The Problem with Calculus Textbooks (Reprise)

Simply put, most calculus textbooks are written in the wrong order.

Unfortunately, this includes the most popular textbooks used in colleges and universities today.

This problem has a long history, and will not be quickly solved for a variety of reasons. I think the solution lies ultimately with high quality, open source e-modules (that is, stand-alone tutorials on all calculus-related topics), but that discussion is for another time. Today, I want to address a more pressing issue: since many of us (including myself) must teach from such textbooks — now, long before the publishing revolution — how might we provide students a more engaging, productive calculus experience?

To be specific, I’ll describe some strategies I’ve used in calculus over the past several years. Once you get the idea, you’ll be able to look through your syllabus and find ways to make similar adaptations. There are so many different versions of calculus taught, there is no “one size fits all” solution. So here goes.

1. I now teach differentiation before limits. The reason is that very little intuition about limits is needed to differentiate quadratics, for example — but the idea of limits is naturally introduced in terms of slopes of secant lines. Once students have the general idea, I give them a list of the usual functions to differentiate. Now they generate the limits we need to study — completely opposite of introducing various limits out of context that “they will need later.”

Students routinely ask, “When am I ever going to use this?” At one time, I dismissed the question as irrelevant — surely students should know that the learning process is not one of immediate gratification. But when I really understood what they were asking — “How do I make sense of what you’re telling me when I have nothing to relate it to except the promise of some unknown future problem?” — I started to rethink how I presented concepts in calculus.

I also didn’t want to write my own calculus textbook from scratch — so I looked for ways to use the resources I already had. Simply doing the introductory section on differentiation before the chapter on limits takes no additional time in the classroom, and not much preparation on the part of the teacher. This point is crucial for the typical teacher — time is precious. What I’m advocating is just a reshuffling of the topics we (have to) teach anyway.

2. I no longer teach the chapter on techniques of integration as a “chapter.” In the typical textbook, nothing in this chapter is sufficiently motivated. So here’s what I do.

I teach the section on integration by parts when I discuss volumes. Finding volumes using cylindrical shells naturally gives rise to using integration by parts, so why wait? Incidentally, I also bring center of mass and Pappus’ theorem into play, as they also fit naturally here. The one-variable formulation of the center of mass gives rise to squares of functions, so I introduce integrating powers of trigonometric functions here. (Though I omit topics such as using integration by parts to integrate unfriendly powers of tangent and secant — I do not feel this is necessary given any mathematician I know would jump to Mathematica or similar software to evaluate such integrals.)

I teach trigonometric substitution (hyperbolic as well — that for another blog post) when I cover arc length and surface area — again, since integrals involving square roots arise naturally here.

Partial fractions can either be introduced when covering telescoping series, or when solving the logistic equation. (A colleague recommended doing series in the middle of the course rather then the end (where it would have naturally have fallen given the order of chapters in our text), since she found that students’ minds were fresher then — so I introduced partial fractions when doing telescoping series. I found this rearrangement to be a good suggestion, by the way. Thanks, Cornelia!)

3. I no longer begin Taylor series by introducing sequences and series in the conventional way. First, I motivate the idea by considering limits like

\displaystyle\lim_{x\to0}\dfrac{\sin x-x}{x^3}=-\dfrac16.

This essentially means that near 0, we can approximate \sin(x) by the cubic polynomial

\sin(x)\approx x-\dfrac{x^3}6.

In other words, the limits we often encounter while studying L’Hopital’s rule provide a good motivation for polynomial approximations. Once the idea is introduced, higher-order — eventually “infinite-order” — approximations can be brought in. Some algorithms approximate transcendental functions with polynomials — this provides food for thought as well. Natural questions arise: How far do we need to go to get a given desired accuracy? Will the process always work?

I won’t say more about this approach here, since I’ve written up a complete set of Taylor series notes. They were written for an Honors-level class, so some sections won’t be appropriate for a typical calculus course. They were also intended for use in an inquiry-based learning environment, and so are not in the usual “text, examples, exercise” order. But I hope they at least convey an approach to the subject, which I have adapted to a more traditional university setting as well. For the interested instructor, I also have compiled a complete Solutions Manual.

I think this is enough to give you the idea of my approach to using a traditional textbook. Every calculus teacher has their own way of thinking about the subject — as it should be. There is no reason to think that every teacher should teach calculus in the same way — but there is every reason to think that calculus teachers should be contemplating how to make this beautiful subject more accessible to their students.

Enumerating the Platonic Solids

The past few weeks, I outlined my approach to a series of lectures on polyhedra.  One of my constraints is that students will not have seen a lot of trigonometry yet, and will not have been exposed to three-dimensional Cartesian coordinates.  But there is Euler’s Formula!  I just finished a pair of lectures on the algebraic enumeration of the Platonic solids using Euler’s Formula, and I thought others might be interested as well.

As a reminder, Euler’s Formula states that if V, E, and F are the number of vertices, edges, and faces, respectively, on a convex polyhedron, then

V-E+F=2.

How might we use this formula to enumerate the Platonic Solids?  We need to make sure we agree on what a Platonic Solid is:  a convex polyhedron with all the same regular polygon for faces, and with the same number meeting at each vertex.

To use this definition, we will define a few more variables:  let p denote the number of sides on the regular polygons, and let q denote the number of polygons meeting at each vertex of the Platonic solid.  (Those familiar with polyhedra will recognize these as the usual variables.)

The trick is to count the number of sides and vertices on all the polygons in two different ways.  For example, since there are F polygons on the Platonic solid, each having p sides, there are a total of pF sides on all of the polygons.

CubeFaces.png

But notice that when we build a cube from six squares, two sides of the squares meet at each edge of the cube.  This implies that 2E also counts all of the sides on the polygons.  Since we are counting the same thing in two different ways, we have

pF=2E.

We may similarly count all the vertices on the polygons as well.  Of course since a regular polygon with p sides also has p vertices, there are pF vertices on all of the polygons.

But notice that when we put the squares together, three vertices from the squares meet at a vertex of the cube.  Thus, if there are V vertices on a Platonic Solid, and if q vertices of the polygons come together at each one, then it must be that qV is the total number of vertices on all of the polygons.  Again, having counted the same thing in two different ways, we have

pF=qV.

Thus, so far we have

V-E+F=2,\quad pF=2E,\quad pF=qV.

Note that we have three equations in five variables here; in general, such a system has infinitely many solutions.  But we have additional constraints here — note that all variables are counting some feature of a Platonic Solid, and so all must be positive integers.

Also, since a regular polygon has at least three sides, we must have p\ge3, and since at least three polygons must come together at the vertex of a convex polyhedron, we must also have q\ge3.

These additional constraints will guarantee a finite (as we know!) number of solutions.  So let’s go about solving this system.  The simplest approach is to solve the last two equations above for E and V and substitute into Euler’s Formula, yielding

\dfrac{pF}q-\dfrac{pF}2+F=2.

Now divide through by F and observe that F>0, so that

\dfrac pq-\dfrac p2+1>0.

Multiply through by 2q and rearrange terms, giving

pq-2p-2q<0.

How should we go about solving this inequality?  There’s a nice trick here:  add 4 to both sides so that the left-hand side factors nicely:

(p-2)(q-2)<4.

Now we are almost done!  Since p,q\ge3, then p-2 and q-2 must both be integers at least 1; but since their product must be less than 4, they can be at most 3.

This directly implies that p and q must be 3, 4, or 5.

This leaves only nine possibilities — but of course, not all options need be considered.  For example, if p=q=5, then

(p-2)(q-2)=9>4,

and so does not represent a valid solution.  But when p=3 and q=4, we have the octahedron, since p=3 means that the polygons on the Platonic Solid are equilateral triangles, and q=4 means that four triangles meet at each vertex.

So out of these nine possibilities to consider, there are just five options for p and q which satisfy the inequality (p-2)(q-2)<4.  And since each pair corresponds to a Platonic Solid, this implies that there are just five of them, as enumerated in the following table:

PlatonicChart

Actually, this implies that there are at most five Platonic Solids.  How do we know that twelve pentagons actually fit together exactly to form a regular dodecahedron?  A further argument is necessary here to be complete.  But for the purposes of my lectures, I just show images of these Platonic Solids, with the presumption that they do, in fact, exist.

Now keep in mind that in an earlier lecture, I enumerated the Platonic Solids using a geometrical approach; that is, by looking at those with triangular faces, square faces, etc.  I like the problem of enumerating the Platonic Solids since the geometric and algebraic methods are so different, and emphasize different aspects of the problem.  Further, both methods are fairly accessible to good algebra students.  The question of when to take an algebraic approach rather than a geometric approach to a geometry problem is frequently difficult for students to answer; hopefully, looking at this problem from both perspectives will give students more insight into this question.

 

Teaching Three-Dimensional Geometry, III

This is the last of a three-part series on teaching three-dimensional geometry.  A few weeks ago, I had begun describing how I would go about putting together a series of about 20 online videos on 3D geometry, each lasting 5–7 minutes.  I just finished a discussion of buckyballs, and why regardless of the number of hexagonal faces on a buckyball, there are always exactly 12 pentagonal faces.

Euler’s Formula was key.  We’ll look at another application of Euler’s Formula, but before doing so, I’d like to point out that students at this level have not encountered Cartesian coordinates in three dimensions, and so I need to find things to talk about at an accessible level.

On to the truncation of polyhedra!  Again, we can apply Euler’s Formula, but it helps to think about the process systematically.  You can count the number of vertices, edges, and faces on a truncated cube, for example, one at a time — but little is gained from a brute force approach.  By thinking more geometrically, we would notice that each edge of the original cube contributes two vertices to the truncated cube, giving a total of 24 vertices.

We can continue on in this fashion, counting as efficiently as possible.  This sets the stage for a discussion of Archimedean solids in general.  A proof of the enumeration of the Archimedean solids is beyond the scope of a single lecture, but the important geometrical ideas can still be addressed.

This concludes the set of lectures on polyhedra in three dimensions.  Of course there is a lot more that can be said, but I need to make sure I get to some other topics.

Like spherical geometry, for instance, next on the slate.  There are two approaches one typically takes, depending how you define a point in spherical geometry.  There is a nice duality of theorems if you define a Point in this new geometry as a pair of antipodal points on a sphere, and a Line as a great circle on a sphere.  Thus two distinct Lines uniquely determine a Point, and two distinct Points uniquely determine a line.

This is a bit abstract for a first go at spherical geometry, so I plan to define a Point as just an ordinary point on a sphere, and a Line as a great circle.  Two points no longer uniquely determine a Line, since there are infinitely many Lines through two antipodal Points.

But still, there are lots of interesting things to discuss.  For example, there is no such thing as a pair parallel lines on a sphere:  two distinct Lines always intersect.

Triangles are also intriguing.  On the sphere, the sides are also angles, measured by the angle subtended at the center of the sphere.  So all together, there are six angular measures in any triangle.

Since students will not have had a lot of exposure to trigonometry at this point, I won’t discuss many of the neat spherical trigonometric formulas.  But still, there is the fact the angle sum of a spherical triangle is always greater than 180^\circ.  And the fact that similarity and congruence on the sphere are the same concept, unlike in Euclidean geometry.  For example, if the angles in a Euclidean triangle are the same in pairs, the triangles are similar.  But on a sphere, if the angles of two spherical triangles measured the same in pairs, they would necessarily have to be congruent.

In other words, students are getting further exposure to non-Euclidean geometries.  (I did a lecture on inversive geometry in a previous section.)  One nice and accessible proof in spherical geometry is the proof that the area of a spherical triangle is proportional to its spherical excess — that is, how much the angle sum is greater than 180^\circ.  So there will be something  I can talk about without needing to say the proof is too complicated to include….

The final topic I plan to address is higher-dimensional geometry.  The first natural go-to here is the hypercube.  Students are always intrigued by a fourth spatial dimension.  Ask a typical student who hasn’t been exposed to these ideas what the fourth dimension is, and the answer you invariably get is “time.”  So you have to do some work getting them to think outside of that box they’ve lived in for so long.

One thing I like about hypercubes is the different ways you can visualize them in two dimensions.

Hypercube1

Viewed this way, you can see the black cube being moved along a direction perpendicular to itself to obtain the blue cube.  Of course the process is necessarily distorted since we’re looking at a static image.

Hypercube2

This perspective highlights a pair of opposite cubes — the green one in the middle, and the outer shell — and the six cubes adjacent to both.

Hypercube4

And this perspective is just aesthetically very pleasing, and also has the nice property that every one of the eight cubes looks exactly the same, except for a rotation.  Again, there won’t be any four-dimensional Cartesian coordinates, but still, there will be plenty to talk about.

I plan to wrap up the series with a discussion of volumes in higher dimensions.  As I mentioned last week, I’d like to discuss why you should avoid peeling a 100-dimensional potato….

Thinking by analogy, it is not difficult to motivate the fact that the volume of a sphere n dimensions is of the form

Kr^n.

Now let’s look at peeling a potato in three dimensions, assuming it’s roughly spherical.  If you were a practiced potato peeler, maybe you could get away with the thickness of your potato peels being, say, just 1% of the radius of your potato.  This leaves the radius of your peeled potato as 0.99r, and calculating a simple ratio reveals that you’ve got 0.99^3\approx0.97 of your potato left.

Extend this idea into higher dimensions.  If your potato-peeling expertise is as good in higher dimensions, you’ll have 0.99^n of your potato left, where n is the number of dimensions of your potato.  Now 0.99^{100}\approx 0.366, so after you’ve peeled your potato, you’ve only got about one-third of it left!

What’s happening here is that as you go up in dimension, there is more volume near the surface of objects than there is near the center.  This is difficult to intuit from two and three dimensions, where it seems the opposite is the case.  Nonetheless, this discussion gives at least some intuition about volumes in higher dimensions.

And that’s it!  I’m looking forward to making these videos; I actually made my first set of slides today.  As usual, if I come across anything startling or unusual during the process, I’ll be sure to post about it!

Teaching Three-Dimensional Geometry, II

A few weeks ago, I began a discussion of what I’d be presenting in a series of twenty (or so) 5—7 minute videos on three-dimensional geometry. I didn’t get very far then, so it’s time to continue….

So to recap a bit, I’ll begin with the usual cones/cylinders/spheres, looking at surface areas and contrasting flat surfaces with the surface of a sphere. Then on to a prelude to calculus by looking at the volume of a cone as a limiting case of a stack of circular disks.

Next, it’s on to polyhedra! A favorite topic of mine, certainly. Polyhedra are interesting, even from the very beginning, since there is still no accepted definition of what a polyhedron actually is. The exception is for convex polyhedra; a perfectly good definition of a convex polyhedron is the convex hull of a finite set of points not all lying in a single plane. Easy enough.

But once you move on to nonconvexity, uncertainties abound. For example, from a historical perspective, sometimes the object below was a polyhedron, and sometimes it wasn’t. Sounds odd, but whether or not you consider this object a polyhedron depends on how you look at the top “face,” which is a square with a smaller square removed from the center. Now is this “face” a polygon, or not? Many definitions of a polygon would exclude this geometrical object – which is problematic if you want to say that a polyhedron has polygons as faces.

toroid.png

So this brings us to a definition of a polygon, which is problematic in its own way – to see why, you can look at a previous post of mine on the definition of a polygon.  Now the point here is not to resolve the issue in an elementary lecture, but rather point out that mathematics is not “black-and-white,” as students tend to believe. Also, it provides a nice example of the importance of definitions in mathematics.

Now this would be discussed briefly in just one video. Next would be the (obligatory) Platonic solids – where else is there to begin? The simplest starting point is the geometric enumeration by looking at what types of polygons – and how many – can appear at any given vertex of a Platonic solid. This enumeration is straightforward enough.

Next, I plan on computing the volume of a regular tetrahedron using the usual Bh/3 formula. This is not really exciting in and of itself, but in the next lecture, I plan to find the volume of a regular tetrahedron by inscribing it in the usual way in the cube by joining alternate vertices.

Of course you get the same result. But for those of us who work a lot in three-dimensional space, we deeply understand the simple algebraic equation, 2 \times 4=8. What I’m referring to, specifically, is that the number of vertices on a three-dimensional simplex is half the number of vertices of a three-dimensional hypercube.

This simple fact is at the heart of any number of intriguing geometrical relationships between polyhedra in three dimensions. In particular, and quite importantly, the simplex and the cross-polytope together fill space. This relationship is at the heart of many architectural constructions in additional to generating other tilings of space with Archimedean solids. But most students have never seen this illustrated before, so I think it is important to include.

Then on to a geometry/algebra relationship: having enumerated the Platonic solids geometrically, how do we proceed to take an algebraic approach? A fairly direct way is to use Euler’s formula to find an algebraic enumeration.

No, I don’t intend to prove Euler’s formula; by far my favorite (and best!) is Legendre’s proof which involves projecting a polyhedron onto a sphere and looking at the areas of the spherical polygons created. This is a bit beyond the scope of this series of videos; there simply isn’t time for everything. But it is important to note the role that convexity plays here; yes, there are other formulas for polyhedra which are not essentially “spheres,” but this is not the place to discuss them.

Next, I want to talk about “buckyballs.” I still have somewhat of a pet peeve about the nomenclature – Buckminster Fuller did not invent the truncated icosahedron – and so the physicists who named this molecule were, in my opinion, polyhedrally rather naïve. But, sadly (as is the case so many times), they did not come to me first before making such a decision…

The polyhedrally interesting fact about buckyballs is this: if a polyhedron has just pentagonal and hexagonal faces, three meeting at every vertex, then there must be exactly twelve pentagons. Always.

Now I know that the polyhedrally savvy among you are well aware of this – but for those who aren’t, I’ll show you the beautiful and very short proof. Once you’ve seen the idea, I don’t think you’ll ever be able to forget it. It’s just remarkable – even with 123,456,789 hexagons, just 12 pentagons.

So let P represent the number of pentagons on the buckyball, and H represent the number of hexagons. Then the number of vertices V is given by

V=\dfrac{5P+6H}{3},

since each pentagon contributes five vertices, each hexagon contributes six, and three vertices of the polygons meet at each vertex of the buckyball.

Moreover, the number of edges is given by

E=\dfrac{5P+6H}{2},

since the polygons on the buckyball meet edge-to-edge. Of course, F=P+H, since the faces are just the pentagons and hexagons. Substitute these expressions into Euler’s formula

V-E+F=2,

and what happens? It turns out that H cancels out, leaving P=12!

Amazes me every time. But what I like about this fact is that it is accessible just knowing Euler’s formula – no more advanced concepts are necessary.

And yes, there’s more! This is now Lecture #12 of my series, so I have a few more to describe to you. Until next time, when I caution you (rather strongly) against peeling a 100-dimensional potato….

Bay Area Mathematical Artists Seminars, XI

This past weekend marked the eleventh meeting of the Bay Area Mathematical Artists Seminars.  Our host this month was Scott Vorthmann, the mastermind behind vZome.  Scott lives in Saratoga, and so those participants who live in the San Jose area were glad of the short commute.

It seems that the content of our seminars is limited only by the creativity of the artists involved, meaning fairly limitless….  Scott invited anyone interested to come early — 1:00 instead of our usual 3:00 — and be involved in a Zome “build;” that is, the construction of a large and intricate model using Zome tools.  Today’s model?  The omnitruncated 24-cell!

This is not the place to have a lengthy discussion of polytopes in four dimensions.  In a nutshell, the 24-cell is a polytope in four dimensions with 24 octahedral facets.  This polytope is truncated in a particular way (called omintruncation), and then projected into three-dimensional space.

But there is just one problem with the projection Scott wanted to build.  You can’t build it with the standard Zome kit!  No matter.  Scott designed and 3D-printed his own struts — olive, maroon, and lavender.  If you’ve ever played around with ZomeTools, you’ll understand what a remarkable feat of design and engineering this is.

The building process is a modular one — six pieces like the one shown below needed to be built and painstakingly assembled together.

IMAG9535

Scott built two of the modules before anyone arrived, so we had something to work from.  That left just four more to complete….

The modules were almost done, but we needed to move on.  In addition to the Zome build, we had two other short presentations.   Andrea and Andy were planning to present a workshop at Bridges 2018 in Stockholm, but at the last minute, were unable to attend.  So they brought their ideas to present to us.

The basic idea is to encode a two-dimensional image using two overlays, as shown here.

Day159crypt1.png

Your friend has an apparently random grid (pad) of black and white squares.  You want to send him a secret message; only you and he have the pad.  So you send him a second grid of black and white squares so that when correctly overlaid on the pad, an image is produced.

This is a great activity for younger students, too, since it can be done with premade templates and graph paper.  And even though Andrea and Andy were not able to attend Bridges, their workshop paper was accepted, and so it is in the Bridges archives.  So if you want to learn more about this method of encryption, you can read all the details about the process in their paper in the Bridges archives.

Our next short presentation was by pianist Hans Boepple, a colleague of Frank Farris at Santa Clara University.  Frank happened to have a very stimulating conversation with Hans about a mathematics/music phenomenon, and thought he might like to present his idea at our meeting.

The idea came from a time when Hans happened to look down a metal cylinder of tubing, like you would find at a hardware store.  It seemed that there was an interesting pattern of reflections along the sides of the tubing, and knowing about music and the overtone series, he wondered if there was any connection with music.

Here is part of a computer-generated image of what Hans produced using paper and pencil many years ago:

IMAG9553

How was this picture generated?  Below is how you’d start making the image.

Day159Hans.png

You can see that the red lines take two zigzags to move from one corner of the rectangle to another, the blue lines take three zigzags, the green four, and the gold lines take five.  If you keep adding more and more lines, you get rather complex and beautiful patterns like the one shown above.  Those familiar with the overtone series will see an immediate connection.

Of course, the mathematical question is about proving various properties of this pattern.  It turns out that the patterns are related to the Ford circles; BAMAS participant Jacob Rus has created an interactive version of this diagram.  Feel free to explore!

In any case, we were delighted that Hans could join us and share his fascination with the relationship between mathematics and music.  You can  learn more about Hans in this interview in The Santa Clara, which is Santa Clara University’s school newspaper.

When Hans finished his presentation, it was time to finish building the omnitruncated 24-cell.  It was quite amazing, as Scott is certainly one of the foremost experts on ZomeTools in the world.  Here is the finished sculpture, suspended from the ceiling in his home.  Just getting the model up there was an engineering feat in its own right!

IMAG9586.jpg

It is difficult to describe the intricacy of this model from just a few pictures.

IMAG9599.jpg

Here is an intriguing perspective of the model, highlighting the parallelism of the blue Zome struts.  It seems there is no end to the geometrical relationships you can find hidden within this model.

And, as usual, the afternoon didn’t end there.  Scott arranged to have Thai food — one of our favorites! — catered in, and we all chipped in our fair share.  We all were having such a great time, the last of us didn’t leave until about 8:30 in the evening.  Another successful seminar!

It is quite heartwarming to see so many so willing to take on hosting our Bay Area Mathematical Artists Seminars.  We have all enjoyed these meetings so much, and we are so glad they continue to happen.  I am confident there will be many, many more delightful Saturday afternoons to experience….

Teaching Three-Dimensional Geometry, I

I have recently had a rather unusual opportunity.  I’ve talked a bit over the last few months about my consulting work producing online videos for a flipped classroom; I’ve been working busily on the Geometry unit.

Now the last section of this unit is on three-dimensional geometry, and I’ve been given pretty free reign as to what to cover in this 20-lectures series of 5-7 minute videos.  And given my interest in polyhedra (which I could focus on exclusively with no shortage of things to discuss!), I felt I had a good start.

But the challenge was also to cover some traditional topics (cones, cylinders, spheres, etc.) — as well as more advanced topics — while not using mathematics beyond what I’ve used in the first several sections of the Geometry unit.

There is, of course, no “correct” answer to this problem.  But I thought I’d share how I’d approach this series of lectures, since geometry is such a passion of mine — and I know it is for many readers as well.  The process of reforming high school geometry courses is now well underway; I hope to contribute to this discussion with today’s post.

Where to start?  Cones and cylinders — a very traditional beginning.  But I thought I’d start with surface areas.  Now for cylinders, this is pretty straightforward.  It’s not much more difficult for cones, but the approach is less obvious than for cylinders.

Earlier in the unit, we derived the formula for the area of a sector of a circle, so finding the lateral surface area of a cone is a nice opportunity to revisit this topic.  And of course, finding the lateral surface area of a cylinder involves just finding the area of a rectangle.

Now what do both of these problems have in common?  Their solution implies that cones and cylinders are flat.  In other words, we reduce what is apparently a three-dimensional problem (the surface area of a three-dimensional object) to a two-dimensional problem.

This is in sharp contrast to finding the surface area of a sphere — you can’t flatten out a sphere.  In fact, the entire science of cartography has evolved specifically in response to this inability.

So this is a nice chance to introduce a little differential geometry!  And no, I don’t really intend to go into differential geometry in any detail — but why not take just a minute in a lecture involving spheres to comment on why the formulas for the surface areas of cones and cylinders are fairly easy to derive, and why — at this level — we’re just given the formula for the surface area of a sphere.

I try to mention such ideas as frequently as I can — pointing out contrasts and connections which go beyond the usual presentation.  Sure, it may be lost on many or most students, but it just may provide that small spark for another.

I think such comments also get at the idea that mathematics is not a series of problems with answers at the back of the book…on the face of it, there is no apparent reason for a student to think that finding the surface area of a cone would be simpler than finding the surface area of a sphere.  This discussion gets them thinking.

Next, I’m planning to discuss Archimedes’ inscription of a sphere in a cylinder (which involves the relative volumes).  This is a bit more straightforward, and it’s a nice way to bring in a little history.

I also plan to look at inscribing a sphere in a right circular cone whose slant height is the same as the diameter of the base, so that we can look at a two-dimensional cross-section to solve the problem.  In particular, this revisits the topic of incircles of triangles in a natural way — I find it more difficult to motivate why you’d want to find an incircle when looking at a strictly two-dimensional problem.

Now on to calculus!  Yes, calculus.  One great mystery for students is the presence of “1/3” in so many volume formulas.  There is always the glib response — the “3” is for “three” dimensions, like the “2” in “1/2 bh” is for “two” dimensions.

When deriving these formulas using integration, this is actually exactly a fairly solid explanation.  But for high school students who have yet to take calculus?

It is easy to approximate the volume of a right circular cone by stacking thin circular disks on top of each other.  If we let the disks get thinner and take more and more of them, we find the volume of the cone as limit of these approximations.  All you need is the sum

\displaystyle\sum_{k=1}^n k^2=\dfrac{n(n+1)(2n+1)}6.

I plan to prove that

\displaystyle\sum_{k=1}^nk=\dfrac{n(n+1)}2,

and then prove (or perhaps just suggest — I’m not sure yet) the formula for the sum of squares.

I think a fairly informal approach could be successful here.  But I do think such discussions are necessary — in calculus, I’ve routinely asked students why certain formulas they remember are true, and they struggle.  As a simple example, students can rarely tell me why the hypotenuse of a 30-60-90 triangle is twice as long as the shorter leg.

When teachers just give students formulas and ask them to plug numbers in to get answers to oversimplified word problems, of course there is a sense of mystery/confusion — where did these formulas come from?  I’m hoping that this discussion suggests that there is a lot more to mathematics than just a bunch of formulas to memorize.

As usual, I realize I have much more to say on this topic than I had originally supposed…I’ve only discussed up to the fifth lecture so far!  Since I have not had extensive experience teaching more traditional topics at the high school, it has been an interesting challenge to tackle the usual geometry topics in a way that grabs students’ attention.  It’s a challenge I enjoy, and of course I’ll have much more to say about it next week….

 

Guest Blogger: Scott Kim, IV

Well, this is the last installment of Scott Kim’s blog post on transforming mathematics education!  These are all important issues, and when you think about them all at once, they seem insurmountable.  It takes each of us working one at a time in our local communities, as well as groups of us working together in broader communities, to effect a change.  What is crucial is that we not only discuss these issues, but we do something about them.  Those of us who participated in the discussion a month ago at the Bay Area Mathematical Artists Seminar are definitely interested in both discussing and doing.

Scott suggests we need to move past our differences and find constructive ways to act.  No, this isn’t easy.  But we need to do this to solve any problem, not just those surrounding mathematics education.  It’s time for some of us to start working on these issues, and many others of us to continue working.  We can’t just sit and watch, passively, any more.  It’s time to act.  What are you waiting for?

Level 4. Resistance from SOCIETY (quarreling crew)

Sailing is a team sport. You can’t get where you want to go without a cooperative crew. Similarly, math education reform is a social issue. You can’t change how math is taught unless parents, teachers, administrators and policy makers are on board. Most adults cling to the way they were taught as if it were the only way to teach math, largely out of ignorance — they simply aren’t aware of other approaches.

Here are three ways society needs to change the way it thinks about math and math education in order for change to happen.

4a. Attitude. The United States has an attitude problem when it comes to math teachers. First, we underpay and under-respect teachers. And the situation is only getting worse as math graduates flock to lucrative high-tech jobs instead of the teaching profession. The book The Smartest Kids in the World and How They Got That Way describes how FInland turned their educational system around — they decided to pay teachers well, set high qualification standards, and give teachers considerable autonomy to teach however they think is best, with the remarkable result that student respect for teachers is extremely high.

Second, it is socially acceptable, even a badge of honor, to say that you were never good at math. You would never say the same thing about reading. Many people do not in fact read books, but no one would publicly brag that they were never good at reading. Our society supports the idea that parents should read to their kids at night, but perpetuates the idea that being no good at math is just fine.

Solution: respect teachers by paying them well, and value math literacy as much as we value reading literacy.

4b. Vision. The national conversation about math education in the United States is locked in a debate about whether we should teach the basics, or the concepts. As a result we see over the decades that the pendulum swings back and forth between No Child Left Behind and standardized testing on one extreme, and New Math and Common Core Math on the other extreme. As long as the pendulum keeps swinging, we will never settle on stable solution. The resolution, of course, is that we need both. In practice, schools that overemphasize rote math find that they must supplement with conceptual exercises, and schools that overemphasize conceptual understanding find that they must supplement with mechanical drill. We need both rote skills and conceptual understanding, just as kids learning to read need both the mechanical skills of grammar and vocabulary, and the conceptual skills of comprehension and argument construction.

Solution: We need a vision of math education that seamlessly integrates mechanical skills and conceptual understanding, in a way that works within the practical realities of teacher abilities and schoolday schedules. To form a vision, don’t just ask people what they want. A vision should go further than conventional wisdom. As Henry Ford is reported to have said (but probably didn’t), “If I had asked people what they wanted, they would have said faster horses.” Or as Steve Jobs did say, “It’s really hard to design products by focus groups. A lot of times, people don’t know what they want until you show it to them.”

4c. The will to act. As a child I grumbled about the educational system I found myself in. As a young adult I started attending math education conferences (regional meetings of the National Council of Teachers of Mathematics), and was astonished to find that all the thousands of teachers at the conference knew perfectly well what math education should look like — full of joyful constructive activities that challenged kids to play with ideas and think deeply. Yet they went back to their schools and largely continued business as usual. They knew what to do, but were unwilling or unable to act, except at a very small scale.

Solution: Yes, a journey of a thousand miles starts with a single step. And change is slow. But if we’re to get where we want to go, we need to think bigger. Assume that big long lasting change is possible, and in the long term, inevitable. As Margaret Mead said, “Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it’s the only thing that ever has.” I’m starting my small group. Others I know are starting theirs. What about you?