Last modified 26 December 2000 by

Math 222, Sections 365 and 367

  1. News
  2. Introduction
  3. Course Requirements
  4. Final Exam
  5. Contact Information
  6. Syllabus
  7. Exam and Quiz Results
  8. Friendly Advice
  9. Extra Stuff



Mathematics 222 is second-semester calculus. Our textbook is Calculus: Early Transcendentals, 3rd Edition, by James Stewart. The main topics to be covered are: Professor Oh will lecture three days a week on new material, and I will lead discussion on the other two days. In discussion sections, students will be expected to ask questions on homework and on concepts covered in lecture. Often the answers to these questions will come from other students. My job is to encourage participation and keep us on track. In short, "discussion" will be a discussion.

Course Requirements

Your grade will be based on the following four components: Attendance will not be taken in lecture or discussion, but all students will have to attend Thursday discussions in order to take the quizzes. Also, your homework will not be graded (unless you ask me to look at your work specifically). The weekly quizzes are intended to ensure that you are doing your work. They will require you to do problems similar to those on the homework, but under time pressure. So it is recommended that you practice by doing all of the homework problems (and even some extra). The exam questions will also be similar to those in the book, except that there might be one or two more challenging problems on each exam.

On the exams and quizzes, students are allowed to use any non-graphing calculator; on the other hand, use of a graphing calculator during an exam or quiz will be considered cheating, and will be dealt with severely. Students should familiarize themselves with the university's academic misconduct policies.

The first midterm exam will be given on 27 September 2000, 5:30-7:00 PM, in 1310 Ingraham Hall (next to Van Vleck). Students who do not have any conflict with this time must take this exam. Students who do have a conflict must arrange with me to attend the make-up exam, which is on 27 September 2000, from 12:00-1:00, in B130 Van Vleck. (This is our normal lecture room, but note the times carefully!) The exam will cover sections 7.1-7.4 and 7.8-7.9. No graphing calculators or "cheat sheets" are allowed. There will be no quiz on the next day, 28 September.

The second midterm exam will be given on 6 November 2000, 5:30-7:00 PM, in 165 Bascom Hall (next to Van Vleck). The exam will cover all course material since our previous exam, up to and including section 10.3. No graphing calculators or "cheat sheets" are allowed. I will be hosting a review session on 5 November, 4:00-5:30 in B113 Van Vleck.

For information on the final exam, see the next section.

Final Exam

The final exam will be given on 20 December 2000, 5:05-7:05 PM, in room 125 of the "old" Biochemistry Building (near where Campus Dr. splits off University Ave.). The exam will be cumulative, but weighted heavily toward the material covered after the second midterm exam. No graphing calculators are permitted. Each student is allowed to bring in a "cheat sheet" with formulas, examples, etc. It should be a single, 8.5 x 11 (or smaller) piece of paper; you are allowed to use both sides. That's a lot of room!

I will be holding two review sessions for the final exam. The first will be Saturday 16 December, 2:45-4:45 PM, B113 Van Vleck. In it I hope to wrap up our discussion of the course material. That is, I envision it as an extra discussion section, in which students will ask questions from the recent homework, lectures, and reading. The second review session will be Monday 18 December, 7:25-9:25 PM, B113 Van Vleck. In it I will be happy to answer questions from Prof. Oh's review sheet, as well as any other questions you have encountered while studying. After this review session, I will take a walk down to the biochemistry buildings, to make sure I know where the final is. Students are welcome to accompany me on this remarkable journey of discovery.

Here is a schedule of review sessions being held by the various teaching assistants. You are welcome to attend any of them that you like. All times are PM.

Wed 13 Dec5:00-8:003331 SterlingJoaquin
Sat 16 Dec2:45-4:45B 113 VVJosh
Sat 16 Dec2:45-4:45B 321 VVRob
Sun 17 Dec2:45-4:45B 219 VVGeircovering series only
Mon 18 Dec12:25-2:25B 135 VVBret
Mon 18 Dec7:25-9:25B 113 VVJoshwalk to 125 Biochemistry afterward
Tue 19 Dec7:25-9:25B 219 VVGeir
Tue 19 Dec12:25-2:25B 135 VVBret
Tue 19 Dec5:05-7:05B 321 VVRob

Here are my recommendations for studying for the final:

Contact Information

Here's how to get in touch with the professor:
Yong-Geun Oh (pronounced "yong gun oh")
Office: Van Vleck Hall, Room 715
Office Phone: (608) 262-2883
Office Hours: Tuesday and Thursday, 2:30-3:30 PM, or by appointment
Here's how to get ahold of me:
Joshua Davis (pronounced "josh")
Office: Van Vleck Hall, Room 518
Office Phone: (608) 262-3860
Office Hours: Monday 9:55-10:45, Thursday 1:20-2:10, Friday 2:25-3:15, or by appointment

If you can't attend my office hours, then consult my weekly schedule and make an appointment with me. If you have a quick question, then try catching me after discussion or lecture; I usually sit in the back of the room at lecture.


15 Sep7.1-7.2integration by parts,
trigonometric functions
7.1: 1, 3, 4, 7, 13, 20, 29, 31, 38, 39, 62;
7.2: 2, 8, 17, 18, 29, 59-61
211 Sep7.3-7.4trigonometric substitutions,
partial fractions
7.3: 1, 4, 8, 10, 15, 21, 27, 35;
7.4: 2, 5, 9, 10, 17, 23, 26, 31, 52, 62
318 Sep7.8-7.9approximate integrals,
improper integrals
7.8: 1-4, 7, 15, 21, 29, 32;
7.9: 3, 11, 12, 17, 18, 23, 24, 41-43, 60, 66
425 Sep8.1-8.2review,
arc length
8.2: 4, 5, 9, 12, 27, 32, 36Midterm I
52 Oct15.1-15.2basics,
first order linear equations
15.1: 1, 3, 4, 5, 7, 11-14, 17-20, 33;
15.2: 1, 3, 7, 13, 17, 19, 23, 25, 33
69 Oct9.1, 9.3parametrized curves,
arc length
9.1: 1, 3, 4, 8, 11, 14, 26, 28;
9.3: 3, 5, 6, 13-16
716 Oct9.4-9.6polar coordinates,
conic sections
9.4: 13-17, 20, 24, 27, 30, 31, 34, 37, 43;
9.5: 1, 2, 5, 7, 10, 15, 19, 21, 23;
9.6: 1, 3, 6, 9, 13, 17, 20, 21, 25, 28, 31, 35
823 Oct10.1-10.2sequences,
10.1: 2, 4-6, 10, 15, 18;
10.2: 7, 8, 11, 14, 21, 30, 52, 54, 55, 58
930 Oct10.3-10.4integral test,
comparison test,
10.3: 1, 4, 7, 9, 14, 15, 21, 31, 32;
10.4: 1, 3, 6, 11, 14, 21, 27, 30, 37, 42
106 Nov10.6absolute convergence,
root test,
ratio test
10.6: 1, 3-5, 7, 8, 13, 17, 29 (just determine whether the series are absolutely convergent or not) Midterm II
1113 Nov10.8-10.10power series10.8: 3, 4, 7, 9, 12, 17, 22, 29, 33;
10.9: 1, 2, 5, 6, 11, 12, 21, 25, 27, 36
1220 Nov10.10-10.12Taylor series10.10: 2, 5, 7, 11, 17, 19, 33, 38, 53;
10.12: 9, 10, 13, 15, 17
1327 Nov11.1-11.3vectors,
dot products
11.1: 5, 7, 9, 10, 15, 17, 18, 22, 27;
11.2: 1, 3, 7, 9, 13, 14, 19, 24, 25, 29, 35;
11.3: 1-8, 12, 13, 19, 21, 25, 28, 53, 61, 63
144 Dec11.4-11.5cross products,
equations of planes
11.4: 1-7, 10, 12, 19, 21, 23, 27, 35;
11.5: 1, 3, 7, 11, 14, 15, 18, 20, 23, 27, 32, 35, 39
1511 Dec11.7motions in space,
11.7: 1-7, 10, 13, 23, 25, 28

Exam and Quiz Results

Grades are not assigned according to a strict 90-80-70-60 percentage scheme. In order to estimate how well you are doing in comparison to your classmates, you should inspect the following data. Here are the approximate mean (average) scores for our in-class quizzes so far:
114 Sep7.1, 7.28.4/10
221 Sep7.3, 7.48.5/10
35 Oct7.9, 8.26.5/10
412 Oct15.1, 15.27.7/10
519 Oct9.1, 9.36.8/10
626 Oct9.4, 9.57.1/10
72 Nov10.1, 10.28.0/10
89 Nov10.3, 10.45.6/10
916 Nov10.69.2/10
1030 Nov10.8 - 10.109.8/14
117 Dec10.12, 11.18.9/10
1214 Dec11.2 - 11.48.8/10

Here are the rough letter grade equivalents for scores (out of 100) on the two midterm exams:
Letter GradeScore Range, Exam 1Score Range, Exam 2

Friendly Advice

You have been given freedom over attendance and homework because it is recognized that you sometimes have other things to do - big papers, stressful days, dental appointments, etc. - and because you know your learning style and needs better than the teachers do. You will have to be self-motivated. Sometimes this is difficult for students in their first semester of college, who are accustomed to more structure. It can also be dangerous for students who have seen some of the material before, because the class suddenly becomes much harder when it starts covering new material.

Be careful not to fall behind. Attend class and do the homework. Have questions about homework problems and lecture material ready for discussion. Review each lecture before the next lecture arrives. Study with your fellow students. Visit me and the professor during our office hours. If you feel that you need additional help, you have a variety of options:

Most importantly, if you're not understanding the material, then get help from me, the professor, or your fellow students as soon as possible. In math courses each concept builds on earlier ones, so it's important to understand the concepts as we go along.

Extra Stuff

Section 7.3, Problem 35

I now see two ways to do this pesky lune problem. The first expresses the area as a difference of two integrals, and then computes these using trig substitutions. This is undoubtedly how the author intended us to do the problem. I went through this solution laboriously in the 11:00 discussion today. The second method of solution requires no calculus and is much faster and easier. (In the following explanation, "*" denotes multiplication and "^" denotes exponentiation. For example, 3*3 = 3^2 = 9. Also, "sqrt" denotes square root.)

Here is a sketch of the first method. Place the center of the circle of radius r at the origin in the x-y plane. Then the function y = sqrt(r^2 - x^2) describes the top edge of the lune. Now the center of the circle of radius R lies at (0, -sqrt(R^2 - r^2)). This means that the function y = sqrt(R^2 - x^2) - sqrt(R^2 - r^2) describes the bottom of the lune. To get the lune's area, just compute the integral, from -r to r, of sqrt(r^2 - x^2) - (sqrt(R^2 - x^2) - sqrt(R^2 - r^2)) dx. This requires substitutions like x = r*sin(theta) and x = R*sin(theta).

Here is the second method. Let P and Q be the two points where the two circles intersect. Let C be the center of the circle of radius R. Draw the line segments PQ, PC, and QC. You end up with an "ice cream cone" consisting of the upper half of the radius r circle and the isosceles triangle PQC. Since you know how to compute the area of semicircles and triangles, you can easily verify that the area of this ice cream cone is pi*(r^2)/2 + r*sqrt(R^2 - r^2).

To get the area of the lune, we just need to subtract out the region of the ice cream cone that lies inside the big circle of radius R. Call this region A. Note that if theta is the measure (in radians) of the angle PCQ (at the bottom of the cone), then the area of A is theta/2pi * pi(R^2). To complete the computation, we just need to find theta. Splitting the triangle into two congruent right triangles, we see that sin(theta/2) = r/R, so theta = 2arcsin(r/R). Thus the final answer is p*(r^2)/2 + r*sqrt(R^2 - r^2) - (R^2)*arcsin(r/R).

Section 10.8, Problem 33

We are asked to find the interval of convergence and an explicit formula for the series f(x) = 1 + 2x + x2 + 2x3 + x4 + 2x5 + .... To be more precise, the nth term in the series is cnxn, where cn is 1 for even n and 2 for odd n. It is possible to write down a formula for cn; in fact,

cn = (3 - (-1)n)/2.

But this formula does us little good in answering the problem. (Try it yourself; the ratio test fails.) The fact that the coefficients cn alternate between 1 and 2 makes this series difficult to analyze using our standard techniques. So we'll use an unusual approach. We will begin by analyzing the odd-numbered partial sums

s1, s3, s5, ..., s2n-1, ....

We'll show that this sequence of odd-numbered partial sums converges. Then we will analyze the even-numbered partial sums

s0, s2, s4, ..., s2n, ...,

and show that they converge to the same thing. Then we can conclude that the entire sequence of partial sums converges to this one thing. (You should stop to convince yourself of this fact. It requires you to understand well what it means for a sequence to converge.) Thus the series will converge. In the process of figuring this out, we'll also compute the interval of convergence.

So let's get started. First, we write down an odd-numbered partial sum s2n-1 and factor it:

s2n-1 = 1 + 2x + x2 + 2x3 + ... + x2n-2 + 2x2n-1

= (1 + 2x)(1 + x2 + x4 + x6 + ... + x2n-2).

Now (1 + 2x) is simple enough, but we need to understand the second factor, (1 + x2 + ... + x2n-2). We're going to need a bit more algebra. You've probably seen the following fact before; if not, then multiply it out yourself:

(1 - x)(1 + x + x2 + x3 + ... + xn-1) = 1 - xn.

In this equation we replace x with x2 to obtain the following identity:

(1 - x2)(1 + x2 + x4 + x6 + ... + x2n-2) = 1 - x2n.

Dividing both sides by (1 - x2), we obtain

1 + x2 + x4 + x6 + ... + x2n-2 = (1 - x2n)/(1 - x2).

The left-hand side of this equation is exactly what we were looking for. Combining this equation with our first equation, we get

s2n-1 = (1 + 2x)(1 - x2n)/(1 - x2).

So we have a nice formula for the odd-numbered partial sum s2n-1. Now the series converges if and only if the sequence of partial sums converges. For the sequence of partial sums to converge, we need at least the sequence of odd-numbered partial sums to converge. So when does it converge?

In our formula for s2n-1, n occurs only in the term x2n. So for our sequence of odd-numbered partial sums to converge as n goes to infinity, we just need x2n to converge as n goes to infinity. Thus we need |x| < 1. In fact, when |x| < 1, x2n goes to 0 as n goes to infinity. Thus, assuming that -1 < x < 1, we have that s2n-1 goes to (1 + 2x)/(1 - x2) as n goes to infinity.

Now let's analyze the even-numbered partial sums. We know that

s2n = s2n-1 + a2n = s2n-1 + x2n.

As before, if -1 < x < 1, then x2n goes to 0 as n goes to infinity. Therefore, as n goes to infinity, s2n goes to s2n-1. This means that the sequence of even-numbered partial sums converges to the same thing as the sequence of odd-numbered partial sums, which was (1 + 2x)/(1 - x2). Therefore the series converges to (1 + 2x)/(1 - x2) when -1 < x < 1.

A careful student will remember to check the endpoints -1 and 1. It is easy to check that the series diverges at both of these endpoints. (It diverges to negative infinity at -1, and to positive infinity at 1.) Our final answer is that the series converges to (1 + 2x)/(1 - x2) on the interval (-1, 1).

Section 10.9, Problem 25

We want to compute the integral, from 0 to 1/5, of 1/(1 + x4), with respect to x, using a power series. First, from Problem 21 we already know that the antiderivative of 1/(1 + x4) for -1 < x < 1 is

C + sum(0 to infinity) (-1)nx4n+1/(4n + 1).

The definite integral we desire equals the antiderivative evaluated at 1/5, minus the antiderivative evaluated at 0. Fortunately, evaluation at 0 kills all terms (except C) in the power series, so the definite integral equals

sum(0 to infinity) (-1)n1/54n+1/(4n + 1).

This series converges to the exact value of the definite integral. We have been asked to approximate the value to six decimal places. I take this to mean "approximate the value, so that the first six decimal places in your answer are definitely correct". Regardless of what it means, I'm just going to approximate it to within 10-7 and consider that a good answer. So, anyway, how many terms of the series do we need to add up, and how do we know that this many makes the error small enough?

The key thing to notice now is that the series in question is alternating. This suggests that we should use the Alternating Series Remainder Theorem from Section 10.5. This theorem says that if you have an alternating series whose terms decrease in magnitude to zero, and you truncate the series somewhere (say, after the nth term, bn), then your error will be smaller than the first term you ignored (bn+1).

Our series satisfies the hypotheses of the theorem. Thus, if we just find a term bn+1 in the series that is smaller than 10-7, then we can cut off the series after bn, and our error in doing so is guaranteed to be smaller than bn+1, which is smaller than 10-7. Applying this idea to the series in question, we start computing the first few terms, which go (ignoring minus signs)

(1/5)1/1, (1/5)5/5, (1/5)9/9, ...

The third term is already less than 10-7, so we can take as our approximation the sum of the first two terms, which is 1/5 + 1/56.