# College Math Teaching

## March 5, 2014

### Comparing calculus exams…..and university students

Filed under: academia, calculus, pedagogy — Tags: — collegemathteaching @ 2:04 am

Some people tried to argue with me about calculus; they seemed to think that calculus at one institution is the same as at another one.

Hardly.

Not only can courses vary in terms of topic and difficulty, but so can exams…and the difference might be very subtle to those who are unfamiliar with giving and grading exams.

Here is one example: suppose you want to examine the students on the Mean Value Theorem. How might you do this?

1. State the Mean Value Theorem (yes, the bad students usually can’t even do this).

2. State and prove the Mean Value Theorem (prove using what?)

3. Let $0 < x < y < 1$. Show that there exists a $c$ between $x$ and $y$ so that $y^2 - x^2 = 2c(y-x)$.

4. Use the Mean Value Theorem to show that $|cos(x) - cos(y)| \leq |x - y |$

5. Show that for all real $x, y, |cos(x) - cos(y) | \leq | x-y|$

Not only do these questions vary in difficulty, they may or may not have been covered directly in class prior to the exam; that makes a big difference.

We are doing a job search. We have someone who is interviewing; he currently teaches at a school whose student population has a median ACT that is about 3 points higher than ours. BUT his institution is “technical majors only”; they don’t have much (any) of a humanities, communication or education program. So, if you compared their calculus student ACT to our “engineering/science calculus” ACT, the difference shrinks considerably, if it remains at all. But our department does teach the “business calculus”, “baby stats” and “math for poets” courses…and he will NOT be used to that type of student.

## July 12, 2013

### Just for fun: NO PROFESSORS!

Filed under: calculus, media — Tags: , , — collegemathteaching @ 9:55 pm

This photo appeared in this College Misery post.
So, I ask you the following questions (professors are disqualified from answering):

1. What is the math on the top line “supposed” to be about?
2. What transcription error is there on the board?
3. Why are the “equals signs”, as they appear, nonsense?

Bonus question (more subtle):
if one erases the first equals sign, and move the “work” in the “vector like” object off to the side where it belongs, why is the resulting equality still (technically) incorrect?

## December 13, 2012

### Domains and Anti Derivatives (Indefinite Integration)

Grading student exams sometimes inspires me to revisit elementary topics. For example, I recently spoke about some unusual (but mostly correct) integration techniques used by students on a final exam.

I’ll recap (and adjust the example slightly): on a recent exam, a student encountered $\int \frac{2}{1-x^2} dx$. I had expected the student to use the usual partial fractions expansion to obtain $\int \frac{1}{1+x} dx + \int \frac{1}{1-x} dx = ln|1+x| - ln|1-x| + C$ which is valid when $x \ne \pm 1$. I admit to being a bad professor and not being picky about domains.

But one student noticed the $1 - x^2$ in the denominator of the fraction and so used the trig substitution $x = sin(\theta), dx = cos(\theta) d\theta$ which leads to the following integral: $\int \frac{2}{cos(\theta)} d\theta = 2ln|sec(\theta) + tan(\theta)| + C$ which leads to $2ln|\frac{1}{\sqrt{1-x^2}} + \frac{x}{\sqrt{1-x^2}}| + C = 2ln|\frac{1+x}{\sqrt{1-x^2}}| = ln|1+x| - ln|1-x| + C$ for $x \in (-1,1)$. Note that, strictly speaking, the “final answer” is really defined for all $x \ne \pm 1$ though the equalities do not hold outside of the domain for $x$ used in the original trig substitution.

And yes, I was a bad professor; I gave full credit to this answer even though we “lost domain” during the string of equalities.

But that got me to wondering: is there a trig substitution that works for $|x| > 1$? Answer: of course:

$\int \frac{2}{1-x^2} dx = -\int \frac{2}{x^2 -1} dx$. Now use $x = sec(\theta), dx = sec(\theta) tan(\theta) d\theta$ which leads to $-2\int csc(\theta) d\theta = 2ln|csc(\theta) + cot(\theta)| + C = 2ln|\frac{x}{\sqrt{x^2-1}} + \frac{1}{\sqrt{x^2 -1}}| + C$ which leads us to our ultimate solution for $|x| > 1$

So, if one REALLY wanted to use trig substituions for this problem, one could and do it in a way to cover the entire domain.

But…as our existence and uniqueness theorems imply, once we get a candidate for an anti-derivative that “works” or the domain, it really doesn’t matter if we did “illegal” steps to get it; we need only show that it is an anti derivative and is valid for the entire domain for the integrand.

Now if one wants a more detailed discussion on domain issues for anti-derivatives, I can recommend the article The Importance of Being Continuous by D. J. JEFFREY which appeared in Mathematics Magazine, Vol 67, pp 294 – 300. (reprint can be found here, scroll down a bit; this mathematician has written quite a bit!). Note: I can recommend this little paper as it talks about the domains of the anti derivatives themselves and not just the domains assumed in doing the calculations along the way or the domains of validity of the substitutions. Note: integral tables and computer algebra systems don’t always give the anti derivative with the “largest” possible domain. One has to watch for that.

## December 7, 2012

### “Unusual” Student Integral Tricks….that are correct!

Filed under: academia, calculus, editorial, elementary mathematics, integrals, integration by substitution, pedagogy — Tags: — collegemathteaching @ 2:52 am

To make this list, the student has to do the integral correctly, but choose a painfully inefficient way of doing it.

On today’s final exam alone (from the most innocent to the most unusual and inefficient….)

1. $\int x\sqrt{x+1} dx =$
Ok, there are two standard methods. The first (and easiest) is to do the change of variable $u = x+1$ which transforms this to $\int (u-1)\sqrt{u} du$ which is very easy to do. The second method: parts, let $u = x, dv = \sqrt{x+1}$ etc. It is an algebraic exercise to see that one gets the same answer either way, though the answers look different at first.

One answer that I saw: $u = \sqrt{x+1}, u^2-1 = x, 2udu = dx$ which leads to $\int 2(u^2 -1)u^2 du$ which of course is doable. So this isn’t that far off of the easiest path, hence this entry only gets an “honorable mention”.

2. $\int \frac{1}{9-x^2} dx$. of course, I thought that I was testing “partial fractions” which leads to an answer of $\frac{1}{6}(ln|3+x| - ln|3-x|)+C$. Fair enough. But what did one of my students do? Well, this looked like trig substitution to him so: $x = 3sin(t), dx = 3cos(t)$ so this was transformed to $\int \frac{3cos(t)}{9cos(t)}dt = \frac{1}{3}\int sec(t) dt = \frac{1}{3}ln|sec(t) + tan(t)|+C$ which transforms back to $\frac{1}{3}ln|\frac{3}{\sqrt{9-x^2}} + \frac{x}{\sqrt{9-x^2}}| = \frac{1}{3}(ln|3+x| - \frac{1}{2}(ln|3-x|+ln|3+x|))+C$ which is, of course, the correct answer.

Yes, I know that there are domain issues with the trig substitution (that is, the integral exists for all values of $x \ne \pm 3$ but I wasn’t being that picky. Besides, this trig substitution is really setting $t = arcsin{\frac{x}{3}}$ and we are really just choosing a convenient “branch” (meaning: viewing the domain “mod (-1,1)”) of the $arcsin(x)$ function.

3. $\int \frac{arcsin(x))^2}{\sqrt{1-x^2}} dx$. Easy, you say? Why not let $u = arcsin(x), du = \frac{1}{\sqrt{1-x^2}},$ etc. Yes, most did it that way. But then we had a couple do the following: $x = sin(t), dx = cos(t)dt, arcsin(x) = t$ which lead to $\int t^2 dt = \frac{t^3}{3} + C$ which transforms to $\frac{(arcsin(x))^3}{3} + C$ which is the correct answer. 🙂

Well, I tell my classes that “this isn’t a gymnastics meet; there are no “degree of difficulty points”” but some insist on trying to entertain me anyway. 🙂

## March 6, 2010

### Implicit Differentiation and Differential Equations

In my “business calculus class”, we were studying implicit differentiation.
Find $dy/dx$ if $x/y - x^2 = 1$
I showed them three ways to do the problem, all of which yield different looking answers:

$x - x^2y = y$ Differentiate both sides: $1 - 2xy -(x^2)dy/dx = dy/dx$ which yields:
$dy/dx = (1-2xy)/(1 + x^2)$

Method 2: do directly:
$(y - (x)dy/dx)/y^2 -2x= 0$  This leads to $dy/dx = (y - 2xy^2)/x$

Of course that looks different; but we can always solve for $y$ and do it directly:
$x/y = 1+ x^2$ which yields $y = x/(x^2+1)$ which yields the easy solution:
$dy/dx = (1 - x^2)/((x^2 + 1)^2$

Now one can check that all three solutions are in fact equal on the domain $y \neq 0$

But here is the question that came to mind:  in the first two methods we had two different two variable equations: $(y - 2xy^2)/x^2 = dy/dx = (1-2xy)/(1 + x^2)$

So what does this mean for $y$?  Is it uniquely determined?

Answer:  of course it is: what we really have is $f(x,y)=g(x,y) = dy/dx$ whose solution IS uniquely determined on an open rectangle so long as $f$ and $g$ are continuous and  $\partial f/y$ and $\partial g/y$ are continuous also. 🙂

But I didn’t realize that until I took my morning swim. 🙂

This is the value of talking to a friend who knows what he/she is doing: I was reminded that $f(x,y)=g(x,y) = dy/dx$ means that $f = dy/dx$ and $g = dy/dx$ indeed have unique solutions that have the same slope at a common point, but with just this there is no reason that the solutions coincide over a whole interval (at least without some other condition).

So we have something to think about and to explore; I don’t like being wrong but I love having stuff to think about!

Now, of course, we have “different” differential equations with the same solution; yes, there is a theory for that.  I’ve got some reading to do!