Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

6 Pages

sola6

Course: MATH 1052, Fall 2009
School: Allan Hancock College
Rating:

Word Count: 1741

Document Preview

Semester MATH1052 2 - 2008 Solutions to Problem Sheet 6 1. Since V = IR = R2 . So I IR1 R2 = it follows that V is a function of I, R1 and 1 1 R1 + R2 + R1 R2 dV V dI V dR1 V dR2 = + + dt I dt R1 dt R2 dt Now V R1 R2 = I R1 + R2 V = IR2 R1 V = IR1 R2 So 1(R1 + R2 ) R1 (R1 + R2 )2 1(R1 + R2 ) R2 (R1 + R2 )2 = 15 8 = 2 IR2 (R1 + R2 )2 2 IR1 . (R1 + R2 )2 = V (2, 3, 5) I V 2.52 50 (2, 3, 5) = = 2 2 R1 (3 + 5) 8...

Register Now

Unformatted Document Excerpt

Coursehero >> California >> Allan Hancock College >> MATH 1052

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Semester MATH1052 2 - 2008 Solutions to Problem Sheet 6 1. Since V = IR = R2 . So I IR1 R2 = it follows that V is a function of I, R1 and 1 1 R1 + R2 + R1 R2 dV V dI V dR1 V dR2 = + + dt I dt R1 dt R2 dt Now V R1 R2 = I R1 + R2 V = IR2 R1 V = IR1 R2 So 1(R1 + R2 ) R1 (R1 + R2 )2 1(R1 + R2 ) R2 (R1 + R2 )2 = 15 8 = 2 IR2 (R1 + R2 )2 2 IR1 . (R1 + R2 )2 = V (2, 3, 5) I V 2.52 50 (2, 3, 5) = = 2 2 R1 (3 + 5) 8 V 2.9 18 (2, 3, 5) = = 2. 2 R2 8 8 Now to put it all together dV 15 1 50 1 18 1 = + 2 + 2 = 0.38125 dt 8 100 8 2 8 10 amps ohms/sec. 2) a) i) Since u(x, y, z) = xy + yz + zx, where x(s, t) = st, y(s, t) = est , z(s, t) = t2 u x u y u z u = + + s x s y s z s Now So u u u = y + z, = x + z, = y + x. x y z u = (est + t2 )t + (st + t2 )test + (est + st). s And evaluated at s = 0, t = 1. u s = (1 + 1).1 + (0 + 1).1 = 3 1 (0,1) ii) If z(x, y) = then x , where x(r, s, t) = rest and y(r, s, t) = rset y z z x z y = + r x r y r Now So z 1 z = , = x y y x y2 , x y = est and = set . r r z et est rest set = r rs (rset )2 z r 1 1 1 = . 2 4 4 1 4 at (1,2,0) = Similarily z s and z t (1,2,0) (1,2,0) = et rtest rest ret rs (rset )2 (1,2,0) = = 1 . b) Dierentiate 2 d(xcos(y) + ycos(x)) d(1) = = 0. dx dx Then thinking of y as a function of x, y(x) and using the chain and product rules cosy xsiny dy dy + cosx ysinx = 0 dx dx dy This implies that (cosx xsiny) = ysinx cosy dx So that dy ysinx cosy = . dx cosx xsiny c) Stewart 6th Ed - Section 15.5 Q38 The volume of a right circular cone is r2 h V = , where r is the base radius and h is the height. 3 We are given that dh dr = 1.8 in/s and = 2.5 in/s. when r = 120 ins and h = 140 ins. dt dt By the chain rule dV V dr V dh = (120, 140) + (120, 140) dt r dt h dt 2rh V (120, 140) = r 3 V r2 (120, 140) = h 3 2 . Now (120,140) = 11200 and (120,140) = 4800 Finally 3) dV = 11200.1.8 + 4800.(2.5) = 8160. dt i) a) First method: dw w dx w dy w w = + ; since = y, =x dt x dt y dt x y dx dy dx dy = y +x since = 2 cos t, = sin t dt dt dt dt = cos t.2 cos t + 2 sin t( sin t) = 2(cos2 t sin2 t) = 2 cos 2t. (b) Second method: w = 2 sin t cos t implies that: dw = 2 cos t cos t 2 sin t sin t = 2(cos2 t sin2 t) = 2 cos 2t. dt ii) (a) First method: Note that z = arctan t implies that tan z = t dz = 1. and sec2 z dt dw w dx w dy w dz = + + dt x dt y dt z dt dx dy dz = y tan z + x tan z xy sec2 z dt dt dt = t2 .t.1 + t.t.2t + t.t2 .1 = 4t3 (b) Second method: w = t3 .t = t4 iii) (a) First method: dw dx dy dz = yz + xz + xy dt dt dt dt = 2t.et .2t + t2 et .2 + t2 .2t(et ) = (6t2 2t3 )et . (b) Second method: w = 2t3 et 4. a) dw = 6t2 et 2t3 et = (6t2 2t3 )et . dt dw = 4t3 . dt w w r w = + . Now x = r cos , y = r sin , r = x2 + y 2 and x r x x r x y sin x y = sec2 = 2 = and = = cos , tan = . So 2 + y2 x x r x x r cos2 x sin = . x r w w w sin So = cos . x r r w r w r y 1 cos w = + , = = sin and = = . So b) 2 y r y y y r y x sec r w w cos w = sin + . y r r 3 c) w x 2 + w y 2 = = = w w sin cos r r w r w r 2 2 2 2 + w w cos sin + r r w 2 2 (cos + sin ) + 2 (cos2 + sin2 ) r2 + w 2 1 r2 5. a) f = (1 2x2 )ye(x 2 +y 2 ) i + (1 2y 2 )xe(x 2 +y 2 ) j So if f = 0 then (1 2x2 )y = 0 and (1 2y 2 )x = 0. 1 Now if (1 2x2 )y = 0 then y = 0 or x = 2 . If y = 0 then (1 2y 2 )x = 0 implies that x = 0 and 1 1 if x = 2 then (12y 2 )x = implies 0 that y = 2 . So the critical points are 1 1 1 1 1 1 1 1 (0, 0), ( , ), ( , ), ( , ), ( , ) 2 2 2 2 2 2 2 2 Now 2f 2 x2 y 2 , 2 = 2x(3 2x )ye x 2f 2 x2 y 2 , 2 = 2y(3 2y )xe y 2f 2 2 = (1 2x2 )(1 2y 2 )ex y . yx and D = 2f 2f 2f 2 ( ) so at (0, 0) D(0, 0) = 1 < 0 which implies a xy x2 y 2 saddle. 1 1 1 1 Now at ( 2 , 2 ) and at ( 2 , 2 ) D= 2(3 1)(3 1)e1 >0 4 2f and < 0, which implies a local max. x2 1 1 1 1 Now at ( 2 , 2 ) and at ( 2 , 2 ) 2f D > 0 and > 0, which implies a local min. x2 b) f = (8y (x+y)3 )i+(8x(x+y)3 )j. So if f = 0 then 8y = (x+y)3 = 8x. This implies x = y. So 8x3 = 8x. Therefore, x = 0 or x = 1. Moreover, 2f = 3(x + y)2 , x2 Now at (0, 0) 2f = 0, x2 2f = 8, xy 4 and 2f = 0. y 2 2f = 8 3(x + y)2 , xy and 2f = 3(x + y)2 . y 2 This leads to D < 0. So the point (0, 0) is a saddle point. At (1, 1) or (1, 1) 2f = 3.4 = 12, x2 2f = 8 3.4 = 4, xy and 2f = 3.4 = 12. y 2 This leads to D = 122 42 > 0. Since (1, 1) are both local maxima. c) 2f = 12 < 0 the points (1, 1) and x2 f = cos x sin yi + sin x cos yj. So if f = 0 then either cos x = cos y = 0 or sin x = sin y = 0. This implies either x = n + /2 and y = m + /2 or x = n and y = m. Moreover, 2f = sin x sin y, x2 2f = cos x cos y, xy and 2f = sin x sin y. y 2 Now at x = n + /2 and y = m + /2 2f = 1, x2 2f = 0, xy and 2f = 1. y 2 This leads to D = 1 > 0. Now depends on n or m to be even or odd we will nd minimum or maximum points. Now at x = n and y = m 2f = 0, x2 2f = 1, xy and 2f = 0. y 2 This leads to D = 1 < 0. So the points x = n and y = m are saddle points. d) f = ex (1 cos y)i + ex sin yj. So if f = 0 then cos y = 1 and sin y = 0. This implies y = 0, 2, 4, . . . but x is unspecied. Moreover, 2f x 2 = e (1 cos y), x At y = 0, 2, 4, . . . we see 2f = ex sin y, xy and 2f x 2 = e cos y. y 2f 2f 2f x = 0, = 0, and 2 2 = e . This xy x y implies D = 0. In fact there is a minimum at y = 0, 2, 4, . . . . But it is not a local min as it occu...

Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Allan Hancock College - MATH - 1052
MATH1052 Semester 1 - 2009 Solutions to Problem Sheet 41. f x 3 3 = 2x+e fx (3, 2) = 2.3+e = 6+e y2 4 4 x g b) = 2y cos(x + y) y 2 sin(x + y). So y gy ( , 2) = 4 cos( + 2) 4 2 sin( + 2) = 4 2 . 2 2 2 a) c) h = 2x2 z z 1 y+z 1 = 2x2 2 1
Allan Hancock College - MATH - 1052
MATH1052 Semester 1 - 2009 Solutions to Assignment 71. a) Let g(x, y) = x2 + 2y 2 so the constraint is g(x, y) = 44. Now we have f = g 3i 2j = (2xi + 4yj). 2x 3 x = or y = . Now using 4y 2 3So 2x = 3 and 4y = 2. This leads to22x2 the constrai
East Los Angeles College - C - 5354
TutorialOpen Source BreakoutA model of software development called open source is gaining acceptance in the software world. Although the exact denition of open source remains debatable, the basic premise is that the source code is freely available
East Los Angeles College - C - 5354
Genetic Algorithmslearning with reference to evolutionDr. Toby P. Breckon School of Engineering Cranfield University, UK.Reading(s) : Mitchell (ch. 9) Course home page : http:/www.cranfield.ac.uk/~toby.breckon/teaching/ml/Toby BreckonPortions
Laurentian - PSYC - 4000
Psychology 4000Evolutionary Psychology Dr Darren HannessonLecture 1IntroductionsMe, youCourse Outline Instructional Methods1MEMeYOUWhats your background?Bio, psych, reason for taking this course2What is evolution?Where are all
Laurentian - PSYC - 4000
Psychology 4000Evolutionary Psychology Dr Darren HannessonLecture 1IntroductionsMe, youCourse Outline Instructional MethodsMEMeYOUWhats your background?Bio, psych, reason for taking this courseWhat is evolution?Where are all the Cy
Laurentian - PSYC - 5850
Behaviour &amp; Evolution Graduate Seminar Psychology 5850/7850 Spring 2006 Co-ordinator: Dr. Paul L. Vasey Office: D 852 E-mail: paul.vasey@uleth.ca Time: 15:05-17:45 Classroom: AH116 Seminar Description: In this seminar we critique the arguments laid o
Laurentian - BIOL - 4110
Biology 41101Human genomeThe Human genome complete DNA sequence What is the human genome? Why is the human genome important? Who gets the human genome? What are important features of the human genome? What are uses of the information from
Laurentian - BIOL - 3520
LAB # 1. THE PROTOZOANS 1. Overview Almost everything about Protozoan phylogeny and biology is controversial. No two biologists can agree whether this group should be alligned with the plants, the animals or even the fungi. We wont go into the detail
Laurentian - BIOL - 3520
LAB 7. Introduction to the Arthropodous Phyla 1. Overview The Ecdysozoa are represented by at least 7 phyla (Nematoda, Nematomorpha, Onychophora, Tardigrada, Kinorhyncha, Priapulida, and Arthropoda) that are protostomous and that possess a cuticle th
Laurentian - GEOG - 1010
GEOG 1010Introduction to Geography1. 2. 3. 4. 5.LECTURE STRUCTUREThe universe and the geologic time scale Inside the earth Crustal Plates: Isostacy and Plate tectonics Relief and topography Crustal deformation OrogenesisAlso consult p. 4 and 5
UCLA - M - 164
Math164 General Course OutlineBy Andrea Brose and Bob Brown164 Optimization Formerly called &quot;Linear Programming&quot;. Three hours lecture and one hour discussion. Prerequisites: course 115A. Text Stephen G. Nash and Ariela Sofer, Linear and Nonlinear
UCLA - M - 167
HW6 - ProblemsAndrea Brose, Math 167/1, F05 Due 14. November 2005AP(x) Let P Rn and Q Rm be convex sets. Show that P Q = {(p, q) : p P, q Q} Rn+m is a convex set. AP(xi) Consider f : [0, 1] R x (a) Dene the correspondence Rf : [0, 1] R x R
UCLA - M - 167
HW10 - Additional ProblemsAndrea Brose, Math 167/1, F05 Due 9. December 2005AP(xxiv) Consider a two-person strategic game whose payo matrix is given by: M= (1, 1) (1, 3) (3, 0) . (1, 0) (0, 1) (0, 3)(a) Sketch the feasible payo region X if player
East Los Angeles College - C - 5354
Visualisation:Lecture18Visualisation:Lecture18Visualisation:CourseReviewVisualisation.VisualisationLecture18 TobyBreckon toby.breckon@ed.ac.uk ComputerVisionLab. InstituteforPerception,Action&amp;Behaviour SchoolofInformaticsTobyBreckon Visualis
East Los Angeles College - C - 5354
Learning Disjunctive Sets of RulesLearning Sets of RulesDr. Toby P. Breckon School of Engineering Cranfield University, UK.Reading(s) : Mitchell (ch. 10) Course home page : http:/www.cranfield.ac.uk/~toby.breckon/teaching/ml/IF parent(x,y) THEN
East Los Angeles College - C - 5354
RL the concept Reinforcement LearningDr. Toby P. Breckon School of Engineering Cranfield University, UK.Reading(s) : Mitchell Ch. 13 Course home page : http:/www.cranfield.ac.uk/~toby.breckon/teaching/ml/Portions: Mitchell, CMU 1997 / Cranfield
East Los Angeles College - C - 5354
Thinking a little bit more about the learning problemComputational Learning TheoryDr. Toby P. Breckon School of Engineering Cranfield University, UK.Reading(s) : Mitchell (ch. 7) Course home page : http:/www.cranfield.ac.uk/~toby.breckon/teaching
East Los Angeles College - C - 5354
Combing Inductive and Analytical LearningDr. Toby P. Breckon School of Engineering Cranfield University, UK.Reading(s) : Mitchell Ch. 12 Course home page : http:/www.cranfield.ac.uk/~toby.breckon/teaching/ml/Portions: Mitchell, CMU 1997 / Vital,
East Los Angeles College - C - 5354
Visualisation:Lecture2Visualisation:Lecture2ComputerGraphics: IntroductiontotheVisualisation ToolkitVisualisationLecture2 TobyBreckon toby.breckon@ed.ac.uk ComputerVisionLab. InstituteforPerception,Action&amp;BehaviourTobyBreckonLastlecture.Vi
East Los Angeles College - C - 5354
Visualisation:Lecture12Visualisation:Lecture12VectorFieldVisualisation:local&amp;globalVectorFieldVisualisation: globalviewVectorFieldsspecifyflowsthroughthefieldaim:visualiseflowinfieldTwopropertiesofvectorfieldstovisualiselocalview
East Los Angeles College - C - 5354
Visualisation:Lecture8Visualisation:Lecture8BriefRecapContouring &amp;ImplicitModellingContouringlecture6 MarchingCubesImplicitFunctionsVisualisationLecture8 TobyBreckon toby.breckon@ed.ac.uk ComputerVisionLab. InstituteforPerception,
East Los Angeles College - C - 5354
Machine Learning ? Brief Introduction to Machine Learning(as an aspect of Pattern Recognition) Why Machine Learning?Dr. Toby P. Breckon School of Engineering Cranfield University, UK.Reading(s) : Mitchell (ch1 + ch2) Course home page : http:/ww
East Los Angeles College - C - 5354
Visualisation Course 2006, UG4 L10 (ONLY)Assignment 2 Visualisation of Glider Flight PathThis assignment is the second of two assessed practical exercises for the visualisation module. The aim of the practical is to visualise a set of GPS coordin
East Los Angeles College - C - 5354
UWO - HISTORY - 2173
The University of Western Ontario Department of History 2008 - 2009HISTORY 2173 (001) UWWAR IN THE ANCIENT AND MEDIAEVAL WORLDMonday 12:30 - 2:30 pm HSB 9 Dr. B. Murison, SSC 4417 Office phone: 661-2111 ext. 84985 Office hours: Tuesdays: 10:30-11
UWO - HISTORY - 588
1UNIVERSITY OF WESTERN ONTARIO DEPARTMENT OF HISTORY Fall 2007HIS 588 The United States and Empire Professor Frank Schumacher Wednesday 2:30 4:30 PMOffice: Phone: Office Hours: tba tba Wednesday 5:00 6:00 PM Thursday 1:00 2:00 and 5:00 6:00
Laurentian - CHEM - 200803
Chemistry 2000 (Fall 2008) Problem Set #1: Molecular Orbital Theory and Diatomic Molecules SolutionsTextbook Questions Answers in Solution Guide that came with text. Additional Practice Problems 1. Draw the potential energy curve for a diatomic mol
Laurentian - CHEM - 200901
University of Lethbridge Department of Chemistry &amp; BiochemistryChemistry 2740 Laboratory Experiment 6THE MUTAROTATION OF -D-GLUCOSEThe rate constant for a reaction that follows a first-order rate equation can be determined relatively easily by m
Laurentian - CHEM - 200901
University of Lethbridge Department of Chemistry &amp; BiochemistryChemistry 2740 Laboratory ContentsCONTENTSLaboratory Rotation and Schedule Introduction General Comments Laboratory Grading Laboratory Reports Laboratory Notebook Laboratory Etiquett
Laurentian - BCHM - 200901
Department of Chemistry and Biochemistry University of LethbridgeBiochemistry 3300III. Metabolism Glucose Catabolism Part IIMetabolic Fates of NADH and Pyruvate1Metabolic Fates of NADH and PyruvatePyruvate is a central branch point in Met
Laurentian - CHEM - 4000
Chemistry 4000/5000/6000PART IISYMMETRY AND THE CRYSTALLINE SOLID STATEMolecular Symmetry: The Schnflies System Symmetry operations and elements Symmetry operation: The movement of a molecule relative to some symmetry element which generates an
Laurentian - CHEM - 4000
Chem4000B Assignment 6 - AnswersFall 20071. (a) Draw the MO diagram for N2 with the correct symmetry labels for the molecular orbitals. Derive the symmetry labels from the appropriate character table. (b) Calculate the bond order for N2. (c) Give
Laurentian - HIST - 200101
History 2800/Week 4/Women &amp; Religion/Prof. WilliamsWomen &amp; Religion Two major themes: 1) Religion, and monastic orders, as socially empowering for women (ie: convents as educational system for women availing skills and opportunities for them.) 2) R
Laurentian - GEOG - 4740
Neumont - STT - 6515
Ramus bone length at four ages for 20 boysVariables are y1=length at 8yr, y2=length at 8.5yr,y3=length at 9 yr and y4=length at 9.5 yr.enfant y1 y2 y3 y41 47.8 48.8 49.0 49.72 46.4 47.3 47.7 48.43 46.3 46.8 47.8 48.54 45.1 45.3 46.1 47.2
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Neumont - MAT - 6470
10.90.80.70.60.50.40.30.20.100510152025303540
Laurentian - PHYS - 2130
Neumont - MAT - 6470
0.250.20.150.10.05000.10.20.30.40.50.60.70.80.91
Laurentian - PHYS - 2130
Neumont - MAT - 6470
Solution of van der Pol Equation, = 1 2.521.510.5 solution y100.511.522.50246810 time t1214161820
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130
Laurentian - PHYS - 2130