Good Calculus Textbook?

My department will be looking for a new calculus textbook over the next few months. We used to use Stewart, but there was some discontent, and we switched in part due to the high price. We've been using Briggs for about two years, and are very unhappy with it. So we want to switch again.

I asked a month and a half ago for a good discrete math text, and Josh suggested Discrete Math With Ducks. It looks fun, and was inexpensive compared to what we had been using. I'm excited that my request for suggestions panned out. (Thanks, Josh!) I'll have more to say about that next semester when I start teaching from it.

That was a decision I got to make on my own. The calculus textbook will be a group decision. The rest of the department will want a more conventional textbook than what I might want. I'm willing to work with whatever textbook we use, but I'm dreaming now of writing my own. (That will take a few years...)

What we didn't like in Briggs:

• The exercises often jumped too quickly to very hard problems
• There's nothing on centroids (until multivariable)

Hmm, I know there's more - I'll have to add to that list next week after our department meeting. I'd like to bring suggestions to the meeting, though. Have any of you used a calculus textbook that you love? Do any of you know of a complete textbook (for Calc I, II, and III, ie going through multivariable calculus) that's under $100?

My department doesn't seem interested in open source textbooks, and the two I used this semester weren't impressive enough for me to want to push it. I love the projects in Boelkins, but that only works if you want to teach through projects. The Guichard made some odd choices. I think any open source book will have more of its own personality than the commercial books. That could be fine, but I haven't seen one yet that will cover all the bases for us.

Ideas?

Centroid (Center of Mass)

This is a topic covered in Calculus II. The textbook explanation is inadequate, and I found nothing good online. So I wrote my own explanation. I now understand it better than I ever did before. (Not surprising, huh? If you're a student, this is an important principle of learning. After you think you understand something, try to write an explanation of it and see how much deeper your understanding can get.)

I'd love to improve this, so please let me know where it's unclear.


Imagine a thin sheet of metal cut in an artistic shape. Is there always a spot where you could hold it balanced on your finger? If there is, can we find that spot? We’ll assume the metal has uniform density. This allows us to treat area as equivalent to weight.

The 1-dimensional Case
To think about this, we first imagine a teeter-totter. We know that two people of the same weight must sit the same distance from the fulcrum (balance point) to balance. We also know that a heavier person must move inward if they want to balance with a lighter person. Experiments show that the weights times the distances from the fulcrum must be equal on the two sides for the teeter-totter to balance. (I wonder if there’s a thought experiment we could do that would convince us this must be true, without the actual experimental evidence.)

Example 1: I weigh 170 pounds and sit 5 feet from the center. My son weighs 75 pounds and sits in front of me, 4 feet from the center. Weights times distances = 170*5 + 75*4 = 1150 feet-pounds. We need someone who weighs 230 pounds to sit 5 feet from the center on the other side. We could write this as:  w₁*d₁ + w₂*d₂ = w₃*d₃

If we change our perspective to a number line below the teeter-totter, with 0 at the fulcrum, then the values on the left will be negative. We won’t have the same equality – we’ll have

 w₁*p₁ + w₂*p₂ = -(w₃*p₃), 

where each d (for distance, always positive) was replaced with a p (for position). This becomes

 w₁*p₁ + w₂*p₂ + w₃*p₃ = 0, 

given that the fulcrum is at 0. But suppose we don’t know where the fulcrum is? Let’s just put our 0 at the left end, and let the former proper place for the 0 - at the fulcrum - be f. Then the equation becomes

 w₁*(p₁ - f) + w₂*(p₂ - f) + w₃*(p₃ - f) = 0, 

or  w₁*p₁ + w₂*p₂ + w₃*p₃ =  w₁*f + w₂*f + w₃*f, 

or w₁*p₁ + w₂*p₂ + w₃*p₃ =  f(w₁ + w₂ + w₃). 

Let W = the sum of all the weights, then we have

f = (w₁*p₁ + w₂*p₂ + w₃*p₃)/W,

which of course extends from 3 weights and positions to n weights and positions.

On to 2 Dimensions
If we use areas instead of weights, we can look for the fulcrum of the x-values (written as an x with a bar over it) and the fulcrum of the y-values (written as a y with a bar over it). For a finite number of small areas, we would get (the same as above)

xbar = (a₁*x₁ + a₂*x₂ + ... + a_n*x_n)/A. 

If we imagine a shape formed by the area under a function f (where f has positive y values), between x=a and x=b, sliced into infinitely many vertical strips, with the area of each vertical strip given by height times width = f(x)^. Δ x, then taking the limit as Δ x goes to 0 gives us

For the y value, we need to notice that the vertical center of each vertical strip is at 1/2 *f(x), and we use this instead of x for the position. So we get

 or

Many times the area we're interested in will not be touching the x-axis, and so we need area between a top function, f(x), and a bottom function, g(x). The height of each slice will be (f (x)- g(x)), making the area  (f(x) - g(x))^. Δ x. The vertical center is now given by averaging f and g. We get:

 

 or

Now if only I could describe this bird shape with functions...

Top Ten Fun Math Books

I wrote about my favorite math books for the Nerdy Book Club site. It's up on their site now.

Have I left out anyone's favorite?

A *Math* Petition?! Yep.

Did you know that the U.S. Federal Government has a website called We the People, where you can post petitions?

This may be the first math-related petition I've ever 'signed'. 

Implement a Policy for Declassifying Discoveries by NSA Mathematicians

The NSA is the largest employer of mathematicians in the United States. Currently, the discoveries of those mathematicians in their official areas of research, being deemed potentially critical to national security, are indiscriminately classified for an indefinite period, with limited circumstances for declassification.
It is requested the White House press the NSA for an expiration policy for the classification status of non-applied discoveries and instituting an expiration for gag order patents in the interest of furthering American academia and industry advancement and in the interest of crediting the discoveries of our nation's talented NSA employees.

Sign here.

Calculus: Anti-derivatives and Area Under a Curve

The textbook we use (Briggs), and I think most of the textbooks I've used in the past (including Stewart and Thomas), introduce anti-derivatives before area under a curve. So they show students the notation for indefinite integrals before showing them the notation for definite integrals. I think this is a BIG mistake.

Here's what happens...

∫ f(x) dx (aka indefinite integral) means find all functions F(x) so that F'(x)=f(x). 

(Why does it use that funny symbol? Why does it have that dx part at the end? Hard to explain without referencing a connection that hasn't been made yet, isn't it?)

And then we start thinking about areas under curves and use a notation that's almost the same.

∫₁² f(x) dx (aka definite integral) means find the area which is between f(x) and the x-axis (area below the axis counts as negative), and between x=1 and x=2.

Seeing almost identical notation and names, we're going to assume that these two act the same in some ways. Students are going to expect anti-derivatives, even though it's area we're talking about here. So it's not much surprise when the Fundamental Theorem of Calculus tells us that to find area we can use anti-derivatives.

Wait! That should be a surprise. It's kind of amazing, isn't it? Derivatives give us slope. Why would going backwards in that process give us area?! Seems to me that's a big one we need to meditate on for a while.

This semester I knew I wanted to connect the new ideas with the position, velocity, and acceleration problems, so I introduced anti-derivatives first. And, I showed the indefinite integral symbol. Oops! I shouldn't have. If I had held off, I believe the meaning of the definite integral would have taken hold better in my student's minds.

Until this semester, I've followed the textbook pretty closely, so my way around this problem has been to introduce the 'Area Function' without using this notation. I found this idea/project in a book put out by the MAA.  I've revised it a lot over the years, but the original author, Charles Jones (of Grinnell College) still deserves credit for getting me started in this direction. (I wish I could figure out how to thank him personally, but he doesn't seem to be at Grinnell College these days, and google gives me lots of people with that name.)

I've put a pdf of the project here. If you'd like my Word file, just email me (mathanthologyeditor on gmail).

We've started that project, and it's going well enough, but I realized that if I hadn't introduced the indefinite integral, we'd be better off. Next semester I'll get that right.

Tomorrow we wrap up the project, and I clarify the implications of the Fundamental Theorem. Cool stuff!

Monk Climbing Mountain Puzzle

Have you seen this puzzle?

A monk climbs from the base of a mountain to its top on the one narrow path up and down, sleeps in a hut at the top, and then descends again to her monastery the next day. She leaves at about 6am on both days, and arrives around 6pm on both days. She stops for a break whenever she feels like it.

Will there be a time of day where she’s at the same spot on both days?

Factor Diagrams

A while back I reviewed You Can Count on Monsters, a delightful book showing monsters built from the prime factorizations of each number 2 through 100. (1 is in the book, but is sad, since it can't be made from primes.)

There are now lots of other takes on this idea. Brent, at The Math Less Traveled, made some gorgeous factor diagrams back in early October. When he posted them, many of his readers took them as inspiration to do more. One made a factor tango. Brent was then inspired to  improve on his own diagrams.  Here's a partial picture of what's he done:

He says he'll be making posters and t-shirts. I think this would make a great poster for a math classroom. Maybe I'll get copies for some of my colleagues.

Proving the Pythagorean Theorem

In a right triangle, where the lengths of the legs are given by a and b, and the length of the hypotenuse is given by c, a²+b²=c². 

We use this so much in math, I have no idea where I first saw it. And it's so simple that I never had trouble remembering it. (The quadratic formula, on the other hand, did not make it into my memory banks until after I had started teaching college. For the first few courses I taught, I had to have it written at the top of my notes.) So I've known and used the Pythagorean Theorem for longer than I can remember.

It comes up in beginning algebra, and for years I showed students how to use it to solve ridiculously artificial algebra problems, never once addressing the issue of proof. This seems terribly wrong to me now. Perhaps about 15 years ago, I realized I'd been 'teaching' this to students for about a decade without even knowing its proof. I tried to come up with a proof on my own and had no idea how to start. Since this was before google became a verb (or even a word), I had to search for a book that would show it. I eventually found it in a high school geometry textbook. Luckily it showed a visually simple proof that stuck with me. (There are hundreds of proofs, many of them hard to follow.)

One of the reasons Pythagoras is held in high esteem by mathematicians is his proof of this idea. It had been used long before Pythagoras and the Greeks, most famously by the Egyptians. Egyptian 'rope-pullers' surveyed the land and helped build the pyramids, using a taut circle of rope with 12 equally-spaced knots to create a 3-4-5 triangle (since 3²+4²=5² this is a right triangle, which is pretty important for building and surveying). But the first evidence we have that it was proven comes from Pythagoras. Ever since the Greeks, proof has been the basis of all mathematics. To do math without understanding why something is true really makes no sense.

Pam Sorooshian is a homeschooler who trusts kids' natural instinct for learning. So she unschooled her kids (who are now grown and doing very well). That means she never required them to learn something they weren't interested in, and never pushed her own interests on them. She has a story in Playing With Math* that really stuck with me. In a talk to other unschooling parents, she said:

Relax and let them develop conceptual understanding slowly, over time. Don't encourage them to memorize anything - the problem is that once people memorize a technique or a 'fact', they have the feeling that they 'know it' and they stop questioning it or wondering about it. Learning is stunted.

It took me decades to wonder about how we know that a²+b²=c². Now I feel that one of my main jobs as a math teacher is to get students to wonder. But my own math education left me with lots of 'knowledge' that has nothing to do with true understanding. (I wonder what else I have yet to question...) And beginning algebra students are still using textbooks that 'give' the Pythagorean Theorem with no justification. No wonder my Calc II students last year didn't know the difference between an example and a proof.

Just this morning I came across an even simpler proof of the Pythagorean Theorem than the one I have liked best over the past 10 to 15 years. I was amazed that I hadn't seen it before. (Maybe I did see it, but wasn't ready to appreciate it.)


My old favorite goes like this:

• Draw a square. 
• Put a dot on one side (not at the middle). 
• Put dots at the same place on each of the other 3 sides. 
• Connect them. 
• You now have a tilted square inside the bigger square, along with 4 triangles. At this point, you can proceed algebraically or visually.

Algebraic version:

• big square = small tilted square + 4 triangles
• (a+b)² = c² + 4*1/2*ab
• a²+2ab+b² = c² + 2ab
• a²+b² = c²



www.cut-the-knot.org/pythag

Visual version:

• Move the triangles around.
• What was c² is now a²+b².
• Also check out Vi Hart's video showing a paper-folding proof (with a bit of ripping). It's pretty similar to this one.

To me, that seemed as simple as it gets. Until I saw this:

from The Step to Rationality, by R. N. Shepard

This is an even more visual proof, although it might take a few geometric remarks to make it clear. In any right triangle, the two acute (less than 90 degrees) angles add up to 90 degrees. Is that enough to see that the original triangle, triangle A, and triangle B are all similar? (Similar means they have exactly the same shape, though they may be different sizes.) Which makes the 'houses with asymmetrical roofs' also all similar. Since the big 'house' has an 'attic' equal in size to the two other 'attics', its 'room' must also be equal in area to the two other 'rooms'.^# Wow!

Added note (6-9-13): I've been asked to clarify why the big house must be equal in size to the two smaller ones added together. Since all three houses are similar (exact same shape, different sizes), the size of the room is some given multiple of the size of the attic. More properly, area(square) = k*area(triangle), where k is the same for all three figures. The square attached to triangle A (whose area we will say is also A) has area kA, similarly for the square attached to triangle B. kA+kB=k(A+B), which is the area of the square attached to the triangle labeled A+B. But kA = a², and kB = b². So k(A+B) = a²+b². And it also equals c², giving us what we sought, a²+b² = c².

I stumbled on the article in which this appeared (The Step to Rationality, by R. N. Shepard) while searching on 'thought experiment weight times distance must equal to balance'. I'm working on a handout for my Calc II students to explain centroid (since the Briggs textbook leaves this topic out). I was wondering if we need experimental evidence to show that the two sides of a teeter-totter will balance only when the weights times distances from the fulcrum are equal on the two sides. I thought maybe we could come up with a thought experiment that would convince us it must be true. I wasn't having any bright ideas, and turned to google. It hasn't solved my centroid question yet, but I love what I discovered.

I think that, even though this proof is simpler in terms of steps, it's a bit harder to see conceptually. So I may stick with my old favorite when explaining to students. Or maybe there's a way to test out which one is more helpful for a deep understanding of both the notion of proof and this theorem in particular.

What do you think? 





____________
*Playing With Math: Stories from Math Circles, Homeschoolers, and Passionate Teachers is a collection of great writing about math education (often outside the classroom), from over 30 authors. I've been working on for 4 years now; it will be available within 2 to 4 months. 
^#I got this language (of houses, attics, and rooms) from a similar description of this proof which I found on Cut-the-Knot.

Maths and Stats by Email (CSIRO)

About two and a half years ago, I signed up for this online newsletter. I have enjoyed most of the issues quite a bit. There's no link to head you over to the latest issue. They don't do it that way. (I have no idea why not.) You pretty much have to sign up for it if you want it.

This month's issue is typical. A simple topic, explained well for young people:

Take a sheet of A4 paper and measure its sides. A4 is 210 millimetres wide and 297 millimetres long. It’s probably the most common size of paper and it’s used in most countries. However, A4 side lengths aren’t simple numbers like 200 or 300 millimetres. So why don’t we use something easier to measure?

If you take a sheet of paper and cut it halfway down the longer side, you end up with two new pieces of paper. These pieces of paper each have half the area of the original sheet, but they are the same proportions as the original sheet! There’s only one type of rectangle that has this ability. Because these half sheets have the same proportions as A4, they also have a name – A5.  If you cut an A5 sheet in half, you get two pieces of A6 paper, with the same proportions as A5 and A4. All these paper sizes are part of a set called the A series.

This pattern also works if you want to go bigger instead of smaller.  If you take two sheets of A4 paper and stick the long sides together, you’ll end up with a sheet of paper that has the same proportions as A4, but is twice as big. This size is called A3. You can use the same process to make A3 sheets into A2, and even A2 sheets into A1 paper.

So why is A4 paper called A4? A4 is half an A3, or one quarter of A2, but more importantly, it’s one sixteenth of A0. A0 has an area of one square metre (but it isn’t a square), and every other paper size in the A series is based on A0. We use A4 for writing on because it is a lot more convenient than trying to write on a square metre sheet of paper!

After the introductory article, they always have a 'try this' activity. This month's is on tangrams (and the activity relates tangrams to the paper sizes described).

Let me know if you decide to sign up.

Seeking Math-Poets for a Reading (JMM, 1-11-13)

JoAnne Growney just posted this on her blog (Intersections - Poetry with Mathematics):

Call for Readers:
     The Journal of Humanistic Mathematics will host a reading of poetry-with-mathematics at the annual Joint Mathematics Meetings (JMM) on Friday, January 11, 5-6:30 PM in Room 1B, Upper Level, San Diego Convention Center.  If you wish to attend the reading and participate, please send,  by December 1, 2012 (via e-mail, to Gizem Karaali (gizem.karaali@pomona.edu)) up to 3 poems that involve mathematics (in content or structure, or both) -- no more than 3 pages -- and a 25 word bio.

There's more at her post, check it out.

I'll be co-hosting this event, and would love to meet you there. (I hope to read a poem or two of my own. I'd better get my submission in.)