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Course: M 307, Fall 2009
School: Iowa State
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307 FALL MATH 2002 QUIZ 2 Name: 1. (5 points) Compute Ax if 1 3 A= 0 2 1 4 0 2 1 -1 1 2 1 1 x= -2 5 6 1 2. (5 points) What is the rank of the matrix A = 2 0 0 2 2 4 ? 4 3. (10 points) Let T (x) = a) Find the matrix of T . 3x1 - 5x2 . -x1 + 7x2 b) Find the inverse transformation. c) Use the result of b) to solve T (x) = y where y = 1 . 1

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307 FALL MATH 2002 QUIZ 2 Name: 1. (5 points) Compute Ax if 1 3 A= 0 2 1 4 0 2 1 -1 1 2 1 1 x= -2 5 6 1 2. (5 points) What is the rank of the matrix A = 2 0 0 2 2 4 ? 4 3. (10 points) Let T (x) = a) Find the matrix of T . 3x1 - 5x2 . -x1 + 7x2 b) Find the inverse tran...

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Iowa State - M - 267
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Iowa State - M - 267
Assignment 1 answers Page 28, #10: Interval of existence is11 6< t < .Page 28, #14: Interval of existence is - < t < . Page 39, #4: y = tan (ex + C) Page 39, #8: y = C(x - 1)ex Page 39, #16: y = - ln (2 - ex ) for x < ln 2 Page 39, #36: T 56.18
Iowa State - M - 267
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Iowa State - M - 267
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MATH 267 FALL 2004 HOMEWORK ASSIGNMENTSWEEK 1August 23: Page 6, #1-4, page 16, #1-5,15-17August 24: page 28, #3,6,11,13,14August 26: page 39, #2-5,9-11,13-16,35b,36August 27: page 39, #26,27WEEK 2August 30: Page 50, #2,3,6,10,12; Page 134,
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MATH 166 REFERENCE SHEETINTEGRATION TRICKS 1. Find a direct formula. Consider ln, inv. trig. Formulas, f ( x) d 2nd Fnd. Thrm. Of Calc./Corollary f (t ) dt dx a 2. Try Substitution 3. Expand. (1 + ex 2) dx = (1 + 2e x + e 2 x )dx4. Separ
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ANCTIL, MICHAEL MATH 166 FINAL REVIEW W ORKSHEET 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 for
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Name:Math 166 Calculus II Dec 13, 2004 FinalAnswers1. Let u = 1 + x2 , then du = 2xdx. Now du -1 x=a = lim |x=1 2 a 2u x=1 u -1 1 = lim = a 2(1 + x2 ) 4 1 = lim a 2x=a 1x 1 dx = lim 2 )2 a 2 (1 + xa 12x dx (1 + x2 )22. 0.125 =125 10
Iowa State - M - 166
Iowa State University Math 166 Instructor:Final Exam May 6, 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 reasons tha
Iowa State - M - 165
MIKE ANCTIL MATH 165: LIMITS PRACTICE1.limx2x2 - 4 x-2Direct substitution gives0 , which is an indeterminate. 0But the numerator can be factored:limx2( x - 2 )( x + 2 )x-2Simplifying the limit gives a form that is solvable by dire
Iowa State - M - 166
Improper IntegralsIndeterminate Forms00 - f (x )dx = llim l f (x )dx - - Taylor polynomial for f about x = ab b+af (x )dx = liml + a f (x )dxl0 0 0p n (x ) = f (a ) + f ' (a )(x - a ) +Taylor series for k =0f(k
Iowa State - M - 165
MIKE ANCTIL MATH 165, SOLVING MOTION PROBLEMSOne of the important applications of calculus is solving motion problems, that is, problems involving position, velocity, and acceleration of an object. There is a general procedure that can normally be
Iowa State - M - 165
MIKE ANCTIL MATH 165, MAKING THE PRODUCT RULE WITH IMPLICIT DIFFERENTIATION EASIERM AKING THE PRODUCT RULE WITH IMPLICIT DIFFERENTIATION EASIER When the Product Rule is combined with Implicit Differentiation, it often causes headaches. Derivative p
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MIKE ANCTIL MATH 165, CURVE SKETCHINGCurve sketching is an important skill to know for quick analysis of an equation. By looking at a few different aspects of the equation, even obscure functions can be sketched with some accuracy. There are six ge
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MIKE ANCTIL MATH 165 SECTION 2.22.4 PRACTICE: EARLY DERIVATIVE RULESFind the derivative: 1)y = 2 x sin x x 2 cos xThis problem uses the product rule twice (once for each product), the power rule, and definitions of trigonometric derivatives. Pr
Iowa State - M - 166
Wangsness Math 166Practice for Test 4 SolutionsName:In order to obtain full credit on this exam, you must show all necessary work when solving each problem. In each problem that requires graphing an equation, you must indicate the vertices, di
Iowa State - M - 166
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 _ A
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
0o 30 o 45o 60 o 90 o 120 o 135o 150 o 180 o 210 o 225o 240 o 270 o 300 o 315o 330 o 360 o 0o 30 o 45o 60 o 90 o 120 o 135o 150 o 180 o 210 o 225o 240 o 270 o 300 o 315o 330 o 360 oJefferson Davis Learning Center, Sandra Petersonrr
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
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Iowa State - M - 166
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MATH 166 IN FOXTROT5/22/05Paige says: "RELAX. IT WAS How much of a discount did she receive? We can use Calculus II to find out!1 3k k =1OFF."continued on next page.MATH 166 IN FOXTROT1 3k k =1is a geometric series. But it is no
Iowa State - M - 166
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Iowa State - M - 166
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Iowa State - M - 166
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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
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Iowa State - M - 166
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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 & 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) > 0 for x < -1, f ( x) < 0 for x > -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 < 0, f(x) continuous. If x > 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