Posts Tagged ‘tutoring’

My Brown Big Spiders

March 21, 2011

Professor: You have to learn to be able to play it blindfolded. The page, for God’s sake! The notes!

David: I’m sorry I was, uh, forgetting them, Professor.

Professor: Would it be asking too much to learn them first?

David: And-And then forget them?

Professor: Precisely.

from the movie Shine

If I want to find the volume and surface area of a sphere, I do it with calculus:

$V = \int_{r = 0}^R\int^{2\pi}_{\phi = 0}\int_{\theta = 0}^\pi r^2\sin\theta \textrm{d}\theta \textrm{d}\phi \textrm{d}r = \frac{4}{3}\pi R^3$

.

$S = \int_{\theta = 0}^\pi\int_{\phi = 0}^{2\pi} R^2 \sin\theta\textrm{d}\theta\textrm{d}\phi = 4\pi R^2$

This is correct, but I can’t use it with high school geometry students because they don’t know what an integral is, much less a Jacobian.

However, Archimedes came up with a beautiful way of discovering the volume and surface area of a sphere. He did it by relating the sphere to a known shape – a cylinder with a cone cut out of it.

He drew a picture like this:

On the left there’s a hemisphere with radius $R$. On the right, there’s a cylinder with radius and height both also $R$, so that the hemisphere would fit perfectly inside the cylinder. The cylinder has had a cone cut out from the top down tapering down to the center of the bottom. First, we’ll show that these two shapes have the same volume.

We imagine slicing the hemisphere horizontally at some certain height $h$. This would reveal a circle as seen in the picture. Call its radius $r$.

At the same height, we also slice the cylinder, leaving us with a disk. We’ll find the areas of this circle and disk.

The area of the circle is $\pi r^2$, which by the Pythagorean theorem is also $\pi (R^2 - h^2)$.

Looking at the cylinder, the outer edge of the disk has radius $R$ and the inner edge has radius $h$, so the area of the disk is also $\pi (R^2 - h^2)$.

Because every horizontal slice of the hemisphere has the same area as the corresponding horizontal slice of the drilled-out cylinder, they must have the same volume. The volume of the cylinder is its original volume minus the volume of the cone, or $\pi R^3 - 1/3 \pi R^3 = 2/3 \pi R^3$. Hence, the volume of a full sphere is

$V = 4/3 \pi R^3$

Next, we’ll show that the hemisphere has the same surface area as the outside of the cylinder (the cone is now unimportant).

Take a slice of the outside of the cylinder at height $h$ and of thickness $\textrm{d}h$. This forms a band around the cylinder whose area is

$\textrm{d}S = 2 \pi R \textrm{d}h$

Now slice the sphere at the same height with the same $\textrm{d}h$. This also forms a band. The band is a shorter distance around, but due to the slant of the edge of the circle, it’s also thicker. Let’s call the thickness of this band $\textrm{d}x$.

The area of the band around the hemisphere is the circumference at height $h$ multiplied by the thickness $\textrm{d}x$.

$\textrm{d}S = 2\pi\sqrt{R^2 - h^2}\textrm{d}x$

If we draw a tangent line on the sphere, it’s perpendicular to the radius. This gives us similar triangles.

So

$\frac{\textrm{d}x}{\textrm{d}h} = \frac{R}{\sqrt{R^2 - h^2}}$

Plugging back into the previous expression,

$\textrm{d}S = 2\pi\sqrt{R^2 - h^2}*\textrm{d}h * \frac{R}{\sqrt{R^2 - h^2}} = 2\pi R \textrm{d}h$

So the band around the outside of the cylinder and sphere have the same surface area, so the entire shapes have the same surface area. That makes the surface area of a sphere

$S = 4 \pi R^2$

This is a really lovely argument. The problem is pretty hard, but the solution is simple. (I’m not sure if this is quite how Archimedes did it. To be honest I never even met the guy. I learned the idea from this animation).

I was reviewing solid geometry with a high school junior the other day, so I showed her this argument (but only the volume part). I was proud of myself for offering this little example of how interesting mathematical ideas can be. At least, I was as we began.

“It’s all so complicated!” she moaned a few minutes later when I asked her to identify a certain quantity in our sketch.

Complicated? I had thought the argument was remarkably simple – just draw a sphere and a cylinder next to each other and you’re practically done. What could be simpler? Somehow my student was getting entangled in brambles I couldn’t even see.

I did not draw quite the same picture for her that I drew earlier in this post. I didn’t want to give it all away, so I drew something more like this and asked for $r$:

Finding $r$ is a simple application of something she knew well – the Pythagorean theorem. She didn’t see it, though, so I showed her this right triangle:

But then she didn’t see how long the new line I just drew was. It’s just $R$ because it’s a radius of the sphere, but although she knew that all radii of a sphere have the same length, she couldn’t easily identify the two lines as radii and call up the relevant information. So I showed her that step, too.

After a bit more prodding, she wrote down $r = \sqrt{R^2 + h^2}$, a mistake that comes from applying the Pythagorean theorem incorrectly. She knows better, and should have found $r^2 = R^2 - h^2$, but by this point she was already flustered from her earlier mistakes, confused about what we were trying to do, self-conscious, and generally unable to approach the problem equanimously.

When she realized she had applied the Pythagorean theorem wrong, her frustration mounted, and moments later, at my next question, I was shocked with, “It’s all so complicated!”

Why did this happen? Why did I so horribly misjudge the difficulty of the exercise?

The other day I read this comment on an essay on teaching

I used to teach English as a second language. It was a mind trip.

I remember one of my students saying something like “I saw a brown big spider”. I responded “No, it should be ‘big brown spider'”. He asked why. Not only did I not know the rule involved, I had never even imagined that anyone would ever say it the other way until that moment.

Tutoring has been exposing my own brown big spiders – the little steps and bits of knowledge that I take for granted – for years. I’ve rarely stopped to notice it.

Just to follow each step in the Archimedes argument, you must make an enormous number of mathematical connections behind the scenes in your mind. Here’s a partial list:

• A “sphere” is a round three-dimensional object like, a ball
• Every point on the surface of a sphere is the same distance from the center
• The “surface” of the sphere means its outside edge, or skin
• A “point” is a little dot with no size at all. It simply marks a place.
• You can represent three-dimensional figures in two dimensions with certain types of drawing.
• The point of doing this drawing is to make things easier to visualize.
• A “hemisphere” is half a sphere – the top half in this case
• A “cylinder” is basically a tube with constant width.
• The center of the bottom of the hemisphere is the same point as the center of the sphere it came from.
• The height of the hemisphere is the same as the distance from the center to the edge horizontally.
• This means that the cylinder drawn is twice as wide as it is tall.
• The volume of a cone is one third the area of its base times its height.
• The volume of a cylinder is its base times its height
• The area of a circle is $\pi$ times the square of its radius

And so on. I only stopped writing so that I’d eventually finish the rest of this post. Each item I added to that list sparked off several new ones I hadn’t considered.

Try writing your own list and you’ll quickly be overwhelmed by the exponentially-proliferating leaves on your conceptual tree. We didn’t even get close to things like the Cavalieri’s principle.

The items on my brown big spider list are not supposed to be mathematical facts so much as cognitive patterns the reader is required to have. For example, mathematically a point is not, “a little dot with no size at all,” as I called it. It’s a primitive notion and has no definition. The list still calls a point a dot, though, because the mathematically-accurate description isn’t helpful to a student, and isn’t they way most people think of it even when they’ve already learned geometry well.

When I started writing the list, I found myself wanting to say, “A sphere is a set of all points equidistant…”, but that’s no good. It uses the significant brown big spiders of “set” and “equidistant”, as well as the general idea of giving mathematical definitions, something most high schoolers don’t yet understand well. Then I wanted to say, “A sphere is a shape that’s symmetric with respect to rotations about any axis…” but this has all the same problems.

Ultimately, I chose “a sphere is a ball.” It’s imprecise, but it’s the way you think about a sphere before you’ve packaged the concept away so tightly you don’t need to think about it any more. Anyone who tells you a sphere is the two-dimensional manifold $S^2$ is someone who has forgotten how much they actually know about spheres. They’ve forgotten it in the good way, of course – the way David was supposed to forget the notes to Rachmaninoff. Unfortunately, I experience a crippling side effect when I forget things this way. I forget that other people haven’t yet forgotten them.

This forgetting is the psychological phenomenon of “chunking“. The most famous example involves chess players. Give expert chess players a position from a game between grandmasters and they can easily memorize the positions of thirty pieces. Give them pieces strewn randomly about the board and they’ll remember just a few – no more, in fact, than your average Joe who knows little more about chess than what the real name of the horsey is.

A position from a real game has lots of meaning, if you’re an expert. If you’re an expert you extract order from the position automatically, without consciously processing every detail. The entire task must seem quite simple to a grandmaster. Similarly, the experienced mathematician sees all the important properties of the sphere and the cylinder and the cone without having to list them out one by one, and the process is so automatic they don’t even realize they’re doing it.

In “Simple” Isn’t “Easy”, I learned not to judge the difficulty of new ideas by how simple they are, but by how familiar to the student. Despite this, I have continued to make a similar mistake when dealing with ideas the students have already learned.

“Learned” isn’t “chunked”. My student understood the meaning of “hemisphere” and the formula for the volume of a cone, but she still needed conscious effort to recall and wield those bits of knowledge. Each sat in its own corner in her mind, accessible only by dint of concerted effort, and certainly not ready to flow into a flood of beautiful ideas.

I was trying to dictate a soliloquy for her to transcribe, but I was assuming that because she could see the letters on her keyboard, should could touch-type. It turned out that the effort to hunt-and-peck was so great, all the artistry of the speech was lost.

I want to watch out for my brown big spiders in the future. I want to be more patient when they are discovered and more studious in cataloging, remembering, and working with them. Most of all, I want to look back later, and remember my students forgetting them.

‘Simple’ Isn’t ‘Easy’

November 7, 2010

You are probably aware that $3^{1/2} = \sqrt{3}$. Sometimes when I’m tutoring I wind up teaching this to young students. Here is the story I use:

You already know that $3^4*3^2 = 3^6$ for a very simple reason.

Forget the reason for a moment, and just focus on the rule. When you multiply exponents with the same base, you can add the powers.

That means

$3^{1/2}*3^{1/2} = 3^1 = 3$

Evidently, $3^{1/2}$ is a number such that if you multiply it by itself, you get three. But that is exactly the meaning of the square root! Hence $3^{1/2} = \sqrt{3}$.

This is a very simple idea, but when I try it on students, it usually fails.

After going through the story, I ask what $16^{1/2}$ is. I’m hoping to hear “four”, but that’s not what happens. Sometimes they say it’s eight. Sometimes they say they don’t know. But the most common response is to go through the whole thing again. The student writes down

$16^{1/2}*16^{1/2} = 16^1 = 16$.

They stare it at for a while. Then they look up at me and say, “Is that right?” We discuss it a bit further to clarify. Circuitously, we stumble upon $16^{1/2}=4$. After that we do a few more half-powers and they get it right. Then I ask what $8^{1/3}$ is. The student will write down

$8^{1/3}*8^{1/3} = 8^{2/3}$.

“It doesn’t work for that one,” they say. “You just get a 2/3 power, and we can’t do that.” So we talk about it some more, until after some time the student can go between roots and exponents.

Then I ask what $4^{3/2}$ is, but they struggle with this, too. Once that’s down we try for $6^{-1}$, but that is also impenetrable (I usually hear that it’s -6). When I suggest trying to figure it out based on the rule of exponent addition, the student feels frustrated and defeated.

It’s curious that I have such difficulty teaching this idea. It is not too complicated or too difficult, even for a young child. It is far simpler than long division and far less abstract than “set the unknown variable equal to x”. The problem is not the sophistication of the idea, but a more fundamental error in communication. When I give my little presentation, the students simply have no idea what I’m doing.

An analogy: I’m teaching someone how to lift weights (this is very hypothetical). I take a dumbbell and I start doing some bicep curls. It’s only a 5-lb dumbbell, and the motion is very simple, so I figure the guy I’m teaching will get it for sure. I hand him the weight and say, “You try.”

When I hand over the weight and the student starts yanking it up and down. He purposely mimics the way I grunt in exertion and copies my facial expressions. He remembers how I looked over my shoulder to talk to him while I demonstrated the exercise, so he looks over his shoulder when trying it out. The weight ultimately does go up and down, but only with a great deal of extraneous commotion. I straighten him out with some effort, but when we move over to the bench press we’ll repeat the whole confused process.

The problem is that before we began, my student didn’t know what weight-lifting is. He didn’t know the point is to make your muscles stronger, or the counter-intuitive idea that to make your muscles stronger, you first have to tire them out by working them hard.

Similarly, my math students watch me do this strange algebraic exercise with exponents not knowing that the goal is to discover new things. They think, instead, that I was simply teaching a new procedure, as in, “This is how you solve problems where the exponent is one half.”

This is not really a big problem. Students can learn new things; that’s what being a student is about. The problem is that students’ ineptitude at this task frustrates me. At times, when watching a student struggle with a problem, I’ve felt ironic wonder at the student’s remarkable creativity – how do they find so many unexpected ways to get everything totally wrong? I wind up concluding that the student is “stupid”, and the student leaves the lesson with only the impression that they have somehow failed at a task they never even understood.

I make these grievous errors in judgment because I assume that since I’ve seen the student handle far more complicated tasks, they should master this one right away. That is not so. ‘Simple’ isn’t ‘easy’. Computing a determinant of a 4×4 matrix isn’t simple, but my students can blaze through it. Showing that the determinant will be zero by noticing that the last row is equal to first row is very simple, but I’ve never had a student use that method.

The things we’re good at are not what’s simplest, but what’s most familiar. The converse also holds: things that are unfamiliar are difficult, even if they’re simple. I personally find it much easier to solve geometry problems using coordinates, algebra, and calculus than using Euclidean geometry, even when the Euclidean approach may be just a few lines of sketching and finding a similar triangle.

When I first noticed that students were having a hard time with problems because they required unfamiliar thinking, and not because they were too hard or because the students were bad, I tried to remedy the situation with speeches. I would talk about how interesting it is to figure out where a formula comes from. I would say over and over that no, I don’t have all the formulas memorized, because as long as I know most of it, I can figure the rest out. I would prove my point by waiting until they embarked on a difficult calculation, and then solving it quickly in my head using some trick or other, supposedly demonstrating how useful it is to be able to approach a problem many different ways. Then I would describe how it’s done. “You’ll like this thing I’m about to show you,” I would say. “It’ll make your life easier.”

This backfired. It mostly led the students to believe that I either gained some ineffable voodoo skills in college or that I am in possession of an extraordinary native intellect that they could never hope to emulate.

I still don’t know quite how to handle the “simple isn’t easy problem”. I have become far more patient when trying to push students’ boundaries, and far less ambitious. I regret the many times I compromised a student’s chance at learning and my own at equanimity by failing to recognize “simple isn’t easy” in practice. I continue to search for simpler and simpler teaching stories, but I don’t spend enough time searching for ways to make the unfamiliar territory easier to navigate. I don’t know how complicated a task that is – to figure out how to build a stepladder to a new level cognition – but I know it isn’t yet easy.

Multiples Rule For 3, 9, and 27

October 23, 2008

I’m sorry I started writing this post. The point was to talk about some simple number theory, but I decided to “build into it” with a pointless anecdote about my pointless childhood. This anecdote happened to involve the TV show “Square One”. Thanks to my simian brain’s stupid ability to make connections between various stores of knowledge, I realized that although I haven’t seen an episode of the show since 1993, now we have YouTube, and therefore I can go watch people roller skating while tied to pickup trucks.  Also I can find that show. It’s probably the greatest thing since Spirograph.

And there’s plenty more where that came from.

But the actual impetus for this story was that I was tutoring some intelligent algebra students ($x$ of them), in the prime factorization of numbers. I was surprised they hadn’t heard the following rules to check whether a number is divisible by three or nine without doing the division problem.

• A number is divisible by three if and only if the sum of its digits is divisible by three.
• Likewise for nine.

The reason I couldn’t imagine someone not familiar with this oh-so-basic fact is that I learned it at age six from a singing cowboy.

But these aren’t theorems about the numbers, exactly.  They’re theorems about numbers and the numbers’ digits.  The distinction is important because numbers themselves are pretty fundamental, while the digits are a consequence of the base of the numeral system you’re using.  We use base ten, which is where these theorems hold.  They clearly fail in base three, for example, where we can write $three = 10$.  Consequently, this quick-factoring trick is one little gem of knowledge unknown to the ancient Pythagoreans or modern Canadians.

Note that three goes into 9, to 99, to 999, etc.  So take a number with some digits in it, say 5832.  (The great thing about being a math dabbler rather than a math student is that proofs by example, which are totally easy, don’t bother you one bit).

$\frac{5832}{3} = \frac{5*1000}{3} + \frac{8*100}{3} + \frac{3*10}{3} + \frac{2}{3}$

Now we’ll write 1000 as 999+1, and pull the same trick for the other powers of ten.  Then break up the fractions using $\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$.

$\frac{5832}{3} = \frac{5*999}{3} + \frac{5*1}{3} + \frac{8*99}{3} + \frac{8}{3} + \frac{3*9}{3} + \frac{3}{3} + \frac{2}{3}$

Since we only care whether this gives an integer, we can drop all those ones with the 9, 99, 999 etc. in the numerator.  They cleary do give integers.  This leaves us to evaluate whether

$\frac{5}{3} + \frac{8}{3} + \frac{3}{3} + \frac{2}{3} = \frac{5+8+3+2}{3}$

is an integer.   So to test whether a number is divisible by three, add its digits.  Likewise for nine, since nine also divides 9, 99, 999…

What got me looking at this was that my student, curious young mathematician that he is, asked if the same holds true for 27.  We can say right off the bat that because every multiple of 27 is also a multiple of nine, the digits of a multiple of 27 must sum to a multple of nine.  However, there is no requirement that the multiple of nine they sum to is 27.  For example, 54 is a multiple of 27, but its digits sum to 9.  2727 is a multiple of 27 (27*101), but its digits sum to 18.

But does a one-way version of the theorem still hold?  If the digits of a number sum to 27, is that number divisible by 27?  The first such number you’ll run into is 999 = 27 * 37.  I tried six or seven other examples.  They all worked.  So I set out to prove it and was running into trouble.  But the nice thing was that although I couldn’t prove it, my attempt to find a proof showed me an easy way to find a counterexample.  I won’t go into the details, but you can go searching for yourself.  Here’s a hint: $\frac{818172}{27} = 30302 \frac{2}{3}$.

Higher powers of three are an obvious generalization.

If you followed the proof closely, then two things:

• wtf?  it’s not like it’s your homework or something.  or like i know what i’m talking about
• you can find similar properties in base n

In base 7, for example. a number’s digits will sum to a multiple of six if and only if the number is itself a multiple of six.  For example, in base 7

$213 = 2*7^2 + 1*7 + 3 = 2*(7^2 - 1) + 1*(7-1) + 2 + 1 + 3$

So all the same stuff will work as long as 6 goes evenly into $7-1, 7^2-1, 7^3-1,...$  It does, because $\frac{7^5 - 1}{6} = \frac{66666}{6} = 11111$ in base 7.  So if you want to see whether $p$ divides $q$ evenly, you could always just convert $q$ to base $p+1$ and add the digits.  However this is quite a bit more work than just doing the straight up division problem.

Also note that the reason the theorem works for three is that it goes evenly into nine. In base 13 the trick works for 12, 6, 4, 3, and 2. And on a practical note, in hexadecimal it works for f (which is what you use for 15), 5, and 3. Hexadecimal is actually quite nice for factoring, because in addition to easy rules for those numbers, you also have easy rules for 2, 4, and 8 – you can just check the last digit of the number the same way you can to see if something is a multiple of 2 or 5 in base 10.