University of Illinois, Urbana Champaign - AE - 321

AE 321 - Equation SheetN0orderzero (scalar)1one (vector)2two3three4fourTensor operations:transformation lawA! = AAi! = " ij A jAi!j = " ik" jl AklAi!jk = " il" jm" kn AlmnAi!jkl = " im" jn " kp " lq Amnpq!a b = aibi(! A )i = Aij, j[

University of Illinois, Urbana Champaign - AE - 321

AE 352: Aerospace Dynamics II, Fall 2011Homework 9Due 5.00 pm, Friday, October 28Problem 1. Working with a general expression for kinetic energy in R3 for a systemof N particles, if T = T2 + T1 + T0 where T2 is quadratic, T1 is linear, and T0 is const

University of Illinois, Urbana Champaign - AE - 321

AE 352: Aerospace Dynamics II, Fall 2011Homework 9Due 5.00 pm, Friday, October 28Problem 1.Solution: Problem 1 Solution is simply the derivation presented on pages 262-264of Greenwood.Problem 2. Greenwood Problem 6-16Solution: This problem caused s

University of Illinois, Urbana Champaign - AE - 321

AE 352: Aerospace Dynamics II, Fall 2011Homework 10Due 5.00 pm, Friday, November 11Problem 1. Greenwood Problem 9-7Problem 2. For the compound pendulum (Greenwood Problem 6-7),1T = ml2 (3 + 2 cos )2 + 2 + 2(1 + cos )2V = mgl 2 cos + cos( + )(a) D

University of Illinois, Urbana Champaign - AE - 321

AE 352: Aerospace Dynamics II, Fall 2010Homework 10Due 5.00 pm, Friday, November 19Problem 1. For the compound pendulum (Greenwood Problem 6-7),1T = ml2 (3 + 2 cos )2 + 2 + 2(1 + cos )2V = mgl 2 cos + cos( + )(a) Determine all the static equilibri

University of Illinois, Urbana Champaign - AE - 321

AE 352: Aerospace Dynamics II, Fall 2011Homework 10Due 5.00 pm, Friday, November 11Problem 1. greenwood 9-7Solution: The position of the center of the disk isrp = ler + rcos(2 1 )er + rsin(2 1 )eUsing R.A.T, we have the velocityvp = r2 sin(2 1 )er

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 1Consider the following system of three linear equations in three unknowns x, y, z :x + y 2z = 12x + y + 2z = 0x y + 3z = 1Use some appropriate method to determine if there are any solutionsto this equation. If there are sol

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 2Consider the following system of three linear equations in three unknowns x, y, z :x y 2z = 1x y + 2z = 02x y + 3z = 1a) Using row operations, nd the echelon form for the augmented matrix that corresponds to this system.b)

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 3Consider the following system of two linear equations in three unknowns x, y , and z :3x 4y + z = 12x 3y z = 2a) Find two solutions to this linear system by taking two dierentparticular choices of z and solving the equations

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 4Consider the following equation of 2 2 matrices A, B, C , and X :A + 2BC + X = B X.a) Write an equation that solves for X in terms of A, B , and C .b) Compute X given thatA=210 1,B=112 1,C=1011.

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 5Given an n n matrix A, we know that the following are equivalent:a) A is invertible (i.e. it has an inverse matrix A1 ),b) Whenever Av = 0, we know that v = 0.c) For any b, there exists v such that Av = b.Now consider the sp

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 6Let a be a real number and consider2a1 2A=0 101the 4 4 matrix1 20 10 223What is det(A)? For what value (or values) of a is A invertible?

University of Illinois, Urbana Champaign - MATH - 225

Math 225, Fall 2010Quiz 7120 and 2 .Consider the subspace S of R4 spanned by 1 1 13844 8 Which of the two vectors 5 and 2 is in this subspace?210

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 111.1Logistic equation and population modelsLet B (t) and D (t) denote the number of births and deaths that have occurred since t = 0 by the time t.B (t+ t)P (t)D(t+ t)t!0P (t)The birth rate (t) = limThe death rate (t) = limt!0B (t)

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 12Example 1 Consider a population P (t) of unsophisticated animals that relysolely on chance of encounter to meet mates for reproductive purposes. Wecan assume that such encounter occurs at a rate that is proportional to theproduct P=2 of ma

University of Illinois, Urbana Champaign - MATH - 285

11.1Lecture 13Variable gravitational accelerationNewton law of gravitation states that the gravitational force between twosmasses M and m is given byGM m; where G is an empirical constant:r2m:G = 6:6726 10 11 Nkg 2F=Let ME be the mass of Ea

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 14We observe from the previous example how one can try to solve the initialvalue problemy 00 + p (x) y 0 + q (x) y = 0;0y (x0 ) = y0 ; y 0 (x0 ) = y0 :We need to know that the particular solution satisfying given initialvalue problem is u

University of Illinois, Urbana Champaign - MATH - 285

11.1Lecture 15Linear independence and WronskianRecall that the determinant of a 2deta11 a12a21 a22=2 matrix is dened bya11 a12a21 a22= a11 a22a12 a21 :Consider a system of two linear equations in two unknowns:a11 x1 + a12 x2 = b1a21 x1 + a

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 161.1Linear ODE of order nIn a similar way, we can consider a linear ODE of order n :y (n) + p1 (x) y (n1)+ p2 (x) y (n2)+ : + pn (x) y (x) = f (x)with the initial conditions:y (x0 ) = b0 ; y 0 (x0 ) = b1 ; :; y (n1)(x0 ) = bn 1 .We

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 17Denition 1 A set of solutions y1 (x) ; y2 (x) ; :; yn (x) ; a < x < b, of thelinear ODE of order ny (n) + p1 (x) y (n1)+ p2 (x) y (n2)+ : + pn (x) y (x) = 0is called the fundamental set of solutions if W (y1 ; y2 ; :; yn ) 6= 0 at leas

University of Illinois, Urbana Champaign - MATH - 285

11.1Lecture 18Non-homogeneous linear ODE of order nLetL = Dn + p1 (x) Dn1+ p2 (x) Dn2+ : + pn1(x) D + pn (x) D0 :Consider non-homogeneous ODEy (n) + p1 (x) y (n1)+ p2 (x) y (n2)+ : + pn (x) y (x) = f (x) , orL [y ] = f:The complementary

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 19For a number a, consider the following operator:a) [f ]jx = f 0 (x)(Daf (x) :First, we prove the following property of some operators. We claim that,for every pair of real numbers,(Da) (Db) = (Db) (Da) :Indeed,(Da) (Db) [f ] = (

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 20Finally, we remark that, as in the case of a real r, for complex r;D [erx ] = rerx :Indeed, let r = a + bi. Thend ax ibxd (a+bi)x=eeedxdxd ax=[e (cos bx + i sin bx)]dx= aeax (cos bx + i sin bx) + eax ( b sin bx + bi cos bx)= ae

University of Illinois, Urbana Champaign - MATH - 285

11.1Lecture 21Free undamped motionmx00 + kx = 0:Setk= !2.0mThen we havex00 + ! 2 x = 0:0Characteristic equation:r2 + !2 = 0 ) r =0! 0 i.General solution:x (t) = A cos ! 0 t + B sin ! 0 t:We can rewrite the foregoing formula as follows:

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 22Example 1 Two pendulums are of length L1 and L2 , and located at therespected distances R1 and R2 from the center of Earth, have periods p1 andp2 . Prove thatpR 1 L1p1p:=p2R 2 L2Solution.00g= 0 ) !0 =+LBecauserg:LMEarth;

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 24c < ccr underdamped case1.0.1In this case, characteristic equation has only complex conjugate roots:qr1 ; r2 = p i ! 2 p2 :0General solution:x (t) = eLet ! 1 =p!20ptC1 cosq!20p2 t + C2 sinq!20p2 t :p2 . Thenx (t) = e pt

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 25Example 1 (Continue)2y 00 + 3y 0 + y = t2 + 3 sin t:I. complementary solution.2y 00 + 3y 0 + y = 0:Characteristic equation:2r2 + 3r + 1 = 0Solution is:r1 =1; r2yc (t) = C1 et1:2+ C2 et2:II. We see that the function t2 + 3 si

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 26Example 1y 004y = cosh 2x:Recall thatet e tet + e t; sinh t =;22cosh0 t = sinh t; sinh0 t = cosh t:cosh t =Complimentary equation:y 004y = 0:Characteristic equation:r24 = 0 ) r1 = 2 and r2 =2.Because r = 2 is a root of mult

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 27Example 1 (Continue)ZZZd (sin x)cos x2=u2 (x) = cos x csc xdx =2 xdx =sin xsin2 xIV. Particular solution:x2x= ln cot2yp (x) = ln cot1sin xcos x +cos xsin x1:General solution:y (x) = yp (x) + yc (x)xcos x= ln cot2

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 28Example 1 (Continue) Considermx00 + kx = F0 cos !twith the initial conditionsx (0) = 0; x0 (0) = 0:x (t) = A (t) sin! + !0t;2whereA (t) =If !2 F0m!0!t.2! 0 , then A (t) is a slow varying amplitude.!2!20sinA rapid oscilla

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 291.1Damped forced oscillationRecall the equation of the damped forced oscillation:mx00 + cx0 + kx = F (t) :We assume that F (t) = F0 cos !t. Recall also that for complimentary solutionmx00 + cx0 + kx = 0the following cases are possible:

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 30Example 1x00 + 3x0 + 5x =4 cos 5t:Complimentary equation:x00 + 3x0 + 5x = 0:Characteristic equation:r2 + 3r + 5 = 0Solution is:r=3 1p+ i 11; r =22Transient solution:xtr (t) = e321pi 11:2!p1111t + C2 sint:C1 cos22p

University of Illinois, Urbana Champaign - MATH - 285

1Lecture 31Example 1 (Continue)y 00 + y = 0;y 0 (0) = y 0 ( ) = 0:III.2=> 0.y 00 + 2 y = 0 ) y (x) = A cos x + B sin x;y 0 (0) = 0 and y 0 ( ) = 0:y 0 (x) =A sin x + B cos x:y 0 (0) = 0 ) B = 0. So,y (x) = A cos x ) y 0 (x) =y0 ( ) = 0 )si

University of Illinois, Urbana Champaign - MATH - 285

11.1Lecture 32Discussion of general non-homogeneous problemConsider ODEx00 + ! 2 x = f (t)0that models the behavior of a mass-spring system with natural circular frequency ! 0 , under inuence of an external force of magnitude f (t) per unitof mas

University of Illinois, Urbana Champaign - MATH - 285

11.1Lecture 33Fourier coe cients: more formulaeNotice that because f (t) ; sin nt and cos nt are periodic functions,R+f (t) cos mtdt; m = 0; 1; 2; :,am = 1+R+1bm =f (t) sin mtdt; m = 1; 2; : .+In particular, for= , we getR2am = 1 0 f (t)

University of Illinois, Urbana Champaign - MATH - 285

HW11. Sec. 1.1: 10. Let y (x) = xln x. Then1xdy (x)=1dxandd2 y (x)1= 2:2dxxWe havex2 y 00 + xy 0y = x21x2Now let1xy (x) =Thendy (x)=dxand1x+x 1(xln x) = ln x:ln x:1x21xd2 y (x)21= 3 + 2:2dxxxWe havex2 y 00 +

University of Illinois, Urbana Champaign - MATH - 285

HW 21. Sec. 1.3: 13.dyp= 3 y ; y (0) = 1:dxThe rate function is given byf (x; y ) =p3y:pIt is known from calculus that f (x; y ) = 3 y is a continuous function forall x and y , in particular in a neighborhood of the point x0 = 0 and y0 = 1.S

University of Illinois, Urbana Champaign - MATH - 285

HW 311. Sec.1.6:3.pxy 0 = y + 2 xy:Sety; y = vx; y 0 = v 0 x + v:xv=Thenpppx(v 0 x + v ) = vx + 2 xvx = vx + 2x v ) v 0 x + v = v + 2 vp) v 0 x = 2 v:This is a separable equation:ZZdvdv1dxp=p = dx ):2x2vv1p2 v = ln jxj + C;

University of Illinois, Urbana Champaign - MATH - 285

HW 411. Sec. 3.1: 25. f (x) = ex sin x; g (x) = ex cos x. Wronskian:ex sin xex cos x:ex sin x + ex cos x ex cos x ex sin xW [f; g ] =Now calculate0111W [f; g ] jx=0 ==1 6= 0,hence f and g are linearly independent.2. Sec. 3.1: 37.2y 00y0y

University of Illinois, Urbana Champaign - MATH - 285

HW 51. Sec. 3.3:11.1y (4)8y (3) + 16y 00 = 0:Characteristic equation:r48r3 + 16r2 = 0 ) r2 r2 8r + 16 = 0 )r1 = r2 = 0; r3 = r4 = 4:General solution:y (x) = (C0 + C1 x) + (C3 + C4 x) e4x :2. Sec. 3.3: 25.3y (3) + 2y 00 = 0; y (0) =1; y 0 (0) =

University of Illinois, Urbana Champaign - MATH - 285

HW 61151. Sec.3.4:3. We have 15 = k 0:2 ) k = 0:2 = 75. x0 = 0 and v0 =10m= sec. Then we have the following problem:rrk75002== 5; x0 = 0; v0 = 10.x + ! 0 x = 0; ! 0 =m3Then x (t) = A cos 5t + B sin 5t. x (0) = 0 = A. Hence x (t) = B sin 5t.

University of Illinois, Urbana Champaign - MATH - 285

HW 711. Sec. 3.6: 3. We havex00 + 100x = 225 cos 5t + 300 sin 5t; x(0) = 375; x0 (0) = 0:The characteristic equation is r2 + 100 = 0 =) r =mentary solution is10i: The compli-xc (t) = c1 cos 10t + c2 sin 10t:r = 5i is not a root of the characteristi

University of Illinois, Urbana Champaign - MATH - 285

HW 811. Sec. 3.8:1. We have y 00 + y = 0; y 0 (0) = 0; y (1) = 0: Ifhavey 00 = 0 =) y 0 = a =) y (x) = ax + b:= 0 then weNow we use the endpoint conditions:y 0 (0) = a = 0 =) a = 0;y (1) = a + b = 0 =) b = 0:Thus, if = 0 =) y (x) = 0 =) = 0 is not

University of Illinois, Urbana Champaign - MATH - 285

HW 911. Sec. 9.1 :21. We have f (t) = t2 ; t 2 [a0 =an =====Z; ] ; see the sketch:Z2 t32322t2 dt =j0 ==:3330ZZZ12222t cos ntdt =t cos ntdt =t2 d sin ntn00Z222t sin ntdt][t sin nt j0n0ZZ44td cos nt =[t cos nt j

University of Illinois, Urbana Champaign - MATH - 285

HW 101. Sec. 9.3: 5. We have f (t) = 0 if t 2 (0; 1) ; f (t) = 1 if t 2 (1; 2) ; andf (t) = 0; if t 2 (2; 3) :1 case: cosine series. See the sketch.1ZZ223222a0 =f (t)dt =dt = t j2 = :1303133Z2223nt22n2ntdt =sinj1 =(sinan =co

North Carolina State University - MA - 241

Extra Problems Math 241 January 23, 2012 1. Evaluate the following integrals (hint: they are improper.) 2a. dx! x 3 0 1b. !0dx3x !dx"xc. 3 2 !d. "1dx 3x2. Evaluate the following integrals

North Carolina State University - MA - 241

Fridays ExamplesI.Partial Fractions#II.Integration by Parts"e!!III.IV.sin xdxdxx 2 " 8x " 9Long division needed and u-substitution#!xCompleting the square and trigonometric substitution#!2x 3 " 5x 2 + 6x " 3dxx 2 ( x 2 + 3)x 3 " 8

North Carolina State University - MA - 241

North Carolina State University - MA - 241

Math 241 Class Information for sections 1 and 2- Spring 2012Instructor: Elizabeth J. DempsterOffice :SAS 3240Email:ejdempst@ncsu.eduOffice Hours Monday and Wednesday 12:30-1:30Text & Materials: Calculus Concepts & Contexts by James StewartMaple 10

North Carolina State University - MA - 241

Math 241 Approximation Methods Section 5.9Outline of LectureWe use approximation methods for one of two reasons,-we cannot find (it is impossible to find for some functions) an antiderivativefor the integrand function- we have data instead of an alge

North Carolina State University - MA - 241

Math 241 Review of integration topics from Calculus IPlease try all of these and well go over them on Wednesday.U-substitutionExample 1: Integrate#sin x1 " cos2 xdx!Partial FractionsExample 2: Integrate11x " 15dx2" 3x# 4x!Integration by p

North Carolina State University - MA - 241

Review Integrals Math 2411. Evaluate the following indefinite integrals.a.!2" x + cos x + 2#d.!xb.+ 5 dx4 " x 2 dxe.!"15" (1 + 3x )x3dxx2 + 2dxc.!!2. Evaluate the following definite integrals.5a.#4x"4dx2x " 5x + 63b.2x

North Carolina State University - MA - 241

Section 5.7 More techniques of IntegrationI.Trigonometric IntegralsII.Partial FractionsIII. Trigonometric Substitution ( well cover this after the test)I.Integrals involving TrigonometryMust play around with to work out; remember your trig.identi

Kaplan University - BUSINESS - Accounting

GB518: Financial Accounting Principles and AnalysisChapter 1- Accounting as a Form of Communication(Decision Case 1.6 Identification of Errors in Financial Statements and Preparation of Revised Statements)Travis DorsoGB518- Unit 1 Decision CaseProfes

Kaplan University - BUSINESS - Accounting

Exercise 12-15 Income Statement, Statement of Cash Flows(Direct Method)1)-Income Statement For First YearHandsome Hounds Grooming CompanyGrooming service revenueExpensesOperating expensesAmortization, patentRent, storage buildingNet incomeIncome

Kaplan University - BUSINESS - Accounting

GB518: Financial Accounting Principles and Analysis(Decision Case 13.5 Acquisition Decision)Travis DorsoGB518- Unit 4 Decision CaseProfessor Wendy W. Achilles, PhD, CPA1.The company seems to be in a state of decent liquidity however each of the meas

Kaplan University - BUSINESS - Accounting

GB518: Financial Accounting Principles and AnalysisChapter 1- Accounting as a Form of Communication(Decision Case 3.3 Reading and Interpreting Southwest Airlines Balance Sheet)Travis DorsoGB518- Unit 2 Decision CaseProfessor Wendy W. Achilles, PhD, C

Kaplan University - BUSINESS - Accounting

GB518: Financial Accounting Principles and AnalysisChapter 1- Accounting as a Form of Communication(Decision Case 4.4 The Use of Net Income and Cash Flow to Evaluate a Company)Travis DorsoGB518- Unit 2 Decision CaseProfessor Wendy W. Achilles, PhD, C

Kaplan University - BUSINESS - Accounting

Users of Accounting Information and Their NeedsUser GroupNeeds Information AboutCompany ManagementStockholderLabor UnionSecurities and Exchange ComminsionsBankerSupplierInternal Revenue ServiceThe profitability of each division in the companyTh

Kaplan University - BUSINESS - Accounting

1. $850,000 dollars out of the $1,000,000 dollars would be consider under cash basis revenue while the full $1,000,000 would be considered accrual basis.2. The matching principle states that revenues of a particular period should be matched with all the

Kaplan University - BUSINESS - Accounting

Problem 6.5 Internal Control1. Problems with the current accounts receivable system-The problems with the current accounts receivable system at Morris Mart, Inc. is that it is a small business that does not have the appropriate amountof personnel to per