## Hanging Slinky Analysis 2: The Pre-Tension Wrinkle

Last time, we saw how the length of a hanging slinky is quadratic in the the number of links, namely,

$\Delta x = \int \mathrm{d}x = \int_0^M \sigma mg \, \mathrm{d}m = \frac {\sigma M^2 g}{2}$,

where M is the mass of the hanging part of the slinky, g is the acceleration of gravity, and $\sigma$ is the “stretchiness” of the material (related to the spring constant k—but see the previous post for details).

And this almost perfectly fit the data, except when we looked closely and found that the fit was better if we slid the parabola to the right a little bit. Here are the two graphs, with residual plots:

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## Hanging Slinky Analysis 1: Sums to Integrals

Last time, we (re-)introduced the Hanging Slinky problem, designed a few years back as a physics lab but suitable for a math class, say Algebra II or beyond. We looked at the length of the hanging slinky as a function of the number of slinks that hang down, and it looked seriously quadratic.

I claim that knowing that the real-world data is quadratic will help you explain  why the data has that shape. That is, “answer analysis” will guide your calculations.

I beg you to work this out for yourself as much as you can before reading this. I made many many many wrong turns in what is supposed to be an easy analysis, and do not want to deprive you of that—and the learning that comes with it.

Slinkies are great. You can demonstrate waves. You can make them go down stairs. They are super-dynamic physics toys. They make a great sound.

But they are also pretty great when static. Consider, for example, a hanging slinky. How far down does it hang?

Well. It depends.

For this post, I’ll skip the question-posing part of this and go directly to what it mostly depends on: the number of coils (slinks) that are hanging down.

Let’s skip all the way to the data. Here is a graph of the length (in cm) of a hanging slinky as a function of the number of slinks. You should, of course, record your own data, if for no other reason than to experience the glorious difficulty of measuring the distance.

We can pause here and make sure the graph makes sense. What do you see in the slinky itself? How would you describe the spacing of the coils in the hanging slinky? How does that pattern get reflected in the data and in the graph?  Continue reading

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## Does Modeling Mean Using Harder Functions?

Ladder and wall, from the “mathisfun” site. They’re doing the trig problem, but no matter.

Here’s something I’m puzzled about in trying to push this picture of math education, the one where we collect data and try to model it with functions: when I come up with ideas for suitable activities, they often require “harder” functions than students may be used to seeing. Let me give an example based on a super-traditional problem such as

A 5-meter ladder is leaning against a wall. The bottom of the ladder is 2.5 meters from the foot of the wall. How high is the top of the ladder?

We know what the kid is supposed to do, traditionally. Draw the picture, recognize the right triangle, notice that the hypotenuse is 5, and write something like

$(2.5)^2+x^2=5^2$

and solve the equation to find that $x = \sqrt{25-6.25} = \sqrt{18.75} \approx 4.33$.

But in the approach I’m trying to push, we de-emphasize the specific answer and focus on the relationship. We take a 5-meter ladder (or its more practical equivalent, a chair) and set it next to a wall, at different distances (let’s call that distance base), and see how high the ladder reaches (height). We plot height against base, and try to figure out a function that is a good model for the data.  Continue reading

## The Home Plate Area Mystery

A major-league home plate. Dimensions in inches. From Math World. Click the image to go there.

The illustration shows the dimensions, in inches, of a major league home plate, according to the official rules of baseball.

What’s the area of the plate?

Another way to present this question is to note that major-league bases are 15 inches square—and wonder which is bigger.

In either case, the problem of figuring out the area of this pentagon involves taking shapes apart or sticking them together. This skill, of dissecting or composing shapes, is important for students; they need to visualize the easy shapes that are inside (or outside) the hard ones. It even appears in the core standards, obliquely, in 7.G.6. It’s also wonderful that there are different ways to do it.

In this case, it’s not too hard. Students who know how to find the area of a right triangle can be successful without much teacher intervention.

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## The Mercator Saga (part 3): the integral

At 60° North, the red box is roughly 10° by 10°. Notice the 2:1 aspect ratio.

You should already have read or skimmed part 1 and part 2 of this saga. In part 2, we showed the picture of the globe (reproduced here) that purported to show how the aspect ratio of a 10° by 10° box at 60° was 2:1.

Of course, it’s only approximately 2:1, I told the students, because (small pause to get attention) the scale on the globe is changing with every degree. In fact, it’s changing continuously. At this point, the students muttered, “mmm, calculus.” So at least they smelt it coming.

I’d like to be able to say that at this point I turned it over to them: “Indeed! Calculus! We’re trying to find $y(\theta)$, the function that gives us the total y-distance on the map as a function of the latitude. Work in your groups to figure out just what calculus you need to do to find that function, and be ready to present in about five minutes.”

The chart, phase three. What happens at an arbitrary angle?

But I can’t. Being realistic, it was good that they sensed calculus, but in this unusual context—the sphere and everything—it would have been excruciating. So I scaffolded like my life depended on it. I reproduce the chart we developed last time that shows the dimensions of the 1° boxes.

“Indeed! Calculus! After all, we want $y(\theta)$, the y-distance as a function of latitude. So to find the total distance, to, say, 60°, I’m going to start at 0° and use this number we have in the table, sec(0°), times 1° because that’s the height per 1°, right?. Then for the next degree, I have to add sec(1°) times 1°, then for the next one, sec(2°) times 1°, all the way up to 60.”

I write the sum on the board, (still ignoring the constant factor, by the way, but that was OK):

$\displaystyle (\sec 0 \times 1) + (\sec 1 \times 1) + (\sec 2 \times 1) + ... + (\sec 59 \times 1)$

“And of course, within each of these 1° sections, it’s changing continuously too. So what should we do?”

Now, almost in unison, “integrate secant,” and a student bravely writes,

$\displaystyle y(\theta) = \int_0^\theta \sec x \,\mathrm{d}x$

### Doing the Integral—the way cool way

“And of course, you all know the integral of the secant, right?”

They squirm, but I take them off the hook. “Of course you don’t. Nobody should remember that. But what’s the most practical way to find this integral?”

“A substitution?” they ask. “Do we, like, do it by parts somehow?”

“All good suggestions, but it’s really not obvious what to substitute. And for parts, usually there’s more stuff, right? So you can have a v and a du? What I’m really asking is, what’s a really practical way to find this integral?”

“Look it up?”

“Yeah! How?”

“That will work, but I want to show you another way.” I pull out my iPhone and hook it up to the projector. “Siri, integrate secant of x.”

“You’re kidding me,” says one kid. Class chatter rises.

“Hmm, let me think…,” says Siri. She shows me an approximation and a link to Wolfram Alpha, which I tap. And this appears:

Siri integrates sec(x)

At this point, they returned to desmos and their data to see if this function actually worked. And it did, to enormous satisfaction throughout the class. That this obscure function, a log of a combination of trig functions, fit these data, was a bit of a miracle, a triumph of actually using calculus.

Data are red dots. Orange is a bogus tangent function. Blue is based on our integral.

### Actually Finding the Integral

We did not find the integral in class, but in case you care, here is a derivation, which shows just how arcane these can get. This is why you look these up. And why it’s so great that Wolfram Alpha exists, because they give you the answer and the derivation.

It turns out that there is a great substitution! First you multiply top and bottom by $(\tan x + \sec x)$:

$\displaystyle \int \sec x \,\mathrm{d}x = \int \frac{\sec x(\tan x + \sec x)}{\tan x + \sec x}\, \mathrm{d}x$

Now we do a u-substitution, choosing

$u = \tan x + \sec x$.

When you take its derivative, you discover that by some miracle,

$\mathrm{d}u = \sec x(\tan x + \sec x)\,\mathrm{d}x$

Which means that

$\displaystyle \int \sec x \,\mathrm{d}x = \int \frac{du}{u} = \ln u = \ln (\tan x + \sec x)$

Which turns out (after even more mind-numbing algebra than it took to find du above) to be equivalent to the horrible formula with the half-angles that Wolfram gave us at first.

Perhaps there is a way to anticipate that that particular substitution would work out so well, but I sure don’t have that kind of insight.

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## The Mercator Saga (part 2): Finding the Scale

Mercator projection of the World (also a handout students used). Basic grid spacing is 15°, but it includes the tropics of Cancer and Capricorn so don’t be deceived.

You really don’t want to read all the details. But among the crucial steps in getting the lesson outlined last time to work right, the crucialest might be the place where the students figure out that the y-scale goes as the secant.

How do we help them figure that out? (And what do we mean by that?) That’s what we’ll talk about today.

First of all: that y-scale is the number of centimeters per degree in the y-direction, and that depends on latitude. And when you look at a Mercator-projected Earth (like the one in the figure) you can see that this scale increases with latitude. How do we know? Because the lines of latitude get farther apart. So more centimeters per degree.

### Coping with Converging Longitudes

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## The Mercator Saga (part 1)

Mercator projection

Last week, Zoya let me into her calculus class to do a data-rich activity of my choosing. Ideally it would involve calculus, appropriate for these kids who had already taken the AB exam. Most of my activities that use elementary functions to model data we get from the world (The Model Shop) or measure ourselves (EGADs) don’t involve calculus, although I think they suit a wide range of students.

For some weeks I thought about freeways, and the optimization problem of figuring out at what speed freeways have the greatest flow of traffic. It’s yummy because of the optimization, of course (and that gives us calculus, or at least smells that way) and also because you have to wrap your mind around what you mean by flow. I also found public data from CalTrans—but that’s all a story for another time.

Wisely, I think, I backed off that idea and instead went with a problem I faced long ago when I tried to write a program to draw a Mercator-projection map of the world. Namely: what’s the function in the y-direction that relates distance on the map to latitude?

The plan went like this:

## Chord Star in the Classroom

A million thanks to Zoya Voskoboynikov and her two sections of “regular” calculus at Lick for letting me come litter their otherwise pure math class with actual data. Of course, it was after the AP exam, and these are last-quarter seniors, so my being there didn’t interfere with any learning they needed to get through.

It worked great. It had what I most wanted: the aha experience of arriving at the destination by another route. Fortunately (and unsurprisingly), none of these successful math students remembered the theorem from the geometry class they took as frosh.

### What we did

1. I set up the problem and had them predict, informally, what the function would look like. The main purpose of this is to orient students to what we’ll be measuring and to the idea that if you measure two quantities, you can see their relationship in a graph. Continue reading
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## Beating the Modeling Drum

Hoping desperately it’s not also a dead horse…

We just did a three-post sequence about “Chord Stars,” finishing up with how we could use insights from data to find radii of curvature remotely, that is, without ever finding the center of the circle. There’s a lot to discuss about that process; this post is part of that discussion.

In particular, it’s an interesting example of modeling. Quite a while ago I was worrying about the definition of modeling, not simply to get it “right”—many people model in different ways—but rather to try to identify things that we were pretty sure demonstrated modeling. Part of my anxiety, as the Core Standards lumber into classrooms, is that people will carelessly define modeling as “real-world” (or something equally weak) and we will lose a great opportunity to improve math education.

I often think of modeling in terms of using functions to model data. That’s partly because some of the coolest, most wonderful math experiences I’ve had have revolved around finding a function that was a good approximation to data. The process of measurement, improving those measurements, finding a suitable function, getting insight about the function as I wrangled it, and getting insight into the situation and the data from the function—all that together is an intoxicating cocktail of mathy-worldy wonderfulness.

But it’s not all there is to modeling, so I want to pause to point out another modeling genre (one of the ones I listed in this old post) that just appeared in Chord Star 3, namely, modeling real-world stuff with geometrical objects.

In fact, here are a curb with tools, and the relevant part of a Sketchpad sketch:

They clearly resemble each other, but I want to make two observations: Continue reading