the middle circles (either blue or red). Hence, there are colorings fixed under this reflection action. The other two actions are symmetric, so they also have fixed colorings. Hence, by Burnside's lemma, the number of unique colorings up to reflections and rotations is
Solution 6
Call the ball in the top row A, the two balls in the 2nd row from left to right B and C, and the bottom rows balls from left to right D, E, and F. The total amount of paintings is . If we divided this number by (for rotations) and (for reflections), we would not get the right answer, because the paintings that do not change when reflected are over-subtracted. So lets find the number of over subtracted paintings. To be symmetric, ball B and C must be the same color, D and F must be the same color, and A and E can be any color because they are on the line of symmetry. Pair B,C and D,F must be blue and red or red and blue, which is possibilities. Then, Ball A has possibilities, and Ball E has possibility (the one remaining color). That means, the number of paintings that when they are reflected do not change is The total amount of these paintings, after accounting for rotations, is Then there are paintings that when reflected, change. (We divided by since we are accounting for rotations and reflections.) Then, adding back the paintings we subtracted off because they didn't change when reflected, we get □ (to look professional)
~heheman
Solution 7
Notice that the green ball can be placed in 6 different positions. If the ball is placed on the top position, bottom-left position, or bottom-right position, it can be rotated to make the green ball on the top position. Meanwhile, if the green ball is placed on the middle-left, middle-right, or bottom-middle position, it can be rotated to move the green ball to the bottom-middle position. To avoid overcounting due to rotations, we can divide this problem into two cases: the green ball is at the top position, or the green ball is at the bottom-middle position.
Consider the first case. If the green ball is placed in the top position, the other five balls can be placed in ways. (We choose 2 of the 5 positions to be occupied by the red ball.) In the following positions,
G G
R R and B B
B B B R B R
the position is identical under reflection over a vertical line. All other 8 positions have a reflection that is different. Thus, we can count only half of these 8 positions (because otherwise we would be over counting), but we can count both of the two special cases (since they don't make a different image when reflected. Thus, this case has cases.
In the second case, the green ball is placed in the bottom-middle position. Similarly, the other five balls can be placed in ways. In the special cases,
B B
R R and B B
B G B R G R
the position is identical under reflection over a vertical line. Proceeding just like the first case, there are cases.
In total, there are different paintings.
~sid2012
See Also
2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17Followed by
Problem 19
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 10 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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187751 | https://testbook.com/question-answer/into-how-many-compartments-do-the-coordinate-plane--5f56268130c8dbd5b6c5ae9e | [Solved] Into how many compartments do the coordinate planes divide t
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Into how many compartments do the coordinate planes divide the space?
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Answer (Detailed Solution Below)
Option 3 : 8
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Detailed Solution
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Explanation:
Two-dimensional coordinate axes divide the plane into four quadrants;
Three-dimensional coordinate axes will divide the space into eight "compartments" known as octants.
Let (x, y, z) be an ordered triad. Take the point L of coordinate x on X′X↔, the point M of coordinate y on Y′Y↔and the point N of coordinate z on Z′Z↔. Through the point L, M, N draw planes π 1,π 2,π 3 perpendicular to the coordinate axes X′X↔,Y′Y↔,Z′Z↔respectively.
The three mutually perpendicular planes π 1,π 2,π 3 intersect at a unique point of P.
Since P ϵ π 1, the projection of P on X′X↔is L and hence the x - coordinate of P is equal to L. Similarly the y - coordinate of P is equal to the y - coordinate of M and the z -coordinate of p is equal to the z -coordinate of N. The coordinates of P are x, y, z in that order. Thus for the ordered triad (x, y, z) we have a unique point P in space.
Hence a one-to-one correspondence is established between the set of points in space and the set of ordered triads of real numbers. This space is called 3D space or R 3 space.
The three coordinate planes divide the space into eight compartments, each of which is called an octant.
Octant with O X→O X′→O X′→O X→O X→O X′→O X′→O X→
founding O Y→O Y→O Y′→O Y′→O Y→O Y→O Y′→O Y′→
lines O Z→O Z→O Z→O Z→O Z′→O Z′→O Z′→O Z′→
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187752 | https://mathbooks.unl.edu/Calculus/sec-6-2-volume.html | Skip to main content
🔗
Section 6.2 Using Definite Integrals to Find Volume by Rotation and Arc Length
🔗
Motivating Questions
How can we use a definite integral to find the volume of a three-dimensional solid?
How can we use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular axis?
In what circumstances do we integrate with respect to instead of integrating with respect to ?
What adjustments do we need to make if we revolve about a line other than the - or -axis?
How can a definite integral be used to measure the length of a curve?
🔗
Just as we can use definite integrals to add the areas of rectangular slices to find the exact area that lies between two curves, we can also use integrals to find the volume of regions whose cross-sections have a particular shape.
🔗
In particular, we can determine the volume of solids whose cross-sections are all thin cylinders (or washers) by adding up the volumes of these individual slices, in particular when these shapes arise by revolving one or more curves around an axis. We first consider a familiar shape in Example 6.16: a circular cone.
Example 6.16.
Consider a circular cone of radius 3 and height 5, which we view horizontally as pictured in Figure 6.17. Our goal in this example is to use a definite integral to determine the volume of the cone.
Projecting the cone onto the -plane yields a triangle with vertices , , and , as shown in Figure 6.17. Find a formula for the linear function corresponding to the side of the triangle joining vertices and .
For the representative slice of thickness that is located horizontally at a location (somewhere between and ), what is the radius of the representative slice? Note that the radius depends on the value of .
What is the volume of the representative slice you found in (b)?
What definite integral will sum the volumes of the thin slices across the full horizontal span of the cone? What is the exact value of this definite integral?
Compare the result of your work in (d) to the volume of the cone that comes from using the formula cone.
Hint.
The line passes through the points and .
Use your answer to the previous question.
A cylinder of radius and height will have volume , so use the answer you found in part (b.) for the radius.
Your answer to part (c.) is the volume of one cylinder, so this will become the integrand (replacing with ).
The two methods for finding the volume should agree.
Answer.
.
The radius at some value will be .
The two methods for finding the volume agree: the volume formula for the cone gives , the same as the integral method.
Solution.
We are trying to find the formula for a linear function which passes through the points and . The -intercept is therefore 3, and the slope is , so the linear equation is .
The radius at some value will be .
A cylinder of radius and height will have volume . For this cylinder, the radius is and the height is , so the volume of this cylinder is
To find the volume of the cone, we will sum up a bunch of cylinders, and take the limit as the width of the cylinders goes to 0. When we do this, the sum turns into an integral and turns into . Thus the volume of the cone is
The two methods for finding the volume agree: the volume formula for the cone gives , the same as the integral method.
🔗
We will also investigate one more application of definite integrals, to find the length of curves by again splitting the curve into small slices and adding up the smaller lengths.
🔗
Subsection 6.2.1 The Volume of a Solid of Revolution
🔗
A solid of revolution is a three dimensional solid that can be generated by revolving one or more curves around a fixed axis. For example, the circular cone in Figure 6.17 is the solid of revolution generated by revolving the portion of the line from to about the -axis. Notice that if we slice a solid of revolution perpendicular to the axis of revolution, the resulting cross-section is a circle.
🔗
We first consider solids whose slices are thin cylinders. Recall that the volume of a cylinder is given by .
Example 6.18.
Find the volume of the solid of revolution generated when the region bounded by and the -axis is revolved about the -axis.
Solution.
First, we observe that intersects the -axis at the points and . When we revolve the region about the -axis, we get the three-dimensional solid pictured in Figure 6.19.
We slice the solid into vertical slices of thickness between and . A representative slice is a cylinder of height and radius . Hence, the volume of the slice is
slice.
Using a definite integral to sum the volumes of the representative slices, it follows that
.
It is straightforward to evaluate the integral and find that the volume is .
🔗
For a solid such as the one in Example 6.18, where each slice is a cylindrical disk, we first find the volume of a typical slice (noting particularly how this volume depends on ), and then integrate over the range of -values that bound the solid. Often, we will be content with simply finding the integral that represents the volume; if we desire a numeric value for the integral, we typically use a calculator or computer algebra system to find that value.
🔗
This method for finding the volume of a solid of revolution is often called the disk method.
🔗
The Disk Method.
🔗
🔗
If is a nonnegative continuous function on , then the volume of the solid of revolution generated by revolving the curve about the -axis over this interval is given by
.
🔗
A different type of solid can emerge when two curves are involved, as we see in the following example.
Example 6.20.
Find the volume of the solid of revolution generated when the finite region that lies between and is revolved about the -axis.
Solution.
First, we must determine where the curves and intersect. Substituting the expression for from the second equation into the first equation, we find that . Rearranging, it follows that
,
and the solutions to this equation are and . The curves therefore cross at and .
When we revolve the region about the -axis, we get the three-dimensional solid pictured at left in Figure 6.21.
Immediately we see a major difference between the solid in this example and the one in Example 6.18: here, the three-dimensional solid of revolution isn’t “solid” because it has open space in its center along the axis of revolution. If we slice the solid perpendicular to the axis of revolution, we observe that the resulting slice is not a solid disk, but rather a washer, as pictured at right in Figure 6.21. At a given location between and , the small radius of the inner circle is determined by the curve , so . Similarly, the big radius comes from the function , and thus .
To find the volume of a representative slice, we compute the volume of the outer disk and subtract the volume of the inner disk. Since
,
it follows that the volume of a typical slice is
slice.
Using a definite integral to sum the volumes of the respective slices across the integral, we find that
.
Evaluating the integral, we find that the volume of the solid of revolution is .
🔗
This method for finding the volume of a solid of revolution generated by two curves is often called the washer method.
🔗
The Washer Method.
🔗
🔗
If and are nonnegative continuous functions on that satisfy for all in , then the volume of the solid of revolution generated by revolving the region between them about the -axis over this interval is given by
.
Example 6.22.
In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find.
The region bounded by the -axis, the curve , and the line ; revolve about the -axis.
The region bounded by the -axis, the curve , and the line ; revolve about the -axis.
The finite region bounded by the curves and ; revolve about the -axis.
The finite region bounded by the curves and ; revolve about the -axis.
The region bounded by the -axis, the curve , and the line ; revolve about the -axis. How is this problem different from the one posed in part (b)?
Hint.
Use slices perpendicular to the -axis.
Note that slices perpendicular to the -axis will be washers.
Find the points where the two curves intersect and draw a labeled graph.
Think about how the solid looks like a donut; what is the outer radius?
Try using slices perpendicular to the -axis.
Answer.
.
.
.
.
.
Solution.
A typical slice (perpendicular to the -axis) has radius , and thus the volume of such a slice is
slice.
It follows by the disk method that the volume of the solid of revolution is
.
2. A typical slice (perpendicular to the -axis) is a washer with outer radius and inner radius . The volume of such a slice is
slice.
It follows by the washer method that the volume of the solid of revolution is
.
3. The curves and intersect at and ; revolving the region about the -axis and taking slices perpendicular to the -axis, we see that a typical slice is a washer with outer radius and inner radius . Hence the volume of a typical slice is
slice.
By the washer method, the volume of the solid is
.
4. The curves and intersect at the points ; after we revolve the region about the -axis, cross-sections perpendicular to the axis of rotation are washers with outer radius and inner radius . By the washer method, we see that
.
5. The curve and the line intersect at ; when we revolve the region about the -axis, slices perpendicular to the -axis are disks. Here we see that each disk’s radius is a function of and the disks have thickness , so we will need to integrate with respect to and consider the radius function (which comes from solving for in terms of . Note how this is different from the problem in part (b) because of the axis we rotate about and how this forces us to integrate in terms of . A typical slice has volume
slice,
and thus by the disk method the total volume is
.
🔗
Subsection 6.2.2 Revolving About the -axis
🔗
When we revolve a given region about the -axis, the representative slices now have thickness , which means that we must integrate with respect to .
Example 6.23.
Find the volume of the solid of revolution generated when the region that lies between and is revolved about the -axis.
Solution.
These two curves intersect when , hence at the point . When we revolve the region about the -axis, we get the three-dimensional solid pictured at left in Figure 6.24.
Note that the slices are cylindrical washers only if taken perpendicular to the -axis. We slice the solid horizontally, starting at and proceeding up to . The thickness of a representative slice is , so we must express the integrand in terms of . The inner radius is determined by the curve , so we solve for and get . In the same way, we solve the curve (which governs the outer radius) for in terms of , and hence . Therefore, the volume of a typical slice is
slice.
We use a definite integral to sum the volumes of all the slices from to . The total volume is
.
It is straightforward to evaluate the integral and find that .
Example 6.25.
In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find.
The region bounded by the -axis, the curve , and the line ; revolve about the -axis.
The region bounded by the -axis, the curve , and the line ; revolve about the -axis.
The finite region in the first quadrant bounded by the curves and ; revolve about the -axis.
The finite region in the first quadrant bounded by the curves and ; revolve about the -axis.
The finite region bounded by the curves and ; revolve about the -axis
Hint.
Consider slices perpendicular to the -axis.
Since we revolve around the -axis, think about slices of thickness .
Note that lies below on the interval .
The curve can be equivalently expressed as .
Find the intersection points by writing the second curve as and solving .
Answer.
.
.
.
.
.
Solution.
Observe that horizontal slices with thickness are all cylindrical disks, each with radius given by since they are given by the curve , which can be equivalently expressed as . The volume of each slice is thus
slice.
The volume of the solid is thus given by the definite integral
.
2. Slicing perpendicular to the axis of rotation, we get slices of thickness that are shaped like washers. The outer radius, , is given by ; the innner radius is given by the function , but expressed with as a function of , so . It follows that a representative washer slice has volume
slice.
Further, since and intersect at the point , we integrate from to and thus find that the total volume is
.
3. The curves and intersect where , so where . The two nonnegative values of that make this equation true are and , and thus the points and form the boundaries of the region that is being rotated about the -axis. Since we rotate about the horizontal axis, we slice perpendicular to that axis with slices of thickness . A representative slice is in the shape of a washer and has volume
slice,
and thus the total volume of the solid is
.
4. To use the same region as in part (c), but now revolved about the -axis, we shift to horizontal slices of thickness , and thus need to express the curves as functions of : the rightmost curve is , while the leftmost curve is . A representative slice is a washer whose volume is
slice.
It follows that the total volume of the solid is
.
5. For this pair of curves, we see they intersect where and are both true, and hence where . Expanding, moving all terms to the left, and combining like terms, we see that , so . It follows that and therefore or . Plotting the region and revolving about the -axis, we see that horizontal slices of thickness are shaped like washers. The outer radius is given by , while the inner radius is . A representative slice thus has volume
slice.
The volume of the solid of revolution is therefore
.
🔗
Subsection 6.2.3 Revolving About Horizontal and Vertical Lines other than the Coordinate Axes
🔗
It is possible to revolve a region around any horizontal or vertical line. Doing so adjusts the radii of the cylinders or washers involved by a constant value. A careful, well-labeled plot of the solid of revolution will usually reveal how the different axis of revolution affects the definite integral.
Example 6.26.
Find the volume of the solid of revolution generated when the finite region that lies between and is revolved about the line .
Solution.
Graphing the region between the two curves in the first quadrant between their points of intersection ( and ) and then revolving the region about the line , we see the solid shown in Figure 6.27. Each slice of the solid perpendicular to the axis of revolution is a washer, and the radii of each washer are governed by the curves and . But we also see that there is one added change: the axis of revolution adds a fixed length to each radius. The inner radius of a typical slice, , is given by , while the outer radius is .
Therefore, the volume of a typical slice is
slice.
Finally, we integrate to find the total volume, and
.
Example 6.28.
In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated axis. In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid. It is not necessary to evaluate the integrals you find. For each prompt, use the finite region in the first quadrant bounded by the curves and .
Revolve about the line .
Revolve about the line .
Revolve about the line .
Revolve about the line .
Hint.
Note that the two curves intersect in the first quadrant at and .
The outer radius is .
Solve the equations governing the two curves for in terms of .
The inner radius is .
Answer.
.
.
.
.
Solution.
The two curves intersect in the first quadrant at and . Revolving the region about the line , we see that a typical slice is a washer with thickness , outer radius , and inner radius . It follows by the washer method that
.
2. In this situation, revolving about , we see that the outer radius of a typical washer is , while the inner radius is , and thus the volume is
.
3. Because we are revolving about , when we slice perpendicular to this we get washers of thickness , and thus we need to write the two curves as functions of in order to determine the radii of the slices as functions of , we note that the line can be expressed as and the cubic curve can be rewritten in the form . The outer radius of a typical slice is thus and the inner radius is . It follows that the volume of the solid is
.
4. Here, the outer radius is and the inner radius is , so the total volume of the solid is
.
🔗
Subsection 6.2.4 Finding the Length of a Curve
🔗
We can also use the definite integral to find the length of a portion of a curve. We use the same fundamental principle: we slice the curve up into small pieces whose lengths we can easily approximate. Specifically, we subdivide the curve into small approximating line segments, as shown at left in Figure 6.29.
🔗
🔗
We estimate the length slice of each portion of the curve on a small interval of length . We use the right triangle with legs parallel to the coordinate axes and hypotenuse connecting the endpoints of the slice, as seen at right in Figure 6.29. The length, , of the hypotenuse approximates the length, slice, of the curve between the two selected points. Thus,
slice.
🔗
🔗
Next we use algebra to rearrange the expression for the length of the hypotenuse into a form that we can integrate. By removing a factor of , we find
slice.
🔗
🔗
Then, as and , we have that . Thus, we can say that
slice.
🔗
Taking a Riemann sum of all of these slices and letting , we arrive at the following fact.
🔗
Arc Length.
🔗
🔗
Given a differentiable function on an interval , the total arc length, , along the curve from to is given by
.
Example 6.30.
Each of the following questions somehow involves the arc length along a curve.
Use the definition and appropriate computational technology to determine the arc length along from to .
Find the arc length of on the interval . Find this value in two different ways: (a) by using a definite integral, and (b) by using a familiar property of the curve.
Determine the arc length of on the interval .
Will the integrals that arise calculating arc length typically be ones that we can evaluate exactly using the First FTC, or ones that we need to approximate? Why?
A moving particle is traveling along the curve given by , and does so at a constant rate of 7 cm/sec, where both and are measured in cm (that is, the curve is the path along which the object actually travels; the curve is not a “position function”). Find the position of the particle when sec, assuming that when , the particle’s location is .
Hint.
Recall that .
Remember that generates a circle centered at of radius 4.
Apply the arc length formula. Use technology to evaluate the integral.
Most expressions involving square roots are difficult to antidifferentiate.
Here you can determine the arc length and then experiment to find the upper limit of integration, which will help you determine position.
Answer.
.
.
.
We will usually have to estimate the value of using computational technology.
Approximately .
Solution.
Using the formula for arc length, we know
.
Evaluating the integral, we find .
2. The curve is a semi-circle of radius 2, so its length must be . This is confirmed by the arc length formula, since , so
.
It follows1
This integral is actually improper because the integrand is undefined at the endpoints, . We learned how to evaluate such integrals in Section 5.10.
that
.
3. For , we observe that , and that
.
Thus, the arc length of the curve on is
.
4. Because will rarely simplify nicely and rarely have an elementary antiderivative, we will usually have to estimate the value of using computational technology.
5. After 4 seconds traveling at 7 cm/sec, the particle has moved 28 cm. So, we have to find the value of for which the arc length along is 28. Thus, we know that
.
Rewriting the integral, we see
.
Experimenting with different values of in a computational engine like Wolfram|Alpha, we find that for , , and thus the position of the particle when is approximately .
🔗
🔗
The same technique can be applied to parametric curves. For now consider a curve travelling in just two dimensions (but see Example 6.33 for curves in 3D). As before, the length of a small segment will be , but here we will factor out to get:
🔗
Now, taking the limit as , and turn into and , so the length of the segment is
.
🔗
We summarize our result here:
🔗
Arc Length for a Parametric Curve.
🔗
🔗
Given two differentiable functions on an interval , the total arc length along the parametric curve from to is given by
.
Example 6.31.
Recall that the unit circle is traced out by the parametric equations and on the interval . Set up and evaluate the arc length integral to find the circumference of the unit circle.
An ellipse (a stretched circle) can be traced out by the parametric equations and . Set up the arc length integral to find the perimeter of this ellipse. You do not have to evaluate the integral.
Hint.
Use the formula for the arc length for a parametric curve to set up the integral. Recall also the Pythagorean Identity for trig functions: .
Use the formula for the arc length for a parametric curve to set up the integral.
Answer.Solution.
Since and , then
Since and , then
We do not need to evaluate this integral, and in fact it is an example of an “elliptical integral of the second kind”, which are a famous class of integrals without elementary antiderivatives.
Example 6.32.
Alex drives to work, and their route is along a path given by , for . How long is their route to work? (You may wish to use a calculator or computer to numerically evaluate the integral.)
Hint.
Set up the integral using formula for arc length for a parametric curve. Then use a computer to evaluate the integral.
Answer.Solution.
Since and , then
Example 6.33.
Suppose that one has a parametric curve travelling through 3 dimensions given by . How would you change the “Arc Length for a Parametric Curve” formula to include the third dimension?
Hint.
Treat the -coordinate the same way the - and -coordinates are treated.
Answer.Solution.
The length of a 3D line segment is given by , so factoring out a factor of and taking the limit as , we get
🔗
Subsection 6.2.5 Summary
🔗
A definite integral may be used to represent the volume of a solid. The idea is to slice the solid into thin shapes whose volume is easy to calculate. The volume of each slice is found by taking the area of the cross-section and multiplying it by the width. Summing up the volumes of the slices and taking the limit as the width results in an integral representing the volume.
A three-dimensional solid of revolution results from revolving a two-dimensional region about a particular axis or line called the axis of revolution. We can use a definite integral to find the volume of such a solid by taking slices perpendicular to the axis of revolution. The slices will be circular disks or washers.
If we revolve about a horizontal line and slice perpendicular to that line, then our slices are vertical and of thickness . This leads us to integrate with respect to .
If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness . This leads us to integrate with respect to .
If we revolve about a line other than the - or -axis, we need to carefully account for the shift that occurs in the radius of a typical slice. Normally, this shift involves taking a sum or difference of the function along with the constant connected to the equation for the horizontal or vertical line; a well-labeled diagram is usually the best way to decide the new expression for the radius.
The arc length, , along the curve from to is given by
.
The definite integral may also be used to find arc length of parametric curves.
🔗
Exercises 6.2.6 Exercises
🔗
1. Solid of revolution from one function about the -axis.
🔗
The region bounded by is rotated around the -axis. Find the volume.
🔗
volume =
🔗
2. Solid of revolution from one function about the -axis.
🔗
Find the volume of the solid obtained by rotating the region in the first quadrant bounded by , , and the -axis around the -axis.
🔗
Volume =
🔗
3. Solid of revolution from two functions about the -axis.
🔗
Find the volume of the solid obtained by rotating the region in the first quadrant bounded by , , and the -axis around the -axis.
🔗
Volume =
🔗
4. Solid of revolution from two functions about a horizontal line.
🔗
Find the volume of the solid obtained by rotating the region in the first quadrant bounded by , , and the -axis about the line .
🔗
Volume =
🔗
5. Solid of revolution from two functions about a different horizontal line.
🔗
🔗
Find the volume of the solid obtained by rotating the region bounded by the curves
🔗
about the line .
🔗
Answer:
🔗
6. Solid of revolution from two functions about a vertical line.
🔗
Find the volume of the solid obtained by rotating the region bounded by the given curves about the line
🔗
🔗
Answer:
🔗
7. Arc length and area of a region, and volume of its solid of revolution.
Consider the curve and the portion of its graph that lies in the first quadrant between the -axis and the first positive value of for which . Let denote the region bounded by this portion of , the -axis, and the -axis.
Set up a definite integral whose value is the exact arc length of that lies along the upper boundary of . Use technology appropriately to evaluate the integral you find.
Set up a definite integral whose value is the exact area of . Use technology appropriately to evaluate the integral you find.
Suppose that the region is revolved around the -axis. Set up a definite integral whose value is the exact volume of the solid of revolution that is generated. Use technology appropriately to evaluate the integral you find.
Suppose instead that is revolved around the -axis. If possible, set up an integral expression whose value is the exact volume of the solid of revolution and evaluate the integral using appropriate technology. If not possible, explain why.
🔗
8. Solid of revolution from a two functions about multiple horizontal and vertical lines.
Consider the curves given by and . For each of the following problems, you should include a sketch of the region/solid being considered, as well as a labeled representative slice.
Sketch the region bounded by the -axis and the curves and up to the first positive value of at which they intersect. What is the exact intersection point of the curves?
Set up a definite integral whose value is the exact area of .
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving about the -axis.
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving about the -axis.
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving about the line .
Set up a definite integral whose value is the exact volume of the solid of revolution generated by revolving about the line .
🔗
9. Area and perimeter of a region and volume of a solid of revolution around multiple lines.
Consider the finite region that is bounded by the curves , , and .
Determine a definite integral whose value is the area of the region enclosed by the two curves.
Find an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region about the line .
Determine an expression involving one or more definite integrals whose value is the volume of the solid of revolution generated by revolving the region about the -axis.
Find an expression involving one or more definite integrals whose value is the perimeter of the region .
🔗
10. Arc length of a curve.
🔗
Find the arc length of the graph of the function from to .
🔗
arc length =
🔗
11. Length of a parametric curve.
The parametric function given by , traces out a figure-8 pattern for . Find the length of this curve. (You may wish to use a calculator or computer to numerically evaluate the integral.)
Hint.
Set up the integral using the “Arc Length for a Parametric Curve” formula, then use a computer to evaluate the integral.
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Enterocutaneous and Enteroatmospheric Fistulas - UpToDate
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Official reprint from UpToD ate www.uptodate.com© 2023 UpToDate, Inc. and/or its affiliates. All Rights Reserved.
Enterocutaneous andenteroatmos pheric f istulas
INTRODUCTION
A fistula is an abnormal connection between two epithelialized hollow spaces or organs. Strictly speaking, an enterocutaneous fistula connects the small bowel to the skin. A more liberal interpretation of the term, however, also includes enteric fistulas originating f rom the colon,stomach, and esophagus. (See 'Classification'below.)The clinical features, diagnosis, and management of enterocutaneous fistulas are reviewed here. Enteroatmospheric fistulas, which are a subset of enterocutan eous fistulas that occur in the setting of an open abdomen, are also discussed here.Fistulas that occur in other areas of the body are discussed in other topics, including:
®
:
Sharon L Stein, MD, FACS, FASCRS
:
Eileen M Bulger, MD, FACS,J Thomas Lamont, MD, David I Soybel, MD
:
Wenliang Chen, MD, PhD All topics are updated as new evidence becomes available and our peer review process is complete.Literature review current through:
Nov 2023.
This topic last updated:
Dec 03, 2020.
Colovesical fistulas (see "Colovesical fistulas")
●
Urogenital tract fistulas in women (see "Urogenital tract fistulas in females"and"Rectovaginal and anovaginal fistulas")
●
Anorectal fistulas (see "Anorectal fistula: Clinical manifestations and diagnosis"and"Operative management of anorectal fistulas"and "Perianal Crohn disease")
●
Pancreatic fistulas
●
Duodenal fistulas (see "Postgastrectomy duodenal leak")
●
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CLASSIFICATION
Enterocutaneou s fistulas can be classified according to source, output volume, and etiology.
By source
— The most common method of defining a fistula is by its organ of origin (eg,gastro-, duodeno-, entero-, jejuno-, ileo-, colo-, recto-) followed by the point of its termination(eg, -cutaneous, -atmospheric). However, at initial presentation, the specific segment of bowel involved (ie, the enteric origin of the fistula) is often unknown, and a broader designation (eg,enterocutaneo us rather than jejunocutaneous) is used until the anatomy of the fistula can b e delineated. (See 'Diagnosis'below.)Fistulas originating from different organ systems will have very different outputs. Electrolyte and nutritional losses can vary greatly between organs of origin. As an example, gastric output has a higher acidity. This information may prove essential in the management of enterocutaneo us fistulas. (See 'Fluid therapy'below.)
By output volume
— The physiologic consequences of enterocutaneous fistulas may depend on the quantity of fluid put out by the fistula, which may differ by the location (proximal versus distal) and the length or diameter of the fistula. An alternative classification defines fistulas by the quantity of their output:In general, high-output fistulas are less likely to heal spontaneously, and patients with high-output fistulas are at a higher risk for metabolic disturbances, fluid loss, and malnutrition. For such patients, controlling external fistula output is as important as replacing fluid and electrolyte loss during the chronic phase of management. (See 'Fistula output reduction'below.)
By etiology
— Enterocutaneous fistulas can also b e classified according to their etiologies (eg,iatrogenic, spontaneous), which are further discussed below. (See 'Etiology and risk factors'below.)
EPIDEMIOLOGY
Tr acheoesophageal fistulas (see "Trache o- and broncho-esophageal fistulas in adults")
●
Bronchopleur al fistulas (see "Bronchopleu ral fistula in adults")
●
A low-output fistula drains less than 200 mL/day.
●
A moderate-output fistula drains between 200 and 500 mL/day.
●
A high-output fistula drains more than 500 mL/day.
●
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Although a few studies have reported postoperative prevalences of enterocutaneous fistulas in patients undergoing surgery for trauma (1.5 percent ), general surgery (3.6 percent ), and Crohn disease (15 to 35 percent [3,4]), the prevalence of enterocutaneous fistulas in the general population (including both iatrogenic and spontaneous cases) is not well known. The National Inpatient Sample (NIS) noted 317,000 admissions between 2004 and 2014 with a diagnosis of enteric fistula, costing the United States hospital systems more than 500 million dollars annually .
ETIOLOGY AND RISK FACTORS
Various etiologies can lead to fistula f ormation (table 1). Most enterocutaneous fistulas are iatrogenic/postoper ative (75 to 85 percent); a minority (15 to 25 percent) develop s pontaneously.
Iatrogenic fistulas
— Iatrogenic causes of enterocutaneo us fistula include trauma and surgery(ie, postoperative fistulas). Postoperative fistulas develop from either a bowel anastomotic leak(50 percent) or a missed enterotomy (45 percent), with a small percentage from erosion of foreign material (eg, mesh for hernia repair, vascular graft) into adjacent bowel . Preoper ative factors that increase the likelihood of the development of a postoperative fistula include Crohn disease, malnutrition, immunosuppression, traumatic injury, infection, smoking, and emergency procedures [8,9].
Enteroatmospheric fistulas
— Enteroatmospheric or exposed fistulas occur in the midst of an open abdomen with no overlying soft tissue (picture 1). The open abdomen is usually the result of a damage control laparotomy, which leaves the abdomen open after trauma or emergency surgery to prevent intra-abdomin al hypertension/abdominal compartment syndrome [10,11]. (See "Abdominal compartment syndrome in adults".)A review of 517 patients from the American Association for Surgery in Trauma (AAST) registry with open abdomen found that 111 patients (21 percent) developed an enterocutaneous fistula,an enteroatmospheric fistula, or intra-abdominal sepsis. Independent risk factors included large bowel resection, large-volume resuscitation, and an increasing number of abdominal re-exploratio ns . The risks of open abdomen are further discussed s eparately. (See"Management of the open abdomen in adults", section on 'Complications of open abdomen'.)
Spontaneous fistulas
— The mnemonic "FRIEND" describes common etiologies for spontaneous enterocutaneous fistulas, which include
f
oreign body,
r
adiation,
i
nflammation (eg,Crohn disease) or
i
nfection (eg, tuberculosis, actinomycosis),
e
pithelialization,
n
eoplasia, and
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d
istal obstruction . These etiologies also describe conditions in which fistulas are less likely to heal spontaneously. Diverticular disease and appendicitis are also reported as uncommon etiologies of fistulas.Crohn disease is the most common etiology for spontaneous fistulas; the p revalen ce varies by patient populations [10,13]. Between 27 and 35 percent of patients undergoing surgery for Crohn disease had a fistula [3,4], while 15.4 percent of patients with complicated Crohn disease,but who had not had surgery, had a fistula . Crohn-related fistulas are typically treated initially with immunosuppressive drugs (thiopurines), biologic agents (TNF inhibitors), and antibiotics (metronidazole) to reduce inflammation and diarrhea. Medical treatment of Crohn-related fistulas is discussed in another topic. (See "Clinical manifestations, diagnosis, and prognosis of Crohn disease in adults", s ection on 'Features of transmural inflammation'.)
CLINICAL FEATURES
The most common presentation of an enterocutaneous fistula is in a postoperative patient who fails to recover normally from abdominal surgery. The patient often presents first with abdominal symptoms, including increased pain, nausea and vomiting, obstipation, and fullness or induration of the abdominal wall. These may be accompanied by fever and leukocytosis. A wound infection is then typically recognized 7 to 10 days postoperatively, and following incisional drainage, enteric contents appear in the surgical wound . An enteric fistula can be distinguished from a wound infection by the presence of bile in the wound. As opposed to infection or seroma, bilious output will stain gauze and dressings an orange/brow n or green color. Frank stool can also be noted in the wound of a colonic fistula.Enterocutaneou s fistulas are characteri zed by leakage of enteric or bowel contents through the abdominal wall. The leakage is also referred to as effluent. Depending upon the location and origin of the fistula, it can be through a prior abdominal wound, incision, or an area of "virgin"abdomen. Leakage of effluent can cause irritation of the skin, loss of fluids and electrolytes,malnutrition, and infection. In addition, as a fistula is forming, the patient may become acutely ill secondary to leakage of effluent into the abdominal cavity.
DIAGNOSIS
The diagnosis of an enterocutaneous fistula is clinical. It should be suspected in a postoperative patient who fails to recovery normally from abdominal surgery and has bilious wound drainage.The diagnosis of enterocutan eous fistula can be confirmed by the appearance of enteric
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187754 | https://math.stackexchange.com/questions/4996820/differentiability-of-piecewise-function-at-endpoint | calculus - Differentiability of piecewise function at endpoint - Mathematics Stack Exchange
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Differentiability of piecewise function at endpoint
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Let f f be a function such that: f:[π/4,π/2]→R f:[π/4,π/2]→R
f(x)={1 tan x 0 π/4≤x<π/2,x=π/2.f(x)={1 tanx π/4≤x<π/2,0 x=π/2.
How can I show that f f is differentiable on the closed interval [π/4,π/2][π/4,π/2]? I can easily show continuity but I’m not sure what to do to show it’s differentiable on the closed interval.
Thank you!
calculus
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derivatives-of-piecewise-functions
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edited Nov 11, 2024 at 1:40
Enforce
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Do you know one side differentiability? en.wikipedia.org/wiki/Semi-differentiabilityzkutch –zkutch 2024-11-11 00:36:06 +00:00 Commented Nov 11, 2024 at 0:36
The typical approach here would be to calculate the derivative away from your "problem point" (in this case π/2 π/2), then calculate the derivative explicitly via the definition at π/2 π/2 and show that the resulting function is continuous. Have you tried that?Enforce –Enforce 2024-11-11 00:36:57 +00:00 Commented Nov 11, 2024 at 0:36
@zkutch I don't think it matters in this case because the function is only defined on the closed interval, so the derivaitve at π/2 π/2 is the one-sided derivative.Enforce –Enforce 2024-11-11 00:39:27 +00:00 Commented Nov 11, 2024 at 0:39
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f(π 2)=0 f(π 2)=0 is used significantly in the following formula:
f(x)−f(π 2)x−π 2=1 tan x−0 x−π 2=−1⋅sin(π 2−x)(π 2−x)⋅1 sin x f(x)−f(π 2)x−π 2=1 tanx−0 x−π 2=−1⋅sin(π 2−x)(π 2−x)⋅1 sinx
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edited Nov 11, 2024 at 12:21
answered Nov 11, 2024 at 1:51
zkutchzkutch
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Do you know sin y y→1,y→0 siny y→1,y→0?zkutch –zkutch 2024-11-11 02:01:08 +00:00 Commented Nov 11, 2024 at 2:01
Yes, but I don’t understand how it’s helpful since I already know the left side derivative at π/2,π/2, which is equal to −1.−1. The only issue I have is with concluding that the function is differentiable over the closed interval.No Name –No Name 2024-11-11 02:03:17 +00:00 Commented Nov 11, 2024 at 2:03
Consider x→π 2,x<π 2 x→π 2,x<π 2, then it is definition of derivative from left.zkutch –zkutch 2024-11-11 02:56:32 +00:00 Commented Nov 11, 2024 at 2:56
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cot x cotx is a continuous ( not piecewise) function including interval (x,0,π)(x,0,π).
By differentiation its derivative −csc 2(x)=−1−csc 2(x)=−1 at point P (π/2,0)(π/2,0) which is a point anyhow right the curve without any slope change.
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answered Nov 11, 2024 at 12:39
NarasimhamNarasimham
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187755 | https://education.ti.com/-/media/C4950CEE91AB4013862AA3E0FAA39B1D | Critical Points and Local Extrema TEACHER NOTES
MATH NSPIRED
©2019 Texas Instruments Incorporated 1education.ti.com
Math Objectives
Students will identify critical points using the definition.
Students will identify local maxima and minima using the
definition.
Students will understand that local maxima and minima must
occur at critical points but that not every critica l point is the
location of a local maximum or local minimum.
Students will reason abstractly and quantitatively. (CCSS
Mathematical Practice)
Students will construct viable arguments and critique the
reasoning of others. (CCSS Mathematical Practice)
Voc abulary
critical point
local maximum, minimum, extrema
About the Lesson
This lesson involves visualizing the connections between the
critical points and local extrema.
As a result, students will :
Zoom in on function graphs at different types of critical
points (including stationary points, locations of vertical
tangents, “corners,” and cusps) to determine whether the
slope of the tangent line is zero or undefined.
See that a local maximum or minimum occurs at critical
points, but the examples illustrate th at not every critical
point is a local extremum.
Use the first derivative test as a means to identify local
maximum and local minimum.
Build on their familiarity with the concept of the derivative
at a point as the local slope of the function graph at that
point.
TI -Nspire™ Navigator™ System
Use Quick Poll to assess students’ understanding.
Use Screen Capture to share students’ work .
Collect student documents and analyze the results.
Utilize Class Analysis to display students’ answers.
TI -Nspire™ Te chnology Skills:
Download a TI -Nspire document
Open a document
Move between pages
Grab and drag a point
Move a slider bar
Grab and drag points along a
graph
Move between screen panes on
a single page
Tech Tips:
Make sure the font size on your
TI -Nspire handheld is set to
Medium.
You can hide the function entry
line by pressing / G.
Lesson Materials:
Student Activity
Critical_Points_Student. pdf
Critical_Points_Student. doc
TI -Nspire document
Critical_Points.tns
Visit www.mathnspired.com for
lesson updates and tech tip videos. Critical Points and Local Extrema TEACHER NOTES
MATH NSPIRED
©2019 Texas Instruments Incorporated 2education.ti.com
Discussion Points and Possible Answers
Tech Tip: When on a page having multiple screens in its layout, / e
will move from screen to screen. If the pointer arrow cursor is available,
you can simply move to another screen pane using it, but you must click
once inside the pane to make it active. An active screen pane has a bold
outline around it.
Move to page 1.2.
The graph of the differentiable function shown in the left window has a box centered around the point (1, 2) . Drag the point on the line segment at the top to see a “zoomed in” view of this boxed
area of the graph in the right window.
a. This function has a local minimum at x = 1. Using the graph and the definition of local minimum above, explain why.
Sample a nswer: The function has a local minimum at x = 1 because the function values for points near x = 1 are all greater than 2; or x = 1 is a local minimum because the graph “goes up”
on either side of the point (1, 2), so the function value here is less than the function value at neighboring points.
b. What appears to happen to the graph as you zoom in on the point (1, 2)?
Answer: The graph appears to be a horizontal line.
c. What is f′(1)? Explain your answer. Why is c = 1 a critical point of f?
Answer: f′(1) = 0 because the slope of the horizontal tangent line is 0. Using the definition above, c = 1 is a critical point because f′(1) = 0.
Teacher Tip: If students have trouble connecting the ideas in 1b and 1c,
you may wish to review the following: If a function f is differentiable at
x = a, then its graph will appear to become linear as you zoom in on the
point ( a, f (a)). The derivative f′(a) is the “local slope” of the graph of
y = f(x) at the point ( a, f(a)) .Critical Points and Local Extrema TEACHER NOTES
MATH NSPIRED
©2019 Texas Instruments Incorporated 3education.ti.com
Move to page 2.1.
This is the graph of a function having a local maximum at x = –2.
a. What appears to happen to the graph as you zoom in on the
point ( –2, 1)?
Answer: The graph appears to be a horizontal line.
b. What is the value of f′(–2)? Explain your answer. Why is
c = –2 a critical point of f?
Answer: f′(–2) = 0 because the slope of the horizontal tangent line is 0. Using the definition
above, c = –2 is a critical point because f′(–2) = 0.
c. What value could the derivative of a func tion have at the location of a local maximum or
minimum? Explain your answer.
Sample answer : Students may reason that the derivative must be zero at a minimum or
maximum value because the graph will always “flatten out” when you zoom in at a “peak” or “lo w
point.” Or, the derivative must be zero because the tangent line would need to be horizontal at the
local minimum and maximum points.
Teacher Tip: Given the examples thus far, it is reasonable for students to
conclude that the derivative must equal zer o at a local maximum or
minimum. The next two examples are used to illustrate that the derivative
may not exist at a local minimum or maximum and thus just locating points
for which the derivative of a function is equal to zero is not sufficient for
findin g local extrema.
Move to page 3.1.
This is the graph of a function having a local minimum at x = –1.
a. What happens to the graph as you zoom in on the point
(–1, –2)?
Answer: The graph does not appear to change as you zoom in. The V shape of t he graph is
maintained. Critical Points and Local Extrema TEACHER NOTES
MATH NSPIRED
©2019 Texas Instruments Incorporated 4education.ti.com
b. Assuming this behavior persists no matter how far you zoom in, is this function differentiable at
x = –1? Why or why not?
Answer: The function is not differentiable at –1 because the graph does not become linear as
you zoom i n. (Instead, there are “two lines” with different slopes .)
Teacher Tip: The purpose of the next two examples is to illustrate that not
all critical points are local maximum or minimum points of a function. In
other words, critical points must be tested.
Move to page 4.1.
The graph of this increasing function has a horizontal tangent at
the point x = 2.
a. Is x = 2 a critical point? Why or why not?
Answer: x = 2 is a critical point because f′(2) = 0; when you zoom in on the point (2, 3), the
grap h appears horizontal.
b. Does f have either a local minimum or local maximum at x = 2?
Answer: No, x = 2 is neither a local minimum nor a local maximum, since the graph is strictly
increasing everywhere.
Teacher Tip: CAUTION: It is important to no te that a few examples do not
constitute a proof of this statement. The examples given have provided
only an intuitive or visual basis for this relationship.
This might be an appropriate time to formalize student conjectures by
providing a statement of Fe rmat’s theorem: if f has a local maximum or
minimum at c, then c is a critical number of f.
Move to page 5.1.
The graph of this increasing function has a vertical tangent at the
point x = –2.
a. Is x = –2 a critical point? Why or why not? Critical Points and Local Extrema TEACHER NOTES
MATH NSPIRED
©2019 Texas Instruments Incorporated 5education.ti.com
Answer: x = –2 is a critical point because f ′(2) is undefined; when you zoom in on the point
(–2, 3), the graph appears vertical. The slope of the vertical tangent line is undefined.
TI -Nspire Navigator Opportunity: Quick Poll or Screen Capture
See Note 1 at the end of this lesson.
b. Does f have either a local minimum or local maximum at x = –2?
Answer: No, x = –2 is neither a local minimum nor a local maximum, since the function is
increasing everywhere.
c. Does this contradict the statement you made in question 3d? Explain why or why not.
Answer: The answer d epends on student responses to question 3d. This does not contradict the
statement that local maxima and minima occur at critical points of a function. However, if students
equated critical points with local extrema, they need to modify the statement to note that while it is
true that all local maxima and minima of a function occur at critical points, not all critical points are
local extrema.
Teacher Tip: If you introduced Fermat’s theorem earlier, this is a good
time to point out that the converse of this theorem is not a true statement.
Wrap Up
Upon completion of the discussion, the teacher should ensure that students understand:
How to identify the critical points of a function given its graph.
Why the local minima or maxima of a function occur at its critical points.
Why not every critical point is a local minima or maxima.
Note: The assumption of continuity is another idea that should be addressed in the lesson wrap up. All of the functions presented in this lesson were continuous, a condition for the first derivative test. Although most functions students will encounter will be continuous, this is an important condition to note. Students could be challenged to consider what might happen if this condition were relaxed. Is it then possible for f
to have a critical point a such that f(a) is a local minimum, but the derivative f′(x) > 0 for x < a and
f′(x) < 0 for x > a? If so, what might it look like?
TI-Nspire Navigator
Note 1
Question 5a, Quick Poll or Screen Capture : You may want to send a Quick Poll or collect a Screen Capture to verify that students understand critical points and local maximum and minimums. |
187756 | https://www.indiabix.com/verbal-reasoning/seating-arrangement/ | Verbal Reasoning :: Seating Arrangement
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Exercise : Seating Arrangement - Seating Arrangement 1
Seating Arrangement - Introduction
Seating Arrangement - Seating Arrangement 1
Seating Arrangement - Seating Arrangement 2
Seating Arrangement - Seating Arrangement 3
Seating Arrangement - Seating Arrangement 4
Seating Arrangement - Seating Arrangement 5
Seating Arrangement - Seating Arrangement 6
Seating Arrangement - Seating Arrangement 7
Seating Arrangement - Seating Arrangement 8
Seating Arrangement - Seating Arrangement 9
Seating Arrangement - Seating Arrangement 10
Seating Arrangement - Seating Arrangement 11
1.
A, P, R, X, S and Z are sitting in a row. S and Z are in the centre. A and P are at the ends. R is sitting to the left of A. Who is to the right of P ?
A
X
S
Z
Answer: Option
Explanation:
The seating arrangement is as follows:
Therefore, right of P is X.
2.
There are 8 houses in a line and in each house only one boy lives with the conditions as given below:
Jack is not the neighbour Siman.
Harry is just next to the left of Larry.
There is at least one to the left of Larry.
Paul lives in one of the two houses in the middle.
Mike lives in between Paul and Larry.
If at least one lives to the right of Robert and Harry is not between Taud and Larry, then which one of the following statement is not correct ?
Robert is not at the left end.
Robert is in between Simon and Taud.
Taud is in between Paul and Jack.
There are three persons to the right of Paul.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
3.
A, B, C, D and E are sitting on a bench. A is sitting next to B, C is sitting next to D, D is not sitting with E who is on the left end of the bench. C is on the second position from the right. A is to the right of B and E. A and C are sitting together. In which position A is sitting ?
Between B and D
Between B and C
Between E and D
Between C and E
Answer: Option
Explanation:
Therefore, A is sitting in between B and C.
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Alburnus alburnus, Bleak : fisheries, aquaculture, bait
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Related species
Species in Alburnus
Species in Leuciscidae
Classification -
Leuciscinae
Leuciscidae
Cypriniformes
Teleostei
Chordata
Animalia
Alburnus alburnus (Linnaeus, 1758)
Bleak
Upload your photos and videos Pictures | Videos | Google imageAlburnus alburnus Picture by Muséum-Aquarium de Nancy/D. Terver
Classification / Names Common names | Synonyms | Catalog of Fishes(genus, species) | ITIS | CoL | WoRMS | Cloffa
Teleostei (teleosts) > Cypriniformes (Carps) > Leuciscidae (Minnows) > Leuciscinae Etymology: Alburnus: From the city of Al Bura, where the fish was known (Ref. 45335).More on author: Linnaeus.
Environment: milieu / climate zone / depth range / distribution range Ecology
Freshwater; brackish; benthopelagic; pH range: 7.0 - ? ; dH range: 10 - ?; potamodromous (Ref. 51243); depth range 1 - ? m (Ref. 30578). Temperate; 10°C - 20°C (Ref. 2059); 68°N - 35°N, 6°W - 60°E
Distribution Countries | FAO areas | Ecosystems | Occurrences | Point map | Introductions | Faunafri
Europe and Asia: most of Europe north of Caucasus, Pyrénées and Alps, eastward to Ural and Emba. Naturally absent from Iberian Peninsula, Adriatic and Aegean basins (except Maritza drainage), Italy, Ireland, Great Britain (except southeast), Norway and Scandinavia north of 67°N, Caspian basin south of Volga. In Anatolia, Marmara basin. Locally introduced in Spain, Portugal and Italy. At least one country reports adverse ecological impact after introduction.
Length at first maturity / Size / Weight / Age
Maturity: Lm 9.9, range 9 - ? cm Max length : 25.0 cm TL male/unsexed; (Ref. 30578); common length : 15.0 cm TL male/unsexed; (Ref. 30578); max. published weight: 60.00 g (Ref. 30578)
Short description Identification keys | Morphology | Morphometrics
Dorsal spines (total): 2 - 4; Dorsal soft rays (total): 7 - 9; Anal spines: 3; Anal soft rays: 14 - 20; Vertebrae: 41 - 44. Diagnosed from congeners in Europe by the possession of the following characters: origin of anal fin below branched dorsal rays 4-5; lateral line with 45-48 + 3 scales; anal fin with 17-20½ branched rays; 16-22 gill rakers; ventral keel exposed from anus to pelvic base; lateral stripe absent in life, faint or absent in preserved specimens; and mouth slightly superior (Ref. 59043). Caudal fin with 19 rays (Ref. 2196). Also Ref. 40476.Body shape (shape guide): fusiform / normal.
Inhabits open waters of lakes and medium to large rivers. Forms large aggregations in backwaters and other still waters during winter. Adults occur in shoals near the surface. Larvae live in littoral zone of rivers and lakes while juveniles leave shores and occupy a pelagic habitat, feeding on plankton, drifting insects or invertebrates fallen on the water surface (Ref. 59043). Feeds mainly on plankton, including crustaceans (Ref. 30578) and insects (Ref. 9696). Spawns in shallow riffles or along stony shores of lakes, occasionally above submerged vegetation (Ref. 59043). Excellent as bait for carnivorous fishes. May be captured using the smallest hook and a fly as bait. Its flesh is tasty (Ref. 30578). Of little interest to commercial or sport fisheries in its native range because of its small size (Ref. 1739). Scales were previously utilized in making Essence d"Orient, a coating for artificial pearls (Ref. 59043).
Life cycle and mating behavior Maturity | Reproduction | Spawning | Eggs | Fecundity | Larvae
Eggs hatch in about 4 days (Ref. 59043).
Main reference Upload your references | References | Coordinator | Collaborators
Kottelat, M., 1997. European freshwater fishes. An heuristic checklist of the freshwater fishes of Europe (exclusive of former USSR), with an introduction for non-systematists and comments on nomenclature and conservation. Biologia, Bratislava, 52/Suppl. 5:1-271. (Ref. 13696)
IUCN Red List Status (Ref. 130435: Version 2025-1)
Least Concern (LC) ; Date assessed: 05 December 2023
CITES
Not Evaluated
CMS (Ref. 116361)
Not Evaluated
Threat to humans
Human uses
Fisheries: minor commercial; aquaculture: commercial; bait: usually
FAO - Aquaculture systems: production; Fisheries: landings; Publication: search | FishSource |
More information
Trophic ecology Food items (preys) Diet composition Food consumption Food rations Predators
Ecology Ecology Home ranges
Population dynamics Growth parameters Max. ages / sizes Length-weight rel. Length-length rel. Length-frequencies Mass conversion Recruitment Abundance
Life cycle Reproduction Maturity Maturity/Gills rel. Fecundity Spawning Spawning aggregations Eggs Egg development Larvae Larval dynamics
Distribution Countries FAO areas Ecosystems Occurrences Introductions BRUVS - Videos
Anatomy Gill area Brain Otolith
Physiology Body composition Nutrients Oxygen consumption Swimming type Swimming speed Visual pigments Fish sound Diseases & Parasites Toxicity (LC50s)
Genetics Genome Genetics Heterozygosity Heritability
Human related Aquaculture systems Aquaculture profiles Strains Ciguatera cases Stamps, coins, misc.
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References References
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Summary page | Point data | Common names | Photos
Internet sources
AFORO (otoliths) | Alien/Invasive Species database | Aquatic Commons | BHL | Cloffa | Websites from users | Check FishWatcher | CISTI | Catalog of Fishes: genus, species | DiscoverLife | DORIS | ECOTOX | FAO - Aquaculture systems: production; Fisheries: landings; Publication: search | Faunafri | Fishipedia | Fishtrace | GloBI | Google Books | Google Scholar | Google | IGFA World Record | National databases | OneZoom | Open Tree of Life | Otolith Atlas of Taiwan Fishes | Public aquariums | PubMed | Reef Life Survey | Socotra Atlas | TreeBase | Tree of Life | Wikipedia: Go, Search | Zoological Record
Estimates based on models
Phylogenetic diversity index (Ref. 82804): PD50 = 0.5000 [Uniqueness, from 0.5 = low to 2.0 = high].
Bayesian length-weight: a=0.00562 (0.00483 - 0.00654), b=3.12 (3.09 - 3.15), in cm total length, based on LWR estimates for this species (Ref. 93245).
Trophic level (Ref. 69278): 2.7 ±0.29 se; based on food items.
Resilience (Ref. 120179): Medium, minimum population doubling time 1.4 - 4.4 years (tm=2-3; K=0.19; tmax=7; Fec=1,500).
Fishing Vulnerability (Ref. 59153): Low to moderate vulnerability (31 of 100). Legend
0 < value ≤ 25: Low vulnerability
25 < value ≤ 35: Low to moderate vulnerability
35 < value ≤ 45: Moderate vulnerability
45 < value ≤ 55: Moderate to high vulnerability
55 < value ≤ 65: High vulnerability
65 < value ≤ 75: High to very high vulnerability
value > 75: Very high vulnerability
">🛈
Price category (Ref. 80766): Unknown.
Nutrients (Ref. 124155): Calcium = 80.3 [38.4, 133.1] mg/100g; Iron = 0.46 [0.27, 0.84] mg/100g; Protein = 16.7 [15.1, 18.1] %; Omega3 = 0.697 [0.310, 1.491] g/100g; Selenium = 9.19 [4.33, 19.16] μg/100g; VitaminA = 40.5 [14.0, 121.6] μg/100g; Zinc = 0.965 [0.689, 1.367] mg/100g (wet weight);
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Maximum area of rectangle with fixed perimeter.
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Asked
Modified 9 years, 7 months ago
Viewed 195k times
11
$\begingroup$
How can you, with polynomial functions, determine the maximum area of a rectangle with a fixed perimeter.
Here's the exact problem
You have 28 feet of rabbit-proof fencing to install around your vegetable garden. What are the dimensions of the garden with the largest area?
I've looked around this Stack Exchange and haven't found an answer to this sort of problem (I have, oddly, found a similar one for concave pentagons).
If you can't give me the exact answer, any hints to get the correct answer would be much appreciated.
geometry
polynomials
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asked Apr 6, 2012 at 21:30
Ethan TurkeltaubEthan Turkeltaub
24322 gold badges33 silver badges99 bronze badges
$\endgroup$
Add a comment |
4 Answers 4
Reset to default
13
$\begingroup$
The result you need is that for a rectangle with a given perimeter the square has the largest area. So with a perimeter of 28 feet, you can form a square with sides of 7 feet and area of 49 square feet.
This follows since given a positive number $A$ with $xy = A$ the sum $x + y$ is smallest when $x = y = \sqrt{A}$.
You have $2x + 2y = P \implies x + y = P/2$, and you want to find the maximum of the area, $A = xy$.
Since $x + y = P/2 \implies y = P/2 - x$, you substitute to get $A = x(P/2-x) = (P/2)x - x^2$. In your example $P = 28$, so you want to find the maximum of $A = 14x - x^2$.
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edited Apr 6, 2012 at 21:57
answered Apr 6, 2012 at 21:46
user23784user23784
$\endgroup$
4
$\begingroup$ So would it be fair to say that the vertex of that function would give the maximum side length and maximum area (x = max side length; y = max area)? $\endgroup$
Ethan Turkeltaub
– Ethan Turkeltaub
2012-04-06 22:29:37 +00:00
Commented Apr 6, 2012 at 22:29
$\begingroup$ @EthanTurkeltaub Well, no. It gives the maximum area. Maximum side length doesn't really make sense. For a rectangle height and width are not necessarily the same, so there would be two different measurements for side length. Since the perimeter is fixed and we know the perimeter, $28 = 2 height + 2 width$ any time you increase the height or the width, you must decrease the other. Also, if you maximize either one, then you would have one of them equal to 14 feet, but that forces the other to be 0 feet (so the total perimeter stays 28ft). $\endgroup$
user23784
– user23784
2012-04-06 22:40:45 +00:00
Commented Apr 6, 2012 at 22:40
$\begingroup$ @EthanTurkeltaub It would be fair to say that graph shows the area ($y$ on the graph) in terms of either the height or width ($x$ on the graph). As you can see the area is at a maximum when $x = 7$, not when $x$ is at it's maximum possible value of 14. $\endgroup$
user23784
– user23784
2012-04-06 22:42:36 +00:00
Commented Apr 6, 2012 at 22:42
$\begingroup$ I see what you're saying. That was just a mistake on my part. Thank you very much for your help on this. $\endgroup$
Ethan Turkeltaub
– Ethan Turkeltaub
2012-04-06 23:36:44 +00:00
Commented Apr 6, 2012 at 23:36
Add a comment |
5
$\begingroup$
Here is a slightly different approach. Let us see what happens if we use a rectangle with base $x$ and height $y$.
Then the perimeter (amount of fencing) used is $2x+2y$. This is $28$, so $2x+2y=28$, or more simply $x+y=14$.
Note that $$4xy=(x+y)^2-(x-y)^2.$$ Since $x+y=14$, it follows that $$4xy=(14)^2-(x-y)^2.$$ To make $4xy$ (and hence $xy$) as large as possible, we must subtract as little as possible from $(14)^2$. So we must make $(x-y)^2$ as small as possible. Since $(x-y)^2$ is a square, it is always $\ge 0$, and it is smallest when $x=y$, that is, when our rectangle is a square.
Share
answered Apr 6, 2012 at 22:30
André NicolasAndré Nicolas
515k4747 gold badges584584 silver badges1k1k bronze badges
$\endgroup$
5
$\begingroup$ Ah, that makes a lot of sense. Thank you very much! $\endgroup$
Ethan Turkeltaub
– Ethan Turkeltaub
2012-04-06 23:38:04 +00:00
Commented Apr 6, 2012 at 23:38
$\begingroup$ Hello, could you expand on how you got to the 4xy= (x+y)2-(x-y)2 $\endgroup$
Jamie Twells
– Jamie Twells
2015-03-18 09:47:39 +00:00
Commented Mar 18, 2015 at 9:47
$\begingroup$ @captainjamie: Recall that $(x+y)^2=x^2+2xy+y^2$ and $(x-y)^2=x^2-2xy+y^2$. Subtract. $\endgroup$
André Nicolas
– André Nicolas
2015-03-18 15:04:18 +00:00
Commented Mar 18, 2015 at 15:04
$\begingroup$ @André ok so you're not coming from some fundamental definition of areas and perimeters or anything, just stating a relationship between x and y. I just wondered if you'd got it from differentiating or something. Thanks. $\endgroup$
Jamie Twells
– Jamie Twells
2015-03-18 15:48:29 +00:00
Commented Mar 18, 2015 at 15:48
$\begingroup$ I gave an algebraic version, since most people nowadays find that most comfortable. But in different forms the identity has been known and used since Neo-Babylonian times. It was in effect used by Diophantus in his Arithmetica. A geometric version is used by al-Khwarizmi, in what is arguably the first systematic algebra west of India. $\endgroup$
André Nicolas
– André Nicolas
2015-03-18 16:51:28 +00:00
Commented Mar 18, 2015 at 16:51
Add a comment |
3
$\begingroup$
Does this help?
Edit: Also, is it elsewhere in the problem that it has to be a rectangle? Because, otherwise a rectangle would not be the best choice.
Share
answered Apr 6, 2012 at 21:39
KevinKevin
1941212 bronze badges
$\endgroup$
5
$\begingroup$ Yes, it must be a rectangle. $\endgroup$
Ethan Turkeltaub
– Ethan Turkeltaub
2012-04-06 21:48:07 +00:00
Commented Apr 6, 2012 at 21:48
$\begingroup$ I'm confused as to why you used 14 instead of 28. $\endgroup$
Ethan Turkeltaub
– Ethan Turkeltaub
2012-04-06 21:49:33 +00:00
Commented Apr 6, 2012 at 21:49
$\begingroup$ because to get the area, you multiply length times width. If you only have 28 feet longest a side could be would be 14 $\endgroup$
Kevin
– Kevin
2012-04-06 21:51:51 +00:00
Commented Apr 6, 2012 at 21:51
$\begingroup$ So, in this instance, the answer would be the vertex of the parabola (i.e., x = max side length; y = max area)? $\endgroup$
Ethan Turkeltaub
– Ethan Turkeltaub
2012-04-06 21:53:57 +00:00
Commented Apr 6, 2012 at 21:53
$\begingroup$ Yep. Although I must say that rar's is the better answer in that it gives much better detail. $\endgroup$
Kevin
– Kevin
2012-04-06 22:00:23 +00:00
Commented Apr 6, 2012 at 22:00
Add a comment |
-2
$\begingroup$
Put the perimeter into the vertex formula.
(Find the P=a+b equation and put it into the A=ab equation.)
The answer is the maximum point.
Share
answered Feb 12, 2016 at 9:21
YvonneYvonne
1
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187759 | https://ilyas.jp/post?id=218 | What is Stiffness of Springs in Parallel and Series? Explained in 1-Minute
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What is Stiffness of Springs in Parallel and Series? Explained in 1-Minute
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Structural Engineering
Structural Analysis
Post ID: 218
Sun 19th Mar 2023 by ilyas
Springs in Parallel
Springs in parallel can be considered to work together. The equivalent stiffness of the springs is the sum of all the springs in parallel. We can consider the following example:
In the above case, the equivalent stiffness will be:
Equivalent stiffness,
Springs in Series
For springs in series, they do not work directly together compared to those in parallel.
For the case above, the equivalent stiffness can be calculated as follows:
Equivalent stiffness,
Applications in Structural Analysis
Let's consider the building frame below...
The columns in the first floor can be considered as have their lateral stiffness in parallel. Similarly for the second floor. The floor slabs can be considered as assembled masses. Therefore we can idealise the building as being made of equivalent "lumped" springs, as follows:
Where...
1F equivalent lateral stiffness:
2F equivalent lateral stiffness:
With this new lumped-mass model, we can find the independent deflection of each floor, or we can perform modal analysis to find the natural frequencies of vibration.
By the way, in order to do such an analysis (by hand), you need to calculate the lateral stiffness of each individual element. Read more about it at the following link:
※ What is Lateral Stiffness? Examples and Formulas
Last Update 19/03/23 05:42 JST
Was this Post Helpful?
1 person found this article helpful.
ANALYSIS
STIFFNESS
BASICS
SIMILAR ARTICLES
Stiffness of Structural Elements
What is In-Plane and Out-of-Plane Stiffness? Differences Explained with Examples
What is Lateral Stiffness? Examples and Formulas
Boundary Conditions for Structural Analysis: Explained in 1-minute.
Floor Vibration
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DbVisualizer
SQL
Understanding the SQL UNIQUE Constraint
Author:
Leslie S. Gyamfi
Length:
6 min
Type:
Guide
Published:
2025-01-07
intro
Let’s see what the SQL UNIQUE constraint is and how it works to ensure the wholeness of our data.
Tools used in the tutorial
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DBVISUALIZER
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THE POSTGRESQL DATABASE
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Constraints in SQL are a set of rules that are responsible for ensuring that the data in a database is whole. Among these set of rules, is the SQL UNIQUE constraint — notable for making sure that the values in a column(s) remain unique across a table in a database.
In this guide, you will find out what the UNIQUE constraint in SQL is, where you can use it, and how to use it in popular database technologies.
Let’s get right into it!
What Is the SQL UNIQUE Constraint?
When it comes to SQL, the UNIQUE constraint is there to ensure that the columns in a database are unique. It comes in handy when you are building a product where the uniqueness of your data matters.
At its core, it enforces that values in a group of columns are distinct from each other. Upon the application of the UNIQUE constraint on a set of columns, SQL will disallow any new or updated row from hosting a duplicate record in those columns.
It is important to note that
Both the UNIQUE and PRIMARY KEY constraints assure uniqueness for a column(s) as a PRIMARY KEY automatically has a UNIQUE constraint.
In some instances, the SQL UNIQUE constraint can be applied to multiple columns (composite unique constraints). For example, in a college database where a student has the ability to take multiple courses, but cannot be enrolled in the same course twice.
Time to explore the syntax of this useful clause!
UNIQUE Constraint in SQL: Operator Syntax
Use the following syntax to apply a UNIQUE constraint when creating a table:
Copy
```
1
CREATE TABLE demo_table (
2
column1 data_type UNIQUE,
3
column2 data_type
4
);
```
But, if you need your constraint to span across multiple columns, do this:
Copy
```
1
CREATE TABLE demo_table (
2
column1 data_type,
3
column2 data_type,
4
UNIQUE (column1, column2)
5
);
```
Instead, to add a UNIQUE constraint to an existing table, you can write:
Copy
```
1
ALTER TABLE customers
2
ADD CONSTRAINT unique_email UNIQUE (email);
```
Before you implement this constraint, verify that the data does not contain duplicates:
Copy
```
1
SELECT email, COUNT()
2
FROM customers
3
GROUP BY email
4
HAVING COUNT() > 1;
```
The above query retrieves all duplicate emails in the records of the customers table (if any).
SQL UNIQUE: Use Cases
Time to explore some use cases in real-world PostgreSQL examples. Keep in mind that you can easily extend them to other popular DBMS technologies, like MySQL, SQL Server, and Oracle
Note: The queries below will be executed in DbVisualizer, the world’s leading database client with the highest user satisfaction. In addition to being able to connect to several DBMSs, it offers great features and full support for all database PostgreSQL capabilities. Try DbVisualizer out now!
Example #1 — Applying UNIQUE on a Single Column
Consider a Customers table where we want to ensure that each customer has a unique phone number.
Copy
```
1
CREATE TABLE Customers (
2
id INT PRIMARY KEY,
3
name VARCHAR(50),
4
phone_number VARCHAR(15) UNIQUE
5
);
```
Creating the ‘customers’ table in DbVisualizer
Now, in attempting to populate our database table with a duplicate value, PostgreSQL will fail with an error:
Copy
```
1
INSERT INTO Customers (id, name, phone_number)
2
VALUES
3
(1, 'Billy Jane', '674-2134')
4
(2, 'Joe Rogan', '674-2134');
```
DbVisualizer throws a unique constraint violation
Example #2: Composite UNIQUE Constraint on Multiple Columns
In this example, we will create a Bookings table with a composite UNIQUE constraint on the combination of columns: room_number and booking_date ensuring that each room can only have one booking per date:
Copy
```
1
CREATE TABLE Bookings (
2
booking_id INT PRIMARY KEY,
3
room_number INT,
4
booking_date DATE,
5
customer_id INT,
6
UNIQUE (room_number, booking_date)
7
);
```
Creating the ‘Bookings’ table with a composite unique constraint
Let’s attempt to insert bookings that conflict with this constraint:
Copy
```
1
INSERT INTO Bookings (booking_id, room_number, booking_date, customer_id)
2
VALUES
3
(1, 101, '2024-11-01', 1)
4
(2, 102, '2024-11-01', 2)
5
(3, 101, '2024-11-02', 3);
```
Violating the composite unique constraint.
The UNIQUE constraint on (room_number, booking_date) ensures that the same room cannot be booked more than once on a specific date.
Best Practices and Considerations for Using SQL UNIQUE
Time to look at some of the best practices that should be taken into consideration when working with SQL UNIQUE:
Although constraints help maintain data quality, too many constraints can also slow down database performance. Learn to apply UNIQUE constraints where they are necessary.
In cases where you want to rename or modify an existing UNIQUE constraint, drop and recreate the constraint with the desired changes as it is generally the most straightforward approach.
Choose between single-column and composite constraints based on your business rules.
Use appropriate naming conventions.
It’s a wrap!
Conclusion
UNIQUE constraints in SQL are essential tools for maintaining data integrity in relational databases. They help ensure data consistency, improve query performance through automatic indexing, and provide a foundation for reliable data relationships.
Database developers should remember that while UNIQUE constraints add overhead to write operations, the benefits of guaranteed data consistency usually outweigh the performance costs and that, they should always test the impact of constraints in their specific use case.
Now that you have a grasp of this concept, put your newfound knowledge to the test. However, do note that SQL UNIQUE is not everything you need to know. While SQL is a powerful tool for querying and manipulating data, there are many other aspects of database management that are equally important, including data modeling, database design, and performance tuning, which can all be tackled effortlessly using SQL clients like DbVisualizer.
See you in the next blog, but until then, follow us on Dev.to to learn more about the newest trends in the database space!
FAQ
What is a UNIQUE constraint in SQL, and why do I need it?
A UNIQUE constraint is a rule you can set on a column(s) in a table to ensure that all values in that column are different from each other. For example, if you add a UNIQUE constraint to an email column in a users table, it will prevent two users from having the same email address. This helps keep your data accurate and prevents accidental duplicates in critical columns.
Can I have multiple UNIQUE constraints in the same table?
Yes, you can have multiple UNIQUE constraints in the same table. For example, you could set a UNIQUE constraint on both a username column and an email column. This way, each username and email in the table will be unique.
How do I add a UNIQUE constraint to multiple columns already containing data?
When adding a UNIQUE constraint to existing data, you need to first check for existing duplicates, clean up any duplicates, and then add the constraint using the ALTER TABLE command:
Copy
```
1
ALTER TABLE demo_table
2
ADD CONSTRAINT constraint_name UNIQUE (column_name);
```
How can I remove a UNIQUE constraint from a table if I no longer need it?
You can remove a UNIQUE constraint using the ALTER TABLE command with the DROP CONSTRAINT clause. Note that you must know the name of the UNIQUE constraint:
Copy
```
1
ALTER TABLE table_name
2
DROP CONSTRAINT constraint_name;
```
About the author
Leslie S. Gyamfi
Leslie Gyamfi is a mobile/web app developer with a passion for creating innovative solutions. He is dedicated to delivering high-quality products and technical articles. You can connect with him on LinkedIn
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187761 | https://discourse.peacefulscience.org/t/birds-and-wallace-s-line/11184 | Birds and Wallace’s Line
Continuing the discussion from Herman Mays Accuses Joshua Swamidass:
Let’s just say I have yet to have any conversations about the distribution of island passerines in East Asia or the distribution of birds across Wallace’s line with creationists where it wasn’t discussed in reference to some Bible beliefs
I’m pretty interested in Wallace’s line. What do you find in the distribution of birds here?
Observers might need a quick refresher on what exactly Wallace’s line actually is.
It’s a biogeographical break in the Indonesian archipelago that separates Asia from Australasia (New Guinea/Australia) named after Alfred Russel Wallace. It’s created by very deep water channels running through the archipelago and is why there are no tigers and rhinos in Australia and no kangaroos in Asia.
It also affects the distribution of birds. Some have managed to cross it (pigeons and cockatoos for example). Most others (birds of paradise for example) not so much.
IIRC there are (at least) two lines - Wallace’s and Weber’s - for different types of fauna.
I forgot about Weber.
Can you link some papers please?
Wallace, Alfred Russel (1863). “On the Physical Geography of the Malay Archipelago”. Royal Geographical Society. 7: 205–212.
Wallace Line
The Wallace Line or Wallace's Line is a faunal boundary line drawn in 1859 by the British naturalist Alfred Russel Wallace and named by English biologist Thomas Henry Huxley that separates the biogeographical realms of Asia and Wallacea, a transitional zone between Asia and Australia. West of the line are found organisms related to Asiatic species; to the east, a mixture of species of Asian and Australian origin is present. Wallace noticed this clear division during his travels through the East I...
Max Carl Wilhelm Weber
Max Carl Wilhelm Weber van Bosse or Max Wilhelm Carl Weber (5 December 1852, in Bonn – 7 February 1937, in Eerbeek) was a German-Dutch zoologist and biogeographer.
Weber studied at the University of Bonn, then at the Humboldt University in Berlin with the zoologist Eduard Carl von Martens (1831–1904). He obtained his doctorate in 1877. Weber taught at the University of Utrecht then participated in an expedition to the Barents Sea. He became Professor of Zoology, Anatomy and Physiology at the Uni...
If I recall, oscine passerines have escaped from Australasia (i.e. crossed Wallace’s line), based on phylogeny, only three times in earth history. Though a few have made it back across the line also.
How many individuals have to cross in order to establish a sustainable population?
Wallace’s Malay Archipelago itself is an interesting read if you want to go old school. Just imagine the collecting permit issues he would have today!
How many individuals have to cross in order to establish a sustainable population?
No idea. The “three times” are of course not individual birds but three instances in which enough birds crossed of one species to establish a clade that survived until the present day.
Yep that indeed appears to be the case.
I believe this is the one: Barker F.K., Cibois A., Schikler P., Feinstein J., Cracraft J. Phylogeny and diversification of the largest avian radiation. Proceedings of the National Academy of Sciences 2004; 101:11040-11045.
Jon Fjeldså, Les Christitis, and Per Ericsson have a new book out on the evolution of the passerines that I need.
Teach me. What is a passarine?
The perching birds. About half of all avian species are passerines.
Yeah it’s a stochastic thing. The more immigrants there are the more likely a new population is founded. Lots of these avian taxa that are good at speciating on oceanic islands aren’t solitary.
Member of the order Passeriformes, or perching birds, which is a bit over half of all extant bird species. Oscines is a suborder that’s about 2/3 of passerines. The evidence is quite good that it originated in Australia.
It’s an order of birds, the passeriformes. That cardinal and finch and wren at your feeder are all passerines as is the robin pulling up worms in the yard and the crows flying overhead. They have some synapomorphies in regards to the anatomy of their feet and the palette among other things but they are a VERY diverse group.
What are the other orders of birds?
There’s like 40 or so. Off the top of my head. This is something where Wikipedia can give you a decent answer.
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187762 | https://newtum.com/converters/power/kilowatt-to-kilojoule-second-converter | Convert Kilowatts to Kilojoules per Second Easily with Our Converter
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Kilowatt To Kilojoule/second Converter
(kW to kJ/s converter)
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Convert Kilowatts to Kilojoules/Second
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(Last Updated On: 2025-08-12)
Explore the seamless process of converting Kilowatts to Kilojoules per second with our user-friendly tool. Designed by Newtum, this converter simplifies your energy calculations, making it an essential resource for engineers, students, and professionals alike. Delve into its features and discover how it can enhance your understanding of energy conversion.
What are Kilowatt and Kilojoule/second
Definition of Kilowatt
A kilowatt (kW) is a unit of power equivalent to 1,000 watts, which measures the rate at which energy is produced or consumed. It is commonly used to express the output power of engines and the power consumption of tools and appliances. Understanding kilowatts is crucial for calculating energy efficiency and cost, as it helps individuals and businesses manage their energy usage effectively. In various sectors, from residential to industrial, kilowatts serve as a standard measurement, facilitating the comparison of energy outputs.
Definition of Kilojoule/second
Kilojoule per second (kJ/s) is a unit of power that represents the rate of energy transfer equivalent to one kilojoule of energy per second. It is a derived unit of power in the International System of Units (SI) and is commonly used in scientific and engineering contexts to quantify the rate at which energy is converted, consumed, or transferred. This unit provides a direct correlation between energy and time, allowing for precise calculations in processes where energy efficiency and consumption are critical, such as in thermal systems and electrical circuits.
Kilowatt to Kilojoule/second Conversion Table
| Kilowatt (kW) | Kilojoule/second (kJ/s) |
--- |
| 0.1 kW | 0.1 kJ/s |
| 0.5 kW | 0.5 kJ/s |
| 1 kW | 1 kJ/s |
| 2 kW | 2 kJ/s |
| 3 kW | 3 kJ/s |
| 4 kW | 4 kJ/s |
| 5 kW | 5 kJ/s |
| 6 kW | 6 kJ/s |
| 7 kW | 7 kJ/s |
| 8 kW | 8 kJ/s |
Conversion of Kilowatt to Kilojoule/second
1 kW = 1 kJ/s
1 kJ/s = 1 kW
Example 1:
convert 5 kW to kJ/s:
5 kW = 5 × 1 kJ/s = 5 kJ/s
Example 2:
convert 3.5 kW to kJ/s:
3.5 kW = 3.5 × 1 kJ/s = 3.5 kJ/s
History of Kilowatt and Kilojoule/second
The Kilowatt to Kilojoule/second Converter has evolved from basic manual calculations to sophisticated digital tools accessible to everyone. Initially used in industrial settings to optimize energy consumption, the need for precise power-to-energy conversions grew with technological advancements. As digital technology progressed, online converters emerged, offering instant, accurate results. Today, these tools are indispensable in various fields, ensuring efficient energy management and fostering a deeper understanding of power and energy dynamics.
How to use Kilowatt to Kilojoule/second Converter
Navigate to the Kilowatt to Kilojoule/second Converter page.
Enter the value in kilowatts (kW) that you wish to convert.
Click the 'Convert' button to obtain the result in kilojoules per second (kJ/s).
Review the calculated value displayed for accuracy.
Repeat the process as needed for multiple conversions.
Real Life Applications of Kilowatt to Kilojoule/second
In various industries and scientific research, converting Kilowatts to Kilojoules per second plays a crucial role in optimizing energy usage. Below, we explore real-life applications where this conversion is essential.
In renewable energy systems, converting kW to kJ/s helps in analyzing energy efficiency and output over time.
In electric vehicle charging, understanding the energy transfer rate is vital for optimizing charge times and energy consumption.
In industrial manufacturing, this conversion assists in evaluating the power requirements of machinery and reducing energy costs.
Common 'Kilowatt to Kilojoule second Converter' Conversion Errors to Avoid
Misunderstanding Units: Ensure you understand that kilowatts measure power, while kilojoules per second measure energy per unit time.
Incorrect Calculations: Always double-check your calculations to avoid errors in results. Remember that 1 kilowatt equals 1000 kilojoules per second.
Ignoring Conversion Factors: Pay attention to conversion factors and constants, as even minor errors can lead to significant discrepancies.
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187763 | https://wayground.com/admin/quiz/5dd6f22b9f2e0c001b3270cc/systems-of-inequalities-test | Systems of Inequalities Test 9th Grade Quiz | Wayground (formerly Quizizz)
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Systems of Inequalities Test
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Systems of Inequalities Test
Quiz
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Mathematics
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9th Grade
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Medium
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CCSS
HSA.REI.D.12, HSA.CED.A.3, 7.EE.B.4B
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Tom Giles
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[x]
1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Preview
Edit
What are the two linear inequalities graphed here?
y ≤ - x + 4
y < x - 2
y ≤ - x + 4
y> x - 2
y ≤ - 2x + 4
y> 2x - 2
y ≤ - x + 2
y> x - 4
Tags
CCSS.HSA.REI.D.12
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Preview
Edit
Which points are solutions to the system shown?
(5, 3) (6, -3)
(3, 3) (5, -3)
(-2, 2) (-4, 4)
(1, 2) (1, 5)
Tags
CCSS.HSA.REI.D.12
3.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Will my line be solid?
5 ≥ x
Yes
No
Tags
CCSS.HSA.CED.A.1
CCSS.HSA.CED.A.3
CCSS.HSA.REI.B.3
4.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Which point below is NOT part of the solution set?
(0, 0)
(5, 1)
(-1, -3)
(20, -5)
Tags
CCSS.HSA.REI.D.12
5.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Select the correct graph for the linear inequality.
A
B
C
D
Tags
CCSS.HSA.REI.D.12
6.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Is (-2,4) a solution to the system?
Solution
Not a solution
Tags
CCSS.HSA.REI.D.12
7.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Which of the following is not a solution to this system of inequalities?
(0,-1)
(0,3)
(4,0)
(6,-2)
Tags
CCSS.HSA.REI.D.12
8.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Is (0,0) a solution to the system?
Solution
Not a solution
Tags
CCSS.HSA.REI.D.12
9.
MULTIPLE CHOICE QUESTION
2 mins • 1 pt
Preview
Edit
Which system of inequalities matches the graph?
y > 2x - 4
y < 2
y < 2x - 4
y < 2
y > 2x - 4
y< 2
y < 2x - 4
y<-2
Tags
CCSS.HSA.REI.D.12
10.
MULTIPLE CHOICE QUESTION
5 mins • 1 pt
Preview
Edit
Write the inequalities
y≥-2x+3
y≤x-3
y>2x+3
y≥x-3
y>-2x+3
y≥x-3
y<-2x+3
y≤x-3
Tags
CCSS.HSA.REI.D.12
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187764 | https://philosophy.stackexchange.com/questions/114323/if-a-predicate-doesnt-determine-a-set-does-that-predicate-even-exist-in-the-fi | philosophy of mathematics - If a predicate doesn't determine a set, does that predicate even exist in the first place? - Philosophy Stack Exchange
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If a predicate doesn't determine a set, does that predicate even exist in the first place?
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I thought of asking this in the Math Stack Exchange, but then I thought this stack exchange is better. Certain predicates define sets, such as "x is not equal to x". Other predicates do not, such as "x is equal to x" and the infamous "x is not a member of x". But, I wonder, if a predicate does not determine a corresponding set, does it even exist in the first place? Of course, to answer this question, we need a formal definition of predicate, and such a definition would also need to distinguish predicates from sets. Maybe a predicate is simply a first-order formula that has exactly one free variable x? Anyway, I would love to know what philosophers and mathematicians have thought of this topic.
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asked Jun 22, 2024 at 22:50
user107952user107952
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1 "Equivalence classes of formulae with one free variable" is basically how classes are usually formulated in ZF set theory.Naïm Camille Favier –Naïm Camille Favier 2024-06-22 22:58:57 +00:00 Commented Jun 22, 2024 at 22:58
If we needed formal definitions of words to answer questions about them we could hardly answer any questions we encounter every day. "Red" is a predicate and it does not define a set, "all red things" is too vague for that. But if you want first order formulas, x = x is a predicate of ZFC and it does not define a set either. If it did, it would be the inconsistent set of all sets and ZFC blocks formation of such sets. So yes, a predicate does not need to define a set to 'exist'.Conifold –Conifold 2024-06-23 08:13:58 +00:00 Commented Jun 23, 2024 at 8:13
1 If X is a predicate and X exists if and only if X determines a set, then the predicate "does not determine a set" would not itself exist, and so instead every predicate would determine a set (since cases of non-determination wouldn't exist, because the predicate of non-determination doesn't exist). So either every predicate can determine a set, or the existence of predicates (what does that even mean?) is not absolutely dependent on their use in determining sets.user40843 –user40843 2024-06-23 09:20:22 +00:00 Commented Jun 23, 2024 at 9:20
1 Exist where? Predicates are objects in the metatheory, and are n-ary relations over the Universe. They may or may not be elements in the universe.Michael Carey –Michael Carey 2024-06-23 20:49:06 +00:00 Commented Jun 23, 2024 at 20:49
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You write "to answer this question, we need a formal definition of predicate" but the main problem is not the definition of predicate but rather the definition of "existing" in your question. To avoid getting into the zillionth debate over Platonism, I would adopt the practical definition of the "existence" of a mathematical entity as having a coherent and well-developed branch of mathematics practicing the said entity, without further ontological commitments.
With that definition, I will argue that predicates do "exist", based on the following branch of mathematics: Edward Nelson's Internal Set Theory (IST). Here IST is a conservative extension of ZFC where infinitesimals have free reign, and more pertinently for our purposes, the "standardness" predicate (indeed a one-place predicate) is part of the language alongside the traditional membership relation of ZFC. It is seen immediately that the standardness predicate does not correspond to a set; indeed, the collection of "standard" natural numbers, which is a proper "subcollection" of N, is not a set.
There are numerous variants of IST and many related publications, establishing the "existence" part in accordance with the above interpretation of "existence".
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answered Jun 24, 2024 at 7:47
Mikhail KatzMikhail Katz
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In ZF(C), yes. Generally, ZFC proceeds by setting out a formal language of logic and then introduces the axiom schema of specification as a way to specify subsets.
So the formula x = x is totally acceptable, and we may use it to predicate sets. B = {x in A: x = x} is a valid set-builder notation, and B = A. Likewise, x not in x is a well-formed formula, and C = {x in A: x not in x} we know to be empty by axiom of regularity.
What isn't valid in ZFC is the general construction S = {x: pred(x)}, because that's (implicitly) referencing a universal set-of-all-sets U, i.e. S = {x in U: pred(x)}, which is prohibited.
Because of Godel's, we know that we can't a (sufficiently complex) formal logic system like ZF(C) within itself, so predicates are not generally regarded as sets and not representable by sets, or if they are we need to be careful to distinguish predicates outside the model and predicates within the model.
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answered Jun 24, 2024 at 19:20
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(I am not a mathematician, you should post this on Math.SE for proper answers)user67129 –user67129 2024-06-24 19:22:26 +00:00 Commented Jun 24, 2024 at 19:22
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Commonly, a logical predicate is an element that provides information about the subject. But that is very subjective. The notion is easier to grasp from the General Systems Theory perspective.
A system is a group of interrelated parts.
In simpler words, such is essentially the definition of a group. A system is essentially a group, which has a particularity: its parts are interrelated, which is a very subjective quality. Take this group of parts: apples, a bottle of wine and a phone call. Are they related? Perhaps not for you, but they are written in my "things to do" list over my fridge. So, in this case, the system (my list of things to do) has three parts.
Notice that the definition of system is recursive, ergo, fractal. A system is a set of parts, so, parts are also systems (e.g. "a bottle of wine" is also a system, which parts can be the bottle itself, the label, and the liquid), and systems are also parts (my list is part of the system house). Notice that the decomposition of a system has multiple approaches: a house can be a set of floors, or a set of walls.
Regarding relationships, think of this family system: a father, a mother and two kids. The father and the mother are also a system, which parts are interrelated. But the relationship between them is not unique: they can be understood to share multiple types of relationships (sex, experiences, friendship, etc.).
Now, applying such perspective:
a proposition is just a system of two interrelated parts: the subject and the predicate.
As discussed above, the predicate is just a kind of relationship between the subject and some specific object.
For example:
Sheila [subject] baked [relationship] a delicious cake [object].
The old house on the corner [subject] needs [relationship] renovation [object].
I walk. This proposition seems complex, but it is just that such structure is a linguistic simplification of what should be I [subject] walk [relationship] the walk [object]. Same happens with I sleep or I go. Linguistically, the object is taken for granted.
Go. This imperative is equivalent, it essentially means I [subject] command [relationship] you [object] to go [complement of the relationship].
Notice there is always an object. The fact that the language could hide or take the object for granted does not imply the object does not exist. Any proposition has always a subject, a predicate and a relationship.
So, a predicate is essentially a logical component that attributes a system a relationship with an object. Linguistically, the relationship is called the verb, but from a systemic standpoint, it is just a relationship.
If a predicate doesn't determine a set
Now, the term set has sense: a system is not only a set (a collection of unrelated parts), but a group of interrelated parts. What is the essential relationship that makes them a system? Metaphysically, its belonging to the set, which does not mean that they have a logical relationship. But this goes way farther: all systems are sets, so no predicate "doesn't determine a set".
does that predicate even exist in the first place?
For that, you need to define existence. This is a common definition of existence, from a systemic perspective: Does black matter exists? No, since we're still not able to interact with it.
From a systemic perspective, an object exists for a subject if the subject is able to interact (physically or rationally) with the object.
Predicates are metaphysical objects (they don't exist physically but rationally). Ergo, if you can interact rationally with such object -essentially, think of it-, it exists. In other words, any metaphysical object exists from the instant you think of it.
Think of a teenage mutant ninja elephant. Since reason abstracts everything, just thinking of the name makes it existing. You [subject] think of [relationship] a teenage mutant ninja elephant [object].
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edited Jun 23, 2024 at 2:52
answered Jun 23, 2024 at 2:33
RodolfoAPRodolfoAP
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"if you can think of it it exists" is imo not a particularly helpful way to reason about math; I am able to think about the properties that a 'lowest counterexample to the collatz conjecture' (it must be odd, if it exists, for instance.), but its existence is still very much up for debate. "a set of all sets" is something we can interact with rationally but it "doesn't exist" in most axiomatic formulations user67129 –user67129 2024-06-24 19:27:26 +00:00 Commented Jun 24, 2024 at 19:27
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Yes. Sometimes predicates can exist even if they don’t determine a well defined set. The relationship between predicates and sets are context dependent. In classical logical framework and set theory , predicates do determine sets but in intuitionistic logic and fuzzy logic the predicates do not always determine a set. For instance , “x is tall” , “ x is somewhat cold “ , “x is beautiful “ etc …
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answered Jun 23, 2024 at 3:24
SacrificialEquationSacrificialEquation
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187765 | https://www.baseball-almanac.com/dictionary-term.php?term=official+at-bat | official at-bat — Baseball Dictionary
A Definition of Official At-bat | Baseball Almanac
The Dickson Baseball Dictionary is an
absolutely invaluable resource for those who love the game of baseball. Referred to as "a staggering piece of scholarship" (Wall Street Journal) and
"an indispensable guide to the language of baseball" (San Diego Union-Tribune), the Baseball Dictionary is the definitive source for baseball terms.
The complete definition of official at-bat, from the Dickson Baseball Dictionary, appears below, along with (where applicable) first usage, synonyms, historical details and more official at-bat research, courtesy of
Baseball Almanac.
"If you're a baseball fan and...you're curious about the etymology of the phrase out in the left field or you need a
refresher on the infield fly rule. Well for 20 years, the authority, the baseball Bible, if you will, has been The Dickson Baseball Dictionary." — Block, Melissa. NPR Radio Host. All Things Considered. 4 March 2014.
"If you're a baseball fan and...you're curious about the etymology of the phrase out in the left field or you need a
refresher on the infield fly rule. Well for 20 years, the authority, the baseball Bible, if you will, has been The Dickson Baseball Dictionary." — Block, Melissa. NPR Radio Host. All Things Considered. 4 March 2014.
| |
| Official At-bat from the Baseball Dictionary A Definition of Official At-bat |
| Term Definition |
| official at-batDefinition An at-bat that is entered in the official record of the game and used as the basis for batting statistics. Four types of plate appearances are not counted as official at-bats: when the batter is awarded a base on balls, is hit by a pitched ball, is awarded first base because of catcher's interference, and hits a sacrifice bunt or sacrifice fly. 1st Use. 1901. "Officially, he was at bat only four times, as his eye was so good yesterday that he drew a base on balls twice and made a sacrifice hit. The 'official' times at bat were judiciously spent on three clean singles and a fly to Hickman." (St. Louis Republic, July 15; Ken Liss). |
| Term Definition |
| Baseball Dictionary | Baseball Almanac |
Official At-bat from the Baseball Dictionary
A Definition of Official At-bat
official at-bat
Definition
An at-bat that is entered in the official record of the game and used as the basis for batting statistics. Four types of plate appearances are not counted as official at-bats: when the batter is awarded a base on balls, is hit by a pitched ball, is awarded first base because of catcher's interference, and hits a sacrifice bunt or sacrifice fly. 1st Use. 1901. "Officially, he was at bat only four times, as his eye was so good yesterday that he drew a base on balls twice and made a sacrifice hit. The 'official' times at bat were judiciously spent on three clean singles and a fly to Hickman." (St. Louis Republic, July 15; Ken Liss).
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187766 | https://physics.uwo.ca/~mhoude2/courses/PDF%20files/astro9701/Ch5-Geometrical_symmetry.pdf | 92 Chapter 5. Geometrical Symmetry Notes: • Most of the material presented in this chapter is taken from Bunker and Jensen (2005), Chap. 3, and Atkins and Friedman, Chap. 5. 5.1 Symmetry Operations We have already encountered many operators that can act on a quantum mechanical system (or any kind of system for that matter). Sometimes the operator will transform the state of the system, but it may also happen that it will leave it unaltered in the case where the state corresponds to an eigenvector of the operator. The latter situation happens when the operator in question is a so-called symmetry operator. This concept can be extended to other situations. We will consider in Chapter 6 sets of symmetry operations that leave the Hamiltonian of a system unchanged. Alternatively, one can investigate ensembles of operations that leave the geometrical shape of a system or an object (e.g., a molecule at equilibrium) unaffected. These are the types of operators with which we are concerned in this chapter. As we are concerned with rigid objects of different geometrical shapes anchored at a given point (the centre of symmetry), we do not take into account any possible translational symmetry and only consider symmetry operators that compose the so-called point groups (see below). There are five types of such operators that can leave the geometrical appearance of objects unchanged, with five corresponding kinds of symmetry elements. An element either is a point, a line, or a plane with respect to which the symmetry operation is effected. The different operations and elements are listed in Table 5-1. Table 5-1 – The five type of symmetry operations and elements. Operation Description Element E The identity operation The object itself Cn An n -fold rotation, a rotation by 2! n about an axis of symmetry The axis of symmetry ! A reflection in a mirror plane The plane of mirror symmetry i An inversion through a centre of symmetry The centre of symmetry Sn An n -fold improper rotation about an axis of improper rotation The axis of improper rotation An object, or a molecule, can have more than one axis of symmetry. For example, a molecule such as the H3 + ion has one three-fold axis of symmetry (C3) perpendicular to the plane of the molecule, and three two-fold axes (C2) going through one nucleus and the centre of symmetry (see Figure 5-1). In a case such as this one, the axis associated with the operation Cn of largest value n (C3 in this case) is called the principal axis. An n-fold symmetry axis will generate n !1 ( ) rotations Cn, Cn 2, …, Cn n!1. A rotation Cn k is right-handed in the usual sense (i.e., the right thumb points in the direction of the corresponding symmetry axis and the rotation sense follows the other fingers). 93 Figure 5-1 – The H3 + ion (allowing for the projection) with its symmetries and symmetry elements. There is one three-fold axis of symmetry C3 (the principal axis), three two-fold axes C2, and three mirror or reflection planes ! v; other symmetry operations are discussed in the text. Note that all the symmetry elements intersect at one point; hence the name point groups. As is apparent from Figure 5-1, the H3 + ion also possesses three mirror (or reflection) planes denoted by ! v (other symmetry operations for this molecule will be discussed later). A mirror plane is called a vertical plane ! v when it contains the principal axis, and it is called a horizontal plane ! h when it is perpendicular to the principal axis. A dihedral plane ! d is a vertical plane that bissects two C2 axes that are perpendicular to the principal axis. The inversion i operation consists of taking each point through the centre of symmetry to an equal distance in the opposite direction. The H3 + ion is obviously not symmetric under this transformation, as the inverted position of a given nucleus is not the original position of another nucleus composing the molecule. An improper rotation Sn is the composite of a Cn rotation followed by a horizontal reflection! h in a plane perpendicular to the n -fold axis, or vice-versa. That is, Sn = Cn! h = ! hCn. (5.1) It is often the case (specifically when n is even) that neither of the latter two operators alone will be a symmetry operation, but their aforementioned combination will be. The reader should verify, for example, that the methane molecule (CH4 ) possesses three S4 axes but no C4 and ! h operators. It can be verified that the improper rotations S1 and S2 are equivalent to the horizontal reflection and inversion operations, respectively. 94 5.2 The Classification of Molecules through Point Groups Upon studying a molecule such as the H3 + ion of Figure 5-1, it will be advantageous to list all of its symmetry operations into an ensemble. When this is accomplished, for any molecule, it is found that the resulting ensemble can be associated with the so-called point group to which the molecule “belongs”. The different point groups are often identified with their dominant symmetry features, as follows 1. The groups C1, Cs , and Ci . These groups respectively consist of the identity E alone, the identity and a reflection ! , and the identity and the inversion i . 2. The groups Cn . These groups consist of the identity and an n-fold rotation. 3. The groups Cnv . Each of these groups contains the elements of the corresponding Cn group, as well as n vertical reflections ! v. A special case is that of the C!v group of heteronuclear diatomic molecules (e.g., CO). 4. The groups Cnh . Each of these groups contains the elements of the corresponding Cn group, as well as a horizontal reflection ! h and whatever other new operations brought about by its multiplication with the elements of Cn (see the definition of a point group below). 5. The groups Dn . Each of these groups contains the elements of the corresponding Cn group, as well as n C2 rotations perpendicular to the n-fold principal axis and whatever other new operations brought about by multiplications between these elements. 6. The groups Dnh . Each of these groups contains the elements of the corresponding Dn group, as well as a horizontal reflection ! h and whatever other new operations brought about by multiplications between these elements. A special case is that of the D!h group which includes homonuclear diatomic molecules. 7. The groups Dnd . Each of these groups contains the elements of the corresponding Dn group, as well as n dihedral reflections ! d and whatever other new operations brought about by multiplications between these elements. 8. The groups Sn . These groups consist of the identity and an n-fold improper rotation Sn, as well as whatever other new operations brought about by multiplications between these elements. Only even values of n need to be considered, as a Sn group with an odd n is equivalent to the corresponding Cnh group. 9. The cubic and icosahedral groups. Each of these groups contains more than one n-fold rotation with n ! 3. The cubic groups are either tetrahedral (labeled T ) or octahedral (labeled O ). The point group Td is that of the regular tetrahedron and of the methane molecule CH4 ; T is the same group but without the reflections; Th is a tetrahedral group with an inversion. The point group Oh is that of the regular octahedron and of the sulphur hexafluoride molecule SF6 , for example; O is the same group but without the reflections. The point group Ih is that of the regular octahedron and of the buckminsterfullerene molecule C60 ; I is the same group but without the inversion. 95 10. The full rotation group R3 or K . This group contains all the possible rotations (an infinite number of them) through any axis that passes through the centre of a molecule. It is also the symmetry group of the sphere. The point group of a given molecule will be determined by first identifying all of its symmetry operations, and then comparing against the list of known point groups. This will be easily accomplished with the help of the algorithm presented in Figure 5-2. Let us work out, for example, the point group of the H3 + ion. Beside the identity element, this molecule possesses not only, as stated before, a C3 principal axis of symmetry with its associated C3 and C3 2 elements, three C2 rotations with their respective axis perpendicular to the principal axis of symmetry (and connecting the centre of symmetry to one hydrogen nucleus), but also three ! v reflections (each associated with a plane containing a hydrogen nucleus), one horizontal reflection ! h , as well as two improper rotations S3 (= C3! h ). The point group of the H3 + ion therefore consists of the following ensemble point group of H3 + = E,C3,C3 2,C2, !
C2, !!
C2," h,S3,S3 2," v, !
" v, !!
" v { }. (5.2) Simply answering the questions contained in the chart of Figure 5-2 quickly reveals that the point group of this molecule is D3h . It is in principle possible to do the same for any molecule. One must, however, be careful that in certain cases some symmetry operations may not be obvious at first sight and could be omitted. It is often preferable to verify that the products of every pair of symmetry operations result in another operation also present in the ensemble. The reason for this becomes apparent when one considers the formal definition of a group; which we now do. 5.2.1 The Definition of a Group Although we will postpone the study of the theory of groups to the next chapter, it will be to our advantage to introduce the concept of a mathematical group. A careful examination of the symmetry operations associated with the geometry of a given molecule would reveal that they fulfill the conditions set forth by the mathematical theory of groups, or group theory. That is, the set composed of all the symmetry operations of a molecule forms a group, in the mathematical sense. The formal definition of a group is as follows 1. The identity E is an operator of the set. 2. The operators multiply associatively; i.e., given three operators R, S and T , then it is true that RS ( )T = R ST ( ). 3. If R and S are two operators of the set, then RS is also an operator contained in the set. 4. The inverse of each operator is a member of the set. 96 Figure 5-2 – An algorithm to determine the point group of a molecule (from Atkins and Friedman). 97 For example, we can verify that the ensemble of equation (5.2) associated with the H3 + ion indeed satisfies these conditions. Although the first condition is obviously met, the second and third conditions can be verified more easily if we produce the so-called multiplication table for the group. This table is obtained by listing the results of every product between two operators of the group, as shown in Table 5-2. Table 5-2 – The multiplication table for the point group of the H3 + ion (i.e., D3h ). The table is calculated by first applying the operator of the top row and then the operator of the left column. E C3 C3 2 C2 !
C2 !!
C2 ! h S3 S3 2 ! v !
" v !!
" v E E C3 C3 2 C2 !
C2 !!
C2 ! h S3 S3 2 ! v !
" v !!
" v C3 C3 C3 2 E !!
C2 C2 !
C2 S3 S3 2 ! h !!
" v ! v !
" v C3 2 C3 2 E C3 !
C2 !!
C2 C2 S3 2 ! h S3 !
" v !!
" v ! v C2 C2 !
C2 !!
C2 E C3 C3 2 ! v !
" v !!
" v ! h S3 S3 2 !
C2 !
C2 !!
C2 C2 C3 2 E C3 !
" v !!
" v ! v S3 2 ! h S3 !!
C2 !!
C2 C2 !
C2 C3 C3 2 E !!
" v ! v !
" v S3 S3 2 ! h ! h ! h S3 S3 2 ! v !
" v !!
" v E C3 C3 2 C2 !
C2 !!
C2 S3 S3 S3 2 ! h !!
" v ! v !
" v C3 C3 2 E !!
C2 C2 !
C2 S3 2 S3 2 ! h S3 !
" v !!
" v ! v C3 2 E C3 !
C2 !!
C2 C2 ! v ! v !
" v !!
" v ! h S3 S3 2 C2 !
C2 !!
C2 E C3 C3 2 !
" v !
" v !!
" v ! v S3 2 ! h S3 !
C2 !!
C2 C2 C3 2 E C3 !!
" v !!
" v ! v !
" v S3 S3 2 ! h !!
C2 C2 !
C2 C3 C3 2 E The fact that every cell in Table 5-2 contains an operator originally contained in the set tells us that condition 3 is indeed satisfied. Condition 2 can be verified using this same table. For example, let us consider the following operation C3! v ( )S3 2 = "" ! v S3 2 = " C2, (5.3) but since C3 ! vS3 2 ( ) = C3 "" C2 = " C2, (5.4) then C3! v ( )S3 2 = C3 ! vS3 2 ( ), as required. The same can be proved for any other association of three operators from the point group. The last condition concerning the existence of the inverse for every symmetry operation in the group can be asserted by the one-time appearance of the identity operation in every row and column. It is in fact possible to show that every operator of the group should appear once and only once in every row and column of the multiplication table. One last comment should be made concerning the results shown in Table 5-2. It is the fact that the different symmetry 98 operations are effected using space-fixed axes. That is, in the product, say, C2! v the axes relative to which the molecule is oriented are not changed by the first operation (i.e., ! v), only the orientation of the molecule in space is affected. Therefore, the following operation (i.e., ! v) is also done with respect to the same set of axes. Finally, although we know from the chart of Figure 5-2 that linear molecules in their equilibrium configuration can either belong to the C!v or D!h , it can also be shown that • A molecule with one, and only one, Cn- or Sn-axis with a finite n ! 3 is a symmetric top. • A molecule with more than one Cn-axis with a finite n ! 3 is a spherical top. • A molecule with no rotational symmetry axis or only C2 axes only is an asymmetric top. It follows from these that the ammonia molecule NH3 (one C3-axis ), the methane molecule CH4 (four C3-axis ), and the water molecule (one C2-axis) are symmetric, spherical, and asymmetric tops, respectively. |
187767 | https://www.khanacademy.org/math/arithmetic/addition-subtraction/addition_carrying/v/carrying-when-adding-three-digit-numbers?utm_content=3&utm_medium=email&utm_campaign=khanacademy&utm_source=digest_html&utm_term=thumbnail | Using place value to add 3-digit numbers: part 1 (video) | Khan Academy
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3rd grade math
Course: 3rd grade math>Unit 3
Lesson 4: Adding with regrouping within 1000
Using place value to add 3-digit numbers: part 1
Using place value to add 3-digit numbers: part 2
Adding 3-digit numbers
Add within 1000
Math>
3rd grade math>
Addition, subtraction, and estimation>
Adding with regrouping within 1000
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Using place value to add 3-digit numbers: part 1
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Let's explore adding two 3-digit numbers using place value and carrying. We walk through the process step by step, starting with the ones place, then the tens place, and finally the hundreds place. We emphasize the importance of understanding what carrying means and how it relates to place value.Created by Sal Khan.
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Eoghan Breslin 11 years ago Posted 11 years ago. Direct link to Eoghan Breslin's post “Why do we have to do carr...” more Why do we have to do carrying?
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Nathan E. Fairchild 11 years ago Posted 11 years ago. Direct link to Nathan E. Fairchild's post “When adding 2 numbers, ca...” more When adding 2 numbers, carrying is required when the numbers in the 1s place add up to 10 or more, or when the numbers in the 10s place add up to 100 or more, or when the numbers in the 100s place add up to 1000 or more, and so on.
If we want to add 98 and 3, we have a several steps to take:
98 + 3 =
90 + 8 + 3 = -- because 98 = 90 + 8
90 + 11 = -- because 8 + 3 = 11
90 + 10 + 1 = -- because 11 = 10 + 1
100 + 1 = -- because 90 + 10 = 100and this the same as our carry step
101 -- our answer
This is a lot to do just to add 3 to 98, so we have this shortcut method called carrying to speed things up:
1) Add 8 and 3 to get 11. This gives us 1 in the 1s place and we carry 10 to the 10s place. So we add a 1 over the 10s place.
2) Add the carried 1 to 9 to get 10. This is equivalent to adding 10 to 90 to get 100. This gives us 0 in the 10s place and we carry 100 to the 100s place by adding a 1 over the 100s place.
3) Add the carried 100 to the 0 to give us 1 in the 100s place
4) We end up with 101
1 <- carried from 10 + 90 1 <- carried from 8 + 3 98 + 3 --- 101
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Blooming_Rosé 892 a year ago Posted a year ago. Direct link to Blooming_Rosé 892's post “What is 536 add 398, I do...” more What is 536 add 398, I don't really understand?
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s178011 9 months ago Posted 9 months ago. Direct link to s178011's post “It is ok! you can add the...” more It is ok! you can add them with many different strategies! you can stack the numbers on top of each on top of each other like Sal did, or you could use a number line. BUT I DONT RECCOMEND USING YOUR FINGERS CAUSE THAT WOULD TAKE A REALLY LONG TIME
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639216 7 months ago Posted 7 months ago. Direct link to 639216's post “why do we have to do math...” more why do we have to do math?
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ashadzz 7 months ago Posted 7 months ago. Direct link to ashadzz's post “it's important.... like w...” more it's important.... like what do you mean "why do we have to do math?". like.... it'll help us in the future.
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4806841910 5 days ago Posted 5 days ago. Direct link to 4806841910's post “why is my thing not loadi...” more why is my thing not loading
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Blooming_Rosé 892 a year ago Posted a year ago. Direct link to Blooming_Rosé 892's post “Oh yay the answer is 934,...” more Oh yay the answer is 934, none needed to help me!!
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Hisham a year ago Posted a year ago. Direct link to Hisham's post “Can he please do the this...” more Can he please do the this is 13 tens this is 13 tens again? If you're wondering what time it is at here it is 0:45
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Jessica Kadavy 2 months ago Posted 2 months ago. Direct link to Jessica Kadavy's post “Bruh he confused me so mu...” more Bruh he confused me so much until I realized that’s just another way of wording how I was doing it.
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migr2578 4 months ago Posted 4 months ago. Direct link to migr2578's post “is there a reason why we ...” more is there a reason why we do carrying?
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starparker a year ago Posted a year ago. Direct link to starparker's post “notice how when he said 5...” more notice how when he said 5+1 he paused for a sec
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Video transcript
Let's add 536 to 398. I'm going to do it two different ways so that we really understand what this carrying is all about. So first we'll do it in the more traditional way. We start in the ones place. We say, well, what's 6 plus 8? Well, we know that 6 plus 8 is equal to 14. And so when we write it down here in the sum, we could say, well, look, the four is in the ones place, so it's equal to 4 plus one 10 So let's write that one 10 in the tens place. And now we focus on the tens place. We have 1 ten, plus 3 tens, plus 9 tens. So what's that going to get us? 1 plus 3 plus 9 is equal to 13. Now, we have to remind ourselves, this is 13 tens. This is 13 tens. Or another way of thinking about it, this is 3 tens. This is 3 tens and 100. You might say, wait, wait. How does that make sense? Remember, this is in the tens place. When we're adding a 1 ten plus the 3 tens plus 9 tens, we're actually adding 10 plus 30 plus 90, and we're getting 130. And so we're putting the 30. The 3 in the tens place represents the 30. So this is the 3. The 3 represents the 30. And then we're placing this 1 in the hundreds place. 10 tens is equal to 100. And now we're adding up the hundreds place. 1 plus 5 plus 3 is equal to-- let's see, 1 plus 5 is equal to 6, plus 3 is equal to 9. But we have to remind ourselves, this is 900. This is in the hundreds place. So this is actually 100-- I want to make all colors the same. So this is actually 100 plus 500 plus 300. Let me get the color coding right. Plus 300 is equal to 900. And that's exactly what we got here. 100 plus 500 plus 300 is equal to 900, and we're done. This is equal to 934.
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187769 | https://fiveable.me/ap-chem/unit-3/ideal-gas-law/study-guide/XINb2AUU6e3c1rGlhBXg | printables
🧪AP Chemistry Unit 3 Review
3.4 Ideal Gas Law
🧪AP Chemistry Unit 3 Review
3.4 Ideal Gas Law
Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
🧪AP Chemistry
Unit & Topic Study Guides
Unit 3 Overview: Properties of Substances and Mixtures
3.1 Intermolecular and Interparticle Forces
3.2 Properties of Solids
3.3 Solids, Liquids, and Gases
3.4 Ideal Gas Law
3.5 Kinetic Molecular Theory
3.6 Deviation from Ideal Gas Law
3.7 Solutions and Mixtures
3.8 Representations of Solutions
3.9 Separation of Solutions and Mixtures
3.10 Solubility
3.11 Spectroscopy and the Electromagnetic Spectrum
3.12 Properties of Photons
3.13 Beer-Lambert Law
What do you currently know about temperature and pressure? Let's expand on those definitions and start thinking about how gases respond to changes in pressure and temperature.
Pressure and Temperature
Gases exert pressure on their surroundings. The key phrase you want to associate with pressure is "the number of times particles hit the walls of the container." If you are ever asked to explain any of the relationships in this key topic, always mention that phrase if pressure is involved!
Below is the value for standard pressure in four different units. Standard pressure is a measure of the pressure exerted by the weight of the Earth's atmosphere at a particular location, and you may know it better as "atmospheric pressure."
Standard Pressure: 1.00 atm = 760 mm Hg = 760 torr = 101.3 kPa
Temperature is a function of mass and velocity and it is most often seen as the average kinetic energy of particles in a substance on this AP exam. Therefore, the higher the temperature of a substance, the greater the average kinetic energy of its particles. We can think of temperature as how fast particles are moving in a substance.
Standard Temperature: 0 degrees C° = 273.15 K (°C + 273.15 = K)
STP, or Standard Temperature and Pressure, are conditions at 1 atm and 273.15 K. You will often see this term on AP questions. Now let's get into it. Don't worry, we're providing some perfect responses to AP questions too
more resources to help you study
practice questionscheatsheetscore calculator
Gas Laws & Relationships
Gas laws are a set of laws that describe the relationship between pressure, volume, temperature, and moles of gas. The following relationships hold true when the amount of gas is constant.
Boyle’s Law
Boyle's law describes the relationship between the pressure and volume of an ideal gas under constant temperature. It indicates an inverse relationship between pressure and volume: P1V1 = P2V2, where P stands for pressure, V stands for volume, and the numbers refer to the initial and final pressures and volumes.
When volume decreases, the particles collide with the side of the container more often, thereby increasing pressure.
Charles’ Law
Charles' law describes the relationship between the volume and temperature of an ideal gas under constant pressure. Charles' law indicates that there is a direct relationship between volume and temperature: V1/T1 = V2/T2, where V stands for volume, T stands for temperature, and the numbers refer to the initial and final conditions.
When temperature or average kinetic energy increases, particles move faster causing more and stronger collisions with the walls of the container. The volume increases to keep the pressure constant.
Gay-Lussac's Law
Gay-Lussac’s law describes the relationship between the temperature and pressure of an ideal gas under constant volume. His law indicates a direct relationship between pressure and temperature: P1/T1 = P2/T2, where P stands for pressure, T stands for temperature, and the numbers refer to initial and final conditions.
As temperature increases, particles move faster causing collisions with the sides of the container to happen more often and to be stronger. This increases the pressure.
Avogadro's Law
Avogadro's law describes the relationship between volume and the moles of gas in the sample. His law indicates a direct relationship between these two variables: V1/n1 = V2/n2, where V stands for volume, n stands for moles of gas, and the numbers refer to initial and final conditions.
Adding more particles to a container causes more collisions with the walls of the container and the volume increases to keep the pressure constant.
Avogadro also found that equal volumes of gases at the same temperature and pressure contain the same number of particles. For example, 5 L of H and 5 L of He at STP contain the same number of particles.
The Combined Gas Law
Three of these laws can be expressed in the combined gas law: P1V1/T1 = P2V2/T2. When solving problems, you can ignore any of the variables that aren’t addressed (ex. in a pressure change problem where you find volume, ignore T and do P1V1 = P2V2 which is really Boyle's Law).
In reality, you only have to remember this equation out of the ones we learned so far but you should understand the reasons behind every relationship.
Ideal Gas Law
The last equation on that chart above is the ideal gas law: PV=nRT. The following information can be found on the AP Chemistry reference table, but it's quick and easy to memorize!
P = pressure in atm
V = volume in L
n = moles of gas
R = universal gas constant (0.08206 Latm/molK)
T = temperature in Kelvin
In every question, we must remember to convert temperature to the Kelvin scale, volume to liters, and pressure to atm. If you ever forget which units you need for these variables, look at the given units of R on the reference sheet!
The ideal gas law is on almost every single AP Examination, so make sure you nail it. Don't worry, it just requires the plugging in of some numbers and sometimes, stoichiometry which you probably mastered by now.
Dalton's Law of Partial Pressure
"In a sample containing a mixture of ideal gases, the pressure exerted by each component (the partial pressure) is independent of the other components."
According to Dalton's law of partial pressure, the sum of all the partial pressures of each gas in a mixture of gases is equal to the total pressure. In mathematical notation, this is expressed by saying: P = Pa + Pb + Pc... where a, b, and c are different gases. You may be wondering then, how do we calculate partial pressure?
Image Courtesy of Science Notes
To do this, we use the mole fraction of that gas. Mole fraction is denoted by Xa and equals moles A/total moles. For example, if we have a mixture of 3 mol O2 and 4 mol H2, the mole fraction of O2 = 3/(3+4) = 3/7. Then, partial pressure = Xa total pressure.
Another way to represent Dalton's Law of Partial Pressures is:
All you have to do here is plug in the values you have. Px represents partial pressure and the fraction on the right is the mole fraction itself. Memorizing the latter representation is much simpler than going through the mole fraction and then calculating pressure.
AP Chemistry Reference Table
As you progress through the course, it is good to familiarize yourself with what information is given on the reference sheet. The following section is what you are given for this chapter and the highlighted information is what you should know thus far from this unit:
🎥 Watch Serene Fang discuss gas behavior under different pressures, temperatures, and volumes. The ideal gas law is reviewed and practice is done with it as well.
Frequently Asked Questions
What is the ideal gas law and what does each letter in PV = nRT stand for?
The ideal gas law is the equation that links a gas sample’s macroscopic properties: pressure, volume, amount (moles), and temperature. It’s written PV = nRT. - P = pressure of the gas (use atm for the common gas-constant value below) - V = volume of the gas (L) - n = amount of substance in moles (mol) - R = ideal gas constant. A common value is 0.08206 L·atm·mol⁻¹·K⁻¹ (other R values exist for different units) - T = absolute temperature (Kelvin, K) Use Kelvin for T and match your R units to P and V. On the AP exam you’ll often use PV = nRT for volume or moles problems and combine it with Dalton’s law for mixtures (partial pressure P_A = P_total × X_A). For a concise review and practice problems on this Topic 3.4, see the Fiveable study guide ( for extra practice, check the AP Chem practice set (
How do I use the ideal gas law to solve problems step by step?
Step-by-step using PV = nRT (CED 3.4.A.1): 1. Identify what you need (P, V, n, or T). Label units. 2. Convert to proper units for your R. Common choice: R = 0.08206 L·atm·mol⁻¹·K⁻¹, so use P in atm, V in L, T in K, n in mol. Convert temperature to Kelvin (T(K) = T(°C) + 273.15). If pressure is given in mmHg or kPa, convert to atm (760 mmHg = 1.00 atm; 101.325 kPa = 1.00 atm) or use R matching those units. 3. If you need moles from mass, use n = mass / molar mass. For gas mixtures, find partial pressures: P_A = X_A × P_total where X_A = moles A / total moles (CED 3.4.A.2). 4. Rearrange PV = nRT to solve for the unknown (e.g., n = PV/RT; V = nRT/P). Plug values with units, calculate, and report with correct units and sig figs. 5. Check reasonableness (e.g., V positive, T > 0 K). For worked examples and AP-aligned tips, see the Topic 3.4 study guide ( and practice problems (
What's the difference between partial pressure and total pressure in gas mixtures?
Partial pressure is the pressure one gas would exert if it alone occupied the container. In a mixture of ideal gases each component acts independently (Dalton’s law): P_total = P_A + P_B + P_C + … and P_A = X_A × P_total, where X_A = moles A / total moles. So partial pressure depends on how many moles of that gas are present (and on T and V through PV = nRT). For example, in a 2.0 L container at 300 K with 1.0 mol A and 1.0 mol B, total n = 2.0 mol so P_total = nRT/V and P_A = 0.5 × P_total. On the AP exam you should be ready to use PV = nRT for mixtures and to convert mole fraction to partial pressure (CED 3.4.A.1–3.4.A.2). If you want a quick refresher, check the Topic 3.4 study guide ( and practice problems (
Why does the ideal gas law work for some gases but not others?
The ideal gas law (PV = nRT) works best when gases behave like “ideal” particles: point-sized, no intermolecular forces, and elastic collisions—assumptions from the kinetic molecular theory. Real gases deviate when molecules have significant size or attractive forces (large, polar, or at high pressure/low temperature). Those factors lower pressure (attractions) or reduce free volume (finite size). Chemists quantify deviation with the compressibility factor Z (= PV/nRT) or with corrections like the van der Waals equation. If Z ≈ 1, PV = nRT is a good model; if Z differs from 1, use corrected models. For AP Chem, connect this to LO 3.4.A and Topic 3.6 on deviations—expect questions about when PV = nRT applies and why (KMT, Z, van der Waals). For a quick review, see the Topic 3.4 study guide ( and grab practice problems at (
I'm confused about when to use Celsius vs Kelvin in gas law problems - which one do I use?
Always use Kelvin for gas-law equations (PV = nRT, combined gas law, Charles’s/Gay-Lussac’s relationships). The ideal gas law depends on absolute temperature—pressure and volume are proportional to the absolute (not relative) thermal energy of molecules. Convert Celsius to Kelvin by adding 273.15 (0.0 °C = 273.15 K). If you plug °C into PV = nRT you’ll get wrong answers and unit mismatches; same for molar volume at STP (STP uses 273.15 K). One subtle point: if you’re using temperature differences in some contexts (like ΔT in a linear expansion equation where units cancel), Celsius increments can work, but not in gas-law formulas. For AP problems, always show the Kelvin conversion when using PV = nRT or related laws. For extra practice and worked examples, see the Topic 3.4 study guide ( and the practice problem bank (
How do I calculate the mole fraction of a gas in a mixture?
Mole fraction (X) is just the fraction of moles of one gas out of the total moles in the mixture. Formula: X_A = n_A / n_total, where n_total = n_A + n_B + n_C + ... . Use PV = nRT to find moles if you’re given pressures, volumes, and temperature (n = PV/RT). By Dalton’s law, a component’s partial pressure equals the total pressure times its mole fraction: P_A = X_A × P_total (so X_A = P_A / P_total). Quick example: if you have 2.0 mol A and 3.0 mol B, X_A = 2.0 / (2.0+3.0) = 0.40. If P_total = 1.50 atm, P_A = 0.40×1.50 = 0.60 atm. This is exactly what the CED expects (3.4.A: PV = nRT and P_A = P_total × X_A). For more worked examples and practice, check the Topic 3.4 study guide ( and thousands of practice problems (
What happens to pressure when I increase temperature but keep volume constant?
Use the ideal gas law: PV = nRT. If n and V are constant, P = (nR/V) × T, so pressure is directly proportional to temperature (in Kelvin). That means if you increase T, P increases linearly—double T (Kelvin) → about double P. Always convert to Kelvin before comparing temperatures (e.g., 300 K → 600 K doubles pressure). This relationship is sometimes called Gay-Lussac’s law and is exactly what AP CED LO 3.4.A expects you to explain using PV = nRT. On the exam, show the equation, state which variables are held constant, and note the Kelvin requirement. For more review and practice on ideal gases, see the Topic 3.4 study guide ( and extra problems (
Can someone explain partial pressure in simple terms with an example?
Partial pressure is the pressure one gas would exert if it alone filled the container. In a gas mixture each component acts independently (Dalton’s law). So the partial pressure of A is P_A = X_A × P_total, where X_A = moles A / total moles (this is in the CED under 3.4.A.2). Example: you have a 1.0 L container at some T with 2.0 mol N2 and 1.0 mol O2. Total moles = 3.0. Mole fraction X_N2 = 2/3, X_O2 = 1/3. If the total pressure is 3.0 atm, then P_N2 = (2/3)(3.0 atm) = 2.0 atm and P_O2 = (1/3)(3.0 atm) = 1.0 atm. Note PV = nRT relates n and P for each gas if you treat them separately. For AP prep, know the equations PV = nRT and P_A = X_A·P_total and that total pressure = sum of partial pressures (3.4.A.1–3.4.A.2). More review and practice problems are in the Topic 3.4 study guide ( and thousands of practice questions at (
How do I know which gas law equation to use on the AP exam?
Pick the law based on what’s given and what changes. - If you’re given P, V, n, T and asked to relate them, use the ideal gas law PV = nRT (R on formula sheet; T must be in K)—this is the CED Essential Knowledge 3.4.A.1. - If two states of the same gas are compared (P, V, T change but n constant), use the combined gas law: P1V1/T1 = P2V2/T2 (it’s just PV = nRT rearranged). - If only two variables change and one is constant, use the simple proportional laws: Boyle’s (P vs. V at constant T, n), Charles’s (V vs. T at constant P, n), Gay-Lussac’s (P vs. T at constant V, n). - For gas mixtures, use Dalton’s law and mole fractions: Ptotal = ΣPi and Pi = X_i·Ptotal (CED 3.4.A.2). - Always convert T to Kelvin and check if n is constant. For quick review, see the Topic 3.4 study guide ( and more practice problems (
What's the relationship between mole fraction and partial pressure?
Mole fraction (X_A) is the fraction of total moles that gas A contributes: X_A = n_A / n_total. By Dalton’s law of partial pressures, each gas in a mixture behaves independently, so its partial pressure is proportional to its mole fraction: P_A = X_A × P_total So if gas A is half the moles (X_A = 0.50) and the total pressure is 2.0 atm, P_A = 1.0 atm. This follows directly from PV = nRT applied to each component at the same V and T (P ∝ n). Also P_total = ΣP_i. This relation is an essential-knowledge point on the AP CED for Topic 3.4 (ideal gas law / Dalton’s law). For more examples and practice problems tied to the CED, check the Topic 3.4 study guide ( and the 1000+ practice questions (
Why do we assume gases are "ideal" when real gases exist?
We call a gas “ideal” because PV = nRT is a simple, useful model that relates macroscopic properties (P, V, T, n) and makes calculations and reasoning easy on the AP (CED EK 3.4.A.1). The ideal-gas assumption comes from the kinetic molecular theory: point particles with no volume and no intermolecular forces. Real gases deviate when molecules have significant size or attractions (high pressure, low temperature). Those deviations are described by corrections like the van der Waals equation and the compressibility factor Z (Topic 3.6). Why assume ideal? Because under typical lab and many exam conditions (low P, high T) the ideal law is accurate enough and lets you focus on relationships (Boyle’s, Charles’s, Dalton’s partial pressures). When a problem asks about limits or deviations, mention intermolecular forces or finite volume and use van der Waals/Z. For more review and AP-style practice on this topic, see the Ideal Gas Law study guide ( and practice problems (
How do I read and interpret PV diagrams for gas behavior?
A PV diagram plots pressure (y) vs volume (x). Read it by noting the axes, the path, and direction (arrow): area under the curve = work done by the gas (W = ∫P dV), positive for expansion. Common paths you’ll see on AP questions (use PV = nRT): - Isobaric: horizontal line (P constant)—volume changes at constant pressure. - Isochoric: vertical line (V constant)—pressure changes, no work done. - Isothermal (T constant): hyperbola (PV = constant)—P drops as V rises (Boyle’s law). - If n changes (reaction or leak), the curve can shift (total P scales with n via PV = nRT). For mixtures remember Dalton’s law: total P = sum of partial pressures (3.4.A.2). On the exam they may ask you to identify process type, calculate work (area), or relate changes using PV = nRT (3.4.A.1). For more examples and practice, see the Topic 3.4 study guide ( and the practice bank (
I don't understand how to find the number of moles when I'm only given pressure, volume, and temperature.
Use the ideal gas law: PV = nRT. If you’re given pressure (P), volume (V), and temperature (T), solve for n: n = PV / (RT) Make sure: - T is in kelvin (K = °C + 273.15). - Use an R that matches P and V units: common choice R = 0.08206 L·atm·mol⁻¹·K⁻¹ when P is in atm and V in L. - Show units and substitution on the AP exam (they expect algebra, numbers, units, and reasonable sig figs). Quick example: P = 1.50 atm, V = 2.00 L, T = 27 °C (300.15 K ≈ 300 K) n = (1.50 atm)(2.00 L) / (0.08206 L·atm·mol⁻¹·K⁻¹ × 300 K) n ≈ 3.00 / 24.6 ≈ 0.122 mol If you have a gas mixture, find the partial pressure of the gas first (P_A = P_total × X_A or use Dalton’s law) then use PV = nRT for that component (CED 3.4.A.1–2). For more review and practice, see the Topic 3.4 study guide ( and practice problems (
What's R in the ideal gas law and what units should I use for it?
R is the gas constant in PV = nRT—it ties pressure, volume, amount, and temperature together. You must use a value of R whose units match the units in your problem so the equation is consistent. Common values you’ll see on the AP formula sheet / in problems: - R = 0.08206 L·atm·mol⁻¹·K⁻¹ (use when P is in atm, V in L, T in K, n in mol) - R = 8.314 J·mol⁻¹·K⁻¹ (use with SI units: P in Pa, V in m³; note 1 L = 1×10⁻³ m³ and 1 atm = 1.013×10⁵ Pa) Always convert temperature to Kelvin (T in K). On the AP exam you’ll be expected to pick the matching R and show correct unit labels (CED Topic 3.4: PV = nRT). For practice and quick review, check the Topic 3.4 study guide ( and more practice problems (
How do partial pressures add up to total pressure in gas mixtures?
Partial pressures add like money in a wallet: each gas in a mixture behaves independently (Dalton’s law), so the pressure one gas would exert alone at the same T and V is its partial pressure. For ideal gases P_A = X_A · P_total, where X_A = n_A / n_total, and P_total = P_A + P_B + P_C + … (this comes from PV = nRT applied to each component). Example: if a container has 2.0 mol A and 3.0 mol B (total 5.0 mol) and P_total = 1.50 atm, then X_A = 0.40 and P_A = 0.60 atm, P_B = 0.90 atm (they sum to 1.50 atm). This idea is tested in Topic 3.4 items on the AP exam—be ready to use mole fractions with PV = nRT or to add partial pressures directly. For a quick review, see the Topic 3.4 study guide ( and try practice problems ( |
187770 | https://unitimesofficial.wordpress.com/wp-content/uploads/2020/07/schaums-outline-theory-and-problems-of-introduction-to-mathematical-economics.pdf | SCHAUM’S OUTLINE OF Theory and Problems of INTRODUCTION TO MATHEMATICAL ECONOMICS Third Edition EDWARD T. DOWLING, Ph.D.
Professor and Former Chairman Department of Economics Fordham University Schaum’s Outline Series McGRAW-HILL New York San Francisco Washington, D.C.
Auckland Bogota ´ Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto To the memory of my parents, Edward T. Dowling, M.D. and Mary H. Dowling EDWARD T. DOWLING is professor of Economics at Fordham University. He was Dean of Fordham College from 1982 to 1986 and Chairman of the Economics Department from 1979 to 1982 and again from 1988 to 1994. His Ph.D. is from Cornell University and his main areas of professional interest are mathematical economics and economic development. In addition to journal articles, he is the author of Schaum’s Outline of Calculus for Business, Economics, and the Social Sciences, and Schaum’s Outline of Mathematical Methods for Business and Economics. A Jesuit priest, he is a member of the Jesuit Community at Fordham.
Copyright © 2001, 1992 by The McGraw-Hill Companies, Inc. All rights reserved. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher.
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PREFACE The mathematics needed for the study of economics and business continues to grow with each passing year, placing ever more demands on students and faculty alike. Introduction to Mathematical Economics, third edition, introduces three new chapters, one on comparative statics and concave programming, one on simultaneous differential and difference equations, and one on optimal control theory. To keep the book manageable in size, some chapters and sections of the second edition had to be excised. These include three chapters on linear programming and a number of sections dealing with basic elements such as factoring and completing the square. The deleted topics were chosen in part because they can now be found in one of my more recent Schaum books designed as an easier, more detailed introduction to the mathematics needed for economics and business, namely, Mathematical Methods for Business and Economics.
The objectives of the book have not changed over the 20 years since the introduction of the first edition, originally called Mathematics for Economists. Introduction to Mathematical Economics, third edition, is designed to present a thorough, easily understood introduction to the wide array of mathematical topics economists, social scientists, and business majors need to know today, such as linear algebra, differential and integral calculus, nonlinear programming, differential and difference equations, the calculus of variations, and optimal control theory. The book also offers a brief review of basic algebra for those who are rusty and provides direct, frequent, and practical applications to everyday economic problems and business situations.
The theory-and-solved-problem format of each chapter provides concise explanations illustrated by examples, plus numerous problems with fully worked-out solutions. The topics and related problems range in difficulty from simpler mathematical operations to sophisticated applications. No mathematical proficiency beyond the high school level is assumed at the start. The learning-by-doing pedagogy will enable students to progress at their own rates and adapt the book to their own needs.
Those in need of more time and help in getting started with some of the elementary topics may feel more comfortable beginning with or working in conjunction with my Schaum’s Outline of Mathematical Methods for Business and Economics, which offers a kinder, gentler approach to the discipline. Those who prefer more rigor and theory, on the other hand, might find it enriching to work along with my Schaum’s Outline of Calculus for Business, Economics, and the Social Sciences, which devotes more time to the theoretical and structural underpinnings.
Introduction to Mathematical Economics, third edition, can be used by itself or as a supplement to other texts for undergraduate and graduate students in economics, business, and the social sciences.
It is largely self-contained. Starting with a basic review of high school algebra in Chapter 1, the book consistently explains all the concepts and techniques needed for the material in subsequent chapters.
Since there is no universal agreement on the order in which differential calculus and linear algebra should be presented, the book is designed so that Chapters 10 and 11 on linear algebra can be covered immediately after Chapter 2, if so desired, without loss of continuity.
This book contains over 1600 problems, all solved in considerable detail. To get the most from the book, students should strive as soon as possible to work independently of the solutions. This can be done by solving problems on individual sheets of paper with the book closed. If difficulties arise, the solution can then be checked in the book.
iii For best results, students should never be satisfied with passive knowledgethe capacity merely to follow or comprehend the various steps presented in the book. Mastery of the subject and doing well on exams requires active knowledgethe ability to solve any problem, in any order, without the aid of the book.
Experience has proved that students of very different backgrounds and abilities can be successful in handling the subject matter presented in this text if they apply themselves and work consistently through the problems and examples.
In closing, I would like to thank my friend and colleague at Fordham, Dr. Dominick Salvatore, for his unfailing encouragement and support over the past 25 years, and an exceptionally fine graduate student, Robert Derrell, for proofreading the manuscript and checking the accuracy of the solutions.
I am also grateful to the entire staff at McGraw-Hill, especially Barbara Gilson, Tina Cameron, Maureen B. Walker, and Deborah Aaronson.
EDWARD T. DOWLING iv PREFACE CONTENTS CHAPTER 1 Review 1 1.1 Exponents.
1.2 Polynomials.
1.3 Equations: Linear and Quadratic.
1.4 Simultaneous Equations.
1.5 Functions.
1.6 Graphs, Slopes, and Intercepts.
CHAPTER 2 Economic Applications of Graphs and Equations 14 2.1 Isocost Lines.
2.2 Supply and Demand Analysis.
2.3 Income Determination Models.
2.4 IS-LM Analysis.
CHAPTER 3 The Derivative and the Rules of Differentiation 32 3.1 Limits.
3.2 Continuity.
3.3 The Slope of a Curvilinear Function.
3.4 The Derivative.
3.5 Differentiability and Continuity.
3.6 Derivative Notation.
3.7 Rules of Differentiation.
3.8 Higher-Order Derivatives.
3.9 Implicit Differentiation.
CHAPTER 4 Uses of the Derivative in Mathematics and Economics 58 4.1 Increasing and Decreasing Functions.
4.2 Concavity and Convexity.
4.3 Relative Extrema.
4.4 Inflection Points.
4.5 Optimization of Functions.
4.6 Successive-Derivative Test for Optimization.
4.7 Marginal Concepts.
4.8 Optimizing Economic Functions.
4.9 Relationship among Total, Marginal, and Average Concepts.
CHAPTER 5 Calculus of Multivariable Functions 82 5.1 Functions of Several Variables and Partial Derivatives.
5.2 Rules of Partial Differentiation.
5.3 Second-Order Partial Derivatives.
5.4 Optimization of Multivariable Functions.
5.5 Constrained Optimization with Lagrange Multipliers.
5.6 Significance of the Lagrange Multiplier.
5.7 Differentials.
5.8 Total and Partial Differentials.
5.9 Total Derivatives.
5.10 Implicit and Inverse Function Rules.
CHAPTER 6 Calculus of Multivariable Functions in Economics 110 6.1 Marginal Productivity.
6.2 Income Determination Multipliers and Comparative Statics.
6.3 Income and Cross Price Elasticities of Demand.
6.4 Differentials and Incremental Changes.
6.5 Optimization of Multivariable Functions in Economics.
6.6 Constrained Optimization of Multivariable v Functions in Economics.
6.7 Homogeneous Production Functions.
6.8 Returns to Scale.
6.9 Optimization of Cobb-Douglas Production Functions.
6.10 Optimization of Constant Elasticity of Substitution Production Functions.
CHAPTER 7 Exponential and Logarithmic Functions 146 7.1 Exponential Functions.
7.2 Logarithmic Functions.
7.3 Properties of Exponents and Logarithms.
7.4 Natural Exponential and Logarithmic Functions.
7.5 Solving Natural Exponential and Logarithmic Functions.
7.6 Logarithmic Transformation of Nonlinear Functions.
CHAPTER 8 Exponential and Logarithmic Functions in Economics 160 8.1 Interest Compounding.
8.2 Effective vs. Nominal Rates of Interest.
8.3 Discounting.
8.4 Converting Exponential to Natural Exponential Functions.
8.5 Estimating Growth Rates from Data Points.
CHAPTER 9 Differentiation of Exponential and Logarithmic Functions 173 9.1 Rules of Differentiation.
9.2 Higher-Order Derivatives.
9.3 Partial Derivatives.
9.4 Optimization of Exponential and Logarithmic Functions.
9.5 Logarithmic Differentiation.
9.6 Alternative Measures of Growth.
9.7 Optimal Timing.
9.8 Derivation of a Cobb-Douglas Demand Function Using a Logarithmic Transformation.
CHAPTER 10 The Fundamentals of Linear (or Matrix) Algebra 199 10.1 The Role of Linear Algebra.
10.2 Definitions and Terms.
10.3 Addition and Subtraction of Matrices.
10.4 Scalar Multiplication.
10.5 Vector Multiplication.
10.6 Multiplication of Matrices.
10.7 Commutative, Associative, and Distributive Laws in Matrix Algebra.
10.8 Identity and Null Matrices.
10.9 Matrix Expression of a System of Linear Equations.
CHAPTER 11 Matrix Inversion 224 11.1 Determinants and Nonsingularity.
11.2 Third-Order Determinants.
11.3 Minors and Cofactors.
11.4 Laplace Expansion and Higher-Order Determinants.
11.5 Properties of a Determinant.
11.6 Cofactor and Adjoint Matrices.
11.7 Inverse Matrices.
11.8 Solving Linear Equations with the Inverse.
11.9 Cramer’s Rule for Matrix Solutions.
vi CONTENTS CHAPTER 12 Special Determinants and Matrices and Their Use in Economics 254 12.1 The Jacobian.
12.2 The Hessian.
12.3 The Discriminant.
12.4 Higher-Order Hessians.
12.5 The Bordered Hessian for Constrained Optimization.
12.6 Input-Output Analysis.
12.7 Characteristic Roots and Vectors (Eigenvalues, Eigenvectors).
CHAPTER 13 Comparative Statics and Concave Programming 284 13.1 Introduction to Comparative Statics.
13.2 Comparative Statics with One Endogenous Variable.
13.3 Comparative Statics with More Than One Endogenous Variable.
13.4 Comparative Statics for Optimization Problems.
13.5 Comparative Statics Used in Constrained Optimization.
13.6 The Envelope Theorem.
13.7 Concave Programming and Inequality Constraints.
CHAPTER 14 Integral Calculus: The Indefinite Integral 326 14.1 Integration.
14.2 Rules of Integration.
14.3 Initial Conditions and Boundary Conditions.
14.4 Integration by Substitution.
14.5 Integration by Parts.
14.6 Economic Applications.
CHAPTER 15 Integral Calculus: The Definite Integral 342 15.1 Area Under a Curve.
15.2 The Definite Integral.
15.3 The Fundamental Theorem of Calculus.
15.4 Properties of Definite Integrals.
15.5 Area Between Curves.
15.6 Improper Integrals.
15.7 L’Ho ˆpital’s Rule.
15.8 Consumers’ and Producers’ Surplus.
15.9 The Definite Integral and Probability.
CHAPTER 16 First-Order Differential Equations 362 16.1 Definitions and Concepts.
16.2 General Formula for First-Order Linear Differential Equations.
16.3 Exact Differential Equations and Partial Integration.
16.4 Integrating Factors.
16.5 Rules for the Integrating Factor.
16.6 Separation of Variables.
16.7 Economic Applications.
16.8 Phase Diagrams for Differential Equations.
CHAPTER 17 First-Order Difference Equations 391 17.1 Definitions and Concepts.
17.2 General Formula for First-Order Linear Difference Equations.
17.3 Stability Conditions.
17.4 Lagged Income Determination Model.
17.5 The Cobweb Model.
17.6 The Harrod Model.
17.7 Phase Diagrams for Difference Equations.
vii CONTENTS CHAPTER 18 Second-Order Differential Equations and Difference Equations 408 18.1 Second-Order Differential Equations.
18.2 Second-Order Difference Equations.
18.3 Characteristic Roots.
18.4 Conjugate Complex Numbers.
18.5 Trigonometric Functions.
18.6 Derivatives of Trigonometric Functions.
18.7 Transformation of Imaginary and Complex Numbers.
18.8 Stability Conditions.
CHAPTER 19 Simultaneous Differential and Difference Equations 428 19.1 Matrix Solution of Simultaneous Differential Equations, Part 1.
19.2 Matrix Solution of Simultaneous Differential Equations, Part 2.
19.3 Matrix Solution of Simultaneous Difference Equations, Part 1.
19.4 Matrix Solution of Simultaneous Difference Equations, Part 2.
19.5 Stability and Phase Diagrams for Simultaneous Differential Equations.
CHAPTER 20 The Calculus of Variations 460 20.1 Dynamic Optimization.
20.2 Distance Between Two Points on a Plane.
20.3 Euler’s Equation and the Necessary Condition for Dynamic Optimization.
20.4 Finding Candidates for Extremals.
20.5 The Sufficiency Conditions for the Calculus of Variations.
20.6 Dynamic Optimization Subject to Functional Constraints.
20.7 Variational Notation.
20.8 Applications to Economics.
CHAPTER 21 Optimal Control Theory 493 21.1 Terminology.
21.2 The Hamiltonian and the Necessary Conditions for Maximization in Optimal Control Theory.
21.3 Sufficiency Conditions for Maximization in Optimal Control.
21.4 Optimal Control Theory with a Free Endpoint.
21.5 Inequality Constraints in the Endpoints.
21.6 The Current-Valued Hamiltonian.
Index 515 viii CONTENTS CHAPTER 1 Review 1.1 EXPONENTS Given n a positive integer, xn signifies that x is multiplied by itself n times. Here x is referred to as the base and n is termed an exponent. By convention an exponent of 1 is not expressed: x1 x, 81 8. By definition, any nonzero number or variable raised to the zero power is equal to 1: x0 1, 30 1. And 00 is undefined. Assuming a and b are positive integers and x and y are real numbers for which the following exist, the rules of exponents are outlined below and illustrated in Examples 1 and 2 and Problem 1.1.
1.
xa(xb) xab 6.
1 xa xa 2.
xa xb xab 7.
x x1/2 3.
(xa)b xab 8.
a x x1/a 4.
(xy)a xaya 9.
b xa xa/b (x1/b)a 5. x y a xa ya 10.
x(a/b) 1 xa/b EXAMPLE 1.
From Rule 2, it can easily be seen why any variable or nonzero number raised to the zero power equals 1. For example, x3/x3 x33 x0 1; 85/85 855 80 1.
EXAMPLE 2.
In multiplication, exponents of the same variable are added; in division, the exponents are subtracted; when raised to a power, the exponents are multiplied, as indicated by the rules above and shown in the examples below followed by illustrations in brackets.
a) x2(x3) x23 x5 x6 Rule 1 [x2(x3) (x · x)(x · x · x) x · x · x · x · x x5] b) x6 x3 x63 x3 x2 Rule 2 x6 x3 x · x · x · x · x · x x · x · x x · x · x x3 c) (x4)2 x4 · 2 x8 x16 or x6 Rule 3 [(x4)2 (x · x · x · x)(x · x · x · x) x8] 1 d) (xy)4 x4y4 xy4 Rule 4 [(xy)4 (xy)(xy)(xy)(xy) (x · x · x · x)(y · y · y · y) x4y4] e) x y 5 x5 y5 x5 y or x y5 Rule 5 x y 5 (x) (y) (x) (y) (x) (y) (x) (y) (x) (y) x5 y5 f) x3 x4 x34 x1 1 x x3/4 Rules 2 and 6 x3 x4 x · x · x x · x · x · x 1 x g) x x1/2 Rule 7 Since x · x x and from Rule 1 exponents of a common base are added in multiplication, the exponent of x, when added to itself, must equal 1. With 1 – 2 1 – 2 1, the exponent of x is 1 – 2. Thus, x · x x1/2 · x1/2 x1/21/2 x1 x.
h) 3 x x1/3 Rule 8 Just as 3 x · 3 x · 3 x x, so x1/3 · x1/3 · x1/3 x1/31/31/3 x1 x.
i) x3/2 (x1/2)3 or (x3)1/2 Rule 9 [43/2 (41/2)3 (4)3 (2)3 8, or equally valid, 43/2 (43)1/2 (64)1/2 64 8] j) x2/3 1 x2/3 1 (x1/3)2 or 1 (x2)1/3 Rule 10 272/3 1 (271/3)2 1 (3)2 1 9, or equally valid, 272/3 1 (272)1/3 1 (729)1/3 1 9 See Problem 1.1.
1.2 POLYNOMIALS Given an expression such as 5x3, x is called a variable because it can assume any number of different values, and 5 is referred to as the coefficient of x. Expressions consisting simply of a real number or of a coefficient times one or more variables raised to the power of a positive integer are called monomials. Monomials can be added or subtracted to form polynomials. Each of the monomials comprising a polynomial is called a term. Terms that have the same variables and exponents are called like terms. Rules for adding, subtracting, multiplying, and dividing polynomials are explained in Examples 3 through 5 and treated in Problems 1.2 to 1.4.
EXAMPLE 3.
Like terms in polynomials can be added or subtracted by adding their coefficients. Unlike terms cannot be so added or subtracted. See Problems 1.2 and 1.3.
a) 4x5 9x5 13x5 b) 12xy 3xy 9xy c) (7x3 5x2 8x) (11x3 9x2 2x) 18x3 4x2 10x d) (24x 17y) (6x 5z) 30x 17y 5z EXAMPLE 4.
Like and unlike terms can be multiplied or divided by multiplying or dividing both the coefficients and variables.
a) (5x)(13y2) 65xy2 b) (7x3y5)(4x2y4) 28x5y9 c) (2x3y)(17y4z2) 34x3y5 z2 d) 15x4y3z6 3x2y2z3 5x2yz3 2 REVIEW [CHAP . 1 e) 4x2y5z3 8x5y3z4 y2 2x3z EXAMPLE 5.
In multiplying two polynomials, each term in the first polynomial must be multiplied by each term in the second and their products added. See Problem 1.4.
(6x 7y)(4x 9y) (2x 3y)(8x 5y 7z) 24x2 54xy 28xy 63y2 24x2 82xy 63y2 16x2 10xy 14xz 24xy 15y2 21yz 16x2 14xy 14xz 21yz 15y2 1.3 EQUATIONS: LINEAR AND QUADRATIC A mathematical statement setting two algebraic expressions equal to each other is called an equation. An equation in which all variables are raised to the first power is known as a linear equation.
A linear equation can be solved by moving the unknown variable to the left-hand side of the equal sign and all the other terms to the right-hand side, as is illustrated in Example 6. A quadratic equation of the form ax2 bx c 0, where a, b, and c are constants and a 0, can be solved by factoring or using the quadratic formula: x b b2 4ac 2a (1.1) Solving quadratic equations by factoring is explained in Example 7 and by the quadratic formula in Example 8 and Problem 1.6.
EXAMPLE 6.
The linear equation given below is solved in three easy steps.
x 4 3 x 5 1 1.
Move all terms with the unknown variable x to the left, here by subtracting x/5 from both sides of the equation.
x 4 3 x 5 1 2.
Move any term without the unknown variable to the right, here by adding 3 to both sides of the equation.
x 4 x 5 1 3 4 3.
Simplify both sides of the equation until the unknown variable is by itself on the left and the solution is on the right, here by multiplying both sides of the equation by 20 and subtracting.
20 · x 4 x 5 4 · 20 5x 4x x 80 80 EXAMPLE 7.
Factoring is the easiest way to solve a quadratic equation, provided the factors are easily recognized integers. Given x2 13x 30 0 by factoring, we have (x 3)(x 10) 0 3 REVIEW CHAP . 1] For (x 3)(x 10) to equal 0, x 3 or x 10 must equal 0. Setting each in turn equal to 0 and solving for x, we have x 3 x 0 3 x 10 x 0 10 Those wishing a thorough review of factoring and other basic mathematical techniques should consult another of the author’s books, Schaum’s Outline of Mathematical Methods for Business and Economics, for a gentler, more gradual approach to the discipline.
EXAMPLE 8.
The quadratic formula is used below to solve the quadratic equation 5x2 55x 140 0 Substituting a 5, b 55, c 140 from the given equation in (1.1) gives x (55) (55)2 4(5)(140) 2(5) 55 3025 2800 10 55 225 10 55 15 10 Adding 15 and then 15 to find each of the two solutions, we get x 55 15 10 7 x 55 15 10 4 See Problem 1.6.
1.4 SIMULTANEOUS EQUATIONS To solve a system of two or more equations simultaneously, (1) the equations must be consistent (noncontradictory), (2) they must be independent (not multiples of each other), and (3) there must be as many consistent and independent equations as variables. A system of simultaneous linear equations can be solved by either the substitution or elimination method, explained in Example 9 and Problems 2.11 to 2.16, as well as by methods developed later in linear algebra in Sections 11.8 and 11.9.
EXAMPLE 9.
The equilibrium conditions for two markets, butter and margarine, where Pb and Pm are the prices of butter and margarine, respectively, are given in (1.2) and (1.3): 8Pb 3Pm 7 Pb 7Pm 19 (1.2) (1.3) The prices that will bring equilibrium to the model are found below by using the substitution and elimination methods.
Substitution Method 1.
Solve one of the equations for one variable in terms of the other. Solving (1.3) for Pb gives Pb 7Pm 19 2.
Substitute the value of that term in the other equation, here (1.2), and solve for Pm.
8Pb 3Pm 7 8(7Pm 19) 3Pm 7 56Pm 152 3Pm 7 53Pm 159 Pm 3 4 REVIEW [CHAP . 1 3.
Then substitute Pm 3 in either (1.2) or (1.3) to find Pb.
8Pb 3(3) 7 8Pb 16 Pb 2 Elimination Method 1.
Multiply (1.2) by the coefficient of Pb (or Pm) in (1.3) and (1.3) by the coefficient of Pb (or Pm) in (1.2).
Picking Pm, we get 7(8Pb 3Pm 7) 3(Pb 7Pm 19) 56Pb 21Pm 49 3Pb 21Pm 57 (1.4) (1.5) 2.
Subtract (1.5) from (1.4) to eliminate the selected variable.
53Pb 106 Pb 2 3.
Substitute Pb 2 in (1.4) or (1.5) to find Pm as in step 3 of the substitution method.
1.5 FUNCTIONS A function f is a rule which assigns to each value of a variable (x), called the argument of the function, one and only one value [f(x)], referred to as the value of the function at x. The domain of a function refers to the set of all possible values of x; the range is the set of all possible values for f(x).
Functions are generally defined by algebraic formulas, as illustrated in Example 10. Other letters, such as g, h, or the Greek letter , are also used to express functions. Functions encountered frequently in economics are listed below.
Linear function: f(x) mx b Quadratic function: f(x) ax2 bx c (a 0) Polynomial function of degree n: f(x) anxn an1xn1 · · · a0 (n nonnegative integer; an 0) Rational function: f(x) g(x) h(x) where g(x) and h(x) are both polynomials and h(x) 0. (Note: Rational comes from ratio.) Power function: f(x) axn (n any real number) EXAMPLE 10.
The function f(x) 8x 5 is the rule that takes a number, multiplies it by 8, and then subtracts 5 from the product. If a value is given for x, the value is substituted for x in the formula, and the equation solved for f(x). For example, if x 3, f(x) 8(3) 5 19 f(x) 8(4) 5 27 If x 4, See Problems 1.7 to 1.9.
5 REVIEW CHAP . 1] EXAMPLE 11.
Given below are examples of different functions: Linear: f(x) 7x 4 g(x) 3x h(x) 9 Quadratic: f(x) 5x2 8x 2 g(x) x2 6x h(x) 6x2 Polynomial: f(x) 4x3 2x2 9x 5 g(x) 2x5 x3 7 Rational: f(x) x2 9 x 4 (x 4) g(x) 5x x 2 (x 2) Power: f(x) 2x6 g(x) x1/2 h(x) 4x3 1.6 GRAPHS, SLOPES, AND INTERCEPTS In graphing a function such as y f(x), x is placed on the horizontal axis and is known as the independent variable; y is placed on the vertical axis and is called the dependent variable. The graph of a linear function is a straight line. The slope of a line measures the change in y ( y) divided by a change in x ( x). The slope indicates the steepness and direction of a line. The greater the absolute value of the slope, the steeper the line. A positively sloped line moves up from left to right; a negatively sloped line moves down. The slope of a horizontal line, for which y 0, is zero. The slope of a vertical line, for which x 0, is undefined, i.e., does not exist because division by zero is impossible. The y intercept is the point where the graph crosses the y axis; it occurs when x 0. The x intercept is the point where the line intersects the x axis; it occurs when y 0. See Problem 1.10.
EXAMPLE 12.
To graph a linear equation such as y 1 – 4x 3 one need only find two points which satisfy the equation and connect them by a straight line. Since the graph of a linear function is a straight line, all the points satisfying the equation must lie on the line.
To find the y intercept, set x 0 and solve for y, getting y 1 – 4(0) 3, y 3. The y intercept is the point (x, y) (0, 3). To find the x intercept, set y 0 and solve for x. Thus, 0 1 – 4x 3, 1 – 4x 3, x 12. The x intercept is the point (x, y) (12, 0). Then plot the points (0, 3) and (12, 0) and connect them by a straight line, as in Fig.
1-1, to complete the graph of y 1 – 4x 3. See Examples 13 and 14 and Problems 1.10 to 1.12.
EXAMPLE 13.
For a line passing through points (x1, y1) and (x2, y2), the slope m is calculated as follows: m y x y2 y1 x2 x1 x1 x2 For the line in Fig. 1-1 passing through (0, 3) and (12, 0), m y x 0 3 12 0 1 4 and the vertical intercept can be seen to be the point (0, 3).
6 REVIEW [CHAP . 1 Fig. 1-1 EXAMPLE 14.
For a linear equation in the slope-intercept form y mx b m, b constants the slope and intercepts of the line can be read directly from the equation. For such an equation, m is the slope of the line; (0, b) is the y intercept; and, as seen in Problem 1.10, (b/m, 0) is the x intercept. One can tell immediately from the equation in Example 12, therefore, that the slope of the line is 1 – 4, the y intercept is (0, 3), and the x intercept is (12, 0).
Solved Problems EXPONENTS 1.1.
Simplify the following, using the rules of exponents: a) x4 · x5 x4 · x5 x45 x9 b) x7 · x3 x7 · x3 x7(3) x4 x7 · x3 x7 · 1 x3 x · x · x · x · x · x · x · 1 x · x · x x4 c) x2 · x4 x2 · x4 x2(4) x6 1 x6 x2 · x4 1 x · x · 1 x · x · x · x 1 x6 d) x2 · x1/2 x2 · x1/2 x2(1/2) x5/2 x5 [x2 · x1/2 (x · x)(x) (x · x · x · x)(x) (x1/2)5 x5/2] e) x9 x3 x9 x3 x93 x6 f) x4 x7 x4 x7 x47 x3 1 x3 x4 x7 x · x · x · x x · x · x · x · x · x · x 1 x3 7 REVIEW CHAP . 1] g) x3 x4 x3 x4 x3(4) x34 x7 x3 x4 x3 1/x4 x3 · x4 x7 h) x3 x x3 x x3 x1/2 x3(1/2) x5/2 x5 i) (x2)5 (x2)5 x2 · 5 x10 j) (x4)2 (x4)2 x4 · (2) x8 1 x8 k) 1 x5 · 1 y5 1 x5 · 1 y5 x5 · y5 (xy)5 1 (xy)5 l) x3 y3 x3 y3 x y 3 POLYNOMIALS 1.2.
Perform the indicated arithmetic operations on the following polynomials: a) 3xy 5xy b) 13yz2 28yz2 c) 36x2y3 25x2y3 d) 26x1x2 58x1x2 e) 16x2y3z5 37x2y3z5 a) 8xy, b) 15yz2, c) 11x2y3, d) 84x1x2, e) 21x2y3z5 1.3.
Add or subtract the following polynomials as indicated. Note that in subtraction the sign of every term within the parentheses must be changed before corresponding elements are added.
a) (34x 8y) (13x 12y) (34x 8y) (13x 12y) 47x 4y b) (26x 19y) (17x 50y) (26x 19y) (17x 50y) 9x 31y c) (5x2 8x 23) (2x2 7x) (5x2 8x 23) (2x2 7x) 3x2 15x 23 d) (13x2 35x) (4x2 17x 49) (13x2 35x) (4x2 17x 49) 9x2 18x 49 8 REVIEW [CHAP . 1 1.4.
Perform the indicated operations, recalling that each term in the first polynomial must be multiplied by each term in the second and their products summed.
a) (2x 9)(3x 8) (2x 9)(3x 8) 6x2 16x 27x 72 6x2 11x 72 b) (6x 4y)(3x 5y) (6x 4y)(3x 5y) 18x2 30xy 12xy 20y2 18x2 42xy 20y2 c) (3x 7)2 (3x 7)2 (3x 7)(3x 7) 9x2 21x 21x 49 9x2 42x 49 d) (x y)(x y) (x y)(x y) x2 xy xy y2 x2 y2 SOLVING EQUATIONS 1.5.
Solve each of the following linear equations by moving all terms with the unknown variable to the left, moving all other terms to the right, and then simplifying.
a) 5x 6 9x 10 b) 26 2x 8x 44 5x 6 5x 9x 4x x 9x 10 10 6 16 4 26 2x 2x 8x 10x x 8x 44 44 26 70 7 c) 9(3x 4) 2x 11 5(4x 1) d) x 3 16 x 12 14 9(3x 4) 2x 27x 36 2x 27x 2x 20x 5x x 11 5(4x 1) 11 20x 5 11 5 36 30 6 x 3 16 x 12 14 x 3 x 12 14 16 Multiplying both sides of the equation by the least common denominator (LCD), here 12, gives 12 · x 3 x 12 30 · 12 4x x x 360 120 e) 5 x 3 x 4 7 x [x 0, 4] 5 x 3 x 4 7 x Multiplying both sides by the LCD, we get x(x 4) · 5 x 3 x 4 7 x · x(x 4) 5(x 4) 3x 8x 20 x 7(x 4) 7x 28 8 9 REVIEW CHAP . 1] 1.6.
Solve the following quadratic equations, using the quadratic formula: a) 5x2 23x 12 0 Using (1.1) and substituting a 5, b 23, and c 12, we get x b b2 4ac 2a 23 (23)2 4(5)(12) 2(5) 23 529 240 10 23 289 10 23 17 10 x 23 17 10 0.6 x 23 17 10 4 b) 3x2 41x 26 0 x (41) (41)2 4(3)(26) 2(3) 41 1681 312 6 41 1369 6 41 37 6 x 41 37 6 13 x 41 37 6 2 3 FUNCTIONS 1.7.
a) Given f(x) x2 4x 5, find f(2) and f(3).
Substituting 2 for each occurrence of x in the function gives f(2) (2)2 4(2) 5 7 Now substituting 3 for each occurrence of x, we get f(3) (3)2 4(3) 5 8 b) Given f(x) 2x3 5x2 8x 20, find f(5) and f(4).
f(5) 2(5)3 5(5)2 8(5) 20 145 f(4) 2(4)3 5(4)2 8(4) 20 260 1.8.
In the following graphs (Fig. 1-2), where y replaces f(x) as the dependent variable in functions, indicate which graphs are graphs of functions and which are not.
For a graph to be the graph of a function, for each value of x, there can be one and only one value of y. If a vertical line can be drawn which intersects the graph at more than one point, then the graph is not the graph of a function. Applying this criterion, which is known as the vertical-line test, we see that (a), (b), and (d) are functions; (c), (e), and (f) are not.
1.9.
Which of the following equations are functions and why?
a) y 2x 7 y 2x 7 is a function because for each value of the independent variable x there is one and only one value of the dependent variable y. For example, if x 1, y 2(1) 7 5. The graph would be similar to (a) in Fig. 1-2.
10 REVIEW [CHAP . 1 b) y2 x y2 x, which is equivalent to y x, is not a function because for each positive value of x, there are two values of y. For example, if y2 9, y 3. The graph would be similar to that of (c) in Fig. 1-2, illustrating that a parabola whose axis is parallel to the x axis cannot be a function.
c) y x2 y x2 is a function. For each value of x there is only one value of y. For instance, if x 5, y 25. While it is also true that y 25 when x 5, it is irrelevant. The definition of a function simply demands that for each value of x there be one value of y, not that for each value of y there be only one value of x. The graph would be like (d) in Fig. 1-2, demonstrating that a parabola with axis parallel to the y axis is a function.
d) y x2 6x 15 y x2 6x 15 is a function. For each value of x there is a unique value of y. The graph would be like (b) in Fig. 1-2.
e) x2 y2 64 x2 y2 64 is not a function. If x 0, y2 64, and y 8. The graph would be a circle, similar to (e) in Fig. 1-2. A circle does not pass the vertical-line test.
f) x 4 x 4 is not a function. The graph of x 4 is a vertical line. This means that at x 4, y has many values. The graph would look like (f) in Fig. 1-2.
GRAPHS, SLOPES, AND INTERCEPTS 1.10.
Find the x intercept in terms of the parameters of the slope-intercept form of a linear equation y mx b.
Setting y 0, 0 mx b mx b x b m Thus, the x intercept of the slope-intercept form is (b/m, 0).
11 REVIEW CHAP . 1] Fig. 1-2 1.11.
Graph the following equations and indicate their respective slopes and intercepts: a) 3y 15x 30 b) 2y 6x 12 c) 8y 2x 16 0 d) 6y 3x 18 0 To graph an equation, first set it in slope-intercept form by solving it for y in terms of x. From Example 14, the slope and two intercepts can then be read directly from the equation, providing three pieces of information, whereas only two are needed to graph a straight line. See Fig. 1-3 and Problems 2.1 to 2.10.
a) 3y 15x 30 3y 15x 30 y 5x 10 Slope m 5 y intercept: (0, 10) x intercept: (2, 0) b) 2y 6x 12 2y 6x 12 y 3x 6 Slope m 3 y intercept: (0, 6) x intercept: (2, 0) c) 8y 2x 16 0 8y 2x 16 y 1 – 4x 2 Slope m 1 – 4 y intercept: (0, 2) x intercept: (8, 0) d) 6y 3x 18 0 6y 3x 18 y 1 – 2x 3 Slope m 1 – 2 y intercept: (0, 3) x intercept: (6, 0) 12 REVIEW [CHAP . 1 Fig. 1-3 1.12.
Find the slope m of the linear function passing through: a) (4, 12), (8, 2); b) (1, 15), (3, 6); c) (2, 3), (5, 18).
a) Substituting in the formula from Example 13, we get m y2 y1 x2 x1 2 12 8 4 10 4 2 1 2 b) m 6 15 3 (1) 9 4 2 1 4 c) m 18 (3) 5 2 21 3 7 1.13.
Graph (a) the quadratic function y 2x2 and (b) the rational function y 2/x.
Fig. 1-3 13 REVIEW CHAP . 1] CHAPTER 2 Economic Applications of Graphs and Equations 2.1 ISOCOST LINES An isocost line represents the different combinations of two inputs or factors of production that can be purchased with a given sum of money. The general formula is PKK PLL E, where K and L are capital and labor, PK and PL their respective prices, and E the amount allotted to expenditures.
In isocost analysis the individual prices and the expenditure are initially held constant; only the different combinations of inputs are allowed to change. The function can then be graphed by expressing one variable in terms of the other, as seen in Example 1 and Problems 2.5 and 2.6.
EXAMPLE 1. Given: PKK PLL E PKK E PLL K E PLL PK K E PK PL PK L This is the familiar linear function of the form y mx b, where b E/PK the vertical intercept and m PL/PK the slope. The graph is given by the solid line in Fig. 2-1.
From the equation and graph, the effects of a change in any one of the parameters are easily discernible. An increase in the expenditure from E to E will increase the vertical intercept and cause the isocost line to shift out to the right (dashed line) parallel to the old line. The slope is unaffected because the slope depends on the relative prices (PL/PK) and prices are not affected by expenditure changes. A change in PL will alter the slope of the line but leave the vertical intercept unchanged. A change in PK will alter the slope and the vertical intercept. See Problems 2.5 and 2.6.
14 2.2 SUPPLY AND DEMAND ANALYSIS Equilibrium in supply and demand analysis occurs when Qs Qd. By equating the supply and demand functions, the equilibrium price and quantity can be determined. See Example 2 and Problems 2.1 to 2.4 and 2.11 to 2.16.
EXAMPLE 2.
Given: Qs 5 3P Qd 10 2P In equilibrium, Qs Qd Solving for P, 5 3P 10 2P 5P 15 P 3 Substituting P 3 in either of the equations, Qs 5 3P 5 3(3) 4 Qd 2.3 INCOME DETERMINATION MODELS Income determination models generally express the equilibrium level of income in a four-sector economy as Y C I G (X Z) where Y income, C consumption, I investment, G government expenditures, X exports, and Z imports. By substituting the information supplied in the problem, it is an easy matter to solve for the equilibrium level of income. Aggregating (summing) the variables on the right allows the equation of be graphed in two-dimensional space. See Example 3 and Problems 2.7 to 2.10 and 2.17 to 2.22.
EXAMPLE 3.
Assume a simple two-sector economy where Y C I, C C0 bY, and I I0. Assume further that C0 85, b 0.9, and I0 55. The equilibrium level of income can be calculated in terms of (1) the general parameters and (2) the specific values assigned to these parameters.
1.
The equilibrium equation is Y C I Substituting for C and I, Y C0 bY I0 Solving for Y, Y bY (1 b)Y C0 I0 C0 I0 Y C0 I0 1 b 15 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] Fig. 2-1 The solution in this form is called the reduced form. The reduced form (or solution equation) expresses the endogenous variable (here Y) as an explicit function of the exogenous variables (C0, I0) and the parameters (b).
2.
The specific equilibrium level of income can be calculated by substituting the numerical values for the parameters in either the original equation (a) or the reduced form (b): a) Y C0 bY I0 Y 0.9Y 0.1Y Y 85 0.9Y 55 140 140 1400 b) Y C0 I0 1 b 85 55 1 0.9 140 0.1 1400 The term 1/(1 b) is called the autonomous expenditure multiplier in economics. It measures the multiple effect each dollar of autonomous spending has on the equilibrium level of income. Since b MPC in the income determination model, the multiplier 1/(1 MPC).
Note: Decimals may be converted to fractions for ease in working with the income determination model. For example, 0.1 1 –– 10, 0.9 9 –– 10, 0.5 1 – 2, 0.2 1 – 5, etc.
2.4 IS-LM ANALYSIS The IS schedule is a locus of points representing all the different combinations of interest rates and income levels consistent with equilibrium in the goods (commodity) market. The LM schedule is a locus of points representing all the different combinations of interest rates and income levels consistent with equilibrium in the money market. IS-LM analysis seeks to find the level of income and the rate of interest at which both the commodity market and the money market will be in equilibrium. This can be accomplished with the techniques used for solving simultaneous equations. Unlike the simple income determination model in Section 2.3, IS-LM analysis deals explicitly with the interest rate and incorporates its effect into the model. See Example 4 and Problems 2.23 and 2.24.
EXAMPLE 4.
The commodity market for a simple two-sector economy is in equilibrium when Y C I. The money market is in equilibrium when the supply of money (Ms) equals the demand for money (Md), which in turn is composed of the transaction-precautionary demand for money (Mt) and the speculative demand for money (Mz). Assume a two-sector economy where C 48 0.8Y, I 98 75i, Ms 250, Mt 0.3Y, and Mz 52 150i.
Commodity equilibrium (IS) exists when Y C I. Substituting into the equation, Y Y 0.8Y 48 0.8Y 98 75i 146 75i 0.2Y 75i 146 0 (2.1) Monetary equilibrium (LM) exists when Ms Mt Mz. Substituting into the equation, 250 0.3Y 52 150i 0.3Y 150i 198 0 (2.2) A condition of simultaneous equilibrium in both markets can be found, then, by solving (2.1) and (2.2) simultaneously: 0.2Y 75i 146 0 (2.1) 0.3Y 150i 198 0 (2.2) 16 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 Multiply (2.1) by 2, add the result (2.3) to (2.2) to eliminate i, and solve for Y 0.4Y 150i 292 0 (2.3) 0.3Y 150i 198 0 0.7Y 490 0 Y 700 Substitute Y 700 in (2.1) or (2.2) to find i.
0.2Y 75i 146 0.2(700) 75i 146 140 75i 146 75i i 0 0 0 6 6 –– 75 0.08 The commodity and money markets will be in simultaneous equilibrium when Y 700 and i 0.08. At that point C 48 0.8(700) 608, I 98 75(0.08) 92, Mt 0.3(700) 210, and Mz 52 150(0.08) 40.
C I 608 92 700 and Mt Mz 210 40 250 Ms.
Solved Problems GRAPHS 2.1.
A complete demand function is given by the equation Qd 30P 0.05Y 2Pr 4T where P is the price of the good, Y is income, Pr is the price of a related good (here a substitute), and T is taste. Can the function be graphed?
Since the complete function contains five different variables, it cannot be graphed as is. In ordinary demand analysis, however, it is assumed that all the independent variables except price are held constant so that the effect of a change in price on the quantity demanded can be measured independently of the influence of other factors, or ceteris paribus. If the other variables (Y, Pr, T) are held constant, the function can be graphed.
2.2 (a) Draw the graph for the demand function in Problem 2.1, assuming Y 5000, Pr 25, and T 30. (b) What does the typical demand function drawn in part (a) show? (c) What happens to the graph if the price of the good changes from 5 to 6? (d) What happens if any of the other variables change? For example, if income increases to 7400?
a) By adding the new data to the equation in Problem 2.1, the function is easily graphable. See Fig. 2-2.
Qd 30P 0.05Y 2Pr 4T 30P 0.05(5000) 2(25) 4(30) 30P 420 b) The demand function graphed in part (a) shows all the diferent quantities of the good that will be demanded at different prices, assuming a given level of income, taste, and prices of substitutes (here 5000, 30, 25) which are not allowed to change.
c) If nothing changes but the price of the good, the graph remains exactly the same since the graph indicates the different quantities that will be demanded at all the possible prices. A simple change in the price of the good occasions a movement along the curve which is called a change in quantity demanded. When the price goes from 5 to 6, the quantity demanded falls from 270 [420 30(5)] to 240 [420 30(6)], a movement from A to B on the curve.
17 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] d) If any of the other variables change, there will be a shift in the curve. This is called a change in demand because it results in a totally new demand function (and curve) in response to the changed conditions.
If income increases to 7400, the new demand function becomes Qd 30P 0.05(7400) 2(25) 4(30) 30P 540 This is graphed as a dashed line in Fig. 2-2.
2.3.
In economics the independent variable (price) has traditionally been graphed on the vertical axis in supply and demand analysis, and the dependent variable (quantity) has been graphed on the horizontal. (a) Graph the demand function in Problem 2.2 according to the traditional method. (b) Show what happens if the price goes from 5 to 6 and income increases to 7400.
a) The function Qd 420 30P is graphed according to traditional economic practice by means of the inverse function, which is obtained by solving the original function for the independent variable in terms of the dependent variable. Solving algebraically for P in terms of Qd, therefore, the inverse function of Qd 420 30P is P 14 1 –– 30Qd. The graph appears as a solid line in Fig. 2-3.
b) If P goes from 5 to 6, Qd falls from 270 to 240.
P 5 1 –– 30Qd Qd 14 1 –– 30Qd 14 1 –– 30Qd 9 270 P 6 1 –– 30Qd Qd 14 1 –– 30Qd 14 1 –– 30Qd 8 240 The change is represented by a movement from A to B in Fig. 2-3.
If Y 7400, as in Problems 2.2(d), Qd 540 30P. Solving algebraically for P in terms of Q, the inverse function is P 18 1 –– 30Qd. It is graphed as a dashed line in Fig. 2-3.
2.4.
Graph the demand function Qd 4P 0.01Y 5Pr 10T when Y 8000, Pr 8, and T 4. (b) What type of good is the related good? (c) What happens if T increases to 8, indicating greater preference for the good? (d) Construct the graph along the traditional economic lines with P on the vertical axis and Q on the horizontal axis.
a) Qd 4P 0.01(8000) 5(8) 10(4) 4P 80 This is graphed as a solid line in Fig. 2-4(a).
b) The related good has a negative coefficient. This means that a rise in the price of the related good will lead to a decrease in demand for the original good. The related good is, by definition, a complementary good.
18 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 Fig. 2-2 Fig. 2-3 c) If T 8, indicating greater preference, there will be a totally new demand.
Qd 4P 0.01(8000) 5(8) 10(8) 4P 120 See the dashed line in Fig. 2-4(a).
d) Graphing P on the vertical calls for the inverse function. Solving for P in terms of Qd, the inverse of Qd 80 4P is P 20 1 – 4Qd and is graphed as a solid line in Fig. 2-4(b). The inverse of Qd 120 4P is P 30 1 – 4Qd. It is the dashed line in Fig. 2-4(b).
2.5 A person has $120 to spend on two goods (X, Y) whose respective prices are $3 and $5. (a) Draw a budget line showing all the different combinations of the two goods that can be bought with the given budget (B). What happens to the original budget line (b) if the budget falls by 25 percent, (c) if the price of X doubles, (d) if the price of Y falls to 4?
a) The general function for a budget line is PxX PYY B If Px 3, PY 5, and B 120, 3X 5Y 120 Solving for Y in terms of X in order to graph the function, Y 24 3 – 5X The graph is given as a solid line in Fig. 2-5(a).
19 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] Fig. 2-4 Fig. 2-5 b) If the budget falls by 25 percent, the new budget is 90 [120 1 – 4(120) 90]. The equation for the new budget line is 3X 5Y 90 Y 18 3 – 5X The graph is a dashed line in Fig. 2-5(a). Lowering the budget causes the budget line to shift parallel to the left.
c) If PX doubles, the original equation becomes 6X 5Y 120 Y 24 6 – 5X The vertical intercept remains the same, but the slope changes and becomes steeper. See the dashed line in Fig. 2-5(b). With a higher price for X, less X can be bought with the given budget.
d) If PY now equals 4, 3X 4Y 120 Y 30 3 – 4X With a change in PY, both the vertical intercept and the slope change. This is shown in Fig. 2-5(c) by the dashed line.
2.6.
Either coal (C) or gas (G) can be used in the production of steel. The cost of coal is 100, the cost of gas 500. Draw an isocost curve showing the different combinations of gas and coal that can be purchased (a) with an initial expenditure (E) of 10,000, (b) if expenditures increase by 50 percent, (c) if the price of gas is reduced by 20 percent, (d) if the price of coal rises by 25 percent. Always start from the original equation.
a) PCC PGG E 100C 500G 10,000 C 100 5G The graph is a solid line in Fig. 2-6(a).
b) A 50 percent increase in expenditures makes the new outlay 15,000 [10,000 0.5(10,000)]. The new equation is 100C 500G 15,000 C 150 5G The graph is the dashed line in Fig. 2-6(a).
20 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 (a) Increase in budget (b) Reduction in PG (c) Increase in PC Fig. 2-6 c) If the price of gas is reduced by 20 percent, the new price is 400 [500 0.2(500)], and the new equation is 100C 400G 10,000 C 100 4G The graph is the dashed line in Fig. 2-6(b).
d) A 25 percent rise in the price of coal makes the new price 125 [100 0.25(100)].
125C 500G 10,000 C 80 4G The graph appears as a dashed line in Fig. 2-6(c).
GRAPHS IN THE INCOME DETERMINATION MODEL 2.7.
Given: Y C I, C 50 0.8Y, and I0 50. (a) Graph the consumption function. (b) Graph the aggregate demand function, C I0. (c) Find the equilibrium level of income from the graph.
a) Since consumption is a function of income, it is graphed on the vertical axis; income is graphed on the horizontal. See Fig. 2-7. When other components of aggregate demand such as I, G, and X Z are added to the model, they are also graphed on the vertical axis. It is easily determined from the linear form of the consumption function that the vertical intercept is 50 and the slope of the line (the MPC or C/ Y) is 0.8.
b) Investment in the model is autonomous investment. This means investment is independent of income and does not change in response to changes in income. When considered by itself, the graph of a constant is a horizontal line; when added to a linear function, it causes a parallel shift in the original function by an amount equal to its value. In Fig. 2-7, autonomous investment causes the aggregate demand function to shift up by 50 parallel to the initial consumption function.
c) To obtain the equilibrium level of income from a graph, a 45 dashed line is drawn from the origin.
If the same scale of measurement is used on both axes, a 45 line has a slope of 1, meaning that as the line moves away from the origin, it moves up vertically ( Y) by one unit for every unit it moves across horizontally ( X). Every point on the 45 line, therefore, has a horizontal coordinate (abscissa) 21 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] Fig. 2-7 exactly equal to its vertical coordinate (ordinate). Consequently, when the aggregate demand function intersects the 45 line, aggregate demand (as graphed on the vertical) will equal national income (as graphed on the horizontal). From Fig. 2-7 it is clear that the equilibrium level of income is 500, since the aggregate demand function (C I) intersects the 45 line at 500.
2.8.
Given: Y C I G, C 25 0.75Y, I I0 50, and G G0 25. (a) Graph the aggregate demand function and show its individual components. (b) Find the equilibrium level of income.
(c) How can the aggregate demand function be graphed directly, without having to graph each of the component parts?
a) See Fig. 2-8.
b) Equilibrium income 400.
c) To graph the aggregate demand function directly, sum up the individual components, Agg. D C I G 25 0.75Y 50 25 100 0.75Y The direct graphing of the aggregate demand function coincides exactly with the graph of the summation of the individual graphs of C, I, and G above.
2.9.
Use a graph to show how the addition of a lump-sum tax (a tax independent of income) influences the parameters of the income determination model. Graph the two systems indi-vidually, using a solid line for (1) and a dashed line for (2).
1) Y C I C 100 0.6Y I0 40 2) Y C I C 100 0.6Yd I0 40 Yd Y T T 50 The first system of equations presents no problems; the second requires that C first be converted from a function of Yd to a function of Y.
1) Agg. D C I 100 0.6Y 40 140 0.6Y 2) Agg. D C I 100 0.6Yd 40 140 0.6(Y T) 140 0.6(Y 50) 110 0.6Y 22 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 Fig. 2-8 A lump-sum tax has a negative effect on the vertical intercept of the aggregate demand function equal to MPC(T). Here 0.6(50) 30. The slope is not affected (note the parallel lines for the two graphs in Fig. 2-9). Income falls from 350 to 275 as a result of the tax. See Fig. 2-9.
2.10.
Explain with the aid of a graph how the incorporation of a proportional tax (a tax depending on income) influences the parameters of the income determination model. Graph the model without the tax as a solid line and the model with the tax as a dashed line.
1) Y C I C 85 0.75Y I0 30 2) Y C I C 85 0.75Yd I0 30 Yd Y T T 20 0.2Y 1) Agg. D C I 85 0.75Y 30 115 0.75Y 2) Agg. D C I 85 0.75Yd 30 115 0.75(Y T) 115 0.75(Y 20 0.2Y) 115 0.75Y 15 0.15Y 100 0.6Y Incorporation of a proportional income tax into the model affects the slope of the line, or the MPC. In this case it lowers it from 0.75 to 0.6. The vertical intercept is also lowered because the tax structure includes a lump-sum tax of 20. Because of the tax structure, the equilibrium level of income falls from 460 to 250.
See Fig. 2-10.
EQUATIONS IN SUPPLY AND DEMAND ANALYSIS 2.11.
Find the equilibrium price and quantity for the following markets: a) Qs 20 3P Qd 220 5P c) Qs 32 7P 0 Qd 128 9P 0 b) Qs 45 8P Qd 125 2P d) 13P Qs 27 Qd 4P 24 0 23 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] Fig. 2-9 Each of the markets will be in equilibrium when Qs Qd.
a) Qs 20 3P 8P P Qs Qs c) Qs Qd 7P 32 16P P Qs Qs Qd 220 5P 240 30 20 3P 20 3(30) 70 Qd 7P 32 128 9P 128 9P 160 10 7P 32 7(10) 32 38 Qd b) Qs 45 8P 10P P Qd Qd d) Qs Qd 27 13P 17P P Qd Qd Qd 125 2P 170 17 125 2P 125 2(17) 91 Qs 27 13P 24 4P 24 4P 51 3 24 4P 24 4(3) 12 Qs 2.12.
Given the following set of simultaneous equations for two related markets, beef (B) and pork (P), find the equilibrium conditions for each market, using the substitution method.
1) QdB 82 3PB PP QsB 5 15PB 2) QdP 92 2PB 4PP QsP 6 32PP Equilibrium requires that Qs Qd in each market.
1) QsB 5 15PB 18PB PP QdB 82 3PB PP 87 2) QsP 6 32PP 36PP 2PB QdP 92 2PB 4PP 98 This reduces the problem to two equations and two unknowns: 18PB PP 87 2PB 36PP 98 (2.4) (2.5) 24 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 Fig. 2-10 Solving for PP in (2.4) gives PP 18PB 87 Substituting the value of this term in (2.5) gives 2PB 36(18PB 87) 98 2PB 648PB 3132 98 646PB 3230 PB 5 Substituting PB 5 in (2.5), or (2.4), 2(5) 36PP 98 36PP 108 PP 3 Finally, substituting the values for PB and PP in either the supply or the demand function for each market, 1) QdB 82 3PB PP 82 3(5) (3) QdB 70 QsB 2) QdP 92 2PB 4PP 92 2(5) 4(3) QdP 90 QsP 2.13.
Find the equilibrium price and quantity for two complementary goods, slacks (S) and jackets (J), using the elimination method.
1) QdS 410 5PS 2PJ QsS 60 3PS 2) QdJ 295 PS 3PJ QsJ 120 2PJ In equilibrium, 1) QdS 410 5PS 2PJ 470 8PS 2PJ QsS 60 3PS 0 2) QdJ 295 PS 3PJ 415 PS 5PJ QsJ 120 2PJ 0 This leaves two equations 470 8PS 2PJ 0 415 PS 5PJ 0 (2.6) (2.7) Multiplying (2.7) by 8 gives (2.8). Subtract (2.6) from (2.8) to eliminate PS, and solve for PJ.
(2.8) 3320 8PS 40PJ 0 (470 8PS 2PJ 0) 2850 38PJ 0 PJ 75 Substituting PJ 75 in (2.6), 470 8PS 2(75) 0 320 8PS PS 40 Finally, substituting PJ 75 and PS 40 into Qd or Qs for each market, 1) QdS 410 5PS 2PJ 410 5(40) 2(75) Qds 60 QsS 2) QdJ 295 PS 3PJ 295 40 3(75) QdJ 30 QsJ 2.14.
Supply and demand conditions can also be expressed in quadratic form. Find the equilibrium price and quantity, given the demand function P Q2 3Q 20 0 (2.9) and the supply function P 3Q2 10Q 5 (2.10) 25 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] Either the substitution method or the elimination method can be used, since this problem involves two equations and two unknowns. Using the substitution method, (2.10) is solved for P in terms of Q.
P 3Q2 10Q 5 P 3Q2 10Q 5 Substituting P 3Q2 10Q 5 in (2.9), (3Q2 10Q 5) Q2 3Q 20 0 4Q2 7Q 15 0 Using the quadratic formula Q1, Q2 (b b2 4ac)/(2a), where a 4, b 7, and c 15, Q1 3, and Q2 1.25. Since neither price nor quantity can be negative, Q 3. Substitute Q 3 in (2.9) or (2.10) to find P.
P (3)2 3(3) 20 0 P 2 2.15.
Use the elimination method to find the equilibrium price and quantity when the demand function is 3P Q2 5Q 102 0 (2.11) and the supply function is P 2Q2 3Q 71 0 (2.12) Multiply (2.12) by 3 to get (2.13) and subtract it from (2.11) to eliminate P.
3P Q2 5Q 102 0 (3P 6Q2 9Q 213 0) 7Q2 4Q 315 0 (2.13) Use the quadratic formula (see Problem 2.14) to solve for Q, and substitute the result, Q 7, in (2.12) or (2.11) to solve for P.
P 2(7)2 3(7) 71 0 P 6 2.16.
Supply and demand analysis can also involve more than two markets. Find the equilibrium price and quantity for the three substitute goods below.
Qd1 23 5P1 P2 P3 Qd2 15 P1 3P2 2P3 Qd3 19 P1 2P2 4P3 Qs1 8 6P1 Qs2 11 3P2 Qs3 5 3P3 For equilibrium in each market, Qd1 23 5P1 P2 P3 31 11P1 P2 P3 Qs1 8 6P1 0 Qd2 15 P1 3P2 2P3 26 P1 6P2 2P3 Qs2 11 3P2 0 Qd3 19 P1 2P2 4P3 24 P1 2P2 7P3 Qs3 5 3P3 0 This leaves three equations with three unknowns: 31 11P1 P2 P3 0 26 P1 6P2 2P3 0 24 P1 2P2 7P3 0 (2.14) (2.15) (2.16) Start by eliminating one of the variables (here P2). Multiply (2.14) by 2 to get 62 22P1 2P2 2P3 0 26 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 From this subtract (2.16).
62 22P1 2P2 2P3 0 (24 P1 2P2 7P3) 0 38 23P1 9P3 0 (2.17) Multiply (2.16) by 3.
72 3P1 6P2 21P3 0 Add the result to (2.15).
26 P1 6P2 2P3 0 72 3P1 6P2 21P3 0 98 4P1 19P3 0 (2.18) Now there are two equations, (2.17) and (2.18), and two unknowns. Multiply (2.17) by 19 and (2.18) by 9; then add to eliminate P3.
722 437P1 171P3 0 882 36P1 171P3 0 1604 401P1 0 P1 4 Substitute P1 4 in (2.18) to solve for P3.
98 4(4) 19P3 0 19P3 114 P3 6 Substitute P1 4 and P3 6 into (2.14), (2.15), or (2.16), to solve for P2.
31 11(4) P2 (6) 0 P2 7 EQUATIONS IN THE INCOME DETERMINATION MODEL 2.17.
Given: Y C I G, C C0 bY, I I0, and G G0, where C0 135, b 0.8, I0 75, and G0 30. (a) Find the equation for the equilibrium level of income in the reduced form. (b) Solve for the equilibrium level of income (1) directly and (2) with the reduced form.
a) From Section 2.3, Y Y bY (1 b)Y C I G C0 bY I0 G0 C0 I0 G0 C0 I0 G0 Y C0 I0 G0 1 b b) 1) Y Y 0.8Y 0.2Y Y C I G 135 0.8Y 75 30 240 240 1200 2) Y C0 I0 G0 1 b 135 75 30 1 0.8 5(240) 1200 2.18.
Find the equilibrium level of income Y C I, when C 89 0.8Y and I0 24.
Y C0 I0 1 b 5(89 24) 565 27 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] From Problem 2.17, the value of the multiplier [1/(1 b)] is already known for cases when b 0.8. Use of the reduced form to solve the equation in this instance is faster, therefore, although the other method is also correct.
2.19.
(a) Find the reduced form of the following income determination model where investment is not autonomous but is a function of income. (b) Find the numerical value of the equilibrium level of income (Ye). (c) Show what happens to the multiplier.
Y C I C C0 bY I I0 aY where C0 65, I0 70, b 0.6, and a 0.2.
a) Y Y bY aY (1 b a)Y Y C I C0 bY I0 aY C0 I0 C0 I0 C0 I0 1 b a b) Y Y 0.6Y 0.2Y 0.2Y Y C I 65 0.6Y 70 0.2Y 65 70 135 675 c) When investment is a function of income, and no longer autonomous, the multiplier changes from 1/(1 b) to 1/(1 b a). This increases the value of the multiplier because it reduces the denominator of the fraction and makes the quotient larger, as substitution of the values of the parameters in the problem shows: 1 1 b 1 1 0.6 1 0.4 2.5 1 1 b a 1 1 0.6 0.2 1 0.2 5 2.20.
Find (a) the reduced form, (b) the numerical value of Ye, and (c) the effect on the multiplier when a lump-sum tax is added to the model and consumption becomes a function of disposable income (Yd).
Y C I C C0 bYd I I0 Yd Y T where C0 100, b 0.6, I0 40, and T 50.
a) Y C I C0 bYd I0 C0 b(Y T) I0 C0 bY bT I0 Y bY C0 I0 bT Y C0 I0 bT 1 b b) Y 100 0.6Yd 40 140 0.6(Y T) or Y 100 40 0.6(50) 1 0.6 110 0.4 140 0.6(Y 50) 140 0.6Y 30 275 Y 0.6Y 110 0.4Y 110 Y 275 The graph of this function is given in Problem 2.9.
c) As seen in part a), incorporation of a lump-sum tax into the model leaves the multiplier at 1/(1 b).
Only the aggregate value of the exogenous variables is reduced by an amount equal to bT.
Incorporation of other autonomous variables such as G0, X0, or Z0 will not affect the value of the multiplier either.
28 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 2.21.
Find (a) the reduced form, (b) the numerical value of Ye, and (c) the effect on the multiplier if a proportional income tax (t) is incorporated into the model.
Y C I C C0 bYd T T0 tY Yd Y T where I I0 30, C0 85, b 0.75, t 0.2, and T0 20.
a) Y C I C0 bYd I0 C0 b(Y T) I0 C0 b(Y T0 tY) I0 C0 bY bT0 btY I0 Y bY btY C0 I0 bT0 (1 b bt)Y C0 I0 bT0 Y C0 I0 bT0 1 b bt b) Once the reduced form is found, its use speeds the solution. But sometimes the reduced form is not available, making it necessary to be familiar with the other method.
Y C I 85 0.75Yd 30 115 0.75(Y T) 115 0.75(Y 20 0.2Y) 115 0.75Y 15 0.15Y Y 0.75Y 0.15Y 0.4Y Y 100 100 250 The graph of this function is given in Problem 2.10.
c) The multiplier is changed from 1/(1 b) to 1/(1 b bt). This reduces the size of the multiplier because it makes the denominator larger and the fraction smaller: 1 1 b 1 1 0.75 1 0.25 4 1 1 b bt 1 1 0.75 0.75(0.2) 1 1 0.75 0.15 1 0.4 2.5 2.22.
If the foreign sector is added to the model and there is a positive marginal propensity to import (z), find (a) the reduced form, (b) the equilibrium level of income, and (c) the effect on the multiplier.
Y C I G (X Z) C C0 bY Z Z0 zY where I I0 90, G G0 65, X X0 80, C0 70, Z0 40, b 0.9, and z 0.15.
a) Y C I G (X Z) Y bY zY (1 b z)Y C0 bY I0 G0 X0 Z0 zY C0 I0 G0 X0 Z0 C0 I0 G0 X0 Z0 Y C0 I0 G0 X0 Z0 1 b z b) Using the reduced form above, Y 70 90 65 80 40 1 0.9 0.15 265 0.25 1060 29 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] c) Introduction of the marginal propensity to import (z) into the model reduces the size of the multiplier.
It makes the denominator larger and the fraction smaller: 1 1 b 1 1 0.9 1 0.1 10 1 1 b z 1 1 0.9 0.15 1 0.25 4 IS-LM EQUATIONS 2.23.
Given: C 102 0.7Y, I 150 100i, Ms 300, Mt 0.25Y, and Mz 124 200i. Find (a) the equilibrium level of income and the equilibrium rate of interest and (b) the level of C, I, Mt, and Mz when the economy is in equilibrium.
a) Commodity market equilibrium (IS) exists where Y Y 0.7Y 0.3Y C I 102 0.7Y 150 100i 252 100i 100i 252 0 Monetary equilibrium (LM) exists where Ms 300 Mt Mz 0.25Y 124 200i 0.25Y 200i 176 0 Simultaneous equilibrium in both markets requires that 0.3Y 100i 252 0 0.25Y 200i 176 0 (2.19) (2.20) Multiply (2.19) by 2, and add the result to (2.20) to eliminate i: 0.6Y 200i 504 0 0.25Y 200i 176 0 0.85Y 680 Y 800 Substitute Y 800 in (2.19) or (2.20): 0.25Y 200i 176 0.25(800) 200i 176 200i i 0 0 24 0.12 b) At Y 800 and i 0.12, C 102 0.7(800) 662 Mt 0.25(800) 200 I 150 100(0.12) 138 Mz 124 200(0.12) 100 and C I 662 138 Y Mt Mz 800 200 100 Ms 300 2.24.
Find (a) the equilibrium income level and interest rate and (b) the levels of C, I, Mt, and Mz in equilibrium when C 89 0.6Y I 120 150i Ms 275 Mt 0.1Y Mz 240 250i 30 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS [CHAP . 2 a) For IS: Y 89 0.6Y 120 150i Y 0.6Y 209 150i 0.4Y 150i 209 0 For LM: Ms Mt Mz 275 0.1Y 240 250i 0.1Y 250i 35 0 In equilibrium, 0.4Y 150i 209 0 0.1Y 250i 35 0 (2.21) (2.22) Multiply (2.22) by 4, and subtract the result from (2.21) to eliminate Y.
0.4Y 150i 209 0 (0.4Y 1000i 140 0) 1150i 69 i 0.06 Substitute i 0.06 in (2.21) or (2.22).
0.4Y 150(0.06) 209 0 0.4Y 200 Y 500 b) At Y 500 and i 0.06, C 89 0.6(500) 389 Mt 0.1(500) 50 I 120 150(0.06) 111 Mz 240 250(0.06) 225 and C I 389 111 Y Mt Mz 500 50 225 Ms 275 31 ECONOMIC APPLICATIONS OF GRAPHS AND EQUATIONS CHAP . 2] CHAPTER 3 The Derivative and the Rules of Differentiation 3.1 LIMITS If the functional values f(x) of a function f draw closer to one and only one finite real number L for all values of x as x draws closer to a from both sides, but does not equal a, L is defined as the limit of f(x) as x approaches a and is written lim x→a f(x) L Assuming that limx→a f(x) and limx→ag(x) both exist, the rules of limits are given below, explained in Example 2, and treated in Problems 3.1 to 3.4.
1.
lim x→a k k (k a constant) 2.
lim x→a xn an (n a positive integer) 3.
lim x→a kf(x) k lim x→a f(x) (k a constant) 4.
lim x→a [f(x) g(x)] lim x→a f(x) lim x→a g(x) 5.
lim x→a [f(x) · g(x)] lim x→a f(x) · lim x→a g(x) 6.
lim x→a [f(x) g(x)] lim x→a f(x) lim x→a g(x) lim x→a g(x) 0 7.
lim x→a [f(x)]n lim x→a f(x) n (n 0) 32 EXAMPLE 1.
a) From the graph of the function f(x) in Fig. 3-1, it is clear that as the value of x approaches 3 from either side, the value of f(x) approaches 2. This means that the limit of f(x) as x approaches 3 is the number 2, which is written lim x→3 f(x) 2 As x approaches 7 from either side in Fig. 3-1, where the open circle in the graph of f(x) signifies there is a gap in the function at that point, the value of f(x) approaches 4 even though the function is not defined at that point.
Since the limit of a function as x approaches a number depends only on the values of x close to that number, the limit exists and is written lim x→7 f(x) 4 b) In Fig. 3-2, as x approaches 4 from the left (from values less than 4), written x →4, g(x) approaches 3, called a one-sided limit; as x approaches 4 from the right (from values greater than 4), written x →4, g(x) approaches 4. The limit does not exist, therefore, since g(x) does not approach a single number as x approaches 4 from both sides.
EXAMPLE 2.
In the absence of a graph, limits can be found by using the rules of limits enumerated above.
a) lim x→5 9 9 Rule 1 b) lim x→6 x2 (6)2 36 Rule 2 c) lim x→3 2x3 2 lim x→3 x3 2(3)3 54 Rules 2 and 3 d) lim x→2 (x4 3x) lim x→2 x4 3 lim x→2 x Rule 4 (2)4 3(2) 22 e) lim x→4 [(x 8)(x 5)] lim x→4 (x 8) · lim x→4 (x 5) Rule 5 (4 8) · (4 5) 12 3.2 CONTINUITY A continuous function is one which has no breaks in its curve. It can be drawn without lifting the pencil from the paper. A function f is continuous at x a if: 1.
f(x) is defined, i.e., exists, at x a 2.
lim x→a f(x) exists, and 3.
lim x→a f(x) f(a) 33 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] Fig. 3-1 Fig. 3-2 All polynomial functions are continuous, as are all rational functions, except where undefined, i.e., where their denominators equal zero. See Problem 3.5.
EXAMPLE 3.
Given that the graph of a continuous function can be sketched without ever removing pencil from paper and that an open circle means a gap in the function, it is clear that f(x) is discontinuous at x 4 in Fig. 3-3(a) and g(x) is discontinuous at x 5 in Fig. 3-3(b), even though limx→5g(x) exists.
3.3 THE SLOPE OF A CURVILINEAR FUNCTION The slope of a curvilinear function is not constant. It differs at different points on the curve. In geometry, the slope of a curvilinear function at a given point is measured by the slope of a line drawn tangent to the function at that point. A tangent line is a straight line that touches a curve at only one point. Measuring the slope of a curvilinear function at different points requires separate tangent lines, as in Fig. 3-4(a).
34 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 Fig. 3-3 Fig. 3-4 The slope of a tangent line is derived from the slopes of a family of secant lines. A secant line S is a straight line that intersects a curve at two points, as in Fig. 3-4(b), where Slope S y2 y1 x2 x1 By letting x2 x1 x and y2 f(x1 x), the slope of the secant line can also be expressed by a difference quotient: Slope S f(x1 x) f(x1) (x1 x) x1 f(x1 x) f(x1) x If the distance between x2 and x1 is made smaller and smaller, i.e., if x →0, the secant line pivots back to the left and draws progressively closer to the tangent line. If the slope of the secant line approaches a limit as x →0, the limit is the slope of the tangent line T, which is also the slope of the function at the point. It is written Slope T lim x→0 f(x1 x) f(x1) x (3.1) Note: In many texts h is used in place of x, giving Slope T lim h→0 f(x1 h) f(x1) h (3.1a) EXAMPLE 4.
To find the slope of a curvilinear function, such as f(x) 2x2, (1) employ the specific function in the algebraic formula (3.1) or (3.1a) and substitute the arguments x1 x (or x1 h) and x1, respectively, (2) simplify the function, and (3) evaluate the limit of the function in its simplified form. From (3.1), Slope T lim x→0 f(x x) f(x) x 1) Employ the function f(x) 2x2 and substitute the arguments.
Slope T lim x→0 2(x x)2 2x2 x 2) Simplify the result.
Slope T lim x→0 2[x2 2x( x) ( x)2] 2x2 x lim x→0 4x( x) 2( x)2 x Divide through by x.
Slope T lim x→0 (4x 2 x) 3) Take the limit of the simplified expression.
Slope T 4x Note: The value of the slope depends on the value of x chosen. At x 1, slope T 4(1) 4; at x 2, slope T 4(2) 8.
35 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] 3.4 THE DERIVATIVE Given a function y f(x), the derivative of the function f at x, written f(x) or dy/dx, is defined as f(x) lim x→0 f(x x) f(x) x if the limit exists (3.2) or from (3.1a), f(x) lim h→0 f(x1 h) f(x1) h (3.2a) where f(x) is read ‘‘the derivative of f with respect to x’’ or ‘‘f prime of x.’’ The derivative of a function f(x), or simply f, is itself a function which measures both the slope and the instantaneous rate of change of the original function f(x) at a given point.
3.5 DIFFERENTIABILITY AND CONTINUITY A function is differentiable at a point if the derivative exists (may be taken) at that point. To be differentiable at a point, a function must (1) be continuous at that point and (2) have a unique tangent at that point. In Fig. 3-5, f(x) is not differentiable at a and c because gaps exist in the function at those points and the derivative cannot be taken at any point where the function is discontinuous.
Continuity alone, however, does not ensure (is not a sufficient condition for) differentiability. In Fig. 3-5, f(x) is continuous at b, but it is not differentiable at b because at a sharp point or kink, called a cusp, an infinite number of tangent lines (and no one unique tangent line) can be drawn.
3.6 DERIVATIVE NOTATION The derivative of a function can be written in many different ways. If y f(x), the derivatives can be expressed as f(x) y dy dx df dx d dx [f(x)] or Dx[f(x)] If y (t), the derivative can be written (t) y dy dt d dt d dt [(t)] or Dt[(t)] If the derivative of y f(x) is evaluated at x a, proper notation includes f(a) and dy dxa .
36 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 Fig. 3-5 EXAMPLE 5.
If y 5x2 7x 12, the derivative can be written y dy dx d dx (5x2 7x 12) or Dx(5x2 7x 12) If z 8t 3, the derivative can be expressed as z dz dt d dt (8t 3) or Dt(8t 3) See Problems 3.6 to 3.8.
3.7 RULES OF DIFFERENTIATION Differentiation is the process of finding the derivative of a function. It involves nothing more complicated than applying a few basic rules or formulas to a given function. In explaining the rules of differentiation for a function such as y f(x), other functions such as g(x) and h(x) are commonly used, where g and h are both unspecified functions of x. The rules of differentiation are listed below and treated in Problems 3.6 to 3.21. Selected proofs are found in Problems 3.24 to 3.26.
3.7.1 The Constant Function Rule The derivative of a constant function f(x) k, where k is a constant, is zero.
Given f(x) k, f(x) 0 EXAMPLE 6.
Given f(x) 8, f(x) 0 Given f(x) 6, f(x) 0 3.7.2 The Linear Function Rule The derivative of a linear function f(x) mx b is equal to m, the coefficient of x. The derivative of a variable raised to the first power is always equal to the coefficient of the variable, while the derivative of a constant is simply zero.
Given f(x) mx b, f(x) m EXAMPLE 7.
Given f(x) 3x 2, f(x) 3 Given f(x) 5 1 – 4x, f(x) 1 – 4 Given f(x) 12x, f(x) 12 3.7.3 The Power Function Rule The derivative of a power function f(x) kxn, where k is a constant and n is any real number, is equal to the coefficient k times the exponent n, multiplied by the variable x raised to the n 1 power.
Given f(x) kxn f(x) k · n · xn1 EXAMPLE 8.
Given f(x) 4x3 f(x) 4 · 3 · x31 12x2 Given f(x) 5x2, f(x) 5 · 2 · x21 10x Given f(x) x4, f(x) 1 · 4 · x41 4x3 See also Problem 3.7.
37 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] 3.7.4 The Rules for Sums and Differences The derivative of a sum of two functions f(x) g(x) h(x), where g(x) and h(x) are both differentiable functions, is equal to the sum of the derivatives of the individual functions. Similarly, the derivative of the difference of two functions is equal to the difference of the derivatives of the two functions.
Given f(x) g(x) h(x), f(x) g(x) h(x) EXAMPLE 9.
Given f(x) 12x5 4x4, f(x) 60x4 16x3 Given f(x) 9x2 2x 3, f(x) 18x 2 See Problem 3.8. For derivation of the rule, see Problem 3.24.
3.7.5 The Product Rule The derivative of a product f(x) g(x) · h(x), where g(x) and h(x) are both differentiable functions, is equal to the first function multiplied by the derivative of the second plus the second function multiplied by the derivative of the first. Given f(x) g(x) · h(x), f(x) g(x) · h(x) h(x) · g(x) (3.3) EXAMPLE 10.
Given f(x) 3x4(2x 5), let g(x) 3x4 and h(x) 2x 5. Taking the individual derivatives, g(x) 12x3 and h(x) 2. Then by substituting these values in the product-rule formula (3.3), f(x) 3x4(2) (2x 5)(12x3) and simplifying algebraically gives f(x) 6x4 24x4 60x3 30x4 60x3 See Problems 3.9 to 3.11; for the derivation of the rule, see Problem 3.25.
3.7.6 The Quotient Rule The derivative of a quotient f(x) g(x) h(x), where g(x) and h(x) are both differentiable functions and h(x) 0, is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the denominator squared. Given f(x) g(x)/h(x), f(x) h(x) · g(x) g(x) · h(x) [h(x)]2 (3.4) EXAMPLE 11.
Given f(x) 5x3 4x 3 where g(x) 5x3 and h(x) 4x 3, we know that g(x) 15x2 and h(x) 4. Substituting these values in the quotientrule formula (3.4), f(x) (4x 3)(15x2) 5x3(4) (4x 3)2 Simplifying algebraically, f(x) 60x3 45x2 20x3 (4x 3)2 40x3 45x2 (4x 3)2 5x2(8x 9) (4x 3)2 See Problems 3.12 and 3.13; for the derivation of the rule, see Problem 3.26.
38 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 3.7.7 The Generalized Power Function Rule The derivative of a function raised to a power, f(x) [g(x)]n, where g(x) is a differentiable function and n is any real number, is equal to the exponent n times the function g(x) raised to the n 1 power, multiplied in turn by the derivative of the function itself g(x). Given f(x) [g(x)]n, f(x) n[g(x)]n1 · g(x) (3.5) EXAMPLE 12.
Given f(x) (x3 6)5, let g(x) x3 6, then g(x) 3x2. Substituting these values in the generalized power function formula (3.5) gives f(x) 5(x3 6)51 · 3x2 Simplifying algebraically, f(x) 5(x3 6)4 · 3x2 15x2(x3 6)4 Note: The generalized power function rule is derived from the chain rule which follows below. See Problems 3.14 and 3.15.
3.7.8 The Chain Rule Given a composite function, also called a function of a function, in which y is a function of u and u in turn is a function of x, that is, y f(u) and u g(x), then y f[g(x)] and the derivative of y with respect to x is equal to the derivative of the first function with respect to u times the derivative of the second function with respect to x: dy dx dy du · du dx (3.6) See Problems 3.16 and 3.17.
EXAMPLE 13.
Consider the function y (5x2 3)4. To use the chain rule, let y u4 and u 5x2 3. Then dy/du 4u3 and du/dx 10x. Substitute these values in (3.6): dy dx 4u3 · 10x 40xu3 Then to express the derivative in terms of a single variable, substitute 5x2 3 for u.
dy dx 40x(5x2 3)3 For more complicated functions, different combinations of the basic rules must be used. See Problems 3.18 and 3.19.
3.8 HIGHER-ORDER DERIVATIVES The second-order derivative, written f
(x), measures the slope and the rate of change of the first derivative, just as the first derivative measures the slope and the rate of change of the original or primitive function. The third-order derivative f(x) measures the slope and rate of change of the second-order derivative, etc. Higher-order derivatives are found by applying the rules of differentia-tion to lower-order derivatives, as illustrated in Example 14 and treated in Problems 3.20 and 3.21.
EXAMPLE 14.
Given y f(x), common notation for the second-order derivative includes f
(x), d 2y/dx2, y
, and D2y; for the third-order derivative, f(x), d3y/dx3, y, and D3y; for the fourth-order derivatives, f (4)(x), d4y/dx4, y(4), and D4y; etc.
39 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] Higher-order derivatives are found by successively applying the rules of differentiation to derivatives of the previous order. Thus, if f(x) 2x4 5x3 3x2, f(x) f
(x) f(x) f (4)(x) 8x3 15x2 6x 24x2 30x 6 48x 30 48 f (5)(x) 0 See Problems 3.20 and 3.21.
3.9 IMPLICIT DIFFERENTIATION Introductory economics deals most often with explicit functions in which the dependent variable appears to the left of the equal sign and the independent variable appears to the right. Frequently encountered in more advanced economics courses, however, are implicit functions in which both variables and constants are to the left of the equal sign. Some implicit functions can be easily converted to explicit functions by solving for the dependent variable in terms of the independent variable; others cannot. For those not readily convertible, the derivative may be found by implicit differentiation. See Example 16 and Problems 3.22, 3.23, 4.25, and 4.26; also see Section 5.10 and Problems 5.20, 5.21, 6.51, and 6.52.
EXAMPLE 15.
Samples of explicit and implicit functions include: Explicit: y 4x y x2 6x 7 y x4 9x3 x2 13 Implicit: 8x 5y 21 0 3x2 8xy 5y 49 0 35x3y7 106 0 EXAMPLE 16.
Given 3x4 7y5 86 0, the derivative dy/dx is found by means of implicit differentiation in two easy steps.
1) Differentiate both sides of the equation with respect to x while treating y as a function of x, d dx (3x4 7y5 86) d dx (0) (3.7) d dx (3x4) d dx (7y5) d dx (86) d dx (0) where d dx (3x4) 12x3, d dx (86) 0, d dx (0) 0. Using the generalized power function rule for d dx (7y5) and noting that d dx (y) dy dx, we get d dx (7y5) 7 · 5 · y51 · d dx (y) 35y4 dy dx Substitute the above values in (3.7).
12x3 35y4 dy dx 0 (3.8) (2) Now simply solve (3.8) algebraically for dy/dx: 35y4 dy dx 12x3 dy dx 12x3 35y4 Compare this answer to that in Example 16 of Chapter 5.
40 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 Solved Problems LIMITS AND CONTINUITY 3.1.
Use the rules of limits to find the limits for the following functions: a) lim x→2 [x3(x 4)] lim x→2 [x3(x 4)] lim x→2 x3 · lim x→2 (x 4) Rule 5 (2)3 · (2 4) 8 · 6 48 b) lim x→4 3x2 5x x 6 lim x→4 3x2 5x x 6 lim x→4 (3x2 5x) lim x→4 (x 6) Rule 6 3(4)2 5(4) 4 6 48 20 10 2.8 c) lim x→2 6x3 1 lim x→2 6x3 1 lim x→2 (6x3 1)1/2 [lim x→2 (6x3 1)]1/2 Rule7 [6(2)3 1]1/2 (49)1/2 7 3.2.
Find the limits for the following polynomial and rational functions.
a) lim x→3 (5x2 4x 9) From the properties of limits it can be shown that for all polynomial functions and all rational functions, where defined, limx→a f(x) f(a). The limits can be taken, therefore, by simply evaluating the functions at the given level of a.
lim x→3 (5x2 4x 9) 5(3)2 4(3) 9 42 b) lim x→4 (3x2 7x 12) lim x→4 (3x2 7x 12) 3(4)2 7(4) 12 8 c) lim x→6 4x2 2x 8 5x2 6 lim x→6 4x2 2x 8 5x2 6 4(6)2 2(6) 8 5(6)2 6 124 186 2 3 3.3.
Find the limits of the following rational functions. If the limit of the denominator equals zero, neither Rule 6 nor the generalized rule for rational functions used above applies.
a) lim x→7 x 7 x2 49 41 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] The limit of the denominator is zero, so Rule 6 cannot be used. Since we are only interested in the function as x draws near to 7, however, the limit can be found if by factoring and canceling, the problem of zero in the denominator is removed.
lim x→7 x 7 x2 49 lim x→7 x 7 (x 7)(x 7) lim x→7 1 x 7 1 14 b) lim x→7 x 7 x2 49 lim x→7 x 7 x2 49 lim x→7 x 7 (x 7)(x 7) lim x→7 1 x 7 The limit does not exist.
c) lim x→6 x2 2x 24 x 6 With the limit of the denominator equal to zero, factor.
lim x→6 x2 2x 24 x 6 lim x→6 (x 4)(x 6) x 6 lim x→6 (x 4) 10 3.4.
Find the limits of the following functions, noting the role that infinity plays.
a) lim x→0 2 x (x 0) As seen in Fig. 1-3(b), as x approaches 0 from the right (x →0), f(x) approaches positive infinity; as x approaches 0 from the left (x →0), f(x) approaches negative infinity. If a limit approaches either positive or negative infinity, the limit does not exist and is written lim x→0 2 x lim x→0 2 x The limit does not exist.
b) lim x→ 2 x lim x→ 2 x As also seen in Fig. 1-3(b), as x approaches , f(x) approaches 0; as x approaches , f(x) also approaches 0. The limit exists in both cases and is written lim x→ 2 x 0 lim x→ 2 x 0 c) lim x→ 3x2 7x 4x2 21 As x →, both numerator and denominator become infinite, leaving matters unclear. A trick in such circumstances is to divide all terms by the highest power of x which appears in the function. Here dividing all terms by x2 leaves lim x→ 3x2 7x 4x2 21 lim x→ 3 (7/x) 4 (21/x2) 3 0 4 0 3 4 42 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 3.5.
Indicate whether the following functions are continuous at the specified points by determining whether at the given point all the following conditions from Section 3.2 hold: (1) f(x) is defined, (2) limx→a f(x) exists, and (3) limx→af(x) f(a).
a) f(x) 5x2 8x 9 at x 3 1) f(3) 5(3)2 8(3) 9 30 2) lim x→3 (5x2 8x 9) 5(3)2 8(3) 9 30 3) lim x→3 f(x) 30 f(3) f(x) is continuous.
b) f(x) x2 3x 12 x 3 at x 4 1) f(4) (4)2 3(4) 12 4 3 40 1 40 2) lim x→4 x2 3x 12 x 3 40 3) lim x→4 f(x) 40 f(4) f(x) is continuous.
c) f(x) x 3 x2 9 at x 3 1) f(3) 3 3 (3)2 9 With the denominator equal to zero, f(x) is not defined at x 3 and so cannot be continuous at x 3 even though the limit exists at x 3. See steps 2 and 3.
2) lim x→3 x 3 x2 9 lim x→3 x 3 (x 3)(x 3) lim x→3 1 x 3 1 6 3) lim x→3 f(x) 1 – 6 f(3). So f(x) is discontinuous at x 3.
DERIVATIVE NOTATION AND SIMPLE DERIVATIVES 3.6.
Differentiate each of the following functions and practice the use of the different notations for a derivative.
a) f(x) 17 b) y 12 f(x) 0 (constant rule) dy dx 0 c) y 5x 12 d) f(x) 9x 6 y 5 (linear function rule) f 9 3.7.
Differentiate each of the following functions using the power function rule. Continue to use the different notations.
a) y 8x3 b) f(x) 6x5 d dx (8x3) 24x2 f 30x4 43 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] c) f(x) 5x2 f(x) 5(2) · x[2(1)] 10x3 10 x3 d) y 9x4 dy dx 9(4) · x[4(1)] 36x5 36 x5 e) y 7 x 7x1 Dx(7x1) 7(1)x2 7x2 7 x2 f) f(x) 18x 18x1/2 df dx 18 1 2 · x1/21 9x1/2 9 x 3.8.
Use the rule for sums and differences to differentiate the following functions. Treat the dependent variable on the left as y and the independent variable on the right as x.
a) R 8t2 5t 6 b) C 4t3 9t2 28t 68 dR dt 16t 5 C 12t2 18t 28 c) p 6q5 3q3 d) q 7p4 15p3 dp dq 30q4 9q2 Dp(7p4 15p3) 28p3 45p4 THE PRODUCT RULE 3.9.
Given y f(x) 5x4(3x 7), (a) use the product rule to find the derivative. (b) Simplify the original function first and then find the derivative. (c) Compare the two derivatives.
a) Recalling the formula for the product rule from (3.3), f(x) g(x) · h(x) h(x) · g(x) let g(x) 5x4 and h(x) 3x 7. Then g(x) 20x3 and h(x) 3. Substitute these values in the product rule formula.
y f(x) 5x4(3) (3x 7)(20x3) Simplify algebraically.
y 15x4 60x4 140x3 75x4 140x3 b) Simplify the original function by multiplication.
y 5x4(3x 7) 15x5 35x4 Take the derivative.
y 75x4 140x3 c) The derivatives found in parts (a) and (b) are identical. The derivative of a product can be found by either method, but as the functions grow more complicated, the product rule becomes more useful.
Knowledge of another method helps to check answers.
44 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 3.10.
Redo Problem 3.9, given y f(x) (x8 8)(x6 11).
a) Let g(x) x8 8 and h(x) x6 11. Then g(x) 8x7 and h(x) 6x5. Substituting these values in (3.3), y f(x) (x8 8)(6x5) (x6 11)(8x7) 6x13 48x5 8x13 88x7 14x13 88x7 48x5 b) Simplifying first through multiplication, y (x8 8)(x6 11) x14 11x8 8x6 88 Then y 14x13 88x7 48x5 c) The derivatives are identical.
3.11.
Differentiate each of the following functions using the product rule. Note: The choice of problems is purposely kept simple in this and other sections of the book to enable students to see how various rules work. While it is proper and often easier to simplify a function algebraically before taking the derivative, applying the rules to the problems as given in the long run will help the student to master the rules more efficiently.
a) y (4x2 3)(2x5) dy dx(4x2 3)(10x4) 2x5(8x) 40x6 30x4 16x6 56x6 30x4 b) y 7x9(3x2 12) dy dx 7x9(6x) (3x2 12)(63x8) 42x10 189x10 756x8 231x10 756x8 c) y (2x4 5)(3x5 8) dy dx (2x4 5)(15x5) (3x5 8)(8x3) 30x8 75x4 24x8 64x3 54x8 75x4 64x2 d) z (3 12t3)(5 4t6) dz dt (3 12t3)(24t5) (5 4t6)(36t2) 72t5 288t8 180t2 144t8 432t8 72t5 180t2 QUOTIENT RULE 3.12.
Given y 10x8 6x7 2x (a) Find the derivative directly, using the quotient rule. (b) Simplify the function by division and then take its derivative. (c) Compare the two derivatives.
a) From (3.4), the formula for the quotient rule is f(x) h(x) · g(x) g(x) · h(x) [h(x)]2 where g(x) the numerator 10x8 6x7 and h(x) the denominator 2x. Take the individual derivatives.
g(x) 80x7 42x6 h(x) 2 45 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] Substitute in the formula, y 2x(80x7 42x6) (10x8 6x7)(2) (2x)2 160x8 84x7 20x8 12x7 4x2 140x8 72x7 4x2 35x6 18x5 b) Simplifying the original function first by division, y 10x8 6x7 2x 5x7 3x6 y 35x6 18x5 c) The derivatives will always be the same if done correctly, but as functions grow in complexity, the quotient rule becomes more important. A second method is also a way to check answers.
3.13.
Differentiate each of the following functions by means of the quotient rule. Continue to apply the rules to the functions as given. Later, when all the rules have been mastered, the functions can be simplified first and the easiest rule applied.
a) y 3x8 4x7 4x3 Here g(x) 3x8 4x7 and h(x) 4x3. Thus, g(x) 24x7 28x6 and h(x) 12x2. Substituting in the quotient formula, y 4x3(24x7 28x6) (3x8 4x7)(12x2) (4x3)2 96x10 112x9 36x10 48x9 16x6 60x10 64x9 16x6 3.75x4 4x3 b) y 4x5 1 3x (x 1 – 3) (Note: The qualifying statement is added because if x 1 – 3, the denominator would equal zero and the function would be undefined.) dy dx (1 3x)(20x4) 4x5(3) (1 3x)2 20x4 60x5 12x5 (1 3x)2 20x4 48x5 (1 3x)2 c) y 15x2 2x2 7x 3 dy dx (2x2 7x 3)(30x) 15x2(4x 7) (2x2 7x 3)2 60x3 210x2 90x 60x3 105x2 (2x2 7x 3)2 105x2 90x (2x2 7x 3)2 d) y 6x 7 8x 5 (x 5 – 8) dy dx (8x 5)(6) (6x 7)(8) (8x 5)2 48x 30 48x 56 (8x 5)2 26 (8x 5)2 46 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 e) y 5x2 9x 8 x2 1 dy dx (x2 1)(10x 9) (5x2 9x 8)(2x) (x2 1)2 10x3 9x2 10x 9 10x3 18x2 16x (x2 1)2 9x2 6x 9 (x2 1)2 THE GENERALIZED POWER FUNCTION RULE 3.14.
Given y (5x 8)2, (a) use the generalized power function rule to find the derivative; (b) simplify the function first by squaring it and then take the derivative; (c) compare answers.
a) From the generalized power function rule in (3.5), if f(x) [g(x)]n, f(x) n[g(x)]n1 · g(x) Here g(x) 5x 8, g(x) 5, and n 2. Substitute these values in the generalized power function rule, y 2(5x 8)21 · 5 10(5x 8) 50x 80 b) Square the function first and then take the derivative, y (5x 8)(5x 8) 25x2 80x 64 y 50x 80 c) The derivatives are identical. But for higher, negative, and fractional values of n, the generalized power function rule is faster and more practical.
3.15.
Find the derivative for each of the following functions with the help of the generalized power function rule.
a) y (6x3 9)4 Here g(x) 6x3 9, g(x) 18x2, and n 4. Substitute in the generalized power function rule, y 4(6x3 9)41 · 18x2 4(6x3 9)3 · 18x2 72x2(6x3 9)3 b) y (2x2 5x 7)3 y 3(2x2 5x 7)2 · (4x 5) (12x 15)(2x2 5x 7)2 c) y 1 7x3 13x 3 First convert the function to an easier equivalent form, y (7x3 13x 3)1 then use the generalized power function rule, y 1(7x3 13x 3)2 · (21x2 13) (21x2 13)(7x3 13x 3)2 (21x2 13) (7x3 13x 3)2 47 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] d) y 34 6x2 Convert the radical to a power function, then differentiate.
y y (34 6x2)1/2 1 – 2(34 6x2)1/2 · (12x) 6x(34 6x2)1/2 6x 34 6x2 e) y 1 4x3 94 Convert to an equivalent form; then take the derivative.
y y (4x3 94)1/2 1 – 2(4x3 94)3/2 · (12x2) 6x2(4x3 94)3/2 6x2 (4x3 94)3/2 6x2 (4x3 94)3 CHAIN RULE 3.16.
Use the chain rule to find the derivative dy/dx for each of the following functions of a function.
Check each answer on your own with the generalized power function rule, noting that the generalized power function rule is simply a specialized use of the chain rule.
a) y (3x4 5)6 Let y u6 and u 3x4 5. Then dy/du 6u5 and du/dx 12x3. From the chain rule in (3.6), dy dx dy du du dx Substituting, dy dx 6u5 · 12x3 72x3u5 But u 3x4 5. Substituting again, dy dx 72x3(3x4 5)5 b) y (7x 9)2 Let y u2 and u 7x 9, then dy/du 2u and du/dx 7. Substitute these values in the chain rule, dy dx 2u · 7 14u Then substitute 7x 9 for u.
dy dx 14(7x 9) 98x 126 c) y (4x5 1)7 Let y u7 and u 4x5 1; then dy/du 7u6, du/dx 20x4, and dy dx 7u6 · 20x4 140x4u6 48 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 Substitute u 4x5 1.
dy dx 140x4(4x5 1)6 3.17.
Redo Problem 3.16, given: a) y (x2 3x 1)5 Let y u5 and u x2 3x 1, then dy/du 5u4 and du/dx 2x 3. Substitute in (3.6).
dy dx 5u4(2x 3) (10x 15)u4 But u x2 3x 1. Therefore, dy dx (10x 15)(x2 3x 1)4 b) y 3(x2 8x 7)4 Let y 3u4 and u x2 8x 7. Then dy/du 12u3, du/dx 2x 8, and dy dx 12u3(2x 8) (24x 96)u3 (24x 96)(x2 8x 7)3 COMBINATION OF RULES 3.18.
Use whatever combination of rules is necessary to find the derivatives of the following functions. Do not simplify the original functions first. They are deliberately kept simple to facilitate the practice of the rules.
a) y 3x(2x 1) 5x 2 The function involves a quotient with a product in the numerator. Hence both the quotient rule and the product rule are required. Start with the quotient rule from (3.4).
y h(x) · g(x) g(x) · h(x) [h(x)]2 where g(x) 3x(2x 1), h(x) 5x 2, and h(x) 5. Then use the product rule from (3.3) for g(x).
g(x) 3x · 2 (2x 1) · 3 12x 3 Substitute the appropriate values in the quotient rule.
y (5x 2)(12x 3) [3x(2x 1)] · 5 (5x 2)2 Simplify algebraically.
y 60x2 15x 24x 6 30x2 15x (5x 2)2 30x2 24x 6 (5x 2)2 Note: To check this answer one could let y 3x · 2x 1 5x 2 or y 3x 5x 2 · (2x 1) and use the product rule involving a quotient.
49 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] b) y 3x(4x 5)2 The function involves a product in which one function is raised to a power. Both the product rule and the generalized power function rule are needed. Starting with the product rule, y g(x) · h(x) h(x) · g(x) where g(x) 3x h(x) (4x 5)2 and g(x) 3 Use the generalized power function rule for h(x).
h(x) 2(4x 5) · 4 8(4x 5) 32x 40 Substitute the appropriate values in the product rule, y 3x · (32x 40) (4x 5)2 · 3 and simplify algebraically, y 96x2 120x 3(16x2 40x 25) 144x2 240x 75 c) y (3x 4) · 5x 1 2x 7 Here we have a product involving a quotient. Both the product rule and the quotient rule are needed. Start with the product rule, y g(x) · h(x) h(x) · g(x) where g(x) 3x 4 h(x) 5x 1 2x 7 and g(x) 3 and use the quotient rule for h(x).
h(x) (2x 7)(5) (5x 1)(2) (2x 7)2 33 (2x 7)2 Substitute the appropriate values in the product rule, y (3x 4) · 33 (2x 7)2 5x 1 2x 7 · 3 99x 132 (2x 7)2 15x 3 2x 7 99x 132 (15x 3)(2x 7) (2x 7)2 30x2 210x 111 (2x 7)2 One could check this answer by letting y (3x 4)(5x 1)/(2x 7) and using the quotient rule involving a product.
d) y (8x 5)3 (7x 4) Start with the quotient rule, where g(x) (8x 5)3 h(x) 7x 4 h(x) 7 and use the generalized power function rule for g(x), g(x) 3(8x 5)2 · 8 24(8x 5)2 Substitute these values in the quotient rule, y (7x 4) · 24(8x 5)2 (8x 5)3 · 7 (7x 4)2 (168x 96)(8x 5)2 7(8x 5)3 (7x 4)2 50 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 To check this answer, one could let y (8x 5)3 · (7x 4)1 and use the product rule involving the generalized power function rule twice.
e) y 3x 4 2x 5 2 Start with the generalized power function rule, y 2 3x 4 2x 5 · d dx 3x 4 2x 5 (3.9) Then use the quotient rule, d dx 3x 4 2x 5 (2x 5)(3) (3x 4)(2) (2x 5)2 7 (2x 5)2 and substitute this value in (3.9), y 2 3x 4 2x 5 · 7 (2x 5)2 14(3x 4) (2x 5)3 42x 56 (2x 5)3 To check this answer, let y (3x 4)2 · (2x 5)2, and use the product rule involving the generalized power function rule twice.
3.19.
Differentiate each of the following, using whatever rules are necesary: a) y (5x 1)(3x 4)3 Using the product rule together with the generalized power function rule, dy dx (5x 1)[3(3x 4)2(3)] (3x 4)3(5) Simplifying algebraically, dy dx (5x 1)(9)(3x 4)2 5(3x 4)3 (45x 9)(3x 4)2 5(3x 4)3 b) y (9x2 2)(7x 3) 5x Using the quotient rule along with the product rule, y 5x[(9x2 2)(7) (7x 3)(18x)] (9x2 2)(7x 3)(5) (5x)2 Simplifying algebraically, y 5x(63x2 14 126x2 54x) 5(63x3 27x2 14x 6) 25x2 630x3 135x2 30 25x2 c) y 15x 23 (3x 1)2 Using the quotient rule plus the generalized power function rule, y (3x 1)2(15) (15x 23)[2(3x 1)(3)] (3x 1)4 Simplifying algebraically, y 15(3x 1)2 (15x 23)(18x 6) (3x 1)4 135x2 414x 123 (3x 1)4 51 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] d) y (6x 1) 4x 9x 1 Using the product rule and the quotient rule, Dx (6x 1) (9x 1)(4) 4x(9) (9x 1)2 4x 9x 1 (6) Simplifying algebraically, Dx (6x 1)(36x 4 36x) (9x 1)2 24x 9x 1 216x2 48x 4 (9x 1)2 e) y 3x 1 2x 5 3 Using the generalized power function rule and the quotient rule, y 3 3x 1 2x 5 2 (2x 5)(3) (3x 1)(2) (2x 5)2 Simplifying algebraically, y 3(3x 1)2 (2x 5)2 17 (2x 5)2 51(3x 1)2 (2x 5)4 HIGHER-ORDER DERIVATIVES 3.20.
For each of the following functions, (1) find the second-order derivative and (2) evaluate it at x 2. Practice the use of the different second-order notations.
a) y 7x3 5x2 12 1) dy dx 21x2 10x 2) At x 2, d2y dx2 42(2) 10 d2y dx2 42x 10 94 b) f(x) x6 3x4 x 1) f(x) 6x5 12x3 1 f
(x) 30x4 36x2 2) At x 2, f
(x) 30(2)4 36(2)2 624 c) y (2x 3)(8x2 6) 1) Dy (2x 3)(16x) (8x2 6)(2) 32x2 48x 16x2 12 48x2 48x 12 D2y 96x 48 2) At x 2, D2y 96(2) 48 240 d) f(x) (x4 3)(x3 2) 1) f (x4 3)(3x2) (x3 2)(4x3) 3x6 9x2 4x6 8x3 7x6 8x3 9x2 f 42x5 24x2 18x 2) At x 2, f 42(2)5 24(2)2 18(2) 1212 52 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 e) y 5x 1 3x 1) y (1 3x)(5) 5x(3) (1 3x)2 2) At x 2, y 30 90(2) [1 3(2)]4 5 15x 15x (1 3x)2 5 (1 3x)2 150 (5)4 y (1 3x)2(0) 5[2(1 3x)(3)] (1 3x)4 6 25 5(6 18x) (1 3x)4 30 90x (1 3x)4 30 (1 3x)3 f) y 7x2 x 1 1) y (x 1)(14x) 7x2(1) (x 1)2 2) At x 2, y 14 (2 1)3 14x2 14x 7x2 (x 1)2 7x2 14x (x 1)2 14 y (x 1)2(14x 14) (7x2 14x)[2(x 1)(1)] (x 1)4 (x2 2x 1)(14x 14) (7x2 14x)(2x 2) (x 1)4 14(x 1) (x 1)4 14 (x 1)3 g) f(x) (8x 4)3 1) f 3(8x 4)2(8) 2) At x 2, f 384[8(2) 4] 24(8x 4)2 4608 f 2(24)(8x 4)(8) 384(8x 4) h) y (5x3 7x2)2 1) Dy 2(5x3 7x2)(15x2 14x) 2) At x 2, D2y 750(2)4 1400(2)3 588(2)2 150x5 350x4 196x3 3152 D2y 750x4 1400x3 588x2 3.21.
For each of the following functions, (1) investigate the successive derivatives and (2) evaluate them at x 3.
a) y x3 3x2 9x 7 1) y 3x2 6x 9 y 6x 6 y 6 y(4) 0 2) At x 3, y 3(3)2 6(3) 9 54 y 6(3) 6 24 y 6 y(4) 0 b) y (4x 7)(9x 2) 1) y (4x 7)(9) (9x 2)(4) 36x 63 36x 8 72x 55 y 72 y 0 2) At x 3, y 72(3) 55 161 y 72 y 0 53 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] c) y (5 x)4 1) Dx 4(5 x)3(1) 4(5 x)3 Dx 2 12(5 x)2(1) 12(5 x)2 Dx 3 24(5 x)(1) 24(5 x) 24x 120 Dx 4 24 Dx 5 0 2) At x 3, Dx 4(5 3)3 32 Dx 2 12(5 3)2 48 Dx 3 24(3) 120 48 Dx 4 24 Dx 5 0 IMPLICIT DIFFERENTIATION 3.22.
Use implicit differentiation to find the derivative dy/dx for each of the following equations.
a) 4x2 y3 97 Take the derivative with respect to x of both sides, d dx (4x2) d dx (y3) d dx (97) (3.10) where d dx(4x2) 8x, d dx (97) 0, and use the generalized power function rule because y is con-sidered a function of x, d dx (y3) 3 · y2 · d dx (y) Set these values in (3.10) and recall that d dx (y) dy dx.
8x 3y2 dy dx 0 3y2 dy dx 8x dy dx 8x 3y2 b) 3y5 6y4 5x6 243 Taking the derivative with respect to x of both sides, d dx (3y5) d dx (6y4) d dx (5x6) d dx (243) 15y4 dy dx 24y3 dy dx 30x5 0 Solve for dy/dx, (15y4 24y3) dy dx 30x5 dy dx 30x5 15y4 24y3 c) 2x4 7x3 8y5 136 d dx (2x4) d dx (7x3) d dx (8y5) d dx (136) 54 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 8x3 21x2 40y4 dy dx 0 40y4 dy dx (8x3 21x2) dy dx (8x3 21x2) 40y4 3.23.
Use the different rules of differentiation in implicit differentiation to find dy/dx for each of the following: a) x4y6 89 d dx (x4y6) d dx (89) Use the product rule and the generalized power function rule.
x4 · d dx (y6) y6 · d dx (x4) d dx (89) x4 · 6y5 dy dx y6 · 4x3 0 Solve algebraically for dy/dx.
6x4y5 dy dx 4x3y6 dy dx 4x3y6 6x4y5 2y 3x b) 2x3 5xy 6y2 87 d dx (2x3 5xy 6y2) d dx (87) Note that the derivative of 5xy requires the product rule.
6x2 5x · dy dx y · (5) 12y dy dx 0 Solving algebraically for dy/dx (5x 12y) dy dx 6x2 5y dy dx (6x2 5y) 5x 12y c) 7x4 3x3y 9xy2 496 28x3 3x3 · dy dx y · 9x2 9x · 2y dy dx y2 · 9 0 28x3 3x3 dy dx 9x2y 18xy dy dx 9y2 0 (3x3 18xy) dy dx 28x3 9x2y 9y2 dy dx (28x3 9x2y 9y2) 3x3 18xy 55 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] d) (5y 21)3 6x5 d dx [(5y 21)3] d dx (6x5) Use the generalized power function rule.
3(5y 21)2 · 5 dy dx 30x4 15(5y 21)2 dy dx 30x4 dy dx 30x4 15(5y 21)2 e) (2x3 7y)2 x5 d dx (2x3 7y)2 d dx (x5) 2(2x3 7y) · d dx (2x3 7y) 5x4 (4x3 14y)6x2 7 dy dx 5x4 24x5 28x3 dy dx 84x2y 98y dy dx 5x4 (28x3 98y) dy dx 5x4 24x5 84x2y dy dx 5x4 24x5 84x2y 28x3 98y See also Problems 4.24, 4.25, 5.20, 5.21, 6.51, and 6.52.
DERIVATION OF THE RULES OF DIFFERENTIATION 3.24.
Given f(x) g(x) h(x), where g(x) and h(x) are both differentiable functions, prove the rule of sums by demonstrating that f(x) g(x) h(x).
From (3.2) the derivative of f(x) is f(x) lim x→0 f(x x) f(x) x Substituting f(x) g(x) h(x), f(x) lim x→0 [g(x x) h(x x)] [g(x) h(x)] x Rearrange terms.
f(x) lim x→0 g(x x) g(x) h(x x) h(x) x Separate terms, and take the limits.
f(x) lim x→0 g(x x) g(x) x h(x x) h(x) x lim x→0 g(x x) g(x) x lim x→0 h(x x) h(x) x g(x) h(x) 56 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION [CHAP . 3 3.25.
Given f(x) g(x) · h(x), where g(x) and h(x) both exist, prove the product rule by demonstrat-ing that f(x) g(x) · h(x) h(x) · g(x).
f(x) lim x→0 f(x x) f(x) x Substitute f(x) g(x) · h(x).
f(x) lim x→0 g(x x) · h(x x) g(x) · h(x) x Add and subtract g(x x) · h(x), f(x) lim x→0 g(x x)h(x x) g(x x)h(x) g(x x)h(x) g(x)h(x) x Partially factor out g(x x) and h(x).
f(x) lim x→0 g(x x)[h(x x) h(x)] h(x)[g(x x) g(x)] x lim x→0 g(x x)[h(x x) h(x)] x lim x→0 h(x)[g(x x) g(x)] x lim x→0 g(x x) · lim x→0 h(x x) h(x) x lim x→0 h(x) · lim x→0 g(x x) g(x) x g(x) · h(x) h(x) · g(x) 3.26.
Given f(x) g(x)/h(x), where g(x) and h(x) both exist and h(x) 0, prove the quotient rule by demonstrating f(x) h(x) · g(x) g(x) · h(x) [h(x)]2 Start with f(x) g(x)/h(x) and solve for g(x), g(x) f(x) · h(x) Then take the derivative of g(x), using the product rule, g(x) f(x) · h(x) h(x) · f(x) and solve algebraically for f(x).
h(x) · f(x) g(x) f(x) · h(x) f(x) g(x) f(x) · h(x) h(x) Substitute g(x)/h(x) for f(x).
f(x) g(x) g(x) · h(x) h(x) h(x) Now multiply both numerator and denominator by h(x), f(x) h(x) · g(x) g(x) · h(x) [h(x)]2 57 THE DERIVATIVE AND THE RULES OF DIFFERENTIATION CHAP . 3] CHAPTER 4 Uses of the Derivative in Mathematics and Economics 4.1 INCREASING AND DECREASING FUNCTIONS A function f(x) is said to be increasing (decreasing) at x a if in the immediate vicinity of the point [a, f(a)] the graph of the function rises (falls) as it moves from left to right. Since the first derivative measures the rate of change and slope of a function, a positive first derivative at x a indicates the function is increasing at s; a negative first derivative indicates it is decreasing. In short, as seen in Fig. 4-1, f(a) 0: increasing function at x a f(a) 0: decreasing function at x a A function that increases (or decreases) over its entire domain is called a monotonic function. It is said to increase (decrease) monotonically. See Problems 4.1 to 4.3.
4.2 CONCA VITY AND CONVEXITY A function f(x) is concave at x a if in some small region close to the point [a, f(a)] the graph of the function lies completely below its tangent line. A function is convex at x a if in an area very close to [a, f(a)] the graph of the function lies completely above its tangent line. A positive second derivative at x a denotes the function is convex at x a; a negative second derivative at x a denotes the function is concave at a. The sign of the first derivative is irrelevant for concavity. In brief, as seen in Fig. 4-2 and Problems 4.1 to 4.4, f
(a) 0: f(x) is convex at x a f
(a) 0: f(x) is concave at x a 58 If f
(x) 0 for all x in the domain, f(x) is strictly convex. If f
(x) 0 for all x in the domain, f(x) is strictly concave.
4.3 RELATIVE EXTREMA A relative extremum is a point at which a function is at a relative maximum or minimum. To be at a relative maximum or minimum at a point a, the function must be at a relative plateau, i.e., neither increasing nor decreasing at a. If the function is neither increasing nor decreasing at a, the first derivative of the function at a must equal zero or be undefined. A point in the domain of a function where the derivative equals zero or is undefined is called a critical point or value.
59 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] Fig. 4-1 Fig. 4-2 To distinguish mathematically between a relative maximum and minimum, the second-derivative test is used. Assuming f(a) 0, 1.
If f
(a) 0, indicating that the function is convex and the graph of the function lies completely above its tangent line at x a, the function is at a relative minimum at x a.
2.
If f
(a) 0, denoting that the function is concave and the graph of the function lies completely below its tangent line at x a, the function is at a relative maximum at x a.
3.
If f
(a) 0, the test is inconclusive.
For functions which are differentiable at all values of x, called differentiable or smooth functions, one need only consider cases where f(x) 0 in looking for critical points. To summarize, f(a) 0 f
(a) 0: relative minimum at x a f(a) 0 f
(a) 0: relative maximum at x a See Fig. 4-3 and Problems 4.5 and 4.6.
4.4 INFLECTION POINTS An inflection point is a point on the graph where the function crosses its tangent line and changes from concave to convex or vice versa. Inflection points occur only where the second derivative equals zero or is undefined. The sign of the first derivative is immaterial. In sum, for an inflection point at a, as seen in Fig. 4-4 and Problems 4.6 and 4.7(c), 1.
f
(a) 0 or is undefined.
2.
Concavity changes at x a.
3.
Graph crosses its tangent line at x a.
4.5 OPTIMIZATION OF FUNCTIONS Optimization is the process of finding the relative maximum or minimum of a function. Without the aid of a graph, this is done with the techniques developed in Sections 4.3 through 4.4 and outlined below. Given the usual differentiable function, 1.
Take the first derivative, set it equal to zero, and solve for the critical point(s). This step represents a necessary condition known as the first-order condition. It identifies all the points at which the function is neither increasing nor decreasing, but at a plateau. All such points are candidates for a possible relative maximum or minimum.
60 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 Fig. 4-3 2.
Take the second derivative, evaluate it at the critical point(s), and check the sign(s). If at a critical point a, f
(a) 0, the function is concave at a, and hence at a relative maximum.
f
(a) 0, the function is convex at a, and hence at a relative minimum.
f
(a) 0, the test is inconclusive. See Section 4.6.
Assuming the necessary first-order condition is met, this step, known as the second-order derivative test, or simply the second-order condition, represents a sufficiency condition. In sum, Relative maximum f(a) 0 f
(a) 0 Relative minimum f(a) 0 f
(a) 0 Note that if the function is strictly concave (convex), there will be only one maximum (minimum), called a global maximum (minimum). See Example 1 and Problems 4.7 to 4.9.
EXAMPLE 1.
Optimize f(x) 2x3 30x2 126x 59.
a) Find the critical points by taking the first derivative, setting it equal to zero, and solving for x.
f(x) 6x2 60x 126 0 6(x 3)(x 7) 0 x 3 x 7 critical points (b) Test for concavity by taking the second derivative, evaluating it at the critical points, and checking the signs to distinguish beteween a relative maximum and minimum.
f
(x) 12x 60 f
(3) 12(3) 60 24 0 concave, relative maximum f
(7) 12(7) 60 24 0 convex, relative minimum The function is maximized at x 3 and minimized at x 7.
4.6 SUCCESSIVE-DERIVATIVE TEST FOR OPTIMIZATION If f
(a) 0, as in Fig. 4-4(a) through (d), the second-derivative test is inconclusive. In such cases, without a graph for guidance, the successive-derivative test is helpful: 61 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] Fig. 4-4 1.
If the first nonzero value of a higher-order derivative, when evaluated at a critical point, is an odd-numbered derivative (third, fifth, etc.), the function is at an inflection point. See Problems 4.6(b) and (d) and 4.7(c).
2.
If the first nonzero value of a higher-order derivative, when evaluated at a critical point a, is an even-numbered derivative, the function is at a relative extremum at a, with a negative value of the derivative indicating that the function is concave and at a relative maximum and a positive value signifying the function is convex and at a relative minimum. See Problems 4.6(a) and (c), 4.7(d), and 4.9(c) and (d).
4.7 MARGINAL CONCEPTS Marginal cost in economics is defined as the change in total cost incurred from the production of an additional unit. Marginal revenue is defined as the change in total revenue brought about by the sale of an extra good. Since total cost (TC) and total revenue (TR) are both functions of the level of output (Q), marginal cost (MC) and marginal revenue (MR) can each be expressed mathematically as derivatives of their respective total functions. Thus, if TC TC(Q), then MC dTC dQ and if TR TR(Q), then MR dTR dQ In short, the marginal concept of any economic function can be expressed as the derivative of its total function. See Examples 2 and 3 and Problems 4.10 to 4.16.
EXAMPLE 2.
1.
If TR 75Q 4Q2, then MR dTR/dQ 75 8Q.
2.
If TC Q2 7Q 23, then MC dTC/dQ 2Q 7.
EXAMPLE 3.
Given the demand function P 30 2Q, the marginal revenue function can be found by first finding the total revenue function and then taking the derivative of that function with respect to Q. Thus, TR PQ (30 2Q)Q 30Q 2Q2 Then MR dTR dQ 30 4Q If Q 4, MR 30 4(4) 14; if Q 5, MR 30 4(5) 10.
4.8 OPTIMIZING ECONOMIC FUNCTIONS The economist is frequently called upon to help a firm maximize profits and levels of physical output and productivity, as well as to minimize costs, levels of pollution, and the use of scarce natural resources. This is done with the help of techniques developed earlier and illustrated in Example 4 and Problems 4.17 to 4.23.
EXAMPLE 4.
Maximize profits for a firm, given total revenue R 4000Q 33Q2 and total cost C 2Q3 3Q2 400Q 5000, assuming Q 0.
a) Set up the profit function: R C.
4000Q 33Q2 (2Q3 3Q2 400Q 5000) 2Q3 30Q2 3600Q 5000 62 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 b) Take the first derivative, set it equal to zero, and solve for Q to find the critical points.
6Q2 60Q 3600 0 6(Q2 10Q 600) 0 6(Q 30)(Q 20) 0 Q 30 Q 20 critical points c) Take the second derivative; evaluate it at the positive critical point and ignore the negative critical point, which has no economic significance and will prove mathematically to be a relative minimum. Then check the sign for concavity to be sure of a relative maximum.
(20) 12Q 60 12(20) 60 300 0 concave, relative maximum Profit is maximized at Q 20 where (20) 2(20)3 30(20)2 3600(20) 5000 39,000 4.9 RELATIONSHIP AMONG TOTAL, MARGINAL, AND A VERAGE CONCEPTS A total product (TP) curve of an input is derived from a production function by allowing the amounts of one input (say, capital) to vary while holding the other inputs (labor and land) constant.
A graph showing the relationship between the total, average, and marginal products of an input can easily be sketched by using now familiar methods, as demonstrated in Example 5.
EXAMPLE 5.
Given TP 90K2 K3, the relationship among the total, average, and marginal products can be illustrated graphically as follows.
1.
Test the first-order condition to find the critical values.
TP 180K 3K2 0 3K(60 K) 0 K 0 K 60 critical values Check the second-order conditions.
TP 180 6K TP
(0) 180 0 convex, relative minimum TP
(60) 180 0 concave, relative maximum Check for inflection points.
TP 180 6K 0 K 30 K 30 TP 0 convex K 30 TP 0 concave Since, at K 30, TP 0 and concavity changes, there is an inflection point at K 30.
2.
Find and maximize the average product of capital APK.
APK TP K 90K K2 AP K 90 2K 0 K 45 critical value AP K 2 0 concave, relative maximum 63 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] 3.
Find and maximize the marginal product of capital MPK, recalling that MPK TP 180K 3K2: MP K 180 6K 0 K 30 critical value MP K 6 0 concave, relative maximum 4.
Sketch the graphs, as in Fig. 4-5.
Note that (a) MPK increases when TP is convex and increasing at an increasing rate, is at a maximum where TP is at an inflection point, and decreases when TP is concave and increasing at a decreasing rate; (b) TP increases over the whole range where MPK is positive, is at a maximum where MPK 0, and declines when MPK is negative; (c) APK is at a maximum where the slope of a line from the origin to the TP curve is tangent to the TP curve, i.e., where MPK APK; (d) MPK APK when APK is increasing, MPK APK when APK is at a maximum, and MPK APK when APK decreases; and (e) MPK is negative when TP declines. See also Problem 4.26.
Solved Problems INCREASING AND DECREASING FUNCTIONS, CONCA VITY AND CONVEXITY 4.1.
From the graphs in Fig. 4-6, indicate which graphs (1) are increasing for all x, (2) are decreasing for all x, (3) are convex for all x, (4) are concave for all x, (5) have relative maxima or minima, and (6) have inflection points.
64 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 Fig. 4-5 1) a, d: increasing for all x.
2) b, f: decreasing for all x.
3) b, c: convex for all x.
4) d, e: concave for all x.
5) c, e: exhibit a relative maximum or minimum.
6) a, f: have an inflection point.
4.2.
Indicate with respect to the graphs in Fig. 4-7 which functions have (1) positive first derivatives for all x, (2) negative first derivatives for all x, (3) positive second derivatives for all x, (4) negative second derivatives for all x, (5) first derivatives equal to zero or undefined at some point, and (6) second derivatives equal to zero or undefined at some point.
1) a, b, h: the graphs all move up from left to right.
2) d, f, g: the graphs all move down from left to right.
3) d, e, h: the graphs are all convex.
4) a, c, f: the graphs are all concave.
5) c, e: the graphs reach a plateau (at an extreme point).
6) b, g: the graphs have inflection points.
65 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] Fig. 4-6 Fig. 4-7 4.3.
Test to see whether the following functions are increasing, decreasing, or stationary at x 4.
a) y 3x2 14x 5 y 6x 14 y(4) 6(4) 14 10 0 Function is increasing.
b) y x3 7x2 6x 2 y 3x2 14x 6 y(4) 3(4)2 14(4) 6 2 0 Function is decreasing.
c) y x4 6x3 4x2 13 y 4x3 18x2 8x y(4) 4(4)3 18(4)2 8(4) 0 Function is stationary.
4.4.
Test to see if the following functions are concave or convex at x 3.
a) y 2x3 4x2 9x 15 y y y
(3) 6x2 8x 9 12x 8 12(3) 8 28 0 concave b) y (5x2 8)2 y y y
(3) 2(5x2 8)(10x) 20x(5x2 8) 100x3 160x 300x2 160 300(3)2 160 2540 0 convex RELATIVE EXTREMA 4.5.
Find the relative extrema for the following functions by (1) finding the critical value(s) and (2) determining if at the critical value(s) the function is at a relative maximum or minimum.
a) f(x) 7x2 126x 23 1) Take the first derivative, set it equal to zero, and solve for x to find the critical value(s).
f(x) 14x 126 0 x 9 critical value 2) Take the second derivative, evaluate it at the critical value(s), and check for concavity to distinguish between a relative maximum and minimum.
f
(x) 14 f
(9) 14 0 concave, relative maximum b) f(x) 3x3 36x2 135x 13 1) f(x) 9x2 72x 135 0 9(x2 8x 15) 0 9(x 3)(x 5) 0 x 3 x 5 critical values 2) f
(x) 18x 72 f
(3) 18(3) 72 18 0 concave, relative maximum f
(5) 18(5) 72 18 0 convex, relative minimum 66 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 c) f(x) 2x4 16x3 32x2 5 1) f(x) 8x3 48x2 64x 0 8x(x2 6x 8) 0 8x(x 2)(x 4) 0 x 0 x 2 x 4 critical values 2) f
(x) 24x2 96x 64 f
(0) 24(0)2 96(0) 64 64 0 convex, relative minimum f
(2) 24(2)2 96(2) 64 32 0 concave, relative maximum f
(4) 24(4)2 96(4) 64 64 0 convex, relative minimum 4.6.
For the following functions, (1) find the critical values and (2) test to see if at the critical values the function is at a relative maximum, minimum, or possible inflection point.
a) y (x 8)4 1) Take the first derivative, set it equal to zero, and solve for x to obtain the critical value(s).
y 4(x 8)3 0 x 8 0 x 8 critical value 2) Take the second derivative, evaluate it at the critical value(s), and check the sign for concavity to distinguish between a relative maximum, minimum, or inflection point.
y 12(x 8)2 y
(8) 12(8 8)2 0 test inconclusive If the second-derivative test is inconclusive, continue to take successively higher derivatives and evaluate them at the critical values until you come to the first higher-order derivative that is nonzero: y y(8) y(4) y(4)(8) 24(x 8) 24(8 8) 0 test inconclusive 24 24 0 As explained in Section 4.6, with the first nonzero higher-order derivative an even-numbered derivative, y is at a relative extremum. With that derivative negative, y is concave and at a relative maximum. See Fig. 4-8(a).
b) y (5 x)3 1) y 3(5 x)2(1) 3(5 x)2 0 x 5 critical value 2) y 6(5 x) y
(5) 6(5 5) 0 test inconclusive Continuing to take successively higher-order derivatives and evaluating them at the critical value(s) in search of the first higher-order derivative that does not equal zero, we get y 6 y(5) 6 0 As explained in Section 4.6, with the first nonzero higher-order derivative an odd-numbered derivative, y is at an inflection point and not at an extreme point. See Fig.
4-8(b).
67 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] c) y 2(x 6)6 1) y 12(x 6)5 0 x 6 critical value 2) y 60(x 6)4 y
(6) 60(0)4 0 test inconclusive Continuing, we get y y(4) y(5) y(6) 240(x 6)3 720(x 6)2 1440(x 6) 1440 y(6) y(4)(6) y(5)(6) y(6)(6) 0 test inconclusive 0 test inconclusive 0 test inconclusive 1440 0 With the first nonzero higher-order derivative an even-numbered derivative, y is at an extreme point; with y(6)(6) 0, y is concave and at a relative maximum.
d) y (4 x)5 1) y 5(4 x)4(1) 5(4 x)4 0 x 4 critical value 2) y 20(4 x)3 y
(4) 20(0)3 0 test inconclusive Moving on to the third- and higher-order derivatives, we get y y(4) y(5) 60(4 x)2 120(4 x) 120 y(4) y(4)(4) y(5)(4) 0 test inconclusive 0 test inconclusive 120 0 With the first nonzero higher-order derivative an odd-numbered derivative, y is at an inflection point.
OPTIMIZATION 4.7.
For the following functions, (1) find the critical values, (2) test for concavity to determine relative maxima or minima, (3) check for inflection points, and (4) evaluate the function at the critical values and inflection points.
68 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 Fig. 4-8 a) f(x) x3 18x2 96x 80 1) f(x) 3x2 36x 96 0 3(x 4)(x 8) 0 x 4 x 8 critical values 2) f
(x) 6x 36 f
(4) 6(4) 36 12 0 concave, relative maximum f
(8) 6(8) 36 12 0 convex, relative minimum 3) f 6x 36 0 x 6 With f
(6) 0 and concavity changing between x 4 and x 8, as seen in step 2, there is an inflection point at x 6.
4) f(4) (4)3 18(4)2 96(4) 80 80 (4, 80) relative maximum f(6) (6)3 18(6)2 96(6) 80 64 (6, 64) inflection point f(8) (8)3 18(8)2 96(8) 80 48 (8, 48) relative minimum b) f(x) x3 6x2 15x 32 1) f(x) 3x2 12x 15 0 3(x 1)(x 5) 0 x 1 x 5 critical values 2) f
(x) 6x 12 f
(1) 6(1) 12 18 0 convex, relative minimum f
(5) 6(5) 12 18 0 concave, relative maximum 3) f
(x) 6x 12 0 x 2 inflection point at x 2 4) f(1) 40 f(2) 14 f(5) 68 (1, 40) (2, 14) (5, 68) relative minimum inflection point relative maximum c) f(x) (2x 7)3 1) f(x) 3(2x 7)2(2) 6(2x 7)2 0 x 3.5 critical value 2) f
(x) 12(2x 7)(2) 24(2x 7) f
(3.5) 24[2(3.5) 7] 0 test inconclusive Continuing on to successively higher-order derivatives, we find f 48 f(3.5) 48 0 3) As explained in Section 4.6, with the first nonzero higher-order derivative an odd-numbered derivative, the function is at an inflection point at x 3.5. With an inflection point at the only critical value, there is no relative maximum or minimum.
4) f(3.5) 0 (3.5, 0) inflection point Testing for concavity to the left (x 3) and right (x 4) of x 3.5 gives f
(3) 24[2(3) 7] 24 0 concave f
(4) 24[2(4) 7] 24 0 convex 69 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] d) f(x) (x 2)4 1) f(x) 4(x 2)3 0 x 2 critical value 2) f
(x) 12(x 2)2 f
(2) 12(2 2)2 0 test inconclusive Continuing, as explained in Section 4.6, we get f(x) f(2) f (4)(x) f (4)(2) 24(x 2) 24(2 2) 0 test inconclusive 24 24 0 relative minimum With the first nonzero higher-order derivative even-numbered and greater than 0, f(x) is minimized at x 2.
3) There is no inflection point.
4) f(2) 0 (2, 0) relative minimum 4.8.
Optimize the following quadratic and cubic functions by (1) finding the critical value(s) at which the function is optimized and (2) testing the second-order condition to distinguish between a relative maximum or minimum.
a) y 7x2 112x 54 1) Take the first derivative, set it equal to zero, and solve for x to find the critical value(s).
y 14x 112 0 x 8 critical value 2) Take the second derivative, evaluate it at the critical value, and check the sign for a relative maximum and minimum.
y 14 y
(8) 14 0 convex, relative minimum Here, with y a constant greater than zero, y is strictly convex and so we can draw the further conclusion that y is at a global minimum at x 8.
b) y 9x2 72x 13 1) y 18x 72 0 x 4 critical value 2) y 18 y
(4) 18 0 concave, relative maximum Here, with y a constant less than zero, y is strictly concave and so we can also conclude that y is at a global maximum at x 4.
c) y x3 6x2 135x 4 1) y 3x2 12x 135 0 3(x2 4x 45) 0 3(x 5)(x 9) 0 x 5 x 9 critical values 2) y 6x 12 y
(5) 6(5) 12 42 0 concave, relative maximum y
(9) 6(9) 12 42 0 convex, relative minimum 70 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 d) y 2x3 15x2 84x 25 1) y 6x2 30x 84 0 6(x 2)(x 7) 0 x 2 x 7 critical values 2) y 12x 30 y
(2) 12(2) 30 54 0 convex, relative minimum y
(7) 12(7) 30 54 0 concave, relative maximum 4.9.
Optimize the following higher-order polynomial functions, using the same procedure as in Problem 4.8.
a) y x4 8x3 80x2 15 1) y 4x3 24x2 160x 0 4x(x 4)(x 10) 0 x 0 x 4 x 10 critical values 2) y 12x2 48x 160 y
(4) 12(4)2 48(4) 160 224 0 convex, relative minimum y
(0) 12(0)2 48(0) 160 160 0 concave, relative maximum y
(10) 12(10)2 48(10) 160 560 0 convex, relative minimum b) y 3x4 20x3 144x2 17 1) y 12x3 60x2 288x 0 12x(x 3)(x 8) 0 x 0 x 3 x 8 critical values 2) y 36x2 120x 288 y
(8) 36(8)2 120(8) 288 1056 0 concave, relative maximum y
(0) 36(0)2 120(0) 288 288 0 convex, relative minimum y
(3) 36(3)2 120(3) 288 396 0 concave, relative maximum c) y (x 13)4 1) y 4(x 13)3 0 x 13 0 x 13 critical value 2) y 12(x 13)2 y
(13) 12(13 13)2 0 test inconclusive Continuing as explained in Section 4.6 and Problem 4.6, we get y y(13) y(4) y(4)(13) 24(x 13) 24(0) 0 test inconclusive 24 24 0 concave, relative maximum d) y (9 4x)4 1) y 4(9 4x)3(4) 16(9 4x)3 0 9 4x 0 x 21 – 4 critical value 71 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] 2) y y
(21 – 4) y y(21 – 4) y(4) y(4)(21 – 4) 48(9 4x)2(4) 192(9 4x)2 192(0)2 0 test inconclusive 384(9 4x)(4) 1536(9 4x) 1536(0) 0 test inconclusive 6144 6144 0 convex, relative minimum MARGINAL, A VERAGE, AND TOTAL CONCEPTS 4.10.
Find (1) the marginal and (2) the average functions for each of the following total functions.
Evaluate them at Q 3 and Q 5.
a) TC 3Q2 7Q 12 1) MC dTC dQ 6Q 7 At Q 3, MC 6(3) 7 25 At Q 5, MC 6(5) 7 37 2) AC TC Q 3Q 7 12 Q At Q 3, AC 3(3) 7 12 –– 3 20 At Q 5, AC 3(5) 7 12 –– 5 24.4 Note: When finding the average function, be sure to divide the constant term by Q.
b) Q2 13Q 78 1) d dQ 2Q 13 2) A Q Q 13 78 Q At Q 3, A 3 13 78 –– 3 16 At Q 5, A 5 13 78 –– 5 7.6 At Q 3, d dQ 2(3) 13 7 At Q 5, d dQ 2(5) 13 3 c) TR 12Q Q2 1) MR dTR dQ 12 2Q At Q 3, MR 12 2(3) 6 At Q 5, MR 12 2(5) 2 2) AR TR Q 12 Q At Q 3, AR 12 3 9 At Q 5, AR 12 5 7 d) TC 35 5Q 2Q2 2Q3 1) MC dTC dQ 5 4Q 6Q2 At Q 3, MC 5 4(3) 6(3)2 47 At Q 5, MC 5 4(5) 6(5)2 135 2) AC TC Q 35 Q 5 2Q 2Q2 At Q 3, AC 35 –– 3 5 2(3) 2(3)2 28.67 At Q 5, AC 35 –– 5 5 2(5) 2(5)2 52 4.11.
Find the marginal expenditure (ME) functions associated with each of the following supply functions. Evaluate them at Q 4 and Q 10.
a) P Q2 2Q 1 To find the ME function, given a simple supply function, find the total expenditure (TE) function and take its derivative with respect to Q.
TE PQ (Q2 2Q 1)Q Q3 2Q2 Q ME dTE dQ 3Q2 4Q 1 At Q 4, ME 3(4)2 4(4) 1 65. At Q 10, ME 3(10)2 4(10) 1 341.
72 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 b) P Q2 0.5Q 3 TE PQ (Q2 0.5Q 3)Q Q3 0.5Q2 3Q ME 3Q2 Q 3 At Q 4, ME 3(4)2 4 3 55. At Q 10, ME 3(10)2 10 3 313.
4.12.
Find the MR functions for each of the following demand functions and evaluate them at Q 4 and Q 10.
a) Q 36 2P b) 44 4P Q 0 P 18 0.5Q TR (18 0.5Q)Q 18Q 0.5Q2 P 11 0.25Q TR (11 0.25Q)Q 11Q 0.25Q2 MR dTR dQ 11 0.5Q MR dTR dQ 18 Q At Q 4, MR 18 4 14 At Q 10, MR 18 10 8 At Q 4, MR 11 0.5(4) 9 At Q 10, MR 11 0.5(10) 6 4.13.
For each of the following consumption functions, use the derivative to find the marginal propensity to consume MPC dC/dY.
a) C C0 bY b) C 1500 0.75Y MPC dC dY b MPC dC dY 0.75 4.14.
Given C 1200 0.8Yd, where Yd Y T and T 100, use the derivative to find the MPC.
When C f(Yd), make C f(Y) before taking the derivative. Thus, C 1200 0.8(Y 100) 1120 0.8Y MPC dC dY 0.8 Note that the introduction of a lump-sum tax into the income determination model does not affect the value of the MPC (or the multiplier).
4.15.
Given C 2000 0.9Yd, where Yd Y T and T 300 0.2Y, use the derivative to find the MPC.
C 2000 0.9(Y 300 0.2Y) 2000 0.9Y 270 0.18Y 1730 0.72Y MPC dC dY 0.72 The introduction of a proportional tax into the income determination model does affect the value of the MPC and hence the multiplier.
4.16.
Find the marginal cost functions for each of the following average cost functions.
a) AC 1.5Q 4 46 Q 73 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] Given the average cost function, the marginal cost function is determined by first finding the total cost function and then taking its derivative, as follows: TC (AC)Q 1.5Q 4 46 Q Q 1.5Q2 4Q 46 MC dTC dQ 3Q 4 b) AC 160 Q 5 3Q 2Q2 TC 160 Q 5 3Q 2Q2 Q 160 5Q 3Q2 2Q3 MC dTC dQ 5 6Q 6Q2 OPTIMIZING ECONOMIC FUNCTIONS 4.17.
Maximize the following total revenue TR and total profit functions by (1) finding the critical value(s), (2) testing the second-order conditions, and (3) calculating the maximum TR or .
a) TR 32Q Q2 1) TR 32 2Q 0 Q 16 critical value 2) TR 2 0 concave, relative maximum 3) TR 32(16) (16)2 256 Note that whenever the value of the second derivative is negative over the whole domain of the function, as in (2) above, we can also conclude that the function is strictly concave and at a global maximum.
b) Q2 11Q 24 1) 2Q 11 0 Q 5.5 critical value 2) 2 0 concave, relative maximum 3) (5.5)2 11(5.5) 24 6.25 c) 1 – 3Q3 5Q2 2000Q 326 1) Q2 10Q 2000 0 1(Q2 10Q 2000) 0 (Q 50)(Q 40) 0 Q 50 Q 40 critical values (4.1) (4.2) 2) 2Q 10
(40) 2(40) 10 90 0 concave, relative maximum
(50) 2(50) 10 90 0 convex, relative minimum Negative critical values will subsequently be ignored as having no economic significance.
3) 1 – 3(40)3 5(40)2 2000(40) 326 50,340.67 Note: In testing the second-order conditions, as in step 2, always take the second derivative from the original first derivative (4.1) before any negative number has been factored out. Taking the second derivative from the first derivative after a negative has been factored out, as in (4.2), will 74 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 reverse the second-order conditions and suggest that the function is maximized at Q 50 and minimized at Q 40. Test it yourself.
d) Q3 6Q2 1440Q 545 1) 3Q2 12Q 1440 0 3(Q 20)(Q 24) 0 Q 20 Q 24 critical values 2) 6Q 12
(20) 6(20) 12 132 0 concave, relative maximum 3) (20)3 6(20)2 1440(20) 545 17,855 4.18.
From each of the following total cost TC functions, find (1) the average cost AC function, (2) the critical value at which AC is minimized, and (3) the minimum average cost.
a) TC Q3 5Q2 60Q 1) AC TC Q Q3 5Q2 60Q Q Q2 5Q 60 2) AC 2Q 5 0 Q 2.5 AC 2 0 convex, relative minimum 3) AC(2.5) (2.5)2 5(2.5) 60 53.75 Note that whenever the value of the second derivative is positive over the whole domain of the function, as in (2) above, we can also conclude that the function is strictly convex and at a global minimum.
b) TC Q3 21Q2 500Q 1) AC Q3 21Q2 500Q Q Q2 21Q 500 2) AC 2Q 21 0 Q 10.5 AC 2 0 convex, relative minimum 3) AC (10.5)2 21(10.5) 500 389.75 4.19.
Given the following total revenue and total cost functions for different firms, maximize profit for the firms as follows: (1) Set up the profit function TR TC, (2) find the critical value(s) where is at a relative extremum and test the second-order condition, and (3) calculate the maximum profit.
a) TR 1400Q 6Q2 TC 1500 80Q 1) 1400Q 6Q2 (1500 80Q) 6Q2 1320Q 1500 2) 12Q 1320 0 Q 110 critical value 12 0 concave, relative maximum 3) 6(110)2 1320(110) 1500 71,000 75 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] b) TR 1400Q 7.5Q2 TC Q3 6Q2 140Q 750 1) 1400Q 7.5Q2 (Q3 6Q2 140Q 750) Q3 1.5Q2 1260Q 750 (4.3) 2) 3Q2 3Q 1260 0 3(Q2 Q 420) 0 3(Q 21)(Q 20) 0 Q 21 Q 20 critical values Take the second derivative directly from (4.3), as explained in Problem 4.17(c), and ignore all negative critical values.
6Q 3
(20) 6(20) 3 123 0 concave, relative maximum 3) (20)3 1.5(20)2 1260(20) 750 15,850 c) TR 4350Q 13Q2 TC Q3 5.5Q2 150Q 675 1) 4350Q 13Q2 (Q3 5.5Q2 150Q 675) Q3 7.5Q2 4200Q 675 2)
(35) 3Q2 15Q 4200 0 3(Q2 5Q 1400) 0 3(Q 40)(Q 35) 0 Q 40 Q 35 critical values 6Q 15 6(35) 15 225 0 concave, relative maximum 3) (35)3 7.5(35)2 4200(35) 675 94,262.50 d) TR 5900Q 10Q2 TC 2Q3 4Q2 140Q 845 1) 5900Q 10Q2 (2Q3 4Q2 140Q 845) 2Q3 6Q2 5760Q 845 2)
(30) 6Q2 12Q 5760 0 6(Q2 2Q 960) 0 6(Q 32)(Q 30) 0 Q 32 Q 30 critical values 12Q 12 12(30) 12 372 0 concave, relative maximum 3) 2(30)3 6(30)2 5760(30) 845 112,555 4.20.
Prove that marginal cost (MC) must equal marginal revenue (MR) at the profit-maximizing level of output.
TR TC To maximize , d/dQ must equal zero.
d dQ dTR dQ dTC dQ 0 dTR dQ dTC dQ MR MC Q.E.D.
76 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 4.21.
A producer has the possibility of discriminating between the domestic and foreign markets for a product where the demands, respectively, are Q1 21 0.1P1 (4.4) Q2 50 0.4P2 (4.5) Total cost 2000 10Q where Q Q1 Q2. What price will the producer charge in order to maximize profits (a) with discrimination between markets and (b) without discrimination? (c) Compare the profit differential between discrimination and nondiscrimination.
a) To maximize profits under price discrimination, the producer will set prices so that MC MR in each market. Thus, MC MR1 MR2. With TC 2000 10Q, MC dTC dQ 10 Hence MC will be the same at all levels of output. In the domestic market, Q1 21 0.1P1 P1 210 10Q1 Hence, TR1 (210 10Q1)Q1 210Q1 10Q1 2 and MR1 dTR1 dQ1 210 20Q1 When MR1 MC, 210 20Q1 10 Q1 10 When Q1 10, P1 210 10(10) 110 In the foreign market, Q2 50 0.4P2 Hence, P2 125 2.5Q2 TR2 (125 2.5Q2)Q2 125Q2 2.5Q2 2 Thus, MR2 dTR2 dQ2 125 5Q2 When MR2 MC, 125 5Q2 10 Q2 23 When Q2 23, P2 125 2.5(23) 67.5 The discriminating producer charges a lower price in the foreign market where the demand is relatively more elastic and a higher price (P1 110) in the domestic market where the demand is relatively less elastic.
b) If the producer does not discriminate, P1 P2 and the two demand functions (4.4) and (4.5) may simply be aggregated. Thus, Q Q1 Q2 21 0.1P 50 0.4P 71 0.5P Hence, P 142 2Q TR (142 2Q)Q 142Q 2Q2 and MR dTR dQ 142 4Q When MR MC, 142 4Q 10 Q 33 When Q 33, P 142 2(33) 76 When no discrimination takes place, the price falls somewhere between the relatively high price of the domestic market and the relatively low price of the foreign market. Notice, however, that the quantity sold remains the same: at P 76, Q1 13.4, Q2 19.6, and Q 33.
77 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] c) With discrimination, TR TR1 TR2 P1Q1 P2Q2 110(10) 67.5(23) 2652.50 TC 2000 10Q, where Q Q1 Q2.
TC 2000 10(10 23) 2330 Thus, TR TC 2652.50 2330 322.50 Without discrimination, TR PQ 76(33) 2508 TC 2330 since costs do not change with or without discrimination. Thus, 2508 2330 178. Profits are higher with discrimination (322.50) than without discrimination.
4.22.
Faced with two distinct demand functions Q1 24 0.2P1 Q2 10 0.05P2 where TC 35 40Q, what price will the firm charge (a) with discrimination and (b) without discrimination?
a) With Q1 24 0.2P1, P1 TR1 MR1 120 5Q1 (120 5Q1)Q1 120Q1 5Q2 1 120 10Q1 The firm will maximize profits where MC MR1 MR2 TC 35 40Q MC 40 When MC MR1, 40 120 10Q1 Q1 8 When Q1 8, P1 120 5(8) 80 In the second market, with Q2 10 0.05P2, P2 TR2 MR2 200 20Q2 (200 20Q2)Q2 200Q2 20Q2 2 200 40Q2 When MC MR2, 40 200 40Q2 Q2 4 When Q2 4, P2 200 20(4) 120 b) If the producer does not discriminate, P1 P2 P and the two demand functions can be combined, as follows: Q Q1 Q2 24 0.2P 10 0.05P 34 0.25P Thus, P TR MR 136 4Q (136 4Q)Q 136Q 4Q2 136 8Q At the profit-maximizing level, MC MR.
40 136 8Q Q 12 At Q 12, P 136 4(12) 88 78 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 4.23.
Use the MR MC method to (a) maximize profit and (b) check the second-order conditions, given TR 1400Q 7.5Q2 TC Q3 6Q2 140Q 750 a) MR TR 1400 15Q, MC TC 3Q2 12Q 140 Equate MR MC.
1400 15Q 3Q2 12Q 140 Solve for Q by moving everything to the right.
3Q2 3Q 1260 0 3(Q 21)(Q 20) 0 Q 21 Q 20 critical values b) TR 15 TC 6Q 12 Since TR TC and the objective is to maximize , be sure to subtract TC from TR
, or you will reverse the second-order conditions and select the wrong critical value.
(20) TR TC 15 6Q 12 6Q 3 6(20) 3 123 0 concave, relative maximum Compare these results with Problem 4.19(b).
THE MARGINAL RATE OF TECHNICAL SUBSTITUTION 4.24.
An isoquant depicts the different combinations of inputs K and L that can be used to produce a specific level of output Q. One such isoquant for the output level Q 2144 is 16K1/4L3/4 2144 (a) Use implicit differentiation from Section 3.9 to find the slope of the isoquant dK/dL which in economics is called the marginal rate of technical substitution (MRTS). (b) Evaluate the marginal rate of technical substitution at K 256, L 108.
a) Take the derivative of each term with respect to L and treat K as a function of L.
d dL (16K1/4 L3/4) d dL (2144) Use the product rule since K is being treated as a function of L.
16K1/4 · d dL (L3/4) L3/4 · d dL (16K1/4) d dL (2144) 16K1/4 · 3 4 L1/4 L3/4 · 16 · 1 4 K3/4 · dK dL 0 12K1/4L1/4 4K3/4L3/4 · dK dL 0 Solve algebraically for dK/dL.
dK dL 12K1/4L1/4 4K3/4L3/4 3K L b) At K 256 and L 108.
MRTS dK dL 3(256) 108 7.11 This means that if L is increased by 1 relatively small unit, K must decrease by 7.11 units in order to remain on the production isoquant where the production level is constant. See also Problem 6.51.
79 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] 4.25.
The equation for the production isoquant is 25K3/5L2/5 5400 (a) Find the MRTS and (b) evaluate it at K 243, L 181.
a) Treat K as a function of L and use the product rule to find dK/dL, which is the MRTS.
25K3/5 · 2 5 L3/5 L2/5 · 25 · 3 5 K2/5 · dK dL 0 10K3/5L3/5 15K2/5L2/5 · dK dL 0 Solve algebraically for dK/dL.
dK dL 10K 3/5L3/5 15K2/5L2/5 2K 3L MRTS b) At K 243 and K 181, MRTS dK dL 2(243) 3(181) 0.895 This means that if L is increased by 1 relatively small unit, K must decrease by 0.895 unit in order to remain on the production isoquant where the production level is constant. See Problem 6.52.
RELATIONSHIP BETWEEN FUNCTIONS AND GRAPHS 4.26.
Given the total cost function C Q3 18Q2 750Q, use your knowledge of calculus to help sketch a graph showing the relationship between total, average, and marginal costs.
a) Take the first and second derivatives of the total cost function C 3Q2 36Q 750 C 6Q 36 and check for (1) concavity and (2) inflection points.
1) For Q 6, C 0 concave For Q 6, C 0 convex 2) 6Q 36 0 Q 6 C(6) (6)3 18(6)2 750(6) 4068 With C(Q) changing from concave to convex at Q 6, (6, 4068) inflection point b) Find the average cost function AC and the relative extrema.
AC TC Q Q2 18Q 750 AC 2Q 18 0 Q 9 critical value AC 2 0 convex, relative minimum 80 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS [CHAP . 4 c) Do the same thing for the marginal cost function MC C MC MC 3Q2 36Q 750 6Q 36 0 Q 6 critical value 6 0 convex, relative minimum d) Sketch the graph as in Fig. 4-9, noting that (1) MC decreases when TC is concave and increasing at a decreasing rate, increases when TC is convex and increasing at an increasing rate, and is at a minimum when TC is at an inflection point and changing concavity; and (2) AC decreases over the whole region where MC AC, is at a minimum when MC AC, and increases when MC AC.
81 USES OF THE DERIVATIVE IN MATHEMATICS AND ECONOMICS CHAP . 4] Fig. 4-9 CHAPTER 5 Calculus of Multivariable Functions 5.1 FUNCTIONS OF SEVERAL VARIABLES AND PARTIAL DERIVATIVES Study of the derivative in Chapter 4 was limited to functions of a single independent variable such as y f(x). Many economic activities, however, involve functions of more than one independent variable. z f(x, y) is defined as a function of two independent variables if there exists one and only one value of z in the range of f for each ordered pair of real numbers (x, y) in the domain of f. By convention, z is the dependent variable; x and y are the independent variables.
To measure the effect of a change in a single independent variable (x or y) on the dependent variable (z) in a multivariable function, the partial derivative is needed. The partial derivative of z with respect to x measures the instantaneous rate of change of z with respect to x while y is held constant.
It is written z/x, f/x, fx(x, y), fx, or zx. The partial derivative of z with respect to y measures the rate of change of z with respect to y while x is held constant. It is written z/y, f/y, fy(x, y), fy, or zy.
Expressed mathematically, z x lim x→0 f(x x, y) f(x, y) x (5.1a) z y lim y→0 f(x, y y) f(x, y) y (5.1b) Partial differentiation with respect to one of the independent variables follows the same rules as ordinary differentiation while the other independent variables are treated as constant. See Examples 1 and 2 and Problems 5.1 and 5.23.
EXAMPLE 1.
The partial derivatives of a multivariable function such as z 3x2y3 are found as follows: a) When differentiating with respect to x, treat the y term as a constant by mentally bracketing it with the coefficient: z [3y3] · x2 82 Then take the derivative of the x term, holding the y term constant, z x zx [3y3] · d dx (x2) [3y3] · 2x Recalling that a multiplicative constant remains in the process of differentiation, simply multiply and rearrange terms to obtain z x zx 6xy3 b) When differentiating with respect to y, treat the x term as a constant by bracketing it with the coefficient; then take the derivative as was done above: z [3x2] · y3 z y zy [3x2] · d dy (y3) [3x2] · 3y2 9x2y2 EXAMPLE 2.
To find the partial derivatives for z 5x3 3x2y2 7y5: a) When differentiating with respect to x, mentally bracket all y terms to remember to treat them as constants: z 5x3 [3y2]x2 [7y5] Then take the derivative of each term, remembering that in differentiation multiplicative constants remain but additive constants drop out, because the derivative of a constant is zero.
z x d dx (5x3) [3y2] · d dx (x2) d dx [7y5] 15x2 [3y2] · 2x 0 15x2 6xy2 b) When differentiating with respect to y, block off all the x terms and differentiate as above.
z [5x3] [3x2]y2 7y5 z y d dy [5x3] [3x2] · d dy (y2) d dy (7y5) 0 [3x2] · 2y 35y4 6x2y 35y4 See Problem 5.1.
5.2 RULES OF PARTIAL DIFFERENTIATION Partial derivatives follow the same basic patterns as the rules of differentiation in Section 3.7. A few key rules are given below, illustrated in Examples 3 to 5, treated in Problems 5.2 to 5.5, and verified in Problem 5.23.
83 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] 5.2.1 Product Rule Given z g(x, y) · h(x, y), z x g(x, y) · h x h(x, y) · g x (5.2a) z y g(x, y) · h y h(x, y) · g y (5.2b) EXAMPLE 3.
Given z (3x 5)(2x 6y), by the product rule, z x (3x 5)(2) (2x 6y)(3) 12x 10 18y z y (3x 5)(6) (2x 6y)(0) 18x 30 5.2.2 Quotient Rule Given z g(x, y)/h(x, y) and h(x, y) 0, z x h(x, y) · g/x g(x, y) · h/x [h(x, y)]2 (5.3a) z y h(x, y) · g/y g(x, y) · h/y [h(x, y)]2 (5.3b) EXAMPLE 4.
Given z (6x 7y)/(5x 3y), by the quotient rule, z x (5x 3y)(6) (6x 7y)(5) (5x 3y)2 30x 18y 30x 35y (5x 3y)2 17y (5x 3y)2 z y (5x 3y)(7) (6x 7y)(3) (5x 3y)2 35x 21y 18x 21y (5x 3y)2 17x (5x 3y)2 5.2.3 Generalized Power Function Rule Given z [g(x, y)]n, z x n[g(x, y)]n1 · g x (5.4a) z y n[g(x, y)]n1 · g y (5.4b) EXAMPLE 5.
Given z (x3 7y2)4, by the generalized power function rule, z x 4(x3 7y2)3 · (3x2) 12x2(x3 7y2)3 z y 4(x3 7y2)3 · (14y) 56y(x3 7y2)3 84 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 5.3 SECOND-ORDER PARTIAL DERIVATIVES Given a function z f(x, y), the second-order (direct) partial derivative signifies that the function has been differentiated partially with respect to one of the independent variables twice while the other independent variable has been held constant: fxx (fx)x x z x 2z x2 fyy (fy)y y z y 2z y2 In effect, fxx measures the rate of change of the first-order partial derivative fx with respect to x while y is held constant. And fyy is exactly parallel. See Problems 5.6 and 5.8.
The cross (or mixed) partial derivatives fxy and fyx indicate that first the primitive function has been partially differentiated with respect to one independent variable and then that partial derivative has in turn been partially differentiated with respect to the other independent variable: fxy (fx)y y z x 2z y x fyx (fy)x x z y 2z x y In brief, a cross partial measures the rate of change of a first-order partial derivative with respect to the other independent variable. Notice how the order of independent variables changes in the different forms of notation. See Problems 5.7 and 5.9.
EXAMPLE 6.
The (a) first, (b) second, and (c) cross partial derivatives for z 7x3 9xy 2y5 are taken as shown below.
a) z x zx 21x2 9y z y zy 9x 10y4 b) 2z x2 zxx 42x 2z y2 zyy 40y3 c) 2z yx y z x y (21x2 9y) zxy 9 2z xy x z y x (9x 10y4) zyx 9 EXAMPLE 7.
The (a) first, (b) second, and (c) cross partial derivatives for z 3x2y3 are evaluated below at x 4, y 1.
a) zx zx(4, 1) 6xy3 6(4)(1)3 24 zy zy(4, 1) 9x2y2 9(4)2(1)2 144 b) zxx zxx(4, 1) 6y3 6(1)3 6 zyy zyy(4, 1) 18x2y 18(4)2(1) 288 c) zxy y (6xy3) 18xy2 zyx x (9x2y2) 18xy2 zxy(4, 1) 18(4)(1)2 72 zyx(4, 1) 18(4)(1)2 72 ByYoung’s theorem, if both cross partial derivatives are continuous, they will be identical. See Problems 5.7 to 5.9.
5.4 OPTIMIZATION OF MULTIVARIABLE FUNCTIONS For a multivariable function such as z f(x, y) to be at a relative minimum or maximum, three conditions must be met: 1.
The first-order partial derivatives must equal zero simultaneously. This indicates that at the 85 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] given point (a, b), called a critical point, the function is neither increasing nor decreasing with respect to the principal axes but is at a relative plateau.
2.
The second-order direct partial derivatives, when evaluated at the critical point (a, b), must both be negative for a relative maximum and positive for a relative minimum. This ensures that from a relative plateau at (a, b) the function is concave and moving downward in relation to the principal axes in the case of a maximum and convex and moving upward in relation to the principal axes in the case of a minimum.
3.
The product of the second-order direct partial derivatives evaluated at the critical point must exceed the product of the cross partial derivatives also evaluated at the critical point. This added condition is needed to preclude an inflection point or saddle point.
In sum, as seen in Fig. 5-1, when evaluated at a critical point (a, b), Relative maximum 1. fx, fy 0 2. fxx, fyy 0 3. fxx · fyy (fxy)2 Relative minimum 1. fx, fy 0 2. fxx,fyy 0 3. fxx · fyy (fxy)2 Note the following: 1) Since fxy fyx by Young’s theorem, fxy · fyx (fxy)2. Step 3 can also be written fxx · fyy (fxy)2 0.
2) If fxx · fyy (fxy)2, when fxx and fyy have the same signs, the function is at an inflection point; when fxx and fyy have different signs, the function is at a saddle point, as seen in Fig. 5-2, where the function is at a maximum when viewed from one axis but at a minimum when viewed from the other axis.
86 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 Fig. 5-1 Fig. 5-2 z z zx 0 zy 0 y x O x O zx 0 zy 0 y 3) If fxx · fyy (fxy)2, the test is inconclusive. See Example 8 and Problems 5.10 and 5.11; for inflection points, see Problems 5.10(c) and 5.11(b) and (c); for saddle points see Problems 5.10(d) and 5.11(a) and (d).
4) If the function is strictly concave (convex) in x and y, as in Fig. 5-1, there will be only one maximum (minimum), called an absolute or global maximum (minimum). If the function is simply concave (convex) in x and y on an interval, the critical point is a relative or local maximum (minimum).
EXAMPLE 8.
(a) Find the critical points. (b) Test whether the function is at a relative maximum or minimum, given z 2y3 x3 147x 54y 12 a) Take the first-order partial derivatives, set them equal to zero, and solve for x and y: zx 3x2 147 x2 x 0 49 7 zy 6y2 54 y2 y 0 9 3 (5.5) With x 7, y 3, there are four distinct sets of critical points: (7, 3), (7, 3), (7, 3), and (7, 3).
b) Take the second-order direct partials from (5.5), evaluate them at each of the critical points, and check the signs: 1) 2) 3) 4) zxx zxx(7, 3) zxx(7, 3) zxx(7, 3) zxx(7, 3) 6x 6(7) 42 0 6(7) 42 0 6(7) 42 0 6(7) 42 0 zyy zyy(7, 3) zyy(7, 3) zyy(7, 3) zyy(7, 3) 12y 12(3) 36 0 12(3) 36 0 12(3) 36 0 12(3) 36 0 Since there are different signs for each of the second direct partials in (1) and (4), the function cannot be at a relative maximum or minimum at (7, 3) or (7, 3). When fxx and fyy are of different signs, fxx · fyy cannot be greater than (fxy)2, and the function is at a saddle point.
With both signs of the second direct partials negative in (2) and positive in (3), the function may be at a relative maximum at (7, 3) and at a relative minimum at (7, 3), but the third condition must be tested first to ensure against the possibility of an inflection point.
c) From (5.5). take the cross partial derivatives and check to make sure that zxx(a, b) · zyy(a, b) [zxy(a, b)]2.
zxy 0 zyx 0 zxx(a, b) · zyy(a, b) [zxy(a, b)]2 From (2), From (3), (42) · (36) 1512 (42) · (36) 1512 (0)2 0 (0)2 0 The function is maximized at (7, 3) and minimized at (7, 3); for inflection points, see Problems 5.10(c) and 5.11(b) and (c).
5.5 CONSTRAINED OPTIMIZATION WITH LAGRANGE MULTIPLIERS Differential calculus is also used to maximize or minimize a function subject to constraint. Given a function f(x,y) subject to a constraint g(x, y) k (a constant), a new function F can be formed by 87 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] (1) setting the constraint equal to zero, (2) multiplying it by (the Lagrange multiplier), and (3) adding the product to the original function: F(x, y, ) f(x, y) [k g(x, y)] (5.6) Here F(x, y, ) is the Lagrangian function, f(x, y) is the original or objective function, and g(x, y) is the constraint. Since the constraint is always set equal to zero, the product [k g(x, y)] also equals zero, and the addition of the term does not change the value of the objective function. Critical values x0, y0, and 0, at which the function is optimized, are found by taking the partial derivatives of F with respect to all three independent variables, setting them equal to zero, and solving simultaneously: Fx(x,y, ) 0 Fy(x, y, ) 0 F(x, y, ) 0 Second-order conditions differ from those of unconstrained optimization and are treated in Section 12.5. See Example 9; Problems 5.12 to 5.14; Sections 6.6, 6.9, and 6.10; and Problems 6.28 to 6.39 and 6.41 to 6.44.
For constraints involving inequalities, see Chapter 13 for concave programming.
EXAMPLE 9.
Optimize the function z 4x2 3xy 6y2 subject to the constraint x y 56.
1.
Set the constraint equal to zero by subtracting the variables from the constant as in (5.6), for reasons to be explained in Section 5.6.
56 x y 0 Multiply this difference by and add the product of the two to the objective function in order to form the Lagrangian function Z.
Z 4x2 3xy 6y2 (56 x y) (5.7) 2.
Take the first-order partials, set them equal to zero, and solve simultaneously.
Zx 8x 3y 0 (5.8) Zy 3x 12y 0 (5.9) 56 x y 0 (5.10) Subtracting (5.9) from (5.8) to eliminate gives 5x 9y 0 x 1.8y Substitute x 1.8y in (5.10), 56 1.8y y 0 y0 20 From which we find x0 36 0 348 Substitute the critical values in (5.7), Z 4(36)2 3(36)(20) 6(20)2 (348)(56 36 20) 4(1296) 3(720) 6(400) 348(0) 9744 In Chapter 12, Example 5, it will be shown that Z is at a minimum. Notice that at the critical values, the Lagrangian function Z equals the objective function z because the constraint equals zero. See Problems 5.12 to 5.14 and Sections 6.6, 6.9, and 6.10.
88 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 5.6 SIGNIFICANCE OF THE LAGRANGE MULTIPLIER The Lagrange multiplier approximates the marginal impact on the objective function caused by a small change in the constant of the constraint. With 348 in Example 9, for instance, a 1-unit increase (decrease) in the constant of the constraint would cause Z to increase (decrease) by approximately 348 units, as is demonstrated in Example 10. Lagrange multipliers are often referred to as shadow prices. In utility maximization subject to a budget constraint, for example, will estimate the marginal utility of an extra dollar of income. See Problem 6.36.
Note: Since in (5.6) above [k g(x, y)] [g(x, y) k] 0, either form can be added to or subtracted from the objective function without changing the critical values of x and y. Only the sign of will be affected. For the interpretation of given in Section 5.6 to be valid, however, the precise form used in Equation (5.6) should be adhered to. See Problems 5.12 to 5.14.
EXAMPLE 10.
To verify that a 1-unit change in the constant of the constraint will cause a change of approximately 348 units in Z from Example 9, take the original objective function z 4x2 3xy 6y2 and optimize it subject to a new constraint x y 57 in which the constant of the constraint is 1 unit larger.
Z Zx Zy Z 4x2 3xy 6y2 (57 x y) 8x 3y 0 3x 12y 0 57 x y 0 When solved simultaneously this gives x0 36.64 y0 20.36 0 354.2 Substituting these values in the Lagrangian function gives Z 10,095 which is 351 larger than the old constrained optimum of 9744, close to the approximation of the 348 increment suggested by .
5.7 DIFFERENTIALS In Section 3.4 the derivative dy/dx was presented as a single symbol denoting the limit of y/ x as x approaches zero. The derivative dy/dx may also be treated as a ratio of differentials in which dy is the differential of y and dx the differential of x. Given a function of a single independent variable y f(x), the differential of y, dy, measures the change in y resulting from a small change in x, written dx.
Given y 2x2 5x 4, the differential of y is found by first taking the derivative of y with respect to x, which measures the rate at which y changes for a small change in x, dy dx 4x 5 a derivative or rate of change and then multiplying that rate at which y changes for a small change in x by a specific change in x(dx) to find the resulting change in y(dy).
dy (4x 5)dx a differential or simple change Change in y rate at which y changes for a small change in x · a small change in x.
EXAMPLE 11.
1.
If y 4x3 5x2 7, then dy/dx 12x2 10x and the differential is dy (12x2 10x) dx 2.
If y (2x 5)2, then dy/dx 2(2x 5)(2) 8x 20 and the differential is dy (8x 20) dx See Problem 5.15.
89 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] 5.8 TOTAL AND PARTIAL DIFFERENTIALS For a function of two or more independent variables, the total differential measures the change in the dependent variable brought about by a small change in each of the independent variables. If z f(x, y), the total differential dz is expressed mathematically as dz zx dx zy dy (5.11) where zx and zy are the partial derivatives of z with respect to x and y respectively, and dx and dy are small changes in x and y. The total differential can thus be found by taking the partial derivatives of the function with respect to each independent variable and substituting these values in the formula above.
EXAMPLE 12.
The total differential is found as follows: 1.
Given: z x4 8xy 3y3 zx 4x3 8y zy 8x 9y2 which, when substituted in the total differential formula, gives dz (4x3 8y) dx (8x 9y2) dy 2.
Given: z (x y)/(x 1) zx (x 1)(1) (x y)(1) (x 1)2 y 1 (x 1)2 zy (x 1)(1) (x y)(0) (x 1)2 1(x 1) (x 1)2 1 x 1 The total differential is dz y 1 (x 1)2 dx 1 x 1 dy If one of the independent variables is held constant, for example, dy 0, we then have a partial differential: dz zx dx A partial differential measures the change in the dependent variable of a multivariate function resulting from a small change in one of the independent variables and assumes the other independent variables are constant. See Problems 5.16 and 5.17 and 6.45 to 6.52.
5.9 TOTAL DERIVATIVES Given a case where z f(x, y) and y g(x), that is, when x and y are not independent, a change in x will affect z directly through the function f and indirectly through the function g. This is illustrated in the channel map in Fig. 5-3. To measure the effect of a change in x on z when x and y are not independent, the total derivative must be found. The total derivative measures the direct effect of x 90 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 Fig. 5-3 on z, z/x, plus the indirect effect of x on z through y, z y dy dx. In brief, the total derivative is dz dx zx zy dy dx (5.12) See Examples 13 to 15 and Problems 5.18 and 5.19.
EXAMPLE 13.
An alternative method of finding the total derivative is to take the total differential of z dz zx dx zy dy and divide through mentally by dx. Thus, dz dx zx dx dx zy dy dx Since dx/dx 1, dz dx zx zy dy dx EXAMPLE 14.
Given z f(x, y) 6x3 7y where y g(x) 4x2 3x 8, the total derivative dz/dx with respect to x is dz dx zx zy dy dx where zx 18x2, zy 7, and dy/dx 8x 3. Substituting above, dz dx 18x2 7(8x 3) 18x2 56x 21 To check the answer, substitute y 4x2 3x 8 in the original function to make z a function of x alone and then take the derivative as follows: z 6x3 7(4x2 3x 8) 6x3 28x2 21x 56 Thus, dz dx 18x2 56x 21 EXAMPLE 15.
The total derivative can be expanded to accommodate other interconnections as well. Given z 8x2 3y2 x 4t y 5t the total derivative of z with respect to t then becomes dz dt zx dx dt zy dy dt where zx 16x, zy 6y, dx/dt 4, and dy/dt 5. Substituting above, dz dt 16x(4) 6y(5) 64x 30y Then substituting x 4t and y 5t immediately above, dz dt 64(4t) 30(5t) 406t 91 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] 5.10 IMPLICIT AND INVERSE FUNCTION RULES As seen in Section 3.9, functions of the form y f(x) express y explicitly in terms of x and are called explicit functions. Functions of the form f(x, y) 0 do not express y explicitly in terms of x and are called implicit functions. If an implicit function f(x, y) 0 exists and fy 0 at the point around which the implicit function is defined, the total differential is simply fx dx fy dy 0.
Recalling that a derivative is a ratio of differentials, we can then rearrange the terms to get the implicit function rule: dy dx fx fy (5.13) Notice that the derivative dy/dx is the negative of the reciprocal of the corresponding partials.
dy dx fx fy 1 fy/fx Given a function y f(x), an inverse function x f 1(y) exists if each value of y yields one and only one value of x. Assuming the inverse function exists, the inverse function rule states that the derivative of the inverse function is the reciprocal of the derivative of the original function. Thus, if Q f(P) is the original function, the derivative of the original function is dQ/dP, the derivative of the inverse function [P f 1(Q)] is dP/dQ, and dP dQ 1 dQ/dP provided dQ dP 0 (5.14) See Examples 16 and 17 and Problems 5.20 to 5.22, 6.51, and 6.52.
EXAMPLE 16.
Given the implicit functions: (a) 7x2 y 0 (b) 3x4 7y5 86 0 the derivative dy/dx is found as follows: a) From (5.13), dy dx fx fy Here fx 14x and fy 1. Substituting above, dy dx 14x (1) 14x The function in this case was deliberately kept simple so the answer could easily be checked by solving for y in terms of x and then taking the derivative directly. Since y 7x2, dy/dx 14x.
b) dy dx fx fy 12x3 35y4 12x3 35y4 Compare this answer with that in Example 16 of Chapter 3.
EXAMPLE 17.
Find the derivative for the inverse of the following functions: 1.
Given Q 20 2P, dP dQ 1 dQ/dP where dQ/dP 2. Thus, dP dQ 1 2 1 2 92 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 2.
Given Q 25 3P3, dP dQ 1 dQ/dP 1 9P2 (P 0) Solved Problems FIRST-ORDER PARTIAL DERIVATIVES 5.1.
Find the first-order partial derivatives for each of the following functions: a) z 8x2 14xy 5y2 b) z 4x3 2x2y 7y5 zx zy 16x 14y 14x 10y zx zy 12x2 4xy 2x2 35y4 c) z 6w3 4wx 3x2 7xy 8y2 d) z 2w2 8wxy x2 y3 zw zx zy 18w2 4x 4w 6x 7y 7x 16y zw zx zy 4w 8xy 8wy 2x 8wx 3y2 5.2.
Use the product rule from (5.2) to find the first-order partials for each of the following functions: a) z 3x2(5x 7y) zx and zy 3x2(5) (5x 7y)(6x) 45x2 42xy 3x2(7) (5x 7y)(0) 21x2 b) z (9x 4y)(12x 2y) zx and zy (9x 4y)(12) (12x 2y)(9) 108x 48y 108x 18y 216x 30y (9x 4y)(2) (12x 2y)(4) 18x 8y 48x 8y 30x 16y c) z (2x2 6y)(5x 3y3) zx and zy (2x2 6y)(5) (5x 3y3)(4x) 10x2 30y 20x2 12xy3 30x2 30y 12xy3 (2x2 6y)(9y2) (5x 3y3)(6) 18x2y2 54y3 30x 18y3 72y3 18x2y2 30x d) z (w x y)(3w 2x 4y) zw zx and zy (w x y)(3) (3w 2x 4y)(1) 6w x 7y (w x y)(2) (3w 2x 4y)(1) w 4x 2y (w x y)(4) (3w 2x 4y)(1) 7w 2x 8y 93 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] 5.3.
Use the quotient rule from (5.3) to find the first-order partials of the following functions: a) z 5x 6x 7y b) z x y 3y zx (6x 7y)(5) (5x)(6) (6x 7y)2 zx 3y(1) (x y)(0) (3y)2 35y (6x 7y)2 1 3y and zy (6x 7y)(0) (5x)(7) (6x 7y)2 and zy 3y(1) (x y)(3) (3y)2 35x (6x 7y)2 3x (3y)2 x 3y2 c) z 4x 9y 5x 2y d) z x2 y2 3x 2y zx (5x 2y)(4) (4x 9y)(5) (5x 2y)2 zx (3x 2y)(2x) (x2 y2)(3) (3x 2y)2 53y (5x 2y)2 3x2 4xy 3y2 (3x 2y)2 and zy (5x 2y)(9) (4x 9y)(2) (5x 2y)2 and zy (3x 2y)(2y) (x2 y2)(2) (3x 2y)2 53x (5x 2y)2 2x2 6xy 2y2 (3x 2y)2 5.4.
Find the first-order partial derivatives for each of the following functions by using the generalized power function rule from (5.4): a) z (x y)2 b) z (2x 5y)3 zx and zy 2(x y)(1) 2(x y) 2(x y)(1) 2(x y) zx and zy 3(2x 5y)2(2) 6(2x 5y)2 3(2x 5y)2(5) 15(2x 5y)2 c) z (7x2 4y3)5 d) z (5w 4x 7y)3 zx and zy 5(7x2 4y3)4(14x) 70x(7x2 4y3)4 5(7x2 4y3)4(12y2) 60y2(7x2 4y3)4 zw zx and zy 3(5w 4x 7y)2(5) 15(5w 4x 7y)2 3(5w 4x 7y)2(4) 12(5w 4x 7y)2 3(5w 4x 7y)2(7) 21(5w 4x 7y)2 5.5.
Use whatever combination of rules is necessary to find the first-order partials for the following functions: a) z (5x2 7y)(3x2 8y) 4x 2y 94 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 Using the quotient rule and product rule, zx (4x 2y)[(5x2 7y)(6x) (3x2 8y)(10x)] (5x2 7y)(3x2 8y)(4) (4x 2y)2 (4x 2y)(30x3 42xy 30x3 80xy) (5x2 7y)(12x2 32y) (4x 2y)2 (4x 2y)(60x3 38xy) (5x2 7y)(12x2 32y) (4x 2y)2 and zy (4x 2y)[(5x2 7y)(8) (3x2 8y)(7)] (5x2 7y)(3x2 8y)(2) (4x 2y)2 (4x 2y)(40x2 56y 21x2 56y) (5x2 7y)(6x2 16y) (4x 2y)2 (4x 2y)(19x2 112y) (5x2 7y)(6x2 16y) (4x 2y)2 b) z (5x2 4y)2(2x 7y3) Using the product rule and the generalized power function rule, zx and zy (5x2 4y)2(2) (2x 7y3)[2(5x2 4y)(10x)] 2(5x2 4y)2 (2x 7y3)(100x3 80xy) (5x2 4y)2(21y2) (2x 7y3)[2(5x2 4y)(4)] 21y2(5x2 4y)2 (2x 7y3)(40x2 32y) c) z (3x 11y)3 2x 6y Using the quotient rule and the generalized power function rule, zx (2x 6y)[3(3x 11y)2(3)] (3x 11y)3(2) (2x 6y)2 (18x 54y)(3x 11y)2 2(3x 11y)3 (2x 6y)2 and zy (2x 6y)[3(3x 11y)2(11)] (3x 11y)3(6) (2x 6y)2 (66x 198y)(3x 11y)2 6(3x 11y)3 (2x 6y)2 d) z 8x 7y 5x 2y 2 Using the generalized power function rule and the quotient rule, zx 2 8x 7y 5x 2y (5x 2y)(8) (8x 7y)(5) (5x 2y)2 16x 14y 5x 2y 19y (5x 2y)2 (266y2 304xy) (5x 2y)3 and zy 2 8x 7y 5x 2y (5x 2y)(7) (8x 7y)(2) (5x 2y)2 16x 14y 5x 2y 19x (5x 2y)2 304x2 266xy (5x 2y)3 95 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] SECOND-ORDER PARTIAL DERIVATIVES 5.6.
Find the second-order direct partial derivatives zxx and zyy for each of the following functions: a) z x2 2xy y2 zx zxx 2x 2y 2 zy 2x 2y zyy 2 b) z x3 9xy 3y3 zx zxx 3x2 9y 6x zy 9x 9y2 zyy 18y c) z 2xy4 7x3y zx zxx 2y4 21x2y 42xy zy 8xy3 7x3 zyy 24xy2 d) z x4 x3y2 3xy3 2y3 zx zxx 4x3 3x2y2 3y3 12x2 6xy2 zy 2x3y 9xy2 6y2 zyy 2x3 18xy 12y e) z (12x 7y)2 zx zxx 2(12x 7y)(12) 288x 168y 288 zy 2(12x 7y)(7) 168x 98y zyy 98 f) z (7x 3y)3 zx zxx 3(7x 3y)2(7) 21(7x 3y)2 42(7x 3y)(7) 2058x 882y zy zyy 3(7x 3y)2(3) 9(7x 3y)2 18(7x 3y)(3) 378x 162y g) z (x2 2y)4 zx zxx 4(x2 2y)3(2x) 8x(x2 2y)3 8x[3(x2 2y)2(2x)] (x2 2y)3(8) 48x2(x2 2y)2 8(x2 2y)3 zy 4(x2 2y)3(2) 8(x2 2y)3 zyy 24(x2 2y)2(2) 48(x2 2y)2 5.7.
Find the cross partial derivatives zxy and zyx for each of the following functions: a) z 3x2 12xy 5y2 zx zxy 6x 12y 12 zy 12x 10y zyx 12 b) z x3 xy 2y3 zx zxy 3x2 y 1 zy x 6y2 zyx 1 c) z 8x2y 11xy3 zx zxy 16xy 11y3 16x 33y2 zy 8x2 33xy2 zyx 16x 33y2 By the product rule.
96 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 d) z (8x 4y)5 zx zxy 5(8x 4y)4(8) 40(8x 4y)4 160(8x 4y)3(4) 640(8x 4y)3 zy zyx 5(8x 4y)4(4) 20(8x 4y)4 80(8x 4y)3(8) 640(8x 4y)3 In items (a) through (d) above, notice how, in accord with Young’s theorem, zxy zyx no matter which first partial is taken initially.
5.8.
Find the first-order and second-order direct partial derivatives for the following functions: a) z x0.4y0.6 zx zxx 0.4x0.6y0.6 0.24x1.6y0.6 zy zyy 0.6x0.4y0.4 0.24x0.4y1.4 b) f(x, y) x0.7y0.2 fx fxx 0.7x0.3y0.2 0.21x1.3y0.2 fy fyy 0.2x0.7y0.8 0.16x0.7y1.8 c) z 2w6x5y3 zw zww 12w5x5y3 60w4x5y3 zx zxx 10w6x4y3 40w6x3y3 zy zyy 6w6x5y2 12w6x5y d) f(x, y, z) 10x3y2z4 fx fxx 30x2y2z4 60xy2z4 fy fyy 20x3yz4 20x3z4 fz fzz 40x3y2z3 120x3y2z2 5.9.
Find the cross partials for each of the following functions: a) z x0.3y0.5 zx zxy 0.3x0.7y0.5 0.15x0.7y0.5 zy zyx 0.5x0.3y0.5 0.15x0.7y0.5 b) f(x, y) x0.1y0.8 fx fxy 0.1x0.9y0.8 0.08x0.9y0.2 fy fyx 0.8x0.1y0.2 0.08x0.9y0.2 c) z w3x4y3 zw zwx zwy 3w2x4y3 12w2x3y3 9w2x4y2 zx zxw zxy 4w3x3y3 12w2x3y3 12w3x3y2 zy zyw zyx 3w3x4y2 9w2x4y2 12w3x3y2 d) f(x, y, z) x3y4z5 fx fxy fxz 3x2y4z5 12x2y5z5 15x2y4z6 fy fyx fyz 4x3y5z5 12x2y5z5 20x3y5z6 fz fzx fzy 5x3y4z6 15x2y4z6 20x3y5z6 Note how by Young’s theorem in (c) zwx zxw, zyw zwy, and zxy zyx and in (d) fxy fyx, fxz fzx, and fyz fzy.
97 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] OPTIMIZING MULTIVARIABLE FUNCTIONS 5.10.
For each of the following quadratic functions, (1) find the critical points at which the function may be optimized and (2) determine whether at these points the function is maximized, is minimized, is at an inflection point, or is at a saddle point.
a) z 3x2 xy 2y2 4x 7y 12 1) Take the first-order partial derivatives, set them equal to zero, and solve simultaneously, using the methods of Section 1.4.
zx zy z 1 6x y 4 0 x 4y 7 0 y 2 (1, 2) critical point (5.15) (5.16) 2) Take the second-order direct partial derivatives from (5.15) and (5.16), evaluate them at the critical point, and check signs.
zxx 6 zyy 4 zxx(1, 2) 6 0 zyy(1, 2) 4 0 With both second-order direct partial derivatives everywhere positive, the function is possibly at a global minimum. Now take the cross partial from (5.15) or (5.16), zxy 1 zyx evaluate it at the critical point and test the third condition: zxy(1, 2) 1 zyx(1, 2) zxx(1, 2) · zyy(1, 2) [zxy(1, 2)]2 6 · 4 (1)2 With zxxzyy (zxy)2 and zxx, zyy 0, the function is at a global minimum at (1, 2).
b) f(x, y) 60x 34y 4xy 6x2 3y2 5 1) Take the first-order partials, set them equal to zero, and solve.
fx fy x 4 y 60 4y 12x 0 34 6y 4x 0 3 (4, 3) critical point (5.17) (5.18) 2) Take the second-order direct partials, evaluate them at the critical point, and check their signs.
fxx fxx(4, 3) 12 12 0 fyy fyy(4, 3) 6 6 0 Take the cross partial from (5.17) or (5.18), fxy 4 fyx evaluate it at the critical point and test the third condition: fxy(4, 3) fxx(4, 3) · fyy(4, 3) 12 · 6 4 fyx(4, 3) [ fxy(4, 3)]2 (4)2 With fxx fyy (fxy)2 and fxx, fyy 0, the function is at a global maximum at (4, 3).
98 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 c) z 48y 3x2 6xy 2y2 72x 1) zx 6x 6y 72 0 zy 6x 4y 48 0 x 0 y 12 (0, 12) critical point 2) Test the second-order direct partials at the critical point.
zxx zxx(0, 12) 6 6 0 zyy zyy(0, 12) 4 4 0 With zxx and zyy 0 for all values, the function may be at a global maximum. Test the cross partials to be sure.
zxy zxy(0, 12) zxx(0, 12) · zyy(0, 12) 6 · 4 6 zyx 6 zyx(0, 12) [zxy(0, 12)]2 (6)2 With zxx and zyy of the same sign and zxxzyy (zxy)2, the function is at an inflection point at (0, 12).
d) f(x, y) 5x2 3y2 30x 7y 4xy 1) fx fy 10x 4y 30 0 4x 6y 7 0 x 2 y 2.5 (2, 2.5) critical point 2) fxx fxx(2, 2.5) 10 10 0 fyy fyy(2, 2.5) 6 6 0 Testing the cross partials, fxy 4 fyx fxx(2, 2.5) · fyy(2, 2.5) [fxy(2, 2.5)]2 10 · 6 42 Whenever fxx and fyy are of different signs, fxx fyy cannot be greater than (fxy)2, and the function will be at a saddle point.
5.11.
For the following cubic functions, (1) find the critical points and (2) determine if at these points the function is at a relative maximum, relative minimum, inflection point, or saddle point.
a) z(x, y) 3x3 5y2 225x 70y 23 1) Take the first-order partials and set them equal to zero.
zx 9x2 225 0 (5.19) zy 10y 70 0 (5.20) Solve for the critical points.
9x2 x2 x 225 25 5 10y y 70 7 (5, 7) (5, 7) critical points 99 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] 2) From (5.19) and (5.20), take the second-order direct partials, zxx 18x zyy 10 evaluate them at the critical points and note the signs.
zxx(5, 7) zxx(5, 7) 18(5) 90 0 18(5) 90 0 zyy(5, 7) zyy(5, 7) 10 0 10 0 Then take the cross partial from (5.19) or (5.20), zxy 0 zyx evaluate it at the critical points and test the third condition.
At (5, 7), At (5, 7), zxx(a, b) 90 90 · · · zyy(a, b) 10 10 [zxy(a, b)]2 0 0 With zxxzyy (zxy)2 and zxx, zyy 0 at (5, 7), z(5, 7) is a relative maximum. With zxxzyy (zxy)2 and zxx and zyy of different signs at (5, 7), z(5, 7) is a saddle point.
b) f(x,y) 3x3 1.5y2 18xy 17 1) Set the first-order partial derivatives equal to zero, fx 9x2 18y 0 (5.21) fy 3y 18x 0 (5.22) and solve for the critical values: 18y y 9x2 1 – 2x2 3y y 18x 6x (5.23) Setting y equal to y, 1 – 2x2 6x x2 12x x(x 12) 0 0 x 0 x 12 Substituting x 0 and x 12 in y 6x from (5.23), y 6(0) 0 y 6(12) 72 Therefore, (0, 0) (12, 72) critical points 2) Take the second-order direct partials from (5.21) and (5.22), fxx 18x fyy 3 evaluate them at the critical points and note the signs.
fxx(0, 0) fxx(12, 72) 18(0) 0 18(12) 216 0 fyy(0, 0) 3 0 fyy(12, 72) 3 0 Then take the cross partial from (5.21) or (5.22), fxy 18 fyx evaluate it at the critical points and test the third conditon.
At (0, 0), At (12, 72), fxx(a, b) 0 216 · · · fyy(a, b) 3 3 648 [fxy(a, b)]2 (18)2 (18)2 324 100 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 With fxx fyy (fxy)2 and fxx, fyy 0 at (12, 72), f(12, 72) is a relative minimum. With fxx fyy (fxy)2 and fxx and fyy of the same sign at (0, 0), f(0, 0) is an inflection point.
c) f 3x3 9xy 3y3 1) fx 9x2 9y 0 (5.24) fy 9y2 9x 0 (5.25) From (5.24), 9y 9x2 y x2 Substitute y x2 in (5.25), 9(x2)2 9x 0 9x4 9x 0 9x(x3 1) 0 9x 0 x 0 or x3 1 0 x3 1 x 1 Substituting these values in (5.24), we find that if x 0, y 0, and if x 1, y 1.
Therefore, (0, 0) (1, 1) critical points 2) Test the second-order conditons from (5.24) and (5.25).
fxx fxx(0, 0) fxx(1, 1) 18x 18(0) 0 18(1) 18 0 fyy fyy(0, 0) fyy(1, 1) 18y 18(0) 0 18(1) 18 0 fxy 9 fyx At (0, 0), At (1, 1), fxx(a, b) 0 18 · · · fyy(a,b) 0 18 [fxy(a,b)]2 (9)2 (9)2 With fxx and fyy 0 and fxx fyy (fxy)2 at (1, 1), the function is at a relative minimum at (1, 1).
With fxx and fyy of the same sign at (0, 0) and fxx fyy (fxy)2, the function is at an inflection point at (0, 0).
d) f(x, y) x3 6x2 2y3 9y2 63x 60y 1) fx 3x2 12x 63 0 3(x2 4x 21) 0 (x 3)(x 7) 0 x 3 x 7 fy 6y2 18y 60 0 6(y2 3y 10) 0 (y 2)(y 5) 0 y 2 y 5 (5.26) Hence (3, 2) (3, 5) (7, 2) (7, 5) critical points 2) Test the second-order direct partials at each of the critical points. From (5.26), (i) (ii) (iii) (iv) fxx fxx(3, 2) fxx(3, 5) fxx(7, 2) fxx(7, 5) 6x 12 30 0 30 0 30 0 30 0 fyy fyy(3, 2) fyy(3, 5) fyy(7, 2) fyy(7, 5) 12y 18 42 0 42 0 42 0 42 0 101 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] With different signs in (i) and (iv), (3, 2) and (7, 5) can be ignored, if desired, as saddle points. Now take the cross partial from (5.26) and test the third condition.
fxy 0 fyx From (ii), From (iii), fxx(a, b) · (30) · (30) · fyy(a, b) (42) (42) [fxy(a, b)]2 (0)2 (0)2 The function is at a relative maximum at (3, 5), at a relative minimum at (7, 2), and at a saddle point at (3, 2) and (7, 5).
CONSTRAINED OPTIMIZATION AND LAGRANGE MULTIPLIERS 5.12.
(1) Use Lagrange multipliers to optimize the following functions subject to the given constraint, and (2) estimate the effect on the value of the objective function from a 1-unit change in the constant of the constraint.
a) z 4x2 2xy 6y2 subject to x y 72 1) Set the constraint equal to zero, multiply it by , and add it to the objective function, to obtain Z 4x2 2xy 6y2 (72 x y) The first-order conditions are Zx 8x 2y 0 (5.27) Zy 2x 12y 0 (5.28) Z 72 x y 0 (5.29) Subtract (5.28) from (5.27) to eliminate .
10x 14y 0 x 1.4y Substitute x 1.4y in (5.29) and rearrange.
1.4y y 72 y0 30 Substitute y0 30 in the previous equations to find that at the critical point x0 42 y0 30 0 276 Thus, Z 4(42)2 2(42)(30) 6(30)2 276(72 42 30) 9936.
2) With 276, a 1-unit increase in the constant of the constraint will lead to an increase of approximately 276 in the value of the objective function, and Z 10,212.
b) f(x, y) 26x 3x2 5xy 6y2 12y subject to 3x y 170 1) The Lagrangian function is F 26x 3x2 5xy 6y2 12y (170 3x y) Thus, Fx 26 6x 5y 3 0 (5.30) Fy 5x 12y 12 0 (5.31) F 170 3x y 0 (5.32) Multiply (5.31) by 3 and subtract from (5.30) to eliminate .
21x 41y 10 0 (5.33) Multiply (5.32) by 7 and subtract from (5.33) to eliminate x.
48y 1200 0 y0 25 102 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 Then substituting y0 25 into the previous equations shows that at the critical point x0 145 3 48 1 3 y0 25 0 139 3 46 1 3 Using x0 145 3 , y0 25, and 0 139 3 , F 3160.
2) With 461 – 3, a 1-unit increase in the constant of the constraint will lead to a decrease of approximately 461 – 3 in the value of the objective function, and F 3206.33.
c) f(x, y, z) 4xyz2 subject to x y z 56 1) F 4xyz2 (56 x y z) Fx 4yz2 0 (5.34) Fy 4xz2 0 (5.35) Fz 8xyz 0 (5.36) F 56 x y z 0 (5.37) Equate ’s from (5.34) and (5.35).
4yz2 4xz2 y x Equate ’s from (5.34) and (5.36) 4yz2 8xyz z 2x Substitute y x and z 2x in (5.37).
56 x x 2x 0 4x 56 x0 14 Then substituting x0 14 in the previous equations gives x0 14 y0 14 z0 28 0 43,904 F0 614,656 2) F1 F0 2 614,656 43,904 658,560 See Problem 12.28 for the second-order conditions.
d) f(x, y, z) 5xy 8xz 3yz subject to 2xyz 1920 1) F 5xy 8xz 3yz (1920 2xyz) Fx 5y 8z 2yz 0 (5.38) Fy 5x 3z 2xz 0 (5.39) Fz 8x 3y 2xy 0 (5.40) F 1920 2xyz 0 (5.41) Solve (5.38), (5.39), and (5.40) for .
5y 8z 2yz 2.5 z 4 y (5.42) 5x 3z 2xz 2.5 z 1.5 x (5.43) 8x 3y 2xy 4 y 1.5 x (5.44) 103 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] Equate ’s in (5.42) and (5.43) to eliminate 2.5/z, 4 y 1.5 x 4x 1.5y x 1.5 4 y and ’s in (5.43) and (5.44) to eliminate 1.5/x.
2.5 z 4 y 4z 2.5y z 2.5 4 y Then substitute x (1.5/4)y and z (2.5/4)y in (5.41).
1920 2 · 1.5 4 y · y · 2.5 4 y y3 1920 · 16 7.5 4096 y0 16 and the critical values are x0 6, y0 16, z0 10, and 0 0.5.
F0 1440 2) F1 F0 0 1440 0.5 1440.5 5.13.
In Problem 5.12(a) it was estimated that if the constant of the constraint were increased by 1 unit, the constrained optimum would increase by approximately 276, from 9936 to 10,212.
Check the accuracy of the estimate by optimizing the original function z 4x2 2xy 6y2 subject to a new constraint x y 73.
Z 4x2 2xy 6y2 (73 x y) Zx 8x 2y 0 Zy 2x 12y 0 Z 73 x y 0 Simultaneous solution gives x0 42.58, y0 30.42, 0 279.8. Thus, Z0 10,213.9, compared to the 10,212 estimate from the original , a difference of 1.9 units or 0.02 percent.
5.14.
Constraints can also be used simply to ensure that the two independent variables will always be in constant proportion, for example, x 3y. In this case measuring the effect of has no economic significance since a 1-unit increase in the constant of the constraint would alter the constant proportion between the independent variables. With this in mind, optimize the following functions subject to the constant proportion constraint: a) z 4x2 3x 5xy 8y 2y2 subject to x 2y With x 2y 0, the Lagrangian function is Z 4x2 3x 5xy 8y 2y2 (x 2y) Zx 8x 3 5y 0 Zy 5x 8 4y 2 0 Z x 2y 0 When solved simultaneously, x0 0.5, y0 0.25, and 0 2.25. Thus, Z0 1.75.
104 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 b) z 5x2 7x 10xy 9y 2y2 subject to y 5x The Lagrangian function is Z 5x2 7x 10xy 9y 2y2 (5x y) Zx 10x 7 10y 5 0 Zy 10x 9 4y 0 Z 5x y 0 Solving simultaneously, x0 5.2, y0 26, and 0 43. Thus, Z0 135.2.
See also Problems 12.19 to 12.28.
DIFFERENTIALS 5.15.
Find the differential dy for each of the following functions: a) y 7x3 5x2 6x 3 dy dx 21x2 10x 6 Thus, dy (21x2 10x 6) dx b) y (4x 3)(3x 8) dy dx (4x 3)(3) (3x 8)(4) 24x 23 Thus, dy (24x 23) dx c) y 9x 4 5x dy dx 5x(9) (9x 4)(5) (5x)2 20 25x2 dy 4 5x2 dx d) y (11x 9)3 dy dx 3(11x 9)2(11) dy 33(11x 9)2 dx 5.16.
Find the total differential dz zx dx zy dy for each of the following functions: a) z 5x3 12xy 6y5 zx 15x2 12y zy 12x 30y4 dz (15x2 12y) dx (12x 30y4) dy b) z 7x2y3 zx 14xy3 zy 21x2y2 dz 14xy3 dx 21x2y2 dy c) z 3x2(8x 7y) zx 3x2(8) (8x 7y)(6x) zy 3x2(7) (8x 7y)(0) dz (72x2 42xy) dx 21x2 dy d) z (5x2 7y)(2x 4y3) zx (5x2 7y)(2) (2x 4y3)(10x) zy (5x2 7y)(12y2) (2x 4y3)(7) dz (30x2 40xy3 14y) dx (112y3 60x2y2 14x) dy 105 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] e) z 9y3 x y zx (x y)(0) 9y3(1) (x y)2 zy (x y)(27y2) 9y3(1) (x y)2 dz 9y3 (x y)2 dx 27xy2 18y3 (x y)2 dy f) z (x 3y)3 zx 3(x 3y)2(1) zy 3(x 3y)2(3) dz 3(x 3y)2 dx 9(x 3y)2 dy 5.17.
Find the partial differential for a small change in x for each of the functions given in Problem 5.16, assuming dy 0.
a) dz (15x2 12y) dx b) dz 14xy3 dx c) dz (72x2 42xy) dx d) dz (30x2 40xy3 14y) dx e) dz 9y3 (x y)2 dx f) dz 3(x 3y)2 dx TOTAL DERIVATIVES 5.18.
Find the total derivative dz/dx for each of the following functions: a) z 6x2 15xy 3y2 where y 7x2 dz dx zx zy dy dx (12x 15y) (15x 6y)(14x) 210x2 84xy 12x 15y b) z (13x 18y)2 where y x 6 dz dx zx zy dy dx 26(13x 18y) 36(13x 18y)(1) 10(13x 18y) c) z 9x 7y 2x 5y where y 3x 4 dz dx zx zy dy dx 59y (2x 5y)2 59x (2x 5y)2 (3) 59(y 3x) (2x 5y)2 d) z 8x 12y where y (x 1)/x2 dz dx zx zy dy dx 8 12(x2 2x) x4 8 12(x 2) x3 106 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 5.19.
Find the total derivative dz/dw for each of the following functions: a) z 7x2 4y2 where x 5w and y 4w dz dw zx dx dw zy dy dw 14x(5) 8y(4) 70x 32y b) z 10x2 6xy 12y2 where x 2w and y 3w dz dw zx dx dw zy dy dw (20x 6y)(2) (6x 24y)(3) 22x 84y IMPLICIT AND INVERSE FUNCTION RULES 5.20.
Find the derivatives dy/dx and dx/dy for each of the following implicit functions: a) y 6x 7 0 dy dx fx fy (6) 1 6 dx dy fy fx (1) 6 1 6 b) 3y 12x 17 0 dy dx fx fy (12) 3 4 dx dy fy fx (3) 12 1 4 c) x2 6x 13 y 0 dy dx fx fy (2x 6) 1 2x 6 dx dy fy fx (1) 2x 6 1 2x 6 (x 3) Notice that in each of the above cases, one derivative is the inverse of the other.
5.21.
Use the implicit function rule to find dy/dx and, where applicable, dy/dz.
a) f(x, y) 3x2 2xy 4y3 b) f(x, y) 12x5 2y dy dx fx fy 6x 2y 12y2 2x dy dx fx fy 60x4 2 30x4 c) f(x, y) 7x2 2xy2 9y4 d) f(x, y) 6x3 5y dy dx fx fy 14x 2y2 36y3 4xy dy dx fx fy 18x2 5 3.6x2 e) f(x, y, z) x2y3 z2 xyz f) f(x,y, z) x3z2 y3 4xyz dy dx fx fy 2xy3 yz 3x2y2 xz dy dx fx fy 3x2z2 4yz 3y2 4xz dy dz fz fy 2z xy 3x2y2 xz dy dz fz fy 2x3z 4xy 3y2 4xz 5.22.
Find the derivative for the inverse function dP/dQ.
a) Q 210 3P dP dQ 1 dQ/dP 1 3 107 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] b) Q 35 0.25P dP dQ 1 0.25 4 c) Q 14 P2 dP dQ 1 2P (P 0) d) Q P3 2P2 7P dP dQ 1 3P2 4P 7 VERIFICATION OF RULES 5.23.
For each of the following functions, use (1) the definition in (5.1a) to find z/x and (2) the definition in (5.1b) to find z/y in order to confirm the rules of differentiation.
a) z 38 7x 4y 1) From (5.1a), z x lim x→0 f(x x, y) f(x, y) x Substituting, z x lim x→0 [38 7(x x) 4y] (38 7x 4y) x lim x→0 38 7x 7 x 4y 38 7x 4y x lim x→0 7 x x lim x→0 7 7 2) From (5.1b), z y lim y→0 f(x,y y) f(x, y) y Substituting, z y lim y→0 [38 7x 4(y y)] (38 7x 4y) y lim y→0 38 7x 4y 4 y 38 7x 4y y lim y→0 4 y y lim y→0 (4) 4 b) z 18x 5xy 14y 1) z x lim x→0 [18(x x) 5(x x)y 14y] (18x 5xy 14y) x lim x→0 18x 18 x 5xy 5 xy 14y 18x 5xy 14y x lim x→0 18 x 5 xy x lim x→0 (18 5y) 18 5y 108 CALCULUS OF MULTIVARIABLE FUNCTIONS [CHAP . 5 2) z y lim y→0 [18x 5x(y y) 14(y y)] (18x 5xy 14y) y lim y→0 18x 5xy 5x y 14y 14 y 18x 5xy 14y y lim y→0 5x y 14 y y lim y→0 (5x 14) 5x 14 c) z 3x2y 1) z x lim x→0 [3(x x)2y] 3x2y x lim x→0 3x2y 6x xy 3( x)2y 3x2y x lim x→0 6x xy 3( x)2y x lim x→0 (6xy 3 xy) 6xy 2) z y lim y→0 [3x2(y y)] 3x2y y lim y→0 3x2y 3x2 y 3x2y y lim y→0 3x2 y y lim y→0 3x2 3x2 d) z 4x2y2 1) z x lim x→0 [4(x x)2y2] 4x2y2 x lim x→0 4x2y2 8x xy2 4( x)2y2 4x2y2 x lim x→0 8x xy2 4( x)2y2 x lim x→0 (8xy2 4 xy2) 8xy2 2) z y lim y→0 [4x2(y y)2] 4x2y2 y lim y→0 4x2y2 8x2y y 4x2( y)2 4x2y2 y lim y→0 8x2y y 4x2( y)2 y lim y→0 (8x2y 4x2 y) 8x2y 109 CALCULUS OF MULTIVARIABLE FUNCTIONS CHAP . 5] CHAPTER 6 Calculus of Multivariable Functions in Economics 6.1 MARGINAL PRODUCTIVITY The marginal product of capital (MPK) is defined as the change in output brought about by a small change in capital when all the other factors of production are held constant. Given a production function such as Q 36KL 2K 2 3L2 the MPK is measured by taking the partial derivative Q/K. Thus, MPK Q K 36L 4K Similarly, for labor, MPL Q/L 36K 6L. See Problems 6.1 to 6.3 6.2 INCOME DETERMINATION MULTIPLIERS AND COMPARATIVE STATICS The partial derivative can also be used to derive the various multipliers of an income determination model. In calculating how the equilibrium level of the endogenous variable can be expected to change in response to a change in any of the exogenous variables or parameters, income determination multipliers provide an elementary exercise in what is called comparative static analysis or, more simply, comparative statics, which we shall study later in greater detail in Chapter 13.
Given Y C I G (X Z) 110 where C C0 bY G G0 Z Z0 I I0 aY X X0 Using simple substitution as in Problem 2.19, the equilibrium level of income is Y ¯ 1 1 b a (C0 I0 G0 X0 Z0) (6.1) Taking the partial derivative of (6.1) with respect to any of the variables or parameters gives the multiplier for that variable or parameter. Thus, the government multiplier is given by Y ¯ G0 1 1 b a The import multiplier is given by Y ¯ Z0 1 1 b a And the multiplier for a change in the marginal propensity to invest is given by Y ¯/a, where, by means of the quotient rule, Y ¯ a (1 b a)(0) (C0 I0 G0 X0 Z0)(1) (1 b a)2 C0 I0 G0 X0 Z0 (1 b a)2 This can alternately be expressed as Y ¯ a 1 1 b a (C0 I0 G0 X0 Z0) 1 1 b a which from (6.1) reduces to Y ¯ a Y ¯ 1 b a See Problems 6.4 to 6.8.
6.3 INCOME AND CROSS PRICE ELASTICITIES OF DEMAND Income elasticity of demand Y measures the percentage change in the demand for a good resulting from a small percentage change in income, when all other variables are held constant. Cross price elasticity of demand c measures the relative responsiveness of the demand for one product to changes in the price of another, when all other variables are held constant. Given the demand function Q1 a bP1 cP2 mY where Y income and P2 the price of a substitute good, the income elasticity of demand is Y Q1 Q1 Y Y Q1 Y Y Q1 and the cross price elasticity of demand is c Q1 Q1 P2 P2 Q1 P2 P2 Q1 See Examples 1 and 2 and Problems 6.18 to 6.21.
111 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] EXAMPLE 1.
Given the demand for beef Qb 4850 5Pb 1.5Pp 0.1Y (6.2) with Y 10,000, Pb 200, and the price of pork Pp 100. The calculations for (1) the income elasticity and (2) the cross price elasticity of demand for beef are given below.
1) Y Qb Qb Y Y Qb Y Y Qb (6.3) From (6.2), Qb Y 0.1 and Qb 4850 5(200) 1.5(100) 0.1(10,000) 5000 (6.4) Substituting in (6.3), Y 0.1(10,000/5000) 0.2.
With Y 1, the good is income-inelastic. For any given percentage increase in national income, demand for the good will increase less than proportionately. Hence the relative market share of the good will decline as the economy expands. Since the income elasticity of demand suggests the growth potential of a market, the growth potential in this case is limited.
2) c Qb Qb Pp Pp Qb Pp Pp Qb From (6.2), Qb/Pp 1.5; from (6.4), Qb 5000. Thus, c 1.5 100 5000 0.03 For substitute goods, such as beef and pork, Q1/P2 0 and the cross price elasticity will be positive. For complementary goods, Q1/P2 0 and the cross price elasticity will be negative. If Q1/P2 0, the goods are unrelated.
EXAMPLE 2.
Continuing with Example 1, the percentage change in the demand for beef resulting from a 10 percent increase in the price of pork is estimated as follows: c Qb Qb Pp Pp Rearranging terms and substituting the known parameters, Qb Qb c Pp Pp (0.03)(0.10) 0.003 The percentage change in the demand for beef Qb/Qb will be 0.3 percent.
6.4 DIFFERENTIALS AND INCREMENTAL CHANGES Frequently in economics we want to measure the effect on the dependent variable (costs, revenue, profit) of a change in an independent variable (labor hired, capital used, items sold). If the change is a relatively small one, the differential will measure the effect. Thus, if z f(x, y), the effect on z of a small change in x is given by the partial differential dz zx dx The effect of larger changes can be approximated by multiplying the partial derivative by the proposed change. Thus, z zx x 112 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 If the original function z f(x, y) is linear, dz dx z x and the effect of the change will be measured exactly: z zx x See Examples 3 and 4 and Problems 6.9 to 6.17.
EXAMPLE 3.
A firm’s costs are related to its output of two goods x and y. The functional relationship is TC x2 0.5xy y2 The additional cost of a slight increment in output x will be given by the differential dTC (2x 0.5y) dx The costs of larger increments can be approximated by multiplying the partial derivative with respect to x by the change in x. Mathematically, TC TC x x (6.5) Since TC/x the marginal cost (MCx) of x, we can also write (6.5) as TC MCx x If initially x 100, y 60, and x 3, then TC [2(100) 0.5(60)] 510 EXAMPLE 4.
Assume in Section 6.2 that b 0.7, a 0.1, and Y 1200. The differential can then be used to calculate the effect of an increase in any of the independent variables. Given the partial derivative Y ¯ G0 1 1 b a the partial differential is dY ¯ 1 1 b a dG0 In a linear model such as this, where the slope is everywhere constant, Y ¯ G0 Y ¯ G0 Hence Y ¯ 1 1 b a G0 If the government increases expenditures by $100, Y ¯ 1 1 0.7 0.1 (100) 500 6.5 OPTIMIZATION OF MULTIVARIABLE FUNCTIONS IN ECONOMICS Food processors frequently sell different grades of the same product: quality, standard, economy; some, too, sell part of their output under their own brand name and part under the brand name of a large chain store. Clothing manufacturers and designers frequentlyhaveatop brand and cheaper imitations for discount department stores. Maximizing profits or minimizing costs under these conditions involve 113 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] functions of more than one variable. Thus, the basic rules for optimization of multivariate functions (see Section 5.4) are required. See Examples 5 and 6 and Problems 6.22 to 6.27.
EXAMPLE 5.
A firm producing two goods x and y has the profit function 64x 2x2 4xy 4y2 32y 14 To find the profit-maximizing level of output for each of the two goods and test to be sure profits are maximized: 1.
Take the first-order partial derivatives, set them equal to zero, and solve for x and y simultaneously.
x 64 4x 4y 0 (6.6) y 4x 8y 32 0 (6.7) When solved simultaneously, x ¯ 40 and y ¯ 24.
2.
Take the second-order direct partial derivatives and make sure both are negative, as is required for a relative maximum. From (6.6) and (6.7), xx 4 yy 8 3.
Take the cross partials to make sure xxyy (xy)2. From (6.6) and (6.7), xy 4 yx. Thus, xxyy (4)(8) 32 (xy)2 (4)2 16 Profits are indeed maximized at x ¯ 40 and y ¯ 24. At that point, 1650.
EXAMPLE 6.
In monopolistic competition producers must determine the price that will maximize their profit.
Assume that a producer offers two different brands of a product, for which the demand functions are Q1 14 0.25P1 (6.8) Q2 24 0.5P2 (6.9) and the joint cost function is TC Q1 2 5Q1Q2 Q2 2 (6.10) The profit-maximizing level of output, the price that should be charged for each brand, and the profits are determined as follows: First, establish the profit function in terms of Q1 and Q2. Since total revenue (TR) minus total cost (TC) and the total revenue for the firm is P1Q1 P2Q2, the firm’s profit is P1Q1 P2Q2 TC Substituting from (6.10), P1Q1 P2Q2 (Q1 2 5Q1Q2 Q2 2) (6.11) Next find the inverse functions of (6.8) and (6.9) by solving for P in terms of Q. Thus, from (6.8), P1 56 4Q1 (6.12) and from (6.9), P2 48 2Q2 (6.13) Substituting in (6.11), (56 4Q1)Q1 (48 2Q2)Q2 Q1 2 5Q1Q2 Q2 2 56Q1 5Q1 2 48Q2 3Q2 2 5Q1Q2 (6.14) Then maximize (6.14) by the familiar rules: 1 56 10Q1 5Q2 0 2 48 6Q2 5Q1 0 which, when solved simultaneously, give Q ¯ 1 2.75 and Q ¯ 2 5.7.
114 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 Take the second derivatives to be sure is maximized: 11 10 22 6 12 5 21 With both second direct partials negative and 1122 (12)2, the function is maximized at the critical values.
Finally, substitute Q ¯ 1 2.75 and Q ¯ 2 5.7 in (6.12) and (6.13), respectively, to find the profit-maximizing price.
P1 56 4(2.75) 45 P2 48 2(5.7) 36.6 Prices should be set at $45 for brand 1 and $36.60 for brand 2, leading to sales of 2.75 of brand 1 and 5.7 of brand 2. From (6.11) or (6.14), the maximum profit is 45(2.75) 36.6(5.7) (2.75)2 5(2.75)(5.7) (5.7)2 213.94 6.6 CONSTRAINED OPTIMIZATION OF MULTIVARIABLE FUNCTIONS IN ECONOMICS Solutions to economic problems frequently have to be found under constraints (e.g., maximizing utility subject to a budget constraint or minimizing costs subject to some such minimal requirement of output as a production quota). Use of the Lagrangian function (see Section 5.5) greatly facilitates this task. See Example 7 and Problems 6.28 to 6.39. For inequality constraints, see concave programming (Section 13.7) in Chapter 13.
EXAMPLE 7.
Find the critical values for minimizing the costs of a firm producing two goods x and y when the total cost function is c 8x2 xy 12y2 and the firm is bound by contract to produce a minimum combination of goods totaling 42, that is, subject to the constraint x y 42.
Set the constraint equal to zero, multiply it by , and form the Lagrangian function, C 8x2 xy 12y2 (42 x y) Take the first-order partials, Cx 16x y 0 Cy x 24y 0 C 42 x y 0 Solving simultaneously, x ¯ 25, y ¯ 17, and ¯ 383. With ¯ 383, a 1-unit increase in the constraint or production quota will lead to an increase in cost of approximately $383. For second-order conditions, see Section 12.5 and Problem 12.27(a).
6.7 HOMOGENEOUS PRODUCTION FUNCTIONS A production function is said to be homogeneous if when each input factor is multiplied by a positive real constant k, the constant can be completely factored out. If the exponent of the factor is 1, the function is homogeneous of degree 1; if the exponent of the factor is greater than 1, the function is homogeneous of degree greater than 1; and if the exponent of the factor is less than 1, the function is homogeneous of degree less than 1. Mathematically, a function z f(x, y) is homogeneous of degree n if for all positive real values of k, f(kx, ky) kn f(x, y). See Example 8 and Problem 6.40.
EXAMPLE 8.
The degree of homogeneity of a function is illustrated below.
1.
z 8x 9y is homogeneous of degree 1 because f(kx, ky) 8kx 9ky k(8x 9y) 2.
z x2 xy y2 is homogeneous of degree 2 because f(kx, ky) (kx)2 (kx)(ky) (ky)2 k2(x2 xy y2) 115 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 3.
z x0.3y0.4 is homogeneous of degree less than 1 because f(kx, ky) (kx)0.3(ky)0.4 k0.30.4(x0.3y0.4) k0.7(x0.3y0.4) 4.
z 2x/y is homogeneous of degree 0 because f(kx, ky) 2kx ky 1 2x y since k k k0 1 5.
z x3 2xy y3 is not homogeneous because k cannot be completely factored out: f(kx, ky) (kx)3 2(kx)(ky) (ky)3 k3x3 2k2xy k3y3 k2(kx3 2xy ky3) 6.
Q AK L is homogeneous of degree because Q(kK, kL) A(kK)(kL) AkK kL k(AK L) 6.8 RETURNS TO SCALE A production function exhibits constant returns to scale if when all inputs are increased by a given proportion k, output increases by the same proportion. If output increases by a proportion greater than k, there are increasing returns to scale; and if output increases by a proportion smaller than k, there are diminishing returns to scale. In other words, if the production function is homogeneous of degree greater than, equal to, or less than 1, returns to scale are increasing, constant, or diminishing.
See Problem 6.40.
6.9 OPTIMIZATION OF COBB-DOUGLAS PRODUCTION FUNCTIONS Economic analysis frequently employs the Cobb-Douglas production function q AKL (A 0; 0 , 1), where q is the quantity of output in physical units, K is the quantity of capital, and L is the quantity of labor. Here (the output elasticity of capital) measures the percentage change in q for a 1 percent change in K while L is held constant; (the output elasticity of labor) is exactly parallel; and A is an efficiency parameter reflecting the level of technology.
A strict Cobb-Douglas function, in which 1, exhibits constant returns to scale. A generalized Cobb-Douglas function, in which 1, exhibits increasing returns to scale if 1 and decreasing returns to scale if 1. A Cobb-Douglas function is optimized subject to a budget constraint in Example 10 and Problems 6.41 and 6.42; second-order condtions are explained in Section 12.5. Selected properties of Cobb-Douglas functions are demonstrated and proved in Problems 6.53 to 6.58.
EXAMPLE 9.
The first and second partial derivatives for (a) q AK L and (b) q 5K 0.4L0.6 are illustrated below.
a) qK qKK qKL AK 1L ( 1)AK 2L AK 1L1 qL qLL qLK AK L1 ( 1)AK L2 AK 1L1 b) qK qKK qKL 2K0.6L0.6 1.2K1.6L0.6 1.2K0.6L0.4 qL qLL qLK 3K 0.4L0.4 1.2K 0.4L1.4 1.2K0.6L0.4 EXAMPLE 10.
Given a budget constraint of $108 when PK 3 and PL 4, the generalized Cobb-Douglas production function q K 0.4L0.5 is optimized as follows: 1.
Set up the Lagrangian function.
Q K 0.4L0.5 (108 3K 4L) 116 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 2.
Using the simple power function rule, take the first-order partial derivatives, set them equal to zero, and solve simultaneously for K0 and L0 (and 0, if desired).
Q K QK 0.4K 0.6L0.5 3 0 (6.15) Q L QL 0.5K 0.4L0.5 4 0 (6.16) Q Q 108 3K 4L 0 (6.17) Rearrange, then divide (6.15) by (6.16) to eliminate .
0.4K0.6L0.5 0.5K 0.4L0.5 3 4 Remembering to subtract exponents in division, 0.8K1L1 0.75 L K 0.75 0.8 L 0.9375K Substitute L 0.9375K in (6.17).
108 3K 4(0.9375K) 0 K0 16 Then by substituting K0 16 in (6.17), L0 15 EXAMPLE 11.
The problem in Example 10 where q K 0.4L0.5, PK 3, PL 4, and B 108 can also be solved using the familiar condition from microeconomic theory for output maximization MUK MUL PK PL as demonstrated in (a) below and illustrated in (b).
a) MUK q K 0.4K0.6L0.5 MUL q L 0.5K 0.4L0.5 Substituting in the equality of ratios above, 0.4K0.6L0.5 0.5K 0.4L0.5 3 4 and solving as in Example 10, 0.8K1L1 0.75 L 0.9375K Then substituting in the budget constraint, 3K 4L 3K 4(0.9375K) K0 108 108 16 L0 15 which is exactly what we found in Example 10 using the calculus. See Fig. 6-1.
117 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] b) 6.10 OPTIMIZATION OF CONSTANT ELASTICITY OF SUBSTITUTION PRODUCTION FUNCTIONS The elasticity of substitution measures the percentage change in the least-cost (K/L) input ratio resulting from a small percentage change in the input-price ratio (PL/PK).
d(K/L) K/L d(PL/PK) PL/PK d(K/L) d(PL/PK) K/L PL/PK (6.18) where 0 . If 0, there is no substitutability; the two inputs are complements and must be used together in fixed proportions. If , the two goods are perfect substitutes. A Cobb-Douglas production function, as shown in Problem 6.57, has a constant elasticity of substitution equal to 1. A constant elasticity of substitution (CES) production function, of which a Cobb-Douglas function is but one example, has an elasticity of substitution that is constant but not necessarily equal to 1.
A CES production function is typically expressed in the form q A[K (1 )L]1/ (6.19) where A is the efficiency parameter, is the distribution parameter denoting relative factor shares, is the substitution parameter determining the value of the elasticity of substitution, and the parameters are restricted so that A 0, 0 1, and 1. CES production functions are optimized subject to budget constraints in Example 12 and Problems 6.43 and 6.44. Various important properties of the CES production function are demonstrated and proved in Problems 6.59 to 6.69.
EXAMPLE 12.
The CES production function q 75[0.3K0.4 (1 0.3)L0.4]1/0.4 is maximized subject to the constraint 4K 3L 120 as follows: 1.
Set up the Lagrangian function.
Q 75(0.3K0.4 0.7L0.4)2.5 (120 4K 3L) 2.
Test the first-order conditions, using the generalized power function rule for QK and QL.
QK 187.5(0.3K0.4 0.7L0.4)3.5(0.12K1.4) 4 0 22.5K1.4(0.3K0.4 0.7L0.4)3.5 4 0 (6.20) QL 187.5(0.3K0.4 0.7L0.4)3.5(0.28L1.4) 3 0 52.5L1.4(0.3K0.4 0.7L0.4)3.5 3 0 (6.21) Q 120 4K 3L 0 (6.22) 118 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 Fig. 6-1 Slope K L MPK/MPL Slope K L PK/PL Rearrange, then divide (6.20) by (6.21) to eliminate .
22.5K1.4(0.3K0.4 0.7L0.4)3.5 52.5L1.4(0.3K0.4 0.7L0.4)3.5 4 3 22.5K1.4 52.5L1.4 4 3 Cross multiply.
67.5K1.4 210L1.4 K1.4 3.11L1.4 Take the 1.4 root, K (3.11)1/1.4L (3.11)0.71L and use a calculator.
K 0.45L Substitute in (6.22).
120 4(0.45L) 3L 0 L0 25 K0 11.25 Note: To find (3.11)0.71 with a calculator, enter 3.11, press the yx key, then enter 0.71 followed by the / key to make it negative, and hit the key to find (3.11)0.71 0.44683.
Solved Problems MARGINAL CONCEPTS 6.1.
Find the marginal productivity of the different inputs or factors of production for each of the following production functions Q: a) Q 6x2 3xy 2y2 MPx Q x 12x 3y MPy Q y 3x 4y b) Q 0.5K 2 2KL L2 MPK K 2L MPL 2L 2K c) Q 20 8x 3x2 0.25x3 5y 2y2 0.5y3 MPx 8 6x 0.75x2 MPy 5 4y 1.5y2 d) Q x2 2xy 3y2 1.5yz 0.2z2 MPx 2x 2y MPy 2x 6y 1.5z MPz 1.5y 0.4z 119 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 6.2.
(a) Assume y ¯ 4 in Problem 6.1(a) and find the MPx for x 5 and x 8. (b) If the marginal revenue at x ¯ 5, y ¯ 4 is $3, compute the marginal revenue product for the fifth unit of x.
a) MPx 12x 3y At x 5, y ¯ 4, MPx 12(5) 3(4) 72.
At x 8, y ¯ 4, MPx 12(8) 3(4) 108.
b) MRPx MPx(MR) At x ¯ 5, y ¯ 4, MRPx (72)(3) 216.
6.3.
(a) Find the marginal cost of a firm’s different products when the total cost function is c 3x2 7x 1.5xy 6y 2y2. (b) Determine the marginal cost of x when x 5, y ¯ 3.
a) MCx 6x 7 1.5y MCy 1.5x 6 4y b) The marginal cost of x when x 5 and y is held constant at 3 is MCx 6(5) 7 1.5(3) 41.5 INCOME DETERMINATION MULTIPLIERS AND COMPARATIVE STATICS 6.4.
Given a three-sector income determination model in which Y C I0 G0 C C0 bYd Yd T Y T C0, I0, G0, T0 0 0 b, t 1 T0 tY determine the magnitude and direction of a 1-unit change in (a) government spending, (b) lump-sum taxation, and (c) the tax rate on the equilibrium level of income. In short, perform the comparative-static exercise of determining the government multiplier, the autonomous tax multiplier, and the tax rate multiplier.
To find the different multipliers, first solve for the equilibrium level of income, as follows: Y C0 bY bT0 btY I0 G0 Y ¯ 1 1 b bt (C0 bT0 I0 G0) (6.23) Then take the appropriate partial derivatives.
a) Y ¯ G0 1 1 b bt Since 0 b 1, Y ¯/G0 0. A 1-unit increase in government spending will increase the equilibrium level of income by 1/(1 b bt).
b) Y ¯ T0 b 1 b bt 0 A 1-unit increase in autonomous taxation will cause national income to fall by b/(1 b bt).
c) Since t appears in the denominator in (6.23), the quotient rule is necessary.
Y ¯ t (1 b bt)(0) (C0 bT0 I0 G0)(b) (1 b bt)2 b(C0 bT0 I0 G0) (1 b bt)2 b 1 b bt C0 bT0 I0 G0 1 b bt Thus, from (6.23), Y ¯ t bY ¯ 1 b bt 0 A 1-unit increase in the tax rate will cause national income to fall by an amount equal to the tax rate multiplier.
120 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 6.5.
Given a simple model Y C I0 G0 C C0 bYd Yd Y T T T0 where taxation does not depend on income, calculate the effect on the equilibrium level of income of a 1-unit change in government expenditure exactly offset by a 1-unit change in taxation. That is, do the comparative-static analysis of finding the balanced-budget multiplier for an economy in which there is only autonomous taxation.
Y C0 b(Y T0) I0 G0 Y ¯ 1 1 b (C0 bT0 I0 G0) Thus, the government multiplier is Y ¯ G0 1 1 b (6.24) and the tax multiplier is Y ¯ T0 b 1 b (6.25) The balanced-budget effect of a 1-unit increase in government spending matched by a 1-unit increase in taxation is the sum of (6.24) and (6.25). Therefore, Y ¯ 1 1 b b 1 b 1 1 b b 1 b 1 b 1 b 1 A change in government expenditure matched by an equal change in government taxation will have a positive effect on the equilibrium level of income exactly equal to the change in government expenditure and taxation. The multiplier in this case is 1.
6.6.
Given Y C I0 G0 Yd Y T C C0 bYd T T0 tY where taxation is now a function of income, demonstrate the effect on the equilibrium level of income of a 1-unit change in government expenditure offset by a 1-unit change in autonomous taxation T0. That is, demonstrate the effect of the balanced-budget multiplier in an economy in which taxes are a positive function of income.
From (6.23), Y ¯ 1/(1 b bt). Thus, Y ¯ G0 1 1 b bt (6.26) and Y ¯ T0 b 1 b bt (6.27) The combined effect on Y ¯ of a 1-unit increase in government spending and an equal increase in autonomous taxation is the sum of (6.26) and (6.27). Thus, Y ¯ 1 1 b bt b 1 b bt 1 b 1 b bt which is positive but less than 1 because 1 b 1 b bt. A change in government expenditures equaled by a change in autonomous taxes when taxes are positively related to income in the model, will have a positive effect on the equilibrium level of income, but the effect is smaller than the initial change in government expenditure. Here the multiplier is less than 1 because the total change in taxes ( T T0 t Y) is greater than the change in G0.
121 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 6.7.
Given Y C I0 G0 X0 Z C C0 bYd T T0 tY Z Z0 zYd where all the independent variables are positve and 0 b, z, t 1. Determine the effect on the equilibrium level of income of a 1-unit change in (a) exports, (b) autonomous imports, and (c) autonomous taxation. In short, perform the comparative-static analysis of finding the export, autonomous import, and autonomous taxation multipliers. [Note that Z f(Yd)].
From the equilibrium level of income, Y C0 b(Y T0 tY) I0 G0 X0 Z0 z(Y T0 tY) Y ¯ 1 1 b bt z zt (C0 bT0 I0 G0 X0 Z0 zT0) a) Y ¯ X0 1 1 b bt z zt 0 because 0 b, z 1. A 1-unit increase in exports will have a positive effect on Y ¯, which is given by the multiplier.
b) Y ¯ Z0 1 1 b bt z zt 0 An increase in autonomous imports will lead to a decrease in Y ¯.
c) Y ¯ T0 z b 1 b bt z zt 0 because a country’s marginal propensity to import z is usually smaller than its marginal propensity to consume b. With z b, z b 0. An increase in autonomous taxes will lead to a decrease in national income, as in (6.27), but the presence of z in the numerator has a mitigating effect on the decrease in income. When there is a positive marginal propensity to import, increased taxes will reduce cash outflows for imports and thus reduce the negative effect of increased taxes on the equilibrium level of income.
6.8.
Determine the effect on Y ¯ of a 1-unit change in the marginal propensity to import z in Problem 6.7.
Y ¯ z (1 b bt z zt)(T0) (C0 bT0 I0 G0 X0 Z0 zT0)(1 t) (1 b bt z zt)2 T0 1 b bt z zt Y ¯(1 t) 1 b bt z zt [Y ¯ (T0 tY ¯)] 1 b bt z zt Y ¯d 1 b bt z zt 0 DIFFERENTIALS AND COMPARATIVE STATICS 6.9.
Y C I0 G0 C C0 bYd Yd Y T T T0 tY C0 100 G0 330 I0 90 T0 240 b 0.75 t 0.20 (a) What is the equilibrium level of income Y ¯? What is the effect on Y ¯ of a $50 increase in (b) government spending and (c) autonomous taxation T0?
122 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 a) From (6.23), Y ¯ 1 1 b bt (C0 bT0 I0 G0) 1 1 0.75 0.75(0.20) [100 0.75(240) 90 330] 1 0.40 (100 180 90 330) 2.5(340) 850 b) If government increases spending by 50, Y ¯ Y ¯ G0 G0 1 1 b bt (50) 2.5(50) 125 c) If autonomous taxation T0 increases by 50, Y ¯ Y ¯ T0 T0 b 1 b bt (50) 0.75 1 0.75 0.75(0.20) (50) 1.875(50) 93.75 6.10.
If the full-employment level of income Yfe in Problem 6.9(a) is 1000 and the government wishes to achieve it, by how much should it change (a) government spending or (b) autonomous taxation?
a) The desired increase in economic activity is the difference between the full-employment level of income (1000) and the present level (850). Thus, the desired Y ¯ 150. Substituting in the formula from Problem 6.9(b), Y ¯ Y ¯ G0 G0 150 2.5 G0 G0 60 Increased government expenditure of 60 will increase Y ¯ by 150.
b) If the government wishes to alter autonomous taxes to achieve full employment, from Problem 6.9(c), Y ¯ Y ¯ T0 T0 150 1.875 T0 T0 80 The government should cut autonomous taxes by 80.
6.11.
Explain the effect on the government deficit (a) if policy a in Problem 6.10 is adopted and (b) if policy b is adopted instead.
a) The government’s financial condition is given by the difference between receipts T and expenditures G. At the initial 850 level of income, T 240 0.2(850) 410 G0 330 T G0 410 330 80 The government has a surplus of 80.
If the government increases spending by 60, expenditures rise by 60. But tax revenues also increase as a result of the increase in income. With Y ¯ 150, T 0.2(150) 30. With expenditures rising by 60 and receipts increasing by 30, the net cost to the government of stimulating the economy to full employment is only $30. At the new Y ¯ 1000, T 240 0.2(1000) 440 G0 330 60 390 T G0 440 390 50 The government surplus is reduced to $50 from the previous $80 surplus.
123 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] b) If the government reduces T0 by 80, tax revenue falls initially by 80. But the $150 stimulatory effect on income has a positive effect on total tax collections, since T 0.2(150) 30. Thus, the net cost of reducing autonomous taxation to stimulate the economy to full employment is $50. The government surplus is reduced to $30: T 160 0.2(1000) 360 G0 330 T G0 360 330 30 6.12.
(a) If the proportional tax in Problem 6.9 is increased by 10 percent, what is the effect on Y ¯?
(b) If the government wants to alter the original marginal tax rate of 20 percent to achieve Yfe 1000, by how much should it change t?
a) If the proportional tax is increased by 10 percent, t 0.10(0.20) 0.02 The resulting change in income is Y ¯ Y ¯ t t Substituting from Problem 6.4(c), Y ¯ bY ¯ 1 b bt (0.02) Since a change in one of the parameters, unlike a change in one of the independent variables, will alter the value of the multiplier, the multiplier will only approximate the effect of the change. Y ¯ 0.75(850) 0.4 (0.02) 31.88 b) The government wants to raise Y ¯ by 150. Substituting Y ¯ 150 in the equation above, 150 0.75(850) 0.4 t t 0.09 The tax rate should be reduced by approximately 0.09. The new tax rate should be around 11 percent (0.20 0.09 0.11).
6.13.
Given Y C C I0 G0 X0 Z C0 bYd T Z T0 tY Z0 zYd with b X0 z 0.9 150 0.15 t Z0 T0 0.2 55 150 C0 I0 G0 125 92.5 600 Calculate (a) the equilibrium level of income, (b) the effect on Y ¯ of an increase of 60 in autonomous exports X0, and (c) the effect on Y ¯ of an increase of 30 in autonomous imports Z0.
a) From Problem 6.7, Y ¯ 1 1 b bt z zt (C0 bT0 I0 G0 X0 Z0 zT0) 1 1 0.9 0.9(0.2) 0.15 0.15(0.2) [125 0.9(150) 92.5 600 150 55 22.5] 2.5(800) 2000 124 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 b) Y ¯ Y ¯ X0 X0 1 1 b bt z zt (60) 2.5(60) 150 c) Y ¯ Y ¯ Z0 Z0 1 1 b bt z zt (30) 2.5(30) 75 6.14.
If the full-employment level of income in Problem 6.13 is 2075, (a) by how much should the government increase expenditures to achieve it? (b) By how much should it cut autonomous taxes to have the same effect?
a) The effect of government spending on national income is Y ¯ Y ¯ G0 G0 substituting Y ¯ 75, 75 1 1 b bt z zt G0 2.5 G0 G0 30 b) Y ¯ Y ¯ T0 T0 75 z b 1 b bt z zt T0 1.875 T0 T0 40 The government should cut autonomous taxation by 40.
6.15.
Calculate the effect on the government deficit if the government in Problem 6.14 achieves full employment through (a) increased expenditures or (b) a tax cut.
a) If the government increases expenditures by 30, the government deficit increases initially by 30.
However, income is stimulated by 75. With Y ¯ 75, T 0.2(75) 15. Tax revenue increases by 15.
Thus the net cost to the government from this policy, and the effect on the deficit, is $15 (30 15 15).
b) If the government cuts autonomous taxation by 40, tax revenues fall initially by 40. But income increases by 75, causing tax revenues to increase by 15. Thus the net cost to the government of this policy is 25 (40 15 25), and the government deficit worsens by 25.
6.16.
Calculate the effect on the balance of payments (B/P) from (a) government spending and (b) the tax reduction in Problem 6.14.
a) Since B/P X Z, substituting from Problem 6.13, B/P X0 (Z0 zYd) X0 Z0 zY zT0 ztY (6.28) With an increase of 30 in government spending, Y ¯ 75. Since Y is the only variable on the right-hand side of (6.28) to change, (B/P) z(75) zt(75) Substituting z 0.15, zt 0.03, (B/P) 9.
b) When the government cuts autonomous taxes by 40, Y ¯ 75. Adjusting (6.28), (B/P) z(75) z(40) zt(75) 15 The reduction in taxes leads to a greater increase in disposable income than the increased government spending, resulting in a higher level of imports and a more serious balance of payments deficit.
125 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 6.17.
Estimate the effect on Y ¯ of a one-percentage-point decrease in the marginal propensity to import from Problem 6.13. Y ¯ Y ¯ z z Substituting from Problem 6.8, Y ¯ Y ¯d 1 b bt z zt z where Y ¯d Y ¯ T0 tY ¯ 2000 150 0.2(2000) 1450 Thus, Y ¯ 1450 0.4 (0.01) 36.25 PARTIAL ELASTICITIES 6.18.
Given Q 700 2P 0.02Y, where P 25 and Y 5000. Find (a) the price elasticity of demand and (b) the income elasticity of demand.
a) d Q P P Q where Q/P 2 and Q 700 2(25) 0.02(5000) 750. Thus, d 2 25 750 0.067 b) Y Q Y Y Q 0.02 5000 750 0.133 6.19.
Given Q 400 8P 0.05Y, where P 15 and Y 12,000. Find (a) the income elasticity of demand and (b) the growth potential of the product, if income is expanding by 5 percent a year.
(c) Comment on the growth potential of the product.
a) Q 400 8(15) 0.05(12,000) 880 and Q/Y 0.05. Thus, Y Q Y Y Q 0.05 12,000 880 0.68 b) Y Q Q Y Y Rearranging terms and substituting the known parameters, Q Q Y Y Y 0.68(0.05) 0.034 The demand for the good will increase by 3.4 percent.
c) Since 0 Y 1, it can be expected that demand for the good will increase with national income, but the increase will be less than proportionate. Thus, while demand grows absolutely, the relative market share of the good will decline in an expanding economy. If Y 1, the demand for the product would grow faster than the rate of expansion in the economy, and increase its relative market share. And if Y 0, demand for the good would decline as income increases.
126 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 6.20.
Given Q1 100 P1 0.75P2 0.25P3 0.0075Y.
At P1 10, P2 20, P3 40, and Y 10,000, Q1 170. Find the different cross price elasticities of demand.
12 Q1 P2 P2 Q1 0.75 20 170 0.088 13 Q1 P3 P3 Q1 0.25 40 170 0.059 6.21.
Given Q1 50 4P1 3P2 2P3 0.001Y. At P1 5, P2 7, P3 3, and Y 11,000, Q1 26.
(a) Use cross price elasticities to determine the relationship between good 1 and the other two goods. (b) Determine the effect on Q1 of a 10 percent price increase for each of the other goods individually.
a) 12 3( 7 –– 26) 0.81 13 2( 3 –– 26) 0.23 With 12 negative, goods 1 and 2 are complements. An increase in P2 will lead to a decrease in Q1.
With 13 positive, goods 1 and 3 are substitutes. An increase in P3 will increase Q1.
b) 12 Q1 Q1 P2 P2 Rearranging terms and substituting the known parameters, Q1 Q1 12 P2 P2 0.81(0.10) 0.081 If P2 increases by 10 percent, Q1 decreases by 8.1 percent.
13 Q1 Q1 P3 P3 Q1 Q1 13 P3 P3 0.23(0.10) 0.023 If P3 increases by 10 percent, Q1 increases by 2.3 percent.
OPTIMIZING ECONOMIC FUNCTIONS 6.22.
Given the profit function 160x 3x2 2xy 2y2 120y 18 for a firm producing two goods x and y, (a) maximize profits, (b) test the second-order condition, and (c) evaluate the function at the critical values x ¯ and y ¯.
a) x 160 6x 2y 0 y 2x 4y 120 0 When solved simultaneously, x ¯ 20 and y ¯ 20.
b) Taking the second partials, xx 6 yy 4 xy 2 With both direct second partials negative, and xxyy (xy)2, is maximized at x ¯ y ¯ 20.
c) 2782 6.23.
Redo Problem 6.22, given 25x x2 xy 2y2 30y 28.
a) x 25 2x y 0 y x 4y 30 0 Thus, x ¯ 10 and y ¯ 5.
127 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] b) xx 2 yy 4 xy 1 With xx and yy both negative and xxyy (xy)2, is maximized.
c) 172 6.24.
A monopolist sells two products x and y for which the demand functions are x 25 0.5Px (6.29) y 30 Py (6.30) and the combined cost function is c x2 2xy y2 20 (6.31) Find (a) the profit-maximizing level of output for each product, (b) the profit-maximizing price for each product, and (c) the maximum profit.
a) Since TRx TRy TC, in this case, Pxx Pyy c (6.32) From (6.29) and (6.30), Px 50 2x (6.33) Py 30 y (6.34) Substituting in (6.32), (50 2x)x (30 y)y (x2 2xy y2 20) 50x 3x2 30y 2y2 2xy 20 (6.35) The first-order condition for maximizing (6.35) is x 50 6x 2y 0 y 30 4y 2x 0 Solving simultaneously, x ¯ 7 and y ¯ 4. Testing the second-order conditon, xx 6, yy 4, and xy 2. With both direct partials negative and xxyy (xy)2, is maximized.
b) Substituting x ¯ 7, y ¯ 4 in (6.33) and (6.34), Px 50 2(7) 36 Py 30 4 26 c) Substituting x ¯ 7, y ¯ 4 in (6.35), 215.
6.25.
Find the profit-maximizing level of (a) output, (b) price, and (c) profit for a monopolist with the demand functions x 50 0.5Px (6.36) y 76 Py (6.37) and the total cost function c 3x2 2xy 2y2 55.
a) From (6.36) and (6.37), Px 100 2x (6.38) Py 76 y (6.39) 128 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 Substituting in Pxx Pyy c, (100 2x)x (76 y)y (3x2 2xy 2y2 55) 100x 5x2 76y 3y2 2xy 55 (6.40) Maximizing (6.40), x 100 10x 2y 0 y 76 6y 2x 0 Thus, x ¯ 8 and y ¯ 10. Checking the second-order condition, xx 10, yy 6, and xy 2.
Since xx, yy 0 and xxyy (xy)2, is maximized at the critical values.
b) Substituting x ¯ 8, y ¯ 10 in (6.38) and (6.39), Px 100 2(8) 84 Py 76 10 66 c) From (6.40), 725.
6.26.
Find the profit-maximizing level of (a) output, (b) price, and (c) profit for the monopolistic producer with the demand functions Q1 491 – 3 2 – 3P1 (6.41) Q2 36 1 – 2P2 (6.42) and the joint cost function c Q1 2 2Q1Q2 Q2 2 120.
a) From (6.41) and (6.42), P1 74 1.5Q1 (6.43) P2 72 2Q2 (6.44) Substituting in P1Q1 P2Q2 c, (74 1.5Q1)Q1 (72 2Q2)Q2 (Q1 2 2Q1Q2 Q2 2 120) 74Q1 2.5Q1 2 72Q2 3Q2 2 2Q1Q2 120 (6.45) The first-order condition for maximizing (6.45) is 1 74 5Q1 2Q2 0 2 72 6Q2 2Q1 0 Thus, Q ¯ 1 11.54 and Q ¯ 2 8.15. Testing the second-order condition, 11 5, 22 6, and 12 2. Thus, 11, 22 0; 1122 (12)2, and is maximized.
b) Substituting the critical values in (6.43) and (6.44), P1 74 1.5(11.54) 56.69 P2 72 2(8.15) 55.70 c) 600.46 6.27.
Find the profit-maximizing level of (a) output, (b) price, and (c) profit when Q1 5200 10P1 (6.46) Q2 8200 20P2 (6.47) and c 0.1Q1 2 0.1Q1Q2 0.2Q2 2 325 a) From (6.46) and (6.47) P1 520 0.1Q1 (6.48) P2 410 0.05Q2 (6.49) 129 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] Thus, (520 0.1Q1)Q1 (410 0.05Q2)Q2 (0.1Q1 2 0.1Q1Q2 0.2Q2 2 325) 520Q1 0.2Q1 2 410Q2 0.25Q2 2 0.1Q1Q2 325 (6.50) Maximizing (6.50), 1 520 0.4Q1 0.1Q2 0 2 410 0.5Q2 0.1Q1 0 Thus, Q ¯ 1 1152.63 and Q ¯ 2 589.47. Checking the second-order condition, 11 0.4, 22 0.5, and 12 0.1 21. Since 11, 22 0 and 1122 (12)2, is maximized at Q ¯ 1 1152.63 and Q ¯ 2 589.47.
b) Substituting in (6.48) and (6.49), P1 520 0.1(1152.63) 404.74 P2 410 0.05(589.47) 380.53 c) 420,201.32 CONSTRAINED OPTIMIZATION IN ECONOMICS 6.28.
(a) What combination of goods x and y should a firm produce to minimize costs when the joint cost function is c 6x2 10y2 xy 30 and the firm has a production quota of x y 34? (b) Estimate the effect on costs if the production quota is reduced by 1 unit.
a) Form a new function by setting the constraint equal to zero, multiplying it by , and adding it to the original or objective function. Thus, C 6x2 10y2 xy 30 (34 x y) Cx 12x y 0 Cy 20y x 0 C 34 x y 0 Solving simultaneously, x ¯ 21, y ¯ 13, and ¯ 239. Thus, C 4093. Second-order conditions are discussed in Section 12.5.
b) With 239, a decrease in the constant of the constraint (the production quota) will lead to a cost reduction of approximately 239.
6.29.
(a) What output mix should a profit-maximizing firm produce when its total profit function is 80x 2x2 xy 3y2 100y and its maximum output capacity is x y 12? (b) Estimate the effect on profits if output capacity is expanded by 1 unit.
a) 80x 2x2 xy 3y2 100y (12 x y) x 80 4x y 0 y x 6y 100 0 12 x y 0 When solved simultaneously, x ¯ 5, y ¯ 7, and ¯ 53. Thus, 868.
b) With ¯ 53, an increase in output capacity should lead to increased profits of approximately 53.
6.30.
A rancher faces the profit function 110x 3x2 2xy 2y2 140y where x sides of beef and y hides. Since there are two sides of beef for every hide, it follows that output must be in the proportion x 2 y x 2y 130 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 At what level of output will the rancher maximize profits?
110x 3x2 2xy 2y2 140y (x 2y) x 110 6x 2y 0 y 2x 4y 140 2 0 x 2y 0 Solving simultaneously, x ¯ 20, y ¯ 10, ¯ 30, and 1800.
6.31.
(a) Minimize costs for a firm with the cost function c 5x2 2xy 3y2 800 subject to the production quota x y 39. (b) Estimate additional costs if the production quota is increased to 40.
a) C 5x2 2xy 3y2 800 (39 x y) Cx 10x 2y 0 Cy 2x 6y 0 C 39 x y 0 When solved simultaneously, x ¯ 13, y ¯ 26, ¯ 182, and c 4349.
b) Since ¯ 182, an increased production quota will lead to additional costs of approximately 182.
6.32.
A monopolistic firm has the following demand functions for each of its products x and y: x 72 0.5Px (6.51) x 120 Py (6.52) The combined cost function is c x2 xy y2 35, and maximum joint production is 40. Thus x y 40. Find the profit-maximizing level of (a) output, (b) price, and (c) profit.
a) From (6.51) and (6.52), Px 144 2x (6.53) Py 120 y (6.54) Thus, (144 2x)x (120 y)y (x2 xy y2 35) 144x 3x2 xy 2y2 120y 35.
Incorporating the constraint, 144x 3x2 xy 2y2 120y 35 (40 x y) Thus, x 144 6x y 0 y x 4y 120 0 40 x y 0 and, x ¯ 18, y ¯ 22, and ¯ 14.
b) Substituting in (6.53) and (6.54), Px 144 2(18) 108 Py 120 22 98 c) 2861 6.33.
A manufacturer of parts for the tricycle industry sells three tires (x) for every frame (y).
Thus, x 3 y x 3y 131 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] If the demand functions are x 63 0.25Px (6.55) y 60 1 – 3Py (6.56) and costs are c x2 xy y2 190 find the profit-maximizing level of (a) output, (b) price, and (c) profit.
a) From (6.55) and (6.56), Px 252 4x (6.57) Py 180 3y (6.58) Thus, (252 4x)x (180 3y)y (x2 xy y2 190) 252x 5x2 xy 180y 190 4y2 Forming a new, constrained function, 252x 5x2 xy 4y2 180y 190 (x 3y) Hence, x 252 10x y 0 y x 8y 180 3 0 x 3y 0 and x ¯ 27, y ¯ 9, and ¯ 27.
b) From (6.57) and (6.58), Px 144 and Py 153.
c) 4022 6.34.
Problem 4.22 dealt with the profit-maximizing level of output for a firm producing a single product that is sold in two distinct markets when it does and does not discriminate. The functions given were Q1 Q2 c 21 0.1P1 50 0.4P2 2000 10Q where Q Q1 Q2 (6.59) (6.60) (6.61) Use multivariable calculus to check your solution to Problem 4.22.
From (6.59), (6.60), and (6.61), P1 210 10Q1 (6.62) P2 125 2.5Q2 (6.63) c 2000 10Q1 10Q2 With discrimination P1 P2 since different prices are charged in different markets, and therefore (210 10Q1)Q1 (125 2.5Q2)Q2 (2000 10Q1 10Q2) 200Q1 10Q1 2 115Q2 2.5Q2 2 2000 Taking the first partials, 1 200 20Q1 0 2 115 5Q2 0 Thus, Q ¯ 1 10 and Q ¯ 2 23. Substituting in (6.62) and (6.63), P ¯ 1 110 and P ¯ 2 67.5.
If there is no discrimination, the same price must be charged in both markets. Hence P1 P2.
Substituting from (6.62) and (6.63), 210 10Q1 125 2.5Q2 2.5Q2 10Q1 85 132 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 Rearranging this as a constraint and forming a new function, 200Q1 10Q1 2 115Q2 2.5Q2 2 2000 (85 10Q1 2.5Q2) Thus, 1 200 20Q1 10 0 2 115 5Q2 2.5 0 85 10Q1 2.5Q2 0 and Q ¯ 1 13.4, Q ¯ 2 19.6, and ¯ 6.8. Substituting in (6.62) and (6.63), P1 210 10(13.4) 76 P2 125 2.5(19.6) 76 Q 13.4 19.6 33 6.35.
Check your answers to Problem 4.23, given Q1 24 0.2P1 Q2 10 0.05P2 c 35 40Q where Q Q1 Q2 From the information given, P1 120 5Q1 (6.64) P2 200 20Q2 (6.65) c 35 40Q1 40Q2 With price discrimination, (120 5Q1)Q1 (200 20Q2)Q2 (35 40Q1 40Q2) 80Q1 5Q1 2 160Q2 20Q2 2 35 Thus, 1 80 10Q1 0 2 160 40Q2 0 and Q ¯ 1 8, Q ¯ 2 4, P1 80, and P2 120.
If there is no price discrimination, P1 P2. Substituting from (6.64) and (6.65), 120 5Q1 200 20Q2 20Q2 5Q1 80 (6.66) Forming a new function with (6.66) as a constraint, 80Q1 5Q1 2 160Q2 20Q2 2 35 (80 5Q1 20Q2) Thus, 1 80 10Q1 5 0 2 160 40Q2 20 0 80 5Q1 20Q2 0 and Q ¯ 1 6.4, Q ¯ 2 5.6, and ¯ 3.2. Substituting in (6.64) and (6.65), P1 120 5(6.4) 88 P2 200 20(5.6) 88 Q 6.4 5.6 12 6.36.
(a) Maximize utility u Q1Q2 when P1 1, P2 4, and one’s budget B 120. (b) Estimate the effect of a 1-unit increase in the budget.
a) The budget constraint is Q1 4Q2 120. Forming a new function to incorporate the constraint, U Q1Q2 (120 Q1 4Q2) Thus, U1 Q2 0 U2 Q1 4 0 U 120 Q1 4Q2 0 and Q ¯ 1 60, Q ¯ 2 15, and ¯ 15.
b) With ¯ 15, a $1 increase in the budget will lead to an increase in the utility function of approximately 15. Thus, the marginal utility of money (or income) at Q ¯ 1 60 and Q ¯ 2 15 is approximately 15.
133 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 6.37.
(a) Maximize utility u Q1Q2, subject to P1 10, P2 2, and B 240. (b) What is the marginal utility of money?
a) Form the Lagrangian function U Q1Q2 (240 10Q1 2Q2).
U1 Q2 10 0 U2 Q1 2 0 U 240 10Q1 2Q2 0 Thus, Q ¯ 1 12, Q ¯ 2 60, and ¯ 6.
b) The marginal utility of money at Q ¯ 1 12 and Q ¯ 2 60 is approximately 6.
6.38.
Maximize utility u Q1Q2 Q1 2Q2, subject to P1 2, P2 5, and B 51.
Form the Lagrangian function U Q1Q2 Q1 2Q2 (51 2Q1 5Q2).
U1 Q2 1 2 0 U2 Q1 2 5 0 U 51 2Q1 5Q2 0 Thus, Q ¯ 1 13, Q ¯ 2 5, and ¯ 3.
6.39.
Maximize utility u xy 3x y subject to Px 8, Py 12, and B 212.
The Lagrangian function is U xy 3x y (212 8x 12y).
Ux y 3 8 0 Uy x 1 12 0 U 212 8x 12y 0 Thus, x ¯ 15, y ¯ 72 – 3, and ¯ 11 – 3.
HOMOGENEITY AND RETURNS TO SCALE 6.40.
Determine the level of homogeneity and returns to scale for each of the following production functions: a) Q x2 6xy 7y2 here Q is homogeneous of degree 2, and returns to scale are increasing because f(kx, ky) (kx)2 6(kx)(ky) 7(ky)2 k2(x2 6xy 7y2) b) Q x3 xy2 3y3 x2y here Q is homogeneous of degree 3, and returns to scale are increasing because f(kx, ky) (kx)3 (kx)(ky)2 3(ky)3 (kx)2(ky) k3(x3 xy2 3y3 x2y) c) Q 3x2 5y2 here Q is homogeneous of degree 0, and returns to scale are decreasing because f(kx, ky) 3(kx)2 5(ky)2 3x2 5y2 and k0 1 d) Q 0.9K 0.2L0.6 here Q is homogeneous of degree 0.8 and returns to scale are decreasing because Q(kK, kL) 0.9(kK)0.2(kL)0.6 Ak0.2K 0.2k0.6L0.6 k0.20.6(0.9K 0.2L0.6) k0.8(0.9K 0.2L0.6) Note that the returns to scale of a Cobb-Douglas function will always equal the sum of the exponents , as is illustrated in part 6 of Example 8.
134 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 CONSTRAINED OPTIMIZATION OF COBB-DOUGLAS FUNCTIONS 6.41.
Optimize the following Cobb-Douglas production functions subject to the given constraints by (1) forming the Lagrange function and (2) finding the critical values as in Example 10.
a) q K 0.3L0.5 subject to 6K 2L 384 1) Q K 0.3L0.5 (384 6K 2L) 2) QK 0.3K 0.7L0.5 6 0 (6.67) QL 0.5K 0.3L0.5 2 0 (6.68) Q 384 6K 2L 0 (6.69) Rearrange, then divide (6.67) by (6.68) to eliminate .
0.3K0.7L0.5 0.5K 0.3L0.5 6 2 Subtracting exponents in division, 0.6K1L1 3 L K 3 0.6 L 5K Substitute L 5K in (6.69).
384 6K 2(5K) 0 K0 24 L0 120 Second-order conditions are tested in Problem 12.27(b).
b) q 10K 0.7L0.1, given PK 28, PL 10, and B 4000 1) Q 10K 0.7L0.1 (4000 28K 10L) 2) QK 7K0.3L0.1 28 0 (6.70) QL 1K 0.7L0.9 10 0 (6.71) Q 4000 28K 10L 0 (6.72) Divide (6.70) by (6.71) to eliminate .
7K0.3L0.1 1K 0.7L0.9 28 10 7K1L1 2.8 L K 2.8 7 L 0.4K Substituting in (6.72), K0 125 L0 50 See Problem 12.27(c) for the second-order conditions.
6.42.
Maximize the following utility functions subject to the given budget constraints, using the same steps as above.
a) u x0.6y0.25, given Px 8, Py 5, and B 680 1) U x0.6y0.25 (680 8x 5y) 2) Ux 0.6x0.4y0.25 8 0 (6.73) Uy 0.25x0.6y0.75 5 0 (6.74) U 680 8x 5y 0 (6.75) 135 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] Divide (6.73) by (6.74).
0.6x0.4y0.25 0.25x0.6y0.75 8 5 2.4x1y1 y 1.6 2 – 3x Substitute in (6.75).
x0 60 y0 40 b) u x0.8y0.2, given Px 5, Py 3, and B 75 1) U x0.8y0.2 (75 5x 3y) 2) Ux 0.8x0.2y0.2 5 0 (6.76) Uy 0.2x0.8y0.8 3 0 (6.77) U 75 5x 3y 0 (6.78) Divide (6.76) by (6.77).
0.8x0.2y0.2 0.2x0.8y0.8 5 3 4x1y1 y 5 – 3 5 –– 12x Substitute in (6.78).
x0 12 y0 5 CONSTRAINED OPTIMIZATION OF CES PRODUCTION FUNCTIONS 6.43.
Optimize the following CES production function subject to the given constraint by (1) forming the Lagrange function and (2) finding the critical values as in Example 12: q 80[0.4K0.25 (1 0.4)L0.25]1/0.25 subject to 5K 2L 150 1) Q 80(0.4K0.25 0.6L0.25)4 (150 5K 2L) 2) Using the generalized power function rule for QK and QL, QK 320(0.4K0.25 0.6L0.25)5(0.1K1.25) 5 0 32K1.25(0.4K0.25 0.6L0.25)5 5 0 (6.79) QL 320(0.4K0.25 0.6L0.25)5(0.15L1.25) 2 0 48L1.25(0.4K0.25 0.6L0.25)5 2 0 (6.80) Q 150 5K 2L 0 (6.81) Rearrange, then divide (6.79) by (6.80) to eliminate .
32K1.25(0.4K0.25 0.6L0.25)5 48L1.25(0.4K0.25 0.6L0.25)5 5 2 32K1.25 48L1.25 2.5 K1.25 3.75L1.25 Take the 1.25 root.
K (3.75)1/1.25L (3.75)0.8L To find (3.75)0.8, enter 3.75 on a calculator, press the yx key, then enter 0.8 followed by the / key to make it negative, and hit the key to find (3.75)0.8 0.34736.
136 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 Thus, K 0.35L Substitute in (6.81).
150 5(0.35L) 2L 0 L0 40 K0 14 6.44.
Optimize the CES production function q 100[0.2K(0.5) (1 0.2)L(0.5)]1/(0.5) subject to the constraint 10K 4L 4100, as in Problem 6.43.
1) Q 100(0.2K 0.5 0.8L0.5)2 (4100 10K 4L) 2) QK 200(0.2K 0.5 0.8L0.5)(0.1K0.5) 10 0 20K0.5(0.2K 0.5 0.8L0.5) 10 0 (6.82) QL 200(0.2K 0.5 0.8L0.5)(0.4L0.5) 4 0 80L0.5(0.2K 0.5 0.8L0.5) 4 0 (6.83) Q 4100 10K 4L 0 (6.84) Divide (6.82) by (6.83) to eliminate .
20K0.5(0.2K 0.5 0.8L0.5) 80L0.5(0.2K 0.5 0.8L0.5) 10 4 20K0.5 80L0.5 2.5 K0.5 10L0.5 Take the 0.5 root.
K (10)1/0.5L (10)2L K 0.01L Substitute in (6.84).
L0 1000 K0 10 PARTIAL DERIVATIVES AND DIFFERENTIALS 6.45.
Given Q 10K 0.4L0.6, (a) find the marginal productivity of capital and labor and (b) determine the effect on output of an additional unit of capital and labor at K 8, L 20.
a) MPK Q K 0.4(10)K0.6L0.6 4K0.6L0.6 MPL Q L 0.6(10)K 0.4L0.4 6K 0.4L0.4 b) Q (Q/K) K. For a 1-unit change in K, at K 8, L 20, Q 4K0.6L0.6 4(8)0.6(20)0.6.
Using a calculator, Q 4K0.6L0.6 4(8)0.6(20)0.6 4(0.28717)(6.03418) 6.93 Note: to find (8)0.6 on a calculator, enter 8, press the yx key, then enter 0.6 followed by the / key to make it negative, and hit the key to find (8)0.6 0.28717. To find (20)0.6, enter 20, press the yx key, then enter 0.6, and hit the key to find (20)0.6 6.03418.
For a 1-unit change in L, Q 6K 0.4L0.4 6(8)0.4(20)0.4 6(2.29740)(0.30171) 4.16 6.46.
Redo Problem 6.45, given Q 12K 0.3L0.5 at K 10, L 15.
a) MPK 3.6K0.7L0.5 MPL 6K 0.3L0.5 137 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] b) For a 1-unit change in K, at K 10, L 15, Q 3.6K0.7L0.5. Q 3.6(10)0.7(15)0.5 3.6(0.19953)(3.87298) 2.78 For a 1-unit change in L, Q 6(10)0.3(15)0.5 6(1.99526)(0.25820) 3.09 6.47.
Given Q 4KL, find (a) MPK and MPL, and (b) determine the effect on Q of a 1-unit change in K and L, when K 50 and L 600.
a) Q 4KL 4(KL)1/2. By the generalized power function rule, MPK QK 2(KL)1/2(L) 2L KL MPL QL 2(KL)1/2(K) 2K KL b) For a 1-unit change in K at K 50, L 600, Q 2[50(600)]1/2(600) 2(0.00577)(600) 6.93 For a 1-unit change in L, Q 2[50(600)]1/2(50) 2(0.00577)(50) 0.58 6.48.
Redo Problem 6.47, given Q 2KL, where K 100 and L 1000.
a) Q 2(KL)1/2 MPK (KL)1/2(L) L KL MPL (KL)1/2(K) K KL b) For a 1-unit change in K at K 100, L 1000, Q [100(1000)]1/2(1000) (0.00316)(1000) 3.16 For a 1-unit change in L, Q [100(1000)]1/2(100) (0.00316)(100) 0.316 6.49.
A company’s sales s have been found to depend on price P, advertising A, and the number of field representatives r it maintains.
s (12,000 900P)A1/2r1/2 Find the change in sales associated with (a) hiring another field representative, (b) an extra $1 of advertising, (c) a $0.10 reduction in price, at P $6, r 49, and A $8100.
a) s s r r 1 2 (12,000 900P)A1/2r1/2 r 1 – 212,000 900(6)1/2(49)1/2(1) 1 – 2(6600)(90)(1 – 7) 42,429 b) s s A A 1 2 (12,000 900P)A1/2r1/2 A 1 2 (6600) 1 90 (7)(1) 256.67 c) s s P P 900A1/2r1/2 P 900(90)(7)(0.10) 56,700 6.50.
Given the sales function for a firm similar to the one in Problem 6.49: s (15,000 1000P)A2/3r1/4, estimate the change in sales from (a) hiring an extra field representative, (b) a $1 increase in advertising, and (c) a $0.01 reduction in price, when P 4, A $6000, and r 24.
138 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 a) s 1 – 4(15,000 1000P)A2/3r3/4 r 1 – 4(11,000)(6000)2/3(24)3/4(1) 2750(330.19)(0.09222) 83,740 b) s 2 – 3(15,000 1000P)A1/3r1/4 A 2 – 3(11,000)(6000)1/3(24)1/4(1) (7333.33)(0.05503)(2.21336) 893 c) s 1000A2/3r1/4 P 1000(6000)2/3(24)1/4(0.01) 10(330.19)(2.21336) 7308 6.51.
Given the equation for a production isoquant 16K 1/4L3/4 2144 use the implicit function rule from Section 5.10 to find the slope of the isoquant dK/dL which is the marginal rate of technical substitution (MRTS).
Set the equation equal to zero to get F(K, L) 16K 1/4L3/4 2144 0 Then from the implicit function rule in Equation (5.13), dK dL FL FK 12K 1/4L1/4 4K3/4L3/4 3K L MRTS Compare this answer with that in Problem 4.24.
6.52.
Given the equation for the production isoquant 25K 3/5L2/5 5400 find the MRTS, using the implicit function rule.
Set up the implicit function, F(K, L) 25K 3/5L2/5 5400 0 and use (5.13).
dK dL FL FK 10K 3/5L3/5 15K2/5L2/5 2K 3L MRTS Compare this answer with that in Problem 4.25.
PROOFS 6.53.
Use the properties of homogeneity to show that a strict Cobb-Douglas production function q AK L, where 1, exhibits constant returns to scale.
Multiply each of the inputs by a constant k and factor.
q(kK, kL) A(kK)(kL) AkK kL k(AK L) k(q) As explained in Section 6.9, if 1, returns to scale are constant; if 1, returns to scale are increasing; and if 1, returns to scale are decreasing.
139 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 6.54.
Given the utility function u Axayb subject to the budget constraint Pxx Pyy B, prove that at the point of constrained utility maximization the ratio of prices Px/Py must equal the ratio of marginal utilities MUx/MUy.
U Axayb (B Pxx Pyy) Ux aAx11yb Px 0 (6.85) Uy bAxayb1 Py 0 (6.86) U B Pxx Pyy 0 where in (6.85) aAxa1yb ux MUx, and in (6.86) bAxayb1 uy MUy.
From (6.85), aAxa1yb Px MUx Px From (6.86), bAxayb1 Py MUy Py Equating ’s, MUx Px MUy Py MUx MUy Px Py Q.E.D.
6.55.
Given a generalized Cobb-Douglas production function q AK L subject to the budget constraint PKK PLL B, prove that for constrained optimization the least-cost input ratio is K L PL PK Using the Lagrangian method, Q AK L (B PKK PLL) QK AK 1L PK 0 (6.87) QL AK L1 PL 0 (6.88) Q B PKK PLL 0 From (6.87) and (6.88), AK 1L PK AK L1 PL Rearranging terms, PL PK AK L1 AK 1L where L1 L/L and 1/K 1 K/K . Thus, PL PK K L K L PL PK Q.E.D.
(6.89) 6.56.
Prove that for a linearly homogeneous Cobb-Douglas production function Q AK L, the output elasticity of capital (QK) and the output elasticity of labor (QL).
From the definition of output elasticity, QK Q/K Q/K and QL Q/L Q/L 140 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 Since 1, let 1 and let k K/L. Then Q AK L1 A K L L AkL Find the marginal functions.
Q K AK 1L1 AK 1L(1) A K L 1 Ak1 Q L (1 )AK L (1 )A K L (1 )Ak Find the average functions.
Q K AkL K Ak K Ak1 Q L AkL L Ak Then divide the marginal functions by their respective average functions to obtain .
QK Q/K Q/K Ak1 Ak1 Q.E.D.
QL Q/L Q/L (1 )Ak Ak 1 Q.E.D.
6.57.
Equation (6.89) gave the least-cost input ratio for a generalized Cobb-Douglas production function. Prove that the elasticity of substitution of any generalized Cobb-Douglas production function is unitary, i.e., that 1.
In Section 6.10, the elasticity of substitution is defined as the percentage change in the least-cost K/L ratio resulting from a small percentage change in the input-price ratio PL/PK.
d(K/L) K/L d(PL/PK) PL/PK d(K/L) d(PL/PK) K/L PL/PK (6.90) Since and are constants in (6.89) and PK and PL are independent variables, K/L can be considered a function of PL/PK. Noting that in the second ratio of (6.90), the marginal function divided by the average function, first find the marginal function of (6.89).
d(K/L) d(PL/PK) Then find the average function by dividing both sides of (6.89) by PL/PK.
K/L PL/PK Substituting in (6.90), d(K/L) d(PL/PK) K/L PL/PK / / 1 Q.E.D.
141 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] 6.58.
Use the least-cost input ratio for a Cobb-Douglas function given in (6.89) to check the answer to Example 10, where q K 0.4L0.5, PK 3, and PL 4.
With 0.4 and 0.5, from (6.89), K L 0.4(4) 0.5(3) 1.6 1.5 Capital and labor must be used in the ratio of 16K:15L. This confirms the answer found in Example 10 of K0 16, L0 15.
6.59.
Given the CES production function q A[K (1 )L]1/ (6.91) and bearing in mind from Problem 6.54 that the ratio of prices must equal the ratios of marginal products if a function is to be optimized, (a) prove that the elasticity of substitution of a CES production is constant and (b) demonstrate the range that may assume.
a) First-order conditions require that Q/L Q/K PL PK (6.92) Using the generalized power function rule to take the first-order partials of (6.91), Q L 1 A[K (1 )L](1/)1()(1 )L1 Canceling ’s, rearranging 1 , and adding the exponents (1/) 1, we get Q L (1 )A[K (1 )L](1)/L(1) Substituting A1/A A for A, Q L (1 ) A1 A [K (1 )L](1)/L(1) From (6.91), A1[K (1 )L](1)/ Q1 and L(1) 1/L1. Thus, Q L 1 A Q L 1 (6.93) Similarly, Q K A Q K 1 (6.94) Substituting (6.93) and (6.94) in (6.92), which leads to the cancellation of A and Q, 1 K L 1 PL PK K L 1 1 PL PK K ¯ L ¯ 1 1/(1) PL PK 1/(1) (6.95) Since and are constants, by considering K ¯ /L ¯ a function of PL/PK, as in Problem 6.57, we can find 142 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 the elasticity of substitution as the ratio of the marginal and average functions. Simplifying first by letting h 1 1/(1) K ¯ L ¯ h PL PK 1/(1) (6.96) The marginal function is d(K ¯ /L ¯) d(PL/PK) h 1 PL PK 1/(1)1 (6.97) and the average function is K ¯ /L ¯ PL/PK h(PL/PK)1/(1) PL/PK h PL PK 1/(1)1 (6.98) By dividing the marginal function in (6.97) by the average function in (6.98), the elasticity of substitution is d(K/L) d(PL/PK) K/L PL/PK h 1 (PL/PK)1/(1)1 h(PL/PK)1/(1)1 1 1 (6.99) Since is a given parameter, 1/(1 ) is a constant.
b) If 1 0, 1. If 0, 1. If 0 , 1.
6.60.
Prove that the CES production function is homogeneous of degree 1 and thus has constant returns to scale.
From (6.91), Q A[K (1 )L]1/ Multiplying inputs K and L by k, as in Section 6.7, f(kK, kL) A[(kK) (1 )(kL)]1/ A{k[K (1 )L]}1/ A(k)1/[K (1 )L]1/ kA[K (1 )L]1/ kQ Q.E.D.
6.61.
Find the elasticity of substitution for the CES production function, q 75(0.3K0.4 0.7L0.4)2.5, given in Example 12.
From (6.99), 1 1 where 0.4. Thus, 1/(1 0.4) 0.71.
6.62.
Use the optimal K/L ratio in (6.95) to check the answer in Example 12 where q 75(0.3K0.4 0.7L0.4)2.5 was optimized under the constraint 4K 3L 120, giving K ¯ 11.25 and L ¯ 25.
From (6.95), K ¯ L ¯ 1 PL PK 1/(1) 143 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] Substituting 0.3, 1 0.7, and 0.4, K ¯ L ¯ 0.3 0.7 3 4 1/1.4 0.9 2.8 0.71 (0.32)0.71 0.45 With K ¯ 11.25 and L ¯ 25, K ¯ /L ¯ 11.25/25 0.45.
6.63.
Use (6.95) to check the answer to Problem 6.43 where q 80(0.4K0.25 0.6L0.25)4 was optimized subject to the constraint 5K 2L 150 at K ¯ 14 and L ¯ 40.
Substituting 0.4, 1 0.6, and 0.25 in (6.95), K ¯ L ¯ 0.4 0.6 2 5 1/1.25 0.8 3 0.8 0.35 Substituting K ¯ 14 and L ¯ 40, 1 – 4 4 – 0 0.35.
6.64.
Find the elasticity of substitution from Problem 6.63.
From (6.99), 1 1 1 1 0.25 0.8 6.65.
Use (6.95) to check the answer to Problem 6.44 where q 100(0.2K 0.5 0.8L0.5)2 was optimized subject to the constraint 10K 4L 4100 at K ¯ 10 and L ¯ 1000.
With 0.2, 1 0.8, and 0.5, K ¯ L ¯ 0.2 0.8 4 10 1/(10.5) 0.8 8 2 (0.1)2 0.01 Substituting K ¯ 10 and L ¯ 1000, 10 –––– 1000 0.01.
6.66.
Find the elasticity of substitution from Problem 6.65.
From (6.99), 1 1 1 1 0.5 2 6.67.
(a) Use the elasticity of substitution found in Problem 6.64 to estimate the effect on the least-cost (K ¯ /L ¯) ratio in Problem 6.43 if PL increases by 25 percent. (b) Check your answer by substituting the new PL in (6.95).
a) The elasticity of substitution measures the relative change in the K ¯ /L ¯ ratio brought about by a relative change in the price ratio PL/PK. If PL increases by 25 percent, PL 1.25(2) 2.5. Thus, PL/PK 2.5/5 0.5 vs. 2 – 5 0.4 in Problem 6.43.
The percentage increase in the price ratio, therefore, is (0.5 0.4)/0.4 0.25. With the elasticity of substitution 0.8 from Problem 6.64, the expected percentage change in the K ¯ /L ¯ ratio is (K ¯ /L ¯) K ¯ /L ¯ 0.8(0.25) 0.2 or 20% With (K ¯ /L ¯)1 0.35, (K ¯ /L ¯)2 1.2(0.35) 0.42.
b) Substituting PL 2.5 in (6.95), K ¯ L ¯ 0.4 0.6 2.5 5 0.8 1 3 0.8 0.42 144 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS [CHAP . 6 6.68.
(a) Use the elasticity of substitution to estimate the new K ¯ /L ¯ ratio if the price of capital decreases by 20 percent. Assume the initial data of Problem 6.43. (b) Check your answer.
a) If PK decreases by 20 percent, PK 0.8(5) 4. Thus, PL/PK 2 – 4 0.5 which is a 25 percent increase in the PL/PK ratio, as seen above. Therefore, (K ¯ /L ¯) K ¯ /L ¯ 0.8(0.25) 0.2 or 20% and (K ¯ /L ¯)2 1.2(0.35) 0.42.
b) Substituting PK 4 in (6.95), K ¯ L ¯ 0.4 0.6 2 4 0.8 0.8 2.4 0.8 0.42 6.69.
(a) If the price of labor decreases by 10 percent in Problem 6.44, use the elasticity of substitution to estimate the effect on the least-cost K ¯ /L ¯ ratio. (b) Check your answer.
a) If PL decreases by 10 percent, PL 0.9(4) 3.6, and the PL/PK ratio also decreases by 10 percent.
With a 10 percent decrease in PL/PK and an elasticity of substitution 2, (K ¯ /L ¯) K ¯ /L ¯ 2(0.10) 0.20 or 20% With the old K ¯ /L ¯ 0.01, (K ¯ /L ¯)2 (1 0.2)(0.01) 0.8(0.01) 0.008.
b) Substituting PL 3.6 in (6.95), K ¯ L ¯ 0.2 0.8 3.6 10 2 0.72 8 2 (0.09)2 0.0081 145 CALCULUS OF MULTIVARIABLE FUNCTIONS IN ECONOMICS CHAP . 6] CHAPTER 7 Exponential and Logarithmic Functions 7.1 EXPONENTIAL FUNCTIONS Previous chapters dealt mainly with power functions, such as y xa, in which a variable base x is raised to a constant exponent a. In this chapter we introduce an important new function in which a constant base a is raised to a variable exponent x. It is called an exponential function and is defined as y ax a 0 and a 1 Commonly used to express rates of growth and decay, such as interest compounding and depreciation, exponential functions have the following general properties. Given y ax, a 0, and a 1: 1.
The domain of the function is the set of all real numbers; the range of the function is the set of all positive real numbers, i.e., for all x, even x 0, y 0.
2.
For a 1, the function is increasing and convex; for 0 a 1, the function is decreasing and convex.
3.
At x 0, y 1, independently of the base.
See Example 1 and Problems 7.1 and 7.2; for a review of exponents, see Section 1.1 and Problem 1.1.
EXAMPLE 1.
Given (a) y 2x and (b) y 2x (1 – 2)x, the above properties of exponential functions can readily be seen from the tables and graphs of the functions in Fig. 7-1. More complicated exponential functions are estimated with the help of the yx key on pocket calculators.
146 7.2 LOGARITHMIC FUNCTIONS Interchanging the variables of an exponential function f defined by y ax gives rise to a new function g defined by x ay such that any ordered pair of numbers in f will also be found in g in reverse order. For example, if f(2) 4, then g(4) 2; if f(3) 8, then g(8) 3. The new function g, the inverse of the exponential function f, is called a logarithmic function with base a. Instead of x ay, the logarithmic function with base a is more commonly written y loga x a 0, a 1 Loga x is the exponent to which a must be raised to get x. Any positive number except 1 may serve as the base for a logarithm. The common logarithm of x, written log10 x or simply log x, is the exponent to which 10 must be raised to get x. Logarithms have the following properties. Given y loga x, a 0, a 1: 1.
The domain of the function is the set of all positive real numbers; the range is the set of all real numbersthe exact opposite of its inverse function, the exponential function.
2.
For base a 1, f(x) is increasing and concave. For 0 a 1, f(x) is decreasing and convex.
3.
At x 1, y 0 independent of the base.
See Examples 2 to 4 and Problems 7.5 and 7.6.
EXAMPLE 2.
A graphoftwofunctionsfandginwhichxandyareinterchanged,suchasy 2xandx 2yinFig.7-2, reveals that one function is a mirror image of the other along the 45 line y x, such that if f(x) y, then g(y) x.
Recall that x 2y is equivalent to and more commonly expressed as y log2 x.
a) y 2x b) y log2 x ↔x 2y x y x y 3 2 1 0 1 2 3 1 – 8 1 – 4 1 – 2 1 2 4 8 1 – 8 1 – 4 1 – 2 1 2 4 8 3 2 1 0 1 2 3 Fig. 7-2 147 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] Fig. 7-1 a) y 2x b) y 2x (1 – 2)x (3, 1 – 8) x 2y or y log2 x (1 – 8, 3) y x y 2x y 2x y 2x (1/2)x EXAMPLE 3.
Knowing that the common logarithm of x is the power to which 10 must be raised to get x, it follows that log 10 1 log 100 2 log 1000 3 since 101 10 since 102 100 since 103 1000 log 1 log 0.1 log 0.01 0 1 2 since 100 1 since 101 0.1 since 102 0.01 EXAMPLE 4.
For numbers that are exact powers of the base, logs are easily calculated without the aid of calculators.
log7 49 log36 6 log3 1 – 9 2 1 – 2 2 since 72 49 since 361/2 6 since 32 1 – 9 log2 16 log16 2 log2 1 – 8 4 1 – 4 3 since 24 16 since 161/4 2 since 23 1 – 8 For numbers that are not exact powers of the base, log tables or calculators are needed.
7.3 PROPERTIES OF EXPONENTS AND LOGARITHMS Assuming a, b 0; a, b 1; and x and y are any real numbers: 1.
ax · ay axy 4.
(ax)y axy 2.
1 ax ax 5.
ax · bx (ab)x 3.
ax ay axy 6.
ax bx a b x For a, x, and y positive real numbers, n a real number, and a 1: 1.
loga xy loga x loga y 3.
loga xn n loga x 2.
loga x y loga x loga y 4.
loga n x 1 n loga x Properties of exponents were treated in Section 1.1 and Problem 1.1. Properties of logarithms are treated in Example 5 and Problems 7.12 to 7.16.
Table 7.1 x log x x log x x log x x log x 1 2 3 4 5 0.0000 0.3010 0.4771 0.6021 0.6990 6 7 8 9 10 0.7782 0.8451 0.9031 0.9542 1.0000 11 12 13 14 15 1.0414 1.0792 1.1139 1.1461 1.1761 16 17 18 19 20 1.2041 1.2304 1.2553 1.2788 1.3010 EXAMPLE 5.
The problems below are kept simple and solved by means of logarithms to illustrate the properties of logarithms.
a) x 7 · 2 log x log 7 log 2 log x 0.8451 0.3010 log x 1.1461 x 14 b) x 18 3 log x log 18 log 3 log x 1.2553 0.4771 log x 0.7782 x 6 148 EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 7 c) x 32 log x 2 log 3 log x 2(0.4771) log x 0.9542 x 9 d) x 3 8 log x 1 – 3 log 8 log x 1 – 3(0.9031) log x 0.3010 x 2 7.4 NATURAL EXPONENTIAL AND LOGARITHMIC FUNCTIONS The most commonly used base for exponential and logarithmic functions is the irrational number e. Expressed mathematically, e lim n→1 1 n n 2.71828 (7.1) Exponentialfunctionstobaseearecallednaturalexponentialfunctionsandarewritteny ex;logarithmic functions to base e are termed natural logarithmic functions and are expressed as y loge x or, more commonly, ln x. Thus ln x is simply the exponent or power to which e must be raised to get x.
As with other exponential and logarithmic functions to a common base, one function is the inverse of the other, such that the ordered pair (a, b) will belong to the set of ex if and only if (b, a) belongs to the set of ln x. Natural exponential and logarithmic functions follow the same rules as other exponential and logarithmic functions and are estimated with the help of tables or the ex and ln x keys on pocket calculators. See Problems 7.3, 7.4, and 7.6.
7.5 SOLVING NATURAL EXPONENTIAL AND LOGARITHMIC FUNCTIONS Since natural exponential functions and natural logarithmic functions are inverses of each other, one is generally helpful in solving the other. Mindful that ln x signifies the power to which e must be raised to get x, it follows that: 1.
e raised to the natural log of a constant (a 0), a variable (x 0), or a function of a variable [f(x) 0] must equal that constant, variable, or function of the variable: eln a a eln x x eln f (x) f(x) (7.2) 2.
Conversely, the natural log of e raised to the power of a constant, variable, or function of a variable must also equal that constant, variable, or function of the variable: ln ea a ln ex x ln ef(x) f(x) (7.3) See Example 6 and Problems 7.18 to 7.22.
EXAMPLE 6.
The principles of (7.2) and (7.3) are used below to solve the given equations for x.
a) 5ex2 120 1) Solve algebraically for ex2, 5ex2 120 ex2 24 2) Take the natural log of both sides to eliminate e.
ln ex2 ln 24 From (7.3), x 2 ln 24 x ln 24 2 Enter 24 on your calculator and press the ln x key to find ln 24 3.17805. Then substitute and solve.
x 3.17805 2 1.17805 149 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] b) 6 ln x 7 12.2 1) Solve algebraically for ln x, 6 ln x 19.2 ln x 3.2 2) Set both sides of the equation as exponents of e to eliminate the natural log expression, eln x e3.2 From (7.2), x e3.2 Enter 3.2 on your calculator and press the ex key to find e3.2 24.53253 and substitute.
x 24.53253 Note: On many calculators the ex key is the inverse (shift, or second function) of the ln x key, and to activate the ex key, one must first press the INV ( Shift , or 2ndF ) key followed by the ln x key.
7.6 LOGARITHMIC TRANSFORMATION OF NONLINEAR FUNCTIONS Linear algebra and regression analysis involving ordinary or two-stage least squares which are common tools in economic analysis assume linear functions or equations. Some nonlinear functions, such as Cobb-Douglas production functions, can easily be converted to linear functions through simple logarithmic transformation; others, such as CES production functions, cannot. For example, from the properties of logarithms, it is clear that given a generalized Cobb-Douglas production function q AKL ln q ln A ln K ln L (7.4) which is log-linear. But given the CES production function, q A[K (1 )L]1/ ln q ln A 1 ln [K (1 )L] which is not linear even in logarithms because of K and L. Ordinary least-square estimation of the coefficients in a log transformation of a Cobb-Douglas production function, such as in (7.4), has the nice added feature that estimates for and provide direct measures of the output elasticity of K and L, respectively, as was proved in Problem 6.56.
Solved Problems GRAPHS 7.1.
Make a schedule for each of the following exponential functions with base a 1 and then sketch them on the same graph to convince yourself that (1) the functions never equal zero; (2) they all pass through (0, 1); and (3) they are all positively sloped and convex.
150 EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 7 a) y 3x b) y 4x c) y 5x a) b) c) x 3 2 1 0 1 2 3 y 1 –– 27 1 – 9 1 – 3 1 3 9 27 x 3 2 1 0 1 2 3 y 1 –– 64 1 –– 16 1 – 4 1 4 16 64 x 3 2 1 0 1 2 3 y 1 ––– 125 1 –– 25 1 – 5 1 5 25 125 Fig. 7-3 7.2.
Make a schedule for each of the following exponential functions with 0 a 1 and then sketch them on the same graph to convince yourself that (1) the functions never equal zero; (2) they all pass through (0, 1), and (3) they are all negatively sloped and convex.
a) y (1 – 3)x 3x b) y (1 – 4)x 4x c) y (1 – 5)x 5x a) b) c) x 3 2 1 0 1 2 3 y 27 9 3 1 1 -3 1 – 9 1 –– 27 x 3 2 1 0 1 2 3 y 64 16 4 1 1 – 4 1 –– 16 1 –– 64 x 3 2 1 0 1 2 3 y 125 25 5 1 1 – 5 1 –– 25 1 ––– 125 Fig. 7-4 7.3.
Using a calculator or tables, set up a schedule for each of the following natural exponential functions y ekx where k 0, noting (1) the functions never equal zero; (2) they all pass through (0, 1), and (3) they are all positively sloped and convex.
151 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] y 5x y 4x y 3x y 5x y 4x y 3x a) y e0.5x b) y ex c) y e2x a) b) c) x 2 1 0 1 2 y 0.37 0.61 1.00 1.65 2.72 x 2 1 0 1 2 y 0.14 0.37 1.00 2.72 7.39 x 2 1 0 1 2 y 0.02 0.14 1.00 7.39 54.60 Fig. 7-5 7.4.
Set up a schedule, rounding to two decimal places, for the following natural exponential functions y ekx where k 0, noting (1) the functions never equal zero; (2) they all pass through (0, 1); and (3) they are all negatively sloped and convex.
a) y e0.5x b) y ex c) y e2x a) b) c) x 2 1 0 1 2 y 2.72 1.65 1.00 0.61 0.37 x 2 1 0 1 2 y 7.39 2.72 1.00 0.37 0.14 x 2 1 0 1 2 y 54.60 7.39 1.00 0.14 0.02 Fig. 7-6 7.5.
Construct a schedule and draw a graph for the following functions to show that one is the mirror image and hence the inverse of the other, noting that (1) the domain of (a) is the range of (b) and the range of (a) is the domain of (b); and (2) a logarithmic function with 0 a 1 is a decreasing function and convex.
a) y (1 – 2)x 2x b) x (1 – 2)y or y log1/2 x a) b) x 3 2 1 0 1 2 3 y 8 4 2 1 1 – 2 1 – 4 1 – 8 x 8 4 2 1 1 – 2 1 – 4 1 – 8 y 3 2 1 0 1 2 3 Fig. 7-7 152 EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 7 y (1 – 2)x y x y log1/2x y e2x y ex y e0.5x y e2x y ex y e0.5x 7.6.
Given (a) y ex and (b) y ln x, and using a calculator or tables, construct a schedule and draw a graph for each of the functions to show that one function is the mirror image or inverse of the other, noting that (1) the domain of (a) is the range of (b) while the range of (a) is the domain of (b), (2) ln x is negative for 0 x 1 and positive for x 1; and (3) ln x is an increasing function and concave.
a) y ex b) y ln x x 2 1 0 1 2 y 0.13534 0.36788 1.00000 2.71828 7.38906 x 0.13534 0.36788 1.00000 2.71828 7.38906 y 2 1 0 1 2 Fig. 7-8 EXPONENTIAL-LOGARITHMIC CONVERSION 7.7.
Change the following logarithms to their equivalent exponential forms: a) log8 64 2 64 82 c) log7 1 – 7 1 1 – 7 71 b) log5 125 3 125 53 d) log3 1 –– 81 4 1 –– 81 34 e) log36 6 1 – 2 6 361/2 f) log16 2 1 – 4 2 161/4 g) loga y 6x y a6x h) log2 y 7x y 27x 7.8.
Convert the following natural logarithms to natural exponential functions: a) ln 32 3.46574 32 e3.46574 b) ln 0.8 0.22314 0.8 e0.22314 c) ln 20 2.99573 20 e2.99573 d) ln 2.5 0.91629 2.5 e0.91629 e) ln y 4x y e4x f) ln y 2t 1 y e2t1 7.0.
Change the following exponential forms to logarithmic forms: a) 81 92 log9 81 2 b) 32 25 log2 32 5 153 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] y ex y x y ln x c) 1 – 9 32 log3 1 – 9 2 (d) 1 –– 16 24 log2 1 –– 16 4 e) 5 1251/3 log125 5 1 – 3 f) 11 1211/2 log121 11 1 – 2 g) 27 93/2 log9 27 3 – 2 h) 64 2563/4 log256 64 3 – 4 7.10.
Convert the following natural exponential expressions to equivalent natural logarithmic forms: a) 4.8 e1.56862 ln 4.8 1.56862 b) 15 e2.70805 ln 15 2.70805 c) 0.6 e0.51083 ln 0.6 0.51083 d) 130 e4.86753 ln 130 4.86753 e) y e(1/2)t ln y 1 – 2t f) y et5 ln y t 5 7.11.
Solve the following for x, y, or a by finding the equivalent expression: a) y log30 900 900 30y y 2 b) y log2 1 –– 32 1 –– 32 2y y 5 c) log4 x 3 x 43 x 64 d) log81 x 3 – 4 x 813/4 x 27 e) loga 27 3 27 a3 a 271/3 a 3 f) loga 4 2 – 3 4 a2/3 a 43/2 a 8 g) loga 125 3 – 2 125 a3/2 a 1252/3 a 25 h) loga 8 3 – 4 8 a3/4 a 84/3 a 16 PROPERTIES OF LOGARITHMS AND EXPONENTS 7.12.
Use the properties of logarithms to write the following expressions as sums, differences, or products: a) loga 56x loga 56x loga 56 logax b) loga 33x4 loga 33x4 loga 33 4 loga x c) loga x2 y3 loga x2 y3 2 loga x 3 loga y d) loga u5 v4 logau5v4 5 logau 4 loga v 154 EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 7 e) loga 6x 7y loga 6x 7y loga 6x loga 7y loga 6 loga x (loga 7 loga y) loga 6 loga x loga 7 loga y f) loga x7 y4 g) loga 3 x loga x7 y4 7 loga x 4 loga y loga 3 x 1 – 3 loga x 7.13.
Use of the properties of logarithms to write the following natural logarithmic forms as sums, differences, or products: a) ln 76x3 ln 763 ln 76 3 ln x b) ln x5y2 ln x5 y2 5 ln x 2 ln y c) ln x4 y6 d) ln 8x 9y ln x4 y6 4 ln x 6 ln y ln 8x 9y ln 8x ln 9y ln 8 ln x (ln 9 ln y) ln 8 ln x ln 9 ln y e) ln 4 x ln 4 x 1 – 4 ln x f) ln(x 5 y) ln(x 5 y) 5 ln x 1 – 2 ln y g) ln 3 5 x y h) ln x7 y4 ln 3 5 x y ln 3 1 – 5 ln x 1 – 2 ln y ln x7 y4 1 – 2(7 ln x 4 ln y) 7.14.
Use the properties of exponents to simplify the following exponential expressions, assuming a, b 0 and a b: a) ax · ay ax · ay axy b) a4x · a5y a4x · a5y a4x5y c) a2x a3y d) ax bx a2x a3y a2x3y ax bx a b x e) a7x f) (ax)4y a7x (a7x)1/2 a(7/2)x (ax)4y a4xy 7.15.
Simplify the following natural exponential expressions: a) e5x · e2y b) (e3x)5 e5x · e2y e5x2y (e3x)5 e15x 155 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] c) e8x e6x d) e4x e7x e8x e6x e8x6x e2x e4x e7x e4x7x e3x 1 e3x 7.16.
Simplify the following natural logarithmic expressions: a) ln 8 ln x b ln x5 ln x3 ln 8 ln x ln 8x ln x5 ln x3 ln x5 x3 ln x2 2 ln x c) ln 12 ln 5 ln 6 d) ln 7 ln x ln 9 ln 12 ln 5 ln 6 ln 12 · 5 6 ln 10 ln 7 ln x ln 9 ln 7 · 9 x ln 63 x e) 1 – 2 ln 81 f) 5 ln 1 – 2 1 – 2 ln 81 ln 811/2 ln 9 5 ln 1 – 2 ln (1 – 2)5 ln 1 –– 32 g) 1 – 3 ln 27 4 ln 2 1 – 3 ln 27 4 ln 2 ln 271/3 ln 24 ln (3 · 16) ln 48 h) 2 ln 4 1 – 3 ln 8 2 ln 4 1 – 3 ln 8 ln 42 ln 81/3 ln 16 –– 2 ln 8 7.17.
Simplify each of the following exponential expressions: a) e3 ln x e3 ln x eln x3 But from (7.2), eln f(x) f(x), so e3 ln x x3 b) e4 ln x5 ln y e4 ln x5 ln y eln x4 · eln y5 x4y5 c) e(1/2) ln 6x e(1/2) ln 6x eln (6x)1/2 (6x)1/2 6x d) e4 ln x9 ln y e4 ln x9 ln y eln x4 eln y9 x4 y9 SOLVING EXPONENTIAL AND LOGARITHMIC FUNCTIONS 7.18.
Use the techniques from Section 7.5 to solve the following natural exponential functions for x: a) 3e5x 8943 1) Solve algebraically for e5x.
e5x 2981 156 EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 7 2) Then take the natural log of both sides to eliminate e.
ln e5x ln 2981 From (7.3), 5x ln 2981 To find the value of ln 2981, enter 2981 on a calculator and press the ln x key to find ln 2981 8.00001 8. Then substitute and solve algebraically.
5x 8 x 1.6 b) 4e3x1.5 360 1) Solve for e3x1.5.
e3x1.5 90 2) Take the natural log.
ln e3x1.5 ln 90 From (7.3), 3x 1.5 ln 90 For ln 90, enter 90 on a calculator and press the ln x key to find ln 90 4.49981 4.5. Substitute and solve.
3x 1.5 4.5 x 2 c) 1 – 2ex2 259 1) Solve for ex2 ex2 518 2) Take the natural log.
ln ex2 ln 518 From (7.3), x2 6.24998 6.25 x 2.5 7.19.
Using the techniques of Section 7.5, solve the following natural logarithmic functions for x: a) 5 ln x 8 14 1) Solve algebraically for ln x.
5 ln x 6 ln x 1.2 2) Set both sides of the equation as exponents of e to eliminate the natural log.
eln x e1.2 From (7.2), x e1.2 To find the value of e1.2, enter 1.2 on a calculator, press the ex key to find e1.2 3.32012, and substitute. If the ex key is the inverse of the ln x key, enter 1.2, then press the lNV key followed by the ln x key.
x 3.32012 3.32 b) ln (x 4)2 3 1) Simplify with the laws of logs, then solve for ln x.
2 ln (x 4) 3 ln (x 4) 1.5 2) eln(x4) e1.5 From (7.2), x 4 e1.5 Using a calculator, x 4 4.48169 x 4.48169 4 0.48169 157 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] c) In x 34 2.55 1) Simplify and solve, 1 – 2 ln (x 34) 2.55 ln (x 34) 5.1 2) From (7.2) x 34 e5.1 164 x 130 7.20.
Solve each of the following equations for x in terms of y: a) loga x y3 b) loga x loga 3 loga y x ay3 x 3y since addition in logs multiplication in algebra c) ln x 3y d) ln x loga y e) loga x ln y x e3y x eloga y x aln y f) y gehx To solve for x when x is an exponent in a natural exponential function, take the natural log of both sides and solve algebraically, as follows: ln y ln g hx ln e ln g hx x ln y ln g h g) y aex1 ln y ln a (x 1) ln e ln a x 1 x ln y ln a 1 h) y p(1 i)x When x is an exponent in an exponential function with a base other than e, take the common log of both sides and solve algebraically.
log y log p x log(1 i) x log y log p log(1 i) 7.21.
Use common logs to solve each of the following equations: a) y 625(0.8) 1) Take the common log of both sides of the equation, using the properties of logarithms from Section 7.3.
log y log 625 log 0.8 To find the logs of 625 and 0.8, enter each number individually on the calculator, press the log x key to get the common log of each number, and perform the required arithmetic.
log y 2.79588 (0.09691) 2.69897 2) Since log y 2.69897 indicates that 10 must be raised to the 2.69897 power to get y, to find the antilogarithm of 2.69897 and solve for y, enter 2.69897 on a calculator, press the 10x key to find that 102.69897 500, and substitute. If the 10x key is the inverse of the log x key, enter 2.69897 and press the INV key followed by the log x key.
y antilog 2.69897 102.69897 500 158 EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 7 b) y 40 ––– 100 1) log y log 40 log 100 2) log y 1.60206 2 0.39794 y antilog (0.39794) 100.39794 0.4 c) y 130 0.25 1) log y log 130 log 0.25 2) log y 2.11394 (0.60206) 2.71600 y antilog 2.71600 102.71600 519.996 520 d) y (1.06)10 1) log y 10 log 1.06 2) log y 10(0.02531) 0.2531 y antilog 0.2531 100.2531 1.791 e) y 10240.2 1) log y 0.2 log 1024 2) log y 0.2(3.0103) 0.60206 y antilog 0.60206 100.60206 4 f) y 5 1024 1) log y 1 – 5 log 1024 2) log y 1 – 5(3.0103) 0.60206 y antilog 0.60206 100.60206 4 The answer is the same as in part (e) because y 10241/5 10240.2 4. Taking the fifth root is the same thing as raising to the 0.2 or one-fifth power. in one case the log is divided by 5; in the other, it is multiplied by 0.2.
7.22.
Use natural logs to solve the following equations: a) y 12.53 1) ln y 3 ln 12.5 To find the natural log of 12.5, enter 12.5 on a calculator and press the ln x key to find ln 12.5 2.52573. Then substitute.
2) ln y 3(2.52573) 7.57719 Since ln y 7.57719 indicates that e must be raised to the 7.57719 power to get y, to find the antilogarithme of 7.57719 and solve for y, enter 7.57719 on a calculator press the ex key to find that e7.57719 1953.1 and substitute. If the ex key is the inverse of the ln x key, enter 7.57719 and press the INV key followed by the ln x key.
y antiloge 7.57719 e7.57719 1953.1 b) y 4 28,561 1) ln y 1 – 4 ln 28,561 2) ln y 1 – 4(10.25980) 2.56495 y antiloge 2.56495 e2.56495 13 159 EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 7] CHAPTER 8 Exponential and Logarithmic Functions in Economics 8.1 INTEREST COMPOUNDING A given principal P compounded annually at an interest rate i for a given number of years t will have a value S at the end of that time given by the exponential function S P(1 i)t (8.1) If compounded m times a year for t years, S P1 i m mt (8.2) If compounded continuously at 100 percent interest for 1 year, S P lim m→1 1 m m P(2.71828) Pe For interest rates r other than 100 percent and time periods t other than 1 year, S Pert (8.3) For negative growth rates, such as depreciation or deflation, the same formulas apply, but i and r are negative. See Example 1 and Problems 8.1 to 8.6 and 8.9 to 8.17.
EXAMPLE 1.
Find the value of $100 at 10 percent interest for 2 years compounded: 1.
Annually, S P(1 i)t.
S 100(1 0.10)2 121 160 2.
Semiannually, S P[1 (i/m)]mt, where m 2 and t 2.
S 1001 0.10 2 2(2) 100(1 0.05)4 To find the value of (1.05)4, enter 1.05 on a calculator, press the yx key, enter 4, and hit the key to find (1.05)4 1.2155. Then substitute.
S 100(1.2155) 121.55 3.
Continuously, S Pert.
S 100e0.1(2) 100e0.2 For e0.2, enter 0.2 on a calculator, press the ex key to find that e0.2 1.2214, and substitute. If the ex key is the inverse (shift, or second function) of the ln x key, enter 0.2 and press the INV ( Shift , or 2nd F ) key followed by the ln x key.
S 100(1.2214) 122.14 8.2 EFFECTIVE VS. NOMINAL RATES OF INTEREST As seen in Example 1, a given principal set out at the same nominal rate of interest will earn different effective rates of interest which depend on the type of compounding. When compounded annually for 2 years, $100 will be worth $121; when compounded semiannually, S $121.55; when compounded continuously, S $122.14.
To find the effective annual rate of interest ie for multiple compounding: P(1 ie)t P1 i m mt Dividing by P and taking the tth root of each side, 1 ie 1 i m m ie 1 i m m 1 (8.4) To find the effective annual rate of interest for continuous compounding: 1 ie er ie er 1 (8.5) See Example 2 and Problems 8.7 and 8.8.
EXAMPLE 2.
Find the effective annual rate of interest for a nominal interest rate of 10 percent when compounded for 2 years (1) semiannually and (2) continuously.
1.
Semiannually, ie 1 i m m 1 (1.05)2 1 For (1.05)2, enter 1.05 on a calculator, press the x2 key, or press the yx key, and then enter 2 followed by the key, to find (1.05)2 1.1025, and substitute.
ie 1.1025 1 0.1025 10.25% 161 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS CHAP . 8] 2.
Continuously, ie er 1 e0.1 1 To find the value of e0.1, enter 0.1 on a calculator, press the ex key to learn e0.1 1.10517, and substitute.
ie 1.10517 1 0.10517 10.52% 8.3 DISCOUNTING A sum of money to be received in the future is not worth as much as an equivalent amount of money in the present, because the money on hand can be lent at interest to grow to an even larger sum by the end of the year. If present market conditions will enable a person to earn 8 percent interest compounded annually, $100 will grow to $108 by the end of the year. And $108 1 year from now, therefore, is worth (has a present value of) only $100 today.
Discounting is the process of determining the present value P of a future sum of money S. If under annual compounding S P(1 i)t then P S (1 i)t S(1 i)t (8.6) Similarly, under multiple compoundings P S[1 (i/m)]mt, and under continuous compounding P Sert. When finding the present value, the interest rate is called the discount rate. See Example 3 and Problems 8.18 to 8.22.
EXAMPLE 3.
The present value of a 5-year bond with a face value of $1000 and no coupons is calculated below.
It is assumed that comparable opportunities offer interest rates of 9 percent under annual compounding.
P S(1 i)t 1000(1 0.09)5 To find the value of (1.09)5, enter 1.09 on a calculator, press the yx key, and enter 5 by first entering 5 and then pressing the / key followed by the key to find (1.09)5 0.64993, and substitute.
P 1000(0.64993) 649.93 Thus, a bond with no coupons promising to pay $1000 5 years from now is worth approximately $649.93 today since $649.93 at 9 percent interest will grow to $1000 in 5 years.
8.4 CONVERTING EXPONENTIAL TO NATURAL EXPONENTIAL FUNCTIONS In Section 8.1 we saw that (1) exponential functions are used to measure rates of discrete growth, i.e., growth that takes place at discrete intervals of time, such as the end of the year or the end of the quarter as in ordinary interest compounding or discounting; and (2) natural exponential functions are used to measure rates of continuous growth, i.e., growth that takes place constantly rather than at discrete intervals, as in continuous compounding, animal development, or population growth. An exponential function S P(1 i/m)mt expressing discrete growth can be converted to an equivalent natural exponential function S Pert measuring continuous growth, by setting the two expressions equal to each other, and solving for r, as follows: P1 i m mt Pert By canceling P’s, 1 i m mt ert 162 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS [CHAP . 8 Taking the natural log of each side, ln1 i m mt ln ert mt ln1 i m rt Dividing both sides by t, r m ln1 i m Thus, S P1 i m mt Pem ln (1i/m)t (8.7) See Examples 4 and 5 and Problems 8.37 to 8.42.
EXAMPLE 4.
A natural exponential function can be used to determine the value of $100 at 10 percent interest compounded semiannually for 2 years, as shown below.
S Pert where r m ln(1 i/m). Thus, r 2 ln1 0.10 2 2 ln 1.05 2(0.04879) 0.09758 Substituting above, S 100e(0.09758)2 100e0.19516 Using a calculator here and throughout, S 100(1.2155) 121.55 as was found in Example 1.
Note that with natural exponential functions, the continuous growth is given by r in Pert. Thus, the continuous growth rate of $100 at 10 percent interest compounded semiannually is 0.09758, or 9.758 percent a year. That is to say, 9.758 percent interest at continuous compounding is equivalent to 10 percent interest when compounded semiannually.
EXAMPLE 5.
A small firm with current annual sales of $10,000 projects a 12 percent growth in sales annually.
Its projected sales in 4 years are calculated below in terms of an ordinary exponential function.
S 10,000(1 0.12)4 10,000(1.5735) 15,735 EXAMPLE 6.
The sales projections specified in Example 5 are recalculated below, using a natural exponential function with r m ln(1 i/m) and m 1.
r ln 1.12 0.11333 S 10,000e0.11333(4) 10,000(1.5735) 15,735 8.5 ESTIMATING GROWTH RATES FROM DATA POINTS Given two sets of data for a functionsales, costs, profitsgrowing consistently over time, annual growth rates can be measured and a natural exponential function estimated through a system of simultaneous equations. For example, if sales volume equals 2.74 million in 1996 and 4.19 million in 163 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS CHAP . 8] 2001, let t 0 for the base year 1996, then t 5 for 2001. Express the two sets of data points in terms of a natural exponential function S Pert, recalling that e0 1.
2.74 Per(0) P (8.8) 4.19 Per(5) (8.9) Substitute P 2.74 from (8.8) in (8.9) and simplify algebraically.
4.19 2.74e5r 1.53 e5r Take the natural log of both sides.
ln 1.53 0.42527 r S ln e5r 5r 5r 0.08505 8.5% 2.74e0.085t Substituting, With r 0.085, the rate of continuous growth per year is 8.5 percent. To find the rate of discrete growth i, recall that r m ln1 i m Thus, for annual compounding with m 1, 0.085 1 i i ln(1 i) antiloge 0.085 e0.085 1.08872 1.08872 1 0.08872 8.9% See Example 7 and Problems 8.43 to 8.45.
EXAMPLE 7.
Given the original information above, an ordinary exponential function for growth in terms of S P(1 i)t can also be estimated directly from the data.
Set the data in ordinary exponential form.
2.74 P(1 i)0 P (8.10) 4.19 P(1 i)5 (8.11) Substitute P 2.74 from (8.10) in (8.11) and simplify.
4.19 2.74(1 i)5 1.53 (1 i)5 Take the common log of both sides.
log 1.53 1 – 5(0.18469) log(1 i) 5 log(1 i) log(1 i) 0.03694 1 i antilog 0.03694 100.03694 1.08878 i 1.08878 1 0.08878 8.9% Substituting, S 2.74(1 0.089)t 164 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS [CHAP . 8 Solved Problems COMPOUNDING INTEREST 8.1.
Given a principal P of $1000 at 6 percent interest i for 3 years, find the future value S when the principal is compounded (a) annually, (b) semiannually, and (c) quarterly.
a) From (8.1), S P(1 i)t 1000(1 0.06)3 For (1.06)3, enter 1.06 on a calculator, press the yx key, enter 3 followed by the key to find (1.06)3 1.19102, then substitute.
S 1000(1.19102) 1191.02 b) From (8.2), S P1 i m mt 10001 0.06 2 2(3) 1000(1.03)6 For (1.03)6, enter 1.03, hit the yx key and then 6, and substitute.
S 1000(1.19405) 1194.05 c) S 10001 0.06 4 4(3) 1000(1.015)12 Enter 1.015, hit the yx key then 12, and substitute.
S 1000(1.19562) 1195.62 8.2.
Redo Problem 8.1, given a principal of $100 at 8 percent for 5 years.
a) S 100(1.08)5 100(1.46933) 146.93 b) S 1001 0.08 2 2(5) 100(1.04)10 100(1.48024) 148.02 c) S 1001 0.08 4 4(5) 100(1.02)20 100(1.48595) 148.60 8.3.
Redo Problem 8.1, given a principal of $1250 at 12 percent for 4 years.
a) S 1250(1.12)4 1250(1.57352) 1966.90 b) S 1250(1.06)8 1250(1.59385) 1992.31 c) S 1250(1.03)16 1250(1.60471) 2005.89 8.4.
Find the future value of a principal of $100 at 5 percent for 6 years when compounded (a) annually and (b) continually.
a) S 100(1.05)6 100(1.34010) 134.01 b) From (8.3), S Pert 100e0.05(6) 100e0.3 For e0.3, enter 0.3, hit the ex key, and substitute.
S 100(1.34986) 134.99 165 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS CHAP . 8] 8.5.
Redo Problem 8.4, given a principal of $150 at 7 percent for 4 years.
a) S 150(1.07)4 150(1.31080) 196.62 b) S 150e0.07(4) 150e0.28 150(1.32313) 198.47 8.6.
From Problems 8.4 and 8.5, use natural logs to find (a) S 100e0.3 and (b) S 150e0.28.
a) ln S ln 100 0.3 ln e 4.60517 0.3(1) 4.90517 S antiloge 4.90517 e4.90517 134.99 b) ln S ln 150 0.28 ln e 5.01064 0.28 5.29064 S antiloge 5.29064 e5.29064 198.47 8.7.
Find the effective annual interest rate on $100 at 6 percent compounded (a) semiannually and (b) continuously.
a) From (8.4), ie 1 i m m 1 1.0609 1 0.06090 6.09% b) From (8.5), ie er 1 e0.06 1 1.06184 1 0.06184 6.18% 8.8.
Calculate the rate of effective annual interest on $1000 at 12 percent compounded (a) quarterly and (b) continuously.
a) ie 1 i m m 1 (1.03)4 1 1.12551 1 0.12551 12.55% b) ie er 1 e0.12 1 1.12750 1 0.12750 12.75% TIMING 8.9.
Determine the interest rate needed to have money double in 10 years under annual compounding.
S P(1 i)t If money doubles, S 2P. Thus, 2P P(1 i)10.
Dividing by P, and taking the tenth root of each side, 2 (1 i)10 (1 i) 10 2 For 10 2, enter 2, press the x y key, then 10 followed by the key, and substitute. If the x y is the inverse (shift, or second function) of the yx key, enter 2, hit the INV ( Shift , or 2ndF ) key followed by the yx key, and then enter 10 and hit the key.
1 i i 1.07177 1.07177 1 0.07177 7.18% Note that since 10 2 21/10 20.1, 10 2 or any root can also be found with the yx key.
166 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS [CHAP . 8 8.10.
Determine the interest rate needed to have money double in 6 years when compounded semiannually.
S P1 i m mt 2P P1 i 2 2(6) 2 1 0.5i ln (1 0.5i) 1 0.5i 0.5i i 1(1 0.5i)12 12 2 1 –– 12 ln 2 1 –– 12(0.69315) 0.05776 e0.05776 1.05946 1.05946 1 0.05946 0.11892 11.89% 8.11.
What interest rate is needed to have money treble in 10 years when compounded quarterly?
S P1 i 4 4(10) If money trebles, 3P P1 i 4 40 3 1 0.25i (1 0.25i)40 40 3 ln (1 0.25i) 1 40 ln 3 0.02747 1 0.25i e0.02747 1.02785 i 0.1114 11.14% 8.12.
At what interest rate will money treble if compounded continuously for 8 years?
S Pert 3P Per(8) ln 3 ln e8r 1.09861 8r r 0.1373 13.73% 8.13.
At what interest rate will money quintuple if compounded continuously for 25 years?
S Pert 5 e25r ln 5 25r 1.60944 25r r 0.0644 6.44% 8.14.
How long will it take money to double at 12 percent interest under annual compounding?
Round answers to two decimal places.
S P(1 i)t 2 (1 0.12)t ln 2 t ln 1.12 0.69315 0.11333t t 6.12 years 167 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS CHAP . 8] 8.15.
How long will it take money to increase to 21 – 2 times its present value when compounded semiannually at 8 percent?
S P1 0.08 2 2t 2.5 (1.04)2t ln 2.5 2t ln 1.04 0.91629 2(0.03922)t t 11.68 years 8.16.
How long will it take money to double at 5 percent interest when compounded quarterly?
S P1 0.05 4 4t 2 (1.0125)4t ln 2 4t ln 1.0125 0.69315 4(0.01242)t t 13.95 years 8.17.
How long will it take money (a) to quadruple when compounded continuously at 9 percent and (b) to treble at 12 percent?
a) S Pert ln 4 0.09t t 15.4 years 4 e0.09t 1.38629 0.09t b) S Pert ln 3 0.12t t 9.16 years 3 e0.12t 1.09861 0.12t DISCOUNTING 8.18.
Find the present value of $750 to be paid 4 years from now when the prevailing interest rate is 10 percent if interest is compounded (a) annually and (b) semiannually.
a) Using (8.6) and its modifications throughout, P S(1 i)t 750(1.10)4 For (1.10)4, enter 1.10, hit the yx key, enter 4, then press the / key to find (1.10)4 0.68301, and substitute.
P 750(0.68301) 512.26 b) P S1 i m mt 750(1.05)8 750(0.67684) 507.63 8.19.
Redo Problem 8.18, for $600 to be paid 7 years hence at a prevailing interest rate of 4 percent.
a) P 600(1.04)7 600(0.75992) 455.95 b) P 600(1.02)14 600(0.75788) 454.73 8.20.
Find the present value of $500 in 3 years at 8 percent when interest is compounded (a) annually and (b) continuously.
a) P 500(1.08)3 500(0.79383) 396.92 b) P Sert 500e0.08(3) 500e0.24 500(0.78663) 393.32 168 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS [CHAP . 8 8.21.
Redo Problem 8.20, for $120 in 5 years at 9 percent.
a) P 120(1.09)5 120(0.64993) 77.99 b) P 120e0.09(5) 120e0.45 120(0.63763) 76.52 8.22.
Use natural logs to solve Problem 8.21(b).
P 120e0.45 ln P ln 120 (0.45) 4.78749 0.45 4.33749 P antiloge 4.33749 e4.33749 76.52 EXPONENTIAL GROWTH FUNCTIONS 8.23.
A firm with sales of 150,000 a year expects to grow by 8 percent a year. Determine the expected level of sales in 6 years.
S 150,000(1.08)6 150,000(1.58687) 238,031 8.24.
Profits are projected to rise by 9 percent a year over the decade. With current profits of 240,000, what will the level of profits be at the end of the decade?
240,000(1.09)10 240,000(2.36736) 568,166 8.25.
The cost of food has been increasing by 3.6 percent a year. What can a family with current food expenditures of $200 a month be expected to pay for food each month in 5 years?
F 200(1.036)5 200(1.19344) 238.69 8.26.
If the cost of living had continued to increase by 12.5 percent a year from a base of 100 in 1993, what would the cost-of-living index be in 2000?
C 100(1.125)7 100(2.28070) 228.07 8.27.
A discount clothing store reduces prices by 10 percent each day until the goods are sold. What will a $175 suit sell for in 5 days?
P 175(1 0.10)5 175(0.9)5 175(0.59049) 103.34 8.28.
A new car depreciates in value by 3 percent a month for the first year. What is the book value of a $6000 car at the end of the first year?
B 6000(0.97)12 6000(0.69384) 4163.04 8.29.
If the dollar depreciates at 2.6 percent a year, what will a dollar be worth in real terms 25 years from now?
D 1.00(0.974)25 1.00(0.51758) 0.5176 or 51.76¢ 169 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS CHAP . 8] 8.30.
The cost of an average hospital stay was $500 at the end of 1989. The average cost in 1999 was $1500. What was the annual rate of increase?
1500 3 1 i 500(1 i)10 (1 i)10 10 3 For 10 3, enter 3, press the x y key then 10, and substitute.
1 i 1.11612 i 0.11612 11.6% 8.31.
A 5-year development plan calls for boosting investment from 2.6 million a year to 4.2 million.
What average annual increase in investment is needed each year?
4.2 1.615 1 i i 2.6(1 i)5 (1 i)5 5 1.615 1.10061 0.10061 10% 8.32.
A developing country wishes to increase savings from a present level of 5.6 million to 12 million.
How long will it take if it can increase savings by 15 percent a year?
12 5.6(1.15)t To solve for an exponent, use a logarithmic transformation ln 12 2.48491 0.13976t ln 5.6 t ln 1.15 1.72277 0.13976t 0.76214 t 5.45 years 8.33.
Population in many third-world countries is growing at 3.2 percent. Calculate the population 20 years from now for a country with 1,000,000 people.
Since population increases continually over time, a natural exponential function is needed.
P 1,000,000e0.032(20) 1,000,000e0.64 1,000,000(1.89648) 1,896,480 8.34.
If the country in Problem 8.33 reduces its population increase to 2.4 percent, what will the population be in 20 years?
P 1,000,000e0.024(20) 1,000,000e0.48 1,000,000(1.61607) 1,616,070 8.35.
If world population grows at 2.6 percent, how long will it take to double?
2 e0.026t ln 2 0.026t 0.69315 0.026t t 26.66 years 8.36.
If arable land in the Sahel is eroding by 3.5 percent a year because of climatic conditions, how much of the present arable land A will be left in 12 years?
P Ae0.035(12) Ae0.42 0.657047A or 66% 170 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS [CHAP . 8 CONVERTING EXPONENTIAL FUNCTIONS 8.37.
Find the future value of a principal of $2000 compounded semiannually at 12 percent for 3 years, using (a) an exponential function and (b) the equivalent natural exponential function.
a) S P1 i m mt 20001 0.12 2 2(3) 2000(1.06)6 2000(1.41852) 2837.04 b) S Pert where r m ln (1 i/m) 2 ln 1.06 2(0.05827) 0.11654.
Thus, S 2000e0.11654(3) 2000e0.34962 2000(1.41853) 2837.06 8.38.
Redo Problem 8.37 for a principal of $600 compounded annually at 9 percent for 5 years.
a) S 600(1.09)5 600(1.53862) 923.17 b) S Pert, where r ln 1.09 0.08618. Thus, S 600e0.08618(5) 600e0.4309 600(1.53864) 923.18 8.39.
Redo Problem 8.37 for a principal of $1800 compounded quarterly at 8 percent interest for 21 – 2 years.
a) S 1800(1.02)10 1800(1.21899) 2194.18 b) r 4 ln 1.02 4(0.01980) 0.07920 S 1800e0.07920(2.5) 1800e0.19800 1800(1.21896) 2194.13 8.40.
Find the equivalent form under annual discrete compounding for S Pe0.07696t.
r m ln1 i m Since compounding is annual, m 1 0.07696 ln (1 i) 1 i antiloge 0.07696 1.08 i 0.08 S P(1.08)t Thus, 8.41.
Find the equivalent form under semiannual discrete compounding for Pe0.09758t.
r 2 ln (1 0.5i) 0.09758 2 ln (1 0.5i) 1 0.5i antiloge 0.04879 1.05 0.5i 0.05 i 0.10 S P(1.05)2t Thus, 8.42.
Find the equivalent form for S Pe0.15688t under quarterly compounding.
r 4 ln (1 0.25i) 1 – 4(0.15688) ln (1 0.25i) 1 0.25i antiloge 0.03922 1.04 i 0.16 S P(1.04)4t 171 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS CHAP . 8] Slight discrepancy is due to earlier rounding.
ESTABLISHING EXPONENTIAL FUNCTIONS FROM DATA 8.43.
An animal population goes from 3.5 million in 1997 to 4.97 million in 2001. Express population growth P in terms of a natural exponential function and determine the rate of growth.
3.50 P0er(0) P0 (8.12) 4.97 P0er(4) (8.13) Substitute P0 3.50 from (8.12) in (8.13) and simplify.
4.97 3.50e4r 1.42 e4r Take the natural log of both sides.
ln 1.42 0.35066 r ln e4r 4r 4r 0.08767 8.8% Thus, P 3.50e0.088t r 8.8% 8.44.
Costs C of a government program escalate from 5.39 billion in 1995 to 10.64 billion in 2001.
Express costs in terms of an ordinary exponential function, and find the annual rate of growth.
5.39 C0(1 i)0 C0 (8.14) 10.64 C0(1 i)6 (8.15) Substitute C0 5.39 from (8.14) in (8.15) and simplify.
10.64 5.39(1 i)6 1.974 (1 i)6 Take the common log of both sides.
log 1.974 1 – 6(0.29535) log (1 i) 6 log (1 i) log (1 i) 0.04923 1 i antilog 0.04923 100.04923 1.12 i 1.12 1 0.12 12% C 5.39(1 0.12)t i 12% Hence, 8.45.
Redo problem 8.44, given C 2.80 in 1991 and C 5.77 in 2001.
2.80 C0(1 i)0 C0 (8.16) 5.77 C0(1 i)10 (8.17) Substitute C0 2.80 in (8.17) and simplify.
5.77 2.80(1 i)10 2.06 (1 i)10 Take the logs.
log 2.06 1 –– 10(0.31387) log (1 i) 10 log (1 i) log (1 i) 0.03139 1 i antilog 0.03139 100.03139 1.07495 i 1.07495 1 0.07495 7.5% Hence, C 2.80(1 0.075)t i 7.5% 172 EXPONENTIAL AND LOGARITHMIC FUNCTIONS IN ECONOMICS [CHAP . 8 CHAPTER 9 Differentiation of Exponential and Logarithmic Functions 9.1 RULES OF DIFFERENTIATION The rules of exponential and logarithmic differentiation are presented below, illustrated in Examples 1 to 4, and treated in Problems 9.1 to 9.8. Selected proofs for the rules are offered in Problems 9.35 to 9.40.
9.1.1 The Natural Exponential Function Rule Given f(x) eg(x), where g(x) is a differentiable function of x, the derivative is f(x) eg(x) · g(x) (9.1) In short, the derivative of a natural exponential function is equal to the original natural exponential function times the derivative of the exponent.
EXAMPLE 1.
The derivatives of each of the natural exponential functions below are found as follows: 1.
f(x) ex Let g(x) x, then g(x) 1. Substituting in (9.1), f(x) ex · 1 ex The derivative of ex is simply ex, the original function itself.
2.
f(x) ex2 Since g(x) x2, then g(x) 2x. Substituting in (9.1), f(x) ex2 · 2x 2xex2 173 3.
f(x) 3e72x Here g(x) 7 2x, so g(x) 2. From (9.1), f(x) 3e72x · 2 6e72x Evaluating the slope of this function at x 4, f(4) 6e72(4) 6e1 6 1 2.71828 2.2 See also Problem 9.1.
9.1.2 The Exponential Function Rule for Base a Other Than e Given f(x) ag(x), where a 0, a 1, and g(x) is a differentiable function of x, the derivative is f(x) ag(x) · g(x) · ln a (9.2) The derivative is simply the original function times the derivative of the exponent times the natural log of the base.
EXAMPLE 2.
The exponential function rule for base a is demonstrated in the following cases: 1.
f(x) a12x. Let g(x) 1 2x, then g(x) 2. Substituting in (9.2), f(x) a12x · 2 · ln a 2a12x ln a 2.
y ax. Here g(x) x and g(x) 1. From (9.2), y ax · 1 · ln a ax ln a Remember that a may also assume a numerical value. See Problem 9.2(c) through (g).
3.
y x2a3x. With y a product of x2 and a3x, the product rule is necessary.
y x2(a3x · 3 · ln a) a3x(2x) xa3x(3x ln a 2) See also Problem 9.2.
9.1.3 The Natural Logarithmic Function Rule Given f(x) ln g(x), where g(x) is positive and differentiable, the derivative is f(x) 1 g(x) · g(x) g(x) g(x) (9.3) See Example 3 and Problems 9.3 to 9.5.
EXAMPLE 3.
Finding the derivative of a natural logarithmic function is demonstrated below: 1.
f(x) ln 6x2. Let g(x) 6x2, then g(x) 12x. Substituting in (9.3), f(x) 1 6x2 · 12x 2 x 2.
y ln x. Since g(x) x, g(x) 1. From (9.3), y 1 x · 1 1 x 174 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 3.
y ln (x2 6x 2). The derivative is y 1 x2 6x 2 · (2x 6) 2x 6 x2 6x 2 Evaluating the slope of this function at x 4, y(4) 1 – 4 4 – 2 1 – 3 9.1.4 The Logarithmic Function Rule for Base a Other Than e Given f(x) loga g(x), where a 0, a 1, and g(x) is positive and differentiable, the derivative is f(x) 1 g(x) · g(x) · loga e or f(x) 1 g(x) · g(x) · 1 ln a (9.4) since loga e 1/ln a. See Example 4 and Problems 9.6 and 9.40.
EXAMPLE 4.
Derivatives of logarithmic functions to base a are found as shown below.
1.
f(x) loga (2x2 1). Let g(x) 2x2 1; then g(x) 4x. Substituting in (9.4), f(x) 1 2x2 1 · 4x · loga e 4x 2x2 1 loga e or, from (9.4), f(x) 4x (2x2 1) ln a 2.
y loga x. Here g(x) x, and g(x) 1. From (9.4), y 1 x · 1 · loga e loga e x or y 1 x ln a 9.2 HIGHER-ORDER DERIVATIVES Higher-order derivatives are found by taking the derivative of the previous derivative, as illustrated in Example 5 and Problems 9.9 and 9.10.
EXAMPLE 5.
Finding the first and second derivatives of exponential and logarithmic functions is illustrated below: 1.
Given y e5x. The first and second derivatives are dy dx e5x(5) 5e5x d 2y dx2 5e5x(5) 25e5x 2.
Given y ax. The first derivative is dy dx ax(1) ln a ax ln a 175 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] where ln a is a constant. Thus, the second derivative is d2y dx2 ax(ln a)(1)(ln a) ax(ln a)2 ax ln2 a 3.
Given y ln 2x. The first derivative is dy dx 1 2x (2) 1 x or x1 By the simple power function rule, d2y dx2 x2 or 1 x2 4.
Given y loga 3x. The first derivative is dy dx 1 3x (3) 1 ln a 1 x ln a By the quotient rule, where ln a is a constant, the second derivative is d2y dx2 x ln a(0) 1 ln a (x ln a)2 ln a x2 ln2 a 1 x2 ln a 9.3 PARTIAL DERIVATIVES Partial derivatives are found by differentiating the function with respect to one variable, while keeping the other independent variables constant. See Example 6 and Problem 9.11.
EXAMPLE 6.
Finding all the first and second partial derivatives for a function is illustrated below: 1.
Given z e(3x2y). The first and second partials are zx e(3x2y)(3) 3e(3x2y) zxx 3e(3x2y)(3) 9e(3x2y) zy e(3x2y)(2) 2e(3x2y) zyy 2e(3x2y)(2) 4e(3x2y) zxy 6e(3x2y) zyx 2.
Given z ln (5x 9y), the partial derivatives are zx 5 5x 9y zy 9 5x 9y By the simple quotient rule, zxx (5x 9y)(0) 5(5) (5x 9y)2 zyy (5x 9y)(0) 9(9) (5x 9y)2 25 (5x 9y)2 81 (5x 9y)2 zxy 45 (5x 9y)2 zyx 9.4 OPTIMIZATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exponential and logarithmic functions follow the general rules for optimization presented in Sections 4.5 and 5.4. The method is demonstrated in Example 7 and treated in Problems 9.12 to 9.21.
176 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 EXAMPLE 7.
The procedure for finding critical values and determining whether exponential and logarithmic functions are maximized or minimized is illustrated below: 1.
Given y 2xe4x. Using the product rule and setting the derivative equal to zero, we get dy dx 2x(4e4x) 2(e4x) 0 2e4x(4x 1) 0 For the derivative to equal zero, either 2e4x 0 or 4x 1 0. Since 2e4x 0 for any value of x, 4x 1 0 x ¯ 1 – 4 Testing the second-order condition, d2y dx2 2e4x(4) (4x 1)(2e4x)(4) 8e4x(4x 2) Evaluated at the critical value, x ¯ 1 – 4, d 2y/dx2 8e1(1 2) 8/e 0. The function is thus at a minimum, since the second derivative is positive.
2.
Given y ln (x2 6x 10). By the natural log rule, dy dx 2x 6 x2 6x 10 0 Multiplying both sides by the denominator x2 6x 10, 2x 6 0 x ¯ 3 Using the simple quotient rule for the second derivative, d2y dx2 (x2 6x 10)(2) (2x 6)(2x 6) (x2 6x 10)2 Evaluating the second derivative at x ¯ 3, d2y/dx2 2 0. The function is minimized.
3.
Given z e(x22xy26y).
zx (2x 2)e(x22xy26y) 0 zy (2y 6)e(x22xy26y) 0 Since e(x22xy26x) 0 for any value of x or y, 2x 2 0 x ¯ 1 2y 6 0 y ¯ 3 Testing the second-order conditions, using the product rule, zxx (2x 2)(2x 2)e(x22xy26y) e(x22xy26y)(2) zyy (2y 6)(2y 6)e(x22xy26y) e(x22xy26y)(2) When evaluated at x ¯ 1, y ¯ 3, zxx 0 2e10 0 zyy 0 2e10 0 since e to any power is positive. Then testing the mixed partials, zxy (2x 2)(2y 6) e(x22xy26y) zyx Evaluated at x ¯ 1, y ¯ 3, zxy 0 zyx. Thus, the function is at a minimum at x ¯ 1 and y ¯ 3 since zxx and zyy 0 and zxxzyy (zxy)2.
9.5 LOGARITHMIC DIFFERENTIATION The natural logarithm function and its derivative are frequently used to facilitate the differen-tiation of products and quotients involving multiple terms. The process is called logarithmic differentiation and is demonstrated in Example 8 and Problem 9.22.
177 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] EXAMPLE 8.
To find the derivative of a function such as g(x) (5x3 8)(3x4 7) (9x5 2) (9.5) use logarithmic differentiation as follows: a) Take the natural logarithm of both sides.
ln g(x) ln (5x3 8)(3x4 7) (9x5 2) ln (5x3 8) ln (3x4 7) ln (9x5 2) b) Take the derivative of ln g(x).
d dx [ln g(x)] g(x) g(x) 15x2 5x3 8 12x3 3x4 7 45x4 9x5 2 (9.6) c) Solve algebraically for g(x) in (9.6).
g(x) 15x2 5x3 8 12x3 3x4 7 45x4 9x5 2 · g(x) (9.7) d) Then substitute (9.5) for g(x) in (9.7).
g(x) 15x2 5x3 8 12x3 3x4 7 45x4 9x5 2 · (5x3 8)(3x4 7) (9x5 2) 9.6 ALTERNATIVE MEASURES OF GROWTH Growth G of a function y f(t) is defined as G dy/dt y f(t) f(t) y y From Section 9.1.3 this is exactly equivalent to the derivative of ln y. The growth of a function, therefore, can be measured (1) by dividing the derivative of the function by the function itself or (2) by taking the natural log of the function and then simply differentiating the natural log function. This latter method is sometimes helpful with more complicated functions. See Example 9 and Problems 9.23 to 9.30.
EXAMPLE 9.
Finding the growth rate of V Pert, where P is a constant, is illustrated below by using the two methods outlined above.
1.
By the first method, G V V where V Pert(r) rPert. Thus, G rPert Pert r 2.
For the second method, take the natural log of the function.
ln V ln P ln ert ln P rt and then take the derivative of the natural log function with respect to t.
G 1 V dV dt d dt (ln V) d dt (ln P rt) 0 r r 178 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 9.7 OPTIMAL TIMING Exponential functions are used to express the value of goods that appreciate or depreciate over time. Such goods include wine, cheese, and land. Since a dollar in the future is worth less than a dollar today, its future value must be discounted to a present value. Investors and speculators seek to maximize the present value of their assets, as is illustrated in Example 10 and Problems 9.31 to 9.34.
EXAMPLE 10.
The value of cheese that improves with age is given by V 1400(1.25)– t. If the cost of capital under continuous compounding is 9 percent a year and there is no storage cost for aging the cheese in company caves, how long should the company store the cheese?
The company wants to maximize the present value of the cheese: P Vert. Substituting the given values of V and r, P 1400(1.25)– t e0.09t. Taking the natural log, ln P ln 1400 t1/2 ln 1.25 0.09t Then taking the derivative and setting it equal to zero to maximize P, 1 P dP dt 0 1 2 (ln 1.25)t1/2 0.09 0 dP dt P 1 2 (ln 1.25)t1/2 0.09 0 (9.8) Since P 0, 1 – 2(ln 1.25)t1/2 0.09 0 t1/2 0.18 ln 1.25 t ln 1.25 0.18 2 0.22314 0.18 2 1.54 years Using the product rule when taking the second derivative from (9.8), because P f(t), we get d2P dt2 P 1 4 (ln 1.25)t3/2 1 2 (ln 1.25)t1/2 0.09 dP dt Since dP/dt 0 at the critical point, d2P dt2 P 1 4 (ln 1.25)t3/2 P(0.05579t3/2) With P, t 0, d2P/dt2 0 and the function is at a maximum.
9.8 DERIVATION OF A COBB-DOUGLAS DEMAND FUNCTION USING A LOGARITHMIC TRANSFORMATION A demand function expresses the amount of a good a consumer will purchase as a function of commodity prices and consumer income. A Cobb-Douglas demand function is derived by maximizing a Cobb-Douglas utility function subject to the consumer’s income. Given u xy and the budget constraint pxx pyy M, begin with a logarithmic transformation of the utility function ln u ln x ln y 179 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] Then set up the Lagrangian function and maximize.
U ln x ln y (M pxx pyy) Ux · 1 x px 0 pxx Uy · 1 y py 0 pyy U M pxx pyy 0 Add from Ux and Uy, recalling that pxx pyy M.
(pxx pyy) M Thus, M Now substitute ( )/M back in Ux and Uy to get x M px 0 x ¯ M px (9.9a) y M py 0 y ¯ M py (9.9b) For a strict Cobb-Douglas function where 1, x ¯ M px and y ¯ M py (9.9c) EXAMPLE 11.
Given the utility function u x0.3y0.7 and the income constraint M 200, from the information derived in Section 9.8, the demand functions for x and y are (a) derived and (b) evaluated at px 5, py 8 and px 6, py 10, as follows: a) From (9.9c), x ¯ M px and y ¯ M py b) At px 5, py 8, x ¯ 0.3(200) 5 12 and y ¯ 0.7(200) 8 17.5 At px 6, py 10, x ¯ 0.3(200) 6 10 and y ¯ 0.7(200) 10 14 Solved Problems DERIVATIVES OF NATURAL EXPONENTIAL FUNCTIONS 9.1.
Differentiate each of the following natural exponential functions according to the rule d/dx[eg(x)] eg(x) · g(x): a) y e2x b) y e(1/3)x Letting g(x) 2x, then g(x) 2, and y e2x(2) 2e2x g(x) 1 – 3x, g(x) 1 – 3, and y e(1/3)x(1 – 3) 1 – 3e(1/3)x 180 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 c) y ex3 d) y 3ex2 y ex3(3x2) 3x2ex3 y 3ex2(2x) 6xex2 e) y e2x1 f) y e14x y e2x1(2) 2ex1 y 4e14x g) y 5e1x2 h) y 2xex y 10xe1x2 By the product rule, y 2x(ex) ex(2) 2ex(x 1) i) y 3xe2x j) y x2e5x y 3x(2e2x) e2x(3) 3e2x(2x 1) y x2(5e5x) e5x(2x) xe5x(5x 2) k) y e5x 1 e5x 1 By the quotient rule, y (e5x 1)(5e5x) (e5x 1)(5e5x) (e5x 1)2 10e5x (e5x 1)2 l) y e2x 1 e2x 1 y (e2x 1)(2e2x) (e2x 1)(2e2x) (e2x 1)2 4e2x (e2x 1)2 DIFFERENTIATION OF EXPONENTIAL FUNCTIONS WITH BASES OTHER THAN e 9.2.
Differentiate each of the following exponential functions according to the rule d/dx[ag(x)] ag(x) · g(x) · ln a: a) y a2x Letting g(x) 2x, then g(x) 2, and y a2x(2) ln a 2a2x ln a b) y a5x2 y a5x2(10x) ln a 10xa5x2 ln a c) y 42x7 y 42x7(2) ln 4 2(4)2x7 ln 4 Using a calculator, y 2(1.38629)(4)2x7 2.77258(4)2x7 d) y 2x y 2x(1) ln 2 2x ln 2 0.69315(2)x e) y 7x2 y 7x2(2x) ln 7 2x(7)x2 ln 7 2x(7)x2(1.94591) 3.89182x(7)x2 f) y x32x By the product rule, recalling that x3 is a power function and 2x is an exponential function, y x3[2x(1) ln 2] 2x(3x2) x22x(x ln 2 3) g) y x225x y x2[25x(5) ln 2] 25x(2x) x25x(5x ln 2 2) 181 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] DERIVATIVES OF NATURAL LOGARITHMIC FUNCTIONS 9.3.
Differentiate each of the following natural log functions according to the rule d/dx[ln g(x)] 1/[g(x)] · g(x): a) y ln 2x3 b) y ln 7x2 Let g(x) 2x3, then g(x) 6x2, and y 1 7x2 (14x) 2 x y 1 2x3 (6x2) 3 x c) y ln (1 x) d) y ln (4x 7) y 1 1 x y 1 4x 7 (4) 4 4x 7 e) y ln 6x f) y 6 ln x y 1 6x (6) 1 x y 6 1 x 6 x Notice how a multiplicative constant within the log expression in part (e) drops out in differentiation, whereas a multiplicative constant outside the log expression in part (f) remains.
9.4.
Redo Problem 9.3 for each of the following functions: a) y ln2x (ln x)2 By the generalized power function rule, y 2 ln x d dx (ln x) 2 ln x 1 x 2 ln x x b) y ln2 8x (ln 8x)2 y 2 ln 8x 1 8x(8) 2 ln 8x x c) y ln2(3x 1) [ln (3x 1)]2 y 2 ln (3x 1) 1 3x 1(3) 6 ln (3x 1) 3x 1 d) y ln2(5x 6) y 2 ln (5x 6) 1 5x 6(5) 10 ln (5x 6) 5x 6 e) y ln3(4x 13) y 3[ln (4x 13)]2 1 4x 13(4) 3 ln2(4x 13) 4 4x 13 12 4x 13 ln2(4x 13) f) y ln (x 5)2 [ln (x 5)]2 Letting g(x) (x 5)2, then g(x) 2(x 5), and y 1 (x 5)2 [2(x 5)] 2 x 5 182 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 g) y ln (x 8)2 y 1 (x 8)2 [2(x 8)] 2 x 8 h) y 3 ln (1 x)2 y 3 1 (1 x)2[2(1 x)] 6 1 x 9.5.
Use the laws of logarithms in Section 7.3 to simplify the differentiation of each of the following natural log functions: a) y ln (x 5)2 From the rules for logs, ln (x 5)2 2 ln (x 5). Thus, as in Problem 9.4(f), y 2 1 x 5(1) 2 x 5 b) y ln (2x 7)2 c) y ln [(3x 7)(4x 2)] y 2 ln (2x 7) y ln (3x 7) ln (4x 2) y 2 1 2x 7(2) 4 2x 7 y 3 3x 7 4 4x 2 d) y ln [5x2(3x3 7)] e) y ln 3x2 x2 1 y ln 5x2 ln (3x3 7) y ln 3x2 ln (x2 1) y 10x 5x2 9x2 3x3 7 y 2 x 2x x2 1 2 x 9x2 3x3 7 f) y ln x3 (2x 5)2 g) y ln 2x2 3 x2 9 y ln x3 ln (2x 5)2 y 1 – 2[ln (2x2 3) ln (x2 9)] y 1 x3 (3x2) 2 1 2x 5(2) y 1 2 4x 2x2 3 2x x2 9 3 x 4 2x 5 2x 2x2 3 x x2 9 DERIVATIVES OF LOGARITHMIC FUNCTIONS WITH BASES OTHER THAN e 9.6.
Differentiate each of the following logarithmic functions according to the rule d dx [loga g(x)] 1 g(x) · g(x) · loga e 1 g(x) · g(x) · 1 ln a a) y loga (4x2 3) b) y log4 9x3 y 1 4x2 3 (8x) 1 ln a y 1 9x3 (27x2) 1 ln 4 8x (4x2 3) ln a 3 x ln 4 183 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] c) y log2 (8 x) d) y x3 log6 x y 1 (8 x) (1) 1 ln 2 1 (8 x) ln 2 By the product rule, y x3 1 x (1) 1 ln 6 log6 x(3x2) x2 ln 6 3x2 log6 x e) y loga x2 7 From the law of logs, y 1 – 2 loga (x2 7). Thus, y 1 2 1 x2 7 (2x) 1 ln a x (x2 7) ln a COMBINATION OF RULES 9.7.
Use whatever combination of rules are necessary to differentiate the following functions: a) y x2 ln x3 By the product rule, y x2 1 x3(3x2) ln x3(2x) 3x 2x ln x3 3x 6x ln x 3x(1 2 ln x) b) y x3 ln x2 y x3 1 x2(2x) ln x2(3x2) 2x2 6x2 ln x 2x2(1 3 ln x) c) y ex ln x By the product rule, y ex 1 x (ln x)(ex) ex 1 x ln x d) y e2x ln 2x y e2x 1 2x(2) ln 2x(2e2x) e2x 1 x 2 ln 2x e) y ln e3x2 y 1 e3x2 (3e3x2) 3 since ln e3x2 3x 2, d/dx(ln e3x2) d/dx(3x 2) 3.
f) y eln x y eln x 1 x x 1 x 1 since eln x x.
g) y eln (2x1) y eln (2x1) 1 2x 1(2) 2 since eln (2x1) 2x 1.
184 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 h) y ex ln x y ex ln x d dx (x ln x) Then using the product rule for d/dx(x ln x), y ex ln xx 1 x (ln x)(1) ex ln x(1 ln x) i) y ex2 ln 3x y ex2 ln 3xx2 1 3x(3) ln 3x(2x) ex2 ln 3x(x 2x ln 3x) xex2 ln 3x(1 2 ln 3x) SLOPES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 9.8.
Evaluate the slope of each of the following functions at the point indicated: a) y 3e0.2x at x 5.
y 0.6e0.2x At x 5, y 0.6e0.2(5) 0.6(2.71828) 1.63097.
b) y 2e1.5x at x 4.
y 3e1.5x At x 4, y 3e1.5(4) 3e6 3(0.00248) 0.00744.
c) y ln (x2 8x 4) at x 2.
y 2x 8 x2 8x 4 At x 2, y 1 – 2 2 – 4 0.5.
d) y ln2 (x 4) at x 6.
At x 6, y [2 ln (x 4)] 1 x 4(1) 2 ln (x 4) x 4 y 2 ln 10 10 2(2.30259) 10 0.46052 SECOND DERIVATIVES 9.9.
Find the first and second derivatives of the following functions: a) y e3x b) y e(1/2)x y 3e3x y 9e3x y 1 – 2e(1/2)x y 1 – 4e(1/2)x c) y 3e5x1 d) y 2xex y 15e5x1 y 75e5x1 By the product rule, y 2x(ex) ex(2) 2ex(x 1) y 2ex(1) (x 1)(2ex) 2ex(x 2) 185 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] e) y ln 2x5 f) y 4 ln x y 1 2x5 (10x4) 5 x 5x1 y 4 1 x(1) 4 x 4x1 y 5x2 5 x2 y 4x2 9.10.
Take the first and second derivatives of each of the following functions: a) y a3x y a3x(3) ln a 3a3x ln a where ln a a constant. Thus, y (3a3x ln a)(3) ln a 9a3x(ln a)2 b) y a5x1 y a5x1(5) ln a 5a5x1 ln a y (5a5x1 ln a)(5) ln a 25a5x1(ln a)2 c) y loga 5x y 1 5x (5) 1 ln a 1 x ln a (x ln a)1 Using the generalized power function rule, y 1(x ln a)2 ln a ln a x2 ln2 a 1 x2 ln a d) y log3 6x y 1 6x (6) 1 ln 3 1 x ln 3 (x ln 3)1 y 1(x ln 3)2 (ln 3) ln 3 x2 ln2 3 1 x2 ln 3 e) y 3xex By the product rule, y 3x(ex) ex(3) 3ex(x 1) y 3ex(1) (x 1)(3ex) 3ex(x 2) f) y 4x 3 ln x By the quotient rule, y (3 ln x)(4) 4x(3)(1/x) 9 ln2 x 12 ln x 12 9 ln2 x 12(ln x 1) 9 ln2 x y (9 ln2 x)[12(1/x)] 12(ln x 1){9(2) ln x} 81 ln4 x (108/x)(ln2 x) (216/x)(ln x 1)(ln x) 81 ln4 x 4 ln x 8(ln x 1) 3x ln3 x 4 ln x 8 3x ln3 x 4(2 ln x) 3x ln3 x 186 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 PARTIAL DERIVATIVES 9.11.
Find all the first and second partial derivatives for each of the following functions: a) z ex2y2 zx 2xex2y2 zy 2yex2y2 By the product rule, zxx 2x(2xex2y2) ex2y2(2) 2ex2y2(2x2 1) zyy 2y(2yex2y2) ex2y2(2) 2ex2y2(2y2 1) zxy 4xyex2y2 zyx b) z e2x23y zx zxx 4xe2x23y 4x(4xe2x23y) e2x23y(4) 4e2x23y(4x2 1) zy zyy 3e2x23y 9e2x23y zxy 12xe2x23y zyx c) z a2x3y zx zxx a2x3y(2) ln a 2a2x3y ln a 2a2x3y(ln a)(2)(ln a) 4a2x3y ln2 a zy zyy a2x3y(3) ln a 3a2x3y ln a 3a2x3y(ln a)(3)(ln a) 9a2x3y ln2 a zxy 6a2x3y ln2 a zyx d) z 43x5y zx zxx 43x5y(3) ln 4 3(4)3x5y ln 4 3(4)3x5y(ln 4)(3)(ln 4) 9(4)3x5y ln2 4 zy zyy 43x5y(5) ln 4 5(4)3x5y ln 4 5(4)3x5y(ln 4)(5)(ln 4) 25(4)3x5y ln2 4 zxy 15(4)3x5y ln2 4 zyx e) z ln (7x 2y) zx 7 7x 2y zy 2 7x 2y By the quotient rule, zxx (7x 2y)(0) 7(7) (7x 2y)2 49 (7x 2y)2 zyy (7x 2y)(0) 2(2) (7x 2y)2 4 (7x 2y)2 zxy 14 (7x 2y)2 zyx f) z ln (x2 4y2) zx 2x x2 4y2 zy 8y x2 4y2 zxx (x2 4y2)(2) 2x(2x) (x2 4y2)2 8y2 2x2 (x2 4y2)2 zyy (x2 4y2)(8) 8y(8y) (x2 4y2)2 8x2 32y2 (x2 4y2)2 zxy 16xy (x2 4y2)2 zyx 187 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] g) z loga (x 2y) zx 1 (x 2y) ln a zy 2 (x 2y) ln a zxx 1 ln a (x 2y)2 ln 2 a 1 (x 2y)2 ln a zyy 4 ln a (x 2y)2 ln2 a 4 (x 2y)2 ln a zxy 2 (x 2y)2 ln a zyx h) z loga (3x2 y2) zx 6x (3x2 y2) ln a zy 2y (3x2 y2) ln a zxx (3x2 y2)(ln a)(6) 6x(6x ln a) (3x2 y2)2 ln2 a zyy (3x2 y2)(ln a)(2) 2y(2y ln a) (3x2 y2)2 ln2 a 6y2 18x2 (3x2 y2)2 ln a 6x2 2y2 (3x2 y2)2 ln a zxy 12xy (3x2 y2)2 ln a zyx OPTIMIZATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS 9.12.
Given y 4xe3x, (a) find the critical values and (b) determine whether the function is maximized or minimized.
a) By the product rule, y 4x(3e3x) e3x(4) 0 4e3x(3x 1) 0 Since there is no value of x for which 4e3x 0, or for which ex 0, 3x 1 0 x ¯ 1 – 3 b) y 4e3x(3) (3x 1)(12e3x) 12e3x(3x 2) At x ¯ 1 – 3, y 12e1(1). And y 12(0.36788) 0. The function is minimized.
9.13.
Redo Problem 9.12, given y 5xe0.2x a) y 5x(0.2e0.2x) e0.2x(5) 0 5e0.2x(1 0.2x) 0 Since 5e0.2x 0, (1 0.2x) 0 x ¯ 5 b) y 5e0.2x(0.2) (1 0.2x)(1e0.2x) e0.2x(0.2x 2) At x ¯ 5, y e1(1 2). And y (0.36788)(1) 0. The function is at a maximum.
9.14.
Redo Problem 9.12, given y ln (x2 8x 20).
a) y 2x 8 x2 8x 20 0 Multiplying both sides by x2 8x 20 gives 2x 8 0 and x ¯ 4.
b) y (x2 8x 20)(2) (2x 8)(2x 8) (x2 8x 20)2 At x ¯ 4, y 8 –– 16 0. The function is at a minimum.
188 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 9.15.
Redo Problem 9.12, given y ln (2x2 20x 5).
a) y 4x 20 2x2 20x 5 0 4x 20 0 x ¯ 5 b) y (2x2 20x 5)(4) (4x 20)(4x 20) (2x2 20x 5)2 At x ¯ 5, y 180/2025 0. The function is at a maximum.
9.16.
Given the function z ln (2x2 12x y2 10y), (a) find the critical values and (b) indicate whether the function is at a maximum or minimum.
a) zx 4x 12 2x2 12x y2 10y 0 zy 2y 10 2x2 12x y2 10y 0 4x 12 0 x ¯ 3 2y 10 0 y ¯ 5 b) zxx (2x2 12x y2 10y)(4) (4x 12)(4x 12) (2x2 12x y2 10y)2 zyy (2x2 12x y2 10y)(2) (2y 10)(2y 10) (2x2 12x y2 10y)2 Evaluated at x ¯ 3, y ¯ 5, zxx (43)(4) 0 (43)2 172 1849 0 zyy (43)(2) 0 (43)2 86 1849 0 zxy (4x 12)(2y 10) (2x2 12x y2 10y)2 zyx At x ¯ 3, y ¯ 5, zxy 0 zyx. With zxx, zyy 0 and zxx zyy (zxy)2, the function is at a maximum.
9.17.
Redo Problem 9.16, given z ln (x2 4x 3y2 6y).
a) zx 2x 4 x2 4x 3y2 6y 0 zy 6y 6 x2 4x 3y2 6y 0 2x 4 0 x ¯ 2 6y 6 0 y ¯ 1 b) zxx (x2 4x 3y2 6y)(2) (2x 4)(2x 4) (x2 4x 3y2 6y)2 zyy (x2 4x 3y2 6y)(6) (6y 6)(6y 6) (x2 4x 3y2 6y)2 At x ¯ 2, y ¯ 1, zxx (7)(2) 0 (7)2 14 49 0 zyy (7)(6) 0 (7)2 42 49 0 zxy (2x 4)(6y 6) (x2 4x 3y2 6y)2 zyx At x ¯ 2, y ¯ 1, zxy 0 zyx. With zxx, zyy 0 and zxx zyy (zxy)2, the function is at a maximum.
9.18.
Redo Problem 9.16, given z e(3x26xy28y).
a) zx (6x 6)e(3x26xy28y) 0 zy (2y 8) e(3x26xy28y) 0 6x 6 0 x ¯ 1 2y 8 0 y ¯ 4 189 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] b) Using the product rule, zxx (6x 6)(6x 6)e(3x26xy28y) e(3x26xy28y)(6) zyy (2y 8)(2y 8)e(3x26xy28y) e(3x26xy28y)(2) Evaluated at x ¯ 1, y ¯ 4, zxx 0 6e19 0 zyy 0 2e19 0 Then testing the cross partials, zxy (6x 6)(2y 8)e(3x26xy28y) zyx At x ¯ 1, y ¯ 4, zxy 0 zyx. The function is at a minimum since zxx, zyy 0 and zxxzyy (zxy)2.
9.19.
Given z e(2x212x2xyy24y), redo Problem 9.16.
a) zx (4x 12 2y)e(2x212x2xyy24y) 0 4x 2y 12 0 (9.10) zy (2x 2y 4)e(2x212x2xyy24y) 0 2x 2y 4 0 (9.11) Solving (9.10) and (9.11) simultaneously, x ¯ 8, y ¯ 10.
b) zxx (4x 12 2y)(4x 12 2y)e(2x212x2xyy24y) e(2x212x2xyy24y)(4) zyy (2x 2y 4)(2x 2y 4)e(2x212x2xyy24y) e(2x212x2xyy24y)(2) Evaluated at x ¯ 8, y ¯ 10, zxx 0 4e68 0 zyy 0 2e68 0 Testing the mixed partials, by the product rule, zxy (4x 12 2y)(2x 2y 4)e(2x212x2xyy24y) e(2x212x2xyy24y)(2) zyx Evaluated at x ¯ 8, y ¯ 10, zxy 0 2e68 zyx. Since zxx,zyy 0 and zxxzyy (zxy)2, the function is at a minimum.
9.20.
Given the demand function P 8.25e0.02Q (9.12) (a) Determine the quantity and price at which total revenue will be maximized and (b) test the second-order condition.
a) TR PQ (8.25e0.02Q)Q By the product rule, dTR dQ (8.25e0.02Q)(1) Q(0.02)(8.25e0.02Q) 0 (8.25e0.02Q)(1 0.02Q) 0 Since (8.25e0.02Q) 0 for any value of Q, 1 0.02Q 0; Q ¯ 50.
Substituting Q ¯ 50 in (9.12), P 8.25e0.02(50) 8.25e1. And P 8.25(0.36788) 3.04.
b) By the product rule, d2TR dQ2 (8.25e0.02Q)(0.02) (1 0.02Q)(0.02)(8.25e0.02Q) (0.02)(8.25e0.02Q)(2 0.02Q) Evaluated at Q ¯ 50, d2TR/dQ2 (0.02)(8.25e1)(1) 0.165(0.36788) 0. TR is at a maxi-mum.
190 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 9.21.
(a) Find the price and quantity that will maximize total revenue, given the demand function P 12.50e0.005Q. (b) Check the second-order condition.
a) TR (12.50e0.005Q)Q dTR dQ (12.50e0.005Q)(1) Q(0.005)(12.50e0.005Q) (12.50e0.005Q)(1 0.005Q) 0 1 0.005Q 0 Q ¯ 200 Thus, P 12.50e0.005(200) 12.50e1 12.50(0.36788) 4.60 b) d2TR dQ2 (12.50e0.005Q)(0.005) (1 0.005Q)(0.005)(12.50e0.005Q) (0.005)(12.50e0.005Q)(2 0.005Q) Evaluated at Q ¯ 200, d2TR/dQ2 (0.005)(12.50e1)(1) 0.0625(0.36788) 0. The function is maximized.
LOGARITHMIC DIFFERENTIATION 9.22.
Use logarithmic differentiation to find the derivatives for the following functions: a) g(x) (x3 2)(x2 3)(8x 5) (9.13) 1) Take the natural logarithm of both sides.
ln g(x) ln (x3 2) ln (x2 3) ln (8x 5) 2) Take the derivative of ln g(x).
d dx [ln g(x)] g(x) g(x) 3x2 x3 2 2x x2 3 8 8x 5 (9.14) 3) Solve algebraically for g(x) in (9.14).
g(x) 3x2 x3 2 2x x2 3 8 8x 5 · g(x) (9.15) 4) Then substitute (9.13) for g(x) in (9.15).
g(x) 3x2 x3 2 2x x2 3 8 8x 5[(x3 2)(x2 3)(8x 5)] b) g(x) (x4 7)(x5 6)(x3 2) (9.16) 1) ln g(x) ln (x4 7) ln (x5 6) ln (x3 2) 2) d dx [ln g(x)] g(x) g(x) 4x3 x4 7 5x4 x5 6 3x2 x3 2 3) g(x) 4x3 x4 7 5x4 x5 6 3x2 x3 2 · g(x) (9.17) 4) Finally, substituting (9.16) for g(x) in (9.17), g(x) 4x3 x4 7 5x4 x5 6 3x2 x3 2 · (x4 7)(x5 6)(x3 2) c) g(x) (3x5 4)(2x3 9) (7x4 5) 1) ln g(x) ln (3x5 4) ln (2x3 9) ln (7x4 5) 191 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] 2) d dx [ln g(x)] g(x) g(x) 15x4 3x5 4 6x2 2x3 9 28x3 7x4 5 3) g(x) 15x4 3x5 4 6x2 2x3 9 28x3 7x4 5 · g(x) 4) g(x) 15x4 3x5 4 6x2 2x3 9 28x3 7x4 5 · (3x5 4)(2x3 9) (7x4 5) GROWTH 9.23.
The price of agricultural goods is going up by 4 percent each year, the quantity by 2 percent.
What is the annual rate of growth of revenue R derived from the agricultural sector?
Converting the revenue formula R PQ to natural logs, ln R ln P ln Q The derivative of the natural log function equals the instantaneous rate of growth G of the function (see Section 9.6). Thus, G d dt (ln R) d dt (ln P) d dt (ln Q) But d dt (ln P) growth of P 4% d dt (ln Q) growth of Q 2% Thus, G d dt (ln R) 0.04 0.02 0.06 The rate of growth of a function involving a product is the sum of the rates of growth of the individual components.
9.24.
A firm experiences a 10 percent increase in the use of inputs at a time when input costs are rising by 3 percent. What is the rate of increase in total input costs?
C ln C PQ ln P ln Q G d dt (ln C) d dt (ln P) d dt (ln Q) 0.03 0.10 0.13 9.25.
Employment opportunities E are increasing by 4 percent a year and population P by 2.5 percent. What is the rate of growth of per capita employment PCE?
PCE E P ln PCE ln E ln P Taking the derivative to find the growth rate, G d dt (ln PCE) d dt (ln E) d dt (ln P) 0.04 0.025 0.015 1.5% The rate of growth of a function involving a quotient is the difference between the rate of growth of the numerator and denominator.
192 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 9.26.
National income Y is increasing by 1.5 percent a year and population P by 2.5 percent a year.
What is the rate of growth of per capita income PCY?
PCY Y P ln PCY ln Y ln P G d dt (ln PCY) d dt (ln Y) d dt (ln P) 0.015 0.025 0.01 1% Per capita income is falling by 1 percent a year.
9.27.
A country exports two goods, copper c and bananas b, where earnings in terms of million dollars are c c(t0) 4 b b(t0) 1 If c grows by 10 percent and b by 20 percent, what is the rate of growth of export earnings E?
E ln E c b ln (c b) G d dt (ln E) d dt ln (c b) From the rules of derivatives in Section 9.1.3, GE 1 c b [c(t) b(t)] (9.18) From Section 9.6, Gc c(t) c(t) Gb b(t) b(t) Thus, c(t) Gcc(t) b(t) Gbb(t) Substituting in (9.18), GE 1 c b [Gcc(t) Gbb(t)] Rearranging terms, GE c(t) c b Gc b(t) c b Gb Then substituting the given values, GE 4 4 1 (0.10) 1 4 1 (0.20) 4 5 (0.10) 1 5 (0.20) 0.12 or 12% The growth rate of a function involving the sum of other functions is the sum of the weighted average of the growth of the other functions.
9.28.
A company derives 70 percent of its revenue from bathing suits, 20 percent from bathing caps, and 10 percent from bathing slippers. If revenues from bathing suits increase by 15 percent, from caps by 5 percent, and from slippers by 4 percent, what is the rate of growth of total revenue?
GR 0.70(0.15) 0.20(0.05) 0.10(0.04) 0.105 0.01 0.004 0.119 or 11.9% 193 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] 9.29.
Find the relative growth rate G of sales at t 4, given S(t) 100,000e0.5– t ln S(t) ln 100,000 ln e0.5– t ln 100,000 0.5t Take the derivative and recall that 0.5t 0.5t1/2, G d dt ln S(t) S(t) S(t) 0.5 1 2 t1/2 0.25 t At t 4, G 0.25 4 0.125 12.5% 9.30.
Find the relative growth of profits at t 8, given (t) 250,000e1.2t1/3.
ln (t) ln 250,000 1.2t1/3 and G d dt ln (t) (t) (t) 1.2 1 3 t2/3 0.4 t2/3 At t 8, G 0.4 82/3 0.4 4 0.1 10% OPTIMAL TIMING 9.31.
Cut glass currently worth $100 is appreciating in value according to the formula V 100e– t 100et1/2 How long should the cut glass be kept to maximize its present value if under continuous compounding (a) r 0.08 and (b) r 0.12?
a) The present value P is P Vert. Substituting for V and r, P 100e– te0.08t 100e– t0.08t Converting to natural logs, ln P ln 100 ln e– t0.08t ln 100 t1/2 0.08t. Taking the derivative, setting it equal to zero, and recalling that ln 100 is a constant, d dt (ln P) 1 P dP dt 1 2 t1/2 0.08 dP dt P 1 2 t1/2 0.08 0 (9.19) Since P 0, 1 – 2t1/2 t1/2 0.08 0.16 t (0.16)2 1 0.0256 39.06 Testing the second-order condition, and using the product rule since P f(t), d2P dt2 P 1 4 t3/2 1 2 t1/2 0.08 dP dt Since dP/dt 0 at the critical value, d2P dt2 P 4t3 which is negative, since P and t must both be positive. Thus, t 39.06 maximizes the function.
194 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 b) If r 0.12, substituting 0.12 for 0.08 in (9.19) above, dP dt P 1 2 t1/2 0.12 0 1 – 2t1/2 0.12 t (0.24)2 1 0.0576 17.36 The second-order condition is unchanged. Note that the higher the rate of discount r, the shorter the period of storage.
9.32.
Land bought for speculation is increasing in value according to the formula V 1000e 3 – t The discount rate under continuous compounding is 0.09. How long should the land be held to maximize the present value?
P 1000e 3 – te0.09t 1000e 3 – t0.09t Convert to natural logs.
ln P ln 1000 t1/3 0.09t Take the derivative.
d dt (ln P) 1 P dP dt 1 3 t2/3 0.09 0 dP dt P 1 3 t2/3 0.09 0 1 – 3t2/3 0.09 t 0.273/2 7.13 years The second-order condition, recalling dP/dt 0 at the critical value, is d2P dt2 P 2 9 t5/3 1 3 t2/3 0.09 dP dt 2P 9 3 t5 0 9.33.
The art collection of a recently deceased painter has an estimated value of V 200,000(1.25) 3 t2 How long should the executor of the estate hold on to the collection before putting it up for sale if the discount rate under continuous compounding is 6 percent?
Substituting the value of V in P Vert, P ln P 200,000(1.25)t2/3e0.06t ln 200,000 t2/3 ln 1.25 0.06t d dt (ln P) 1 P dP dt 2 3 (ln 1.25)t1/3 0.06 0 dP dt P 2 3 (ln 1.25)t1/3 0.06 0 t1/3 3(0.06) 2 ln 1.25 t 0.18 2(0.22314) 3 (0.403)3 15.3 years 9.34.
The estimated value of a diamond bought for investment purposes is V 250,000(1.75) 4 – t 195 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] If the discount rate under continuous compounding is 7 percent, how long should the diamond be held?
P ln P 250,000(1.75)t1/4e0.07t ln 250,000 t1/4 (ln 1.75) 0.07t d dt (ln P) 1 P dP dt 1 4 (ln 1.75)t3/4 0.07 0 dP dt P 1 4 (ln 1.75)t3/4 0.07 0 1 4 (ln 1.75) t3/4 0.07 0 t 0.28 ln 1.75 4/3 (0.50)4/3 2.52 years SELECTED PROOFS 9.35.
Derive the derivative for ln x.
From the definition of a derivative in Equation (3.2), f(x) lim x→0 f(x x) f(x) x Specifying f(x) ln x, d dx (ln x) lim x→0 ln (x x) ln x x From the properties of logs, where ln a ln b ln (a/b), d dx (ln x) lim x→0 ln [(x x)/x] x Rearranging first the denominator and then the numerator, d dx (ln x) lim x→0 1 x ln x x x lim x→0 1 x ln 1 x x Multiply by x/x.
d dx (ln x) lim x→0 1 x · x x ln 1 x x From the properties of logs, where a ln x ln xa, d dx (ln x) lim x→0 1 x ln 1 x x x/ x 1 x lim x→0 ln 1 x x x/ x Since the logarithmic function is continuous, d dx (ln x) 1 x ln lim x→01 x x x/ x Let n x/ x and note that as x →0, n →. Then d dx (ln x) 1 x ln lim n→ 1 1 n n But from Section 7.4, e the limit as n → of (1 1/n)n, so d dx (ln x) 1 x ln e 1 x · 1 1 x 196 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 9.36.
Derive the derivative for y ln g(x), assuming g(x) is positive and differentiable.
Using the chain rule notation from (3.6), let y ln u and u g(x). Then dy dx dy du · du dx where dy du 1 u and du dx g(x) Substituting, dy dx 1 u · g(x) Then replacing u with g(x), dy dx 1 g(x) · g(x) Hence, d dx [ln g(x)] 1 g(x) · g(x) g(x) g(x) 9.37.
Show that the derivative of the function y ex is ex.
Take the natural logarithm of both sides, ln y ln ex ln y x From Equation (7.3), Use implicit differentiation and recall that y is a function of x and so requires the chain rule, 1 y · dy dx 1 dy dx y Replace y with ex, d dx (ex) ex 9.38.
Given y eg(x), prove that dy/dx eg(x) · g(x).
Use the chain rule, letting y eu and u g(x); then dy dx dy du · du dx where dy du eu and du dx g(x) Substituting, dy dx eu · g(x) Then replace y with eg(x) on the left-hand side and u with g(x) on the right-hand side, d dx (eg(x)) eg(x) · g(x) 197 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS CHAP . 9] 9.39.
Prove that loga (x/y) loga x loga y.
Let s loga x and t loga y. Then from the definition of a logarithm in Section 7.2 as x and at y and as at x y Substituting from the property of exponents, where as/at ast, ast x y Again using the definition of a logarithm, loga x y s t But s loga x and t loga y, so loga x y loga x loga y 9.40.
From Equation (9.4), prove that loga e 1/ln a.
Set each side of the given equation as an exponent of a.
aloga e a1/ln a But aloga e e. Substituting, e a1/ln a Then taking the log of both sides, loga e loga a1/ln a 1 ln a 198 DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS [CHAP . 9 CHAPTER 10 The Fundamentals of Linear (or Matrix) Algebra 10.1 THE ROLE OF LINEAR ALGEBRA Linear algebra (1) permits expression of a complicated system of equations in a succinct, simplified way, (2) provides a shorthand method to determine whether a solution exists before it is attempted, and (3) furnishes the means of solving the equation system. Linear algebra however, can be applied only to systems of linear equations. Since many economic relationships can be approximated by linear equations and others can be converted to linear relationships, this limitation can in part be averted.
See Example 2 and Section 7.6.
EXAMPLE 1.
For a company with several different outlets selling several different products, a matrix provides a concise way of keeping track of stock.
Outlet 1 2 3 4 Skis 120 200 175 140 Poles 110 180 190 170 Bindings 90 210 160 180 Outfits 150 110 80 140 By reading across a row of the matrix, the firm can determine the level of stock in any of its outlets. By reading down a column of the matrix, the firm can determine the stock of any line of its products.
EXAMPLE 2.
A nonlinear function, such as the rational function z x0.3/y0.6, can be easily converted to a linear function by a simple rearrangement z x0.3 y0.6 x0.3y0.6 followed by a logarithmic transformation ln z 0.3 ln x 0.6 ln y 199 which is log-linear. In similar fashion, many exponential and power functions are readily convertible to linear functions and then handled by linear algebra. See Section 7.6.
10.2 DEFINITIONS AND TERMS A matrix is a rectangular array of numbers, parameters, or variables, each of which has a carefully ordered place within the matrix. The numbers (parameters, or variables) are referred to as elements of the matrix. The numbers in a horizontal line are called rows; the numbers in a vertical line are called columns. The number of rows r and columns c defines the dimensions of the matrix (r c), which is read ‘‘r by c.’’ The row number always precedes the column number. In a square matrix, the number of rows equals the number of columns (that is, r c). If the matrix is composed of a single column, such that its dimensions are r 1, it is a column vector; if a matrix is a single row, with dimensions 1 c, it is a row vector. A matrix which converts the rows of A to columns and the columns of A to rows is called the transpose of A and is designated by A (or AT). See Example 3 and Problems 10.1 to 10.3.
EXAMPLE 3.
Given A a11 a21 a31 a12 a22 a32 a13 a23 a33 33 B 3 9 8 4 2 7 23 C 7 4 5 31 D [3 0 1]13 Here A is a general matrix composed of 3 3 9 elements, arranged in three rows and three columns. It is thus a square matrix. Note that no punctuation separates the elements of a matrix. The elements all have double subscripts which give the address or placement of the element in the matrix; the first subscript identifies the row in which the element appears, and the second identifies the column. Positioning is precise within a matrix. Thus, a23 is the element which appears in the second row, third column; a32 is the element which appears in the third row, second column. Since row always precedes column in matrix notation, it might be helpful to think of the subscripts in terms of RC cola or some other mnemonic device. To determine the number of rows, always count down; to find the number of columns, count across.
Here B is a 2 3 matrix. Its b12 element is 9, its b21 element is 4. And C is a column vector with dimensions 3 1; D is a row vector with dimensions 1 3.
The transpose of A is A a11 a21 a31 a12 a22 a32 a13 a23 a33 and the transpose of C is C [7 4 5] 10.3 ADDITION AND SUBTRACTION OF MATRICES Addition (and subtraction) of two matrices A B (or A B) requires that the matrices be of equal dimensions. Each element of one matrix is then added to (subtracted from) the corresponding element of the other matrix. Thus, a11 in A will be added to (subtracted from) b11 in B; a12 to b12; etc.
See Examples 4 and 5 and Problems 10.4 to 10.8.
EXAMPLE 4.
The sum A B is calculated below, given matrices A and B: A 8 9 7 3 6 2 4 5 10 33 B 1 3 6 5 2 4 7 9 2 33 A B 8 1 9 3 7 6 3 5 6 2 2 4 4 7 5 9 10 2 33 9 12 13 8 8 6 11 14 12 33 200 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 The difference C D, given matrices C and D, is found as follows: C 4 9 2 6 22 D 1 7 5 422 C D 4 1 9 7 2 5 6 4 22 3 2 3 2 22 EXAMPLE 5.
Suppose that deliveries D are made to the outlets of the firm in Example 1. What is the new level of stock?
D 40 20 50 10 25 30 10 60 15 0 40 70 60 40 10 50 To find the new level of stock, label the initial matrix S and solve for S D. Adding the corresponding elements of each matrix, S D 120 40 110 20 90 50 150 10 200 25 180 30 210 10 110 60 175 15 190 0 160 40 80 70 140 60 170 40 180 10 140 50 160 130 140 160 225 210 220 170 190 190 200 150 200 210 190 190 10.4 SCALAR MULTIPLICATION In matrix algebra, a simple number such as 12, 2, or 0.07 is called a scalar. Multiplication of a matrix by a number or scalar involves multiplication of every element of the matrix by the number.
The process is called scalar multiplication because it scales the matrix up or down according to the size of the number. See Example 6 and Problems 10.10 to 10.12.
EXAMPLE 6.
The result of scalar multiplication kA, given k 8 and A 6 9 2 7 8 4 32 is shown below kA 8(6) 8(9) 8(2) 8(7) 8(8) 8(4) 32 48 72 16 56 64 32 32 10.5 VECTOR MULTIPLICATION Multiplication of a row vector A by a column vector B requires as a precondition that each vector have precisely the same number of elements. The product is then found by multiplying the individual elements of the row vector by their corresponding elements in the column vector and summing the products: AB (a11 b11) (a12 b21) (a13 b31) etc.
The product of row-column multiplication will thus be a single number or scalar. Row-column vector multiplication is of paramount importance. It serves as the basis for all matrix multiplication. See Example 7 and Problems 10.13 to 10.18.
201 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] EXAMPLE 7.
The product AB of the row vector A and the column vector B, given A [4 7 2 9]14 B 12 1 5 6 41 is calculated as follows: AB 4(12) 7(1) 2(5) 9(6) 48 7 10 54 119 The product of vectors C [3 6 8]13 D 2 4 5 31 is CD (3 2) (6 4) (8 5) 6 24 40 70 Note that since each set of vectors above has the same number of elements, multiplication is possible.
Reversing the order of multiplication in either of the above and having column-row vector multiplication (BA or DC) will give a totally different answer. See Problem 10.36.
EXAMPLE 8.
The meaning of vector multiplication is perhaps easiest understood in terms of the following example. Assume Q is a row vector of the physical quantities of hamburgers, fries, and sodas sold, respectively, on a given day and P is a column vector of the corresponding prices of hamburgers, fries, and sodas.
Q [12 8 10] P 1.25 0.75 0.50 Then by vector multiplication the total value of sales (TVS) for the day is TVS QP [12(1.25) 8(0.75) 10(0.50)] 26.00 10.6 MULTIPLICATION OF MATRICES Multiplication of two matrices with dimensions (r c)1 and (r c)2 requires that the matrices be conformable, i.e., that c1 r2, or the number of columns in 1, the lead matrix, equal the number of rows in 2, the lag matrix. Each row vector in the lead matrix is then multiplied by each column vector of the lag matrix, according to the rules for multiplying row and column vectors discussed in Section 10.5. The row-column products, called inner products or dot products, are then used as elements in the formation of the product matrix, such that each element cij of the product matrix C is a scalar derived from the multiplication of the ith row of the lead matrix and the jth column of the lag matrix. See Examples 9 to 11 and Problems 10.19 to 10.33.
EXAMPLE 9.
Given A 3 6 7 12 9 11 23 B 6 12 5 10 13 2 32 C 1 7 8 2 4 323 A shorthand test for conformability, which should be applied before undertaking any matrix multiplication, is to place the two sets of dimensions in the order in which the matrices are to be multiplied, then mentally circle the last number of the first set and the first number of the second set. If the two numbers are equal, the number of columns in the lead matrix will equal the number of rows in the lag matrix, and the two matrices will be 202 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 conformable for multiplication in the given order. Moreover, the numbers outside the circle will provide, in proper order, the dimensions of the resulting product matrix. Thus, for AB 2 3 3 2 2 2 The number of columns in the lead matrix equals the number of rows in the lag matrix, 3 3; the matrices are conformable for multiplication; and the dimensions of the product matrix AB will be 2 2. When two matrices such as A and B are conformable for multiplication, the product AB is said to be defined.
For BC, 3 2 2 3 3 3 The number of columns in the lead matrix equals the number of rows in the lag matrix, 2 2; hence B and C are conformable. The product BC is defined, and BC will be a 3 3 matrix.
For AC, 2 3 2 3 A and C are not conformable for multiplication. Thus, AC is not defined.
EXAMPLE 10.
Having determined that A and B in Example 9 are conformable, the product AB D can be found. Remembering to use only rows R from the lead matrix and only columns C from the lag matrix, multiply the first row R1 of the lead matrix by the first column C1 of the lag matrix to find the first element d11 ( R1C1) of the product matrix D. Then multiply the first row R1 of the lead matrix by the second column C2 of the lag matrix to get d12 ( R1 C2). Since there are no more columns left in the lag matrix to be multiplied by the first row of the lead matrix, move to the second row of the lead matrix. Multiply the second row R2 of the lead matrix by the first column C1 of the lag matrix to get d21 (R2C1). Finally, multiply the second row R2 of the lead matrix by the second column C2 of the lag matrix to get d22 ( R2 C2). Thus, AB D R1 C1 R1 C2 R2 C1 R2 C2 3(6) 6(5) 7(13) 3(12) 6(10) 7(2) 12(6) 9(5) 11(13) 12(12) 9(10) 11(2) 22 139 110 260 256 22 The product of BC is calculated below, using the same method: BC E R1C1 R1C2 R1C3 R2C1 R2C2 R2C3 R3C1 R3C2 R3C3 6(1) 12(2) 6(7) 12(4) 6(8) 12(3) 5(1) 10(2) 5(7) 10(4) 5(8) 10(3) 13(1) 2(2) 13(7) 2(4) 13(8) 2(3) 33 30 90 84 25 75 70 17 99 110 33 EXAMPLE 11.
Referring to Example 1, suppose that the price of skis is $200, poles $50, bindings $100, and outfits $150. To find the value V of the stock in the different outlets, express the prices as a column vector P, and multiply S by P.
V SP 120 110 90 150 200 180 210 110 175 190 160 80 140 170 180 140 44 200 50 100 150 41 203 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] The matrices are conformable, and the product matrix will be 4 1 since 4 4 4 1 4 1 Thus, V R1C1 R2C1 R3C1 R4C1 120(200) 110(50) 90(100) 150(150) 200(200) 180(50) 210(100) 110(150) 175(200) 190(50) 160(100) 80(150) 140(200) 170(50) 180(100) 140(150) 41 61,000 86,500 72,500 75,500 41 10.7 COMMUTATIVE, ASSOCIATIVE, AND DISTRIBUTIVE LAWS IN MATRIX ALGEBRA Matrix addition is commutative (that is, A B B A) since matrix addition merely involves the summing of corresponding elements of two matrices and the order in which the addi-tion takes place is inconsequential. For the same reason, matrix addition is also associative, (A B) C A (B C). The same is true of matrix subtraction. Since matrix subtraction A B can be converted to matrix addition A (B), matrix subtraction is also commutative and associative.
Matrix multiplication, with few exceptions, is not commutative (that is, AB BA). Scalar multiplication, however, is commutative (that is, kA Ak). If three or more matrices are conformable, that is, Xab, Ycd, Zef, where b c and d e, the associative law will apply as long as the matrices are multiplied in the order of conformability. Thus (XY)Z X(YZ). Subject to these same conditions, matrix multiplication is also distributive: A(B C) AB AC. See Examples 12 to 14 and Problems 10.34 to 10.48.
EXAMPLE 12.
Given A 4 11 17 6 B 3 7 6 2 To show that matrix addition and matrix subtraction are commutative, demonstrate that (1) A B B A and (2) A B B A. The calculations are shown below.
1) A B 4 3 11 7 17 6 6 2 7 18 23 8 B A 3 4 7 11 6 17 2 6 7 18 23 8 2) A B 4 3 11 7 17 6 6 2 1 4 11 4 B A 3 4 7 11 6 17 2 6 1 4 11 4 EXAMPLE 13.
Given A 3 6 7 12 9 11 23 B 6 12 5 10 13 2 32 It can be demonstrated that matrix multiplication is not commutative, by showing AB BA, as follows: Matrix AB is conformable.
2 3 3 2 AB will be 2 2 AB 3(6) 6(5) 7(13) 3(12) 6(10) 7(2) 12(6) 9(5) 11(13) 12(12) 9(10) 11(2) 22 139 110 260 25622 204 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 Matrix BA is conformable.
3 2 2 3 BA will be 3 3 BA 6(3) 12(12) 6(6) 12(9) 6(7) 12(11) 5(3) 10(12) 5(6) 10(9) 5(7) 10(11) 13(3) 2(12) 13(6) 2(9) 13(7) 2(11) 33 162 144 174 135 120 145 63 96 113 33 Hence AB BA. Frequently matrices will not even be conformable in two directions.
EXAMPLE 14.
Given A 7 5 1 3 8 6 32 B 4 9 10 2 6 5 23 C 2 6 7 31 To illustrate that matrix multiplication is associative, that is, (AB)C A(BC), the calculations are as follows: AB 7(4) 5(2) 7(9) 5(6) 7(10) 5(5) 1(4) 3(2) 1(9) 3(6) 1(10) 3(5) 8(4) 6(2) 8(9) 6(6) 8(10) 6(5) 33 38 93 95 10 27 25 44 108 110 33 (AB)C 38 93 95 10 27 25 44 108 110 33 2 6 7 31 38(2) 93(6) 95(7) 10(2) 27(6) 25(7) 44(2) 108(6) 110(7) 31 1299 357 1506 31 BC 4(2) 9(6) 10(7) 2(2) 6(6) 5(7) 21 132 75 21 A(BC) 7 5 1 3 8 6 32 132 75 21 7(132) 5(75) 1(132) 3(75) 8(132) 6(75) 31 1299 357 1506 31 Q.E.D.
10.8 IDENTITY AND NULL MATRICES An identity matrix I is a square matrix which has 1 for every element on the principal diagonal from left to right and 0 everywhere else. See Example 15. When a subscript is used, as in In, n denotes the dimensions of the matrix (n n). The identity matrix is similar to the number 1 in algebra since multiplication of a matrix by an identity matrix leaves the original matrix unchanged (that is, AI IA A). Multiplication of an identity matrix by itself leaves the identity matrix unchanged: I I I2 I. Any matrix for which A A is a symmetric matrix. A symmetric matrix for which A A A is an idempotent matrix. The identity matrix is symmetric and idempotent.
A null matrix is composed of all 0s and can be of any dimension; it is not necessarily square.
Addition or subtraction of the null matrix leaves the original matrix unchanged; multiplication by a null matrix produces a null matrix. See Example 15 and Problems 10.49 to 10.51.
EXAMPLE 15.
Given A 7 10 14 9 2 6 1 3 7 B 5 12 20 4 N 0 0 0 0 I 1 0 0 0 1 0 0 0 1 it is possible to show that (1) multiplication by an identity matrix leaves the original matrix unchanged, that is, AI A, (2) multiplication by a null matrix produces a null matrix, that is, BN N, and (3) addition or subtraction of a null matrix leaves the original matrix unchanged, that is, B N B. The calculations are shown below.
205 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] 1) AI 7 10 14 9 2 6 1 3 7 1 0 0 0 1 0 0 0 1 7(1) 10(0) 14(0) 7(0) 10(1) 14(0) 7(0) 10(0) 14(1) 9(1) 2(0) 6(0) 9(0) 2(1) 6(0) 9(0) 2(0) 6(1) 1(1) 3(0) 7(0) 1(0) 3(1) 7(0) 1(0) 3(0) 7(1) 7 10 14 9 2 6 1 3 7 Q.E.D.
2) BN 5(0) 12(0) 5(0) 12(0) 20(0) 4(0) 20(0) 4(0) 0 0 0 0 Q.E.D.
3) B N 5 0 12 0 20 0 4 0 5 12 20 4 Q.E.D.
10.9 MATRIX EXPRESSION OF A SYSTEM OF LINEAR EQUATIONS Matrix algebra permits the concise expression of a system of linear equations. As a simple illustration, note that the system of linear equations 7x1 3x2 45 4x1 5x2 29 can be expressed in matrix form AX B where A 7 3 4 5 X x1 x2 and B 45 29 Here A is the coefficient matrix, X is the solution vector, and B is the vector of constant terms. And X and B will always be column vectors. See Examples 16 and 17.
EXAMPLE 16.
To show that AX B accurately represents the given system of equations above, find the product AX. Multiplication is possible since AX is conformable, and the product matrix will be 2 1.
2 2 2 1 (2 1) Thus, AX 7 3 4 5 x1 x2 7x1 3x2 4x1 5x221 and AX B: 7x1 3x2 4x1 5x2 45 29 Q.E.D.
Here, despite appearances, AX is a 2 1 column vector since each row is composed of a single element which cannot be simplified further through addition.
EXAMPLE 17.
Given 8w 12x 7y 2z 139 3w 13x 4y 9z 242 206 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 To express this system of equations in matrix notation, mentally reverse the order of matrix multiplication: 8 12 7 2 3 13 4 9 24 w x y z 41 139 242 21 Then, letting A matrix of coefficients, W the column vector of variables, and B the column vector of constants, the given system of equations can be expressed in matrix form A24W41 B21 Solved Problems MATRIX FORMAT 10.1.
(a) Give the dimensions of each of the following matrices. (b) Give their transposes and indicate the new dimensions.
A 6 7 9 2 8 4 B 12 9 2 6 7 5 8 3 9 1 0 4 C 12 19 25 D 2 1 7 8 3 0 9 5 E [10 2 9 6 8 1] F 1 2 5 5 9 3 6 7 6 3 8 9 a) Recalling that dimensions are always listed row by column or rc, A 2 3, B 3 4, C 3 1, D 4 2, E 1 6, and F 4 3. C is also called a column vector; E, a row vector.
b) The transpose of A converts the rows of A to columns and the columns of A to rows.
A 6 2 7 8 9 4 32 B 12 7 9 9 5 1 2 8 0 6 3 4 43 C [12 19 25]13 D 2 7 3 9 1 8 0 5 24 E 10 2 9 6 8 1 61 F 1 5 6 3 2 9 7 8 5 3 6 9 34 10.2.
Given a21 4, a32 5, a13 3, a23 6, a12 10, and a31 5, use your knowledge of subscripts to complete the following matrix: A 6 — — — 7 — — — 9 207 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] Since the subscripts are always given in row-column order, a21 4 means that 4 is located in the second row, first column; a32 5 means that 5 appears in the third row, second column; etc. Thus, A 6 10 3 4 7 6 5 5 9 10.3.
A firm with five retail stores has 10 TVs t, 15 stereos s, 9 tape decks d, and 12 recorders r in store 1; 20t, 14s, 8d, and 5r in store 2; 16t, 8s, 15d, and 6r in store 3; 25t, 15s, 7d, and 16r in store 4; and 5t, 12s, 20d, and 18r in store 5. Express present inventory in matrix form.
Retail store 1 2 3 4 5 t 10 20 16 25 5 s 15 14 8 15 12 d 9 8 15 7 20 r 12 5 6 16 18 MATRIX ADDITION AND SUBTRACTION 10.4.
Find the sums A B of the following matrices: a) A 8 9 12 7 B 13 4 2 6 A B 8 13 9 4 12 2 7 6 21 13 14 13 b) A 7 10 8 2 B 8 12 4 6 A B 7 (8) 10 4 8 12 2 (6) 1 6 4 4 c) A [12 16 2 7 8] B [0 1 9 5 6] A B [12 17 11 12 14] d) A 9 4 2 7 3 5 8 6 B 1 3 6 5 2 8 9 2 A B 10 7 8 12 5 13 17 8 10.5.
Redo Problem 10.4, given A 0 1 6 2 3 5 8 7 2 9 1 6 B 7 2 12 6 5 4 3 8 10 6 1 0 5 11 9 208 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 Matrices A and B are not conformable for addition because they are not of equal dimensions; A 3 4, B 3 5.
10.6.
The parent company in Problem 10.3 sends out deliveries D to its stores: D 4 3 5 2 0 9 6 1 5 7 2 6 12 2 4 8 9 6 3 5 What is the new level of inventory?
I2 I1 D 10 15 9 12 20 14 8 5 16 8 15 6 25 15 7 16 5 12 20 18 4 3 5 2 0 9 6 1 5 7 2 6 12 2 4 8 9 6 3 5 14 18 14 14 20 23 14 6 21 15 17 12 37 17 11 24 14 18 23 23 10.7.
Find the difference A B for each of the following: a) A 3 7 11 12 9 2 B 6 8 1 9 5 8 A B 3 6 7 8 11 1 12 9 9 5 2 8 3 1 10 3 4 6 b) A 16 2 15 9 B 7 11 3 8 A B 16 7 2 11 15 3 9 8 9 9 12 1 c) A 13 5 8 4 9 1 10 6 2 B 14 9 3 2 6 13 5 8 11 A B 1 5 13 7 3 7 13 7 13 10.8.
A monthly report R on sales for the company in Problem 10.6 indicates R 8 12 6 9 10 11 8 3 15 6 9 7 21 14 5 18 6 11 13 9 209 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] What is the inventory left at the end of the month?
I2 R 14 18 14 14 20 23 14 6 21 15 17 12 37 17 11 24 14 18 23 23 8 12 6 9 10 11 8 3 15 6 9 7 21 14 5 18 6 11 13 9 6 6 8 5 10 12 6 3 6 9 8 5 16 3 6 6 8 7 10 14 CONFORMABILITY 10.9.
Given A 7 2 6 5 4 8 3 1 9 B 6 2 5 0 C 11 4 13 D 14 4 E [8 1 10] F [13 3] Determine for each of the following whether the products are defined, i.e., conformable for multiplication. If so, indicate the dimensions of the product matrix. (a) AC, (b) BD, (c) EC, (d) DF, (e) CA, (f) DE, (g) DB, (h) CF, (i) EF.
a) The dimensions of AC, in the order of multiplication, are 3 3 3 1. Matrix AC is defined since the numbers within the dashed circle indicate that the number of columns in A equals the number of rows in C. The numbers outside the circle indicate that the product matrix will be 3 1.
b) The dimensions of BD are 2 2 2 1. Matrix BD is defined; the product matrix will be 2 1.
c) The dimensions of EC are 1 3 3 1. Matrix EC is defined; the product matrix will be 1 1, or a scalar.
d) The dimensions of DF are 2 1 1 2. Matrix DF is defined; the product matrix will be 2 2.
e) The dimensions of CA are 3 1 3 3. Matrix CA is undefined. The matrices are not conformable for multiplication in that order. [Note that AC in part (a) is defined. This illustrates that matrix multiplication is not commutative: AC CA.] f) The dimensions of DE are 2 1 1 3. Matrix DE is defined; the product matrix will be 2 3.
g) The dimensions of DB are 2 1 2 2. The matrices are not conformable for multiplication.
Matrix DB is not defined.
h) The dimensions of CF are 3 1 1 2. The matrices are conformable; the product matrix will be 3 2.
i) The dimensions of EF are 1 3 1 2. The matrices are not conformable, and EF is not defined.
210 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 SCALAR AND VECTOR MULTIPLICATION 10.10. Determine Ak, given A 3 2 9 5 6 7 k 4 Here k is a scalar, and scalar multiplication is possible with a matrix of any dimension. Hence the product is defined.
Ak 3(4) 2(4) 9(4) 5(4) 6(4) 7(4) 12 8 36 20 24 28 10.11. Find kA, given k 2 A 7 5 2 3 6 7 2 8 9 kA 2(7) 2(5) 2(2) 2(3) 2(6) 2(7) 2(2) 2(8) 2(9) 14 10 4 6 12 14 4 16 18 10.12. A clothing store discounts all its slacks, jackets, and suits by 20 percent at the end of the year.
If V1 is the value of stock in its three branches prior to the discount, find the value V2 after the discount, when V1 5,000 4,500 6,000 10,000 12,000 7,500 8,000 9,000 11,000 A 20 percent reduction means that the clothing is selling for 80 percent of its original value. Hence V2 0.8V1, V2 0.8 5,000 4,500 6,000 10,000 12,000 7,500 8,000 9,000 11,000 4,000 3,600 4,800 8,000 9,600 6,000 6,400 7,200 8,800 10.13. Find AB, given A [9 11 3] B 2 6 7 Matrix AB is defined; 1 3 3 1; the product will be a scalar, derived by multiplying each element of the row vector by its corresponding element in the column vector and then summing the products.
AB 9(2) 11(6) 3(7) 18 66 21 105 211 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] 10.14. Find AB, given A [12 5 6 11] B 3 2 8 6 Matrix AB is defined; 1 4 4 1.
AB 12(3) (5)(2) 6(8) 11(6) 44 10.15. Find AB, given A [9 6 2 0 5] B 2 13 5 8 1 Matrix AB is defined; 1 5 5 1.
AB 9(2) 6(13) 2(5) 0(8) (5)(1) 101 10.16. Find AB, given A [12 9 2 4] B 6 1 2 Matrix AB is undefined; 1 4 3 1. Multiplication is not possible.
10.17. If the price of a TV is $300, the price of a stereo is $250, the price of a tape deck is $175, and the price of a recorder is $125, use vectors to determine the value of stock for outlet 2 in Problem 10.3.
The value of stock is V QP. The physical volume of stock in outlet 2 in vector form is Q [20 14 8 5]. The price vector P can be written P 300 250 175 125 Matrix QP is defined; 1 4 4 1. Thus V QP 20(300) 14(250) 8(175) 5(125) 11,525 10.18. Redo Problem 10.17 for outlet 5 in Problem 10.3.
Here Q [5 12 20 18], P remains the same. Matrix QP is defined. Thus, V 5(300) 12(250) 20(175) 18(125) 10,250 MATRIX MULTIPLICATION 10.19. Determine whether AB is defined, indicate what the dimensions of the product matrix will be, and find the product matrix AB, given A 12 14 20 5 B 3 9 0 2 212 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 Matrix AB is defined; 2 2 2 2; the product matrix will be 2 2. Matrix multiplication is nothing but a series of row-column vector multiplications in which the a11 element of the product matrix is determined by the product of the first row R1 of the lead matrix and the first column C1 of the lag matrix; the a12 element of the product matrix is determined by the product of the first row R1 of the lead matrix and the second column C2 of the lag matrix; the aij element of the product matrix is determined by the product of the ith row Ri of the lead matrix and the jth column Cj of the lag matrix, etc. Thus, AB R1C1 R1C2 R2C1 R2C2 12(3) 14(0) 12(9) 14(2) 20(3) 5(0) 20(9) 5(2) 36 136 60 190 10.20. Redo Problem 10.19, given A 4 7 9 1 B 3 8 5 2 6 7 Matrix AB is defined; 2 2 2 3; the product matrix will be 2 3.
AB R1C1 R1C2 R1C3 R2C1 R2C2 R2C3 4(3) 7(2) 4(8) 7(6) 4(5) 7(7) 9(3) 1(2) 9(8) 1(6) 9(5) 1(7) 26 74 69 29 78 52 10.21. Redo Problem 10.19, given A 3 1 8 2 B 2 9 4 6 7 5 Matrix AB is not defined; 2 2 3 2. The matrices cannot be multiplied because they are not conformable in the given order. The number of columns (2) in A does not equal the number of rows (3) in B.
10.22. Redo Problem 10.19 for BA in Problem 10.21.
Matrix BA is defined; 3 2 2 2; the product matrix will be 3 2.
BA 2 9 4 6 7 5 3 1 8 2 R1C1 R1C2 R2C1 R2C2 R3C1 R3C2 2(3) 9(8) 2(1) 9(2) 4(3) 6(8) 4(1) 6(2) 7(3) 5(8) 7(1) 5(2) 78 20 60 16 61 17 10.23. Redo Problem 10.19 for AB in Problem 10.21, where B is the transpose of B: B 2 4 7 9 6 5 Matrix AB is defined: 2 2 2 3; the product will be a 2 3 matrix.
AB 3 1 8 2 2 4 7 9 6 5 3(2) 1(9) 3(4) 1(6) 3(7) 1(5) 8(2) 2(9) 8(4) 2(6) 8(7) 2(5) 15 18 26 34 44 66 (Note from Problems 10.21 to 10.23 that AB BA AB. The noncommutative aspects of matrix multiplication are treated in Problems 10.36 to 10.41.) 213 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] 10.24. Redo Problem 10.19, given A 7 11 2 9 10 6 B 12 4 5 3 6 1 Matrix AB is defined; 3 2 2 3; the product matrix will be 3 3.
AB R1C1 R1C2 R1C3 R2C1 R2C2 R2C3 R3 C1 R3C2 R3C3 7(12) 11(3) 7(4) 11(6) 7(5) 11(1) 2(12) 9(3) 2(4) 9(6) 2(5) 9(1) 10(12) 6(3) 10(4) 6(6) 10(5) 6(1) 117 94 46 51 62 19 138 76 56 10.25. Redo Problem 10.19, given A 6 2 5 7 9 4 B 10 1 11 3 2 9 Matrix AB is defined; 2 3 3 2; the product matrix will be 2 2.
AB R1 C1 R1C2 R2C1 R2C2 6(10) 2(11) 5(2) 6(1) 2(3) 5(9) 7(10) 9(11) 4(2) 7(1) 9(3) 4(9) 92 57 177 70 10.26. Redo Problem 10.19, given A [2 3 5] B 7 1 6 5 2 4 9 2 7 Matrix AB is defined; 1 3 3 3. The product matrix will be 1 3.
AB [R1C1 R1C2 R1C3] [2(7) 3(5) 5(9) 2(1) 3(2) 5(2) 2(6) 3(4) 5(7)] [74 18 59] 10.27. Redo Problem 10.19, given A 5 1 10 B 3 9 4 2 1 8 5 6 1 Matrix AB is not defined; 3 1 3 3. Multiplication is impossible in the given order.
10.28. Find BA from Problem 10.27.
Matrix BA is defined; 3 3 3 1. The product matrix will be 3 1.
BA 3 9 4 2 1 8 5 6 1 5 1 10 R1C1 R2C1 R3C1 3(5) 9(1) 4(10) 2(5) 1(1) 8(10) 5(5) 6(1) 1(10) 64 91 41 214 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 10.29. Redo Problem 10.19, given A 2 1 5 3 2 6 1 4 3 B 10 1 2 5 3 6 2 1 2 Matrix AB is defined; 3 3 3 3. The product matrix will be 3 3.
AB R1C1 R1C2 R1C3 R2C1 R2C2 R2C3 R3 C1 R3C2 R3C3 2(10) 1(5) 5(2) 2(1) 1(3) 5(1) 2(2) 1(6) 5(2) 3(10) 2(5) 6(2) 3(1) 2(3) 6(1) 3(2) 2(6) 6(2) 1(10) 4(5) 3(2) 1(1) 4(3) 3(1) 1(2) 4(6) 3(2) 35 10 20 52 15 30 36 16 32 10.30. Redo Problem 10.19, given A 3 1 4 5 B [2 6 5 3] Matrix AB is defined; 4 1 1 4. The product matrix will be 4 4.
AB R1C1 R1C2 R1C3 R1C4 R2C1 R2C2 R2C3 R2C4 R3 C1 R3C2 R3C3 R3C4 R4C1 R4C2 R4C3 R4C4 3(2) 3(6) 3(5) 3(3) 1(2) 1(6) 1(5) 1(3) 4(2) 4(6) 4(5) 4(3) 5(2) 5(6) 5(5) 5(3) 6 18 15 9 2 6 5 3 8 24 20 12 10 30 25 15 10.31.
Find AB when A [3 9 8 7] B 2 5 3 Matrix AB is undefined and cannot be multiplied as given; 1 4 3 1.
10.32. Find BA from Problem 10.31.
Matrix BA is defined; 3 1 1 4. The product matrix will be 3 4.
BA R1C1 R1C2 R1C3 R1C4 R2C1 R2C2 R2C3 R2C4 R3 C1 R3C2 R3C3 R3C4 2 5 3 [3 9 8 7] 2(3) 2(9) 2(8) 2(7) 5(3) 5(9) 5(8) 5(7) 3(3) 3(9) 3(8) 3(7) 6 18 16 14 15 45 40 35 9 27 24 21 215 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] 10.33. Use the inventory matrix for the company in Problem 10.3 and the price vector from Problem 10.17 to determine the value of inventory in all five of the company’s outlets.
V QP.
QP is defined; 5 4 4 1; V will be 5 1.
V 10 20 16 25 5 15 14 8 15 12 9 8 15 7 20 12 5 6 16 18 300 250 175 125 R1C1 R2C1 R3C1 R4C1 R5C1 10(300) 15(250) 9(175) 12(125) 20(300) 14(250) 8(175) 5(125) 16(300) 8(250) 15(175) 6(125) 25(300) 15(250) 7(175) 16(125) 5(300) 12(250) 20(175) 18(125) 9,825 11,525 10,175 14,475 10,250 THE COMMUTATIVE LAW AND MATRIX OPERATIONS 10.34. To illustrate the commutative or noncommutative aspects of matrix operations (that is, A B B A, but in general, AB BA), find (a) A B and (b) B A, given A 7 3 2 1 4 6 2 5 4 B 2 0 5 3 4 1 7 9 6 a) A B 7 2 3 0 2 5 1 3 4 4 6 1 2 7 5 9 4 6 9 3 7 4 8 7 9 14 10 b) B A 2 7 0 3 5 2 3 1 4 4 1 6 7 2 9 5 6 4 9 3 7 4 8 7 9 14 10 A B B A. This illustrates that the commutative law does apply to matrix addition.
Problems 10.35 to 10.42 illustrate the application of the commutative law to other matrix operations.
10.35. Find (a) A B and (b) B A given A 5 3 4 9 10 8 6 12 B 3 13 7 9 2 1 8 6 a) A B 5 3 4 7 10 2 6 8 3 13 9 9 8 1 12 6 2 10 3 0 8 7 2 6 b) B A 3 5 7 4 2 10 8 6 13 3 9 9 1 8 6 12 2 10 3 0 8 7 2 6 A B B A. This illustrates that matrix subtraction is commutative.
216 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 10.36. Find (a) AB and (b) BA, given A [4 12 9 6] B 13 5 2 7 Check for conformability first and indicate the dimensions of the product matrix.
a) Matrix AB is defined; 1 4 4 1. The product will be a 1 1 matrix or scalar.
AB [4(13) 12(5) 9(2) 6(7)] 136 b) Matrix BA is also defined; 4 1 1 4; the product will be a 4 4 matrix.
BA 13(4) 5(4) 2(4) 7(4) 13(12) 5(12) 2(12) 7(12) 13(9) 5(9) 2(9) 7(9) 13(6) 5(6) 2(6) 7(6) 52 20 8 28 156 60 24 84 117 45 18 63 78 30 12 42 AB BA. This illustrates the noncommutative aspect of matrix multiplication. Products generally differ in dimensions and elements if the order of multiplication is reversed.
10.37. Find (a) AB and (b) BA, given A 7 4 6 2 1 8 B 3 9 1 2 12 7 a) Matrix AB is defined; 3 2 2 3; the product will be 3 3.
AB 7(3) 4(2) 7(9) 4(12) 7(1) 4(7) 6(3) 2(2) 6(9) 2(12) 6(1) 2(7) 1(3) 8(2) 1(9) 8(12) 1(1) 8(7) 13 111 35 14 78 20 13 105 57 b) Matrix BA is also defined; 2 3 3 2; the product will be 2 2.
BA 3(7) 9(6) 1(1) 3(4) 9(2) 1(8) 2(7) 12(6) 7(1) 2(4) 12(2) 7(8) 34 14 93 88 AB BA. Matrix multiplication is not commutative. Here the products again differ in dimensions and elements.
10.38. Find (a) AB and (b) BA, given A 4 9 8 7 6 2 1 5 3 B 1 2 0 5 3 1 0 2 4 a) Matrix AB is defined; 3 3 3 3; the product will be 3 3.
AB 4(1) 9(5) 8(0) 4(2) 9(3) 8(2) 4(0) 9(1) 8(4) 7(1) 6(5) 2(0) 7(2) 6(3) 2(2) 7(0) 6(1) 2(4) 1(1) 5(5) 3(0) 1(2) 5(3) 3(2) 1(0) 5(1) 3(4) 49 51 41 37 36 14 26 23 17 217 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] b) Matrix BA is also defined and will result in a 3 3 matrix.
BA 1(4) 2(7) 0(1) 1(9) 2(6) 0(5) 1(8) 2(2) 0(3) 5(4) 3(7) 1(1) 5(9) 3(6) 1(5) 5(8) 3(2) 1(3) 0(4) 2(7) 4(1) 0(9) 2(6) 4(5) 0(8) 2(2) 4(3) 18 21 12 42 68 49 18 32 16 AB BA. The dimensions are the same but the elements differ.
10.39. Find (a) AB and (b) BA, given A 7 5 2 6 1 3 9 4 B 1 0 1 3 a) Matrix AB is defined; 2 4 4 1; the product will be 2 1.
AB 7(1) 5(0) 2(1) 6(3) 1(1) 3(0) 9(1) 4(3) 23 4 b) Matrix BA is not defined; 4 1 2 4. Multiplication is impossible. This is but another way in which matrix multiplication is noncommutative.
10.40. Find (a) AB and (b) BA, given A 11 14 2 6 B 7 6 4 5 1 3 a) Matrix AB is not defined; 2 2 3 2 and so cannot be multiplied.
b) Matrix BA is defined; 3 2 2 2 and will produce a 3 2 matrix.
BA 7(11) 6(2) 7(14) 6(6) 4(11) 5(2) 4(14) 5(6) 1(11) 3(2) 1(14) 3(6) 89 134 54 86 17 32 BA AB, because AB does not exist.
10.41. Find (a) AB and (b) BA, given A 2 4 7 B [3 6 2] a) Matrix AB is defined; 3 1 1 3; the product will be a 3 3 matrix.
AB 2(3) 4(3) 7(3) 2(6) 4(6) 7(6) 2(2) 4(2) 7(2) 6 12 21 12 24 42 4 8 14 b) Matrix BA is also defined; 1 3 3 1, producing a 1 1 matrix or scalar.
BA [3(2) 6(4) (2)(7)] 4 Since matrix multiplication is not commutative, reversing the order of multiplication can lead to widely different answers. Matrix AB results in a 3 3 matrix, BA results in a scalar.
218 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 10.42. Find (a) AB and (b) BA, for a case where B is an identity matrix, given A 23 6 14 18 12 9 24 2 6 B 1 0 0 0 1 0 0 0 1 a) Matrix AB is defined; 3 3 3 3. The product matrix will also be 3 3.
AB 23(1) 6(0) 14(0) 23(0) 6(1) 14(0) 23(0) 6(0) 14(1) 18(1) 12(0) 9(0) 18(0) 12(1) 9(0) 18(0) 12(0) 9(1) 24(1) 2(0) 6(0) 24(0) 2(1) 6(0) 24(0) 2(0) 6(1) 23 6 14 18 12 9 24 2 6 b) Matrix BA is also defined; 3 3 3 3. The product matrix will also be 3 3.
BA 1(23) 0(18) 0(24) 1(6) 0(12) 0(2) 1(14) 0(9) 0(6) 0(23) 1(18) 0(24) 0(6) 1(12) 0(2) 0(14) 1(9) 0(6) 0(23) 0(18) 1(24) 0(6) 0(12) 1(2) 0(14) 0(9) 1(6) 23 6 14 18 12 9 24 2 6 Here AB BA. Premultiplication or postmultiplication by an identity matrix gives the original matrix. Thus in the case of an identity matrix, matrix multiplication is commutative. This will also be true of a matrix and its inverse. See Section 11.7.
ASSOCIATIVE AND DISTRIBUTIVE LAWS 10.43. To illustrate whether the associative and distributive laws apply to matrix operations [that is, (A B) C A (B C), (AB)C A(BC), and A(B C) AB AC, subject to the condi-tions in Section 10.7], find (a) (A B) C and (b) A (B C), given A 6 2 7 9 5 3 B 9 1 3 4 2 6 C 7 5 1 10 3 8 a) A B 6 9 2 1 7 3 9 4 5 2 3 6 15 3 10 13 7 9 (A B) C 15 7 3 5 10 1 13 10 7 3 9 8 22 8 11 23 10 17 b) B C 9 7 1 5 3 1 4 10 2 3 6 8 16 6 4 14 5 14 A (B C) 6 16 2 6 7 4 9 14 5 5 3 14 22 8 11 23 10 17 Thus, (A B) C A (B C). This illustrates that matrix addition is associative. Other aspects of these laws are demonstrated in Problems 10.44 to 10.47.
10.44. Find (a) (A B) C and (b) A (B C), given A 7 6 12 B 3 8 5 C 13 2 6 a) A B 7 3 6 8 12 5 4 2 7 b) B C 3 13 8 2 5 6 10 6 1 (A B) C 4 13 2 2 7 6 17 0 13 A (B C) 7 10 6 (6) 12 1 17 0 13 Matrix subtraction is also associative.
219 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] 10.45. Find (a) (AB)C and (b) A(BC), given A [7 1 5] B 6 5 2 4 3 8 C 9 4 3 10 a) Matrix AB is defined; 1 3 3 2, producing a 1 2 matrix.
AB [7(6) 1(2) 5(3) 7(5) 1(4) 5(8)] [59 79] Matrix (AB)C is defined; 1 2 2 2, leaving a 1 2 matrix.
(AB)C [59(9) 79(3) 59(4) 79(10)] [768 1026] Matrix BC is defined; 3 2 2 2, creating a 3 2 matrix.
BC 6(9) 5(3) 6(4) 5(10) 2(9) 4(3) 2(4) 4(10) 3(9) 8(3) 3(4) 8(10) 69 74 30 48 51 92 Matrix A(BC) is also defined; 1 3 3 2, producing a 1 2 matrix.
A(BC) [7(69) 1(30) 5(51) 7(74) 1(48) 5(92)] [768 1026] Matrix multiplication is associative, provided the proper order of multiplication is maintained.
10.46. Find (a) A(B C) and (b) AB AC, given A [4 7 2] B 6 5 1 C 9 5 8 a) B C 6 9 5 5 1 8 15 10 9 Matrix A(B C) is defined; 1 3 3 1. The product matrix will be 1 1.
A(B C) [4(15) 7(10) 2(9)] 148 b) Matrix AB is defined; 1 3 3 1, producing a 1 1 matrix.
AB [4(6) 7(5) 2(1)] 61 Matrix AC is defined; 1 3 3 1, also producing a 1 1 matrix.
AC [4(9) 7(5) 2(8)] 87 Thus, AB AC 61 87 148. This illustrates the distributive law of matrix multiplication.
10.47. A hamburger chain sells 1000 hamburgers, 600 cheeseburgers, and 1200 milk shakes in a week.
The price of a hamburger is 45¢, a cheeseburger 60¢, and a milk shake 50¢. The cost to the chain of a hamburger is 38¢, a cheeseburger 42¢, and a milk shake 32¢. Find the firm’s profit for the week, using (a) total concepts and (b) per-unit analysis to prove that matrix multiplication is distributive.
220 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 a) The quantity of goods sold Q, the selling price of the goods P, and the cost of goods C, can all be represented in matrix form: Q 1000 600 1200 P 0.45 0.60 0.50 C 0.38 0.42 0.32 Total revenue TR is TR PQ 0.45 0.60 0.50 1000 600 1200 which is not defined as given. Taking the transpose of P or Q will render the vectors conformable for multiplication. Note that the order of multiplication is all-important. Row-vector multiplication (PQ or QP) will produce the scalar required; vector-row multiplication (PQ or QP) will produce a 3 3 matrix that has no economic meaning. Thus, taking the transpose of P and premultiplying, we get TR PQ [0.45 0.60 0.50] 1000 600 1200 where PQ is defined; 1 3 3 1, producing a 1 1 matrix or scalar.
TR [0.45(1000) 0.60(600) 0.50(1200)] 1410 Similarly, total cost TC is TC CQ: TC [0.38 0.42 0.32] 1000 600 1200 [0.38(1000) 0.42(600) 0.32(1200)] 1016 Profits, therefore, are TR TC 1410 1016 394 b) Using per-unit analysis, the per-unit profit U is U P C 0.45 0.60 0.50 0.38 0.42 0.32 0.07 0.18 0.18 Total profit is per-unit profit times the number of items sold UQ 0.07 0.18 0.18 1000 600 1200 which is undefined. Taking the transpose of U, UP [0.07 0.18 0.18] 1000 600 1200 [0.07(1000) 0.18(600) 0.18(1200)] 394 Q.E.D.
10.48. Crazy Teddie’s sells 700 CDs, 400 cassettes, and 200 CD players each week. The selling price of CDs is $4, cassettes $6, and CD players $150. The cost to the shop is $3.25 for a CD, $4.75 for a cassette, and $125 for a CD player. Find weekly profits by using (a) total and (b) per-unit concepts.
221 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] a) Q 700 400 200 P 4 6 150 C 3.25 4.75 125.00 TR PQ [4 6 150] 700 400 200 [4(700) 6(400) 150(200)] 35,200 TC CQ [3.25 4.75 125] 700 400 200 [3.25(700) 4.75(400) 125(200)] 29,175 TR TC 35,200 29,175 6025 b) Per-unit profit U is U P C 4 6 150 3.25 4.75 125.00 0.75 1.25 25.00 Total profit is UQ [0.75 1.25 25] 700 400 200 [0.75(700) 1.25(400) 25(200)] 6025 UNIQUE PROPERTIES OF MATRICES 10.49. Given A 6 12 3 6 B 12 6 6 3 (a) Find AB. (b) Why is the product unique?
a) AB 6(12) 12(6) 6(6) 12(3) 3(12) 6(6) 3(6) 6(3) 0 0 0 0 b) The product AB is unique to matrix algebra in that, unlike ordinary algebra in which the product of two nonzero numbers can never equal zero, the product of two non-null matrices may produce a null matrix. The reason for this is that the two original matrices are singular. A singular matrix is one in which a row or column is a multiple of another row or column (see Section 11.1). In this problem, row 1 of A is 2 times row 2, and column 2 is 2 times column 1. In B, row 1 is 2 times row 2, and column 1 is 2 times column 2. Thus, in matrix algebra, multiplication involving singular matrices may, but need not, produce a null matrix as a solution. See Problem 10.50.
10.50. (a) Find AB and (b) comment on the solution, given A 6 12 3 6 B 12 6 6 3 a) AB 6(12) 12(6) 6(6) 12(3) 3(12) 6(6) 3(6) 6(3) 144 72 72 36 b) While both A and B are singular, they do not produce a null matrix. The product AB, however, is also singular.
222 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA [CHAP . 10 10.51. Given A 4 8 1 2 B 2 1 2 2 C 2 1 4 2 a) Find AB and AC. (b) Comment on the unusual property of the solutions.
a) AB 4(2) 8(2) 4(1) 8(2) 1(2) 2(2) 1(1) 2(2) 24 20 6 5 AC 4(2) 8(4) 4(1) 8(2) 1(2) 2(4) 1(1) 2(2) 24 20 6 5 b) Even though B C, AB AC. Unlike algebra, where multiplication of one number by two different numbers cannot give the same product, in matrix algebra multiplication of one matrix by two different matrices may, but need not, produce identical matrices. In this case, A is a singular matrix.
223 THE FUNDAMENTALS OF LINEAR (OR MATRIX) ALGEBRA CHAP . 10] CHAPTER 11 Matrix Inversion 11.1 DETERMINANTS AND NONSINGULARITY The determinant A of a 2 2 matrix, called a second-order determinant, is derived by taking the product of the two elements on the principal diagonal and subtracting from it the product of the two elements off the principal diagonal. Given a general 2 2 matrix A a11 a12 a21 a22 the determinant is A a11 a12 a21 a22 () () a11a22 a12a21 The determinant is a single number or scalar and is found only for square matrices. If the determinant of a matrix is equal to zero, the determinant is said to vanish and the matrix is termed singular. A singular matrix is one in which there exists linear dependence between at least two rows or columns. If A 0, matrix A is nonsingular and all its rows and columns are linearly independent.
If linear dependence exists in a system of equations, the system as a whole will have an infinite number of possible solutions, making a unique solution impossible. Hence we want to preclude linearly dependent equations from our models and will generally fall back on the following simple determinant test to spot potential problems. Given a system of equations with coefficient matrix A, If A 0, the matrix is singular and there is linear dependence among the equations. No unique solution is possible.
If A 0, the matrix is nonsingular and there is no linear dependence among the equations. A unique solution can be found.
The rank of a matrix is defined as the maximum number of linearly independent rows or columns in the matrix. The rank of a matrix also allows for a simple test of linear dependence which follows immediately. Assuming a square matrix of order n, If (A) n, A is nonsingular and there is no linear dependence.
If (A) n, A is singular and there is linear dependence.
See Example 1 and Problems 11.1, 11.3, and 11.17. For proof of nonsingularity and linear independence, see Problem 11.16.
224 ↙ ↘ EXAMPLE 1.
Determinants are calculated as follows, given A 6 4 7 9 B 4 6 6 9 From the rules stated above, A 6(9) 4(7) 26 Since A 0, the matrix is nonsingular, i.e., there is no linear dependence between any of its rows or columns.
The rank of A is 2, written (A) 2. By way of contrast, B 4(9) 6(6) 0 With B 0, B is singular and linear dependence exists between its rows and columns. Closer inspection reveals that row 2 and column 2 are equal to 1.5 times row 1 and column 1, respectively. Hence (B) 1.
11.2 THIRD-ORDER DETERMINANTS The determinant of a 3 3 matrix A a11 a12 a13 a21 a22 a23 a31 a32 a33 is called a third-order determinant and is the summation of three products. To derive the three products: 1.
Take the first element of the first row, a11, and mentally delete the row and column in which it appears. See (a) below. Then multiply a11 by the determinant of the remaining elements.
2.
Take the second element of the first row, a12, and mentally delete the row and column in which it appears. See (b) below. Then multiply a12 by 1 times the determinant of the remaining elements.
3.
Take the third element of the first row, a13, and mentally delete the row and column in which it appears. See (c) below. Then multiply a13 by the determinant of the remaining elements.
a11 a12 a13 a21 a22 a23 a31 a32 a33 a11 a12 a13 a21 a22 a23 a31 a32 a33 a11 a12 a13 a21 a22 a23 a31 a32 a33 (a) (b) (c) Thus, the calculations for the determinant are as follows: A a11 a22 a23 a32 a33 a12(1) a21 a23 a31 a33 a13 a21 a22 a31 a32 a11(a22a33 a23a32) a12(a21a33 a23a31) a13(a21a32 a22a31) (11.1) a scalar See Examples 2 and 3 and Problems 11.2, 11.3, and 11.17.
In like manner, the determinant of a 4 4 matrix is the sum of four products; the determinant of a 5 5 matrix is the sum of five products; etc. See Section 11.4 and Example 5.
225 MATRIX INVERSION CHAP . 11] EXAMPLE 2.
Given A 8 3 2 6 4 7 5 1 3 the determinant A is calculated as follows: A 8 4 7 1 3 3(1) 6 7 5 3 2 6 4 5 1 8[4(3) 7(1)] 3[6(3) 7(5)] 2[6(1) 4(5)] 8(5) 3(17) 2(14) 63 With A 0, A is nonsingular and (A) 3.
11.3 MINORS AND COFACTORS The elements of a matrix remaining after the deletion process described in Section 11.2 form a subdeterminant of the matrix called a minor. Thus, a minor Mij is the determinant of the submatrix formed by deleting the ith row and jth column of the matrix. Using the matrix from Section 11.2, M11 a22 a23 a32 a33 M12 a21 a23 a31 a33 M13 a21 a22 a31 a32 where M11 is the minor of a11, M12 the minor of a12, and M13 the minor of a13. Thus, the determinant in (11.1) can be written A a11 M11 a12(1) M12 a13 M13 (11.2) A cofactor Cij is a minor with a prescribed sign. The rule for the sign of a cofactor is Cij (1)ij Mij Thus if the sum of the subscripts is an even number, Cij Mij , since 1 raised to an even power is positive. If i j is equal to an odd number, Cij Mij , since 1 raised to an odd power is negative.
See Example 3 and Problems 11.18 to 11.24.
EXAMPLE 3.
The cofactors (1) C11 , (2) C12 , and (3) C13 for the matrix in Section 11.2 are found as follows: 1) C11 (1)11 M11 Since (1)11 (1)2 1, C11 M11 a22 a23 a32 a33 2) C12 (1)12 M12 Since (1)12 (1)3 1, 3) C13 (1)13 M13 Since (1)13 (1)4 1, C12 M12 a21 a23 a31 a33 C13 M13 a21 a22 a31 a32 226 MATRIX INVERSION [CHAP . 11 11.4 LAPLACE EXPANSION AND HIGHER-ORDER DETERMINANTS Laplace expansion is a method for evaluating determinants in terms of cofactors. It thus simplifies matters by permitting higher-order determinants to be established in terms of lower-order deter-minants. Laplace expansion of a third-order determinant can be expressed as A a11 C11 a12 C12 a13 C13 (11.3) where Cij is a cofactor based on a second-order determinant. Here, unlike (11.1) and (11.2), a12 is not explicitly multiplied by 1, since by the rule of cofactors C12 will automatically be multiplied by 1.
Laplace expansion permits evaluation of a determinant along any row or column. Selection of a row or column with more zeros than others simplifies evaluation of the determinant by eliminating terms. Laplace expansion also serves as the basis for evaluating determinants of orders higher than three. See Examples 4 and 5 and Problem 11.25.
EXAMPLE 4.
Given A 12 7 0 5 8 3 6 7 0 the determinant is found by Laplace expansion along the third column, as demonstrated below: A a13 C13 a23 C23 a33 C33 Since a13 and a33 0 A a23 C23 (11.4) Deleting row 2 and column 3 to find C23 , C23 (1)23 12 7 6 7 (1)[12(7) 7(6)] 42 Then substituting in (11.4) where a23 3, A 3(42) 126. So A is nonsingular and (A) 3.
The accuracy of this answer can be readily checked by expanding along the first row and solving for A .
EXAMPLE 5.
Laplace expansion for a fourth-order determinant is A a11 C11 a12 C12 a13 C13 a14 C14 where the cofactors are third-order subdeterminants which in turn can be reduced to second-order subdeter-minants, as above. Fifth-order determinants and higher are treated in similar fashion. See Problem 11.25(d) to (e).
11.5 PROPERTIES OF A DETERMINANT The following seven properties of determinants provide the ways in which a matrix can be manipulated to simplify its elements or reduce part of them to zero, before evaluating the determinant: 1.
Adding or subtracting any nonzero multiple of one row (or column) from another row (or column) will have no effect on the determinant.
2.
Interchanging any two rows or columns of a matrix will change the sign, but not the absolute value, of the determinant.
3.
Multiplying the elements of any row or column by a constant will cause the determinant to be multiplied by the constant.
227 MATRIX INVERSION CHAP . 11] 4.
The determinant of a triangular matrix, i.e., a matrix with zero elements everywhere above or below the principal diagonal, is equal to the product of the elements on the principal diagonal.
5.
The determinant of a matrix equals the determinant of its transpose: A A .
6.
If all the elements of any row or column are zero, the determinant is zero.
7.
If two rows or columns are identical or proportional, i.e., linearly dependent, the determinant is zero.
These properties and their use in matrix manipulation are treated in Problems 11.4 to 11.15.
11.6 COFACTOR AND ADJOINT MATRICES A cofactor matrix is a matrix in which every element aij is replaced with its cofactor Cij . An adjoint matrix is the transpose of a cofactor matrix. Thus, C C11 C12 C13 C21 C22 C23 C31 C32 C33 Adj A C C11 C21 C31 C12 C22 C32 C13 C23 C33 EXAMPLE 6.
The cofactor matrix C and the adjoint matrix Adj A are found below, given A 2 3 1 4 1 2 5 3 4 Replacing the elements aij with their cofactors Cij according to the laws of cofactors, C 1 2 3 4 3 1 3 4 3 1 1 2 4 2 5 4 2 1 5 4 2 1 4 2 4 1 5 3 2 3 5 3 2 3 4 1 2 6 7 9 3 9 5 0 10 The adjoint matrix Adj A is the transpose of C, Adj A C 2 9 5 6 3 0 7 9 10 11.7 INVERSE MATRICES An inverse matrix A1, which can be found only for a square, nonsingular matrix A, is a unique matrix satisfying the relationship AA1 I A1A Multiplying a matrix by its inverse reduces it to an identity matrix. Thus, the inverse matrix in linear algebra performs much the same function as the reciprocal in ordinary algebra. The formula for deriving the inverse is A1 1 A Adj A See Example 7 and Problem 11.25.
228 MATRIX INVERSION [CHAP . 11 EXAMPLE 7.
Find the inverse for A 4 1 5 2 3 1 3 1 4 1.
Check that it is a square matrix, here 3 3, since only square matrices can have inverses.
2.
Evaluate the determinant to be sure A 0, since only nonsingular matrices can have inverses. A 4[3(4) 1(1)] 1[(2)(4) 1(3)] (5)[(2)(1) 3(3)] 52 11 35 98 0 Matrix A is nonsingular; (A) 3.
3.
Find the cofactor matrix of A, C 3 1 1 4 1 5 1 4 1 5 3 1 2 1 3 4 4 5 3 4 4 5 2 1 2 3 3 1 4 1 3 1 4 1 2 3 13 11 7 1 31 7 16 6 14 Then transpose the cofactor matrix to get the adjoint matrix.
Adj A C 13 1 16 11 31 6 7 7 14 4.
Multiply the adjoint matrix by 1/ A 1 –– 98 to get A1.
A1 1 98 13 1 16 11 31 6 7 7 14 13 –– 98 11 –– 98 1 –– 14 1 –– 98 31 –– 98 1 –– 14 16 –– 98 6 –– 98 1 – 7 0.1327 0.0102 0.1633 0.1122 0.3163 0.0612 0.0714 0.0714 0.1429 5.
To check your answer, multiply AA1 or A1A. Both products will equal I if the answer is correct. An inverse is checked in Problem 11.26(a).
11.8 SOLVING LINEAR EQUATIONS WITH THE INVERSE An inverse matrix can be used to solve matrix equations. If AnnXn1 Bn1 and the inverse A1 exists, multiplication of both sides of the equation by A1, following the laws of conformability, gives A1 nnAnnXn1 A1 nnBn1 From Section 11.7, A1A I. Thus, InnXn1 A1 nnBn1 From Section 10.8, IX X. Therefore, Xn1 (A1B)n1 The solution of the equation is given by the product of the inverse of the coefficient matrix A1 and the column vector of constants B.
See Problems 11.27 to 11.33.
229 MATRIX INVERSION CHAP . 11] EXAMPLE 8.
Matrix equations and the inverse are used below to solve for x1, x2, and x3, given 4x1 x2 5x3 8 2x1 3x2 x3 12 3x1 x2 4x3 5 First, express the system of equations in matrix form, AX B 4 1 5 2 3 1 3 1 4 x1 x2 x3 8 12 5 From Section 11.8, X A1B Substituting A1 from Example 7 and multiplying, X 13 –– 98 11 –– 98 1 –– 14 1 –– 98 31 –– 98 1 –– 14 16 –– 98 6 –– 98 1 – 7 8 12 5 104 ––– 98 88 –– 98 8 –– 14 12 –– 98 372 ––– 98 12 –– 14 80 –– 98 30 –– 98 5 – 7 196 ––– 98 490 ––– 98 14 –– 14 2 5 1 Thus, x ¯1 2, x ¯2 5, and x ¯3 1.
11.9 CRAMER’S RULE FOR MATRIX SOLUTIONS Cramer’s rule provides a simplified method of solving a system of linear equations through the use of determinants. Cramer’s rule states x ¯i Ai A where xi is the ith unknown variable in a series of equations, A is the determinant of the coefficient matrix, and Ai is the determinant of a special matrix formed from the original coefficient matrix by replacing the column of coefficients of xi with the column vector of constants. See Example 9 and Problems 11.34 to 11.37. Proof for Cramer’s rule is given in Problem 11.38.
EXAMPLE 9.
Cramer’s rule is used below to solve the system of equations 6x1 5x2 49 3x1 4x2 32 1.
Express the equations in matrix form.
AX B 6 5 3 4 x1 x2 49 32 2.
Find the determinant of A A 6(4) 5(3) 9 3.
Then to solve for x1, replace column 1, the coefficients of x1, with the vector of constants B, forming a new matrix A1.
A1 49 5 32 4 230 MATRIX INVERSION [CHAP . 11 Find the determinant of A1, A1 49(4) 5(32) 36 and use the formula for Cramer’s rule, x ¯1 A1 A 36 9 4 4.
To solve for x2, replace column 2, the coefficients of x2, from the original matrix, with the column vector of constants B, forming a new matrix A2.
A2 6 49 3 32 Take the determinant, A2 6(32) 49(3) 45 and use the formula x ¯2 A2 A 45 9 5 For a system of three linear equations, see Problem 11.35(b) to (e).
Solved Problems DETERMINANTS 11.1.
Find the determinant A for the following matrices: a) A 9 13 15 18 b) A 40 10 25 5 A 9(18) 13(15) 33 A 40(5) (10)(25) 50 c) A 7 6 9 5 2 12 The determinant does not exist because A is a 3 2 matrix and only a square matrix can have a determinant.
11.2.
Find the determinant A for the following matrices. Notice how the presence of zeros simplifies the task of evaluating a determinant.
a) A 3 6 5 2 1 8 7 9 1 A 3 1 8 9 1 6 2 8 7 1 5 2 1 7 9 3[1(1) 8(9)] 6[2(1) 8(7)] 5[2(9) 1(7)] 3(71) 6(54) 5(11) 166 231 MATRIX INVERSION CHAP . 11] b) A 12 0 3 9 2 5 4 6 1 A 12 2 5 6 1 0 9 5 4 1 3 9 2 4 6 12(2 30) 0 3(54 8) 198 c) A 0 6 0 3 5 2 7 6 9 A 0 5 2 6 9 6 3 2 7 9 0 3 5 7 6 0 6(27 14) 0 78 RANK OF A MATRIX 11.3.
Determine the rank of the following matrices: a) A 3 6 2 1 5 4 4 8 2 A 3 5 4 8 2 6(1) 1 4 4 2 2 1 5 4 8 3[10 (32)] 6(2 16) 2(8 20) 98 With A 0, A is nonsingular and the three rows and columns are linearly independent. Hence, (A) 3.
b) B 5 2 3 9 12 18 3 4 6 B 5 12 4 18 6 9(1) 2 4 3 6 3 2 12 3 18 5[72 (72)] 9[12 (12)] 3[36 (36)] 5(0) 9(0) 3(0) 0 With B 0, B is singular and the three rows and columns are not linearly independent. Hence, (B) 3. Now test to see if any two rows or columns are independent. Starting with the submatrix in the upper left corner, take the 2 2 determinant.
5 9 2 12 60 (18) 78 0 Thus, (B) 2. There are only two linearly independent rows and columns in B. Row 3 is 1.5 times row 2, and column 3 is 1 – 3 times column 2.
232 MATRIX INVERSION [CHAP . 11 c) C 8 10 24 2 2.5 6 6 7.5 18 C 8 2.5 7.5 6 18 2(1) 10 7.5 24 18 6 10 2.5 24 6 8[45 (45)] 2(180 180) 6[60 (60)] 0 With C 0, (C) 3. Trying various 2 2 submatrices, 8 10 10 24 2 2.5 2.5 6 20 20 0 60 (60) 0 2 2.5 2.5 6 6 7.5 7.5 18 15 15 0 45 (45) 0 With all the determinants of the different 2 2 submatrices equal to zero, no two rows or columns of C are linearly independent. So (C) 2 and (C) 1. Row 2 is 1.25 times row 1, row 3 is 3 times row 1, column 2 is 1 – 4 times column 1, and column 3 is 3 – 4 times column 1.
d) D 2 5 7 11 3 1 Since the maximum number of linearly independent rows (columns) must equal the maximum number of linearly independent columns (rows), the rank of D cannot exceed 2. Testing a submatrix, 2 5 7 11 22 35 13 0 (D) 2 While it is clear that there are only two linearly independent columns, there are also only two linearly independent rows because row 2 2 times row 1 plus row 3.
PROPERTIES OF DETERMINANTS 11.4.
Given A 2 5 1 3 2 4 1 4 2 Compare (a) the determinant of A and (b) the determinant of the transpose of A. (c) Specify which property of determinants the comparison illustrates.
a) A 2(4 16) 5(6 4) 1(12 2) 24 b) A 2 3 1 5 2 4 1 4 2 A 2(4 16) 3(10 4) 1(20 2) 24 c) This illustrates that the determinant of a matrix equals the determinant of its transpose. See Section 11.5.
11.5.
Compare (a) the determinant of A and (b) the determinant of A, given A a11 a12 a21 a22 233 MATRIX INVERSION CHAP . 11] a) A a11a22 a12a21 b) A a11 a21 a12 a22 A a11a22 a21a12 11.6.
Given A 1 4 2 3 5 4 2 3 2 (a) Find the determinant of A. (b) Form a new matrix B by interchanging row 1 and row 2 of A, and find B . (c) Form another matrix C by interchanging column 1 and column 3 of A, and find C . (d) Compare determinants and specify which property of determinants is illustrated.
a) A 1(10 12) 4(6 8) 2(9 10) 4 b) B 3 5 4 1 4 2 2 3 2 B 3(8 6) 5(2 4) 4(3 8) 4 c) C 2 4 1 4 5 3 2 3 2 C 2(10 9) 4(8 6) 1(12 10) 4 d) B A . Interchanging any two rows or columns will affect the sign of the determinant, but not the absolute value of the determinant.
11.7.
Given W w y x z (a) Find the determinant of W. (b) Interchange row 1 and row 2 of W, forming a new matrix Y, and compare the determinant of Y with that of W.
a) W wz yx b) Y y z w x Y yx wz (wz yx) W 11.8.
Given A 3 5 7 2 1 4 4 2 3 (a) Find the determinant of A. (b) Form a new matrix B by multiplying the first row of A by 2, and find the determinant of B. (c) Compare determinants and indicate which property of determinants this illustrates.
a) A 3(3 8) 5(6 16) 7(4 4) 35 b) B 6 10 14 2 1 4 4 2 3 B 6(3 8) 10(6 16) 14(4 4) 70 c) B 2 A . Multiplying a single row or column of a matrix by a scalar will cause the value of the determinant to be multiplied by the scalar. Here doubling row 1 doubles the determinant.
234 MATRIX INVERSION [CHAP . 11 11.9.
Given A 2 5 8 3 10 1 1 15 4 (a) Find A . (b) Form a new matrix B by multiplying column 2 by 1 – 5 and find B . (c) Compare determinants.
a) A 2(40 15) 5(12 1) 8(45 10) 275 b) Recalling that multiplying by 1 – 5 is the same thing as dividing by or factoring out 5, B 2 1 8 3 2 1 1 3 4 B 2(8 3) 1(12 1) 8(9 2) 55 c) B 1 – 5 A 11.10. Given A a11 a12 a21 a22 B a11 ka12 a21 ka22 Compare (a) the determinant of A and (b) the determinant of B.
a) A a11a22 a12a21 b) B a11ka22 ka12a21 k(a11a22) k(a12a21) k(a11a22 a12a21) k A 11.11. Given A 5 1 4 3 2 5 4 1 6 (a) Find A . (b) Subtract 5 times column 2 from column 1, forming a new matrix B, and find B . (c) Compare determinants and indicate which property of determinants is illustrated.
a) A 5(12 5) 1(18 20) 4(3 8) 17 b) B 0 1 4 7 2 5 1 1 6 B 0 1(42 5) 4(7 2) 17 c) B A . Addition or subtraction of a nonzero multiple of any row or column to or from another row or column does not change the value of the determinant.
11.12. (a) Subtract row 3 from row 1 in A of Problem 11.11, forming a new matrix C, and (b) find C .
a) C 1 0 2 3 2 5 4 1 6 b) C 1(12 5) 0 (2)(3 8) 17 11.13. Given the upper-triangular matrix A 3 0 0 2 5 0 6 1 4 235 MATRIX INVERSION CHAP . 11] which has zero elements everywhere above the principal diagonal, (a) find A . (b) Find the product of the elements along the principal diagonal and (c) specify which property of determinants this illustrates.
a) A 3(20 0) 0 0 60 b) Multiplying the elements along the principal diagonal, (3)(5)(4) 60.
c) The determinant of a triangular matrix is equal to the product of the elements along the principal diagonal.
11.14. Given the lower-triangular matrix A 2 5 1 0 3 6 0 0 7 which has zero elements everywhere below the principal diagonal, find (a) A and (b) the product of the diagonal elements.
a) A 2(21 0) (5)(0 0) 1(0 0) 42 b) 2(3)(7) 42 11.15. Given A 12 16 13 0 0 0 15 20 9 (a) Find A . (b) What property of determinants is illustrated?
a) A 12(0 0) 16(0 0) 13(0 0) 0 b) If all the elements of a row or column equal zero, the determinant will equal zero. With all the elements of row 2 in A equal to zero, the matrix is, in effect, a 2 3 matrix, not a 3 3 matrix. Only square matrices have determinants.
SINGULAR AND NONSINGULAR MATRICES 11.16. Using a 2 2 coefficient matrix A, prove that if A 0, there is linear independence between the rows and columns of A and a unique solution exists for the system of equations.
Start with two linear equations in two unknowns a11x a12y b1 (11.5) a21x a22y b2 (11.6) and solve for x by multiplying (11.5) by a22 and (11.6) by a12 and then adding to eliminate y.
a11a22x a12a22y a12a21x a12a22y (a11a22 a12a21)x x a22b1 a12b2 a22b1 a12b2 a22b1 a12b2 a11a22 a12a21 (11.7) where a11a22 a12a21 A . If, in (11.7), A a11a22 a12a21 0, x has no unique solution, indicating linear dependence between the equations; if A a11a22 a12a21 0, x has a unique solution and the equations must be linearly independent.
236 MATRIX INVERSION [CHAP . 11 11.17. Use determinants to determine whether a unique solution exists for each of the following systems of equations: a) 12x1 7x2 147 15x1 19x2 168 To determine whether a unique solution exists, find the coefficient matrix A and take the determinant A . If A 0, the matrix is nonsingular and a unique solution exists. If A 0, the matrix is singular and there is no unique solution. Thus, A 12 7 15 19 A 12(19) (7)15 123 Since A 0, A is nonsingular and a unique solution exists.
b) 2x1 3x2 27 6x1 9x2 81 A 2 3 6 9 A 2(9) 6(3) 0 There is no unique solution. The equations are linearly dependent. The second equation is 3 times the first equation.
c) 72x1 54x2 216 64x1 48x2 192 A 72 54 64 48 A 72(48) (54)(64) 3456 3456 0 A unique solution does not exist because the equations are linearly dependent. Closer inspection reveals the second equation is 8 – 9 times the first equation.
d) 4x1 3x2 5x3 27 x1 6x2 2x3 19 3x1 x2 3x3 15 A 4 3 5 1 6 2 3 1 3 A 4(18 2) 3(3 6) 5(1 18) 12 A unique solution exists.
e) 4x1 2x2 6x3 28 3x1 x2 2x3 20 10x1 5x2 15x3 70 A 4 2 6 3 1 2 10 5 15 A 4(15 10) 2(45 20) 6(15 10) 0 237 MATRIX INVERSION CHAP . 11] There is no unique solution because the equations are linearly dependent. Closer examination reveals the third equation is 2.5 times the first equation.
f) 56x1 47x2 8x3 365 84x1 39x2 12x3 249 28x1 81x2 4x3 168 A 56 47 8 84 39 12 28 81 4 Factoring out 28 from column 1 and 4 from column 3 before taking the determinant, A 28(4) 2 47 2 3 39 3 1 81 1 The linear dependence between column 1 and column 3 is now evident. The determinant will therefore be zero, and no unique solution exists. A 112[2(39 243) 47(0) 2(243 39)] 112(0) 0 MINORS AND COFACTORS 11.18. Find (a) the minor Mij and (b) the cofactor Cij for each of the elements in the first row, given A a11 a12 a21 a22 a) To find the minor of a11, mentally delete the row and column in which it appears. The remaining element is the minor. Thus, M11 a22. Similarly, M12 a21.
b) From the rule of cofactors, C11 (1)11 M11 1(a22) a22 C12 (1)12 M12 1(a21) a21 11.19. Find (a) the minors and (b) the cofactors for the elements of the second row, given A 13 17 19 15 a) M21 17 M22 13 b) C21 (1)21 M21 1(17) 17 C22 (1)22 M22 1(13) 13 11.20. Find (a) the minors and (b) the cofactors for the elements of the second column, given A 6 7 12 9 a) M12 12 M22 6 b) C12 (1)12 M12 12 C22 (1)22 M22 6 238 MATRIX INVERSION [CHAP . 11 11.21. Find (a) the minors and (b) the cofactors for the elements of the first row, given A 5 2 4 6 3 7 1 2 4 a) Deleting row 1 and column 1, M11 3 7 2 4 26 Similarly, M12 6 7 1 4 17 M13 6 3 1 2 15 b) C11 (1)2 M11 26 C12 (1)3 M12 17 C13 (1)4 M13 15 11.22. Find (a) the minors and (b) the cofactors for the elements of the third row, given A 9 11 4 3 2 7 6 10 4 a) Deleting row 3 and column 1, M31 11 4 2 7 69 Similarly, M32 9 4 3 7 51 M33 9 11 3 2 15 b) C31 (1)4 M31 69 C32 (1)5 M32 51 C33 (1)6 M33 15 11.23. Find (a) the minors and (b) the cofactors for the elements in the second column, given A 13 6 11 12 9 4 7 10 2 a) M12 12 4 7 2 4 M22 13 11 7 2 51 M32 13 11 12 4 80 b) C12 (1)3 M12 1(4) 4 C22 (1)4 M22 51 C32 (1)5 M32 1(80) 80 239 MATRIX INVERSION CHAP . 11] 11.24. Find (1) the cofactor matrix C and (2) the adjoint matrix Adj A for each of the following: a) A 7 12 4 3 1) C C11 C12 C21 C22 M11 M12 M21 M22 3 4 12 7 2) Adj A C 3 12 4 7 b) A 2 5 13 6 1) C 6 13 5 2 2) Adj A 6 5 13 2 c) A 9 16 20 7 1) C 7 20 16 9 2) Adj A 7 16 20 9 d) A 6 2 7 5 4 9 3 3 1 1) C C11 C12 C13 C21 C22 C23 C31 C32 C33 4 9 3 1 2 7 3 1 2 7 4 9 5 9 3 1 6 7 3 1 6 7 5 9 5 4 3 3 6 2 3 3 6 2 5 4 23 22 3 19 15 12 10 19 14 2) Adj A C 23 19 10 22 15 19 3 12 14 e) A 13 2 8 9 6 4 3 2 1 1) C 6 4 2 1 2 8 2 1 2 8 6 4 9 4 3 1 13 8 3 1 13 8 9 4 9 6 3 2 13 2 3 2 13 2 9 6 2 3 0 14 11 20 40 20 60 2) Adj A C 2 3 0 14 11 20 40 20 60 240 MATRIX INVERSION [CHAP . 11 LAPLACE EXPANSION 11.25. Use Laplace expansion to find the determinants for each of the following, using whatever row or column is easiest: a) A 15 7 9 2 5 6 9 0 12 Expanding along the second column, A a12 C12 a22 C22 a32 C32 7(1) 2 6 9 12 5 15 9 9 12 0 7(30) 5(99) 705 b) A 23 35 0 72 46 10 15 29 0 Expanding along the third column, A a13 C13 a23 C23 a33 C33 0 10(1) 23 35 15 29 0 10(142) 1420 c) A 12 98 15 0 25 0 21 84 19 Expanding along the second row, A a21 C21 a22 C22 a23 C23 0 25 12 15 21 19 0 25(87) 2175 d) A 2 4 1 5 3 2 5 1 1 2 1 4 3 4 3 2 Expanding along the first row, A a11 C11 a12 C12 a13 C13 a14 C14 2(1)11 2 5 1 2 1 4 4 3 2 4(1)12 3 5 1 1 1 4 3 3 2 1(1)13 3 2 1 1 2 4 3 4 2 5(1)14 3 2 5 1 2 1 3 4 3 Then expanding each of the 3 3 subdeterminants along the first row, A 22 1 4 3 2 5 2 4 4 2 1 2 1 4 3 43 1 4 3 2 5 1 4 3 2 1 1 1 3 3 13 2 4 4 2 2 1 4 3 2 1 1 2 3 4 53 2 1 4 3 2 1 1 3 3 5 1 2 3 4 2[2(10) 5(12) 1(2)] 4[3(10) 5(10) 1(0)] 1[3(12) 2(10) 1(2)] 5[3(2) 2(0) 5(2)] 2(42) 4(20) 1(18) 5(4) 6 241 MATRIX INVERSION CHAP . 11] e) A 5 0 1 3 4 2 6 0 3 0 1 5 0 1 4 2 Expanding along the second column, A a12 C12 a22 C22 a32 C32 a42 C42 0 2(1)22 5 1 3 3 1 5 0 4 2 0 1(1)42 5 1 3 4 6 0 3 1 5 Then substituting the values for the 3 3 subdeterminants, A 2(60) 1(88) 32 INVERTING A MATRIX 11.26. Find the inverse A1 for the following matrices. Check your answer to part (a).
a) A 24 15 8 7 A1 1 A Adj A Evaluating the determinant, A 24(7) 15(8) 48 Then finding the cofactor matrix to get the adjoint, C 7 8 15 24 and Adj A C 7 15 8 24 Thus, A1 1 48 7 15 8 24 7 –– 48 1 – 6 5 –– 16 1 – 2 (11.8) Checking to make sure A1A I, and using the unreduced form of A1 from (11.8) for easier computation, A1A 1 48 7 15 8 24 24 15 8 7 1 48 7(24) 15(8) 7(15) 15(7) 8(24) 24(8) 8(15) 24(7) 1 48 48 0 0 48 1 0 0 1 b) A 7 9 6 12 A 7(12) 9(6) 30 The cofactor matrix is C 12 6 9 7 and Adj A C 12 9 6 7 242 MATRIX INVERSION [CHAP . 11 Thus, A1 1 30 12 9 6 7 2 – 5 1 – 5 3 –– 10 7 –– 30 c) A 7 16 9 13 A 7(13) 16(9) 53 C 13 9 16 7 Adj A C 13 16 9 7 A1 1 53 13 16 9 7 13 –– 53 9 –– 53 16 –– 53 7 –– 53 d) A 4 2 5 3 1 8 9 6 7 A 4(7 48) 2(21 72) 5(18 9) 17 The cofactor matrix is C 1 8 6 7 2 5 6 7 2 5 1 8 3 8 9 7 4 5 9 7 4 5 3 8 3 1 9 6 4 2 9 6 4 2 3 1 41 51 9 16 17 6 11 17 2 and Adj A C 41 16 11 51 17 17 9 6 2 Thus, A1 1 17 41 16 11 51 17 17 9 6 2 41 –– 17 3 9 –– 17 16 –– 17 1 6 –– 17 11 –– 17 1 2 –– 17 e) A 14 0 6 9 5 0 0 11 8 A 14(40) 0 6(99) 1154 The cofactor matrix is C 40 72 99 66 112 154 30 54 70 The adjoint is Adj A 40 66 30 72 112 54 99 154 70 243 MATRIX INVERSION CHAP . 11] Then A1 1 1154 40 66 30 72 112 54 99 154 70 20 ––– 577 36 ––– 577 99 –––– 1154 33 ––– 577 56 ––– 577 77 ––– 577 15 ––– 577 27 ––– 577 35 ––– 577 MATRIX INVERSION IN EQUATION SOLUTIONS 11.27. Use matrix inversion to solve the following systems of linear equations. Check your answers on your own by substituting into the original equations.
a) 4x1 3x2 28 2x1 5x2 42 4 3 2 5 x1 x2 28 42 where from Section 11.8, X A1B. Find first the inverse of A, where A 4(5) 3(2) 14. The cofactor matrix of A is C 5 2 3 4 and Adj A C 5 3 2 4 Thus, A1 1 14 5 3 2 4 5 –– 14 1 – 7 3 –– 14 2 – 7 Then substituting in X A1B and simply multiplying matrices, X 5 –– 14 1 – 7 3 –– 14 2 – 7 22 28 42 21 10 9 4 12 21 1 8 21 Thus, x ¯ 1 and x ¯2 8.
b) 6x1 7x2 56 2x1 3x2 44 6 7 2 3 x1 x2 56 44 where A 6(3) 7(2) 4.
C 3 2 7 6 Adj A C 3 7 2 6 and A1 1 4 3 7 2 6 3 – 4 1 – 2 7 – 4 3 – 2 Thus, X 3 – 4 1 – 2 7 – 4 3 – 2 22 56 44 21 42 77 28 66 21 35 38 21 and x ¯1 35 and x ¯2 38.
11.28. The equilibrium conditions for two related markets (pork and beef) are given by 18Pb Pp 87 2Pb 36Pp 98 244 MATRIX INVERSION [CHAP . 11 Find the equilibrium price for each market.
18 1 2 36 Pb Pp 87 98 where A 18(36) (1)(2) 646.
C 36 2 1 18 Adj A 36 1 2 18 and A1 1 646 36 1 2 18 18 ––– 323 1 ––– 323 1 ––– 646 9 ––– 323 Thus, X 18 ––– 323 1 ––– 323 1 ––– 646 9 ––– 323 87 98 1615 –––– 323 969 ––– 323 5 3 and P ¯ b 5 and P ¯ p 3.
This is the same solution as that obtained by simultaneous equations in Problem 2.12. For practice try the inverse matrix solution for Problem 2.13.
11.29. The equilibrium condition for two substitute goods is given by 5P1 2P2 15 P1 8P2 16 Find the equilibrium prices.
5 2 1 8 P1 P2 15 16 where A 5(8) (1)(2) 38.
C 8 1 2 5 Adj A 8 2 1 5 and A1 1 38 8 2 1 5 4 –– 19 1 –– 38 1 –– 19 5 –– 38 Thus, X 4 –– 19 1 –– 38 1 –– 19 5 –– 38 15 16 60 16 19 15 80 38 4 2.5 and P ¯ 1 4 and P ¯ 2 2.5.
11.30. Given: the IS equation 0.3Y 100i 252 0 and the LM equation 0.25Y 200i 176 0.
Find the equilibrium level of income and rate of interest.
The IS and LM equations can be reduced to the form 0.3Y 100i 252 0.25Y 200i 176 and then expressed in matrix form where A 0.3 100 0.25 200 X Y i B 252 176 245 MATRIX INVERSION CHAP . 11] Thus, A 0.3(200) 100(0.25) 85 C 200 0.25 100 0.3 Adj A 200 100 0.25 0.3 and A1 1 85 200 100 0.25 0.3 40 –– 17 0.05 17 20 –– 17 0.06 17 Thus, X 40 –– 17 0.05 17 20 –– 17 0.06 17 252 176 10,080 3520 17 12.6 10.56 17 800 0.12 In equilibrium Y ¯ 800 and i ¯ 0.12 as found in Problem 2.23 where simultaneous equations were used.
On your own, practice with Problem 2.24.
11.31. Use matrix inversion to solve for the unknowns in the system of linear equations given below.
2x1 4x2 3x3 12 3x1 5x2 2x3 13 x1 3x2 2x3 17 2 4 3 3 5 2 1 3 2 x1 x2 x3 12 13 17 where A 2(16) 4(8) 3(4) 76.
C 5 2 3 2 4 3 3 2 4 3 5 2 3 2 1 2 2 3 1 2 2 3 3 2 3 5 1 3 2 4 1 3 2 4 3 5 16 8 4 17 1 10 7 13 22 Adj A 16 17 7 8 1 13 4 10 22 A1 1 76 16 8 4 17 1 10 7 13 22 16 –– 76 8 –– 76 4 –– 76 17 –– 76 1 –– 76 10 –– 76 7 –– 76 13 –– 76 22 –– 76 where the common denominator 76 is deliberately kept to simplify later calculations.
Thus, X 16 –– 76 8 –– 76 4 –– 76 17 –– 76 1 –– 76 10 –– 76 7 –– 76 13 –– 76 22 –– 76 12 13 17 192 221 119 76 96 13 221 76 48 130 374 76 7 4 6 x ¯1 x ¯2 x ¯3 246 MATRIX INVERSION [CHAP . 11 11.32. The equilibrium condition for three related markets is given by 11P1 P2 P3 31 P1 6P2 2P3 26 P1 2P2 7P3 24 Find the equilibrium price for each market.
11 1 1 1 6 2 1 2 7 P1 P2 P3 31 26 24 where A 11(38) 1(9) 1(8) 401.
C 6 2 2 7 1 1 2 7 1 1 6 2 1 2 1 7 11 1 1 7 11 1 1 2 1 6 1 2 11 1 1 2 11 1 1 6 38 9 8 9 76 23 8 23 65 Adj A 38 9 8 9 76 23 8 23 65 A1 1 401 38 9 8 9 76 23 8 23 65 38 ––– 401 9 ––– 401 8 ––– 401 9 ––– 401 76 ––– 401 23 ––– 401 8 ––– 401 23 ––– 401 65 ––– 401 X 38 ––– 401 9 ––– 401 8 ––– 401 9 ––– 401 76 ––– 401 23 ––– 401 8 ––– 401 23 ––– 401 65 ––– 401 31 26 24 1178 234 192 401 279 1976 552 401 248 598 1560 401 4 7 6 P ¯ 1 P ¯ 2 P ¯ 3 See Problem 2.16 for the same solution with simultaneous equations.
11.33. Given Y C I0, where C C0 bY. Use matrix inversion to find the equilibrium level of Y and C.
The given equations can first be rearranged so that the endogenous variables C and Y, together with their coefficients b, are on the left-hand side of the equation and the exogenous variables C0 and I0 are on the right.
Y C I0 bY C C0 Thus, 1 1 b 1 Y C I0 C0 The determinant of the coefficient matrix is A 1(1) 1(b) 1 b. The cofactor matrix is C 1 b 1 1 Adj A 1 1 b 1 247 MATRIX INVERSION CHAP . 11] and A1 1 1 b 1 1 b 1 Letting X Y C, X 1 1 b 1 1 b 1 I0 C0 1 1 b I0 C0 bI0 C0 Thus, Y ¯ 1 1 b (I0 C0) C ¯ 1 1 b (C0 bI0) Example 3 in Chapter 2 was solved for the equilibrium level of income without matrices.
CRAMER’S RULE 11.34. Use Cramer’s rule to solve for the unknowns in each of the following: a) 2x1 6x2 22 x1 5x2 53 From Cramer’s rule, x ¯i Ai A where Ai is a special matrix formed by replacing the column of coefficients of xi with the column of constants. Thus, from the original data, 2 6 1 5 x1 x2 22 53 where A 2(5) 6(1) 16.
Replacing the first column of the coefficient matrix with the column of constants, A1 22 6 53 5 where A1 22(5) 6(53) 208. Thus, x ¯1 A1 A 208 16 13 Replacing the second column of the original coefficient matrix with the column of constants, A2 2 22 1 53 where A2 2(53) 22(1) 128. Thus, x ¯2 A2 A 128 16 8 b) 7p1 2p2 60 p1 8p2 78 A 7 2 1 8 where A 7(8) 2(1) 54.
A1 60 2 78 8 248 MATRIX INVERSION [CHAP . 11 where A1 60(8) 2(78) 324.
A2 7 60 1 78 where A2 7(78) 60(1) 486.
p ¯1 A1 A 324 54 6 and p ¯2 A2 A 486 54 9 c) 18Pb Pp 87 2Pb 36Pp 98 A 18 1 2 36 where A 18(36) (1)(2) 646.
A1 87 1 98 36 where A1 87(36) 1(98) 3230.
A2 18 87 2 98 where A2 18(98) 87(2) 1938.
P ¯ b A1 A 3230 646 5 and P ¯ p A2 A 1938 646 3 Compare the work involved in this method of solution with the work involved in Problems 2.12 and 11.28 where the same problem is treated first with simultaneous equations and then with matrix inversion.
11.35. Redo Problem 11.34 for each of the following: a) 0.4Y 150i 209 0.1Y 250i 35 A 0.4 150 0.1 250 where A 0.4(250) 150(0.1) 115.
A1 209 150 35 250 where A1 209(250) 150(35) 57,500.
A2 0.4 209 0.1 35 where A2 0.4(35) 209(0.1) 6.9.
Y ¯ A1 A 57,500 115 500 and i ¯ A2 A 6.9 115 0.06 Compare this method of solution with Problem 2.24.
249 MATRIX INVERSION CHAP . 11] b) 5x1 2x2 3x3 16 2x1 3x2 5x3 2 4x1 5x2 6x3 7 A 5 2 4 2 3 5 3 5 6 5(18 25) 2(12 20) 3(10 12) 37 A1 16 2 7 2 3 5 3 5 6 16(18 25) 2(12 35) 3(10 21) 111 A2 5 2 4 16 2 7 3 5 6 5(12 35) 16(12 20) 3(14 8) 259 A3 5 2 4 2 3 5 16 2 7 5(21 10) 2(14 8) 16(10 12) 185 x ¯1 A1 A 111 37 3 x ¯2 A2 A 259 37 7 x ¯3 A3 A 185 37 5 c) 2x1 4x2 x3 52 x1 5x2 3x3 72 3x1 7x2 2x3 10 A 2 1 3 4 5 7 1 3 2 2(31) 4(11) 1(8) 114 A1 52 72 10 4 5 7 1 3 2 52(31) 4(114) 1(554) 1710 A2 2 1 3 52 72 10 1 3 2 2(114) 52(11) 1(226) 1026 A3 2 1 3 4 5 7 52 72 10 2(554) 4(226) 52(8) 1596 x ¯1 A1 A 1710 114 15 x ¯2 A2 A 1026 114 9 x ¯3 A3 A 1596 114 14 d) 11p1 p2 p3 31 p1 6p2 2p3 26 p1 2p2 7p3 24 A 11 1 1 1 6 2 1 2 7 11(38) 1(9) 1(8) 401 A1 31 26 24 1 6 2 1 2 7 31(38) 1(230) 1(196) 1604 250 MATRIX INVERSION [CHAP . 11 A2 11 1 1 31 26 24 1 2 7 11(230)31(9) 1(2) 2807 A3 11 1 1 1 6 2 31 26 24 11(196) 1(2) 31(8) 2406 Thus, p ¯1 A1 A 1604 401 4 p ¯2 A2 A 2807 401 7 p ¯3 A3 A 2406 401 6 Compare the work involved in this type of solution with the work involved in Problems 2.16 and 11.32.
11.36. Use Cramer’s rule to solve for x and y, given the first-order conditions for constrained optimization from Example 7 of Chapter 6: TC x 16x y 0 TC y 24y x 0 TC 42 x y 0 Rearrange the equations, 16x y 0 x 24y 0 x y 42 and set them in matrix form.
16 1 1 1 24 1 1 1 0 x y 0 0 42 Expanding along the third column, A (1)(1 24) (1)(16 1) 0 42 A1 0 0 42 1 24 1 1 1 0 Expanding along the first column, A 1 42(1 24) 1050.
A2 16 1 1 0 0 42 1 1 0 Expanding along the second column, A2 (42)(16 1) 714.
A3 16 1 1 1 24 1 0 0 42 Expanding along the third column, A3 42(384 1) 16,086.
251 MATRIX INVERSION CHAP . 11] Thus, x ¯ A1 A 1050 42 25 y ¯ A2 A 714 42 17 and ¯ A3 A 16,086 42 383 11.37. Use Cramer’s rule to find the critical values of Q1 and Q2, given the first-order condi-tions for constrained utility maximization in Problem 6.37: Q2 10 0, Q1 2 0, and 240 10Q1 2Q2 0.
0 1 10 1 0 2 10 2 0 Q1 Q2 0 0 240 Thus, A (1)(20) (10)(2) 40.
A1 0 0 240 1 0 2 10 2 0 240(2) 480 A2 0 1 10 0 0 240 10 2 0 (240)(10) 2400 A3 0 1 10 1 0 2 0 0 240 240(1) 240 Thus, Q ¯ 1 A1 A 480 40 12 Q ¯ 2 A2 A 2400 40 60 and ¯ A3 A 240 40 6 11.38. Given ax1 bx2 g (11.9) cx1 dx2 h (11.10) Prove Cramer’s rule by showing x ¯1 g b h d a b c d A1 A x ¯2 a g c h a b c d A2 A Dividing (11.9) by b, a b x1 x2 g b (11.11) 252 MATRIX INVERSION [CHAP . 11 Multiplying (11.11) by d and subtracting (11.10), ad b x1 dx2 dg b cx1 dx2 h ad cb b x1 dg hb b x ¯1 dg hb ad cb g b h d a b c d A1 A Similarly, dividing (11.9) by a, x1 b a x2 g a (11.12) Multiplying (11.12) by c and adding to (11.10), cx1 bc a x2 cg a cx1 dx2 h ad bc a x2 ah cg a x ¯2 ah cg ad bc a g c h a b c d A2 A Q.E.D.
253 MATRIX INVERSION CHAP . 11] CHAPTER 12 Special Determinants and Matrices and Their Use in Economics 12.1 THE JACOBIAN Section 11.1 showed how to test for linear dependence through the use of a simple determinant.
In contrast, a Jacobian determinant permits testing for functional dependence, both linear and nonlinear. A Jacobian determinant J is composed of all the first-order partial derivatives of a system of equations, arranged in ordered sequence. Given y1 f1(x1, x2, x3) y2 f2(x1, x2, x3) y3 f3(x1, x2, x3) J y1, y2, y3 x1, x2, x3 y1 x1 y2 x1 y3 x1 y1 x2 y2 x2 y3 x2 y1 x3 y2 x3 y3 x3 Notice that the elements of each row are the partial derivatives of one function yi with respect to each of the independent variables x1, x2, x3, and the elements of each column are the partial derivatives 254 of each of the functions y1, y2, y3 with respect to one of the independent variables xj. If J 0, the equations are functionally dependent; if J 0, the equations are functionally independent. See Example 1 and Problems 12.1 to 12.4.
EXAMPLE 1.
Use of the Jacobian to test for functional dependence is demonstrated below, given y1 5x1 3x2 y2 25x1 2 30x1x2 9x2 2 First, take the first-order partials, y1 x1 5 y1 x2 3 y2 x1 50x1 30x2 y2 x2 30x1 18x2 Then set up the Jacobian, J 5 50x1 30x2 3 30x1 18x2 and evaluate, J 5(30x1 18x2) 3(50x1 30x2) 0 Since J 0, there is functional dependence between the equations. In this, the simplest of cases, (5x1 3x2)2 25x2 1 30x1x2 9x2 2.
12.2 THE HESSIAN Given that the first-order conditions zx zy 0 are met, a sufficient condition for a multivariable function z f(x, y) to be at an optimum is 1) zxx, zyy 0 for a minimum zxx, zyy 0 for a maximum zxxzyy (zxy)2 2) See Section 5.4. A convenient test for this second-order condition is the Hessian. A Hessian H is a determinant composed of all the second-order partial derivatives, with the second-order direct partials on the principal diagonal and the second-order cross partials off the principal diagonal. Thus, H zxx zxy zyx zyy where zxy zyx. If the first element on the principal diagonal, the first principal minor, H1 zxx is positive and the second principal minor H2 zxx zxy zxy zyy zxxzyy (zxy)2 0 the second-order conditions for a minimum are met. When H1 0 and H2 0, the Hessian H is called positive definite. A positive definite Hessian fulfills the second-order conditions for a minimum.
If the first principal minor H1 zxx 0 and the second principal minor H2 zxx zxy zxy zyy 0 the second-order conditions for a maximum are met. When H1 0, H2 0, the Hessian H is negative definite. A negative definite Hessian fulfills the second-order conditions for a maximum. See Example 2 and Problems 12.10 to 12.13.
255 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] EXAMPLE 2.
In Problem 5.10(a) it was found that z 3x2 xy 2y2 4x 7y 12 is optimized at x0 1 and y0 2. The second partials were zxx 6, zyy 4, and zxy 1. Using the Hessian to test the second-order conditions, H zxx zxy zyx zyy 6 1 1 4 Taking the principal minors, H1 6 0 and H2 6 1 1 4 6(4) (1)(1) 23 0 With H1 0 and H2 0, the Hessian H is positive definite, and z is minimized at the critical values.
12.3 THE DISCRIMINANT Determinants may be used to test for positive or negative definiteness of any quadratic form. The determinant of a quadratic form is called a discriminant D . Given the quadratic form z ax2 bxy cy2 the discriminant is formed by placing the coefficients of the squared terms on the principal diagonal and dividing the coefficients of the nonsquared term equally between the off-diagonal positions.
Thus, D a b 2 b 2 c Then evaluate the principal minors as in the Hessian test, where D1 a and D2 a b 2 b 2 c ac b2 4 If D1 , D2 0, D is positive definite and z is positive for all nonzero values of x and y. If D1 0 and D2 0, z is negative definite and z is negative for all nonzero values of x and y. If D2 0, z is not sign definite and z may assume both positive and negative values. See Example 3 and Problems 12.5 to 12.7.
EXAMPLE 3.
To test for sign definiteness, given the quadratic form z 2x2 5xy 8y2 form the discriminant as explained in Section 12.3. D 2 2.5 2.5 8 Then evaluate the principal minors as in the Hessian test. D1 2 0 D2 2 2.5 2.5 8 16 6.25 9.75 0 Thus, z is positive definite, meaning that it will be greater than zero for all nonzero values of x and y.
256 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 12.4 HIGHER-ORDER HESSIANS Given y f(x1, x2, x3), the third-order Hessian is H y11 y12 y13 y21 y22 y23 y31 y32 y33 where the elements are the various second-order partial derivatives of y: y11 2y x1 2 y12 2y x2x1 y23 2y x3x2 etc.
Conditions for a relative minimum or maximum depend on the signs of the first, second, and third principal minors, respectively. If H1 y11 0, H2 y11 y12 y21 y22 0 and H3 H 0 where H3 is the third principal minor, H is positive definite and fulfills the second-order conditions for a minimum. If H1 y11 0, H2 y11 y12 y21 y22 0 and H3 H 0 H is negative definite and will fulfill the second-order conditions for a maximum. Higher-order Hessians follow in analogous fashion. If all the principal minors of H are positive, H is positive definite and the second-order conditions for a relative minimum are met. If all the principal minors of H alternate in sign between negative and positive, H is negative definite and the second-order conditions for a relative maximum are met. See Example 4 and Problems 12.8, 12.9, and 12.14 to 12.18.
EXAMPLE 4.
The function y 5x1 2 10x1 x1x3 2x2 2 4x2 2x2x3 4x2 3 is optimized as follows, using the Hessian to test the second-order conditions.
The first-order conditions are y x1 y1 10x1 10 x3 0 y x2 y2 4x2 2x3 4 0 y x3 y3 x1 2x2 8x3 0 which can be expressed in matrix form as 10 0 1 0 4 2 1 2 8 x1 x2 x3 10 4 0 (12.1) Using Cramer’s rule (see Section 11.9) and taking the different determinants, A 10(28) 1(4) 276 0.
Since A in this case is the Jacobian and does not equal zero, the three equations are functionally independent. A1 10(28) 1(8) 288 A2 10(32) (10)(2) 1(4) 336 A3 10(8) 10(4) 120 257 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] Thus, x ¯1 A1 A 288 276 1.04 x ¯2 A2 A 336 276 1.22 x ¯3 A3 A 120 276 0.43 taking the second partial derivatives from the first-order conditions to prepare the Hessian, y11 10 y21 0 y31 1 y12 0 y22 4 y32 2 y13 1 y23 2 y33 8 Thus, H 10 0 1 0 4 2 1 2 8 which has the same elements as the coefficient matrix in (12.1) since the first-order partials are all linear. Finally, applying the Hessian test, by checking the signs of the first, second, and third principal minors, respectively, H1 10 0 H2 10 0 0 4 40 0 H3 H A 276 0 Since the principal minors alternate correctly in sign, the Hessian is negative definite and the function is maximized at x ¯1 1.04, x ¯2 1.22, and x ¯3 0.43.
12.5 THE BORDERED HESSIAN FOR CONSTRAINED OPTIMIZATION To optimize a function f(x, y) subject to a constraint g(x, y), Section 5.5 showed that a new function could be formed F(x, y, ) f(x, y) [k g(x, y)], where the first-order conditions are Fx Fy F 0.
The second-order conditions can now be expressed in terms of a bordered Hessian H ¯ in either of two ways: H ¯ Fxx Fyx gx Fxy Fyy gy gx gy 0 or 0 gx gy gx Fxx Fyx gy Fxy Fyy which is simply the plain Hessian Fxx Fxy Fyx Fyy bordered by the first derivatives of the constraint with zero on the principal diagonal. The order of a bordered principal minor is determined by the order of the principal minor being bordered. Hence H ¯ above represents a second bordered principal minor H ¯ 2 , because the principal minor being bordered is 2 2.
For a function in n variables f(x1, x2, ..., xn), subject to g(x1, x2,..., xn), H ¯ F11 F21 – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – Fn1 g1 F12 F22 Fn2 g2 · · · · · · · · · · · · F1n F2n Fnn gn g1 g2 gn 0 or 0 g1 g2 – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – gn g1 F11 F21 Fn1 g2 F12 F22 Fn2 · · · · · · · · · · · · gn F1n F2n Fnn where H ¯ H ¯ n , because of the n n principal minor being bordered.
If all the principal minors are negative, i.e., if H ¯ 2 , H ¯ 3 , . . ., H ¯ n 0, the bordered Hessian is positive definite, and a positive definite Hessian always satisfies the sufficient condition for a relative minimum.
If the principal minors alternate consistently in sign from positive to negative, i.e., if H ¯ 2 0, 258 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 H ¯ 3 0, H ¯ 4 0, etc., the bordered Hessian is negative definite, and a negative definite Hessian always meets the sufficient condition for a relative maximum. Further tests beyond the scope of the present book are needed if the criteria are not met, since the given criteria represent sufficient conditions, and not necessary conditions. See Examples 5 and 6 and Problems 12.19 to 12.27. For a 4 4 bordered Hessian, see Problem 12.28.
EXAMPLE 5.
Refer to Example 9 in Chapter 5. The bordered Hessian can be used to check the second-order conditions of the optimized function and to determine if Z is maximized or minimized, as demonstrated below.
From Equations (5.8) and (5.9), Zxx 8, Zyy 12, Zxy Zyx 3. From the constraint, x y 56, gx 1, and gy 1. Thus, H ¯ 8 3 1 3 12 1 1 1 0 Starting with the second principal minor H ¯ 2 , H ¯ 2 H ¯ 8(1) 3(1) 1(3 12) 14 With H ¯ 2 0, H ¯ is positive definite, which means that Z is at a minimum. See Problems 12.19 to 12.22.
EXAMPLE 6.
The bordered Hessian is applied below to test the second-order condition of the generalized Cobb-Douglas production function maximized in Example 10 of Chapter 6.
From Equations (6.15) and (6.16), QKK 0.24K1.6L0.5, QLL 0.25K0.4L1.5, QKL QLK 0.2K0.6L0.5; and from the constraint, 3K 4L 108, gK 3, gL 4, H ¯ 0.24K1.6L0.5 0.2K0.6L0.5 3 0.2K0.6L0.5 0.25K0.4L1.5 4 3 4 0 Starting with H ¯ 2 and expanding along the third row, H ¯ 2 3(0.8K0.6L0.5 0.75K0.4L1.5) 4(0.96K1.6L0.5 0.6K 0.6L0.5) 2.25K0.4L1.5 4.8K0.6L0.5 3.84K1.6L0.5 2.25K0.4 L1.5 4.8 K0.6L0.5 3.84L0.5 K1.6 0 With H ¯ 2 0, H ¯ is negative definite and Q is maximized. See Problems 12.23 to 12.28.
12.6 INPUT-OUTPUT ANALYSIS In a modern economy where the production of one good requires the input of many other goods as intermediate goods in the production process (steel requires coal, iron ore, electricity, etc.), total demand x for product i will be the summation of all intermediate demand for the product plus the final demand b for the product arising from consumers, investors, the government, and exporters, as ultimate users. If aij is a technical coefficient expressing the value of input i required to produce one dollar’s worth of product j, the total demand for product i can be expressed as xi ai1x1 ai2x2 · · · ainxn bi for i 1, 2, ..., n. In matrix form this can be expressed as X AX B (12.2) where X x1 x2 · · · xn A a11 a21 – – – – – – – – – – – – – – – – – an1 a12 a22 an2 · · · · · · · · · a1n a2n ann B b1 b2 · · · bn 259 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] and A is called the matrix of technical coefficients. To find the level of total output (intermediate and final) needed to satisfy final demand, we can solve for X in terms of the matrix of technical coefficients and the column vector of final demand, both of which are given. From (12.2), X AX (I A)X X B B (I A)1B (12.3) Thus, for a three-sector economy x1 x2 x3 1 a11 a12 a13 a21 1 a22 a23 a31 a32 1 a33 1 b1 b2 b3 where the I A matrix is called the Leontief matrix. In a complete input-output table, labor and capital would also be included as inputs, constituting value added by the firm. The vertical summation of elements along column j in such a model would equal 1: the input cost of producing one unit or one dollar’s worth of the commodity, as seen in Problem 12.39. See Example 7 and Problems 12.29 to 12.39.
EXAMPLE 7.
Determine the total demand x for industries 1, 2, and 3, given the matrix of technical coefficients A and the final demand vector B.
A 0.3 0.4 0.1 0.5 0.2 0.6 0.1 0.3 0.1 B 20 10 30 From (12.3), X (I A)1B, where I A 1 0 0 0 1 0 0 0 1 0.3 0.4 0.1 0.5 0.2 0.6 0.1 0.3 0.1 0.7 0.4 0.1 0.5 0.8 0.6 0.1 0.3 0.9 Taking the inverse, (I A)1 1 0.151 0.54 0.39 0.32 0.51 0.62 0.47 0.23 0.25 0.36 and substituting in (12.3), X 1 0.151 0.54 0.39 0.32 0.51 0.62 0.47 0.23 0.25 0.36 20 10 30 1 0.151 24.3 30.5 17.9 160.93 201.99 118.54 x1 x2 x3 12.7 CHARACTERISTIC ROOTS AND VECTORS (EIGENVALUES, EIGENVECTORS) To this point, the sign definiteness of a Hessian and a quadratic form has been tested by using the principal minors. Sign definiteness can also be tested by using the characteristic roots of a matrix.
Given a square matrix A, if it is possible to find a vector V 0 and a scalar c such that A V cV (12.4) the scalar c is called the characteristic root, latent root, or eigenvalue; and the vector is called the characteristic vector, latent vector, or eigenvector. Equation (12.4) can also be expressed A V cIV 260 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 which can be rearranged so that A V cIV 0 (A cI)V 0 (12.5) where A cI is called the characteristic matrix of A. Since by assumption V 0, the characteristic matrix A cI must be singular (see Problem 10.49) and thus its determinant must vanish. If A 3 3 matrix, then A cI a11 c a21 a31 a12 a22 c a32 a13 a23 a33 c 0 With A cI 0 in (12.5), there will be an infinite number of solutions for V. To force a unique solution, the solution may be normalized by requiring of the elements vi of V that vi 2 1, as shown in Example 9.
If 1) All characteristic roots (c) are positive, A is positive definite.
2) All c’s are negative, A is negative definite.
3) All c’s are nonnegative and at least one c 0, A is positive semidefinite.
4) All c’s are nonpositive and at least one c 0, A is negative semidefinite.
5) Some c’s are positive and others negative, A is sign indefinite.
See Examples 8 and 9 and Problems 12.40 to 12.45, and Section 9.3.
EXAMPLE 8.
Given A 6 3 3 6 To find the characteristic roots of A, the determinant of the characteristic matrix A cI must equal zero. Thus, A cI 6 c 3 3 6 c 0 (12.6) (6 c)(6 c) (3)(3) 0 c2 12c 27 0 (c 9)(c 3) 0 c1 9 c2 3 Testing for sign definiteness, since both characteristic roots are negative, A is negative definite. Note (1) that c1 c2 must equal the sum of the elements on the principal diagonal of A and (2) c1c2 must equal the determinant of the original matrix A .
EXAMPLE 9.
Continuing with Example 8, the first root c1 9 is now used to find the characteristic vector.
Substituting c 9 in (12.6), 6 (9) 3 3 6 (9) v1 v2 0 3 3 3 3 v1 v2 0 (12.7) Since the coefficient matrix is linearly dependent, (12.7) is capable of an infinite number of solutions. The product of the matrices gives two equations which are identical.
3v1 3v2 0 Solving for v2 in terms of v1, v2 v1 (12.8) 261 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] Then, normalizing the solution in (12.8) so that v1 2 v2 2 1 (12.9) v2 v1 is substituted in (12.9), getting v1 2 (v1)2 1 Thus, 2v1 2 1, v1 2 1 – 2. Then taking the positive square root, v1 1 – 2 0.5. From (12.8), v2 v1. Thus, v2 0.5, and the first characteristic vector is V1 0.5 0.5 When the second characteristic root c2 3 is used, 6 (3) 3 3 6 (3) v1 v2 3 3 3 3 v1 v2 0 Multiplying the 2 2 matrix by the column vector, 3v1 3v2 0 3v1 3v2 0 Thus, v1 v2. Normalizing, v1 2 v2 2 1 (v2)2 v2 2 1 2v2 2 1 v2 0.5 v1 0.5 Thus, V2 0.5 0.5 Solved Problems THE JACOBIAN 12.1.
Use the Jacobian to test for functional dependence in the following system of equations: y1 6x1 4x2 y2 7x1 9x2 Taking the first-order partials to set up the Jacobian J , y1 x1 6 y1 x2 4 y2 x1 7 y2 x2 9 Thus, J 6 4 7 9 6(9) 7(4) 26 Since J 0, there is no functional dependence. Notice that in a system of linear equations the Jacobian J equals the determinant A of the coefficient matrix, and all their elements are identical. See Section 11.1, where the determinant test for nonsingularity of a matrix is nothing more than an application of the Jacobian to a system of linear equations.
262 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 12.2.
Redo Problem 12.1, given y1 3x1 4x2 y2 9x1 2 24x1x2 16x2 2 The first-order partials are y1 x1 3 y1 x2 4 y2 x1 18x1 24x2 y2 x2 24x1 32x2 Thus, J 3 18x1 24x2 4 24x1 32x2 3(24x1 32x2) 4(18x1 24x2) 0 There is functional dependence: (3x1 4x2)2 9x1 2 24x1x2 16x2 2.
12.3.
Redo Problem 12.1, given y1 x1 2 3x2 5 y2 x1 4 6x1 2x2 9x2 2 y1 x1 2x1 y1 x2 3 y2 x1 4x1 3 12x1x2 y2 x2 6x1 2 18x2 J 2x1 4x1 3 12x1x2 3 6x1 2 18x2 2x1(6x1 2 18x2) 3(4x1 3 12x1x2) 0 There is functional dependence: y2 (y1 5)2, where y1 5 (x1 2 3x2)2 x1 2 3x2 5 5 x1 2 3x2 x1 4 6x2 1x2 9x2 2 and 12.4.
Test for functional dependence in each of the following by means of the Jacobian: a) y1 4x1 x2 y2 16x1 2 8x1x2 x2 2 J 4 32x1 8x2 1 8x1 2x2 4(8x1 2x2) 1(32x1 8x2) 64x1 16x2 0 The equations are functionally independent.
b) y1 1.5x1 2 12x1x2 24x2 2 y2 2x1 8x2 J 3x1 12x2 2 12x1 48x2 8 8(3x1 12x2) 2(12x1 48x2) 0 There is functional dependence between the equations.
c) y1 4x1 2 3x2 9 y2 16x1 4 24x1 2x2 9x2 2 12 J 8x1 64x1 3 48x1x2 3 24x1 2 18x2 8x1(24x1 2 18x2) 3(64x1 3 48x1x2) 0 The equations are functionally dependent.
263 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] DISCRIMINANTS AND SIGN DEFINITENESS OF QUADRATIC FUNCTIONS 12.5.
Use discriminants to determine whether each of the following quadratic functions is positive or negative definite: a) y 3x1 2 4x1x2 4x2 2 Since the coefficients of the squared terms are placed on the principal diagonal and the coefficient of the nonsquared term x1x2 is divided evenly between the a12 and a21 positions, in this case, D 3 2 2 4 where D1 3 0 D2 3 2 2 4 (3)(4) (2)(2) 8 0 With D1 0 and D2 0, y is negative definite and y will be negative for all nonzero values of x1 and x2.
b) y 5x1 2 2x1x2 7x2 2 The discriminant is D 5 1 1 7 where D1 5 0 and D2 D 5(7) (1)(1) 34 0. With D1 0 and D2 0, y is positive definite and y will be positive for all nonzero values of x1 and x2.
12.6.
Redo Problem 12.5 for y 5x1 2 6x1x2 3x2 2 2x2x3 8x3 2 3x1x3.
For a quadratic form in three variables, the coefficients of the squared terms continue to go on the principal diagonal, while the coefficient of x1x3 is divided evenly between the a13 and a31 positions, the coefficient of x2x3 is divided between the a23 and a32 positions, etc. Thus, D 5 3 1.5 3 3 1 1.5 1 8 where D1 5 0 D2 5 3 3 3 6 0 and D3 D 5(23) 3(25.5) 1.5(7.5) 27.25 0. Therefore, y is positive definite.
12.7.
Use discriminants to determine the sign definiteness of the following functions: a) y 2x1 2 4x1x2 5x2 2 2x2x3 3x3 2 2x1x3 D 2 2 1 2 5 1 1 1 3 where D1 2 0 D2 2 2 2 5 6 0 and D3 D 2(14) 2(7) 1(7) 7 0. So y is negative definite.
b) y 7x1 2 2x2 2 2x2x3 4x3 2 6x1x3 D 7 0 3 0 2 1 3 1 4 264 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 where D1 7 D2 7 0 0 2 14 and D3 D 7(7) 3(6) 31. And y is negative definite.
THE HESSIAN IN OPTIMIZATION PROBLEMS 12.8.
Optimize the following function, using (a) Cramer’s rule for the first-order condition and (b) the Hessian for the second-order condition: y 3x1 2 5x1 x1x2 6x2 2 4x2 2x2x3 4x3 2 2x3 3x1x3 a) The first-order conditions are y1 6x1 5 x2 3x3 0 y2 x1 12x2 4 2x3 0 (12.10) y3 2x2 8x3 2 3x1 0 which in matrix form is 6 1 3 1 12 2 3 2 8 x1 x2 x3 5 4 2 Using Cramer’s rule, A 6(92) 1(2) 3(34) 448. Since A also equals J , the equations are functionally independent. A1 5(92) 1(36) 3(32) 400 A2 6(36) 5(2) 3(14) 184 A3 6(32) 1(14) 5(34) 8 Thus, x ¯1 400 448 0.89 x ¯2 184 448 0.41 x ¯3 8 448 0.02 b) Testing the second-order condition by taking the second-order partials of (12.10) to form the Hessian, y11 6 y21 1 y31 3 y12 1 y22 12 y32 2 y13 3 y23 2 y33 8 Thus, H 6 1 3 1 12 2 3 2 8 where H1 6 0 H2 6 1 1 12 71 0 and H3 H A 448 0. With H positive definite, y is minimized at the critical values.
12.9.
Redo Problem 12.8, given y 5x1 2 10x1 x1x3 2x2 2 4x2 2x2x3 4x3 2.
a) y1 10x1 10 x3 0 y2 4x2 4 2x3 0 (12.11) y3 x1 2x2 8x3 0 In matrix form, 10 0 1 0 4 2 1 2 8 x1 x2 x3 10 4 0 265 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] Using Cramer’s rule, A A1 A2 A3 10(28) 1(4) 276 10(28) 1(8) 288 10(32) 10(2) 1(4) 336 10(8) 10(4) 120 Thus, x ¯1 288 276 1.04 x ¯2 336 276 1.22 x ¯3 120 276 0.43 b) Taking the second partials of (12.11) and forming the Hessian, H 10 0 1 0 4 2 1 2 8 where H1 10 0, H2 40 0, and H3 A 276 0. Thus, H is negative definite, and y is maximized.
12.10. A firm produces two goods in pure competition and has the following total revenue and total cost functions: TR 15Q1 18Q2 TC 2Q1 2 2Q1Q2 3Q2 2 The two goods are technically related in production, since the marginal cost of one is dependent on the level of output of the other (for example, TC/Q1 4Q1 2Q2). Maximize profits for the firm, using (a) Cramer’s rule for the first-order condition and (b) the Hessian for the second-order condition.
a) TR TC 15Q1 18Q2 2Q1 2 2Q1Q2 3Q2 2 The first-order conditions are 1 15 4Q1 2Q2 0 2 18 2Q1 6Q2 0 In matrix form, 4 2 2 6 Q1 Q2 15 18 Solving by Cramer’s rule, A 24 4 20 A1 90 36 54 A2 72 30 42 Thus, Q ¯ 1 54 20 2.7 Q ¯ 2 42 20 2.1 b) Using the Hessian to test for the second-order condition, H 4 2 2 6 where H1 4 and H2 20. With H negative definite, is maximized.
12.11. Using the techniques of Problem 12.10, maximize profits for the competitive firm whose goods are not technically related in production. The firm’s total revenue and total cost functions are TR 7Q1 9Q2 TC Q1 2 2Q1 5Q2 2Q2 2 266 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 a) 7Q1 9Q2 Q1 2 2Q1 5Q2 2Q2 2 1 7 2Q1 2 0 Q ¯ 1 2.5 2 9 5 4Q2 0 Q ¯ 2 1 b) H 2 0 0 4 where H1 2 and H2 8. So H is negative definite, and is maximized.
12.12. Maximize profits for a monopolistic firm producing two related goods, i.e., P1 f(Q1, Q2) when the goods are substitutes and the demand and total cost functions are P1 80 5Q1 2Q2 P2 50 Q1 3Q2 TC 3Q1 2 Q1Q2 2Q2 2 Use (a) Cramer’s rule and (b) the Hessian, as in Problem 12.10.
a) TR TC, where TR P1Q1 P2Q2.
(80 5Q1 2Q2)Q1 (50 Q1 3Q2)Q2 (3Q1 2 Q1Q2 2Q2 2) 80Q1 50Q2 4Q1Q2 8Q1 2 5Q2 2 1 80 4Q2 16Q1 0 2 50 4Q1 10Q2 0 In matrix form, 16 4 4 10 Q1 Q2 80 50 A 160 16 144 A1 800 200 600 A2 800 320 480 and Q ¯ 1 600 144 4.17 Q ¯ 2 480 144 3.33 b) H 16 4 4 10 where H1 16 and H2 144. So is maximized.
12.13. Maximize profits for a producer of two substitute goods, given P1 130 4Q1 Q2 P2 160 2Q1 5Q2 TC 2Q1 2 2Q1Q2 4Q2 2 Use (a) Cramer’s rule for the first-order condition and (b) the Hessian for the second-order condition.
a) (130 4Q1 Q2)Q1 (160 2Q1 5Q2)Q2 (2Q1 2 2Q1Q2 4Q2 2) 130Q1 160Q2 5Q1Q2 6Q1 2 9Q2 2 1 130 5Q2 12Q1 0 2 160 5Q1 18Q2 0 Thus, 12 5 5 18 Q1 Q2 130 160 A 191 A1 1540 Q ¯ 1 1540 191 8.06 A2 1270 Q ¯ 2 1270 191 6.65 b) H 12 5 5 18 H1 12 and H2 191. So is maximized.
267 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] 12.14. Redo Problem 12.13 for a monopolistic firm producing three related goods, when the demand functions and the cost function are P1 180 3Q1 Q2 2Q3 P2 200 Q1 4Q2 P3 150 Q2 3Q3 TC Q1 2 Q1Q2 Q2 2 Q2Q3 Q3 2 a) (180 3Q1 Q2 2Q3)Q1 (200 Q1 4Q2)Q2 (150 Q2 3Q3)Q3 (Q1 2 Q1Q2 Q2 2 Q2Q3 Q3 2) 180Q1 200Q2 150Q3 3Q1Q2 2Q2Q3 2Q1Q3 4Q1 2 5Q2 2 4Q3 2 1 180 3Q2 2Q3 8Q1 0 2 200 3Q1 2Q3 10Q2 0 3 150 2Q2 2Q1 8Q3 0 In matrix form, 8 3 2 3 10 2 2 2 8 Q1 Q2 Q3 180 200 150 A A1 A2 A3 8(76) 3(20) 2(14) 520 180(76) 3(1300) 2(1100) 7580 8(1300) 180(20) 2(50) 6900 8(1100) 3(50) 180(14) 6130 Thus, Q ¯ 1 7580 520 14.58 Q ¯ 2 6900 520 13.27 Q ¯ 3 6130 520 11.79 b) H 8 3 2 3 10 2 2 2 8 where H1 8, H2 71, and H3 H A 520. And is maximized.
12.15. Maximize profits as in Problem 12.14, given P1 70 2Q1 Q2 Q3 P2 120 Q1 4Q2 2Q3 P3 90 Q1 Q2 3Q3 TC Q1 2 Q1Q2 2Q2 2 2Q2Q3 Q3 2 Q1Q3 a) 70Q1 120Q2 90Q3 3Q1Q2 5Q2Q3 3Q1Q3 3Q1 2 6Q2 2 4Q3 2 1 70 3Q2 3Q3 6Q1 0 2 120 3Q1 5Q3 12Q2 0 3 90 3Q1 5Q2 8Q3 0 Thus, 6 3 3 3 12 5 3 5 8 Q1 Q2 Q3 70 120 90 A A1 A2 A3 336 2000 Q ¯ 1 2160 Q ¯ 2 1680 Q ¯ 3 2000 336 2160 336 1680 336 5.95 6.43 5 268 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 b) H 6 3 3 3 12 5 3 5 8 where H1 6, H2 63, and H3 H A 336. is maximized.
12.16. Given that Q F(P), maximize profits by (a) finding the inverse function P f(Q), (b) using Cramer’s rule for the first-order condition, and (c) using the Hessian for the second-order condition. The demand functions and total cost function are Q1 100 3P1 2P2 Q2 75 0.5P1 P2 TC Q1 2 2Q1Q2 Q2 2 where Q1 and Q2 are substitute goods, as indicated by the opposite signs for P1 and P2 in each equation (i.e., an increase in P2 will increase demand for Q1 and an increase in P1 will increase demand for Q2).
a) Since the markets are interrelated, the inverse functions must be found simultaneously. Rearranging the demand functions to get P f(Q), in order ultimately to maximize as a function of Q alone, 3P1 2P2 Q1 100 0.5P1 P2 Q2 75 In matrix form, 3 0.5 2 1 P1 P2 Q1 100 Q2 75 Using Cramer’s rule, A 2 A1 Q1 100 Q2 75 2 1 Q1 100 2Q2 150 250 Q1 2Q2 P1 A1 A 250 Q1 2Q2 2 125 0.5Q1 Q2 A2 3 0.5 Q1 100 Q2 75 3Q2 225 0.5Q1 50 275 0.5Q1 3Q2 P2 A2 A 275 0.5Q1 3Q2 2 137.5 0.25Q1 1.5Q2 b) (125 0.5Q1 Q2)Q1 (137.5 0.25Q1 1.5Q2)Q2 (Q1 2 2Q1Q2 Q2 2) 125Q1 137.5Q2 3.25Q1Q2 1.5Q1 2 2.5Q2 2 1 125 3.25Q2 3Q1 0 2 137.5 3.25Q1 5Q2 0 Thus, 3 3.25 3.25 5 Q1 Q2 125 137.5 A A1 A2 4.4375 178.125 Q ¯ 1 6.25 Q ¯ 2 178.125 4.4375 6.25 4.4375 40.14 1.4 c) H 3 3.25 3.25 5 H1 3, H2 H A 4.4375, and is maximized.
269 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] 12.17. Redo Problem 12.16 by maximizing profits for Q1 90 6P1 2P2 Q2 80 2P1 4P2 TC 2Q1 2 3Q1Q2 2Q2 2 where Q1 and Q2 are complements, as indicated by the same sign for P1 and P2 in each equation.
a) Converting the demand functions to functions of Q, 6 2 2 4 P1 P2 Q1 90 Q2 80 A 20 A1 Q1 90 2 Q2 80 4 4Q1 360 2Q2 160 200 4Q1 2Q2 P1 200 4Q1 2Q2 20 10 0.2Q1 0.1Q2 A2 6 Q1 90 2 Q2 80 6Q2 480 2Q1 180 300 6Q2 2Q1 P2 300 6Q2 2Q1 20 15 0.3Q2 0.1Q1 b) (10 0.2Q1 0.1Q2)Q1 (15 0.3Q2 0.1Q1)Q2 (2Q1 2 3Q1Q2 2Q1 2) 10Q1 15Q2 2.8Q1Q2 2.2Q1 2 2.3Q2 2 1 10 2.8Q2 4.4Q1 0 2 15 2.8Q1 4.6Q2 0 Thus, 4.4 2.8 2.8 4.6 Q1 Q2 10 15 A 12.4 A1 4 Q ¯ 1 4 12.4 0.32 A2 38 Q ¯ 2 38 12.4 3.06 c) H 4.4 2.8 2.8 4.6 H1 4.4, H2 A 12.4, and is maximized.
12.18. Redo Problem 12.16, given Q1 150 3P1 P2 P3 Q2 180 P1 4P2 2P3 Q3 200 2P1 P2 5P3 TC Q1 2 Q1Q2 2Q2 2 Q2Q3 Q3 2 Q1Q3 a) Finding the inverses of the demand functions 3 1 2 1 4 1 1 2 5 P1 P2 P3 Q1 150 Q2 180 Q3 200 A 36 A1 (Q1 150)(20 2) 1(5Q2 900 2Q3 400) 1(Q2 180 4Q3 800) 4980 18Q1 6Q2 6Q3 P1 4980 18Q1 6Q2 6Q3 36 138.33 0.5Q1 0.17Q2 0.17Q3 270 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 A2 3(5Q2 900 2Q3 400) (Q1 150)(5 4) 1(Q3 200 2Q2 360) 5090 9Q1 13Q2 7Q3 P2 5090 9Q1 13Q2 7Q3 36 141.39 0.25Q1 0.36Q2 0.19Q3 A3 3(4Q3 800 Q2 180) 1(Q3 200 2Q2 360) (Q1 150)(1 8) 4450 9Q1 5Q2 11Q3 P3 4450 9Q1 5Q2 11Q3 36 123.61 0.25Q1 0.14Q2 0.31Q3 b) P1Q1 P2Q2 P3Q3 TC 138.33Q1 141.39Q2 123.61Q3 1.42Q1Q2 1.33Q2Q3 1.42Q1Q3 1.5Q1 2 2.36Q2 2 1.31Q3 2 1 138.33 1.42Q2 1.42Q3 3Q1 0 2 141.39 1.42Q1 1.33Q3 4.72Q2 0 3 123.61 1.33Q2 1.42Q1 2.62Q3 0 Thus, 3 1.42 1.42 1.42 4.72 1.33 1.42 1.33 2.62 Q1 Q2 Q3 138.33 141.39 123.61 A 22.37 A1 612.27 Q ¯ 1 612.27 22.37 27.37 A2 329.14 Q ¯ 2 329.14 22.37 14.71 A3 556.64 Q ¯ 3 556.64 22.37 24.88 c) H1 3 1.42 1.42 1.42 4.72 1.33 1.42 1.33 2.62 H1 3, H2 12.14, and H3 A 22.37. And is maximized.
THE BORDERED HESSIAN IN CONSTRAINED OPTIMIZATION 12.19. Maximize utility u 2xy subject to a budget constraint equal to 3x 4y 90 by (a) finding the critical values x ¯, y ¯, and ¯ and (b) using the bordered Hessian H ¯ to test the second-order condition.
a) The Lagrangian function is U 2xy (90 3x 4y) The first-order conditions are Ux 2y 3 0 Uy 2x 4 0 U 90 3x 4y 0 In matrix form, 0 2 3 2 0 4 3 4 0 x y 0 0 90 (12.12) Solving by Cramer’s rule, A 48, A1 720, A2 540, and A3 360. Thus, x ¯ 15, y ¯ 11.25, and ¯ 7.5.
271 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] b) Taking the second partials of U with respect to x and y and the first partials of the constraint with respect to x and y to form the bordered Hessian, Uxx 0 Uyy 0 Uxy 2 Uyx cx 3 cy 4 From Section 12.5, H ¯ 0 2 3 2 0 4 3 4 0 or H ¯ 0 3 4 3 0 2 4 2 0 H ¯ 2 H ¯ 2(12) 3(8) 48 0 H ¯ 2 H ¯ 3(8) 4(6) 48 0 The bordered Hessian can be set up in either of the above forms without affecting the value of the principal minor. With H ¯ A 0, from the rules of Section 12.5 H ¯ is negative definite, and U is maximized.
12.20. Maximize utility u xy x subject to the budget constraint 6x 2y 110, by using the techniques of Problem 12.19.
a) U xy x (110 6x 2y) Ux y 1 6 0 Uy x 2 0 U 110 6x 2y 0 In matrix form, 0 1 6 1 0 2 6 2 0 x y 1 0 110 Solving by Cramer’s rule, x ¯ 91 – 3, y ¯ 27, and ¯ 42 – 3.
b) Since Uxx 0, Uyy 0, Uxy 1 Uyx, cx 6, and cy 2, H ¯ 0 1 6 1 0 2 6 2 0 H ¯ 2 H ¯ 24 With H ¯ 2 0, H ¯ is negative definite, and U is maximized.
12.21. Minimize a firm’s total costs c 45x2 90xy 90y2 when the firm has to meet a production quota g equal to 2x 3y 60 by (a) finding the critical values and (b) using the bordered Hessian to test the second-order conditions.
a) C 45x2 90xy 90y2 (60 2x 3y) Cx 90x 90y 2 0 Cy 90x 180y 3 0 C 60 2x 3y 0 In matrix form, 90 90 2 90 180 3 2 3 0 x y 0 0 60 Solving by Cramer’s rule, x ¯ 12, y ¯ 12, and ¯ 1080.
b) Since Cxx 90, Cyy 180, Cxy 90 Cyx, gx 2, and gy 3, H ¯ 90 90 2 90 180 3 2 3 0 H ¯ 2 450. With H ¯ 2 0, H ¯ is positive definite and C is minimized.
272 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 12.22. Minimize a firm’s costs c 3x2 5xy 6y2 when the firm must meet a production quota of 5x 7y 732, using the techniques of Problem 12.21.
a) C 3x2 5xy 6y2 (732 5x 7y) Cx 6x 5y 5 0 Cy 5x 12y 7 0 C 732 5x 7y 0 Solving simultaneously, x ¯ 75 y ¯ 51 ¯ 141 b) With Cxx 6, Cyy 12, Cxy 5 Cyx, gx 5, and gy 7, H ¯ 6 5 5 5 12 7 5 7 0 H ¯ 2 5(35 60) 7(42 25) 244. Thus, H ¯ is positive definite, and C is minimized.
12.23. Redo Problem 12.21 by maximizing utility u x0.5y0.3 subject to the budget constraint 10x 3y 140.
a) U x0.5y0.3 (140 10x 3y) Ux 0.5x0.5y0.3 10 0 Uy 0.3x0.5y0.7 3 0 U 140 10x 3y 0 Solving simultaneously, as shown in Example 10 of Chapter 6, x ¯ 8.75 y ¯ 17.5 and ¯ 0.04 b) With Uxx 0.25x1.5y0.3, Uyy 0.21x0.5y1.7, Uxy Uyx 0.15x0.5y0.7, gx 10, and gy 3, H ¯ 0.25x1.5y0.3 0.15x0.5y0.7 10 0.15x0.5y0.7 0.21x0.5y1.7 3 10 3 0 Expanding along the third column, H ¯ 2 10(0.45x0.5y0.7 2.1x0.5y1.7) 3(0.75x1.5y0.3 1.5x0.5y0.7) 21x0.5y1.7 9x0.5y0.7 2.25x1.5y0.3 0 since x and y 0, and a positive number x raised to a negative power n equals 1/xn, which is also positive. With H ¯ 2 0, H ¯ is negative definite, and U is maximized.
12.24. Maximize utility u x0.25y0.4 subject to the budget constraint 2x 8y 104, as in Problem 12.23.
a) U x0.25y0.4 (104 2x 8y) Ux 0.25x0.75y0.4 2 0 Uy 0.4x0.25y0.6 8 0 U 104 2x 8y 0 Solving simultaneously, x ¯ 20, y ¯ 8, and ¯ 0.03.
b) H ¯ 0.1875x1.75y0.4 0.1x0.75y0.6 2 0.1x0.75y0.6 0.24x0.25y1.6 8 2 8 0 Expanding along the third row, H ¯ 2 2(0.8x0.75y0.6 0.48x0.25y1.6) 8(1.5x1.75y0.4 0.2x0.75y0.6) 0.96x0.25y1.6 3.2x0.75y0.6 12x1.75y0.4 0 Thus, H ¯ is negative definite, and U is maximized.
273 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] 12.25. Minimize costs c 3x 4y subject to the constraint 2xy 337.5, using the techniques of Problem 12.21(a) and (b). (c) Discuss the relationship between this solution and that for Problem 12.19.
a) C 3x 4y (337.5 2xy) Cx 3 2y 0 1.5 y (12.13) Cy 4 2x 0 2 x (12.14) C 337.5 2xy 0 (12.15) Equate ’s in (12.13) and (12.14).
1.5 y 2 x y 0.75x Substitute in (12.15).
337.5 2x(0.75x) 1.5x2 x2 225 x ¯ 15 Thus, y ¯ 11.25 and ¯ 0.133.
b) With Cxx 0, Cyy 0, and Cxy Cyx 2 and from the constraint 2xy 337.5, gx 2y, and gy 2x, H ¯ 0 2 2y 2 0 2x 2y 2x 0 H ¯ 2 (2)(4xy) 2y(4x) 16xy. With ¯, x ¯, y ¯ 0, H ¯ 2 0. Hence H ¯ is positive definite, and C is minimized.
c) This problem and Problem 12.19 are the same, except that the objective functions and constraints are reversed. In Problem 12.19, the objective function u 2xy was maximized subject to the constraint 3x 4y 90; in this problem the objective function c 3x 4y was minimized subject to the constraint 2xy 337.5. Therefore, one may maximize utility subject to a budget constraint or minimize the cost of achieving a given level of utility.
12.26. Minimize the cost of 434 units of production for a firm when Q 10K0.7L0.1 and PK 28, PL 10 by (a) finding the critical values and (b) using the bordered Hessian. (c) Check the answer with that of Problem 6.41(b).
a) The objective function is c 28K 10L, and the constraint is 10K0.7L0.1 434. Thus, C CK CL C 28K 10L (434 10K0.7L0.1) 28 7K0.3L0.1 0 10 K0.7L0.9 0 434 10K0.7L0.1 0 (12.16) (12.17) (12.18) Rearranging and dividing (12.16) by (12.17) to eliminate , 28 10 7K0.3L0.1 K0.7L0.9 2.8 7L K K 2.5L 274 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 Substituting in (12.18) and using a calculator, 434 10(2.5)0.7L0.7L0.1 434 19L0.8 L ¯ (22.8)1/0.8 (22.8)1.25 50 Thus, K ¯ 125 and ¯ 11.5.
b) With CKK 2.1K 1.3L0.1, CLL 0.9K0.7L1.9, and CKL 0.7K0.3L0.9 CLK and from the constraint gK 7K0.3L0.1 and gL K0.7L0.9, H ¯ 2.1K1.3L0.1 0.7K0.3L0.9 7K0.3L0.1 0.7K0.3L0.9 0.9K0.7L1.9 K0.7L0.9 7K0.3L0.1 K0.7L0.9 0 Expanding along the third row, H ¯ 2 7K0.3L0.1(0.7K 0.4L1.8 6.3K 0.4L1.8) K0.7L0.9(2.1K0.6L0.8 4.9K0.6L0.8) 49K0.1L1.7 7K0.1L1.7 56K0.1L1.7 With K, L, 0, H ¯ 2 0; H ¯ is positive definite, and C is minimized.
c) The answers are identical with those in Problem 6.41(b), but note the difference in the work involved when the linear function is selected as the objective function and not the constraint. See also the bordered Hessian for Problem 6.41(b), which is calculated in Problem 12.27(c).
12.27. Use the bordered Hessian to check the second-order conditions for (a) Example 7 of Chapter 6, (b) Problem 6.41(a), and (c) Problem 6.41(b).
a) H ¯ 16 1 1 1 24 1 1 1 0 H ¯ 2 1(1 24) 1(16 1) 42. With H ¯ 2 0, H ¯ is positive definite and C is minimized.
b) H ¯ 0.21K1.7L0.5 0.15K0.7L0.5 6 0.15K0.7L0.5 0.25K0.3L1.5 2 6 2 0 H ¯ 2 6(0.30K0.7L0.5 1.5K0.3L1.5) 2(0.42K1.7L0.5 0.9K0.7L0.5) 9K0.3L1.5 3.6K0.7L0.5 0.84K1.7L0.5 0 With H ¯ 2 0, H ¯ is negative definite, and Q is maximized.
c) H ¯ 2.1K1.3L0.1 0.7K0.3L0.9 28 0.7K0.3L0.9 0.9K0.7L1.9 10 28 10 0 H ¯ 2 28(7K0.3L0.9 25.2K0.7L1.9) 10(21K1.3L0.1 19.6K0.3L0.9) 705.6K0.7L1.9 392K0.3L0.9 210K1.3L0.1 0 With H ¯ 2 0, H ¯ is negative definite, and Q is maximized.
12.28. Use the bordered Hessian to check the second-order conditions in Problem 5.12(c), where 4xyz2 was optimized subject to the constraint x y z 56; the first-order conditions were Fx 4yz2 0, Fy 4xz2 0, and Fz 8xyz 0; and the critical values were x ¯ 14, y ¯ 14, and z ¯ 28.
275 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] Take the second partial derivatives of F and the first partials of the constraint, and set up the bordered Hessian, as follows: H ¯ 0 gx gy gz gx Fxx Fyx Fzx gy Fxy Fyy Fzy gz Fxz Fyz Fzz 0 1 1 1 1 0 4z2 8yz 1 4z2 0 8xz 1 8yz 8xz 8xy Start with H ¯ 2 , the 3 3 submatrix in the upper left-hand corner. H ¯ 2 0 1(4z2) 1(4z2) 8z2 0 Next evaluate H ¯ 3 , which here equals H ¯ . Expanding along the first row, H ¯ 3 0 1 1 1 1 4z2 0 8xz 8yz 8xz 8xy 1 1 1 1 0 4z2 8yz 8yz 8xz 8xy 1 1 1 1 0 4z2 8yz 4z2 0 8xz H ¯ 3 1[1(0 8xz · 8xz) 4z2(8xy 8xz) 8yz(8yz 0)] 1[1(4z2 · 8xy 8yz · 8xz) 0 8yz(8yz 4z2)] 1[1(4z2 · 8xz 0) 0 4z2(8yz 4z2)] H ¯ 3 1(64x2z2 32xyz2 32xz3 64xyz2) 1(32xyz2 64xyz2 64y2z2 32yz3) 1(32xz3 32yz3 16z4) H ¯ 3 16z4 64xz3 64yz3 64xyz2 64x2z2 64y2z2 Evaluated at x ¯ 14, y ¯ 14, z ¯ 28, H ¯ 3 19,668,992 0 With H ¯ 2 0 and H ¯ 3 0, H ¯ is negative definite and the function is maximized.
INPUT-OUTPUT ANALYSIS 12.29. Determine the total demand for industries 1, 2, and 3, given the matrix of technical coefficients A and the final demand vector B below.
A Output industry 1 0.2 0.4 0.3 2 0.3 0.1 0.5 3 0.2 0.3 0.2 1 2 3 Input industry B 150 200 210 From (12.3), the total demand vector is X (I A)1B, where I A 0.8 0.4 0.3 0.3 0.9 0.5 0.2 0.3 0.8 Taking the inverse of I A, (I A)1 1 0.239 0.57 0.34 0.27 0.41 0.58 0.32 0.47 0.49 0.60 Substituting in X (I A)1B, X 1 0.239 0.57 0.34 0.27 0.41 0.58 0.32 0.47 0.49 0.60 150 200 210 1 0.239 210.2 244.7 294.5 879.50 1023.85 1232.22 x1 x2 x3 276 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 12.30. Determine the new level of total demand X2 for Problem 12.29 if final demand increases by 40 in industry 1, 20 in industry 2, and 25 in industry 3. X (I A)1 B X 1 0.239 0.57 0.34 0.27 0.41 0.58 0.32 0.47 0.49 0.60 40 20 25 1 0.239 36.35 36.00 43.60 152.09 150.63 182.43 X2 X1 X 879.50 1023.85 1232.22 152.09 150.63 182.43 1031.59 1174.48 1414.65 12.31. Determine the total demand for industries 1, 2, and 3, given the matrix of technical coefficients A and the final demand vector B below.
A Output industry 1 0.4 0.2 0.2 2 0.3 0.2 0.4 3 0.1 0.3 0.2 1 2 3 Input industry B 140 220 180 X (I A)1B where I A 0.6 0.2 0.2 0.3 0.8 0.4 0.1 0.3 0.8 and the inverse (I A)1 1 0.222 0.52 0.28 0.17 0.22 0.46 0.20 0.24 0.30 0.42 Thus, X 1 0.222 0.52 0.28 0.17 0.22 0.46 0.20 0.24 0.30 0.42 140 220 180 743.24 756.76 789.19 x1 x2 x3 12.32. Determine the new total demand X2 if final demand increases by 30 for industry 1 and decreases by 15 and 35 for industries 2 and 3, respectively, in Problem 12.31. X (I A)1 B X 1 0.222 0.52 0.28 0.17 0.22 0.46 0.20 0.24 0.30 0.42 30 15 35 1 0.222 5.45 7.30 12.00 24.55 32.88 54.05 X2 X1 X 743.24 756.76 789.19 24.55 32.88 54.05 767.79 723.88 735.14 277 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] 12.33. Given the interindustry transaction demand table in millions of dollars below, find the matrix of technical coefficients.
Sector of Destination Sector of Origin Steel Coal Iron Auto Final Demand Total Demand Steel Coal Iron Auto Value added Gross production 80 200 220 60 40 600 20 50 110 140 280 600 110 90 30 160 10 400 230 120 40 240 370 1000 160 140 0 400 600 600 400 1000 Thetechnicalcoefficientaijexpressesthenumberofunitsordollarsofinputirequiredtoproduceoneunit or one dollar of product j. Thus a11 the percentage of steel in one dollar ofsteel, a21 thepercentageofcoal in one dollar of steel, a31 the percentage of iron in one dollar of steel, and a41 the percentage of autos in one dollar of steel. To find the technical coefficients, simply divide every element in each column by the value of gross production at the bottom of the column, omitting value added. Thus, A 80 ––– 600 200 ––– 600 220 ––– 600 60 ––– 600 20 ––– 600 50 ––– 600 110 ––– 600 140 ––– 600 110 ––– 400 90 ––– 400 30 ––– 400 160 ––– 400 230 –––– 1000 120 –––– 1000 40 –––– 1000 240 –––– 1000 0.133 0.033 0.275 0.23 0.333 0.083 0.225 0.12 0.367 0.183 0.075 0.04 0.10 0.233 0.40 0.24 12.34. Check the matrix of technical coefficients A in Problem 12.33.
To check matrix A, multiply it by the column vector of total demand X. The product should equal the intermediate demand which is total demand X final demand B. Allow for slight errors due to rounding.
AX 0.133 0.033 0.275 0.23 0.333 0.083 0.225 0.12 0.367 0.183 0.075 0.04 0.10 0.233 0.40 0.24 600 600 400 1000 439.6 459.6 400 599.8 X B 600 600 400 1000 160 140 0 400 440 460 400 600 12.35. Given the interindustry transaction demand table below, (a) find the matrix of technical coefficients and (b) check your answer.
Sector of Destination Sector of Origin 1 2 3 Final Demand Total Demand 1 2 3 Value added Gross production 20 50 40 30 140 60 10 30 50 150 10 80 20 20 130 50 10 40 140 150 130 278 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 a) A 20 ––– 140 50 ––– 140 40 ––– 140 60 ––– 150 10 ––– 150 30 ––– 150 10 ––– 130 80 ––– 130 20 ––– 130 0.143 0.357 0.286 0.4 0.067 0.2 0.077 0.615 0.154 b) AX 0.143 0.357 0.286 0.4 0.067 0.2 0.077 0.615 0.154 140 150 130 90 140 90 X B 140 150 130 50 10 40 90 140 90 12.36. Find the new level of total demand in Problem 12.35 if in year 2 final demand is 70 in industry 1, 25 in industry 2, and 50 in industry 3.
X (I A)1B where I A 0.857 0.357 0.286 0.4 0.933 0.2 0.077 0.615 0.846 and (I A)1 1 0.354 0.666 0.354 0.318 0.478 0.703 0.555 0.338 0.286 0.657 Thus, X 1 0.354 0.666 0.354 0.318 0.478 0.703 0.555 0.338 0.286 0.657 70 25 50 1 0.354 71.37 78.79 63.66 201.61 222.57 179.83 12.37. Having found the inverse of I A, use it to check the accuracy of the matrix of coefficients derived in Problem 12.35; i.e., check to see if (I A)1B X.
(I A)1B 1 0.354 0.666 0.354 0.318 0.478 0.703 0.555 0.338 0.286 0.657 50 10 40 1 0.354 49.56 53.13 46.04 140 150 130 12.38. Assume in Problem 12.35 that value added is composed entirely of the primary input labor.
How much labor would be necessary to obtain the final demand (a) in Problem 12.35 and (b) in Problem 12.36? (c) If the amount of labor available in the economy is 100, is the output mix feasible?
a) To get the technical coefficient of labor aLj in Problem 12.35, simply divide the value added in each column by the gross production. Thus, aL1 30 ––– 140 0.214, aL2 50 ––– 150 0.333, and aL3 20 ––– 130 0.154. The amount of labor needed to meet the final demand will then equal the row of technical coefficients for labor times the column vector of total demand, since labor must also be used to produce the intermediate products. Thus, L1 [0.214 0.333 0.154] 140 150 130 99.93 b) L2 [0.214 0.333 0.154] 201.61 222.57 179.83 144.95 c) Final demand in Problem 12.35 is feasable since 99.93 100. Final demand in Problem 12.36 is not feasible since society does not have sufficient labor resources to produce it.
279 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] 12.39. Check the accuracy of the technical coefficients found in Problem 12.38.
Having found the technical coefficients of labor for Problem 12.35, where value added was due totally to labor inputs, the accuracy of the technical coefficients can be easily checked. Since each dollar of output must be completely accounted for in terms of inputs, simply add each column of technical coefficients to be sure it equals 1.
1 2 3 1 2 3 Value added (labor) 0.143 0.357 0.286 0.214 1.000 0.4 0.067 0.2 0.333 1.000 0.077 0.615 0.154 0.154 1.000 EIGENVALUES, EIGENVECTORS 12.40. Use eigenvalues (characteristic roots, latent roots) to determine sign definiteness for A 10 3 3 4 To find the characteristic roots of A, the determinant of the characteristic matrix A cI must equal zero. Thus, A cI 10 c 3 3 4 c 0 40 c2 14c 9 0 c2 14c 31 0 Using the quadratic formula, c 14 196 4(31) 2 14 8.485 2 c1 11.2425 c2 2.7575 With both characteristic roots positive, A is positive definite.
12.41. Redo Problem 12.40, given A 4 2 2 6 A cI 4 c 2 2 6 c 0 24 c2 10c 4 0 c2 10c 20 0 c 10 100 4(20) 2 10 4.4721 2 c1 5.5279 2 2.764 c2 14.4721 2 7.236 With both characteristic roots negative, A is negative definite.
280 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 12.42. Redo Problem 12.40, given A 6 2 2 2 A cI 6 c 2 2 2 c 0 12 c2 8c 4 0 c2 8c 8 0 c 8 64 4(8) 2 8 5.66 2 c1 1.17 c2 6.83 A is positive definite.
12.43. Redo Problem 12.40, given A 4 6 3 0 2 5 0 1 3 A cI 4 c 0 0 6 2 c 1 3 5 3 c 0 Expanding along the first column, A cI (4 c)[(2 c)(3 c) 5] 0 (12.19) c3 9c2 21c 4 0 (12.20) To solve (12.20), we may use a standard formula for finding cube roots or note that (12.20) will equal zero if in (12.19) 4 c 0 or (2 c)(3 c) 5 0 Thus, the characteristic roots are 4 c 0 (2 c)(3 c) 5 0 c1 4 c2 5c 1 0 c 5 25 4 2 5 4.58 2 c2 4.79 c3 0.21 With all three characteristic roots positive, A is positive definite.
12.44. Redo Problem 12.40, given A 6 13 5 1 4 1 0 0 9 A cI 6 c 13 5 1 4 c 1 0 0 9 c 0 Expanding along the third column, A cI (9 c)[(6 c)(4 c) 13] 0 (12.21) c3 19c2 101c 99 0 (12.22) 281 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] which will equal zero if in (12.21) 9 c 0 or (6 c)(4 c) 13 0 Thus, c1 9 c2 10c 11 0 c 10 100 4(11) 2 10 7.48 2 c2 8.74 c3 1.26 With all latent roots positive, A is positive definite.
12.45. Redo Problem 12.40, given A 5 0 4 1 2 2 2 0 3 A cI 5 c 0 4 1 2 c 2 2 0 3 c 0 Expanding along the second row, A cI (2 c)[(5 c)(3 c) 8] 0 Thus, 2 c c1 0 or (5 c)(3 c) 8 0 2 c2 8c 7 0 (c 7)(c 1) 0 c2 7 c3 1 With all latent roots negative, A is negative definite.
12.46. Given A 6 6 6 3 Find (a) the characteristic roots and (b) the characteristic vectors.
a) A cI 6 c 6 6 3 c 0 18 c2 3c 36 0 c2 3c 54 0 (c 9)(c 6) 0 c1 9 c2 6 With one root positive and the other negative, A is sign indefinite.
b) Using c1 9 for the first characteristic vector V1, 6 9 6 6 3 9 v1 v2 3 6 6 12 v1 v2 0 v1 2v2 Normalizing, as in Example 9, (2v2)2 v2 2 1 5v2 2 1 v2 0.2 v1 2v2 20.2 282 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS [CHAP . 12 Thus, V1 20.2 0.2 Using c2 6 for the second characteristic vector, 6 (6) 6 6 3 (6) v1 v2 12 6 6 3 v1 v2 0 v2 2v1 Normalizing, v1 2 (2v1)2 1 5v1 2 1 v1 0.2 v2 2v1 20.2 Thus, V2 0.2 20.2 12.47. Redo Problem 12.46, given A 6 3 3 2 a) A cI 6 c 3 3 2 c 0 c2 4c 21 0 c1 7 c2 3 With c1 0 and c2 0, A is sign indefinite.
b) Using c1 7 to form the first characteristic vector, 6 7 3 3 2 7 v1 v2 1 3 3 9 v1 v2 0 v1 3v2 Normalizing, (3v2)2 v2 2 1 9v2 2 v2 2 1 10v2 2 1 v2 0.1 and v1 3v2 30.1 Thus, V1 30.1 0.1 Using c2 3, 6 (3) 3 3 2 (3) v1 v2 9 3 3 1 v1 v2 0 v2 3v1 Normalizing, v1 2 (3v1)2 1 10v1 2 1 v1 0.1 and v2 3v1 30.1 Thus, V2 0.1 30.1 283 SPECIAL DETERMINANTS AND MATRICES AND THEIR USE IN ECONOMICS CHAP . 12] CHAPTER 13 Comparative Statics and Concave Programming 13.1 INTRODUCTION TO COMPARATIVE STATICS Comparative-static analysis, more commonly known as comparative statics, compares the different equilibrium values of the endogenous variables resulting from changes in the values of the exogenous variables and parameters in the model. Comparative statics allows economists to estimate such things as the responsiveness of consumer demand to a projected excise tax, tariff, or subsidy; the effect on national income of a change in investment, government spending, or the interest rate; and the likely price of a commodity given some change in weather conditions, price of inputs, or availability of transportation. Comparative statics essentially involves finding the appropriate derivative, as we saw earlier in Section 6.2.
13.2 COMPARATIVE STATICS WITH ONE ENDOGENOUS VARIABLE Comparative statics can be used both with specific and general functions. Example 1 provides a specific function illustration; Example 2 demonstrates the method with a general function. In the case of specific functions the prerequisite derivatives can be derived from either explicit or implicit functions. In the case of general functions, implicit functions must be used. Whenever there is more than one independent variable (Problem 13.2), partial derivatives are the appropriate derivatives and are found in a similar fashion (Problem 13.3).
EXAMPLE 1.
Assume the demand QD and supply QS of a commodity are given by specific functions, here expressed in terms of parameters.
QD m nP kY m, n, k 0 QS a bP a, b 0 284 where P price and Y consumers’ income. The equilibrium condition is QD QS Substituting from above and solving for the equilibrium price level P, we have m nP kY a bP (13.1) m a kY (b n)P P m a kY b n (13.2) Using comparative statics we can now determine how the equilibrium level of the endogenous variable P will change for a change in the single exogenous variable (Y) or any of the five parameters (a, b, m,n, k), should the latter be of interest. Comparative-static analysis simply involves taking the desired derivative and determining its sign. To gauge the responsiveness of the equilibrium price to changes in income, we have from the explicit function (13.2), dP dY k b n 0 (13.3) This means that an increase in consumers’ income in this model will lead to an increase in the equilibrium price of the good. If the values of the parameters are known, as in Problem 13.1, the specific size of the price increase can also be estimated.
Comparative statics can be applied equally well to implicit functions. By moving everything to the left in (13.1), so that QD QS 0, or excess demand equals zero, we can derive the implicit function F for the equilibrium condition: F m nP kY a bP 0 (13.4) Then the implicit function rule (Section 5.10) can be employed to find the desired comparative-static derivative.
Assuming Fp 0, dP dY FY FP where from (13.4), FY k and FP (n b). Substituting and simplifying, we have dP dY k (n b) k b n 0 Comparative statics can also be used to estimate the effect on P of a change in any of the parameters (m, n, k, a, b), but since these merely represent intercepts and slopes in demand and supply analysis, such as we have above, they generally have little practical relevance for economics. In other instances, however, such as income determination models (Problem 13.3), the parameters will frequently have economic significance and may warrant comparative-static derivatives of their own.
EXAMPLE 2.
Now assume a general model in which the supply and demand of a commodity are given solely by general functions: Demand D(P, Y) Dp 0, DY 0, Supply S(P) SP 0 The equilibrium price level P can be found where demand equals supply: D(P, Y) S(P) or equivalently where excess demand equals zero, D(P, Y) S(P) 0 (13.5) 285 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] With general functions, only implicit forms such as (13.5) are helpful in finding the comparative-static derivatives.
Assuming FP 0, dP dY FY FP where from (13.5), FY DY and FP DP SP. Substituting, dP dY DY DP SP From theory, we always expect SP 0. If the good is a normal good, then DY 0 and DP 0.
Substituting above, we have in the case of a normal good dP dY () () () 0 If the good is an inferior good but not a Giffen good, then DY 0 and DP 0 so that dP/dY 0; and if the good is a Giffen good, then DY 0 and DP 0 and the sign of the derivative will be indeterminate and depend on the sign of the denominator. See Problems 13.1 to 13.7.
13.3 COMPARATIVE STATICS WITH MORE THAN ONE ENDOGENOUS VARIABLE In a model with more than one endogenous variable, comparative statics requires that there be a unique equilibrium condition for each of the endogenous variables. A system of n endogenous variables must have n equilibrium conditions. Measuring the effect of a particular exogenous variable on any or all of the endogenous variables involves taking the total derivative of each of the equilibrium conditions with respect to the particular exogenous variable and solving simultaneously for each of the desired partial derivatives. If the functions have continuous derivatives and the Jacobian consisting of the partial derivatives of all the functions with respect to the endogenous variables does not equal zero, then from the implicit function theorem the optimal values of the endogenous variables can be expressed as functions of the exogenous variables, as outlined in Problem 13.8, and the comparative-static derivatives can be estimated with the help of Cramer’s rule, as demonstrated in Example 3 below.
The method is also illustrated in terms of a typical economic problem in Example 4.
EXAMPLE 3.
For simplicity of exposition, assume a model with only two endogenous variables and two exogenous variables, expressed in terms of implicit general functions in which the endogenous variables are listed first, followed by the exogenous variables, with a semicolon separating the former from the latter. The model can be easily expanded to any number of endogenous variables (n) and any number of exogenous variables (m), where n need not equal m.
F 1(y1, y2; x1, x2) 0 F 2(y1, y2; x1, x2) 0 To find the comparative-static partial derivatives of the system with respect to one of the independent variables, say x1, we take the total derivative (Section 5.9) of both functions with respect to x1.
F 1 y1 · y1 x1 F 1 y2 · y2 x1 F 1 x1 0 F 2 y1 · y1 x1 F 2 y2 · y2 x1 F 2 x1 0 When evaluated at the equilibrium point, which is also frequently indicated by a bar, all the partial derivatives will 286 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 have fixed values and so can be expressed in matrix notation. Moving the partial derivatives of both functions with respect to x1 to the right, we have F 1 y1 F 2 y1 F 1 y2 F 2 y2 y ¯1 x1 y ¯2 x1 F 1 x1 F 2 x1 JX B If both functions have continuous first and second derivatives and the Jacobian J , consisting of all the first-order partial derivatives of both functions (Fi) with respect to both endogenous variables (yj), does not equal zero, J F 1 y1 · F 2 y2 F 2 y1 · F 1 y2 0, then by making use of the implicit function theorem we can express the optimal values of the endogenous values as implicit functions of the exogenous variables and solve for the desired comparative-static derivatives in X using Cramer’s rule. Specifically, assuming J 0, y ¯i x1 Ji J (13.6) Thus, to solve for the first derivative, y ¯1/x1, we form a new matrix J1 by replacing the first column of J with the column vector B and then substitute it above in (13.6).
y ¯1 x1 J1 J F 1 x1 F 2 x1 F 1 y2 F 2 y2 F 1 x1 · F 2 y2 F 2 x1 · F 1 y2 F 1 y1 F 2 y1 F 1 y2 F 2 y2 F 1 y1 · F 2 y2 F 2 y1 · F 1 y2 Similarly, y ¯2 x1 J2 J F 1 y1 F 2 y1 F 1 x1 F 2 x1 F 1 y1 · F 2 x1 F 2 y1 · F 1 x1 F 1 y1 F 2 y1 F 1 y2 F 2 y2 F 1 y1 · F 2 y2 F 2 y1 · F 1 y2 The partials with respect to x2 are found in like fashion, after starting with the total derivatives of both functions with respect to x2. See Problem 13.9.
EXAMPLE 4.
Assume that equilibrium in the goods and services market (IS curve) and the money market (LM curve) are given, respectively, by F 1(Y, i; C0, M0, P) Y C0 C(Y, i) 0 0 CY 1, Ci 0 (13.7) F 2(Y, i; C0, M0, P) L(Y, i) M0/P 0 LY 0, Li 0 (13.8) where L(Y, i) the demand for money, M0 the supply of money, C0 autonomous consumption, and P the price level, which makes M0/P the supply of real rather than nominal money. For simplicity, we will hold P 287 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] constant. The effect on the equilibrium levels of Y and i of a change in C0 using comparative statics is demonstrated below.
a) Take the total derivative of the equilibrium conditions, (13.7) and (13.8), with respect to the desired exogenous variable, here C0.
Y C0 1 CY · Y C0 Ci · i C0 0 LY · Y C0 Li · i C0 0 b) Rearranging and setting in matrix form, 1 CY LY Ci Li Y ¯ C0 i ¯ C0 1 0 JX B c) Then check to make sure the Jacobian J 0 so the implicit function theorem holds. J (1 CY)Li CiLY Applying the signs, J ()() ()() () 0 Therefore, J 0.
d) Solve for the first derivative, Y ¯/C0 by forming a new matrix J1 in which the first column of J is replaced with the column vector B and substituted in (13.6). J1 1 0 Ci Li Li Thus, Y ¯ C0 J1 J Li (1 CY)Li CiLY () () 0 An increase in autonomous consumption C0 will lead to an increase in the equilibrium level of income.
e) Solve for the second derivative, i ¯/C0 by forming J2 in which the second column of J is replaced with the column vector B and substituted in (13.6). J2 1 CY LY 1 0 LY and i ¯ C0 J2 J LY (1 CY)Li CiLY () () 0 An increase in C0 will also lead to an increase in the equilibrium level of interest. The effect on Y ¯ and i ¯ of a change in M0 is treated in Problem 13.10. See also Problems 13.8 to 13.18.
13.4 COMPARATIVE STATICS FOR OPTIMIZATION PROBLEMS In addition to their general interest in the effects of changes in exogenous variables on the equilibrium values of their models, economists frequently also want to study the effects of changes in exogenous variables on the solution values of optimization problems. This is done by applying comparative-static techniques to the first-order conditions from which the initial optimal values are determined. Since first-order conditions involve first-order derivatives, comparative-static analysis of optimization problems, as we shall see, is intimately tied up with second-order derivatives and Hessian determinants. The methodology is set forth in Example 5.
288 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 EXAMPLE 5.
A price-taking firm has a strictly concave production function Q(K, L). Given P output price, r rental rate of capital, and w wage, its profit function is PQ(K, L) rK wL If we take the derivatives, /K and /L, for the first-order optimization conditions, and express them as implicit functions, we have F 1(K, L; r, w, P) PQK(K ¯ , L ¯) r 0 F 2(K, L; r, w, P) PQL(K ¯ , L ¯) w 0 where the bars indicate that the first derivatives QK and QL are evaluated at the optimal values of the profit function. From these first-order conditions we can determine the effects of a change in the exogenous variables (r, w) on the optimal values of the endogenous variables (K ¯ , L ¯) by use of comparative statics as follows: a) Take the total derivatives of the first-order conditions with respect to either of the exogenous variables and set them in the now familiar matrix form. Starting with the rental rate of capital r and noting that each of the first derivatives QK and QL is a function of both K and L, we have F 1 K F 2 K F 1 L F 2 L K ¯ r L ¯ r F 1 r F 2 r or specifically, PQKK PQLK PQKL PQLL K ¯ r L ¯ r 1 0 JX B J P2(QKKQLL QLKQKL) Provided the second-order sufficient conditions, as expressed within the parentheses above, are met, J 0. Here we observe that when finding the comparative-static derivatives from the first derivatives of the first-order optimization conditions, J H , the Hessian (Section 12.2). For optimization of a (2 2) system, we also recall H 0.
b) Since J H 0, and assuming continuous first and second derivatives, the conditions of the implicit function theorem are met and we can use Cramer’s rule to find the desired derivatives.
K ¯ r J1 J J 1 0 PQKL PQLL PQLL P2(QKKQLL QKLQLK) 0 where K ¯ /r 0 because we are assuming strictly concave production functions, which means QLL 0, QKK 0, and QKKQLL QKLQLK over the entire domain of the function. We also know from microtheory that a profit-maximizing firm will only produce where the marginal productivity of inputs (QL, QK) is declining. Hence at the optimal level of production, QLL 0 and QKK 0. Similarly, we can find L ¯ r J2 J J PQKK PQLK 1 0 PQLK P2(QKKQLL QKLQLK) To be able to sign this comparative-static derivative, we need to know the sign of the cross partial QLK, which is the effect of a change in capital on the marginal productivity of labor QL. If we assume it is positive, which is likely, an increase in the interest rate will cause a decrease in the use of labor due to the negative sign in the numerator. For the effects of a change in wage w on K ¯ , L ¯, see Problem 13.19.
See also Problems 13.20 to 13.24.
289 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] 13.5 COMPARATIVE STATICS USED IN CONSTRAINED OPTIMIZATION Comparative-static analysis can also be applied to constrained optimization problems. In constrained optimization, the Lagrangian multiplier is an endogenous variable and in comparative-static analysis it is evaluated at its optimal value ( ¯). If the second-order sufficient condition is satisfied, the bordered Hessian H ¯ may be positive or negative, depending on the type of optimization, but it will never equal zero. If H ¯ 0, the Jacobian will not equal zero since J H ¯ , as seen in Example 6. When J 0 and assuming continuous first and second derivatives, we know from the implicit function theorem that the optimal values of the endogenous variables can be expressed as implicit functions of the exogenous variables and the desired comparative-static derivatives found by means of Cramer’s rule. An illustration is provided in Example 6.
EXAMPLE 6.
Assume a firm operating in perfectly competitive input and output markets wants to maximize its output q(K, L) subject to a given budgetary constraint rK wL B The Lagrangian function to be maximized is Q q(K, L) (B rK wL) and the three first-order derivatives (Q/K, Q/L, Q/) representing the first-order conditions can be expressed as the following implicit functions: F 1(K ¯ , L ¯, ¯; r, w, B) QK(K ¯ , L ¯) r ¯ 0 F 2(K ¯ , L ¯, ¯; r, w, B) QL(K ¯ , L ¯) w ¯ 0 F 3(K ¯ , L ¯, ¯; r, w, B) B rK ¯ wL ¯ 0 From these first-order conditions for constrained optimization, assuming continuous derivatives and satisfaction of the second-order sufficient condition, we can determine the effects of a change in any of the exogenous variables (r, w, B) on the optimal values of the three endogenous variables (K ¯ , L ¯, ¯) with comparative-static analysis.
To find the effect of a change in the budget B on the optimal values of the endogenous variables, we take the total derivative of each of the three functions with respect to B.
F 1 K ¯ F 2 K ¯ F 3 K ¯ F 1 L ¯ F 2 L ¯ F 3 L ¯ F 1 ¯ F 2 ¯ F 3 ¯ K ¯ B L ¯ B ¯ B F 1 B F 2 B F 3 B or specifically, QKK QLK r QKL QLL w r w 0 K ¯ B L ¯ B ¯ B 0 0 1 J QKK(w2) QKL(rw) r(rQLL wQLK) J w2QKK rwQKL r2QLL rwQLK 0 since J H ¯ (Section 12.5) and, if the second-order sufficient condition is met, H ¯ 0 for constrained maximization. Since profit-maximizing firms in perfect competition operate only in the area of decreasing marginal productivity of inputs (QKK, QLL 0), the second-order condition will be fulfilled whenever K and L are complements (QKL, QLK 0) and will depend on the relative strength of the direct and cross partials when K and L are substitutes (QKL, QLK 0).
290 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 With J H ¯ 0, and assuming continuous first- and second-order derivatives, we can now use Cramer’s rule to find the desired derivatives.
1.
K ¯ B J1 J J 0 0 1 QKL QLL w r w 0 wQKL rQLL J K ¯ /B 0 when K and L are complements and indeterminate when K and L are substitutes.
2.
L ¯ B J2 J J QKK QLK r 0 0 1 r w 0 rQLK wQKK J L ¯/B 0 when K and L are complements and indeterminate when K and L are substitutes.
3.
¯ B J3 J J QKK QLK r QKL QLL w 0 0 1 1(QKKQLL QKLQLK) J ¯/B is indeterminate. See also Problems 13.25 to 13.29.
13.6 THE ENVELOPE THEOREM The envelope theorem enables us to measure the effect of a change in any of the exogenous variables on the optimal value of the objective function by merely taking the derivative of the Lagrangian function with respect to the desired exogenous variable and evaluating the derivative at the values of the optimal solution. The rationale is set forth in Example 7 and an illustration is offered in Example 8. The envelope theorem also provides the rationale for our earlier description of the Lagrange multiplier as an approximation of the marginal effect on the optimized value of the objective function due to a small change in the constant of the constraint (Section 5.6). One important implication of the envelope theorem for subsequent work in concave programming is that if ¯ 0 at the point at which the function is optimized, the constraint must be nonbinding. Conversely, if the constraint is nonbinding, ¯ 0.
EXAMPLE 7.
Assume one wishes to maximize the function z(x, y; a, b) subject to f(x, y; a, b) The Lagrangian function is Z(x,y, ; a, b) z(x, y; a, b) f(x, y; a, b) and the first-order conditions are Zx zx(x ¯, y ¯; a, b) ¯fx(x ¯, y ¯; a, b) 0 Zy zy(x ¯, y ¯; a, b) ¯fy(x ¯, y ¯; a, b) 0 Z f(x ¯, y ¯; a, b) 0 If we assume all the functions have continuous first- and second-order derivatives and if J zxx ¯fxx zyx ¯fyx fx zxy ¯fxy zyy ¯fyy fy fx fy 0 0 291 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] we know from the implicit function theorem that we can express the optimal values of the endogenous variables as functions of the exogenous variables.
Z(x ¯, y ¯, ¯; a, b) z[x ¯(a, b), y ¯(a, b); a, b] ¯(a, b)f[x ¯(a, b), y ¯(a, b); a, b] The objective function, when evaluated at the values of the optimal solution, is known as the indirect objective function.
z ¯(a, b) z[x ¯(a, b), y ¯(a, b); a, b] z(a, b) The envelope theorem states that the partial derivative of the indirect objective function with respect to any one of the exogenous variables, say b, equals the partial derivative of the Lagrangian function with respect to the same exogenous variable. To prove the envelope theorem, then, we need to show z ¯ b Z b Making use of the chain rule to take the derivative of the indirect objective function, and recalling that it is always evaluated at the optimal solution, we have z ¯ b zx x ¯ b zy y ¯ b zb Then substituting zx ¯fx, zy ¯fy from the first two first-order conditions, z ¯ b ¯ fx x ¯ b fy y ¯ b zb (13.9) Also from the third first-order condition, we know f[x ¯(a, b), y ¯(a, b), a, b] 0 Taking the derivative with respect to b and rearranging, fx x ¯ b fy y ¯ b fb Then substituting fb in (13.9) and rearranging, z ¯ b (zb ¯fb) Z b Q.E.D.
The derivative of the Lagrangian function with respect to a specific exogenous variable, when evaluated at the optimal values of the problem, is a reliable measure of the effect of that exogenous variable on the optimal value of the objective function.
EXAMPLE 8.
Assume a utility maximization problem subject to a budget constraint: maximize u(x, y) subject to pxx pyy B If there are continuous first and second derivatives and the Jacobian determinant consisting of the derivatives of the first-order conditions with respect to the endogenous variables does not equal zero (or vanish), then the Lagrangian function can be written U(x ¯, y ¯, ¯; px, py, B) u(x ¯, y ¯) ¯(B pxx ¯ pyy ¯) and the indirect objective function is V(x ¯, y ¯; px, py, B) u[x ¯(px, py, B), y ¯(px, py, B)] 292 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 Then using the envelope theorem to estimate the effect on the optimal value of the objective function of a change in any of the three exogenous variables, we have a) V px U px ¯x ¯ b) V py U py ¯y ¯ c) V B U B ¯ In (c), ¯ can be called the marginal utility of money, i.e., the extra utility the consumer would derive from a small change in his or her budget or income. Notice that with the budget constraint (B) appearing only in the constraint, the derivative comes more easily from the Lagrangian function. In (a) and (b), assuming positive utility for income from (c), a change in the price of the good will have a negative impact weighted by the quantity of the good consumed when utility is being maximized. In both cases, with prices appearing only in the constraint, the derivatives once again come more readily from the Lagrangian function. See also Problems 13.30 to 13.32.
13.7 CONCA VE PROGRAMMING AND INEQUALITY CONSTRAINTS In the classical method for constrained optimization seen thus far, the constraints have always been strict equalities. Some economic problems call for weak inequality constraints, however, as when individuals want to maximize utility subject to spending not more than x dollars or business seeks to minimize costs subject to producing no less than x units of output. Concave programming, so called because the objective and constraint functions are all assumed to be concave, is a form of nonlinear programming designed to optimize functions subject to inequality constraints. Convex functions are by no means excluded, however, because the negative of a convex function is concave. Typically set up in the format of a maximization problem, concave programming can nevertheless also minimize a function by maximizing the negative of that function.
Given an optimization problem subject to an inequality constraint with the following differenti-able concave objective and constraint functions, maximize f(x1, x2) subject to g(x1, x2) 0 x1, x2 0 and the corresponding Lagrangian function, F(x1, x2, ) f(x1, x2) g(x1, x2) the first-order necessary and sufficient conditions for maximization, called the Kuhn-Tucker conditions, are 1.
a) F xi fi(x ¯1, x ¯2) ¯gi(x ¯1, x ¯2) 0 b) xi 0 c) x ¯i F xi 0, i 1, 2 2.
a) F g(x ¯1, x ¯2) 0 b) ¯ 0 c) ¯ F 0 where the conditions in (c) are called the complementary-slackness conditions, meaning that both x ¯ and f(x ¯) cannot simultaneously both be nonzero. Since a linear function is concave and 293 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] convex, though not strictly concave or strictly convex, a concave-programming problem consisting solely of linear functions that meet the Kuhn-Tucker conditions will always satisfy the necessary and sufficient conditions for a maximum.
Note: (1) condition 1(a) requires that the Lagrangian function be maximized with respect to x1 and x2, while condition 2(a) calls for the Lagrangian to be minimized with respect to . This means concave programming is designed to seek out a saddle point in the Lagrangian function in order to optimize the objective function subject to an inequality constraint.
(2) in the Kuhn-Tucker conditions the constraint is always expressed as greater than or equal to zero. This means that unlike equality constraints set equal to zero, where it makes no difference whether you subtract the constant from the variables in the constraint or the variables from the constant, the order of subtraction is important in concave programming (see Problem 13.33).
The rationale for the tripartite conditions (a)–(c) can be found in Example 9 and a demonstration of the basic maximization method is offered in Example 10. For minimization, see Problem 13.34. For multiple constraints, see Problem 13.39. For other applications, see Problems 13.33 to 13.42.
EXAMPLE 9.
Consider a single-variable function for which we seek a local maximum in the first quadrant where x 0. Three scenarios are possible, each with slightly different conditions, as seen in Fig. 13-1.
a) For the maximum at F, an interior solution, f(x) 0 and x 0 b) For the maximum at G, a boundary solution, f(x) 0 and x 0 c) For the maximum at H or J, both boundary solutions, f(x) 0 and x 0 All the possibilities for a maximum in the first quadrant can be summarized more succinctly, however, as f(x) 0 x 0 and xf(x) 0 which we recognize as part of the Kuhn-Tucker conditions. Note that the conditions automatically exclude a point like K in (a) which is not a maximum, because f(K) 0.
EXAMPLE 10.
A consumer wishing to maximize utility while spending no more than a predetermined budget faces the following concave-programming problem, maximize u(x, y) subject to B pxx pyy 0 x, y 0 294 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 Fig. 13-1 K f(x) F x (a) f(x) G x (b) J f(x) H x (c) f(x) (a) (b) (c) and Lagrangian function, U u(x, y) (B pxx pyy) Using the Kuhn-Tucker conditions, the Lagrangian function is first maximized with respect to the choice variables x and y and the related conditions checked.
1.
a) U x ux ¯px 0 U y uy ¯py 0 b) x ¯ 0 y ¯ 0 c) x ¯(ux ¯px) 0 y ¯(uy ¯py) 0 Then the Lagrangian is minimized with respect to the constraint variable and the related conditions checked.
2.
a) U B pxx ¯ pyy ¯ 0 b) ¯ 0 c) ¯(B pxx ¯ pyy ¯) 0 This leaves three categories of solutions that are nontrivial: (a) x ¯, y ¯ 0, (b) x ¯ 0, y ¯ 0, and (c) x ¯ 0, y ¯ 0. We deal with the first two below and leave the third to you as a private exercise.
a) First scenario.
If x ¯, y ¯ 0, then from 1(c), ux ¯px 0 uy ¯py 0 Therefore, ¯ ux px ¯ uy py (13.10) With px, py 0, and assuming nonsatiation of the consumer, i.e., ux, uy 0, ¯ 0 If ¯ 0, then from 2(c), B pxx ¯ pyy ¯ 0 and the budget constraint holds as an exact equality, not a weak inequality. This means the optimal point, x ¯, y ¯, will lie somewhere on the budget line, and not below it.
By reconfiguring (13.10), we can also see ux uy px py Since ux/uy the slope of the indifference curve and px/py the slope of the budget line, whenever both x ¯, y ¯ 0, the indifference curve will be tangent to the budget line at the point of optimization and we have an interior solution. With the budget constraint functioning as an exact equality, this first scenario, in which both x ¯, y ¯ 0, exactly parallels the classical constrained optimization problem, as seen in Fig.
13-2(a).
b) Second scenario.
If x ¯ 0, y ¯ 0, then from 1(c), ux ¯px 0 uy ¯py 0 and ux px ¯ ¯ uy py (13.11) Assuming px, py, ux, uy 0, then ¯ 0. From 2(c), therefore, the budget constraint holds as an exact equality, not a weak inequality, even though only one variable is greater than zero and the other equals zero. This means that once again the optimal point, x ¯, y ¯, will lie on the budget line, and not below it.
295 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] Substituting ¯ uy/py from the equality on the right in (13.11) for ¯ in the inequality on the left in (13.11), we have ux px uy py or ux uy px py This means that along the budget line the indifference curves are everywhere flatter than the budget line, leading to a corner solution in the upper left, as seen in Fig. 13-2(b). At the corner solution, the slope of the highest indifference curve that just touches the budget line may be flatter than or equal to the slope of the budget line.
Solved Problems COMPARATIVE STATICS WITH ONE EXOGENOUS VARIABLE 13.1.
Given the model from Example 1, QD m nP kY m, n, k 0 QS a bP a, b 0 and now assuming we know the value of income and the parameters where Y 100, m 60, n 2, k 0.1, a 10, and b 0.5 (a) find the equilibrium price and quantity, (b) use comparative statics to estimate the effect on the equilibrium price P of a $1 change in income, and (c) confirm the comparative statics results by reestimating the equilibrium price.
a) QD QS 60 2P 0.1(100) 2.5P P 10 0.5P 60 24 Q 22 b) From (13.3), dP dY k b n 0 Substituting k 0.1, b 0.5, n 2, dP dY 0.1 0.5 2 0.04 296 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 Fig. 13-2 (a) y y x x 0 (b) y x 0 (0, y) A $1 increase in income can be expected to cause a 4¢ increase in the equilibrium price of the commodity.
c) Reestimating the equilibrium equation for Y 101, 60 2P 0.1(101) 2.5P P 10 0.5P 60.1 24.04 Q.E.D.
13.2.
Assume QD m nP cPc sPs m, n, c, s 0 QS a bP iPi a, b, i 0 where P price of the good, Pc price of a complement, Ps price of a substitute, and Pi price of an input. (a) Find the equilibrium price P. (b) Find the comparative-static derivatives resulting from a change in Pc and Pi. (c) Find the implicit function for the equilibrium condition and (d) from it derive the comparative-static derivative for a change in Ps.
a) The equilibrium condition is m nP cPc sPs a bP iPi (13.12) m a cPc sPs iPi (b n)P P m a cPc sPs iPi b n b) Whenever there is more than one independent variable, we have to work in terms of partial derivatives which will keep the other independent variables constant.
P Pc c b n P Pi i b n c) From (13.12), m nP cPc sPs a bP iPi 0 d) P Ps FPs FP s b n 13.3.
Assume a two-sector income determination model where consumption depends on income and investment is autonomous, so that C bY, I I0, 0 b 1 and equilibrium occurs where Y C I. (a) Solve for the equilibrium level of income Y explicitly. (b) Use comparative statics to estimate the effect on Y of a change in autonomous investment I0. (c) Find the same comparative-static derivative from the implicit function. (d) Evaluate the effect on Y of a change in the marginal propensity to consume b explicitly and (e) implicitly.
a) Y C I Substituting, Y bY I0 (13.13) Y bY I0 Y I0 1 b (13.14) 297 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] b) The effect on Y of a change in autonomous investment I0 is dY dI0 1 1 b 0 since 0 b 1.
c) Moving everything to the left in (13.13), we obtain the implicit function, Y bY I0 0 (13.15) From the implicit function rule, always under the usual assumptions, dY dI0 FI0 FY where FI0 1 and FY 1 b. So, dY dI0 1 1 b 1 1 b 0 d) If we treat the marginal propensity to consume b as an exogenous variable instead of as a parameter, we have a function of more than one independent variable and must find the partial derivative, which will hold I0 constant. First applying the quotient rule on the explicit function in (13.14), Y b I0 (1 b)2 Substituting from (13.14) where Y I0/(1 b), we have Y b Y (1 b) e) Next using the implicit function rule on (13.15) when evaluated at Y, Y b Fb FY Y 1 b Though seemingly difficult at first, implicit functions are frequently faster and easier to work with in the long run. See Problems 13.5 to 13.6.
13.4.
From (a) the explicit and (b) the implicit function, find the effect on the profit-maximizing level of output of a per-unit tax t placed on a monopoly with total revenue and total cost functions: TR mQ nQ2, TC kQ m, n, k 0 a) Profit for the monopoly is mQ nQ2 kQ tQ d dQ m 2nQ k t 0 (13.16) Q m k t 2n profit-maximizing level of output Then from the explicit function above, the comparative-static derivative estimating the effect on Q of a change in the per-unit tax t is dQ dt 1 2n 0 298 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 b) From (13.16), the implicit function for the optimum condition is m 2nQ k t 0 By the implicit function rule, dQ dt Ft FQ 1 2n 0 13.5.
Assume a two-sector income determination model expressed in general functions: C C(Y), I I0 with equilibrium when Y C I. (a) Determine the implicit function for the equilibrium level of income Y. (b) Estimate the effect on Y of a change in autonomous investment I0.
a) Y C(Y) I0 0 b) dY dI0 FI0 FY 1 1 CY 13.6.
In the model of Problem 6.4 we found the equilibrium condition and explicit function for the equilibrium level of income Y ¯ derived from it were, respectively, Y C0 bY bT0 btY I0 G0 (13.17) and Y ¯ 1 1 b bt (C0 bT0 I0 G0) We also found, after considerable simplification, the comparative-static derivative for the effect on the equilibrium level of income Y ¯ of a change in the tax rate t was Y ¯ t bY ¯ 1 b bt (a) Find the implicit function for the equilibrium condition in (13.17) and (b) from it derive the same derivative Y ¯/t to convince yourself of the convenience of working with implicit functions.
a) From (13.17), the implicit function for the equilibrium condition is Y ¯ C0 bY ¯ bT0 btY ¯ I0 G0 0 b) From the implicit function rule, Y ¯ t Ft FY ¯ bY ¯ 1 b bt 13.7.
In Problems 6.7 and 6.8 we found the equilibrium condition and explicit function for the equilibrium level of income derived from it were, respectively, Y C0 b(Y T0 tY) I0 G0 X0 Z0 z(Y T0 tY) (13.18) Y ¯ 1 1 b bt z zt (C0 bT0 I0 G0 X0 Z0 zT0) 299 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] We also found, after considerable simplification, the comparative-static derivative for the effect on Y ¯ of a change in the marginal propensity to import z was Y ¯ z Y ¯ d 1 b bt z zt (a) Find the implicit function for the equilibrium condition in (13.18) and (b) from it derive the same derivative Y ¯/z to see once again the convenience of working with implicit functions.
a) From (13.18), the implicit function for the equilibrium condition is Y ¯ C0 b(Y ¯ T0 tY ¯) I0 G0 X0 Z0 z(Y ¯ T0 tY ¯) b) From the implicit function rule, Y ¯ z Fz FY – Y ¯ T0 tY ¯ 1 b bt z zt Y ¯ d 1 b bt z zt COMPARATIVE STATICS WITH MORE THAN ONE ENDOGENOUS VARIABLE 13.8.
Set forth the implicit function theorem.
Given the set of simultaneous equations, f 1(y1, y2, . . ., yn; x1, x2, . . ., xm) 0 f 2(y1,y2, . . ., yn; x1, x2, . . ., xm) 0 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · f n(y1, y2, . . ., yn; x1, x2, . . ., xm) 0 if all the above functions in f have continuous partial derivatives with respect to all the x and y variables, and if at a point (y10, y20, . . ., yn0; x10, x20, . . ., xm0), the Jacobian consisting of the partial derivatives of all the functions fi with respect to all of the dependent variables yi is nonzero, as indicated below: J f 1 y1 f 2 y1 · · · f n y1 f 1 y2 f 2 y2 · · · f n y2 · · · · · · · · · · · · f 1 yn f 2 yn · · · f n yn 0 then there exists an m-dimensional neighborhood N in which the variables y1, y2, . . ., yn are implicit functions of the variables x1, x2, . . ., xm in the form y10 f 1(x10, x20, . . ., xm0) y20 f 2(x10, x20, . . ., xm0) · · · · · · · · · · · · · · · · · · · · · · · yn0 f n(x10, x20, . . ., xm0) The implicit functions, f 1, f 2, . . ., f n, are continuous and have continuous partial derivatives with respect to all the independent variables. Derivation of the comparative-static derivatives is explained in Example 3.
13.9.
Working with the model from Example 3 in which F 1(y1, y2; x1, x2) 0 F 2(y1, y2; x1, x2) 0 find the comparative-static partial derivatives of the system with respect to x2.
300 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 Taking the total derivative of both functions with respect to x2, F 1 y1 · y1 x2 F 1 y2 · y2 x2 F 1 x2 0 F 2 y1 · y1 x2 F 2 y2 · y2 x2 F 2 x2 0 Moving the partials of the functions with respect to x2 to the right, we have F 1 y1 F 2 y1 F 1 y2 F 2 y2 y1 x2 y2 x2 F 1 x2 F 2 x2 JX B where J F 1 y1 · F 2 y2 F 2 y1 · F 1 y2 0 To solve for the first derivative, y1/x2, form a new matrix J1 , replacing the first column of J with the column vector B and substitute in (13.6).
y1 x2 J1 J F 1 x2 F 2 x2 F 1 y2 F 2 y2 F 1 x2 · F 2 y2 F 2 x2 · F 1 y2 F 1 y1 F 2 y1 F 1 y2 F 2 y2 F 1 y1 · F 2 y2 · F 2 y1 · F 1 y2 Similarly, y2 x2 J2 J F 1 y1 F 2 y1 F 1 x2 F 2 x2 F 1 y1 · F 2 x2 F 2 y1 · F 1 x2 F 1 y1 F 2 y1 F 1 y2 F 2 y2 F 1 y1 · F 2 y2 · F 2 y1 · F 1 y2 13.10. Assume the model from Example 4, Y C0 C(Y, i) 0 0 CY 1, Ci 0 L(Y, i) M0/P 0 LY 0, Li 0 and use comparative-static analysis to find the effect on the equilibrium levels of Y and i of a change in the money supply M0, recalling that P is constant.
Taking the total derivatives with respect to M0, Y M0 CY · Y M0 Ci · i M0 0 LY · Y M0 Li · i M0 1 P 0 301 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] and setting them in matrix form, 1 CY LY Ci Li Y ¯ M0 i ¯ M0 0 1/P JX B where J (1 CY)Li CiLY J ()() ()() () 0 We then solve for the first derivative, Y ¯/M0 by forming the new matrix J1 . J1 0 1/P Ci Li Ci P and Y ¯ M0 J1 J Ci P[(1 CY)Li CiLY] () () 0 An increase in the money supply M0 will cause the equilibrium level of income to increase. For i ¯/M0, J2 1 CY LY 0 1/P 1 CY P and i ¯ M0 J2 J 1 CY P[(1 CY)Li CiLY] () () 0 An increase in M0 will cause the equilibrium interest rate to fall.
13.11. Working with a system of three simultaneous equations similar to Problem 6.4, Y C I0 G0 C C0 b(Y T) T T0 tY (a) Express the system of equations as both general and specific implicit functions. (b) Use the simultaneous equations approach to comparative statics to find the Jacobian for both the general functions and the specific functions. (c) Express in matrix form the total derivatives of both the general and the specific functions with respect to G0. Then find and sign (d) Y ¯/G0, (e) C ¯/G0, and (f) T ¯/G0.
a) F 1(Y, C, T; C0, I0, G0, T0, b, t) Y C I0 G0 0 F 2(Y, C, T; C0, I0, G0, T0, b, t) C C0 b(Y T) 0 F 3(Y, C, T; C0, I0, G0, T0, b, t) T T0 tY 0 b) The Jacobian consists of the partial derivatives of all the equations with respect to all the endogenous or dependent variables. J F 1 Y F 2 Y F 3 Y F 1 C F 2 C F 3 C F 1 T F 2 T F 3 T 1 b t 1 1 0 0 b 1 Expanding along the first row, J 1(1) (1)(b bt) 1 b bt 0 302 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 c) Taking the total derivatives of the general functions with respect to G0, setting them in the now familiar matrix form, and using bars over the endogenous variables to indicate that they are to be evaluated at the point of equilibrium in the model, we have F 1 Y F 2 Y F 3 Y F 1 C F 2 C F 3 C F 1 T F 2 T F 3 T Y ¯ G0 C ¯ G0 T ¯ G0 F 1 G0 F 2 G0 F 3 G0 The same derivatives in terms of the specific functions are 1 b t 1 1 0 0 b 1 Y ¯ G0 C ¯ G0 T ¯ G0 1 0 0 (13.19) JX B where the signs in the B matrix change as the matrix is moved to the right of the equal sign.
d) To find Y ¯/G0, the a11 element in the X matrix in (13.19), we create a new matrix J1 by substituting B in the first column of J and using (13.6).
Y ¯ G0 J1 J J 1 0 0 1 1 0 0 b 1 1 1 b bt 0 e) To find C ¯/G0, the a21 element in the X matrix in (13.19), we create J2 by substituting B in the second column of J and using (13.6).
C ¯ G0 J2 J J 1 b t 1 0 0 0 b 1 b(1 t) 1 b bt 0 f) To find T ¯/G0, the a31 element in the X matrix in (13.19), we create J3 by substituting B in the third column of J and using (13.6).
T ¯ G0 J3 J J 1 b t 1 1 0 1 0 0 1 1 b bt 0 13.12. Using the previous model in Problem 13.11, where the original Jacobian J will remain the same, (a) express in matrix form the total derivatives of both the general and the specific functions with respect to T0. Then find and sign (b) Y ¯/T0, (c) C ¯/T0, and (d) T ¯/T0.
303 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] a) In matrix form, the total derivatives of the general functions with respect to T0 are F 1 Y F 2 Y F 3 Y F 1 C F 2 C F 3 C F 1 T F 2 T F 3 T Y ¯ T0 C ¯ T0 T ¯ T0 F 1 T0 F 2 T0 F 3 T0 The same derivatives in terms of the specific function are 1 b t 1 1 0 0 b 1 Y ¯ T0 C ¯ T0 T ¯ T0 0 0 1 b) To find Y ¯/T0, the a11 element in the X matrix, create J1 by substituting B in the first column of J and using (13.6).
Y ¯ T0 J1 J J 0 0 1 1 1 0 0 b 1 b 1 b bt 0 c) For C ¯/T0, the a21 element in the X matrix, create J2 by substituting B in the second column of J.
C ¯ T0 J2 J J 1 b t 0 0 1 0 b 1 b 1 b bt 0 d) For T ¯/T0, substitute B in the third column of J and use (13.6).
T ¯ T0 J3 J J 1 b t 1 1 0 0 0 1 1 b 1 b bt 0 13.13. Retaining the same model from Problem 13.11, (a) express in matrix form the total derivatives of both the general and the specific functions with respect to the tax rate t. Then find and sign (b) Y ¯/t, (c) C ¯/t, and (d) T ¯/t.
a) The total derivatives of the general functions with respect to t are F 1 Y F 2 Y F 3 Y F 1 C F 2 C F 3 C F 1 T F 2 T F 3 T Y ¯ t C ¯ t T ¯ t F 1 t F 2 t F 3 t 304 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 The same derivatives in terms of the specific functions are 1 b t 1 1 0 0 b 1 Y ¯ t C ¯ t T ¯ t 0 0 Y ¯ b) For Y ¯/t, Y ¯ t J1 J J 0 0 Y ¯ 1 1 0 0 b 1 bY ¯ 1 b bt 0 c) For C ¯/t, C ¯ t J2 J J 1 b t 0 0 Y ¯ 0 b 1 bY ¯ 1 b bt 0 d) For T ¯/t, T ¯ t J3 J J 1 b t 1 1 0 0 0 Y ¯ (1 b)Y ¯ 1 b bt 0 13.14. Given the income determination model Y C I0 G0 X0 Z C C0 bY Z Z0 zY where X exports, Z imports, and a zero subscript indicates an exogenously fixed variable, (a) express the system of equations as both general and specific implicit functions. (b) Express in matrix form the total derivatives of both the general and the specific functions with respect to exports X0. Then find and sign (c), Y ¯/X0, (d) C ¯/X0, and (e) Z ¯/X0.
a) F1(Y, C, Z; C0, I0, G0, X0, Z0, b, z) Y C I0 G0 X0 Z 0 F 2(Y, C, Z; C0, I0 G0, X0, Z0, b, z) C C0 bY 0 F 3(Y, C, Z; C0, I0, G0, X0, Z0, b, z) Z Z0 zY 0 b) F 1 Y F 2 Y F 3 Y F 1 C F 2 C F 3 C F 1 Z F 2 Z F 3 Z Y ¯ X0 C ¯ X0 Z ¯ X0 F 1 X0 F 2 X0 F 3 X0 1 b Z 1 1 0 1 0 1 Y ¯ X0 C ¯ X0 Z ¯ X0 1 0 0 J 1 b z 0 305 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] c) For Y ¯/X0, Y ¯ X0 J1 J J 1 0 0 1 1 0 1 0 1 1 1 b z 0 d) For C ¯/X0, C ¯ X0 J2 J J 1 b z 1 0 0 1 0 1 b 1 b z 0 e) For Z ¯/X0, Z ¯ X0 J3 J J 1 b z 1 1 0 1 0 0 z 1 b z 0 13.15. Using the model from Problem 13.14, (a) express in matrix form the total derivatives of the specific functions with respect to the marginal propensity to consume b. Then find and sign (b) Y ¯/b, (c) C ¯/b, and (d) Z ¯/b.
a) 1 b z 1 1 0 1 0 1 Y ¯ b C ¯ b Z ¯ b 0 Y ¯ 0 b) Y ¯ b J1 J J 0 Y ¯ 0 1 1 0 1 0 1 Y ¯ 1 b z 0 c) C ¯ b J2 J J 1 b z 0 Y ¯ 0 1 0 1 (1 z)Y ¯ 1 b z 0 d) Z ¯ b J3 J J 1 b z 1 1 0 0 Y ¯ 0 zY ¯ 1 b z 0 13.16. Continuing with the model from Problem 13.14, (a) express in matrix form the total derivatives of the specific functions with respect to the marginal propensity to import z. Then find and sign (b) Y ¯/z, (c) C ¯/z, and (d) Z ¯/z.
a) 1 b z 1 1 0 1 0 1 Y ¯ z C ¯ z Z ¯ z 0 0 Y ¯ 306 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 b) For Y ¯/z, Y ¯ z J1 J J 0 0 Y ¯ 1 1 0 1 0 1 Y ¯ 1 b z 0 c) For C ¯/z, C ¯ z J2 J J 1 b z 0 0 Y ¯ 1 0 1 bY ¯ 1 b z 0 d) For Z ¯/z, Z ¯ z J3 J J 1 b z 1 1 0 0 0 Y ¯ (1 b)Y ¯ 1 b z 0 13.17. Having introduced the foreign trade market to the goods market in our national income model, let us now combine them with the money market. Assume The goods market: I S I(i) S(Y, i) (Ii 0) (0 SY 1; Si 0) The foreign trade market: Z X Z(Y, i) X0 0 ZY 1; Zi 0 and the money market: MD MS L(Y, i) M0 LY 0, Li 0 where Z imports, S savings, X0 autonomous exports, MD demand for money, MS supply of money, and all the other symbols are familiar. (a) Express the equilibrium conditions for the combined goods market and the money market. (b) Express the equilibrium conditions as general and specific implicit functions. (c) Express in matrix form the total derivatives of these functions with respect to M0. Find and sign (d) the Jacobian, (e) Y ¯/M0, and (f) i ¯/M0.
a) The combined goods market is in equilibrium when injections equal leakages: I(i) X0 S(Y, i) Z(Y, i) The money market is in equilibrium when the demand for money equals the money supply: L(Y, i) M0 b) F 1(Y, i; M0, X0) I(i) X0 S(Y, i) Z(Y, i) F 2(Y, i; M0, X0) L(Y, i) M0 c) F 1 Y F 2 Y F 1 i F 2 i Y ¯ M0 i ¯ M0 F 1 M0 F 2 M0 SY ZY LY Ii Si Zi Li Y ¯ M0 i ¯ M0 0 1 JX B 307 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] d) J Li(SY ZY) LY(Ii Si Zi) 0 e) Using Cramer’s rule for Y ¯/M0, Y ¯ M0 J1 J J 0 1 Ii Si Zi Li (Ii Si Zi) Li(SY ZY) LY(Ii Si Zi) 0 f) For i ¯/M0, i ¯ M0 J2 J J SY ZY LY 0 1 (SY ZY) Li(SY ZY) LY(Ii Si Zi) 0 13.18. Using the model in Problem 13.17, (a) express in matrix form the total derivatives of the specific functions with respect to exports X0. Then find and sign (b) Y ¯/X0 and (c) i ¯/X0.
a) SY ZY LY Ii Si Zi Li Y ¯ X0 i ¯ X0 1 0 b) For Y ¯/X0, Y ¯ X0 J1 J J 1 0 Ii Si Zi Li Li Li(SY ZY) LY(Ii Si Zi) 0 c) For i ¯/X0, i ¯ X0 J2 J J SY ZY LY 1 0 LY Li(SY ZY) LY(Ii Si Zi) 0 COMPARATIVE STATIC ANALYSIS IN OPTIMIZATION PROBLEMS 13.19. Returning to the model in Example 5, where the first-order conditions were F 1(K, L; r, w, P) PQK(K ¯ , L ¯) r 0 F 2(K, L; r, w, P) PQL(K ¯ , L ¯) w 0 (a) express the total derivatives of the functions with respect to the wage w in matrix form. Then find and sign (b) K ¯ /w and (c) L ¯/w.
a) PQKK PQLK PQKL PQLL K ¯ w L ¯ w 0 1 J P2(QKKQLL QLKQKL) 0 b) For K ¯ /w, K ¯ w J1 J J 0 1 PQKL PQLL PQKL P2(QKKQLL QKLQLK) Without specific knowledge of the sign of the cross partial QKL, it is impossible to sign the derivative.
Assuming the marginal productivity of capital will increase for an increase in labor, QKL 0 and 308 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 K ¯ /w 0, meaning that the optimal level of capital will likely fall in response to an increase in wage.
c) For L ¯/w, L ¯ w J2 J J PQKK PQLK 0 1 PQKK P2(QKKQLL QKLQLK) 0 The optimal level of labor will decrease for an increase in wage since QKK 0.
13.20. Staying with the model in Example 5, where the first-order conditions were F 1(K, L; r, w, P) PQK(K ¯ , L ¯) r 0 F 2(K, L; r, w, P) PQL(K ¯ , L ¯) w 0 (a) express the total derivatives of the functions with respect to the commodity price P in matrix form. Then find and sign (b) K ¯ /P and (c) L ¯/P.
a) PQKK PQLK PQKL PQLL K ¯ P L ¯ P QK QL b) For K ¯ /P, K ¯ P J1 J J QK QL PQKL PQLL P(QKQLL QLQKL) P2(QKKQLL QKLQLK) (QLQKL QKQLL) P(QKKQLL QKLQLK) Since QK MPK 0, QL MPL 0, QLL 0 for maximization, and J 0 in the denominator, the sign depends completely on the cross partial QKL. If K and L are complements, so that an increased use of one input will lead to an increase in the MP of the other input, QKL 0 and the comparative-static derivative K ¯ /P 0. If QKL 0, the sign of K ¯ /P is indeterminate.
c) For L ¯/P, L ¯ P J2 J J PQKK PQLK QK QL P(QLQKK QKQLK) P2(QKKQLL QKLQLK) (QKQLK QLQKK) P(QKKQLL QKLQLK) and the sign will depend on the cross partial QLK, exactly as in (b).
13.21. Assume now a firm seeks to optimize the discounted value of its profit function, P0Q(X, Y)ert PxX PyY where the first derivatives of the first-order conditions, /X and /Y, when expressed as implicit functions, are F 1(X, Y; P0, Px, Py, r, t) P0Qx(X ¯ , Y ¯)ert Px 0 F 2(X, Y; P0, Px, Py, r, t) P0Qy(X ¯ , Y ¯)ert Py 0 (a) Express the total derivatives of the functions with respect to P0 in matrix form, recalling that r and t are constants. Then find and sign (b) the Jacobian, (c) X ¯ /P0 and (d) Y ¯/P0.
a) P0Qxxert P0Qyxert P0Qxyert P0Qyyert X ¯ P0 Y ¯ P0 Qxert Qyert 309 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] b) J P0 2e2rt(QxxQyy QyxQxy) 0 With P0 2e2rt 0 and (QxxQyy QyxQxy) 0 from the second-order sufficient condition, J 0.
c) For X ¯ /P0, X ¯ P0 J1 J J Qxert Qyert P0Qxyert P0Qyyert P0e2rt(QyQxy QxQyy) P0 2e2rt(QxxQyy QyxQxy) (QyQxy QxQyy) P0(QxxQyy QyxQxy) As in Problem 13.20(b), the sign will depend on the cross partial Qxy.
d) For Y ¯/P0, Y ¯ P0 J2 J J P0Qxxert P0Qyxert Qxert Qyert P0e2rt(QxQyx QyQxx) P0 2e2rt(QxxQyy QyxQxy) (QxQyx QyQxx) P0(QxxQyy QyxQxy) 13.22. Using the same model in Problem 13.21, (a) express the total derivatives of the functions with respect to time t in matrix form. Then find and sign (b) X ¯ /t and (c) Y ¯/t.
a) P0Qxxert P0Qyxert P0Qxyert P0Qyyert X ¯ t Y ¯ t rP0Qxert rP0Qyert b) X ¯ t J1 J J rP0Qxert rP0Qyert P0Qxyert P0Qyyert rP0 2e2rt(QxQyy QyQxy) P0 2e2rt(QxxQyy QyxQxy) r(QxQyy QyQxy) (QxxQyy QyxQxy) c) Y ¯ t J2 J J P0Qxxert P0Qyxert rP0Qxert rP0Qyert rP0 2e2rt(QyQxx QxQyx) P0 2e2rt(QxxQyy QyxQxy) r(QyQxx QxQyx) (QxxQyy QyxQxy) In both cases, if the cross partials are positive, the comparative-static derivatives will be negative and if the cross partials are negative, the comparative-static derivatives will be indeterminate.
13.23. Assume the production function in Example 5 is specified as a Cobb-Douglas function with decreasing returns to scale (Section 6.9), so that the competitive firm’s profit function is PAK L rK wL and the first derivatives /K and /L from the first-order conditions are F 1(K, L; r, w, P, A, , ) PAK 1L r 0 F 2(K, L; r, w, P, A, , ) PAK L1 w 0 (a) Express the total derivatives of the functions with respect to the wage w in matrix form.
Then find and sign (b) the Jacobian, (c) K ¯ /w, and (d) L ¯/w.
a) ( 1)PAK 2L PAK 1L1 PAK 1L1 ( 1)PAK L2 K ¯ w L ¯ w 0 1 b) J ( 1)PAK 2L · ( 1)PAK L2 (PAK 1 L1)2 J (1 )P2A2K 22L22 0 310 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 since J H and H H2 0 in unconstrained optimization problems. This condition implies that a profit-maximizing firm in perfect competition operates under decreasing returns to scale since J 0 requires ( ) 1.
c) For K ¯ /w, K ¯ w J1 J J 0 1 PAK 1L1 ( 1)PAK L2 PAK 1L1 J 0 since the numerator, independent of the negative sign, is unambiguously positive and the denominator is positive, the comparative-static derivative K ¯ /w is unquestionably negative. An increase in the wage will decrease the demand for capital. Through further simplification, if desired, we can also see K ¯ w PAK 1L1 (1 )P2A2K 22L22 KL (1 )TR 0 d) For L ¯/w, L ¯ w J2 J J ( 1)PAK 2L PAK 1L1 0 1 ( 1)PAK 2L J 0 since 1, making ( 1) 0. An increase in the wage will lead to a reduction in the optimal level of labor used. Through further simplification, we can also see L ¯ w ( 1)PAK 2L (1 )P2A2K 22L22 (1 )L2 (1 )TR 0 13.24. Working with the same model in Problem 13.23, (a) express the total derivatives of the functions with respect to output price P in matrix form. Then find and sign (b) K ¯ /P and (c) L ¯/P.
a) ( 1)PAK 2L PAK 1L1 PAK 1L1 ( 1)PAK L2 K ¯ P L ¯ P AK 1L AK L1 b) For K ¯ /P, K ¯ P J1 J J AK 1L AK L1 PAK 1L1 ( 1)PAK L2 PA2K 21L22 (1 )P2A2K 22L22 K ¯ P K (1 )P 0 c) For L ¯/P, L ¯ P J2 J J ( 1)PAK 1L PAK 1L1 AK 1L AK L1 PA2K 22L21 (1 )P2A2K 22L22 L ¯ P L (1 )P 0 311 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] COMPARATIVE STATICS IN CONSTRAINED OPTIMIZATION 13.25. A consumer wants to maximize utility u(a, b) subject to the constraint paa pbb Y, a constant. Given the Lagrangian function U u(a, b) (Y paa pbb) and assuming the second-order sufficient condition is met so that H ¯ J 0, the endogenous variables in the first-order conditions can be expressed as implicit functions of the exogenous variables, such that F 1(a, b, ; pa, pb, Y) Ua pa 0 F 2(a, b, ; pa, pb, Y) Ub pb 0 (13.20) F 3(a, b, ; pa, pb, Y) Y paa pbb 0 (a) Express the total derivatives of the functions with respect to pa in matrix form and (b) find a ¯/pa.
a) Uaa Uba pa Uab Ubb pb pa pb 0 a ¯ pa b ¯ pa ¯ pa ¯ 0 a ¯ J pa 2Ubb papbUab papbUba pb 2Uaa 0 since J H ¯ 0 from the second-order sufficient condition for constrained maximization. But theory leaves unspecified the signs of the individual second partials.
b) For a ¯/pa, a ¯ pa J1 J J ¯ 0 a ¯ Uab Ubb pb pa pb 0 a ¯(paUbb pbUab) ¯pb 2 J (13.21) where the sign is indeterminate because the signs of the second partials are unknown.
13.26. Working with the same model in Problem 13.25, (a) express the total derivatives of the functions with respect to pb in matrix form and (b) find b ¯/pb.
a) Uaa Uba pa Uab Ubb pb pa pb 0 a ¯ pb b ¯ pb ¯ pb 0 ¯ b ¯ b) For b ¯/pb, b ¯ pb J2 J J Uaa Uba pa 0 ¯ b ¯ pa pb 0 b ¯(pbUaa paUba) ¯pa 2 J (13.22) which is also indeterminate.
312 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 13.27. Continuing with the same model in Problem 13.25, (a) express the total derivatives of the functions with respect to Y in matrix form and find (b) a ¯/Y and (c) b ¯/Y.
a) Uaa Uba pa Uab Ubb pb pa pb 0 a ¯ Y b ¯ Y ¯ Y 0 0 1 b) For a ¯/Y, a ¯ Y J1 J J 0 0 1 Uab Ubb pb pa pb 0 (paUbb pbUab) J (13.23) which cannot be signed from the mathematics but can be signed from the economics. If a is a normal good, a ¯/Y 0; if a is a weakly inferior good, a ¯/Y 0; and if a is a strictly inferior good, a ¯/Y 0.
c) For b ¯/Y, b ¯ Y J2 J J Uaa Uba pa 0 0 1 pa pb 0 (pbUaa paUba) J (13.24) which can also be signed according to the nature of the good, as in (b) above.
13.28. Derive the Slutsky equation for the effect of a change in pa on the optimal quantity of the good demanded a ¯ and determine the sign of the comparative-static derivative a ¯/pa from the information gained in Problems 13.25 to 13.27.
From (13.21), with slight rearrangement, a ¯ pa ¯pb 2 J a ¯(paUbb pbUab) J (13.25) But from (13.23), a ¯ Y (paUbb PbUab) J Substituting a ¯/Y in (13.25), we get the Slutsky equation for a ¯, where the first term on the right is the substitution effect and the second term is the income effect.
a ¯ pa ¯pb 2 J a ¯ a ¯ Y Substitution effect Income effect Since J H ¯ 0 for constrained maximization and from (13.20), ¯ Ua pa MUa pa 0 the substitution effect in the first term is unambiguously negative. The income effect in the second term will depend on the nature of the good. For a normal good, a ¯/Y 0 and the income effect above will be negative, making a ¯/pa 0. For a weakly inferior good, a ¯/Y 0, and a ¯/pa 0. For a strictly inferior good, a ¯/Y 0 and the sign of a ¯/pa will depend on the relative magnitude of the different effects. If the income effect overwhelms the substitution effect, as in the case of a Giffen good, a ¯/pa 0 and the demand curve will be positively sloped.
313 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] 13.29. Derive the Slutsky equation for the effect of a change in pb on the optimal quantity of the good demanded b ¯ and determine the sign of the comparative-static derivative b ¯/pb.
From (13.22), b ¯ pb ¯pa 2 J b ¯(pbUaa paUba) J But from (13.24), b ¯ Y (pbUaa paUba) J Substituting above, b ¯ pb ¯pa 2 J b ¯ b ¯ Y where the substitution effect in the first term on the right is unquestionably negative and the income effect in the second term will depend on the nature of the good, as in Problem 13.28.
THE ENVELOPE THEOREM 13.30. A firm in perfect competition with the production function Q f(K, L) and a production limit of Q0 seeks to maximize profit PQ rK wL Assuming conditions are satisfied for the implicit function theorem, the Lagrangian function and the indirect objective function can be written, respectively, (K, L, Q, ; r, w, P, Q0) PQ(r, w, P, Q0) rK(r, w, P, Q0) wL(r, w, P, Q0) [Q0 f(K, L)] ¯(K ¯ , L ¯, Q ¯ ; r, w, P, Q0) PQ ¯ (r, w, P, Q0) rK ¯ (r, w, P, Q0) wL ¯(r, w, P, Q0) Use the envelope theorem to find and comment on the changes in the indirect objective function signified by (a) ¯/r, (b) ¯/w, (c) ¯/P, (d) ¯/Q0.
a) ¯ r r K ¯ (r, w, P, Q0) b) ¯ w w L ¯(r, w, P, Q0) Differentiating the profit function with respect to input prices gives the firm’s demand for inputs.
Notice here where input prices (r, w) appear in the objective function and not in the constraint, the desired derivatives can be readily found from either function.
c) ¯ P P Q ¯ (r, w, P, Q0) Differentiating the profit function with respect to output prices gives the firm’s supply function. Since output price (P) appears only in the objective function, the derivative can once again easily be found from either function.
d) ¯ Q0 Q0 ¯ Differentiating the profit function with respect to an output constraint gives the marginal value of relaxing that constraint, i.e., the extra profit the firm could earn if it could increase output by one unit.
Notice here where the output limit (Q0) appears only in the constraint, the derivative can be derived more quickly and readily from the Lagrangian function.
314 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 13.31. A consumer wants to minimize the cost of attaining a specific level of utility: c pxx pyy subject to u(x, y) U0 If the implicit function theorem conditions are fulfilled, the Lagrangian and indirect object functions are C(x, y, ; px, py, U0) pxx(px, py,U0) pyy(px, py, U0) [U0 u(x, y)] c ¯(x ¯, y ¯; px, py, U0) pxx ¯(px, py, U0) pyy ¯(px, py, U0) Use the envelope theorem to find and comment on the changes in the indirect objective function signified by (a) c ¯/px, (b) c ¯/py, (c) c ¯/U0.
a) c ¯ px C px x ¯(px, py, U0) b) c ¯ py C py y ¯(px, py, U0) In both cases, a change in the price of a good has a positive effect on the cost that is weighted by the amount of the good consumed. Since prices appear only in the objective function and not the constraint, the desired derivatives can be easily taken from either function.
c) c ¯ U0 C U0 ¯ Here ¯ measures the marginal cost of changing the given level of utility. Since the utility limit U0 appears only in the constraint, the derivative is more easily found from the Lagrangian function.
13.32. Assume the model in Example 7 is a function with only a single exogenous variable a. Show that at the optimal solution the total derivative of the Lagrangian function with respect to a is equal to the partial derivative of the same Lagrangian function with respect to a.
The new Lagrangian function and first-order conditions are Z(x, y, ; a) z[x(a), y(a); a] (a)f[x(a), y(a); a] Zx zx(x ¯, y ¯; a) ¯fx(x ¯, y ¯; a) 0 Zy zy(x ¯, y ¯; a) ¯fy(x ¯, y ¯; a) 0 Z f(x ¯, y ¯; a) 0 Taking the total derivative of the original Lagrangian function with respect to a, dZ da (zx ¯fx) dx ¯ da (zy ¯fy) dy ¯ da f d ¯ da (za ¯fa) But from the first-order conditions, zx ¯fx 0, zy ¯fy 0, and f 0 so the first three terms cancel and the total derivative of the function with respect to the exogenous variable a ends up equal to the partial derivative of the function with respect to the exogenous variable a: dZ da (za ¯fa) Z a This suggests we can find the total effect of a change in a single exogenous variable on the optimal value of the Lagrangian function by simply taking the partial derivative of the Lagrangian function with respect to that exogenous variable.
315 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] CONCA VE PROGRAMMING 13.33. Given the typical format for constrained optimization of general functions below, (a) maximize f(x, y) subject to g(x, y) B (b) minimize f(x, y) subject to g(x, y) B express them in suitable form for concave programming and write out the Lagrangians.
a) For less than or equal to constraints in maximization problems, subtract the variables in the constraint from the constant of the constraint.
Maximize f(x, y) subject to B g(x, y) 0 x ¯, y ¯ 0 Max F f(x, y) [B g(x, y)] b) For minimization, multiply the objective function by 1 to make it negative and then maximize the negative of the original function. For the corresponding greater than or equal to constraints in minimization problems, subtract the constant of the constraint from the variables in the constraint.
Maximize f(x, y) subject to g(x, y) B 0 x ¯, y ¯ 0 Max F f(x, y) [g(x, y) B] 13.34. Assume a firm with the production function Q(K, L) and operating in a purely competitive market for inputs wishes to minimize cost while producing no less than a specific amount of output, given by minimize c rK wL subject to Q(K, L) Q0 (a) Express the problem in concave-programming format, (b) write out the Lagrangian function, and (c) solve the problem.
a) Multiplying the objective function by 1 to make it negative and maximizing it, Maximize rK wL subject to Q(K, L) Q0 0 K, L 0 b) Maximize C rK wL [Q(K, L) Q0] c) Applying the Kuhn-Tucker conditions, we first maximize the Lagrangian function with respect to the choice variables K and L and check the related conditions.
1.
a) C K r ¯QK 0 C L w ¯QL 0 b) c) K ¯ 0 K ¯ (r ¯QK) 0 L ¯ 0 L ¯(w ¯QL) 0 We then minimize the Lagrangian with respect to the constraint variable and check the related conditions.
2.
a) b) c) C ¯ 0 ¯[Q(K ¯ , L ¯) Q0] 0 Q(K ¯ , L ¯) Q0 0 Assuming production depends on both inputs, K ¯ , L ¯ 0, the two expressions within the parentheses in 1(c) have to be equalities.
r ¯QK 0 w ¯QL 0 Rearranging, we see ¯ r QK w QL 0 (13.26) 316 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 since input prices r, w 0 and marginal productivities QK, QL 0. With ¯ 0, from 2(c) the budget constraint binds as an equality, with Q(K ¯ , L ¯) Q0 Rearranging (13.26), we have an interior solution with the isoquant tangent to the isocost line, QL QK w r 13.35. Maximize profits 64x 2x2 96y 4y2 13 subject to the production constraint x y 20 We first set up the Lagrangian function 64x 2x2 96y 4y2 13 (20 x y) and set down the Kuhn-Tucker conditions.
1.
a) x 64 4x ¯ ¯ 0 y 96 8y ¯ ¯ 0 b) x ¯ 0 y ¯ 0 c) x ¯(64 4x ¯ ¯) 0 y ¯(96 8y ¯ ¯) 0 2.
a) 20 x ¯ y ¯ 0 b) ¯ 0 c) ¯(20 x ¯ y ¯) 0 We then test the Kuhn-Tucker conditions methodically.
1.
Check the possibility that ¯ 0 or ¯ 0.
If ¯ 0, then from 1(a), 64 4x ¯ 0 96 8y ¯ 0 Therefore, 4x ¯ 64 8y ¯ 96 x ¯ 16 y ¯ 12 But this violates the initial constraint since x ¯ y ¯ 28 20. Hence ¯ 0 and from 2(b) we conclude ¯ 0.
2.
If ¯ 0, from 2(c), the constraint holds as an equality, with 20 x ¯ y ¯ 0 3.
Next check to see if either of the choice variables x ¯ or y ¯ can equal zero.
a) If x ¯ 0, y ¯ 20 and the second condition in 1(c) is violated.
20[96 8(20) ( ¯ 0)] 0 b) If y ¯ 0, x ¯ 20 and the first condition in 1(c) is violated.
20[64 4(20) ( ¯ 0)] 0 So neither choice variable can equal zero and from 1(b), if x ¯ 0, x ¯ 0 and if y ¯ 0, y ¯ 0 317 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] 4.
If x ¯, y ¯, ¯ 0, then from 1(c) and 2(c), the following equalities hold, 64 4x ¯ ¯ 0 96 8y ¯ ¯ 0 20 x ¯ y ¯ 0 Setting them down in matrix form, 4 0 1 0 8 1 1 1 0 x ¯ y ¯ ¯ 64 96 20 and solving by Cramer’s rule where A 12, A1 128, A2 112, and A3 256, we get the solution: x ¯ 10.67, y ¯ 9.33, and ¯ 21.33 which we know to be optimal because none of the Kuhn-Tucker conditions is violated. With ¯ 21.33, a unit increase in the constant of the production constraint will cause profits to increase by approximately 21.33.
13.36. Maximize the profit function in Problem 13.35, 64x 2x2 96y 4y2 13 subject to the new production constraint x y 36 The Lagrangian function and the Kuhn-Tucker conditions are 64x 2x2 96y 4y2 13 (36 x y) 1.
a) x 64 4x ¯ ¯ 0 y 96 8y ¯ ¯ 0 b) x ¯ 0 y ¯ 0 c) x ¯(64 4x ¯ ¯) 0 y ¯(96 8y ¯ ¯) 0 2.
a) 36 x ¯ y ¯ 0 b) ¯ 0 c) ¯(36 x ¯ y ¯) 0 We then check the Kuhn-Tucker conditions systematically.
1.
Test the possibility of ¯ 0 or ¯ 0.
If ¯ 0, then from 1(a), 64 4x ¯ 0 96 8y ¯ 0 Therefore, x ¯ 16 y ¯ 12 Since x ¯ y ¯ 28 36, no condition is violated. Therefore, it is possible that ¯ 0 or ¯ 0.
2.
Now check to see if either of the choice variables x ¯ or y ¯ can equal zero.
a) If x ¯ 0, y ¯ 36, and the second condition in 1(c) is violated.
36[96 8(36) ( ¯ 0)] 0 b) If y ¯ 0, x ¯ 36, and the first condition in 1(c) is violated.
36[64 4(36) ( ¯ 0)] 0 Therefore, neither choice variable can equal zero and from 1(b), x ¯ 0 and y ¯ 0 318 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 3.
Check the solutions when (a) ¯ 0 and (b) ¯ 0.
a) If ¯, x ¯, y ¯ 0, then from the Kuhn-Tucker conditions listed under (c), 64 4x ¯ ¯ 0 96 8y ¯ ¯ 0 36 x y ¯ 0 In matrix form, 4 0 1 0 8 1 1 1 0 x ¯ y ¯ ¯ 64 96 36 Using Cramer’s rule where A 12, A1 256, A2 176, and A3 256, we get x ¯ 21.33, y ¯ 14.67, and ¯ 21.33 which cannot be optimal because ¯ 0 in violation of 2(b) of the Kuhn-Tucker conditions. With ¯ 0, from 2(c), the constraint is a strict equality and decreasing the level of output will increase the level of profit.
b) If ¯ 0 and x ¯, y ¯ 0, then from 1(c), 64 4x ¯ 0, x ¯ 16 96 8y ¯ 0, y ¯ 12 This gives us the optimal solution, x ¯ 16, y ¯ 12, and ¯ 0, which we know is optimal because it violates none of the Kuhn-Tucker conditions. With ¯ 0, the constraint is nonbinding as we see from the optimal solution x ¯ y ¯ 28 36.
13.37. Minimize cost c 5x2 80x y2 32y subject to x y 30 Multiplying the objective function by 1 and setting up the Lagrangian, we have Max C 5x2 80x y2 32y (x y 30) where the Kuhn-Tucker conditions are, 1.
a) Cx 10x ¯ 80 ¯ 0 Cy 2y ¯ 32 ¯ 0 b) x ¯ 0 y ¯ 0 c) x ¯(10x ¯ 80 ¯) 0 y ¯(2y ¯ 32 ¯) 0 2.
a) C x ¯ y ¯ 30 0 b) ¯ 0 c) ¯(x ¯ y ¯ 30) 0 1.
Check the possibility of ¯ 0.
If ¯ 0, then from 1(a), 10x ¯ 80 0 2y ¯ 32 0 Therefore, x ¯ 8 y ¯ 16 But this is a violation because x ¯ y ¯ 24 fails to satisfy the initial constraint x ¯ y ¯ 30. So ¯ 0.
2.
Check to see if x ¯ or y ¯ can equal zero.
From 1(a), if x ¯ 0, ¯ 80, and if y ¯ 0, ¯ 32, both of which violate the nonnegativity constraint on variables. So x ¯, y ¯ 0.
319 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] 3.
Now check the Kuhn-Tucker conditions when x ¯, y ¯ 0 and ¯ 0. If ¯ 0 and x ¯, y ¯ 0, all the first partials are strict equalities and we have 10 0 1 0 2 1 1 1 0 x ¯ y ¯ ¯ 80 32 30 where A 12, A1 109, A2 252, and A3 440, giving the optimal solution, which violates none of the Kuhn-Tucker conditions: x ¯ 9, y ¯ 21, ¯ 36.67 13.38. Minimize the same function as above, 5x2 80x y2 32y subject to a new constraint, x y 20 The Lagrangian function and Kuhn-Tucker conditions are, Max C 5x2 80x y2 32y (x y 20) 1.
a) Cx 10x ¯ 80 ¯ 0 Cy 2y ¯ 32 ¯ 0 b) x ¯ 0 y ¯ 0 c) x ¯(10x ¯ 80 ¯) 0 y ¯(2y ¯ 32 ¯) 0 2.
a) C x ¯ y ¯ 20 0 b) ¯ 0 c) ¯(x ¯ y ¯ 20) 0 1.
Check the possibility of ¯ 0.
If ¯ 0, then from 1(a), 10x ¯ 80 0 2y ¯ 32 0 Therefore, x ¯ 8 y ¯ 16 This violates no constraint because x ¯ y ¯ 24 satisfies x ¯ y ¯ 20. So ¯ 0 or ¯ 0.
2.
By the same reasoning as in step 2 above, we can show x ¯, y ¯ 0.
3.
So we are left with two possibilities, depending on whether ¯ 0, or ¯ 0.
a) If ¯ 0 and x ¯, y ¯ 0, all the derivatives are strict equalities and we have 10 0 1 0 2 1 1 1 0 x ¯ y ¯ ¯ 80 32 20 With A 12, A1 88, A2 152, and A3 80, the solution is x ¯ 7.33, y ¯ 12.67, ¯ 6.67 Since ¯ is negative, condition 2(b) is violated and the solution is nonoptimal. The solution suggests we can reduce the cost by increasing output.
b) If ¯ 0 and x ¯, y ¯ 0, from 1(c), 10x ¯ 80 0 2y ¯ 32 0 x ¯ 8 y ¯ 16 320 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 This satisfies the new constraint x y 20 and violates none of the Kuhn-Tucker conditions, so the optimal solution is x ¯ 8, y ¯ 16, ¯ 0 With ¯ 0, the constraint is nonbinding. The firm can minimize costs while exceeding its production quota.
13.39. Maximize the utility function, u xy subject to the following budget and dietary constraints, 3x 4y 144 budget constraint 5x 2y 120 dietary constraint The Lagrangian function and Kuhn-Tucker conditions are Max U xy 1(144 3x 4y) 2(120 5x 2y) 1.
a) Ux y ¯ 3 ¯1 5 ¯2 0 Uy x ¯ 4 ¯1 2 ¯2 0 b) x ¯ 0 y ¯ 0 c) x ¯(y ¯ 3 ¯1 5 ¯2) 0 y ¯(x ¯ 4 ¯1 2 ¯2) 0 2.
a) U1 144 3x ¯ 4y ¯ 0 U2 120 5x ¯ 2y ¯ 0 b) ¯1 0 ¯2 0 c) ¯1(144 3x ¯ 4y ¯) 0 ¯2(120 5x ¯ 2y ¯) 0 Given the nature of the objective function, we assume neither of the choice variables, x ¯, y ¯, can equal zero. Otherwise the utility function, u xy 0. If x ¯, y ¯ 0, from 1(c), y ¯ 31 52 0 x ¯ 41 22 0 Noting that the MUx ux y ¯, MUy uy x ¯, and assuming MU’s 0, both ’s cannot equal zero, so at least one of the constraints must be binding. This leaves us with three possibilities: (a) ¯1 0, ¯2 0, (b) ¯1 0, ¯2 0, and (c) ¯1 0, ¯2 0. We examine each in turn.
a) ¯1 0, ¯2 0, x ¯, y ¯ 0. From the 1(c) and 2(c) conditions we have four equalities, three of which still call for solution.
y ¯ 3 ¯1 5 ¯2 0 x ¯ 4 ¯1 2 ¯2 0 144 3x ¯ 4y ¯ 0 ¯2 0 Putting the latter in matrix form, 0 1 3 1 0 4 3 4 0 x ¯ y ¯ ¯1 0 0 144 and solving by Cramer’s rule, where A 24, A1 576, A2 432, and A3 144, we have x ¯ 24, y ¯ 18, ¯1 6 But checking the answer for internal consistency, we find U2 120 5(24) 2(18) 36 0 This is clearly a violation of the Kuhn-Tucker conditions which require U/i 0 and hence the solution is not optimal.
321 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] b) ¯1 0, ¯2 0, x ¯, y ¯ 0. From 1(c) and 2(c), we have four equalities, three of which must still be solved.
y ¯ 3 ¯1 5 ¯2 0 x ¯ 4 ¯1 2 ¯2 0 ¯1 0 120 5x ¯ 2y ¯ 0 Setting the latter in matrix form and solving by Cramer’s rule, 0 1 5 1 0 2 5 2 0 x ¯ y ¯ ¯2 0 0 120 we have A 20, A1 240, A2 600, and A3 120, and x ¯ 12, y ¯ 30, ¯2 6 But checking for internal consistency once again, we find U1 144 3(12) 4(30) 12 0 This violates the Kuhn-Tucker conditions and so the solution cannot be optimal.
c) ¯1 0, ¯2 0, x ¯,y ¯ 0. From 1(c) and 2(c), all four derivatives are strict equalities which we set down immediately in matrix form.
0 1 3 5 1 0 4 2 3 4 0 0 5 2 0 0 x ¯ y ¯ ¯1 ¯2 0 0 144 120 From Cramer’s rule, where A 196, A1 2688, A2 5040, A3 240, and A4 864, we find the optimal solution, which violates none of the conditions: x ¯ 13.71, y ¯ 25.71, ¯1 1.22, ¯2 4.41 13.40. Confirm the results of Problem 13.39 with (a) a graph and (b) explain what the graph illustrates.
a) See Fig. 13-3.
322 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 Fig. 13-3 B y A C 60 48 36 30 5.71 24 18 12 0 12 13.71 24 36 48 5x + 2y < 120 Dietary Constraint 3x + 4y < 144 Income Constraint x b) At point A, x ¯ 24, y ¯ 18, the income constraint is exactly fulfilled, but the dietary constraint is violated. Hence A cannot be optimal. At point B, x ¯ 12, y ¯ 30, the dietary constraint is exactly fulfilled, but the income constraint is violated. Hence B cannot be optimal. At point C, x ¯ 13.71, y ¯ 25.71, and both the income constraint and the dietary constraint are binding with no Kuhn-Tucker condition violated. Hence C is optimal and note that it occurs exactly at the intersection of the two constraints.
13.41. Following a fair-rate-of-return policy, regulators of natural monopolies, such as utility com-panies, restrict profits to a certain fixed proportion of the capital employed. The policy, however, leads to a distortion of inputs on the part of regulated industries that has been termed the Averch-Johnson effect. Assuming a firm wishes to maximize profits (K, L) E(K, L) rK wL subject to the fair-rate-of-return constraint, E(K, L) wL mK where E earnings and EK, EL 0, EKK, ELL 0, and EKKELL EKLELK, r cost of capital, w wage, m maximum rate of return on capital, and m r 0, use concave programming to demonstrate the possibility of a distorting effect predicted by Averch-Johnson.
The Lagrangian function and the Kuhn-Tucker conditions are (K, L) E(K, L) rK wL [mK E(K, L) wL] 1.
a) K EK r ¯m ¯EK 0 L EL w ¯EL ¯w 0 b) K ¯ 0 L ¯ 0 c) K ¯ (EK r ¯m ¯EK) 0 L ¯(EL w ¯EL ¯w) 0 2.
a) mK ¯ wL ¯ E(K ¯ , L ¯) 0 b) ¯ 0 c) ¯[mK ¯ wL ¯ E(K ¯ , L ¯)] 0 Making the common sense assumption that K ¯ , L ¯ 0, adding and subtracting ¯r within the parentheses of the equation on the left in 1(c), and rearranging we have, (1 ¯)(EK r) ¯(m r) 0 (13.27) (1 ¯)(EL w) 0 (13.28) mK ¯ wL ¯ E(K ¯ , L ¯) 0 ¯[mK ¯ wL ¯ E(K ¯ , L ¯)] 0 If ¯ 1 in (13.27), m r 0 and this would contradict the assumption that the maximum allowable rate of return is greater than the cost of capital m r. So ¯ 1. But if ¯ 1, from (13.27), EK r ¯(m r) (1 ¯) (13.29) and from (13.28), EL w (13.30) Dividing (13.29) by (13.30), EK EL r ¯(m r) (1 ¯) w r w ¯(m r) (1 ¯)w (13.31) 323 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] If ¯ 0, the constraint is not binding and we have the unregulated optimum, EK EL r w where EK MRPK and EL MRPL. Dividing numerator and denominator on the left by the common output price, the above expression is equivalent to the familiar, MPK MPL r w But if ¯ 0 and the constraint is binding, regulation interferes with the economically optimal solution in which the ratio of marginal products exactly equals the ratio of respective prices. Thus if ¯ 0, there will be a distorting effect on the profit-maximizing level of output.
13.42. (a) Specify the direction of the distortion the Averch-Johnson effect predicts and (b) demonstrate the conditions necessary to verify it in terms of the previous model.
a) The Averch-Johnson effect predicts the distortion will lead to a higher K/L ratio than in an unregulated market. If more than the optimal amount of capital is used, the marginal productivity of capital will be diminished and the result predicted by the Averch-Johnson effect will be MPK MPL r w b) In terms of (13.31), the bias towards greater than optimal capital intensity will be true whenever ¯(m r) (1 ¯)w 0 (13.32) Since we know r 0, w 0, and by the assumption from common practice, m r 0, (13.32) will be positive whenever ¯ 1. To determine the sign of ¯, we revert to comparative-static techniques.
Having determined that x ¯, y ¯, ¯ 0, we know that all three partial derivatives in the Kuhn-Tucker conditions must hold as equalities: (1 ¯)EK r ¯m 0 (1 ¯)(EL w) 0 mK ¯ wL ¯ E(K ¯ , L ¯) 0 These equations are the same as the first-order conditions we would obtain if we had maximized the original function subject to the equality constraint: Max (K, L) E(K, L) rK wL subject to E(K, L) wL mK The second-order conditions for maximization require that the Bordered Hessian be positive: H ¯ (1 ¯)EKK (1 ¯)ELK m EK (1 ¯)EKL (1 ¯)ELL w EL m EK w EL 0 0 Expanding along the third row, H (m EK)[(w EL)(1 ¯)EKL (m EK)(1 ¯)ELL] (w EL)[(w EL)(1 ¯)EKK (m EK)(1 ¯)ELK] Since w EL at the optimum, all the (w EL) terms 0, leaving H ¯ (m EK)2(1 ¯)ELL 324 COMPARATIVE STATICS AND CONCAVE PROGRAMMING [CHAP . 13 For H 0, (1 ¯)ELL 0 Since ELL 0 from our earlier assumption of strict concavity, it follows that ¯ 1 With ¯ 1, (13.32) 0 and from (13.31), MPK MPL r w Q.E.D.
325 COMPARATIVE STATICS AND CONCAVE PROGRAMMING CHAP . 13] CHAPTER 14 Integral Calculus: The Indefinite Integral 14.1 INTEGRATION Chapters 3 to 6 were devoted to differential calculus, which measures the rate of change of functions. Differentiation, we learned, is the process of finding the derivative F(x) of a function F(x).
Frequently in economics, however, we know the rate of change of a function F(x) and want to find the original function. Reversing the process of differentiation and finding the original function from the derivative is called integration, or antidifferentiation. The original function F(x) is called the integral, or antiderivative, of F(x).
EXAMPLE 1.
Letting f(x) F(x) for simplicity, the antiderivative of f(x) is expressed mathematically as f(x) dx F(x) c Here the left-hand side of the equation is read, ‘‘the indefinite integral of f of x with respect to x.’’ The symbol is an integral sign, f(x) is the integrand, and c is the constant of integration, which is explained in Example 3.
14.2 RULES OF INTEGRATION The following rules of integration are obtained by reversing the corresponding rules of differentiation. Their accuracy is easily checked, since the derivative of the integral must equal the integrand. Each rule is illustrated in Example 2 and Problems 14.1 to 14.6.
Rule 1.
The integral of a constant k is k dx kx c 326 Rule 2.
The integral of 1, written simply as dx, not 1 dx, is dx x c Rule 3.
The integral of a power function xn, where n 1, is given by the power rule: xn dx 1 n 1 xn1 c n 1 Rule 4.
The integral of x1 (or 1/x) is x1 dx ln x c x 0 The condition x 0 is added because only positive numbers have logarithms. For negative numbers, x1 dx ln x c x 0 Rule 5.
The integral of an exponential function is akx dx akx k ln a c Rule 6.
The integral of a natural exponential function is ekx dx ekx k c since ln e 1 Rule 7.
The integral of a constant times a function equals the constant times the integral of the function. kf(x) dx k f(x) dx Rule 8.
The integral of the sum or difference of two or more functions equals the sum or difference of their integrals. [f(x) g(x)] dx f(x) dx g(x) dx Rule 9.
The integral of the negative of a function equals the negative of the integral of that function. f(x) dx f(x) dx EXAMPLE 2.
The rules of integration are illustrated below. Check each answer on your own by making sure that the derivative of the integral equals the integrand.
i) 3 dx 3x c (Rule 1) ii) x2 dx 1 2 1 x21 c 1 3 x3 c (Rule 3) 327 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] iii) 5x4 dx 5 x4 dx (Rule 7) 5 1 5 x5 c1 (Rule 3) x5 c where c1 and c are arbitrary constants and 5c1 c. Since c is an arbitrary constant, it can be ignored in the preliminary calculation and included only in the final solution.
iv) (3x3 x 1) dx 3 x3 dx x dx dx (Rules 7, 8, and 9) 3(1 – 4x4) 1 – 2x2 x c (Rules 2 and 3) 3 – 4x4 1 – 2x2 x c v) 3x1 dx 3 x1 dx (Rule 7) 3 ln x c (Rule 4) vi) 23x dx 23x 3 ln 2 c (Rule 5) vii) 9e3x dx 9e3x 3 c (Rule 6) 3e3x c EXAMPLE 3.
Functions which differ by only a constant have the same derivative. The function F(x) 2x k has the same derivative, F(x) f(x) 2, for any infinite number of possible values for k. If the process is reversed, it is clear that 2 dx must be the antiderivative or indefinite integral for an infinite number of functions differing from each other by only a constant. The constant of integration c thus represents the value of any constant which was part of the primitive function but precluded from the derivative by the rules of differentiation.
The graph of an indefinite integral f(x) dx F(x) c, where c is unspecified, is a family of curves parallel in the sense that the slope of the tangent to any of them at x is f(x). Specifying c specifies the curve; changing c shifts the curve. This is illustrated in Fig. 14-1 for the indefinite integral 2 dx 2x c where c 7, 3, 1, and 5, respectively. If c 0, the curve begins at the origin.
328 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 Fig. 14-1 14.3 INITIAL CONDITIONS AND BOUNDARY CONDITIONS In many problems an initial condition (y y0 when x 0) or a boundary condition (y y0 when x x0) is given which uniquely determines the constant of integration. By permitting a unique determination of c, the initial or boundary condition singles out a specific curve from the family of curves illustrated in Example 3 and Problems 14.3 to 14.5.
EXAMPLE 4.
Given the boundary condition y 11 when x 3, the integral y 2 dx is evaluated as follows: y 2 dx 2x c Substituting y 11 when x 3, 11 2(3) c c 5 Therefore, y 2x 5. Note that even though c is specified, 2 dx remains an indefinite integral because x is unspecified. Thus, the integral 2x 5 can assume an infinite number of possible values.
14.4 INTEGRATION BY SUBSTITUTION Integration of a product or quotient of two differentiable functions of x, such as 12x2(x3 2) dx cannot be done directly by using the simple rules above. However, if the integrand can be expressed as a constant multiple of another function u and its derivative du/dx, integration by substitution is possible. By expressing the integrand f(x) as a function of u and its derivative du/dx and integrating with respect to x, f(x) dx
u du dx dx f(x) dx u du F(u) c The substitution method reverses the operation of the chain rule and the generalized power function rule in differential calculus. See Examples 5 and 6 and Problems 14.7 to 14.18.
EXAMPLE 5.
The substitution method is used below to determine the indefinite integral 12x2(x3 2) dx 1.
Be sure that the integrand can be converted to a product of another function u and its derivative du/dx times a constant multiple. (a) Let u equal the function in which the independent variable is raised to the higher power in terms of absolute value; here let u x3 2. (b) Take the derivative of u; du/dx 3x2. (c) Solve algebraically for dx; dx du/3x2. (d) Then substitute u for x3 2 and du/3x2 for dx in the original integrand: 12x2(x3 2) dx 12x2 · u · du 3x2 4u du 4 u du where 4 is a constant multiple of u.
2.
Integrate with respect to u, using Rule 3 and ignoring c in the first step of the calculation.
4 u du 4(1 – 2u2) 2u2 c 329 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] 3.
Convert back to the terms of the original problem by substituting x3 2 for u. 12x2(x3 2) dx 2u2 c 2(x3 2)2 c 4.
Check the answer by differentiating with the generalized power function rule or chain rule.
d dx [2(x3 2)2 c] 4(x3 2)(3x2) 12x2(x3 2) See also Problems 14.7 to 14.18.
EXAMPLE 6.
Determine the integral 4x(x 1)3 dx.
Let u x 1. Then du/dx 1 and dx du/1 du. Substitute u x 1 and dx du in the original integrand. 4x(x 1)3 dx 4xu3 du 4 xu3 du Since x is a variable multiple which cannot be factored out, the original integrand cannot be transformed to a constant multiple of u du/dx. Hence the substitution method is ineffectual. Integration by parts (Section 14.5) may be helpful.
14.5 INTEGRATION BY PARTS If an integrand is a product or quotient of differentiable functions of x and cannot be expressed as a constant multiple of u du/dx, integration by parts is frequently useful. The method is derived by reversing the process of differentiating a product. From the product rule in Section 3.7.5, d dx [f(x) g(x)] f(x) g(x) g(x) f(x) Taking the integral of the derivative gives f(x) g(x) f(x) g(x) dx g(x) f(x) dx Then solving algebraically for the first integral on the right-hand side, f(x)g(x) dx f(x)g(x) g(x) f(x) dx (14.1) See Examples 7 and 8 and Problems 14.19 to 14.24.
For more complicated functions, integration tables are generally used. Integration tables provide formulas for the integrals of as many as 500 different functions, and they can be found in mathematical handbooks.
EXAMPLE 7.
Integration by parts is used below to determine 4x(x 1)3 dx 1.
Separate the integrand into two parts amenable to the formula in (14.1). As a general rule, consider first the simpler function for f(x) and the more complicated function for g(x). By letting f(x) 4x and g(x) (x 1)3, then f(x) 4 and g(x) (x 1)3 dx, which can be integrated by using the simple power rule (Rule 3): g(x) (x 1)3 dx 1 – 4(x 1)4 c1 330 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 2.
Substitute the values for f(x), f(x), and g(x) in (14.1); and note that g(x) is not used in the formula. 4x(x 1)3 dx f(x) · g(x) [g(x) · f(x)] dx 4x[1 – 4(x 1)4 c1] 1 – 4(x 1)4 c1 dx x(x 1)4 4c1x [(x 1)4 4c1] dx 3.
Use Rule 3 to compute the final integral and substitute. 4x(x 1)3 dx x(x 1)4 4c1x 1 – 5(x 1)5 4c1x c x(x 1)4 1 – 5(x 1)5 c Note that the c1 term does not appear in the final solution. Since this is common to integration by parts, c1 will henceforth be assumed equal to 0 and not formally included in future problem solving.
4.
Check the answer by letting y(x) x(x 1)4 1 – 5(x 1)5 c and using the product and generalized power function rules.
y(x) [x · 4(x 1)3 (x 1)4 · 1] (x 1)4 4x(x 1)3 EXAMPLE 8.
The integral 2xex dx is determined as follows: Let f(x) 2x and g(x) ex; then f(x) 2, and by Rule 6, g(x) ex dx ex. Substitute in (14.1). 2xex dx f(x) · g(x) g(x) · f(x) dx 2x · ex ex · 2 dx 2xex 2 ex dx Apply Rule 6 again and remember the constant of integration. 2xex dx 2xex 2ex c Then let y(x) 2xex 2ex c and check the answer.
y(x) 2x · ex ex · 2 2ex 2xex 14.6 ECONOMIC APPLICATIONS Net investment I is defined as the rate of change in capital stock formation K over time t. If the process of capital formation is continuous over time, I(t) dK(t)/dt K(t). From the rate of investment, the level of capital stock can be estimated. Capital stock is the integral with respect to time of net investment: Kt I(t) dt K(t) c K(t) K0 where c the initial capital stock K0.
Similarly, the integral can be used to estimate total cost from marginal cost. Since marginal cost is the change in total cost from an incremental change in output, MC dTC/dQ, and only variable costs change with the level of output TC MC dQ VC c VC FC 331 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] since c the fixed or initial cost FC. Economic analysis which traces the time path of variables or attempts to determine whether variables will converge toward equilibrium over time is called dynamics. For similar applications, see Example 9 and Problems 14.25 to 14.35.
EXAMPLE 9.
The rate of net investment is given by I(t) 140t3/4, and the initial stock of capital at t 0 is 150.
Determining the function for capital K, the time path K(t), K 140t3/4 dt 140 t3/4 dt By the power rule, K 140(4 – 7t7/4) c 80t7/4 c But c K0 150. Therefore, K 80t7/4 150.
Solved Problems INDEFINITE INTEGRALS 14.1.
Determine the following integrals. Check the answers on your own by making sure that the derivative of the integral equals the integrand.
a) 3.5 dx 3.5 dx 3.5x c (Rule 1) b) 1 – 2 dx 1 – 2 dx 1 – 2 dx 1 – 2x c (Rules 1 and 9) c) dx dx x c (Rule 2) d) x5 dx x5 dx 1 – 6x6 c (Rule 3) e) 4x3 dx 4x3 dx 4 x3 dx (Rule 7) 4(1 – 4x4) c x4 c (Rule 3) f) x2/3 dx x2/3 dx 3 – 5x5/3 c (Rule 3) g) x1/5 dx x1/5 dx 5 – 4x4/5 c (Rule 3) 332 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 h) 4x2 dx 4x2 dx 4x1 c 4 x c (Rule 3) i) x5/2 dx x5/2 dx 2 3 x3/2 c 2 3x3 c (Rule 3) 14.2.
Redo Problem 14.1 for each of the following: a) dx x dx x 1 x dx ln x c (Rule 4) b) 5x1 dx 5x1 dx 5 ln x c (Rules 7 and 4) c) 1 3x dx 1 3x dx 1 3 1 x dx 1 3 ln x c (Rules 7 and 4) d) x dx x dx x1/2 dx 2 – 3x3/2 c (Rule 3) e) dx x4 dx x4 x4 dx 1 3 x3 c (Rule 3) f) dx 3 x dx 3 x x1/3 dx 3 2 x2/3 c (Rule 3) g) (5x3 2x2 3x) dx (5x3 2x2 3x) dx 5 x3 dx 2 x2 dx 3 x dx (Rules 7 and 8) 5(1 – 4x4) 2(1 – 3x3) 3(1 – 2x2) c (Rule 3) 5 – 4x4 2 – 3x3 3 – 2x2 c 333 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] h) (2x6 3x4) dx (2x6 3x4) dx 2 – 7x7 3 – 5x5 c (Rules 3, 7, 8, and 9) 14.3.
Find the integral for y (x1/2 3x1/2) dx, given the initial condition y 0 when x 0.
y (x1/2 3x1/2) dx 2 – 3x3/2 6x1/2 c Substituting the initial condition y 0 when x 0 above, c 0. Hence, y 2 – 3x3/2 6x1/2.
14.4.
Find the integral for y (2x5 3x1/4) dx, given the initial condition y 6 when x 0.
y (2x5 3x1/4) dx 1 – 3x6 4x3/4 c Substituting y 6 and x 0, c 6. Thus, y 1 – 3x6 4x3/4 6.
14.5.
Find the integral for y (10x4 3) dx, given the boundary condition y 21 when x 1.
y (10x4 3) dx 2x5 3x c Substituting y 21 and x 1, 21 2(1)5 3(1) c c 22 y 2x5 3x 22 14.6.
Redo Problem 14.1 for each of the following: a) 24x dx b) 8x dx 24x dx 24x 4 ln 2 c (Rule 5) 8x dx 8x ln 8 c c) e5x dx d) 16e4x dx e5x dx e5x 5 c (Rule 6) 16e4x dx 16e4x 4 c 4e4x c 1 – 5e5x c e) (6e3x 8e2x) dx (6e3x 8e2x) dx 6e3x 3 8e2x 2 c 2e3x 4e2x c INTEGRATION BY SUBSTITUTION 14.7.
Determine the following integral, using the substitution method. Check the answer on your own. Given 10x(x2 3)4 dx.
334 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 Let u x2 3. Then du/dx 2x and dx du/2x. Substituting in the original integrand to reduce it to a function of u du/dx, 10x(x2 3)4 dx 10xu4 du 2x 5 u4 du Integrating by the power rule, 5 u4 du 5(1 – 5u5) u5 c Substituting u x2 3, 10x(x2 3)4 dx u5 c (x2 3)5 c 14.8.
Redo Problem 14.7, given x4(2x5 5)4 dx.
Let u 2x5 5, du/dx 10x4, and dx du/10x4. Substituting in the original integrand, x4(2x5 5)4 dx x4u4 du 10x4 1 10 u4 du Integrating, 1 10 u4 du 1 10 1 5 u5 1 50 u5 c Substituting, x4(2x5 5)4 dx 1 50 u5 c 1 50 (2x5 5)5 c 14.9.
Redo Problem 14.7, given (x 9)7/4 dx.
Let u x 9. Then du/dx 1 and dx du. Substituting, (x 9)7/4 dx u7/4 du Integrating, u7/4 du 4 11 u11/4 c Substituting, (x 9)7/4 dx 4 11 (x 9)11/4 c Whenever du/dx 1, the power rule can be used immediately for integration by substitution.
14.10. Redo Problem 14.7, given (6x 11)5 dx.
Let u 6x 11. Then du/dx 6 and dx du/6. Substituting, (6x 11)5 dx u5 du 6 1 6 u5 du Integrating, 1 6 u5 du 1 6 1 4 u4 1 24 u4 c Substituting, (6x 11)5 dx 1 –– 24(6x 11)4 c Notice that here du/dx 6 1, and the power rule cannot be used directly.
14.11. Redo Problem 14.7, given x2 (4x3 7)2 dx x2 (4x3 7)2 dx x2(4x3 7)2 dx 335 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] Let u 4x3 7, du/dx 12x2, and dx du/12x2. Substituting, x2u2 du 12x2 1 12 u2 du Integrating, 1 12 u2 du 1 12 u1 c Substituting, x2 (4x3 7)2 dx 1 12(4x3 7) c 14.12. Redo Problem 14.7, given 6x2 4x 10 (x3 x2 5x)3 dx Let u x3 x2 5x. Then du/dx 3x2 2x 5 and dx du/(3x2 2x 5). Substituting, (6x2 4x 10)u3 du 3x2 2x 5 2 u3 du Integrating, 2 u3 du u2 c Substituting, 6x2 4x 10 (x3 x2 5x)3 dx 1 (x3 x2 5x)2 c 14.13. Redo Problem 14.7, given dx 9x 5 dx 9x 5 (9x 5)1 dx Let u 9x 5, du/dx 9, and dx du/9. Substituting, u1 du 9 1 9 u1 du Integrating with Rule 4, 1 – 9 u1 du 1 – 9 ln u c. Since u may be 0, and only positive numbers have logs, always use the absolute value of u. See Rule 4. Substituting, dx 9x 5 1 9 ln 9x 5 c 14.14. Redo Problem 14.7, given 3x2 2 4x3 8x dx Let u 4x3 8x, du/dx 12x2 8, and dx du/(12x2 8). Substituting, (3x2 2)u1 du 12x2 8 1 4 u1 du Integrating, 1 4 u1 du 1 4 ln u c 336 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 Substituting, 3x2 2 4x3 8x dx 1 4 ln 4x3 8x c 14.15. Use the substitution method to find the integral for x3ex4 dx. Check your answer.
Let u x4. Then du/dx 4x3 and dx du/4x3. Substituting, and noting that u is now an exponent, x3eu du 4x3 1 4 eu du Integrating with Rule 6, 1 4 eu du 1 4 eu c Substituting, x3ex4 dx 1 4 ex4 c 14.16. Redo Problem 14.15, given 24xe3x2 dx.
Let u 3x2, du/dx 6x, and dx du/6x. Substituting, 24xeudu 6x 4 eu du Integrating, 4 eu du 4eu c Substituting, 24xe3x2 dx 4e3x2 c 14.17. Redo Problem 14.15, given 14e2x7 dx.
Let u 2x 7; then du/dx 2 and dx du/2. Substituting, 14eu du 2 7 eu du 7eu c Substituting, 14e2x7 dx 7e2x7 c 14.18. Redo Problem 14.15, given 5xe5x23 dx.
Let u 5x2 3, du/dx 10x, and dx du/10x. Substituting, 5xeu du 10x 1 2 eu du Integrating, 1 2 eu du 1 2 eu c Substituting, 5xe5x23 dx 1 2 e5x23 c INTEGRATION BY PARTS 14.19. Use integration by parts to evaluate the following integral. Keep in the habit of checking your answers. Given 15x(x 4)3/2 dx.
337 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] Let f(x) 15x, then f(x) 15. Let g(x) (x 4)3/2, then g(x) (x 4)3/2 dx 2 – 5(x 4)5/2. Sub-stituting in (14.1), 15x(x 4)3/2 dx f(x)g(x) g(x) f(x) dx 15x[2 – 5(x 4)5/2] 2 – 5(x 4)5/215 dx 6x(x 4)5/2 6 (x 4)5/2 dx Evaluating the remaining integral, 15x(x 4)3/2 dx 6x(x 4)5/2 12 –– 7 (x 4)7/2 c 14.20. Redo Problem 14.19, given 2x (x 8)3 dx Let f(x) 2x, f(x) 2, and g(x) (x 8)3; then g(x) (x 8)3 dx 1 – 2(x 8)2. Substituting in (14.1), 2x (x 8)3 dx 2x 1 2 (x 8)2 1 2 (x 8)2 2 dx x(x 8)2 (x 8)2 dx Integrating for the last time, 2x (x 8)3 dx x(x 8)2 (x 8)1 c x (x 8)2 1 x 8 c 14.21. Redo Problem 14.19, given 5x (x 1)2 dx Let f(x) 5x, f(x) 5, and g(x) (x 1)2; then g(x) (x 1)2 dx (x 1)1. Substituting in (14.1), 5x (x 1)2 dx 5x[(x 1)1] (x 1)15 dx 5x(x 1)1 5 (x 1)1 dx Integrating again, 5x (x 1)2 dx 5x(x 1)1 5 ln x 1 c 5x x 1 5 ln x 1 c 14.22. Redo Problem 14.19, given 6xex7 dx.
Let f(x) 6x, f(x) 6, g(x) ex7, and g(x) ex7 dx ex7. Using (14.1), 6xex7 dx 6xex7 ex76 dx 6xex7 6
ex7 dx Integrating again, 6xex7 dx 6xex7 6ex7 c 338 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 14.23. Use integration by parts to evaluate 16xe(x9) dx.
Let f(x) 16x, f(x) 16, g(x) e(x9), and g(x) e(x9) dx e(x9). Using (14.1), 16xe(x9) dx 16xe(x9) e(x9)16 dx 16xe(x9) 16 e(x9) dx Integrating once more, 16xe(x9) dx 16xe(x9) 16e(x9) c 14.24. Redo Problem 14.23, given x2e2x dx.
Let f(x) x2, f(x) 2x, g(x) e2x, and g(x) e2x dx 1 – 2e2x. Substituting in (14.1). x2e2x dx x2(1 – 2e2x) 1 – 2e2x(2x) dx 1 – 2x2e2x xe2x dx (14.2) Using parts again for the remaining integral, f(x) x, f(x) 1, g(x) e2x, and g(x) e2x dx 1 – 2e2x.
Using (14.1), xe2x dx x(1 – 2e2x) 1 – 2e2x dx 1 – 2xe2x 1 – 2(1 – 2e2x) Finally, substituting in (14.2), x2e2x dx 1 – 2x2e2x 1 – 2xe2x 1 – 4e2x c ECONOMIC APPLICATIONS 14.25. The rate of net investment is I 40t3/5, and capital stock at t 0 is 75. Find the capital function K.
K I dt 40t3/5 dt 40(5 – 8t8/5) c 25t8/5 c Substituting t 0 and K 75, 75 0 c c 75 Thus, K 25t8/5 75.
14.26. The rate of net investment is I 60t1/3, and capital stock at t 1 is 85. Find K.
K 60t1/3 dt 45t4/3 c At t 1 and K 85, 85 45(1) c c 40 Thus, K 45t4/3 40.
14.27. Marginal cost is given by MC dTC/dQ 25 30Q 9Q2. Fixed cost is 55. Find the (a) total cost, (b) average cost, and (c) variable cost functions.
a) TC MC dQ (25 30Q 9Q2) dQ 25Q 15Q2 3Q3 c With FC 55, at Q 0, TC FC 55. Thus, c FC 55 and TC 25Q 15Q2 3Q3 55.
339 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] b) AC TC Q 25 15Q 3Q2 55 Q c) VC TC FC 25Q 15Q2 3Q3 14.28. Given MC dTC/dQ 32 18Q 12Q2, FC 43. Find the (a) TC, (b) AC, and (c) VC functions.
a) TC MC dQ (32 18Q 12Q2) dQ 32Q 9Q2 4Q3 c At Q 0, TC FC 43, TC 32Q 9Q2 4Q3 43.
b) AC TC Q 32 9Q 4Q2 43 Q c) VC TC FC 32Q 9Q2 4Q3 14.29. Marginal revenue is given by MR dTR/dQ 60 2Q 2Q2. Find (a) the TR function and (b) the demand function P f(Q).
a) TR MR dQ (60 2Q 2Q2) dQ 60Q Q2 2 – 3Q3 c At Q 0, TR 0. Therefore c 0. Thus, TR 60Q Q2 2 – 3Q3.
b) TR PQ. Therefore, P TR/Q, which is the same as saying that the demand function and the average revenue function are identical. Thus, P AR TR/Q 60 Q 2 – 3Q2.
14.30. Find (a) the total revenue function and (b) the demand function, given MR 84 4Q Q2 a) TR MR dQ (84 4Q Q2) dQ 84Q 2Q2 1 – 3Q3 c At Q 0, TR 0. Therefore c 0. Thus, TR 84Q 2Q2 1 – 3Q3.
b) P AR TR Q 84 2Q 1 3 Q2 14.31. With C f(Y), the marginal propensity to consume is given by MPC dC/dY f(Y). If the MPC 0.8 and consumption is 40 when income is zero, find the consumption function.
C f(Y) dY 0.8 dY 0.8Y c At Y 0, C 40. Thus, c 40 and C 0.8Y 40.
14.32. Given dC/dY 0.6 0.1/ 3 Y MPC and C 45 when Y 0. Find the consumption function.
C 0.6 0.1 3 Y dY (0.6 0.1Y1/3) dY 0.6Y 0.15Y2/3 c At Y 0, C 45. Thus, C 0.6Y 0.15Y2/3 45.
340 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL [CHAP . 14 14.33. The marginal propensity to save is given by dS/dY 0.5 0.2Y1/2. There is dissaving of 3.5 when income is 25, that is, S 3.5 when Y 25. Find the savings function.
S (0.5 0.2Y1/2) dY 0.5Y 0.4Y 1/2 c At Y 25, S 3.5.
3.5 0.5(25) 0.4(25) c c 14 Thus, S 0.5Y 0.4Y 1/2 14.
14.34. Given MC dTC/dQ 12e0.5Q and FC 36. Find the total cost.
TC 12e0.5Q dQ 12 1 0.5 e0.5Q c 24e0.5Q c With FC 36, TC 36 when Q 0. Substituting, 36 24e0.5(0) c. Since e0 1, 36 24 c, and c 12.
Thus, TC 24e0.5Q 12. Notice that c does not always equal FC.
14.35. Given MC 16e0.4Q and FC 100. Find TC.
TC 16e0.4Q dQ 16 1 0.4 e0.4Q c 40e0.4Q c At Q 0, TC 100.
100 40e0 c c 60 Thus, TC 40e0.4Q 60.
341 INTEGRAL CALCULUS: THE INDEFINITE INTEGRAL CHAP . 14] CHAPTER 15 Integral Calculus: The Definite Integral 15.1 AREA UNDER A CURVE There is no geometric formula for the area under an irregularly shaped curve, such as y f(x) between x a and x b in Fig. 15-1(a). If the interval [a, b] is divided into n subintervals [x1, x2], [x2, x3], etc., and rectangles are constructed such that the height of each is equal to the smallest value of the function in the subinterval, as in Fig. 15-1(b), then the sum of the areas of the rectangles n i1 f(xi) xi, called a Riemann sum, will approximate, but underestimate, the actual area under the curve. The smaller the subintervals (the smaller the xi), the more rectangles are created and the closer the combined area of the rectangles n i1 f(xi) xi approaches the actual area under the curve. If 342 Fig. 15-1 the number of subintervals is increased so that n →, each subinterval becomes infinitesimal ( xi dxi dx) and the area A under the curve can be expressed mathematically as A lim n→ n i1 f(xi) xi 15.2 THE DEFINITE INTEGRAL The area under a graph of a continuous function such as that in Fig. 15-1 from a to b (a b) can be expressed more succinctly as the definite integral of f(x) over the interval a to b. Put mathematically, b a f(x) dx lim n→ n i1 f(xi) xi Here the left-hand side is read, ‘‘the integral from a to b of f of x dx.’’ Here a is called the lower limit of integration and b the upper limit of integration. Unlike the indefinite integral which is a set of functions containing all the antiderivatives of f(x), as explained in Example 3 of Chapter 14, the definite integral is a real number which can be evaluated by using the fundamental theorem of calculus (Section 15.3).
15.3 THE FUNDAMENTAL THEOREM OF CALCULUS The fundamental theorem of calculus states that the numerical value of the definite integral of a continuous function f(x) over the interval from a to b is given by the indefinite integral F(x) c evaluated at the upper limit of integration b, minus the same indefinite integral F(x) c evaluated at the lower limit of integration a. Since c is common to both, the constant of integration is eliminated in subtraction. Expressed mathematically, b a f(x) dx F(x) b a F(b) F(a) where the symbol b a, ]b a, or [ · · · ] b a indicates that b and a are to be substituted successively for x. See Examples 1 and 2 and Problems 15.1 to 15.10.
EXAMPLE 1.
The definite integrals given below (1) 4 1 10x dx (2) 3 1 (4x3 6x) dx are evaluated as follows: 1) 4 1 10x dx 5x2 4 1 5(4)2 5(1)2 75 2) 3 1 (4x3 6x) dx [x4 3x2]3 1 [(3)4 3(3)2] [(1)4 3(1)2] 108 4 104 EXAMPLE 2.
The definite integral is used below to determine the area under the curve in Fig. 15-2 over the interval 0 to 20 as follows: A 20 0 1 – 2x dx 1 – 4x2 20 0 1 – 4(20)2 1 – 4(0)2 100 343 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] The answer can be checked by using the geometric formula A 1 – 2xy: A 1 – 2xy 1 – 2(20)(10) 100 15.4 PROPERTIES OF DEFINITE INTEGRALS 1.
Reversing the order of the limits changes the sign of the definite integral.
b a f(x) dx a b f(x) dx (15.1) 2.
If the upper limit of integration equals the lower limit of integration, the value of the definite integral is zero.
a a f(x) dx F(a) F(a) 0 (15.2) 3.
The definite integral can be expressed as the sum of component subintegrals.
c a f(x) dx b a f(x) dx c b f(x) dx a b c (15.3) 4.
The sum or difference of two definite integrals with identical limits of integration is equal to the definite integral of the sum or difference of the two functions.
b a f(x) dx b a g(x) dx b a [ f(x) g(x)] dx (15.4) 5.
The definite integral of a constant times a function is equal to the constant times the definite integral of the function.
b a kf(x) dx k b a f(x) dx (15.5) See Example 3 and Problems 15.11 to 15.14.
344 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 Fig. 15-2 EXAMPLE 3.
To illustrate a sampling of the properties presented above, the following definite integrals are evaluated: 1.
3 1 2x3 dx 1 3 2x3 dx 3 1 2x3 dx 1 – 2x4 3 1 1 – 2(3)4 1 – 2(1)4 40 Checking this answer, 1 3 2x3 dx 1 – 2x4 1 3 1 – 2(1)4 1 – 2(3)4 40 2.
5 5 (2x 3) dx 0 Checking this answer, 5 5 (2x 3) dx [x2 3x]5 5 [(5)2 3(5)] [(5)2 3(5)] 0 3.
4 0 6x dx 3 0 6x dx 4 3 6x dx 4 0 6x dx 3x2 4 0 3(4)2 3(0)2 48 3 0 6x dx 3x2 3 0 3(3)2 3(0)2 27 4 3 6x dx 3x2 4 3 3(4)2 3(3)2 21 Checking this answer, 48 27 21 15.5 AREA BETWEEN CURVES The area of a region between two or more curves can be evaluated by applying the properties of definite integrals outlined above. The procedure is demonstrated in Example 4 and treated in Problems 15.15 to 15.18.
EXAMPLE 4.
Using the properties of integrals, the area of the region between two functions such as y1 3x2 6x 8 and y2 2x2 4x 1 from x 0 to x 2 is found in the following way: a) Draw a rough sketch of the graph of the functions and shade in the desired area as in Fig. 15-3.
345 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] Fig. 15-3 b) Note the relationship between the curves. Since y1 lies above y2, the desired region is simply the area under y1 minus the area under y2 between x 0 and x 2. Hence, A 2 0 (3x2 6x 8) dx 2 0 (2x2 4x 1) dx From (15.4), A 2 0 [(3x2 6x 8) (2x2 4x 1)] dx 2 0 (5x2 10x 7) dx (5 – 3x3 5x2 7x) 2 0 71 – 3 0 71 – 3 15.6 IMPROPER INTEGRALS The area under some curves that extend infinitely far along the x axis, as in Fig. 15-4(a), may be estimated with the help of improper integrals. A definite integral with infinity for either an upper or lower limit of integration is called an improper integral.
a f(x) dx and b f(x) dx are improper integrals because is not a number and cannot be substituted for x in F(x). They can, however, be defined as the limits of other integrals, as shown below.
a f(x) dx lim b→ b a f(x) dx and b f(x) dx lim a→ b a f(x) dx If the limit in either case exists, the improper integral is said to converge. The integral has a definite value, and the area under the curve can be evaluated. If the limit does not exist, the improper integral diverges and is meaningless. See Example 5 and Problems 15.19 to 15.25.
EXAMPLE 5.
The improper integrals given below (a) 1 3 x2 dx (b) 1 6 x dx are sketched in Fig. 15-4(a) and (b) and evaluated as follows: a) 1 3 x2 dx lim b→ b 1 3 x2 dx lim b→ 3 x b 1 lim b→ 3 b (3) 1 lim b→ 3 b 3 3 346 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 Fig. 15-4 because as b →, 3/b →0. Hence the improper integral is convergent and the area under the curve in Fig. 15-4(a) equals 3.
b) 1 6 x dx lim b→ b 1 6 x dx lim b→ [6 ln x ]b 1 lim b→ [6 ln b 6 ln 1 ] lim b→ [6 ln b ] since ln 1 0 As b →, 6 ln b →. The improper integral diverges and has no definite value. The area under the curve in Fig. 15-4(b) cannot be computed even though the graph is deceptively similar to the one in (a).
15.7 L’HO ˆ PITAL’S RULE If the limit of a function f(x) g(x)/h(x) as x →a cannot be evaluated, such as (1) when both numerator and denominator approach zero, giving rise to the indeterminate form 0/0, or (2) when both numerator and denominator approach infinity, giving rise to the indeterminate form /, L’Ho ˆpital’s rule can often be helpful. L’Ho ˆpital’s rule states: lim x→a g(x) h(x) lim x→a g(x) h(x) (15.6) It is illustrated in Example 6 and Problem 15.26.
EXAMPLE 6.
The limits of the functions given below are found as follows, using L’Ho ˆpital’s rule. Note that numerator and denominator are differentiated separately, not as a quotient.
(a) lim x→4 x 4 16 x2 (b) lim x→ 6x 2 7x 4 a) As x →4, x 4 and 16 x2 →0. Using (15.6), therefore, and differentiating numerator and denominator separately, lim x→4 x 4 16 x2 lim x→4 1 2x 1 8 b) As x →, both 6x 2 and 7x 4 →. Using (15.6), lim x→ 6x 2 7x 4 lim x→ 6 7 6 7 15.8 CONSUMERS’ AND PRODUCERS’ SURPLUS A demand function P1 f1(Q), as in Fig. 15-5(a), represents the different prices consumers are willing to pay for different quantities of a good. If equilibrium in the market is at (Q0, P0), then the consumers who would be willing to pay more than P0 benefit. Total benefit to consumers is represented by the shaded area and is called consumers’ surplus. Mathematically, Consumers’ surplus Q0 0 f1(Q) dQ Q0P0 (15.7) A supply function P2 f2(Q), as in Fig. 15-5(b), represents the prices at which different quantities of a good will be supplied. If market equilibrium occurs at (Q0, P0), the producers who would supply at a lower price than P0 benefit. Total gain to producers is called producers’ surplus and is designated by the shaded area. Mathematically, Producers’ surplus Q0P0 Q0 0 f2(Q) dQ (15.8) See Example 7 and Problems 15.27 to 15.31.
347 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] EXAMPLE 7.
Given the demand function P 42 5Q Q2. Assuming that the equilibrium price is 6, the consumers’ surplus is evaluated as follows: At P0 6, 42 5Q Q2 6 36 5Q Q2 0 (Q 9)(Q 4) 0 So Q0 4, because Q 9 is not feasible. Substituting in (15.7), Consumers’ surplus 4 0 (42 5Q Q2) dQ (4)(6) [42Q 2.5Q2 1 – 3Q3]4 0 24 (168 40 211 – 3) 0 24 822 – 3 15.9 THE DEFINITE INTEGRAL AND PROBABILITY The probability P that an event will occur can be measured by the corresponding area under a probability density function. A probability density or frequency function is a continuous function f(x) such that: 1.
f(x) 0. Probability cannot be negative.
2.
f(x) dx 1. The probability of the event occurring over the entire range of x is 1.
3.
P(a x b) b a f(x) dx. The probability of the value of x falling within the interval [a, b] is the value of the definite integral from a to b.
See Example 8 and Problems 15.32 and 15.33.
EXAMPLE 8.
The time in minutes between cars passing on a highway is given by the frequency function f(t) 2e2t for t 0. The probability of a car passing in 0.25 minute is calculated as follows: P 0.25 0 2e2t dt e2t 0.25 0 e0.5 (e0) 0.606531 1 0.393469 348 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 Fig. 15-5 Solved Problems DEFINITE INTEGRALS 15.1.
Evaluate the following definite integrals: a) 6 0 5x dx 6 0 5x dx 2.5x2 6 0 2.5(6)2 2.5(0)2 90 b) 10 1 3x2 dx 10 1 3x2 dx x3 10 1 (10)3 (1)3 999 c) 64 1 x2/3 dx 64 1 x2/3 dx 3x1/3 64 1 3 3 64 3 3 1 9 d) 3 1 (x3 x 6) dx 3 1 (x3 x 6) dx (1 – 4x4 1 – 2x2 6x) 3 1 1 – 4(3)4 1 – 2(3)2 6(3) [1 – 4(1)4 1 – 2(1)2 6(1)] 36 e) 4 1 (x1/2 3x1/2) dx 4 1 (x1/2 3x1/2) dx (2x1/2 2x3/2) 4 1 24 243 (21 213) 16 f) 3 0 4e2x dx 3 0 4e2x dx 2e2x 3 0 2(e2(3) e2(0)) 2(403.4 1) 804.8 g) 10 0 2e2x dx 10 0 2e2x dx e2x 10 0 e2(10) (e2(0)) e20 e0 1 SUBSTITUTION METHOD 15.2.
Use the substitution method to integrate the following definite integral: 3 0 8x(2x2 3) dx 349 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] Let u 2x2 3. Then du/dx 4x and dx du/4x. Ignore the limits of integration for the moment, and treat the integral as an indefinite integral. Substituting in the original integrand, 8x(2x2 3) dx 8xu du 4x 2 u du Integrating with respect to u, 2 u du 2 u2 2 c u2 c (15.9) Finally, by substituting u 2x2 3 in (15.9) and recalling that c will drop out in the integration, the definite integral can be written in terms of x, incorporating the original limits: 3 0 8x(2x2 3) dx (2x2 3)2 3 0 [2(3)2 3]2 [2(0)2 3]2 441 9 432 Because in the original substitution u x but 2x2 3, the limits of integration in terms of x will differ from the limits of integration in terms of u. The limits can be expressed in terms of u, if so desired. Since we have set u 2x2 3 and x ranges from 0 to 3, the limits in terms of u are u 2(3)2 3 21 and u 2(0)2 3 3. Using these limits with the integral expressed in terms of u, as in (15.9), 2 21 3 u du u2 21 3 441 9 432 15.3.
Redo Problem 15.2, given 2 1 x2(x3 5)2 dx.
Let u x3 5, du/dx 3x2, and dx du/3x2. Substituting independently of the limits, x2(x3 5)2 dx x2u2 du 3x2 1 3 u2 du Integrating with respect to u and ignoring the constant, 1 – 3 u2 du 1 – 3(1 – 3u3) 1 – 9u3 Substituting u x3 5 and incorporating the limits for x, 2 1 x2(x3 5)2 dx [1 – 9(x3 5)3]2 1 1 – 9[(2)3 5]3 1 – 9[(1)3 5]3 1 – 9(27) 1 – 9(64) 10.11 Since u x3 5 and the limits for x are x 1 and x 2, by substitution the limits for u are u (1)3 5 4 and u (2)3 5 3. Incorporating these limits for the integral with respect to u, 1 – 3 3 4 u2 du [1 – 9u3]3 4 1 – 9(3)3 1 – 9(4)3 10.11 15.4.
Redo Problem 15.2, given 2 0 3x2 (x3 1)2 dx Let u x3 1. Then du/dx 3x2 and dx du/3x2. Substituting, 3x2 (x3 1)2 dx 3x2u2 du 3x2 u2 du 350 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 Integrating with respect to u and ignoring the constant, u2 du u1 Substituting u x3 1 with the original limits, 2 0 3x2 (x3 1)2 dx (x3 1)1 2 0 1 23 1 1 03 1 1 9 1 8 9 With u x3 1, and the limits of x ranging from 0 to 2, the limits of u are u (0)3 1 1 and u (2)3 1 9. Thus, 9 1 u2 du u1 9 1 (1 – 9) (1 – 1) 8 – 9 15.5.
Integrate the following definite integral by means of the substitution method: 3 0 6x x2 1 dx Let u x2 1, du/dx 2x, and dx du/2x. Substituting, 6x x2 1 dx 6xu1 du 2x 3 u1 du Integrating with respect to u, 3 u1 du 3 ln u Substituting u x2 1, 3 0 6x x2 1 dx 3 ln x2 1 3 0 3 ln 32 1 3 ln 02 1 3 ln 10 3 ln 1 3 ln 10 6.9078 Since ln 1 0, The limits of u are u (0)2 1 1 and u (3)2 1 10. Integrating with respect to u, 3 10 1 u1 du 3 ln u 10 1 3 ln 10 3 ln 1 3 ln 10 6.9078 15.6.
Redo Problem 15.5, given 2 1 4xex22 dx.
Let u x2 2. Then du/dx 2x and dx du/2x. Substituting, 4xex22 dx 4xeu du 2x 2 eu du Integrating with respect to u and ignoring the constant, 2 eu du 2eu Substituting u x2 2, 2 1 4xex22 dx 2ex22 2 1 2(e(2)22 e(1)22) 2(e6 e3) 2(403.43 20.09) 766.68 351 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] With u x2 2, the limits of u are u (1)2 2 3 and u (2)2 2 6.
2 6 3 eu du 2eu 6 3 2(e6 e3) 766.68 15.7.
Redo Problem 15.5, given 1 0 3x2e2x31 dx.
Let u 2x3 1, du/dx 6x2, and dx du/6x2. Substituting, 3x2e2x31 dx 3x2eu du 6x2 1 2 eu du Integrating with respect to u, 1 – 2 eu du 1 – 2eu Substituting u 2x3 1, 1 0 3x2e2x31 dx 1 – 2e2x31 1 0 1 – 2(e3 e1) 1 – 2(20.086 2.718) 8.684 With u 2x3 1, the limits of u are u 2(0)3 1 1 and u 2(1)3 1 3. Thus 1 – 2 3 1 eu du 1 – 2eu 3 1 1 – 2(e3 e1) 8.68 INTEGRATION BY PARTS 15.8.
Integrate the following definite integral, using the method of integration by parts: 5 2 3x (x 1)2 dx Let f(x) 3x; then f(x) 3. Let g(x) (x 1)2; then g(x) (x 1)2 dx (x 1)1. Sub-stituting in (14.1), 3x (x 1)2 dx 3x[(x 1)1] (x 1)13 dx 3x(x 1)1 3 (x 1)1 dx Integrating and ignoring the constant, 3x (x 1)2 dx 3x(x 1)1 3 ln x 1 Applying the limits, 5 2 3x (x 1)2 dx [3x(x 1)1 3 ln x 1 ]5 2 3(5) 5 1 3 ln 5 1 3(2) 2 1 3 ln 2 1 5 – 2 3 ln 6 2 3 ln 3 3(ln 6 ln 3) 1 – 2 3(1.7918 1.0986) 0.5 1.5796 352 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 15.9.
Redo Problem 15.8, given 3 1 4x (x 2)3 dx Let f(x) 4x, f(x) 4, g(x) (x 2)3, and g(x) (x 2)3 dx 1 – 2(x 2)2. Substituting in (14.1), 4x (x 2)3 dx 4x 1 2 (x 2)2 1 2 (x 2)24 dx 2x(x 2)2 2 (x 2)2 dx Integrating, 4x (x 2)3 dx 2x(x 2)2 2(x 2)1 Applying the limits, 3 1 4x (x 2)3 dx [2x(x 2)2 2(x 2)1]3 1 [2(3)(3 2)2 2(3 2)1] [2(1)(1 2)2 2(1 2)1] 6 –– 25 2 – 5 2 – 9 2 – 3 56 ––– 225 15.10. Redo Problem 15.8, given 3 1 5xex2 dx.
Let f(x) 5x, f(x) 5, g(x) ex2, and g(x) ex2 dx ex2. Applying (14.1), 5xex2 dx 5xex2 ex25 dx 5xex2 5 ex2 dx Integrating, 5xex2 dx 5xex2 5ex2 Applying the limits, 3 1 5xex2 dx [5xex2 5ex2]3 1 (15e5 5e5) (5e3 5e3) 10e5 10(148.4) 1484 PROPERTIES OF DEFINITE INTEGRALS 15.11. Show 4 4 (8x3 9x2) dx 0 4 (8x3 9x2) dx 4 0 (8x3 9x2) dx.
4 4 (8x3 9x2) dx 2x4 3x3 4 4 704 320 384 0 4 (8x3 9x2) dx 2x4 3x3 0 4 0 320 320 4 0 (8x3 9x2) dx 2x4 3x3 4 0 704 0 704 Checking this answer, 320 704 384 353 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] 15.12. Show 16 0 (x1/2 3x) dx 4 0 (x1/2 3x) dx 9 4 (x1/2 3x) dx 16 9 (x1/2 3x) dx.
16 0 (x1/2 3x) dx 2x1/2 1.5x2 16 0 392 0 392 4 0 (x1/2 3x) dx 2x1/2 1.5x2 4 0 28 0 28 9 4 (x1/2 3x) dx 2x1/2 1.5x2 9 4 127.5 28 99.5 16 9 (x1/2 3x) dx 2x1/2 1.5x2 16 9 392 127.5 264.5 Checking this answer, 28 99.5 264.5 392 15.13. Show 3 0 6x x2 1 dx 1 0 6x x2 1 dx 2 1 6x x2 1 dx 3 2 6x x2 1 dx From Problem 15.5, 3 0 6x x2 1 dx 3 ln x2 1 3 0 3 ln 10 1 0 6x x2 1 dx 3 ln x2 1 1 0 3 ln 2 0 3 ln 2 2 1 6x x2 1 dx 3 ln x2 1 2 1 3 ln 5 3 ln 2 3 2 6x x2 1 dx 3 ln x2 1 3 2 3 ln 10 3 ln 5 Checking this answer, 3 ln 2 3 ln 5 3 ln 2 3 ln 10 3 ln 5 3 ln 10 15.14. Show 3 1 5xex2 dx 2 1 5xex2 dx 3 2 5xex2 dx.
From Problem 15.10, 3 1 5xex2 dx [5xex2 5ex2]3 1 10e5 2 1 5xex2 dx [5xex2 5ex2]2 1 (10e4 5e4) (5e3 5e3) 5e4 3 2 5xex2 dx [5xex2 5ex2]3 2 (15e5 5e5) (10e4 5e4) 10e5 5e4 Checking this answer, 5e4 10e5 5e4 10e5 AREA BETWEEN CURVES 15.15. (a) Draw the graphs of the following functions, and (b) evaluate the area between the curves over the stated interval: y1 7 x and y2 4x x2 from x 1 to x 4 a) See Fig. 15-6.
b) From Fig. 15-6, the desired region is the area under the curve specified by y1 7 x from x 1 to 354 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 x 4 minus the area under the curve specified by y2 4x x2 from x 1 to x 4. Using the properties of definite integrals, A 4 1 [1 – 3x3 2.5x2 7x]4 1 [1 – 3(4)3 2.5(4)2 7(4)] [1 – 3(1)3 2.5(1)2 7(1)] 4.5 (7 x) dx 4 1 (4x x2) dx 4 1 (x2 5x 7) dx 15.16. Redo Problem 15.15, given y1 6 x and y2 4 from x 0 to x 5 Notice the shift in the relative positions of the curves at the point of intersection.
a) See Fig. 15-7.
b) From Fig. 15-7, the desired area is the area between y1 6 x and y2 4 from x 0 to x 2 plus the area between y2 4 and y1 6 x from x 2 to x 5. Mathematically, A 2 0 [(6 x) 4] dx 5 2 [4 (6 x)] dx 2 0 (2 x) dx 5 2 (x 2) dx [2x 1 – 2x2]2 0 [1 – 2x2 2x]5 2 2 0 2.5 (2) 6.5 15.17. Redo Problem 15.15, given y1 x2 4x 8 and y2 2x from x 0 to x 3 a) See Fig. 15-8.
b) A 2 0 [(x2 4x 8) 2x] dx 3 2 [2x (x2 4x 8)] dx 2 0 (x2 6x 8) dx 3 2 (x2 6x 8) dx [1 – 3x3 3x2 8x]2 0 [1 – 3x3 3x2 8x]3 2 71 – 3 15.18. Redo Problem 15.15, given y1 x2 4x 12 and y2 x2 from x 0 to x 4 a) See Fig. 15-9.
355 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] Fig. 15-6 Fig. 15-7 b) A 3 0 [(x2 4x 12) x2] dx 4 3 [x2 (x2 4x 12)] dx 3 0 (12 4x) dx 4 3 (4x 12) dx [12x 2x2]3 0 [2x2 12x]4 3 20 IMPROPER INTEGRALS AND L’HO ˆ PITAL’S RULE 15.19. (a) Specify why the integral given below is improper and (b) test for convergence. Evaluate where possible.
1 2x (x2 1)2 dx a) This is an example of an improper integral because the upper limit of integration is infinite.
b) 1 2x (x2 1)2 dx lim b→ b 1 2x (x2 1)2 dx Let u x2 1, du/dx 2x, and dx du/2x. Substituting, 2x (x2 1)2 dx 2xu2 du 2x u2 du Integrating with respect to u and ignoring the constant, u2 du u1 Substituting u x2 1 and incorporating the limits of x, 1 2x (x2 1)2 dx lim b→ b 1 2x (x2 1)2 dx (x2 1)1 b 1 1 b2 1 1 (1)2 1 1 2 1 b2 1 As b →, 1/(b2 1) →0. The integral converges and has a value of 1 – 2..
356 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 Fig. 15-8 Fig. 15-9 15.20. Redo Problem 15.19, given 1 dx x 7 a) This is an improper integral because one of its limits of integration is infinite.
b) 1 dx x 7 lim b→ b 1 dx x 7 ln x 7 b 1 ln b 7 ln 1 7 As b →, ln b 7 →. The integral diverges and is meaningless.
15.21. Redo Problem 15.19, given 0 e3x dx.
a) The lower limit is infinite.
b) 0 e3x dx lim a→ 0 a e3x dx 1 – 3e3x 0 a 1 – 3e3(0) 1 – 3e3a 1 – 3 1 – 3e3a As a →, 1 – 3e3a →0. The integral converges and has a value of 1 – 3.
15.22. (a) Specify why the integral given below is improper and (b) test for convergence. Evaluate where possible: 0 (5 x)2 dx a) The lower limit is infinite.
b) 0 (5 x)2 dx lim a→ 0 a (5 x)2 dx Let u 5 x, du/dx 1, and dx du. Substituting, (5 x)2 dx u2(du) u2 du Integrating with respect to u, u2 du u1 Substituting u 5 x and incorporating the limits of x, 0 (5 x)2 dx lim a→ 0 a (5 x)2 dx (5 x)1 0 a 1 5 0 1 5 a 1 5 1 5 a As a →, 1/(5 a) →0. The integral converges and equals 1 – 5.
15.23. Redo Problem 15.22, given 0 2xex dx.
a) The lower limit is infinite.
b) 0 2xex dx lim a→ 0 a 2xex dx 357 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] Using integration by parts, let f(x) 2x, f(x) 2, g(x) ex, and g(x) ex dx ex. Substituting in (14.1), 2xex dx 2xex ex2 dx Integrating once again, 2xex dx 2xex 2ex Incorporating the limits, 0 2xex dx [2(0)e0 2e0] (2aea 2ea) 2 2aea 2ea since e0 1 lim a→ 0 a 2xex dx (2xex 2ex) 0 a As a →, ea →0. Therefore the integral converges and has a value of 2.
15.24. Redo Problem 15.22, given 6 0 dx x 6 a) This is also an improper integral because, as x approaches 6 from the left (x →6), the integrand →.
b) 6 0 dx x 6 ln b 6 ln 0 6 lim b→6 b 0 dx x 6 ln x 6 b 0 As b →6, b 6 →0 and ln 0 is undefined. Therefore, the integral diverges and is meaningless.
15.25. Redo Problem 15.22, given 8 0 (8 x)1/2 dx.
a) As x →8, the integrand approaches infinity.
b) 8 0 (8 x)1/2 dx (28 b) (28 0) 28 28 b lim b→8 b 0 (8 x)1/2 dx 2(8 x)1/2 b 0 As b →8, 28 b →0. The integral converges and has a value of 28 42.
15.26. Use L’Ho ˆpital’s rule to evaluate the following limits: a) lim x→ 5x 9 ex As x →, both 5x 9 and ex tend to , giving rise to the indeterminate form /. Using (15.6), therefore, and differentiating numerator and denominator separately, lim x→ 5x 9 ex lim x→ 5 ex 5 0 b) lim x→ 1 e1/x 1/x 358 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 As x →, 1 e1/x and 1/x →0. Using (15.6), therefore, and recalling that 1/x x1, lim x→ 1 e1/x 1/x lim x→ (1/x2)e1/x 1/x2 Simplifying algebraically, lim x→ 1 e1/x 1/x lim x→ (e1/x) e0 1 c) lim x→ ln 2x e5x As x →, ln 2x and e5x →. Again using (15.6), lim x→ ln 2x e5x lim x→ 1/x 5e5x 0 0 since 0 is not an indeterminate form.
d) lim x→ 6x3 7 3x2 9 lim x→ 6x3 7 3x2 9 lim x→ 18x2 6x lim x→ 3x e) lim x→ 3x2 7x 4x2 21 lim x→ 3x2 7x 4x2 21 lim x→ 6x 7 8x Whenever application of L’Ho ˆpital’s rule gives rise to a new quotient whose limit is also an indeterminate form, L’Ho ˆpital’s rule must be applied again. Thus, lim x→ 6x 7 8x lim x→ 6 8 3 4 See Problem 3.4(c).
f) lim x→ 8x3 5x2 13x 2x3 7x2 18x Using L’Ho ˆpital’s rule repeatedly, lim x→ 8x3 5x2 13x 2x3 7x2 18x lim x→ lim x→ 24x2 10x 13 6x2 14x 18 4 – 1 8 – 2 4 lim x→ 48x 10 12x 14 CONSUMERS’ AND PRODUCERS’ SURPLUS 15.27. Given the demand function P 45 0.5Q, find the consumers’ surplus CS when P0 32.5 and Q0 25.
Using (15.7), CS 25 0 [45(25) 0.25(25)2] 0 812.5 156.25 (45 0.5Q) dQ (32.5)(25) [45Q 0.25Q2]0 25 812.5 359 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] 15.28. Given the supply function P (Q 3)2, find the producers’ surplus PS at P0 81 and Q0 6.
From (15.8), PS (81)(6) 486 [1 – 3(6 3)3 1 – 3(0 3)3] 252 6 0 (Q 3)2 dQ 486 [1 – 3(Q 3)3]6 0 15.29. Given the demand function Pd 25 Q2 and the supply function Ps 2Q 1. Assuming pure competition, find (a) the consumers’ surplus and (b) the producers’ surplus.
For market equilibrium, s d. Thus, 2Q 1 25 Q2 Q2 2Q 24 0 (Q 6) (Q 4) 0 Q0 4 P0 9 since Q0 cannot equal 6.
a) CS 4 0 [25(4) 1 – 3(4)3] 0 36 42.67 (25 Q2) dQ (9) (4) [25Q 1 – 3Q3]4 0 36 b) PS (9) (4) 36 [Q2 Q]4 0 16 4 0 (2Q 1) dQ 15.30. Given the demand function Pd 113 Q2 and the supply function Ps (Q 1)2 under pure competition, find (a) CS and (b) PS.
Multiplying the supply function out and equating supply and demand, Q2 2Q 1 113 Q2 2(Q2 Q 56) 0 (Q 8) (Q 7) 0 Q0 7 P0 64 a) CS 7 0 (113 Q2) dQ (64) (7) [113Q 1 – 3Q3]7 0 448 228.67 b) PS (64) (7) 7 0 (Q 1)2 dQ 448 [1 – 3(Q 1)3]0 7 448 (170.67 0.33) 277.67 15.31. Under a monopoly, the quantity sold and market price are determined by the demand function.
If the demand function for a profit-maximizing monopolist is P 274 Q2 and MC 4 3Q, find the consumers’ surplus.
Given P 274 Q2, TR PQ (274 Q2)Q 274Q Q3 and MR dTR dQ 274 3Q2 The monopolist maximizes profit at MR MC. Thus, 274 3Q2 4 3Q 3(Q2 Q 90) 0 (Q 10) (Q 9) 0 Q0 9 P0 193 and CS 9 0 (274 Q2) dQ (193) (9) [274Q 1 – 3Q3]9 0 1737 486 360 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL [CHAP . 15 FREQUENCY FUNCTIONS AND PROBABILITY 15.32. The probability in minutes of being waited on in a large chain restaurant is given by the frequency function f(t) 4 –– 81t3 for 0 t 3. What is the probability of being waited on between 1 and 2 minutes?
P 2 1 4 –– 81t3 dt 1 –– 81t4 2 1 1 –– 81(16) 1 –– 81(1) 0.1852 15.33. The proportion of assignments completed within a given day is described by the probability density function f(x) 12(x2 x3) for 0 x 1. What is the probability that (a) 50 percent or less of the assignments will be completed within the day and (b) 50 percent or more will be completed?
a) Pa 0.5 0 12(x2 x3) dx 12 x3 3 x4 4 0.5 0 12 0.125 3 0.0625 4 0 0.3125 b) Pb 1 0.5 12(x2 x3) dx 12 x3 3 x4 4 1 0.5 12 1 3 1 4 0.125 3 0.0625 4 0.6875 As expected, Pa Pb 0.3125 0.6875 1.
OTHER ECONOMIC APPLICATIONS 15.34. Given I(t) 9t1/2, find the level of capital formation in (a) 8 years and (b) for the fifth through the eighth years (interval [4, 8]).
a) K 8 0 9t1/2 dt 6t3/2 8 0 6(8)3/2 0 962 135.76 b) K 8 4 9t1/2 dt 6t3/2 8 4 6(8)3/2 6(4)3/2 135.76 48 87.76 361 INTEGRAL CALCULUS: THE DEFINITE INTEGRAL CHAP . 15] CHAPTER 16 First-Order Differential Equations 16.1 DEFINITIONS AND CONCEPTS A differential equation is an equation which expresses an explicit or implicit relationship between a function y f(t) and one or more of its derivatives or differentials. Examples of differential equations include dy dt 5t 9 y 12y and y 2y 19 0 Equations involving a single independent variable, such as those above, are called ordinary differential equations. The solution or integral of a differential equation is any equation, without derivative or differential, that is defined over an interval and satisfies the differential equation for all the values of the independent variable(s) in the interval. See Example 1.
The order of a differential equation is the order of the highest derivative in the equation. The degree of a differential equation is the highest power to which the derivative of highest order is raised.
See Example 2 and Problem 16.1.
EXAMPLE 1.
To solve the differential equation y
(t) 7 for all the functions y(t) which satisfy the equation, simply integrate both sides of the equation to find the integrals.
y(t) 7 dt 7t c1 y(t) (7t c1)dt 3.5t2 c1t c This is called a general solution which indicates that when c is unspecified, a differential equation has an infinite number of possible solutions. If c can be specified, the differential equation has a particular or definite solution which alone of all possible solutions is relevant.
362 EXAMPLE 2.
The order and degree of differential equations are shown below.
1.
dy dt 2x 6 first-order, first-degree 2. dy dt 4 5t5 0 first-order, fourth-degree 3.
d2y dt2 dy dt 3 x2 0 second-order, first-degree 4. d2y dt2 7 d3y dt3 5 75y third-order, fifth-degree 16.2 GENERAL FORMULA FOR FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS For a first-order linear differential equation, dy/dt and y must be of the first degree, and no product y(dy/dt) may occur. For such an equation dy dt vy z where v and z may be constants or functions of time, the formula for a general solution is y(t) ev dtA zev dt dt (16.1) where A is an arbitrary constant. A solution is composed of two parts: ev dtA is called the complementary function, and ev dt zev dt dt is called the particular integral. The particular integral yp equals the intertemporal equilibrium level of y(t); the complementary function yc represents the deviation from the equilibrium. For y(t) to be dynamically stable, yc must approach zero as t approaches infinity (that is, k in ekt must be negative). The solution of a differential equation can always be checked by differentiation. See Examples 3 to 5, Problems 16.2 to 16.12, and Problem 20.33.
EXAMPLE 3.
The general solution for the differential equation dy/dt 4y 12 is calculated as follows. Since v 4 and z 12, substituting in (16.1) gives y(t) e 4 dtA 12e 4 dt dt From Section 14.2, 4 dt 4t c. When (16.1) is used, c is always ignored and subsumed under A. Thus, y(t) e4tA 12e4t dt (16.2) Integrating the remaining integral gives 12e4t dt 3e4t c. Ignoring the constant again and substituting in (16.2), y(t) e4t(A 3e4t) Ae4t 3 (16.3) since e4te4t e0 1. As t →, yc Ae4t →0 and y(t) approaches yp 3, the intertemporal equilibrium level.
y(t) is dynamically stable.
To check this answer, which is a general solution because A has not been specified, start by taking the derivative of (16.3).
dy dt 4Ae4t 363 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] From the original problem, dy dt 4y 12 dy dt 12 4y Substituting y Ae4t 3 from (16.3), dy dt 12 4(Ae4t 3) 4Ae4t EXAMPLE 4.
Given dy/dt 3t2y t2 where v 3t2 and z t2. To find the general solution, first substitute in (16.1), y(t) e 3t2 dtA t2e 3t2 dt dt (16.4) Integrating the exponents, 3t2 dt t3. Substituting in (16.4), y(t) et3A t2et3 dt (16.5) Integrating the remaining integral in (16.5) calls for the substitution method. Letting u t3, du/dt 3t2, and dt du/3t2, t2et3 dt t2 eudu 3t2 1 3 eu du 1 3 eu 1 3 et3 Finally, substituting in (16.5), y(t) et3(A 1 – 3et3) Aet3 1 – 3 (16.6) As t →, yc Aet3 →0 and y(t) approaches 1 – 3. The equilibrium is dynamically stable.
Differentiating (16.6) to check the general solution, dy/dt 3t2Aet3. From the original problem, dy dt 3t2y t2 dy dt t2 3t2y Substituting y from (16.6), dy dt t2 3t2Aet3 1 3 3t2Aet3 EXAMPLE 5.
Suppose that y(0) 1 in Example 4. The definite solution is calculated as follows: From (16.6), y Aet3 1 – 3. At t 0, y(0) 1. Hence, 1 A 1 – 3 since e0 1, and A 2 – 3. Substituting A 2 – 3 in (16.6), the definite solution is y 2 – 3et3 1 – 3.
16.3 EXACT DIFFERENTIAL EQUATIONS AND PARTIAL INTEGRATION Given a function of more than one independent variable, such as F(y, t) where M F/y and N F/t, the total differential is written dF(y, t) M dy N dt (16.7) Since F is a function of more than one independent variable, M and N are partial derivatives and Equation (16.7) is called a partial differential equation. If the differential is set equal to zero, so that M dy N dt 0, it is called an exact differential equation because the left side exactly equals the differential of the primitive function F(y, t). For an exact differential equation, M/t must equal N/y, that is, 2F/(ty) 2F/(yt). For proof of this proposition, see Problem 16.49.
Solution of an exact differential equation calls for successive integration with respect to one independent variable at a time while holding constant the other independent variable(s). The 364 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 procedure, called partial integration, reverses the process of partial differentiation. See Example 6 and Problems 16.13 to 16.17.
EXAMPLE 6.
Solve the exact nonlinear differential equation (6yt 9y2) dy (3y2 8t) dt 0 (16.8) 1.
Test to see if it is an exact differential equation. Here M 6yt 9y2 and N 3y2 8t. Thus, M/t 6y and N/y 6y. If M/t N/y, it is not an exact differential equation.
2.
Since M F/y is a partial derivative, integrate M partially with respect to y by treating t as a constant, and add a new function Z(t) for any additive terms of t which would have been eliminated by the original differentiation with respect to y. Note that y replaces dy in partial integration.
F(y, t) (6yt 9y2) y Z(t) 3y2t 3y3 Z(t) (16.9) This gives the original function except for the unknown additive terms of t, Z(t).
3.
Differentiate (16.9) with respect to t to find F/t (earlier called N). Thus, F t 3y2 Z(t) (16.10) Since F/t N and N 3y2 8t from (16.8), substitute F/t 3y2 8t in (16.10).
3y2 8t 3y2 Z(t) Z(t) 8t 4.
Next integrate Z(t) with respect to t to find the missing t terms.
Z(t) Z(t) dt 8t dt 4t2 (16.11) 5.
Substitute (16.11) in (16.9), and add a constant of integration.
F(y, t) 3y2t 3y3 4t2 c This is easily checked by differentiation.
16.4 INTEGRATING FACTORS Not all differential equations are exact. However, some can be made exact by means of an integrating factor. This is a multiplier which permits the equation to be integrated. See Example 7 and Problems 16.18 to 16.22.
EXAMPLE 7.
Testing the nonlinear differential equation 5yt dy (5y2 8t) dt 0 reveals that it is not exact.
With M 5yt and N 5y2 8t, M/t 5y N/y 10y. Multiplying by an integrating factor of t, however, makes it exact: 5yt2 dy (5y2t 8t2) dt 0. Now M/t 10yt N/y, and the equation can be solved by the procedure outlined above. See Problem 16.22.
To check the answer to a problem in which an integrating factor was used, take the total differential of the answer and then divide by the integrating factor.
16.5 RULES FOR THE INTEGRATING FACTOR Two rules will help to find the integrating factor for a nonlinear first-order differential equation, if such a factor exists. Assuming M/t N/y, Rule 1.
If 1 N M t N y f(y) alone, then e f (y) dy is an integrating factor.
365 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] Rule 2.
If 1 M N y M t g(t) alone, then e g(t) dt is an integrating factor.
See Example 8 and Problems 16.23 to 16.28.
EXAMPLE 8.
To illustrate the rules above, find the integrating factor given in Example 7, where 5yt dy (5y2 8t) dt 0 M 5yt N 5y2 8t M t 5y N y 10y Applying Rule 1, 1 5y2 8t (5y 10y) 5y 5y2 8t which is not a function of y alone and will not supply an integrating factor for the equation. Applying Rule 2, 1 5yt (10y 5y) 5y 5yt 1 t which is a function of t alone. The integrating factor, therefore, is e (1/t) dt eln t t.
16.6 SEPARATION OF VARIABLES Solution of nonlinear first-order first-degree differential equations is complex. (A first-order first-degree differential equation is one in which the highest derivative is the first derivative dy/dt and that derivative is raised to a power of 1. It is nonlinear if it contains a product of y and dy/dt, or y raised to a power other than 1.) If the equation is exact or can be rendered exact by an integrating factor, the procedure outlined in Example 6 can be used. If, however, the equation can be written in the form of separated variables such that R(y) dy S(t) dt 0, where R and S, respectively, are functions of y and t alone, the equation can be solved simply by ordinary integration. The procedure is illustrated in Examples 9 and 10 and Problems 16.29 to 16.37.
EXAMPLE 9.
The following calculations illustrate the separation of variables procedure to solve the nonlinear differential equation dy dt y2t (16.12) First, separating the variables by rearranging terms, dy y2 t dt where R 1/y2 and S t. Then integrating both sides, y2 dy t dt y1 c1 t2 2 c2 1 y t2 2c2 2c1 2 Letting c 2c2 2c1, y 2 t2 c (16.13) Since the constant of integration is arbitrary until it is evaluated to obtain a particular solution, it will be treated generally and not specifically in the initial steps of the solution. ec and ln c can also be used to express the constant.
366 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 This solution can be checked as follows: Taking the derivatives of y 2(t2 c)1 by the generalized power function rule, dy dt (1)(2)(t2 c)2(2t) 4t (t2 c)2 From (16.12), dy/dt y2t. Substituting into (16.12) from (16.13), dy dt 2 t2 c 2 t 4t (t2 c)2 EXAMPLE 10.
Given the nonlinear differential equation t2 dy y3 dt 0 (16.14) where M f(y) and N f(t). But multiplying (16.14) by 1/(t2y3) to separate the variables gives 1 y3 dy 1 t2 dt 0 (16.14a) Integrating the separated variables, y3 dy t2 dt 1 – 2y2 t1 c and F(y, t) 1 – 2y2 t1 c 1 2y2 1 t c For complicated functions, the answer is frequently left in this form. It can be checked by differentiating and comparing with (16.14a), which can be reduced to (16.14) through multiplication by y3t2. For other forms in which an answer can be expressed, see Problems 16.20 to 16.22 and 16.29 to 16.35.
16.7 ECONOMIC APPLICATIONS Differential equations serve many functions in economics. They are used to determine the conditions for dynamic stability in microeconomic models of market equilibria and to trace the time path of growth under various conditions in macroeconomics. Given the growth rate of a function, differential equations enable the economist to find the function whose growth is described; from point elasticity, they enable the economist to estimate the demand function (see Example 11 and Problems 16.38 to 16.47). In Section 14.6 they were used to estimate capital functions from investment functions and total cost and total revenue functions from marginal cost and marginal revenue functions.
EXAMPLE 11.
Given the demand function Qd c bP and the supply function Qs g hP, the equilibrium price is P ¯ c g h b (16.15) Assume that the rate of change of price in the market dP/dt is a positive linear function of excess demand Qd Qs such that dP dt m(Qd Qs) m a constant 0 (16.16) The conditions for dynamic price stability in the market [i.e., under what conditions P(t) will converge to P ¯ as t →] can be calculated as shown below.
367 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] Substituting the given parameters for Qd and Qs in (16.16), dP dt m[(c bP) (g hP)] m(c bP g hP) Rearranging to fit the general format of Section 16.2, dP/dt m(h b)P m(c g). Letting v m(h b) and z m(c g), and using (16.1), P(t) e v dtA ze v dt dt evtA zevt dt evtA zevt v Aevt z v (16.17) At t 0, P(0) A z/v and A P(0) z/v.
Substituting in (16.17), P(t) P(0) z v evt z v Finally, replacing v m(h b) and z m(c g), P(t) P(0) c g h b em(hb)t c g h b and making use of (16.15), the time path is P(t) [P(0) P ¯ ]em(hb)t P ¯ (16.18) Since P(0), P ¯ , m 0, the first term on the right-hand side will converge toward zero as t →, and thus P(t) will converge toward P ¯ only if h b 0. For normal cases where demand is negatively sloped (b 0) and supply is positively sloped (h 0), the dynamic stability condition is assured. Markets with positively sloped demand functions or negatively sloped supply functions will also be dynamically stable as long as h b.
16.8 PHASE DIAGRAMS FOR DIFFERENTIAL EQUATIONS Many nonlinear differential equations cannot be solved explicitly as functions of time. Phase diagrams, however, offer qualitative information about the stability of equations that is helpful in determining whether the equations will converge to an intertemporal (steady-state) equilibrium or not.
A phase diagram of a differential equation depicts the derivative which we now express as y · for simplicity of notation as a function of y. The steady-state solution is easily identified on a phase diagram as any point at which the graph crosses the horizontal axis, because at that point y · 0 and the function is not changing. For some equations there may be more than one intersection and hence more than one solution.
Diagrammatically, the stability of the steady-state solution(s) is indicated by the arrows of motion.
The arrows of motion will point to the right (indicating y is increasing) any time the graph of y · is above the horizontal axis (indicating y · 0) and to the left (indicating y is decreasing) any time the graph of y · is below the horizontal axis (indicating y · 0). If the arrows of motion point towards a steady-state solution, the solution is stable; if the arrows of motion point away from a steady-state solution, the solution is unstable.
Mathematically, the slope of the phase diagram as it passes through a steady-state equilibrium point tells us if the equilibrium point is stable or not. When evaluated at a steady-state equilibrium point, if dy · dy 0, the equilibrium is stable if dy · dy 0, the point is unstable 368 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 Phase diagrams are illustrated in Example 12. The derivative test for stability is used in Example 13.
See also Problems 16.48 to 16.50.
EXAMPLE 12.
Given the nonlinear differential equation y · 8y 2y2 a phase diagram can be constructed and employed in six easy steps.
1.
The intertemporal or steady-state solution(s), where there is no pressure for change, is found by setting y · 0 and solving algebraically.
y · 8y 2y2 0 2y(4 y) 0 y ¯ 0 y ¯ 4 steady-state solutions The phase diagram will pass through the horizontal axis at y 0, y 4.
2.
Since the function passes through the horizontal axis twice, it has one turning point. We next determine whether that point is a maximum or minimum.
dy · dy 8 4y 0 y 2 is a critical value d2y · dy2 4 0 concave, relative maximum 3.
A rough, but accurate, sketch of the phase diagram can then easily be drawn. See Fig. 16-1.
4.
The arrows of motion complete the graph. As explained above, where the graph lies above the horizontal axis, y · 0 and the arrows must point to the right; where the graph lies below the horizontal axis, y · 0 and the arrows must point to the left.
5.
The stability of the steady-state equilibrium points can now be read from the graph. Since the arrows of motion point away from the first intertemporal equilibrium y ¯1 0, y ¯1 is an unstable equilibrium. With the arrows of motion pointing toward the second intertemporal equilibrium y ¯2 4, y ¯2 is a stable equi-librium.
6.
The slope of the phase diagram at the steady-state solutions can also be used to test stability independently of the arrows of motion. Since the slope of the phase diagram is positive at y ¯1 0, we can conclude y ¯1 is an unstable equilibrium. Since the slope of the phase diagram is negative at y ¯2 4, we know y ¯2 must be stable.
EXAMPLE 13.
Without even resorting to a phase diagram, we can use the simple first-derivative evaluated at the intertemporal equilibrium level(s) to determine the stability of differential equations. Given y · 8y 2y2, dy · dy 8 4y 369 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] y y .
0 2 4 Fig. 16-1 Evaluated at the steady-state levels, y ¯1 0 and y ¯2 4, dy · dy (0) 8 4(0) 8 0 dy · dy (4) 8 4(4) 8 0 y ¯1 0 is unstable y ¯2 4 is stable Solved Problems ORDER AND DEGREE 16.1.
Specify the order and degree of the following differential equations: a) d2y dx2 dy dx 3 12x b) dy dx 3x2 c) d3y dx3 4 d2y dx2 6 4 y d) d2y dx2 3 d4y dx4 75y 0 e) d3y dx3 x2y d2y dx2 4y4 0 (a) Second order, first degree; (b) first order, first degree; (c) third order, fourth degree; (d) fourth order, first degree; (e) third order, first degree.
FIRST-ORDER FIRST-DEGREE LINEAR DIFFERENTIAL EQUATIONS 16.2.
(a) Use the formula for a general solution to solve the following equation. (b) Check your answer.
dy dt 5y 0 (16.19) a) Here v 5 and z 0. Substituting in (16.1), y(t) e 5 dtA 0e 5 dtdt Integrating the exponents, 5 dt 5t c, where c can be ignored because it is subsumed under A.
Thus, y(t) e5t(A 0 dt). And 0 dt k, a constant, which can also be subsumed under A.
Hence, y(t) e5tA Ae5t (16.20) b) Taking the derivative of (16.20), dy/dt 5Ae5t. From (16.19), dy/dt 5y. Substituting y from (16.20), dy dt 5(Ae5t) 5Ae5t 16.3.
Redo Problem 16.2, given dy dt 3y y(0) 2 (16.21) 370 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 a) Rearranging to obtain the general format, dy dt 3y 0 Here v 3 and z 0. Substituting in (16.1), y(t) e3 dtA 0e3 dt dt Substituting 3 dt 3t, y(t) e3t(A 0 dt) Ae3t. At t 0, y 2. Thus, 2 Ae3(0), A 2.
Substituting, y(t) 2e3t (16.22) b) Taking the derivative of (16.22), dy/dt 6e3t. From (16.21), dy/dt 3y. Substituting y from (16.22), dy/dt 3(2e3t) 6e3t.
16.4.
Redo Problem 16.2, given dy dt 15 (16.23) a) Here v 0 and z 15. Thus, y(t) e 0 dtA 15e 0 dt dt where 0 dt k, a constant. Substituting and recalling that ek is also a constant, y(t) ekA 15ek dt ek(A 15tek) Aek 15t 15t A (16.24) where A is an arbitrary constant equal to Aek or simply c. Whenever the derivative is equal to a constant, simply integrate as in Example 1.
b) Taking the derivative of (16.24), dy/dt 15. From (16.23), dy/dt 15.
16.5.
Redo Problem 16.2, given dy dt 6y 18 (16.25) a) Here v 6, z 18, and 6 dt 6t. Substituting in (16.1), y(t) e6tA 18e6t dt where 18e6t dt 3e6t. Thus, y(t) e6t(A 3e6t) Ae6t 3 (16.26) b) Taking the derivative of (16.26), dy/dt 6Ae6t. From (16.25), dy/dt 18 6y. Substituting y from (16.26), dy/dt 18 6(Ae6t 3) 6Ae6t.
16.6.
Redo Problem 16.2, given dy dt 4y 20 y(0) 10 (16.27) 371 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] a) Here v 4, z 20, and 4 dt 4t. Thus, y(t) e4tA 20e4t dt where 20e4t dt 5e4t. Substituting, y(t) e4t(A 5e4t) Ae4t 5. At t 0, y 10. Thus, 10 Ae4(0) 5, and A 15. Substituting, y(t) 15e4t 5 5(3e4t 1) (16.28) b) The derivative of (16.28) is dy/dt 60e4t. From (16.27), dy/dt 20 4y. Substituting from (16.28) for y, dy/dt 20 4(15e4t 5) 60e4t.
16.7.
Redo Problem 16.2, given dy dt 4ty 6t (16.29) a) v 4t, z 6t, and 4t dt 2t2. Thus, y(t) e2t2A 6te2t2 dt (16.30) Using the substitution method for the remaining integral, let u 2t2, du/dt 4t, and dt du/4t.
Thus, 6te2t2 dt 6teu du 4t 1.5 eu du 1.5e2t2 Substituting back in (16.30), y(t) e2t2(A 1.5e2t2) Ae2t2 1.5 (16.31) b) The derivative of (16.31) is dy/dt 4tAe2t2. From (16.29), dy/dt 6t 4ty. Substituting from (16.31), dy/dt 6t 4t(Ae2t2 1.5) 4tAe2t2.
16.8.
(a) Solve the equation below using the formula for a general solution. (b) Check your answer.
2 dy dt 2t2y 9t2 y(0) 2.5 (16.32) a) Dividing through by 2, dy/dt t2y 4.5t2. Thus, v t2, z 4.5t2, and t2 dt 1 – 3t3. Substituting, y(t) e(1/3)t3A 4.5t2e(1/3)t3 dt (16.33) Let u 1 – 3t3, du/dt t2, and dt du/t2. Thus, 4.5t2e(1/3)t3 dt 4.5t2eu du t 2 4.5 eu du 4.5e(1/3)t3 Substituting in (16.33), y(t) e(1/3)t3(A 4.5e(1/3)t3) Ae(1/3)t3 4.5 At t 0, 2.5 A 4.5; A 2. Thus, y(t) 2e(1/3)t3 4.5 (16.34) b) Taking the derivative of (16.34), dy/dt 2t2e(1/3)t3. From (16.32), dy/dt 4.5t2 t2y. Substituting from (16.34), dy/dt 4.5t2 t2(2e(1/3)t3 4.5) 2t2e(1/3)t3.
372 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 16.9.
Redo Problem 16.8, given dy dt 2ty et2 (16.35) a) v 2t, z et2, and 2t dt t2. Thus, y(t) et2A et2et2 dt et2A e0 dt where e0 1 and 1 dt t. Substituting back, y(t) et2(A t) (16.36) b) The derivative of (16.36), by the product rule, is dy/dt 2tet2(A t) et2(1) 2tAet2 2t2et2 et2.
From (16.35), dy/dt et2 2ty. Substituting from (16.36), dy dt et2 2t[et2(A t)] et2 2tAet2 2t2et2 16.10. Redo Problem 16.8, given dy dt 3y 6t y(0) 1 3 (16.37) a) v 3, z 6t, and 3 dt 3t. Then, y(t) e3tA 6te3t dt (16.38) Using integration by parts for the remaining integral, let f(t) 6t, then f(t) 6; let g(t) e3t, then g(t) e3t dt 1 – 3e3t. Substituting in (14.1), 6te3t dt 6t(1 – 3e3t) 1 – 3e3t6 dt 2te3t 2 e3t dt 2te3t 2 – 3e3t Substituting back in (16.38), y(t) e3t(A 2te3t 2 – 3e3t) Ae3t 2t 2 – 3 At t 0, 1 – 3 Ae3(0) 2(0) 2 – 3; A 1. Thus, y(t) e3t 2t 2 – 3 (16.39) b) Taking the derivative of (16.39), dy/dt 3e3t 2. From (16.37), dy/dt 6t 3y. Substituting (16.39) directly above, dy/dt 6t 3(e3t 2t 2 – 3) 3e3t 2.
16.11. Redo Problem 16.8, given dy dt y t 0 y(3) 12 (16.40) a) v 1/t, z 0, and (1/t) dt ln t. Thus, y(t) eln tA 0 dt At since eln t t. At t 3, 12 A(3); A 4. Thus, y(t) 4t (16.41) 373 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] b) The derivative of (16.41) is dy/dt 4. From (16.40), dy/dt y/t. Substituting from (16.41), dy/dt 4t/t 4.
16.12. Redo Problem 16.8, given dy dt y y(3) 20 (16.42) a) With rearranging, dy/dt y 0. Therefore, v 1, z 0, and 1 dt t. Thus, y(t) etA 0 dt Aet At t 3, 20 Ae3; 20 A(0.05), so A 400. Thus, y(t) 400et (16.43) b) Taking the derivative of (16.43), dy/dt 400et. From (16.42), dy/dt y. Substituting from (16.43), dy/dt (400et) 400et.
EXACT DIFFERENTIAL EQUATIONS AND PARTIAL INTEGRATION 16.13. Solve the following exact differential equation. Check the answer on your own.
(4y 8t2) dy (16yt 3) dt 0 As outlined in Example 6, 1.
Check to see if it is an exact differential equation. Letting M 4y 8t2 and N 16yt 3, M/t 16t N/y.
2.
Integrate M partially with respect to y and add Z(t) to get F(y, t).
F(y, t) (4y 8t2) y Z(t) 2y2 8t2y Z(t) (16.44) 3.
Differentiate F(y, t) partially with respect to t and equate with N above.
F t 16ty Z(t) But F/t N 16yt 3, so 16ty Z(t) 16yt 3 Z(t) 3 4.
Integrate Z(t) with respect to t to get Z(t).
Z(t) Z(t) dt 3 dt 3t (16.45) 5.
Substitute (16.45) in (16.44) and add a constant of integration.
F(y, t) 2y2 8t2y 3t c 16.14. Redo Problem 16.13, given (12y 7t 6) dy (7y 4t 9) dt 0.
1.
M/t 7 N/y.
2.
F(y, t) (12y 7t 6) y Z(t) 6y2 7yt 6y Z(t) 3.
F/t 7y Z(t). But F/t N 7y 4t 9, so 7y Z(t) 7y 4t 9 Z(t) 4t 9 374 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 4.
Z(t) (4t 9) dt 2t2 9t 5.
F(y, t) 6y2 7yt 6y 2t2 9t c 16.15. Redo Problem 16.13, given (12y2t2 10y) dy (8y3t) dt 0.
1.
M/t 24y2t N/y.
2.
F(y, t) (12y2t2 10y) y Z(t) 4y3t2 5y2 Z(t) 3.
F/t 8y3t Z(t). But N 8y3t, so 8y3t 8y3t Z(t) Z(t) 0 4.
Z(t) 0 dt k, which will be subsumed under c.
5.
F(y,t) 4y3t2 5y2 c 16.16. Redo Problem 16.13, given 8tyy (3t2 4y2).
By rearranging, 8ty dy (3t2 4y2) dt 8ty dy (3t2 4y2) dt 0 1.
M/t 8y N/y.
2.
F(y, t) 8ty y Z(t) 4ty2 Z(t) 3.
F/t 4y2 Z(t). But F/t N 3t2 4y2, so 4y2 Z(t) 3t2 4y2 Z(t) 3t2 4.
Z(t) 3t2 dt t3 5.
F(y, t) t3 4ty2 c 16.17. Redo Problem 16.13, given 60ty2y (12t3 20y3).
By rearranging, 60ty2 dy (12t3 20y3) dt 0 1.
M/t 60y2 N/y.
2.
F(y, t) 60ty2 y Z(t) 20ty3 Z(t) 3.
F/t 20y3 Z(t). But F/t N 12t3 20y3, so 20y3 Z(t) 12t3 20y3 Z(t) 12t3 4.
Z(t) 12t3 dt 3t4 5.
F(y, t) 3t4 20ty3 c INTEGRATING FACTORS 16.18. Use the integrating factors provided in parentheses to solve the following differential equation.
Check the answer on your own (remember to divide by the integrating factor after taking the total differential of the answer).
6t dy 12y dt 0 (t) 375 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] 1.
M/t 6 N/y 12. But multiplying by the integrating factor t, 6t2 dy 12yt dt 0 where M/t 12t N/y. Continuing with the new function, 2.
F(y, t) 6t2 y Z(t) 6t2y Z(t) 3.
F/t 12ty Z(t). But F/t N 12ty, so Z(t) 0.
4.
Z(t) 0 dt k, which will be subsumed under the c below.
5.
F(y, t) 6t2y c 16.19. Redo Problem 16.18, given t2 dy 3yt dt 0 (t) 1.
M/t 2t N/y 3t. But multiplying by t, t3 dy 3yt2 dt 0 where M/t 3t2 N/y.
2.
F(y, t) t3 y Z(t) t3y Z(t) 3.
F/t 3t2y Z(t). But F/t N 3t2y, so Z(t) 0 and F(y, t) t3y c.
16.20. Redo Problem 16.18, given dy dt y t 1 ty Rearranging, t dy y dt t dy y dt 0 1.
M/t 1 N/y 1. Multiplying by 1/(ty), dy y dt t 0 where M/t 0 N/y, since neither function contains the variable with respect to which it is being partially differentiated.
2.
F(y, t) 1 y y Z(t) ln y Z(t) 3.
F/t Z(t). But F/t N 1/t, so Z(t) 1/t.
4.
Z(t) 1 t dt ln t 5.
F(y, t) ln y ln t c which can be expressed in different ways. Since c is an arbitrary constant, we can write ln y ln t c. Making use of the laws of logs (Section 7.3), ln y ln t ln (y/t). Thus, ln (y/t) c. Finally, expressing each side of the equation as exponents of e, and recalling that eln x x, eln (y/t) ec y t ec or y tec For other treatments of c, see Problems 16.29 to 16.37.
376 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 16.21. Redo Problem 16.18, given 4t dy (16y t2) dt 0 (t3) 1.
M/t 4 N/y 16. Multiplying by t3, 4t4 dy (16t3y t5) dt 0 where M/t 16t3 N/y.
2.
F(y, t) 4t4 y Z(t) 4t4y Z(t) 3.
F/t 16t3y Z(t). But F/t N 16t3 y t5, so 16t3y Z(t) 16t3y t5 Z(t) t5 4.
Z(t) t5 dt 1 – 6t6 5.
F(y, t) 4t4y 1 – 6t6 c 24t4y t6 c or 24t4y t6 c 16.22. Redo Problem 16.18, given 5yt dy (5y2 8t) dt 0 (t) 1.
M/t 5y N/y 10y. Multiplying by t, as in Example 7, 5yt2 dy (5y2t 8t2) dt 0 where M/t 10yt N/y.
2.
F(y, t) 5yt2 y Z(t) 2.5y2t2 Z(t) 3.
F/t 5y2t Z(t). But F/t N 5y2t 8t2, so 5y2t Z(t) 5y2t 8t2 Z(t) 8t2 4.
Z(t) 8t2 dt 8 – 3t3 5.
F(y, t) 2.5y2t2 8 – 3t3 c 7.5y2t2 8t3 c FINDING THE INTEGRATING FACTOR 16.23. (a) Find the integrating factor for the differential equation given below, and (b) solve the equation, using the five steps from Example 6.
(7y 4t2) dy 4ty dt 0 (16.46) a) M/t 8t N/y 4t. Applying Rule 1 from Section 16.5, since M 7y 4t2 and N 4ty, 1 4ty (8t 4t) 4t 4ty 1 y f(y) alone Thus the integrating factor is e (1/y) dy eln y y b) Multiplying (16.46) by the integrating factor y, (7y2 4yt2) dy 4ty2 dt 0.
1.
M/t 8yt N/y. Thus, 2.
F(y, t) (7y2 4yt2) y Z(t) 7 – 3y3 2y2t2 Z(t) 3.
F t 4y2t Z(t) 377 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] 4.
F/t N 4y2t, so Z(t) 0 and Z(t) is a constant. Thus, 5.
F(y, t) 7 – 3y3 2y2t2 c 7y3 6y2t2 c 16.24. Redo Problem 16.23, given y3t dy 1 – 2y4 dt 0 (16.47) a) M/t y3 N/y 2y3. Applying Rule 1, 1 1 – 2y4 (y3 2y3) 2 y4 (y3) 2 y f(y) alone Thus, e2y1 dy e2 ln y eln y2 y2 b) Multiplying (16.47) by y2, yt dy 1 – 2y2 dt 0.
1.
M/t y N/y. Thus, 2.
F(y, t) yt y Z(t) 1 – 2y2t Z(t) 3.
F t 1 2 y2 Z(t) 4.
F/t N 1 – 2y2, so Z(t) 0, and Z(t) is a constant. Thus, 5.
F(y, t) 1 – 2y2t c 16.25. Redo Problem 16.23, given 4t dy (16y t2) dt 0 (16.48) a) M 4t, N 16y t2, and M/t 4 N/y 16. Applying Rule 1, 1 16y t2 (4 16) 12 16y t2 f(y) alone Applying Rule 2, 1 4t (16 4) 3 t g(t) alone Thus, e 3t1 dt e3 ln t eln t3 t3 b) Multiplying (16.48) by t3, 4t4 dy (16yt3 t5) dt 0 which was solved in Problem 16.21.
16.26. Redo Problem 16.23, given t2 dy 3yt dt 0 (16.49) a) Here M t2, N 3yt, and M/t 2t N/y 3t. Applying Rule 1, 1 3yt (2t 3t) t 3yt 1 3y f(y) alone Thus, e (1/3y) dy e(1/3) ln y eln y1/3 y1/3 Consequently, y1/3 is an integrating factor for the equation, although in Problem 16.19 t was given as an integrating factor. Let us check y1/3 first.
b) Multiplying (16.49) by y1/3, t2y1/3 dy 3ty2/3 dt 0.
1.
M/t 2ty1/3 N/y. Thus, 378 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 2.
F(y, t) t2y1/3 y Z(t) 1.5t2y2/3 Z(t) 3.
F t 3ty2/3 Z(t) 4.
F/t N 3ty2/3, so Z(t) 0 and Z(t) is a constant. Hence, 5.
F1(y, t) 1.5t2y2/3 c (16.50) Here F1 is used to distinguish this function from the function F2 below.
16.27. Test to see if t is a possible integrating factor in Problem 16.26.
Applying Rule 2 to the original equation, 1 t2 (3t 2t) t t2 1 t Thus, e (1/t) dt eln t t Hence t is also a possible integrating factor, as demonstrated in Problem 16.19, where the solution was F2(y, t) t3y c. This differs from (16.50) but is equally correct, as you can check on your own.
16.28. Redo Problem 16.23, given (y t) dy dt 0 (16.51) a) M y t, N 1, and M/t 1 N/y 0. Applying Rule 1, 1 1 (1 0) 1 f(y) alone Thus, e 1 dy ey b) Multiplying (16.51) by ey, (y t)ey dy ey dt 0 (16.52) 1.
M/t ey N/y. Thus.
2.
F(y, t) (y t)ey y Z(t) (16.53) which requires integration by parts. Let f(y) y t f(y) 1 g(y) ey g(y) ey dy ey Substituting in (14.1), (y t)ey y (y t)ey ey1dy (y t)ey ey Substituting in (16.53), F(y, t) (y t)ey ey Z(t).
3.
F t ey Z(t) 4.
F/t N ey in (16.52), so Z(t) 0 and Z(t) is a constant. Thus, 5.
F(y, t) (y t)ey ey c or (y 1)ey tey c 379 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] SEPARATION OF VARIABLES 16.29. Solve the following differential equation, using the procedure for separating variables described in Section 16.6.
dy dt 5t y Separating the variables, y dy 5t dt y dy 5t dt 0 Integrating each term separately, y2 2 5t2 2 c1 y2 5t2 Letting c 2c1, y2 5t2 2c1 c 16.30. Redo Problem 16.29, given a) dy dt t5 y4 b) t2 dy y2 dt 0 y4 dy t5 dt 0 dy y2 dt t2 0 Integrating, y5 5 t6 6 c1 Integrating, 1 y 1 t c 6y5 5t6 30c1 y t cty Letting c 30c1, 6y5 5t6 c 16.31. Redo Problem 16.29, given t dy y dt 0.
dy y dt t 0 Integrating, ln y ln t ln c (an arbitrary constant) By the rule of logs, ln yt ln c yt c 16.32. Use separation of variables to solve the following differential equation.
dy dt y Separating the variables, dy y dt Integrating both sides and using ln c for the constant of integration, ln y t ln c Then playing with the constant of integration for ultimate simplicity of the solution, ln y ln c t ln y c t 380 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 Setting both sides as exponents of e, y c et y cet 16.33. Redo Problem 16.32, given dy dt b ay Separating the variables and then multiplying both sides by 1, dy b ay dt dy ay b dt Integrating both sides and being creative once again with the constant of integration, 1 a ln(ay b) t 1 a ln c Multiplying both sides by a and rearranging, ln (ay b) at ln c ln ay b c at ay b c eat ay b ceat y Ceat b a where C c a 16.34. Redo Problem 16.32, given (t 5) dy (y 9) dt 0.
dy y 9 dt t 5 0 Integrating, ln (y 9) ln (t 5) ln c By the rule of logs, ln y 9 t 5 ln c y 9 t 5 c or y 9 c(t 5) 16.35. Using the procedure for separating variables, solve the differential equation dy 3t2y dt.
dy y 3t2 dt 0 Integrating, ln y t3 ln c 381 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] Expressing each side of the equation as an exponent of e, eln yt3 eln yet3 yet3 y eln c eln c c cet3 16.36. Redo Problem 16.35, given y2(t3 1) dy t2(y3 5) dt 0.
y2 y3 5 dy t2 t3 1 dt 0 Integrating by substitution, 1 – 3 ln (y3 5) 1 – 3 ln (t3 1) ln c ln [(y3 5)(t3 1)] ln c (y3 5)(t3 1) c 16.37. Redo Problem 16.35, given 3 dy t t2 1 dt 0 Integrating, 3y 1 – 2 ln (t2 1) c Setting the left-hand side as an exponent of e and ignoring c, because, as an arbitrary constant, it can be expressed equally well as c or ec, e3y(1/2) ln (t21) c e3yeln (t21)1/2 c e3y(t2 1)1/2 c USE OF DIFFERENTIAL EQUATIONS IN ECONOMICS 16.38. Find the demand function Q f(P) if point elasticity is 1 for all P 0.
dQ dP P Q 1 dQ dP Q P Separating the variables, dQ Q dP P 0 Integrating, ln Q ln P ln c QP c Q c P 16.39. Find the demand function Q f(P) if k, a constant.
dQ dP P Q k dQ dP kQ P Separating the variables, dQ Q k P dP 0 ln Q k ln P c QPk c Q cPk 382 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 16.40. Find the demand function Q f(P) if (5P 2P2)/Q and Q 500 when P 10.
dQ dP P Q (5P 2P2) Q dQ dP (5P 2P2) Q Q P (5 2P) Separating the variables, dQ (5 2P) dP 0 Integrating, Q 5P P2 c Q P2 5P c At P 10 and Q 500, 500 100 50 c c 650 Thus, Q 650 5P P2.
16.41. Derive the formula P P(0)eit for the total value of an initial sum of money P(0) set out for t years at interest rate i, when i is compounded continuously.
If i is compounded continuously, dP dt iP Separating the variables, dP P i dt 0 Integrating, ln P it c Setting the left-hand side as an exponent of e, eln Pit c Peit c P ceit At t 0, P P(0). Thus P(0) ce0, c P(0), and P P(0)eit.
16.42. Determine the stability condition for a two-sector income determination model in which C ˆ , I ˆ, Y ˆ are deviations of consumption, investment, and income, respectively, from their equilibrium values Ce, Ie, Ye. That is, C ˆ C(t) Ce, etc., where C ˆ is read ‘‘C hat.’’ Income changes at a rate proportional to excess demand C I Y, and C ˆ (t) gY ˆ (t) I ˆ(t) bY ˆ (t) dY ˆ (t) dt a(C ˆ I ˆ Y ˆ ) 0 a, b, g 1 Substituting the first two equations in the third, dY ˆ dt a(g b 1)Y ˆ Separating the variables and then integrating, dY ˆ Y ˆ a(g b 1) dt ln Y ˆ a(g b 1)t c eln Y ˆ ea(gb1)tc 383 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] Letting the constant ec c, Y ˆ cea(gb1)t At t 0, Y ˆ Y(0) Ye c.
Substituting above, Y ˆ [Y(0) Ye]ea(gb1)t.
Since Y ˆ Y(t) Ye, Y(t) Ye Y ˆ . Thus, Y(t) Ye [Y(0) Ye]ea(gb1)t As t →, Y(t) →Ye only if g b 1. The sum of the marginal propensity to consume g and the marginal propensity to invest b must be less than 1.
16.43. In Example 11 we found P(t) [P(0) P ¯ ]em(hb)t P ¯ . (a) Explain the time path if (1) the initial price P(0) P ¯, (2) P(0) P ¯, and (3) P(0) P ¯ . (b) Graph your findings.
a) 1) If the initial price equals the equilibrium price, P(0) P ¯ , the first term on the right disappears and P(t) P ¯ . The time path is a horizontal line, and adjustment is immediate. See Fig. 16-2.
2) If P(0) P ¯ , the first term on the right is positive. Thus P(t) P ¯ and P(t) approaches P ¯ from above as t → and the first term on the right →0.
3) If P(0) P ¯ , the first term on the right is negative. And P(t) P ¯ and approaches it from below as t → and the first term →0.
b) See Fig. 16-2.
16.44. A change in the rate of investment will affect both aggregate demand and the productive capability of an economy. The Domar model seeks to find the time path along which an economy can grow while maintaining full utilization of its productive capacity. If the marginal propensity to save s and the marginal capital-output ratio k are constant, find the investment function needed for the desired growth.
The change in aggregate demand is equal to the change in investment times the multiplier 1/s, dY dt 1 s dI dt (16.54) The change in productive capacity is equal to the change in the capital stock times the reciprocal of the marginal capital-output ratio, dQ dt 1 k dK dt 1 k I since dK dt I (16.55) Equating (16.54) and (16.55) for fully utilized capacity, 1 s dI dt 1 k I 1 s dI 1 k I dt 384 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 Fig. 16-2 Separating the variables, dI I s k dt 0 Integrating, ln I s k t c Ie(s/k)t c I ce(s/k)t At t 0, I(0) c, and I I(0)e(s/k)t.
Investment must grow at a constant rate determined by s/k: the savings rate divided by the capital-output ratio.
16.45. The Solow model examines equilibrium growth paths with full employment of both capital and labor. Based on the assumptions that 1.
Output is a linearly homogeneous function of capital and labor exhibiting constant returns to scale, Y f(K, L) (16.56) 2.
A constant proportion s of output is saved and invested, dK dt K · sY (16.57) 3.
The supply of labor is growing at a constant rate r, L L0ert (16.58) derive the differential equation in terms of the single variable K/L, which serves as the basis of the model.
Substituting Y from (16.56) in (16.57), dK dt sf(K, L) (16.59) Substituting L from (16.58) in (16.59), dK dt sf(K, L0ert) (16.60) This is the time path capital formation (dK/dt) must follow for full employment of a growing labor force.
Preparing to convert to a function of K/L, let z K/L, then K zL. Making use of (16.58), K zL0ert (16.61) Taking the derivative of (16.61) and using the product rule since z is a function of t, dK dt z(rL0ert) L0ert dz dt zr dz dt L0ert (16.62) Equating (16.60) and (16.62), sf(K, L0ert) zr dz dt L0ert (16.63) Since the left-hand side of (16.63) is a linearly homogeneous production function, we may divide both inputs by L0ert and multiply the function itself by L0ert without changing its value. Thus, sf(K, L0ert) sL0ertf K L0ert , 1 (16.64) 385 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] Substituting (16.64) in (16.63) and dividing both sides by L0ert sf K L0ert , 1 zr dz dt (16.65) Finally, substituting z for K/L0ert and subtracting zr from both sides, dz dt sf(z, 1) zr (16.66) which is a differential equation in terms of the single variable z and two parameters r and s, where z K/L, r the rate of growth of the labor force, and s the savings rate.
16.46. Assume that the demand for money is for transaction purposes only. Thus, Md kP(t)Q (16.67) where k is constant, P is the price level, and Q is real output. Assume Ms Md and is exogenously determined by monetary authorities. If inflation or the rate of change of prices is proportional to excess demand for goods in society and, from Walras’ law, an excess demand for goods is the same thing as an excess supply of money, so that dP(t) dt b(Ms Md) (16.68) find the stability conditions, when real output Q is constant.
Substituting (16.67) in (16.68), dP(t) dt bMs bkP(t)Q (16.69) If we let P ˆ P(t) Pe (16.70) where P ˆ is the deviation of prices from the equilibrium price level Pe, then taking the derivative of (16.70), dP ˆ dt dP(t) dt dPe dt But in equilibrium dPe/dt 0. Hence, dP ˆ dt dP(t) dt (16.71) Substituting in (16.69), dP ˆ dt bMs bkP(t)Q (16.72) In equilibrium, Ms Md kPeQ. Hence Ms kPeQ 0 and b(Ms kPeQ) 0. Subtracting this from (16.72), dP ˆ dt bMs bkP(t)Q bMs bkPeQ bkQ[P(t) Pe] bkQP ˆ (16.73) which is a differential equation. Separating the variables, dP ˆ P ˆ bkQ dt Integrating, ln P ˆ bkQt c, P ˆ AebkQt, where ec A.
386 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 Since b, k, Q 0, P ˆ →0 as t →, and the system is stable. To find the time path P(t) from P ˆ , see the conclusion of Problem 16.42, where Y(t) was derived from Y ˆ .
16.47. If the expectation of inflation is a positive function of the present rate of inflation dP(t) dt E h dP(t) dt (16.74) and the expectation of inflation reduces people’s desire to hold money, so that Md kP(t)Q g dP(t) dt E (16.75) check the stability conditions, assuming that the rate of inflation is proportional to the excess supply of money as in (16.68).
Substituting (16.74) in (16.75), Md kP(t)Q gh dP(t) dt (16.76) Substituting (16.76) in (16.68), dP(t) dt bMs bkP(t)Q gh dP(t) dt By a process similar to the steps involving (16.70) to (16.73), dP ˆ dt bMs bkP(t)Q bgh dP(t) dt bMs bkPeQ bkQP ˆ bgh dP(t) dt (16.77) Substituting (16.71) for dP(t)/dt in (16.77), dP ˆ dt bkQP ˆ bgh dP ˆ dt bkQP ˆ 1 bgh Separating the variables, dP ˆ P ˆ bkQ 1 bgh dt Integrating, ln P ˆ bkQt/(1 bgh) P ˆ AebkQt/(1bgh) Since b, k, Q 0, P ˆ →0 as t →, if bgh 1. Hence even if h is greater than 1, meaning people expect inflation to accelerate, the economy need not be unstable, as long as b and g are sufficiently small.
PHASE DIAGRAMS FOR DIFFERENTIAL EQUATIONS 16.48. (a) Construct a phase diagram for the following nonlinear differential equation and test the dynamic stability using (b) the arrows of motion, (c) the slope of the phase line, and (d) the derivative test.
y · 3y2 18y a) By setting y · 0, we find the intertemporal equilibrium solution(s) where the phase diagram crosses the horizontal axis.
3y(y 6) 0 y ¯1 0 y ¯2 6 387 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] We then find the critical value and whether it represents a maximum or minimum.
dy · dy 6y 18 0 y 3 critical value d2y · dy2 6 0 relative minimum Armed with this information, we can then draw a rough but accurate sketch of the graph, as in Fig. 16-3.
b) Above the horizontal axis, where y · 0, the arrows of motion point to the right; below the horizontal axis, y · 0 and the arrows of motion point to the left. Since the arrows of motion point towards y ¯1 0 and away from y ¯2 6, y ¯1 is stable and y ¯2 is unstable.
c) With the slope of the phase diagram negative as it passes through y ¯1 0, we know y ¯1 must be stable.
With a positive slope at y ¯2 6, y ¯2 must be unstable.
d) Taking the derivative of the equation, independently of the graph, and evaluating it at the critical values, we see dy · dy 6y 18 dy · dy (0) 6(0) 18 18 0 y ¯1 0 is stable dy · dy (6) 6(6) 18 18 0 y ¯2 6 is unstable 16.49. Repeat the exercise in Problem 16.48 for y · y2 6y 5 a) Setting y · 0, (y 1)(y 5) 0 y ¯1 1 y ¯2 5 Optimizing, dy · dy 2y 6 0 y 3 critical value d2y · dy2 2 0 relative maximum Then sketching the graph, as in Fig. 16-4.
388 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 Fig. 16-3 y y .
0 3 6 The phase diagram for y = 3y2 – 18y .
b) Since the arrows of motions point away from y ¯1 1 and towards y ¯2 5, y ¯1 is unstable while y ¯2 is a stable intertemporal equilibrium.
c) The positive slope at y ¯1 1 and the negative slope at y ¯2 5 indicate that y ¯1 is an unstable equilibrium and y ¯2 is a stable equilibrium.
d) dy · dy 2y 6 dy · dy (1) 2(1) 6 4 0 y ¯1 1 is unstable dy · dy (5) 2(5) 6 4 0 y ¯2 5 is stable 16.50. Repeat Problem 16.49, given y · y2 10y 16 a) (y 2)(y 8) 0 y ¯1 2 y ¯2 8 dy · dy 2y 10 0 y 5 critical value d2y · dy2 2 0 relative minimum Then sketching the graph, as in Fig. 16.5.
389 FIRST-ORDER DIFFERENTIAL EQUATIONS CHAP . 16] Fig. 16-4 y y .
0 2 5 The phase diagram for y = y2 – 10y + 16 .
8 Fig. 16-5 y y .
0 1 3 5 The phase diagram for y = –y2 + 6y – 5 .
b) Since the arrows of motion point towards y ¯1 2 and away from y ¯2 8, y ¯1 is stable while y ¯2 is an unstable intertemporal equilibrium.
c) The negative slope at y ¯1 2 and the positive slope at y ¯2 8 indicate that y ¯1 is a stable equilibrium and y ¯2 is an unstable equilibrium.
d) dy · dy 2y 10 dy · dy (2) 2(2) 10 6 0 y ¯1 2 is stable dy · dy (8) 2(8) 10 6 0 y ¯2 8 is unstable 390 FIRST-ORDER DIFFERENTIAL EQUATIONS [CHAP . 16 CHAPTER 17 First-Order Difference Equations 17.1 DEFINITIONS AND CONCEPTS A difference equation expresses a relationship between a dependent variable and a lagged independent variable (or variables) which changes at discrete intervals of time, for example, It f(Yt1), where I and Y are measured at the end of each year. The order of a difference equation is determined by the greatest number of periods lagged. A first-order difference equation expresses a time lag of one period; a second-order, two periods; etc. The change in y as t changes from t to t 1 is called the first difference of y. It is written y t yt yt1 yt (17.1) where is an operator replacing d/dt that is used to measure continuous change in differential equations. The solution of a difference equation defines y for every value of t and does not contain a difference expression. See Examples 1 and 2.
EXAMPLE 1.
Each of the following is a difference equation of the order indicated.
It Qs yt3 9yt2 2yt1 6yt 8 order 3 yt a(Yt1 Yt2) a bPt1 5yt order 2 order 1 order 1 Substituting from (17.1) for yt above, yt1 yt 5yt yt1 6yt order 1 EXAMPLE 2.
Given that the initial value of y is y0, in the difference equation yt1 byt (17.2) 391 a solution is found as follows. By successive substitutions of t 0, 1, 2, 3, etc. in (17.2), y1 by0 y2 by1 b(by0) b2y0 y3 by2 b(b2y0) b3y0 y4 by3 b(b3y0) b4y0 Thus, for any period t, yt bty0 This method is called the iterative method. Since y0 is a constant, notice the crucial role b plays in determining values for y as t changes.
17.2 GENERAL FORMULA FOR FIRST-ORDER LINEAR DIFFERENCE EQUATIONS Given a first-order difference equation which is linear (i.e., all the variables are raised to the first power and there are no cross products), yt byt1 a (17.3) where b and a are constants, the general formula for a definite solution is yt y0 a 1 b bt a 1 b when b 1 (17.4) yt y0 at when b 1 (17.4a) If no initial condition is given, an arbitrary constant A is used for y0 a/(1 b) in (17.4) and for y0 in (17.4a). This is called a general solution. See Example 3 and Problems 17.1 to 17.13.
EXAMPLE 3.
Consider the difference equation yt 7yt1 16 and y0 5. In the equation, b 7 and a 16.
Since b 1, it is solved by using (17.4), as follows: yt 5 16 1 7 (7)t 16 1 7 3(7)t 2 (17.5) To check the answer, substitute t 0 and t 1 in (17.5).
y0 3(7)0 2 5 since (7)0 1 y1 3(7)1 2 19 Substituting y1 19 for yt and y0 5 for yt1 in the original equation, 19 7(5) 16 35 16 17.3 STABILITY CONDITIONS Equation (17.4) can be expressed in the general form yt Abt c (17.6) where A y0 a/(1 b) and c a/(1 b). Here Abt is called the complementary function and c is the particular solution. The particular solution expresses the intertemporal equilibrium level of y; the complementary function represents the deviations from that equilibrium. Equation (17.6) will be dynamically stable, therefore, only if the complementary function Abt →0, as t →. All depends on the base b. Assuming A 1 and c 0 for the moment, the exponential expression bt will generate seven different time paths depending on the value of b, as illustrated in Example 4. As seen there, if b 1, the time path will explode and move farther and farther away from equilibrium; if b 1, the time path will be damped and move toward equilibrium. If b 0, the time path will oscillate between positive and negative values; if b 0, the time path will be nonoscillating. If A 1, the value of the 392 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 multiplicative constant will scale up or down the magnitude of bt, but will not change the basic pattern of movement. If A 1, a mirror image of the time path of bt with respect to the horizontal axis will be produced. If c 0, the vertical intercept of the graph is affected, and the graph shifts up or down accordingly. See Examples 4 and 5 and Problems 17.1 to 17.13.
EXAMPLE 4.
In the equation yt bt, b can range from to . Seven different time paths can be generated, each of which is explained below and graphed in Fig. 17-1.
1.
If b 1, bt increases at an increasing rate as t increases, thus moving farther and farther away from the horizontal axis. This is illustrated in Fig. 17-1(a), which is a step function representing changes at discrete intervals of time, not a continuous function. Assume b 3. Then as t goes from 0 to 4, bt 1, 3, 9, 27, 81.
2.
If b 1, bt 1 for all values of t. This is represented by a horizontal line in Fig. 17-1(b).
3.
If 0 b 1, then b is a positive fraction and bt decreases as t increases, drawing closer and closer to the horizontal axis, but always remaining positive, as illustrated in Fig. 17-1(c). Assume b 1 – 3. Then as t goes from 0 to 4, bt 1, 1 – 3, 1 – 9, 1 –-27, 1 –– 81.
4.
If b 0, then bt 0 for all values of t. See Fig. 17-1(d).
5.
If 1 b 0, then b is a negative fraction; bt will alternate in sign and draw closer and closer to the horizontal axis as t increases. See Fig. 17-1(e). Assume b 1 – 3. Then as t goes from 0 to 4, bt 1, 1 – 3, 1 – 9, 1 –– 27, 1 –– 81.
393 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] Fig. 17-1 Time path of bt 6.
If b 1, then bt oscillates between 1 and 1. See Fig. 17-1(f).
7.
If b 1, then bt will oscillate and move farther and farther away from the horizontal axis, as illustrated in Fig. 17-1(g). Assume b 3. Then bt 1, 3, 9, 27, 81, as t goes from 0 to 4.
In short, if b 1 b 1 b 0 b 0 the time path explodes the time path converges the time path is nonoscillating the time path oscillates EXAMPLE 5.
In the equation yt 6(1 – 4)t 6, since b 1 – 4 0, the time path oscillates. Since b 1, the time path converges.
When yt 5(6)t 9 and b 6 0, there is no oscillation. With b 1, the time path explodes.
17.4 LAGGED INCOME DETERMINATION MODEL In the simple income determination model of Section 2.3 there were no lags. Now assume that consumption is a function of the previous period’s income, so that Ct C0 cYt1 Yt Ct It where It I0. Thus, Yt C0 cYt1 I0. Rearranging terms to conform with (17.3), Yt cYt1 C0 I0 (17.7) where b c and a C0 I0. Substituting these values in (17.4), since the marginal propensity to consume c cannot equal 1, and assuming Yt Y0 at t 0, Yt Y0 C0 I0 1 c (c)t C0 I0 1 c (17.8) The stability of the time path thus depends on c. Since 0 MPC 1, c 1 and the time path will converge. Since c 0, there will be no oscillations. The equilibrium is stable, and as t →, Yt →(C0 I0)/(1 c), which is the intertemporal equilibrium level of income. See Example 6 and Problems 17.14 to 17.20.
EXAMPLE 6.
Given Yt Ct It, Ct 200 0.9Yt1, It 100, and Y0 4500. Solving for Yt, Yt 200 0.9Yt1 100 0.9Yt1 300 (17.9) Using (17.4), Yt 4500 300 1 0.9 (0.9)t 300 1 0.9 1500(0.9)t 3000 (17.10) With 0.9 1, the time path converges; with 0.9 0, there is no oscillation. Thus, Yt is dynamically stable. As t →, the first term on the right-hand side goes to zero, and Yt approaches the intertemporal equilibrium level of income: 300/(1 0.9) 3000.
To check this answer, let t 0 and t 1 in (17.10). Thus, Y0 1500(0.9)0 3000 4500 Y1 1500(0.9)1 3000 4350 Substituting Y1 4350 for Yt and Y0 4500 for Yt1 in (17.9), 4350 0.9(4500) 300 4350 4050 300 394 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 17.5 THE COBWEB MODEL For many products, such as agricultural commodities, which are planted a year before marketing, current supply depends on last year’s prices. This poses interesting stability questions. If Qdt c bPt and Qst g hPt1 in equilibrium, c bPt g hPt1 (17.11) bPt hPt1 g c (17.12) Dividing (17.12) by b to conform to (17.3), Pt h b Pt1 g c b Since b 0 and h 0 under normal demand and supply conditions, h/b 1. Using (17.4), Pt P0 (g c)/b 1 h/b h b t (g c)/b 1 h/b P0 g c b h h b t g c b h (17.13) When the model is in equilibrium, Pt Pt1. Substituting Pe for Pt and Pt1 in (17.11) Pe g c b h (17.13a) Substituting in (17.13), Pt (P0 Pe) h b t Pe With an ordinary negative demand function and positive supply function, b 0 and h 0.
Therefore, h/b 0 and the time path will oscillate.
If h b , h/b 1, and the time path Pt explodes.
If h b , h/b 1, and the time oscillates uniformly.
If h b , h/b 1, and the time path converges, and Pt approaches Pe.
In short, when Q f(P) in supply-and-demand analysis, as is common in mathematics, the supply curve must be flatter than the demand curve for stability. See Example 7 and Problems 17.21 to 17.25.
But if P f(Q), as is typical in economics, the reverse is true. The demand curve must be flatter, or more elastic, than the supply curve if the model is to be stable.
EXAMPLE 7.
Given Qdt 86 0.8Pt and Qst 10 0.2Pt1, the market price Pt for any time period and the equilibrium price Pe can be found as follows. Equating demand and supply, 86 0.8Pt 10 0.2Pt1 0.8Pt 0.2Pt1 96 Dividing through by 0.8 to conform to (17.3), Pt 0.25Pt1 120. Using (17.4), Pt P0 120 1 0.25 (0.25)t 120 1 0.25 (P0 96)(0.25)t 96 which can be checked by substituting the appropriate values in (17.13). From (17.13a), Pe (10 86)/ (0.8 0.2) (96)/(1) 96.
With the base b 0.25, which is negative and less than 1, the time path oscillates and converges. The equilibrium is stable, and Pt will converge to Pe 96 as t →.
395 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] 17.6 THE HARROD MODEL The Harrod model attempts to explain the dynamics of growth in the economy. It assumes St sY t where s is a constant equal to both the MPS and APS. It also assumes the acceleration principle, i.e., investment is proportional to the rate of change of national income over time.
It a(Y t Y t1) where a is a constant equal to both the marginal and average capital-output ratios. In equilibrium, It St. Therefore, a(Y t Y t1) sY t (a s)Y t aY t1 Dividing through by a s to conform to (17.3), Y t [a/(a s)]Y t1. Using (17.4) since a/(a s) 1, Y t (Y 0 0) a a s t 0 a a s t Y 0 (17.14) The stability of the time path thus depends on a/(a s). Since a the capital-output ratio, which is normally larger than 1, and since s MPS which is larger than 0 and less than 1, the base a/(a s) will be larger than 0 and usually larger than 1. Therefore, Y t is explosive but nonoscillating. Income will expand indefinitely, which means it has no bounds. See Examples 8 and 9 and Problems 17.26 and 17.27. For other economic applications, see Problems 17.28 to 17.30.
EXAMPLE 8.
The warranted rate of growth (i.e., the path the economy must follow to have equilibrium between saving and investment each year) can be found as follows in the Harrod model.
From (17.14) Y t increases indefinitely. Income in one period is a/(a s) times the income of the previous period.
Y 1 a a s Y 0 (17.15) The rate of growth G between the periods is defined as G Y1 Y0 Y0 Substituting from (17.15), G [a/(a s)]Y 0 Y 0 Y 0 [a/(a s) 1]Y 0 Y 0 a a s 1 a a s a s a s s a s The warranted rate of growth, therefore, is Gw s a s (17.16) EXAMPLE 9.
Assume that the marginal propensity to save in the Harrod model above is 0.12 and the capital-output ratio is 2.12. To find Y t from (17.14), Y t 2.12 2.12 0.12 t Y 0 (1.06)tY 0 The warranted rate of growth, from (17.16), is Gw 0.12 2.12 0.12 0.12 2 0.06 396 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 17.7 PHASE DIAGRAMS FOR DIFFERENCE EQUATIONS While linear difference equations can be solved explicitly, nonlinear difference equations in general cannot. Important information about stability conditions, however, can once again be gleaned from phase diagrams. A phase diagram of a difference equation depicts yt as a function of yt1. If we restrict the diagram to the first quadrant for economic reasons so all the variables will be nonnegative, a 45 line from the origin will capture all the possible steady-state equilibrium points where yt yt1.
Consequently, any point at which the phase diagram intersects the 45 line will indicate an intertemporal equilibrium solution. The stability of a solution can be tested diagrammatically (Example 10) and mathematically (Example 11). Mathematically, the test depends on the following criteria for the first derivative of the phase line when it is evaluated at a steady-state point.
1.
If dyt dyt1 (y ¯) 1, y ¯ is locally stable. If dyt dyt1 (y ¯) 1, y ¯ is locally unstable.
2.
If dyt dyt1 (y ¯) 0, no oscillation. If dyt dyt1 (y ¯) 0, oscillation.
EXAMPLE 10.
Given a nonlinear difference equation, such as yt y0.5 t1 yt1 we can construct a phase diagram in a few easy steps.
1.
Find the steady-state solution(s), where yt yt1, by setting both yt and yt1 y ¯ and solving algebraically for y ¯.
y ¯ y ¯0.5 y ¯ y ¯0.5 0 y ¯ y ¯0.5 y ¯ 1 y ¯(y ¯0.5 1) 0 y ¯1 0 y ¯2 1 steady-state solutions The phase diagram must intersect the 45 line at y ¯1 0 and y ¯2 1.
2.
Take the first-derivative to see if the slope is positive or negative.
dyt dyt1 0.5yt1 0.5 0.5 yt1 0 Assuming yt, yt1 0, the phase diagram must be positively sloped.
3.
Take the second derivative to see if the phase line is concave or convex.
d2yt dy2 t1 0.25yt1 1.5 0.25yt1 3/2 0.25 y3 t1 0 concave 4.
Draw a rough sketch of the graph, as in Fig. 17-2.
397 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] Fig. 17-2 yt yt–1 0 0.25 1 2 45˚ line: yt = yt–1 (Slope = 1) 1 Phase line 5.
To analyze the phase diagram for stability conditions, we pick an arbitrary value for yt1, say yt1 0.25, and in a repeated sequence of steps (shown as dotted arrows) move up vertically to the phase diagram, then across horizontally to the 45 line, to see if the process converges to an equilibrium point or diverges from an equilibrium point. Here, starting from a value 0 yt1 1, the process converges towards y ¯2 1 and diverges from y ¯1 0. We conclude, therefore, that starting from a value 0 yt1 1, y ¯2 1 is a locally stable equilibrium and y ¯1 0 is locally unstable.
We next pick a value yt1 1, here yt1 2, and repeat the process. From the pattern of dotted arrows that emerges, we conclude that starting from a value yt1 1, y ¯2 1 is also locally stable when approached from a value 1.
EXAMPLE 11.
Here we confirm the results of the phase diagram test with the simple calculus test described above. Assuming once again, yt y0.5 t1, dyt dyt1 0.5yt1 0.5 Evaluated in absolute value at y ¯2 1, 0.5(1)0.5 0.5 1 0.5 1 locally stable.
Evaluated simply as y ¯2 1, 0.5(1)0.5 0.5 1 0.5 0 no oscillation.
Evaluated at y ¯1 0, the derivative is undefined but approaches infinity as yt1 →0. Therefore, y ¯1 0 is locally unstable. See Problems 17.31 to 17.33.
Solved Problems USE OF GENERAL FORMULA FOR FIRST-ORDER LINEAR DIFFERENCE EQUATIONS 17.1.
(a) Solve the difference equation given below; (b) check your answer, using t 0 and t 1; and (c) comment on the nature of the time path.
yt 6yt1 a) Here b 6 and a 0. Using (17.4) for all cases in which b 1, yt (y0 0)(6)t 0 y0(6)t A(6)t (17.17) where A, as a more generally used unspecified constant, replaces y0.
b) Estimating (17.17) at t 0 and t 1, y0 A(6)0 A y1 A(6) 6A Substituting y0 A for yt1 and y1 6A for yt in the original problems, 6A 6(A).
c) With the base b 6 in (17.17) positive and greater than 1, that is, b 0 and b 1, the time path is nonoscillating and explosive.
17.2.
Redo Problem 17.1 for yt 1 – 8yt1.
a) Using (17.4), yt (y0 0)(1 – 8)t 0 y0(1 – 8)t A(1 – 8)t 398 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 b) At t 0, y0 A(1 – 8)0 A. At t 1, y1 A(1 – 8) 1 – 8A. Substituting y0 A for yt1 and y1 1 – 8A for yt in the original equation, 1 – 8A 1 – 8(A).
c) With b 1 – 8, b 0 and b 1. The time path is nonoscillating and converging.
17.3.
Redo Problem 17.1, given yt 1 – 4yt1 60 and y0 8.
a) yt 8 60 1 1 – 4 1 4 t 60 1 1 – 4 40 1 4 t 48 b) At t 0, y0 40(1 – 4)0 48 8. At t 1, y1 40(1 – 4) 48 58. Substituting in the original equation, 58 1 – 4(8) 60 58.
c) With b 1 – 4, b 0 and b 1. The time path oscillates and converges.
17.4.
Redo Problem 17.1, given xt 3xt1 8 0 and x0 16.
a) Rearranging to conform with (17.3), xt 3xt1 8 Thus, b 3 and a 8. Substituting in (17.4), xt 16 8 1 3(3)t 8 1 3 18(3)t 2 b) At t 0, x0 18(3)0 2 16. At t 1, x1 18(3) 2 56. Substituting in the original, 56 3(16) 8 0.
c) With b 3, b 0 and b 1. The time path oscillates and explodes.
17.5.
Redo Problem 17.1, given yt yt1 17.
a) Rearranging, yt yt1 17. Here b 1. Using (17.4a), therefore, yt y0 17t A 17t.
b) At t 0, y0 A. At t 1, y1 A 17. Substituting in the original, A 17 A 17.
c) Here b 1. Thus b 0 and yt will not oscillate. But with b 1, 1 b 1. This presents a special case. With a 0, unless y0 A 0, the time path is divergent because the complementary function A does not approach 0 as t →. Thus, yt approaches A at, and not the particular solution, at, itself.
For b 1 and a 0, see Problem 17.17.
17.6.
Redo Problem 17.1, given gt gt1 25 and g0 40.
a) Using (17.4a), gt 40 25t.
b) At t 0, g0 40. At t 1, gt 15. Substituting in the original, 15 40 25.
c) With b 1, a 0 and A g0 0. The time path is nonoscillatory and divergent.
17.7.
Redo Problem 17.1, given 2yt yt1 18.
a) Dividing through by 2 to conform to (17.3) and then using (17.4), yt 1 2 yt1 9 y0 9 1 1 – 2 1 2 t 9 1 1 – 2 A 1 2 t 18 where A is an arbitrary constant for y0 18.
b) At t 0, y0 A 18. At t 1, y1 1 – 2A 18. Substituting in the original, 2(1 – 2A 18) A 18 18; A 36 A 36.
c) With b 1 – 2, b 0 and b 1. So yt is nonoscillating and convergent.
399 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] 17.8.
(a) Solve the following difference equation; (b) check the answer, using t 0 and t 1; and (c) comment on the nature of the time path.
5yt 2yt1 140 0 y0 30 a) Dividing by 5, rearranging terms, and using (17.4), yt 0.4yt1 28 30 28 1 0.4 (0.4)t 28 1 0.4 10(0.4)t 20 b) At t 0, y0 30. At t 1, y1 16. Substituting in the original, 5(16) 2(30) 140 0.
c) With b 0.4, b 0 and b 1. So yt oscillates and converges.
17.9.
Redo Problem 17.8, given xt1 4xt 36.
a) Shifting the time periods back one period to conform with (17.3), xt 4xt1 36. Using (17.4) and allowing A to replace x0 a/(1 b) as in Problem 17.7.
xt A(4)t 36 1 4 A(4)t 12 b) At t 0, x0 A 12. At t 1, x1 4A 12. Substituting x1 4A 12 for xt1 and x0 A 12 for xt in the original equation, 4A 12 4(A 12) 36; 4A 12 4A 12.
c) With b 4, b 0 and b 1. So xt does not oscillate but it explodes.
17.10. Redo Problem 17.8, given yt5 2yt4 57 0 and y0 11.
a) Moving the time periods back 5 periods, rearranging terms, and using (17.4), yt 2yt1 57 11 57 1 2(2)t 57 1 2 30(2)t 19 b) At t 0, y0 11. At t 1, y1 79. Substituting y1 for yt5 and y0 for yt4 in the original equation, 79 2(11) 57 0.
c) With b 2, b 0 and b 1. yt oscillates and explodes.
17.11. Redo Problem 17.8, given 8yt2 2yt3 120 and y0 28.
a) Divide through by 8, shift the time periods ahead by 2, and rearrange terms.
yt 1 4 yt1 15 28 15 1 1 – 4 1 4 t 15 1 1 – 4 8 1 4 t 20 b) At t 0, y0 28. At t 1, y1 22. Substitute y1 for yt2 and y0 for yt3.
8(22) 2(28) 120 120 120 c) With b 1 – 4, b 0 and b 1. So yt is nonoscillating and convergent.
17.12. Redo Problem 17.8, given gt 14.
a) Substituting (17.1) for gt, gt1 gt 14 (17.18) Set the time periods back 1 and rearrange terms.
gt gt1 14 Using (17.4a), gt g0 14t A 14t.
400 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 b) At t 0, g0 A. At t 1, g1 A 14. Substituting g1 for gt1 and g0 for gt in (17.18), A 14 A 14.
c) With b 1, gt is nonoscillatory. If A 0, gt is divergent.
17.13. Redo Problem 17.8, given yt yt 13 and y0 45.
a) Substituting from (17.1), moving the time periods back 1, and rearranging terms, yt 2yt1 13 (17.19) Using (17.4), yt 45 13 1 2 (2)t 13 1 2 58(2)t 13 b) At t 0, y0 45. At t 1, y1 103. Substituting in (17.19), 103 2(45) 13; 103 103.
c) With b 2, b 0 and b 1. Thus yt is nonoscillatory and explosive.
LAGGED INCOME DETERMINATION MODELS 17.14. Given the data below, (a) find the time path of national income Yt; (b) check your answer, using t 0 and t 1; and (c) comment on the stability of the time path.
Ct 90 0.8Y t1 It 50 Y 0 1200 a) In equilibrium, Y t Ct It. Thus, Y t 90 0.8Y t1 50 0.8Y t1 140 (17.20) Using (17.4), Y t 1200 140 1 0.8 (0.8)t 140 1 0.8 500(0.8)t 700 b) Y 0 1200; Y 1 1100. Substituting in (17.20), 1100 0.8(1200) 140 1100 1100 c) With b 0.8, b 0 and b 1. The time path Y t is nonoscillating and convergent. Y t converges to the equilibrium level of income 700.
17.15. Redo Problem 17.14, given Ct 200 0.75Y t1, It 50 0.15Y t1, and Y 0 3000.
a) Y t 200 0.75Y t1 50 0.15Y t1 0.9Yt1 250 Using (17.4), Y t 3000 250 1 0.9 (0.9)t 250 1 0.9 500(0.9)t 2500 b) Y 0 3000; Y 1 2950. Substituting above, 2950 0.9(3000) 250; 2950 2950.
c) With b 0.9, the time path Y t is nonoscillatory and converges toward 2500.
17.16. Redo Problem 17.14, given Ct 300 0.87Y t1, It 150 0.13Y t1, and Y0 6000.
a) Y t 300 0.87Y t1 150 0.13Y t1 Y t1 450 (17.21) Using (17.4a), Y t 6000 450t.
b) Y 0 6000; Y 1 6450. Substituting in (17.21) above, 6450 6000 450.
c) With b 1 and A 0, the time path Y t is nonoscillatory but divergent. See Problem 17.5.
401 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] 17.17. Redo Problem 17.14, given Ct 0.92Y t1, It 0.08Y t1, and Y 0 4000.
a) Y t 0.92Y t1 0.08Y t1 Y t1 Using (17.4a), Y t 4000 0 4000.
b) Y 0 4000 Y 1.
c) When b 1 and a 0, Y t is a stationary path.
17.18. Redo Problem 17.14, given Ct 400 0.6Y t 0.35Y t1, It 240 0.15Y t1, and Y 0 7000.
a) Y t 400 0.6Y t 0.35Y t1 240 0.15Y t1 0.4Y t 0.5Y t1 640 Divide through by 0.4 and then use (17.4).
Y t 1.25Y t1 1600 7000 1600 1 1.25 (1.25)t 1600 1 1.25 13,400(1.25)t 6400 b) Y 0 7000; Y 1 10,350. Substituting in the initial equation, 10,350 400 0.6(10,350) 0.35(7000) 240 0.15(7000) 10,350 c) With b 1.25, the time path Y t is nonoscillatory and explosive.
17.19. Redo Problem 17.14, given Ct 300 0.5Yt 0.4Y t1, It 200 0.2Y t1, and Y 0 6500.
a) Y t 300 0.5Y t 0.4Y t1 200 0.2Y t1 0.5Y t 0.6Y t1 500 Dividing through by 0.5 and then using (17.4), Y t 1.2Y t1 1000 6500 1000 1 1.2 (1.2)t 1000 1 1.2 11,500(1.2)t 5000 b) Y 0 6500; Y 1 8800. Substituting in the initial equation, 8800 300 0.5(8800) 0.4(6500) 200 0.2(6500) 8800 c) With b 1.2, Y t is nonoscillatory and explosive.
17.20. Redo Problem 17.14, given Ct 200 0.5Y t, It 3(Y t Y t1), and Y 0 10,000.
a) Y t 200 0.5Y t 3(Yt Y t1) 2.5Y t 3Y t1 200 Dividing through by 2.5 and then using (17.4), Y t 1.2Y t1 80 10,000 80 1 1.2(1.2)t 80 1 1.2 9600(1.2)t 400 b) Y 0 10,000; Y 1 11,920. Substituting in the initial equation, 11,920 200 0.5(11,920) 3(11,920 10,000) 11,920 c) With b 1.2, the time path Y t explodes but does not oscillate.
THE COBWEB MODEL 17.21. For the data given below, determine (a) the market price Pt in any time period, (b) the equilibrium price Pe, and (c) the stability of the time path.
Qdt 180 0.75Pt Qst 30 0.3Pt1 P0 220 402 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 a) Equating demand and supply, 180 0.75Pt 30 0.3Pt1 (17.22) 0.75Pt 0.3Pt1 210 Dividing through by 0.75 and using (17.4), Pt 0.4P t1 280 220 280 1 0.4 (0.4)t 280 1 0.4 20(0.4)t 200 (17.23) b) If the market is in equilibrium, Pt Pt1. Substituting Pe for Pt and Pt1 in (17.22), 180 0.75Pe 30 0.3Pe Pe 200 which is the second term on the right-hand side of (17.23).
c) With b 0.4, the time path Pt will oscillate and converge.
17.22. Check the answer to Problem 17.21(a), using t 0 and t 1.
From (17.23), P0 20(0.4)0 200 220 and P1 20(0.4) 200 192. Substituting P1 for Pt and P0 for Pt1 in (17.22), 180 0.75(192) 30 0.3(220) 36 36 17.23. Redo Problem 17.21, given Qdt 160 0.8Pt, Qst 20 0.4Pt1, and P0 153.
a) 160 0.8Pt 20 0.4Pt1 (17.24) 0.8Pt 0.4Pt1 180 Dividing through by 0.8 and using (17.4), Pt 0.5P t1 225 153 225 1 0.5 (0.5)t 225 1 0.5 3(0.5)t 150 (17.25) b) As shown in Problem 17.21(b), Pe 150. See also Section 17.5.
c) With b 0.5, Pt oscillates and converges toward 150.
17.24. Check the answer to Problem 17.23(a), using t 0 and t 1.
From (17.25), P0 3(0.5)0 150 153 and P1 3(0.5) 150 148.5. Substituting in (17.24), 160 0.8(148.5) 20 0.4(153) 41.2 41.2 17.25. Redo Problem 17.21, given Qdt 220 0.4Pt, Qst 30 0.6Pt1, and P0 254.
a) 220 0.4Pt 30 0.6Pt1 0.4Pt 0.6Pt1 250 Dividing through by 0.4 and then using (17.4), Pt 1.5P t1 625 254 625 1 1.5 (1.5)t 625 1 1.5 4(1.5)t 250 b) Pe 250 c) With b 1.5, Pt oscillates and explodes.
403 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] THE HARROD GROWTH MODEL 17.26. For the following data, find (a) the level of income Y t for any period and (b) the warranted rate of growth.
It 2.66(Y t Y t1) St 0.16Y t Y 0 9000 a) In equilibrium, 2.66(Y t Y t1) 0.16Y t 2.5Y t 2.66Y t1 Dividing through by 2.5 and then using (17.4), Y t 1.064Y t1 (9000 0)(1.064)t 0 9000(1.064)t b) From (17.16), Gw 0.16/(2.66 0.16) 0.064.
17.27. Redo Problem 17.26, given It 4.2(Y t Y t1), St 0.2Y t, and Y 0 5600.
a) 4.2(Y t Y t1) 4Yt Y t 0.2Yt 4.2Y t1 1.05Y t1 Using (17.4), Y t 5600(1.05)t.
b) Gw 0.2 4.2 0.2 0.05 OTHER ECONOMIC APPLICATIONS 17.28. Derive the formula for the value Pt of an initial amount of money P0 deposited at i interest for t years when compounded annually.
When interest is compounded annually, Pt1 Pt iPt (1 i)Pt Moving the time periods back one to conform with (17.3), Pt (1 i)Pt1 Using (17.4) since i 0, Pt (P0 0)(1 i)t 0 P0(1 i)t 17.29. Assume that Qdt c zPt, Qst g hPt, and Pt1 Pt a(Qst Qdt) (17.26) i.e., price is no longer determined by a market-clearing mechanism but by the level of inventory Qst Qdt. Assume, too, that a 0 since a buildup in inventory (Qst Qdt) will tend to reduce price and a depletion of inventory (Qst Qdt) will cause prices to rise. (a) Find the price Pt for any period and (b) comment on the stability conditions of the time path.
a) Substituting Qst and Qdt in (17.26), Pt1 Pt a(g hPt c zPt) [1 a(h z)]Pt a(g c) [1 a(z h)]Pt a(g c) Shifting the time periods back 1 to conform to (17.3) and using (17.4), Pt P0 a(g c) 1 [1 a(z h)] [1 a(z h)]t a(g c) 1 [1 a(z h)] P0 g c z h [1 a(z h)]t g c z h (17.27) 404 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 Substituting as in (17.13a), Pt (P0 Pe)[1 a(z h)]t Pe (17.28) b) The stability of the time path depends on b 1 a(z h). Since a 0 and under normal conditions z 0 and h 0, a(z h) 0. Thus, If 0 a(z h) 1, If a(z h) 1, If 2 a(z h) 1, If a(z h) 2, If a(z h) 2, 0 b 1; b 0; 1 b 0; b 1; b 1; Pt converges and is nonoscillatory.
Pt remains in equilibrium (Pt P0).
Pt converges with oscillation.
uniform oscillation takes place.
Pt oscillates and explodes.
17.30. Given the following data, (a) find the price Pt for any time period; (b) check the answer, using t 0 and t 1; and (c) comment on the stability conditions.
Qdt 120 0.5Pt Qst 30 0.3Pt Pt1 Pt 0.2(Qst Qdt) P0 200 a) Substituting, Pt1 Pt 0.2(30 0.3Pt 120 0.5Pt) 0.84Pt 30 Shifting time periods back 1 and using (17.4), Pt 0.84P t1 30 200 30 1 0.84 (0.84)t 30 1 0.84 12.5(0.84)t 187.5 b) P0 200; P1 198. Substituting in the first equation of the solution, 198 200 0.2[30 0.3(200) 120 0.5(200)] 198.
c) With b 0.84, Pt converges without oscillation toward 187.5.
PHASE DIAGRAMS FOR DIFFERENCE EQUATIONS 17.31. (a) Construct a phase diagram for the nonlinear difference equation below, (b) use it to test for dynamic stability, and (c) confirm your results with the derivative test.
yt y3 t1 a) Setting yt yt1 y ¯ for the intertemporal equilibrium solution, y ¯ y ¯3 y ¯(1 y ¯2) 0 y ¯1 0 y ¯2 1 intertemporal equilibrium levels The phase diagram will intersect the 45 line at y 0 and y 1. We next take the first and second derivative to determine the slope and concavity of the phase line.
dyt dyt1 3y2 t1 0 positive slope d2yt dy2 t1 6yt1 0 convex With the above information we then draw a rough sketch of the graph, as in Fig. 17-3.
405 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] Fig. 17-3 yt yt–1 0 0.75 1 1.5 The phase diagram for yt = y3 t–1 1 yt = yt–1 b) Starting from a value of yt 0.75 and following the series of moves indicated by the dotted arrows we see the function converges to y ¯1 0 and diverges from y ¯2 1. From a starting point of yt 1.5, the function also diverges from y ¯2 1. We conclude, therefore, that y ¯2 1 is an unstable equilibrium when approached from either side and y ¯1 0 is a stable equilibrium when approached from a positive value.
c) Independently of the phase diagram, we can also test the stability conditions by taking the first derivative of the equation and evaluating it at the steady-state solutions.
dyt dyt1 3y2 t1 dyt dyt1 (0) 3(0) 0 1 locally stable dyt dyt1 (1) 3 1 locally unstable With the derivative positive at y ¯1 0 and y ¯2 1, there is no oscillation.
17.32. Repeat the steps in Problem 17.31 for the nonlinear difference equation, yt yt1 0.25 1 4 yt1 a) Setting yt yt1 y ¯, and substituting above, y ¯ y ¯0.25 y ¯ y ¯0.25 0 y ¯(1 y ¯1.25) 0 y ¯ 0 y ¯ 1 Since y ¯ 0 is undefined at y ¯1.25, there is only one intertemporal equilibrium, y ¯ 1. Taking the derivatives, dyt dyt1 0.25yt1 1.25 0.25 y1.25 t1 0 negative slope d2yt dy2 t1 0.3125yt1 2.25 0 convex We can then sketch the graph, as in Fig. 17-4.
b) Starting at a value less than 1, say yt1 0.75, the function oscillates between values larger and smaller than one but converges to y ¯ 1. If we start at a value larger than 1, we also get the same results.
406 FIRST-ORDER DIFFERENCE EQUATIONS [CHAP . 17 Fig. 17-4 yt yt–1 0 0.75 1 The phase diagram for yt = yt–1 1 Convergent –0.25 c) Working simply with the derivative and evaluating it in absolute value at the steady-state equilibrium solution, we have dyt dyt1 0.25yt1 1.25 dyt dyt1 (1) 0.25(1) 1 locally stable dyt dyt1 (1) 0.25(1) 0 oscillation 17.33. Redo Problem 17.31 for the equation yt yt1 1.5 a) The intertemporal equilibrium solution is again y ¯ 1.
dyt dyt1 1.5yt1 2.5 0 negative slope d2yt dy2 t1 3.75yt1 3.5 0 convex See Fig. 17-5.
b) Starting at a value less than 1, say yt1 0.7, the function oscillates between values larger and smaller than one and ultimately diverges from y ¯ 1.
c) Working solely with the first derivative of the function, dyt dyt1 (1) 1.5 1 locally unstable dyt dyt1 (1) 1.5 0 oscillation 407 FIRST-ORDER DIFFERENCE EQUATIONS CHAP . 17] Fig. 17-5 yt yt–1 0 0.7 1 The phase diagram for yt = yt–1 1 Divergent –1.5 CHAPTER 18 Second-Order Differential Equations and Difference Equations 18.1 SECOND-ORDER DIFFERENTIAL EQUATIONS Second-order differential equations require separate solutions for the complementary function yc and the particular integral yp. The general solution is the sum of the two: y(t) yc yp. Given the second-order linear differential equation y
(t) b1y(t) b2y(t) a (18.1) where b1, b2, and a are constants, the particular integral will be yp a b2 b2 0 (18.2) yp a b1 t b2 0 b1 0 (18.2a) yp a 2 t2 b1 b2 0 (18.2b) The complementary function is yc y1 y2 (18.3) 408 where y1 A1er1t (18.3a) y2 A2er2t (18.3b) and r1, r2 b1 b1 2 4b2 2 (18.4) Here, A1 and A2 are arbitrary constants, and b1 2 4b2. r1 and r2 are referred to as characteristic roots, and (18.4) is the solution to the characteristic or auxiliary equation: r2 b1r b2 0. See Examples 1 to 4 and Problems 18.1 to 18.11, 20.9, 20.10, 20.13, 20.14 and 20.16 to 20.20.
EXAMPLE 1.
The particular integral for each of the following equations (1) y
(t) 5y(t) 4y(t) 2 (2) y
(t) 3y(t) 12 (3) y
(t) 16 is found as shown below.
For (1), using (18.2), For (2), using (18.2a), For (3), using (18.2b), yp 2 – 4 1 – 2 yp 12 –– 3t 4t yp 16 –– 2t2 8t2 (18.5) (18.5a) (18.5b) EXAMPLE 2.
The complementary functions for equations (1) and (2) in Example 1 are calculated below.
Equation (3) will be treated in Example 9.
For (1), from (18.4), r1, r2 5 (5)2 4(4) 2 5 3 2 1, 4 Substituting in (18.3a) and (18.3b), and finally in (18.3), yc A1et A2e4t (18.6) For (2), r1, r2 3 (3)2 4(0) 2 3 3 2 0, 3 Thus, yc A1e0 A2e3t A1 A2e3t (18.6a) EXAMPLE 3.
The general solution of a differential equation is composed of the complementary function and the particular integral (Section 16.2), that is, y(t) yc yp. As applied to the equations in Example 1, For (1), from (18.6) and (18.5), For (2), from (18.6a) and (18.5a), y(t) A1et A2e4t 1 – 2 (18.7) y(t) A1 A2e3t 4t (18.7a) EXAMPLE 4.
The definite solution for (1) in Example 3 is calculated below. Assume y(0) 51 – 2 and y(0) 11.
From (18.7), y(t) A1et A2e4t 1 – 2 (18.8) Thus, y(t) A1et 4A2e4t (18.8a) Evaluating (18.8) and (18.8a) at t 0, and setting y(0) 51 – 2 and y(0) 11 from the initial conditions, y(0) A1e0 A2e4(0) 1 – 2 51 – 2 thus A1 A2 5 y(0) A1e0 4A2e4(0) 11 thus A1 4A2 11 Solving simultaneously, A1 3 and A2 2. Substituting in (18.7), y(t) 3et 2e4t 1 – 2 (18.9) To check this solution, from (18.9), y(t) 3et 2e4t 1 – 2 409 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] Thus, y(t) 3et 8e4t y
(t) 3et 32e4t Substituting in the original equation [(1) in Example 1], (3et 32e4t) 5(3et 8e4t) 4(3et 2e4t 1 – 2) 2 18.2 SECOND-ORDER DIFFERENCE EQUATIONS The general solution of a second-order difference equation is composed of a complementary function and a particular solution: y(t) yc yp. Given the second-order linear difference equation yt b1yt1 b2yt2 a (18.10) where b1, b2, and a are constants, the particular solution is yp a 1 b1 b2 b1 b2 1 (18.11) yp a 2 b1 t b1 b2 1 b1 2 (18.11a) yp a 2 t2 b1 b2 1 b1 2 (18.11b) The complementary function is yc A1rt 1 A2rt 2 (18.12) where A1 and A2 are arbitrary constants and the characteristic roots r1 and r2 are found by using (18.4), assuming b1 2 4b2. See Examples 5 to 8 and Problems 18.12 to 18.20.
EXAMPLE 5.
The particular solution for each of the following equations: 1) yt 10yt1 16yt2 14 2) yt 6yt1 5yt2 12 3) yt 2yt1 yt2 8 is found as shown below.
For (1), using (18.11), yp 14 1 10 16 2 (18.13) For (2), using (18.11a), yp 12 2 6 t 3t (18.13a) For (3), using (18.11b), yp 8 – 2t2 4t2 (18.13b) EXAMPLE 6.
From Example 5, the complementary functions for (1) and (2) are calculated below. For (3), see Example 9.
For (1), using (18.4) and then substituting in (18.12), r1, r2 10 100 4(16) 2 10 6 2 2, 8 Thus, yc A1(2)t A2(8)t (18.14) For (2), r1, r2 6 36 4(5) 2 6 4 2 1, 5 Thus, yc A1(1)t A2(5)t A1 A2(5)t (18.14a) 410 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 EXAMPLE 7.
The general solutions for (1) and (2) from Example 5 are calculated below.
For (1), y(t) yc yp. From (18.14) and (18.13), y(t) A1(2)t A2(8)t 2 (18.15) For (2), from (18.14a) and (18.13a), y(t) A1 A2(5)t 3t (18.15a) EXAMPLE 8.
Given y(0) 10 and y(1) 36, the definite solution for (1) in Example 7 is calculated as follows: Letting t 0 and t 1 successively in (18.15), y(0) A1(2)0 A2(8)0 2 A1 A2 2 y(1) A1(2) A2(8) 2 2A1 8A2 2 Setting y(0) 10 and y(1) 36 from the initial conditions, A1 A2 2 10 2A1 8A2 2 36 Solving simultaneously, A1 5 and A2 3. Finally, substituting in (18.15), y(t) 5(2)t 3(8)t 2 (18.16) This answer is checked by evaluating (18.16) at t 0, t 1, and t 2, y(0) 5 3 2 10 y(1) 10 24 2 36 y(2) 20 192 2 214 Substituting y(2) for yt, y(1) for yt1, and y(0) for yt2 in yt 10yt1 16yt2 14 of Equation (1) in Example 5, 214 10(36) 16(10) 14.
18.3 CHARACTERISTIC ROOTS A characteristic equation can have three different types of roots.
1.
Distinct real roots. If b1 2 4b2, the square root in (18.4) will be a real number, and r1 and r2 will be distinct real numbers as in (18.6) and (18.6a).
2.
Repeated real roots. If b1 2 4b2, the square root in (18.4) will vanish, and r1 and r2 will equal the same real number. In the case of repeated real roots, the formulas for yc in (18.3) and (18.12) must be changed to yc A1ert A2tert (18.17) yc A1rt A2trt (18.18) 3.
Complex roots. If b1 2 4b2, (18.4) contains the square root of a negative number, which is called an imaginary number. In this case r1 and r2 are complex numbers. A complex number contains a real part and an imaginary part; for example, (12 i) where i 1.
[As a simple test to check your answers when using (18.4), assuming the coefficient of the y
(t) term is 1, r1 r2 must equal b1; r1 r2 must equal b2.] EXAMPLE 9.
The complementary function for Equation (3) in Example 1, where y
(t) 16, is found as follows: From (18.4), r1, r2 0 0 4(0) 2 0 Using (18.17) since r1 r2 0, which is a case of repeated real roots, yc A1e0 A2te0 A1 A2t.
In Equation (3) of Example 5, yt 2yt1 yt2 8. Solving for the complementary function, from (18.4), r1, r2 2 4 4(1) 2 2 0 2 1 Using (18.18) because r1 r2 1, yc A1(1)t A2t(1)t A1 A2t.
411 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] 18.4 CONJUGATE COMPLEX NUMBERS If b1 2 4b2 in (18.4), factoring out 1 gives r1, r2 b1 1 4b2 b1 2 2 b1 i4b2 b1 2 2 Put more succinctly, r1, r2 g hi where g 1 – 2b1 and h 1 – 24b2 b1 2 (18.19) g hi are called conjugate complex numbers because they always appear together. Substituting (18.19) in (18.3) and (18.12) to find yc for cases of complex roots, yc A1e(ghi)t A2e(ghi)t egt(A1ehit A2ehit) (18.20) yc A1(g hi)t A2(g hi)t (18.21) See Example 10 and Problems 18.28 to 18.35, 20.11 and 20.12.
EXAMPLE 10.
The complementary function for y
(t) 2y(t) 5y(t) 18 is calculated as shown below. Using (18.19) since b1 2 4b2, g 1 – 2(2) 1 h 1 – 24(5) (2)2 1 – 2(4) 2 Thus, r1, r2 1 2i. Substituting in (18.20), yc et(A1e2it A2e2it).
18.5 TRIGONOMETRIC FUNCTIONS Trigonometric functions are often used in connection with complex numbers. Given the angle ! in Fig. 18-1, which is at the center of a circle of radius k and measured counterclockwise, the trigono-metric functions of ! are sine (sin) ! tangent (tan) ! secant (sec) ! h k h g k g cosine (cos) ! cotangent (cot) ! cosecant (csc) ! g k g h k h 412 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 Fig. 18-1 The signs of the trigonometric functions in each of the four quadrants are sin, csc cos, sec tan, cot The angle ! is frequently measured in radians. Since there are 2 radians in a circle, 1 /180 radian. Thus 360 2 radians, 180 radians, 90 /2 radians, and 45 /4 radians.
EXAMPLE 11.
If the radius OL in Fig. 18-1 starts at A and moves counterclockwise 360 , sin ! h/k goes from 0 at A, to 1 at B, to 0 at C, to 1 at D, and back to 0 at A. Cosine ! g/k goes from 1 at A, to 0 at B, to 1 at C, to 0 at D, and back to 1 at A. This is summarized in Table 18-1 and graphed in Fig. 18-2. Notice that both functions are periodic with a period of 2 (i.e., they repeat themselves every 360 or 2 radians). Both have an amplitude of fluctuation of 1 and differ only in phase or location of their peaks.
Table 18-1 Degrees 0 90 180 270 360 Radians 0 – 2 3 – 2 2 sin !
0 1 0 1 0 cos !
1 0 1 0 1 18.6 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS Given that u is a differentiable function of x, 1) d dx (sin u) cos u du dx 2) d dx (cos u) sin u du dx 3) d dx (tan u) sec2u du dx 4) d dx (cot u) csc2u du dx 5) d dx (sec u) sec u tan u du dx 6) d dx (csc u) csc u cot u du dx See Example 12 and Problems 18.21 to 18.27.
413 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] Fig. 18-2 EXAMPLE 12.
The derivatives for the trigonometric functions 1) y sin (3x2 6) 2) y 4 cos 2x 3) y (1 tan x)2 are calculated as follows: 1) dy dx 6x cos (3x2 6) 2) dy dx 8 sin 2x 3) dy dx 2(1 tan x)(sec2 x) 2 sec2 x(1 tan x) 18.7 TRANSFORMATION OF IMAGINARY AND COMPLEX NUMBERS Three rules are helpful in transforming imaginary and complex numbers to trigonometric functions.
1.
g and h in Fig. 18-1, which are Cartesian coordinates, can be expressed in terms of ! and k, which are called polar coordinates, by the simple formula g k cos !
h k sin !
k 0 Thus for the conjugate complex number (g hi), g hi k cos ! ik sin ! k(cos ! i sin !) (18.22) 2.
By what are called Euler relations, ei! cos ! i sin !
(18.23) Thus, by substituting (18.23) in (18.22) we can also express (g hi) as g hi kei!
(18.23a) 3.
From (18.23a), raising a conjugate complex number to the nth power means (g hi)n (kei!)n knein!
(18.24) Or, by making use of (18.23) and noting that n! replaces !, we have De Moivre’s theorem: (g hi)n kn(cos n! i sin n!) (18.25) See Examples 13 to 15 and Problems 18.28 to 18.35, 20.11 and 20.12.
EXAMPLE 13.
The value of the imaginary exponential function e2i is found as follows. Using (18.23), where ! 2, e2i cos 2 i sin 2 From Table 18-1, cos 2 1 and sin 2 0. Thus, e2i 1 i(0) 1.
EXAMPLE 14.
The imaginary exponential expressions in (18.20) and (18.21) are transformed to trigonometric functions as shown below.
From (18.20), yc egt(A1ehit A2ehit). Using (18.23) where ! ht, yc egt[A1(cos ht i sin ht) A2(cos ht i sin ht)] egt[(A1 A2) cos ht (A1 A2) i sin ht] egt(B1 cos ht B2 sin ht) (18.26) where B1 A1 A2 and B2 (A1 A2)i.
From (18.21), yc A1(g hi)t A2(g hi)t. Using (18.25) and substituting t for n, yc A1kt(cos t! i sin t!) A2kt(cos t! i sin t!) kt[(A1 A2) cos t! (A1 A2)i sin t!] kt(B1 cos t! B2 sin t!) (18.27) where B1 A1 A2 and B2 (A1 A2)i.
414 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 EXAMPLE 15.
The time paths of (18.26) and (18.27) are evaluated as follows: Examining each term in (18.26), 1.
Here B1 cos ht is a cosine function of t, as in Fig. 18-2, with period 2/h instead of 2 and amplitude of the multiplicative constant B1 instead of 1.
2.
Likewise B2 sin ht is a sine function of t with period 2/h and amplitude of B2.
3.
With the first two terms constantly fluctuating, stability depends on egt.
If g 0, egt gets increasingly larger as t increases. This increases the amplitude and leads to explosive fluctuations of yc, precluding convergence.
If g 0, egt 1 and yc displays uniform fluctuations determined by the sine and cosine functions. This also precludes convergence.
If g 0, egt approaches zero as t increases. This diminishes the amplitude, produces damped fluctuations, and leads to convergence. See Fig. 18-3.
Since (18.27) concerns a difference equation in which t can only change at discrete intervals, yc is a step function rather than a continuous function (see Fig. 17-1). Like (18.26), it will fluctuate, and stability will depend on kt. If k 1, yc will converge. See Problems 18.8 to 18.11, 18.18 to 18.20, and 18.31 to 18.35.
18.8 STABILITY CONDITIONS For a second-order linear differential equation with distinct or repeated real roots, both roots must be negative for convergence. If one of the roots is positive, the exponential term with the positive root approaches infinity as t approaches infinity, thereby precluding convergence. See Problems 18.8 to 18.11. In the case of complex roots, g in egt of (18.26) must be negative, as illustrated in Example 15.
For a second-order linear difference equation with distinct or repeated real roots, the root with the largest absolute value is called the dominant root because it dominates the time path. For convergence, the absolute value of the dominant root must be less than 1. See Problems 18.18 to 18.20. In the case of complex roots, the absolute value of k in (18.27) must be less than 1, as explained in Example 15.
For economic applications, see Problems 18.36 and 18.37.
415 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] Fig. 18-3 Time path of yc(t).
Solved Problems SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS Distinct Real Roots 18.1.
For the following equation, find (a) the particular integral yp, (b) the complementary function yc, and (c) the general solution y(t).
y
(t) 9y(t) 14y(t) 7 a) Using (18.2), yp 7 –– 14 1 – 2.
b) Using (18.4), r1, r2 9 81 4(14) 2 9 5 2 2, 7 Substituting in (18.3), yc A1e2t A2e7t.
c) y(t) yc yp A1e2t A2e7t 1 – 2 (18.28) 18.2.
Redo Problem 18.1, given y
(t) 12y(t) 20y(t) 100.
a) From (18.2), yp 100 ––– 20 5.
b) From (18.4), r1, r2 12 144 4(20) 2 12 8 2 2, 10 Thus, yc A1e2t A2e10t.
c) y(t) yc yp A1e2t A2e10t 5 (18.29) 18.3.
Redo Problem 18.1, given y
(t) 4y(t) 5y(t) 35.
a) From (18.2), yp 35 5 7.
b) From (18.4) r1, r2 4 16 4(5) 2 4 6 2 5, 1 Thus, yc A1e5t A2et c) y(t) A1e5t A2et 7 (18.30) 18.4.
Redo Problem 18.1, given y
(t) 7y(t) 28.
a) Using (18.2a), yp 28 –– 7 t 4t.
b) From (18.4), r1, r2 7 49 4(0) 2 7 7 2 0, 7 Thus, yc A1e(0)t A2e7t A1 A2e7t c) y(t) A1 A2e7t 4t (18.31) 416 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 18.5.
Redo Problem 18.1, given y
(t) 1 – 2y(t) 13.
a) From (18.2a), yp 13 1 – 2 t 26t.
b) r1, r2 1 – 2 1 – 4 4(0) 2 1 – 2 1 – 2 2 0, 1 – 2 Thus, yc A1 A2e(1/2)t c) y(t) A1 A2e(1/2)t 26t (18.32) Repeated Real Roots 18.6.
Find (a) the particular integral yp, (b) the complementary function yc, and (c) the general solution y(t), given y
(t) 12y(t) 36y(t) 108.
a) yp 108 ––– 36 3 b) r1, r2 12 144 4(36) 2 12 0 2 6 Using (18.17) since r1 r2 6, yc A1e6t A2te6t.
c) y(t) A1e6t A2te6t 3 (18.33) 18.7.
Redo Problem 18.6, given y
(t) y(t) 1 – 4y(t) 9.
a) yp 9 1 – 4 36 b) r1, r2 1 1 4(1 – 4) 2 1 0 2 1 2 Using (18.17) since r1 r2 1 – 2, yc A1e(1/2)t A2te(1/2)t c) y(t) A1e(1/2)t A2te(1/2)t 36 (18.34) DEFINITE SOLUTIONS AND STABILITY CONDITIONS 18.8.
Find (a) the definite solution for the following equation, (b) check your answer, and (c) comment on the dynamic stability of the time path, given y
(t) 9y(t) 14y(t) 7, y(0) 21 – 2, and y(0) 31.
a) From (18.28), y(t) A1e2t A2e7t 1 – 2 (18.35) Thus, y(t) 2A1e2t 7A2e7t (18.35a) Evaluating (18.35) and (18.35a) at t 0, y(0) A1 A2 1 – 2 y(0) 2A1 7A2 Setting y(0) 21 – 2 and y(0) 31 from the initial conditions, A1 A2 1 – 2 21 – 2 2A1 7A2 31 Solving simultaneously, A1 2 and A2 5, which when substituted in (18.35) gives y(t) 2e2t 5e7t 1 – 2 (18.36) 417 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] b) From (18.36), y(t) 2e2t 5e7t 1 – 2. Thus, y(t) 4e2t 35e7t y
(t) 8e2t 245e7t Substituting these values in the original problem, where y 9y(t) 14y(t) 7, 8e2t 245e7t 9(4e2t 35e7t) 14(2e2t 5e7t 1 – 2) 7 c) With both characteristic roots negative, (18.36) will approach 1 – 2 as t →. Therefore y(t) is convergent.
Any time both characteristic roots are negative, the time path will converge.
18.9.
Redo Problem 18.8, given y
(t) 4y(t) 5y(t) 35, y(0) 5, and y(0) 6.
a) From (18.30), y(t) A1e5t A2et 7 (18.37) Thus, y(t) 5A1e5t A2et (18.37a) Evaluating (18.37) and (18.37a) at t 0 and setting them equal to the initial conditions where y(0) 5 and y(0) 6, y(0) A1 A2 7 5 thus A1 A2 12 y(0) 5A1 A2 6 Solving simultaneously, A1 3 and A2 9, which when substituted in (18.37) gives y(t) 3e5t 9et 7 (18.38) b) From (18.38), y(t) 3e5t 9et 7. Thus, y(t) 15e5t 9et and y
(t) 75e5t 9et.
Substituting these values in the original problem, where y
(t) 4y(t) 5y(t) 35, 75e5t 9et 4(15e5t 9et) 5(3e5t 9et 7) 35 c) With one characteristic root positive and the other negative, the time path is divergent. The positive root dominates the negative root independently of their relative absolute values because as t →, the positive root → and the negative root →0.
18.10. Redo Problem 18.8, given y
(t) 1 – 2y(t) 13, y(0) 17, and y(0) 191 – 2.
a) From (18.32), y(t) A1 A2e(1/2)t 26t (18.39) Thus, y(t) 1 – 2A2e(1/2)t 26 (18.39a) Evaluating (18.39) and (18.39a) at t 0 and setting them equal to the initial conditions, y(0) A1 A2 17 y(0) 1 – 2A2 26 191 – 2 A2 13 With A2 13, A1 4. Substituting in (18.39) and rearranging terms, y(t) 13e(1/2)t 26t 4 (18.40) b) From (18.40), y(t) 13e(1/2)t 26t 4. Thus, y(t) 6.5e(1/2)t 26 y
(t) 3.25e(1/2)t Substituting these values in the original equation, 3.25e(1/2)t 1 – 2(6.5e(1/2)t 26) 13 c) With both characteristic roots positive, the time path will diverge.
418 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 18.11. Redo Problem 18.8, given y
(t) y(t) 1 – 4y(t) 9, y(0) 30, and y(0) 15.
a) From (18.34), y(t) A1e(1/2)t A2te(1/2)t 36 (18.41) Using the product rule for the derivative of the second term, y(t) 1 – 2A1e(1/2)t 1 – 2A2te(1/2)t A2e(1/2)t (18.41a) Evaluating (18.41) and (18.41a) at t 0 and equating to the initial conditions, y(0) A1 36 30 A1 6 y(0) 1 – 2A1 A2 15 With A1 6, A2 12. Substituting in (18.41), y(t) 12te(1/2)t 6e(1/2)t 36 (18.42) b) From (18.42), y(t) 12te(1/2)t 6e(1/2)t 36. By the product rule, y(t) 6te(1/2)t 12e(1/2)t 3e(1/2)t y
(t) 3te(1/2)t 6e(1/2)t 6e(1/2)t 1.5e(1/2)t 3te(1/2)t 13.5e(1/2)t Substituting in the original equation, (3te(1/2)t 13.5e(1/2)t) (6te(1/2)t 15e(1/2)t) 1 – 4(12te(1/2)t 6e(1/2)t 36) 9 c) With the repeated characteristic roots negative, the time path will converge since tert follows basically the same time path as ert.
SECOND-ORDER LINEAR DIFFERENCE EQUATIONS Distinct Real Roots 18.12. Find (a) the particular solution, (b) the complementary function, and (c) the general solution, given yt 7yt1 6yt2 42.
a) From (18.11), yp 42 1 7 6 3 b) From (18.4), r1, r2 7 49 4(6) 2 7 5 2 1, 6 From (18.12), yc A1(1)t A2(6)t c) y(t) yc yp A1(1)t A2(6)t 3 (18.43) 18.13. Redo Problem 18.12, given yt 12yt1 11yt2 6.
a) From (18.11), yp 6 1 12 11 1 4 b) r1, r2 12 144 4(11) 2 12 10 2 1, 11 Thus, yc A1(1)t A2(11)t c) y(t) A1(1)t A2(11)t 1 – 4 (18.44) 419 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] 18.14. Redo Problem 18.12, given yt2 11yt1 10yt 27.
a) Shifting the time periods back 2 to conform with (18.10), yt 11yt1 10yt2 27. Then from (18.11a), yp 27 2 11 t 3t b) r1, r2 11 121 4(10) 2 11 9 2 1, 10 yc A1 A2(10)t c) y(t) A1 A2(10)t 3t (18.45) 18.15. Redo Problem 18.12, given yt 7yt1 8yt2 45.
a) From (18.11a), yp 45 2 7 t 5t b) r1, r2 7 49 4(8) 2 7 9 2 1, 8 yc A1 A2(8)t c) y(t) A1 A2(8)t 5t (18.46) Repeated Real Roots 18.16. Redo Problem 18.12, given yt 10yt1 25yt2 8.
a) yp 8 1 10 25 1 2 b) r1, r2 10 100 4(25) 2 10 0 2 5 Using (18.18) because r1 r2 5, yc A1(5)t A2t(5)t.
c) y(t) A1(5)t A2t(5)t 1 – 2 (18.47) 18.17. Redo Problem 18.12, given yt 14yt1 49yt2 128.
a) yp 128 1 14 49 2 b) r1, r2 14 196 4(49) 2 14 0 2 7 From (18.18), yc A1(7)t A2t(7)t.
c) y(t) A1(7)t A2t(7)t 2 (18.48) DEFINITE SOLUTIONS AND STABILITY CONDITIONS 18.18. (a) Find the definite solution, (b) check the answer, and (c) comment on dynamic stability, given yt 7yt1 6yt2 42, y(0) 16, and y(1) 35.
a) From (18.43), y(t) A1(1)t A2(6)t 3 (18.49) Letting t 0 and t 1 successively in (18.49) and making use of the initial conditions, y(0) A1 A2 3 16 y(1) A1 6A2 3 35 420 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 Solving simultaneously, A1 8 and A2 5. Substituting in (18.49), y(t) 8(1)t 5(6)t 3 (18.50) b) Evaluating (18.50) at t 0, t 1, and t 2 to check this answer, y(0) 8 5 3 16 y(1) 8 30 3 35 y(2) 8 180 3 191 Substituting in the initial equation with y(2) yt y(1) yt1 y(0) yt2 191 7(35) 6(16) 42 c) The characteristic roots are 1 and 6. The characteristic root with the largest absolute value is called the dominant root because it dominates the time path. For convergence, the absolute value of the dominant root must be less than 1. Since 6 1, the time path is divergent.
18.19. (a) Find the definite solution and (b) comment on dynamic stability, given yt2 11yt1 10yt 27 y(0) 2 y(1) 53 a) From (18.45), y(t) A1 A2(10)t 3t (18.51) Letting t 0 and t 1, and using the initial conditions, y(0) A1 A2 2 y(1) A1 10A2 3 53 Solving simultaneously, A1 4 and A2 6. Substituting in (18.51), y(t) 6(10)t 3t 4 b) The time path is divergent because the dominant root 10 is greater than 1.
18.20. Redo Problem 18.19, given yt 10yt1 25yt2 8, y(0) 1, and y(1) 5.
a) From (18.47), y(t) A1(5)t A2t(5)t 1 – 2 (18.52) Letting t 0 and t 1, and using the initial conditions, y(0) A1 1 – 2 1 A1 1 – 2 y(1) 5A1 5A2 1 – 2 5 With A1 1 – 2, A2 2 – 5. Substituting in (18.52), y(t) 1 – 2(5)t 2 – 5t(5)t 1 – 2 (18.53) b) Convergence in the case of repeated real roots likewise depends on r 1 since the effect of rt dominates the effect of t in the second term A2trt. Here with r 5 1, the time path is divergent.
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 18.21. Find the first-order derivative for the following trigonometric functions. Note that they are also called circular functions or sinusoidal functions.
a) y sin 7x b) y cos (5x 2) dy dx 7 cos 7x dy dx 5 sin (5x 2) c) y tan 11x d) y csc (8x 3) dy dx 11 sec2 11x dy dx 8[csc (8x 3) cot (8x 3)] 421 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] e) y sin (3 x2) f) y sin (5 x)2 dy dx 2x cos (3 x2) dy dx 2(5 x) cos (5 x)2 (chain rule) 18.22. Redo Problem 18.21, given y x2 tan x.
By the product rule, dy dx x2 (sec2 x) (tan x)(2x) x2 sec2 x 2x tan x 18.23. Redo Problem 18.21, given y x3 sin x.
dy dx x3 (cos x) (sin x)(3x2) x3 cos x 3x2 sin x 18.24. Redo Problem 18.21, given y (1 cos x)2.
By the chain rule, dy dx 2(1 cos x)(sin x) (2 sin x)(1 cos x) 18.25. Redo Problem 18.21, given y (sin x cos x)2.
dy dx 2(sin x cos x)(cos x sin x) 2(cos2 x sin2 x) 18.26. Redo Problem 18.21, given y sin2 5x, where sin2 5x (sin 5x)2.
By the chain rule, dy dx 2 sin 5x cos 5x(5) 10 sin 5x cos 5x 18.27. Redo Problem 18.21, given y csc2 12x.
dy dx (2 csc 12x)[csc 12x cot 12x(12)] 24 csc2 12x cot 12x COMPLEX ROOTS IN SECOND-ORDER DIFFERENTIAL EQUATIONS 18.28. Find (a) the particular integral, (b) the complementary function, and (c) the general solution, given the second-order linear differential equation y
(t) 2y(t) 10y(t) 80.
a) From (18.2), yp 8 – 1 0 – 0 8 b) Using (18.19) since b1 2 4b2, that is, (2)2 4(10), g 1 – 2(2) 1 h 1 – 24(10) (2)2 3 Thus, r1, r2 1 3i. Substituting g and h in (18.26), yc et(B1 cos 3t B2 sin 3t) c) y(t) yc yp et(B1 cos 3t B2 sin 3t) 8 (18.54) 18.29. Redo Problem 18.28, given y
(t) 6y(t) 25y(t) 150.
a) yp 150 ––– 25 6 b) From (18.19), g 1 – 2(6) 3 and h 1 – 24(25) (6)2 4. Substituting in (18.26), yc e3t(B1 cos 4t B2 sin 4t) c) y(t) e3t(B1 cos 4t B2 sin 4t) 6 (18.55) 422 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 18.30. Redo Problem 18.28, given y
(t) 4y(t) 40y(t) 10.
a) yp 1 – 4 0 – 0 1 – 4 b) From (18.19), g 2 and h 1 – 2160 16 6. Thus, yc e2t(B1 cos 6t B2 sin 6t).
c) y(t) e2t(B1 cos 6t B2 sin 6t) 1 – 4 (18.56) 18.31. (a) Find the definite solution for the following data. (b) Comment on the dynamic stability.
y
(t) 2y(t) 10y(t) 80 y(0) 10 y(0) 13 a) From (18.54), y(t) et(B1 cos 3t B2 sin 3t) 8 (18.57) By the product rule, y(t) et(3B1 sin 3t 3B2 cos 3t) (B1 cos 3t B2 sin 3t)(et) et(3B2 cos 3t 3B1 sin 3t) et(B1 cos 3t B2 sin 3t) (18.57a) Evaluating (18.57) and (18.57a) at t 0 and equating them to the initial conditions, y(0) e0(B1 cos 0 B2 sin 0) 8 10 From Table 18-1, cos 0 1 and sin 0 0. Thus, y(0) B1 0 8 10 B1 2 Similarly, y(0) e0(3B2 cos 0 3B1 sin 0) e0(B1 cos 0 B2 sin 0) 13.
y(0) 3B2 B1 13 Since B1 2 from above, B2 5. Finally, substituting in (18.57), y(t) et(2 cos 3t 5 sin 3t) 8 b) With g 1, the time path converges, as it does in Fig. 18-3 (see Example 15).
18.32. Redo Problem 18.31, given y
(t) 6y(t) 25y(t) 150, y(0) 13, and y(0) 25.
a) From (18.55), y(t) e3t(B1 cos 4t B2 sin 4t) 6 (18.58) Thus, y(t) e3t(4B1 sin 4t 4B2 cos 4t) 3e3t(B1 cos 4t B2 sin 4t) (18.58a) Evaluating (18.58) and (18.58a) at t 0 and equating them to the initial conditions, y(0) e0(B1 cos 0 B2 sin 0) 6 13 y(0) B1 0 6 13 B1 7 and y(0) e0(4B1 sin 0 4B2 cos 0) 3e0(B1 cos 0 B2 sin 0).
y(0) 4B2 3B1 25 B2 1 Substituting in (18.58), y(t) e3t(7 cos 4t sin 4t) 6.
b) With g 3, the time path is divergent.
18.33. Redo Problem 18.31, given y
(t) 4y(t) 40y(t) 10, y(0) 1 – 2, and y(0) 21 – 2.
a) From (18.56), y(t) e2t=(B1cos 6t B2 sin 6t) 1 – 4 (18.59) Thus, y(t) e2t(6B1 sin 6t 6B2 cos 6t) 2e2t(B1 cos 6t B2 sin 6t) (18.59a) 423 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] Evaluating (18.59) and (18.59a) at t 0 and equating them to the initial conditions, y(0) B1 1 – 4 1 – 2 B1 1 – 4 y(0) 6B2 2B1 21 – 2 B2 1 – 2 Thus, y(t) e2t(1 – 4 cos 6t 1 – 2 sin 6t) 1 – 4 1 – 4e2t(cos 6t 2 sin 6t) 1 – 4.
b) With g 2, the time path is convergent.
COMPLEX ROOTS IN SECOND-ORDER DIFFERENCE EQUATIONS 18.34. Find (a) the particular solution, (b) the complementary function, (c) the general solution, and (d) the definite solution. (e) Comment on the dynamic stability of the following second-order linear difference equation: yt 4yt2 15 y(0) 12 y(1) 11 a) From (18.11), yp 15 1 0 4 3 b) From (18.19), g 1 – 2(0) 0 and h 1 – 24(4) 0 2. For second-order difference equations we now need k and !. Applying the Pythagorean theorem to Fig. 18-1, k2 g2 h2 k g2 h2 Substituting with the parameters of (18.19) for greater generality, k b1 2 4b2 b1 2 4 b2 (18.60) Thus, k 4 2. From the definitions of Section 18.5, sin ! h k cos ! g k (18.61) Substituting the values from the present problem, sin ! 2 – 2 1 cos ! 0 – 2 0 From Table 18-1, the angle with sin ! 1 and cos ! 0 is /2. Thus, ! /2. Substituting in (18.27), yc 2tB1 cos 2 t B2 sin 2 t c) y(t) 2tB1 cos 2 t B2 sin 2 t 3 (18.62) d) Using Table 18-1 to evaluate (18.62) at t 0 and t 1 from the initial conditions, y(0) (B1 0) 3 12 B1 9 y(1) 2(0 B2) 3 11 B2 4 Thus, y(t) 2t9 cos 2 t 4 sin 2 t 3 e) With k 2, the time path is divergent, as explained in Example 15.
18.35. Redo Problem 18.34, given yt 2yt2 24, y(0) 11, and y(1) 18.
a) From (18.11), yp 24 1 0 2 8 424 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 b) From (18.60), k b2 2. From (18.19) g 1 – 2(0) 0 h 1 – 24(2) 0 2 From (18.61), sin ! 2 2 1 cos ! 0 From Table 18-1, ! /2. Substituting in (18.27), yc (2)tB1 cos 2 t B2 sin 2 t c) y(t) (2)tB1 cos 2 t B2 sin 2 t 8 (18.63) d) y(0) B1 8 11 B1 3 y(1) 2B2 8 18 B2 7.07 Thus, y(t) (2)t3 cos 2 t 7.07 sin 2 t 8 e) With k 2 1, the time path is divergent.
ECONOMIC APPLICATIONS 18.36. In many markets supply and demand are influenced by current prices and price trends (i.e., whether prices are rising or falling and whether they are rising or falling at an increasing or decreasing rate). The economist, therefore, needs to know the current price P(t), the first derivative dP(t)/dt, and the second derivative d2P(t)/dt2. Assume Qs c1 w1P u1P v1P Qd c2 w2P u2P v2P (18.64) Comment on the dynamic stability of the market if price clears the market at each point in time.
In equilibrium, Qs Qd. Therefore, c1 w1P u1P v1P c2 w2P u2P v2P (v1 v2)P (u1 u2)P (w1 w2)P (c1 c2) Letting v v1 v2, u u1 u2, w w1 w2, c c1 c2, and dividing through by v to conform to (18.1), P u v P w v P c v (18.65) Using (18.2) to find the particular integral, which will be the intertemporal equilibrium price P ¯ , P ¯ Pp c/v w/v c w Since c c1 c2 and w w1 w2 where under ordinary supply conditions, c1 0, w1 0, and under ordinary demand conditions, c2 0, w2 0, c/w 0, as is necessary for P ¯ . Using (18.4) to find the characteristic roots for the complementary function, r1, r2 u/v (u/v)2 4w/v 2 (18.66) which can assume three different types of solutions, depending on the specification of w, u, and v: 1.
If (u/v)2 4w/v, r1 and r2 will be distinct real roots solvable in terms of (18.66); and P(t) A1er1t A2er2t c/w.
425 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] 2.
If (u/v)2 4w/v, r1 and r2 will be repeated real roots. Thus, (18.66) reduces to (u/v)/2 or u/2v.
Then from (18.17), P(t) A1e(u/2v)t A2te(u/2v)t c/w.
3.
If (u/v)2 4w/v, r1 and r2 will be complex roots and from (18.26), P(t) egt(B1 cos ht B2 sin ht) c/w, where from (18.19), g u/(2v) and h 1 – 24w/v (u/v)2.
Specification of w, u, v depends on expectations. If people are bothered by inflationary psychology and expect prices to keep rising, u2 in (18.64) will be positive; if they expect prices to ultimately fall and hold off buying because of that expectation, u2 will be negative; and so forth.
18.37. In a model similar to Samuelson’s interaction model between the multiplier and the accelerator, assume Yt Ct It Gt (18.67) Ct C0 cYt1 (18.68) It I0 w(Ct Ct1) (18.69) where 0 c 1, w 0, and Gt G0. (a) Find the time path Y(t) of national income and (b) comment on the stability conditions.
a) Substituting (18.68) in (18.69), It I0 cw(Yt1 Yt2) (18.70) Substituting Gt G0, (18.70), and (18.68) into (18.67), and then rearranging to conform with (18.10), Yt C0 cYt1 I0 cw(Yt1 Yt2) G0 Yt c(1 w)Yt1 cwYt2 C0 I0 G0 (18.71) Using (18.11) for the particular solution, Yp C0 I0 G0 1 c(1 w) cw C0 I0 G0 1 c which is the intertemporal equilibrium level of income Y ¯ . Using (18.4) to find the characteristic roots for the complementary function, r1, r2 c(1 w) [c(1 w)]2 4cw 2 (18.72) which can assume three different types of solutions depending on the values assigned to c and w: 1.
If c2(1 w)2 4cw, or equivalently, if c(1 w)2 4w, r1 and r2 will be distinct real roots solvable in terms of (18.72) and Y(t) A1r1 t A2r2 t C0 I0 G0 1 c 2.
If c(1 w)2 4w, r1 and r2 will be repeated real roots, and from (18.72) and (18.18), Y(t) A1 1 2c(1 w) t A2t 1 2 c(1 w) t C0 I0 G0 1 c 3.
If c(1 w)2 4w, r1 and r2 will be complex roots; from (18.27), Y(t) kt(B1 cos t! B2 sin t!) C0 I0 G0 1 c where from (18.60), k cw, and from (18.61) ! must be such that sin ! h k cos ! g k where from (18.19), g 1 – 2c(1 w) and h 1 – 24cw c2(1 w)2.
426 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS [CHAP . 18 b) For stability in the model under all possible initial conditions, the necessary and sufficient conditions are (1) c 1 and (2) cw 1. Since c MPC with respect to the previous year’s income, c will be less than 1; for cw 1, the product of the MPC and the marginal capital-output ratio must also be less than 1. If the characteristic roots are conjugate complex, the time path will oscillate.
427 SECOND-ORDER DIFFERENTIAL EQUATIONS AND DIFFERENCE EQUATIONS CHAP . 18] CHAPTER 19 Simultaneous Differential and Difference Equations 19.1 MATRIX SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS, PART 1 Assume a system of n first-order, autonomous, linear differential equations in which no derivative is a function of another derivative and which we limit here to n 2 for notational simplicity.
Autonomous simply means all aij and bi are constant.
y ˙1 a11y1 a12y2 b1 y ˙2 a21y1 a22y2 b2 (19.1) Expressed in matrix form, y ˙1 y ˙2 a11 a21 a12 a22 y1 y2 b1 b2 or Y ˙ A Y B The complete solution to such a system will consist of n equations, each in turn composed of (1) a complementary solution yc and (2) a particular solution yp.
1.
a) From our earlier work with single differential equations, we can expect the complemen-tary solution of the system of equations, given distinct real roots, to take the general form, yc n i1 kiCierit k1C1er1t k2C2er2t (19.2) where ki a scalar or constant, Ci (2 1) column vector of constants called an eigenvector, and ri a scalar called the characteristic root. See Section 12.8.
428 b) As demonstrated in Problem 19.13, the characteristic roots, also called eigenvalues, can be found by solving the quadratic equation ri Tr(A) [Tr(A)]2 4 A 2 (19.3) where A determinant of A, Tr(A) trace of A, and Tr(A) of all the elements on the principal diagonal of A. Here where A (2 2), Tr(A) a11 a22 c) As explained in Problem 19.12, the solution of the system of simultaneous equations requires that (A riI)Ci 0 (19.4) where (A riI) a11 a21 a12 a22 ri 1 0 0 1 a11 ri a21 a12 a22 ri ri is a scalar, and I is an identity matrix, here I2. Equation (19.4) is called the eigenvalue problem. The eigenvectors are found by solving (19.4) for Ci. To preclude trivial, i.e., null-vector, solutions for Ci, the matrix (A riI) must be constrained to be singular.
2.
The particular integral, yp, is simply the intertemporal or steady-state solution. As demonstrated in Problem 19.14, yp Y ¯ A1B (19.5) where A1 the inverse of A and B the column of constants.
The stability of the model depends on the characteristic roots.
If all ri 0, the model is dynamically stable.
If all ri 0, the model is dynamically unstable.
If the ri are of different signs, the solution is at a saddle-point equilibrium and the model is unstable, except along the saddle path. See Section 19.5 and Examples 10 and 12.
EXAMPLE 1.
Solve the following system of first-order, autonomous, linear differential equations, y ˙1 5y1 0.5y2 12 y1(0) 12 y ˙2 2y1 5y2 24 y2(0) 4 1.
Convert them to matrices for ease of computation.
y ˙1 y ˙2 5 2 0.5 5 y1 y2 12 24 Y ˙ A Y B 2.
Then find the complementary functions. From (19.2), assuming distinct real roots, yc k1C1er1t k2C2er2t But from (19.3), the characteristic roots are r1, r2 Tr(A) [Tr(A)]2 4 A 2 where Tr(A) a11 a22 5 5 10 and A 25 1 24 429 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] Substituting, r1, r2 10 (10)2 4(24) 2 10 2 2 r1 4 r2 6 characteristic roots or eigenvalues 3.
We next find the eigenvectors. Using (19.4) and recalling (A riI) is singular, (A riI)Ci 0 where (A riI)Ci a11 ri a21 a12 a22 ri c1 c2 0 a) Substituting first for r1 4, 5 4 2 0.5 5 4 c1 c2 1 2 0.5 1 c1 c2 0 Then, by simple multiplication of row by column, we have c1 0.5c2 0 c1 0.5c2 2c1 c2 0 c1 0.5c2 Since (A riI) is constrained to be singular, there will always be linear dependence between the equations and we can work with either one. With linear dependence, there is also an infinite number of eigenvectors that will satisfy the equation. We can normalize the equation by choosing a vector whose length is unity, i.e., c1 2 c2 2 1, which is called the Euclidian distance condition, or we can simply pick any arbitrary value for one element while maintaining the relationship between elements.
Opting for the latter, let c1 1.
If c1 1, then c2 1 0.5 2 Thus, the eigenvector C1 corresponding to r1 4 is C1 c1 c2 1 2 and the first elements of the complementary function of the general solution are y1 c k1 1 2 e4t k1e4t 2k1e4t b) Substituting next for r2 6, 5 6 2 0.5 5 6 c1 c2 1 2 0.5 1 c1 c2 0 Multiplying row by column, c1 0.5c2 0 c1 0.5c2 2c1 c2 0 c1 0.5c2 If c1 1, then c2 1 0.5 2 Thus, the eigenvector C2 corresponding to r2 6 is C2 c1 c2 1 2 430 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 and the second elements of the complementary function of the general solution are yc 2 k2 1 2 e6t k2e6t 2k2e6t Putting them together for the complete complementary solution to the system, y1(t) k1e4t k2e6t y2(t) 2k1e4t 2k2e6t 4.
Now we find the intertemporal or steady-state solutions for yp. From (19.5), yp Y ¯ A1B where B 12 24 , A 5 2 0.5 5 , A 25 1 24, the cofactor matrix is C 5 0.5 2 5 , the adjoint matrix is Adj. A C 5 2 0.5 5 , and the inverse is A1 1 24 5 2 0.5 5 Substituting in (19.5), Y ¯ 1 24 5 2 0.5 5 12 24 Multiplying row by column, Y ¯ y ¯1 y ¯2 1 24 72 144 3 6 Thus the complete general solution, y(t) yc yp, is y1(t) k1e4t k2e6t 3 (19.6) y2(t) 2k1e4t 2k2e6t 6 With r1 4 0 and r2 6 0, the equilibrium is unstable. See also Problems 19.1 to 19.3.
EXAMPLE 2.
To find the definite solution for Example 1, we simply employ the initial conditions, y1(0) 12, y2(0) 4. When (19.6) is evaluated at t 0, we have y1(0) k1 k2 3 12 y2(0) 2k1 2k2 6 4 Solved simultaneously, k1 4 k2 5 Substituting in (19.6), we have the definite solution, y1(t) 4e4t 5e6t 3 y2(t) 8e4t 10e6t 6 which remains dynamically unstable because of the positive roots.
19.2 MATRIX SOLUTION OF SIMULTANEOUS DIFFERENTIAL EQUATIONS, PART 2 Assume a system of n first-order, autonomous, linear differential equations in which one or more derivatives is a function of another derivative, and which we limit here to n 2 simply for notational simplicity, a11y ˙1 a12y ˙2 a13y1 a14y2 b1 a21y ˙1 a22y ˙2 a23y1 a24y2 b2 (19.7) 431 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] In matrix form, a11 a12 a21 a22 y ˙1 y ˙2 a13 a14 a23 a24 y1 y2 b1 b2 A1Y ˙ A2Y B The general solution y(t) will consist of a complementary function yc and a particular integral or solution yp. As in previous examples, for distinct real roots we can expect the complementary function to assume the general form yc k1C1er1t k2C2er2t (19.8) As explained in Problems 19.15 to 19.16, the eigenvalue problem here is (A2 riA1)Ci 0 (19.9) where (A2 riA1) a13 a14 a23 a24 ri a11 a12 a21 a22 a13 a11ri a14 a12ri a23 a21ri a24 a22ri The characteristic equation is A2 riA1 0 (19.10) the particular integral is Y ¯ A2 1B (19.11) and the stability conditions are the same as in Section 19.1.
EXAMPLE 3.
Solve the following system of first-order, autonomous, nonlinear differential equations.
y ˙1 3y1 1.5y2 2.5y ˙2 2.4 y1(0) 14 y ˙2 2y1 5y2 16 y2(0) 15.4 1.
First rearrange the equations to conform with (19.7) and set them in matrix form, 1 0 2.5 1 y ˙1 y ˙2 3 2 1.5 5 y1 y2 2.4 16 2.
Assuming distinct real roots, find the complementary function.
yc k1C1er1t k2C2er2t a) Start with the characteristic equation to find the characteristic roots. From (19.10), A2 riA1 0 b) Substituting and dropping the i subscript for simplicity, A2 rA1 3 2 1.5 5 r 1 0 2.5 1 3 r 2 1.5 2.5r 5 r 0 (3 r)(5 r) 2(1.5 2.5r) 0 r2 13r 12 0 r1 1 r2 12 characteristic roots 3.
Find the eigenvectors, Ci. From (19.9), (A2 riA1)Ci 0 where (A2 riA1)Ci 3 ri 2 1.5 2.5ri 5 ri c1 c2 432 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 a) Substituting first for r1 1, 3 (1) 2 1.5 2.5(1) 5 (1) c1 c2 2 2 4 4 c1 c2 0 By simple matrix multiplication, 2c1 4c2 0 c1 2c2 2c1 4c2 0 c1 2c2 If we let c1 2, c2 1. Thus, C1 c1 c2 2 1 and the first elements of the general complementary function for r1 1 are yc 1 k1 2 1 et 2k1 k1 et b) Now substituting for r2 12, 3 (12) 2 1.5 2.5(12) 5 (12) c1 c2 9 2 31.5 7 c1 c2 0 Multiplying row by column, 9c1 31.5c2 0 c1 3.5c2 2c1 7c2 0 c1 3.5c2 Letting c1 3.5, c2 1. So, C2 c1 c2 3.5 1 and the second elements of the complementary function for r2 12 are yc 2 k2 3.5 1 e12t 3.5k2 k2 e12t Adding the two together, the complementary functions are y1(t) 2k1et 3.5k2e12t y2(t) k1et k2e12t (19.12) 4.
Find the particular integral yp which is simply the intertemporal equilibrium Y ¯ . From (19.11), Y ¯ A2 1B where B 2.4 16 , A2 3 2 1.5 5 , A2 15 3 12 the cofactor matrix is C 5 1.5 2 3 , the adjoint matrix is Adj. A C 5 2 1.5 3 and the inverse is A2 1 1 12 5 2 1.5 3 Substituting in (19.11), Y ¯ y ¯1 y ¯2 1 12 5 2 1.5 3 2.4 16 3 4.4 433 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] 5.
By adding the particular integrals or steady-state solutions to the complementary function in (19.12), we derive the complete general solution.
y1(t) 2k1et 3.5k2e12t 3 y2(t) k1et k2e12t 4.4 (19.13) With r1 1 0, r2 12 0, the system of equations is dynamically stable. See also Problems 19.4 to 19.6.
EXAMPLE 4.
To find the definite solution for Example 3, we simply employ the initial conditions, y1(0) 14, y2(0) 15.4. When (19.13) is evaluated at t 0, we have y1(0) 2k1 3.5k2 3 14 y2(0) k1 k2 4.4 15.4 Solved simultaneously, k1 9 k2 2 Substituting in (19.13) for the definite solution, y1(t) 18et 7e12t 3 y2(t) 9et 2e12t 4.4 19.3 MATRIX SOLUTION OF SIMULTANEOUS DIFFERENCE EQUATIONS, PART 1 Assume a system of n linear first-order difference equations in which no difference is a function of another difference, the coefficients are constants, and we again set n 2 for notational simplicity.
xt a11xt1 a12yt1 b1 yt a21xt1 a22yt1 b2 In matrix form, xt yt a11 a12 a21 a22 xt1 yt1 b1 b2 Letting Yt xt yt , Yt1 xt1 yt1 , B b1 b2 , and A the coefficient matrix, we have Yt A Yt1 B (19.14) The complete solution will consist of n equations, each in turn composed of the complementary solution yc and the particular solution yp. Based on our earlier work with single difference equations and assuming distinct real roots, we can expect the complementary function will take the general form, yc n i1 kiCiri t k1C1r1 t k2C2r2 t (19.15) As demonstrated in Problem 19.17, the eigenvalue problem breaks down to (A riI)Ci 0 (19.16) By similar steps to the demonstrations in Problems 19.13 to 19.14, it can be shown that the characteristic equation is A riI 0 where the characteristic roots can be found with (19.3), and the particular solution is yp (I A)1B (19.17) 434 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 The stability conditions require that each of the n roots be less than 1 in absolute value for dynamic stability. If even one root is greater than 1 in absolute value, it will dominate the other(s) and the time path will be divergent.
EXAMPLE 5.
Solve the following system of first-order linear difference equations, xt 4xt1 yt1 12 x0 16 yt 2xt1 3yt1 6 y0 8 (19.18) 1.
Set them in matrix form, xt yt 4 2 1 3 xt1 yt1 12 6 Yt A Yt1 B 2.
We next find the complementary functions. Assuming a case of distinct real roots, yc k1C1r1 t k2C2r2 t and the characteristic roots are r1, r2 Tr(A) [Tr(A)]2 4 A 2 where Tr(A) 4 3 7 and A 12 2 10 Substituting, r1, r2 7 (7)2 4(10) 2 7 3 2 r1 2 r2 5 characteristic roots or eigenvalues 3.
We then find the eigenvectors. Using (19.16) and recalling (A riI) is singular, (A riI)Ci a11 ri a21 a12 a22 ri c1 c2 0 a) Substituting first for r1 2, 4 (2) 2 1 3 (2) c1 c2 2 2 1 1 c1 c2 0 Then by multiplying row by column, we find 2c1 c2 0 c2 2c1 2c1 c2 0 c2 2c1 Letting c1 1, c2 2 Hence the eigenvector corresponding to r1 2 is C1 1 2 and the ai1 elements of the complementary function are k1 1 2 (2)t k1(2)t 2k1(2)t b) Then substituting for r2 5, 4 (5) 2 1 3 (5) c1 c2 1 1 2 2 c1 c2 0 435 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] Multiplying row by column, c1 c2 0 c2 c1 2c1 2c2 0 c2 c1 If c1 1, c2 1 The eigenvector for r2 5 is C2 1 1 and the ai2 elements of the complementary function are k2 1 1 (5)t k2(5)t k2(5)t Combining the two for the general complementary function, we have xc k1(2)t k2(5)t yc 2k1(2)t k2(5)t 4.
Now we find the particular solution for the steady-state solutions x ¯, y ¯. From (19.17), yp (I A)1B where (I A) 1 0 0 1 4 2 1 3 5 2 1 4 Following a series of steps similar to those in Example 3, yp x ¯ y ¯ 1 18 4 1 2 5 12 6 3 3 This makes the complete general solution, xt k1(2)t k2(5)t 3 yt 2k1(2)t k2(5)t 3 With 2 1 and 5 1, the time path is divergent. See also Problems 19.7 to 19.8.
EXAMPLE 6.
For the specific solution, we need only employ the initial conditions. Given x0 16, y0 8, at t 0, the general functions reduce to k1 k2 3 16 2k1 k2 3 8 Solving simultaneously, k1 6, k2 7 By simple substitution, we then find the specific solution, xt 6(2)t 7(5)t 3 yt 12(2)t 7(5)t 3 (19.19) To check the answer, substitute t 1 and t 0 in (19.19).
x1 6(2)1 7(5)1 3 44 x0 6(2)0 7(5)0 3 16 y1 12(2)1 7(5)1 3 14 y0 12(2)0 7(5)0 3 8 Then go back to (19.18) and substitute x1, y1 for xt and yt, and x0, y0 for xt1 and yt1.
44 4(16) 8 12 44 14 2(16) 3(8) 6 14 436 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 19.4 MATRIX SOLUTION OF SIMULTANEOUS DIFFERENCE EQUATIONS, PART 2 Assume a system of n linear first-order difference equations in which one or more differences is a function of another difference, the coefficients are constant, and we again set n 2 for notational simplicity.
a11xt a12yt a13xt1 a14yt1 b1 a21xt a22yt a23xt1 a24yt1 b2 (19.20) or a11 a12 a21 a22 xt yt a13 a14 a23 a24 xt1 yt1 b1 b2 A1Yt A2Yt1 B From previous sections we can expect the general solution yt will consist of a complementary function yc and a particular solution yp, where the complementary function for distinct real roots will take the general form yc k1C1(r1)t k2C2(r2)t As demonstrated in Problems 19.18 to 19.19, the eigenvalue problem here reduces to (A2 riA1)Ci 0 (19.21) where (A2 riA1) a13 a14 a23 a24 ri a11 a12 a21 a22 a13 a11ri a14 a12ri a23 a21ri a24 a22ri and the particular integral is Y ¯ (A1 A2)1B (19.22) The stability conditions remain the same as in Section 19.3.
EXAMPLE 7.
Solve the following system of linear first-order difference equations.
xt 4xt1 2yt1 yt 10 x0 20 yt 3xt1 6yt1 4 y0 3 (19.23) 1.
Rearrange to conform with (19.20) and set in matrix form.
1 0 1 1 xt yt 4 3 2 6 xt1 yt1 10 4 A1Yt A2Yt1 B 2.
For the complementary function, begin with the characteristic equation derived from (19.21) A2 riA1 0 and substitute the parameters of the problem, A2 riA1 4 r 3 2 r 6 r 0 (4 r)(6 r) 3(2 r) r2 13r 30 0 0 r1 3 r2 10 3.
Find the nontrivial solutions for the eigenvectors Ci. From (19.21), (A2 riA1)Ci 0 437 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] where (A2 riA1)Ci 4 r 3 2 r 6 r c1 c2 a) Substituting for r1 3, 4 3 3 2 3 6 3 c1 c2 1 1 3 3 c1 c2 0 Multiplying row by column, c1 c2 0 c1 c2 3c1 3c2 0 c1 c2 If c1 1, c2 1, the eigenvector is C1 1 1 making the ai1 elements of the complementary function for r1 3, k1 1 1 (3)t k1(3)t k1(3)t b) For r2 10, 4 10 3 2 10 6 10 c1 c2 6 3 8 4 c1 c2 0 By matrix multiplication, 6c1 8c2 0 c2 0.75c1 3c1 4c2 0 c2 0.75c1 Letting c1 1, c2 0.75, the corresponding eigenvector becomes C1 1 0.75 and the ai2 elements of the complementary function for r2 10 is k2 1 0.75 (10)t k2(10)t 0.75k2(10)t Combining the two, the complete complementary function becomes xc k1(3)t k2(10)t yc k1(3)t 0.75k2(10)t (19.24) 4.
Now find the particular or steady-state solution yp Y ¯ . From (19.22), Y ¯ (A1 A2)1B where A1 A2 1 0 1 1 4 3 2 6 3 3 1 5 Hence Y ¯ x ¯ y ¯ 1 18 5 3 1 3 10 4 3 1 5.
The complete general solution, y0 yc yp, then becomes xt k1(3)t k2(10)t 3 yt k1(3)t 0.75k2(10)t 1 (19.25) Since r1 3, r2 10 1 , the system of equations is dynamically unstable. See also Problems 19.9 to 19.10.
438 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 EXAMPLE 8.
The specific solution is found with the help of the initial conditions. With x0 20 and y0 3, at t 0, (19.24) reduces to k1 k2 3 20 k1 0.75k2 1 3 Solving simultaneously, k1 5, k2 12 and by substituting into (19.25), we come to the definite solution, xt 5(3)t 12(10)t 3 yt 5(3)t 9(10)t 1 (19.26) To check the answer, substitute t 1 and t 0 in (19.26).
x1 5(3) 12(10) 3 138 x0 5 12 3 20 y1 5(3) 9(10) 1 74 y0 5 9 1 3 Then substitute x1, y1 for xt and yt, and x0, y0 for xt1 and yt1 back in (19.23).
138 4(20) 2(3) 74 10 138 74 3(20) 6(3) 4 74 19.5 STABILITY AND PHASE DIAGRAMS FOR SIMULTANEOUS DIFFERENTIAL EQUATIONS Given a system of linear autonomous differential equations, the intertemporal equilibrium level will be asymptotically stable, i.e., y(t) will converge to y ¯ as t →, if and only if all the characteristic roots are negative. In the case of complex roots, the real part must be negative. If all the roots are positive, the system will be unstable. A saddle-point equilibrium, in which roots assume different signs, will generally be unstable. If, however, the initial conditions for y1 and y2 satisfy the condition y2 r1 a11 a12 (y1 y ¯1) y ¯2 where r1 the negative root, we have what is called a saddle path, and y1(t) and y2(t) will converge to their intertemporal equilibrium level (see Example 10).
A phase diagram for a system of two differential equations, linear or nonlinear, graphs y2 on the vertical axis and y1 on the horizontal axis. The y1, y2 plane is called the phase plane. Construction of a phase diagram is easiest explained in terms of an example.
EXAMPLE 9.
Given the system of linear autonomous differential equations, y ˙1 4y1 16 y ˙2 5y2 15 a phase diagram is used below to test the stability of the model. Since neither variable is a function of the other variable in this simple model, each equation can be graphed separately.
1.
Find the intertemporal equilibrium level, y ¯1, i.e., the locus of points at which y ˙1 0.
y ˙1 4y1 16 0 y ¯1 4 The graph of y ¯1 4, a vertical line at y1 4, is called the y1 isocline. The y1 isocline divides the phase plane into two regions called isosectors, one to the left of the y1 isocline and one to the right.
2.
Find the intertemporal equilibrium level, y ¯2, i.e., the locus of points at which y ˙2 0.
y ˙2 5y2 15 0 y ¯2 3 439 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] The graph of y ¯2 3 is a horizontal line at y2 3, called the y2 isocline. The y2 isocline divides the phase plane into two isosectors, one above the y2 isocline and the other below it. See Fig. 19-1.
The intersection of the isoclines demarcates the intertemporal equilibrium level, (y ¯1, y ¯2) (4, 3) 3.
Determine the motion around the y1 isocline, using arrows of horizontal motion.
a) To the left of the y1 isocline, y1 4.
b) To the right of the y1 isocline, y1 4.
By substituting these values successively in y ˙1 4y1 16, we see If y1 4, y ˙ 1 0, and there will be motion to right If y1 4, y ˙1 0, and there will be motion to left.
4.
Determine the motion around the y2 isocline, using arrows of vertical motion.
a) Above the y2 isocline, y2 3.
b) Below the y2 isocline, y2 3.
Substitution of these values successively in y ˙2 5y2 15, shows If y2 3, y ˙ 2 0, and the motion will be downward.
If y2 3, y ˙2 0, and the motion will be upward.
The resulting arrows of motion in Fig. 19-1, all pointing to the intertemporal equilibrium, suggest the system of equations is convergent. Nevertheless, trajectory paths should be drawn because the arrows by themselves can be deceiving, as seen in Fig. 19-2. Starting from an arbitrary point, such as (3, 2) in the southwest quadrant, or any point in any quadrant, we can see that the dynamics of the model will lead to the steady-state solution (4, 3). Hence the time path converges to the steady-state solution, making that solution stable.
Since the equations are linear, the answer can be checked using the techniques of Chapter 16 or 19, getting y1(t) k1e4t 4 y2(t) k2e5t 3 With both characteristic roots negative, the system must be stable.
440 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 Fig. 19-1 Fig. 19-2 2 y1 4 3 y1 = 0 y2 = 0 (3, 2) .
.
y2 = 3 (a) y1 y1 = 0 y2 = 0 y1 = 2 (d) (c) (b) y2 .
.
EXAMPLE 10.
A phase diagram is constructed in Fig. 19-2 and used below to test the dynamic stability of a saddle point equilibrium for the system of equations, y ˙1 2y2 6 y ˙2 8y1 16 1.
Find the y1 isocline on which y ˙1 0.
y ˙1 2y2 6 0 y ¯2 3 Here the y1 isocline is a horizontal line at y ¯2 3.
2.
Find the y2 isocline on which y ˙2 0.
y ˙2 8y1 16 y ¯1 2 The y2 isocline is a vertical line at y ¯1 2. See Fig. 19-2.
3.
Determine the motion around the y1 isocline, using arrows of horizontal motion.
a) Above the y1 isocline, y2 3.
b) Below the y1 isocline, y2 3.
Substitution of these values successively in y ˙1 2y2 6, shows If y2 3, y ˙ 1 0, and the arrows of motion point to the right.
If y2 3, y ˙1 0, and the arrows of motion point to the left.
4.
Determine the motion around the y2 isocline, using arrows of vertical motion.
a) To the left of the y2 isocline, y1 2.
b) To the right of the y2 isocline, y1 2.
By substituting these values successively in y ˙2 8y1 16, we see If y1 2, y ˙ 2 0, and there will be motion downward.
If y1 2, y ˙2 0, and there will be motion upward.
Despite appearances in Fig. 19-2, the system is unstable even in the northwest and southeast quadrants. As explained in Example 11, we can show by simply drawing trajectories that the time paths diverge in all four quadrants, whether we start at point a, b, c, or d.
EXAMPLE 11.
The instability in the model in Fig. 19-2 is made evident by drawing a trajectory from any of the quadrants. We do two, one from a and one from b, and leave the other two for you as a practice exercise. In each case the path of the trajectory is best described in four steps.
1.
Departure from point a.
a) The trajectory moves in a southeasterly direction.
b) But as the time path approaches the y1 isocline where y ˙1 0, the y1 motion eastward slows down while the y2 motion southward continues unabated.
c) At the y1 isocline, y ˙1 0. Consequently, the trajectory must cross the y1 isocline vertically.
d) Below the y1 isocline, the arrows of motion point in a southwesterly direction, taking the time path away from the equilibrium and hence indicating an unstable equilibrium.
2.
Departure from point b.
a) The trajectory once again moves in a southeasterly direction.
b) But as the time path approaches the y2 isocline where y ˙2 0, the y2 motion southward ebbs while the y1 motion eastward continues unaffected.
c) Since y ˙2 0 at the y2 isocline, the time path must cross the y2 isocline horizontally.
d) To the right of the y2 isocline, the arrows of motion point in a northeasterly direction, taking the time path away from the equilibrium and belying the appearance of a stable equilibrium.
EXAMPLE 12.
The dotted line in Fig. 19-2 is a saddle path. Only if the initial conditions fall on the saddle path will the steady-state equilibrium prove to be stable. The equation for the saddle path is y2 r1 a11 a12 (y1 y ¯1) y ¯2 441 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] where we already know all but r1, the negative root. From the original equations, A a11 a12 a21 a22 0 2 8 0 16 Any time A 0, we have a saddle-point equilibrium. Substituting in (19.3), ri Tr(A) [Tr(A)]2 4 A 2 r1, r2 0 0 4(16) 2 4, 4 Then substituting in the saddle-path equation above, y2 4 0 2 (y1 2) 3 y2 7 2y1 saddle path Note that the intertemporal equilibrium (2, 3) falls on the saddle path. Only if the initial conditions satisfy the saddle-path condition will the intertemporal equilibrium be stable.
Solved Problems SIMULTANEOUS DIFFERENTIAL EQUATIONS 19.1.
Solve the following system of first-order, autonomous, linear differential equations, y ˙1 8y1 5y2 4 y1(0) 7 y ˙2 3.25y1 4y2 22 y2(0) 21.5 1.
Putting them in matrix form for ease of computation, y ˙1 y ˙2 8 3.25 5 4 y1 y2 4 22 Y ˙ A Y B 2.
Then find the complementary functions. Assuming distinct real roots, yc k1C1er1t k2C2er2t where r1, r2 Tr(A) [Tr(A)]2 4 A 2 Tr(A) 12, and A 15.75.
r1, r2 12 (12)2 4(15.75) 2 12 9 2 r1 1.5 r2 10.5 3.
Next we find the eigenvectors Ci from (A riI)Ci a11 ri a21 a12 a22 ri c1 c2 0 442 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 a) For r1 1.5, 8 (1.5) 3.25 5 4 (1.5) c1 c2 6.5 3.25 5 2.5 c1 c2 0 By simple multiplication of row by column, 6.5c1 5c2 0 c2 1.3c1 3.25c1 2.5c2 0 c2 1.3c1 If c1 1, then c2 1.3. Thus, the eigenvector C1 corresponding to r1 1.5 is C1 c1 c2 1 1.3 and the first elements of the complementary function of the general solution are yc 1 k1 1 1.3 e1.5t k1e1.5t 1.3k1e1.5t b) Substituting next for r2 10.5, 8 (10.5) 3.25 5 4 (10.5) c1 c2 2.5 3.25 5 6.5 c1 c2 0 Then simply multiplying the first row by the column, since the final results will always be the same due to the singularity of the (A riI) matrix 2.5c1 5c2 0 c1 2c2 If c2 1, then c1 2; the eigenvector C2 for r2 10.5 is C2 c1 c2 2 1 and the second elements of the general complementary function are yc 2 k2 2 1 e10.5t 2k2e10.5t k2e10.5t This makes the complete complementary solution, y1(t) k1e1.5t 2k2e10.5t y2(t) 1.3k1e1.5t k2e10.5t 4.
Now find the intertemporal equilibrium solutions for yp, yp Y ¯ A1B where A 8 3.25 5 4, C 4 5 3.25 8 , Adj. A 4 3.25 5 8 , A1 1 15.75 4 3.25 5 8 Substituting above, Y ¯ y ¯1 y ¯2 1 15.75 4 3.25 5 8 4 22 8 12 Thus the complete general solution, y(t) yc yp, is y1(t) k1e1.5t 2k2e10.5t 8 y2(t) 1.3k1e1.5t k2e10.5t 12 (19.27) 443 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] 5.
To find the definite solution, simply evaluate the above equations at t 0 and use the initial conditions, y1(0) 7, y2(0) 21.5.
y1(0) k1 2k2 8 7 y2(0) 1.3k1 k2 12 21.5 Solved simultaneously, k1 5 k2 3 Substituting in (19.27), y1(t) 5e1.5t 6e10.5t 8 y2(t) 6.5e1.5t 3e10.5t 12 With r1 1.5 0, r2 10.5 0, the equilibrium is dynamically stable.
19.2.
Solve the following system of differential equations, y ˙1 2y2 6 y1(0) 1 y ˙2 8y1 16 y2(0) 4 1.
In matrix form, y ˙1 y ˙2 0 2 8 0 y1 y2 6 16 Y ˙ A Y B 2.
Finding the characteristic roots, r1, r2 Tr(A) [Tr(A)]2 4 A 2 r1, r2 0 (0)2 4(16) 2 8 2 r1 4 r2 4 3.
Now determine the eigenvectors.
a) For r1 4, 0 (4) 8 2 0 (4) c1 c2 4 2 8 4 c1 c2 0 4c1 2c2 0 c2 2c1 If c1 1, then c2 2, and the first elements of the complementary function are yc 1 k1 1 2 e4t k1e4t 2k1e4t b) For r1 4, 0 4 8 2 0 4 c1 c2 4 8 2 4 c1 c2 0 4c1 2c2 0 c2 2c1 If c1 1, then c2 2, and the second elements of the complementary function are yc 2 k2 1 2 e4t k2e4t 2k2e4t 444 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 This means the general complementary functions are y1(t) k1e4t k2e4t y2(t) 2k1e4t 2k2e4t 4.
For the steady-state solutions yp, yp Y ¯ A1B Y ¯ y ¯1 y ¯2 1 16 0 8 2 0 6 16 2 3 Thus the complete general solution, y(t) yc yp, is y1(t) k1e4t k2e4t 2 y2(t) 2k1e4t 2k2e4t 3 (19.28) 5.
We then find the definite solution from the initial conditions, y1(0) 1, y2(0) 4.
y1(0) k1 k2 2 1 y2(0) 2k1 2k2 3 4 Solved simultaneously, k1 0.75 k2 0.25 Substituting in (19.28), we have the final solution.
y1(t) 0.75e4t 0.25e4t 2 y2(t) 1.5e4t 0.5k2e4t 3 With r1 4 0 and r2 4 0, we have a saddle-point solution. Saddle-point solutions are generally unstable unless the initial conditions fall on the saddle path: y2 r1 a11 a12 (y1 y ¯1) y ¯2 Substituting, y2 y2 7 2y1 4 0 2 (y1 2) 3 This is the equation for the saddle path, which was graphed in Fig. 19-2 of Example 10. Substituting the initial conditoins, y1(0) 1, y2(0) 4, we see 4 7 2(1) 5 Since the initial conditions do not fall on the saddle path, the system is unstable.
19.3.
Solve the following system of equations.
y ˙1 4y1 7y2 3 y1(0) 7 y ˙2 y1 2y2 4 y2(0) 10 1.
Converting to matrices, y ˙1 y ˙2 4 1 7 2 y1 y2 31 4 Y ˙ A Y B 2.
The characteristic roots are r1,r2 2 (2)2 4(15) 2 2 8 2 r1 3 r2 5 445 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] 3.
The eigenvector for r1 3, 4 (3) 1 7 2 (3) c1 c2 7 7 1 1 c1 c2 0 7c1 7c2 0 c1 c2 If c2 1, then c1 1, and yc 1 k1 1 1 e3t k1e3t k1e3t For r2 5, 4 5 1 7 2 5 c1 c2 1 1 7 7 c1 c2 0 c1 7c2 0 c1 7c2 If c2 1, then c1 7, and yc 2 k2 7 1 e5t 7k2e5t k2e5t The general complementary functions are y1(t) k1e3t 7k2e5t y2(t) k1e3t k2e5t 4.
The steady-state solutions yp are yp Y ¯ A1B Y ¯ y ¯1 y ¯2 1 15 2 1 7 4 31 4 6 1 and the complete general solution is y1(t) k1e3t 7k2e5t 6 y2(t) k1e3t k2e5t 1 (19.29) 5.
The definite solution, given y1(0) 7, y2(0) 10, is y1(0) k1 7k2 6 7 y2(0) k1 k2 1 10 k1 8 k2 3 Substituting in (19.29) for the final solution, y1(t) 8e3t 21e5t 6 y2(t) 8e3t 3e5t 1 With r1 3 0 and r2 5 0, we again have a saddle-point solution which will be unstable unless the initial conditions fulfill the saddle-path equation: y2 r1 a11 a12 (y1 y ¯1) y ¯2 Substituting, y2 y2 7 y1 3 4 7 [y1 (6)] (1) Employing the initial conditions, y1(0) 7, y2(0) 10, 10 7 (7) 14 446 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 Since the initial conditions do not satisfy the saddle-path equation, the system of equations is unstable.
19.4.
Solve the following system of nonlinear, autonomous, first-order differential equations in which one or more derivative is a function of another derivative.
y ˙1 4y1 y2 6 y1(0) 9 y ˙2 8y1 5y2 y ˙1 6 y2(0) 10 1.
Rearranging the equations to conform with (19.7) and setting them in matrix form, 1 0 1 1 y ˙1 y ˙2 4 1 8 5 y1 y2 6 6 A1Y ˙ A2Y B 2.
Find the characteristic roots from the characteristic equation, A2 riA1 0 where, dropping the i subscript for simplicity, A2 rA1 4 1 8 5 r 1 0 1 1 4 r 8 r 1 5 r 0 r2 8r 12 0 r1 2 r2 6 3.
Find the eigenvectors Ci where (A2 riA1)Ci 0 and (A2 riA1)Ci 4 ri 8 ri 1 5 ri c1 c2 Substituting for r1 2, 4 2 8 2 1 5 2 c1 c2 2 1 6 3 c1 c2 0 2c1 c2 0 c2 2c1 If c1 1, c2 2, and yc 1 k1 1 2 e2t k1 2k1 e2t Now substituting for r2 6, 4 6 8 6 1 5 6 c1 c2 2 2 1 1 c1 c2 0 2c1 c2 0 c2 2c1 If c1 1, c2 2, and yc 2 k2 1 2 e6t k2 2k2 e6t Adding the two components of the complementary functions, y1(t) k1e2t k2e6t y2(t) 2k1e2t 2k2e6t 447 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] 4.
For the particular integral yp, Y ¯ A2 1B where B 6 6 , A2 4 1 8 5 , A2 20 8 12, and A2 1 1 12 5 8 1 4 .
Substituting, Y ¯ y ¯1 y ¯2 1 12 5 8 1 4 6 6 3 6 Adding the particular integrals to the complementary functions, y1(t) k1e2t k2e6t 3 y2(t) 2k1e2t 2k2e6t 6 (19.30) 5.
For the definite solution, set t 0 in (19.30) and use y1(0) 9, y2(0) 10.
y1(0) k1 k2 3 9 y2(0) 2k1 2k2 6 10 k1 5 k2 7 Then substituting back in (19.30), y1(t) 5e2t 7e6t 3 y2(t) 10e2t 14e6t 6 With r1 2 0 and r2 6 0, the system of equations will be dynamically unstable.
19.5.
Solve the following system of differential equations.
y ˙1 y1 4y2 0.5y ˙2 1 y1(0) 4.5 y ˙2 4y1 2y2 10 y2(0) 16 1.
Rearranging and setting in matrix form, 1 0 0.5 1 y ˙1 y ˙2 1 4 4 2 y1 y2 1 10 A1Y ˙ A2Y B 2.
From the characteristic equation, A2 rA1 1 r 4 4 0.5r 2 r 0 we find the characteristic roots, r2 5r 14 0 r1 7 r2 2 3.
We next find the eigenvectors Ci from the eigenvalue problem, (A2 riA1)Ci 1 r 4 4 0.5r 2 r c1 c2 0 Substituting for r1 7, 1 (7) 4 4 0.5(7) 2 (7) c1 c2 6 4 7.5 5 c1 c2 0 6c1 7.5c2 0 c1 1.25c2 448 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 If c2 1, c1 1.25, and yc 1 k1 1.25 1 e7t 1.25k1e7t k1e7t Substituting for r1 2, 1 2 4 4 0.5(2) 2 2 c1 c2 3 4 3 4 c1 c2 0 3c1 3c2 0 c1 c2 If c2 1, c1 1, and yc 2 k2 1 1 e2t k2e2t k2e2t This makes the complete general complementary functions, y1(t) 1.25k1e7t k2e2t y2(t) k1e7t k2e2t 4.
Finding the particular integral Y ¯ A2 1B, where A2 1 4 4 2 , A2 1 1 14 2 4 4 1 , and Y ¯ y ¯1 y ¯2 1 14 2 4 4 1 1 10 3 1 By adding the particular integrals to the complementary functions, we get y1(t) 1.25k1e7t k2e2t 3 y2(t) k1e7t k2e2t 1 (19.31) 5.
For the definite solution, we set t 0 in (19.31) and use y1(0) 4.5, y2(0) 16.
y1(0) 1.25k1 k2 3 4.5 y2(0) k1 k2 1 16 k1 6 k2 9 Finally, substituting in (19.31), y1(t) 7.5e7t 9e2t 3 y2(t) 6e7t 9e2t 1 With characteristic roots of different signs, we have a saddle-point equilibrium which will be unstable unless the initial conditions happen to coincide with a point on the saddle path.
19.6.
Solve the following.
y ˙1 3y1 y2 0.5y ˙2 5 y1(0) 22.2 y ˙2 2y1 4y2 y ˙1 10 y2(0) 3.9 1.
In matrix form, 1 1 0.5 1 y ˙1 y ˙2 3 2 1 4 y1 y2 5 10 A1Y ˙ A2Y B 449 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] 2.
The characteristic equation is A2 rA1 3 r 2 r 1 0.5r 4 r 0 0.5r2 5r 10 0 Multiplying by 2 and using the quadratic formula, the characteristic roots are r1 7.235 r2 2.765 3.
The eigenvector for r1 7.235 is 3 (7.235) 2 (7.235) 1 0.5(7.235) 4 (7.235) c1 c2 4.235 2.6175 5.235 3.235 c1 c2 0 4.235c1 2.6175c2 0 c2 1.62c1 If c1 1, c2 1.62, and yc 1 k1 1 1.62 e7.235t k1e7.235t 1.62k1e7.235t For r2 2.765, 3 (2.765) 2 (2.765) 1 0.5(2.765) 4 (2.765) c1 c2 0.235 0.3825 0.765 1.235 c1 c2 0 0.235c1 0.3825c2 0 c1 1.62c2 If c2 1, c1 1.62, and yc 2 k2 1.62 1 e2.765t 1.62k2e2.765t k2e2.765t The complete complementary function, then, is y1(t) k1e7.235t 1.62k2e2.765t y2(t) 1.62k1e7.235t k2e2.765t 4.
The particular integral Y ¯ A2 1B is Y ¯ y ¯1 y ¯2 1 10 4 2 1 3 5 10 1 2 and the general solution is y1(t) k1e7.235t 1.62k2e2.765t 1 y2(t) 1.62k1e7.235t k2e2.765t 2 (19.32) 5.
Using y1(0) 22.2, y2(0) 3.9 to solve for k1 and k2, k1 1.62k2 1 22.2 1.62k1 k2 2 3.9 k1 5 k2 10 Substituting in (19.32) for the definite solution, y1(t) 5e7.235t 16.2e2.765t 1 y2(t) 8.1e7.235t 10e2.765t 2 With both characteristic roots negative, the intertemporal equilibrium is stable.
450 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 SIMULTANEOUS DIFFERENCE EQUATIONS 19.7.
Solve the following system of first-order linear difference equations in which no difference is a function of another difference.
xt 0.4xt1 0.6yt1 6 x0 14 yt 0.1xt1 0.3yt1 5 y0 23 1.
Setting them in matrix form, xt yt 0.4 0.6 0.1 0.3 xt1 yt1 6 5 Yt A Yt1 B 2.
Using (19.3) on the characteristic equation A riI 0, find the characteristic roots.
r1, r2 0.7 (0.7)2 4(0.06) 2 0.7 0.5 2 r1 0.6 r2 0.1 3.
The eigenvector for r1 0.6 is 0.4 0.6 0.1 0.6 0.3 0.6 c1 c2 0.2 0.1 0.6 0.3 c1 c2 0 0.2c1 0.6c2 0 c1 3c2 If c2 1, c1 3, and we have k1 3 1 (0.6)t 3k1(0.6)t k1(0.6)t For r2 0.1, 0.4 0.1 0.1 0.6 0.3 0.1 c1 c2 0.3 0.6 0.1 0.2 c1 c2 0 0.3c1 0.6c2 0 c1 2c2 If c2 1, c1 2, and k2 2 1 (0.1)t 2k2(0.1)t k2(0.1)t Combining the two for the general complementary functions, xc 3k1(0.6)t 2k2(0.1)t yc k1(0.6)t k2(0.1)t 4.
For the particular solution, yp (I A)1B where (I A) 1 0 0 1 0.4 0.6 0.1 0.3 0.6 0.1 0.6 0.7 and yp x ¯ y ¯ 1 0.36 0.7 0.6 0.1 0.6 6 5 20 10 This makes the complete general solution, xt 3k1(0.6)t 2k2(0.1)t 20 yt k1(0.6)t k2(0.1)t 10 (19.33) 451 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] 5.
Employing the initial conditions, x0 14, y0 23, (19.33) reduces to 3k1 2k2 20 14 k1 k2 10 23 Solved simultaneously, k1 4, k2 9 Substituting in (19.33), xt 12(0.6)t 18k2(0.1)t 20 yt 4(0.6)t 9(0.1)t 10 With 0.6 1 and 0.1 1, the time path is convergent. With both roots positive, there will be no oscillation.
19.8.
Solve the following system of first-order linear difference equations.
xt 0.6xt1 0.1yt1 9 x0 7.02 yt 0.5xt1 0.2yt1 42 y0 57.34 1.
In matrix form, xt yt 0.6 0.5 0.1 0.2 xt1 yt1 9 42 Yt A Yt1 B 2.
The characteristic roots are r1, r2 0.8 (0.8)2 4(0.07) 2 0.8 0.6 2 r1 0.1 r2 0.7 3.
The eigenvector for r1 0.1 is 0.6 (0.1) 0.5 0.1 0.2 (0.1) c1 c2 0.5 0.5 0.1 0.1 c1 c2 0 0.5c1 0.1c2 0 c2 5c1 If c1 1, c2 5, and k1 1 5 (0.1)t k1(0.1)t 5k1(0.1)t For r2 0.7, 0.6 (0.7) 0.5 0.1 0.2 (0.7) c1 c2 0.1 0.1 0.5 0.5 c1 c2 0 0.1c1 0.1c2 0 c1 c2 If c2 1, c1 1, and k2 1 1 (0.7)t k2(0.7)t k2(0.7)t This makes the general complementary functions, xc k1(0.1)t k2(0.7)t yc 5k1(0.1)t k2(0.7)t 452 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 4.
For the particular solution, yp (I A)1B where (I A) 1 0 0 1 0.6 0.1 0.5 0.2 1.6 0.5 0.1 1.2 and yp x ¯ y ¯ 1 1.87 1.2 0.1 0.5 1.6 9 42 8.02 38.34 This makes the complete general solution, xt k1(0.1)t k2(0.7)t 8.02 yt 5k1(0.1)t k2(0.7)t 38.34 (19.34) 5.
Using x0 7.02, y0 57.34, (19.34) reduces to k1 k2 8.02 7.02 5k1 k2 38.34 57.34 Solved simultaneously, k1 3, k2 4 Substituting in (19.34), xt 3(0.1)t 4(0.7)t 8.02 yt 15(0.1)t 4(0.7)t 38.34 With both characteristic roots in absolute value less than 1, the system of equations will approach a stable intertemporal equilibrium solution. With the roots negative, there will be oscillation.
19.9.
Solve the following system of first-order linear difference equations in which one difference is a function of another difference.
xt 0.7xt1 0.4yt1 40 x0 24 yt 0.575xt1 0.5yt1 xt 6 y0 32 1.
Rearranging and setting in matrix form, 1 0 1 1 xt yt 0.7 0.4 0.575 0.5 xt1 yt1 40 6 A1Yt A2Yt1 B 2.
We then find the characteristic roots from the characteristic equation, A2 riA1 0 0.7 0.4 0.575 0.5 ri 1 0 1 1 0.7 ri 0.575 ri 0.4 0.5 ri 0 r2 0.8r 0.12 0 r1 0.6 r2 0.2 3.
The eigenvector for r1 0.6 is 0.7 (0.6) 0.575 (0.6) 0.4 0.5 (0.6) c1 c2 0.1 0.025 0.4 0.1 c1 c2 0 0.1c1 0.4c2 0 c1 4c2 If c2 1, c1 4 and the eigenvector is k1 4 1 (0.6)t 4k1(0.6)t k1(0.6)t 453 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] For r2 0.2, 0.7 (0.2) 0.575 (0.2) 0.4 0.5 (0.2) c1 c2 0.5 0.375 0.4 0.3 c1 c2 0 0.5c1 0.4c2 0 c2 1.25c1 If c1 1, c2 1.25, and the eigenvector for r2 0.2 is k2 1 1.25 (0.2)t k2(0.2)t 1.25k2(0.2)t Adding the two eigenvectors, the complementary functions are xc 4k1(0.6)t k2(0.2)t yc k1(0.6)t 1.25k2(0.2)t 4.
For the particular solution yp Y ¯ , Y ¯ (A1 A2)1B where (A1 A2) 1 0 1 1 0.7 0.4 0.575 0.5 1.7 0.4 1.575 1.5 and Y ¯ x ¯ y ¯ 1 1.92 1.5 1.575 0.4 1.7 40 6 30 27.5 Adding yc and yp, the complete general solution is xt 4k1(0.6)t k2(0.2)t 30 yt k1(0.6)t 1.25k2(0.2)t 27.5 (19.35) 5.
Finally, we apply the initial conditions, x0 24 and y0 32, to (19.35), 4k1 k2 30 24 k1 1.25k2 27.5 32 k1 3 k2 6 and substitute these values back in (19.35) for the definite solution.
xt 12(0.6)t 6(0.2)t 30 yt 3(0.6)t 7.5(0.2)t 27.5 With both characteristic roots less than 1 in absolute value, the solution is stable.
19.10. Solve the following system of first-order linear difference equations.
xt 0.6xt1 0.85yt1 yt 15 x0 27 yt 0.2xt1 0.4yt1 6 y0 38 1.
In matrix form, 1 1 0 1 xt yt 0.6 0.85 0.2 0.4 xt1 yt1 15 6 A1Yt A2Yt1 B 2.
For the characteristic roots, A2 riA1 0.
0.6 0.85 0.2 0.4 ri 1 1 0 1 0.6 ri 0.2 0.85 ri 0.4 ri 0 r2 0.8r 0.07 0 r1 0.7 r2 0.1 454 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 3.
For r1 0.7, 0.6 0.7 0.2 0.85 0.7 0.4 0.7 c1 c2 0.1 0.2 0.15 0.3 c1 c2 0 0.1c1 0.15c2 0 c1 1.5c2 If c2 1, c1 1.5 and the eigenvector is k1 1.5 1 (0.7)t 1.5k1(0.7)t k1(0.7)t For r2 0.1, 0.6 0.1 0.2 0.85 0.1 0.4 0.1 c1 c2 0.5 0.75 0.2 0.3 c1 c2 0 0.5c1 0.75c2 0 c1 1.5c2 If c2 1, c1 1.5 and the eigenvector is k2 1.5 1 (0.1)t 1.5k2(0.1)t k2(0.1)t 4.
For the particular solution, Y ¯ (A1 A2)1B Here (A1 A2) 1 1 0 1 0.6 0.85 0.2 0.4 0.4 0.2 0.15 0.6 and Y ¯ x ¯ y ¯ 1 0.27 0.6 0.2 0.15 0.4 15 6 30 20 Adding yc and yp, xt 1.5k1(0.7)t 1.5k2(0.1)t 30 yt k1(0.7)t k2(0.1)t 20 (19.36) 5.
For the definite solution, we apply x0 27 and y0 38 to (19.36), 1.5k1 1.5k2 30 27 k1 k2 20 38 k1 8 k2 10 Substituting back in (19.36), xt 12(0.7)t 15(0.1)t 30 yt 8(0.7)t 10(0.1)t 20 With both characteristic roots less than 1 in absolute value, the solution is stable.
PHASE DIAGRAMS FOR SIMULTANEOUS DIFFERENTIAL EQUATIONS 19.11. Use a phase diagram to test the stability of the system of equations, y ˙1 3y1 18 y ˙2 2y2 16 1.
Determine the steady-state solutions y ¯i where y ˙i 0 to find the isoclines.
y ˙1 3y1 18 0 y ˙2 2y2 16 0 y ¯1 6 the y1 isocline y ¯2 8 the y2 isocline 455 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] As seen in Fig. 19-3, the intersection of the isoclines demarcates the intertemporal equilibrium level, (y ¯1, y ¯2) (6, 8).
2.
Determine the motion around the y1 isocline using arrows of horizontal motion.
a) To the left of the y1 isocline, y1 6.
b) To the right of the y1 isocline, y1 6.
Substituting these values successively in y ˙1 3y1 18, we see If y1 6, y ˙1 0, and there will be motion to the left.
If y1 6, y ˙1 0, and there will be motion to the right.
3.
Determine the motion around the y2 isocline, using arrows of vertical motion.
a) Above the y2 isocline, y2 8.
b) Below the y2 isocline, y2 8.
Substitution of these values successively in y ˙2 2y2 16 shows If y2 8, y ˙2 0, and the motion will be downward.
If y2 8, y ˙2 0, and the motion will be upward.
The resulting arrows of motion in Fig. 19-3, all pointing away from the intertemporal equilibrium, suggest the system of equations is divergent. Drawing trajectory paths to be sure confirms that the system is indeed divergent.
Proofs and Demonstrations 19.12. Given y ˙1 y ˙2 a11 a12 a21 a22 y1 y2 b1 b2 or Y ˙ A Y B (19.37) show in terms of Section 19.1 and Example 1 that (A riI)Ci 0 Starting with the homogeneous form of the system of equations in which B 0, or a null vector, and assuming distinct real roots, we can expect the solution to be in the form Y kiCieri t (19.38) where ki a scalar, Ci (2 1) column vector of constants, and ri a scalar. Taking the derivative of (19.38) with respect to t, we have Y ˙ rikiCierit (19.39) 456 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 Fig. 19-3 y2 = 8 y1 y1 = 0 y2 = 0 y1 = 6 y2 .
.
Substituting (19.38) and (19.39) in the homogeneous form of (19.37) where B 0, rikiCierit AkiCierit Canceling the common ki and erit terms, we have riCi ACi ACi riCi 0 Factoring out Ci and recalling that A is a (2 2) matrix while ri is a scalar, we multiply ri by a (2 2) identity matrix I2, or simply I, to get (A riI)Ci 0 Q.E.D.
(19.40) 19.13. Continuing with the model in Problem 19.12, show that r1, r2 Tr(A) [Tr(A)]2 4 A 2 If (A riI) is nonsingular in (19.40), meaning it contains no linear dependence, then Ci must be a null column vector, making the solution trivial. To find a nontrivial solution, (A riI) must be singular. A necessary condition for a nontrivial solution (Ci 0), then, is that the determinant A riI 0 (19.41) where equation (19.41) is called the characteristic equation or characteristic polynomial for matrix A.
Dropping the subscript for simplicity and substituting from above, we have a11 r a21 a12 a22 r 0 a11a22 a11r a22r r2 a12a21 0 Rearranging, r2 (a11 a22)r (a11a22 a12a21) 0 Or, using matrix notion, r2 Tr(A)r A 0 which is a quadratic equation that can be solved for r with the quadratic formula, r1, r2 Tr(A) [Tr(A)]2 4 A 2 Q.E.D.
19.14. Continuing with the model in Problem 19.13, show that the particular integral or solution is yp Y ¯ A1B (19.42) The particular integral is simply the intertemporal or steady-state solution Y ¯ . To find the steady-state solution, we simply set the column vector of derivatives equal to zero such that Y ˙ 0. When Y ˙ 0, there is no change and Y Y ¯ . Substituting in (19.37), Y ˙ A Y ¯ Y ¯ A Y ¯ B 0 B A1B Q.E.D.
19.15. Given a11 a12 a21 a22 y ˙1 y ˙2 a13 a14 a23 a24 y1 y2 b1 b2 or A1 Y ˙ A2Y B (19.43) 457 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] show in terms of Section 19.2 and Example 3 that to find the complementary function, one must solve the specific eigenvalue problem (A2 riA1)Ci 0 Starting with the homogeneous form of (19.43) in which B 0, or a null vector, and assuming distinct real roots, we can expect the solution and its derivative to take the forms Y kiCierit Y ˙ rikiCierit (19.44) Substituting from (19.44) into the homogeneous form of (19.43) where B 0, A1rikiCierit A2kiCierit Canceling the common ki and erit terms, we have A1riCi A2Ci (A2 riA1)Ci 0 Q.E.D.
19.16. In terms of the model in Problem 19.15, show that the particular integral is yp Y ¯ A2 1B (19.45) The particular integral is the steady-state solution Y ¯ when Y ˙ 0. Substituting in (19.43), A1Y ˙ A2Y ¯ Y ¯ A2Y ¯ B 0 B A2 1B Q.E.D.
19.17. Given xt yt a11 a12 a21 a22 xt1 yt1 b1 b2 or Yt A Yt1 B (19.46) show in terms of Section 19.3 and Example 5 that the eigenvalue problem for a system of simultaneous first-order linear difference equations when no difference is a function of another difference is (A riI)Ci 0 Starting with the homogeneous form of the system of equations in which B 0, and assuming a case of distinct real roots, from what we know of individual difference equations, we can expect that Yt kiCi(ri)t and Yt1 kiCi(ri)t1 (19.47) where ki and ri are scalars, and Ci (2 1) column vector of constants. Substituting in (19.46) when B 0, we have kiCi(ri)t AkiCi(ri)t1 Canceling the common ki terms and rearranging, ACi(ri)t1 Ci(ri)t 0 Evaluated at t 1, (A riI)Ci 0 Q.E.D.
19.18. Given a11 a12 a21 a22 xt yt a13 a14 a23 a24 xt1 yt1 b1 b2 A1Yt A2Yt1 B (19.48) 458 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS [CHAP . 19 show in terms of Section 19.4 and Example 9 that for a system of simultaneous first-order linear difference equations when one or more differences is a function of another difference the eigenvalue problem is (A2 riA1)Ci 0 From earlier work, assuming distinct real roots, we can anticipate Yt kiCi(ri)t and Yt1 kiCi(ri)t1 (19.49) Substituting in the homogeneous form of (19.48) where B 0, we have A1kiCi(ri)t A2kiCi(ri)t1 A2Ci(ri)t1 A1Ci(ri)t 0 Evaluated at t 1, (A2 riA1)Ci 0 Q.E.D.
19.19. Remaining with the same model as in Problem 19.18, show that the particular solution is yp Y ¯ (A1 A2)1B (19.50) For the particular or steady-state solution, xt xt1 x ¯ and yt yt1 y ¯ In matrix notation, Yt Yt1 Y ¯ Substituting in (19.48), A1Y ¯ A2Y ¯ B Solving for Y ¯ , Y ¯ (A1 A2)1B Q.E.D.
459 SIMULTANEOUS DIFFERENTIAL AND DIFFERENCE EQUATIONS CHAP . 19] CHAPTER 20 The Calculus of Variations 20.1 DYNAMIC OPTIMIZATION In the static optimization problems studied in Chapters 4 and 5, we sought a point or points that would maximize or minimize a given function at a particular point or period of time. Given a function y y(x), the first-order condition for an optimal point x is simply y(x) 0. In dynamic optimization we seek a curve x(t) which will maximize or minimize a given integral expression. The integral to be optimized typically defines the area under a curve F which is a function of the independent variable t, the function x(t), and its derivative dx/dt. In brief, assuming a time period from t0 0 to t1 T and using x · for the derivative dx/dt, we seek to maximize or minimize T 0 F[t, x(t), x ·(t)] dt (20.1) where F is assumed continuous for t, x(t), and x ·(t) and to have continuous partial derivatives with respect to x and x ·. An integral such as (20.1) which assumes a numerical value for each of the class of functions x(t) is called a functional. A curve that maximizes or minimizes the value of a functional is called an extremal. Acceptable candidates for an extremal are the class of functions x(t) which are continuously differentiable on the defined interval and which typically satisfy some fixed endpoint conditions. In our work with extremals, we start with the classical approach, called the calculus of variations, pioneered by Isaac Newton and James and John Bernoulli toward the end of the seventeenth century.
EXAMPLE 1.
A firm wishing to maximize profits from time t0 0 to t1 T finds that demand for its product depends on not only the price p of the product but also the rate of change of the price with respect to time dp/dt.
By assuming that costs are fixed and that both p and dp/dt are functions of time, and employing p · for dp/dt, the firm’s objective can be expressed mathematically as max T 0 [t, p(t), p ·(t)] dt A second firm has found that its total cost C depends on the level of production x(t) and the rate of change 460 of production dx/dt x ·, due to start-up and tapering-off costs. Assuming that the firm wishes to minimize costs and that x and x · are functions of time, the firm’s objective might be written min t1 t0 C[t, x(t), x ·(t)] dt subject to x(t0) x0 and x(t1) x1 These initial and terminal constraints are known as endpoint conditions.
20.2 DISTANCE BETWEEN TWO POINTS ON A PLANE The length S of any nonlinear curve connecting two points on a plane, such as the curve connecting the points (t0, x0) and (t1, x1) in Fig. 20-1(a), can be approximated mathematically as follows. Subdivide the curve mentally into subintervals, as in Fig. 20-1(b), and recall from the Pythagorean theorem that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. Accordingly, the length of an individual subsegment ds is (ds)2 (dt)2 (dx)2 Simplifying mathematically, ds (dt)2 (dx)2 Dividing, then multiplying, both sides by (dt)2, or simply dt, ds dt 1 dx dt 2 ds 1 dx dt 2 dt or, using the more compact symbol, ds 1 (x ·)2 dt from which the length of the total curve S from t0 to t1 can be estimated by simple integration to get S t1 t0 1 (x ·)2 dt See Problems 20.1 to 20.3.
461 THE CALCULUS OF VARIATIONS CHAP . 20] Fig. 20-1 20.3 EULER’S EQUATION: THE NECESSARY CONDITION FOR DYNAMIC OPTIMIZATION For a curve X x(t) connecting points (t0, x0) and (t1, x1) to be an extremal for (i.e., to optimize) a functional t1 t0 F[t, x(t), x ·(t)] dt the necessary condition, called Euler’s equation, is F x d dt F x · (20.2a) Although it is the equivalent of the first-order necessary conditions in static optimization, Euler’s equation is actually a second-order differential equation which can perhaps be more easily understood in terms of slightly different notation. Using subscripts to denote partial derivatives and listing the arguments of the derivatives, which are themselves functions, we can express Euler’s equation in (20.2a) as Fx(t, x, x ·) d dt [Fx ·(t, x, x ·)] (20.2b) Then using the chain rule to take the derivative of Fx · with respect to t and omitting the arguments for simplicity, we get Fx Fx ·t Fx ·x(x ·) Fx ·x ·(x ¨) (20.2c) where x ¨ d2x/dt2.
Proof that Euler’s equation is a necessary condition for an extremal in dynamic optimization is offered in Example 2. See also Problems 20.26 to 20.33.
EXAMPLE 2.
To prove that Euler’s equation in (20.2a) is a necessary condition for an extremal, let X x(t) be the curve connecting points (t0, x0) and (t1, x1) in Fig. 20-2 which optimizes the functional (i.e., posits the optimizing function for) t1 t0 F[t, x(t), x ·(t)] dt (20.3) Let X ˆ x(t) mh(t) be a neighboring curve joining these points, where m is an arbitrary constant and h(t) is an arbitrary function. In order for the curve X ˆ to also pass through the points (t0, x0) and (t1, x1), that is, for X ˆ to also satisfy the endpoint conditions, it is necessary that h(t0) 0 and h(t1) 0 (20.4) By holding both x(t) and h(t) fixed, the value of the integral becomes a function of m alone and can be written g(m) t1 t0 F[t, x(t) mh(t), x ·(t) mh ·(t)] dt (20.5) 462 THE CALCULUS OF VARIATIONS [CHAP . 20 Fig. 20-2 Since x(t) by definition optimizes the functional in (20.3), the function g(m) in (20.5) can be optimized only when m 0 and dg dm m0 0 (20.6) To differentiate under the integral sign in (20.5), we use Leibnitz’s rule which states that given g(m) t1 t0 f(t, m) dt where t0 and t1 are differentiable functions of m, dg dm t1 t0 F m dt f(t1, m) t1 m f(t0, m) t0 m (20.7) Since the boundaries of integration t0 and t1 are fixed in the present example, t0/m t1/m 0, and we have to consider only the first term in Leibnitz’s rule. Applying the chain rule to (20.5) to find F/m, because F is a function of x and x ·, which in turn are functions of m, and substituting in (20.7), we have dg dm t1 t0 F x (x mh) m F x · (x · mh ·) m dt With (x mh)/m h and (x · mh ·)/m h ·, and using (20.6), dg dm m0 t1 t0 F x h(t) F x · h ·(t) dt 0 (20.8) Leaving the first term in the brackets in (20.8) untouched and integrating the second term by means of parts, dg dm m0 t1 t0 F x h(t) dt F x · h(t) t1 t0 t1 t0 d dt F x · h(t) dt 0 With h(t0) h(t1) 0 from (20.4), the second term above drops out. Combining the other two terms and rearranging, dg dm m0 t1 t0 F x d dt F x · h(t) dt 0 (20.9) Since h(t) is an arbitrary function that need not equal zero, it follows that a necessary condition for an extremal is that the integrand within the brackets equal zero, namely, F x d dt F x · 0 or F x d dt F x · which is Euler’s equation. See also Problems 20.26 to 20.33.
20.4 FINDING CANDIDATES FOR EXTREMALS Finding candidates for extremals to maximize or minimize a given integral subject to fixed endpoint conditions in dynamic optimization problems is facilitated by the following five steps: 1.
Let the integrand equal F. Normally F F(t, x, x ·).
2.
Take the partial derivatives of F with respect to x and x · to find F/x Fx and F/x · Fx ·.
3.
Substitute in Euler’s equation from (20.2a) or (20.2b).
4.
Take the derivative with respect to t of Fx ·, recalling that the chain rule may be necessary because Fx · can be a function of t, x, and x ·, and x and x · are functions of t.
5.
If there are no derivative terms (x · or x ¨), solve immediately for x; if there are x · or x ¨ terms, integrate until all the derivatives are gone and then solve for x.
463 THE CALCULUS OF VARIATIONS CHAP . 20] Illustrations of this technique are provided in Examples 3 and 4 and Problems 20.4 to 20.18.
EXAMPLE 3.
Given T 0 (6x2e3t 4tx ·) dt the functional is optimized by using the procedure outlined in Section 20.4 and the notation from (20.2a), as follows: 1.
Let F 6x2e3t 4tx · 2.
Then F x 12xe3t and F x · 4t 3.
Substituting in Euler’s equation from (20.2a), 12xe3t d dt (4t) 4.
But d(4t)/dt 4. Substituting above, 12xe3t 4 5.
Solving for x directly since there are no x · or x ¨ terms, and expressing the solution as x(t), x(t) 1 – 3e3t This satisfies the necessary condition for dynamic optimization, which only makes the solution a candidate for an extremal. The sufficiency conditions, which follow in Section 20.5, must also be applied.
EXAMPLE 4.
The functional 2 0 (4x · 2 12xt 5t) dt subject to x(0) 1 x(2) 4 is optimized as above, but now with the notation from (20.2b).
1.
Let F 4x · 2 12xt 5t 2.
Then Fx 12t and Fx · 8x · 3.
Substituting in Euler’s equation from (20.2b), 12t d dt (8x ·) 4.
Recalling that x · dx dt and that d dt dx dt d2x dt2 x ¨, 12t 8x ¨ 5.
Since an x ¨ term remains, integrate both sides of the equation successively twice, using only one constant of integration term at each step. 12t dt 8x ¨ dt 6t2 c1 8x · Integrating again, (6t2 c1) dt 8x · dt 2t3 c1t c2 8x 464 THE CALCULUS OF VARIATIONS [CHAP . 20 Solving for x, x(t) 1 4 t3 c1 8 t c2 8 Applying the boundary conditions, x(0) c2 8 1 c2 8 x(2) 1 – 4(2)3 1 – 8(2)c1 1 4 c1 4 Substituting, x(t) 1 – 4t3 1 – 2t 1 20.5 THE SUFFICIENCY CONDITIONS FOR THE CALCULUS OF VARIATIONS Assuming the necessary conditions for an extremal are satisfied, 1.
If the functional F[t, x(t), x ·(t)] is jointly concave in x(t), x ·(t), then the necessary conditions are sufficient for a maximum.
2.
If the functional F[t, x(t), x ·(t)] is jointly convex in x(t), x ·(t), the necessary conditions are sufficient for a minimum.
Joint concavity and convexity are easily determined in terms of the sign definiteness of the quadratic form of the second derivatives of the functional. Given the discriminant, D Fxx Fx ·x Fxx · Fx ·x · 1.
a) If D1 Fxx 0 and D2 D 0, D is negative definite and F is strictly concave, making the extremal a global maximum.
b) If D1 Fxx · 0 and D2 D 0, when tested for all possible orderings of the variables, D is negative semidefinite and F is simply concave, which is sufficient for a local maximum.
2.
a) If D1 Fxx 0 and D2 D 0, D is positive definite and F is strictly convex, making the extremal a global minimum.
b) If D1 Fxx 0 and D2 D 0, when tested for all possible orderings of the variables, D is positive semidefinite and F is simply convex, which is sufficient for a local minimum. See Example 5 and Problems 20.4 to 20.18.
EXAMPLE 5.
The sufficiency conditions are illustrated below for Example 3 where the functional was F 6x2e3t 4tx ·, Fx 12xe3t, and Fx · 4t. D1 Fxx Fx ·x Fxx · Fx ·x · 12e3t 0 0 0 D1 1 12e3t 0 D1 2 0 D1 fails to meet the positive definite criteria for a global minimum, but may prove to be positive semidefinite for a local minimum if the discriminant for the reversed order of variables is also positive semidefinite. D2 Fx ·x · Fxx · Fx ·x Fxx 0 0 0 12e3t D2 1 0, D2 2 0 With D1 0 and D2 0 for both possible orderings of variables, D is positive semidefinite, which is sufficient to establish that the functional is at a local minimum. The sufficiency conditions for Example 4 test out in a perfectly analogous fashion.
465 THE CALCULUS OF VARIATIONS CHAP . 20] 20.6 DYNAMIC OPTIMIZATION SUBJECT TO FUNCTIONAL CONSTRAINTS To find an extremal that maximizes or minimizes a given integral T 0 F[t, x(t), x ·(t)] dt (20.10) under a constraint that keeps the integral T 0 G[t, x(t), x ·(t)] dt k (20.11) where k is a constant, the Lagrangian multiplier method may be used. Multiply the constraint in (20.11) by , and add it to the objective function from (20.10) to form the Lagrangian function: T 0 (F G) dt (20.12) The necessary, but not sufficient, condition to have an extremal for dynamic optimization is the Euler equation H x d dt H x · where H F G (20.13) See Example 6 and Problem 20.25.
EXAMPLE 6.
Constrained optimization of functionals is commonly used in problems to determine a curve with a given perimeter that encloses the largest area. Such problems are called isoperimetric problems and are usually expressed in the functional notation of y(x) rather than x(t). Adjusting for this notation, to find the curve Y of given length k which encloses a maximum area A, where A 1 2 (xy · y) dx and the length of the curve is x1 x0 1 y · 2 dx k set up the Lagrangian function, as explained in Section 20.6.
x1 x0 [1 – 2(xy · y) 1 y · 2] dx (20.14) Letting H equal the integrand in (20.14), the Euler equation is H y d dx H y · where from (20.14), H y 1 2 and H y · 1 2 x y · 1 y · 2 Substituting in Euler’s equation, 1 2 d dx 1 2 x y · 1 y · 2 1 2 1 2 d dx y · 1 y · 2 1 d dx y · 1 y · 2 466 THE CALCULUS OF VARIATIONS [CHAP . 20 Integrating both sides directly and rearranging, y · 1 y · 2 (x c1) Squaring both sides of the equation and solving algebraically for y ·, 2y · 2 (x c1)2(1 y · 2) 2y · 2 (x c1)2y · 2 (x c1)2 y · 2 (x c1)2 2 (x c1)2 y · x c1 2 (x c1)2 Integrating both sides, using integration by substitution on the right, gives, y c2 2 (x c1)2 which, by squaring both sides and rearranging, can be expressed as a circle (x c1)2 (y c2)2 2 where c1, c2, and are determined by x0, x1, and k.
20.7 VARIATIONAL NOTATION A special symbol " is used in the calculus of variations which has properties similar to the differential d in differential calculus.
Given a function F[t, x(t), x ·(t)] and considering t as constant, let F F[t, x(t) mh(t), x ·(t) mh ·(t)] F[t, x(t), x ·(t)] (20.15) where m is an arbitrary constant, h(t) is an arbitrary function as in Example 2, and the arguments are frequently omitted for succinctness. Using the Taylor expansion which approximates a function such as x(t) by taking successive derivatives and summing them in ordered sequence to get x(t) x(t0) x ·(t0)(t t0) x ¨(t0)(t t0)2 2!
· · · we have F(t, x mh, x · mh ·) F(t, x, x ·) F x mh F x · mh · · · · (20.16) Substituting (20.16) in (20.15) and subtracting as indicated, F F x mh F x · mh · · · · (20.17) where the sum of the first two terms in (20.17) is called the variation of F and is denoted by "F.
Thus, "F F x mh F x · mh · (20.18) From (20.18) it is readily seen that if F x, by substituting x for F, we have "x mh (20.19) 467 THE CALCULUS OF VARIATIONS CHAP . 20] Similarly, if F x · "x · mh · (20.20) Hence an alternative expression for (20.18) is "F F x "x F x · "x · (20.21) and the necessary condition for finding an extremal in dynamic optimization can aslso be expressed as " t1 t0 F[t, x(t), x ·(t)] dt 0 For proof, see Problems 20.34 and 20.35.
20.8 APPLICATIONS TO ECONOMICS A firm wishes to minimize the present value at discount rate i of an order of N units to be delivered at time t1. The firm’s costs consist of production costs a[x ·(t)]2 and inventory costs bx(t), where a and b are positive constants; x(t) is the accumulated inventory by time t; the rate of change of inventory is the production rate x ·(t), where x ·(t) 0; and ax ·(t) is the per unit cost of production.
Assuming x(t0) 0 and the firm wishes to achieve x(t1) N, in terms of the calculus of variations the firm must min t1 t0 eit(ax · 2 bx) dt subject to x(t0) 0 x(t1) N To find a candidate for the extremal that will minimize the firm’s cost, let F[t, x(t), x ·(t)] eit(ax · 2 bx) then Fx beit and Fx · 2aeitx · Substituting in Euler’s equation from (20.2b), beit d dt (2aeitx ·) Using the product rule and the chain rule to take the derivative on the right since x · is a function of t, we have beit 2aeit(x ¨) x ·(i2aeit) 2aeitx ¨ 2aieitx · Canceling the eit terms and rearranging to solve for x ¨, x ¨(t) ix ·(t) b 2a (20.22) With x ¨(t) ix ·(t) b/(2a) from (20.22) and x ·(t) 0 by assumption, x ¨(t) in (20.22) must be positive, indicating that the firm should maintain a strictly increasing rate of production over time.
Equation (20.22) is a second-order linear differential equation which can be solved with the method outlined in Section 18.1. Using Equation (18.2a) to find the particular integral, since in terms 468 THE CALCULUS OF VARIATIONS [CHAP . 20 of (18.1) b1 i, b2 0, and a b/2a, and adjusting the functional notation from y y(t) to x x(t), we have xp b/2a i t b 2ai t Using (18.4) to find r1 and r2 for the complementary function, r1, r2 (i) (i)2 4(0) 2 i i 2 r1 i r2 0 Substituting in (18.3) to find xc and adding to xp, x(t) A1eit A2 b 2ai t (20.23) Letting t0 0 and t1 T, from the boundary conditions we have x(0) A1 A2 0 A2 A1 x(T) A1eiT (A1) b 2ai T N Solving x(T) for A1, A1(eiT 1) N b 2ai T A1 N [b/(2ai)]T eiT 1 (20.24) Finally, substituting in (20.23), and recalling that A2 A1, we have as a candidate for an extremal: x(t) N [b/(2ai)]T eiT 1 eit N [b/(2ai)]T eiT 1 b 2ai t x(t) N b 2ai T eit 1 eiT 1 b 2ai t 0 t T Then testing the sufficiency conditions, where Fx beit and Fx · 2aeitx, D1 Fxx Fx ·x Fxx · Fx ·x · 0 0 0 2aeit D1 1 0 D1 2 0 D2 Fx ·x · Fxx · Fx ·x Fxx 2aeit 0 0 0 D2 1 2aeit 0 D2 2 0 With the discriminant of the quadratic form of the second-order derivatives of the functional positive semidefinite when tested for both orderings of the variables, the sufficiency condition for a local minimum is met. For further economic applications, see Problems 20.19 to 20.24.
469 THE CALCULUS OF VARIATIONS CHAP . 20] Solved Problems DISTANCE BETWEEN TWO POINTS ON A PLANE 20.1.
Minimize the length of a curve S connecting the points (t0, x0) and (t1, x1) in Fig. 20-1 from Section 20.2, i.e., min t1 t0 1 x ·2 dt subject to x(t0) x0 x(t1) x1 Using the procedure outlined in Section 20.4 to find a candidate for an extremal to minimize the functional, 1.
Let F 1 x · 2 (1 x · 2)1/2 2.
Take the partial derivatives Fx and Fx ·, noting that there is no x term in F, only an x · term, and that the chain rule or generalized power function rule is necessary for Fx ·.
Fx 0 Fx · 1 2 (1 x · 2)1/2 · 2x · x · 1 x · 2 3.
Substitute in Euler’s equation.
0 d dt x · 1 x · 2 4.
Since there are no variables on the left-hand side, integrate both sides immediately with respect to t. Integrating the derivative on the right-hand side will produce the original function. With 0 dt c, a constant, we have c x · 1 x · 2 Squaring both sides and rearranging to solve for x ·, c2(1 x · 2) x · 2 c2 x · 2 c2x ˙2 (1 c2)x · 2 x · c2 1 c2 k1 a constant 5.
With an x · term remaining, integrate again to get x(t) k1t k2 (20.25) 6.
With only one variable x · in the functional, the sufficiency conditions of concavity or convexity can be determined solely by the sign of the second derivative. From Fx · x ·(1 x · 2)1/2, we have by the product rule, Fx ·x · Fx ·x · Fx ·x · (1 x · 2)1/2 x · 2(1 x · 2)3/2 (1 x · 2)3/2[(1 x ˙2) x · 2] (1 x · 2)3/2 1 (1 x · 2)3 Since the square root of a distance can never be negative, Fx ˙x ˙ 0. The functional is convex and the sufficiency conditions for a minimum are satisfied.
Note the solution in (20.25) is linear, indicating that the shortest distance between two points is a straight line. The parameters k1 (slope) and k2 (vertical intercept) are uniquely determined by the boundary conditions, as is illustrated in Problem 20.2.
470 THE CALCULUS OF VARIATIONS [CHAP . 20 20.2.
Minimize 2 0 1 x · 2 dt subject to x(0) 3 x(2) 8 From (20.25), x(t) k1t k2 Applying the boundary conditions, x(0) k1(0) k2 3 k2 3 x(2) k1(2) 3 8 k1 2.5 Substituting, x(t) 2.5t 3 20.3.
(a) Estimate the distance between the points (t0, x0) and (t1, x1) from Problem 20.2, using the functional; (b) draw a graph and check your answer geometrically.
a) Given 2 0 1 x · 2 dt and x(t) 2.5t 3 by taking the derivative x ·(t) 2.5 and substituting, we have 2 0 1 (2.5)2 dt 2 0 7.25 dt 7.25t 2 0 2.69258(2) 2.69258(0) 5.385 b) Applying the Pythagorean theorem to Fig. 20-3, x2 x 52 22 29 5.385 FINDING CANDIDATES FOR EXTREMALS 20.4.
Optimize t1 t0 (2x · 2 42xt 11t) dt subject to x(t0) x0 x(t1) x1 Using the now familiar six steps to find a candidate for an extremal, 1.
F 2x · 2 42xt 11t 2.
Fx 42t Fx · 4x · 471 THE CALCULUS OF VARIATIONS CHAP . 20] Fig. 20-3 3.
Substituting in Euler’s equation, 42t d dt (4x ·) 42t 4x ¨ 4.
Integrating both sides to eliminate the x ¨ term and using only one constant of integration term at each step throughout, 21t2 c1 4x · Integrating again to eliminate the x · term, 7t3 c1t c2 4x 5.
Solving for x, x(t) 1.75t3 0.25c1t 0.25c2 6.
Testing the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 0 0 0 4 D1 1 0 D1 2 0 D2 Fx ·x · Fxx · Fx ·x Fxx 4 0 0 0 D2 1 4 0 D2 2 0 The discriminant of the second-order derivatives of the functional is positive semidefinite when tested for both orderings of the variables, which satisfies the sufficiency condition for a local minimum.
20.5.
Optimize t1 t0 (x · 2 60t3x) dt subject to x(t0) x0 x(t1) x1 1.
F x · 2 60t3x 2.
Fx 60t3 Fx · 2x · 3.
Substituting in Euler’s equation, 60t3 d dt (2x ·) 60t3 2x ¨ 4.
Integrating both sides and combining constants of integration for each step, 15t4 c1 2x · Integrating again and solving for x, 3t5 c1t c2 2x 5.
x(t) 1.5t5 0.5c1t 0.5c2 6.
The sufficiency conditions, when tested as in step 6 of the previous problem, reveal the functional is at a local minimum.
472 THE CALCULUS OF VARIATIONS [CHAP . 20 20.6.
Optimize 1 0 (13t 3x · 2 36xt) dt subject to x(0) 2 x(1) 4 1.
F 13t 3x · 2 36xt 2.
Fx 36t Fx · 6x · 3.
36t d dt (6x ·) 36t 6x ¨ 4.
Integrating both sides twice but using only one constant of integration each time, 18t2 c1 6x · 6t3 c1t c2 6x 5.
x(t) t3 c1 6 t c2 6 Applying the initial conditions, x(0) c2 6 2 c2 12 x(1) 1 c1 6 2 4 c1 18 Then substituting above, x(t) t3 3t 2 6.
Finally, testing the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 0 0 0 6 D1 1 0 D1 2 0 D2 Fx ·x · Fxx · Fx ·x Fxx 6 0 0 0 D2 1 6 0 D2 2 0 When tested for both orderings of the variables, the discriminant of the second-order derivatives is negative semidefinite. This fulfills the sufficiency conditions for a local maximum.
20.7.
Optimize t1 t0 (3x2e5t 4t3x ·) dt subject to x(t0) x0 x(t1) x1 1.
F 3x2e5t 4t3x · 2.
Fx 6xe5t Fx · 4t3 473 THE CALCULUS OF VARIATIONS CHAP . 20] 3.
Substituting in Euler’s equation, 6xe5t d dt (4t3) 4.
6xe5t 12t2 5.
With no x · terms left, simply solve for x algebraically.
x(t) 2t2e5t 6.
Checking the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 6e5t 0 0 0 D1 1 6e5t 0 D1 2 0 D2 Fx ·x · Fxx · Fx ·x Fxx 0 0 0 6e5t D2 1 0 D2 2 0 For both orderings of the variables, the discriminant of the second-order derivatives is positive semidefinite, fulfilling the sufficiency condition for a local minimum.
20.8.
Optimize t1 t0 3x ·2 8t3 dt subject to x(t0) x0 x(t1) x1 1.
F 3x · 2 8t3 2.
Fx 0 Fx · 6x · 8t3 3x · 4t3 3.
Substituting in Euler’s equation, 0 d dt 3x · 4t3 4.
With Fx 0, integrate immediately.
c1 3x · 4t3 3x · 4c1t3 With an x · still remaining, integrate again.
3x c1t4 c2 5.
x(t) c1 3 t4 c2 3 k1t4 k2 where k1 c1 3 k2 c2 3 6.
When tested, as above, the sufficiency conditions reveal a local minimum.
20.9.
Optimize t1 t0 (5t2x · 4x2e0.7t) dt 474 THE CALCULUS OF VARIATIONS [CHAP . 20 subject to x(t0) x0 x(t1) x1 1.
F 5t2x · 4x2e0.7t 2.
Fx 8xe0.7t Fx · 5t2 3.
8xe0.7t d dt (5t2) 10t 8xe0.7t 10t 4.
With no derivative term remaining and no need to integrate, we solve directly for x.
5.
x(t) 1.25te0.7t 6.
Testing the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 8e0.7t 0 0 0 D1 1 8e0.7t 0 D1 2 0 D2 Fx ·x · Fxx · Fx ·x Fxx 0 0 0 8e0.7t D2 1 0 D2 2 0 With both orderings of the variables, D is negative semidefinite, making F concave and fulfilling the sufficiency condition for a local maximum.
20.10. Optimize t1 t0 (7t2 2x · 2t) dt subject to x(t0) x0 x(t1) x1 1.
F 7t2 2x · 2t 2.
Fx 0 Fx · 4x ·t 3.
0 d dt (4x ·t) 4.
Integrating immediately, c1 4x ·t x · c1 4t Integrating again, x c1 4 ln t c2 5.
x(t) k1 ln t k2 where k1 c1 4 k2 c2 6.
The sufficiency conditions for a relative minimum are satisfied.
20.11. Optimize t1 t0 (15x2 132x 19xx · 12x · 2) dt subject to x(t0) x0 x(t1) x1 1.
F 15x2 132x 19xx · 12x · 2 475 THE CALCULUS OF VARIATIONS CHAP . 20] 2.
Fx 30x 132 19x · Fx · 19x 24x · 3.
30x 132 19x · d dt (19x 24x ·) 4.
30x 132 19x · 19x · 24x ¨ Solving algebraically, x ¨ 1.25x 5.5 (20.26) 5.
Equation (20.26) is a second-order differential equation which can be solved with the techniques of Section 18.1. Using Equation (18.2) to find the particular integral xp, where b1 0, b2 1.25, and a 5.5, xp a b2 5.5 1.25 4.4 Then using (18.4) to find the characteristic roots, r1, r2 0 0 4(1.25) 2 1.25 and substituting in (18.3) to find the complementary function xc, xc A1e1.25t A2e1.25t Finally, by adding xc and xp, we have x(t) A1e1.25t A2e1.25t 4.4 6.
Checking the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 30 19 19 24 D1 1 30 0 D1 2 359 0 With D1 0 and D2 0, D is positive definite, which means F is strictly convex and we have an absolute minimum. There is no need to test in reverse order with D2 .
20.12. Optimize 1 0 (16x2 144x 11xx · 4x · 2) dt subject to x(0) 8 x(1) 8.6 1.
F 16x2 144x 11xx · 4x · 2 2.
Fx 32x 144 11x · Fx · 11x 8x · 3.
32x 144 11x · d dt (11x 8x ·) 32x 144 11x · 11x · 8x ¨ Simplifying and rearranging to conform with (18.1), x ¨ 4x 18 4.
From (18.2), the particular integral is xp 18 4 4.5 476 THE CALCULUS OF VARIATIONS [CHAP . 20 From (18.4), the characteristic roots are r1, r2 0 0 4(4) 2 r1 2 r2 2 Thus, x(t) A1e2t A2e2t 4.5 Applying the initial conditions, x(0) 8, x(1) 8.6, x(0) A1 A2 4.5 8 x(1) 7.3891A1 0.1353A2 4.5 8.6 Solving simultaneously, A1 0.5, A2 3 5.
Substituting, x(t) 0.5e2t 3e2t 4.5 6.
Applying the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 32 11 11 8 D1 1 32 0 D1 2 135 0 Since D1 0 and D2 0, D is negative definite. F is strictly concave and there is an absolute maximum. Hence D2 need not be tested.
20.13. Optimize t1 t0 (16x2 9xx · 8x · 2) dt subject to x(t0) x0 x(t1) x1 1.
F 16x2 9xx · 8x · 2 2.
Fx 32x 9x · Fx · 9x 16x · 3.
32x 9x · d dt (9x 16x ·) 4.
32x 9x · 9x · 16x ¨ x ¨ 2x 0 5.
Using (18.2) where b1 0, b2 2, and a 0 to find xp, xp a b2 0 2 0 Using (18.4) to find r1 and r2, r1, r2 4(2) 2 2 and substituting in (18.3) to find xc, xc A1e2t A2e2t Then since xp 0, x(t) xc A1e2t A2e2t 6.
Sufficiency conditions reveal an absolute minimum.
477 THE CALCULUS OF VARIATIONS CHAP . 20] 20.14. Optimize t1 t0 (7x · 2 4xx · 63x2) dt subject to x(t0) x0 x(t1) x1 1.
F 7x · 2 4xx · 63x2 2.
Fx 4x · 126x Fx · 14x · 4x 3.
4x · 126x d dt (14x · 4x) 4.
4x · 126x 14x ¨ 4x · x ¨ 9x 0 where in terms of (18.1), b1 0, b2 9, and a 0.
5.
Using (18.2) for xp, xp 0 9 0 Using (18.19) since b1 2 4b2, r1, r2 g hi where g 1 – 2b1 1 – 2(0) 0 h 1 – 24b2 b1 2 1 – 236 3 and r1, r2 0 3i 3i Substituting in (18.26), xc B1 cos 3t B2 sin 3t With xp 0, x(t) B1 cos 3t B2 sin 3t 6.
Sufficiency conditions indicate an absolute minimum.
20.15. Optimize t1 t0 (5x2 27x 8xx · x · 2) dt subject to x(t0) x0 x(t1) x1 1.
F 5x2 27x 8xx · x · 2 2.
Fx 10x 27 8x · Fx · 8x 2x · 3.
10x 27 8x · d dt (8x 2x ·) 4.
10x 27 8x · 8x · 2x ¨ x ¨ 5x 13.5 5.
Using (18.2), xp 13.5 5 2.7 Using (18.19), g 1 – 2(0) 0 h 1 – 24(5) 5 r1, r2 0 5i 5i Substituting in (18.26), xc B1 cos 5t B2 sin 5t and x(t) B1 cos 5t B2 sin 5t 2.7 478 THE CALCULUS OF VARIATIONS [CHAP . 20 6.
For the second-order conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 10 8 8 2 D1 1 10 0 D1 2 84 0 With D2 0, D fails the test both for concavity and convexity. F is neither maximized nor minimized. It is at a saddle point.
20.16. Optimize t1 t0 e0.12t(5x · 2 18x) dt subject to x(t0) x0 x(t1) x1 1.
F e0.12t(5x · 2 18x) 2.
Fx 18e0.12t Fx · 10x ·e0.12t 3.
18e0.12t d dt (10x ·e0.12t) 4.
Using the product rule, 18e0.12t 10x ·(0.12e0.12t) e0.12t(10x ¨) Canceling the e0.12t terms and rearranging algebraically, x ¨ 0.12x · 1.8 5.
Using (18.2a) and (18.4), xp 1.8 0.12 t 15t r1, r2 0.12 (0.122 0) 2 0.12, 0 xc A1e0.12t A2 and x(t) A1e0.12t A2 15t 6.
Checking the sufficiency conditions, D1 Fxx Fx ·x Fxx · Fx ·x · 0 0 0 10e0.12t D1 1 0 D1 2 0 D2 Fx ·x · Fxx · Fx ·x Fxx 10e0.12t 0 0 0 D2 1 10e0.12t 0 D2 2 0 D is positive semidefinite, F is convex, and we have a local minimum.
20.17. Optimize t1 t0 e0.05t(4x · 2 15x) dt subject to x(t0) x0 x(t1) x1 1.
F e0.05t(4x · 2 15x) 2.
Fx 15e0.05t Fx · 8x ·e0.05t 479 THE CALCULUS OF VARIATIONS CHAP . 20] 3.
15e0.05t d dt (8x ·e0.05t) 4.
15e0.05t 8x ·(0.05e0.05t) e0.05t(8x ¨) Canceling the e0.05t terms and rearranging, x ¨ 0.05x · 1.875 5.
Using (18.2a) and (18.4), xp 1.875 0.05 t 37.5t r1, r2 (0.05) (0.05)2 0 2 0.05, 0 xc A1e0.05t A2 and x(t) A1e0.05t A2 37.5t 6.
The sufficiency conditions indicate a local minimum.
20.18. Find the curve connecting (t0, x0) and (t1, x1) which will generate the surface of minimal area when revolved around the t axis, as in Fig. 20.4. That is, Minimize 2 t1 t0 x(1 x · 2)1/2 dt subject to x(t0) x0 x(t1) x1 1.
F x(1 x · 2)1/2 2.
Using the chain rule for Fx ·, Fx (1 x · 2)1/2 Fx · xx ·(1 x · 2)1/2 3.
(1 x · 2)1/2 d dt [xx ·(1 x · 2)1/2] 4.
Using the product rule and the chain rule, (1 x · 2)1/2 xx · [1 – 2(1 x · 2)3/2 · 2x ·x ¨] (1 x · 2)1/2(xx ¨ x ·x ·) xx · 2x ¨(1 x · 2)3/2 (xx ¨ x · 2)(1 x · 2)1/2 480 THE CALCULUS OF VARIATIONS [CHAP . 20 Fig. 20-4 Multiplying both sides by (1 x · 2)3/2, (1 x · 2)2 1 2x · 2 x · 4 x ¨ x · 2 x xx · 2x ¨ (xx ¨ x · 2)(1 x · 2) xx · 2x ¨ xx ¨ x · 2 xx · 2x ¨ x · 4 1 x (20.27) 5.
Let x · u dx/dt, then x ¨ u · du dt du dx · dx dt du dx · u u du dx Substituting in (20.27), u du dx u2 x 1 x Separating the variables and integrating, u du dx 1 x (1 u2) u 1 u2 du 1 x dx 1 – 2ln (1 u2) c1 ln x Solving for u, eln1u2c1 eln x c21 u2 x where c2 ec1. Squaring both sides and rearranging algebraically, 1 u2 x2 c2 2 u x2 c2 2 c2 dx dt Separating variables again and integrating, dx x2 c2 2 dt c2 (20.28) Using integral tables for the left-hand side, ln (x x2 c2 2) t c3 c2 (20.29) or by applying trigonometric substitution directly to (20.28), cosh1 x c2 t c3 c2 x(t) c2 cosh t c3 c2 (20.30) The curve in (20.30) is called a catenary from the Latin word for chain because it depicts the shape a chain would assume if hung from points (t0, x0) and (t1, x1). The constants c2 and c3 can be found from either (20.29) or (20.30) by using the initial conditions x(t0) x0 and x(t1) x1.
481 THE CALCULUS OF VARIATIONS CHAP . 20] ECONOMIC APPLICATIONS 20.19. The demand for a monopolist’s product in terms of the number of units x(t) she or he can sell depends on both the price p(t) of the good and the rate of change of price p ·(t): x(t) ap(t) bp ·(t) c (20.31) Production costs z(x) at rate of production x are z(x) mx2 nx k (20.32) Assuming p(0) p0 and the desired price at time T is p(T) p1, find the pricing policy to maximize profits over 0 t T. That is, Maximize T 0 [p(t)x(t) z(x)] dt Substituting from (20.31) and (20.32), T 0 [p(t)x(t) z(x)] dt T 0 [p(ap bp · c) (mx2 nx k)] dt Substituting from (20.31) again, T 0 [p(t)x(t) z(x)] dt T 0 [ap2 bpp · cp m(ap bp · c)2 n(ap bp · c) k] dt T 0 (ap2 bpp · cp ma2p2 mabpp · macp mabpp · mb2p · 2 mbcp · macp mbcp · mc2 nap nbp · nc k) dt T 0 [a(1 ma)p2 (c 2mac na)p (b 2mab)pp · (2mbc nb)p · mb2p · 2 mc2 nc k] dt (20.33) Letting F the integrand in (20.33), Fp 2a(1 ma)p c 2mac na (b 2mab)p · and Fp · (b 2mab)p 2mbc nb 2mb2p · Using the Euler equation, 2a(1 ma)p c 2mac na (b 2mab)p · d dt [(b 2mab)p 2mbc nb 2mb2p ·] 2a(1 ma)p c 2mac na (b 2mab)p · (b 2mab)p · 2mb2p ¨ Rearranging algebraically, 2mb2p ¨ 2a(1 ma)p 2mac na c p ¨ a(1 ma) mb2 p 2mac na c 2mb2 Using (18.2) and (18.4), pp (2mac na c)/(2mb2) a(1 ma)(mb2) 2mac na c 2a(1 ma) r1, r2 0 0 4a(1 ma)/(mb2) 2 a(ma 1) mb2 482 THE CALCULUS OF VARIATIONS [CHAP . 20 Thus, p(t) A1 exp a(ma 1) mb2 t A2 exp a(ma 1) mb2 t 2mac na c 2a(1 ma) If a 0 and m 0, as one would expect from economic theory, independently of the sign for b, a(ma 1)/(mb2) 0 and the time path p(t) will have distinct real roots of equal magnitude and opposite signs.
Note: This is a classic case that has appeared in one form or another in the economic literature over the decades.
20.20. Maximize the stream of instantaneous utility U(t) from the flow of consumption C(t) where C(t) ↓ Flow of consumption G[K(t)] ↓ production K ·(t) ↓ investment (20.34) and the endpoints are fixed at K(0) K0, K(T) KT. That is, Maximize T 0 U[C(t)] dt T 0 U{G[K(t)] K · (t)} dt subject to K(0) K0 K(T) KT Letting F U{G[K(t)] K · (t)}, U dU/dC, and G dG/dK, FK U[C(t)]G[K(t)] FK · U[(C(t)] Substituting in Euler’s equation, U[C(t)]G[K(t)] d dt {U[C(t)]} U
[C(t)] · C · where, upon using the chain rule on (20.34), C · dC dt G[K(t)] · K · K ¨ Substituting above, U[C(t)]G[K(t)] U
[C(t)] · (G[K(t)] · K · K ¨ ) The Euler equation thus yields a second-order ordinary differential equation, the solution of which maximizes the given extremal.
20.21. Maximize the discounted stream of utility from consumption C(t) over 0 t T, that is, Maximize T 0 {eitU[C(t)]} dt (20.35) where C(t) G[K(t)] K · (t) bK(t), 0 b 1, G[K(t)] is the rate of production, K ·(t) bK(t) is investment, and b is a constant rate of capital depreciation.
Substituting in (20.35), we seek to maximize T 0 (eitU{G[K(t)] K · (t) bK(t)} dt With F eitU{G[K(t)] K · (t) bK(t)} 483 THE CALCULUS OF VARIATIONS CHAP . 20] and letting U dU/dC and G dG/dK, while retaining K · dK/dt, FK eitU{G[K(t)] K · (t) bK(t)} · {G[K(t)] b} FK · eitU{G[K(t)] K · (t) bK(t)} Substituting in Euler’s equation and simplifying notation, eit[U · (G b)] d dt (eitU) (20.36) Using the product and chain rules for the derivative on the right, eit[U · (G b)] eitU · (GK · K ¨ bK · ) Uieit and canceling the eit terms, U · (G b) U · (GK · K ¨ bK · ) Ui With U{G[K(t)]} unspecified, we cannot proceed further. Going back to (20.36) and using general notation for the derivative on the right, however, we see that eit[U · (G b)] eit d dt (U) U(ieit) Canceling the eit terms and rearranging, d dt (U) Ui U · (G b) d(U)/dt U i b G where the term on the left, the rate of change of the marginal utility, equals the discount rate plus the depreciation rate minus the marginal product of capital. In brief, if we consider the term on the left as capital gains, the optimal time path suggests that if capital gains are greater than the discount rate plus the depreciation rate minus the marginal product of capital, then more capital and hence more consumption should be forthcoming. If it is less, capital accumulation and consumption should be scaled back.
20.22. Maximize the discounted stream of utility from consumption C(t) over 0 t T, that is, Maximize T 0 eitU[C(t)] dt (20.37) given (a) U[C(t)] [C(t)]n where 0 n 1 (b) C(t) ↓ Flow of consumption G[K(t)] ↓ production I(t) ↓ investment where G[K(t)] aK(t), a linear production function with a 0 and (c) I(t) K ·(t) B bK(t) 0 b 1 B 0 derived from K · (t) ↓ in K stock I(t) ↓ investment [B bK(t)] ↓ linear depreciation Substituting in (20.37), we wish to maximize T 0 eit[aK(t) K · (t) B bK(t)]n dt 484 THE CALCULUS OF VARIATIONS [CHAP . 20 which by rearranging and omitting the arguments for simplicity becomes T 0 eit(mK K · B)n dt where m a b. Letting F eit(mK K · B)n, FK mneit(mK K · B)n1 FK · neit(mK K · B)n1 Using Euler’s equation, mneit(mK K · B)n1 d dt [neit(mK K · B)n1] Using the product rule and the chain rule on the right, mneit(mK K · B)n1 neit(n 1)(mK K · B)n2(mK · K ¨ ) (mK K · B)n1(ineit) Dividing both sides by neit(mK K · B)n1, m (n 1)(mK K · B)1(mK · K ¨ ) i m i (1 n)(mK · K ¨ ) mK K · B Cross-multiplying and simplifying, (1 n)K ¨ (i mn 2m)K · (m2 im)K (m i)B K ¨ i mn 2m 1 n K · m2 im 1 n K m i 1 n B (20.38) Letting Z1 i mn 2m 1 n Z2 m2 im 1 n Z3 m i 1 n B Kp Z3 Z2 B(m i)/(1 n) (m2 im)/(1 n) (m i)B m(m i) 1 mB where m a b, a is the marginal product of capital (dG/dK a), and b is the constant rate of depreciation.
Kc A1er1t A2er2t where r1, r2 Z1 Z2 1 4Z2 2 and A1 and A2 can be computed from the boundary conditions.
20.23. Maximize T 0 eitU[C(t)] dt given the discount rate i 0.12, the endpoints K(0) 320 and K(5) 480, and the utility function U[C(t)] [C(t)]0.5, where C(t) G[K(t)] I(t) G[K(t)] 0.25K(t) I(t) K · 60 0.05K(t) Substituting in the given functional, we seek to maximize 5 0 e0.12t[0.25K(t) K · (t) 60 0.05K(t)]0.5 dt 485 THE CALCULUS OF VARIATIONS CHAP . 20] Rearranging and omitting the arguments for simplicity, 5 0 e0.12t(0.2K K · 60)0.5 dt Letting F e0.12t(0.2K K · 60)0.5 and using the chain rule or the generalized power function rule, FK 0.1e0.12t(0.2K K · 60)0.5 FK · 0.5e0.12t(0.2K K · 60)0.5 Substituting in Euler’s equation, then using the product rule and the chain rule, 0.1e0.12t(0.2K K · 60)0.5 d dt [0.5e0.12t(0.2K K · 60)0.5] 0.5e0.12t[0.5(0.2K K · 60)1.5(0.2K · K ¨ )] (0.2K K · 60)0.5(0.06e0.12t) Dividing both sides by 0.5e0.12t(0.2K K · 60)0.5 and rearranging, 0.2 0.5(0.2K · K ¨ ) 0.2K K · 60 0.12 0.08(0.2K K · 60) 0.1K · 0.5K ¨ K ¨ 0.36K · 0.032K 9.6 (20.39) Using (18.2) and (18.4) Kp 9.6 0.032 300 r1, r2 (0.36) (0.36)2 4(0.032) 2 0.36 0.0016 2 r1 0.2 r2 0.16 Thus, K(t) A1e0.2t A2e0.16t 300 Applying the endpoint conditions, K(0) A1 A2 300 320 A2 20 A1 K(5) A1e0.2(5) (20 A1)e0.16(5) 300 480 A1(2.71828) (20 A1)(2.22554) 180 A1 274.97 275 A2 20 275 255 Substituting, K(t) 275e0.2t 255e0.16t 300 20.24. Since Problem 20.23 is a specific application of Problem 20.22, check the accuracy of the answer in Problem 20.23 by substituting the given values of a 0.25, b 0.05, B 60, i 0.12, m 0.2, and n 0.5 in Equation (20.38) to make sure it yields the same answer as Equation (20.39).
Substituting the specific values in (20.38), K ¨ 0.12 (0.2)(0.5) 2(0.2) 1 0.5 K · (0.2)2 (0.12)(0.2) 1 0.5 K 0.2 0.12 1 0.5 60 K ¨ 0.36K · 0.032K 9.6 On your own, check the values of r1, r2, Kc, and K(t) by substituting the specific values in the equations immediately following (20.38) in Problem 20.19 and comparing them with the solutions found in Problem 20.20.
486 THE CALCULUS OF VARIATIONS [CHAP . 20 CONSTRAINED OPTIMIZATION 20.25. Minimize t1 t0 eit(ax · 2 bx) dt subject to t1 t0 x ·(t) dt N where x(t0) 0 and x(t1) N Setting up the Lagrangian function, as in Section 20.6, t1 t0 [eit(ax · 2 bx) x ·] dt Letting H equal the integrand, the Euler equation is H x d dt H x · Taking the needed partial derivatives, Hx beit Hx · 2ax ·eit Substituting in Euler’s equation, beit d dt (2ax ·eit ) Using the product and generalized power function rules, beit 2aix ·eit 2ax ¨eit Canceling the eit terms and rearranging, x ¨ ix · b 2a which is identical to what we found in (20.22) without using constrained dynamic optimization. This is another example of an isoperimetric problem, and it can be solved as it was in Section 20.8.
PROOFS AND DEMONSTRATIONS 20.26. In seeking an extremal for t2 t0 F(t, x, x ·) dt show that Euler’s equation can also be expressed as d dt F x · F x · F t 0 (20.40) Taking the derivative with respect to t of each term within the parentheses and using the chain rule, dF dt F t F x dx dt F x · dx · dt F t F x x · F x · x ¨ d dt x · F x · x · d dt F x · F x · x ¨ 487 THE CALCULUS OF VARIATIONS CHAP . 20] and substituting in (20.40), F t F x x · F x · x ¨ x · d dt F x · F x · x ¨ F t 0 x · F x d dt F x · 0 F x d dt F x · or Fx d dt (Fx ·) 20.27. Show that if F is not an explicit function of t, the Euler equation can be expressed as F x · F x · c a constant If t is not an explicit argument of F, F/t 0 and Equation (20.40) reduces to d dt F x · F x · 0 Integrating both sides with respect to t, F x · F x · c 20.28. (a) Show that if F F(t, x ·), with x not one of the arguments, the Euler equation reduces to Fx · c a constant (b) Explain the significance.
a) With F F(t, x ·) Fx 0 Fx · Fx · Substituting in Euler’s equation, 0 d dt (Fx ·) Integrating both sides with respect to t, Fx · c (20.41) b) Equation (20.41) is a first-order differential equation with arguments of t and x · alone which, when solved, provides the desired extremal. See Problems 20.7 and 20.8.
20.29. (a) Show that if F F(x ·), that is, a function of x · alone, the Euler equatiuon reduces to Fx ·x · x ¨ 0 and (b) explain the significance.
a) Given F F(x ·) Fx 0 Fx · Fx · Using Euler’s equation, 0 d dt [Fx ·(x ·)] Fx ·x · x ¨ (20.42) 488 THE CALCULUS OF VARIATIONS [CHAP . 20 b) From (20.42), either x ¨ 0 or Fx ·x · 0. If x ¨ 0, integrating twice yields x(t) c1t c2, which is linear.
If Fx ·x · 0, Fx · c, a constant, which means that F is linear in x ·. If F is linear in x ˙, the solution is trivial.
See Problems 20.30 and 20.31.
20.30. Find extremals for t1 t0 e3x ·2 dt subject to x(t0) x0 x(t1) x1 Fx 0 Fx · 6x ·e3x ·2 0 d dt (6x ·e3x ·2) By the product rule, 0 6x ·(6x ·e3x ·2x ¨) e3x ·2(6x ¨) 6x ¨e3x ·2(6x · 2 1) 0 (20.43) which is a nonlinear second-order differential equation, not easily solved. However, since x ¨ in (20.43) must equal zero for the equation to equal zero, from Problem 20.29 we know that the solution must be linear.
Thus, x(t) c1t c2 20.31. Find extremals for t1 t0 (27 5x ·) dt subject to x(t0) x0 x(t1) x1 Fx 0 Fx · 5 0 d dt (5) 0 The Euler equation is an identity which any admissible value of x satisfies trivially, as was indicated in Problem 20.29. This becomes clear upon direct integration of the extremal in this problem: t1 t0 (27 5x ·) dt 27(t1 t0) 5[x(t1) x(t0)] and any x satisfying the endpoint conditions yields the same value for this integrand.
20.32. (a) Show that if F F(t, x), with x · not one of the arguments, the Euler equation reduces to Fx 0 (b) Explain the significance.
a) With F F(t, x), Fx Fx Fx · 0 Substituting in Euler’s equation, Fx d dt (0) 0 489 THE CALCULUS OF VARIATIONS CHAP . 20] b) When there is no x · term in F, the optimization problem is static and not dynamic. The condition for optimization, therefore, is the same as that in static optimization, namely, Fx 0 20.33. Show that if application of Euler’s equation results in a second-order differential equation with no x or t terms, the second-order differential equation can be converted to a first-order differential equation and a solution found by means of equation (16.1). Demonstrate in terms of Equation (20.22) from Section 20.8.
From Equation (20.22), x ¨ ix · b 2a Since there are no x or t terms in (20.22), it can be converted to a first-order linear differential equation by letting u x · and u · x ¨ Substituting in (20.22) above and rearranging, u · iu b 2a which can be solved by means of the formula in (16.1). Letting v i and z b/2a.
u e(i) dtA b 2a e(i) dt dt eitA b 2a eit dt Taking the remaining integral, u eitA 1 i b 2a eit u Aeit b 2ai But u x ·(t) by definition, so we must integrate once again to find x(t). Replacing A with c1 for notational consistency with ordinary integration, x(t) c1 i eit b 2ai t c2 (20.44) Letting t0 0 and t1 T, from the boundary conditions we have x(0) c1 i c2 0 c2 c1 i x(T) c1 i eiT b 2ai T c2 N Substituting c2 c1/i in x(T) and solving for c1, c1 i eiT b 2ai T c1 i N c1 i (eiT 1) N b 2ai T c1 i{N [b/(2ai)]T} eiT 1 (20.45) 490 THE CALCULUS OF VARIATIONS [CHAP . 20 Finally, substituting in (20.44) and noting that the i in (20.45) cancels out the i in the denominator of c1/i, we have as a candidate for an extremal: x(t) N [b/(2ai)]T eiT 1 eit b 2ai t N [b/(2ai)]T eiT 1 N b 2ai T eit 1 eiT 1 b 2ai t 0 t T Compare the work done here with the work done in Section 20.8, and note that conversion of a second-order differential equation to a first-order differential equation before integrating does not necessarily reduce the work involved in finding a solution.
VARIATIONAL NOTATION 20.34. Show that the operators " and d/dt are commutative, i.e., show that " dx dt d dt ("x) From (20.19) and (20.20), "x mh and "x · mh · where m is an arbitrary constant and h is an arbitrary function h h(t). Substituting dx/dt for x · above on the right, " dx dt mh · Expressing h · as dh/dt and recalling that m is a constant, " dx dt d dt (mh) Then substituting from (20.19), " dx dt d dt ("x) 20.35. Given t1 t0 F[t, x(t), x ·(t)] dt show that in terms of variational notation a necessary condition for an extremal is " t1 t0 F[t, x(t), x ·(t)] dt 0 Moving " within the integral sign, t1 t0 "F[t, x(t), x ·(t)] dt 0 From (20.21), t1 t0 F x "x F x · "x · dt 0 491 THE CALCULUS OF VARIATIONS CHAP . 20] Substituting from (20.19) and (20.20) where "x mh "x · mh · t1 t0 F x mh F x · mh · dt 0 Dividing by m, an arbitrary constant, t1 t0 F x h F x · h · dt 0 (20.46) Equation (20.46) is identical to Equation (20.18) in the Euler equation proof from Example 2 and can be concluded in the same manner.
492 THE CALCULUS OF VARIATIONS [CHAP . 20 CHAPTER 21 Optimal Control Theory 21.1 TERMINOLOGY Optimal control theory is a mid-twentieth-century advance in the field of dynamic optimization that can handle any problem the calculus of variations was designed for. More importantly, optimal control theory is more powerful than the calculus of variations because it can manage some problems the calculus of variations cannot, such as corner solutions, and other problems the calculus of variations cannot handle readily, such as constraints on the derivatives of the functions being sought.
In optimal control theory, the aim is to find the optimal time path for a control variable, which we shall denote as y. The variable for which we previously sought an optimal time path in the calculus of variations, known as a state variable, we shall continue to designate as x. State variables always have equations of motion or transition set equal to x ˙. The goal of optimal control theory is to select a stream of values over time for the control variable that will optimize the functional subject to the constraint set on the state variable.
Optimal control theory problems involving continuous time, a finite time horizon, and fixed endpoints are generally written: Maximize J T 0 f[x(t), y(t), t] dt subject to x ˙ x(0) g[x(t), y(t), t] x0 x(T) xT (21.1) where J the value of the functional to be optimized; y(t) the control variable, so called because its value is selected or controlled to optimize J; x(t) the state variable, which changes over time according to the differential equation set equal to x ˙ in the constraint, and whose value is indirectly determined by the control variable in the constraint; and t time. The solution to an optimal control problem demarcates the optimal dynamic time path for the control variable y(t).
493 21.2 THE HAMILTONIAN AND THE NECESSARY CONDITIONS FOR MAXIMIZATION IN OPTIMAL CONTROL THEORY Dynamic optimization of a functional subject to a constraint on the state variable in optimal control involves a Hamiltonian function H similar to the Lagrangian function in concave programming.
In terms of (21.1), the Hamiltonian is defined as H[x(t), y(t), (t), t] f[x(t), y(t), t] (t)g[x(t), y(t), t] (21.2) where (t) is called the costate variable. Similar to the Lagrangian multiplier, the costate variable (t) estimates the marginal value or shadow price of the associated state variable x(t). Working from (21.2), formation of the Hamiltonian is easy. Simply take the integrand under the integral sign and add to it the product of the costate variable (t) times the constraint.
Assuming the Hamiltonian is differentiable in y and strictly concave so there is an interior solution and not an endpoint solution, the necessary conditions for maximization are 1.
H y 0 2.
a) t ˙ H x b) x t x ˙ H 3.
a) x(0) x0 b) x(T) xT The first two conditions are known as the maximum principle and the third is called the boundary condition. The two equations of motion in the second condition are generally referred to as the Hamiltonian system or the canonical system. For minimization, the objective functional can simply be multiplied by 1, as in concave programming. If the solution does not involve an end point, H/y need not equal zero in the first condition, but H must still be maximized with respect to y. See Chapter 13, Example 9, and Fig. 13-1(b)–(c), for clarification. We shall generally assume interior solutions.
EXAMPLE 1.
The conditions in Section 21.2 are used below to solve the following optimal control problem: Maximize subject to 3 0 (4x 5y2) dt x ˙ x(0) 8y 2 x(3) 117.2 A.
From (21.1), set up the Hamiltonian.
H 4x 5y2 (8y) B.
Assuming an interior solution, apply the maximum principle.
1.
H y 0 H y 10y 8 0 y 0.8 (21.3) 2.
a) ˙ ˙ 4 H x (21.4) b) x ˙ x ˙ 8y H 494 OPTIMAL CONTROL THEORY [CHAP . 21 But from (21.3), y 0.8. So, x ˙ 8(0.8) 6.4 (21.5) Having employed the maximum principle, we are left with two differential equations, which we now solve for the state variables x(t) and the costate variable (t).
By integrating (21.4) we find the costate variable.
(t) ˙ dt 4 dt 4t c1 (21.6) Substituting (21.6) in (21.5), x ˙ 6.4(4t c1) 25.6t 6.4c1 Integrating, x(t) (25.6t 6.4c1) dt x(t) 12.8t2 6.4c1t c2 (21.7) C.
The boundary conditions can now be used to solve for the constants of integration. Applying x(0) 2, x(3) 117.2 successively to (21.7), x(0) 12.8(0)2 6.4c1(0) c2 2 c2 2 x(3) 12.8(3)2 6.4c1(3) 2 117.2 c1 12 Then by substituting c1 12 and c2 2 in (21.7) and (21.6), we have, x(t) 12.8t2 76.8t 2 state variable (t) 4t 12 costate variable (21.8) D.
Lastly, we can find the final solution for the control variable y(t) in either of two ways.
1.
From (21.3), y(t) 0.8, so y(t) 0.8(4t 12) 3.2t 9.6 control variable 2.
Or taking the derivative of (21.8), x ˙ 25.6t 76.8 we substitute for x ˙ in the equation of motion in the constraint, x ˙ 25.6t 76.8 y(t) 8y 8y 3.2t 9.6 control variable Evaluated at the endpoints, y(0) 3.2(0) 9.6 9.6 y(3) 3.2(3) 9.6 0 The optimal path of the control variable is linear starting at (0, 9.6) and ending at (3, 0), with a slope of 3.2. For similar problems involving fixed endpoints, see also Problems 21.1 to 21.3.
21.3 SUFFICIENCY CONDITIONS FOR MAXIMIZATION IN OPTIMAL CONTROL Assuming the maximum principle representing the necessary conditions for maximization in optimal control is satisfied, the sufficiency conditions will be fulfilled if: 1.
Both the objective functional f[x(t), y(t), t] and the constraint g[x(t), y(t), t] are differentiable and jointly concave in x and y, and 495 OPTIMAL CONTROL THEORY CHAP . 21] 2.
(t) 0, if the constraint is nonlinear in x or y. If the constraint is linear, may assume any sign.
Linear functions are always both concave and convex, but neither strictly concave or strictly convex.
For nonlinear functions, an easy test for joint concavity is the simple discriminant test. Given the discriminant of the second-order derivatives of a function, D fxx fxy fyx fyy a function will be strictly concave if the discriminant is negative definite, D1 fxx 0 and D2 D 0 and simply concave if the discriminant is negative semidefinite, D1 fxx 0 and D2 D 0 A negative definite discriminant indicates a global maximum and is, therefore, always sufficient for a maximum. A negative semidefinite discriminant is indicative of a local maximum and is sufficient for a maximum if the test is conducted for every possible ordering of the variables with similar results.
EXAMPLE 2.
The sufficiency conditions for the problem in Example 1 are demonstrated below. Starting with the objective functional which is nonlinear, we take the second derivatives and apply the discriminant test.
D fxx fxy fyx fyy 0 0 0 10 where D1 0 and D2 D 0 D fails the strict negative-definite criteria but proves to be negative semidefinite with D1 0 and D2 D 0.
However, for the semidefinite test we must also test the variables in reverse order.
D fyy fyx fxy fxx 10 0 0 0 where D1 10 and D2 D 0 With both discriminant tests negative semidefinite, the objective functional f is jointly concave in x and y. Since the constraint is linear, it is also jointly concave and does not need testing. We can conclude, therefore, that the functional is indeed maximized.
21.4 OPTIMAL CONTROL THEORY WITH A FREE ENDPOINT The general format for an optimal control problem involving continuous time with a finite time horizon and a free endpoint is Maximize subject to J x ˙ x(0) T 0 f[x(t), y(t), t] dt g[x(t), y(t), t] x0 x(T) free (21.9) where the upper limit of integration x(T) is free and unrestricted. Assuming an interior solution, the first two conditions for maximization, comprising the maximum condition, remain the same but the third or boundary condition changes: 1.
H y 0 2.
a) t ˙ H x b) x t x ˙ H 496 OPTIMAL CONTROL THEORY [CHAP . 21 3.
a) x(0) x0 b) (T) 0 where the very last condition is called the transversality condition for a free endpoint. The rationale for the transversality condition follows straightforward from what we learned in concave programming. If the value of x at T is free to vary, the constraint must be nonbinding and the shadow price evaluated at T must equal 0, i.e., (T) 0. For problems involving free endpoints, see Examples 3 to 4 and Problems 21.4 to 21.6.
EXAMPLE 3.
The conditions in Section 21.4 are used below to solve the following optimal control problem with a free endpoint: Maximize subject to 2 0 (3x 2y2) dt x ˙ x(0) 8y 5 x(2) free A.
From (21.1), H 3x 2y2 (8y) B.
Assuming an interior solution and applying the maximum principle.
1.
H y 0 H y 4y 8 0 y 2 (21.10) 2.
a) ˙ ˙ 3 H x (21.11) b) x ˙ x ˙ 8y H But from (21.10), y 2. So, x ˙ 8(2) 16 (21.12) From the maximum principle, two differential equations emerge, which can now be solved for the state variable x(t) and the costate variable (t).
Integrating (21.11), (t) ˙ dt 3 dt 3t c1 (21.13) Substituting (21.13) in (21.12), x ˙ 16(3t c1) 48t 16c1 Integrating, x(t) 24t2 16c1t c2 (21.14) C.
We now use the boundary conditions to specify the constants of integration.
1.
Start with the transversality condition (T) 0 for a free endpoint. Here (2) 0 497 OPTIMAL CONTROL THEORY CHAP . 21] Substituting in (21.13), (2) 3(2) c1 0 c1 6 Therefore, (t) 3t 6 costate variable (21.15) 2.
Now substitute c1 6 in (21.14), x(t) 24t2 16(6)t c2 x(t) 24t2 96t c2 and apply the initial boundary condition, x(0) 5.
x(0) 24(0)2 96(0) c2 5 c2 5 So, x(t) 24t2 96t 5 state variable (21.16) D.
The control variable y(t) can then be found in either of two ways.
1.
From (21.10), y(t) 2. Substituting from (21.15) for the final solution, y(t) 2(3t 6) 6t 12 control variable (21.17) 2.
Or take the derivative of (21.16), x ˙ 48t 96 and substitute in the equation of transition in the constraint, x ˙ 48t 96 y(t) 8y 8y 6t 12 control variable Evaluated at the endpoints, y(0) 6(0) 12 12 y(2) 6(2) 12 0 The optimal path of the control variable is linear starting at (0, 12) and ending at (2, 0), with a slope of 6.
EXAMPLE 4.
The sufficiency conditions for Example 3 are found in the same way as in Example 2. Taking the second derivatives of the objective functional and applying the discriminant test, D fxx fxy fyx fyy 0 0 0 4 where D1 0 and D2 D 0 D is not negative-definite but it is negative semidefinite with D1 0 and D2 D 0. For the semidefinite test, however, we must test the variables in reverse order.
D fyy fyx fxy fxx 4 0 0 0 where D1 4 and D2 D 0 With both discriminants negative semidefinite, the objective functional f is jointly concave in x and y. The constraint is linear and so needs no testing. The functional is maximized.
21.5 INEQUALITY CONSTRAINTS IN THE ENDPOINTS If the terminal value of the state variable is subject to an inequality constraint, x(T) xmin, the optimal value x(T) may be chosen freely as long as it does not violate the value set by the constraint xmin. If x(T) xmin, the constraint is nonbinding and the problem reduces to a free endpoint problem. So (T) 0 when x(T) xmin 498 OPTIMAL CONTROL THEORY [CHAP . 21 If x(T) xmin, the constraint is binding and the optimal solution will involve setting x(T) xmin, which is equivalent to a fixed-end problem with (T) 0 when x(T) xmin For conciseness, the endpoint conditions are sometimes reduced to a single statement analogous to the Kuhn-Tucker condition, (T) 0 x(T) xmin [x(T) xmin](T) 0 In practice, solving problems with inequality constraints on the endpoints is straightforward. First solve the problem as if it were a free endpoint problem. If the optimal value of the state variable is greater than the minimum required by the endpoint condition, i.e., if x(T) xmin, the correct solution has been found. If x(T) xmin, set the terminal endpoint equal to the value of the constraint, x(T) xmin, and solve as a fixed endpoint problem. The method is illustrated in Examples 5 and 6 and further explained and developed in Example 7 and Problems 21.7 to 21.10.
EXAMPLE 5.
Maximize subject to 2 0 (3x 2y2) dt x ˙ x(0) 8y 5 x(2) 95 To solve an optimal control problem involving an inequality constraint, solve it first as an unconstrained problem with a free endpoint. This we did in Example 3 where we found the state variable in (21.16) to be x(t) 24t2 96t 5 Evaluating (21.16) at x 2, the terminal endpoint, we have x(2) 101 95 Since the free endpoint solution satisfies the terminal endpoint constraint x(T) 95, the constraint is not binding and we have indeed found the proper solution. From (21.17) in Example 3, y(t) 6t 12 EXAMPLE 6.
Redo the same problem in Example 5 with the new boundary conditions, x(0) 5 x(2) 133 A.
From Example 5 we know that the value for the state variable when optimized under free endpoint conditions is x(2) 101 133 which fails to meet the new endpoint constraints. This means the constraint is binding and we have now to optimize the functional as a fixed endpoint problem with the value of the constraint as the terminal endpoint, x(0) 5 x(2) 133 B.
The first two steps remain the same as when we solved the problem as a free endpoint in Example 3.
Employing the maximum principle, we found: in (21.10), in (21.11), in (21.12), in (21.13), in (21.14), y ˙ x ˙ (t) x(t) 2 3 16 3t c1 24t2 16c1t c2 Now we continue on with the new boundary conditions for a fixed endpoint.
499 OPTIMAL CONTROL THEORY CHAP . 21] C.
Applying x(0) 5 and x(2) 133 successively in (21.14), we have x(0) 24(0)2 16c1(0) c2 5 c2 5 x(2) 24(2)2 16c1(2) 5 133 c1 7 Then, substituting c1 7, c2 5 in (21.13) and (21.14), we derive (t) 3t 7 costate variable x(t) 24t2 112t 5 state variable D.
The control variable can be found in either of the two familiar ways.We opt once again for the first. From (21.10), y(t) 2 2(3t 7) 6t 14 control variable EXAMPLE 7.
With an inequality constraint as a terminal endpoint, in accord with the rules of Section 21.5, we first optimize the Hamiltonian subject to a free endpoint. With a free endpoint, we set (T) 0, allowing the marginal value of the state variable to be taken down to zero. This, in effect, means that as long as the minimum value set by the constraint is met, the state variable is no longer of any value to us. Our interest in the state variable does not extend beyond time T.
Most variables have value, however, and our interest generally extends beyond some narrowly limited time horizon. In such cases we will not treat the state variable as a free good by permitting its marginal value to be reduced to zero. We will rather require some minimum value of the state variable to be preserved for use beyond time T. This means maximizing the Hamiltonian subject to a fixed endpoint determined by the minimum value of the constraint. In such cases, (T) 0, the constraint is binding, and we will not use as much of the state variable as we would if it were a free good.
21.6 THE CURRENT-VALUED HAMILTONIAN Optimal control problems frequently involve discounting, such as Maximize subject to J x ˙ x(0) T 0 e t f[x(t), y(t), t] dt g[x(t), y(t), t] x0 x(T) free The Hamiltonian for the discounted or present value follows the familiar format H e t f[x(t), y(t), t] (t)g[x(t), y(t), t] but the presence of the discount factor e t complicates the derivatives in the necessary conditions. If we let #(t) (t)e t, however, we can form a new, ‘‘current-valued’’ Hamiltonian Hc He t f[x(t), y(t), t] #(t)g[x(t), y(t), t] (21.18) which is generally easier to solve and requires only two adjustments to the previous set of necessary conditions. Converting condition 2(a) from Section 21.2 to correspond to the current-valued Hamiltonian, we have ˙ H x Hc x e t Taking the derivative of (t) #(t)e t, we have ˙ # ˙ e t #e t 500 OPTIMAL CONTROL THEORY [CHAP . 21 Equating the ˙’s, canceling the common e t terms, and rearranging, we derive the adjusted condition for 2(a): # ˙ # Hc x The second adjustment involves substituting (t) #(t)e t in the boundary conditions. The transver-sality condition for a free endpoint then changes from (T) 0 to the equivalent #(T)e t 0.
In short, given the current-valued Hamiltonian in (21.18) and assuming an interior solution, the necessary conditions for optimization are 1.
Hc y 0 2.
a) # t # ˙ # Hc x b) x t x ˙ Hc # 3.
a) x(0) x0 b) #(T)e t 0 If the solution does not involve an end point, Hc/y need not equal zero in the first condition, but Hc must still be maximized with respect to y. With Hc He t, the value of y that will maximize Hc will also maximize H since e t is treated as a constant when maximizing with respect to y. The sufficiency conditions of Section 21.3 remain the same, as shown in Example 9. Maximization of a current-valued Hamiltonian is demonstrated in Examples 8 to 9 and followed up in Problems 21.11 to 21.12.
EXAMPLE 8.
Maximize subject to 2 0 e0.02t(x 3x2 2y2) dt x ˙ x(0) y 0.5x 93.91 x(2) free A.
Set up the current-valued Hamiltonian.
Hc x 3x2 2y2 #(y 0.5x) B.
Assuming an interior solution, apply the modified maximum principle.
1.
Hc y 0 Hc y 4y # 0 y 0.25# (21.19) 2.
a) # ˙ # ˙ 0.02# (1 6x 0.5#) 0.52# 6x 1 (21.20) b) x ˙ Hc # y 0.5x Substituting from (21.19), x ˙ 0.25# 0.5x (21.21) Arranging the two simultaneous first-order differential equations from (21.20) and (21.21) in matrix form and solving with the techniques from Section 19.3, # ˙ x ˙ 0.52 0.25 6 0.5 # x 1 0 501 OPTIMAL CONTROL THEORY CHAP . 21] or Y ˙ A Y B The characteristic equation is A rI 0.52 r 0.25 6 0.5 r 0 From (19.3), the characteristic roots are, r1, r2 0.02 (0.02)2 4(1.76) 2 r1 1.3367 r2 1.3167 For r1 1.3367, the eigenvector is 0.52 1.3367 0.25 6 0.5 1.3367 c1 c2 0.8167 0.25 6 1.8367 c1 c2 0 0.8167c1 6c2 0 c1 7.3466c2 y1 c 7.3466 1 k1e1.3367t For r2 1.3167, the eigenvector is 0.52 1.3167 0.25 6 0.5 1.3167 c1 c2 1.8367 0.25 6 0.8167 c1 c2 0 1.8367c1 6c2 0 c1 3.2667c2 y2 c 3.2667 1 k2e1.3167t From (19.5), the particular solution is Y ¯ A1B Y ¯ # ¯ x ¯ 1 1.76 0.5 0.25 6 0.52 1 0 0.28 0.14 Adding the complementary and particular solutions, we have #(t) 7.3466k1e1.3667t 3.2667k2e1.3167t 0.28 (21.22) x(t) k1e1.3667t k2e1.3167t 0.14 (21.23) C.
Next we apply the boundary conditions.
1.
From the transversality condition for the free endpoint, #(T)e t 0, we have at T 2, #(2)e0.02(2) 0 Substituting for #(2), (7.3466k1e1.3667(2) 3.2667k2e1.3167(2) 0.28)e0.04 0 113.0282k1 0.2350k2 0.2690 0 (21.24) 2.
Evaluating x(t) at x(0) 93.91, k1 k2 0.14 93.91 (21.25) Solving (21.24) and (21.25) simultaneously, k1 0.2 k2 93.57 502 OPTIMAL CONTROL THEORY [CHAP . 21 Then substituting k1 0.2, k2 93.57 in (21.22) and (21.23), we get #(t) 1.4693e1.3667t 305.6651e1.3167t 0.28 costate variable x(t) 0.2e1.3667t 93.57e1.3167t 0.14 state variable D.
The solution for the control variable can now be found in either of the two usual ways. We choose the easier. From (21.19), y(t) 0.25#. Substituting from the costate variable above, y(t) 0.3673e1.3667t 76.41631.3167t 0.07 control variable EXAMPLE 9.
The sufficiency conditions follow the usual rules.
D fxx fxy fyx fyy 6 0 0 4 With D1 6 0, and D2 24 0, D is negative definite, making f strictly concave in both x and y. With g linear in x and y, the sufficiency condition for a global maximum is fulfilled.
Solved Problems FIXED ENDPOINTS 21.1.
Maximize subject to 2 0 (6x 4y2) dt x ˙ x(0) 16y 24 x(2) 408 A.
The Hamiltonian is H 6x 4y2 (16y) B.
The necessary conditions from the maximum principle are 1.
H y 8y 16 0 y 2 (21.26) 2.
a) ˙ H x 6 (21.27) b) x ˙ H 16y From (21.26), x ˙ 16(2) 32 (21.28) Integrating (21.27), (t) 6t c1 (21.29) Substituting in (21.28) and then integrating, x ˙ 32(6t c1) 192t 32c1 x(t) 96t2 32c1t c2 (21.30) C.
Applying the boundary conditions, x(0) 24, x(2) 408, x(0) c2 24 c2 24 x(2) 96(2)2 32c1(2) 24 408 c1 12 503 OPTIMAL CONTROL THEORY CHAP . 21] Substituting c1 12 and c2 24 in (21.29) and (21.30), (t) 6t 12 costate variable (21.31) x(t) 96t2 384t 24 state variable (21.32) D.
For the control variable solution, we then either substitute (21.31) into (21.26), y(t) 2 2(6t 12) 12t 24 control variable or take the derivative of (21.32) and substitute it in the constraint x ˙ 16y.
x ˙ 192t 384 16y y(t) 12t 24 control variable E.
The sufficiency conditions are met in analogous fashion to Example 2.
21.2.
Maximize subject to 1 0 (5x 3y 2y2) dt x ˙ x(0) 6y 7 x(1) 70 A.
The Hamiltonian is H 5x 3y 2y2 (6y) B.
The necessary conditions from the maximum principle are 1.
H y 3 4y 6 0 y 0.75 1.5 (21.33) 2.
a) ˙ H x 5 (21.34) b) x ˙ H 6y From (21.33), x ˙ 6(0.75 1.5) 4.5 9 (21.35) Integrating (21.34), (t) 5t c1 (21.36) Substituting in (21.35) and then integrating, x ˙ 4.5 9(5t c1) 4.5 45t 9c1 x(t) 4.5t 22.5t2 9c1t c2 (21.37) C.
Applying the boundary conditions, x(0) 7, x(1) 70, x(0) c2 7 c2 7 x(1) 4.5 22.5 9c1 7 70 c1 9 Substituting c1 9 and c2 7 in (21.36) and (21.37), (t) 5t 9 costate variable (21.38) x(t) 22.5t2 85.5t 7 state variable (21.39) D.
For the final solution, we then simply substitute (21.38) into (21.33), y(t) 0.75 1.5(5t 9) 7.5t 14.25 control variable E.
The sufficiency conditions are once again similar to Example 2.
504 OPTIMAL CONTROL THEORY [CHAP . 21 21.3.
Maximize subject to 5 0 (8x y2) dt x ˙ x(0) 0.2y 3 x(5) 5.6 A.
H 8x y2 (0.2y) B.
1.
H y 2y 0.2 y 0.1 (21.40) 2.
a) ˙ H x (8) 8 (21.41) b) x ˙ H 0.2y From (21.40), x ˙ 0.2(0.1) 0.02 (21.42) Integrating (21.41), (t) 8t c1 (21.43) Substituting in (21.42) and then integrating, x ˙ 0.02(8t c1) 0.16t 0.02c1 x(t) 0.08t2 0.02c1t c2 (21.44) C.
Applying the boundary conditions, x(0) 3, x(5) 5.6, x(0) c2 3 c2 3 x(5) 0.08(5)2 0.02c1(5) 3 5.6 c1 6 Substituting c1 6 and c2 3 in (21.43) and (21.44), (t) 8t 6 costate variable (21.45) x(t) 0.08t2 0.12t 3 state variable (21.46) D.
Substituting (21.45) into (21.40) for the final solution, y(t) 0.1 0.8t 0.6 control variable or taking the derivative of (21.46) and substituting it in the constraint x ˙ 0.2y, x ˙ 0.16t 0.12 0.2y y(t) 0.8t 0.6 control variable FREE ENDPOINTS 21.4.
Maximize subject to 4 0 (8x 10y2) dt x ˙ x(0) 24y 7 x(4) free A.
H 8x 10y2 (24y) B.
1.
H y 20y 24 0 y 1.2 (21.47) 2.
a) ˙ H x (8) 8 (21.48) 505 OPTIMAL CONTROL THEORY CHAP . 21] b) x ˙ H 24y From (21.47), x ˙ 24(1.2) 28.8 (21.49) Integrating (21.48), (t) 8t c1 Substituting in (21.49), x ˙ 28.8(8t c1) 230.4t 28.8c1 and then integrating, x(t) 115.2t2 28.8c1t c2 (21.50) C.
We now apply the transversality condition (4) 0.
(4) 8(4) c1 0 c1 32 Therefore, (t) 8t 32 costate variable (21.51) Next we substitute c1 32 in (21.50) and apply the initial condition x(0) 7.
So, x(t) x(0) x(t) 115.2t2 921.6t c2 0 0 c2 7 c2 7 115.2t2 921.6t 7 state variable D.
Then substituting (21.51) in (21.47), we have the final solution, y(t) 1.2(8t 32) 9.6t 38.4 control variable E.
The sufficiency conditions are easily confirmed as in Example 2.
21.5.
Maximize subject to 3 0 (2x 18y 3y2) dt x ˙ x(0) 12y 7 5 x(3) free A.
H 2x 18y 3y2 (12y 7) B.
1.
H y 18 6y 12 0 y 3 2 (21.52) 2.
a) ˙ H x (2) 2 (21.53) b) x ˙ H 12y 7 From (21.52), x ˙ 12(3 2) 7 43 24 (21.54) Integrating (21.53), (t) 2t c1 Substituting in (21.54), x ˙ 43 24(2t c1) 43 48t 24c1 and then integrating, x(t) 43t 24t2 24c1t c2 (21.55) C.
We now resort to the transversality condition (3) 0.
(3) 2(3) c1 0 c1 6 Therefore, (t) 2t 6 costate variable (21.56) Next we substitute c1 6 in (21.55) and apply the initial condition x(0) 5.
506 OPTIMAL CONTROL THEORY [CHAP . 21 So, x(t) x(0) x(t) 24t2 187t c2 0 0 c2 5 c2 5 24t2 187t 5 state variable D.
Then substituting (21.56) in (21.52), we have the final solution, y(t) 3 2(2t 6) 4t 15 control variable E.
The sufficiency conditions once again follow Example 2.
21.6.
Maximize subject to 1 0 (4y y2 x 2x2) dt x ˙ x(0) x y 6.15 x(1) free A.
H 4y y2 x 2x2 (x y) B.
1.
H y 4 2y 0 y 2 0.5 (21.57) 2.
a) ˙ H x (1 4x ) 1 4x b) x ˙ H x y From y in (21.57), x ˙ x 2 0.5 In matrix form, ˙ x ˙ 1 0.5 4 1 x 1 2 Y AX B Using (19.3) for the characteristic roots, r1, r2 0 0 4(3) 2 3.464 2 1.732 The eigenvector corresponding to r1 1.732 is A rI 1 1.732 0.5 4 1 1.732 c1 c2 2.732 0.5 4 0.732 c1 c2 0 2.732c1 4c2 0 c1 1.464c2 yc 1 1.464 1 k1e1.732t The eigenvector corresponding to r2 1.732 is A rI 1 1.732 0.5 4 1 1.732 c1 c2 0.732 0.5 4 2.732 c1 c2 0 0.732c1 4c2 0 c1 5.464c2 yc 2 5.464 1 k2e1.732t 507 OPTIMAL CONTROL THEORY CHAP . 21] For the particular solution, Y ¯ A1B.
Y ¯ ¯ x ¯ 1 3 1 0.5 4 1 1 2 2.33 0.83 Combining the complementary and particular solutions, (t) 1.464k1e1.732t 5.464k2e1.732t 2.33 (21.58) x(t) k1e1.732t k2e1.732t 0.83 (21.59) C.
Applying the transversality condition for a free endpoint (1) 0, (1) 8.2744k1 0.9667k2 2.33 0 From the initial condition x(0) 6.15, x(0) k1 k2 0.83 6.15 Solved simultaneously, k1 0.98, k2 5.95 Substituting in (21.58) and (21.59), (t) 1.4347e1.732t 32.511e1.732t 2.33 costate variable (21.60) x(t) 0.98e1.732t 5.95e1.732t 0.83 state variable (21.61) D.
Finally, substituting (21.60) in (21.57), we find the solution.
y(t) 0.7174e1.732t 16.256e1.732t 0.835 E.
For the sufficiency conditions, with f 4y y2 x 2x2, fx 1 4x, and fy 4 2y, we have D fxx fxy fyx fyy 4 0 0 2 D1 4 0 D2 8 0 Therefore, D is negative-definite and f is strictly concave. With the constraint (x y) linear and hence also concave, the sufficiency conditions for a global maximization in optimal control theory are satisfied.
INEQUALITY CONSTRAINTS 21.7.
Maximize subject to 4 0 (8x 10y2) dt x ˙ x(0) 24y 7 x(4) 2000 1.
For inequality constraints, we always start with a free endpoint. This problem was previously solved as a free endpoint in Problem 21.4 where we found (t) x(t) y(t) 8t 32 115.2t2 921.6t 7 1.2(8t 32) 9.6t 38.4 costate variable state variable control variable Evaluating the state variable at t 4, we have x(4) 1850.2 2000 a constraint violation 2.
Faced with a constraint violation, we have to redo the problem with a new fixed terminal endpoint: x(0) 7 x(4) 2000 508 OPTIMAL CONTROL THEORY [CHAP . 21 Solution of this new problem follows along exactly the same as in Problem 21.4 until the end of part B where we found (t) 8t c1 x(t) 115.2t2 28.8c1t c2 Now instead of the transversality condition, we apply each of the boundary conditions.
x(0) 115.2t2 28.8c1t c2 7 c2 7 Substituting c2 7 in the terminal boundary and solving, x(4) 115.2t2 28.8c1t 7 2000 c1 33.3 This gives us (t) x(t) 8t 33.3 115.2t2 959.04t 7 costate variable state variable Then for the final solution, from (21.46) we have y(t) 1.2(8t 33.3) 9.6t 39.96 control variable 21.8.
Maximize subject to 1 0 (5x 3y 2y2) dt x ˙ x(0) 6y 7 x(1) 70 1.
For an inequality constraint, we always start with a free terminal endpoint. Here we can build on the work already done in solving this problem under fixed endpoint conditions in Problem 21.2. There we found in (21.36) and (21.37), (t) x(t) 5t c1 4.5t 22.5t2 9c1t c2 Now applying the transversality condition (T) 0, (1) 5 c1 0 c1 5 and then substituting c1 5 and solving for the initial condition, x(0) 4.5t 22.5t2 45t c2 7 c2 7 This leads to, (t) x(t) 5t 5 22.5t2 49.5t 7 costate variable state variable Then from y 0.75 1.5 in (21.33), the solution is, y(t) 7.5t 8.25 control variable To see if the solution is acceptable, we evaluate the state variable at t 1.
x(1) 22.5t2 49.5t 7 34 70 constraint violation 2.
With the terminal constraint violated, we must now rework the problem with a fixed endpoint x(1) 70. We did this earlier in Problem 21.2 where we found (t) x(t) y(t) 5t 9 22.5t2 85.5t 7 7.5t 14.25 costate variable state variable control variable To determine if the control variable is an acceptable solution, we evaluate the state variable at the terminal endpoint and see that it fulfills the constraint.
x(1) 22.5t2 85.5t 7 70 509 OPTIMAL CONTROL THEORY CHAP . 21] 21.9.
Maximize subject to 1 0 (4y y2 x 2x2) dt x ˙ x(0) x y 6.15 x(1) 5 We have already optimized this function subject to free endpoint conditions in Problem 21.6. There we found, (t) x(t) y(t) 1.4347e1.732t 32.784e1.732t 2.33 0.98e1.732t 6e1.732t 0.83 0.7174e1.732t 16.392e1.732t 0.835 costate variable state variable control variable Evaluating the state variable at the terminal endpoint, we have x(1) 0.98e1.732t 6e1.732t 0.83 5.7705 5 Since the endpoint constraint is satisfied, we have found the solution to the problem.
21.10. Redo Problem 21.9 with a new set of endpoint constraints, x(0) 6.15 x(1) 8 From Problem 21.9 we know the free endpoint solution fails to meet the new terminal endpoint constraint x(1) 8. We must optimize under fixed endpoint conditions, therefore, by setting x(1) 8.
Starting from (21.55) and (21.56) where we found (t) x(t) 1.464k1e1.732t 5.464k2e1.732t 2.33 k1e1.732t k2e1.732t 0.83 we apply the endpoint conditions where we have at x(0) 6.15, at x(1) 8, k1 5.6519k1 k2 0.1769k2 0.83 6.15 0.83 8 When solved simultaneously, k1 1.3873 k2 5.5927 Substituting in (21.55) and (21.56) repeated immediately above, (t) x(t) 2.0310e1.732t 30.5585e1.732t 2.33 1.3873e1.732t 5.5927e1.732t 0.83 costate variable state variable Then from y 2 0.5 in (21.54), we derive the solution, y(t) 1.0155e1.732t 15.2793e1.732t 0.835 control variable Finally, to be sure the solution is acceptable, we evaluate the state variable at t 1, x(1) 1.3873e1.732t 5.5927e1.732t 0.83 8 and see that it satisfies the terminal endpoint constraint.
CURRENT-VALUED HAMILTONIANS 21.11. Maximize subject to 3 0 e0.05t(xy x2 y2) dt x ˙ x(0) x y 134.35 x(3) free A.
Setting up the current-valued Hamiltonian, Hc xy x2 y2 #(x y) 510 OPTIMAL CONTROL THEORY [CHAP . 21 B.
Assuming an interior solution, we first apply the modified maximum principle.
1.
Hc y x 2y # 0 y 0.5(x #) (21.59) 2.
a) # ˙ # Hc x # ˙ # ˙ 0.05# (y 2x #) 0.95# 2x y Using (21.59), # ˙ # ˙ 0.95# 2x 0.5(x #) 1.45# 1.5x b) x ˙ Hc # x y From (21.59), x ˙ 1.5x 0.5# In matrix form, # ˙ x ˙ 1.45 0.5 1.5 1.5 # x 0 0 The characteristic equation is A rI 1.45 r 0.5 1.5 1.5 r 0 and from (19.3), the characteristic roots are r1, r2 0.05 (0.05)2 4(2.925) 2 r1 1.7354 r2 1.6855 For r1 1.7354, the eigenvector is 1.45 1.7354 0.5 1.5 1.5 1.7354 c1 c2 3.1854 0.5 1.5 0.2354 c1 c2 0 3.1854c1 1.5c2 0 c2 2.1236c1 yc 1 1 2.1326 k1e1.7354t For r2 1.6855, the eigenvector is 1.45 1.6855 0.5 1.5 1.5 1.6855 c1 c2 0.2355 0.5 1.5 3.1855 c1 c2 0 0.2355c1 1.5c2 0 c1 6.3694c2 yc 2 6.3694 1 k2e1.6855t With B 0 in Y ¯ A1B, Y ¯ # ¯ x ¯ 0 0 Adding the complementary functions for the general solution, we have #(t) k1e1.7354t 6.3694k2e1.6855t (21.60) x(t) 2.1236k1e1.7354t k2e1.6855t (21.61) 511 OPTIMAL CONTROL THEORY CHAP . 21] C.
Next we apply the transversality condition for the free endpoint, #(T)e t 0, #(3)e0.05(3) (k1e1.7354(3) 6.3694k2e1.6855(3))e0.15 0 156.9928k1 0.0349k2 0 (21.62) and evaluate x(t) at x(0) 134.25, x(0) 2.1236k1 k2 134.25 (21.63) Solving (21.62) and (21.63) simultaneously, k1 0.03 k2 134.1819 Then substituting k1 0.03, k2 134.1819 in (21.60) and (21.61), we find #(t) x(t) 0.03e1.7354t 854.6582e1.6855t 0.0637e1.7354t 134.1819e1.6855t costate variable state variable D.
From (21.59), y(t) 0.5(x #). Substituting from above, y(t) 0.04685e1.7354t 360.23811.6855t control variable E.
For the sufficiency conditions, D fxx fxy fyx fyy 2 1 1 2 With D1 2 0, and D2 3 0, D is negative-definite, making f strictly concave in both x and y. With g linear in x and y, the sufficiency condition for a global maximum is fulfilled.
21.12. Maximize subject to 1 0 e0.08t(10x 4y xy 2x2 0.5y2) dt x ˙ x(0) x 2y 88.52 x(1) free A.
Hc 10x 4y xy 2x2 0.5y2 #(x 2y) B.
Assuming an interior solution, 1.
Hc y 4 x y 2# 0 y x 2# 4 (21.64) 2.
a) # ˙ # Hc x # ˙ # ˙ 0.08# (y 4x # 10) 0.92# 4x y 10 Substituting for y from (21.64) # ˙ 2.92# 3x 14 b) x ˙ Hc # x 2y From (21.64), x ˙ 4# 3x 8 In matrix form, # ˙ x ˙ 2.92 4 3 3 # x 14 8 512 OPTIMAL CONTROL THEORY [CHAP . 21 where the characteristic equation is A rI 2.92 r 4 3 3 r 0 and the characteristic roots are r1, r2 0.08 (0.08)2 4(20.76) 2 r1 4.5965 r2 4.5165 The eigenvector for r1 4.5965 is 2.92 4.5965 4 3 3 4.5965 c1 c2 7.5165 4 3 1.5965 c1 c2 0 7.5165c1 3c2 0 c2 2.5055c1 yc 1 1 2.5055 k1e4.5965t The eigenvector for r2 4.5165 is 2.92 4.5165 4 3 3 4.5165 c1 c2 1.5965 4 3 7.5165 c1 c2 0 1.5965c1 3c2 0 c1 1.8791c2 yc 2 1.8791 1 k2e4.5165t For the particular solutions, Y ¯ A1B.
Y ¯ # ¯ x ¯ 1 20.76 3 4 3 2.92 14 8 3.1792 1.5723 Combining the complementary functions and particular solutions for the general solution, #(t) k1e4.5965t 1.8791k2e4.5165t 3.1792 (21.65) x(t) 2.5055k1e4.5965t k2e4.5165t 1.5723 (21.66) C.
Applying the transversality condition for the free endpoint, #(T)e t 0, #(1)e0.08(1) (k1e4.5965(1) 1.8791k2e4.5165(1) 3.1792)e0.08 0 91.5147k1 0.0189k2 2.9348 0 (21.67) and evaluating x(t) at x(0) 88.52, x(0) 2.5055k1 k2 1.5723 88.52 (21.68) Solving (21.67) and (21.68) simultaneously, k1 0.05 k2 86.8230 Then substituting these values in (21.65) and (21.66), we find #(t) x(t) 0.05e4.5965t 163.1491e4.5165t 3.1792 0.1253e4.5965t 86.8230e4.5165t 1.5723 costate variable state variable D.
From (21.64), y(t) x 2# 4. Substituting for x and # from above, y(t) 0.2253e4.5965t 239.47524.5165t 0.7861 control variable 513 OPTIMAL CONTROL THEORY CHAP . 21] E.
For the sufficiency conditons, D fxx fxy fyx fyy 4 1 1 1 D1 4 0, and D2 3 0. D is negative-definite and f is strictly concave in both x and y. Since the constraint is linear in x and y, the sufficiency condition for a global maximum is satisfied.
514 OPTIMAL CONTROL THEORY [CHAP . 21 INDEX 515 Abscissa, 21p Acceleration principle, 396 Addition of matrices, 200–201, 208–209p Address, 200, 207–208p Adjoint matrix, 228 Aggregation, 15 Amplitude of fluctuation, 413 Antiderivative (see Integral) Antidifferentiation (see Integration) Area: between curves, 345–346, 354–356p under a curve, 342–343 of revolution, 480p Argument of function, 5 Arrows of Motion, 368 Associative law, 204–205, 219–222p Autonomous Equation, 428 Auxiliary equation, 408–409, 457p Average concepts, 63–64, 72–74p relationship to total and marginal concepts, 63–64, 72–74p, 80–81p Averch-Johnson effect, 323–325p Bordered Hessian, 258–259, 271–276p, evaluation of (4 4), 275–276p Boundary condition, 329, 404 Budget: constraint, 115 line, 19p Calculus: fundamental theorem of, 343–344 of variations, 460 (See also Differentiation; Integration) Canonical system, 494 Cartesian coordinate, 414 Catenary, 481 CES production function, 118–119, 136–137p Chain rule, 39, 48–49p Change: incremental, 112–113, 133–134p Characteristic equation, 408–409, 457p Characteristic matrix, 261 Characteristic polynomial, 457p Characteristic root, in differential and difference equations, 409–411, 416–421p, 428–439, 442–455p Characteristic root (Cont.): and sign-definiteness of quadratic form, 260–261, 280–283p Cobb-Douglas function, 116–118, 135–137p derivation of C-D demand function, 179–180 logarithmic transformation of, 150 Cobweb model, 395, 402–403p Coefficient, 2 technical, 259 Coefficient matrix, 206 Cofactor, 226, 238–240p Cofactor matrix, 228–229, 242–248p Column, 200 Column vector, 200 Commutative law, 204–205, 216–219p Comparative static: analysis, 110–111, 120–126p, 284–288, 296–308p derivatives, 110–111, 120–122p in constrained optimization, 290–291, 312–314p in optimization problems, 288–289, 308–311p Complementary function, 363, 392, 409 Complementary-slackness conditions, 293–294 Complex number, 411–412 transformation of, 414–415 Complex root, 411, 422–425p Composite-function rule, 38, 48–49p Compounding interest, 160–161, 165–166p Concave programming, 293–296, 316–323p Concavity, 58–59, 64–66p joint, 465–466 strict, 59 Conformability of matrices, 202, 210p Conjugate complex number, 412 Constant of integration, 326 Constant elasticity of substitution, 118 production functions, 118–119, 136–137p Constant function rule, 37, 43p Constrained optimization, 87–89, 102–105p, 115, 130–134p in calculus of variations, 466–467, 487p of CES production function, 118–119, 136–137p of Cobb-Douglas production function, 116–118, 135–136p in optimal control theory, 493–496, 503–514p and Lagrange multiplier, 87–89, 102–105p, 115, 130–134p The letter p following a page number refers to a Problem.
516 INDEX Demand: analysis, 15, 23–27p expressed in exponential functions, 190–191p final, 259–260 for money, 16–17 DeMoivre’s theorem, 414 Dependence, functional, 254–255, 262–267p linear, 224–225, 231–233p Dependent variable, 6 Depreciation, 160–161 Derivation of: addition rule in differentiation, 56p Cramer’s rule, 252–253p Cobb-Douglas demand function, 179–180 Euler’s equation, 462–463 product rule, 57p quotient rule, 57p Derivative, 36 of composite function, 38, 48–49p of constant function, 37, 43p of exponential function, 173–174, 181p of generalized power function, 38, 47–48p, 94p higher-order, 39–40, 52–54p, 85, 96–97p of an integral, 463 of linear function, 37, 43p of logarithmic function, 174–175, 182–184p notation, 36–37, 43–44p partial, 82–85, 93–105p, 176, 187–188p cross (mixed), 85, 96–97p second-order, 85, 96–97p of power function, 37, 43–44p, 93p of product, 38, 44–45p 93p of quotient, 38, 45–47p, 94p rules of (see Differentiation rules), of sums and differences, 38, 44p total, 90–91, 106–107p of trigonometric function, 413–414, 421–422p Determinant, 224–226, 231–232p Discriminant, 256, 264–265p Hessian, 255–256, 265–271p bordered, 258–259, 271–276p Jacobian, 254–255, 262–263p properties of, 227–228, 233–236p vanishing, 224 Determinant test for: sign–definiteness of quadratic form, 254–255, 264–265p singular matrix, 224–225, 236–238p Deviation from equilibrium, 363, 392, 409 Diagonal (principal), 220 Difference (first), 391 Difference equation: and Cobweb model, 395, 402–403p economic applications of, 404–405p, 425–427p first–order, 391–392, 398–401p Constraint: budget, 115 in dynamic optimization, 466–467, 487p endpoint, in calculus of variations, 461 inequality, 293–296, 316–325p nonnegativity, 293 in static optimization, 87–89, 102–105p, 115, 130–134p Consumers’ surplus, 347–348, 359–360p Continuity, 33–34, 36, 41–43p and differentiability, 36–37 Continuous growth, 162–163 Control variable, 493 Convergence: in difference equations, 392–393, 398–401p, 415, 417–419p in differential equations, 363–364, 368–370, 387–389p in improper integrals, 346–347, 356–359p Conversion, exponential-logarithmic, 162–163, 171p Convexity, 58–59, 64–66p strict, 59 Coordinate: Cartesian, 414 polar, 414 Corner solution, 296 Cosecant function, 412 derivative of, 413–414, 421–422p Cosine function, 412 derivative of, 413–414, 421–422p Costate variable, 493 Cotangent function, 412 derivative of, 413–414, 421–422p Cramer’s rule, 230–231, 248–253p proof of, 252–253p Critical point (value), 59 Cross elasticity of demand, 111–112, 126–127p Cross partial derivative, 85, 96–97p Current-valued Hamiltonian, 500–503, 510–514p Curve: area between, 345–346, 354–356p area under, 342–343 Curvilinear function, slope of, 34–35 Cusp, 36 Decreasing function, 58, 64–66p Defined (product of matrices), 203 Definite: integral, 343, 349p negative-, positive-, 256–257 semi-, 260–261 solution, 363, 392, 409 Deflation, 160 Degree, of differential equation, 362, 370p 517 INDEX Difference equation (Cont): and Harrod model, 396, 404p and lagged income determination model, 394, 401–402p second-order, 410–411, 419–420p, 424–425p Difference quotient, 35 Differentiable function, 36 Differentiability, 36, 60 Differential, 89–90, 105–106p, 112–113, 122–126p and incremental change, 112–113 partial, 90, 106p, 122–126p total, 90, 105–106p Differential equation, 362 economic applications of, 367–368, 382–387p exact, 364–365, 374–375p first-order linear, 363–365, 370–374p and integrating factors, 364–365, 374–375p nonlinear, 366–367, 380–382p ordinary, 362 partial, 364 second-order, 408–410, 416–417p and separation of variables, 366–367, 380–382p Differentiation, 36 of exponential function, 173–174, 181p implicit, 40, 54–56p logarithmic, 177–178, 191–192p of logarithmic function, 174–175, 183–184p partial, 82–85, 93–105p, 176, 187–188p total, 90–91, 106–107p of trigonometric function, 413–414, 421–422p (See also Differentiation rules and Derivative) Differentiation rules, 37–39, 82–85 chain rule, 38, 48–49p combination of rules, 49–52p, 94–95p composite function rule, 38, 48–49p constant function rule, 37, 43p exponential function rule, 173–174, 181p derivation of, 197p function of a function rule, 38, 48–49p generalized power function rule, 39, 47–48p implicit function rule, 92, 107p inverse function rule, 92, 107–108p linear function rule, 37, 43p logarithmic function rule, 174–175, 182–184p derivation of, 196p power function rule, 37, 43–44p product rule, 38, 44–45p derivation of, 57p quotient rule, 38, 45–47p derivation of, 57p sum and difference rule, 38, 44p derivation of, 56p Dimension of matrix, 200, 207–208p Diminishing returns, 116, 134p Discounting, 162, 168–169p Discrete growth, 162 Discriminant, 256, 264–265p Discrimination, in pricing, 77–78p, 132–133p Distance between two points on a plane, 461, 470–471p Distribution parameter, 118 Distributive law, 204–205, 219–222p Divergence, 346–347, 356–359p Domain, 5 Domar model, 384–385p Dominant root, 415 Dynamic optimization, 462–463, 471–481p economic applications, 482–486p under functional constraints, 466–467, 487p Dynamic stability of equilibrium, 363, 391, 428 e, the number, 149 Effective rate of interest, 161–162 Efficiency parameter, 116 Eigenvalue (vector), 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–445p Eigenvalue problem, 429, 458–459p Elasticity: cross, 111–112, 127p of demand, 126–127p of income, 111–112, 126–127p of price, 127p of supply, 126–127 Elasticity of substitution, 118, 140–145p of CES function, 118, 142–145p of Cobb-Douglas function, 140–141p Elements of a matrix, 200 Elimination method for solving simultaneous equations, 4, 26p Endogenous variable, 16 Endpoint condition, 461 Envelope theorem, 291–293, 314–316p Equation, 3 auxiliary, 408–409, 457p characteristic, 408–409 difference, 391–392, 398–401p differential, 362–365, 370–374p equilibrium, 15 in income determination models, 15–16, 27–30p in IS–LM models, 16–17, 30–31p linear, 3 quadratic, 3 of motion, 493 reduced form, 16, 27–30p simultaneous, 4–5, 25–27p solution of, 9–10p using logarithms, 158–159p in supply and demand analysis, 23–27p of transition, 493 Equilibrium: dynamic stability of, 363, 391, 428 inter-temporal, 363, 391, 428 518 INDEX Function (Cont.): optimization of, 60–63, 68–72p, 74–79p, 85–57, 98–102p, 113–115, 127–130p periodic, 413 polynomial, 5 power, 5 primitive, 39 quadratic, 4–5, 10p, 13p, 256, 264–265p, 280–283p rational, 5, 13p sinusoidal, 421 smooth, 60 trigonometric, 412–413, 421–422p Functional, 460 Functional dependence, 254–255, 262–263p Fundamental theorem of calculus, 343 Generalized power function rule, 38, 47–48p, 94p Giffen good, 286 Global maximum (minimum), 61 Goods: complementary (substitute), 111–112 final (intermediate), 259–260 Graph, 6, 11–13p of exponential function, 147, 150–152p in income determination model, 21–23p of linear function, 6, 17–23p of logarithmic function, 147, 152–153p of quadratic function, 13p of rational function, 13p Growth, 192–194p continuous (discrete), 162–163 conversion factor for, 162–163, 171p measures of, 178 Growth functions, exponential, 163–164, 169–179p Growth rates: estimation from data points, 163–164, 172p per capita, 192p warranted, 396 Hamiltonian, 494–495, 503–510p current-valued, 500–503, 510–514p system, 494 Harrod model, 396, 404p Hessian, 255, 256, 265–271p bordered, 258–259, 271–276p Higher-order derivative, 39–40, 52–54p, 85, 96–97p Homogeneity, 115–116, 134p i, the number, 412 Idempotent matrix, 205 Identity matrix, 205–206 Image, mirror, 147, 152–153p Imaginary number, 411–412 transformation of, 414–415 Equilibrium (Cont.): steady-state, 363, 391, 428 Euclidian distance condition, 431 Euler equation, 414 proof of, 462–463 Euler relations, 414 Exact differential equation, 364–365, 374–375p Exogenous variable, 16 Explicit function, 40 Exponent, rules of, 1, 7–8p, 148, 154–156p Exponential function, 146, 150–151p, 160 base conversion of, 162–163, 170–171p derivative of, 173–175, 181p and discounting, 162, 168–169p and growth, 163–164, 169–170p and interest compounding, 160–161, 165–168p integration of, 327, 337p logarithmic solutions of, 149, 156–159p natural, 149–150, 151–153p and optimal timing, 179, 194–196p optimization of, 176–177, 188–191p relationship to logarithmic functions, 149–150, 156–159p slope of, 185p Extremal, 460 finding candidates for, 463–465, 471–481p Extremum (relative), 60, 66–72p Factoring, 3–4 Final demand, 259–260 First difference, 391 First-order condition, 60 Frequency functions and probability, 348, 361p Free endpoint, 496–498, 505–508p Function, 5, 10–11p Cobb-Douglas, 116–118, 135–137p complementary, 363, 392, 409 composite, 39, 48–49p concave (convex), 58–59, 64–66p constant, 37 constant elasticity of substitution (CES), 118 curvilinear, slope of, 34–35 decreasing (increasing), 58, 64–66p differentiable, 60 explicit, 40 exponential, 146, 150–151p, 160 frequency, 348, 361p of a function, 39, 48–49p homogeneous, 115–116, 134p implicit, 40, 54–56p, 92, 107p inverse, 18p linear, 5 logarithmic, 147 monotonic, 58 multivariable, 82, 110 objective, 87–88 519 INDEX Implicit differentiation, 40, 54–56p, 92, 107p Implicit function, 40 rule, 40, 54–56p, 92, 107p theorem, 286, 300p Improper integral, 346–347, 356–359p and L’Ho ˆpital’s rule, 347, 356–359p Income determination model, 15–16 equations in, 27–30p graphs in, 21–23p lagged, 394, 401–402p Income determination multiplier, 15–16, 27–30p, 120–126p Income elasticity of demand, 111–112 Increasing function, 58, 64–66p Indefinite integral, 326 Independence, 4 Independent variable, 6 Indirect objective function, 292 Inequality constraints, 293–296, 316–325p, in endpoints, 498–500, 508–510p Inferior good, 286 Inflection point, 60, 66–68p, 85–87, 98–102p Initial condition, 329 Inner product, 202 Input–output analysis, 259–260, 276–280p Integral, 326, 343 definite, 343, 349p probability and, 348, 361p properties of, 344–345, 353–354p of differential equation, 362 improper, 346–347, 356–359p indefinite, 326, 332–334p particular, 363 sign, 326 Integral calculus, 326, 342 Integrand, 326 Integrating factor, 365–366, 375–379p Integration, 326, 342 constant of, 326–328 and consumers’ and producers’ surplus, 347–348, 359–360p economic applications of, 331–332, 339–341p, 359–361p and initial conditions and boundary conditions, 329, 494 limits of, 343 partial, 364–366, 374–379p by parts, 330–331, 337–339p, 352–353p rules of, 326–328, 332–334p by substitution, 329–330, 334–337p, 349–352p tables, 330 Intercept (x–, y–), 6, 11–13p Interest compounding, 160, 165–168p Interest rate (effective, nominal), 161 and timing, 166–168p Interior solution, 294 Intermediate good, 259 Intertemporal equilibrium level (solution), 363, 392 Inverse function, 18p rule, 92, 107–108p Inverse matrix, 228–229, 242–248p and solution of linear–equation system, 229–230, 242–248p IS-LM analysis, 16–17, 30–31p, 287–288p Isocline, 339–342 Isocost (isoprofit) line, 14–15 Isoperimetric problem, 466–467 Isoquant, 79–80p, 139p Isosector, 439–442 Iterative method, 392 Jacobian, 254–255, 262–263p Joint concavity (convexity), 465–466 Kuhn-Tucker conditions, 293–294, 316–325p Lag, 395 Lag matrix, 202 Lagrange multiplier, 87–89, 102–105p, 115, 130–134p in calculus of variations, 466–467, 487p and inequality constraints, 293–296, 316–325p in optimal control theory, 493–496, 503–514p and shadow prices, 89 Lagrangian function, 87–89, 102–105p, 115, 130–134p Laplace expansion, 227, 241–243p Latent root (vector), 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–445p Lead matrix, 202 Least common denominator (LCD), 9p Leibnitz’s rule, 463 Leontief matrix, 260 L’Ho ˆpital’s rule, 347, 356–359p Like terms, 2 Limits, 32–33, 41–43p lower (upper), 343 one-sided, 33 rules of, 32–33, 41–43p Linear algebra, 199 commutative, associative, distributive laws, 204–205, 216–222p expression of set of linear systems by, 206–207 Linear dependence, test for, 224–225, 236–238p Linear equations: solving with Cramer’s rule, 230–231, 248–253p solving with the inverse matrix, 228–229, 242–248p Linear function, 5 Linear function rule, 37, 43p LM schedule, 16, 287–288p 520 INDEX Minor, 226, 238–240p principal, 255 bordered, 258 Mirror image, 147, 152–153p Mixed partial derivative, 85, 96–97p Modified maximum principle, 500–501 Monomial, 2 Monotonic function, 58 Multiplication of matrices, 202–204, 212–216p scalar and vector, 201–202, 211–212p Multiplier, 15–16, 110–111 autonomous expenditure, 16, 28p autonomous (lump-sum) tax, 28p balanced–budget, 121p foreign–trade, 29–30p, 122p government, 120–121p proportional tax, 29p Multivariate functions, 82, 110 optimization of, 85–87, 98–102p, 113–115, 127–130p Natural exponential function, 149 conversion of, 153–154 derivative of, 173–174, 180–181p graph of, 153p optimization of, 176–177, 188–191p slope of, 174–175 solution of, 149 Natural logarithmic function, 149 conversion of, 153–154 derivative of, 174–175, 183–184p graph of, 153p optimization of, 176–177, 188–191p slope of, 174–175 solution of, 149 Negative definiteness, 256, 260–261, 264–265p, 280–282p Negative semi-definiteness, 260–261, 264–265p, 280–282p Nominal rate of interest, 161 Nonlinear programming, 293–296, 316–323p Nonnegativity constraint, 293 Nonsingularity of matrices, 224–225, 236–238p Nontrivial solutions, 429 Normal good, 286 Normalization, 261, 430 Null matrix, 205–206, 222p Numbers: complex, 411–412 conjugate, 411–412 imaginary, 411–412 Objective function, 87–88 Optimal control theory, 493 fixed endpoints, 494–495, 503–505p free endpoints, 496–498, 505–508p ln, 149–150 Logarithm, 147–148 conversion formulas, 162–163 natural, 149–150 properties of, 148, 154–156p Logarithmic differentiation, 177–178, 191–192p Logarithmic function, 147–148 conversion of, 174–175, 183–184p derivative of, 191, 198–199p optimization of, 176–177, 188–191p relationship to exponential functions, 147 slope of, 185p Logarithmic transformation, 150 Lower–triangular matrix, 236p Marginal concepts, 62–63, 72–74p, 110, 119–120p relationship to total and average concepts, 62–63, 72–74p, 110, 119–120p Marginal product (MP), 110 Marginal rate of technical substitution (MRTS), 79–80p, 139p Marginal utility of money, 133p Matrix(ces), 200 addition and subtraction, 200–201, 208–209p adjoint, 228 associative, commutative, distributive laws, 204–205, 216–222p characteristic, 261 coefficient, 206 cofactor, 228–229, 242–248p column vector, 200 conformability, 202, 210p idempotent, 205 identity, 205–206 inverse, 228–229, 242–248p lag, 202 lead, 202 Leontief, 260 multiplication, 202–204, 212–216p nonsingular, 224, 236–238p null, 205–206, 222p rank of, 224–225, 232–233p row vector, 200 singular, 222p, 224, 236–238p square, 20 symmetric, 205 technical coefficients of, 260 transpose, 200, 207p triangular, 228, 235–236p unique properties of, 222–223p Matrix algebra, (see Linear algebra) Matrix inversion, 228–229, 242–248p Maximum principle, 494 modified, 500–501 Maximization, (Minimization), (see Optimization) 521 INDEX Optimal control theory (Cont.): inequality constraints in endpoints, 498–500, 508–510p Optimal timing, 179, 194–196p Optimization, 60–63, 68–72p, 74–79p, 85–87, 98–102p, 113–115, 127–130p constrained, 87–89, 102–105p, 115, 130–134p, 258–259, 271–276p of CES production function, 118–119, 136–137p of Cobb-Douglas production function, 116–118, 135–137p, 259, 275p dynamic, 462–463, 471–486p constrained, 466–467, 487p of exponential functions, 176–177, 188–191p of logarithmic functions, 176–177, 188–191p of multivariable functions, 85–87, 98–102p, 113–115, 127–130p Order: of difference equation, 391 of differential equation, 362, 370p Ordinate, 22p Oscillation, 392–394 Output elasticity, 116 Parabola, 13p Parameter distribution, 116–118 efficiency, 116–118 substitution, 116–118 Partial derivative, 82–85, 93–105p, 176, 187–188p cross (mixed), 85, 96–97p exponential, 176, 187–188p logarithmic, 176, 187–188p second-order direct, 85, 96–97p, 102 Partial differentiation, 82–85, 93–105p, 176, 187–188p rules of, 83–85, 93–97p Partial differential, 90, 106p, 122–126p Partial integration, 364–365, 374–379p Particular integral, 363, 458p Parts, integration by, 330–331, 337–339p, 352–353p Per capita growth, 192p Period, 413 Periodic function, 413 Phase, 413 diagram, 368–370, 397–398 in difference equation, 397–398, 405–407p in differential equation, 368–370, 387–390p, 439–442, 455–459p Phase plane, 439 Polar coordinate, 414 Polynomial, 2, 8–9p Polynomial function, 5 continuity of, 34 limit of, 41p Population growth, 162 Positive definiteness, 256, 260–261, 264–265p, 280–282p Positive semi–definiteness, 260–261, 264–265p, 280–282p Power function, 5 derivative of, 37, 43–44p generalized, 38, 47–48p Power function rule, 37, 43–44p generalized, 38, 47–48p Precautionary demand for money, 16–17 Present value, 162 Price, shadow, 89 Price discrimination, 77–78, 132–133p Price elasticity, 127p Primitive function, 39 Principal diagonal, 205 Principal minor, 255 bordered, 258 Probability, and definite integral, 348, 361p Producers’ surplus, 347–348, 359–360p Product rule, 38, 44–45p, 84 derivation of, 57p Production function: CES, 118–119, 136–137p Cobb-Douglas, 116–118, 135–136p elasticity of substitution of, 118, 140–145p output elasticity of, 116 homogeneous, 115–116, 134p Production isoquant, 79–80p, 139p Proof of: addition rule, 56p Cramer’s rule, 252–253p ex rule, 197p eg(x) rule, 197p Euler’s equation, 462–463 ln x rule, 196p ln g(x) rule, 197p Px/Py MUx/MUy for optimization, 140p product rule, 57p properties of logarithms, 148, 154–156p quotient rule, 57p Pythagorean theorem, 461 Quadratic equation, 3 graph of, 13p solution by: factoring, 3–4 quadratic formula, 3–4, 10p, 26p Quadratic formula, 3–4, 10p, 26p Quadratic function, 4–5, 10p, 13p, 256, 264–265p discriminant and sign–definiteness of, 256, 264–265p Quotient rule, 38, 45–47p derivation of, 57p 522 INDEX Sine function, 412 derivative of, 413–414, 421–422p Singular matrix, 224 test for, 224–225, 262–263p Slope, 6, 11–13p of curvilinear function, 34–35 of exponential function, 173–174, 185p of linear function, 6, 11–13p of logarithmic function, 174–175, 185p Slope-intercept form, 7, 11–13p Slutsky effect, 313–314p Smooth function, 60 Solow model, 385–386p Solution: definite, 363, 392 general, 363, 392 particular, 363, 392 trivial, 429, 489p vector, 206 Speculative demand for money, 16 Stability conditions: Cobweb model, 395, 402–403p difference equation, 391–398, 398–407p, 415, 420–421p differential equation, 362–370, 370–390p, 415, 417–419p State variables, 493 Steady-state solution, 368, 397 Strictly concave (convex), 58–59, 64–66p Subintegral, 344–345 Substitution: elasticity of, 118, 142–143p integration by, 329–330, 334–337p method for solving simultaneous equations, 4, 26p Subtraction of matrices, 200–201, 209–210p Successive-derivative test, 61–62, 67–72p Sum and difference rule, 38, 44p derivation of, 56p Supply, elasticity of, 126–127p Supply and demand analysis, 15, 23–27p Surplus, consumers’, producers’, 347–348, 359–360p Symmetric matrix, 205 Tables of integration, 330 Tangent function, 412 derivative of, 413–414, 421–422p Tangent line, 34 Taylor expansion, 467 Technical coefficient, 259–260 Terms, 2 Time path, 384p, 392–393, 413, 415 Timing, optimal, 179, 194–196p Total concepts, 63–64, 72–74p, 80–81p relationship to marginal and average concepts, 63–64, 72–74p Radian, 413 Range, 5 Rank, 224–225, 232–233p Rate of change, 36 Rate of discount, 162, 168–169p Rational function, 5, continuity of, 34 graph of, 13p limits of, 32–33 Reduced form equation, 16, 27–30p Relative extremum, 59–60, 66–68p Returns to scale, 116, 134p Riemann sum, 342 Root: characteristic, 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–455p complex, 411 distinct real, 411, 416–417p dominant, 415 latent, 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–455p repeated real, 411, 417, 420p Row, 200 Row vector, 200 Rules: Cramer’s, 230–231, 248–253p De Moivre’s, 414 of differentiation (see Differentiation rules) of exponents, 1, 7–8p for integrating factor, 365–366, 375–379p of integration, 326–328, 322–334p for inverse function, 92, 107–108p L’Ho ˆpital’s, 347, 356–359p Leibnitz’s, 463 of limits, 32–33, 41–43p of logarithms, 148, 154–156p Saddle path, 429, 439 Saddle point, 87 Saddle point solution, 429 Samuelson model, 426p Scalar, 206 multiplication, 201, 211p Secant function, 412 derivative of, 413–414, 421–422p Secant line, 35 Second-derivative test, 60 Second-order condition, 61 Semi-definiteness, 260–261 Shadow price, 89 Separation of variables, 366–367, 380–382p Sign-definiteness of quadratic function, 256, 260–261, 264–265p, 280–282p Simultaneous equations, 4–5, 25–27p difference equations, 434–439, 451–455p differential equations, 428–434, 442–450p 523 INDEX Total derivative, 90–91, 106–107p Total differential, 90, 105–106p Trace, 429 Transaction demand for money, 16 Transformation, of complex and imaginary numbers, 414–415 logarithmic, 150 Transpose matrix, 200 Transversality condition, 497 Triangular matrix, 228 lower (upper), 235p Trigonometric function, 413–414 derivative of, 413–414, 421–422p Upper-triangular matrix, 235p Utility: marginal, of money, 133p optimization of, 133–134p Value, critical, 59 of function, 5 Vanishing determinant, 224 Variable, 2 control, 493 costate, 493 dependent and independent, 6 Variable (Cont.): endogenous and exogenous, 16 separated, 366–367, 380–382p state, 493 Variation of F, 467 Variational notation, 467–468, 491–492p Variations, the calculus of, 460 Vector: characteristic, 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–455p column and row, 200 of constant terms, 229 eigen, 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–455p latent, 260–261, 280–283p, 409–411, 416–421p, 428–439, 442–455p multiplication of, 201–202, 211–212p solution, 206 Vertical line test, 10p Warranted rate of growth, 396 Young’s Theorem, 85, 97p This page intentionally left blank NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES NOTES |
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7th Grade Math Module 1 Ratios and Proportions: Week 1
Created by
Teach How They Learn
This document consists of 5 days of lesson plans as an introduction to ratios and proportions. I used the 7th Grade Math Module 1 from Engageny.org as a guide when I created my notes and activities. Everything you need for the activities are included, along with directions. I used the questions from the module as practice problems in some of my lessons. I have included formal lesson plans with essential questions for each day, guided notes, practice problems, homework, exit tickets and activ
6th - 9th
Math
CCSS
6.RP.A.1
, 6.RP.A.2
, 7.RP.A.2a
+2
$7.00
Original Price $7.00
Rated 4.92 out of 5, based on 12 reviews
4.9 (12)
7th Grade Math Module 1 Ratios and Proportions Week 2
Created by
Teach How They Learn
This document consists of 6 days of lesson plans on ratios and proportions. I used the 7th Grade Math Module 1 from Engageny.org as a guide when I created my notes and activities. Everything you need for the activities are included, along with directions. I used the questions from the module as practice problems in some of my lessons. I have included formal lesson plans with essential questions for each day, guided notes, practice problems, homework, exit tickets and activities. Lesson hi
6th - 9th
Math
CCSS
7.RP.A.2a
, 7.RP.A.2b
, 7.RP.A.2c
+1
$7.00
Original Price $7.00
Rated 5 out of 5, based on 6 reviews
5.0 (6)
7th Grade Math Module 1 Ratios & Proportions Week 4 Pt 1: Review Station Activty
Created by
Teach How They Learn
This is a station activity that reviews ratios and proportions using cooperative learning and different learning styles. I have included a formal lesson plan along with all the materials and directions needed for each station. Lesson highlights: Station #1: Students will be answering NYS questions (from assessments and modules on engageny.org) on proportional and non-proportional relationships. Station #2: Students will be given pictures and they have to work with a partner to create word probl
6th - 9th
Math
CCSS
7.RP.A.2a
, 7.RP.A.2b
, 7.RP.A.2c
+1
$3.00
Original Price $3.00
Rated 5 out of 5, based on 4 reviews
5.0 (4)
BUNDLE Ratio and Proportion POWERPOINT with matching WORKSHEETS
Created by
Teach with Annalise
The perfect pair for teaching students how to represent ratios, simplify ratios, create and identify equivalent ratios, divide quantities into a given ratio AND solve real-world problems involving ratios! You get a worksheets/notes that PERFECTLY match the structure, graphic notes and questions in the PowerPoint Lesson and lesson plan! This 55-slide PowerPoint with lesson plans and matching student worksheets does the hard work for you, providing step-by-by step explicit instructions with wo
6th - 7th
Math
$6.50Original Price $6.50
$5.00
Price $5.00
8th Grade Math Back to School Activities | Fractions Ratios & Proportions Review
Created by
Boldly Inspired Curriculum
Simplify your back to school lesson planning with these 8th grade math back to school activities! Build confidence with this bundle of review activities for fraction operations, simplifying ratios, and solving proportions. This bite sized version of the Algebra 1 Prep Packet includes weeks 3 and 4 of the 8 week packet. Your students will review fraction operations, ratios, and proportions so that their skills are sharp when the first day of school rolls around! Inside the Google Drive folder you
8th
Algebra, Math
CCSS
5.NF.A.1
, 6.NS.A.1
, 7.NS.A.2
+2
$13.25Original Price $13.25
$11.25
Price $11.25
Algebra - Ratios & Proportions Unit Guided Notes, Worksheets, & Quiz Bundle
Created by
Marisa May Education
Are your Algebra students starting their Ratios & Proportions Unit and you're looking for an easy ready to go total unit lesson plan (guided notes) bundle with quiz assessment forms to be able to just hit print & teach? If so, this ready to teach bundle will provide you with just what you need! What topics it covers:Ratios & ProportionsRatios & Proportions in Similar FiguresPercents with Ratios & ProportionsThe Return of PercentsDirect & Inverse VariationProbabilityTheoretical vs Experiment
7th - 10th
Algebra, Applied Math, Math
$14.50Original Price $14.50
$13.05
Price $13.05
7th Grade Math Module 1 Ratios and Proportions Week 3
Created by
Teach How They Learn
This document consists of 5 days of lesson plans on ratios and proportions. I used the 7th Grade Math Module 1 from Engageny.org as a guide when I created my notes and activities. Everything you need for the activities are included, along with directions. I used the questions from the module as practice problems in some of my lessons. I have included formal lesson plans with essential questions for each day, guided notes, practice problems,and activities. Lesson highlights: Days 1-3: not
6th - 9th
Math
CCSS
7.RP.A.2a
, 7.RP.A.2b
, 7.RP.A.2c
+1
$5.00
Original Price $5.00
Rated 5 out of 5, based on 4 reviews
5.0 (4)
Unit Plan - Math 7 Ratios & Proportions
Created by
Mrs Dambach
This Unit Plan lays out the vocabulary needed for the unit, the standards for Ratios and Proportions with IXL lessons to match each standards. There is a place for students to reflect and rate themselves 1-4 for each learning target as well as a place for them to make a goal for the unit. Great resource all on one page.
7th
Math
CCSS
7.RP.A.1
, 7.RP.A.2
, 7.RP.A.2a
+3
$2.00
Original Price $2.00
Math 6, Lesson 6: Ratios and Proportions Complete Lesson Plan
Created by
All Things Math Emporium
Engage and energize your students with a dynamic lesson plan unveiling the wonders of Math 6, Topic 6: Ratios and Proportions! This vibrant presentation comprises 211 animated and illustrated slides, delivering a captivating learning experience. Dive into a multitude of classroom exercises that bring the below topics and objectives to life. Get ready for an interactive journey that transforms abstract concepts into tangible understanding! Topic Objectives Section 6-1: Ratios Define a Ratio.Write
6th - 7th
Algebra, Math, Other (Math)
Also included in: Math 6 Superiority Bundle: 12 Comprehensive Lesson Plans Year-Long Course
$4.99
Original Price $4.99
Ratio and Rates 3 day Mini Unit - STAAR REVIEW LESSON - TEKS 7.4B and 7.4D
Created by
Teaching Math Magicians
This activity can be used multiple ways. It can created into stations, notes, guided practice, and/or independent practice. This mini-unit includes 3 days of lesson plans, starting with solving proportions, setting up proportions, and proportion word problems. Each lesson includes notes and activities. This bundle includes: Day 1 Notes (1 page)12 Task Cards (3 pages)Recording Sheet (2 pages)Exit Ticket (1 page)Answer Key (2 pages) Day 2 Notes (1 page)Activity - 2 options (2 pages each)E
6th - 8th
Math, Math Test Prep, Other (Math)
CCSS
6.RP.A.2
, 6.RP.A.3
, 6.RP.A.3b
+1
Also included in: Ratios, Rates and Percent Proportions - STAAR REVIEW Lessons - 4 day bundle
$5.40
Original Price $5.40
Algebra - Ratios & Proportions Complete Unit Guided Notes & Worksheets
Created by
Marisa May Education
Are your Algebra students starting their Ratios & Proportions Unit and you're looking for an easy ready to go lesson plan bundle complete with guided notes to be able to just hit print & teach? If so, these ready to teach lessons with guided notes will provide you with just what you need! What topics it covers:Ratios & ProportionsRatios & Proportions in Similar FiguresPercents with Ratios & ProportionsThe Return of PercentsDirect & Inverse VariationProbabilityTheoretical vs Experimental Prob
7th - 10th
Algebra, Applied Math, Math
Also included in: Algebra - Ratios & Proportions Unit Guided Notes, Worksheets, & Quiz Bundle
$10.50
Original Price $10.50
Ratio & Proportion - Complete Unit of Work
Created by
Tom O'Toole Math Resources
In this unit, students will learn how to use ratio and proportion in Maths. They will start by learning how to write ratios, including using unitary form and simplest form, before splitting quantities in ratios and using ratios to find unknown amounts. Students will then look at the links between ratio and fractions, and how ratio can be used to write linear functions and scale up quantities in direct proportion. They will then apply this to the specific cases of scaling up recipes, maps, scale
7th - 10th
Decimals, Fractions, Other (Math)
Also included in: Complete Bundle - All units of work
$16.00
Original Price $16.00
Ratio and Proportions | Digital and Print Word Problem Activities
Created by
Room to Discover
Want an easy way to engage your students in studying ratio and proportions word problems? This amazing digital activity for distance learning supports student in making sense of ratio and proportion word problems and using Polya's 4-step problem-solving process. This resource is complete with a teacher's guide, lesson plan, print and digital option. This no-prep math-workshop lesson is designed to be taught in a live or online classroom. Your students will think deeply and learn through colla
6th
Math, Other (Math)
CCSS
6.RP.A.2
, 6.RP.A.3b
$5.00
Original Price $5.00
Ratios and Solving Proportions - 8th Grade Math Activity with Google Slides
Created by
Boldly Inspired Curriculum
Increase student engagement with this no-prep ratios and proportions activity using individual whiteboards! This 8th grade math lesson includes animated Google Slides with a printable version. This activity includes teacher directions, a review of ratios and solving proportions, 12 questions, and lesson slides. The printable version can be used with the slides if you don't have a classroom set of individual whiteboards. Note: You can make a 'whiteboard' using any piece of paper and sheet prote
8th
Algebra, Math
CCSS
7.RP.A.3
$3.00
Original Price $3.00
Number 8 (Standard Form Calculations, Ratio and Proportion)
Created by
Excellent Resources Outstanding Progress
GCSE Mathematics (Grade 8-10) - 2 hour fully interactive and animated lesson planned with answers including student exercise booklet. Ready to be used straight away! FULL GCSE Course - Other lessons available from store: Geometry 1 (Angles Facts, Angles in shapes & Symmetry) Geometry 2 (parallel lines, interior & exterior) Geometry 3 (Circle Theorems) Geometry 4 (Pythagoras) Geometry 5 (Trigonometry) Geometry 6 (Area, Perimeter, Surface Area, Area and Circumference of a Circle) Geometry 7
8th - 12th
Decimals, Numbers
Also included in: KS3 KS4 - GCSEs (Grade 8-11) - NUMBER - ALL Topics Powerpoints and Booklets
$9.95
Original Price $9.95
Proportion & Ratio - Complete Lesson
Created by
Tom O'Toole Math Resources
This is a complete lesson on proportion and ratio that looks at how to scale up quantities in proportion using a ratio, and express a ratio as a linear function. The pack contains a full lesson plan, along with accompanying resources, including a student worksheet and suggested support and extension activities.
6th - 9th
Decimals, Fractions, Numbers
$4.00
Original Price $4.00
Back to School with Ratio and Proportion
Created by
Mo Don
Fall is a great time to buy this 36 page complete unit. This plan includes topics, steps and procedures, tips, 4 good-sized worksheets, and an end of unit test. My students have gained tremendously from this complete lesson plan. They are swift to explain the facts and differences between ratios and proportions, where before the development of this format, they could not. The topics are aligned to the Common Core and PA standards (6th and 7th) and also CMP3 math curriculum. If you need any gra
5th - 8th
Math
$4.99
Original Price $4.99
Ratios and proportions with Ancient Recipes- Review Before 6th Grade
Created by
Easy as Pi Learning Incorporated
Dive into ancient culinary arts with our lesson plan, "Ratios in Ancient Recipes." Perfect for 5th graders preparing for 6th grade, this engaging activity helps students master ratios and proportions through the fascinating lens of historical recipes from around the world. Starting with a comprehensive review, students will explore ratios and proportions before applying their knowledge to adjust recipes like Italian pasta sauce and Indian biryani. The adventure continues with challenging exte
5th - 6th
Math, Other (Math)
CCSS
5.NF.B.3
, 5.NF.B.4
, 6.RP.A.1
+3
Also included in: Review-5th Grade Concepts-Math Adventure - Journey to the Ancient World
$2.99
Original Price $2.99
Ratios, Rates, Proportions and Percents Unit 6th Grade Math Curriculum
Created by
Beyond the Worksheet with Lindsay Gould
⭐️⭐️⭐️ This ultimate 6th grade math ratios, rates, proportions and percent unit is COMPLETELY EDITABLE and is designed to help facilitate the needs of your students! ✅ This unit includes six multi-day lessons that cover the following skills: Ratios and Rates (including tape diagrams)Proportions and ScaleUnits of MeasurePercent of a NumberFractions, Decimals and Percents RelationshipsCompare and Order Fractions, Decimals and Percents✅ This unit includes:Google forms version of all assessmen
6th
Math, Other (Math)
CCSS
6.NS.C.7
, 6.RP.A.1
, 6.RP.A.2
+1
Also included in: 6th Grade Math Curriculum: Guided Notes, Practice, Assessments and More!
$35.99
Original Price $35.99
Rated 4.89 out of 5, based on 103 reviews
4.9 (103)
Ratios, Rates, Proportions and Percents Unit : 7th Grade Math Curriculum
Created by
Beyond the Worksheet with Lindsay Gould
⭐️⭐️⭐️ This COMPLETELY EDITABLE Ratios, Rates, Proportions and Percents Unit for 7th Grade Math is designed to help you best meet the needs of your students! ⭐️⭐️⭐️ ⭐️ This unit includes FOUR multi-day lessons that cover: ⭐️ ✅ Represent and Solve Proportional Relationships ✅ Identify and Compute Unit Rates ✅ Multi-Step Percent Problems ✅ Scale Drawings ⭐️ ALSO included in this resource: ⭐️ ✅ Google Forms versions of the assessments ✅ Weekly warm up recording sheets ✅ Weekly exit ticket s
7th
Algebra, Math
CCSS
7.G.A.1
, 7.RP.A.1
, 7.RP.A.2
+5
Also included in: 7th Grade Math Curriculum: Guided Notes, Practice, Assessments and More!
$22.99
Original Price $22.99
Rated 4.79 out of 5, based on 128 reviews
4.8 (128)
6th Grade Math Ratios and Proportions Curriculum Unit CCSS
Created by
Transforming the Middle
This 12 lesson CCSS aligned Ratios and Proportions Unit includes a full set of lessons/guided notes that cover all 6th grade RP standards, as well as a study guide and assessment. Topics included are ratio language, equivalent ratios, ratio tables, real world ratios, rates, unit rates, real world rates, unit rates involving conversions, percents, and real world percents. This unit has all instructional lessons needed to cover and teach the 6th grade ratios and proportional relationships standa
6th
Algebra, Math, Other (Math)
CCSS
6.RP.A.1
, 6.RP.A.2
, 6.RP.A.3
Also included in: 6th Grade Math Curriculum CCSS Aligned Bundle
$49.00Original Price $49.00
Price $41.00
Rated 5 out of 5, based on 13 reviews
5.0 (13)
6th Grade Ratios & Proportions Relationships Choice Board – Distance Learning
Created by
Simon Says School
Ratios and Proportional Relationships Enrichment Choice Board – Sixth Grade– This enrichment menu project is an amazing differentiation tool that not only empowers students through choice but also meets their individual needs. You will find that the ratio and proportions enrichment board contains three leveled activities for each standard: appetizer, entrée, and dessert. No two activities are alike!Students can either choose to complete activities from the project menu or design their own
6th
Algebra, Arithmetic, Math
CCSS
6.RP.A.1
, 6.RP.A.2
, 6.RP.A.3
+4
Also included in: 6th Grade Math Choice Boards Bundle - ALL STANDARDS - Distance Learning
$2.00
Original Price $2.00
Rated 4.79 out of 5, based on 130 reviews
4.8 (130)
7th Grade Ratios & Proportions Relationships Choice Board - Distance Learning
Created by
Simon Says School
Ratios and Proportions Relationships Enrichment Choice Board – Seventh Grade– This enrichment menu project is an amazing differentiation tool that not only empowers students through choice but also meets their individual needs. You will find that the ratio and proportions enrichment board contains three leveled activities for each standard: appetizer, entrée, and dessert. No two activities are alike!Students can either choose to complete activities from the project menu or design their own
7th
Algebra, Math, Other (Math)
CCSS
7.RP.A.1
, 7.RP.A.2
, 7.RP.A.2a
+4
Also included in: 7th Grade Math Choice Boards Bundle - ALL STANDARDS - Distance Learning
$2.00
Original Price $2.00
Rated 4.85 out of 5, based on 51 reviews
4.9 (51)
Proportional Reasoning with Ratios and Rates - Unit 5-6th Grade + Distance Learn
Created by
Proportional Reasoning with Ratios and Rates - Unit 5 - 6th Grade Math - Curriculum THIS FILE NOW CONTAINS THE PDF VERSION OF THIS PRODUCT PLUS A GOOGLE SLIDES VERSION FOR DISTANCE LEARNINGThis bundle pack contains Lesson Plans, Notes, INB pages, Homework, Quizzes, Activities, Study Guide, and a Unit Test. Topics Covered: • Converting Fractions to Decimals • Ratio Introduction • Ratio Models • Qualitative Reasoning • Quantitative Reasoning • Rates • Unit Rates and Unit Costs • Representing Ratio
6th - 8th
Fractions, Math, Other (Math)
$38.00Original Price $38.00
Price $28.50
Rated 4.95 out of 5, based on 28 reviews
5.0 (28)
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Folding and Unfolding a Net
Author:David Wees
Topic:Solids or 3D Shapes
I found this activity designed by Luxmath and buried in a forum, so I decided to publish it here. This could be an excellent tool to help students develop relationships between the nets of prisms and the appearance of those prisms in 3D space.
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Why do we not name $-a+bi$ in relation to $a+bi$?
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$\begingroup$
Given a complex number $a+bi$, it has a complex conjugate $a-bi$. The product of this complex number with its complex conjugate gives $(a+bi)(a-bi)=a^2+b^2$.
One might imagine flipping the sign of the real part instead of the imaginary part to get a sort of "anticonjugate", resulting in a similar product of $(a+bi)(-a+bi)=-(a^2+b^2)$.
Clearly this "anticonjugate" is the negative of the conjugate. I suspect I've never seen this concept before because of either (1) no finds this anticonjugate useful or (2) the negative of the complex conjugate is considered without giving it a special name.
Is there a different reason why we don't seem to use or consider "anticonjugates"? Or does the above account for this perception?
complex-numbers
terminology
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edited Jan 10, 2022 at 15:34
user21820
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asked Jan 10, 2022 at 6:56
GalenGalen
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$\endgroup$
2
4
$\begingroup$ Yes, as you say, your "anti-conjugate" is just minus the conjugate, I don't think there is any need to introduce a separate concept for that. $\endgroup$
Captain Lama
– Captain Lama
2022-01-10 07:01:24 +00:00
Commented Jan 10, 2022 at 7:01
1
$\begingroup$ Thank you for providing the background and motivation for your question and showing a possible answer to it. $\endgroup$
jimjim
– jimjim
2022-01-10 07:05:02 +00:00
Commented Jan 10, 2022 at 7:05
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2 Answers 2
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$\begingroup$
Complex conjugates are important because $i$ and $-i$ are fundamentally indistinguishable by definition; $i$ is defined to be a number satisfying the equation $i^2 = -1$, but of course $-i$ must satisfy the same equation. So "any fact" which can be stated about complex numbers must remain true if we swap all occurrences of $i$ with $-i$ (though one must take care with "hidden" occurrences). Complex conjugation is therefore a mapping of complex numbers which preserves many algebraic properties.
In contrast, complex anticonjugation as defined in your question does not preserve any useful properties, because $1$ and $-1$ are not fundamentally indistinguishable; $-1$ is not a successor of the number $0$, it is not a multiplicative identity such that $1x \equiv x$, nor does it satisfy any other reasonable definition of the number $1$.
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answered Jan 10, 2022 at 15:30
kaya3kaya3
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$\endgroup$
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$\begingroup$ More specifically and succinctly, complex conjugation is the [sic] non-trivial field automorphism of the complex numbers. (+1) $\endgroup$
Andrew D. Hwang
– Andrew D. Hwang
2022-01-10 17:40:39 +00:00
Commented Jan 10, 2022 at 17:40
$\begingroup$ This in my opinion, is the right answer; the other reply completely fails to explain why the conjugate is important and the anticonjugate isn't. $\endgroup$
MJD
– MJD
2022-01-11 02:06:42 +00:00
Commented Jan 11, 2022 at 2:06
1
$\begingroup$ @MJD This one here is a very good answer, which I had +1'd already. That said, I don't think my "other reply completely fails to explain" - at least not what I had set out to explain. There is a historical perspective to this, too, and complex conjugation first emerged for the purpose of solving cubic equations with real coefficients long before abstract algebra even existed. By the time automorphisms of $\mathbb C$ came to be studied, conjugation was proved to be (the only) one, but it was not brought around because of that. $\endgroup$
dxiv
– dxiv
2022-01-11 03:34:04 +00:00
Commented Jan 11, 2022 at 3:34
$\begingroup$ @dxiv Complex conjugates appear in pairs as roots because if one is a root of a real polynomial then the other must be too - so it's pretty much the same reason, just described at a different level of abstraction. $\endgroup$
kaya3
– kaya3
2022-01-11 08:02:55 +00:00
Commented Jan 11, 2022 at 8:02
2
$\begingroup$ @kaya3 That's correct, of course. However, the question here was not why conjugates are used and useful, but rather why "anti-conjugates" are not. The historical reason is that "anti-conjugates" would have been redundant, because there is only one such transformation needed to describe all complex manipulations, and that place was taken by conjugates from the very beginning. The original reason why conjugates were preferred was merely convenience related to solving the casus irreducibilis of cubics in the XVI$^{th}$ century. Deeper reasons were found later, but that wasn't my point. $\endgroup$
dxiv
– dxiv
2022-01-11 08:14:10 +00:00
Commented Jan 11, 2022 at 8:14
Add a comment |
6
$\begingroup$
There is a tradition in math to avoid (or, at least, minimize) redundancies.
In this case, complex conjugation $\,\overline{a+ib}=a-ib\,$ has been known and used for a long time. It has many applications, from polynomial equations to calculus, abstract algebra, geometry etc.
In contrast, the proposed "anti-conjugate", say we write it as $\,\widetilde{a+ib}=-a+ib\,$, would be a new concept, without any obvious advantage - conceptual or practical. Moreover, it can be easily expressed in terms of the conjugate as $\,\widetilde{a+ib}=-\,\overline{a+ib}=\overline{i\cdot\overline{i \cdot(a+ib)}}\,$. Thus, redundant.
[ EDIT ] $\;$ Side by side summary.
$$ \begin{matrix} & & & \small{\text{conjugate}}\;\overline{\,z\,} & & & \small{\text{anti-conjugate}}\; \widetilde{\,z\,} \ \small{\text{symmetry}} & & & \small{\text{over real axis}} & & & \small{\text{over imaginary axis}} \ \small{\text{involution}} & & \small{\text{yes:}} & \overline{\overline{\,z\,}} = z & & \small{\text{yes:}} & \widetilde{\widetilde{\,z\,}} = z \ \small{\text{distrib over +}} & & \small{\text{yes:}} & \overline{\,z_1+z_2\,} = \overline{\,z_1\,}+\overline{\,z_2\,} & & \small{\text{yes:}} & \widetilde{\,z_1+z_2\,} = \widetilde{\,z_1\,}+\overline{\,z_2\,} \ \small{\text{distrib over }\times} & & \small{\text{yes:}} & \overline{\,z_1\,\cdot\,z_2\,} = \overline{\,z_1\,}\,\cdot\,\overline{\,z_2\,} & & \color{red}{\small{\text{no:}}} & \widetilde{\,z_1\,\cdot\,z_2\,} \ne \widetilde{\,z_1\,}\,\cdot\,\widetilde{\,z_2\,} \end{matrix} $$
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edited Jan 10, 2022 at 18:37
answered Jan 10, 2022 at 8:29
dxivdxiv
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$\endgroup$
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$\begingroup$ I feel that it is quite a stretch to call the last expression "easily expressed" :) $\endgroup$
lisyarus
– lisyarus
2022-01-10 08:44:23 +00:00
Commented Jan 10, 2022 at 8:44
$\begingroup$ @lisyarus Granted ;-) though it could be shortened to $\,\overline{\;z\;} = f(z)\,$, $\,\widetilde{\;z\;}=f(i\,f(i\,z))\,$. $\endgroup$
dxiv
– dxiv
2022-01-10 08:51:57 +00:00
Commented Jan 10, 2022 at 8:51
$\begingroup$ If you're multiplying by $i$, you may as well multiply by $-1$ (which is only $i^2$ anyway), without calling $f$ twice. $\endgroup$
J.G.
– J.G.
2022-01-10 09:33:42 +00:00
Commented Jan 10, 2022 at 9:33
1
$\begingroup$ @lisyarus: It seems that it is not easy because it is wrong! Lol! c(i
c(i
(a+b·i))) = c(i·c(a·ib)) = c(i·(a·ib)) = c(ab·i) = a+b·i and we are back to the start!! In fact, we don't have to do any computation to know that it is wrong, because no sequence comprising rotations and an even number of reflections can ever be equivalent to a single reflection. $\endgroup$
user21820
– user21820
2022-01-11 18:41:42 +00:00
Commented Jan 11, 2022 at 18:41
1
$\begingroup$ @user21820 Oh, indeed! A nice geometric argument, by the way. $\endgroup$
lisyarus
– lisyarus
2022-01-12 09:40:54 +00:00
Commented Jan 12, 2022 at 9:40
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187775 | https://math.answers.com/basic-math/What_is_the_fraction_of_5_eights_of_56 | What is the fraction of 5 eights of 56? - Answers
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What is the fraction of 5 eights of 56?
Anonymous ∙ 12 y ago
Updated: 4/28/2022
fraction of 5/8 of 56 = 35/1
Wiki User ∙ 12 y ago
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### What is six eights as a fraction? Six eights is 48 = 48/1
### What fraction is greater three fifths or five eights? LCD(5, 8) = 40.So 3/5 = 24/40 and 5/8 = 25/40. The second fraction, that is 5/8, is greater.
### How do you know if a fraction is less than another fraction? You don't know!!!! Compare 5/7 & 7/8 ; which greater? To compare , we bring to a common denominator of 56 ( 7 x 8 ) in this case. We then make the numerators the equivalent fraction. Hence 5/7 = 5(8)/7(8) & 7/8 = 7(7)/8(7) 5/7 = 40/56 & 7/8 = 49/56 We now compare numerators, because the denominators are the same value at '56'. (top number) 40 < 49 Hence 40/56 < 49/56 It follows that 5/7 < 7/8
### What is equivalent to the fraction 56? 11
### What is the fraction value of .56? It is 56/100, which can be simplified.
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187776 | https://www.cancerresearchuk.org/about-cancer/gestational-trophoblastic-disease-gtd/about | What is gestational trophoblastic disease | Cancer Research UK
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HomeAbout cancerGestational trophoblastic disease (GTD)What is gestational trophoblastic disease?
What is gestational trophoblastic disease?
Gestational trophoblastic disease (GTD) is the name for abnormal cells or tumours that grow from the tissue that forms in the womb during pregnancy.
GTD is very rare. It can be non cancerous (benign) or cancerous (malignant).
There are different types of GTD. These include:
molar pregnancy (complete or partial)
invasive mole or persistent trophoblastic disease (PTD)
choriocarcinoma
placental site trophoblastic tumour (PSTT)
epithelioid trophoblastic tumour (ETT)
atypical placental site nodule (APSN)
Although GTD starts in the womb, it behaves very differently from cancer of the womb. It is also treated differently.
If you are looking for information about womb cancer go to our womb cancer section. Womb cancer is also known as uterine or endometrial cancer.
Find out about womb cancer
How gestational trophoblastic diseases develop
Gestation means pregnancy. Trophoblasts are the cells that form during the normal development of a baby.
Usually, after a sperm fertilises an egg, new cells grow within the womb to form an embryo. As the embryo grows, its cells start to specialise. Some cells start to form the baby (foetus) and others form the placenta. The placenta protects and nourishes the baby during pregnancy.
The first layer of cells that develops into the placenta is called the trophoblast. The trophoblast produces tiny, finger-like, outgrowths known as villi. These villi attach the placenta to the lining of the womb.
Molar pregnancy
There is a problem when the sperm fertilises the egg. The foetus either doesn’t develop at all, or it partly forms but can't grow normally. The villi may swell up and grow in clusters, a bit like bunches of grapes. Unfortunately a molar pregnancy cannot develop into a healthy baby.
Molar pregnancy is the most common type of GTD and can be either partial or complete. They are not cancerous – they are benign. But rarely, a molar pregnancy can become cancerous and can spread to other parts of the body.
Find out more about molar pregnancy
Invasive mole and choriocarcinoma
Some trophoblastic cells grow abnormally and develop into a tumour. These tumours are cancerous and can sometimes spread outside the womb.
An invasive mole is a cancer that can form in the womb after a molar pregnancy. It is also called persistent trophoblastic disease (PTD).
Choriocarcinoma is a very rare cancer that can occur after a:
normal pregnancy
molar pregnancy
miscarriage
termination of pregnancy (abortion)
Treatment is usually with chemotherapy.
Find out more about invasive mole and choriocarcinoma
Placental site trophoblastic tumours (PSTT) and epithelioid trophoblastic tumours (ETT)
These cancers can occur several months, or even years after a pregnancy. They can happen after any type of pregnancy. This includes molar pregnancy, miscarriage or a full term normal pregnancy.
They develop in the area where the placenta joined the lining of the womb (uterus). They can grow into the muscle layer of the womb and can sometimes spread to other parts of the body.
Treatment is usually surgery to remove the womb. Some people may also have chemotherapy.
Find out more about placental site trophoblastic tumours and epithelioid trophoblastic tumours
Other terms used to describe gestational trophoblastic disease (GTD)
The medical descriptions for gestational trophoblastic disease are all quite long. And they can sound complicated. There are many terms that may be used for these types of tumours. These include:
gestational trophoblastic tumour (GTT)
trophoblastic disease
gestational tumour
gestational trophoblastic neoplasia (GTN)
molar pregnancy (partial or complete)
invasive mole
persistent trophoblastic disease (PTD)
choriocarcinoma
atypical placental site nodule (APSN)
placental site trophoblastic tumour (PSTT)
epithelioid trophoblastic tumour (ETT)
Treatment
Treatment for all types of GTD is very successful and most women are cured.
You can read more about treatment in the section or page about each type of GTD.
Find information about the different types of GTD
Research
Doctors are always looking to improve treatments for GTD, and reduce side effects. As part of your treatment, your doctor may ask you to take part in a clinical trial. This might be to test a new treatment. Or to look at different combinations of existing treatments.
To make sure the research is accurate, trials have certain entry conditions for who can take part. These are different for each trial.
References
Diagnosis and management of gestational trophoblastic disease: 2025 update
H Y S Ngan and others
The International Journal of Gynecology & Obstetrics. 2025
The Management of Gestational Trophoblastic Disease(4th edition)
Royal College of Obstetricians and Gynaecologists, September 2020
Molar pregnancies
Antonio Braga and others
BMJ best practice, Last reviewed:August 2025
Gestational Trophoblastic Disease (5th Edition)
International Society for the Study of Trophoblastic Diseases, 2022
Practical Guidelines for the Treatment of Gestational Trophoblastic Disease: Collaboration of the European Organisation for the Treatment of Trophoblastic Disease (EOTTD)–European Society of Gynaecologic Oncology (ESGO)–Gynecologic Cancer InterGroup (GCIG)–International Society for the Study of Trophoblastic Diseases (ISSTD)
C Lok and others
Journal of Clinical Oncology, 2025
Last reviewed:
26 Sep 2025
Next review due:
26 Sep 2028
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Molar pregnancy
Invasive mole and choriocarcinoma
Placental site trophoblastic tumour and epithelioid trophoblastic tumour
Mental health and cancer
We know that it is common to struggle with your mental health when you have cancer or care for someone with cancer.
We have information to help
Related links
Molar pregnancy
In a molar pregnancy the fertilisation of the egg by the sperm goes wrong and creates abnormal cells or clusters of water filled sacs inside the womb.Molar pregnancies are not cancer (they are benign).
Invasive mole and choriocarcinoma
Invasive mole and choriocarcinoma are very rare types of cancer that can occur after pregnancy. They are types of gestational trophoblastic disease (GTD).
Placental site trophoblastic tumour and epithelioid trophoblastic tumour
Placental site trophoblastic tumours (PSTTs) and epithelioid trophoblastic tumours (ETTs) happen after pregnancy. They are extremely rare and are slow growing.
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187777 | http://users.math.uoc.gr/~pamfilos/eGallery/problems/ParabolaChords.html | Parabola Chords
Consider the parabola with directrix line (a) and focus F. The following properties are valid. The line DG joining D to the middle G of PP' is parallel to the axis of the parabola. The intersection point E of the parabola with DG is the middle of DG. The tangent at E is parallel to the chord PP' and QQ' is the half of PP'. The middle points G of all chords PP', parallel to a given direction e are on a line DG parallel to the axis of the parabola. The previous line DG is the parallel to the axis going through the tangent point E with a line parallel to the direction of PP'. The line HF is orthogonal to the chord direction. The segments on the chord UV, U'V' between the parabola and the tangents are equal. The segments UW, U'W' are also equal. Triangles DFP and DFP' are similar and DF2 = FPFP'. For triangles like DQQ', formed by tangents, their circumcircle passes through the focus F. The Simson-Wallace line of F with respect to triangle DQQ' is the tangent at the vertex of the parabola. The orthocenter of DQQ' is on the directrix of the parabola. DF is the bisector of angle(PFP'), which is twice the angle(PDP'). Tangents DP, DP' are orthogonal exactly when D is on the directrix (a) of the parabola. Then PP' passes through F. Inversely, if PP' passes through F then the tangents at P, P' are orthogonal and meet on (a). DF is the symmedian of triangle DPP' from D.
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Because PD, PD' are medial lines of AF, A'F, D is the circumcenter of the triangle AFA'. D projects on the middle H of AA'. Apply to the tangent pairs QE, QP and Q'E, Q'P'. R, R' are the middles of GP, GP' hence QQ' joins the middles of PD, DP'. Follows from . DG passes through the fixed point E, where the tangent is parallel to the direction e. Its direction is orthogonal to (a). See above. The tangent at E is medial line of HF. UU' and VV' have a common point as their middle. Follows from the previous. Follows from the facts: angle(DFP') = angle(DFP) = angle(DAP). angle(S'DF) = angle(S'SF) = angle(HAF) = angle(APS) = angle(SPF). The second equality because of the cyclicity of DSFS' and the parallelity of SS' to AA'. Follows from the previous, since angle(Q'FS') = angle(QFS) => angle(QFQ')+angle(QDQ') = pi. Notice also that triangle QFQ' is similar to PFD, DFP', since FQ, FQ' are the medians of the two similar triangles. Follows from the fact that the Simson-Wallace line of DQQ' with respect to F passes through points {S,S'}. The locus of these points is the tangent at the vertex. The statement on the orthocenter follows from the characteristic property of the Steiner line (see ). Follows easily from . is a consequence of since angle(ADA') = angle(PFP'). Follows also from , since the ratio FS/FS = DP'/DP, which is characteristic for the symmedian from D. Notice that and identify F with a vertex of the second Brocard triangle and the parabola with one of the Artzt parabolas (first kind) of triangle DPP' These properties are used in the discussion about the determination of the parabola in oblique axes (see ). Look at the file for the Archimedean method to calculate the area of a parabola sector. The file contains a discussion on the parabola tangent to four given lines.
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187778 | https://www.cs.ucr.edu/~neal/Slides/Karger99Rounding/Karger99Rounding_slides.pdf | A 12/11-Approximation Algorithm for Minimum 3-Way Cut David Karger (MIT), Phillip Klein (Brown), Cliff Stein (Columbia) Mikkel Thorup (AT&T), Neal Young (UCR) ``As the field of approximation algorithms matures, methodologies are emerging that apply broadly to many NP-hard optimization problems. One such approach has been the use of metric and geometric embeddings in addressing graph optimization problems. Faced with a discrete graph optimization problem, one formulates a relaxation that maps each graph node into a metric or geometric space, which in turn induces lengths on the graph’s edges. One solves this relaxation optimally and then derives from the relaxed solution a near-optimal solution to the original problem.’’ problem: 3-way cut • input: undirected graph, three terminal nodes • output: three-way cut (subset of edges whose removal separates the terminals) • objective: minimize number of edges cut • NP-HARD 3-way cut 3-way cut 1. Embed graph into triangle.
2. Cut triangle using randomized cutting scheme. ... induces cut of embedded graph.
Approach [Calinescu et al, 1998] goal: Bound expected number of edges cut.
1.
2.
Step 1: embedding a. Assign vertices to points in the triangle.
b. Constrain each terminal to a corner.
c. Minimize sum of edge lengths (L1 metric).
• Optimal embedding via linear program.
• Value of LP is at most |optimal 3-cut| .
LP for finding optimal embedding minimize 1 2 ∑ (u,v)∈E duv (xt1,yt1,zt1)=(1,0,0) (xt2,yt2,zt2)=(0,1,0) (xt3,yt3,zt3)=(0,0,1) (∀u) xu +yu +zu=1 (∀u,v) duv≥|xu −xv|+|yu −yv|+|zu −zv| Each vertex u is mapped to a point (xu, yu, zu), determined by the LP , to minimize sum of embedded edge lengths.
Embedding (animated) Step 2: cutting the triangle (Calinescu et al’s scheme) a. Choose 2 of 3 sides randomly.
b. Choose a random slice parallel to each sides.
Pr[ edge (u,v) cut ] ≤ (4/3) duv a. Pr[ cut by red ] = (2/3) duv b. Pr[ cut by green ] = (2/3) duv c. Pr[ cut ] ≤ 2×(2/3) duv d Expected #edges cut ≤ 4/3 OPT lemma: corollary: Pr[edge (u,v) cut] ≤4 3duv expected number of edges cut ≤4 3 ∑ (u,v)∈E duv = 4 3 |value of LP| ≤4 3 |optimal 3-cut| Better cutting scheme (probability 8/11) (probability 3/11) or ball cut corner cut i. Choose random point on star ii. Choose three of six rays parallel to sides Ball cut density of ball cut slices 3/2 x 1/2 x 2 x 8/11 = 12/11 2/3 -- density of horizontal slice -- only one of two rays (red or green) -- segment can be cut from two orientations -- probability of ball cut distribution of slices made by ball cuts Expected #edges cut ≤ 12/11 OPT Pr[edge (u,v) cut] ≤12 11duv lemma: corollary: expected number of edges cut ≤12 11 ∑ (u,v)∈E duv = 12 11 |value of LP| ≤12 11 |optimal 3-cut| More • Generalizes to K-way cut (ratio < 1.34...) • K=3 case done also by Cunningham and Tang • Meta-problem of finding an optimal cutting scheme can be formulated as an infinite LP!
• For K=3, no better cutting scheme for this LP relaxation is possible. Would need better relaxation to improve result.
• K > 3 much harder. Improve constant?
probability that edge is cut |
187779 | https://math.stackexchange.com/questions/2875142/pigeonhole-principle-based-algorithm | combinatorics - Pigeonhole principle based algorithm - Mathematics Stack Exchange
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Pigeonhole principle based algorithm
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I was trying to solve this problem.
INMO-2015 P6. From a set of 11 11 square integers, show that one can choose 6 6 numbers a 2,b 2,c 2,d 2,e 2,f 2 a 2,b 2,c 2,d 2,e 2,f 2 such that a 2+b 2+c 2≡d 2+e 2+f 2 mod 12 a 2+b 2+c 2≡d 2+e 2+f 2 mod 12.
I came up with the following solution.
We note that there are 4 distinct residues modulo 12 12, namely {0,1,4,9}{0,1,4,9}.
LEMMA:
Among any 5 squares we can find a pair that is congruent modulo 12. Since there are atmost 4 unique residues mod 12, by the Pigeonhole Principle we must have two squares with the same residue.
We describe the following algorithm for finding a,b,c,d,e,f a,b,c,d,e,f.
Label the set of given 11 squares Z={Z 1,Z 2,Z 3…}Z={Z 1,Z 2,Z 3…} and let R={},L={}R={},L={}.
Consider the set of the last 5 elements. Among these elements, we may choose two elements with the same residue. Put one of these in R R and the other in L L
Remove the elements from Z Z.
Repeat step 2 two more times. We can do this because each step removes two elements, and we need remove only 6 elements so at each step we still have 5 elements to work with.
In fact by my count, we can do this one more time and end up with 4 integers in R,L R,L each.
My issues
Is an algorithm a valid proof style?
If it is, any tips on writing it better?
Is the 4th integer possible?
Thanks in advance!
combinatorics
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edited Aug 7, 2018 at 16:31
darij grinberg
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asked Aug 7, 2018 at 16:25
Damien AshwoodDamien Ashwood
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2 Everything you've done is correct as far as I can see, including the fourth pair. A correct algorithm is certain a correct proof. I think your proof is superior to the one given in the link.saulspatz –saulspatz 2018-08-07 16:34:27 +00:00 Commented Aug 7, 2018 at 16:34
"We note that there are 4 distinct residues modulo 12" Nitpick. There are 4 distinct square residues. There are, of course, 12 distinct residues.fleablood –fleablood 2018-08-07 17:08:18 +00:00 Commented Aug 7, 2018 at 17:08
Of course. Definitely not a nitpick.Damien Ashwood –Damien Ashwood 2018-08-07 17:58:03 +00:00 Commented Aug 7, 2018 at 17:58
It's not bad. I might put it this way: Using the set-theorists' notation [C]2[C]2 to denote the set of all two-element subsets of any set C:C: Let S S be the set of the 11 11 squares. Any T⊂S T⊂S with at least 5 5 members has at least 2 2 members congruent to each other mod 12.12. So let A 0=∅A 0=∅ and for 1≤j≤3 1≤j≤3 let A j={u j,v j}∈[S∖∪j−1 i=0 A i]2 A j={u j,v j}∈[S∖∪i=0 j−1 A i]2 with u j≡v j mod 12.u j≡v j mod 12.... (I'm Canadian . "Not bad" means "quite good".)DanielWainfleet –DanielWainfleet 2018-08-07 20:24:02 +00:00 Commented Aug 7, 2018 at 20:24
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Your argument is correct and nice.
I would generalize it: "Let S S be a finite set. Let d d and k k be two nonnegative integers. Let ∼∼ be an equivalence relation on S S such that there are at most d d equivalence classes with respect to ∼∼. Assume that |S|≥d+2 k−1|S|≥d+2 k−1. Then, we can find 2 k 2 k distinct elements s 1,s 2,…,s k,t 1,t 2,…,t k s 1,s 2,…,s k,t 1,t 2,…,t k of S S such that s 1∼t 1 s 1∼t 1, s 2∼t 2 s 2∼t 2, ..., s k∼t k s k∼t k." Now that you have turned the k k into a proper variable, you can do induction over k k; this gives a clearer writeup than the algorithm (but of course is just a recursive rewording of the latter).
Yes, because 11≥4+2⋅4−1 11≥4+2⋅4−1. Note that introducing a 4 4-th integer into the original problem does not make it stronger, because a 2+b 2+c 2+d 2≡e 2+f 2+g 2+h 2 mod 12 a 2+b 2+c 2+d 2≡e 2+f 2+g 2+h 2 mod 12 does not imply a 2+b 2+c 2≡e 2+f 2+g 2 mod 12 a 2+b 2+c 2≡e 2+f 2+g 2 mod 12. Only once you have strengthened a 2+b 2+c 2≡e 2+f 2+g 2 mod 12 a 2+b 2+c 2≡e 2+f 2+g 2 mod 12 to a≡e mod 12,b≡f mod 12,c≡g mod 12 a≡e mod 12,b≡f mod 12,c≡g mod 12 does the 4 4-th integer become an obvious boon.
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answered Aug 7, 2018 at 16:37
darij grinbergdarij grinberg
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Once you know where you are going you can rewrite the proof to be cleaner and crisper.
Point 1: As there are only 4 4 distinct square residues mod 12 mod 12 (0 2≡6 2≡0;1 2≡5 2≡7 2≡11 2≡1;2 2≡4 2≡8 2≡10 2≡4;3 2≡9 2≡9 0 2≡6 2≡0;1 2≡5 2≡7 2≡11 2≡1;2 2≡4 2≡8 2≡10 2≡4;3 2≡9 2≡9) then
Point 2: Given any group of 5 5 or more integers at least 2 2 when squared will belong to the same residue (one of four possible residues) by the pigeon hole principal.
Point 3: Of the eleven integers we can label and sort them as {x 11,x 10,x 9,....,x 1}{x 11,x 10,x 9,....,x 1} so that x 11 x 11 and x 10 x 10 are selected from {x 11,x 10,x 9,....,x 1}{x 11,x 10,x 9,....,x 1}, which has more than 5 5, so that x 2 11≡x 2 10 mod 12 x 11 2≡x 10 2 mod 12. And so long as 2 m+1≥5 2 m+1≥5 (i.e. m≥2 m≥2), x 2 m+1 x 2 m+1 and x 2 m x 2 m are chosen from {x 2 m+1,....,x 1}{x 2 m+1,....,x 1} so that x 2 2 m+1≡x 2 2 m mod 12 x 2 m+1 2≡x 2 m 2 mod 12.
Conclusion: So for m=2,3,4,5 m=2,3,4,5 we have x 2 2 m+1≡x 2 2 m x 2 m+1 2≡x 2 m 2 and therefore:
x 2 4+x 2 6+x 2 8≡x 2 5+x 2 7+x 2 9 x 4 2+x 6 2+x 8 2≡x 5 2+x 7 2+x 9 2;
x 2 4+x 2 6+x 2 8+x 2 10≡x 2 5+x 2 7+x 2 9+x 2 10 x 4 2+x 6 2+x 8 2+x 10 2≡x 5 2+x 7 2+x 9 2+x 10 2
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answered Aug 7, 2018 at 18:16
fleabloodfleablood
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187780 | https://pressbooks.bccampus.ca/physicsforlifesciences1phys1108/chapter/5-6-flagella-and-the-cytoskeleton/ | 5.6 Flagella and the Cytoskeleton from OpenStax Biology and Anatomy and Physiology – x-2019-Douglas College Physics 1108 Physics for the Life Sciences
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Contents
Welcome and Territorial Acknowledgement
Preface to College Physics - the Basis for this Textbook
Introduction to Open Textbooks
Physics 1108 Curriculum Guidelines - Course Outline
Scholarly Research
Chapter 1 The Nature of Science and Physics
1.0 Introduction
1.1 Physics: An Introduction
1.2 Physical Quantities and Units
1.3 Accuracy, Precision, and Significant Figures
1.4 Approximation
1.5 Introduction to Measurements
1.6 Expressing Numbers
1.7 Significant Figures
1.8 Converting Units
1.9 Other Units: Temperature and Density
1.10 Expressing Units
1.11 Additional Exercises
Chapter 2 One-Dimensional Kinematics
2.0 Introduction
2.1 Displacement
2.2 Vectors, Scalars, and Coordinate Systems
2.3 Time, Velocity, and Speed
2.4 Acceleration
2.5 Graphical Analysis of One-Dimensional Motion
2.6 Motion Equations for Constant Acceleration in One Dimension
2.7 Problem-Solving Basics for One-Dimensional Kinematics
2.8 Falling Objects
2.9 New - Motion with a non-constant acceleration - Derivatives
2.10 New - anti-derivatives and motion
Chapter 3 Two-Dimensional Kinematics
3.0 Introduction
3.1 Kinematics in Two Dimensions: An Introduction
3.2 Vector Addition and Subtraction: Graphical Methods
3.3 Vector Addition and Subtraction: Analytical Methods
3.4 Projectile Motion
3.5 Addition of Velocities
Chapter 4 Dynamics: Force and Newton's Laws of Motion
4.0 Introduction
4.1 Development of Force Concept
4.2 Hooke's Law (Originally Section 5.3 Elasticity: Stress and Strain)
4.3 Newton’s First Law of Motion: Inertia
4.4 Newton’s Second Law of Motion: Concept of a System
4.5 Newton’s Third Law of Motion: Symmetry in Forces
4.6 Normal, Tension, and Other Examples of Forces
4.7 Problem-Solving Strategies
4.8 Further Applications of Newton’s Laws of Motion
Chapter 5 Further Applications of Newton's Laws: Friction, Drag
5.0 Introduction
5.1 Friction
5.2 Drag Forces
5.3 Drag Forces on the Human Body
5.4 Lab Activity : Testing a terminal speed hypothesis using coffee filters
5.5 Viscous forces in cells - Flagellum
5.6 Flagella and the Cytoskeleton from OpenStax Biology and Anatomy and Physiology
Chapter 6 Uniform Circular Motion and Gravitation
6.0 Introduction
6.1 Rotation Angle and Angular Velocity
6.2 Centripetal Acceleration
6.3 Centripetal Force
6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
6.5 Newton’s Universal Law of Gravitation
Chapter 7 Work, Energy, and Energy Resources
7.0 Introduction
7.1 Work: The Scientific Definition
7.2 Kinetic Energy and the Work-Energy Theorem
7.3 Gravitational Potential Energy
7.4 Conservative Forces and Potential Energy
7.5 Nonconservative Forces
7.6 Conservation of Energy
7.7 Power
7.8 Work, Energy, and Power in Humans
7.9 World Energy Use
Chapter 8 Linear Momentum and Collisions
8.0 Introduction
8.1 Linear Momentum and Force
8.2 Impulse
8.3 Conservation of Momentum
8.5 Inelastic Collisions in One Dimension
8.6 Collisions of Point Masses in Two Dimensions
Chapter 9 Statics and Torque
9.0 Introduction
9.1 The First Condition for Equilibrium
9.2 The Second Condition for Equilibrium
9.3 Stability
9.4 Applications of Statics, Including Problem-Solving Strategies
9.5 Simple Machines
9.6 Forces and Torques in Muscles and Joints
Chapter 10 Rotational Motion and Angular Momentum
10.0 Introduction
10.1 Angular Acceleration
10.2 Kinematics of Rotational Motion
10.3 Dynamics of Rotational Motion: Rotational Inertia
10.4 Rotational Kinetic Energy: Work and Energy Revisited
10.5 Angular Momentum and Its Conservation
10.6 Collisions of Extended Bodies in Two Dimensions
10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
Chapter 13 Temperature, Kinetic Theory, and the Gas Laws
13.0 Introduction
13.1 Temperature
Chapter 15 Thermodynamics
15.0 Introduction
15.1 The First Law of Thermodynamics
15.2 The First Law of Thermodynamics and Some Simple Processes
Chapter 16 Oscillatory Motion and Waves
16.0 Introduction
16.1 Hooke’s Law: Stress and Strain Revisited
16.2 Period and Frequency in Oscillations
16.3 Simple Harmonic Motion: A Special Periodic Motion
16.4 The Simple Pendulum
16.5 Energy and the Simple Harmonic Oscillator
16.6 Uniform Circular Motion and Simple Harmonic Motion
16.7 Damped Harmonic Motion
16.8 Forced Oscillations and Resonance
16.9 Waves
16.10 Superposition and Interference
16.11 Energy in Waves: Intensity
Chapter 17 Physics of Hearing
17.0 Introduction
17.1 Sound
17.2 Speed of Sound, Frequency, and Wavelength
17.3 Sound Intensity and Sound Level
17.4 Doppler Effect and Sonic Booms
17.5 Sound Interference and Resonance: Standing Waves in Air Columns
17.6 Hearing
17.7 Ultrasound
Floating and Sinking - Buoyancy
Introduction to Floating and Sinking
Definitions - Solids Liquids and Gases
Appendix
Appendix A Useful Information - Constants, Units, Formulae
Appendix B Useful Mathematics (originally from Open Stax Chemistry)
Appendix C: Periodic Table of the Elements from Open Stax Chemistry 1st Canadian Edition
David W. Ball
25. Appendix D Atomic Masses
26. Appendix E: Half-lives for Several Radioactive Isotopes (Originally from OpenStax Chemistry)
27. Appendix F Glossary of Key Symbols and Notation
x-2019-Douglas College Physics 1108 Physics for the Life Sciences
Chapter 5 Further Applications of Newton’s Laws: Friction, Drag
5.6 Flagella and the Cytoskeleton from OpenStax Biology and Anatomy and Physiology
Learning Objectives
By the end of this section, you will be able to do the following:
Describe the cytoskeleton
Compare the roles of microfilaments, intermediate filaments, and microtubules
Compare and contrast cilia and flagella
Summarize the differences among the components of prokaryotic cells, animal cells, and plant cells
If you were to remove all the organelles from a cell, would the plasma membrane and the cytoplasm be the only components left? No. Within the cytoplasm, there would still be ions and organic molecules, plus a network of protein fibers that help maintain the cell’s shape, secure some organelles in specific positions, allow cytoplasm and vesicles to move within the cell, and enable cells within multicellular organisms to move. Collectively, scientists call this network of protein fibers the cytoskeleton. There are three types of fibers within the cytoskeleton: microfilaments, intermediate filaments, and microtubules (Figure 4.22). Here, we will examine each.
Figure 4.22 Microfilaments thicken the cortex around the cell’s inner edge. Like rubber bands, they resist tension. There are microtubules in the cell’s interior where they maintain their shape by resisting compressive forces. There are intermediate filaments throughout the cell that hold organelles in place.
Microfilaments
Of the three types of protein fibres in the cytoskeleton, microfilaments are the narrowest. They function in cellular movement, have a diameter of about 7 nm, and are comprised of two globular protein intertwined strands, which we call actin (Figure 4.23). For this reason, we also call microfilaments actin filaments.
ATP powers actin to assemble its filamentous form, which serves as a track for the movement of a motor protein we call myosin. This enables actin to engage in cellular events requiring motion, such as cell division in eukaryotic cells and cytoplasmic streaming, which is the cell cytoplasm’s circular movement in plant cells. Actin and myosin are plentiful in muscle cells. When your actin and myosin filaments slide past each other, your muscles contract.
Microfilaments also provide some rigidity and shape to the cell. They can depolymerize (disassemble) and reform quickly, thus enabling a cell to change its shape and move. White blood cells (your body’s infection-fighting cells) make good use of this ability. They can move to an infection site and phagocytize the pathogen.
Intermediate Filaments
Several strands of fibrous proteins that are wound together comprise intermediate filaments (Figure 4.24). Cytoskeleton elements get their name from the fact that their diameter, 8 to 10 nm, is between those of microfilaments and microtubules.
Figure 4.24 Intermediate filaments consist of several intertwined strands of fibrous proteins.
Intermediate filaments have no role in cell movement. Their function is purely structural. They bear tension, thus maintaining the cell’s shape, and anchor the nucleus and other organelles in place. Figure 4.22 shows how intermediate filaments create a supportive scaffolding inside the cell.
The intermediate filaments are the most diverse group of cytoskeletal elements. Several fibrous protein types are in the intermediate filaments. You are probably most familiar with keratin, the fibrous protein that strengthens your hair, nails, and the skin’s epidermis.
Microtubules – Flagella
As their name implies, microtubules are small hollow tubes. Polymerized dimers of α-tubulin and β-tubulin, two globular proteins, comprise the microtubule’s walls (Figure 4.25). With a diameter of about 25 nm, microtubules are cytoskeletons’ widest components. They help the cell resist compression, provide a track along which vesicles move through the cell, and pull replicated chromosomes to opposite ends of a dividing cell. Like microfilaments, microtubules can disassemble and reform quickly.
Figure 4.25 Microtubules are hollow. Their walls consist of 13 polymerized dimers of α-tubulin and β-tubulin (right image). The left image shows the tube’s molecular structure.
Microtubules are also the structural elements of flagella, cilia, and centrioles (the latter are the centrosome’s two perpendicular bodies). In animal cells, the centrosome is the microtubule-organizing center. In eukaryotic cells, flagella and cilia are quite different structurally from their counterparts in prokaryotes, as we discuss below.
Flagella and Cilia
The flagella (singular = flagellum) are long, hair-like structures that extend from the plasma membrane and enable an entire cell to move (for example, sperm, Euglena, and some prokaryotes). When present, the cell has just one flagellum or a few flagella. However, when cilia (singular = cilium) are present, many of them extend along the plasma membrane’s entire surface. They are short, hair-like structures that move entire cells (such as paramecia) or substances along the cell’s outer surface (for example, the cilia of cells lining the Fallopian tubes that move the ovum toward the uterus, or cilia lining the cells of the respiratory tract that trap particulate matter and move it toward your nostrils.)
Despite their differences in length and number, flagella and cilia share a common structural arrangement of microtubules called a “9 + 2 array.” This is an appropriate name because a single flagellum or cilium is made of a ring of nine microtubule doublets, surrounding a single microtubule doublet in the center (Figure 4.26).
Figure 4.26This transmission electron micrograph of two flagella shows the microtubules’ 9 + 2 array: nine microtubule doublets surround a single microtubule doublet. (credit: modification of work by Dartmouth Electron Microscope Facility, Dartmouth College; scale-bar data from Matt Russell)
From Anatomy and Physiology
The Cytoskeleton
Much like the bony skeleton structurally supports the human body, the cytoskeleton helps the cells to maintain their structural integrity. The cytoskeleton is a group of fibrous proteins that provide structural support for cells, but this is only one of the functions of the cytoskeleton. Cytoskeletal components are also critical for cell motility, cell reproduction, and transportation of substances within the cell.
The cytoskeleton forms a complex thread-like network throughout the cell consisting of three different kinds of protein-based filaments: microfilaments, intermediate filaments, and microtubules (Figure 3.18). The thickest of the three is the microtubule, a structural filament composed of subunits of a protein called tubulin. Microtubules maintain cell shape and structure, help resist compression of the cell, and play a role in positioning the organelles within the cell. Microtubules also make up two types of cellular appendages important for motion: cilia and flagella. Cilia are found on many cells of the body, including the epithelial cells that line the airways of the respiratory system. Cilia move rhythmically; they beat constantly, moving waste materials such as dust, mucus, and bacteria upward through the airways, away from the lungs and toward the mouth. Beating cilia on cells in the female fallopian tubes move egg cells from the ovary towards the uterus. A flagellum (plural = flagella) is an appendage larger than a cilium and specialized for cell locomotion. The only flagellated cell in humans is the sperm cell that must propel itself towards female egg cells.
Figure 3.18 The Three Components of the Cytoskeleton The cytoskeleton consists of (a) microtubules, (b) microfilaments, and (c) intermediate filaments. The cytoskeleton plays an important role in maintaining cell shape and structure, promoting cellular movement, and aiding cell division.
A very important function of microtubules is to set the paths (somewhat like railroad tracks) along which the genetic material can be pulled (a process requiring ATP) during cell division, so that each new daughter cell receives the appropriate set of chromosomes. Two short, identical microtubule structures called centrioles are found near the nucleus of cells. A centriole can serve as the cellular origin point for microtubules extending outward as cilia or flagella or can assist with the separation of DNA during cell division. Microtubules grow out from the centrioles by adding more tubulin subunits, like adding additional links to a chain.
In contrast with microtubules, the microfilament is a thinner type of cytoskeletal filament (see Figure 3.18b). Actin, a protein that forms chains, is the primary component of these microfilaments. Actin fibers, twisted chains of actin filaments, constitute a large component of muscle tissue and, along with the protein myosin, are responsible for muscle contraction. Like microtubules, actin filaments are long chains of single subunits (called actin subunits). In muscle cells, these long actin strands, called thin filaments, are “pulled” by thick filaments of the myosin protein to contract the cell.
Actin also has an important role during cell division. When a cell is about to split in half during cell division, actin filaments work with myosin to create a cleavage furrow that eventually splits the cell down the middle, forming two new cells from the original cell.
The final cytoskeletal filament is the intermediate filament. As its name would suggest, an intermediate filament is a filament intermediate in thickness between the microtubules and microfilaments (see Figure 3.18c). Intermediate filaments are made up of long fibrous subunits of a protein called keratin that are wound together like the threads that compose a rope. Intermediate filaments, in concert with the microtubules, are important for maintaining cell shape and structure. Unlike the microtubules, which resist compression, intermediate filaments resist tension—the forces that pull apart cells. There are many cases in which cells are prone to tension, such as when epithelial cells of the skin are compressed, tugging them in different directions. Intermediate filaments help anchor organelles together within a cell and also link cells to other cells by forming special cell-to-cell junctions.
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187781 | https://www.hopkinsmedicine.org/health/treatment-tests-and-therapies/brachytherapy | Skip to Main Content
Health
Brachytherapy
What is brachytherapy?
Brachytherapy is radiation treatment that is given directly into your body. It is placed as close to the cancer as possible. The radiation is given using tiny devices, such as wires, seeds, or rods filled with radioactive materials. These devices are called implants.
Brachytherapy lets your healthcare provider use a higher total dose of radiation over a shorter time than is possible with external beam radiation therapy. The radiation dose is focused on the cancer cells. It does less damage to the nearby normal cells.
Brachytherapy implants may be short-term (temporary) or long-lasting (permanent).
Temporary brachytherapy
Temporary implants are removed after the treatment has ended. The implants may be hollow needles, hollow tubes (catheters), or balloons filled with fluid. The implants are inserted into or near the cancer for short time, then removed. Either high-dose or low-dose brachytherapy may be used:
High-dose rate (HDR) brachytherapy.The implants stay in the body for several minutes and then are removed.
Low-dose rate (LDR) brachytherapy.The implants stay in the body for hours or days and then are removed.
Permanent brachytherapy
This type is also called low-dose rate brachytherapy or seed implantation. It uses implants called pellets or seeds. These implants are very small, about the size of a grain of rice. Your healthcare provider inserts the seeds directly into a tumor with thin, hollow needles. The seeds are left in place after the radiation has been used up. Their small size causes little or no pain. With time, the radiation lessens, then stops completely.
Why might I need brachytherapy?
Brachytherapy may be used to treat many types of cancer, such as:
Prostate cancer
Cancer of the vagina, cervix, or uterus
Breast cancer
Eye cancer
Head and neck cancers
It may be done along with external beam radiation therapy to help destroy tumor cells for certain types of cancer.
What are the risks of brachytherapy?
Side effects of brachytherapy often depend on where the therapy is given and the therapy dose. Risks and possible complications include:
Extreme tiredness (fatigue)
Pain from staying in 1 position
Temporary side effects in the area being treated. For example, head and neck brachytherapy can cause sores in the mouth and throat. Brachytherapy to the pelvis can cause urinary or digestive issues, such as urinary frequency or diarrhea.
Failure to affect tumor growth
Damage to healthy tissue and organs
Having another cancer later due to radiation exposure
Risks of anesthesia
How do I get ready for brachytherapy?
Before brachytherapy starts, your healthcare provider will decide which tests you need. These may include blood tests, an electrocardiogram (EKG), chest X-rays, and scans, such as ultrasound, MRI, or CT.
A few days before your brachytherapy starts, you will be given specific directions about how to prepare. Follow these directions closely. Tell your healthcare provider about all medicines you take. This includes both over-the-counter and prescription medicines. It also includes vitamins, herbs, and other supplements. You may need to stop taking some or all of them before the procedure. You may need to stop taking aspirin and over-the-counter pain and fever medicines (called NSAIDS or nonsteroidal anti-inflammatory drugs). These can cause bleeding problems. Also, follow any directions you’re given for not eating or drinking before the procedure.
What happens during brachytherapy?
The type of cancer you have, its location, and other factors will determine your treatment schedule. How long brachytherapy lasts will depend on the type of treatment given. You may go home after treatment. Or you may stay in the hospital for 1 or more nights.
Before the procedure. Right before the brachytherapy starts, you may need anesthesia medicine to keep you free from pain while the implants are placed in your body. This will depend on the size and number of implants, as well as the location of the insertion site. Anesthesia makes you numb, drowsy, or completely asleep. An IV (intravenous) line is put into a vein in your arm or hand to give fluids and medicines.
During the procedure. A delivery device, such as a needle, is placed into the cancer site. The device may be passed through a nearby opening in the body, such as the vagina or rectum. Or a cut (incision) may be made in the skin. Implants are then passed through the delivery device into or near the cancer site. The implants are placed by hand or machine. X-rays, ultrasound, or another imaging test may be used to make sure of correct placement. Here's what happens depending on the type you have:
High-dose rate (HDR) brachytherapy.The implants are put in once a day for several minutes, then removed. The implant tube may be removed after each treatment session. Or it may stay in place. You may go home between treatments. Or you may stay in the hospital until all treatment sessions are done.
Low-dose rate (LDR) brachytherapy.The implants and delivery device stay in place for hours to days. You stay in the hospital during this time.
Permanent brachytherapy. The implants are put in place and not removed. Very low doses of radiation are given, and the radiation stops over time.
What happens after brachytherapy?
You will recover from the procedure, then go to your hospital room. Or you will be released to go home. If you are able to go home, have an adult family member or friend drive you.
If you stay in the hospital
Some types of brachytherapy need you to be in the hospital for a few days. You will need to follow specific rules to protect others from the effects of the radiation while it is active inside your body. Generally, treatment may include:
Staying in a private room
Hospital staff time spending as little time as possible in your room when care is being given
Placing portable shields between you and staff or visitors
Limits for visitors may include:
Not allowing pregnant women or children under a certain age to visit
Limiting how long visitors may stay
Limiting how close visitors can get to you
If you are discharged home, you may have additional visitor limitations. Ask your healthcare provider for guidelines.
If you go home
Follow any directions you are given for caring for yourself between treatments. These may include:
Take any prescribed medicine exactly as directed.
Care for your bandage and implant site as directed.
If you have an implant tube left in place, follow your healthcare provider’s directions for taking care of it.
Follow your healthcare provider’s directions about not putting weight or pressure on the implant site.
Follow any radiation safety measures you are given.
Next steps
Before you agree to the test or procedure, make sure you know:
The name of the test or procedure
The reason you are having the test or procedure
What results to expect and what they mean
The risks and benefits of the test or procedure
What the possible side effects or complications are
When and where you are to have the test or procedure
Who will do the test or procedure and what that person’s qualifications are
What would happen if you did not have the test or procedure
Any alternative tests or procedures to think about
When and how you will get the results
Who to call after the test or procedure if you have questions or problems
How much you will have to pay for the test or procedure
Specializing In:
Brachytherapy
Find Additional Treatment Centers at:
Howard County Medical Center
Sibley Memorial Hospital
Suburban Hospital
Request an Appointment
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187782 | https://spot.pcc.edu/math/orcca/ed2/html/section-scientific-notation.html | ORCCA Scientific Notation
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Open Resources for Community College Algebra
Portland Community College Faculty
Contents
IndexCalcPrevUpNext
ContentsPrevUpNext
Front Matter
Colophon
Acknowledgements
To All
Pedagogical Decisions
Entering WeBWorK Answers
I Linear Equations and Lines
1 Variables, Expressions, and Equations
Variables and Evaluating Expressions
Combining Like Terms
Comparison Symbols and Notation for Intervals
Equations and Inequalities as True/False Statements
Solving One-Step Equations
Solving One-Step Inequalities
Algebraic Properties and Simplifying Expressions
Modeling with Equations and Inequalities
Variables, Expressions, and Equations Chapter Review
2 Linear Equations and Inequalities
Solving Multistep Linear Equations
Solving Multistep Linear Inequalities
Linear Equations and Inequalities with Fractions
Special Solution Sets
Isolating a Linear Variable
Linear Equations and Inequalities Chapter Review
3 Graphing Lines
Cartesian Coordinates
Graphing Equations
Exploring Two-Variable Data and Rate of Change
Slope
Slope-Intercept Form
Point-Slope Form
Standard Form
Horizontal, Vertical, Parallel, and Perpendicular Lines
Summary of Graphing Lines
Graphing Lines Chapter Review
4 Systems of Linear Equations
Solving Systems of Linear Equations by Graphing
Substitution
Elimination
Systems of Linear Equations Chapter Review
II Preparation for STEM
5 Exponents and Polynomials
Adding and Subtracting Polynomials
Introduction to Exponent Rules
Dividing by a Monomial
Multiplying Polynomials
Special Cases of Multiplying Polynomials
More Exponent Rules
Exponents and Polynomials Chapter Review
6 Radical Expressions and Equations
Square and n n th Root Properties
Rationalizing the Denominator
Radical Expressions and Rational Exponents
Solving Radical Equations
Radical Expressions and Equations Chapter Review
7 Solving Quadratic Equations
Solving Quadratic Equations by Using a Square Root
The Quadratic Formula
Complex Solutions to Quadratic Equations
Solving Equations in General
Solving Quadratic Equations Chapter Review
8 Quantities in the Physical World
Scientific Notation
Unit Conversion
Geometry Formulas
Geometry Applications
Quantities in the Physical World Chapter Review
9 Topics in Graphing
Review of Graphing
Key Features of Quadratic Graphs
Graphing Quadratic Expressions
Graphically Solving Equations and Inequalities
Topics in Graphing Chapter Review
III Preparation for College Algebra
10 Factoring
Factoring Out the Common Factor
Factoring by Grouping
Factoring Trinomials with Leading Coefficient One
Factoring Trinomials with a Nontrivial Leading Coefficient
Factoring Special Polynomials
Factoring Strategies
Solving Quadratic Equations by Factoring
Factoring Chapter Review
11 Functions
Function Basics
Domain and Range
Using Technology to Explore Functions
Simplifying Expressions with Function Notation
Technical Definition of a Function
Functions Chapter Review
12 Rational Functions and Equations
Introduction to Rational Functions
Multiplication and Division of Rational Expressions
Addition and Subtraction of Rational Expressions
Complex Fractions
Solving Rational Equations
Rational Functions and Equations Chapter Review
13 Graphs and Equations
Overview of Graphing
Quadratic Graphs and Vertex Form
Completing the Square
Absolute Value Equations
Solving Mixed Equations
Compound Inequalities
Solving Inequalities Graphically
Graphs and Equations Chapter Review
Appendices
A Basic Math Review
Arithmetic with Negative Numbers
Fractions and Fraction Arithmetic
Absolute Value and Square Root
Percentages
Order of Operations
Set Notation and Types of Numbers
B Unit Conversions
C PCC Course Content and Outcome Guides
MTH 60
MTH 65
MTH 95
Index
Authored in PreTeXt
Section 8.1 Scientific Notation
¶
Objectives:PCC Course Content and Outcome Guide
MTH 65 CCOG 1.d
MTH 65 CCOG 1.e
Very large and very small numbers can be awkward to write and calculate with. These kinds of numbers can show in the sciences. For example in biology, a human hair might be as thick as 0.000181 0.000181 meters. And the closest that Mars gets to the sun is 206620000 206620000 meters. Keeping track of the decimal places and extra zeros raises the potential for mistakes to be made. In this section, we discuss a format used for very large and very small numbers called scientific notation that helps alleviate the issues with these numbers.
Figure 8.1.1. Alternative Video Lesson
Subsection 8.1.1 The Basics of Scientific Notation
An October 3, 2016 CBS News headline1 read:
Federal Debt in FY 2016 Jumped $1,422,827,047,452.46$1,422,827,047,452.46—that's $12,036$12,036 Per Household.
The article also later states:
By the close of business on Sept.30, 2016, the last day of fiscal 2016, it had climbed to $19,573,444,713,936.79.$19,573,444,713,936.79.
When presented in this format, trying to comprehend the value of these numbers can be overwhelming. More commonly, such numbers would be presented in a descriptive manner:
The federal debt climbed by 1.42 1.42 trillion dollars in 2016.
The federal debt was 19.6 19.6 trillion dollars at the close of business on Sept.30, 2016.
In science, government, business, and many other disciplines, it's not uncommon to deal with very large numbers like these. When numbers get this large, it can be hard to discern when a number has eleven digits and when it has twelve.
We have descriptive language for all numbers based on the place value of the different digits: ones, tens, thousands, ten thousands, etc. We tend to rely upon this language more when we start dealing with larger numbers. Here's a chart for some of the most common numbers we see and use in the world around us:
Number US English Name Power of 10 10
1 1 one 10 0 10 0
10 10 ten 10 1 10 1
100 100 hundred 10 2 10 2
1,000 1,000 one thousand 10 3 10 3
10,000 10,000 ten thousand 10 4 10 4
100,000 100,000 one hundred thousand 10 5 10 5
1,000,000 1,000,000 one million 10 6 10 6
1,000,000,000 1,000,000,000 one billion 10 9 10 9
Figure 8.1.2. Whole Number Powers of 10 10
Each number above has a corresponding power of ten and this power of ten will be important as we start to work with the content in this section. This descriptive language also covers even larger numbers: trillion, quadrillion, quintillion, sextillion, septillion, and so on. There's also corresponding language to describe very small numbers, such as thousandth, millionth, billionth, trillionth, etc.
Through centuries of scientific progress, humanity became increasingly aware of very large numbers and very small measurements. As one example, the star that is nearest to our sun is Proxima Centauri2imagine.gsfc.nasa.gov/features/cosmic/nearest_star_info.html. Proxima Centauri is about 25,000,000,000,000 25,000,000,000,000 miles from our sun. Again, many will find the descriptive language easier to read: Proxima Centauri is about 25 25 trillion miles from our sun.
To make computations involving such numbers more manageable, a standardized notation called “scientific notation” was established. The foundation of scientific notation is the fact that multiplying or dividing by a power of 10 10 will move the decimal point of a number so many places to the right or left, respectively. So first, let's take a moment to review that level of basic arithmetic.
Checkpoint 8.1.3.
Multiplying a number by 10 n 10 n where n n is a positive integer had the effect of moving the decimal point n n places to the right.
Every number can be written as a product of a number between 1 1 and 10 10 and a power of 10.10. For example, 650=6.5×100.650=6.5×100. Since 100=10 2,100=10 2, we can also write
650=6.5×10 2 650=6.5×10 2
and this is our first example of writing a number in scientific notation.
Definition 8.1.4.
A positive number is written in scientific notation when it has the form a×10 n a×10 n where n n is an integer and 1≤a<10.1≤a<10. In other words, a a has precisely one non-zero digit to the left of the decimal place. The exponent n n used here is called the number's order of magnitude. The number a a is sometimes called the significand or the mantissa.
Some conventions do not require a a to be between 1 1 and 10,10, excluding both values, but that is the convention used in this book.
Some calculators and computer readouts cannot display exponents in superscript. In some cases, these devices will display scientific notation in the form 6.5E2 instead of 6.5×10 2.6.5×10 2.
Subsection 8.1.2 Scientific Notation for Large Numbers
To write a number larger than 10 10 in scientific notation, like 89412,89412, first write the number with the decimal point right after its first digit, like 8.9412.8.9412. Now count how many places there are between where the decimal point originally was and where it is now.
8.49412 8.9412⏞4
Use that count as the power of 10.10. In this example, we have
89412=8.9412×10 4 89412=8.9412×10 4
Scientific notation communicates the “essence” of the number (8.9412 8.9412) and then its size, or order of magnitude (10 4 10 4).
Example 8.1.5.
To get a sense of how scientific notation works, let's consider familiar lengths of time converted to seconds.
Length of Time Length in Seconds Scientific Notation
one second 1 second 1×10 0 1×10 0 second
one minute 60 seconds 6×10 1 6×10 1 seconds
one hour 3600 seconds 3.6×10 3 3.6×10 3 seconds
one month 2,628,000 seconds 2.628×10 6 2.628×10 6 seconds
ten years 315,400,000 seconds 3.154×10 8 3.154×10 8 seconds
79 years (about a lifetime)2,491,000,000 seconds 2.491×10 9 2.491×10 9 seconds
Note that roughly 2.6 2.6 million seconds is one month, while roughly 2.5 2.5 billion seconds is an entire lifetime.
Checkpoint 8.1.6.
Checkpoint 8.1.7.
Subsection 8.1.3 Scientific Notation for Small Numbers
Scientific notation can also be useful when working with numbers smaller than 1.1. As we saw in Figure 8.1.2, we can represent thousands, millions, billions, trillions, etc., with positive integer exponents on 10.10. We can similarly represent numbers smaller than 1 1 (which are written as tenths, hundredths, thousandths, millionths, billionths, trillionths, etc.), with negative integer exponents on 10.10. This relationship is outlined in Figure 8.1.8.
Number English Name Power of 10 10
1 1 one 10 0 10 0
0.1 0.1 one tenth 1 10=10−1 1 10=10−1
0.01 0.01 one hundredth 1 100=10−2 1 100=10−2
0.001 0.001 one thousandth 1 1,000=10−3 1 1,000=10−3
0.0001 0.0001 one ten thousandth 1 10,000=10−4 1 10,000=10−4
0.00001 0.00001 one hundred thousandth 1 100,000=10−5 1 100,000=10−5
0.000001 0.000001 one millionth 1 1,000,000=10−6 1 1,000,000=10−6
0.000000001 0.000000001 one billionth 1 1,000,000,000=10−9 1 1,000,000,000=10−9
Figure 8.1.8. Negative Integer Powers of 10 10
To see how this works with a digit other than 1,1, let's look at 0.005.0.005. When we state 0.005 0.005 as a number, we say “5 thousandths.” Thus 0.005=5×1 1000.0.005=5×1 1000. The fraction 1 1000 1 1000 can be written as 1 10 3,1 10 3, which we know is equivalent to 10−3.10−3. Using negative exponents, we can then rewrite 0.005 0.005 as 5×10−3.5×10−3. This is the scientific notation for 0.005.0.005.
In practice, we won't generally do that much computation. To write a small number in scientific notation we start as we did before and place the decimal point behind the first non-zero digit. We then count the number of decimal places between where the decimal had originally been and where it now is. Keep in mind that negative powers of ten are used to help represent very small numbers (smaller than 1 1) and positive powers of ten are used to represent very large numbers (larger than 1 1). So to convert 0.005 0.005 to scientific notation, we have:
0 3.005=5×10−3 0.005⏞3=5×10−3
Example 8.1.9.
In quantum mechanics, there is an important value called Planck's Constant3en.wikipedia.org/wiki/Planck_constant. Written as a decimal, the value of Planck's constant (rounded to six significant digits) is
0.000 000 000 000 000 000 000 000 000 000 000 662 607.0.000 000 000 000 000 000 000 000 000 000 000 662 607.
In scientific notation, this number will be 6.62607×10?.6.62607×10?. To determine the exponent, we need to count the number of places from where the decimal originally is to where we will move it (following the first “6”):
0 34 places.000 000 000 000 000 000 000 000 000 000 000 6 62 607 0.000 000 000 000 000 000 000 000 000 000 000 6⏞34 places 62 607
So in scientific notation, Planck's Constant is 6.62607×10−34.6.62607×10−34. It will be much easier to use 6.62607×10−34 6.62607×10−34 in a calculation, and an added benefit is that scientific notation quickly communicates both the value and the order of magnitude of Planck's Constant.
Checkpoint 8.1.10.
Checkpoint 8.1.11.
Checkpoint 8.1.12.
Subsection 8.1.4 Multiplying and Dividing Using Scientific Notation
One main reason for having scientific notation is to make calculations involving immensely large or small numbers easier to perform. By having the order of magnitude separated out in scientific notation, we can separate any calculation into two components.
Example 8.1.13.
On Sept.30th, 2016, the US federal debt was about $19,600,000,000,000$19,600,000,000,000 and the US population was about 323,000,000.323,000,000. What was the average debt per person that day?
Calculate the answer using the numbers provided, which are not in scientific notation.
First, confirm that the given values in scientific notation are 1.96×10 13 1.96×10 13 and 3.23×10 8.3.23×10 8. Then calculate the answer using scientific notation.
Explanation
We've been asked to answer the same question, but to perform the calculation using two different approaches. In both cases, we'll need to divide the debt by the population.
We may need to use a calculator to handle such large numbers and we have to be careful that we type the correct number of 0s.
\begin{gather} \frac{19600000000000}{323000000}\approx 60681.11 \end{gather}
2. To perform this calculation using scientific notation, our work would begin by setting up the quotient as (\frac{1.96 \times 10^{13}}{3.23 \times 10^8}\text{.}) Dividing this quotient follows the same process we did with variable expressions of the same format, such as (\frac{1.96 w^{13}}{3.23 w^8}\text{.}) In both situations, we'll divide the coefficients and then use exponent rules to simplify the powers.
\begin{align} \frac{1.96 \times 10^{13}}{3.23 \times 10^8} \amp= \frac{1.96 }{3.23} \times\frac{10^{13}}{ 10^8}\ \amp\approx 0.6068111 \times 10^5\ \amp\approx 60681.11 \end{align}
The federal debt per capita in the US on September 30th, 2016 was about (\$60{,}681.11) per person. Both calculations give us the same answer, but the calculation relying upon scientific notation has less room for error and allows us to perform the calculation as two smaller steps.
Whenever we multiply or divide numbers that are written in scientific notation, we must separate the calculation for the coefficients from the calculation for the powers of ten, just as we simplified earlier expressions using variables and the exponent rules.
Example 8.1.14.
Multiply (2×10 5)(3×10 4).(2×10 5)(3×10 4).
Divide 8×10 17 4×10 2.8×10 17 4×10 2.
Explanation
We will simplify the significand/mantissa parts as one step and then simplify the powers of (10) as a separate step.
(\begin{aligned}[t] \left( 2\times 10^5 \right)\left( 3\times10^4 \right) \amp= \left( 2\times 3 \right)\times \left(10^5 \times 10^4 \right)\ \amp= 6 \times 10^{9} \end{aligned})
(\begin{aligned}[t] \frac{8 \times 10^{17}}{4\times 10^2} \amp= \frac{8}{4} \times \frac{10^{17}}{10^2}\ \amp= 2 \times 10^{15} \end{aligned})
Often when we multiply or divide numbers in scientific notation, the resulting value will not be in scientific notation. Suppose we were multiplying (9.3×10 17)(8.2×10−6)(9.3×10 17)(8.2×10−6) and need to state our answer using scientific notation. We would start as we have previously:
(9.3×10 17)(8.2×10−6)=(9.3×8.2)×(10 17×10−6)=76.26×10 11(9.3×10 17)(8.2×10−6)=(9.3×8.2)×(10 17×10−6)=76.26×10 11
While this is a correct value, it is not written using scientific notation. One way to convert this answer into scientific notation is to turn just the coefficient into scientific notation and momentarily ignore the power of ten:
=76.26×10 11=7.626×10 1×10 11=76.26×10 11=7.626×10 1×10 11
Now that the coefficient fits into the proper format, we can combine the powers of ten and have our answer written using scientific notation.
=7.626×10 1×10 11=7.626×10 12=7.626×10 1×10 11=7.626×10 12
Example 8.1.15.
Multiply or divide as indicated. Write your answer using scientific notation.
(8×10 21)(2×10−7)(8×10 21)(2×10−7)
2×10−6 8×10−19 2×10−6 8×10−19
Explanation
Again, we'll separate out the work for the significand/mantissa from the work for the powers of ten. If the resulting coefficient is not between (1) and (10\text{,}) we'll need to adjust that coefficient to put it into scientific notation.
(\begin{aligned}[t] \left( 8 \times 10^{21} \right)\left( 2 \times 10^{-7} \right) \amp= \left( 8 \times 2 \right)\times\left( 10^{21} \times 10^{-7} \right)\ \amp= \highlight{16} \times 10^{14}\ \amp= \highlight{1.6\times 10^1} \times 10^{14}\ \amp= 1.6 \times 10^{15} \end{aligned})
We need to remember to apply the product rule for exponents to the powers of ten.
(\begin{aligned}[t] \frac{ 2 \times 10^{-6} }{ 8 \times 10^{-19} } \amp= \frac{ 2 }{ 8 }\times\frac{ 10^{-6} }{ 10^{-19} }\ \amp= \highlight{0.25} \times 10^{13}\ \amp= \highlight{2.5\times 10^{-1}} \times 10^{13}\ \amp= 2.5 \times 10^{12} \end{aligned})
There are times where we will have to raise numbers written in scientific notation to a power. For example, suppose we have to find the area of a square whose radius is 3×10 7 3×10 7 feet. To perform this calculation, we first remember the formula for the area of a square, A=s 2 A=s 2 and then substitute 3×10 7 3×10 7 for s:s:A=(3×10 7)2.A=(3×10 7)2. To perform this calculation, we'll need to remember to use the product to a power rule and the power to a power rule:
A=(3×10 7)2=(3)2×(10 7)2=9×10 14 A=(3×10 7)2=(3)2×(10 7)2=9×10 14
Reading Questions 8.1.5 Reading Questions
1.
Which number is very large and which number is very small?
9.99×10−47 1.01×10 23 9.99×10−47 1.01×10 23
2.
Since some computer/calculator screens can't display an exponent, how might a computer/calculator display the number 2.318×10 13?2.318×10 13?
3.
Why do we bother having scientific notation for numbers?
Exercises 8.1.6 Exercises
Converting To and From Scientific Notation
1.
Write the following number in scientific notation.
100000=100000=
2.
Write the following number in scientific notation.
20000=20000=
3.
Write the following number in scientific notation.
300=300=
4.
Write the following number in scientific notation.
400000=400000=
5.
Write the following number in scientific notation.
0.005=0.005=
6.
Write the following number in scientific notation.
0.0006=0.0006=
7.
Write the following number in scientific notation.
0.07=0.07=
8.
Write the following number in scientific notation.
0.008=0.008=
9.
Write the following number in decimal notation without using exponents.
9×10 2=9×10 2=
10.
Write the following number in decimal notation without using exponents.
1.1×10 5=1.1×10 5=
11.
Write the following number in decimal notation without using exponents.
2.02×10 3=2.02×10 3=
12.
Write the following number in decimal notation without using exponents.
3.02×10 2=3.02×10 2=
13.
Write the following number in decimal notation without using exponents.
4.01×10 0=4.01×10 0=
14.
Write the following number in decimal notation without using exponents.
5.01×10 0=5.01×10 0=
15.
Write the following number in decimal notation without using exponents.
6×10−4=6×10−4=
16.
Write the following number in decimal notation without using exponents.
7×10−2=7×10−2=
17.
Write the following number in decimal notation without using exponents.
8×10−4=8×10−4=
18.
Write the following number in decimal notation without using exponents.
8.99×10−2=8.99×10−2=
Arithmetic with Scientific Notation
19.
Multiply the following numbers, writing your answer in scientific notation.
(9×10 2)(7×10 2)=(9×10 2)(7×10 2)=
20.
Multiply the following numbers, writing your answer in scientific notation.
(2×10 4)(4×10 5)=(2×10 4)(4×10 5)=
21.
Multiply the following numbers, writing your answer in scientific notation.
(3×10 2)(9×10 4)=(3×10 2)(9×10 4)=
22.
Multiply the following numbers, writing your answer in scientific notation.
(4×10 3)(6×10 3)=(4×10 3)(6×10 3)=
23.
Multiply the following numbers, writing your answer in scientific notation.
(5×10 5)(3×10 5)=(5×10 5)(3×10 5)=
24.
Multiply the following numbers, writing your answer in scientific notation.
(6×10 3)(9×10 4)=(6×10 3)(9×10 4)=
25.
Divide the following numbers, writing your answer in scientific notation.
4.2×10 5 7×10 3=4.2×10 5 7×10 3=
26.
Divide the following numbers, writing your answer in scientific notation.
2.4×10 3 8×10 2=2.4×10 3 8×10 2=
27.
Divide the following numbers, writing your answer in scientific notation.
7.2×10 5 9×10 2=7.2×10 5 9×10 2=
28.
Divide the following numbers, writing your answer in scientific notation.
5.4×10 6 9×10 4=5.4×10 6 9×10 4=
29.
Divide the following numbers, writing your answer in scientific notation.
6×10 3 2×10−4=6×10 3 2×10−4=
30.
Divide the following numbers, writing your answer in scientific notation.
2.4×10 5 3×10−2=2.4×10 5 3×10−2=
31.
Divide the following numbers, writing your answer in scientific notation.
2×10 2 4×10−3=2×10 2 4×10−3=
32.
Divide the following numbers, writing your answer in scientific notation.
1×10 4 5×10−2=1×10 4 5×10−2=
33.
Divide the following numbers, writing your answer in scientific notation.
4.8×10−5 6×10 2=4.8×10−5 6×10 2=
34.
Divide the following numbers, writing your answer in scientific notation.
3.5×10−3 7×10 5=3.5×10−3 7×10 5=
35.
Divide the following numbers, writing your answer in scientific notation.
1.6×10−2 8×10 4=1.6×10−2 8×10 4=
36.
Divide the following numbers, writing your answer in scientific notation.
6.3×10−4 9×10 3=6.3×10−4 9×10 3=
37.
Simplify the following expression, writing your answer in scientific notation.
(5×10 5)4=(5×10 5)4=
38.
Simplify the following expression, writing your answer in scientific notation.
(2×10 2)2=(2×10 2)2=
39.
Simplify the following expression, writing your answer in scientific notation.
(2×10 8)3=(2×10 8)3=
40.
Simplify the following expression, writing your answer in scientific notation.
(3×10 5)2=(3×10 5)2=
41.
Simplify the following expression, writing your answer in scientific notation.
(3×10 10)3=(3×10 10)3=
42.
Simplify the following expression, writing your answer in scientific notation.
(4×10 7)4=(4×10 7)4=
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187783 | https://parksinchaisri.github.io/files/ta/rec11-2015.pdf | 1.200J, Fall 2015—Recitation 11 1 Massachusetts Institute of Technology 1.200J—Transportation Systems Analysis: Performance and Optimization Fall 2015 — TA: Wichinpong “Park” Sinchaisri Recitation 11 Unit 5 — Advanced Stochastic Modeling of Transportation Applications 1 Probability Refresher X has pdf λe−λx1x≥0 for some λ > 0. Compute the following: (a) The cdf of X (b) The expected value of X (c) The variance of X (d) The probability that X is greater than its expected value.
2 Inverse Transform Method (a) Weibull distribution with parameter (α, β) has the following probability density function (pdf) f(x) = αβxβ−1e−αxβ if x ≥0; 0 otherwise Use the inverse transform method to generate samples from the Weibull distribution.
Write down the pseudo-code.
1.200J, Fall 2015—Recitation 11 2 (b) Write down the pseudo-code for an algorithm to generate X with the following cdf F(x) = 0.4(1 −e−2x) + 0.6(1 −e−2√x) 1x≥0 by using two random variables U1, U2 ∼Uniform(0,1).
3 Discrete Event Simulation (a) Simulate a M/M/1 queueing system that stops admitting new customers after time T.
PDFs for interarrival times and service times are fa and fs, respectively.
(i) Define variables and events for the system.
(ii) Initialization (iii) Event 1: Arrival of a customer (iv) Event 2: Departure of a customer (v) Event 3: Serving customers after system closure (b) Simulate a M/M/1 queuing system up until N customers getting served.
1.200J, Fall 2015—Recitation 11 3 Solution - Probability Refresher (a) The key is to integrate the PDF from the lower bound up to x as the definition of CDF is P(X ≤x).
CDF: FX(x) = P(X ≤x) = R x −∞λe−λx1x≥0dx = R x 0 λe−λx for x ≥0. Thus, FX(x) = 1 −e−λx if x > 0 0 otherwise (b) E[X] = R ∞ 0 xλe−λxdx = 1 λ (c) V ar[X] = E[X2] −(E[X])2 = R ∞ 0 x2λe−λxdx −( 1 λ)2 = 2 λ2 −1 λ2 = 1 λ2 (d) P(X ≥1/λ) = 1 −P(X ≤1/λ) = 1 −(1 −e−1)) = 1/e or you can also achieve the same answer by calculating R ∞ 1/λ λe−λxdx = e−1.
Solution - Inverse Transform Method (a) The inverse transform method relies on the fact that for X ∼FX we have that F −1 X (X) ∼ Uniform(0, 1). For this example, we need to compute the distribution function of the general Weibull distribution and then compute its inverse: FX(x) = Z x 0 f(y)dy = αβ Z x 0 yβ−1e−αyβdy = 1 −e−αxβ so that x = −log(1 −FX(x)) α 1/β and we can define: F −1 X (U) = −log(1 −U) α 1/β as (1 −U) is also of Uniform(0,1), we can say that: F −1 X (U) = −log(U) α 1/β Define Finv(u, alpha, beta) to be exactly this function, then we can use the following pseudo-code to generate random variables: function [sample] = weibull(alpha, beta) u = rand(1); % generate a uniform(0,1) r.v.
sample = Finv(u, alpha, beta); end 1.200J, Fall 2015—Recitation 11 4 (b) The CDF: F(x) = 0.4(1 −e−2x) + 0.6(1 −e−2√x) 1x≥0 Let U1, U2 ∼Uniform(0,1) and p = 0.4. We notice that that (1 −e−2x) is F1 of the exponential distribution at rate λ = 2, while F2 = 1 −e−2√x is of Weibull distribution we solved in part (b) with α = 2, β = 1/2.
We found the inverse functions for both distributions: F −1 1 (U) = −log(U) 2 and F −1 2 (U) = (log(U))2 4 Note that we will use U1 to choose which F −1 to use, and use U2 to generate the sample.
The pseudo-code: (a) Generate U1, U2 ∼Uniform(0,1).
(b) If U1 ≥0.4, set X = −0.5 ln(U2); otherwise, X = 0.25(ln(U2))2.
MATLAB: function [sample] = mixture u1 = rand(1); u2 = rand(1); if (u1 <= 0.4) sample = -0.5log(u2); else sample =0.25(log(u2))$ˆ2$; end end |
187784 | https://www.ebsco.com/research-starters/chemistry/doppler-effect | Doppler effect | Research Starters | EBSCO Research
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Doppler effect
The Doppler effect, also known as Doppler shift, is a phenomenon that describes the change in frequency of waves due to the relative motion between a wave source and an observer. This effect is observable in various types of waves, including sound, light, and water waves. A common example is the change in pitch of a siren from an approaching police car, where the sound appears higher in frequency as the car approaches and lower as it moves away. The actual frequency of the sound remains constant; it is the observer's perception that changes based on their movement relative to the source.
The Doppler effect has significant applications across multiple fields. In meteorology, Doppler radar is utilized to track storm systems, while in astronomy, it helps measure the movement of stars and the expansion of the universe through phenomena like redshift and blueshift. In medicine, echocardiograms leverage the Doppler effect to assess blood flow and cardiac health. Additionally, it plays a critical role in satellite communications and air traffic control, ensuring accurate tracking and signal integrity despite high speeds and movement. Understanding the Doppler effect provides essential insights into both natural phenomena and technological advancements.
Published in:2024
By:Dewey, Joseph
Go to EBSCOhost and sign in to access more content about this topic.
Doppler effect
The Doppler effect, or Doppler shift, measures change in the frequency of waves based on the relative motion of the source and the position of an observer. It applies to any type of wave, including water, electromagnetic, visible light, and sound waves. The most familiar application of the Doppler effect involves sound, specifically sound coming from a source that is in motion (or reaching an observer who is in motion). A constant sound—that is, a single sustained note—can appear to change pitch when the source moves with respect to its listener. As the distance between the source of the sound and the receiver shortens, the frequency of the sound waves increases, causing the sound to become higher pitched; as the distance lengthens, the frequency decreases, causing the sound to become lower pitched.
It is important to note that what changes is the observer’s perception of the sound, not the sound itself. The emitted frequency of the sound remains the same, but the received frequency changes based on the increasing or decreasing speed and distance between the source and the listener.
Background
When a police car passes another car with its siren on, the other driver can hear the pitch of the siren increase as the police car approaches and then drop as it pulls away. The frequency of the emitted siren remains the same; it is only with respect to the driver of the other car that the sound seems to change. As the siren approaches from behind, the sound waves are compressed closer together by the forward movement, and therefore the frequency of the sound waves increases. Once the siren passes, the opposite effect occurs. The familiar pitched engine whine of competing sports cars flying past stationary spectators at a racing event also demonstrates this effect.
The Doppler effect was first described by Austrian physicist and mathematician Christian Doppler (1803–53) in his 1842 paper on the color shifts of binary star systems. Doppler determined that he could calculate the speed of the stars by observing the shift in the frequency of their light waves, although the lower quality of telescopes at the time made confirmation of his hypothesis difficult. Testing by Dutch scientist C. H. D. Buys Ballot (1817–1890) confirmed Doppler’s hypothesis with regard to sound waves in 1845; Buys Ballot placed a horn section playing a single, calibrated note on a train and observed the Doppler effect on the sound as the train repeatedly passed him. If the receiver of the sound is stationary and the source of the sound is moving, the only time that the listener actually hears the emitted sound at the appropriate pitch is the brief moment when the moving source actually passes by.
The effects of the Doppler shift become more complicated when both the source and the observer are in motion. This complexity has given rise to generations of increasingly dense mathematical suppositions that have tested Doppler’s original hypothesis. When both source and observer are in motion, the wave crests assume varied heights and thus affect the perceived wave frequency and pitch.
Overview
The Doppler effect has far-reaching applications. The predictable action of waves across a moving medium led to the development of sonar technology, in which sound waves are projected through water to measure the distance and speed of underwater objects. Doppler radar is an essential element of modern meteorological forecasting, using sound waves to determine the velocity and direction of storm systems.
The Doppler effect also enables astronomers and astrophysicists to calculate the expansion of the universe and the movements of distant star systems. In the spectrum of visible light, the lowest-frequency light waves appear red, while the highest-frequency light waves appear blue. (Light waves just below the frequency range of visible light are called infrared, while those just above the range are called ultraviolet.) When stars are moving away from Earth, the observed frequency of the electromagnetic waves they emit decreases, creating a phenomenon known as redshift, in which the color of the visible light shifts toward the red end of the spectrum. Conversely, if a star or some other light-emitting object is moving toward Earth, the waves shorten and increase in frequency, causing the light to appear blue; this is known as blueshift.
Surgeons using an echocardiogram use the principle of the Doppler effect to examine the cardiac system for valve abnormalities or to measure blood flow, greatly improving preventive cardiac care and reducing surgical risks. Echocardiograms rely on Doppler sonography and ultrasound devices, which produce high-frequency sound waves that bounce off of circulating blood cells, enabling doctors to measure the frequency shift between blood cells moving toward or away from the probe and the velocity of blood flow. Ultrasound technology is also used to listen to fetal heartbeats and observe gestational development, allowing doctors to detect genetic abnormalities and determine the sex of the fetus before birth.
Doppler technology has also been applied to satellite technology and communications systems. When ground stations broadcast radio waves to satellites hurtling through space, it is necessary to ensure that the transmitted signal maintains its integrity and usefulness, despite Earth’s rotation and the great speeds of the satellites themselves. In addition, the mathematical measure of distance and velocity between moving objects using radar waves is the basis of air-traffic-control technology and radar tracking of virtually any object launched into space.
Bibliography
Chen, Jiabi, et al. “Observation of the Inverse Doppler Effect in Negative-Index Materials at Optical Frequencies.” Nature Photonics 5.4 (2011): 239–45. Print.
Chen, Victor C., David Tahmoush, and William J. Miceli. Radar Micro-Doppler Signature: Processing and Applications. London: Inst. of Engineering and Technology, 2014. Print.
Deeg, Karl-Heinz, Thomas Rupprecht, and Michael Hofbeck. Doppler Sonography in Infancy and Childhood. Cham: Springer, 2015. Print.
Denny, Mark. Blip, Ping, and Buzz: Making Sense of Radar and Sonar. Baltimore: Johns Hopkins UP, 2007. Print.
Doviak, Richard J., and Dušan S. Zrnić.Doppler Radar and Weather Observations. 2nd ed. 1993. Minelola: Dover, 2006. Print.
Drake, Samuel Picton, and Alan Purvis. “Everyday Relativity and the Doppler Effect.” American Journal of Physics 82.1 (2014): 52–59. Print.
Fleisher, Paul. Doppler Radar, Satellites, and Computer Models: The Science of Weather Forecasting. Minneapolis: Lerner, 2011. Print.
Poessel, Markus. “Waves, Motion and Frequency: The Doppler Effect.” Einstein Online. Max Planck Inst. for Gravitational Physics, 2011. Web. 26 Sept. 2014.
Temming, Maria. "Scientists Say: Doppler Effect." Science News Exlores, 4 Apr. 2022, www.snexplores.org/article/scientists-say-doppler-effect. Accessed 19 Nov. 2024.
Tyson, Neil deGrasse, and Donald Goldsmith. Origins: Fourteen Billion Years of Cosmic Evolution. New York: Norton, 2004. Print.
Qui, Song, et. al. "Rotational Doppler Effect with Vortex Beams: Fundamental Mechanism and Technical Progress." Frontiers in Physics, 28 June 2022, doi.org/10.3389/fphy.2022.938593. Accessed 29 Dec. 2022.
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Paradoxically split S2 heart sound | Learn the Heart
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Paradoxically Split S2 Heart Sound
BySteven Lome, MD
A paradoxical split S2 heart sound occurs when the splitting is heard during expiration and disappears during inspiration, the opposite of the physiologic split S2.
A paradoxical split S2 occurs in any setting that delays the closure of the aortic valve, such as severe aortic stenosis, hypertrophic obstructive cardiomyopathy (HOCM) or in the setting of a left bundle branch block (LBBB).
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187787 | https://les-mathematiques.net/vanilla/uploads/editor/fr/4beg0ut3r86b.pdf | The D'Alembert Functional Equation Author(s): F. J. Papp Reviewed work(s): Source: The American Mathematical Monthly, Vol. 92, No. 4 (Apr., 1985), pp. 273-275 Published by: Mathematical Association of America Stable URL: .
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1985] NOTES 273 and thus a sum of two squares in F is again a square. In short, F is real closed. We have shown now that any field F whose algebraic closure has prime degree must be real closed. Consider finally an arbitrary K with K/K finite Galois. Let L = K( -1). Then K/L is finite Galois. If L A K, choose-some nontrivial automorphism T of K over L. Let n be its order, and choose some prime p dividing n. Then the field F fixed by T"/P has IK: Fl = p. Hence F should be real closed; but this is impossible, since -1 is in L C F. This contradiction shows that L = K. But then K/K has degree 2, and K must be a real closed field. U References 1. E. Artin and 0. Schreier, Eine Kennzeichnung der reell abgeschlossenen Korper, Abh. Math. Seminar Hamburg, 5 (1927) 225-231. 2. N. Bourbaki, Alg&bre VI: Groupes et Corps Ordonnes, Hermann, Paris, 1952. 3. N. Jacobson, Lectures in Abstract Algebra III: Theory of Fields and Galois Theory, Van Nostrand, New York, 1964. 4. , Basic Algebra I, Freeman, San Francisco, 1974. 5. I. Kaplansky, Fields and Rings, U. Chicago Press, Chicago, 1969. 6. G. Kreisel and J. L. Krivine, Elements of Mathematical Logic, North-Holland, London, 1967. THE D'ALEMBERT FUNCTIONAL EQUATION F. J. PAPP Department of Mathematics, University of Michigan, Dearborn, MI 48128 I. Introduction. D'Alembert's functional equation (1) 2f(x)f(y) = f(x + y) + f(x - y) has a long history going back to d'Alembert , Poisson and , and Picard and [5, Chapter I, Section III]. The equation plays an important role in determining the sum of two vectors in various Euclidean and non-Euclidean geometries; in addition to the references above see and [7, Section 2.4.2]. In equation (1) f is taken to be a continuous real valued function defined on the entire real line. With these assumptions, the only solutions of (1) are f(x)-0, f(x)-1, f(x) = coskx, and f(x) = coshkx, where k is a real constant. The Cauchy method of showing that these are the only solutions to (1) first determines f on a dense set and then uses continuity to pass to the entire real line; see [7, Section 2.4.1] for the details. This paper provides an alternative proof which makes use of a few elementary ideas from calculus and differential equations. This method given below is quite straightforward and will, perhaps, be somewhat more accessible than Cauchy's method. II. Preliminary Properties. We will first obtain several properties which any solution of (1) must satisfy. All of these are derived directly from the functional equation itself. If x = y = 0 in (1), then (2) 2[f(0)]2 = 2f(0) so that, either (3) f(0) =0 or (4) f(0) = 1. If (3) holds, then for any real x, 0 = 2f(x)f(0) =f(x + 0) +f(x - 0) = 2f(x). 274 F. J. PAPP [April In other words, f is the identically zero function. So we will assume from now on that (3) does not hold and that f satisfies (4). Next, any solution of (1) must be an even function. To see this, let x = 0 and y be any real number. It then follows from (1) and (4) that 2f( y) = 2f(O)f( y) = f(0 + y) + f?( - y) = f (y) + f( -y); or (5) f(Y) =f(-Y). III. Construction of the Continuous Solutions of (1). Since f is assumed to be continuous on the entire line, it will be integrable on any finite subinterval. Consequently, for t > 0 we have (6) |2f((x)f(y)dy= tf(x+y)dy+ tf(x-y)dy. With changes of the variables in the two integrals on the right hand side of this last equation, we obtain (7) 2f(x)ft f(y) dy = 2f f(y) dy. Since (4) holds, and since we have assumed that f is continuous, there exists t > 0 such that (8) f f(y) dy> 0. Note that the right hand side of (7) is differentiable with respect to x and thus so is the left hand side. Knowing now that f is differentiable allows us to conclude that f" exists as well. In fact, proceeding step-by-step, we see that any continuous solution of (1) is in fact infinitely differentia- ble. Differentiate both sides of (7) with respect to x, evaluate the resulting expression at x = 0, and choose t > 0 sufficiently small so that (8) holds. It then follows that (9) 2f'(0) f f(y) dy = 2[f(t) - f( - t)] = 0, since f is an even function. Consequently (10) f'(0) = 0. Now, differentiate (1) twice with respect to y and obtain (11) 2f(x)f"(y) = f"( X + y) + f"( X - y) Letting y = 0, we get (12) 2f( x)f"(0) = 2f"( X). Let K = f"(0). So (12) becomes (13) below, and we have now shown that if f is a non-trivial solution of (1) it must also be a solution to the initial value problem (13)-(4)-(10): (13) f"(x) = Kf(x), (4) f(0) =1, (10) f'(O) = 0. Depending on K there are three possibilities. For K = 0, f(X) = C1X + C2; for K > 0, f(x) = C1sinh kx + C2cosh kx 1985] NOTES 275 where k = K1"2; and for K < 0, f(x) = C,sinkx + C2coskx with k (-K)1/2. In all three cases, (4) and (10) lead to C1 = 0 and C2 = 1. Thus the continuous solutions of (1) are (14) f(x) 0, (15) f(x) 1, (16) f x) = cosh kx, and (17) f( x) = cos kx. References 1. J. d'Alembert, M6moire sur les principes de m6canique, Hist. Acad. Sci., Paris, 1769, pp. 278-286. 2. S. Poisson, Du parallelogramme des forces, Correspondance sur lEcole Polytechnique, 1 (1804) 356-360. 3. , Traite de m6canique, Paris, 1811. 4. C.-E. Picard, Deux leqons sur certaines 6quations fonctionnelles et la g6ometrie non-euclidienne, Bull. Soc. Math., XLVI (1922) 404-416, 425-432. 5. -_ , Leqons sur quelques 6quations fonctionnelles avec des applications i divers problWmes d'analyse et de la physique math6matique, Gauthier-Villars, Paris, 1928. 6. R. Bonola, Non-Euclidean Geometry, Dover, New York, 1955. 7. J. Acz6l, Lectures on Functional Equations and Their Applications, Academic Press, New York and London, 1966. THE DISCRIMINATOR (A SIMPLE APPLICATION OF BERTRAND'S POSTULATE) L. K. ARNOLD, S. J. BENKOSKI, AND B. J. MCCABE Daniel H. Wagner, Associates, Station Square Onie, Paoli, PA 19301 1. Introduction. For a positive integer n, what is the smallest value of k such that 12, 22,,,., n2 are all incongruent modulo k? The answer to this question requires only elementary number theory techniques and also provides a rare application of Bertrand's Postulate in an elementary setting. Bertrand postulated in 1845 that for every natural number n > 3, there is a prime p satisfying n < p < 2 n - 2. He verified this for n < 3,000,000. Tchebychef proved this in 1850. Usually, the Postulate is now stated with n < p < 2n. Most modern texts use a proof due to Erdos . 2. Motivation. This particular question arose from consideration of a problem in computer simulation. The problem may be described as developing a method to determine quickly the square roots of a long sequence of integers. Each of these integers is a perfect square and the range of the integers is known. However, the integers themselves are generated by another process and, in fact, it can be assumed that they are generated randomly. The goal is to determine the square roots quickly. The usual square root algorithm is much too slow. An alternative procedure would be to calculate all the required square roots once and then "look up" the appropriate value in an array. More precisely, let S = {1, 22, .. ., n2 } be the set of integers whose square roots must be determined. Let A be a 1 x n2 array with A(x) = x1/2. For any value s E S,A(s) is the square root of s. This is obviously a very quick procedure; however, it requires an array of size I x n2. This is a large array and most of the values in the array are not used. Is there an alternative method which would still allow quick calculation of the square roots but require a much smaller |
187788 | https://www.youtube.com/watch?v=4QRGDovumlU | Problem 2.7 - Electric Field, Continuous Charge Distributions: Introduction to Electrodynamics
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Alright alright alright, so clearly this geometry is a step up in difficulty. The limit cases verify our results but clearly getting there is quite difficult. Notice how messy these integral formulations tend to be! Not only in the setup but in the evaluation of the integrals. If only there was another way to use the geometry and symmetries provided.
Here I draw on the surface of the sphere and redraw the cross section so that you can see how the geometry leads to these vectors and distances. The law of cosines will be used repeated! But all and all, this was really fun!
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187789 | https://blog.mannequinmadness.com/2010/08/mannequin-terminology/ | Mannequin Terminology -
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Mannequin Terminology
August 17, 2010No Comments
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Although mannequins can’t talk, they are very self-expressive. Here are all the terms, defintions and descriptions of mannequins. This information was compiled by Martin M Pelger and the info donated by James ML of www.fashionwindows.com
ABSTRACT MANNEQUIN:A highly stylized, usually non-featured mannequin devoid of wig and / or make-up details. Through based on human measurements and proportions, the shape and sculpting is not realistic and strives instead for a decorative and non-objective effect. A female, male or child mannequin that is ageless, non-ethnic, nonspecific and can be finished in a variety of decorative colors or metallics.
ANKLE ROD:The short, upright bar that extends up from the floor base and inserts into the fitting above the mannequin’s ankle. It is the usual way of keeping male mannequins upright and is also desirable for female mannequins that wear pants. It is almost invisible and usually doesn’t require the opening of any seams for the insertion of the supporting rod. See: Foot Spike.
ARMATURE:The metal framework or skeletal construction upon which the original clay sculpting of a mannequin or form is done. The metal frame inside a molded form or the bendable and shapable skeleton in the stuffed doll type of mannequin which makes possible the repositioning of these soft forms.
ARTICULATED FORMS or MANNEQUINS:Forms similar to, or the life-size variations of, the wooden artist’s mannequins that can be repositioned into myriad human poses. Forms or mannequins with articulated or movable joints (elbows, wrists, knees, hips, etc.) which can be swiveled or turned in different directions.
BASE FLANGE:A flat disk with a perpendicular hollow tube extending up from the plate. The flat plate can be screwed or secured to the floor, onto a base or platform and the hollow tube will accept the mannequin’s butt or ankle rod and thus support the mannequin. See: Ankle Rod, Butt Rod.
BIKINI CUT:A female mannequin with a removable leg and the break or cut line is close to the pubis area and thus will be successfully hidden by the bottom of a bikini swim suit. A more natural look for displaying abbreviated swim wear, sportswear and lingerie.
BLOUSE FORM:An armless and headless, bust defined form which ends just below the waistline. It may be equipped with an adjustable up-and-down rod and a decorative base. It is used to display ladies blouses, sweaters and sometimes jackets.
BODY TRUNK:A torso form, sometimes male, which starts above the waistline and continues down to just below the knees and is used to show walk shorts, underwear, swimwear, etc. See: Torso Form, Trunk Form.
BRA FORM:A headless and armless bust form, with or without shoulders, which ends just below the bustline. For long-line bras and braselettes it is possible to get longer bra forms that continue down to the waistline or slightly below. The form is usually scaled for Misses at 34B and to 32A for Juniors.
BRASELETTE FORM:A bra form that ends at the hips rather than below the bustline or at the waistline. It can also be used to show lingerie and slips. It takes a 348.
BUTT FITTING:A square metal hollow tube with a set screw that is set into the mannequin’s butt or upper thigh. It receives the butt rod which angles up from the metal, glass or plastic mannequin base and holds the mannequin erect. See: Butt Rod.
BUTT ROD:The square metal rod which extends up, at an angle, from the mannequin base. The metal end fits into a square opening, equipped with a set screw, on the mannequin’s butt or upper thigh. The butt rod, when secured in place, keeps the mannequin upright in the position it was meant to hold.
COAT FORM:A headless and usually armless male form which ends above the hips and can be raised or lowered on an adjustable rod. The form will show coats, jackets or sweaters. Most are flatfronted, without chest muscle definition, and the arms, when used, are either bendable rods or sleeve pads.
CUSTOM MANNEQUIN:A mannequin that is especially sculpted to order for a particular customer. It may be a special head which can be used on an existing line body, or it may be an all new form which will express a store’s Image or look. It may also refer to a very individualized makeup, finish, glaze or texture.
DICKEY: A detachable insert which simulates a shirt front, with or without a tie. A bib either of fabric or molded in plastic which fills in for a shirt on a coat or suit form.
DRESS FORM:A headless, armless and legless form which goes from the neck to below the hips. It is often made of papier-mâché and covered in linen, jersey or velvet. There is usually a fitting on the bottom of the form to receive a rod which may be attached to a base, fixture, platform, etc. A dress form averages between 36 and 38 ins., depending upon the neck extension.
DRESSMAKER FORM:A dress form used by designers, tailors, seamstresses, etc. to fit garments in work, or to try out designs or patterns. It is available in a variety of dress sizes and can also be customized for specific bust, waist and hip measurements. The dressmaker form is often seen on a wire basket-like structure supported on an ornate cast-iron base, with or without casters. The dressmaker form is usually associated with custom-made and designer fashions.
DUMMY:Reference for a mannequin most commonly used to describe the headless, legless and armless upholstered dressmaker form.
EGG HEAD:An abstract sculpture of a head which resembles a smooth, oval egg with no defined features or anatomical details. It may be used as a millinery head form or a wig stand, or be the head of an abstract mannequin. The egg head is usually available in a variety of non-skin tones and metallics and in assorted neck lengths. Faces can be painted onto the featureless face.
ETHNIC MANNEQUIN: A mannequin which is realistically portrayed with the skin tone and body and facial physiognomy of a particular ethnic group.
FLEX ARMS: Flexible metal extensions from the arm sockets of coat or suit forms which can be bent or reshaped into various natural arm positions. They may be similar to EX cables or the uprights of goose neck lamps, or they may be metal rods joined with swivel sockets at the shoulder and elbow. With flex arms it is possible to break the rigid dummy look of a coat or suit form and give some semblance of naturalness to a garment.
FOOT BRACKET:A sandal-strap-like device that is attached to a flage or base and will accept and hold a leg form or pantyhose form in an upright position. It is usually made of a clear plastic material. It may also resemble a cup.
FOOT SPIKE:The short metal rod that extends up from the metal, glass or plastic mannequin base and insets into the matching square fitting, with set screw, midway between the heel and the calf of the mannequin leg. It supports and holds the mannequin upright, but can interfere with the use of hosiery or the wearing of pants. See: Ankle Rod.
FORM:A headless mannequin. Specifically a three-dimensional representation of a part or parts of the human anatomy; the torso, bust, shoulder to waist, hips to ankles, etc. See: Torso Form, Bra Form, Coat Form, Dress Form, Blouse Form.
FULLER FIGURE:The larger sized mannequin for the plumper, fuller figure. The female form wears a size 14′/2 and, depending upon the pose, stands about 5 ft. 9 ins. tall. The bust, waist and hips, as well as the arms, legs and head, are proportioned to suit the half-size garment.
GIRDLE FORM:See: Pantie Form.
GLASS BASE PLATE: The heavy piece of glass, equipped with a metal plate or flange and extending rod, which is used to support a mannequin. The base plate may be almost any shape and is usually about 18 ins. wide.
GLASS EYES:Large, artificial, but extremely realistic eyes which fit into the hollowed-out eye sockets of a mannequin. They are usually made so that they can be positioned to look to the right, left, straight ahead, up or down and appear to make contact with other mannequins in a grouping.
GLOVE HAND:An accessory to a mannequin, replacing the regular hand that joins the arm at the wrist or a separate entity, which is designed to wear gloves. The finger arrangement facilitates the putting on and taking off of gloves and yet presents them in a graceful manner.
GROUPING: Two or more mannequins which are designed, arranged or positioned to go together and create a situation or a semblance of belonging in the same place at the same time. Mannequins which are proportioned and posed to be used together.
HEADLESS MANNEQUIN:A complete, life-like mannequin which ends at the neck. Sometimes the neck is straight or flat cut, or it may end in a fanciful swirl. Either way, the decapitated form is a full-size, full-scale, non-personalized representation.
JUNIOR:A mannequin size rather than an age. It is often posed and made-up as a young, active type of woman, but depending upon pose, make-up and wig style, the Junior mannequin can be a college freshman, young executive, or sophisticated, mature lady. It wears a size 7 dress and averages about 5 ft. 8 ins. in height. Hip, bust and waist measurements will vary slightly with the manufacturer and the fashion trends.
JUNIOR PETITE: A special size and type of mannequin which, depending upon the manufacturer, will wear a size 5 or 7 and average about 5 ft. 5 ins. in height. It is often made-up and posed as the superannuated, freckled and braided, saucy and perky teenager, though with the right pose, make-up and wig can represent the smaller woman.
MAKE·UP:Body and facial coloring — the color and art work used on the lips, cheeks and around the eyes. The subtle or dramatic use of rouge, mascara and lipstick — the enhancement of the facial sculpture. The brushwork, the blending or sharpness of line and color. The mannequin type or Image as personalized by art work.
MANNEQUIN, MANNEKIN, MANIKEN:A three-dimensional representation of the human form, somewhat idealized and stylized, to show wearing apparel. It may be a realistic interpretation, semirealistic or abstract. Mannequins will vary in sizes and proportions depending upon type, age group, manufacturer and current fashion look.
METAL BASE:A flat base with perpendicular flange into which the butt rod or ankle rod fits. It is heavy enough to sit on the ground and support the mannequin in an erect position. See: Butt Rod, Base Flange.
MISSY or MISS:A female mannequin that wears a size 8 and varies in height from 5 ft. 8 ins. to 6 ft. tall. There is a great variety of interpretation in the Missy group and depending upon the manufacturer, the pose, make-up and wig, the Missy can be the young college type, active sportswoman, career woman, super-sophisticate or grandest lady at the ball. The Missy mannequin can be personalized to represent the store Image.
NECK BLOCK or NECK CAP:The cap, usually metal, that finishes off the neck top of a suit, coat, shirt or other headless form. It is usually finished in chrome or brass and sometimes topped with a finial or ring.
PADDLE HAND:See: Pocket Hand.
PANTS FORM:A three-dimensional male or female form which starts at the waistline and ends at the feet. In men’s forms, it usually wears a size 30 with a 32 in. length. Boy’s forms take a 28 in. waist with a 30 in. length. Female pants forms wear a size 8. See: Slack Form.
PANTIE or PANTIE FORM:A three-dimensional form that entends from above the waist to the knees and is used for the counter and ledge display of panties, girdles and bikini bottoms. The longer leg girdle form shows more leg and averages 25 to 26 ins, in height to the pantie form’s 23 inches.
PANTIEHOSE or PANTYHOSE FORM:Usually a lightweight, three-dimensional form, flesh colored waist to toes. It may be used toes up and waist as the base, or upright with the toe set into a foot bracket. It may also be used to show stretch tightsand slacks.
PEARLESCENT or PEARLIZED FINISH:A lustrous, milky finish on a mannequin or form reminiscent of the sheen of a pearl. An opalescent or macrous quality often used to tone down a color.
PETITE:A smaller, slighter mannequin that usually wears a size 5 and is not necessarily posed or made-up as the superannuated teenager. A mannequin for the presentation of merchandise cut for the smaller, slighter woman.
PLEXIGLASS BASE:Similar in deisgn and function to the glass base plate, but made of a heavy-weight, heavy-duty, non-breakable plastic material. See: Glass Base Plate.
POCKET HAND:Also called a paddle hand. A hand lacking in anatomical details that resembles a shaped thumbless mitten that fits into a pocket and presents a smooth outline of a hand rather than bumps and knuckles. It is also flatter and more tapered to fit into and out of pockets.
PORCELAIN FINISH:Usually a shiny, white enamel finish which is similar to the gloss of fine china, but may also refer to other pastel lacquer finishes.
PORTRAIT FINISH:The matte, lightly stippled texture and finish of a mannequin’s face and body. The skin appears more porous and thus more realistic. The flat finish adds depth to the skin color which varies as per the customer’s specifications.
POSE:Body position of a mannequin or form. The relationship of the arms and legs to the body, the body line, the angle of the head to the shoulders. Movement and attitude of the body composition.
PRE-TEEN:The ten-to-twelve-year-oId mannequin which is still a child-like mannequin, but has the beginnings of a more mature body. Usually proportioned to wear a girl’s size 8 to 10.
REALISTIC MANNEQUIN:A full round sculpted form that resembles in face, pose and proportions a particular type and size woman, man or child. Not abstract.
RIGGING or RIG:Dressing of a mannequin or form. The padding, pinning and plumping out of merchandise on the inanimate object to make the merchandise look and fit better and to emphasize the best features of the garment.
SEMI-ABSTRACT MANNEQUIN:An abstract or highly stylized mannequin which may suggest either in the sculpting or the artwork a face or semblance to a particular type of individual. Sometimes a mannequin with sculptured features but no make-up or tonal qualities.
SEMI·REALISTIC MANNEQUIN: A fairly realistic sculpted mannequin which is made more stylized by the vagueness of the facial features or the lack or realism in the art work or make-up. Wigs may be replaced by painted hair styles, or the hair may be sculpted in and not detailed.
SHELL FORM:A half-round form. Usually a lightweight, plastic bra, blouse, sweater or dress form with a fully dimensional front but scooped out back for one-sided merchandise presentation.
SHIRT FORM:The male version of the blouse form. See: Blouse Form.
SLACK FORM:A sculpted dimensional form of a pair of legs and hips that end at the waist or slightly above. If the legs are crossed, one leg will be removable. To facilitate standing, the form usually comes with a foot spike that holds the form erect without affecting the fit of the garment. See: Pants Form.
SLEEVE PADS:Flattened, sausage-shaped cushions that are used to pad out or fill in the sleeves of coats and jackets shown on armless coat or suit forms. They add dimension and form to what would otherwise be limp appendages to a contoured garment.
STOCKING FORMS:Flesh-colored, dimensional, plastic leg shaped, usually with hollow tops, available in assorted lengths depending upon the type of merchandise to be displayed: thigh-high; knee-high or calf-high. The forms are usually shown with the instep arched and the weight on the ball of the foot. Some stocking forms are two-dimensional, plastic shapes with a twist that are presented toe up with the thigh as a base.
SUPPORT ROD:See: Ankle Rod, Butt Rod, Foot Spike.
THREE QUARTER FORM:A dimensional form or mannequin, with head, arms and separated legs, that ends at the knee or just below. Usually used on ledges or where there isn’t sufficient height to accommodate a full-standing mannequin.
THREE QUARTER MANNEQUIN:See: Three Quarter Form.
TORSO:See: Torso Form.
TORSO FORM:A full round, headless and armless body form that ends just above the knees and is used for showing bathing suits, lingerie and sportswear. The average height is 38 ins, for the female and 43 ins. for the male figure. It differs from the usual dress form in that the body line is more animated and the legs are parted and defined.
T·ROD: An upside-down, T-shaped metal device. The upright end secures into the butt fitting of the mannequin while the flat, horizontal bar rests on the ground and serves as an easel base for the mannequin.
TRUNK FORM:A masculine body part that extends from above the waistline to the knees or slightly below. It is ideally suited for showing shorts, swim trunks and briefs. This form is usually designed to wear a size 30. Longer forms will show more leg while the shorter forms will cut in mid-length.
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187790 | https://www.calctool.org/fluid-mechanics/darcy-weisbach | Share via
Darcy–Weisbach Equation Calculator
Created by Luis Hoyos
Last updated: Jul 09, 2022
Table of contents:
Darcy–Weisbach equation for pressure drop
Darcy friction factor calculation for pressure drop
Welcome to the Darcy–Weisbach equation calculator, a tool to calculate the pressure drop in a pipe. Input the pipe length, flow velocity, fluid density, friction factor, and pipe diameter, and the pressure drop will appear immediately.
To calculate the pressure drop, you need to know the friction factor (f). You can find it using our darcy friction factor calculator.
Use our flow rate calculator if you need to know the flow velocity but only know the flow rate.
Density is another variable of the pressure drop equation. In our density calculator, you can find the density of many materials, including fluids.
Additionally, you can use this calculator in various ways, and not only to calculate the pressure drop. For example, if you're designing a pipe system, you can use it to know the necessary pipe diameter if pressure drop and flow conditions are known.
Darcy–Weisbach equation for pressure drop
The equation to calculate pressure drop in a pipe is:
ΔP=fLρV2/2D
where:
ΔP — Pressure drop, in Pascals (Pa);
f — Darcy friction factor, dimensionless;
L — Pipe length, in meters (m);
D — Pipe hydraulic diameter (pipe inner diameter), in meters (m);
ρ — Fluid density, in kg/m3; and
V — Average flow velocity through the pipe, in meters per second (m/s).
If you don't know the flow velocity, you can calculate it by remembering that flow velocity equals flow rate divided by the area covered by the hydraulic diameter:
V=Q/A=Q/(πD2/4)=4Q/πD2
where:
Q — Volumetric flow rate, in cubic meters per second (m³/s);
A — Area covered by the hydraulic diameter, in square meters (m²).
Darcy friction factor calculation for pressure drop
To calculate the Darcy–Weisbach equation, calculating the Darcy friction factor is necessary. The friction factor formula will differ depending on whether the flow is laminar or turbulent.
Friction factor for laminar flow
For laminar flow, the friction factor only depends on the dimensionless quantity known as Reynolds number (Re):
f=64/Re
Friction factor for turbulent flow
For turbulent flow, the friction factor is given by the Colebrook-White equation:
f1=−2log(3.7k/D+Ref2.52)
where:
k — Pipe's surface roughness, in meters (m);
D — Pipe hydraulic diameter, in meters (m); and
The ratio k/D is commonly known as relative roughness.
Reynolds number
We say a flow is laminar if its Reynolds number is below 2100 (Re < 2100).
We consider a flow as turbulent if its Reynolds number exceeds 3000 (Re > 3000).
When 2100 < Re < 3000, the flow is transitional, a situation in which it intermittently changes from laminar to turbulent.
For a pipe, the Reynolds number is given by:
Re=ρVL/μ=VL/ν
where:
μ — Dynamic viscosity of the fluid, in kg/(m·s); and
ν — Kinematic viscosity of the fluid (ν=μ/ρ), in m²/s.
Hopefully, now you have a broader sense of the formulas behind this pipe pressure drop calculation. Thanks for reading!
Luis Hoyos
Pipe length (L)
ft
Pipe diameter (D)
ft
Flow velocity (V)
ft/s
Density of fluid (ρ)
lb/cu ft
Darcy friction factor (f)
Pressure drop (ΔP)
psi
To calculate Darcy friction factor (f), try our dedicated friction factor calculator.
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187791 | https://www.quora.com/What-is-the-ratio-of-the-circumference-of-a-circle-with-the-radius-R-to-its-areas | Something went wrong. Wait a moment and try again.
Ratio Explanation
Circle Area
Circumference Theorem
Area Reference
Circle (shape)
5
What is the ratio of the circumference of a circle with the radius R to its areas?
·
To find the ratio of the circumference of a circle to its area, we start with the formulas for both:
Circumference (C) of a circle with radius R:
C=2πR
Area (A) of a circle with radius R:
A=πR2
Now, we can form the ratio of the circumference to the area:
Ratio=CA=2πRπR2
Simplifying this expression:
Ratio=2R
Thus, the ratio of the circumference of a circle with radius R to its area is:
2R
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Manji Lambert
B.Tech in Marine Engineering, Arab Academy for Science, Technology & Maritime Transport (Graduated 2019)
·
6y
Well pi (22/7 or 3.142) is the ratio of the circumference of the circle to its diameter and it is a constant. while the ratio of the circumference to the area is 2/r .. that is as far as i know
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Circumference = 2piR
Area =piR^2
Ratio = (2piR) /(piR^2)
Ratio = 2/R
Howard Medhurst
Former Software Developer
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Author has 6.5K answers and 4.3M answer views
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The ratio of the circumference to the area is meaningless; you can’t compare 1-dimensional length with 2-dimensional area.
For example, a circle of radius 1 yard would have a circumference of 2π yards, and an area of π square yards, giving a circumference:area ratio of 2:1; while a circle of radius 3 feet would have a circumference of 2π×3 feet, and an area of π×32 square feet, giving a circumference:area ratio of 6:9. But there are 3 feet in a yard, so it’s the same circle!
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Mike Raguin
Former Computer Engineer
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The area of a circle is equal to (1/2) circumference times its radius. This was a formula discovered by Archimedes over 2,000 years ago. At that time the number pi was not known yet.
From that formula we see that the ratio of the circumference of a circle to its area is 2/R.
Sanjiv Soni
Former Pvt Tutor
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Author has 9.7K answers and 1.9M answer views
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Circumference=2πR
Area of the circle=πR^2=πRR
Ratio of the circumference to the area of the circle of radius (R)
=2πR/πRR
=2/R
=2:R
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Author has 2.1K answers and 3.3M answer views
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Related
What part of a circle are the radius, diameter, circumference, area, and center?
This is an excellent question so let’s start off by drawing a simple circle.
The black line that makes the circle is called the circumference. The circumference of a circle is the distance all the way round the circle.
The point that is equally far away from every point on the circumference is called the center of the circle.
The distance from the center (usually spelled centre outside the USA) of th
This is an excellent question so let’s start off by drawing a simple circle.
The black line that makes the circle is called the circumference. The circumference of a circle is the distance all the way round the circle.
The point that is equally far away from every point on the circumference is called the center of the circle.
The distance from the center (usually spelled centre outside the USA) of the circle to the circumference is called the radius, it is the same all the way round the circumference. This is the defining property of a circle, everywhere on the circumference is exactly the same distance from the centre and this distance is the radius.
if you draw a line from one part of the circumference to another part of the circumference that line is called a chord. The largest possible chord is the largest possible width of the circle which is called the diameter. Since the radius goes from the center of the circle to the circumference the diameter must be exactly double the length of the radius.
The total space inside the circle is called the area of the circle. The area of the circle is the amount of space the circle takes up.
There are a couple of other terms you should probably be aware of if you are studying circles at school or...
Greg Gruzalski
Former Senior Scientist at Oak Ridge National Laboratory (1980–2011)
·
Author has 222 answers and 545.5K answer views
·
7y
Related
A circle has a circumference of 60.4cm with a diameter of 19cm. What is the ratio?
A circle has a circumference of 60.4cm with a diameter of 19cm. What is the ratio?
I don’t know where those numbers came from, but let’s assume that they came from measurements made in a lab, maybe at a high school. The ratio of circumference to diameter is
cd=60.4 cm19 cm=3.18.
The circle and the diameter most likely were constructed with relatively little error (when compared with the errors in measurement). That is, the error in the ratio is likely due to uncertainties in the measurements of c and d. If we let these uncertainties equal δc and δ
A circle has a circumference of 60.4cm with a diameter of 19cm. What is the ratio?
I don’t know where those numbers came from, but let’s assume that they came from measurements made in a lab, maybe at a high school. The ratio of circumference to diameter is
cd=60.4 cm19 cm=3.18.
The circle and the diameter most likely were constructed with relatively little error (when compared with the errors in measurement). That is, the error in the ratio is likely due to uncertainties in the measurements of c and d. If we let these uncertainties equal δc and δd, respectively, then the experimentally determined ratio cd could be approximated by
3.18 (1±√(δcc)2+(δdd)2 )
If the uncertainties were each about 1%, we might express the ratio as
3.18 (1±√(0.01)2+(0.01)2 )=3.18±0.04
Note that π would be within the stated experimental error.
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Roger Pickering
Spent 6 years at 2 universities doing maths
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Author has 14.9K answers and 5.9M answer views
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3y
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What makes the ratio of the circumference of a circle to its diameter irrational? Is it because the circumference Of a circle has no exact value?
No. It is the ratio of the circumference to the diameter that is the irrational number pi.
The symbol for that is π which is just the Greek letter p. They even understood ways to calculate pretty accurate estimates of the value of π. For example Archimedes proved that π is between 22371 and 227.
[Approximations of π - Wikipedia
Varying methods used to calculate pi This page is about the history of approximations of π ; see also chronology of computation of π for a tabular summary. See also the history of π for other aspects of the evolution of our knowledge about mathematical properties of π . Graph showing the historical evolution of the record precision of numerical approximations to pi, measured in decimal places (depicted on a logarithmic scale; time before 1400 is not shown to scale) Approximations for the mathematical constant pi ( π ) in the history of mathematics reached an accuracy within 0.04% of the true value before the beginning of the Common Era . In Chinese mathematics , this was improved to approximations correct to what corresponds to about seven decimal digits by the 5th century. Further progress was not made until the 14th century, when Madhava of Sangamagrama developed approximations correct to eleven and then thirteen digits. Jamshīd al-Kāshī achieved sixteen digits next. Early modern mathematicians reached an accuracy of 35 digits by the beginning of the 17th century ( Ludolph van Ceulen ), and 126 digits by the 19th century ( Jurij Vega ). The record of manual approximation of π is held by William Shanks , who calculated 527 decimals correctly in 1853. [ 1 ] Since the middle of the 20th century, the approximation of π has been the task of electronic digital computers (for a comprehensive account, see Chronology of computation of π ). On April 2, 2025, the current record was established by Linus Media Group and Kioxia with Alexander Yee's y-cruncher with 300 trillion (3× 10 14 ) digits. [ 2 ] The best known approximations to π dating to before the Common Era were accurate to two decimal places; this was improved upon in Chinese mathematics in particular by the mid-first millennium, to an accuracy of seven decimal places. After this, no further progress was made until the late medieval period. Some Egyptologists [ 3 ] have claimed that the ancient Egyptians used an approximation of π as 22 ⁄ 7 = 3.142857 (about 0.04% too high) from as early as the Old Kingdom (c. 2700–2200 BC). [ 4 ] This claim has been met with skepticism. [ 5 ] [ 6 ] Babylonian mathematics usually approximated π to 3, sufficient for the architectural projects of the time (notably also reflected in the description of Solomon's Temple in the Hebrew Bible ). [ 7 ] The Babylonians were aware that this was an approximation, and one Old Babylonian mathematical tablet excavated near Susa in 1936 (dated to between the 19th and 17th centuries BCE) gives a better approximation of π as 25 ⁄ 8 = 3.125, about 0.528% below the exact value. [ 8 ] [ 9 ] [ 10 ] [ 11 ] At about the same time, the Egyptian Rhind Mathematical Papyrus (dated to the Second Intermediate Period , c. 1600 BCE, although stated to be a copy of an older, Middle Kingdom text) implies an approximation of π as 256 ⁄ 81 ≈ 3.16 (accurate to 0.6 percent) by calculating the area of a circle via approximation with the octagon . 5
If you have a right angled triangle whose two short sides have length 1, then the longer side opposite them is called the hypothenuse and the ratio of that to one of the short sides is also irrational and because that is the square root of 2 it i
No. It is the ratio of the circumference to the diameter that is the irrational number pi.
The symbol for that is π which is just the Greek letter p. They even understood ways to calculate pretty accurate estimates of the value of π. For example Archimedes proved that π is between 22371 and 227.
[Approximations of π - Wikipedia
Varying methods used to calculate pi This page is about the history of approximations of π ; see also chronology of computation of π for a tabular summary. See also the history of π for other aspects of the evolution of our knowledge about mathematical properties of π . Graph showing the historical evolution of the record precision of numerical approximations to pi, measured in decimal places (depicted on a logarithmic scale; time before 1400 is not shown to scale) Approximations for the mathematical constant pi ( π ) in the history of mathematics reached an accuracy within 0.04% of the true value before the beginning of the Common Era . In Chinese mathematics , this was improved to approximations correct to what corresponds to about seven decimal digits by the 5th century. Further progress was not made until the 14th century, when Madhava of Sangamagrama developed approximations correct to eleven and then thirteen digits. Jamshīd al-Kāshī achieved sixteen digits next. Early modern mathematicians reached an accuracy of 35 digits by the beginning of the 17th century ( Ludolph van Ceulen ), and 126 digits by the 19th century ( Jurij Vega ). The record of manual approximation of π is held by William Shanks , who calculated 527 decimals correctly in 1853. [ 1 ] Since the middle of the 20th century, the approximation of π has been the task of electronic digital computers (for a comprehensive account, see Chronology of computation of π ). On April 2, 2025, the current record was established by Linus Media Group and Kioxia with Alexander Yee's y-cruncher with 300 trillion (3× 10 14 ) digits. [ 2 ] The best known approximations to π dating to before the Common Era were accurate to two decimal places; this was improved upon in Chinese mathematics in particular by the mid-first millennium, to an accuracy of seven decimal places. After this, no further progress was made until the late medieval period. Some Egyptologists [ 3 ] have claimed that the ancient Egyptians used an approximation of π as 22 ⁄ 7 = 3.142857 (about 0.04% too high) from as early as the Old Kingdom (c. 2700–2200 BC). [ 4 ] This claim has been met with skepticism. [ 5 ] [ 6 ] Babylonian mathematics usually approximated π to 3, sufficient for the architectural projects of the time (notably also reflected in the description of Solomon's Temple in the Hebrew Bible ). [ 7 ] The Babylonians were aware that this was an approximation, and one Old Babylonian mathematical tablet excavated near Susa in 1936 (dated to between the 19th and 17th centuries BCE) gives a better approximation of π as 25 ⁄ 8 = 3.125, about 0.528% below the exact value. [ 8 ] [ 9 ] [ 10 ] [ 11 ] At about the same time, the Egyptian Rhind Mathematical Papyrus (dated to the Second Intermediate Period , c. 1600 BCE, although stated to be a copy of an older, Middle Kingdom text) implies an approximation of π as 256 ⁄ 81 ≈ 3.16 (accurate to 0.6 percent) by calculating the area of a circle via approximation with the octagon . 5
If you have a right angled triangle whose two short sides have length 1, then the longer side opposite them is called the hypothenuse and the ratio of that to one of the short sides is also irrational and because that is the square root of 2 it is easily proved to be irrational.
So this kind of property is not unique to circles, and π is by no means the only irrational number we encounter in mathematics.
Khalik Shaikh
B E Chemical Engineering
·
Author has 90 answers and 42.5K answer views
·
3y
Related
What is the ratio of the circumference of a circle to its area?
2pi r divided by pi r^2 = 2/r
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Robert Sheraw
Author has 1.3K answers and 215K answer views
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3y
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What is the ratio of the circumference of a circle to its area?
Using C for circumference and A for area, C/A=(2πr)/(πr^2)=2/r.
Huzaifa Ali
.
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 71 answers and 527K answer views
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8y
Related
What is the ratio of the circumference of a circle to its radius?
As we all know
Circumference of circle : 2πr
Radius of circle : r
Ratio = 2πr : r = 2πr/r
Thus we conclude that 2π:1
Roger Pickering
First class degree in mathematics.
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How do you calculate the ratio of the circumference of a circle if the circumference is 27 cm?
As you know the circumference of the circle, I will resume you meant to ask how to calulate the radius.
The formula for the circumference is 2πr so the radius = circumference /2π
To get a very approximate answer we can use a crude estimate of π=3 giving an estimated radius of 27/6=9/2 (or 4.5) cm.
You may have been told to use the value 22/7 for π which is also an approximation. That would give the radius equal to 27/2 X 7/22 which is 189/44, or about 4.3 cm.
That is also only approximate because you can never have an exact value for π that is either the ratio of two whole numbers or a
As you know the circumference of the circle, I will resume you meant to ask how to calulate the radius.
The formula for the circumference is 2πr so the radius = circumference /2π
To get a very approximate answer we can use a crude estimate of π=3 giving an estimated radius of 27/6=9/2 (or 4.5) cm.
You may have been told to use the value 22/7 for π which is also an approximation. That would give the radius equal to 27/2 X 7/22 which is 189/44, or about 4.3 cm.
That is also only approximate because you can never have an exact value for π that is either the ratio of two whole numbers or a decimal with a finite number of digits.
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187792 | https://www.doubtnut.com/qna/644015030 | Find the local maximum and local minimumof f ( x ) = x 3 - 3 x in [-2,4].
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187793 | https://en.wikipedia.org/wiki/Orthographic_map_projection | Orthographic map projection - Wikipedia
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1 History
2 Mathematics
3 Orthographic projections onto cylinders
4 See also
5 References
6 External links
Orthographic map projection
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From Wikipedia, the free encyclopedia
Azimuthal perspective map projection
Orthographic projection (equatorial aspect) of eastern hemisphere 30W–150E
The orthographic projection with Tissot's indicatrix of deformation.
Orthographic projection in cartography has been used since antiquity. Like the stereographic projection and gnomonic projection, orthographic projection is a perspective projection in which the sphere is projected onto a tangent plane or secant plane. The point of perspective for the orthographic projection is at infinite distance. It depicts a hemisphere of the globe as it appears from outer space, where the horizon is a great circle. The shapes and areas are distorted, particularly near the edges.
History
[edit]
The orthographic projection has been known since antiquity, with its cartographic uses being well documented. Hipparchus used the projection in the 2nd century BC to determine the places of star-rise and star-set. In about 14 BC, Roman engineer Marcus Vitruvius Pollio used the projection to construct sundials and to compute sun positions.
Vitruvius also seems to have devised the term orthographic (from the Greek orthos (= “straight”) and graphē (= “drawing”)) for the projection. However, the name analemma, which also meant a sundial showing latitude and longitude, was the common name until François d'Aguilon of Antwerp promoted its present name in 1613.
The earliest surviving maps on the projection appear as crude woodcut drawings of terrestrial globes of 1509 (anonymous), 1533 and 1551 (Johannes Schöner), and 1524 and 1551 (Apian). A highly-refined map, designed by Renaissance polymathAlbrecht Dürer and executed by Johannes Stabius, appeared in 1515.
Photographs of the Earth and other planets from spacecraft have inspired renewed interest in the orthographic projection in astronomy and planetary science.
Mathematics
[edit]
The formulas for the spherical orthographic projection are derived using trigonometry. They are written in terms of longitude (λ) and latitude (φ) on the sphere. Define the radius of the sphereR and the centerpoint (and origin) of the projection (λ 0, φ 0). The equations for the orthographic projection onto the (x, y) tangent plane reduce to the following:
x=R cosφ sin(λ−λ 0)y=R(cosφ 0 sinφ−sinφ 0 cosφ cos(λ−λ 0)){\displaystyle {\begin{aligned}x&=R\,\cos \varphi \sin \left(\lambda -\lambda {0}\right)\y&=R{\big (}\cos \varphi {0}\sin \varphi -\sin \varphi {0}\cos \varphi \cos \left(\lambda -\lambda {0}\right){\big )}\end{aligned}}}
Latitudes beyond the range of the map should be clipped by calculating the angular distancec from the center of the orthographic projection. This ensures that points on the opposite hemisphere are not plotted:
cosc=sinφ 0 sinφ+cosφ 0 cosφ cos(λ−λ 0){\displaystyle \cos c=\sin \varphi {0}\sin \varphi +\cos \varphi {0}\cos \varphi \cos \left(\lambda -\lambda {0}\right)\,}.
The point should be clipped from the map if cos(_c) is negative. That is, all points that are included in the mapping satisfy:
−π 2<c<π 2{\displaystyle -{\frac {\pi }{2}}<c<{\frac {\pi }{2}}}.
The inverse formulas are given by:
φ=arcsin(cosc sinφ 0+y sinc cosφ 0 ρ)λ=λ 0+arctan(x sinc ρ cosc cosφ 0−y sinc sinφ 0){\displaystyle {\begin{aligned}\varphi &=\arcsin \left(\cos c\sin \varphi {0}+{\frac {y\sin c\cos \varphi {0}}{\rho }}\right)\\lambda &=\lambda {0}+\arctan \left({\frac {x\sin c}{\rho \cos c\cos \varphi {0}-y\sin c\sin \varphi _{0}}}\right)\end{aligned}}}
where
ρ=x 2+y 2 c=arcsinρ R{\displaystyle {\begin{aligned}\rho &={\sqrt {x^{2}+y^{2}}}\c&=\arcsin {\frac {\rho }{R}}\end{aligned}}}
For computation of the inverse formulas the use of the two-argument atan2 form of the inverse tangent function (as opposed to atan) is recommended. This ensures that the sign of the orthographic projection as written is correct in all quadrants.
The inverse formulas are particularly useful when trying to project a variable defined on a (λ, φ) grid onto a rectilinear grid in (x, y). Direct application of the orthographic projection yields scattered points in (x, y), which creates problems for plotting and numerical integration. One solution is to start from the (x, y) projection plane and construct the image from the values defined in (λ, φ) by using the inverse formulas of the orthographic projection.
See References for an ellipsoidal version of the orthographic map projection.
Comparison of the Orthographic map projection and some azimuthal projections centred on 90°N at the same scale, ordered by projection altitude in Earth radii. (click for detail)
Orthographic projections onto cylinders
[edit]
In a wide sense, all projections with the point of perspective at infinity (and therefore parallel projecting lines) are considered as orthographic, regardless of the surface onto which they are projected. Such projections distort angles and areas close to the poles.[clarification needed]
An example of an orthographic projection onto a cylinder is the Lambert cylindrical equal-area projection.
See also
[edit]
List of map projections
Stereographic projection in cartography
References
[edit]
^ Jump up to: abSnyder, J. P. (1987). Map Projections—A Working Manual (US Geologic Survey Professional Paper 1395). Washington, D.C.: US Government Printing Office. pp.145–153.
^ Jump up to: abcdSnyder, John P. (1993). Flattening the Earth: Two Thousand Years of Map Projections pp. 16–18. Chicago and London: The University of Chicago Press. ISBN9780226767475.
^Zinn, Noel (June 2011). "Ellipsoidal Orthographic Projection via ECEF and Topocentric (ENU)"(PDF). Retrieved 2011-11-11.
External links
[edit]
Wikimedia Commons has media related to Orthographic projection (cartography).
Orthographic Projection—from MathWorld
| show v t e Map projection |
| History List Portal |
| | show By surface | | | Cylindrical | | Mercator-conformal | Gauss–Krüger Transverse Mercator Oblique Mercator | | Equal-area | Balthasart Behrmann Gall–Peters Hobo–Dyer Lambert Smyth equal-surface Trystan Edwards | | Cassini Central Equirectangular Gall stereographic Gall isographic Miller Space-oblique Mercator Web Mercator | | | Pseudocylindrical | | Equal-area | Collignon Eckert II Eckert IV Eckert VI Equal Earth Goode homolosine Mollweide Sinusoidal Tobler hyperelliptical | | Kavrayskiy VII Wagner VI Winkel I and II | | | Conical | Albers Equidistant Lambert conformal | | Pseudoconical | Bonne Bottomley Polyconic American Chinese Werner | | Azimuthal (planar) | | General perspective | Gnomonic Orthographic Stereographic | | Equidistant Lambert equal-area | | | Pseudoazimuthal | Aitoff Hammer Wiechel Winkel tripel | | |
| | show By metric | | | Conformal | Adams hemisphere-in-a-square Gauss–Krüger Guyou hemisphere-in-a-square Lambert conformal conic Mercator Peirce quincuncial Stereographic Transverse Mercator | | Equal-area | | Bonne | Sinusoidal Werner | | Bottomley | Sinusoidal Werner | | Cylindrical | Balthasart Behrmann Gall–Peters Hobo–Dyer Lambert cylindrical equal-area Smyth equal-surface Trystan Edwards | | Tobler hyperelliptical | Collignon Mollweide | | Albers Briesemeister Eckert II Eckert IV Eckert VI Equal Earth Goode homolosine Hammer Lambert azimuthal equal-area Quadrilateralized spherical cube Strebe 1995 | | | Equidistant in some aspect | Conic Equirectangular Sinusoidal Two-point Werner | | Gnomonic | Gnomonic | | Loxodromic | Loximuthal Mercator | | Retroazimuthal (Mecca or Qibla) | Craig Hammer Littrow | | |
| | show By construction | | | Compromise | Chamberlin trimetric Kavrayskiy VII Miller cylindrical Natural Earth Robinson Van der Grinten Wagner VI Winkel tripel | | Hybrid | Goode homolosine HEALPix | | Perspective | | Planar | Gnomonic Orthographic Stereographic | | Central cylindrical | | | Polyhedral | AuthaGraph Cahill Butterfly Cahill–Keyes M-shape Dymaxion ISEA Quadrilateralized spherical cube Waterman butterfly | | |
| | show See also | | Interruption (map projection) Latitude Longitude Tissot's indicatrix Map projection of the tri-axial ellipsoid | |
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187794 | https://www.medicinenet.com/jock_itch/article.htm | Jock Itch: Causes, Symptoms & Treatment
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health centersskin center
Jock Itch
Medical Author: Gary W. Cole, MD, FAAD
Medical Author: Karthik Kumar, MBBS
Medical Editor: John P. Cunha, DO, FACOEP
Medically Reviewed on 8/1/2024
Introduction
What is jock itch?
Causes
What causes jock itch?
Risk Factors
What are risk factors for jock itch?
Symptoms
What are jock itch symptoms?
Potential to Spread
Can jock itch spread to other parts of the entire body?
When to See a Doctor
When should I see a doctor for jock itch?
Diagnosis
Which healthcare professionals diagnose and treat jock itch?
Doctors Who Treat
How do healthcare professionals diagnose jock itch?
Treatment
What is the treatment for jock itch?
Fungal Jock Itch
How do I treat fungal jock itch?
Bacterial Jock Itch
How do I treat bacterial jock itch?
Best Drug Treatment
What is the best drug for jock itch?
OTC Medications
What over-the-counter medications are effective for jock itch?
Home Remedies
How can you get rid of jock itch on your own?
Fastest Treatment
What is the fastest way to cure jock itch?
Holistic Treatment
What holistic jock itch treatments are available?
Contagious
Is jock itch curable? Is jock itch contagious?
Prognosis
What is the prognosis for jock itch?
Complications
What are possible complications of jock itch?
Prevention
How can you prevent jock itch?
What is jock itch?
Jock itch is a term for any rash that occurs in the male groin area.
Jock itch, or tinea cruris, is a term for any rash that occurs in the male groin area.
Jock itch is a common, itchy rash of the groin. It can produce a very intense itch and is associated with a red or pink rash involving the groin folds and genitals. It is primarily a skin condition in men because of anatomic structures unique to males, the male genitalia.
The symptoms and signs of the condition may come and go, and many cases of jock itch resolve spontaneously without any treatment. Tinea cruris is primarily seen in the groin, although it may spread to the inner thighs, genitals (including the penis, scrotum, labia, and vaginal opening), and anus. The condition causes a red or pink rash on the sides of the groin folds. There may be a dry, scaly, well-demarcated skin rash or a collection of small, pinpoint red or pink bumps at each hair follicle. This form of eruption is often called ringworm because of its well-defined red edge with central clearing. The medical term for ringworm of the groin is tinea cruris, and it is caused by a fungal infection.
While tinea cruris is frequently noted in otherwise healthy people, those with diabetes and/or obesity are more susceptible. Possible causes include irritation from tight-fitting or abrasive underwear, excess moisture, sweating, skin rubbing or friction, allergic problems, fungal infection, Candida (yeast) infection, and bacterial overgrowth.
Treatment of fungal-related jock itch may include one or a combination of antifungal creams and, rarely, antifungal medications. Treatment of jock itch that is not caused by fungus involves proper groin hygiene, keeping the genital area clean and dry, and washing frequently with gentle soap and water (especially after sweating or exercise).
What causes jock itch?
Causes of the condition include the following:
Warmth, skin friction, and moist areas in the groin
Tight, occlusive clothing and undergarments that trap sweat
Infections caused by fungus and yeasts: Candida (yeast), Trichophyton, and Epidermophyton (fungal molds)
Infections by certain types of bacteria
Generally, diet does not seem to affect tinea cruris.
What are risk factors for jock itch?
Jock itch is most common in adult and middle-aged men. Anyone can get it and it is thought to affect nearly all people at some point in their lives.
Certain groups of people may be more prone to the condition. Patients with diabetes, obesity, and those with a compromised immune system such as from HIV/AIDS, hepatitis, chronic illnesses, cancer, systemic chemotherapy, immunosuppressive drugs such as prednisone, and those on biologic immune-system-modifying drugs such as infliximab (Remicade) or etanercept (Enbrel) may be more prone to tinea cruris.
Other risk factors include:
Male genitalia
Heat
Moisture
Humidity
Excess sweating
Exercise
A weakened immune system
Tight, occlusive fabrics and undergarments
Athlete's foot infection or other fungal infections on the body
What are jock itch symptoms?
Jock itch usually begins with mild intermittent itching in the groin. The itching can get worse and become unbearable in some cases. The rash is usually on both sides of the groin and affects the folds.
The rash may become dry, rough, and bumpy, develop pus blisters, or begin to ooze. Sometimes, there is central clearing as the redness of the rash spreads outward to the thighs. The itching and rash can spread to the genitals, including the labia, vagina, scrotum, penis, and anus.
Women may also develop vaginal white discharge and yeast infections. Men may develop infections on the head of the penis (balanitis), especially if they are not circumcised.
Severe cases may be very uncomfortable and develop secondary complications such as breaks in the skin, open sores, ulcers, and rarely cellulitis (bacterial infection of the skin and underlying soft tissues).
Can jock itch spread to other parts of the entire body?
The condition does not affect the entire body. It is usually limited to the groin, inner thigh folds, genitals, and anal area. Itching (pruritus) of the entire body is not typical of tinea cruris.
#### SLIDESHOW Ringworm: Treatment, Pictures, Causes, and Symptoms See Slideshow
When should I see a doctor for jock itch?
How long does it take for jock itch to heal? If jock itch persists for one to two weeks despite proper skin care and the use of over-the-counter medications, it may be necessary to schedule an appointment to see a physician. In addition, if the rash worsens despite medical treatment or if any of the following signs of an advancing skin infection (cellulitis) develop, contact a physician.
Spreading despite treatment
Increasing pain
Rapidly spreading rash
Formation of pus, abscesses, or draining sores
Red streak(s) extending from the groin (called lymphangitis)
Fever or chills
Failing to improve after two weeks of continuous topical treatment
Which healthcare professionals diagnose and treat jock itch?
Most primary care physicians can accurately diagnose and treat tinea cruris. A few other medical conditions may appear just like jock itch and should be examined more closely by a dermatologist.
Other medical conditions can mimic tinea cruris. Some possible mimics include:
Ringworm, also called tinea
Contact dermatitis
Intertrigo
Erythrasma
Impetigo
Diaper rash
Irritant or contact dermatitis
Heat rash
Familial pemphigus
Inverse psoriasis
Jock itch may be associated with athlete's foot, also called tinea pedis. The same fungus that causes athlete's foot in a person may spread to the groin in some cases. It is important to always check the feet for rashes in people with tinea cruris. The spread of the fungus usually occurs when fungal particles pass onto the crotch while putting on underwear. Any foot infection must be treated to avoid the recurrence of the condition.
How do healthcare professionals diagnose jock itch?
The diagnosis of the condition is usually based on the symptoms and skin appearance. A microscopic examination of skin scrapings covered in a drop of potassium hydroxide will confirm a fungal cause of jock itch. Occasionally, a fungal culture of the skin scrapings may be necessary.
Certain bacteria can produce an eruption in the groin indistinguishable from a fungal infection. An examination with a special ultraviolet light, Wood's light, will enable identification.
Occasionally, a small skin biopsy may be used to help the doctor confirm the diagnosis. Rarely, a skin biopsy (surgically taking a small piece of skin using local numbing medicine) that is examined under a microscope may be necessary in atypical or widespread cases. Sometimes skin biopsies help to exclude other possible diagnoses.
A skin swab or culture may be taken and sent to the lab to detect an infectious cause of the condition. A bacterial culture may be useful to detect bacteria, such as Staphylococcus.
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What is the treatment for jock itch?
There are many treatment options and skin care recipes for treating tinea cruris. Since the two primary causes of the condition are excess moisture and fungal infections, treatment depends on the exact cause of the condition.
Treatment of tinea cruris associated with skin irritation and excess moisture should address general measures to keep the groin clean and dry along with the use of zinc oxide ointment.
Treatment of fungal jock itch should include antifungal creams used continuously for two to four weeks. Clotrimazole cream is an effective treatment for both dermatophyte molds and Candida(yeast) and can be purchased without a prescription.
It is important to keep in mind that no therapy is uniformly effective in all people.
How do I treat fungal jock itch?
Keeping the affected areas clean and dry is the mainstay of treatment and prevention of jock itch. Some fungi and bacteria are inhibited by acidic conditions. Patting the area with a solution of 1 part vinegar and 4 parts water and allowing it to dry after washing may inhibit fungi and relieve itching. If sweating is hard to control during an episode of jock itch, the application of drying solutions (astringent) such as vinegar or aluminum acetate (Burow's solution or Domeboro) may be more effective.
Antifungal therapy may also be necessary:
Mild fungal or yeast jock itch may be treated by the following:
washing the groin twice daily with an antifungal shampoo like ketoconazole (Nizoral shampoo) or selenium sulfide (Selsun Blue shampoo).
Moderate fungal or yeast jock itch is often treated by a combination of the following:
washing the groin twice daily with an antifungal shampoo like ketoconazole or selenium sulfide; and
using a topical antifungal cream like miconazole (Monistat, Micatin), clotrimazole (Lotrimin, Mycelex), or terbinafine (Lamisil).
Severe fungal or yeast jock itch is typically treated by a combination of the following:
washing the groin twice daily with an antifungal shampoo like ketoconazole or selenium sulfide;
using a topical antifungal cream like miconazole, clotrimazole, or terbinafine; and
taking an antifungal pill like fluconazole (Diflucan), itraconazole (Sporanox), or terbinafine.
How do I treat bacterial jock itch?
Mild bacterial jock itch may be treated with the following:
antibacterial skin washes and a neomycin-containing ointment.
Moderate bacterial jock itch may be treated with the following:
antibacterial skin washes like chlorhexidine soap twice daily; and
twice-daily application of a topical antibiotic like erythromycin lotion or metronidazole (Flagyl) lotion.
Severe bacterial jock itch may be treated with the following:
antibacterial skin washes like chlorhexidine soap twice daily and
a 5- to 14-day course of an oral antibiotic like cephalexin (Keflex), dicloxacillin, doxycycline, minocycline (Dynacin, Minocin), or erythromycin.
What is the best drug for jock itch?
What is the best cream for jock itch? Overall, the best jock-itch drug is a topical antifungal cream like miconazole, clotrimazole, or terbinafine, assuming the condition is produced by a fungus. If the condition does not improve within two to three weeks of treatment, then a physician should be consulted. The use of the topical antifungal should be suspended for two weeks before seeing the physician.
What over-the-counter medications are effective for jock itch?
Assuming the condition is not caused by an infectious agent, itching from jock itch can be treated with a short course of one of the following:
Noncorticosteroid creams should be tried first, such as over-the-counter 1% pramoxine, which relieves itching and is mildly anesthetic (numbing).
Hydrocortisone cream should be avoided if possible, or given only for as long as needed to reduce inflammation, as they can promote the growth of yeasts.
Use a short five- to seven-day course of a mild to medium potency, topical steroid cream like prescription triamcinolone 0.025% once or twice a day for inflamed or itchy areas.
Use a short five- to seven-day course of a mild over-the-counter topical steroid cream like 1% hydrocortisone (Cortaid) one to three times a day for itching.
How can you get rid of jock itch on your own?
What are the best home remedies for jock itch? Home remedies for mild jock itch include the following:
Wash the groin skin two to three times a day.
Keep the groin area dry.
Avoid excess groin skin irritation by wearing 100% cotton underwear.
Avoid fabric softeners, bleaches, or harsh laundry detergents.
Apply a solution of 1 part vinegar and 4 parts water two to three times a day to the area after washing and allow to dry.
Apply a mix of over-the-counter hydrocortisone cream and clotrimazole (Lotrimin, Mycelex) cream one to two times a day to the affected area and cover this liberally with zinc oxide ointment.
What is the fastest way to cure jock itch?
The fastest way to cure jock itch involves a combination of good hygiene, over-the-counter antifungal treatments, and sometimes prescription antifungal medications.
Keep the area clean and dry:
Wash the affected area with an antifungal soap and water daily. Dry thoroughly after washing and use an antifungal dusting powder to fight fungal infections caused by sweat and moisture accumulation, as they worsen the infection.
Keep the affected area cool. Avoid excessive heat and sweating in the groin area, as it can worsen the jock itch.
Use over-the-counter antifungal treatments:
Apply antifungal creams, powders, or sprays such as clotrimazole, miconazole, or terbinafine as directed on the packaging.
Continue using the antifungal treatment for the full recommended duration, even if symptoms improve earlier.
Wear loose, breathable clothing:Wear loose-fitting, cotton underwear and clothing to keep the area dry and reduce irritation.
Maintain good hygiene:
Change clothes, especially underwear, daily.
Wash workout clothes or any clothing that comes in contact with the infected area regularly.
Avoid scratching, as it can spread the infection and cause additional irritation.
Consider prescription medications: If over-the-counter treatments are not effective, consult a doctor. Prescription-strength antifungal creams or oral medications may be necessary.
Avoid sharing personal items: Do not share towels, clothing, or personal items to prevent spreading the infection to others.
Jock itch is contagious. To prevent the infection from spreading, take precautions such as avoiding intimate contact with others, washing your hands thoroughly after touching the affected area, and covering the area with a clean towel while using communal facilities such as gym showers or changing rooms.
By following these measures and maintaining good hygiene practices, most cases of jock itch can be effectively treated and resolved. If your jock itch isn’t getting better, or if it’s getting worse, see a health care provider.You might need to start a prescription-strength antifungal cream or oral medication to fight off the fungus.
What holistic jock itch treatments are available?
Holistic (nonmedicated) home remedy options for tinea cruris include:
soaking the affected area daily with a washcloth dipped in dilute white vinegar (1 part vinegar to 4 parts water) and drying the skin; and
soaking in a bathtub daily or every other day with very dilute Clorox bleach (1-quarter cup of Clorox bleach in a bathtub full of water) and drying the skin.
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Is jock itch curable? Is jock itch contagious?
Most cases of tinea cruris are easily and fully curable. There are very uncommon, long-standing cases of jock itch that may not be cured. Often these more resistant cases may be controlled with proper treatment and tinea cruris medication. The condition sometimes clears completely by itself without treatment.
Although most cases of tinea cruris are not contagious, cases caused by an infection may be transmitted through skin or sexual contact, sharing of swimwear, or towels. It is possible to transmit fungal jock itch to someone else through close skin contact.
Some people are simply more prone to developing jock itch because of their overall health, activity, anatomy, possible altered immune status, exposure history, and other predisposing skin conditions like eczema. People with athlete's foot (tinea pedis) are more prone to developing jock itch.
What is the prognosis for jock itch?
The prognosis for tinea cruris is very good. Overall, the condition tends to be an easily treated and curable skin condition. Commonly, it is a mild, benign, usually noncontagious, and self-limited skin condition. More widespread, atypical cases of jock itch may be embarrassing, chronically disfiguring, and psychologically distressing.
What are possible complications of jock itch?
Complications are infrequent since jock itch is usually a self-limited skin condition. Rarely, the rash may spread past the groin onto the thighs and genitals. Secondary skin infections from scratching or rubbing can uncommonly deepen, causing cellulitis or abscess formation.
Another potential complication includes temporary skin discoloration called post-inflammatory hypopigmentation (lighter than the regular skin color) or hyperpigmentation (darker than the regular skin color). This altered skin color may occur after the rash has improved or after a temporary flare. It usually resolves over several weeks to months with avoidance of sunlight in the area. Permanent scarring is uncommon.
Why is my groin still discolored?
Residual skin discoloration in the groin may persist for weeks to months after more severe forms of jock itch clear. Because it is in an area that is not generally visible, it does not usually require treatment. Eventually, the skin discoloration will return to a more natural color. Sunscreen or avoidance of sunbathing helps to avoid worsening the pigmentation.
How can you prevent jock itch?
Jock itch prevention efforts include good general skin hygiene and keeping the groin clean and dry.
The following steps will help prevent the itchy condition:
Wash the groin and buttocks with soap and water after exercise and sweating.
Wash workout clothes, underwear, and swimwear after each use.
Minimize groin moisture by using white cotton underwear.
Change underwear frequently, especially after sweating.
It may be useful to apply petroleum jelly or zinc oxide ointment (Desitin) liberally to the groin creases before physical exercise to minimize friction skin damage.
Wash clothes, undergarments, and towels in hot soapy water.
Use loose-fitting cotton underwear and clothing.
Avoid undergarments with polyesters, nylon, or synthetic fibers.
Use an antifungal powder like Lamisil or Zeasorb to keep the groin dry.
Avoid perfumed creams, powders, sprays, or lotions on the groin.
Do not go barefoot, especially at gyms, schools, and public pools.
Treat athlete's foot if you have it.
Cover your feet with socks before you put on your underwear and pants.
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Medically Reviewed on 8/1/2024
References
CDC. Fungal Diseases – Symptoms of Ringworm Infections. January 14, 2021.
Dinulos, JGH. Superficial Fungal Infections. Chapter 13. Habif's Clinical Dermatology. 483-524.e1
Hay, RJ. Dermatophytosis (Ringworm) and Other Superficial Mycoses. Chapter 266. Mandell, Douglas, and Bennett's Principles and Practice of Infectious Diseases, 3201-3210.e1
van Zuuren EJ, Fedorowicz Z, El-Gohary M. Evidence-based topical treatments for tinea cruris and tinea corporis: a summary of a Cochrane systematic review. Br J Dermatol. 2015;172(3):616-641. doi:10.1111/bjd.13441
El-Gohary M, van Zuuren EJ, Fedorowicz Z, et al. Topical antifungal treatments for tinea cruris and tinea corporis. Cochrane Database Syst Rev. 2014;2014(8):CD009992. doi:10.1002/14651858.CD009992.pub2
Kalra MG, Higgins KE, Kinney BS. Intertrigo and secondary skin infections. Am Fam Physician. 2014;89(7):569-573.
King LM. Jock Itch. WebMD. Published May 30, 2024. Accessed July 16, 2024.
Professional CCM. Jock Itch (Tinea Cruris). Cleveland Clinic. Accessed July 16, 2024.
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187795 | https://www2.tntech.edu/leap/murdock/books/v1chap3.pdf | Chapter 3 Motion in Two and Three Dimensions 3.1 The Important Stuff 3.1.1 Position In three dimensions, the location of a particle is specified by its location vector, r: r = xi + yj + zk (3.1) If during a time interval ∆t the position vector of the particle changes from r1 to r2, the displacement ∆r for that time interval is ∆r = r1 −r2 (3.2) = (x2 −x1)i + (y2 −y1)j + (z2 −z1)k (3.3) 3.1.2 Velocity If a particle moves through a displacement ∆r in a time interval ∆t then its average velocity for that interval is v = ∆r ∆t = ∆x ∆t i + ∆y ∆t j + ∆z ∆t k (3.4) As before, a more interesting quantity is the instantaneous velocity v, which is the limit of the average velocity when we shrink the time interval ∆t to zero. It is the time derivative of the position vector r: v = dr dt (3.5) = d dt(xi + yj + zk) (3.6) = dx dt i + dy dt j + dz dt k (3.7) can be written: v = vxi + vyj + vzk (3.8) 51 52 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS where vx = dx dt vy = dy dt vz = dz dt (3.9) The instantaneous velocity v of a particle is always tangent to the path of the particle.
3.1.3 Acceleration If a particle’s velocity changes by ∆v in a time period ∆t, the average acceleration a for that period is a = ∆v ∆t = ∆vx ∆t i + ∆vy ∆t j + ∆vz ∆t k (3.10) but a much more interesting quantity is the result of shrinking the period ∆t to zero, which gives us the instantaneous acceleration, a. It is the time derivative of the velocity vector v: a = dv dt (3.11) = d dt(vxi + vyj + vzk) (3.12) = dvx dt i + dvy dt j + dvz dt k (3.13) which can be written: a = axi + ayj + azk (3.14) where ax = dvx dt = d2x dt2 ay = dvy dt = d2y dt2 az = dvz dt = d2z dt2 (3.15) 3.1.4 Constant Acceleration in Two Dimensions When the acceleration a (for motion in two dimensions) is constant we have two sets of equations to describe the x and y coordinates, each of which is similar to the equations in Chapter 2. (Eqs. 2.6—2.9.) In the following, motion of the particle begins at t = 0; the initial position of the particle is given by r0 = x0i + y0j and its initial velocity is given by v0 = v0xi + v0yj and the vector a = axi + ayj is constant.
vx = v0x + axt vy = v0y + ayt (3.16) x = x0 + v0xt + 1 2axt2 y = y0 + v0yt + 1 2ayt2 (3.17) v2 x = v2 0x + 2ax(x −x0) v2 y = v2 0y + 2ay(y −y0) (3.18) x = x0 + 1 2(v0x + vx)t y = y0 + 1 2(v0y + vy)t (3.19) Though the equations in each pair have the same form they are not identical because the components of r0, v0 and a are not the same.
3.1. THE IMPORTANT STUFF 53 3.1.5 Projectile Motion When a particle moves in a vertical plane during free–fall its acceleration is constant; the acceleration has magnitude 9.80 m s2 and is directed downward. If its coordinates are given by a horizontal x axis and a vertical y axis which is directed upward, then the acceleration of the projectile is ax = 0 ay = −9.80 m s2 = −g (3.20) For a projectile, the horizontal acceleration ax is zero!!!
Projectile motion is a special case of constant acceleration, so we simply use Eqs. 3.16– 3.19, with the proper values of ax and ay.
3.1.6 Uniform Circular Motion When a particle is moving in a circular path (or part of one) at constant speed we say that the particle is in uniform circular motion. Even though the speed is not changing, the particle is accelerating because its velocity v is changing direction.
The acceleration of the particle is directed toward the center of the circle and has mag-nitude a = v2 r (3.21) where r is the radius of the circular path and v is the (constant) speed of the particle.
Because of the direction of the acceleration (i.e. toward the center), we say that a particle in uniform circular motion has a centripetal acceleration.
If the particle repeatedly makes a complete circular path, then it is useful to talk about the time T that it takes for the particle to make one complete trip around the circle. This is called the period of the motion. The period is related to the speed of the particle and radius of the circle by: T = 2πr v (3.22) 3.1.7 Relative Motion The velocity of a particle depends on who is doing the measuring; as we see later on it is perfectly valid to consider “moving” observers who carry their own clocks and coordinate systems with them, i.e. they make measurements according to their own reference frame; that is to say, a set of Cartesian coordinates which may be in motion with respect to another set of coordinates. Here we will assume that the axes in the different system remain parallel to one another; that is, one system can move (translate) but not rotate with respect to another one.
Suppose observers in frames A and B measure the position of a point P. Then then if we have the definitions: rPA = position of P as measured by A rPB = position of P as measured by B 54 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS rBA = position of B’s origin, as measured by A with v’s and a’s standing for the appropriate time derivatives, then we have the relations: rPA = rPB + rBA (3.23) vPA = vPB + vBA (3.24) For the purposes of doing physics, it is important to consider reference frames which move at constant velocity with respect to one another; for these cases, vBA = 0 and then we find that point P has the same acceleration in these reference frames: aPA = aPB Newton’s Laws (next chapter!) apply to such a set of inertial reference frames. Observers in each of these frames agree on the value of a particle’s acceleration.
Though the above rules for translation between reference frames seem very reasonable, it was the great achievement of Einstein with his theory of Special Relativity to understand the more subtle ways that we must relate measured quantities between reference frames. The trouble comes about because time (t) is not the same absolute quantity among the different frames.
Among other places, Eq. 3.24 is used in problems where an object like a plane or boat has a known velocity in the frame of (with respect to) a medium like air or water which itself is moving with respect to the stationary ground; we can then find the velocity of the plane or boat with respect to the ground from the vector sum in Eq. 3.24.
3.2 Worked Examples 3.2.1 Velocity 1. The position of an electron is given by r = 3.0ti −4.0t2j + 2.0k (where t is in seconds and the coefficients have the proper units for r to be in meters). (a) What is v(t) for the electron? (b) In unit–vector notation, what is v at t = 2.0 s?
(c) What are the magnitude and direction of v just then? [HRW5 4-9] (a) The velocity vector v is the time–derivative of the position vector r: v = dr dt = d dt(3.0ti −4.0t2j + 2.0k) = 3.0i −8.0tj where we mean that when t is in seconds, v is given in m s .
3.2. WORKED EXAMPLES 55 (b) At t = 2.00 s, the value of v is v(t = 2.00 s) = 3.0i −(8.0)(2.0)j = 3.0i −16.j that is, the velocity is (3.0i −16.j) m s .
(c) Using our answer from (b), at t = 2.00 s the magnitude of v is v = q v2 x + v2 y + v2 z = q (3.00 m s )2 + (−16. m s )2 + (0)2 = 16. m s we note that the velocity vector lies in the xy plane (even though this is a three–dimensional problem!) so that we can express its direction with a single angle, the usual angle θ measured anti-clockwise in the xy plane from the x axis. For this angle we get: tan θ = vy vx = −5.33 = ⇒ θ = tan−1(−5.33) = −79◦.
When we take the inverse tangent, we should always check and see if we have chosen the right quadrant for θ. In this case −79◦is correct since vy is negative and vx is positive.
3.2.2 Acceleration 2.
A particle moves so that its position as a function of time in SI units is r = i + 4t2j + tk. Write expressions for (a) its velocity and (b) its acceleration as functions of time. [HRW5 4-11] (a) To clarify matters, what we mean here is that when we use the numerical value of t in seconds, we will get the values of r in meters. Since the velocity vector is the time–derivative of the position vector r, we have: v = dr dt = d dt(i + 4t2j + tk) = 0i + 8tj + k That is, v = 8tj + k. Here, we mean that when we use the numerical value of t in seconds, we will get the value of v in m s .
(b) The acceleration a is the time–derivative of v, so using our result from part (a) we have: a = dv dt = d dt(8tj + k) = 8j 56 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS So a = 8j, where we mean that the value of a is in units of m s2. In fact, we should really include the units here and write: a = 8 m s2 j 3. A particle moving with an initial velocity v = (50 m s )j undergoes an acceleration a = [35 m/ s2 + (2 m/ s5)t3)i + [4 m/ s2 −(1 m/ s4)t2]j. What are the particle’s position and velocity after 3.0 s, assuming that it starts at the origin? [FGT2 3-20] In the problem we are given the acceleration at all times, the initial velocity and also the initial position. We know that at t = 0, the velocity components are vx = 0 and vy = 50 m s and the coordinates are x = 0 and y = 0.
From the acceleration a we do know something about the velocity. Since the acceleration is the time derivative of the velocity: a = dv dt , the velocity is the anti-derivative (or “indefinite integral”, “primitive”. . . ) of the accelera-tion. Having learned our calculus well, we immediately write: v = 35t + 1 2t4 + C1 i + 4t −1 3t3 + C2 j Here, for simplicity, I have omitted the units that are supposed to go with the coefficients.
(I’m not supposed to do that!) Just keep in mind that time is supposed to be in seconds, length is in meters. . .
Of course, when we do the integration, we get constants C1 and C2 which (so far) have not been determined. We can determine them using the rest of the information in the problem.
Since vx = 0 at t = 0 we get: 35(0) + 1 2(0)4 + C1 = 0 = ⇒ C1 = 0 and 4(0) −1 3(0)3 + C2 = 50 = ⇒ C2 = 50 so the velocity as a function of time is v = 35t + 1 2t4 i + 4t −1 3t3 + 50 j where t is in seconds and the result is in m s .
We can find r as a function of time in the same way. Since v = dr dt 3.2. WORKED EXAMPLES 57 then r is the anti-derivative of v. We get: r = 35 2 t2 + 1 10t5 + C3 i + 2t2 −1 12t4 + 50t + C4 j and once again we need to solve for the constants. x = 0 at t = 0, so 35 2 (0)2 + 1 10(0)5 + C3 = 0 = ⇒ C3 = 0 and y = 0 at t = 0, so 2(0)2 −1 12(0)4 + 50(0) + C4 = 0 = ⇒ C4 = 0 and so r is fully determined: r = 35 2 t2 + 1 10t5 i + 2t2 −1 12t4 + 50t j Now we can answer the questions.
We want to know the value of r (the particle’s position) at t = 3.0 s. Just plug in!
x(t = 3.0 s) = 35 2 (3.0)2 + 1 10(3.0)5 = 181 m and y(t = 3.0 s) = 2(3.0)2 −1 12(3.0)4 + 50(3.0) = 161 m .
The components of the velocity at t = 3.0 s are vx(t = 3.0 s) = 35(3.0) + 1 2(3.0)4 = 146 m s and vy(t = 3.0 s) = 4(3.0) −1 3(3.0)3 + 50 = 53 m s .
Here we have been careful to include the proper (SI) units in the final answers because coordinates and velocities must have units.
3.2.3 Constant Acceleration in Two Dimensions 4. A fish swimming in a horizontal plane has a velocity v0 = (4.0i + 1.0j) m s at a point in the ocean whose position vector is r0 = (10.0i −4.0j) m relative to a stationary rock at the shore. After the fish swims with constant acceleration for 20.0 s, its velocity is v = (20.0i −5.0j) m s . (a) What are the components of the acceleration? (b) What is the direction of the acceleration with respect to the fixed x axis? (c) Where is the fish at t = 25 s and in what direction is it moving?
[Ser4 4-7] 58 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS (a) Since we are given that the acceleration is constant, we can use Eqs. 3.16: vx = v0x + axt vy = v0y + ayt to get: ax = vx −v0x t = (20.0 m s −4.0 m s ) 20.0 s = 0.80 m s2 and ay = vy −v0y t = (−5.0 m s −1.0 m s ) 20.0 s = −0.30 m s2 and the acceleration vector of the fish is a = (0.80 m s2 )i −(0.30 m s2)j .
(b) With the angle θ measured counterclockwise from the +x axis, the direction of the acceleration a is: tan θ = ay ax = −0.30 0.80 = −0.375 A calculator gives us: θ = tan−1(−0.375) = −20.6◦ Since the y component of the acceleration is negative, this angle is in the proper quadrant.
The direction of the acceleration is given by θ = −20.6◦. (The same as θ = 360◦−20.6◦= 339.4◦.
(c) We can use Eq. 3.17 to find the values of x and y at t = 25 s: x = x0 + v0x + 1 2axt2 = 10 m + 4.0 m s (25 s) + 1 2(0.80 m s2)(25 s)2 = 360 m and y = y0 + v0y + 1 2ayt2 = −4.0 m + 1.0 m s (25 s) + 1 2(−0.30 m s2)(25 s)2 = −72.8 m At t = 25 s the velocity components of the fish are given by: vx = v0x + axt = 4.0 m s + (0.80 m s2)(25 s) = 24 m s and vy = v0y + ayt = 1.0 m s + (−0.30 m s2)(25 s) = −6.5 m s 3.2. WORKED EXAMPLES 59 1.9 cm 30 m y x Figure 3.1: Bullet hits target 1.9 cm below the aiming point.
so that at that time the speed of the fish is v = q v2 x + v2 y = q (24 m s )2 + (−6.5 m s )2 = 24.9 m s and the direction of its motion θ is found from: tan θ = vy vx = −6.5 24 = −0.271 so that θ = −15.2◦.
Again, since vy is negative and vx is positive, this is the correct choice for θ. So the direction of the fish’s motion is −15.2◦from the +x axis.
3.2.4 Projectile Motion 5. A rifle is aimed horizontally at a target 30 m away. The bullet hits the target 1.9 cm below the aiming point. (a) What is the bullet’s time of flight? (b) What is the muzzle velocity? [HRW5 4-19] (a) First, we define our coordinates. I will use the coordinate system indicated in Fig. 3.1, where the origin is placed at the tip of the gun. Then we have x0 = 0 and y0 = 0. We also know the acceleration: ax = 0 and ay = −9.80 m s2 = −g What else do we know? The gun is fired horizontally so that v0y = 0, but we don’t know v0x. We don’t know the time of flight but we do know that when x has the value 30 m then y has the value −1.9 × 10−2 m. (Minus!) 60 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS Our equation for the y coordinate is y = y0 + y0yt + 1 2ayt2 = 0 + 0 + 1 2(−g)t2 = −1 2gt2 We can now ask: “At what time t does y equal −1.9 × 10−2 m?” . Substitute y = −1.9 × 10−2 m and solve: t2 = −2y g = −2(−1.9 × 10−2 m) 9.80 m s2 = 3.9 × 10−3 s2 which gives: t = 6.2 × 10−2 s Since this is the time of impact with the target, the time of flight of the bullet is t = 6.2 × 10−2 s.
(b) The equation for x−motion is x = x0 + v0xt + 1 2axt2 = 0 + v0xt + 0 = v0xt From part (a) we know that when t = 6.2 × 10−2 s then x = 30 m. This allows us to solve for v0x: v0x = x t = 30 m 6.2 × 10−2 s = 480 m s The muzzle velocity of the bullet is 480 m s .
6. In a local bar, a customer slides an empty beer mug on the counter for a refill.
The bartender does not see the mug, which slides offthe counter and strikes the floor 1.40 m from the base of the counter. If the height of the counter is 0.860 m, (a) with what speed did the mug leave the counter and (b) what was the direction of the mug’s velocity just before it hit the floor? [Ser4 4-11] (a) The motion of the beer mug is shown in Fig. 3.2(a). We choose the origin of our xy coordinate system as being at the point where the mug leaves the counter. So the mug’s initial coordinates for its flight are x0 = 0, y0 = 0.
At the very beginning of its motion through the air, the velocity of the mug is horizontal.
(This is because its velocity was horizontal all the time it was sliding on the counter.) So we know that v0y = 0 but we don’t know the value of v0x. (In fact, that’s what we’re trying to figure out!) 3.2. WORKED EXAMPLES 61 vo 1.40 m 0.860 m v q q (a) (b) x y Figure 3.2: (a) Beer mug slides offcounter and strikes floor! (b) Velocity vector of the beer mug at the time of impact.
We might begin by finding the time t at which the mug hit the floor. This is the time t at which y = −0.860 m (recall how we chose the coordinates!), and we will need the y equation of motion for this; since v0y = 0 and ay = −g, we get: y = v0yt + 1 2ayt2 = −1 2gt2 So we solve −0.860 m = −1 2gt2 which gives t2 = 2(0.860 m) g = 2(0.860 m) (9.80 m s2 ) = 0.176 s2 so then t = 0.419 s is the time of impact.
To find v0x we consider the x equation of motion; x0 = 0 and ax = 0, so we have x = v0xt .
At t = 0.419 s we know that the x coordinate was equal to 1.40 m. So 1.40 m = v0x(0.419 s) Solve for v0x: v0x = 1.40 m 0.419 s = 3.34 m s which tells us that the initial speed of the mug was v0 = 3.34 m s .
(b) We want to find the components of the mug’s velocity at the time of impact, that is, at t = 0.419 s. Substitute into our expressions for vx and vy: vx = v0x + axt = v0x = 3.34 m s 62 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS 40o 22.0 m 25.0 m/s Figure 3.3: Ball is thrown toward wall at 40◦above horizontal, in Example 7.
and vy = v0y + ayt = 0 + (−9.80 m s2 )(0.419 s) = −4.11 m s .
So at the time of impact, the speed of the mug was v = q v2 x + v2 y = q (3.34 m s )2 + (−4.11 m s )2 = 5.29 m s and, if as in Fig. 3.2(b) we let θ be the angle below the horizontal at which the velocity vector is pointing, we see that tan θ = 4.11 3.34 = 1.23 = ⇒ θ = tan−1(1.23) = 50.9◦.
At the time of impact, the velocity of the mug was directed at 50.9◦below the horizontal.
7. You throw a ball with a speed of 25.0 m s at an angle of 40.0◦above the horizontal directly toward a wall, as shown in Fig. 3.3. The wall is 22.0 m from the release point of the ball. (a) How long does the ball take to reach the wall? (b) How far above the release point does the ball hit the wall? (c) What are the horizontal and vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory? [HRW5 4-28] (a) We will use a coordinate system which has its origin at the point of firing, which we take to be at ground level.
What is the mathematical condition which determines when the ball hits the wall? It is when the x coordinate of the ball is equal to 22.0 m. Then let’s write out the x−equation of motion for the ball. The ball’s initial x−velocity is v0x = v0 cos θ0 = (25.0 m s ) cos 40.0◦= 19.2 m s and of course ax = 0, so that the x motion is given by x = x0 + voxt + 1 2axt2 = 19.2 m s t 3.2. WORKED EXAMPLES 63 We solve for the time at which x = 22.0 m: x = 19.2 m s t = 22.0 m = ⇒ t = 22.0 m 19.2 m s = 1.15 s The ball hits the wall 1.15 s after being thrown.
(b) We will be able to answer this question if we can find the y coordinate of the ball at the time that it hits the wall, namely at t = 1.15 s.
We need the y equation of motion. The initial y velocity of the ball is v0y = v0 sin θ0 = 25.0 m s sin 40.0◦= 16.1 m s and the y acceleration of the ball is ay = −g giving: y = y0 + v0yt + 1 2ayt2 = 16.1 m s t −1 2gt2 which we use to find the y coordinate at t = 1.15 s: y = (16.1 m s )(1.15 s) −1 2(9.80 m s2 )(1.15 s)2 = 12.0 m which tells us that the ball hits the wall at 12.0 m above the ground level (above the release point).
(c) The x and y components of the balls’s velocity at the time of impact, namely at t = 1.15 s are found from Eqs. 3.16: vx = v0x + axt = 19.2 m s + 0 = 19.2 m s and vy = v0y + ayt = 16.1 m s + (−9.80 m s2)(1.15 s) = +4.83 m s .
(d) Has the ball already passed the highest point on its trajectory? Suppose the ball was on its way downward when it struck the wall. Then the y component of the velocity would be negative, since it is always decreasing and at the trajectory’s highest point it is zero. (Of course, the x component of the velocity stays the same while the ball is in flight.) Here we see that the y component of the ball’s velocity is still positive at the time of impact. So the ball was still climbing when it hit the wall; it had not reached the highest point of its (free) trajectory.
8. The launching speed of a certain projectile is five times the speed it has at its maximum height. Calculate the elevation angle at launching. [HRW5 4-32] We make a diagram of the projectile’s motion in Fig. 3.4. The launch it speed is v0, and the projectile is launched at an angle θ0 upward from the horizontal.
We might start this problem by solving for the time it takes the projectile to get to maxi-mum height, but we can note that at maximum height, there is no y velocity component, and 64 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS v0 Speed is v0 / 5 q0 Figure 3.4: Motion of projectile in Example 8.
the x velocity component is the same as it was when the projectile was launched. Therefore at maximum height the velocity components are vx = v0 cos θ0 and vy = 0 and so the speed of the projectile at maximum height is v0 cos θ0.
Now, we are told that the launching speed (v0) is five times the speed at maximum height.
This gives us: v0 = 5v0 cos θ0 = ⇒ cos θ0 = 1 5 which has the solution θ0 = 78.5◦ So the elevation angle at launching is θ0 = 78.5◦.
9. A projectile is launched from ground level with speed v0 at an angle of θ0 above the horizontal.
Find: (a) the maximum height H attained by the projectile, and (b) the distance from the starting point at which the projectile strikes the ground; this is called the range R of the projectile.
Comment: This problem is worked in virtually every physics text, and it is sometimes simply called “The Projectile Problem”. I include it in this book for the sake of completeness and so that we can use the results if we need them later on. I do not treat it as part of the fundamental material of this chapter because it is a very particular application of free–fall motion. In this problem, the projectile impacts at the same height as the one from which it started, and that is not always the case. We must think about all projectile problems individually and not rely on simple formulae to plug numbers into!
The path of the projectile is shown in Fig. 3.5. The initial coordinates of the projectile are x0 = 0 and y0 = 0 , 3.2. WORKED EXAMPLES 65 R H v0 q0 y x Figure 3.5: The common projectile problem; projectile is shot from ground level with speed v0 and angle θ0 above the horizontal.
the components of the initial velocity are v0x = v0 cos θ0 and v0y = v0 sin θ0 and of course the (constant) acceleration of the projectile is ax = 0 and ay = −g = −9.80 m s2 Then our equations for x(t), vx(t), y(t) and vy(t) are vx = v0 cos θ0 x = v0 cos θ0 t vy = v0 sin θ0 −gt y = v0 sin θ0 t −1 2gt2 (a) What does it mean for the projectile to get to “maximum height”? This is when it is neither increasing in height (rising) nor decreasing in height (falling); the vertical component of the velocity at this point is zero. At this particular time then, vy = v0 sin θ0 −gt = 0 so solving this equation for t, the projectile reaches maximum height at t = v0 sin θ0 g .
How high is the projectile at this time? To answer this, substitute this value of t into the equation for y and get: y = v0 sin θ0 v0 sin θ0 g !
−1 2g v0 sin θ0 g !2 = v2 0 sin2 θ0 g −v2 0 sin2 θ0 2g = v2 0 sin2 θ0 2g 66 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS This is the maximum height attained by the projectile: H = v2 0 sin2 θ0 2g (b) What is the mathematical condition for when the projectile strikes the ground (since that is how we will find the range R)? We know that at this point, the projectile’s y coordinate is zero: y = v0 sin θ0 t −1 2gt2 = 0 We want to solve this equation for t; we can factor out t in this expression to get: t(v0 sin θ0 −1 2gt) = 0 which has two solutions: t = 0 and t = 2v0 sin θ0 g The first of these is just the time when the projectile was fired; yes, y was equal to zero then, but that’s not what we want! The time at which the projectile strikes the ground is t = 2v0 sin θ0 g .
We want to find the value of x at the time of impact. Substituting this value of t into our equation for x(t), we find: x = v0 cos θ0 2v0 sin θ0 g !
= 2v2 0 sin θ0 cos θ0 g This value of x is the range R of the projectile.
We can make this result a little simpler by recalling the trig relation: sin 2θ0 = 2 sin θ0 cos θ0 .
Using this in our result for the range gives: R = 2v2 0 sin θ0 cos θ0 g = v2 0 sin 2θ0 g 10. A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection? [Ser4 4-23] Now, this problem does deal with a projectile which starts and ends its flight at the same height, just as we calculated in the previous example. So we can use the results for the range R and maximum height H that we found there.
3.2. WORKED EXAMPLES 67 A B 9.40 km 3.30 km 35o Figure 3.6: Volcanic bombs away!
The problem tells us that R = 3H. Substituting the expressions for H and R that we found in the last example (we pick the first expression we got for R), we get: R = 2v2 0 sin θ0 cos θ0 g = 3H = 3 v2 0 sin2 θ0 2g !
Cancelling stuff, we get: 2 cos θ0 = 3 2 sin θ0 = ⇒ tan θ0 = 4 3 The solution is: θ0 = tan−1(4/3) = 53.1◦ The projectile was fired at 53.1◦above the horizontal.
11.
During volcanic eruptions, chunks of solid rock can be blasted out of a volcano; these projectiles are called volcanic bombs. Fig. 3.6 shows a cross section of Mt.
Fuji in Japan.
(a) At what initial speed would the bomb have to be ejected, at 35◦to the horizontal, from the vent at A in order to fall at the foot of the volcano at B? (Ignore the effects of air on the bomb’s travel.) (b) What would be the time of flight?
[HRW5 4-42] (a) We use a coordinate system with its origin at point A (the volcano “vent”); then for the flight from the vent at A to point B, the initial coordinates are x0 = 0 and y0 = 0, and the final coordinates are x = 9.40 km and y = −3.30 km. Aside from this, we don’t know the initial speed of the rock (that’s what we’re trying to find) or the time of flight from A to B.
Of course, the acceleration of the rock is given by ax = 0, ay = −g.
We start with the x equation of motion. The initial x−velocity is v0x = v0 cos θ 68 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS where θ = 35◦so the function x(t) is x = x0 + v0xt + 1 2axt2 = 0 + v0 cos θt + 0 = v0 cos θ t Now, we do know that at the time of impact x had the value x = 9.40 km so if we now let t be the time of flight, then (9.40 km) = v0 cos θt or t = (9.40 km) v0 cos θ (3.25) Next we look at the y equation of motion. Since v0y = v0 sin θ we get: y = y0 + v0yt + 1 2ayt2 = 0 + v0 sin θt −1 2gt2 = v0 sin θ t −1 2gt2 But at the time t of impact the y coordinate had the value y = −3.30 km.
If we also substitute for t in this expression using Eq. 3.25 we get: −3.30 km = v0 sin θ 9.40 km v0 cos θ !
−1 2g 9.40 km v0 cos θ !2 = (9.40 km) tan θ −g(9.40 km)2 2v2 0 cos2 θ At this point we are done with the physics problem. The only unknown in this equation is v0, which we can find by doing a little algebra: g(9.40 km)2 2v2 0 cos2 θ = (9.40 km) tan θ + 3.30 km = 9.88 km which gives: v2 0 = g(9.40 km)2 cos2 θ(9.88 km) = g(0.951 km) cos2 θ = (9.80 m s2)(951 m) cos2 35◦ = 1.39 × 104 m2 s2 and finally v0 = 118 m s 3.2. WORKED EXAMPLES 69 v0 f q0 d Path of projectile Figure 3.7: Projectile is fired up an incline, as described in Example 12 (b) Having v0 in hand, finding t is easy. Using our result from part(a) and Eq. 3.25 we find: t = (9.40 km) v0 cos θ = (9400 m) (118 m s ) cos 35◦= 97.2 s The time of flight is 97.2 s.
12. A projectile is fired up an incline (incline angle φ) with an initial speed v0 at an angle θ0 with respect to the horizontal (θ0 > φ) as shown in Fig. 3.7. (a) Show that the projectile travels a distance d up the incline, where d = 2v2 0 cos θ0 sin(θ0 −φ) g cos2 φ (b) For what value of θ0 is d a maximum, and what is the maximum value? [Ser4 4-56] (a) This is a relatively challenging problem, and of course it is completely analytic.
We can start by writing down equations for x and y as functions of time. By now we can easily see that we have: x = v0 cos θ0 t y = v0 sin θ0 t −1 2gt2 We can combine these equations to get a relation between x and y for points on the trajectory; from the first, we have t = x/(v0 cos θ0), and putting this into the second one gives: y = v0 sin θ0 x v0 cos θ0 −1 2g x v0 cos θ0 2 = (tan θ0)x −g 2 x2 v2 0 cos2 θ0 What is the condition for the time that the projectile hits the slope? Unlike the problems where a projectile impacts with the flat ground or a wall, we don’t know the value of x or y 70 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS at impact. But since the incline has a slope of tan φ, the relation between x and y for points on the slope is y = (tan φ)x .
These two relations between x and y allow us to solve for the values of x and y where the impact occurs. Substituting for y above, we find: (tan φ)x = (tan θ0)x −g 2 x2 v2 0 cos2 θ0 A little rearranging gives: g 2 x v2 0 cos2 θ0 + (tan φ −tan θ0) = 0 and the solution for x is: x = 2v2 0 cos2 θ0(tan θ0 −tan φ) g The problem has us solve for the distance d up the slope; this distance is related to the impact value of x by: d = x cos φ and this gives us: d = x cos φ = 2v2 0 cos2 θ0(tan θ0 −tan φ) g cos φ .
Although this is a perfectly good expression for d, it is not the one presented in the problem. (Among other things, it has another factor of cos φ downstairs.) If we multiply top and bottom by cos φ we find: d = 2v2 0 cos2 θ0 cos φ(tan θ0 −tan φ) g cos2 φ = 2v2 0 cos θ0(cos θ0 cos φ tan θ0 −cos θ0 cos φ tan φ) g cos2 φ = 2v2 0 cos θ0(cos φ sin θ0 −cos θ0 sin φ) g cos2 φ And now using an angle–addition identity from trigonometry in the numerator, we arrive at d = 2v2 0 cos θ0 sin(θ0 −φ) g cos2 φ which is the preferred expression for d.
(b) In part (a) we found the up–slope impact distance as a function of launch angle θ0. (The launch speed v0 and the slope angle φ are taken to be fixed.) For a certain value of theta0, 3.2. WORKED EXAMPLES 71 this function d(θ0) will take on a maximum value. To find this value, we differentiate the function d(θ0) and set the derivative equal to zero. We find: d′(θ0) = 2v2 0 g cos2 φ d dθ0 [cos θ0 sin(θ0 −φ)] = 2v2 0 g cos2 φ[−sin θ0 sin(θ0 −φ) + cos θ0 cos(θ0 −φ)] = 2v2 0 g cos2 φ cos(2θ0 −φ) where in the last step we used the trig identity cos α cos β −sin α sin β = cos(α + β).
Now, to satisfy d′(θ0) = 0 we must have cos(2θ0 −φ) = 0. While this equation has infinitely many solutions for θ0, considering the values that θ0 and φ may take on, we see that we need only look at the case where 2θ0 −φ = π 2 which of course, does solve the equation. This gives us: θ0 = π 4 + φ 2 for the value of θ which makes the projectile go the farthest distance d up the slope.
To find what this value of d is, we substitute for θ0 in our function d(θ0). We find: dmax = 2v2 0 g cos2 φ cos π 4 + φ 2 !
sin π 4 + φ 2 −φ !
= 2v2 0 g cos2 φ cos π 4 + φ 2 !
sin π 4 −φ 2 !
This expression is correct but it can be simplified. We use the trig identity which states: sin α cos β = 1 2 sin(α + β) + 1 2 sin(α −β) this gives us: sin π 4 −φ 2 !
cos π 4 + φ 2 !
= 1 2 sin π 2 + 1 2 sin(−φ) = 1 2 −1 2 sin φ = 1 2(1 −sin φ) which is a lot simpler. Using this result in our expression for dmax gives: dmax = 2v2 0 g cos2 φ (1 −sin φ) 2 = v2 0(1 −sin φ) g(1 −sin2 φ) = v2 0 g(1 + sin φ) which is a simple as it’s going to get!
We can check result for a couple well–known cases. If φ = 0 we are dealing with the common projectile problem on level ground for which we know we get maximum range when θ0 = 45◦and from our solution for that problem we get R = v2 0 g . If φ = 90◦we have the problem of a projectile fired straight up; one can show that the maximum height reached is H = v2 0 2g which again agrees with the formula we’ve derived.
72 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS 3.2.5 Uniform Circular Motion 13. In one model of the hydrogen atom, an electron orbits a proton in a circle of radius 5.28 × 10−11 m with a speed of 2.18 × 106 m s . (a) What is the acceleration of the electron in this model? (b) What is the period of the motion? [HRW5 4-57] (a) The electron moves in a circle with constant speed. It is accelerating toward the center of the circle and the acceleration has magnitude acent = v2 r . Substituting the given values, we have: acent = v2 r = (2.18 × 106 m s )2 (5.28 × 10−11 m) = 9.00 × 1022 m s2 The acceleration has magnitude 9.00 × 1022 m s2.
(b) As the electron makes one trip around the circle of radius r, it moves a distance 2πr (the circumference of the circle). If T is the period of the motion, then the speed of the electron is given by the ratio of distance to time, v = 2πr T which gives...
T = 2πr v which shows why Eq. 3.22 is true. Substituting the given values, we get: T = 2π(5.28 × 10−11 m) (2.18 × 106 m s ) = 1.52 × 10−16 s The period of the electron’s motion is 1.52 × 10−16 s.
14. A rotating fan completes 1200 revolutions every minute. Consider a point on the tip of a blade, at a radius of 0.15 m. (a) Through what distance does the point move in one revolution? (b) What is the speed of the point? (c) What is its acceleration? (d) What is the period of the motion? [HRW5 4-63] (a) As the fan makes one revolution, the point in question moves through a circle of radius 0.15 m so the distance it travels is the circumference of that circle, i.e.
d = 2πr = 2π(0.15 m) = 0.94 m The point travels 0.94 m.
(b) If in one minute (60 s) the fan makes 1200 revolutions, the time to make one revolution must be Time for one rev = T = 1 1200 · (1.00 min) = 1 1200 · (60.0 s) = 5.00 × 10−2 s Using our answer from part (a), we know that the point travels 0.94 m in 5.000 × 10−2 s, moving at constant speed. Therefore that speed is: v = d T = 0.94 m 5.000 × 10−2 s = 19 m s 3.2. WORKED EXAMPLES 73 vWG= -0.5 m/s vSW= +1.2 m/s vWG= -0.5 m/s vSW= -1.2 m/s (a) (b) x Figure 3.8: (a) Velocities for case where swimmer swims upstream. (b) Velocities for case where swimmer swims downstream.
(c) The point is undergoing uniform circular motion; its acceleration is always toward the center and has magnitude acent = v2 r . Substituting, acent = v2 r = (19 m s )2 (0.15 m) = 2.4 × 103 m s2 (d) The period of the motion is the time for the fan to make one revolution. And we already found this in part (b)! It is: T = 5.00 × 10−2 s 3.2.6 Relative Motion 15. A river has a steady speed of 0.500 m s . A student swims upstream a distance of 1.00 km and returns to the starting point. If the student can swim at a speed of 1.20 m s in still water, how long does the trip take? Compare this with the time the trip would take if the water were still. [Ser4 4-43] What happens if the water is still? The student swims a distance of 1.00 km “upstream” at a speed of 1.20 m s ; using the simple distance/time formula d = vt the time for the trip is t = d v = 1.0 × 103 m 1.20 m s = 833 s and the same is true for the trip back “downstream”. So the total time for the trip is 833 s + 833 s = 1.67 × 103 s = 27.8 min Good enough, but what about the case where the water is not still? And what does that have to do with relative velocities? In Fig. 3.8, the river is shown; it flows in the −x 74 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS direction. At all times, the velocity of the water with respect to the ground is vWG = −0.500 m s .
When the student swims upstream, as represented in Fig. 3.8(a), his velocity with respect to the water is vSW = +1.20 m s .
We know this because we are given his swimming speed for still water.
Now we are interested in the student’s velocity with respect to the ground, which we will call vSG. It is given by the sum of his velocity with respect to the water and the water’s velocity with respect to the ground: vSG = vSW + vWG = +1.20 m s −0.500 m s = 0.70 m s and so to cover a displacement of ∆x = 1.00 km (measured along the ground!) requires a time ∆t = ∆x vSG = 1.00 × 103 m 0.70 m s = 1.43 × 103 s Then the student swims downstream (Fig. 3.8(b)) and his velocity with respect to the water is vSW = −1.20 m s giving him a velocity with respect to the ground of vSG = vSW + vWG = −1.20 m s −0.500 m s = 1.70 m s so that the time to cover a displacement of ∆x = −1.00 km is ∆t = ∆x vSG = (−1.00 × 103 m) (−1.70 m s ) = 5.88 × 102 s The total time to swim upstream and then downstream is tTotal = tup + tdown = 1.43 × 103 s + 5.88 × 102 s = 2.02 × 103 s = 33.6 min .
16.
A light plane attains an airspeed of 500 km/ hr.
The pilot sets out for a destination 800 km to the north but discovers that the plane must be headed 20.0◦east of north to fly there directly. The plane arrives in 2.00 hr. What was the wind velocity vector? [HRW5 4-83] Whoa! What the Hell is this problem talking about???
When a plane flies in air which itself is moving (i.e. there is a wind velocity) there are three (vector) velocities we need to think about; I will refer to them as: 3.2. WORKED EXAMPLES 75 vPA vPG vPG vAG vPA 20o N S E W N W E S 500 km/hr 400 km/hr (a) (b) Figure 3.9: (a) Vectors for the planes velocity with respect to the ground (vPG) and with respect to the moving air (vPA). (b) The sum of the plane’s velocity relative to the air and the wind velocity gives the plane’s velocity with respect to the ground, vPG.
vPA: Velocity of the plane with respect to the air. The magnitude of this vector is the “airspeed” of the plane. (This is the only thing that a plane’s “speedometer” can really measure.) vAG: Velocity of the air with respect to the ground. This is the wind velocity.
vPG: Velocity of the plane with respect to the ground. This is the quantity which tells us the rate of (ground!) travel of the plane.
These three vectors are related via: vPG = vPA + vAG The first thing we are given in this problem is that the magnitude of vPA is 500 km/ hr.
The plane needs to fly due north and this tells us that vPG (the real direction of motion of the plane) points north (along the y axis). We are then told that the plane’s “heading” is 20.0◦east of north, which tells us that the direction of vPA lies in this direction. These facts are illustrated in Fig. 3.9(a).
Now if the plane travels 800 km in 2.00 hr then its speed (with respect to the ground!) is vPG = 800 km 2.00 hr = 400 km hr .
which we note in Fig. 3.9(b). Since we now have the magnitudes and and directions of vPA and vPG we can compute the wind velocity, vAG = vPG −vPA The x component of this vector is vAG,x = 0 −500 km hr sin 20.0◦= −171 km hr and its y component is vAG,y = 400 −500 km hr cos 20.0◦= −69.8 km hr 76 CHAPTER 3. MOTION IN TWO AND THREE DIMENSIONS So the wind velocity is vAG = −171 km hr i −69.8 km hr j If we want to express the velocity as a magnitude and direction, we find: vAG = q (−171 km hr )2 + (−69.8 km hr )2 = 185 km hr so the wind speed is 185 km hr . The direction of the wind, measured as an angle θ counter-clockwise from the east is found from its components: tan θ = −69.8 −171 = 0.408 = ⇒ θ = tan−1(0.408) = 202◦ (Here we have made sure to get the angle right! Since both components are negative, θ lies in the third quadrant!) Since 180◦would be due West and the wind direction is 22◦larger than that, we can also say that the wind direction is “22◦south of west”. |
187796 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9609269/ | The Hydrophobic Effects: Our Current Understanding - PMC
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Molecules
. 2022 Oct 18;27(20):7009. doi: 10.3390/molecules27207009
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The Hydrophobic Effects: Our Current Understanding
Qiang Sun
Qiang Sun
1 Key Laboratory of Orogenic Belts and Crustal Evolution, Ministry of Education, The School of Earth and Space Sciences, Peking University, Beijing 100871, China; qiangsun@pku.edu.cn
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Editor: Apostolos Avgeropoulos 1
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1 Key Laboratory of Orogenic Belts and Crustal Evolution, Ministry of Education, The School of Earth and Space Sciences, Peking University, Beijing 100871, China; qiangsun@pku.edu.cn
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Apostolos Avgeropoulos: Academic Editor
Received 2022 Aug 13; Accepted 2022 Oct 17; Collection date 2022 Oct.
© 2022 by the author.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC9609269 PMID: 36296602
Abstract
Hydrophobic interactions are involved in and believed to be the fundamental driving force of many chemical and biological phenomena in aqueous environments. This review focuses on our current understanding on hydrophobic effects. As a solute is embedded into water, the interface appears between solute and water, which mainly affects the structure of interfacial water (the topmost water layer at the solute/water interface). From our recent structural studies on water and air-water interface, hydration free energy is derived and utilized to investigate the origin of hydrophobic interactions. It is found that hydration free energy depends on the size of solute. With increasing the solute size, it is reasonably divided into initial and hydrophobic solvation processes, and various dissolved behaviors of the solutes are expected in different solvation processes, such as dispersed and accumulated distributions in solutions. Regarding the origin of hydrophobic effects, it is ascribed to the structural competition between the hydrogen bondings of interfacial and bulk water. This can be applied to understand the characteristics of hydrophobic interactions, such as the dependence of hydrophobic interactions on solute size (or concentrations), the directional natures of hydrophobic interactions, and temperature effects on hydrophobic interactions.
Keywords: water, solute, interface, hydrogen bonding, hydrophobic effects
1. Introduction
Hydrophobic effects refer to the tendency of nonpolar molecules (or parts of molecules) to be aggregated in water. They are involved in and believed to be the fundamental driving force in many chemical and biological phenomena in aqueous solutions, such as molecular recognition, protein folding, formation and stability of micelles, biological membranes and macromolecular complexes, surfactant aggregation, coagulation, complexation, detergency, and the formation of gas clathrates [1,2,3,4,5]. To date, numerous experimental and theoretical works have been carried out to understand the physical origin of hydrophobic effects.
In general, hydrophobicity is expressed through the empirically calculated logarithm of partition coefficient (logP), which is widely used in drug design and medicinal chemistry [6,7]. Historically, the concept of hydrophobicity arose in the context of the low solubility of non-polar solutes in water. Experimentally, it has been found that the entropy change upon transferring a hydrocarbon from a nonpolar environment into water is large and negative. In 1945, Frank and Evans suggested that the observed loss of entropy was related to the structural changes of liquid water as hydrophobic solutes were dissolved into water. They proposed that a kind of “cage”, consisting of water molecules, was formed around the solute. This ordered water structure, similar to the gas clathrate, was known as the “iceberg” model . Since then, many works [9,10,11,12,13,14,15,16,17] are conducted to measure the water structural changes around hydrophobic surfaces. In accordance with the “iceberg” model, some studies [9,10,11,12] support the existence of increased tetrahedral order around small hydrophobic groups in aqueous solutions. However, the decrease of water structure around hydrophobic groups is also found in many works [13,14,15,16,17]. It is well known that the neutron scattering may be sensitive to the structure of water. From the neutron scattering experimental measurements on aqueous solutions containing tetramethylammonium chloride and methane molecules , these do not suggest that water around these hydrophobic solutes may be more ordered than bulk water. To date, there remain strong debates on the “iceberg” structural model.
In 1959, based on the “iceberg” model, Kauzmann introduced the concept of hydrophobic interactions. When two “caged” hydrophobes come together, the “structured” water between solutes may be released into the bulk (Figure 1), which undoubtedly leads to the increase of entropy. Subsequently, the attractive force between these particles may be related to the entropy increase. Therefore, hydrophobic interactions are classically regarded to be entropy driven. However, the “classic” hydrophobic effects may be in contrast with some works. In Diederich et al. work, they found that complexation of benzene in a cyclophane host molecule was enthalpy driven at room temperature, which was also accompanied with a slightly negative entropy change. Additionally, in Baron, Setny, and McCammon works [20,21], molecular dynamics (MD) simulations are used to investigate the binding in a hydrophobic receptor-ligand system. It is found that the association between the non-polar ligand and binding pocket may be driven by enthalpy and opposed by entropy. The “non-classic” hydrophobic effects are ascribed to the release of weakly hydrogen-bonded water molecules into the more strongly hydrogen-bonded bulk water .
Figure 1.
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The “iceberg” structural model of hydrophobic effects. The ordered water structure is expected to form around the solute. As two “caged” solutes become together, the “structured” water in the region between them may be returned to the bulk.
In the “iceberg” structural model, water molecules can rearrange around a small hydrophobic solute, without losing their hydrogen bondings. However, as a large solute is embedded into water, hydrogen bondings of water may be broken at the surface of the solute, which may result in an enthalpic penalty. Therefore, hydrophobic interactions depend on the size of dissolved solute. In recent years, a theoretical approach was developed by Lum, Chandler, and Weeks (LCW) [23,24,25,26] to understand the dependence of hydrophobic interactions on solute size. Both the Gaussian density fluctuations related to small size and the physics of interfacial formation related to large size are incorporated in LCW theory [23,24,25,26]. It can be found that the hydration free energy grows linearly with the solvated volume for small solute, but grows linearly with the solvated surface area for large solute . Therefore, the crossover may be expected between small and large regime, which takes place on the nanometer length scale .
Liquid water is generally believed to play a vital role in the process of hydrophobic interactions. In cell biology, water may be regarded as an active constituent, rather than a bystander [27,28,29]. According to our recent structural works [30,31,32,33,34,35,36,37,38] on liquid water and air-water interface, hydration free energy is determined. This is utilized to understand the nature of hydrophobic interactions. It is found that hydrophobic interactions may be related to the size of dissolved solute. With increasing the solute size, it is reasonably divided into initial and hydrophobic solvation processes . Additionally, different dissolved behaviors of solutes are expected in initial and hydrophobic solvation processes, such as dispersed and accumulated distributions in aqueous environments. In addition, hydrophobic interactions may be ascribed to the structural competition between interfacial and bulk water .
Hydrophobic interactions have long been recognized to be the fundamental driving force in physical chemistry and biochemistry. To understand the origin of hydrophobic interactions, numerous works have been carried out. This review is focused on our current understanding on hydrophobic interactions. Water plays a vital role while solutes are aggregated in solutions. Therefore, it is necessary to study the structure of water, which is included in Section 2. In Section 3, according to the structural works on water and air-water interface, hydration free energy may be determined, and used to investigate the nature of hydrophobic interactions. From this, it is applied to understand the characteristics of hydrophobic interactions, which may be included in Section 4. This work is devoted to studying the dependence of hydrophobic interactions on solute size (or concentrations), the directional natures of hydrophobic interactions, and the temperature effects on hydrophobic interactions.
2. Water Structure
Gibbs free energy (ΔG), related to the changes of enthalpy (ΔH) and entropy (ΔS), may be used to study whether a process is likely to take place. Thermodynamically, it is expressed as,
(1)
where ΔH is enthalpy changes, and ΔS is entropy changes. In general, ΔH quantifies the average potential energy between molecules, ΔS measures the order (or intermolecular) correlations of system.
In fact, various interactions between solutes and water may be expected when solutes are embedded into water. The total Gibbs free energy of system is reasonably described as,
(2)
in which ΔG Water-water, ΔG Solute-water, and ΔG Solute-solute, respectively, mean the Gibbs energies are due to water-water, solute-water, and solute-solute interactions. In fact, the solutes are necessarily attracted to approach each other in aqueous solutions before they may be affected by the interactions between them. This is due to the hydrophobic interactions, which accumulate the solutes in the solutions. Therefore, to understand the molecular mechanism of hydrophobic effects, it is important to study the water structure and the effects of solutes on water structure.
Numerous experimental and theoretical works have been carried out to investigate the structure of water. To date, different structural models have been proposed, which are generally partitioned into the mixture and continuum structural models [39,40]. In the mixture model, two distinct types of structures are regarded to simultaneously exist in ambient water. It is likely that the first mixture model was proposed by W.C. Röntgen in 1892, who suggested that liquid water was a mixture of two components, a low-density fluid and a high-density fluid . Since then, various mixture structural models have been proposed [42,43,44]. For the continuum structural model, water is comprised of a random, three-dimensional hydrogen-bonded network, which may be characterized by a broad distribution of O-H···O hydrogen bond distances and angles. However, the water networks cannot be “broken” (or separated into distinct molecular species) as in the mixture model. To date, liquid water is usually regarded as a tetrahedral fluid, which is based on the first coordination number, , where ρ means the density of water, r min and r max are the lower and upper limits of integration in oxygen-oxygen radial distribution function, g OO(r). For ambient water, Nc is determined to be 4.3 and 4.7 , respectively.
Liquid water is usually regarded as an anomalous fluid, which is due to the formation of hydrogen bondings between neighboring water molecules. To understand the physical nature of hydrogen bondings, various theoretical methods have been developed, such as symmetry-adapted perturbation theory (SAPT) [47,48]. From the theoretical calculations on a water dimer ((H 2 O)2), besides van der Waals interactions between water molecules, obvious electrostatic interactions can also be found between them. Of course, this is reasonably attributed to the formation of hydrogen bondings between water molecules. Therefore, hydrogen bondings may be ascribed to the electrostatic interactions between the neighboring water molecules.
For an H 2 O molecule, the vibrational normal modes may be 2A 1+B 1 , which are all Raman active. When hydrogen bonding is formed between neighboring water molecules, this decreases the O···O distance between them, and weakens the O-H covalent bond . The formation of hydrogen bonding may result in OH vibrational frequencies moving towards a low wavenumber (red shift). Therefore, OH vibrations may be sensitive to hydrogen bondings of liquid water, and widely utilized to investigate the structure of water.
With decreasing temperature from 298 K to 248 K at 0.1 MPa, based on the normalized intensity, an isosbestic point is found around 3330 cm−1 in the Raman OH stretching bands of water (Figure 2). The isosbestic point is the wavelength where a series of spectra cross, and the spectral intensity may keep constant. In mixture structural model, the isosbestic point is generally used to support the two-state behavior of water structure. However, after considering the electric field experienced by the proton projected onto the OH covalent bond, Smith et al. suggested that the isosbestic point was explained through a continuous distribution of local hydrogen-bonded networks, which was due to the increasing distortions around a single-component tetrahedral structural motif. It is noted that, along with the isosbestic point, temperature increase (or adding NaCl) may also lead to the decrease of the second peak at 4.5 Å in g OO(r), which undoubtedly means the breakage of tetrahedral hydrogen bondings of water. Therefore, it can be derived that the isosbestic point indicates the structural transition between tetrahedral and non-tetrahedral hydrogen-bonded networks.
Figure 2.
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The Raman OH stretching bands of water from 298 to 248 K under 0.1 MPa. Based on normalized intensity, an isosbestic point is found around 3330 cm−1.
Many works have been conducted to explain the Raman OH stretching band of water. In fact, water molecular clusters, (H 2 O)n, provide an approach to investigate the dependence of OH vibrational frequencies on hydrogen bondings between water molecules. From the theoretical calculations, the stable configurations of water molecular clusters can be determined. In combination with the experimental measurements on OH vibration frequencies of clusters, these may be utilized to unravel the relationship between OH vibrational frequencies and hydrogen-bonded networks. It is found that, when three-dimensional hydrogen-bonded networks occur in water molecular clusters (n ≥ 6), different OH vibrational frequencies may correspond to various hydrogen bondings in the first shell of a water molecule (local hydrogen bondings), and the effects of hydrogen bondings beyond the first shell on OH vibrational frequencies may be weak (Figure 3). From this, as three-dimensional hydrogen bondings appear, different OH vibrations may be ascribed to OH vibrations engaged into various local hydrogen bondings [30,32].
Figure 3.
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The dependence of the OH stretching frequency on hydrogen bondings of water molecular clusters, (H 2 O)n. Different symbols are used to discriminate OH vibrations engaged in various local hydrogen-bonded networks of a water molecule. Various structures of hexamers are also shown.
For a water molecule, the local hydrogen-bonded network can be differentiated by whether the molecule forms hydrogen bonds as a proton donor (D), proton acceptor (A), or a combination of both with neighboring molecules. Under ambient conditions, the main local hydrogen bonding motifs for a water molecule can be classified as DDAA (double donor-double acceptor), DDA (double donor-single acceptor), DAA (single donor-double acceptor), and DA (single donor-single acceptor) (Figure 4). For ambient water, the Raman OH stretching band may be fitted into five sub-bands, which can be assigned to the ν DAA-OH, ν DDAA-OH, ν DA-OH, ν DDA-OH, and free OH symmetric stretching vibrations, respectively (Figure 4). Therefore, at ambient conditions, different local hydrogen bondings may be expected around a water molecule.
Figure 4.
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The Raman OH stretching band of ambient water may be deconvoluted into five sub-bands, located at 3041, 3220, 3430, 3572, and 3636 cm−1, and assigned to the ν DAA-OH, ν DDAA-OH, ν DA-OH, ν DDA-OH, and free OH symmetric stretching vibrations, respectively. At ambient conditions, the main local hydrogen-bonded networks for a water molecule are expected to be DDAA, DDA, DAA, and DA hydrogen bondings. Hydrogen bondings are drawn with dashed lines.
From the Raman spectroscopic studies [30,31,32] on ambient water, the local statistical model (LSM) is proposed. This suggests that a water molecule interacts with the neighboring water molecules (in the first shell) through different local hydrogen bondings. Of course, it is different from continuum structural models of ambient water. Additionally, according to the mixture structural model, water has been considered as a mixture of two distinct types of structures. Therefore, different spatial distributions may be necessary for various structural types, and sharp boundary is expected between them. In mixture model, water structure may be heterogeneous. Additionally, according to recent X-ray experimental studies [53,54], these mean that liquid water is heterogeneous at ambient conditions, and this becomes enhanced in the supercooled region. However, based on LSM of water structure, various local hydrogen bondings are expected for a water molecule at ambient conditions. This indicates that it is impossible to find a sharp phase boundary between various structural motifs, and the structure of ambient water may be homogeneous. Of course, it is different from the mixture structural model. In addition, based on LSM model, the local hydrogen-bonded networks of a water molecule may be affected by pressure, temperature, dissolved salt, and a confined environment.
According to the explanation on Raman OH stretching band of water [30,31,32], ν DDAA-OH is due to OH vibration engaged in DDAA (tetrahedral or “unbroken”) hydrogen bonding, and ν Free-OH is free OH symmetric stretching vibration. From the van’t Hoff equation, this may be applied to calculate the thermodynamic functions of tetrahedral (DDAA) hydrogen bonding (Figure 5),
(3)
Figure 5.
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The dependence of ln(I Free-OH/I DDAA-OH) on 1/T. The solid line represents linear fit (R 2 = 0.9975) with a slope of −∆H/R. This is used to determine the thermodynamic characteristics of tetrahedral hydrogen bonding.
From the Raman OH stretching bands from 298 K to 248 K, the enthalpy (∆H) and entropy (∆S) from tetrahedral hydrogen bonding to free water are calculated to be 11.35 kJ/mol and 29.66 J/mol, respectively.
Additionally, based on the structural explanation on the Raman OH stretching band of water, it is derived that the isosbestic point (Figure 2) may indicate the structural equilibrium between different local hydrogen bondings around a water molecule. This is expressed as follows,
(4)
With decreasing temperature from 298 to 248 K at 0.1 MPa, this decreases the intensity of the high wavenumber sub-bands (>3330 cm−1), but increases the intensity of the low wavenumber sub-bands (<3330 cm−1) (Figure 2). Therefore, temperature decrease may enhance the probability to form tetrahedral hydrogen-bonded networks around a water molecule, which is related to the nucleation of ice. In fact, according to recent MD simulations [55,56,57] on homogeneous ice nucleation, the ordered ice-like intermediate can be found in the supercooled water, and ice nucleation is found to occur in the low-mobility regions [56,57]. Additionally, this intermediate phase has also been found in other theoretical simulations [58,59,60] during the nucleation from supercooled liquids, such as Lenard-Jones hard spheres and metals. From these, a non-classical pathway, rather than classical nucleation theory (CNT), is proposed in order to understand the nucleation mechanism in supercooled liquids.
Water has many unusual thermodynamic and dynamic properties, both in pure form and as a solvent. Additionally, these anomalous behaviors may be strongly enhanced in the supercooled state, such as thermal expansion, isothermal compressibility, etc. To explain the origin of the anomalous behaviors of supercooled water, many theories are proposed, such as the stability limit (SL) conjecture , the liquid-liquid critical-point (LLCP) hypothesis , the singularity-free (SF) model , and the critical-point free scenario . Recently, numerous works [54,65,66,67,68,69] have been carried out to demonstrate the existence of second critical point in LLCP. In fact, LLCP is based on the mixture model of water structure. In LLCP, two liquid phases are expected in water, such as low-density water (LDW) and high-density water (HDW), which may interconvert through a first-order liquid-liquid transition terminating at the second critical point in the supercooled regime. In this hypothesis, the anomalous behavior of water is due to the fluctuations emanating from the LLCP.
With decreasing temperature from 298 to 248 K at 0.1 MPa (Figure 2), this increases the probability to form the tetrahedral (DDAA) hydrogen bondings in supercooled water. In comparison with bulk water, DDAA hydrogen bondings have a larger volume, and lower entropy . This indicates that the formation of tetrahedral hydrogen bondings in a supercooled regime may be used to explain the anomalies of supercooled water, such as the thermal expansion. From statistical mechanics, the thermal expansion may be related to the correlation of the fluctuations of volume and entropy, . For most fluids, with increasing volume, it is also accompanied with an increase of entropy. However, regarding water below TM (temperature of maximum density), the fluctuations of volume and entropy may be anti-correlated. From the above, it is more reasonable to explain the anomalous properties of supercooled water from the SF model . Further study is necessary.
When a solute is dissolved into liquid water, an interface appears between the solute and water. Therefore, the dissolved solute undoubtedly affects the structure of water. At ambient conditions, the OH vibration is mainly related to the local hydrogen-bonded networks of a water molecule. It is derived that the dissolved solute may mainly affect the structure of topmost water layer at the solute-water interface (interfacial water) (Figure 6). In fact, this is in accordance with other studies [70,71,72,73,74] on the structure and dynamics of water around ions. These mean that the effects of dissolved ions on water structure may be largely limited to the first solvation shell. Therefore, as a solute is embedded into water, it may be divided into interfacial and bulk water (Figure 6). Regarding the effects of solutes on water structure, these may be related to the solute-water interfaces. In other words, it is related to the water in confined environments. In our Raman spectroscopic study on confined water, it can be found that, in comparison with DDAA (tetrahedral) hydrogen bondings, DA hydrogen bondings are expected to form under confined environments. Therefore, it can be derived that the DA hydrogen bonding may tend to form at the solute-water interface.
Figure 6.
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Structural changes across the solute-water interface. The dissolved solute mainly affects the structure of interfacial water (topmost water layer at the interface).
In theory, vibrational sum frequency generation (SFG) spectroscopy provides a unique and powerful method to investigate the surface and interface at the molecular level. Many SFG experimental works have been conducted to study the structure of air-water interface [33,76,77,78,79]. In recent years, phase sensitive sum-frequency generation (PS-SFG) spectroscopy has been developed by Shen et al. [80,81]. From the PS-SFG measurements, Imχ(2) may be directly determined, and the sign of the imaginary part of χ(2) may be utilized to reflect the water molecular dipole direction. Based on our SFG study on the air-water interface , no tetrahedral (DDAA) hydrogen bonding is expected to form in the interfacial water, and an obvious structural difference may be found across the air-water interface.
From the Raman spectroscopic studies [30,31,32], DDAA (tetrahedral) and DA are the predominant hydrogen-bonded networks in ambient water. Additionally, they are related to the structural changes across the solute-water interface (Figure 6). It is necessary to understand the characteristics of DDAA and DA hydrogen bondings. For ambient water, the DDAA-OH, located around 3220 cm−1, lies at a lower frequency than DA-OH (3430 cm−1) (Figure 4). This indicates that, in comparison with DA hydrogen bonding, the formation of DDAA structural motif may result in higher hydrogen bonding energies. Additionally, it is necessary for the interacting molecules to lie in specific relative orientations to form hydrogen bondings between neighboring water molecules, therefore DDAA (tetrahedral) is expected to own the lower entropy rather than DA hydrogen bonding. In addition, based on the experimental measurements and theoretical simulations [83,84], higher water density can be found at the solute-water interface, which may be related to the formation of DA in interfacial water. Therefore, higher density is expected for DA as a structural motif than it is for DDAA hydrogen bonding. In comparison with DDAA (tetrahedral) hydrogen-bonded network, the DA structural motif owns a lower enthalpy, and a higher entropy and density.
The dissolved solute mainly affects the structure of interfacial water. Additionally, DA structural motif tends to form at the interface between solute and water. In other words, this means that the loss of tetrahedral hydrogen bonding may be related to the formation of solute-water interface (Figure 6). Certainly, this is different from the “iceberg” structural model proposed by Frank and Evans . In history, the “iceberg” structural model was proposed in order to understand the large and negative entropy while the simple solutes were dissolved in water. In Frank and Evans’ work , an increase in the structure of water was utilized to explain the negative entropy incurred from the dissolved non-polar molecules. However, they did not explain why the hydrophobic solute could lead to increase of the order of the system. In fact, this may be due to the transition from interfacial to bulk water as the solutes are dissolved in solutions, which leads to the increase of DDAA hydrogen bondings in water.
The effects of dissolved solute on water structure are mainly limited within the interfacial water layer (Figure 6). Additionally, the formation of solute-water interface is due to the loss of tetrahedral hydrogen bonding in interfacial water. Therefore, as the ratio of interfacial water layer to volume is obtained, this may be utilized to calculate the Gibbs free energy of the interface between solute and water. From this, it is reasonably expressed as,
(5)
in which ∆G DDAA means the Gibbs free energy of tetrahedral hydrogen bonding, R Interfacial water/volume is the molecular number ratio of interfacial water layer to volume, and n HB means the average hydrogen bonding number per molecule. For tetrahedral hydrogen bonding, n HB is equal to 2.
3. Hydrophobic Effects
When a solute is embedded in liquid water, it is thermodynamically equivalent to form the solute-water interface. After the solute is treated as a sphere, the R Interfacial water/volume is 4·r H2O/R, in which R means the radius of solute. The hydration free energy is the free energy associated with the transfer of solute from vacuum to water. Therefore, as the sphere solute is dissolved into water, hydration free energy may be described as (Figure 7),
(6)
in which ΔG Water-water is the Gibbs free energy of water, r H2O is the average radius of a H 2 O molecule. For water at 293 K and 0.1 MPa, Gibbs free energy (ΔG Water-water) is −1500 cal/mol . Additionally, the average volume of a water molecule is 3 × 10−29 m 3 at ambient conditions. After the water molecule is regarded as a sphere, the corresponding diameter is 3.8 Å, and r H2O is 1.9 Å. In addition, the Gibbs free energy of tetrahedral hydrogen bonding (ΔG DDAA) is calculated to be −2.66 kJ/mol at 293 K and 0.1 MPa.
Figure 7.
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Hydration free energy at 293 K and 0.1 MPa. Hydration free energy is related to the size of solute, and critical radius (Rc) is expected. With increasing the solute size, it is divided into initial and hydrophobic solvation processes.
In thermodynamics, the lower the hydration free energy, the more stable the system is. Because hydration free energy is the sum of ∆G Water-water and ∆G Solute-water, it may be dominated by the Gibbs energy of bulk water (∆G Water-water) or interfacial water (∆G Solute-water). Of course, it is related to the size of dissolved solute. Therefore, the structural transition may be expected to occur as ∆G Water-water being equal to ∆G Solute-water,
(7)
in which Rc means the critical radius of dissolved solute . In our recent study , Rc is determined to be 6.5 Å at 293 K and 0.1 MPa (Figure 7).
With increasing the solute size, it is reasonably divided into the initial (ΔG Water-water< ΔG Solute-water) and hydrophobic (ΔG Water-water> ΔG Solute-water) solvation processes. The Gibbs free energy between solute and water (ΔG Solute-water) is inversely proportional to the solute size (1/R), which is related to the ratio of surface area to volume. Therefore, various dissolved behaviors of solutes in aqueous solutions may be expected in initial and hydrophobic processes, which may be related to the solute size (or concentrations) (Figure 8).
Figure 8.
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Different dissolved behaviors of solutes in aqueous solutions may be expected in initial (a) and hydrophobic (b) solvation processes.
In the initial solvation process, ΔG Solute-water is lower than ΔG Water-water (both of them are negative), or the solute size is smaller than Rc. Therefore, hydration free energy is dominated by the Gibbs free energy of interfacial water (ΔG Solute-water) [34,36]. To become more thermodynamically stable, this is fulfilled through maximizing the |ΔG Solute-water|. In other words, it is achieved through maximizing the ratio of surface area to volume of dissolved solutes. Therefore, the solutes may be dispersed in aqueous solutions, and water molecules are found between them (Figure 8). In addition, hydration free energy is proportional to the volume (or concentrations) of dissolved solutes.
Additionally, the dissolved solute mainly affects the hydrogen-bonded networks of interfacial water, DA hydrogen bondings tend to form within interfacial water layer . In the initial solvation process, hydration free energy may be related to DA hydrogen bondings. In comparison with DDAA (tetrahedral) structural motif, DA hydrogen bonding owns weaker hydrogen bonding energy, and higher entropy. Therefore, the driving force may be thermodynamically ascribed to the increase of entropy arising from interfacial water .
In hydrophobic solvation process, Gibbs free energy of interfacial water is higher than bulk water (ΔG Solute-water> ΔG Water-water). To be more thermodynamically stable, this may be fulfilled through maximizing |ΔG Water-water|. Indeed, this is accompanied with the minimization of Gibbs free energy of interfacial water (|ΔG Solute-water|). From the above, the ΔG Solute-water is related to the ratio of surface area to volume of solutes. It can be derived that the dissolved solutes may be aggregated in solutions in order to maximize the hydrogen bondings of water [34,36]. Therefore, the “attractive” forces may be expected between solutes in hydrophobic solvation process (Figure 8).
Due to the existence of DDAA (tetrahedral) hydrogen bondings in bulk water, this leads to ΔG Water-water being lower than ΔG Solute-water. In comparison with DA hydrogen bonding, tetrahedral hydrogen-bonded networks own the stronger hydrogen bonding energy, and lower entropy . Regarding the “attractive” force between solutes, it may be ascribed to be an enthalpic process, which is related to DDAA (tetrahedral) hydrogen bondings in bulk water. As the solutes are aggregated in water, they are also accompanied with the loss of entropy, which is related to the transition from interfacial to bulk water (Figure 8). Additionally, hydration free energy is proportional to the surface area of dissolved solutes.
The solutes may be “dispersed” or “accumulated” in water, which is related to the size of the solute. As two same sphere solutes are embedded into water, hydration free energy may be related to the separation between them. Therefore, the corresponding Rc is 3.25 Å at 293 K and 0.1 MPa . At ambient conditions, hydrophobic interactions may be expected as the solute radius is larger than 3.25 Å (>3.25 Å). This is demonstrated by our recent MD simulations on two C 60 fullerenes in water, and a pair of CH 4 molecules in water (Figure 9). Based on the calculated potential mean forces (PMFs), the water-induced contributions may be determined for C 60-C 60 in water, and CH 4-CH 4 in water, respectively. It seems that there exists the “attractive” force between two fullerenes (Figure 9). This is due to the radius of C 60 fullerene is larger than Rc. Because the CH 4 radius is less than 3.25 Å, water molecules are expected to exist in the region between two CH 4 molecules. The dissolved CH 4 molecules tend to be engaged in the solvent-separated conformation (Figure 9). In other words, as the CH 4 molecules are pushed together, this leads to the water molecules between them to be expelled into bulk water, which may be less thermodynamically stable. Therefore, as the CH 4 molecules are associated in water, the “repulsive” forces between them may be expected. This is in agreement with the Ashbaugh et al. MD simulations .
Figure 9.
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The dependence of hydrophobic interactions on solute size. Based on the calculated potential of mean forces (PMFs) between C 60 fullerenes in water and under vacuum (a), the water-induced PMF between C 60 fullerenes is determined (b). It is fitted as, ΔG = −3.35 + γ·3.80/(r − 10). During the H1w process, γ = 1. In the H2s process, solutes become contact in solutions. The fitted results at various γ (0.8, 0.6, 0.4) are drawn in squares. From the PMFs between CH 4 dimers in water and under vacuum (c), these are used to determine the water-induced PMF between CH 4 dimers (d). Stable configurations are also shown.
To investigate the thermodynamic properties of hydrophobic interactions, the PMFs can be determined through MD simulations on C 60-C 60 fullerenes in water, and CH 4-CH 4 in water at different temperature (300 K, 320 K and 340 K) (Unpublished data). Based on the calculated ΔG Water-induced for C 60-C 60 in water, and CH 4-CH 4 in water, these can be applied to determine the thermodynamic functions as the solutes are associated in water (Figure 10). To be more thermodynamically stable, the two CH 4 molecules are engaged in the solvent-separated conformation. In thermodynamics, it is driven by the entropy contributions related to interfacial water (Figure 10). However, the two fullerenes tend to be accumulated in solutions. Thermodynamically, it is dominated by enthalpy related to maximizing the hydrogen bondings of water. Additionally, this is also accompanied with the loss of entropy. Indeed, this is related to the transition from interfacial to bulk water as the fullerenes are aggregated in solutions, especially as the distance between them is less than 13 Å (Figure 10). Therefore, various thermodynamic driving forces may be expected in initial and hydrophobic solvation processes, which may be also in agreement with the above structural studies.
Figure 10.
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The thermodynamic characteristics of hydrophobic interactions. Based on the calculated PMFs of C 60-C 60 fullerenes and CH 4-CH 4 in water at different temperatures (300 K, 320 K and 340 K), water contributions to Gibbs energy (ΔG W) are determined, which are used to calculate the enthalpic (ΔH W) and entropic (-T·ΔS W) contributions. Different thermodynamic characteristics may be expected in initial (a) and hydrophobic (b) solvation processes.
In addition, enthalpy-entropy compensation (EEC) has been attracting attention because it is involved in many research fields [87,88,89,90,91], especially for the understanding of molecular recognition and drug design. In thermodynamics, EEC means that, if, for the particular reaction, ΔH and ΔS are changing in one direction (either increasing or decreasing), their changes that are transformed into ΔG are mutually compensated, and there is little change in the value of ΔG. To understand the nature of EEC, many works have been carried out. This means that EEC is real, very common, and a consequence of the properties of liquid water. Based on the calculated thermodynamics functions (Figure 10), hydrophobic interactions are closely related to EEC. Therefore, water plays a vital role in the process of EEC. Of course, this has been demonstrated in the work of Gilli et al. .
The dissolved solutes mainly affect the structure of interfacial water. Therefore, the effects of solutes on water structure may be related to the surfaces of solutes to be available for interfacial water. Owing to hydrophobic interactions, the dissolved solutes are attracted and tend to be aggregated in aqueous solutions in order to maximize the hydrogen bondings of water. In fact, the solutes coming into contact undoubtedly leads to the decrease of the solute surfaces available for interfacial water. Therefore, the Gibbs free energy of interfacial water may be described as,
(8)
where γ is named as the geometric factor . It is proposed to reflect the changes of solute surfaces while solutes are accumulated in water. Naturally, the solutes are rarely rigid. As they are dissolved in solutions, this may be accompanied by the changes of solute volume. From this, γ may be generally expressed as,
(9)
in which r Separation means the distance between solutes. When the solutes come into contact, the corresponding distance between them is termed ‘the hydrophobic radius’ (R H) . As the solutes are aggregated in water, it may be divided into H1w and H2s hydrophobic solvation processes, respectively (Figure 9).
Water molecules may be found between the dissolved solutes in H1w hydrophobic process, the separation between solutes is larger than R H (>R H), or γ is 1 . Due to hydrophobic interactions, the solutes are attracted to approaching each other. This decreases the distance between the solutes, and the water molecules in the region between them are expelled into bulk water. Therefore, hydrophobic interactions are fulfilled by the rearrangement of water molecules. Additionally, energy barriers may be expected in the H1w process, due to the expelled water molecules. Thermodynamically, the dissolved solutes are expected to approach each other in the direction with the lowest energy barrier, in which less water molecules may be expelled. From the above, the directional nature may be expected in the H1w hydrophobic process .
As the dissolved solutes come into contact in the H2s hydrophobic process (<R H), γ is less than 1 (γ < 1). The aggregation of solute surface undoubtedly leads to the decrease of solute surface to be available for interfacial water. ΔG Solute-water is proportional to the ratio of surface area to volume of the solutes. To become more thermodynamically stable, the solutes may be accumulated to minimize their ratio of surface area to volume. In fact, this may be demonstrated by MD simulations on graphite sheets [92,93] and carbon nanotubes (CNTs) [93,94] in solutions. Therefore, in the H2s solvation process, the directional nature of hydrophobic interactions is also expected. From the above, different directional natures are found in H1w and H2s processes, which may be discussed in the following section.
While the solutes are accumulated in solutions, the dewetting transition process, which is similar to the liquid-gas phase transition, may be observed (Figure 11). In 1995, according to MD simulations, Wallqvist and Berne [95,96] found the depletion of water between two nanoscale hydrophobic particles. Since then, water dewetting has been observed in many MD simulations on nanotubes and plates in water [97,98], water-protein interfaces , and collapsing polymers . Generally, the dewetting is believed to lead to the long-range hydrophobic attraction between solutes. Based on our recent study , dewetting may be closely related to H2s hydrophobic process, in which a single water layer between solutes may be expelled, and solutes become contact in solutions. In addition, dewetting is also related to the intermolecular interactions between solutes.
Figure 11.
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The changes of interfacial (a) and bulk (b) water during two C 60 fullerenes are aggregated in solutions. The dashed line represents the corresponding time of R H.
Hydrophobic effects are usually defined as the tendency of non-polar solutes to be aggregated in aqueous environments. From our recent studies [34,36,37] on hydrophobic effects, this may be reasonably described as the tendency for minimizing the ratio of the surface area to the volume of the solutes in order to maximize the hydrogen bondings of water. In fact, this is due to the dissolved solute mainly affecting the structure of interfacial water, as the hydrogen bondings of interfacial water are weaker than bulk water. Additionally, it is found that hydrophobic interactions are reasonably attributed to the structural competition between interfacial and bulk water. Therefore, it is reasonable to regard the hydrophobicity as “effects” rather than “bonds” [34,36,37].
4. Characteristics of Hydrophobic Interactions
Owing to hydrophobic interactions, the solutes are attracted to and tend to be aggregated in aqueous solutions to maximize the hydrogen bondings of water. The dissolved solutes mainly affect the structure of interfacial water. Therefore, hydrophobic interactions may be closely related to the geometric characteristics of solute, such as the solute size (or concentrations), the shape of solute, and the relative orientation between dissolved solutes. Additionally, hydrophobic interactions are due to the structural competition between interfacial and bulk water. In fact, water structure may be affected by many factors, such as temperature, pressure, dissolved salt, confined environments, etc. Therefore, these factors may also affect the hydrophobic interactions. In this review, it is focused on the dependence of hydrophobic interactions on solute size (or concentrations), the directional natures of hydrophobic interactions, and the temperature effects on hydrophobic interactions.
4.1. Dependence of Hydrophobic Interactions on Solute Size (or Concentrations)
With increasing the solute size, it may be divided into initial and hydrophobic solvation processes. Additionally, different dissolved behaviors of solutes may be expected in initial and hydrophobic solvation processes, such as the “dispersed” or “accumulated” distributions in solutions, which may be related to the solute size (or concentrations). In hydrophobic solvation process, the “attractive” forces are expected between the dissolved solutes. To maximize the hydrogen bondings of water, the dissolved solutes tend to be accumulated to form the aggregate in aqueous environments. Regarding the strength of hydrophobic interactions, this may be related to the size of aggregate. In our recent study , this is utilized to understand the origin of intermediate phase, which may be found before solute nucleation occurs in aqueous solutions.
In general, classical nucleation theory (CNT) is employed to understand the mechanism of nucleation. In the nucleation process, it is assumed that the free energy may be divided into a favorable term, and an unfavorable term. They are respectively related to the number of particles in the nucleus, and the dividing surface between the nucleus and the solution. In CNT, the free energy difference may be expressed as,
(10)
in which n is the molecular number of the crystal phase, Δμ represents the difference of chemical potential between crystal and liquid phase, γ means the surface tension, and s is the surface of the nucleus. From this, the critical nucleus (Nc) is expected in the process of nucleation.
The nucleation of crystals from solution is a ubiquitous process, which plays an important role in many research fields. To unravel the nucleation mechanism, many experimental approaches [101,102,103,104,105] have been developed and applied to study the nucleation mechanism of crystal in aqueous solutions. Different from CNT, the intermediate phase is found before the nucleation takes place in solutions, such as amorphous calcium carbonate (ACC) [104,105]. Therefore, the intermediate phase is regarded to be related to the solidification pathway of dissolved solutes. To understand the non-CNT pathway, various models are put forth, such as the two-step nucleation model [106,107]. According to the two-step nucleation mechanism, amorphous nuclei (intermediate phase) may be formed in the first step. Of course, it is necessary to overcome the energy barrier to form intermediate phases in solutions. In the second step, the nucleation of crystal is expected to take place in the middle of the amorphous phase. In comparison with direct nucleation from solution, the lower energy barrier is needed to overcome in the second step.
Regarding the non-CNT pathway, it is important to understand the origin of intermediate phase. Many theoretical simulations [38,108,109,110,111,112,113,114,115] have been carried out to study the mechanism of NaCl nucleation from the solutions. In a Giberti et al. study , a wurtzite-like polymorph was found in the nucleation process, which was regarded as an intermediate route during NaCl nucleation in the solution. Based on the Chakraborty and Patey MD simulation , an unstructured dense NaCl nucleus was found and transformed into the rock salt structure. These seem to support the two-step nucleation mechanism. According to our recent MD simulations , it is found that the dissolved behaviors of NaCl in water may be related to ion concentrations. With increasing the concentrations, the dissolved Na+ and Cl− ions may be associated to form the aggregate in solutions. However, no energy barrier is necessary to overcome this during the formation of the aggregate, which is different from the two-step nucleation mechanism. Additionally, as the ion aggregate is formed in solution, this is accompanied with the maximization of hydrogen bondings of water. In combination with our recent works [34,36], this is reasonably ascribed to the hydrophobic interactions. Compared with CNT, the formation of solute aggregate may lower the barrier height of nucleation and affect the nucleation mechanism of NaCl crystal in water.
With increasing ion concentrations, the dissolved ions tend to be aggregated to form the intermediate phase in solutions, which may be related to hydrophobic interactions. However, this is not considered in CNT theory. To explain the nucleation mechanism in solutions, it may be necessary to take into account the hydrophobic interactions related to the formation of aggregate. From this, CNT is reasonably revised as (Rev-CNT) (Figure 12),
(11)
where ΔG H represents the hydrophobic interactions related to the formation of intermediate phase. Additionally, different from the Nc in CNT, the critical aggregate (Agg C) may be expected, which is the largest aggregate as the nucleation takes place in the middle of intermediate phase.
Figure 12.
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The homogeneous nucleation mechanism of dissolved solutes in water. The dissolved behaviors of solutes in water are related to the ion concentrations, which affect the nucleation mechanism of crystal in water. Due to the formation of solute aggregate, this lowers the height of nucleation barrier.
Generally, nucleation processes can be classified as homogeneous and heterogeneous . For heterogeneous nucleation, it is formulated within the CNT framework, which is expressed as,
(12)
in which f(θ) is the shape factor in the range of 0 to 1 . Due to the existence of foreign surface, this lowers the nucleation barrier in the heterogeneous nucleation process. Therefore, it is crucial to understand the fundamentals of heterogeneous nucleation in order to achieve control over the nucleation process. In fact, as the foreign substances are embedded into water, they also affect the structure of interfacial water. To maximize the hydrogen bondings of water, the dissolved ions are expected to be accumulated at the surfaces of the substances. Due to the appearance of foreign substances, it is helpful to form the ion aggregate at the foreign surfaces, which may facilitate the phase transition (heterogeneous nucleation). Additionally, in combination with recent study , it is derived that heterogeneous nucleation may be affected by the geometric characteristics of foreign surface, especially geometric shape. In addition, it is also related to the molecular polarity of foreign surface. Further study is necessary.
4.2. Directional Natures of Hydrophobic Interactions
As the solutes are aggregated in aqueous solutions, it may be divided into H1w and H2s hydrophobic processes. In our recent study , different directional natures are found in the H1w and H2s processes. In the H1w process, the dissolved solutes are expected to approach in the specific direction with the lowest energy barrier, in which fewer expelled water molecules are expected. However, as the solutes come into contact in the H2s hydrophobic process (<R H), the solutes may be aggregated to minimize the ratio of surface area to volume of them. Additionally, the necessary pathway, where the solutes tend to pass through, may be expected while they are aggregated in water. Based on our recent study on directional nature of hydrophobic interactions, it is utilized to understand the nature of molecular recognition, especially the molecular specificity.
Molecular recognition is closely related to biological process and drug design. In the process of molecular recognition, two or more binding partners may interact through non-covalent interactions to form a specific complex in solutions. Numerous works have been devoted to understanding the nature of molecular recognition. To date, three structural models have been proposed, such as Fischer’s lock-and-key model , Koshland’s induced fit model , and the conformational selection model [120,121,122,123]. Based on the lock-and-key model, it is assumed that the conformation of the free protein is the same as that of ligand-bound protein. In the induced fit model, the conformational differences are expected after the ligand is associated with the target protein. Therefore, a new conformation, which may be more complementary to the binding partner, is expected for the protein. According to the conformational selection model, this means that all protein conformations pre-exist, and the ligand may select the most favored conformation. After the ligand is bound with the protein, a population shift may be expected, which may lead to redistribution of the conformational states.
In the process of molecular recognition, it is necessary to require the molecular specificity and affinity. Molecular specificity may be defined as why two or more binding partners approach and bind together to form a specific complex in aqueous environment [124,125,126,127]. Regarding the binding affinity, it determines whether the complex will be formed in aqueous environments, which is related to the strength of non-covalent interactions between partners. In other words, the binding affinity may be closely related to the thermodynamic stability of the molecular association in aqueous solutions.
In fact, the molecules are rarely rigid. Therefore, the conformational changes may be expected for partners before and after they are associated in aqueous solutions. However, it is well known that shape complementarity plays an important role as a ligand is associated with a target protein. In fact, this is emphasized in each structural model of molecular recognition. This means that, when a ligand is associated with a given target protein, it must fit into the cavity of the protein without steric conflicts. Based on our studies [34,36,37] on hydrophobic interactions, this may agree with the minimization of the ratio of surface area to volume of the solutes in the H2s process.
With decreasing the separation between partners, the direct interactions between them become stronger, such as van der Waals interactions, etc. In fact, it is necessary for partners to approach each other in solutions so that they are affected by the direct solute-solute interactions. Of course, this is due to hydrophobic interactions, which may lead to the dissolved solutes to approach each other in solutions. In other words, it seems that there exist the “attractive” forces between the solutes. Based on our recent work , molecular recognition is reasonably regarded to be driven by hydrophobic interactions (HDM, hydrophobic-interaction driven model ) (Figure 13).
Figure 13.
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Hydrophobic-interaction-driven model (HDM) of molecular recognition. The solutes mainly affect the structure of interfacial water (dashed line). Due to hydrophobic interactions, the solutes are attracted and approach in the direction with the lowest energy barrier in the H1w process. As the solutes become contact in the H2s process, they are accumulated in a specific direction to minimize the surface area to volume ratio. Additionally, with decreasing separation between the solutes in the H2s process, the solute-solute interactions become stronger. The affinity of molecular recognition is related to both the hydrophobic interactions and the solute-solute interactions.
From our recent studies [34,37], hydrophobic interactions may be closely related to both molecular specificity and affinity, especially the specificity of molecular recognition (Figure 13). According to HDM model, the molecular recognition may be divided into the following stages:
(I) Owing to the directional nature of hydrophobic interactions in H1w process, the dissolved solutes tend to approach in the specific direction with the lowest energy barrier, in which fewer water molecules between solutes may be expelled into bulk.
(II) When the solutes become contact in the H2s process, the solutes may be aggregated to minimize the ratio of surface area to volume of them. This is related to the directional nature of H2s hydrophobic process. In addition, this is utilized to explain the specificity in the process of molecular recognition.
(III) With decreasing distance between the partners, the direct intermolecular interactions between them become stronger, especially in the H2s process. Regarding the affinity of molecular recognition, it may be related to both hydrophobicity and the solute-solute interactions.
4.3. Temperature Effects on on Hydrophobic Interactions
Water plays an important role in the process of hydrophobic interactions. The hydrogen bondings of water may be influenced by temperature. Therefore, the changes of temperature undoubtedly affect the hydrophobic interactions. According to our recent study , the temperature effects on hydrophobic interactions may be related to the size of solute (Figure 14). With increasing temperature, this leads to the decrease of Rc (Figure 14). This may be utilized to investigate the nature of protein unfolding at high temperature (heat unfolding) and low temperature (cold denaturation).
Figure 14.
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The temperature effects on hydration free energy at 0.1 MPa. Regarding the dependence of hydration free energy on temperature, it is related to the size of solute. With increasing temperature, this decreases the Rc.
In thermodynamics, proteins may be stable in a limited range of temperatures. Both low and high temperatures may make the native protein structure become unstable, which leads to the denaturation of protein structure. It is well known that the native structure of protein may be disrupted at high temperature. However, regarding the molecular mechanism of cold denaturation, it is still under debate [129,130,131,132,133,134]. To date, the cold denaturation is usually attributed to the decreasing strength of the hydrophobic interactions [129,130,131].
Hydrophobic interactions play a vital role in the process of protein folding. In general, protein folding may be regarded to be driven by hydrophobic interactions. Due to hydrophobic interactions, this leads to the structural collapse of nonpolar amino acids of protein in solutions, which decreases the separation between the residues of protein. During protein folding, the direct interactions between residues become stronger, such as van der Waals interactions, hydrogen bondings, etc. Additionally, hydrophobic interactions also play an important role to keep the native structure of protein stable. According to Pace et al. works [135,136], hydrophobic interactions may be treated as the dominant contribution to keep protein stable, which may be greater than 50%.
Thermodynamically, protein stability may be evaluated by the energy difference between the folded and unfolded state of the protein in aqueous solutions. This energy difference can be used to determine whether a protein will keep its native folded conformation at a low temperature (or a high temperature). Based on experimental measurements, the free energy difference between these states is usually between 20 and 60 kJ/mol . Of course, this may be sufficient to prevent spontaneous unfolding at ambient conditions. However, it is noted that the protein native structure may be only marginally stable .
The changes of temperature may affect hydrophobic interactions, which undoubtedly affects the stability of protein structure. From our recent studies [34,35,36], hydrophobic interactions may be related to the size of solute. Therefore, it is necessary to understand the structural characteristics of protein. After the protein is globally regarded as a solute, the radii (R) are usually in the range from 12 and 23 Å , which is larger than Rc. Therefore, protein folding is wholly regarded to be driven by hydrophobic interactions. However, regarding the interior of protein, the average radius of amino acids is determined to be 3.3 Å with the standard deviation being 0.3 Å . It is found that the average radius of amino acids (3.3 Å) is slightly larger than 3.25 Å, which is the Rc as two same solutes are embedded into ambient water . Therefore, regarding the protein interior (local structure), it may be roughly engaged in the hydrophobic solvation process. Additionally, it is also engaged in initial solvation process. This is demonstrated by the buried water molecules, which are found in the interior of protein [139,140,141,142].
The dissolved behaviors of solutes in water may be related to the solute size. To understand the dependence of protein stability on temperature, it is necessary to investigate the changes of dissolved behaviors related to different solute size, being larger than Rc (>Rc), or less than Rc (<Rc). Temperature increase may lead to the decrease of Rc (Figure 15). With increasing temperature, the size of residue may change from being less than Rc to larger than Rc, and the obvious structural changes may be expected. In other words, the “repulsive” forces between solutes may be changed into the “attractive” forces. Therefore, with increasing temperature, the water molecules in the interior of protein (buried water) may be released into bulk water (Figure 15). Due to the expelled water, this may result in the accumulation in the interior of protein, and the global volume of protein is expected to become shrinking. Therefore, with increasing temperature, the inner structure of protein will be destroyed, which makes protein to lose the core activity.
Figure 15.
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The mechanism of protein unfolding. Inlet shows the dependence of Rc on temperature as two identical solutes are embedded into water. With increasing (or decreasing) temperature, obvious structural changes are expected as the solute size passing through Rc. This is reflected on the global and local (interior) structures of protein. Regarding the molecular mechanism of protein unfolding, cold denaturation may be different from heat unfolding.
Temperature decrease may lead to the increase of Rc (Figure 14). Therefore, protein stability is also affected by temperature decrease, which is different from the effects related to temperature increase. Due to temperature decrease, this may lead to the transition from hydrophobic to initial processes, which also affects the structure of protein (Figure 15). In other words, with decreasing temperature, water molecules may penetrate the core of protein (Figure 15). Therefore, the “repulsive” forces may be expected in the interior of protein. In addition, the global structure of protein may be changed from folded to extending conformation.
Regarding temperature effects on hydrophobic interactions, these are related to the size of solute. In combination with our recent studies [34,36,128], the obvious changes of dissolved behaviors of solutes are expected as the solute size passing though Rc. Therefore, the obvious structural difference may be expected between cold denaturation and heat unfolding. With decreasing temperature, this may lead to the transition from hydrophobic to initial process, water molecules are expected to penetrate the hydrophobic core of proteins in cold denaturation, which leads to the increase of the solvent accessibility (surface area). Regarding heat unfolding, although global structure of protein will be preserved, this may result in the packing of the hydrophobic core during the transition from initial to hydrophobic processes. Therefore, cold denatured proteins are more expanded, and heat unfolded states become more compact [133,134]. Of course, these can be demonstrated by the experimental measurements [133,134]. From the above, cold denaturation may be different from heat unfolding.
Owing to hydrophobic interactions, the dissolved solutes may be aggregated in solutions in order to minimize their ratio of surface area to volume. Additionally, the dissolved solutes mainly affect the structure of interfacial water. Therefore, hydrophobic interactions may be related to the shape of solute. This may be applied to explain the molecular packing parameter. The parameter was proposed by Israelachvili, Mitchell, and Ninham , which was defined as v 0/aI 0, where v 0 and I 0 were the volume and the length of the surfactant tail and a was the surface area of the hydrophobic core of the aggregate expressed per molecule in the aggregate. The concept of molecular packing parameter is widely used in physics, chemistry, and biology because it allows a simple and intuitive insight into the self-assembly phenomenon. It is well known that there exists the following connection between the molecular packing parameter and the aggregate shape: 0 ≤ v 0/aI 0 ≤ 1/3 for sphere, 1/3 ≤ v 0/aI 0 ≤ 1/2 for cylinder, and 1/2 ≤ v 0/aI 0 ≤ 1 for bilayer . Therefore, after the molecular packing parameter is determined, the shape and size of the equilibrium aggregate may be readily identified as shown above. From this work, it can be found that this parameter may be closely related to the effects of solute shape on hydrophobic interactions. Further study may be covered in our next work.
Hydrophobic interactions are involved in a lot of chemical and biological phenomena in aqueous environments. This review focuses on our current understanding on hydrophobic interactions, such as the dependence of hydrophobic interactions on the solute size, the directional natures of hydrophobic interactions, and the temperature effects on hydrophobic interactions. In fact, hydrophobic interactions may also be extended to investigate the molecular mechanism of protein folding, ion selectivity of nanopore (ion channel), Hofmeister effects, superhydrophobic surface, etc.
To maximize the hydrogen bondings of water, the dissolved solutes are attracted to approach each other until they may be affected by the direct solute-solute interactions. Therefore, hydrophobic interactions are reasonably regarded to be the fundamental driving force in the chemical and biological processes. Based on our recent study on hydrophobic interactions, they are reasonably ascribed to the structural competition between interfacial and bulk water. Therefore, water plays a vital role in the process of hydrophobic effects. From our current understanding, water is reasonably regarded as “the director of a wonderful performance, not the audience”.
5. Conclusions
Hydrophobic interactions may be regarded as the fundamental driving force in numerous biophysical and biochemical processes in aqueous solutions. In this study, we focus on our current understanding on hydrophobic effects. Based on our recent studies, the following conclusions are derived:
(1) According to the structural studies on water and air-water interface, hydration free energy is determined, and applied to unravel the nature of hydrophobic effects. It is found that hydrophobic interactions may be attributed to the structural competition between interfacial and bulk water.
(2) Hydration free energy is related to the size of dissolved solute (or concentrations). With increasing the solute size, it may be divided into initial and hydrophobic solvation processes. In addition, various dissolved behaviors of solutes, such as dispersed and accumulated distributions in solutions, may be expected in initial and hydrophobic solvation processes. This is utilized to understand the origin of intermediate phase, which is found before solute nucleation occurs in solutions.
(3) While the dissolved solutes are accumulated in hydrophobic solvation process, it is divided into H1w and H2s hydrophobic processes. Additionally, different directional natures are expected in H1w and H2s processes. This is applied to understand the mechanism of molecular recognition, especially the specificity.
(4) Temperature changes may affect hydrophobic interactions, which may be related to solute size. With increasing temperature, this leads to the decrease of Rc. This is utilized to understand the nature of protein unfolding at high temperature (heat unfolding) and low temperature (cold denaturation). It can be found that cold denaturation may be different from heat unfolding.
Acknowledgments
The editor and reviewers are greatly appreciated for providing good suggestions to revise the paper.
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Informed Consent Statement
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Data Availability Statement
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Conflicts of Interest
The authors declare no conflict of interest.
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187797 | https://onlinelibrary.wiley.com/doi/10.1002/ccr3.70091 | A Rare and Intriguing Case of Wilson's Disease Initially Suspected of Systemic Lupus Erythematosus - Khodashahi - 2025 - Clinical Case Reports - Wiley Online Library
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Clinical Case Reports
Volume 13, Issue 1 e70091
CASE REPORT
Open Access
A Rare and Intriguing Case of Wilson's Disease Initially Suspected of Systemic Lupus Erythematosus
Mandana Khodashahi,
Mandana Khodashahi
khodashahimn@mums.ac.ir
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Conceptualization, Data curation, Investigation, Project administration, Supervision, Validation, Writing - review & editing
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Najmeh Mohajeri,
Najmeh Mohajeri
mohajerin1@mums.ac.ir
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Data curation, Investigation, Validation, Writing - review & editing
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Moeid Reza Alipour,
Moeid Reza Alipour
moeedrezalipoor@gmail.com
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Data curation, Investigation, Writing - review & editing
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Reza Khademi,
Reza Khademi
khademir982@mums.ac.ir
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Data curation, Investigation, Writing - review & editing
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Nama Mohamadian Roshan,
Nama Mohamadian Roshan
roshann@mums.ac.ir
Department of Pathology, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Investigation, Methodology, Validation, Writing - original draft
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Behzad Aminzadeh,
Behzad Aminzadeh
aminzadehb@mums.ac.ir
orcid.org/0000-0001-6569-4670
Department of Radiology, Faculty of Medicine, Vakil Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Investigation, Methodology, Validation, Writing - original draft
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Muhammed Joghatayi,
Corresponding Author
Muhammed Joghatayi
muhammedjyi@gmail.com
rdrc.mums@gmail.com
orcid.org/0009-0004-9947-4616
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Correspondence:
Muhammed Joghatayi (muhammedjyi@gmail.com)
Contribution: Conceptualization, Data curation, Investigation, Project administration, Validation, Visualization, Writing - original draft, Writing - review & editing
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Mandana Khodashahi,
Mandana Khodashahi
khodashahimn@mums.ac.ir
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Conceptualization, Data curation, Investigation, Project administration, Supervision, Validation, Writing - review & editing
Search for more papers by this author
Najmeh Mohajeri,
Najmeh Mohajeri
mohajerin1@mums.ac.ir
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Data curation, Investigation, Validation, Writing - review & editing
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Moeid Reza Alipour,
Moeid Reza Alipour
moeedrezalipoor@gmail.com
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Data curation, Investigation, Writing - review & editing
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Reza Khademi,
Reza Khademi
khademir982@mums.ac.ir
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Data curation, Investigation, Writing - review & editing
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Nama Mohamadian Roshan,
Nama Mohamadian Roshan
roshann@mums.ac.ir
Department of Pathology, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Investigation, Methodology, Validation, Writing - original draft
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Behzad Aminzadeh,
Behzad Aminzadeh
aminzadehb@mums.ac.ir
orcid.org/0000-0001-6569-4670
Department of Radiology, Faculty of Medicine, Vakil Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Contribution: Investigation, Methodology, Validation, Writing - original draft
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Muhammed Joghatayi,
Corresponding Author
Muhammed Joghatayi
muhammedjyi@gmail.com
rdrc.mums@gmail.com
orcid.org/0009-0004-9947-4616
Rheumatic Diseases Research Center, Ghaem Hospital, Taghi Abad, Mashhad University of Medical Sciences, Mashhad, Iran
Correspondence:
Muhammed Joghatayi (muhammedjyi@gmail.com)
Contribution: Conceptualization, Data curation, Investigation, Project administration, Validation, Visualization, Writing - original draft, Writing - review & editing
Search for more papers by this author
First published: 07 January 2025
Funding: The authors received no specific funding for this work.
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1 Introduction
2 Case History
3 Methods
4 Conclusion and Results
5 Discussion
6 Conclusion
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ABSTRACT
When systematic lupus erythematosus-like lab results (e.g., positive anti-double-stranded DNA antibody, low complement component 3) are inconsistent with physical findings, such as the absence of arthritis or nephritis, clinicians should consider diagnoses such as Wilson's disease, especially in the presence of abnormal liver function and elevated international normalized ratio (INR).
1 Introduction
Wilson's disease (WD) is a rare autosomal-recessive disorder characterized by defective copper metabolism. Copper accumulation starts at birth, and patients usually become symptomatic in adolescence or early adulthood. Symptoms arise due to copper buildup in various organs, particularly the liver and brain .
Systemic lupus erythematosus (SLE) is a multisystem autoimmune disease that can affect any organ. It is also known as the “Great Imitator” due to its vast variability in presentations, signs, and symptoms. Despite significant research, the diagnosis and management of SLE remain complex and challenging. It can manifest in a wide array of clinical forms, ranging from mild fatigue to severe complications such as acute kidney injury, psychosis, and seizures .
Our patient was referred to the rheumatology clinic because of positive paraclinical tests for SLE, pancytopenia, elevated liver enzymes, and fatigue, which can all be seen in possible cases of SLE.
This case report underscores the diagnostic complexities when distinguishing WD from SLE, particularly in patients presenting with hepatic and hematologic abnormalities commonly associated with both conditions.
2 Case History
A 19-year-old girl with no significant previous medical history except for a low mood and depression, with a suicide attempt 6 months prior, medication use, or illicit drug use, was referred to an internist due to a 2-year history of fatigue, asthenia, easy bruising, and occasional epistaxis. These symptoms had been gradually worsening over the past 6 months. She had no significant family history among her first- and second-degree relatives, and her parents were non-consanguineous. Routine tests revealed pancytopenia (white blood cell (WBC) = 3.3 × 10^9/L, red blood cell (RBC) = 3.8 × 10^12/L, hemoglobine (Hb) = 10.9 g/dL, and platelets (PLT) = 59 × 10^9/L), normal serum creatine level of 0.9 mg/dL (normal range = 0.6–1.3), elevated liver enzymes (aspartate aminotransferase (AST) = 214 U/L [normal range: up to 38], alanine aminotransferase (ALT) = 196 U/L [normal range: up to 31], alkaline phosphatase (ALP) = 442 U/L [normal range: up to 480]), total bilirubin of 3 mg/dL, direct bilirubin of 1 mg/dL, and lactate dehydrogenase (LDH) of 418 U/L [normal range: up to 480], along with disturbed coagulation tests (partial thromboplastin time (PTT) = 49.4 s [normal range: 25–35], INR 1.52, prothrombin time (PT) = 22.4 s). She and her parents had no history of jaundice.
The conjugated hyperbilirubinemia increased AST and ALT, and normal ALP suggested hepatocellular damage. Considering her age, the internist ordered a viral hepatitis panel (antibody to hepatitis A virus (Anti-HAV), hepatitis B core antibody (HBc Ab), hepatitis C virus antibody (HCV Ab), and hepatitis B surface antigen (HBs Ag)) and markers for rheumatologic diseases and autoimmune hepatitis. The viral panel results were negative, but fluorescent antinuclear antibody (FANA), Anti-Ro, and anti-cyclic citrullinated peptide antibodies (Anti-CCP) were positive, while anti-mitochondrial antibodies (AMA) was negative. She had an rheumatoid factor (RF) (55 IU/mL [normal range: up to 20]) and erythrocyte sedimentation rate (ESR) (20 mm/h), with C-reactive protein (CRP) within the normal range (0.2 mg/dL [normal range: up to 6]). Urinalysis showed no hematuria or proteinuria. The internist referred the patient to a rheumatology clinic for further investigation.
Except for fatigue, her physical examination and history-taking revealed no significant findings. There was no history of arthritis, arthralgia, seizures, malar rash, skin lesions, or morning stiffness. The physical examination was normal; no findings suggested SLE or rheumatoid arthritis (RA). The pull test was negative, and no mouth ulcers were found.
Considering her coagulation disturbances, positive ANA profile, high liver enzyme levels, and history of easy bruising and epistaxis, the rheumatologist requested another complete blood count (CBC) and ANA profile, serum levels of C3 and C4, anti-smooth muscle antibodies (ASMA), anti-liver kidney microsomal antibodies (anti-LKM1), perinuclear antinuclear neutrophil antibodies (P-ANNA), liver and spleen sonography, and a liver coagulation panel (factors VIII, IX, X, and XI activity levels). The results showed a negative anti-dsDNA and Anti-Ro, low C3 serum level, normal C4 serum level, and confirmed pancytopenia (WBC = 1.7 × 10^9/L, RBC = 3.7 × 10^12/L, Hb = 10.5 g/dL, PLT = 59 × 10^9/L) and elevated liver enzymes (AST = 218 U/L, ALT = 180 U/L, ALP = 201 U/L). While Anti-CCP, RF, ASMA, anti-LKM1, and P-ANNA were negative. Abdominal sonography showed hepatomegaly with coarse and heterogeneous echogenicity and splenomegaly with normal echogenicity. The activity levels of factors IX, X, and X1 were low, while factor VIII had a high activity level with no response to the mixing test. Serum protein electrophoresis showed low albumin and increased immunoglobulin G (IgG) levels (Figure1).
FIGURE 1
Open in figure viewerPowerPoint
Serum protein electrophoresis shows an increase in γ band and a decrease in albumin level.
Due to her physical examinations and lab data incompatibility, the rheumatologist admitted the patient to the rheumatology unit for further investigations.
3 Methods
She received prednisolone 1 mg/kg/day and intravenous immunoglobulin (IVIG) for 6 days. However, her pancytopenia and high liver enzyme levels did not improve. To rule out cirrhosis, an abdominal CT scan with and without injection and color-doppler ultrasound of the portal vein was performed. An endoscopy was also performed to check for esophageal varices; no varices were found. The color-doppler ultrasound showed no thrombosis, normal blood flow, and portal vein diameter. The abdominal CT scan showed hepatomegaly with heterogeneous parenchymal density, an enlarged spleen with normal density, and several lymphadenopathies in para-aortic and retrocaval regions, suggesting lymphoma (Figure2).
FIGURE 2
Open in figure viewerPowerPoint
Non-contrast CT scan with mildly heterogenous liver, which appears normal on the post-contrast image with enlargement of the spleen. (a) sagittal view, (b) axial view.
However, the patient had no history of B symptoms (unintended weight loss, low appetite, or night sweats), and peripheral blood smear (PBS) showed no abnormal white blood cells. Bone marrow aspiration showed normal bone marrow activity, suggesting peripheral pancytopenia.
On the seventh day of her admission, she became gradually agitated and aggressive, had a generalized tonic–clonic seizure, experienced a decrease in consciousness level, and went into a coma (GCS = 3), leading to immediate Intensive Care Unit (ICU) admission. The vital signs were normal. Her pulse rate was 80/min. Her respiratory rate was 22/min. Her blood pressure was 126/71 mmHg, and her body temperature was 36.6°C. Her oxygen saturation was 98% on room air. Lab data showed within normal range serum electrolytes (Na = 138 [normal range: 135–145 mEq/L] K = 3.7 [normal range: 3.5–5.3 mEq/L]) and a mild respiratory alkalosis (pH = 7.49, pCO 2 = 29 mmHg and HCO3 = 23.3). A brain CT scan showed effacement and slit in both lateral ventricles with brain edema and no mass lesion, prompting an MRI due to her acute condition.
A neurology consult requested a brain magnetic resonance imaging (MRI) with and without injection and MRV, revealing hypersignal intensity in both globus pallidi on T1-weighted images without enhancement and restriction, compatible with toxic or metabolic encephalopathy (Figure3). No abnormality was seen after the gadolinium (GD) injection. Magnetic resonance venography (MRV) showed no thrombosis and was reported as normal.
FIGURE 3
Open in figure viewerPowerPoint
Symmetrical high T1 signal is seen within the bilateral globus pallidi with a normal appearance on T2W images. (a) hyperintensity on bilateral globi pallidi on T1W images. (b) hyperintensity in corticospinal (pyramidal) tract.
Ophthalmologic examination was negative for Kayser-Fleischer (KF) rings or sunflower cataracts.
After correcting the INR with vitamin K and receiving multiple fresh frozen plasma (FFP) and platelet transfusions, a lumbar puncture to rule out septic meningoencephalitis, including Tuberculosis (TB) and Herpes simplex virus polymerase chain reaction (HSV PCR), was performed and came back negative. Given the paraclinical results indicating hepatic encephalopathy, she was listed for a liver transplant and was treated with mannitol and lactulose in the ICU. She fully recovered and became conscious and oriented after 24 h. A liver biopsy revealed nodular liver and fibrosis with portal and lobular inflammation, suggesting metabolic disorders such as wilson's disease or autoimmune hepatitis (Figure4).
FIGURE 4
Open in figure viewerPowerPoint
The patient's liver biopsy showed nodular liver and fibrosis with portal and lobular inflammation. (a) Inter-nodular fibrosis, (b) nodular pattern, (c) hepatocytic Rosette, (d) hepatocyte ballooning, (e) lympho-plasma cell infiltration, (f) hepatocyte damage, (g) mild cholestasis.
Further copper concentration measurement in the liver biopsy showed a high level of copper (260 μg/g dry weight [normal range: up to 50 μg/g]). Serum ceruloplasmin was low (157 μg/dL [normal range: 204–407 μg/dL]), and 24-h urine copper excretion was high (168 μg/24 h [normal range: 15–70 μg/24 h]), confirming wilson's disease. The summary of laboratory tests, imaging studies, and their findings, used to evaluate the patient's condition are presented in Table1.
TABLE 1. Diagnostic workup timeline.
| Date | Tests performed | Findings | Normal range |
--- --- |
| Initial lab tests ordered by internist | CBC | WBC = 3.3 × 10^9/L RBC = 3.8 × 10^12/L PLT = 59 × 10^9/L | 4.5–11.0 × 10^9/L 4.2–5.4 × 10^12/L 150–400 × 10^9/L |
| Liver enzymes | AST = 214 U/L ALT = 196 U/L ALP = 442 U/L | Up to 38 Up to 31 Up to 480 |
| Coagulation tests | PTT 49.4 s INR 1.52 PT = 22.4 s | 25–35 s—11–13.5 s |
| Follow-up tests ordered by internist | FANA | Positive | Qualitative |
| Anti-CCP | Positive | Qualitative |
| RF | 55 IU/mL (elevated) | Up to 20 |
| Anti-dsDNA | Negative | Qualitative |
| Anti-Ro | Positive | Qualitative |
| Urinalysis | No hematuria or proteinuria | — |
| Viral hepatitis panel (Anti-HAV, HBc Ab, HCV Ab, HBs Ag) | All negative | — |
| Autoimmune hepatitis biomarkers | All negative | Qualitative |
| Tests ordered by rheumatologis | CBC | WBC = 1.7 × 10^9/L RBC = 3.7 × 10^12/L PLT = 59 × 10^9 | 4.5–11.0 × 10^9/L 4.2–5.4 × 10^12/L 150–400 × 10^9/L |
| Liver enzymes | AST = 214 U/L ALT = 196 U/L ALP = 442 U/L | Up to 38 Up to 31 Up to 480 |
| VIII, IX, X, and XI activity levels | VIII (elevated) IX, X, and XI (decreased) | — |
| ANA profile | All negative | — |
| Abdominal sonography | Hepatomegaly Splenomegaly | — |
| Hospital admission tests | MRI | Hypersignal intensity in globi pallidi | — |
| Endoscopy | No varices | — |
| Color-doppler ultrasound of portal vein | No thrombosis Normal flow | — |
| Abodmianl CT scan | Hepatomegaly Splenomegaly Para-aortic lymphadenopathy | — |
| Bone marrow aspiration | Normal bone marrow activity | — |
| Copper studies and liver biopsy | Urinary copper | 168 μg/24 h (elevated) | 15–70 μg/24 h |
| Serum ceruloplasmin | 57 μg/dL (low) | 204–407 μg/dL |
| Liver biopsy | 260 μg/g (elevated) | Up to 50 μg/g dry weight |
| Follow-up | CBC | WBC = 6.28 × 10^9/L RBC = 3.47 × 10^12/L PLT = 70 × 10^9/L | 4.5–11.0 × 10^9/L 4.2–5.4 × 10^12/L 150–400 × 10^9/L |
| Liver enzymes | AST = 33 U/L ALT = 110 U/L (elevated) ALP = 498 U/L | Up to 38 Up to 31 Up to 480 |
4 Conclusion and Results
Following the confirmation of wilson's disease, treatment was initiated with zinc sulfate (25 mg three times daily) and trientine (1250 mg/day) to reduce copper accumulation and promote its excretion. The patient's pancytopenia and coagulation abnormalities showed progressive improvement, indicating a favorable response to treatment (WBC = 6.28 × 10^9/L, RBC = 3.47 × 10^12/L, PLT = 70 × 10^9/L, PT = 34 s, PTT = 35, INR = 1.11). Her liver enzymes began to normalize (AST = 33 U/L, ALT = 110 U/L, ALP = 498 U/L). She was discharged 5 days later with a good general well-being. During her regular examinations, she showed significant improvement.
5 Discussion
This case report describes a 19-year-old female initially suspected to have SLE due to her symptoms of fatigue, pancytopenia, elevated liver enzymes, and positive autoimmune markers. However, further investigations revealed WD as the correct diagnosis, highlighted by abnormal liver biopsy findings, high hepatic copper content, low serum ceruloplasmin, and elevated 24-h urine copper excretion.
WD primarily targets the liver and brain, similar to SLE, which can present with neurological and hepatic manifestations [1, 3].
While the involvement of SLE in the onset of asymptomatic liver disease remains a topic of debate, many experts acknowledge that SLE frequently results in subclinical liver dysfunction, referred to as lupus hepatitis. This condition is a nonspecific reactive liver disease, primarily driven by complement deposition and vasculitis-induced damage to the liver. Research indicates that hepatomegaly is present in about 20%–40% of SLE patients and is often associated with autoimmune hepatitis, lupus hepatitis, or drug-induced liver injury. However, hepatomegaly is relatively uncommon in SLE and typically signals the coexistence of autoimmune or drug-induced hepatitis .
In this case, specific clinical and laboratory findings did not match with SLE or autoimmune hepatitis. Although PTT can be abnormal in SLE, the presence of high INR, along with a coarsely enlarged liver [5, 6], led the rheumatologist to lean more toward a liver injury rather than SLE: the negative AMA, ASMA, and P-ANNA. Anti-LKM1, coupled with no response to corticosteroids, significantly lowered the probability of autoimmune hepatitis .
This case presented several diagnostic pitfalls. Positive ANA and RF can also be seen in WD, but anti-Ro antibodies and low C3 levels are not typically reported in WD and mislead the diagnosis. The absence of KF rings in the eyes, a hallmark of WD , was another diagnostic challenge. Additionally, symptoms like pancytopenia, fatigue, and a history of depression are common in both WD and SLE, complicating the diagnosis further [1, 9]. Moreover, the patient's age (19 years) posed a diagnostic pitfall since the onset of both diseases typically occurs in early adulthood.
Notably, WD has been associated with a range of immunological abnormalities, including positive ANA, positive anti-neutrophil cytoplasmic antibody (ANCA), and anti-CCP. Such markers, typically linked to autoimmune diseases such as SLE, autoimmune hepatitis, and multiple sclerosis, can complicate diagnosis by mimicking or coexisting with autoimmune conditions .
For example, anti-Ro antibodies, primarily associated with Sjögren's syndrome and occasionally found in SLE, can appear at low titers in about 15% of healthy individuals. Though less common, false positives remain a possibility .
This immunological overlap becomes even more complex in WD patients treated with D-penicillamine, a drug linked to drug-induced lupus , which increases the risk of misdiagnosis. Therefore, the presence of these autoantibodies in patients with WD, especially those receiving D-penicillamine, requires careful interpretation, such as assessing coexisting antibodies like Anti-histone antibodies (AHA), to differentiate between underlying autoimmune processes and drug effects rather than assuming a definitive autoimmune disease.
Low complement levels, particularly C3, in SLE, are primarily due to immune complex formation that activates and consumes complement proteins, and they are mainly suggestive of high disease activity. However, complement levels can also be reduced due to liver dysfunction, as the liver synthesizes these proteins, a key pathology in WD . In the absence of SLE symptoms and signs of high activity, such as lupus nephritis or arthritis, other mechanisms, including liver dysfunction, should be considered. It is crucial not to attribute low complement levels solely to autoimmune processes in the absence of signs and symptoms of SLE.
Several case reports have documented the overlap of SLE and WD, complicating the diagnosis and treatment strategies. For instance, Hadef et al. reported a case of a 12-year-old boy with concurrent WD and SLE, where both conditions were diagnosed simultaneously. The patient initially presented with hemolytic anemia and impaired liver function, leading to the suspicion of WD. Further investigations confirmed both WD and SLE, and the patient responded to treatment with corticosteroids and chelation therapy .
Zhang et al. described a 35-year-old woman with SLE who was later found to have WD during her routine follow-up due to unexplained liver fibrosis. Genetic testing confirmed WD, and the patient was treated with zinc sulfate and medications for SLE. This case underscores the importance of considering WD in patients with unexplained hepatic involvement in SLE .
Yang et al. presented a case of a 9-year-old girl diagnosed with both SLE and WD. Despite effective control of her SLE symptoms, her liver function did not improve, leading to the suspicion of WD. Genetic testing confirmed the diagnosis, and the patient underwent liver transplantation. Unfortunately, she passed away shortly after the surgery, highlighting the severe implications of delayed diagnosis and treatment .
However, the patient did not respond to SLE treatment in this case. In contrast, WD treatment with zinc sulfate and trientine significantly improved pancytopenia, elevated liver enzymes, and impaired INR. This response indicates that the case was not an overlap of SLE and WD but rather a misdiagnosis of SLE in the presence of WD.
This case highlights one key point: positive ANA profile results alone are insufficient to diagnose SLE. Physicians must consider other laboratory data, such as impaired INR and hepatomegaly, which are uncommon in SLE.
The absence of typical physical examination findings should be strongly considered, as emphasized in the 2019 EULAR/ACR (2019 European League Against Rheumatism/American College of Rheumatology) criteria . Physicians should avoid unnecessary lab tests that can mislead the diagnostic approach and increase the burden on the healthcare system . The case emphasizes the importance of comprehensive clinical evaluation and the interpretation of laboratory tests in the context of the patient's overall condition . Highlighting the famous quote in medicine: “Treat the patient, not the disease” .
6 Conclusion
This case report underscores the diagnostic challenges in distinguishing between SLE and WD. It emphasizes the importance of thorough investigation and consideration of WD in patients with hepatic and hematologic abnormalities unresponsive to conventional treatments for autoimmune diseases. Early diagnosis and appropriate treatment of WD are essential to improve patient outcomes and prevent irreversible organ damage.
Author Contributions
Mandana Khodashahi: conceptualization, data curation, investigation, project administration, supervision, validation, writing – review and editing. Najmeh Mohajeri: data curation, investigation, validation, writing – review and editing. Moeid Reza Alipour: data curation, investigation, writing – review and editing. Reza Khademi: data curation, investigation, writing – review and editing. Nama Mohamadian Roshan: investigation, methodology, validation, writing – original draft. Behzad Aminzadeh: investigation, methodology, validation, writing – original draft. Muhammed Joghatayi: conceptualization, data curation, investigation, project administration, validation, visualization, writing – original draft, writing – review and editing.
Acknowledgments
No funding was received for the preparation or publication of this article.
Ethics Statement
This case report was conducted in accordance with the principles of the Declaration of Helsinki. All efforts were made to maintain the patient's privacy and confidentiality.
Consent
Written informed consent was obtained from the patient for the publication of this case report, including clinical details and images.
Conflicts of Interest
The authors declare no conflicts of interest.
Open Research
Data Availability Statement
Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.
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Keywords
hepatomegaly
liver injury
pancytopenia
systemic lupus erythematosus
Wilson's disease
Publication History
Issue Online: 07 January 2025
Version of Record online: 07 January 2025
Manuscript accepted: 22 November 2024
Manuscript revised: 09 November 2024
Manuscript received: 11 September 2024
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187798 | https://brainly.com/question/39131859 | [FREE] The chemical equation for the formation of water (H_2O) from hydrogen gas (H_2) and oxygen gas (O_2) is: - brainly.com
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The chemical equation for the formation of water (H 2O) from hydrogen gas (H 2) and oxygen gas (O 2) is:
2 H 2+O 2→2 H 2O
To balance this equation, the coefficient 2 should be placed in front of H 2 and H 2O.
So, the balanced equation is:
2 H 2+O 2→2 H 2O
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The chemical equation for making water from hydrogen and oxygen gases is: 2H2 + O2 → 2H2O. This and all other chemical reactions should always be balanced according to the principle of conservation of matter. An example of another reaction is 2H2O2 breaking down into 2H2O and O2.
Explanation
The chemical equation for the formation of water from hydrogen gas and oxygen gas is indeed: 2H2 + O2 → 2H2O. This equation is balanced, meaning the number of atoms for each element is the same on both sides of the equation. This complies with the law of conservation of matter, which states that no atoms are created or destroyed in a chemical reaction under normal circumstances.
An example of a simple chemical reaction would be the decomposition of hydrogen peroxide molecules (H2O2), which breaks down into water (H2O) and oxygen (O2). The balanced equation for this reaction would be 2H2O2 → 2H2O + O2.
In both examples, it's crucial to balance the equation to accurately represent the chemical process. In a balanced equation, the number of atoms of each element in the reactants matches the number in the products, adhering to the conservation of matter principle.
Learn more about Balancing Chemical Equations here:
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The balanced chemical equation for forming water from hydrogen and oxygen gases is 2H₂ + O₂ → 2H₂O. This equation ensures that the number of each type of atom is the same on both sides, adhering to the law of conservation of mass. Thus, two molecules of hydrogen react with one molecule of oxygen to form two molecules of water.
Explanation
The chemical equation for the formation of water (H₂O) from hydrogen gas (H₂) and oxygen gas (O₂) is written as:
2 H 2+O 2→2 H 2O
This equation is balanced, which means that there are equal numbers of each type of atom on both sides of the equation. To break it down:
Reactants: On the left side, we have 2 molecules of hydrogen (H₂) and 1 molecule of oxygen (O₂). This gives us:
2 hydrogen atoms (from 2H₂)
2 oxygen atoms (from O₂)
Products: On the right side, we have 2 molecules of water (H₂O). Each water molecule contains:
2 hydrogen atoms
1 oxygen atom
Therefore, with 2 molecules of H₂O, we have:
4 hydrogen atoms (from 2H₂O)
2 oxygen atoms (from 2H₂O)
Balancing the Equation:
The total number of hydrogen atoms on the left (4 from 2H₂) matches the total on the right (4 from 2H₂O).
The total number of oxygen atoms on both sides also matches (2 from O₂ on the left and 2 from 2H₂O on the right).
Law of Conservation of Mass: This equation follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. The balanced equation reflects that the same number of each type of atom is present before and after the reaction.
Also, it's important to note that coefficients in chemical equations indicate the number of molecules involved in the reaction. The balanced equation shows that two molecules of hydrogen gas react with one molecule of oxygen gas to produce two molecules of water. This concept can be applied to other reactions as well by ensuring that the number of each atom on either side is equal.
Examples & Evidence
An example of another balanced chemical reaction is 2H₂O₂ → 2H₂O + O₂, where hydrogen peroxide decomposes into water and oxygen, maintaining the balance of atoms on both sides of the equation.
The chemical equation 2H₂ + O₂ → 2H₂O has been established through fundamental chemical principles and upheld by the law of conservation of mass, as taught in chemistry courses.
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Community Answer The following balanced equation shows the formation of water. 2H2 + O2 Right arrow. 2H2O How many moles of oxygen (O2) are required to completely react with 27.4 mol of H2?
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Predict the products of the following reaction. If no reaction will occur, use the NO REACTION button. Be sure your chemical equation is balanced! C H 3(C H)2C H 3(g)+O 2(g)→
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187799 | https://www.hey.nhs.uk/wp/wp-content/uploads/2018/07/Hodgkin-Lymphoma.pdf | HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Hull and East Yorkshire and North Lincolnshire NHS Trusts Haematology Multidisciplinary Team Guideline and Pathway Hodgkin Lymphoma 1 BACKGROUND The Hull and North Lincolnshire Haematology Multidisciplinary team manages patients with haematological malignancies on three sites, Diana Princess of Wales Hospital Grimsby, the Queens Centre for Haematology and Oncology at Castle Hill Hospital Hull and East Yorkshire Hospitals NHS Trust and Scunthorpe Hospital. Levels of service provided in these organisations is as defined in the NICE guidance “Haematological Cancers: improving outcomes NG47” 25th May 2016. Low-to-intermediate intensity chemotherapy is delivered in Grimsby, the Queens Centre Castle Hill Hospital and Scunthorpe Hospital. High-intensity chemotherapy and autologous stem cell transplantation is delivered at the Queens Centre, Castle Hill Hospital. The following chemotherapy regimens for Hodgkin Lymphoma (HL) can be delivered in centres providing low-to intermediate intensity chemotherapy: ChlVPP Vinblastine single agent Rituximab single agent The following chemotherapy regimens for HL can only be delivered in centres providing high intensity chemotherapy: Escalated BEACOPP ABVD/AVD RCVP RCHOP DHAP IVE ICE BEAM autoPBSC LEAM autoPBSC Brentuximab Nivolumab HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 2 POLICY / PROCEDURE / GUIDELINE DETAILS Diagnosis of Hodgkin Lymphoma. HL will be diagnosed in line with current BSH guidelines: First line management of classical Hodgkin Lymphoma, published 09/04/2014 and Investigation and management of Nodular Lymphocyte Predominant Hodgkin Lymphoma, published 05/11/2015. Patient’s referral with a suspected lymphoproliferative disorder are to be reviewed by a haematologist who decides about timescale of the first appointment - to be offered in accordance with relevant NHS patient pathway. Booking instructions for the first clinic appointment should include a panel of routine blood tests: full blood count, group and screen, DAT, biochemical profile, lactate dehydrogenase, B2 microglobulin, serum protein electrophoresis, coagulation screen, virology screen (hepatitis B - S antigen, hepatitis C antibody, HIV, varicella zoster antibody). Additional blood tests are to be requested at clinician discretion - relevant to the referral details. Diagnostic material (lymph node or soft tissue biopsies, bone marrow) is to be sent directly to HMDS, Leeds. Lymph node excision biopsy is strongly recommended, but if risk of a surgical procedure outweighs the potential benefits of an excision biopsy – multiple needle core biopsies is an alternative. Based on HMDS report - diagnosis is to be defined as Classical Hodgkin Lymphoma (cHL) with specified subtype or Nodular Lymphocyte Predominant Hodgkin Lymphoma (NLPHL). Full body PET/CT scan is an obligatory staging scan modality. Early stage cHL (Ann Arbor IA and IIA) patients may be classified as favorable or unfavorable based on EORTC criteria. Risk stratification and prognostication for advanced stage cHL (Ann Arbor III and IV) is to be assessed in accordance with International Prognostic Score (Hasenclever Index) There is currently no evidence to support a risk‐stratified approach to therapy in early stage Nodular Lymphocyte Predominant Hodgkin Lymphoma (NLPHL) and there is no internationally validated scoring system for advanced stage NLPHL. Staging bone marrow biopsy is not routinely required but may be recommended if clinically appropriate (i.e. ambiguous bone marrow involvement on PET/CT scan or unexplained cytopenias). All patients with reproductive potential requiring anti-cancer treatment should be fully informed prior to treatment about the possible gonadal toxic consequences of any given treatment approach. Fertility-preserving treatments should be offered to eligible patients in accordance with guidelines: “The effects of cancer treatment on reproductive functions:” RCP 2007 All newly diagnosed cases of HL are to be discussed at the network MDT. HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Management of Hodgkin Lymphoma cHL will be managed within the Hull and North Lincolnshire MDT in line with current BSH guidelines: First line management of classical Hodgkin Lymphoma, published 09/04/2014, Investigation and management of Nodular Lymphocyte Predominant Hodgkin Lymphoma, published 05/11/2015 and Guideline on the management of primary resistant and relapsed classical Hodgkin lymphoma, published 01/10/2013. The local management of HL will also take account of the following NICE pathways and guidance: Nivolumab for treating relapsed or refractory classical Hodgkin lymphoma NICE guidance TA462, published 26/07/2017 Brentuximab vedotin for treating CD30-positive Hodgkin lymphoma NICE guidance TA446, published 28/06/2017 First line-treatment of Hodgkin Lymphoma. Classical Hodgkin Lymphoma Treatment decisions are to be based on a risk stratified approach for cHL. Treatment decisions concerning elderly patients are to be based on assessment of fitness/frailty and co-morbidities rather than age alone. Early stage cHL patients IA and IIA without bulk are to be offered ABVD based treatment (doxorubicin, bleomycin, vinblastine, dacarbazine) with interim PET/CT scan assessment. Recent studies indicate that risk stratification of early cHL is less predictive than PET response and therefore we will not routinely use it to guide treatment. In light of the results of the RAPID and other similar trials each patient needs to be assessed and counselled in relation to the risks and benefits of radiotherapy in the treatment of their disease. Standard therapy would be with 2 cycles of ABVD and radiotherapy (RT) with PET assessment pre-treatment and post therapy. If the patient and MDT feel it would be appropriate to try to avoid radiotherapy and accept the small increase in relapse then treatment should proceed as per the RAPID protocol with PET pre-treatment and after 3 ABVD. If the PET is negative (Deauville 2) then no further treatment is required. If the PET is positive then a further cycle of ABVD is given followed by radiotherapy and an end of treatment PET. HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Patients with advanced stage cHL fit for intensive chemotherapy should initiate treatment with 2 cycles of ABVD chemotherapy and then undergo risk stratification based on interim PET/CT scan result:- Patients with complete response (Deauville 2) on interim PET/CT are to continue ABVD chemotherapy up to planned 6 cycles with no need for end of treatment PET/CT. Also patients who achieved complete response on interim PET/CT may discontinue bleomycin from subsequent chemotherapy cycles due to balance of benefit vs risk of lung toxicity. Patients with equivocal (Deauville 3) response on interim PET/CT are to continue ABVD chemotherapy with end of treatment PET/CT reassessment. Patients with unsatisfactory response (Deauville 4 and 5) are to be considered for treatment with escalated BEACOPP with PET/CT scan reassessment after 2 cycles of escalated treatment. An end of treatment PET/CT will be required if no complete response achieved post 2 cycles of escalated BEACOPP chemotherapy. Patients who remain PET‐positive on completion of therapy may require re-biopsy with consideration of radiotherapy to residual disease and PET/CT scan re-assessment 3 months after radiotherapy. Transplant eligible patients with primary refractory/progressive disease are to be offered salvage chemotherapy followed by high dose chemotherapy with autologous stem cell transplant. Patients who are not fit for intensive chemotherapy are to be offered palliative treatment with steroids and/or weekly Vinblastine monotherapy and radiotherapy to symptomatic bulk of disease. Nodular Lymphocyte Predominant Hodgkin Lymphoma Patients with NLPHL stage IA who underwent diagnostic excision biopsy with no residual disease on PET/CT scan are to be considered for active monitoring. Patients with residual but localized NLPHL (stage IA and IIA with ≤2 sites of disease) where the involved nodes are contiguous and can be safely encompassed within a radiation field are to be offered radiotherapy treatment. Patients with stage IIA NLPHL who are not suitable for involved field radiotherapy are to be treated as advanced stage disease. Patients with advanced stage NLPHL are to be offered 8 cycles of R‐CVP chemotherapy. Patients with advanced stage with a high index of suspicion of transformed disease are to be considered for 6 cycles of R‐CHOP chemotherapy. Response to treatment is to be assessed with repeated PET-CT scan after 4 cycles of chemotherapy with aim to continue chemotherapy up to the planned number of cycles - unless there is evidence of primary refractory/progressive disease on interim PET scan. Management primary refractory/progressive NLPHL requires MDT discussion with highly individualized treatment approach. HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Treatment of Relapsed Hodgkin Lymphoma Relapsed disease is to be reassessed with PET/CT scan with repeat biopsy strongly recommended. All relapsed cases of HL are to be discussed at the network MDT. Treatment decisions must be considered in the context of the patient’s goals (palliative vs curative), performance status, transplant eligibility, previous treatments and response. Transplant eligible patients are to be offered salvage chemotherapy (DHAP, IVE or ICE). After two cycles of salvage chemotherapy response to treatment is to be assessed with PET/CT scan. Responding patients are to be offered high dose chemotherapy with autologous stem cell transplantation. Patients with unsatisfactory response to initial salvage chemotherapy are to be considered for alternative salvage regimen, Brentuximab, palliative treatment or clinical trial. If autologous stem cell transplantation has not resulted in long lasting remission or if patient failed stem cell harvesting but being still fit enough for transplantation and suitable donor can be found - allogenic stem cell transplantation should be discussed with patient and transplant centre in Leeds or Sheffield. Brentuximab vedotin and Nivolumab are the treatment options for primary refractory or relapsed cHL to be used in accordance with the products registration characteristic and NICE approval. Clinical Trials All patients should be considered for, and offered entry into clinical trials where available. Palliative care services All Hodgkin Lymphoma patients will have a named key-worker who will undertake a holistic needs assessment and provide support and advice based upon this. Referral to hospital or community palliative care services should be considered for patients unfit for chemotherapy and those with poor prognosis relapsed/refractory disease. HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Patient Pathway, First line treatment, early stage Clasical Hodgkin Lymphoma Early stage cHL Fit for chemotherapy Not fit for chemotherapy Radiotherapy/palliative care Appropriate to include radiotherapy in the treatment plan (by patient and MDT) Appropriate and accepted to try avoid radiotherapy in the treatment plan (by patient and MDT) ABVD X 2 + Radiotherapy. End of treatment PET/CT ABVD x 3 End of treatment PET/CT PET negative Active monitoring PET positive Treat as refractory PET negative Active monitoring PET positive Consider radiotherapy or escalation of chemo HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Patient Pathway, First line treatment, advanced stage Clasical Hodgkin Lymphoma Advanced stage cHL Fit for chemotherapy Not fit for chemotherapy Radiotherapy/palliative care ABVD x 2 Interim PET/CT Complete metabolic response Continue with 4 x AVD, (no need for end of treatment PET/CT) Equivocal response Continue with 4 x ABVD End of treatment PET/CT Unsatisfactory response Escalate treatment to BEACOPP PET/CT reassessment PET negative Active monitoring PET positive Consider radiotherapy or treat as refractory PET negative Active monitoring PET positive Consider radiotherapy or treat as refractory HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 Patient Pathway, Refractory/Relapsed Hodgkin Lymphoma REFRACTORY/RELPSED HODGKIN LYMPHOMA TRANSPLANT ELIGIBLE SALVAGE CHEMOTHERAPY TRANSPLANT INELIGIBLE CONSIDERATION OF BRENTUXIMAB/ NIVOLUMAB/RADIOTHERAPY/ PALLIATIVE TREATMENT OR CLINICAL TRIAL NO RESPONSE SATISFACTORY RESPONSE AUTO PBSCT ALTERNATIVE SALVAGE REGIMEN/ BRENTUXIMAB/NIVOLUMA B/C.TRIAL FOLLOW UP/SURVIVORSHIP HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 3 PROCESS FOR MONITORING COMPLIANCE Compliance will be audited within the MDT audit programme. 4 REFERENCES Hull and North Lincolnshire Haematology MDT operational policy September 2017. Haematological cancers: improving outcomes. NICE guideline [NG47] Published date: May 2016 André MPE, Reman O, Federico M, Girinski T, Brice P, Brusamolino E et al. Interim Analysis of the Randomized EORTC/LYSA/FIL Intergroup H10 Trial On Early PET-Scan Driven Treatment Adaptation in Stage I/II Hodgkin Lymphoma. American Society of Hematology 2012 Annual Meeting. Abstract 549. Hasenclever D and Diehl V. A prognostic score for advanced Hodgkin’s disease. International Prognostic Factors Project on Advanced Hodgkin’s Disease. New Engl J Med 1998; 339: 1506-1514. Collins, G. P., Parker, A. N., Pocock, C. , Kayani, I. , Sureda, A. , Illidge, T. , Ardeshna, K. , Linch, D. C., Peggs, K. S. and , (2014), Guideline on the management of primary resistant and relapsed classical Hodgkin lymphoma. Br J Haematol, 164: 39-52. doi:10.1111/bjh.12582 Follows, G. A., Ardeshna, K. M., Barrington, S. F., Culligan, D. J., Hoskin, P. J., Linch, D. , Sadullah, S. , Williams, M. V., Wimperis, J. Z. and , (2014), Guidelines for the first line management of classical Hodgkin lymphoma. Br J Haematol, 166: 34-49. doi:10.1111/bjh.12878 McKay, P. , Fielding, P. , Gallop‐Evans, E. , Hall, G. W., Lambert, J. , Leach, M. , Marafioti, T. , McNamara, C. and , (2016), Guidelines for the investigation and management of nodular lymphocyte predominant Hodgkin lymphoma. Br J Haematol, 172: 32-43. doi:10.1111/bjh.13842 Radford J., Illidge T, Counsell N Results of a Trial of PET-Directed Therapy for Early-Stage Hodgkin’s Lymphoma N Engl J Med 2015; 372:1598-1607 DOI: 10.1056/NEJMoa1408648, 2015 Royal College of Physicians, The Royal College of Radiologists, Royal College of Obstetricians and Gynaecologists. The effects of cancer treatment on reproductive functions: Guidance on management. Report of a Working Party. London: RCP, 2007 Nivolumab for treating relapsed or refractory classical Hodgkin lymphoma HL Guideline Hull and North Lincs Haematology MDT ver 1.3 18/07/2018 NICE guidance TA462, published 26/07/2017 Brentuximab vedotin for treating CD30-positive Hodgkin lymphoma NICE guidance TA446, published 28/06/2017 Document Control Reference No: First published: 18/07/2018 Version: 1.3 Current Version Published: 18/07/2018 Lead Director: Haematology MDT lead Review Date: 17/07/2021 Document Managed by Name: Haematology MDT lead Ratification Committee: Hull and North Lincs Haematology MDT Document Managed by Title: Date EIA Completed: Version Control Date Version Author Revision description 24/04/2018 1.3 Bailey/Patmore/Frygier New draft. |
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