KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

Size: px
Start display at page:

Download "KEY. Math 334 Midterm I Fall 2008 sections 001 and 003 Instructor: Scott Glasgow"

Transcription

1 1 KEY Mah 4 Miderm I Fall 8 secions 1 and Insrucor: Sco Glasgow Please do NOT wrie on his eam. No credi will be given for such work. Raher wrie in a blue book, or on our own paper, preferabl engineering paper. Wrie our name, course, and secion number on he blue book, or our own pile of papers. Again, do no wrie his or an oher pe of informaion on his eam.

2 1. Solve he following iniial value problem: 1 poins Soluion d 1, ( d = =. (1.1 Also, wha is he value of his soluion a = 4? (I.e. wha is ( 4? The equaion separaes o d ( 1, 1+ = + d (1. which, wih he iniial daa in (1.1, ields he inegral saemen which, afer some work, gives d = ( 1+ d 1+, (1. ( log 1+ = log 1+ log 1+ = log 1+ = d = + 1+ ( 1 d = + = = = ( + 8 = ( ( + 4, (1.4 i.e. 1 = ep ( ( (1.5 1 ( 4 = ep = 1 1 = = (. Thus

3 . Prove ha he following differenial equaion is eac and hen find an epression for is general soluion. 14 poins Soluion d d =. (1.6 The equaion (1.6 is, b definiion, eac if he lef-hand side is he differenial of a (coninuousl differeniable funcion (of wo variables and, in some simplconneced region of he - plane, ec., ec., i.e. if here is a funcion ψ (, such ha dψ (, = + + d d. (1.7 Bu we have, b definiion, dψ (, = ψ (, d + ψ (, d, (1.8 so ha he equaion (1.7 is he (poeniall over-deermined ssem of PDE s ψ (, = + +, and ψ (, = ( This over-deermined pair of equaions is consisen (or inegrable iff ( ψ ( ψ iff ( 4 6 ( =, i.e. + + = (1.1 (1.1 holds rue, so ha he equaion (1.6 is indeed eac, because eiher side of (1.1 5 is As for developing he funcion ψ (,, and hen (an epression for he general soluion of(1.6, one noes ha he equaions (1.9 demand, respecivel, ha ψ ψ (, = + + d = f (, and (, d g(, = = (1.11 for some iniiall raher arbirar funcions f ( and g(. The wo saemens (1.11 are compaible iff f ( = g( f ( = g(, which

4 4 implies boh sides of he las equaion are independen of boh and. As far as finding he general soluion of (1.6 is concerned, wihou loss of generali we can choose 5 f ( = g( = so ha (1.11 becomes ( in eiher case ψ (, = = (1.1 (1.1 is NOT he general soluion o he (eac differenial equaion (1.6. I is no even a specific soluion. Raher (1.1 defines a poenial (funcion for he soluion. Using i one noes ha (1.6 can be wrien as he general soluion o which is clearl ( dψ (, = d 1+ + =, (1.1 ( 1 C + + =. (1.14. B using one of he esimaes from Picard s proof of he Fundamenal Theorem of Firs Order ODE s, show ha here eiss a soluion = φ o he IVP d ( 1, ( d = + =, (1.15 a leas hroughou he inerval [ h, h] [ 1,1] =. 1 (1.16 Hin: If ou do no remember he esimae, do he following o jog our memor. Insead of (1.15, and following Picard, wrie (for some h > o be deermined ha and, so, deduce ha 1, ( φ φ = s + s ds h, ( Noe ha he soluion of (1.15 acuall persiss hroughou he 1/ (, 1/ ( 1.46,1.46 inerval ( π ( π, since he soluion has he formula an ( / π π. since he domain of (he relevan insance of he angen funcion is,, and

5 5 h h φ s 1 + φ ( s ds = s 1 + φ ( s ds s 1 + φ ( s ds.(1.18 Now demand ha h is he bigges number ha is sill small enough so ha, for some Y >, imposing ( s Y s h, h on he righ hand side of φ (for each [ ] (1.18 cerainl ensures ha φ Y (for each [ h, h] on he lef of (1.18. B doing his ou will ge an h ha depends ony, i.e. ou will ge an h( Y. Now find 1 poins Soluion Y > h : = ma h Y. (1.19 This should be he number indicaed in(1.16, i.e. he resul of (1.19 should be 1. Demanding φ( s Y (for each s [ h, h] on he righ hand side of (1.18 gives here ha h h h φ s 1 + ( s ds s 1+ Y ds = 1+ Y ( φ. (1. So now we cerainl ge φ Y (for each [ h, h] provided we choose h small h enough so ha ( 1 Y + Y, he larges such h accomplishing his being h Y Y 1+ Y h = So hen we ge he h indicaed in (1.16 b noing ha 1/. (1.1 1/ 1/ 1/ Y 1 h : = ma h( Y = ma 1. Y > Y > = = = 1+ Y 1+ 1 (1. We could ge he maimum indicaed in (1. b using he relevan ools from calculus.

6 4. Suppose we have coninuous soluions φ and inegral equaion indicaed in (1.17, i.e. suppose ha boh ψ, [ h, h] [ 1,1] =, o he 6 φ = s 1 + ( s ds, and ( φ = 1 + ( ψ ψ s s ds (1. for each [ 1,1] soluions obes. Then we could noe ha he difference φ ψ of hese wo = = φ ψ s φ s s ψ s ds s φ s ψ s ds s s s s s ds = ( φ + ψ ( φ ψ, (1.4 and, for [ 1,1], we could ge he esimae = + φ ψ s φ s ψ s φ s ψ s ds s φ( s ψ ( s φ( s ψ ( s ds = + { τ [,1] } ma τ φ( τ + ψ ( τ φ( s ψ ( s ds = : K φ( s ψ ( s ds K φ( s ψ ( s ds = (1.5 where evidenl K : ma τ φ( τ ψ ( τ τ = + <, (1.6 [,1] he las inequali in (1.6 holding because we have a coninuous funcion on he τ,1. So now define he new funcion bounded inerval [ ] U : = φ( s ψ ( s ds (1.7

7 7 for each [ 1,1], and noe ha [ ] =, U U,,1, (1.8 and also hen noe ha (1.5 can hen be wrien as for [,1], i.e. we ge Use (1.8 and (1. o show ha U = φ ψ K φ( s ψ ( s ds KU = (1.9 U KU, [,1 ]. (1. U ( =, (1.1 for [,1] and, so, deduce ha φ ψ, [,1] coninuous soluion o he inegral equaion (1.17 for [,1]. (You could also show U ( = for [ 1,] b a relaed bu differen argumen. =, i.e. deduce ha here is a mos one 15 poins Soluion From (1. we have ha, for each [,1], and hen ha for [,1]. Bu hen since U KU, (1. K (1. / K / e U e KU

8 8 d = (, (1.4 d / K / K / K e U e KU e U (1. is d e U d / K (, (1.5 again for [,1]. Togeher wih U = (see (1.8, (1.5 gives, for each [,1] / K / K / K / Ki / Ks e U e U e U e U e U ( s / Ks ( U ( s d e ds ds i.e. (1.5 gives or, equivalenl, =, ds, = = = = e / K U s= s= (1.6, (1.7 U (1.8 for each [,1]. Beween (1.8 and U (from (1.8, we mus have U ( =. Bu hen, using (1.7, we have for [,1] d d d = = U = φ( s ψ ( s ds = φ ψ d d d (1.9, which hen gives φ ψ = for [,1]. 5. Afer a ime =, a soluion of consan concenraion of 1gram solue per lier solven eners a (perfec sirring ank a a consan rae of 6liers per minue. The well-sirred miure eis he ank a a consan rae of 4 liers per minue. Suppose he solue akes no volume in soluion. If he ank conains 1liers of fluid a a ime =, wrie down a (self-conained differenial equaion for he ime evoluion of he grams of solue Q accumulaed in he ank a ime, one ha is valid for as long as he ank is no overflowing. Then, assuming here are

9 9 1 grams of solue in he ank a =, give an epression for he grams Q( of solue accumulaed in he ank a ime (b solving he relevan IVP. 15 poins Soluion B soichiomeric / uni-canceling / chain-rule reasoning, one has dq dq dq dq dq dv dq dv = = = d d d d dv d dv d = C R C R oal in ou in in ou ou in in ou ou Q Q Q = RinCin Rou = = 6 4, V V V (1.4 where he fluid ank volume V = V is specified b dv d = R R = 6 4 =, V ( = V = 1, (1.41 in ou he laer (rivial iniial value problem having he unique soluion V = V + ( R R = 1 + = 1 +. (1.4 Thus he required, self-conained differenial equaion is in ou dq Q = RinCin Rou d V + ( Rin Rou 4 = 6 Q 1 + dq + Q = 6. d 5 + (1.4 We solve he iniial value problem which is ODE (1.4 ogeher wih iniial daa Q ( = 1. (1.44 An inegraing facor for he ODE (1.4 is, according o he sandard heor,

10 1 µ = ep d = ep( log( (1.45 = 5 +. Use of he inegraing facor (1.45 in (1.4 gives ( d 5 + Q dq d 5 + = ( ( 5 + Q = 6( 5 + = d d 5 + d. (1.46 Inegraion of (1.46 using relevan limis (and dumm variables gives 5 + Q 5 = 5 + Q 5 1 = 5 + Q 5 + Q( = 5 + s Q( s = or, equivalenl, (( 5 + = ( 5 + d s Q s d s ( s = 5 + = = 5 + 5, (1.47 Q = 5 + = 1 +. ( Show ha he following differenial equaion, equaion(1.49, is no eac, bu can be rendered eac b muliplicaion b an inegraing facor ha is onl a funcion of or onl a funcion of. Find an epression of he general soluion of he differenial equaion. 15 poins d d =. (1.49 Soluion (1.49 is no eac since ψ 4 ( ψ : ( ( : ( ψ = = + + = + +. ( = = = ψ

11 11 B heorem we know ha, wih an inegraing facor µ, (1.49 can be made eac. We noe from (1.5 ha 4 ( + + ( = ( = 4 6 ( = + + which divides he firs erm in (1.49, (which is ( + +, he remaining facor ( / being onl a funcion of. Thus we suspec he eisence of an inegraing facor onl depending on. A an rae, wih he use of such a facor, ODE (1.49 becomes µ ( d µ ( d =, (1.5 and eacness demands ha 4 (( µ (( µ 4 ( µ ( ( µ ( = = µ ( µ ( µ ( = µ ( = µ (. ( ( µ ( µ ( = Thus, as suspeced, here is an inegraing facor depending onl on. A soluion of he las differenial equaion in (1.5 is given b,(1.51 (1.5 µ ( =, (1.54 in which case (1.5 becomes = d d = + + d d, (1.55 which is he eac equaion considered in problem. Thus he soluion is he same as in problem, namel

12 ( C = + + =. ( Find a linear, firs order, ordinar differenial equaion wih he proper ha ever soluion of i approaches he funcion f = 1+ arbiraril closel as +. Noe ha he (oo simple equaion = f = 1+ = (1.57 does no work since he general soluion of (1.57 is, (1.58 = d = C + giving lim f = lim C + 1+ = C 1, ( which is no zero for ever possible choice of C. 1 poins Soluion Inroduce a general soluion of he form 1 Ce a = + + (1.6 wih a > o ge a a lim f = lim 1+ + Ce 1+ = lim Ce = ( for ever choice of C, as demanded b he problem. Thus, o ge a firs order ODE wih he required proper we differeniae (1.6 wih respec o and eliminae C beween (1.6 and his new resul. Differeniaing (1.6 gives = ace a, (1.6 and eliminaion of C beween (1.6 and (1.6 gives he required firs order ODE, namel

13 1. ( a = ace = a 1+ = a + a + + a ( Solve he following iniial value problem. Sae he properies of he soluion as for all possible choices of he iniial value. = + 1 +, ( =. ( poins Soluion The ODE in (1.64 can be wrien as + 1 = 1+. (1.65 (1.65 suggess he inegraing facor µ = ep 1d = e, (1.66 which renders he ODE (1.65 as d e ( = d e + e = ( 1+ e d = ( 1+ e e d d 1+ d = d + + d = d d = d d = d d = (( 1+ e d d (( 1 e e ( 1 d (( 1 e e ( 1 e ( 1 d (( 1 e e ( 1 e (( 1 e e ( 1 e (1.67 which, wih he iniial daa specified in (1.64, inegraes o

14 14 or, equivalenl, s 1 ( e = e = e e = e s = s s de ( s = d ( 1+ s e (1.68 s = 1+ s e = 1+ e 1+ e = 1+ e 1, = + + e. ( Noe hen he differenial equaion in (1.64 gives a soluion o problem 7, which indicaes he desired properies. 9. Carefull sae he (nonlinear eisence and uniqueness heorem for a single firs order ODE. 15 poins Soluion Consider he iniial value problem = f (,, ( =. (1.7 Suppose f (, and f (, are boh coninuous in an open recangle (, (, conaining he poin (, Then here eiss an h > such ha (1.7 has a unique, coninuousl differeniable soluion = φ persising over he inerval ( h, + h (poeniall much smaller han he inerval (, Carefull sae he linear eisence and uniqueness heorem for a single firs order ODE. Eplain in general erms how i is proven. 15 poins Soluion Consider he iniial value problem = p + q, ( =. (1.71

15 15 Suppose p( and q are boh coninuous in an open inerval (, conaining he poin. Then (1.71 has a unique, coninuousl differeniable soluion = φ persising over he inerval (, The heorem is proven b eplicil inegraing (1.71, using he various heorems of calculus, including ha he inegral of a coninuous funcion eiss, ec.

Math 334 Test 1 KEY Spring 2010 Section: 001. Instructor: Scott Glasgow Dates: May 10 and 11.

Math 334 Test 1 KEY Spring 2010 Section: 001. Instructor: Scott Glasgow Dates: May 10 and 11. 1 Mah 334 Tes 1 KEY Spring 21 Secion: 1 Insrucor: Sco Glasgow Daes: Ma 1 and 11. Do NOT wrie on his problem saemen bookle, excep for our indicaion of following he honor code jus below. No credi will be

More information

KEY. Math 334 Midterm III Winter 2008 section 002 Instructor: Scott Glasgow

KEY. Math 334 Midterm III Winter 2008 section 002 Instructor: Scott Glasgow KEY Mah 334 Miderm III Winer 008 secion 00 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering

More information

Solutions to Assignment 1

Solutions to Assignment 1 MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we

More information

KEY. Math 334 Midterm III Fall 2008 sections 001 and 003 Instructor: Scott Glasgow

KEY. Math 334 Midterm III Fall 2008 sections 001 and 003 Instructor: Scott Glasgow KEY Mah 334 Miderm III Fall 28 secions and 3 Insrucor: Sco Glasgow Please do NOT wrie on his exam. No credi will be given for such work. Raher wrie in a blue book, or on your own paper, preferably engineering

More information

MA 366 Review - Test # 1

MA 366 Review - Test # 1 MA 366 Review - Tes # 1 Fall 5 () Resuls from Calculus: differeniaion formulas, implici differeniaion, Chain Rule; inegraion formulas, inegraion b pars, parial fracions, oher inegraion echniques. (1) Order

More information

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities: Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial

More information

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by

More information

( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+

( ) 2. Review Exercise 2. cos θ 2 3 = = 2 tan. cos. 2 x = = x a. Since π π, = 2. sin = = 2+ = = cotx. 2 sin θ 2+ Review Eercise sin 5 cos sin an cos 5 5 an 5 9 co 0 a sinθ 6 + 4 6 + sin θ 4 6+ + 6 + 4 cos θ sin θ + 4 4 sin θ + an θ cos θ ( ) + + + + Since π π, < θ < anθ should be negaive. anθ ( + ) Pearson Educaion

More information

Math 10B: Mock Mid II. April 13, 2016

Math 10B: Mock Mid II. April 13, 2016 Name: Soluions Mah 10B: Mock Mid II April 13, 016 1. ( poins) Sae, wih jusificaion, wheher he following saemens are rue or false. (a) If a 3 3 marix A saisfies A 3 A = 0, hen i canno be inverible. True.

More information

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,

More information

Exam 1 Solutions. 1 Question 1. February 10, Part (A) 1.2 Part (B) To find equilibrium solutions, set P (t) = C = dp

Exam 1 Solutions. 1 Question 1. February 10, Part (A) 1.2 Part (B) To find equilibrium solutions, set P (t) = C = dp Exam Soluions Februar 0, 05 Quesion. Par (A) To find equilibrium soluions, se P () = C = = 0. This implies: = P ( P ) P = P P P = P P = P ( + P ) = 0 The equilibrium soluion are hus P () = 0 and P () =..

More information

Review - Quiz # 1. 1 g(y) dy = f(x) dx. y x. = u, so that y = xu and dy. dx (Sometimes you may want to use the substitution x y

Review - Quiz # 1. 1 g(y) dy = f(x) dx. y x. = u, so that y = xu and dy. dx (Sometimes you may want to use the substitution x y Review - Quiz # 1 (1) Solving Special Tpes of Firs Order Equaions I. Separable Equaions (SE). d = f() g() Mehod of Soluion : 1 g() d = f() (The soluions ma be given implicil b he above formula. Remember,

More information

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n Module Fick s laws of diffusion Fick s laws of diffusion and hin film soluion Adolf Fick (1855) proposed: d J α d d d J (mole/m s) flu (m /s) diffusion coefficien and (mole/m 3 ) concenraion of ions, aoms

More information

t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t...

t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t... Mah 228- Fri Mar 24 5.6 Marix exponenials and linear sysems: The analogy beween firs order sysems of linear differenial equaions (Chaper 5) and scalar linear differenial equaions (Chaper ) is much sronger

More information

Second Order Linear Differential Equations

Second Order Linear Differential Equations Second Order Linear Differenial Equaions Second order linear equaions wih consan coefficiens; Fundamenal soluions; Wronskian; Exisence and Uniqueness of soluions; he characerisic equaion; soluions of homogeneous

More information

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x . 1 Mah 211 Homework #3 February 2, 2001 2.4.3. y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih

More information

Math 333 Problem Set #2 Solution 14 February 2003

Math 333 Problem Set #2 Solution 14 February 2003 Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial

More information

Chapter 2. First Order Scalar Equations

Chapter 2. First Order Scalar Equations Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.

More information

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,

More information

MA Study Guide #1

MA Study Guide #1 MA 66 Su Guide #1 (1) Special Tpes of Firs Order Equaions I. Firs Order Linear Equaion (FOL): + p() = g() Soluion : = 1 µ() [ ] µ()g() + C, where µ() = e p() II. Separable Equaion (SEP): dx = h(x) g()

More information

23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes

23.5. Half-Range Series. Introduction. Prerequisites. Learning Outcomes Half-Range Series 2.5 Inroducion In his Secion we address he following problem: Can we find a Fourier series expansion of a funcion defined over a finie inerval? Of course we recognise ha such a funcion

More information

4.6 One Dimensional Kinematics and Integration

4.6 One Dimensional Kinematics and Integration 4.6 One Dimensional Kinemaics and Inegraion When he acceleraion a( of an objec is a non-consan funcion of ime, we would like o deermine he ime dependence of he posiion funcion x( and he x -componen of

More information

ME 391 Mechanical Engineering Analysis

ME 391 Mechanical Engineering Analysis Fall 04 ME 39 Mechanical Engineering Analsis Eam # Soluions Direcions: Open noes (including course web posings). No books, compuers, or phones. An calculaor is fair game. Problem Deermine he posiion of

More information

Chapter 6. Systems of First Order Linear Differential Equations

Chapter 6. Systems of First Order Linear Differential Equations Chaper 6 Sysems of Firs Order Linear Differenial Equaions We will only discuss firs order sysems However higher order sysems may be made ino firs order sysems by a rick shown below We will have a sligh

More information

AP Calculus BC Chapter 10 Part 1 AP Exam Problems

AP Calculus BC Chapter 10 Part 1 AP Exam Problems AP Calculus BC Chaper Par AP Eam Problems All problems are NO CALCULATOR unless oherwise indicaed Parameric Curves and Derivaives In he y plane, he graph of he parameric equaions = 5 + and y= for, is a

More information

SMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15.

SMT 2014 Calculus Test Solutions February 15, 2014 = 3 5 = 15. SMT Calculus Tes Soluions February 5,. Le f() = and le g() =. Compue f ()g (). Answer: 5 Soluion: We noe ha f () = and g () = 6. Then f ()g () =. Plugging in = we ge f ()g () = 6 = 3 5 = 5.. There is a

More information

APPM 2360 Homework Solutions, Due June 10

APPM 2360 Homework Solutions, Due June 10 2.2.2: Find general soluions for he equaion APPM 2360 Homework Soluions, Due June 10 Soluion: Finding he inegraing facor, dy + 2y = 3e µ) = e 2) = e 2 Muliplying he differenial equaion by he inegraing

More information

SOLUTIONS TO ECE 3084

SOLUTIONS TO ECE 3084 SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no

More information

Chapter 15: Phenomena. Chapter 15 Chemical Kinetics. Reaction Rates. Reaction Rates R P. Reaction Rates. Rate Laws

Chapter 15: Phenomena. Chapter 15 Chemical Kinetics. Reaction Rates. Reaction Rates R P. Reaction Rates. Rate Laws Chaper 5: Phenomena Phenomena: The reacion (aq) + B(aq) C(aq) was sudied a wo differen emperaures (98 K and 35 K). For each emperaure he reacion was sared by puing differen concenraions of he 3 species

More information

Limits at Infinity. Limit at negative infinity. Limit at positive infinity. Definition of Limits at Infinity Let L be a real number.

Limits at Infinity. Limit at negative infinity. Limit at positive infinity. Definition of Limits at Infinity Let L be a real number. 0_005.qd //0 : PM Page 98 98 CHAPTER Applicaions of Differeniaion f() as Secion.5 f() = + f() as The i of f as approaches or is. Figure. Limis a Infini Deermine (finie) is a infini. Deermine he horizonal

More information

THE WAVE EQUATION. part hand-in for week 9 b. Any dilation v(x, t) = u(λx, λt) of u(x, t) is also a solution (where λ is constant).

THE WAVE EQUATION. part hand-in for week 9 b. Any dilation v(x, t) = u(λx, λt) of u(x, t) is also a solution (where λ is constant). THE WAVE EQUATION 43. (S) Le u(x, ) be a soluion of he wave equaion u u xx = 0. Show ha Q43(a) (c) is a. Any ranslaion v(x, ) = u(x + x 0, + 0 ) of u(x, ) is also a soluion (where x 0, 0 are consans).

More information

Y 0.4Y 0.45Y Y to a proper ARMA specification.

Y 0.4Y 0.45Y Y to a proper ARMA specification. HG Jan 04 ECON 50 Exercises II - 0 Feb 04 (wih answers Exercise. Read secion 8 in lecure noes 3 (LN3 on he common facor problem in ARMA-processes. Consider he following process Y 0.4Y 0.45Y 0.5 ( where

More information

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals

More information

LAPLACE TRANSFORM AND TRANSFER FUNCTION

LAPLACE TRANSFORM AND TRANSFER FUNCTION CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Spring 2018 Dep. of Chemical and Biological Engineering 5-1 Road Map of he Lecure V Laplace Transform and Transfer funcions

More information

The fundamental mass balance equation is ( 1 ) where: I = inputs P = production O = outputs L = losses A = accumulation

The fundamental mass balance equation is ( 1 ) where: I = inputs P = production O = outputs L = losses A = accumulation Hea (iffusion) Equaion erivaion of iffusion Equaion The fundamenal mass balance equaion is I P O L A ( 1 ) where: I inpus P producion O oupus L losses A accumulaion Assume ha no chemical is produced or

More information

Math Week 14 April 16-20: sections first order systems of linear differential equations; 7.4 mass-spring systems.

Math Week 14 April 16-20: sections first order systems of linear differential equations; 7.4 mass-spring systems. Mah 2250-004 Week 4 April 6-20 secions 7.-7.3 firs order sysems of linear differenial equaions; 7.4 mass-spring sysems. Mon Apr 6 7.-7.2 Sysems of differenial equaions (7.), and he vecor Calculus we need

More information

Math 2214 Solution Test 1B Fall 2017

Math 2214 Solution Test 1B Fall 2017 Mah 14 Soluion Tes 1B Fall 017 Problem 1: A ank has a capaci for 500 gallons and conains 0 gallons of waer wih lbs of sal iniiall. A soluion conaining of 8 lbsgal of sal is pumped ino he ank a 10 galsmin.

More information

5.1 - Logarithms and Their Properties

5.1 - Logarithms and Their Properties Chaper 5 Logarihmic Funcions 5.1 - Logarihms and Their Properies Suppose ha a populaion grows according o he formula P 10, where P is he colony size a ime, in hours. When will he populaion be 2500? We

More information

15. Vector Valued Functions

15. Vector Valued Functions 1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example,

More information

(π 3)k. f(t) = 1 π 3 sin(t)

(π 3)k. f(t) = 1 π 3 sin(t) Mah 6 Fall 6 Dr. Lil Yen Tes Show all our work Name: Score: /6 No Calculaor permied in his par. Read he quesions carefull. Show all our work and clearl indicae our final answer. Use proper noaion. Problem

More information

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3. Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies

More information

Problem Set 7-7. dv V ln V = kt + C. 20. Assume that df/dt still equals = F RF. df dr = =

Problem Set 7-7. dv V ln V = kt + C. 20. Assume that df/dt still equals = F RF. df dr = = 20. Assume ha df/d sill equals = F + 0.02RF. df dr df/ d F+ 0. 02RF = = 2 dr/ d R 0. 04RF 0. 01R 10 df 11. 2 R= 70 and F = 1 = = 0. 362K dr 31 21. 0 F (70, 30) (70, 1) R 100 Noe ha he slope a (70, 1) is

More information

13.3 Term structure models

13.3 Term structure models 13.3 Term srucure models 13.3.1 Expecaions hypohesis model - Simples "model" a) shor rae b) expecaions o ge oher prices Resul: y () = 1 h +1 δ = φ( δ)+ε +1 f () = E (y +1) (1) =δ + φ( δ) f (3) = E (y +)

More information

CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIRCUITS CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As

More information

Math 115 Final Exam December 14, 2017

Math 115 Final Exam December 14, 2017 On my honor, as a suden, I have neiher given nor received unauhorized aid on his academic work. Your Iniials Only: Iniials: Do no wrie in his area Mah 5 Final Exam December, 07 Your U-M ID # (no uniqname):

More information

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan Ground Rules PC11 Fundamenals of Physics I Lecures 3 and 4 Moion in One Dimension A/Prof Tay Seng Chuan 1 Swich off your handphone and pager Swich off your lapop compuer and keep i No alking while lecure

More information

Solutions from Chapter 9.1 and 9.2

Solutions from Chapter 9.1 and 9.2 Soluions from Chaper 9 and 92 Secion 9 Problem # This basically boils down o an exercise in he chain rule from calculus We are looking for soluions of he form: u( x) = f( k x c) where k x R 3 and k is

More information

Math 334 Fall 2011 Homework 11 Solutions

Math 334 Fall 2011 Homework 11 Solutions Dec. 2, 2 Mah 334 Fall 2 Homework Soluions Basic Problem. Transform he following iniial value problem ino an iniial value problem for a sysem: u + p()u + q() u g(), u() u, u () v. () Soluion. Le v u. Then

More information

EXERCISES FOR SECTION 1.5

EXERCISES FOR SECTION 1.5 1.5 Exisence and Uniqueness of Soluions 43 20. 1 v c 21. 1 v c 1 2 4 6 8 10 1 2 2 4 6 8 10 Graph of approximae soluion obained using Euler s mehod wih = 0.1. Graph of approximae soluion obained using Euler

More information

2. Nonlinear Conservation Law Equations

2. Nonlinear Conservation Law Equations . Nonlinear Conservaion Law Equaions One of he clear lessons learned over recen years in sudying nonlinear parial differenial equaions is ha i is generally no wise o ry o aack a general class of nonlinear

More information

Second-Order Differential Equations

Second-Order Differential Equations WWW Problems and Soluions 3.1 Chaper 3 Second-Order Differenial Equaions Secion 3.1 Springs: Linear and Nonlinear Models www m Problem 3. (NonlinearSprings). A bod of mass m is aached o a wall b means

More information

PROBLEMS FOR MATH 162 If a problem is starred, all subproblems are due. If only subproblems are starred, only those are due. SLOPES OF TANGENT LINES

PROBLEMS FOR MATH 162 If a problem is starred, all subproblems are due. If only subproblems are starred, only those are due. SLOPES OF TANGENT LINES PROBLEMS FOR MATH 6 If a problem is sarred, all subproblems are due. If onl subproblems are sarred, onl hose are due. 00. Shor answer quesions. SLOPES OF TANGENT LINES (a) A ball is hrown ino he air. Is

More information

Math 2214 Solution Test 1A Spring 2016

Math 2214 Solution Test 1A Spring 2016 Mah 14 Soluion Tes 1A Spring 016 sec Problem 1: Wha is he larges -inerval for which ( 4) = has a guaraneed + unique soluion for iniial value (-1) = 3 according o he Exisence Uniqueness Theorem? Soluion

More information

Inventory Analysis and Management. Multi-Period Stochastic Models: Optimality of (s, S) Policy for K-Convex Objective Functions

Inventory Analysis and Management. Multi-Period Stochastic Models: Optimality of (s, S) Policy for K-Convex Objective Functions Muli-Period Sochasic Models: Opimali of (s, S) Polic for -Convex Objecive Funcions Consider a seing similar o he N-sage newsvendor problem excep ha now here is a fixed re-ordering cos (> 0) for each (re-)order.

More information

Math Final Exam Solutions

Math Final Exam Solutions Mah 246 - Final Exam Soluions Friday, July h, 204 () Find explici soluions and give he inerval of definiion o he following iniial value problems (a) ( + 2 )y + 2y = e, y(0) = 0 Soluion: In normal form,

More information

AP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES. Question 1. 1 : estimate = = 120 liters/hr

AP CALCULUS AB/CALCULUS BC 2016 SCORING GUIDELINES. Question 1. 1 : estimate = = 120 liters/hr AP CALCULUS AB/CALCULUS BC 16 SCORING GUIDELINES Quesion 1 (hours) R ( ) (liers / hour) 1 3 6 8 134 119 95 74 7 Waer is pumped ino a ank a a rae modeled by W( ) = e liers per hour for 8, where is measured

More information

Physics 127b: Statistical Mechanics. Fokker-Planck Equation. Time Evolution

Physics 127b: Statistical Mechanics. Fokker-Planck Equation. Time Evolution Physics 7b: Saisical Mechanics Fokker-Planck Equaion The Langevin equaion approach o he evoluion of he velociy disribuion for he Brownian paricle migh leave you uncomforable. A more formal reamen of his

More information

Econ107 Applied Econometrics Topic 7: Multicollinearity (Studenmund, Chapter 8)

Econ107 Applied Econometrics Topic 7: Multicollinearity (Studenmund, Chapter 8) I. Definiions and Problems A. Perfec Mulicollineariy Econ7 Applied Economerics Topic 7: Mulicollineariy (Sudenmund, Chaper 8) Definiion: Perfec mulicollineariy exiss in a following K-variable regression

More information

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4. PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence

More information

ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004

ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004 ODEs II, Lecure : Homogeneous Linear Sysems - I Mike Raugh March 8, 4 Inroducion. In he firs lecure we discussed a sysem of linear ODEs for modeling he excreion of lead from he human body, saw how o ransform

More information

Section 4.4 Logarithmic Properties

Section 4.4 Logarithmic Properties Secion. Logarihmic Properies 59 Secion. Logarihmic Properies In he previous secion, we derived wo imporan properies of arihms, which allowed us o solve some asic eponenial and arihmic equaions. Properies

More information

Section 4.4 Logarithmic Properties

Section 4.4 Logarithmic Properties Secion. Logarihmic Properies 5 Secion. Logarihmic Properies In he previous secion, we derived wo imporan properies of arihms, which allowed us o solve some asic eponenial and arihmic equaions. Properies

More information

Hamilton Jacobi equations

Hamilton Jacobi equations Hamilon Jacobi equaions Inoducion o PDE The rigorous suff from Evans, mosly. We discuss firs u + H( u = 0, (1 where H(p is convex, and superlinear a infiniy, H(p lim p p = + This by comes by inegraion

More information

Solutions of Sample Problems for Third In-Class Exam Math 246, Spring 2011, Professor David Levermore

Solutions of Sample Problems for Third In-Class Exam Math 246, Spring 2011, Professor David Levermore Soluions of Sample Problems for Third In-Class Exam Mah 6, Spring, Professor David Levermore Compue he Laplace ransform of f e from is definiion Soluion The definiion of he Laplace ransform gives L[f]s

More information

ES.1803 Topic 22 Notes Jeremy Orloff

ES.1803 Topic 22 Notes Jeremy Orloff ES.83 Topic Noes Jeremy Orloff Fourier series inroducion: coninued. Goals. Be able o compue he Fourier coefficiens of even or odd periodic funcion using he simplified formulas.. Be able o wrie and graph

More information

Theory of! Partial Differential Equations!

Theory of! Partial Differential Equations! hp://www.nd.edu/~gryggva/cfd-course/! Ouline! Theory o! Parial Dierenial Equaions! Gréar Tryggvason! Spring 011! Basic Properies o PDE!! Quasi-linear Firs Order Equaions! - Characerisics! - Linear and

More information

Differential Equations

Differential Equations Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding

More information

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010 Simulaion-Solving Dynamic Models ABE 5646 Week 2, Spring 2010 Week Descripion Reading Maerial 2 Compuer Simulaion of Dynamic Models Finie Difference, coninuous saes, discree ime Simple Mehods Euler Trapezoid

More information

Math Wednesday March 3, , 4.3: First order systems of Differential Equations Why you should expect existence and uniqueness for the IVP

Math Wednesday March 3, , 4.3: First order systems of Differential Equations Why you should expect existence and uniqueness for the IVP Mah 2280 Wednesda March 3, 200 4., 4.3: Firs order ssems of Differenial Equaions Wh ou should epec eisence and uniqueness for he IVP Eample: Consider he iniial value problem relaed o page 4 of his eserda

More information

INDEX. Transient analysis 1 Initial Conditions 1

INDEX. Transient analysis 1 Initial Conditions 1 INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera

More information

Announcements: Warm-up Exercise:

Announcements: Warm-up Exercise: Fri Apr 13 7.1 Sysems of differenial equaions - o model muli-componen sysems via comparmenal analysis hp//en.wikipedia.org/wiki/muli-comparmen_model Announcemens Warm-up Exercise Here's a relaively simple

More information

Hamilton- J acobi Equation: Explicit Formulas In this lecture we try to apply the method of characteristics to the Hamilton-Jacobi equation: u t

Hamilton- J acobi Equation: Explicit Formulas In this lecture we try to apply the method of characteristics to the Hamilton-Jacobi equation: u t M ah 5 2 7 Fall 2 0 0 9 L ecure 1 0 O c. 7, 2 0 0 9 Hamilon- J acobi Equaion: Explici Formulas In his lecure we ry o apply he mehod of characerisics o he Hamilon-Jacobi equaion: u + H D u, x = 0 in R n

More information

Sterilization D Values

Sterilization D Values Seriliaion D Values Seriliaion by seam consis of he simple observaion ha baceria die over ime during exposure o hea. They do no all live for a finie period of hea exposure and hen suddenly die a once,

More information

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,

More information

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx. . Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.

More information

MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.

MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du. MATH 3B: MIDTERM REVIEW JOE HUGHES. Inegraion by Pars. Evaluae 3 e. Soluion: Firs make he subsiuion u =. Then =, hence 3 e = e = ue u Now inegrae by pars o ge ue u = ue u e u + C and subsiue he definiion

More information

System of Linear Differential Equations

System of Linear Differential Equations Sysem of Linear Differenial Equaions In "Ordinary Differenial Equaions" we've learned how o solve a differenial equaion for a variable, such as: y'k5$e K2$x =0 solve DE yx = K 5 2 ek2 x C_C1 2$y''C7$y

More information

CHAPTER 2: Mathematics for Microeconomics

CHAPTER 2: Mathematics for Microeconomics CHAPTER : Mahemaics for Microeconomics The problems in his chaper are primarily mahemaical. They are inended o give sudens some pracice wih he conceps inroduced in Chaper, bu he problems in hemselves offer

More information

Theory of! Partial Differential Equations-I!

Theory of! Partial Differential Equations-I! hp://users.wpi.edu/~grear/me61.hml! Ouline! Theory o! Parial Dierenial Equaions-I! Gréar Tryggvason! Spring 010! Basic Properies o PDE!! Quasi-linear Firs Order Equaions! - Characerisics! - Linear and

More information

1998 Calculus AB Scoring Guidelines

1998 Calculus AB Scoring Guidelines AB{ / BC{ 1999. The rae a which waer ows ou of a pipe, in gallons per hour, is given by a diereniable funcion R of ime. The able above shows he rae as measured every hours for a {hour period. (a) Use a

More information

MATH 128A, SUMMER 2009, FINAL EXAM SOLUTION

MATH 128A, SUMMER 2009, FINAL EXAM SOLUTION MATH 28A, SUMME 2009, FINAL EXAM SOLUTION BENJAMIN JOHNSON () (8 poins) [Lagrange Inerpolaion] (a) (4 poins) Le f be a funcion defined a some real numbers x 0,..., x n. Give a defining equaion for he Lagrange

More information

ACCUMULATION. Section 7.5 Calculus AP/Dual, Revised /26/2018 7:27 PM 7.5A: Accumulation 1

ACCUMULATION. Section 7.5 Calculus AP/Dual, Revised /26/2018 7:27 PM 7.5A: Accumulation 1 ACCUMULATION Secion 7.5 Calculus AP/Dual, Revised 2019 vie.dang@humbleisd.ne 12/26/2018 7:27 PM 7.5A: Accumulaion 1 APPLICATION PROBLEMS A. Undersand he quesion. I is ofen no necessary o as much compuaion

More information

!!"#"$%&#'()!"#&'(*%)+,&',-)./0)1-*23)

!!#$%&#'()!#&'(*%)+,&',-)./0)1-*23) "#"$%&#'()"#&'(*%)+,&',-)./)1-*) #$%&'()*+,&',-.%,/)*+,-&1*#$)()5*6$+$%*,7&*-'-&1*(,-&*6&,7.$%$+*&%'(*8$&',-,%'-&1*(,-&*6&,79*(&,%: ;..,*&1$&$.$%&'()*1$$.,'&',-9*(&,%)?%*,('&5

More information

Finish reading Chapter 2 of Spivak, rereading earlier sections as necessary. handout and fill in some missing details!

Finish reading Chapter 2 of Spivak, rereading earlier sections as necessary. handout and fill in some missing details! MAT 257, Handou 6: Ocober 7-2, 20. I. Assignmen. Finish reading Chaper 2 of Spiva, rereading earlier secions as necessary. handou and fill in some missing deails! II. Higher derivaives. Also, read his

More information

3.1.3 INTRODUCTION TO DYNAMIC OPTIMIZATION: DISCRETE TIME PROBLEMS. A. The Hamiltonian and First-Order Conditions in a Finite Time Horizon

3.1.3 INTRODUCTION TO DYNAMIC OPTIMIZATION: DISCRETE TIME PROBLEMS. A. The Hamiltonian and First-Order Conditions in a Finite Time Horizon 3..3 INRODUCION O DYNAMIC OPIMIZAION: DISCREE IME PROBLEMS A. he Hamilonian and Firs-Order Condiions in a Finie ime Horizon Define a new funcion, he Hamilonian funcion, H. H he change in he oal value of

More information

04. Kinetics of a second order reaction

04. Kinetics of a second order reaction 4. Kineics of a second order reacion Imporan conceps Reacion rae, reacion exen, reacion rae equaion, order of a reacion, firs-order reacions, second-order reacions, differenial and inegraed rae laws, Arrhenius

More information

Lecture 10: The Poincaré Inequality in Euclidean space

Lecture 10: The Poincaré Inequality in Euclidean space Deparmens of Mahemaics Monana Sae Universiy Fall 215 Prof. Kevin Wildrick n inroducion o non-smooh analysis and geomery Lecure 1: The Poincaré Inequaliy in Euclidean space 1. Wha is he Poincaré inequaliy?

More information

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x WEEK-3 Reciaion PHYS 131 Ch. 3: FOC 1, 3, 4, 6, 14. Problems 9, 37, 41 & 71 and Ch. 4: FOC 1, 3, 5, 8. Problems 3, 5 & 16. Feb 8, 018 Ch. 3: FOC 1, 3, 4, 6, 14. 1. (a) The horizonal componen of he projecile

More information

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle Chaper 2 Newonian Mechanics Single Paricle In his Chaper we will review wha Newon s laws of mechanics ell us abou he moion of a single paricle. Newon s laws are only valid in suiable reference frames,

More information

Math 116 Second Midterm March 21, 2016

Math 116 Second Midterm March 21, 2016 Mah 6 Second Miderm March, 06 UMID: EXAM SOLUTIONS Iniials: Insrucor: Secion:. Do no open his exam unil you are old o do so.. Do no wrie your name anywhere on his exam. 3. This exam has pages including

More information

ASTR415: Problem Set #5

ASTR415: Problem Set #5 ASTR45: Problem Se #5 Curran D. Muhlberger Universi of Marland (Daed: April 25, 27) Three ssems of coupled differenial equaions were sudied using inegraors based on Euler s mehod, a fourh-order Runge-Kua

More information

DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS PAGE # An equaion conaining independen variable, dependen variable & differenial coeffeciens of dependen variables wr independen variable is called differenial equaion If all he

More information

( ) = Q 0. ( ) R = R dq. ( t) = I t

( ) = Q 0. ( ) R = R dq. ( t) = I t ircuis onceps The addiion of a simple capacior o a circui of resisors allows wo relaed phenomena o occur The observaion ha he ime-dependence of a complex waveform is alered by he circui is referred o as

More information

t 2 B F x,t n dsdt t u x,t dxdt

t 2 B F x,t n dsdt t u x,t dxdt Evoluion Equaions For 0, fixed, le U U0, where U denoes a bounded open se in R n.suppose ha U is filled wih a maerial in which a conaminan is being ranspored by various means including diffusion and convecion.

More information

Second Order Linear Differential Equations

Second Order Linear Differential Equations Second Order Linear Differenial Equaions Second order linear equaions wih consan coefficiens; Fundamenal soluions; Wronskian; Exisence and Uniqueness of soluions; he characerisic equaion; soluions of homogeneous

More information

20. Applications of the Genetic-Drift Model

20. Applications of the Genetic-Drift Model 0. Applicaions of he Geneic-Drif Model 1) Deermining he probabiliy of forming any paricular combinaion of genoypes in he nex generaion: Example: If he parenal allele frequencies are p 0 = 0.35 and q 0

More information

TMA4329 Intro til vitensk. beregn. V2017

TMA4329 Intro til vitensk. beregn. V2017 Norges eknisk naurvienskapelige universie Insiu for Maemaiske Fag TMA439 Inro il viensk. beregn. V7 ving 6 [S]=T. Sauer, Numerical Analsis, Second Inernaional Ediion, Pearson, 4 Teorioppgaver Oppgave 6..3,

More information

Predator - Prey Model Trajectories and the nonlinear conservation law

Predator - Prey Model Trajectories and the nonlinear conservation law Predaor - Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories

More information

Class Meeting # 10: Introduction to the Wave Equation

Class Meeting # 10: Introduction to the Wave Equation MATH 8.5 COURSE NOTES - CLASS MEETING # 0 8.5 Inroducion o PDEs, Fall 0 Professor: Jared Speck Class Meeing # 0: Inroducion o he Wave Equaion. Wha is he wave equaion? The sandard wave equaion for a funcion

More information