Creativity in Mathematics

Of course, there is always more to say about hyperbolic trigonometry….  Next, we’ll look at what is usually called the logistic curve, which is the solution to the differential equation

The logistic curve comes up in the usual chapter on differential equations, and is an example of population growth.  Without going into too many details (since the emphasis is on hyperbolic trigonometry), is a constant which influences how fast the population grows, and is called the carrying capacity of the environment.

Note that when is very small, and so the population growth is almost exponential.  But when gets very close to then and so population growth slows down.  And of course when growth stops — hence calling the carrying capacity of the environment.  It represents the largest population the environment can sustain.

Here is an example of such a curve where and

Notice the S shape, obtained from a curve rapidly growing when the population is small. It happens that the population grows fastest at half the carrying capacity, and then growth slows to zero as the carrying capacity is reached.

Skipping the details (simple separation of variables), the solution to this differential equation is given by

I will digress for a moment, however, to mention partial fractions (as I step on my calculus soapbox).  I have mentioned elsewhere that incomprehensible chapter in calculus textbooks:  Techniques of Integration.  Pedagogically a disaster for so many reasons.

The first time I address partial fractions is when summing telescoping series, such as

It really is necessary.  But I only go so far as to be able to sum such series.  (Note:  I do series as the middle third of Calculus II, rather than the end.  A colleague suggested that students are more tired near the end of the course, which is better for a more technique-oriented discussion of the solution to differential equations, which typically comes before series.)

You also need partial fractions to solve the differential equation for the logistic curve, which is when I revisit the topic.  After finding the logistic curve, we talk about partial fractions in more detail.  The point is that students see some motivation for the method of partial fractions — which they decidedly don’t in a chapter on techniques of integration.

OK, time to step off the soapbox and talk about hyperbolic trigonometry….  The punch line is that the logistic curve is actually a scaled and shifted hyperbolic tangent curve!  Of course it looks like a hyperbolic tangent, but let’s take a moment to see why.

We first use the definitions of and to write

This results in

You can see the form of the equation of the logistic curve starting to take shape.  Since the hyperbolic tangent has horizontal tangents at and we need to scale by a factor of so that the asymptotes of the logistic curve are units apart:

Note that this puts the horizontal asymptotes of the function at and

To take into account the initial population, we need a horizontal shift, since otherwise the initial population would be We can accomplish this be replacing with

We’re almost done at this point:  we simply need

Solving and substituting back results in

which, since is an odd function, becomes

And there it is!  The logistic curve as a scaled, shifted hyperbolic tangent.

Now what does showing this accomplish?  I can’t give you a definite answer from the point of view of the students.  But for me, it is a way to tie two seemingly unrelated concepts — hyperbolic trigonometry and solution of differential equations by separation of variables — together in a way that is not entirely contrived (as so many calculus textbook problems are).

I would love to perform the following experiment:  work out the solution to the differential equation together as a guided discussion, and then prompt students to suggest functions this curve “looks like.”  Of course the might be suggested, but how would we relate this to the exponential function?

Eventually we’d tease out the hyperbolic tangent, since this function actually does involve the exponential function.  Then I’d move into an inquiry-based lesson:  give the students the equation of a logistic curve, and have them work out the conversion to the hyperbolic tangent.

And as is typical in such an approach, I would put students into groups, and go around the classroom and nudge them along.  See what happens.

I say that yes, calculus students should be able to do this.  I recently sent an email about pedagogy in calculus which, among other things, addressed the question:  What do calculus students really need to know?

There is no room to adequately address that important question here, but in today’s context, I would say this:  I think it is more important for a student to be able to rewrite as a hyperbolic tangent than it is for them to know how to sketch the graph of

Why?  Because it is trivial to graph functions, now.  Type the formula into Desmos.  But how to interpret the graph?  Rewrite it?  Analyze it?  Draw conclusions from it?  We need to focus on what is no longer necessary, and what is now indispensable.  To my knowledge, no one has successfully done this.

I think it is about time for that to change….

We continue where we left off on the last post about hyperbolic trigonometry.  Recall that we ended by finding an antiderivative for using the hyperbolic trigonometric substitution  Today, we’ll look at this substitution in more depth.

The functional relationship between and is described by the gudermannian function, defined by

This is not at all obvious, so we’ll look at the derivation of this rather surprising-looking formula.  It’s the only formula I’m aware of which involves both the arctangent and the exponential function.  We remark (as we did in the last post) that we restrict to the interval so that this relationship is in fact invertible.

We use a technique similar to that used to derive a formula for the inverse hyperbolic cosine.  First, write

and then multiply through by to obtain the quadratic

This quadratic equation results in

Which sign should we choose?  We note that and increase together, so that because is an increasing function of then must be an increasing function of It is not difficult to see that we must choose “plus,” so that  and consequently

We remark that no absolute values are required here; this point was discussed in the previous post.

Now to solve for  The trick is to use a lesser-known trigonometric identity:

There is such a nice geometrical proof of this identity, I can’t help but include it.  Start with the usual right triangle, and extend the segment of length by in order to form an isosceles triangle.  Thus,

To find observe that is supplementary to both and so that

which easily implies

Therefore

which is precisely what we need to prove the identity.

Now we substitute back into the previous expression for which results in

This may be solved for giving

So let’s see how to use this to relate circular and hyperbolic trigonometric functions.  We have

which after using the usual circular trigonometric identities, becomes

It is also an easy exercise to see that

So revisiting the integral

we may alternatively make the substitution giving

which is the same simple integral we saw in the previous post.

What about the other trigonometric functions?  Certainly we know that  Again using the usual circular trigonometric identities, we can show that

Knowing these three relationships, the rest are easy to find:  and

I think that the gudermannian function should be more widely known.  On the face of it, circular and hyperbolic trigonometric functions are very different beasts — but they relate to each other in very interesting ways, in my opinion.

I will admit that I don’t teach students about the gudermannian function as part of a typical calculus course.  Again, there is the issue of time:  as you are well aware, students finishing one course in the calculus sequence must be adequately prepared for the next course in the sequence.

So what I do is this:  I put the exercises on the gudermannian function as extra challenge problems.  Then, if a student is already familiar with hyperbolic trigonometry, they can push a little further to learn about the gudermannian.

Not many students take on the challenge — but there are always one or two who will visit my office hours with questions.  Such a treat for a mathematics professor!  But I feel it is always necessary to give something to the very best students to chew on, so they’re not bored.  The gudermannian does the trick as far as hyperbolic trigonometry is concerned….

As a parting note, I’d like to leave you with a few more exercises which I include in my “challenge” question on the gudermannian.  I hope you enjoy working them out!

1.  Show that
2. Show that
3. Show that if is the inverse of the gudermannian function, then

Now on to some calculus involving hyperbolic trigonometry!  Today, we’ll look at trigonometric substitutions involving hyperbolic functions.

Let’s start with a typical example:

The usual technique involving circular trigonometric functions is to put so that and the integral transforms to

In general, we note that when taking square roots, a negative sign is sometimes needed if the limits of the integral demand it.

This integral requires integration by parts, and ultimately evaluating the integral

And how is this done?  I shudder when calculus textbooks write

How does one motivate that “trick” to aspiring calculus students?  Of course the textbooks never do.

Now let’s see how to approach the original integral using a hyperbolic substitution.  We substitute so that and  Note well that taking the positive square root is always correct, since is always positive!

This results in the integral

which is quite simple to evaluate:

Now and

Recall from last week that we derived an explicit formula for and so our integral finally becomes

You likely noticed that using a hyperbolic substitution is no more complicated than using the circular substitution  What this means is — no need to ever integrate

again!  Frankly, I no longer teach integrals involving and which involve integration by parts.  Simply put, it is not a good use of time.  I think it is far better to introduce students to hyperbolic trigonometric substitution.

Now let’s take a look at the integral

The usual technique?  Substitute and transform the integral into

Sigh.  Those irksome tangents and secants.  A messy integration by parts again.

But not so using  We get and (here, a negative square root may be necessary).

We rewrite as

This results in

All we need now is a formula for which may be found using the same technique we used last week for

Thus, our integral evaluates to

We remark that the integral

is easily evaluated using the substitution  Thus, integrals of the forms and may be computed by using the substitutions and respectively.  It bears repeating:  no more integrals involving powers of tangents and secants!

One of the neatest applications of hyperbolic trigonometric substitution is using it to find

without resorting to a completely unmotivated trick.  Yes, I saved the best for last….

So how do we proceed?  Let’s think by analogy.  Why did the substitution work above?  For the same reason works: we can simplify using one of the following two identities:

So is playing the role of and is playing the role of  What does that suggest?  Try using the substitution !

No, it’s not the first think you’d think of, but it makes sense.  Comparing the use of circular and hyperbolic trigonometric substitutions, the analogy is fairly straightforward, in my opinion.  There’s much more motivation here than in calculus textbooks.

So with  we have

But notice that — just look at the above identities and compare. We remark that if is restricted to the interval then as a result of the asymptotic behavior, the substitution gives a bijection between the graphs of and and between the graphs of and In this case, the signs are always correct — and always have the same sign.

So this means that

What could be simpler?

Thus, our integral becomes

But

Thus,

Voila!

We note that if is restricted to the interval as discussed above,  then we always have so there is no need to put the argument of the logarithm in absolute values.

Well, I’ve done my best to convince you of the wonder of hyperbolic trigonometric substitutions!  If integrating didn’t do it, well, that’s the best I’ve got.

The next installment of hyperbolic trigonometry?  The Gudermannian function!  What’s that, you ask?  You’ll have to wait until next time — or I suppose you can just google it….

I love hyperbolic trigonometry.  I always include it when I teach calculus, as I think it is important for students to see.  Why?

1.  Many applications in the sciences use hyperbolic trigonometry; for example, the use of Laplace transforms in solving differential equations, various applications in physics, modeling population growth (the logistic model is a hyperbolic tangent curve);
2. Hyperbolic trigonometric substitutions are, in many instances, easier than circular trigonometric substitutions, especially when a substitution involving or is involved;
3. Students get to see another form of trigonometry, and compare the new form with the old;
4. Hyperbolic trigonometry is fun.

OK, maybe that last reason is a bit of hyperbole (though not for me).

Not everyone thinks this way.  I once had a colleague who told me she did not teach hyperbolic trigonometry because it wasn’t on the AP exam.  What do you say to someone who says that?  I dunno….

In any case, I want to introduce the subject here for you, and show you some interesting aspects of hyperbolic trigonometry.  I’m going to stray from my habit of not discussing things you can find anywhere online, since in order to get to the better stuff, you need to know the basics.  I’ll move fairly quickly through the introductory concepts, though.

The hyperbolic cosine and sine are defined by

I will admit that when I introduce this definition, I don’t have an accessible, simple motivation for doing so.  I usually say we’ll learn a lot more as we work with these definitions, so if anyone has a good idea in this regard, I’d be interested to hear it.

The graphs of these curves are shown below.

The graph of is shown in blue, and the graph of is shown in red.  The dashed orange graph is which is easily seen to be asymptotic to both graphs.

Parallels to the circular trigonometric functions are already apparent:   is an even function, just like  Similarly, is odd, just like

Another parallel which is only slight less apparent is the fundamental relationship

Thus, lies on a unit hyperbola, much like lies on a unit circle.

While there isn’t a simple parallel with circular trigonometry, there is an interesting way to characterize and  Recall that given any function we may define

to be the even and odd parts of respectively.  So we might simply say that and are the even and odd parts of respectively.

There are also many properties of the hyperbolic trigonometric functions which are reminiscent of their circular counterparts.  For example, we have

and

None of these are especially difficult to prove using the definitions.  It turns out that while there are many similarities, there are subtle differences.  For example,

That is, while some circular trigonometric formulas become hyperbolic just by changing to and to sometimes changes of sign are necessary.

These changes of sign from circular formulas are typical when working with hyperbolic trigonometry.  One particularly interesting place the change of sign arises is when considering differential equations, although given that I’m bringing hyperbolic trigonometry into a calculus class, I don’t emphasize this relationship.  But recall that is the unique solution to the differential equation

Similarly, we see that is the unique solution to the differential equation

Again, the parallel is striking, and the difference subtle.

Of course it is straightforward to see from the definitions that and  Gone are the days of remembering signs when differentiating and integrating trigonometric functions!  This is one feature of hyperbolic trigonometric functions which students always appreciate….

Another nice feature is how well-behaved the hyperbolic tangent is (as opposed to needing to consider vertical asymptotes in the case of ).  Below is the graph of

The horizontal asymptotes are easily calculated from the definitions.  This looks suspiciously like the curves obtained when modeling logistic growth in populations; that is, finding solutions to

In fact, these logistic curves are hyperbolic tangents, which we will address in more detail in a later post.

One of the most interesting things about hyperbolic trigonometric functions is that their inverses have closed formulas — in striking contrast to their circular counterparts.  I usually have students work this out, either in class or as homework; the derivation is quite nice, so I’ll outline it here.

So let’s consider solving the equation for  Begin with the definition:

The critical observation is that this is actually a quadratic in

All that is necessary is to solve this quadratic equation to yield

and note that is always negative, so that we must choose the positive sign.  Thus,

And this is just the beginning!  At this stage, I also offer more thought-provoking questions like, “Which is larger, or  These get students working with the definitions and thinking about asymptotic behavior.

Next week, I’ll go into more depth about the calculus of hyperbolic trigonometric functions.  Stay tuned!