# 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.

1 Page

### FinalS04ansCorr

Course: M 166, Fall 2009
School: Iowa State
Rating:

#### Document Preview

to Correction the solution to #10: The last two rows of the posted solution be should as follows: 27 27 36 3 9 3 27 3 = + = 9 2 6 2 2 2 1

Register Now

#### Unformatted Document Excerpt

Coursehero >> Iowa >> Iowa State >> M 166

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.
to Correction the solution to #10: The last two rows of the posted solution be should as follows: 27 27 36 3 9 3 ...

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:

Iowa State - M - 166
MIKE ANCTIL TH MATH 166 REVIEW, VARBERG 9 ED., CHAPTERS 5, 75.1: The Area of a Plane Region Find the area bounded by the function y = x and y = x - 2 .25.2: Volumes of Solids: Slabs, Disks, and Washers Find the volume of the solid whose base is
Iowa State - M - 166
MIKE ANCTIL 12.4 PRACTICE PROBLEMSChoose the Best Graph:(x + 3)2 + ( y + 2)24 16A) B)=1A)9 x 2 - 16 y 2 + 54 x + 64 y - 127 = 0B)C)D)C)D)MIKE ANCTIL 12.4 PRACTICE PROBLEMSDetermine the Correct Equation: Horizontal Ellipse Cent
Iowa State - M - 166
12.3: #13 Find points of tangency. Ellipse9x 2 + 4 y 2 = 1y-intercept: (0, 6)x y2 + =1 4 92(A)Equation of a tangent line for an ellipse (given in the book) at (x0, y0):xx0 yy 0 + = 1 (B) 4 9Plugging the y-intercept (0, 6) into equation
Iowa State - M - 166
MIKE ANCTIL MATH 166 PRACTICE PROBLEMThis problem is similar to problem #22 in the Integration by Substitution Section of the Techniques of Integration chapter of the book.x4x dx + 16Initially this problem appears to not match any direct fo
Iowa State - M - 166
Iowa State University Math 166 Instructor:Midterm Exam October 7, 2004 TA:Name: Section:Instructions: Answer each question completely. Show all work. No credit for mere answers with no work shown. Show the steps of calculations. State the reaso
Iowa State - M - 166
MIKE ANCTIL APRIL 28 FINAL REVIEW PROBLEMS1) Find the Sum of the Geometric Series:1 1 1 1 1 + 2 + 3 + 4 + . + n 7 7 7 7 7You should also be able to identify this series if given in this form:1 1 1 1 + + + + . 7 49 343 2401From this we can ea
Iowa State - M - 166
MATH 166 REFERENCE SHEETSUMMARY OF TESTS FOR SERIES Test Series nth - TermConvergesDivergesComment This test cannot be used to show convergence. Sum: S =an =1 n =0limnnan 0Geometric Series ar (bn =1 nnr &lt;1lim n bn =
Iowa State - MATH - 166
1 Exercises 6.2 1. R is the region bounded by the graphs of y = x2+1, y = 0, x = 0, x = 2. Let S be the solid of revolution generated by revolving R about the x-axis. Find the volume of S. y R 2 y = x2 + 1 or x = y 1
Iowa State - MATH - 166
Math 166 Final Exam Spring 2006Name &amp; Section: Instructor:Answer each question completely. Show all work and all steps, and state reasons for conclusions. No credit is allowed for mere answers with no work shown. Give numerical answers exactly, n
Iowa State - MATH - 166
1 Here let f(x) be a real-valued function with continuous derivatives of all orders on an open interval I of the x-axis. Let x0 be a real number in I. The Taylor series for f(x) about x0 is the power series (1)k =0f ( k ) ( x0 ) k ( x - x0 )
Iowa State - MATH - 166
1 Return now to the above power series (1)an =0n ( x x0 ) n .Suppose now that one of the following two cases occurs. (i) The power series (1) converges for all real numbers x. In this case, let R = . (ii) There is a positive real number R
Iowa State - MATH - 166
1Some ObservationsNow let be the curve which is the graph of the quadratic function y = f(x) = ax2 + bx + c where a0. Note as before that 2 b ax2 + bx + c = a x + x + c a 2 2 b b b a x2 + x + + c a = a 2a 2a
Iowa State - MATH - 166
Exercises 6.1 26. ( (1,-1) y (8,4) R (3,-1)xIntersection of x = y 2 - 2 y and x = y + 4 : y 2 - 2 y = y + 4. y 2 - 3 y - 4 = 0. ( y + 1) ( y - 4) = 0. y = -1 y = 4 or x=3 x = 8. Area of R = 4 -1( ( y
Iowa State - MATH - 166
Exercises 8.3 3.tdt 3t + 4Let u = 3t + 4. u 2 = 3t + 4. 2udu = 3dt dt = t= 1 2 ( u - 4). 3 tdt 2 udu. 31 2 ( u - 4) 2 udu 3 3 3t + 4 = u 3 2 ( u - 4u ) = du 9 u 2 = ( u 2 - 4 )du 9 2 1 = u 3 - 4u + C 9 3 3 1 2 8 = ( 3t + 4 ) 2 -
Iowa State - MATH - 166
1 Departmental Final Solution Spring 2006 1. ( x + 1)213dx= lim bb 21 dx ( x + 1) 3b 1 -2 = lim - ( x + 1) b 2 2 1 -2 = ( 2 + 1) 2 1 = . 18 2. k + 5 - k + 6 k =1n kk +1 n n n +1 1 2 2 3 3 4 n - 2 n -1
Iowa State - MATH - 166
1 Exercises 8.3 27. y = f ( x) = f ( x) = = -2 (21 . x + 2x + 5 x 2 + 2 x + 5 0 - 1 ( 2 x + 2)2. 2 + 2x + 5 f ( x) = 0 for x = -1. f ( x) &gt; 0 for x &lt; -1, f ( x) &lt; 0 for x &gt; -1. 1 Maximum value of f(x) is f(-1) = . 4(xx +1(x)2+ 2
Iowa State - MATH - 166
1 Exercises 10.1 11. e 2n an = 2 . n + 3n - 1 e2x Let f(x) = 2 . x + 3x - 1 d 2x e dx lim f ( x) = lim x x d x 2 + 3x - 1 dx 2x 2e = lim x 2 x + 3 d 2e 2 x = lim dx x d ( 2 x + 3) dx 4e 2 x = lim x 2 = . lim a n = .( )()()x 18.1
Iowa State - MATH - 166
1 The Integral Test (Section 10.3)f ( x) 0 for 1 x &lt; 0, f(x) continuous. If x &gt; x , then f(x ) f(x ). a n = f(n) for = 1,2,3,. a 2 f ( x)dx a1 .1 2a3 f ( x)dx a 2 .23a 4 f ( x)dx a3 .34etc. a 2 + a3 + a 4 + + a n f ( x)
Iowa State - MATH - 166
Exercise 10.36.k =100 ( k + 2)b32Let f(x) =3 . ( x + 2) 2 dx = 3 2 1 x + 2 100 b ( x + 2)10031 1 1 + 3 as b . b+2 100 + 2 34 Converges. = 3 1000k 2 10. 3 k =1 1 + k 1000x 2 Let f(x) = . 1+ x3 b 1000 3 1 f ( x)dx = 3 l
Iowa State - MATH - 166
1 Exercises 11.2 11.1.51cos( x) dx.Let f(x) = cos( x) . Let n be a positive integer. 1.5-1 .5 Let h = = . n n Let x 0 = 1, x1 = 1 + 1 h, x 2 = 1 + 2 h, x3 = 1 + 3 h, x 4 = 1 + 4 h, , x n -1 = 1 + (n - 1) h, x n = 1 + n h = 1.5. Let Tn
Iowa State - MATH - 166
( 2n) 6 2 4 3 3 3 3 dx x 13 dx - 1 1 - x + x - x + + ( - 1) n x cos 0 0 2! 4! 6! ( 2n)! 1 ( 2n ) 6 2 4 3 13 x 3 x 3 x 3 n x -dx = cos x - 1 - + - + + ( - 1) 0 2! 4! 6! (2n)! 1 (2n) 6 2 4 3 3 3 3 dx x 13 - 1
Iowa State - MATH - 166
Exercises 8.4 21. arctan t dt 1 Let u = arctan and dv = dt. t 1 1 1 du = 2 dt = 2 dt. 2 t +1 1 t 1+ t v = t.1 arctan t dt = arctan t t + 1 + t 23.11t2dt1 1 = t arctan + ln(1 + t 2 ) + c. t 2 x cos2( x)
Iowa State - MATH - 166
1 Exercises 8.2 5.cos 5 ( ) = cos 4 ( ) cos( ) = ( cos 2 ( ) ) cos( )220cos 5 ( )d2= (1 - sin 2 ( ) ) cos( )9.= (1 - 2 sin 2 ( ) + sin 4 ( ) ) cos( ). 2 3 1 5 5 ^ cos ( )d = sin ( ) - 3 sin ( ) + 5 sin ( ) + c. 2 1 8 2 0 cos( )d
Iowa State - MATH - 166
Exercises 10.3 4. 2kk =132+1 32. 2x + 1 3 f ( x)dx = 2 arctan 2 x . b 3 3 1 f ( x)dx = 2 arctan 2b - 2 arctan 2 3 3 - arctan 2 as b . 2 2 2 The improper integralLet f(x) =()()( )( )dx 2x + 1 converges. Thus the infi
Iowa State - MATH - 166
Exercises 10.2 5. k + 2.k =1k -51 k -5 k -5 k = k+2 k+2 1 k 5 1- k 1 0 as k . = 2 1+ k Diverges. 7. k - k - 1 . k =2 n1 11 1 k - k - 1 . k =21 1 1 1 1 1 1 1 1 1 = - + - + - + - + - 2 1 3 2 4 3 n
Iowa State - MATH - 166
Exercises 6.4 33. : x = a ( t sin(t ) ) , y = a (1 cos(t ) ) for 0 t 2. is one arch of a cycloid. (See Problem 18 and Figure 20.) Revolve about the x - axis. Area of the surface of revolution generated = 2 2 0 2y(t) [ x (t)] 2 + [ y (
Iowa State - MATH - 166
Mathematics 166 Practice Test 3 1. Determine whether or not each of the following infinite series converges. Give a reason for your conclusion in each case. 1 (a) 4 n =1 n 3 1 (b) 4 n =1 n 3 + 1 n (c) 4 n =1 n 3 + 1(d) (e) (f)e (n =1
Iowa State - MATH - 166
1 Solution Final Exam Spring 2006 1.0e -4 x dx= lim e -4 x dxb 0b 1 = lim - e - 4 x b 4 0 1 1 = lim - e - 4b + . b 4 4 1 = . 4 1 1 1 2 3 = 2 1 = 2 2 = 3. 2. n =0 1- 3 3 3. (a) nbnn =11321 x3 21
Iowa State - MATH - 166
Solutions 1. (a). Vertices are ( 5,0 ). a 2 = 25 and b 2 = 144. b 2 = c 2 - a 2 c 2 = a 2 + b 2 = 169. Foci are ( 13,0 ). (b). Asymptotes b y = x. a 12 y = x. 5 (c). yx2. Ellipse, center at (1,2), a focus at (4,2), length of major axis 10. c
Iowa State - LSSU - 111
HOMEWORK 7: SOLUTIONS - MATH 111 INSTRUCTOR: George VoutsadakisProblem 1 Find the present value of the future amount \$10,000 compounded semiannually at 6% for 5 years. Solution: We have A = P (1 + P =10000 . 1.0310 r mt m) ,i.e., P =A r (1+ m )
Iowa State - LSSU - 215
EXAM 1: SOLUTIONS - MATH 215 INSTRUCTOR: George VoutsadakisProblem 1 Let P and Q be propositional forms. Define the new connective following truth table P Q P Q F F F T F T T F T T T F 1. Prove that P Q is equivalent to the negation of P Q. 2. Is P
Iowa State - LSSU - 490
EXAM 2: SOLUTIONS - MATH 490 INSTRUCTOR: George VoutsadakisProblem 1 1. Give the definition of a topological space. 2. Let X be an arbitrary set. Let T be the collection of all subsets of X whose complements are either finite or all of X. Then (X, T
Iowa State - LSSU - 325
HOMEWORK 3 - MATH 325 INSTRUCTOR: George VoutsadakisProblem 1 Assume that ABC and EF G are similar with ratio k. Let AD be an altitude of ABC and EH be an altitude of EF G. Prove that EH = k AD. What can you conclude about the areas of the two tria
Iowa State - LSSU - 325
EXAM 4: SOLUTIONS - MATH 325 INSTRUCTOR: George VoutsadakisProblem 1 (a) State Pierre Varignon's Theorem and give the definition of Varignon parallelogram. (b) Show that, if one diagonal divides a quadrangle into two triangles of equal area, it bise
Iowa State - LSSU - 111
EXAM 1: SOLUTIONS - MATH 111 INSTRUCTOR: George VoutsadakisProblem 1 Find the equation of the line that is perpendicular to 2x - 7y = 21 and passes through the point (-2, 7). Solution: The slope of the given line may be found by solving its equation
Iowa State - LSSU - 111
HOMEWORK 3: SOLUTIONS - MATH 111 INSTRUCTOR: George VoutsadakisProblem 1 The equation of the line perpendicular to y = 5x 2 and passing through (5, 2) is 1 1 1 (a) y = x + 1 (b) y = 5x + 2 (c) y = x + 3 (d) y = x + 1 5 5 5 Solution: Since the unk
Iowa State - LSSU - 490
HOMEWORK 4: SOLUTIONS - MATH 490 INSTRUCTOR: George VoutsadakisProblem 1 Prove that a function f : (X, T ) (Y, T ) is a homeomorphism if and only if 1. f is one-one; 2. f is onto; 3. For each point x X and each subset N of X, N is a neighborhood o
Iowa State - LSSU - 111
HOMEWORK 8: SOLUTIONS - MATH 111 INSTRUCTOR: George Voutsadakis y = x-1 y = 8+z Problem 1 Solve the system of equations using the Gauss-Jordan method z = -3 - x Solution: The given system can be rewritten as = 1 x - y y - z = 8 x + z = -3
Iowa State - LSSU - 152
HOMEWORK 8 - MATH 152 DUE DATE: Monday, November 29 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. One part of each homework problem will be chosen at random and graded. Each question is worth 1 point. It
Iowa State - LSSU - 215
HOMEWORK 4 - MATH 215 DUE DATE: After Chapter 2 has been covered! INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. A few randomly selected problems will be graded for a total of 10 points. It is necessary t
Iowa State - LSSU - 111
HOMEWORK 6: SOLUTIONS - MATH 111 INSTRUCTOR: George Voutsadakisx Problem 1 If ln x = 3 and ln y = 4 find ln ( y ).2Solution: We have x2 ln ( y ) = ln (x2 ) - ln y = = = = 2 ln x - ln (y 2 ) 1 2 ln x - 2 ln y 23- 1 4 2 4.1Problem 2 Solve the
Iowa State - LSSU - 111
HOMEWORK 2 - MATH 111 DUE DATE: Monday, September 19 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 1 point. It is necessary to show your work. Correct answers without explanations
Iowa State - LSSU - 215
HOMEWORK 3 - MATH 215 DUE DATE: After Section 2.3 has been covered! INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. A few randomly selected problems will be graded for a total of 10 points. It is necessary
Iowa State - LSSU - 140
HOMEWORK 2 - MATH 140 DUE DATE: Monday, February 7 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. One part of each homework problem will be chosen at random and graded. Each question is worth 1 point. It
Iowa State - LSSU - 490
HOMEWORK 7 - MATH 490 DUE DATE: Monday, April 28 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 5 points. It is necessary to show your work. GOOD LUCK!1. Let X be a compact space a
Iowa State - LSSU - 111
HOMEWORK 7: SOLUTIONS - MATH 111 INSTRUCTOR: George VoutsadakisProblem 1 Find the present value of the future amount \$8,000 compounded quarterly at 4% for 10 years. Solution: We have A = P (1 + P =8000 . 1.0140 r mt m) ,i.e., P =A r (1+ m )mt ,
Iowa State - LSSU - 111
HOMEWORK 7 - MATH 111 DUE DATE: Friday, November 14 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 1 point. It is necessary to show your work. Correct answers without explanations a
Iowa State - LSSU - 111
HOMEWORK 7 - MATH 111 DUE DATE: Friday, March 21 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 1 point. It is necessary to show your work. Correct answers without explanations are
Iowa State - LSSU - 111
HOMEWORK 8: SOLUTIONS - MATH 111 INSTRUCTOR: George VoutsadakisProblem 1 Solve the systems 3x + 5y = 12 x + 2y = 7 by the substitution method. Solution: For the rst system we have, solving the second equation for x, x = 2y 7. Substituting this valu
Iowa State - LSSU - 140
HOMEWORK 7 - MATH 140 DUE DATE: Monday, October 25 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. One part of each homework problem will be chosen at random and graded. Each question is worth 1 point. It
Iowa State - LSSU - 351
HOMEWORK 4 - MATH 351 INSTRUCTOR: George VoutsadakisProblem 1 Given a vertex v of Cn , find e(v). Show that v has either one or two eccentric vertices, depending on the parity of n. Solution: If n = 2k, for some k, then e(v) = k. If e(v) = 2k + 1, f
Iowa State - LSSU - 111
HOMEWORK 8 - MATH 111 DUE DATE: Monday, April 5 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 1 point. It is necessary to show your work. Correct answers without explanations are w
Iowa State - LSSU - 111
HOMEWORK 6 - MATH 111 DUE DATE: Monday, November 8 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 1 point. It is necessary to show your work. Correct answers without explanations ar
Iowa State - LSSU - 490
HOMEWORK 2 - MATH 490 DUE DATE: Monday, February 10 INSTRUCTOR: George VoutsadakisRead each problem very carefully before starting to solve it. Each question is worth 5 points. It is necessary to show your work. GOOD LUCK!1. Let X be the set of con
Iowa State - RU - 477
Name:7 30 98Final Exam In-Class Portion You are permitted to have one sheet of paper, handwritten on both sides and a calculator.Explain as much of your reasoning as you can it can result in more partial credit. Do not pass Go. Do not collect \$2
Iowa State - CMU - 801
On se of inte rs containing no k e m nts in t ge le e arithm tic progre e ssionE. S m r ze e di Acta Arithm tica, 1975 eS huchi C hawlaTheProble and som History m e [van de Wae n, 1926] Le N = S [S . The e r S or S r rde t n ithe 1 1 2 2conta
Iowa State - MATH - 690
The Hat Game11/19/04 James FiedlerReferencesHendrik W. Lenstra, Jr. and Gadiel Seroussi, On Hats and Other Covers, preprint, 2002, http:/www.hpl.hp.com/research/info_ theory/hats_extsum.pdf . J.P. Buhler, Hat Tricks, The Mathematical Intell
Iowa State - RU - 373
{VERSION 2 3 &quot;SUN SPARC SOLARIS&quot; &quot;2.3&quot; } {USTYLETAB {CSTYLE &quot;Maple Input&quot; -1 0 &quot;Courier&quot; 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{PSTYLE &quot;Normal&quot; -1 0 1 {CSTYLE &quot; -1 -1 &quot; 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {SECT 0 {EXCHG {PARA
Iowa State - RU - 373
{VERSION 2 3 &quot;SUN SPARC SOLARIS&quot; &quot;2.3&quot; } {USTYLETAB {CSTYLE &quot;Maple Input&quot; -1 0 &quot;Courier&quot; 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE &quot;2D Math&quot; -1 2 &quot;Times&quot; 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE &quot;2D Output&quot; 2 20 &quot; 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTY
Iowa State - RU - 373
{VERSION 2 3 &quot;SUN SPARC SOLARIS&quot; &quot;2.3&quot; } {USTYLETAB {CSTYLE &quot;Maple Input&quot; -1 0 &quot;Courier&quot; 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE &quot;2D Math&quot; -1 2 &quot;Times&quot; 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE &quot;2D Output&quot; 2 20 &quot; 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTY
Iowa State - RU - 373
{VERSION 2 3 &quot;SUN SPARC SOLARIS&quot; &quot;2.3&quot; } {USTYLETAB {CSTYLE &quot;Maple Input&quot; -1 0 &quot;Courier&quot; 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE &quot;2D Math&quot; -1 2 &quot;Times&quot; 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE &quot;2D Output&quot; 2 20 &quot; 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTY