Math 221, Fall 2011 (Washington)

  • The syllabus
  • Homework assignments
  • Section 9.5, number 12: answer = .9 Pi
  • Math Department Tutoring room
  • Math Success Tutoring program
  • Past Math 221 Final Exams (you can also find old exams from other math classes at this link)

    Midterm 1: Thursday, October 6 (on Sections 8.1 through 9.6)

  • A sample exam
  • Another sample exam

    Some answers for Sample 1a:

  • 1a: (1/20)(2+5 ln x)^4 + C, 1b: (1/2)x^2 ln x +2x ln x -(1/4)x^2 -2x + C
  • 1c (without the limits of integration): (-1/3)x e^(-3x) - (1/9)e^(-3x) + C; with the limits of integration: 1/9
  • 2a: three values are (2/3)+pi/6, (2/3)-pi/6, (2/3)+pi/2, 2b: 100 tan(pi/3)= 100 sqrt(3), 3a: 2500
  • 4: Use quotient rule: [3x^2 cos(x^3)cos(x) + sin(x^3)sin(x)]/cos^2 (x)
  • 5a: H'(6)=-sin(2 pi/3) (pi/12)= -pi sqrt(3)/24, 5b: Max value is 5 (since the biggest cos can be is 1).
  • 6. (a) .680 (b) 21/20 (c) .0012

    Some answers for Sample 1b:

  • 1a: (1/6)x sin(6x) + (1/36)cos(6x) + C
  • 1b: without limits of integration: (1/7)(x^2+3x)^7 + C; with limits of integration: (1/7)(10^7 - 4^7)
  • 1c: (1/5)x^5 ln x - (1/25)x^5 + C, 1d: (1/3)ln(x^3+7) + C, 2a: 3x^2 cos(x^5) - 5x^7 sin(x^5)
  • 2b: (-3sin(3x)sin(4x)-4cos(3x)cos(4x))/(sin(4x))^2, 2c: (2x+5)cos(x^2+5x+3), 3: sin(t)= 12/13, cos(t)= 5/13, tan(t)= 12/5
  • 4a: (sqrt(1)+2sqrt(9)+2sqrt(65)+sqrt(217))(1)
  • 4c: the actual value of the integral is smaller than the trapezoidal approximation, since the trapezoids lie above the curve.
  • 5a: Longest: 25, Shortest: 51, 5b: decreasing at 3Pi/26 hours per week (the derivative is -3Pi/26)

    Midterm 2: Thursday, November 3

  • A sample exam
  • Another sample exam
  • Another sample exam

    Some answers for Review Exam 2:

  • 1. y = -1/(t^2 + 3t - 1/4), 2. y = t + 1 + 4e^(-t), 4. y' = ky(1-y) (or y' = k(y - y^2) ), 5. y' = -k(y-T) , k > 0

    Some answers for Second Exam Review:

  • 1. y = ln(t^2 + 3t +1), 2. y = (1/2)t - (1/4) + (9/4)e^(4-2t), 4. y' = -ky, 5. y' = .06y - 400

    Answers for the third Sample Exam:

  • 1. y'=ky(500000-y), 2. y'=-ky, 3. y=(1/5)te^(-t) - (1/25)e^(-t) + (76/25)e^(-6t), 4. y' = .06y - 3000,
  • 5. y = (2t^3 + 12t^2 + 36)^(1/2), 6. Starting at y(0)=6: going downwards asymptotic to 0. Starting at y(0)=10: going upwards towards infinity
  • 7. A goes asymptotically to 500. B goes asymptotically to 0. The difference is 500 - 0 = 500, 8. 4/7, .12, .0154

    Answers to even problems:

  • 11.5: 16: x^2 - (1/2)x^4 + (1/3)x^6 - (1/4)x^4 + ...
  • 11.5: 18: 1 - (1/2!)x^4 + (1/4!)x^8 - (1/6!)x^12 + ...
  • 12.1: 8(a): 1/2, 8(b): 1/4, 8(c): 1/100, 8(d): 0
  • 12.2: 32: ln(1.1)/ln(2)
  • 12.4: 28: .0228
  • 12.4: 30: the longer route over city streets
  • 12.5: skip 12, 13, 14, 17

    Midterm 3: Thursday, December 8

  • Sample problems and Answers
  • An old exam 3
    For Midterm 3 and for the final: Know the Taylor series for e^x, sin x, cos x, 1/(1-x), ln(1+x). Know the formula a/(1-r) for the sum of a geometric series.
  • Answers to Old exam 3
  • 1. k=3/7
  • 2. a. 1/2 b. 8/3
  • 3. x - x^2/2 + x^3/3
  • 4. 2
  • 5a. 15/2 b. sin(pi)=0
  • 6. e^(-1.2)
  • 7.a .5328 b. .0228
  • 8. 13 e^(-4)

    Final Exam: Thursday, December 15, 1:30-3:30pm

    Rooms for Final Exam: Lee: Francis Scott Key 0103; Amanda: Armory 0110 (NOTE: These are not the lecture room in the Armory)


  • Final from Fall 2005
  • Answers to Fall 2005 and Spring 2006 exams
  • Final from Spring 2006 and Solutions to a few problems
  • Other final exams are available on Testbank