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!