Northern Illinois University

Department of Mathematical Sciences


MATH 211
Fall 2002
Section C


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

Teaching Assistant: April Galyardt, DuSable 370, 753-1146
Office Hours: TBA


Assignments | Quiz Solutions | 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   APRIL GALYARDT
               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   5.2         5.3          5.4           17 18 19 20 21 22 23
 11/25   5.4         5.5         Holiday        24 25 26 27 28 29 30
 12/2    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

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


REVIEW

Please note: these review sheets will be updated, since they are from two years ago, using a different textbook.

Spring 99:
Review sheet for Test I | Test I Solutions | Test II Solutions
Review sheet for Test III (in .pdf format) | Test III Solutions

I have put some previous tests on the WEB. 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.

Fall 98: Final Exam

Fall 97: Final Exam

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

Fall 96: Tests and quizzes

Tutoring Center: The Math Assistance Center is staffed by teaching assistants. It is located in DU326, and is open during the day, Monday through Friday.
In addition, the LARC provides tutoring during the evening, Sunday through Thursday. It is located on the first floor of Lincoln Hall.


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)

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


QUIZ SOLUTIONS

QUIZ 1, 1/15/99

1. (6 pts; p 14 #19) If f(x) = x2 +2x, find f(a+2) and simplify.

Solution: f(a+2) = (a+2)2 +2(a+2) = a2 + 4a + 4 + 2a + 4 = a2 + 6a + 8

2. (8 pts; p 25 #5) Graph the linear function f(x) = -2x + 3.

3. (6 pts) If f(x) = x2, find f(x+h)-f(x) / h and simplify.

Solution:

f(x+h)-f(x)   (x+h)2-x2   x2+2xh+h2-x2   h(2x+h)
----------- = -------- = ------------ = ------- = 2x+h
     h            h            h           h

QUIZ 2, 1/22/99

1. (5 pts; p 70 #22) Find the equation of the line through the point (-1,-1) parallel to the line 3x + y = 7.

Solution: To find the slope of the given line, solve for y, to get y = -3x + 7. In the the point-slope form of the line, y = m(x - x1) + y1, take m = -3, x1 = -1, and y1 = -1. The equation of the line is
y = -3(x + 1) - 1.

2. (5 pts; p 46 #65) Simplify this expression: (16x8)-3/4 =

Solution: (16x8)-3/4 = 16-3/4(x8)-3/4 = (161/4)-3x-6 = 2-3x-6 = 1 / (8x6)

3. (5 pts; p 46 #68) Simplify this expression: (25xy)3/2 / (x2y) =

Solution: (25xy)3/2 / (x2y) = (25)3/2x3/2y3/2 / x2y = (251/2)3x3/2-2y3/2-1 = 53x-1/2y1/2 = 125y1/2 / x1/2

4. (5 pts; p 56 #11) A rectangular box has no top. Write an equation expressing the fact that the surface area of the box has 65 square inches.

Solution: Use the variables x and y to give the dimensions of the base, and let h be the height of the box. The area of the base is xy, two sides have the area xh, and the other two sides have the area yh. The equation is
xy + 2xh + 2yh = 65.

Note: you can solve #2 and #3 in many different ways. As long as your steps are correct, you should get full credit. In #4 you can use any variables you like, but you must say which lengths correspond to which variables, since it is the base that only gets counted once when you figure out the total surface area.

QUIZ 3, 1/29/99

1. (5pts; p.77 #34) Write the equation of the line tangent to the graph of y = x3 at the point where x = -1. (You may use the fact that the tangent line to the graph of y = x3 at the point (x,y) has slope 3 x2.)

Solution: Use the point-slope form of the equation of a line, with y = m(x - x1) + y1.
We are given x1 = -1, and so the correspond point y1 is found from the equation y = x3, and we get y1 = (-1)3 = -1.
The slope m is found using the formula 3x2, so we get m = 3(-1)2 = 3.
The equation of the tangent line is y = 3(x + 1) - 1.

2. (5pts; p.85 #25) Use the power formula to find the slope of the curve y = x4 at x = 3.

Solution: For the function f(x) = x4, the power formula gives the derivative f'(x) = 4x3. Then the slope of the curve at x = 3 is given by the derivative at x = 3, so the solution is f'(3) = 4(3)3 = 108.

3. (5pts; p.85 #40) If f(x) = x-1/3, use the power formula to find the derivative f'(x).

Solution: If f(x) = xr, then the power formula states that f'(x) = rxr-1. For the function in this problem we have r = -1/3, and so the derivative is f'(x) = (-1/3)x-1/3 - 1 = (-1/3)x-4/3.

4. (5pts; p. 86, #55) For f(x) = x3, compute and simplify the expression (f(x+h)-f(x)) / h.

Solution:

f(x+h)-f(x)   (x+h)3-x3   
----------- = -------- 
     h            h     
  x3+3x2h+3xh2+h3-x3
= -----------------
       h           
  3x2h+3xh2+h3   h(3x2+3xh+h2)
= ----------- = -------------
       h             h
= 3x2+3xh+h2

QUIZ 4, 2/12/99

1. (5pts; p.108 #35) Find the derivative of the function f(x) = 4 / (x2+x)1/2.

Solution: Use the general power formula.
f(x) = 4 (x2+x)-1/2
f'(x) = 4(-1/2)(x2+x)-3/2(2x+1)

2. (5pts; p.109 #53) The tangent line to the curve y = 1/3 x3 - 4x2 + 18 x + 22 is parallel to the line 6x - 2y = 1 at two points on the curve. Find the two points.

Solution: The line is 2y = 6x - 1, or y = 3x - 1/2, so its slope is 3.
We need to solve the equation dy/dx = 3.
dy/dx = x2 - 8x2 + 18
x2 - 8x2 + 18 = 3
x2 - 8x2 + 15 = 0
(x-5)(x-3) = 0
x=5 or x=3
The corresponding points are (5,161/3) and (3,49).

3. (5pts; p.114 #24) Find d/dT ( 1 / (3T+1) ) |T=2.

Solution: d/dT (3T+1)-1 = (-1)(3T+1)-2(3) = -3 / (3T+1)2
Substituting x=2 gives -3 / 49.

4. (5pts; p. 123, #9) An analysis of the daily output of a factory assembly line shows that about 60t + t2 - 1/12 t3 units are produced after t hours of work. What is the rate of production (in units per hour) when t = 2?

Solution: The rate of production is given by the derivative 60 + 2t - 1/4 t2.
Substituting t=2 gives 60 + 4 - 1 = 63.

QUIZ 5, 2/19/99

1. (8 points) For f(x) = -1/9x3+x2+9x, find f '(x) and f ''(x). Use this information to find the relative maximum and relative minimum points of the graph of f(x) (the graph has one of each).

Solution: The first derivative is f '(x) = -1/3x2+2x+9

Set f '(x) = 0
-1/3x2+2x+9 = 0
x2-6x-27 = 0
(x-9)(x+3) = 0
x = 9 or x = 3

Now we need to use the second derivative, f ''(x) = -2/3 x + 2

f ''(9) = (-2/3)(9) + 2 = -6 + 2 = -4, so f(x) is concave down at x=9, which means that f(9) is a relative maximum
f ''(-3) = (-2/3)(-3) + 2 = 2 + 2 = 4, so f(x) is concave up at x=-3, which means that f(-3) is a relative minimum
The corresponding points are (9,81) and (-3,-15).

2. (12 points) In this question you will sketch the curve y = 2/x + 1/2 x + 1 = 2x-1 + 1/2 x + 1.

(a) Find the first derivative: y ' = -2x-2 + 1/2 = -2 / x2 + 1/2

(b) Solve y' = 0 to find the points at which the tangent line is horizontal.

-2 / x2 + 1/2 = 0
-2 / x2 = - 1/2
4 = x2
x = 2 or x = -2

(c) Find the second derivative y'' =

y '' = +4x-3 = 4 / x3

(d) Solve y'' = 0 to find the points of inflection (if there are any).

4 / x3 = 0 has no solution

(e) Use the sign of y'' to find the intervals on which the graph is concave up or concave down.

Since y '' = 4 / x3, y'' is positive when x is positive, and negative when x is negative. Therefore y is concave up when x is positive, and concave down when x is negative.

(f) Using the above information, sketch the curve on the given axes. You should actually plot at least 5 or 6 points, and be sure to include the asymptotes (if there are any).


QUIZ 7, 2/26/99

1. Solve for x: 23x = 4 . 25x
23x = 22 . 25x = 22+5x
Now the exponents must be equal.
3x = 2+5x
-2 = 2x
x = -1

2. Solve for x: ex2-2x = e8
x2-2x = 8
x2-2x - 8 = 0
(x-4)(x+2) = 0
x = 4 or x = -2

3. Find y' if y = (1+5ex)4:
y' = 4(1+5ex)3(5ex)

4. Find y' if y = x3 e2x:
y' = 3x2e2x + x3e2x(2)

5. Find y' if y = (e-3x) / (1 - 4x):
Write y = (e-3x)(1 - 4x)-1, and then
y' = (e-3x)(-3)(1 - 4x)-1 + (e-3x)(-1)(1 - 4x)-2(-4)


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 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.

--Still writing--

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:

    DATE              DUE


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