First Reading Assignment

Since I am not sure when Blackboard is going to be available for the class (I procrastinated asking for the two sections to be merged into a single section), here is the reading assignment and preparation for classwork for Wednesday.

Reading Assignments:

Online Calculus Textbook: Read Sections 1.1 and 1.2 (link below). These sections emphasize the idea that variables (which represent physical quantities) can be related, as independent and dependent variables. We want to think about related quantities throughout this semester. This text specifically asks for the web-browser Firefox version 3.0 or later (this is to render formulas correctly).

• Section 1.1
• Section 1.2

Come prepared to class having prepared answers for Problems 3, 4, 6, 7, 10 from Section 1.2 Problems. We will discuss these problems but will not turn them in.

How Students Learn: This book prepared by the National Academy of Sciences is actually written for teachers to focus on making courses better suited for students to learn. The first 12 pages introduce three concepts that students should be aware of in their own learning. The link below goes to the first page, and then follow the links to read through page 12.

• How Students Learn, page 1 through page 12.

Why is math fun? Why is math hard?

Today in class, I tried to help break the ice and reduce some of the anxiety related to taking a university mathematics course (Math 231). I asked students for examples of why they might find mathematics fun and why they might find mathematics hard. Here are some of the responses.

Why fun?

• It's fun when you struggle with a concept and then it finally clicks and you understand.
• It's fun to see mathematics actually being applied to a real problem.
• It's fun when you are able to spot trends and make predictions based on data.
• It's fun why you really understand why instead of just the "required" steps.
• When you understand, it becomes easy.
• It can be a lot like a game or solving a puzzle.
• It's fun to develop things logically.

Why hard?

• Later material builds on earlier material, so missing something early is permanent hardship.
• It can be really hard when the teacher goes too fast.
• It can be really hard when the teacher is unclear, especially if they can't give alternate ways of thinking about an idea.
• It can be hard if the teaching style is very different from your learning style.
• There are so many formulas, it can be overwhelming to try to memorize them.
• The theorems, rules and definitions are full of little details.
• It can be difficult to understand the many conceptual ideas that interact.
• Learning related technology can be challenging.
• Only one answer, so you can't fake it.
• Very hard to cram for exams.
• It can be really hard to find a little (stupid) mistake when proofing your work.
• Lots of homework, and problems can take a lot of time.

I'd welcome more comments, including examples of when you found mathematics especially exciting or examples of how your relationship with mathematics soured. Feel free to post a comment.

A New Semester --- Two Classes

This semester I am teaching a mathematical models in biology (Math/Bio 342) as well as the first semester of Calculus with Functions (Math 231). So I expect to have entries for both of these courses showing up.

Values, Equations, and Theorems

There is some confusion about theorems. For example, consider the Mean Value Theorem: If f is continuous on [a,b] and differentiable on (a,b), then there is some value c∈(a,b) so that f'(c)=(f(b)-f(a))/(b-a).

Some students think that the ratio (f(b)-f(a))/(b-a) is the Mean Value Theorem. But it is not; it is just a value that is called the average rate of change of f between a and b. You can compute this value as long as both f(a) and f(b) exist. It has nothing to do with derivatives or continuity.

Other students think that the equation f'(c)=(f(b)-f(a))/(b-a). This is closer to the truth, but still is incorrect. First of all, what is c? Second, this statement may not be true. For example, suppose that f(x)=-1 if x<0 and f(x)=1 if x>0. Suppose that a=-2 and b=+2. Then the ratio (f(b)-f(a))/(b-a) is equal to 1/2. But f'(x)=0 everywhere except at x=0, where f'(0) does not exist.

Even closer is to say that f'(c)=(f(b)-f(a))/(b-a) for some c between a and b. This is actually the conclusion of the Mean Value Theorem. It requires the entire statement, particularly the statement that the equation is true for some c and that the value c must be between a and b.

But my example above provides an example where the conclusion of the Mean Value Theorem is false. That does not mean that the Mean Value Theorem itself is false. After all, it is a theorem, and that means that it has been proved to be true always. The part that is missing is the hypothesis for the theorem. The conclusion can only be guaranteed to be true using the theorem if the hypotheses are all satisfied. In this case, you must also check (or give a reason why) the function f is continuous and differentiable on the interval from a to b, including the endpoints for continuity.

Similarly, ∫_a^b f(x) dx/(b-a) computes the average value of a function f on an interval [a,b]. The value can be computed anytime the function is integrable over the interval [a,b]. The Mean Value Theorem for Integrals has nothing to do (in principle) with this calculation.

However, if f is continuous on [a,b], then
f(c)=∫_a^b f(x) dx/(b-a)
for some c∈(a,b). This entire statement comprises the Mean Value Theorem for Integrals. The hypothesis that must be verified to use the theorem is that f is continuous on [a,b]. The conclusion is that you are guaranteed that
f(c)=∫_a^b f(x) dx / (b-a)
for at least one value c between a and b.

Limit or Function Evaluation?

I've noticed that some students are perplexed about when they use a limit or function evaluation. I presume that the cause of this confusion is that students have learned that you evaluate a limit by plugging in a value. But this is only because nearly all functions that they work with are continuous.

You use a limit evaluation when you need to know what the value of the function should be by using information from the side of the point of interest. When using a limit, you must use limit notation: lim_x→c f(x). Then you use the appropriate rules of limits to evaluate (and hopefully, the function is continuous).

You use a function evaluation when you need to know the value of the function at an actual point. There is no limit involved, just function evaluation. You just use function notation, say f(c), and compute the value defined by the function.

For example, suppose you are calculating an instantaneous rate of change as the limit of an average rate of change. The average rate of change only makes sense when the interval of interest includes two points (endpoints of an interval). The instantaneous rate of change is found by seeing what the value of the average rate of change does when the two points move closer to each other, or more particularly, as the second point approaches the first point.

On the other hand, suppose that you know the derivative, which is itself a function. Then the instantaneous rate of change is calculated by function evaluation using the derivative function. (The limit was already used to create this new function.)

In particular, I have noticed this problem when dealing with finding extreme values of a function. When the interval of interest is an open interval, we are acting as though the domain does not include the endpoints. So, with this restricted domain, evaluation of the function is not possible (since the points are not in the domain). So we must use the information about the function immediately adjacent to the endpoints, and this is with a limit. In this context, the value found in the limit is not achieved at the endpoint, although it might be achieved somewhere else in the domain.

On the other hand, if the interval is a closed interval, then the endpoints are included in the restricted domain. If the function is continuous at these points, then evaluating the function directly is appropriate, since the point is in the domain.

Final remark, if the function is discontinuous at some point in the interval, you must also check the limits at that point for consideration when looking for extreme values.

Differential Equations Project Tips

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) Start with the proposed form of X(t).  Compute X'(t) and X''(t) based on that form.  Then use those calculations to discover when X'' + k/m X = 0.

(2) The question "mean physically about the mass on a spring" is not asking you to think about the mass (as in measurement) but is asking you to think about what the statement X(0)=1 means about the state of the mass at time t=0 and what the statement X'(0)=0 means about the state of the mass at time t=0.

(3) X(0) is a constant and has derivative of d/dt[X(0)] = 0.  Recall that dX/dt > 0 implies that X is increasing, dX/dt < 0 implies that X is stationary (instantaneous rate = 0)

(4) An arbitrary quantity A is proportional to some other quantity B if it is always that case that A = k B for some constant value k (the constant of proportionality).  Now interpret the statements to identify what pieces of the equation are proportional to what.

(5) Although you have studied e^x in precalculus, you will not use the logarithm at all in this work.  Instead, I just want you to consider some function that has the special property that exp' = exp.  (This sentence is analogous to sin'=cos and cos'=-sin.)  However, you do need to think about the chain rule: X(t) = A exp(rt)  (Since the argument is not simply t, you must use the chain rule.)  This problem is exactly analogous to Step 1.

(6) You will get something like

dX/dt = "formula involving X and a, b, and m"

X is increasing when "formula" > 0, decreasing when "formula" < 0, and stationary when "formula" = 0.  So use your skills with algebra (think sign analysis) to find conditions when these are the case.

(7) You need to understand the relationship between a rate of change and an actual change.  To understand this as well as possible, see the section we skipped in Chapter 3 (last section).  But what you essentially need is that we will follow the tangent line for the time increment Δt.  How much change is there when the rate of change and the duration of time are both known?

(8) The new version of Excel has some unanticipated differences from what I had when I wrote the project.  The labels are not assigned from a menu anymore.  Instead of the 3-step process that is described, you just click in the label field in the header section of Excel and type in the new label and then hit enter.

The calculations you see in the first few lines should exactly match your hand calculations in part (7).  

Do not print the spreadsheet (it takes WAY too many pages).  That is why I ask you to submit your spreadsheet on Blackboard as part of the project.

(9) I must receive a print out of the graph -- hand drawn figures are not acceptable.   Ideally, this entire project report would be typed (perhaps using Equation Editor for the equations), with the figures naturally fitting in.

(10) Make hypotheses and test your hypotheses.

Exponential Project Tips

As some questions are asked more regularly, I thought I'd provide some general discussion here.

(1) exp is the name of the function, just as sin and cos are names of functions. From calculus, you learn that sin'=cos and cos'=-sin.  This step shows that exp_b' = ln b * exp_b. (That is, it leaves the function alone except for a constant multiple. (But be careful where the chain rule is needed!)

(2) You do not need to use the limit definition (epsilons and deltas).  Instead, for perhaps the easiest solution, you should think about how to finish the statement:

lim b^x = lim [(b^x-1)/x ... ]

That is, if you start with (b^x-1)/x, what do you do to that expression to leave only b^x.  Then use elementary limit rules to compute your resulting limit.

(3) One method is to use the method of substitution for limits (change of variables) and then use an identity for the function so that the result of Step 2 is applied --- this method mimics what is done to show that sin x is continuous everywhere.  A second method is to use a general theorem that makes continuity an obvious conclusion of the results from Step 1.

(4) You must start with a statement like:

ln (1/b) = lim_{x → 0} [(1/b)^x - 1]/x

There are two easy approaches: (1) Find a common denominator to rewrite this as a simple fraction before continuing or (2) Think of (1/b)^x as b to some appropriate power and then use a limit substitution.

(5) Since you do not know the derivative of ln x, it is incorrect to use the Mean Value Theorem applied to the logarithm.  Instead, you should apply the MVT to the function exp_b(x) on an interval so that b^a and b^b are incredibly easy and where it is clear which value is larger (so that you know if the average rate of change is positive or negative).  You may use the fact that b^x is positive for all values of x.

(6) The function fb(x) is a linear function. You should write it in slope-intercept form (e.g., mx+b).

(7) Do not attempt to solve the equation fb(x) = exp_b(x).  There is one obvious solution from the definition: x=0.  But the formulas themselves do not explain where there would not be more solutions.  Instead, you should define a function (perhaps g) so that

g(x) = exp_b(x) - fb(x).

You know that g(0) = 0.  You need to show that g(x)>0 for all x ≠ 0.  My hint suggested Rolle's theorem, but I have since found that the Mean Value Theorem helps even more.  Use the Mean Value Theorem to show that for x>0, the average rate of change between 0 and x must be positive.  What about x<0? x="0?">

(8) You may not use a limit form of the type b^∞. You may take a limit of the function fb(x) because that is of a form we know how to work with. Then you should use the result of (7) to conclude what the limit of exp_b(x) must be.

(9) and (10) put all of the previous steps together to perform analysis similar to Sections 4.2 and 4.4 to understand the graph.