Mathematics II. Tutorial 5 Basic mathematical modelling. Groups: B03 & B08. Ngo Quoc Anh Department of Mathematics National University of Singapore

Similar documents
General Equilibrium. What happens to cause a reaction to come to equilibrium?

MAC Calculus II Summer All you need to know on partial fractions and more

Relative Maxima and Minima sections 4.3

MTH 142 Solution Practice for Exam 2

arxiv:gr-qc/ v2 6 Feb 2004

Chapter 15 Equilibrium. Reversible Reactions & Equilibrium. Reversible Reactions & Equilibrium. Reversible Reactions & Equilibrium 5/27/2014

Chapter 15 Equilibrium. Reversible Reactions & Equilibrium. Reversible Reactions & Equilibrium. Reversible Reactions & Equilibrium 2/3/2014

Acoustic Waves in a Duct

Name Solutions to Test 1 September 23, 2016

3 Tidal systems modelling: ASMITA model

Some facts you should know that would be convenient when evaluating a limit:

Theory. Coupled Rooms

Millennium Relativity Acceleration Composition. The Relativistic Relationship between Acceleration and Uniform Motion

Wavetech, LLC. Ultrafast Pulses and GVD. John O Hara Created: Dec. 6, 2013

finalsol.nb In my frame, I am at rest. So the time it takes for the missile to reach me is just 8µ106 km

Chapter 8 Thermodynamic Relations

Mass Transfer 2. Diffusion in Dilute Solutions

Heat exchangers: Heat exchanger types:

The Laws of Acceleration

Lecture 11 Buckling of Plates and Sections

Modes are solutions, of Maxwell s equation applied to a specific device.

UTC. Engineering 329. Proportional Controller Design. Speed System. John Beverly. Green Team. John Beverly Keith Skiles John Barker.

( ) ( ) Volumetric Properties of Pure Fluids, part 4. The generic cubic equation of state:

REVIEW QUESTIONS Chapter 15

2 How far? Equilibrium Answers

Controller Design Based on Transient Response Criteria. Chapter 12 1

Geometry of Transformations of Random Variables

The Hanging Chain. John McCuan. January 19, 2006

Systems and Matrices VOCABULARY

23.1 Tuning controllers, in the large view Quoting from Section 16.7:

Methods of evaluating tests

Chapter 15 Chemical Equilibrium

Maximum Entropy and Exponential Families

Where as discussed previously we interpret solutions to this partial differential equation in the weak sense: b

2. The Energy Principle in Open Channel Flows

Remark 4.1 Unlike Lyapunov theorems, LaSalle s theorem does not require the function V ( x ) to be positive definite.

5.1 Composite Functions

PY Modern Physics

Chapter-3 PERFORMANCE MEASURES OF A MULTI-EVAPORATOR TYPE COMPRESSOR WITH STANDBY EXPANSION VALVE

Green s function for the wave equation

A NORMALIZED EQUATION OF AXIALLY LOADED PILES IN ELASTO-PLASTIC SOIL

Chapter 15: Chemical Equilibrium

Lecture 3 - Lorentz Transformations

The gravitational phenomena without the curved spacetime

Fiber Optic Cable Transmission Losses with Perturbation Effects

Chapter 2: Solution of First order ODE

7 Max-Flow Problems. Business Computing and Operations Research 608

Strauss PDEs 2e: Section Exercise 3 Page 1 of 13. u tt c 2 u xx = cos x. ( 2 t c 2 2 x)u = cos x. v = ( t c x )u

Intermediate Math Circles November 18, 2009 Solving Linear Diophantine Equations

Chapter 8 Hypothesis Testing

Optimization of Statistical Decisions for Age Replacement Problems via a New Pivotal Quantity Averaging Approach

Study on the leak test technology of spacecraft using ultrasonic

Special and General Relativity

arxiv: v1 [physics.soc-ph] 26 Nov 2014

ON A PROCESS DERIVED FROM A FILTERED POISSON PROCESS

Extra Credit Solutions Math 181, Fall 2018 Instructor: Dr. Doreen De Leon

Discrete Bessel functions and partial difference equations

TENSOR FORM OF SPECIAL RELATIVITY

Simplify each expression. 1. 6t + 13t 19t 2. 5g + 34g 39g 3. 7k - 15k 8k 4. 2b b 11b n 2-7n 2 3n x 2 - x 2 7x 2

Fundamental Theorem of Calculus

4.4 Solving Systems of Equations by Matrices

Quantum Mechanics: Wheeler: Physics 6210

Relativistic Dynamics

A two storage inventory model with variable demand and time dependent deterioration rate and with partial backlogging

PAST YEAR EXAM PAPER SOLUTION SEMESTER II, ACADEMIC YEAR PAP261 Introduction to Lasers (Solution by: Phann Sophearin)

MODELLING THE POSTPEAK STRESS DISPLACEMENT RELATIONSHIP OF CONCRETE IN UNIAXIAL COMPRESSION

UPPER-TRUNCATED POWER LAW DISTRIBUTIONS

Determination of the reaction order

MATHEMATICAL AND NUMERICAL BASIS OF BINARY ALLOY SOLIDIFICATION MODELS WITH SUBSTITUTE THERMAL CAPACITY. PART II

CSC321: 2011 Introduction to Neural Networks and Machine Learning. Lecture 10: The Bayesian way to fit models. Geoffrey Hinton

DIGITAL DISTANCE RELAYING SCHEME FOR PARALLEL TRANSMISSION LINES DURING INTER-CIRCUIT FAULTS

Answers to Coursebook questions Chapter J2

INTERNATIONAL JOURNAL OF CIVIL AND STRUCTURAL ENGINEERING Volume 2, No 4, 2012

Math 220A - Fall 2002 Homework 8 Solutions

F = c where ^ı is a unit vector along the ray. The normal component is. Iν cos 2 θ. d dadt. dp normal (θ,φ) = dpcos θ = df ν

PHY 108: Optical Physics. Solution to Midterm Test

Chapter 26 Lecture Notes

Average Rate Speed Scaling

A Queueing Model for Call Blending in Call Centers

MOLECULAR ORBITAL THEORY- PART I

Design and evaluation of a connection management mechanism for an ATM-based connectionless service

Most results in this section are stated without proof.

Complexity of Regularization RBF Networks

Advanced Computational Fluid Dynamics AA215A Lecture 4

Critical Reflections on the Hafele and Keating Experiment

Non-Markovian study of the relativistic magnetic-dipole spontaneous emission process of hydrogen-like atoms

Differential Equations 8/24/2010

Does P=NP? Karlen G. Gharibyan. SPIRIT OF SOFT LLC, 4-th lane 5 Vratsakan 45, 0051, Yerevan, Armenia

6x and find the. y=2x+5 and y=4 -x 2. gradient of the curve at this point. tangent to the curve y = 4 - x 2? Figure 5.11

22.01 Fall 2015, Problem Set 6 (Normal Version Solutions)

Answer: Easiest way to determine equilibrium concentrations is to set up a table as follows: 2 SO 2 + O 2 2 SO 3 initial conc change

2.2 BUDGET-CONSTRAINED CHOICE WITH TWO COMMODITIES

Q2. [40 points] Bishop-Hill Model: Calculation of Taylor Factors for Multiple Slip

Measuring & Inducing Neural Activity Using Extracellular Fields I: Inverse systems approach

10.2 The Occurrence of Critical Flow; Controls

max min z i i=1 x j k s.t. j=1 x j j:i T j

Control Theory association of mathematics and engineering

Chapter 35. Special Theory of Relativity (1905)

What are the locations of excess energy in open channels?

THE EFFECT OF CONSOLIDATION RATIOS ON DYNAMIC SHEAR MODULUS OF SOIL

Transcription:

Mathematis II Tutorial 5 Basi mathematial modelling Groups: B03 & B08 February 29, 2012 Mathematis II Ngo Quo Anh Ngo Quo Anh Department of Mathematis National University of Singapore 1/13

: The ost of making ans We setup notations first: J is the ost of aluminium per square m, K is the ost per m of welding. Regarding to J, we need to find total area of the an; and for K we need to find the length of welding. From the question, only the top of the an is welded on. Thus, the length of the welding is nothing but the perimeter of the top, whih equals 2πr. For the area of the an, it is simply equal to twie of the area of the top plus the area of the side of the an. Mathematially, we find that 2r = 5 m h = 12.5 m Mathematis II Ngo Quo Anh Total area of the an = 2πr 2 + 2V r. In onlusion, we find the ost as The ost Cr) = J 2πr 2 + 2V r ) + K2πr). 2/13

: The ost of making ans In order to minimize Cr), we have to find its ritial points. By differentiating, we first have C r) = J 4πr 2V ) r 2 + 2πK ) = J 4πr 2πr2 h r 2 + 2πK = J 4πr 2πh) + 2πK. By studying C r) = 0, we get that Jh 2r) = K. Equivalently, we have Mathematis II Ngo Quo Anh h r = 2 + K/J. r This tells us that if r is fixed, both h and K J have the same monotoniity. By using the following fats h = 12.5, r = 2.5, we get that K J = 7.5. 3/13

: The Malthusian growth model Suppose P is the population, B is the birth-rate, and D is the death-rate. The Malthusian growth model The model an be stated as a simple ODE Equivalently, we have d P t) = B D)P t). dt P t) = P 0 e B D)t, Mathematis II Ngo Quo Anh where P 0) = P 0 is the initial population. Using this model with P 0) = 10, 000 and the fat that P 1 + 1 2 ) = 11, 000 we find that 11, 000 = 10, 000e 3 2 B D). 4/13

: The Malthusian growth model Solving the equation above, we get that B D = 2 3 ln 11 10. To measure the number of bateria after 10 hours, we simply alulate P 10) whih is P 10) = 10, 000e 2 3 ln 11 10 )10 14, 600. To find the duration so that the number of bateria reahes 20, 000, we simply solve for t from the following Mathematis II Ngo Quo Anh 20, 000 = 10, 000e 2 3 ln 11 10 )t. Obviously, t 18.18 hours). NB: The Malthusian growth model is the diret anestor of the logisti funtion. 5/13

: The logisti growth model Sine a fixed number of olonists are sent out eah year, the rate of emigration is nothing but onstant. For that reason, let us assume that is the number of emigrants per year. Sine the Malthusian model would hold if there were no emigration, we have the following ODE B D + d P t) = B D)P t). dt By solving, using P t) = P h t) + A, we get that P t) = P 0) B D ) e B D)t, t 0, where P 0) denotes the initial population. Depending on the sign of P 0) B D, we basially have three ases If P 0) < B D, then P t) 0 exponentially. If P 0) = B D, then P t) = B D all the time. If P 0) >, then P t) + exponentially. B D Mathematis II Ngo Quo Anh 6/13

: The logisti growth model If the rate of emigration is proportional to time, say t. The orresponding ODE is now d P t) = B D)P t) t. dt Using the method of undetermined oeffiients, the general solution P t) = P h t) + At + B) is ) P t) = B D t+ B D) 2 + P 0) B D) 2 e B D)t, where t 0 and P 0) is the initial population. Again, we basially have three ases If P 0) <, then P t) and then P t) 0 B D) 2 sine the exponential funtions grow faster than linear funtions. If P 0) =, then P t) = B D) 2 B D t + linearly. If P 0) > B D) 2 + B D) 2, then P t) + exponentially. Mathematis II Ngo Quo Anh 7/13

: The logisti growth model Sine the birth-rate B and death-rate D are onstants, we an adopt the Malthusian growth model to the problem, that is, we have d P t) = B D)P t). dt The general solution is Sine P 20) P 0) P t) = P 0)e B D)t, t 0. = 2, we get that 2 = e B D)20 whih implies that B D = ln 2 20. After the departure of the women, say at t = t, learly B is zero. Therefore, Using the fat that P t) = P t)e Dt t), t t. P t + 10) = 1 2 P t), Mathematis II Ngo Quo Anh 8/13

: The logisti growth model we find that e 10D = 1 ln 2 2. Hene, D = 10. Having D, we an solve for B from the equation B D = ln 2 20. A simple alulation shows that B = 0.10397, that is, 10.397% per year. From the solution above, one an find that the following assumptions have been made: Men and old women have same death rate as young women, whih is not true in reality beause men usually smoke, get into fights, et while on the other hand old women are indestrutible. The death rate of the remaining population is not hanged by the departure of the women. That is D before t and D after t are the same. This helps us to go bak to solve for B one we have D. This is a very questionable assumption sine morale will be affeted, et. Question: Study the presentation of P t) using t = 0, t. Mathematis II Ngo Quo Anh 9/13

: The logisti growth model Let N be the number of neutrons. It is known that D is a positive onstant and B = sn for some onstant s > 0. The model an be stated via the following ODE To solve this, we first write dt = d Nt) = sn D)Nt). dt dn sn D)N = s D By integrating, we get that 1 sn D 1 sn ) dn. t + C = 1 D ln sn D 1 D ln N = 1 D ln sn D N In other words, sn D N = edt+c), or Nt) = D s e Dt+C).. Mathematis II Ngo Quo Anh 10/13

: The logisti growth model Apparently, we have to determine the orret sign for Nt). This proess depends on the sign of sn0) D. Suppose sn0) D > 0. It follows from N0) = D s e DC > D s that is the orret sign. Thus showing that Nt) = D s e Dt+C) Mathematis II Ngo Quo Anh Suppose sn0) D = 0. By the no-rossing priniple, we find that N = D s is a solution. Suppose sn0) D < 0. We then find that the solution is Nt) = D s + e Dt+C). 11/13

: The logisti growth model First, remember the phase portrait of the logisti equation dp dt = BP sp 2, i.e., the lassifiation of solutions. Initially, you have 200 bugs in a bottle but 2 days later, you have 360 bugs. As suh, the solution for the logisti growth model was inreasing. Consequently, the form of suh a solution is B P t) = ), t 0. s + B P 0) s e Bt From the question, there hold B = 3 2, P 0) = 200. We still need to find s. Thanks to P 2) = 360, we find that Therefore, 3/2 360 = ). s + 3/2 200 s e 3/2)2 s = 1 9e 3 5 1200 e 3 1 0.3992 Mathematis II Ngo Quo Anh 12/13

: The logisti growth model Thus, after 3 days, the number of bugs is equal to P 3) whih is P 3) = 3/2 1 9e 3 5 1200 e 3 1 + 3/2 200 1 1200 ) 372. 9e 3 5 e e 3 1 3/2)3 Conerning the number of bugs that eventually have, say P, we atually study the following limit lim P t) = t + lim t + B ) s + B P 0) s e Bt Mathematis II Ngo Quo Anh whih is nothing but B s. Therefore, P = B s = 376. In our ase, the birth rate B = 3 2 seems to be high. 13/13