Department of Mathematical Sciences, Northern Illinois University


MATH 211
Fall 2002
Section C


Professor: John Beachy, Watson 355, 753-6753
Office Hours: 1:00-1:50 M T F (in Watson 355), or by appointment

Teaching Assistant: Lingju Kong, DuSable 374, 753-1724
Office Hours: MWF 1:00 - 2:00 pm



FINAL EXAM

The final exam is on Friday, December 13, 8:00-9:50 AM.
Our section will take the exam in DuSable 148.

Review session: Wednesday, 12/10, 5:00-7:00 PM, in DU 204

If you need copies of the review sheets, you can print them off from these links.
Exam 1   |   Exam 2   |   Exam 3   |   Final

There is also a review page with sample exams.

Optional review problems (Hand in by Tuesday, 4:30 pm)
2.3 #29; 2.4 #19; 2.5 #55, 33; 2.7 #77; 2.8 #19; 3.3 #21; 3.5 #19; 4.1 #29; 4.3 #21
These are worth 2 pts each, and the total will replace your lowest quiz or homework grade.



Assignments | Review | Class notes

Syllabus | Schedule | Homework | WEB resources

MATH 211 Homepage (for all sections)


SCHEDULE OF LECTURES


LECTURES          11:00-11:50  MWF    DU 204   DR. JOHN BEACHY

HELP SESSIONS  C1 10:00-10:50  T      DU 302   LINGJU KONG
               C2 11:00-11:50  T      DU 328       "
               C3 11:00-11:50  TH     FR 238       "

         MONDAY      WEDNESDAY    FRIDAY         S  M Tu  W Th  F  S
Week of                                            
 8/26    1.1,1.2     1.3,1.4      1.5       AUG 25 26 27 28 29 30 31
 9/2    Labor Day    2.1          2.2       SEP  1  2  3  4  5  6  7
 9/9     2.3         2.4          2.4            8  9 10 11 12 13 14
 9/16    2.5         ###          Exam I        15 16 17 18 19 20 21
 9/23    2.6         2.7          2.7           22 23 24 25 26 27 28
 9/30    2.8         2.9          3.1       OCT 29 30  1  2  3  4  5
 10/7    3.2         3.2          3.3            6  7  8  9 10 11 12
 10/14   3.3         ###          Exam II       13 14 15 16 17 18 19
 10/21   3.4         3.5          3.5           20 21 22 23 24 25 26
 10/28   4.1         4.2          4.2           27 28 29 30 31  1  2
 11/4    4.3         4.4          5.1       NOV  3  4  5  6  7  8  9
 11/11   5.1         ###          Exam III      10 11 12 13 14 15 16
 11/18   5.1         5.2          5.2           17 18 19 20 21 22 23
 11/25   5.3         5.4         Holiday        24 25 26 27 28 29 30
 12/2    5.4         5.5          5.5       DEC  1  2  3  4  5  6  7

      FINAL EXAM       Friday, December 13, 8:00-9:50 AM

### These days may be used for review.


ASSIGNMENTS

Current assignments for Section C:
    DATE              DUE

Tuesday, 12/10    Hand in review problems (optional)
                  2.3 #29; 2.4 #19; 2.5 #55, #33; 2.7 #77; 
                  2.8 #19; 3.3 #21; 3.5 #19; 4.1 #29; 4.3 #21

Friday, 12/6      No quiz
                  Hmwk 6    Hand in Section 5.5 #17,26,34,38,40 (p 412)

Wednesday, 11/27  Hmwk 5    Hand in p 385 #38,60, p 397 #14, p372 #44

Friday, 11/22     Quiz 11   Sections 5.1, 5.2
                            In 5.2, only #1 - 14 

Monday, 11/18     Quiz 10   Section 5.1

Friday, 11/15     EXAM III  Sections 3.4-3.5, 4.1-4.4

Friday, 11/8      Quiz 9    Sections 4.1, 4.2, 4.3

Friday, 11/1      Quiz 8    Section 3.5 
                  Hmwk 4    Hand in Section 3.5, #32, 52 (p 262)

Friday, 10/25     Quiz 7    Sections 3.4, 3.5 (thru #22 only)


REVIEW

FALL 2002

EXAM III:   Solutions   |   Exam III   |   Review Sheet   |   Review Problems (Section 3.5) with Answers

                    Quiz 9 solutions   |   Quiz 8 solutions   |   Homework 4 solutions   |   Quiz 7 solutions

EXAM II:   Solutions   |   Exam II   |   Review Sheet   |   Quiz 6 solutions   |   Quiz 5 solutions   |   Quiz 4 solutions

EXAM I:   Solutions   |   Exam I   |   Review sheet   |   Homework 2 solutions   |   Quiz 3 solutions



OLD EXAMS

When these older exams were given, we used a different textbook, so quite a bit has changed!

In past years I've put some tests on the WEB. (You can see that I don't teach this course very often.) It is difficult to write mathematics for WEB pages, so the tests are in Adobe Acrobat (.pdf) format, which is supposed to be supported by the campus computer labs.

Please use the sample tests carefully. Coverage may change, and the emphasis of individual tests may change from semester to semester. Don't start studying by using the previous tests. You should study the book, class notes, homework problems, and quizzes. Only then are you ready to test yourself on an old exam.

Spring 1999:
Review for Test I | Test I Solutions | Test II Solutions
Review for Test III | Test III Solutions

Fall 1998: Final Exam

Fall 1997: Final Exam

Spring 1997: Test I, 2/7/97 | Test II, 3/7/97 | Test III, 4/11/97 | Final Exam, 5/3/97

Fall 1996: Tests and quizzes



SYLLABUS (SECTION C)

Prerequisite: The prerequisite is the equivalent of a good Algebra II course in high school, although Math 211 does not require trig. On the other hand, Math 211 does depend on a knowledge of logarithms and exponential functions. If you have not had any algebra for several semesters, or if you had problems in earlier algebra courses, you will need to do extra work to relearn the algebra while you are learning calculus.

Calculators: You will not be allowed to use a calculator on tests or quizzes.
I have seen too many examples of students who can solve a problem using a calculator, but do not understand the basic principles behind the solution. For example, using a calculator to compute logs and exponentials does not really help to understand the log and exponential functions that we need to use.
I do encourage you to check your homework on a graphing calculator.

Hour tests: I expect to follow the schedule for all sections, with hour exams on September 20, October 18, and November 15. Makeup tests will be allowed only if you make arrangements with me before the scheduled time of the test.

Grading: Final grades will be based on a total of 600 points. I intend to use a curve close to
90% (A), 80% (B), 65% (C), 55% (D).

Quizzes: You should expect a 20 minute quiz each Friday, worth 20 points, unless an hour exam is scheduled. The quizzes are designed to check that you are doing (and understanding) the assigned homework problems. If necessary, additional quizzes may be given in the recitation classes. I do not give makeup quizzes--if you have a valid excuse for missing a quiz you should check with me.

Recommended homework problems: The homework problems are the key to your success in the course. They represent the minimum--do as many additional problems as possible. You should keep all of your homework problems in a notebook so that you can study from it for quizzes, hour tests, and the final exam. You will need to hand in only selected problems from the list of recommended problems, since the quizzes will be designed test whether or not you have been doing the assignments.
You may find it useful to buy the Study Guide for the book, which has solutions to some of the exercises.

General advice: The key ideas of calculus are not the main barrier in the course--they can be explained rather intuitively. In my experience, students have trouble either (1) because of a weak background in algebra or (2) because they fall behind. We can help you review algebra, but you are responsible for keeping up, by attending every class and doing all of the homework problems. You should expect to spend an average of two hours preparing for each hour in class.

Tutoring: The Math Assistance Center is staffed by teaching assistants. It is located in DU 326, and is open during the day, Monday through Friday. In addition, the ACCESS office runs walk-in tutoring centers in Grant North, Grant South, and Lincoln residence halls.

Handouts: The original handout for Section C can be printed from your WEB browser. You can also print the homework assignments, and the general syllabus for all sections.


HOMEWORK PROBLEMS

Recommended problems: These homework problems are the key to your success in the course. They represent the minimum--do as many additional problems as possible. You should keep all of your homework problems in a notebook so that you can study from them for the final exam.
You may find it useful to buy the Study Guide for students. It has solutions to every sixth problem.

Handouts: List of homework problems   |   Homework 1 (in PDF format)   |   Homework 2   |   Homework 3   |   Homework 5

Section
     Page Problems

1.1   13  5, 7, 11, 12, 15, 16
1.2   26  1, 3, 5, 13, 15, 17, 43, 45
1.3   37  1, 3, 5, 7, 8, 9, 11, 13, 15, 16, 19, 21, 27, 31, 33, 35, 37, 53
1.4   51  5, 9, 12, 13, 17, 18, 21, 22, 23, 25, 27, 29, 30, 33-36, 49, 50, 
          53, 57, 65, 72
1.5   71  1, 3, 4, 9, 11, 13, 15, 17, 19-25, 27, 29, 31, 33, 37, 43, 49, 51,
          53, 59, 67-72, 73, 75, 79, 81
2.1  106  1, 4-10, 12, 13-18, 19-22, 25-28
2.2  113  1, 3, 4, 7, 8, 11-14, 17-20, 23-26
2.3  121  1, 2, 5, 7, 8, 11, 13, 14, 17, 18, 20, 21, 23, 29, 31, 32, 34
2.4  137  1, 2, 7, 8, 13, 14, 17, 18, 20, 21, 23, 24, 29-32
2.5  148  1-4, 7-10, 13-16, 19, 20, 23, 24, 27-30, 33-38, 41-44, 47-50, 53-60, 
          65-68, 76-78
               Exam I
2.6  155  1-4, 7, 8, 9, 13, 18, 21
2.7  163  3, 4, 7-10, 13, 14, 19, 20, 23-26, 37, 38, 77-79, 87, 89, 90, 95, 99
2.8  172  1, 2, 5, 6, 9-14, 17-20, 23-28, 33, 34, 37, 38, 39, 41, 42, 47, 48,
          49, 51, 56, 57, 58, 61, 62, 65, 66, 69, 73, 74, 79, 81
2.9  176  1, 3, 6, 7, 8, 9, 10, 15, 19, 20, 27, 28, 35, 36, 41, 45, 56, 48
3.1  199  1-4, 7, 8, 10, 13, 14, 17-26, 57, 59, 61, 62
3.2  214  1, 2, 5, 6, 9-18, 21, 22, 28, 31, 32, 35, 36, 39-42, 85, 87, 88, 89
3.3  232  1-14, 17-20, 25-27, 29, 30-34, 37, 38, 40, 41, 43-49, 54, 57, 60, 61
               Exam II
3.4  245  1-8, 11, 12, 15, 16-20, 28, 31-35, 50-54, 59-62, 65, 69, 70, 83-86, 
          97-100, 103, 104, 107, 108  
3.5  261  7-10, 13, 14, 17-22, 25, 27-32, 37, 38, 39, 40
4.1  298  1-4, 7-32, 35-40, 51-54, 57-68, 71, 72
4.2  314  1, 2, 5, 6, 9, 10, 13, 14, 17, 18, 23, 24, 35-42, 43-64, 67, 69, 71,
          72, 77, 78, 79, 81-86, 88, 90, 93, 94, 98
4.3  327  1-8, 11-14, 18, 22, 24, 25, 33, 39, 41
4.4  340  1-4, 8, 17, 18, 26, 28, 29, 35, 36
5.1  371  1-32, 34, 35, 39, 40, 43-46, 53, 54, 56, 61, 62
               Exam III
5.2  385  3-8, 9, 10, 12, 13, 15, 16, 41-46, 47-50, 53, 54, 55, 57, 62, 66
5.3  396  1, 2, 3-8, 13-18, 23-26, 29, 30, 31, 33, 35, 36, 37
5.4  406  1, 2, 5, 6, 8, 9, 13, 14, 17, 18, 23, 25, 29, 30
5.5  412  1-18, 21-26, 31-34, 35-38, 41-44, 55, 61-65, 69, 
               FINAL EXAM


CLASS NOTES

Introductory lecture

The study of calculus is divided into two parts: (1) derivatives and (2) integrals. The derivative of a function measures its rate of growth. The integral of a function (over a given interval) involves finding an average value of the function (over the interval).

The original motivation for studying derivatives came from physics, at a time when people were trying to understand things in motion. In this context, the derivative measures velocity, which we can think of as an instantaneous rate of change. The acceleration of an object measures how fast its velocity is changing, and so this can also be measured by a derivative (the second derivative).

When working with the graph of a function, the derivative can be very useful. For a straight line, the derivative is just the slope of the line, and a positive slope corresponds to a graph that moves up (from left to right), while a negative slope corresponds to a graph that moves down. For more general functions, just knowing the sign of the derivative is useful, since the graph goes up when the derivative is positive, and down when it is negative.

The very simple idea of measuring growth as either positive or negative makes it possible to find maximum and minimum values for a function. If the growth of a function changes from positive to negative, then you know that it has a (relative) maximum value when it changes. If the growth changes from negative to positive, then the function has a (relative) minimum value when it changes. So if we can find a function to describe a relationship in real life, we can predict where the maximum and minimum values will occur. Some of the applications will involve writing down a function to describe the profit for a company, and then deciding on the strategy that will maximize the profit.

The formula   rate x time = distance   should be familiar to you. This works only when the rate stays constant. Using calculus, we can deal with situations in which the rate is changing, since the concept of an integral allows us to find an average rate and use that to find the total distance traveled. Similarly, if we have a function that describes the rate of output of some process, we can find the total output even if the rate is variable, since we can calculate the average rate of output and multiply this average value by the total time. and multiply this average value by the total time.

There is a longer essay online, if you have the time to read it.

Understanding the product rule

To take the derivative of the sum of two functions, we use the rule that the derivative of a sum is the sum of the derivatives. Unfortunately, the rule for taking the derivative of a product of two functions is more complicated. A simple example for which we already know the answer gives a good illustration of the difficulties.

Suppose that f(x) = x2 and g(x) = x3. Then the individual derivatives are f '(x) = 2x and g '(x) = 3x2. We know that the derivative of the product f(x)g(x) = x2x3 = x5 should be 5x4, but it is difficult to see how to combine f '(x) and g '(x) to get this answer. It turns out that we also have to use f(x) and g(x) in the formula.

One approach to understanding the formula for differentiating products comes from working with tangent lines. Remember that the equation of the line through the point (a,b) with slope m is

y = m(x-a) + b.

At the point (a,f(a)) on the curve y = f(x), we have the y-coordinate b = f(a), and the slope of the tangent line at this point is given by m = f '(a). This allows us to calculate the equation of the line tangent to the curve y = f(x) at x=a as

y = f '(a)(x-a) + f(a).

In the same way, the line tangent to the curve y = g(x) at x=a is

y = g '(a)(x-a) + g(a).

To get the formula for the tangent line to y = f(x)g(x) at x=a, we could try multiplying the two formulas. We would get the following quadratic expression.

f '(a)g '(a)(x-a)2 + ( f '(a)g(a) + f(a)g '(a) )(x-a) + f(a)g(a)

Because the tangent line has to be linear, this doesn't solve the problem, unless we can change it in some way. The tangent line is intended to be the line that gives the best approximation to the curve, near the point (a,f(a)). When x is very close to a, the factor (x-a) is very small. In comparison, the factor (x-a)2 is even smaller; in fact, it is very much smaller. For example, if x-a = .1, then (x-a)2 = .01; if x-a = .01, then (x-a)2 = .0001; if x-a = .001, then (x-a)2 = .000001. Because the tangent line is just an approximation to the curve, it turns out that we can ignore the quadratic term. Notice that the part we had to ignore contains the term f '(a)g '(a), which would have given us a "nice" formula.

Conclusion: the tangent line to y = f(x)g(x) at x=a is

y = ( f '(a)g(a) + f(a)g '(a) )(x-a) + f(a)g(a).

The general formula is best given using x instead of a.


The Product Rule

The product rule for taking the derivative of the product of two functions f(x) and g(x) can be stated this way:

d/dx ( f(x)g(x) ) = f '(x)g(x) + f(x)g '(x).


Besides remembering the product rule in the "visual" form given above, it is good to remember it in words, as an algorithm.

The derivative of the product of two functions is equal to the derivative of the first times the second, plus the first times the derivative of the second.

Example 1. Going back to the first example, we can now compute

d/dx ( x2 x3 ) = (2x)(x3) + (x2)(3x2) = 2x4 + 3x4 = 5x4.

Our formula gives the expected answer, but has to include f(x) and g(x) along with f '(x) and g '(x).

Example 2. It is interesting to see the formula for the derivative of the product of three functions. It clearly shows the pattern in which we add up terms where we differentiate one function at a time.

d/dx ( f(x)g(x)h(x) ) = f '(x)g(x)h(x) + f(x)g '(x)h(x) + f(x)g(x)h '(x)


Understanding the chain rule

The rule for taking the derivative of the composition of two functions is also complicated. Again, a simple example for which we already know the answer gives a good illustration of the difficulties.

Suppose that f(y) = y3 and u(x) = x4. The formula for the composite function is f(u(x)) = (x4)3. The individual derivatives are f '(y) = 3y2 and u '(x) = 4x3. We know that the derivative of the composite function f(u(x)) = x12 should be 12x11. In this case we just need to substitute y=u(x) into f '(y) and multiply by u '(x). We can show this by looking at the tangent lines.

The line tangent to the curve y = u(x) at x=a is

y = u '(a)(x-a) + u(a).

At the point (b,f(b)) on the curve z = f(y), the slope of the tangent line at this point is given by f '(b). Thus the equation of the line tangent to the curve z = f(y) at y=b as

z = f '(b)(y-b) + f(b).

To compute the composite function f(u(x)), we substitute the formula for u(x) into the formula for f(y). If x=a, then we evaluate f(y) at y=u(a). This means that in the equation of the tangent line, we must take b=u(a). The equation becomes

z = f '(u(a))(y-u(a)) + f(u(a)).

To get the formula for the tangent line to z = f(u(x)) at x=a, we can substitute the formula

y = u '(a)(x-a) + u(a)

into the formula

z = f '(u(a))(y-u(a)) + f(u(a)).

This gives us

z = f '(u(a))(u '(a)(x-a) + u(a) -u(a)) + f(u(a))

or

z = f '(u(a))(u '(a)(x-a)) + f(u(a))

The derivative is the slope f '(u(a))(u '(a). The general formula is best given using x instead of a.


The Chain Rule

The chain rule for taking the derivative of the composite of two functions f(y) and u(x) can be stated this way:

d/dx ( f(u(x)) ) = f '(u(x)) u '(x).


Besides remembering the product rule in the "visual" form given above, it is good to remember it in words, as an algorithm.

The derivative of the composite of two functions is found by taking the derivative of the first function and substituting in the second, the multiplying by the derivative of the second.

Example. 1 Going back to the first example, we can now compute

d/dx ( (x4)3 ) = 3(x4)2(4x3) = (3x8)(4x3) = 12x11.

Our formula gives the expected answer. It is important to remember to multiply the term f '(u(x)) by u '(x).

Example. 2 To give another point of view, we can check the chain rule in a case where we can actually apply the product rule. If f(u(x)) = u(x)3, we can write

f(u(x)) = u(x)3 = u(x)u(x)u(x),

and so from the product rule we get

f '(u(x)) = u '(x)u(x)u(x) + u(x)u '(x)u(x) + u(x)u(x)u '(x).

Combining terms gives us the correct answer

f '(u(x)) = 3 u(x)2 u '(x).


Past assignments for Section C:     Homework 3

    DATE              DUE

Friday, 10/18     EXAM II   Sections 2.5-2.9, 3.1-3.3

Monday, 10/14     Hmwk 3    (p 232 #28)

Friday, 10/7      Quiz 6    Sections 3.1, 3.2

Friday, 10/4      Quiz 5    Sections 2.8, 2.9

Friday, 9/27      Quiz 4    Sections 2.6, 2.7

Friday, 9/20      EXAM I    Sections 1.1-1.5 and 2.1-2.5

Monday, 9/16      Homework 2

Friday, 9/13      Quiz 3    Sections 2.2, 2.3

Friday, 9/6       Quiz 2    Sections 1.5, 2.1

Friday, 8/29      Homework 1
                  Quiz 1 over Sections 1.1--1.4


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