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Course: M 166, Fall 2009
School: Iowa State
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Word Count: 102

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Standard Form: __________________________ Parabola: Vertex: ________ Focus: ________ Directrix: _______________________ Circle: Center: ________ Radius: ________ Ellipse: Center: _____ a: _____ b: ______ Foci: ________ and _________ Ends of Major Axis: _____ and ______ Ends of Minor Axis: _____ and ______ Hyperbola: Center: _____ a: _____ b: ______ Foci: ________ and _________ Vertices: _________ and __________...

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Standard Form: __________________________ Parabola: Vertex: ________ Focus: ________ Directrix: _______________________ Circle: Center: ________ Radius: ________ Ellipse: Center: _____ a: _____ b: ______ Foci: ________ and _________ Ends of Major Axis: _____ and ______ Ends of Minor Axis: _____ and ______ Hyperbola: Center: _____ a: _____ b: ______ Foci: ________ and _________ Vertices: _________ and __________ Asymptotes: ____________________ ____________________ Standard Form: __________________________ Parabola: Vertex: ________ Focus: ________ Directrix: _______________________ Circle: Center: ________ Radius: ________ Ellipse: Center: _____ a: _____ b: ______ Foci: ________ and _________ Ends of Major Axis: _____ and ______ Ends of Minor Axis: _____ and ______ Hyperbola: Center: _____ a: _____ b: ______ Foci: ________ and _________ Vertices: _________ and __________ Asymptotes: ____________________ ____________________ Standard Form: __________________________ Parabola: Vertex: ________ Focus: ________ Directrix: _______________________ Circle: Center: ________ Radius: ________ Ellipse: Center: _____ a: _____ b: ______ Foci: ________ and _________ Ends of Major Axis: _____ and ______ Ends of Minor Axis: _____ and ______ Hyperbola: Center: _____ a: _____ b: ______ Foci: ________ and _________ Vertices: _________ and __________ Asymptotes: ____________________ ____________________
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Iowa State - M - 165
MIKE ANCTIL MATH 165 SECTION 2.8 RELATED RATES PRACTICE PROBLEMA train is moving to the left (left on the x-axis) along a curved track. The direction its headlight points is described by the functiony = ( x 2 ) + 1 . Joe is standing at the point
Iowa State - M - 166
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Iowa State - M - 165
MIKE ANCTIL MATH 165 SECTION 2.8 RELATED RATES PRACTICE PROBLEMA train is moving to the left (left on the x-axis) along a curved track. The direction its headlight points is described by the functiony = ( x - 2 ) + 1 . Joe is standing at the poin
Iowa State - M - 166
Iowa State University Math 166 Instructor:Midterm Exam February 26, 2004 TA:AnswersSection:Instructions: Answer each question completely. Show all work. No credit for mere answers with no work shown. Show the steps of calculations. State the r
Iowa State - M - 165
Iowa State - M - 166
MIKE ANCTIL THE UNIT CIRCLEThe Unit Circle
Iowa State - M - 166
MIKE ANCTIL x MATH 166: GRAPHS OF y = eGraphs of y = exy = e x and y = ln x can be confusing functions, especially when evaluating limits of them. Alwaysremember that while it frequently works out to be this simple, a limit is not just pluggin
Iowa State - M - 166
MATH 166 IN FOXTROT5/22/05Paige says: &quot;RELAX. IT WAS How much of a discount did she receive? We can use Calculus II to find out!1 3k k =1OFF.&quot;continued on next page.MATH 166 IN FOXTROT1 3k k =1is a geometric series. But it is no
Iowa State - M - 166
ANCTIL, MICHAEL MATH 166 FINAL REVIEW WORKSHEET Topics to Know: I. Applications of the Integral a) Area of a plane region b) Volume of a solid of revolution c) Length of a plane curve d) Area of a surface of revolution e) Work done by a variable forc
Iowa State - M - 166
Objectives for Math 1661Applications of the IntegralSet up and evaluate integrals to calculate 1. Area of a plane region 2. Volume of a solid of revolution 3. Length of a plane curve 4. Area of a surface of revolution 5. Work done by a variable
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 .2Step 1: GraphStep 2: Find the intercepts This is done by setting the equations equal to eac
Iowa State - M - 166
Correction to the solution to #10: The last two rows of the posted solution should be as follows: 27 27 36 3 9 3 27 3 = + = 9 2 6 2 2 21
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