Department of Mathematical Sciences
Northern Illinois University
MATH 229 Homepage

## MATH 229 H, Fall 2003

### 1:00-1:50 M T W Th F, DU 310

Professor: John Beachy, Watson 355, 753-6753

Office Hours: 11:00-11:50 M W F, or by appointment

Teaching Assistant: Claudine Myers, DuSable 370

Office Hours: 2:00-3:00 M W

Assignments | Quiz Solutions | Review | Class notes | Syllabus | List of homework problems

Handouts (in printable form):
Student Information Sheet for my section | Homework problems

## ASSIGNMENTS

Current assignments for 229 H:
```    DATE            DUE

Friday, 8/29     NO QUIZ
Homework #1: 1.1 #40,53,62  1.2 #12  1.3 #37
Homework #2: 2.1 #5,8  2.2 #6,22,33

Friday, 9/5      Quiz #1: Emphasis on 2.3, but covering everything
we have done so far

Friday, 9/12     Quiz #2: Section 2.4 #1 - 10 (bring a calculator)
Section 2.5

Homework #3: 2.4 #29, 31

Friday, 9/19     Quiz #3: Sections 2.6, 3.1, 3.2

Friday, 9/26     EXAM I  Covering sections 1.1-1.3, 2.1-2.6, 3.1-3.4
Note: no calculators

Friday, 10/3     Quiz #4: Sections 3.5, 3.6
Homework #4: Appendix D #51, 56, 57, 74, 83

```

## MY SECTION

QUIZZES: You should expect a 20 minute quiz each Friday (20 points). I do not give makeup quizzes-if you have a valid excuse for missing a quiz you should check with me.

HONORS: The honors part of the course (Math 297) will use these notes: The Calculus of Polynomials. We will usually cover honors material on Tuesdays.

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 in this course either

(1) because of a weak background in algebra or
(2) because they fall behind.
I can help with your algebra, but you are responsible for keeping up, by attending every class and doing all of the homework problems.
The WEB site Understanding Mathematics: a study guide has a good discussion about studying and learning mathematics. There is also a list of resources on the WEB.

## DEPARTMENT SYLLABUS

CALCULUS I (4) A first course in calculus.
PRQ: A level placement on the Math Placement Exam or MATH 155 (with at least a C)

TEXT: Calculus, 5th Ed., by James Stewart, McMaster University

COURSE OBJECTIVES: You are expected to acquire not only computational facility with the topics introduced, but also a basic understanding of the concepts and theory of calculus. Mathematics is a truly universal language; we want you to become more fluent in reading, writing, and using this precise language.

```    3 one-hour examinations     300 points
Final examination           200 points
Homework/quizzes            100 points
```

FINAL EXAM: The comprehensive departmental final exam is scheduled as a mass exam on Friday, December 12, 9:00-9:50 A.M.

COURSE WITHDRAWAL: The last day to withdraw is Friday, October 17.

CALCULATORS: A graphing calculator is required for the course. with capabilities similar to those of the TI-83 calculator. Graphing calculators will not be permitted on the final exam.

SYLLABUS: The course will cover Chapters 1-5, §6.1, and §8.7.

```       Tentative lecture schedule
Monday    Tuesday   Wednesday  Thursday   Friday           M Tu  W Th  F
1.1-1.4     2.1        2.2     Recitation  2.3        AUG 25 26 27 28 29
Holiday    H1.1        2.3        "        2.4        SEP  1  2  3  4  5
2.4       H1.2        2.5        "        2.6             8  9 10 11 12
3.1       H1.3        3.2        "        3.3            15 16 17 18 19
3.3       H1.4        3.4      Review    EXAM 1          22 23 24 25 26
3.5       H1.5        3.6     Recitation  3.7        OCT 29 30  1  2  3
3.8       H1.6        3.9        "        3.10            6  7  8  9 10
4.1      H Exam       4.2        "        4.3            13 14 15 16 17
4.4       H2.1        4.5      Review    EXAM 2          20 21 22 23 24
4.6       H2.2        4.7     Recitation  4.7            27 28 29 30 31
4.9       H2.3        4.10       "        5.1        NOV  3  4  5  6  7
5.2       H2.4        5.2        "        5.3            10 11 12 13 14
5.3       H2.5        5.4      Review    EXAM 3          16 17 18 19 20
5.5        5.5      Thanksgiving ------- holiday         24 25 26 27 28
6.1      H Exam       8.7      Review     8.7        DEC  1  2  3  4  5
FINAL EXAM   Tuesday, 4-5:50 PM                   8  9 10 11 12
```

ADVICE: The key to success in this course is to work as many problems as you can. The general rule in college is to spend at least two hours preparing for each class period. Above all, you can't afford to fall behind, since it is almost impossible to make up even a few days of lost time.

## 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.
Note: A * indicates additional problems assigned in 229H.

Week   Section Page       Problems

```8/25   1.1   p.22   2,7-9,19,21,24,26,29,35,38,40,48,53,62
1.2   p.35   2,3,6,10,12
1.3   p.46   4,10,12,17,18,24,27,32,36,37
8/25   2.1   p.69   1,3-5,8
2.2   p.80   5,6,12,14,15,22,23,25,33
2.3   p.89   1,4-6,10,12,13,15,21,31,37,46
9/1   *2.3   p.90   ***16,17,18,24,26,29,33,43,44,47,52,53,59
2.4  p.100   2,3,7,15
9/8   *2.4  p.101   ***29,31,37
2.5  p.110   3,4,6,11,12,15,16,20,33,38,39,41,43
2.6  p.119   7,8,12,15,17,18
9/15   3.1  p.132   3,4,7,8,13,19-22,25
*3.2  p.142   4-7,17,24,31    ***39,40,43,44,45
3.3  p.154   4,6,12,14,17,20-22,26,27,32,49,54,56,57,58,61,63,64
9/22  *3.3  p.156   ***65-75,80-88
3.4  p.166   1,2,7,8,11,14,18,20,25,26
EXAM 1

9/29  AppD  p.A32   1,2,7,8,13-15,17-21,23,24,29,30,35,36,74,78,81
3.5  p.174   2,7,8,10,12,13,17,23,30,33,35,36,38,40,41,43,46
3.6  p.181   4,6,7,12,17,20,24,27,28,34,36,37,39,40,45,52,53
3.7  p.188   1,7,14,16,20,23,26,27,31,38,40
10/6   3.8  p.195   1,4,5,13,18,19,23,26,30,35,38,44,51,53
3.9  p.202   1-3,5,7,11,13,17,18,23,24,25,29,31
3.10 p.210   5,6,9,15,17,24,34,39,41,42
10/13  4.1  p.230   3,7-10,14,16,18,25,30,31,34,39,41,47,48,54,56
4.2  p.238   2,6,12,17,19,23,24
4.3  p.247   2,8,11,14,16-18,22,24,25,28,32,34,38,50
10/20  4.4  p.260   4,7-10,13,16,20,32,39,40,47,48,52,54
4.5  p.270   3,5,7,13,14,25,26,31,37
EXAM 2

10/27  4.6  p.277   1,3,7,11,12,20,25
4.7  p.283   2,6,9,12,16,19,20,22,28
4.7  p.284   29,34,35,38,51,52,58
11/3   4.9  p.298   1,5,6,10,11,27,30,35
4.10  p.305   1,2,4,10,12,13,17,22,24,27,31,36,53,55,71
5.1  p.324   1,3,5,11,13,17-20
11/10  5.2  p.336   1,2,5,11,18,20,22-25
5.2  p.338   30,35,37,43,47,48,51
5.3  p.347   1,4,6,8,10,19-21,25,26,28
11/17  5.3  p.348   30,31,33,34,45,46,53,54
5.4  p.356   1,2,5-13(odd),18-38 (even),43,47,48
EXAM 3

11/24  5.5  p.365   1-31(odd),37-51(odd)
THANKSGIVING
12/1   6.1  p.380   5-15(odd),21,22,25
8.7  p.563   7-13(odd),29,33
```

Quiz 1 | Quiz 3

## REVIEW

Review sheets (in pdf format):
Review sheet for Exam 1 | Exam 1 from 2/16/2001 | Solutions | Exam 1 from 9/22/2000 | Solutions
Review sheet for Exam 2 | Exam 2 from 10/27/2000
Review sheet for Exam 3 | Exam 3 from 12/1/2000

## CLASS NOTES

Worksheets (in pdf format): Optimization problems | Antiderivatives | Integrals (by substitution)

## 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 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. This allows to find maximum and minimum values for a function.

The indefinite integral is the opposite of the derivative. It allows us to solve the following problem: given a formula that tells the rate of growth at any point in time, try to reconstruct the original function.

The definite integral is an averaging process. For example, the something is moving at a constant rate, the distance it travels is just the rate multiplied by the time it takes. But what if the rate is variable? That is where the definite integral can be used. As another example, in physics the work done by a force is defined to be the magnitude of the force times the distance through which it acts (provided the force is constant). But even in stretching a spring the force is variable, and this simple formula can't be used. The definite integral of the force finds the average force times the distance, and that is why physicists usually define work in terms of an integral. Finally, the integral allows us to find the area under a curve because it can find an average height, multiplied by the width of the interval.

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 229 H:

```    DATE              DUE

```