Dan,
I really like this lesson, but I want to pick an important nit with you. Did you really measure the size of this doll to +/- 0.01 mm? If so, how did you do this? With a pair of 10 micron calipers and a microscope?

I raise this because as a science teacher, I get infuriated when students tell me that an object that travels 20 meters in 30s is going at 0.666666666 m/s or even worse, 0.6 repeating, when all they did was measure the distance using the marks on a football field and the time using the timer on their iphone.

I know teaching measurement theory isn’t the point of math class, but if we’re going to bring the real world into problems for our kids, the measurements we give them (or have them make) need to be appropriate, right?

I think one element of pseudocontext must also be the made up measurements with fantastical precision that most textbooks hand to students without much explanation.

I have to say Dan, I think this problem as you’ve set it up is a lot more closed than I;m used to from you. Once you give all three pieces of information, I don’t see how this is any different than a traditional textbook problem. A b^n = C

I’d vote that you just have the starting video and let them have at it, have them estimate the size of the dolls just from the video, or put a dollar bill or something standing in the background.

I agree that adding a dollar bill to the mix would only add to estimation (but let’s agree that estimation is important too), but I guess I’m worried that by offering those three particular clues, and only those three clues, you are already suggesting fairly strongly an approach to the problem, giving the average shrink percentage in particular hints strongly at exponentiation. I thought part of the value of ‘be less helpful’ was that it requires the students themselves to formulate the approaches. By giving the average shrink rate, you’ve already short circuited a whole dialog about what sort of governing number would be important for figuring the problem out. A thought process I think could reveal how well the student’s /think/ in terms of exponentiation.

re: post [7]

Measure directly off the screen. Yardstick on the ol’ projection screen, or fancy pixel measurements on a smartboard. Reshoot the video if necessary so that we are sure the height is not distorted by perspective (too much).

Maybe that’s too far removed from real measurement, since we make assumptions about the camera, the angle, and the projection.

> what do you think about me just providing the two heights instead of the percent? Let them calculate the percent?

Both slides could be prepared, and adapted to the class. My last post talked about the idea above about how to measure.

> 75% means (hypothetically) doll #2 is 80% the height of doll #1 while doll #3 is 70% the height of doll #2.

I guess I assumed they were similar under the same ratio. Are they not? If not, can we really answer anything about their progression and how many there are?

Dan: I don’t think you’d want the arithmetic average of .7 and .8

Suppose we have the first three doll heights, a, b, c. And lets say we have two ratios “r” and “s”.

Such that r*a = b and s*b = c.
Thus (rs)*a = c

But you can’t average r and s using (r+s)/2.
(r+s)/2 * (r+s)/2 * a is not equal to (rs)*a.

We need the geometric mean (rs)^(1/2).

Or, if we are finding the geometric mean of all rates, it would be the product of (r_1 * r_2 * … * r_n)^(1/n)

I quickly ran the numbers and it makes such a small difference that it does not affect the outcome, but maybe that would be something to look for in another lesson. When would the arithmetic mean cause us trouble when we really need the geometric mean? And by trouble I mean real trouble.

I’m thinking that if you gave them the height of the first and the second dolls then most students (depending upon year level of course) would assume a linear progression and work from there.

Is giving them the first two heights in an almost deliberate attempt to send them off on a wild goose chase too nasty?

I’m thinking that if you give them those two dimensions then ask if they need anything further. The student who asks for the third dimension is probably thinking along the right lines already?

Personally, I like the idea of not giving exact numbers and letting the students extract them from the video,

But, I’d like to point out that I think that John’s original nit remains to be picked. I think you’ve fallen prey to your instruments Dan. Your calipers may know the distance between their ends to 0.01 mm, but that doesn’t mean you know the height of the doll to 0.01 mm. Grab the largest doll and your calipers and try to measure the height again. I’ll bet you a nickel you won’t get 62.82 mm.

Set the doll down, open the calipers wide and set them down, stand up, walk around your chair, sit down and measure again. Do it ten times. I’ll bet another nickel that the standard deviation of your measurements is greater than 0.1 mm. In fact, I’ll throw another nickel on the measured error in your measurements being within a factor of 3 from 1.0 mm.

I don’t think you should be bothered is the numbers don’t ‘work out’. Real data is messy and thats something we should embrace.

If it helps, Scott, I can’t get any closer than 3 seconds on the escalator problem.

This brings us back to “messy data”. If one of our goals is to help students see mathematics in the world, then messy data is unavoidable. Do we like the stacking dolls because they have an interesting but approximate mathematical pattern? Or do we like them only if we can shoehorn them into an exact pattern? I vote the former.

Yes, this one is slippery. I’m ok with messy data — if its messy due to our collection. But this data is messy inherently.

Compare with the basketball parabola: our predictions could be off because we may not fit the parabola onto the path of the ball precisely enough. But in the case of these dolls, our predictions could be off because the dolls do not automatically follow a predictable pattern. How are we to know if the next ratio will be 43% (14:15), 68% (12:13), or 85% (5:6)?

I mean, the %error by the 4th Doll using geometric ratios is just as large as the %error using a linear model for the 3rd doll. So imagine this scenario:

Call the dolls A, B, C, D, etc…

You have an AB slide and a ABC slide. Especially if students are used to doing linear problems, then have them work based off that assumption from the AB slide. When you reveal the ABC slide we can see that the relationship is not linear. (due to the fact that we are so far off … 7mm (16% error))

But how can we justify using a geometric relationship if we match that error with our geometric calculations? We obtain a measurement for doll D that is 5mm off (15% error) using the first ratio, or 7mm (21%) off again using the “cheater” geomean.

If you consider the geomean of the first TWO ratios (B/A, and C/B) then you get pretty accurate data until the 12th doll when we’re back up to 13% error, and by the 15th doll we’re 200% off.

I think a student that guesses 15 dolls would not have a good reason. Even if its the reason we are hoping her to give.

I’m ok with messy data – if its messy due to our collection. But this data is messy inherently.

I have to strongly disagree with this sentiment, in general. Mathematical models are simplifications that capture only whatever behavior is considered “essential.” In fact, some useful mathematical models are only qualitative, not even attempting to fit the relevant data in a numerical way. One example that comes to mind is the use of Lotka—Volterra equations in population biology.

Even in physics, mathematical models are often only very good approximations. For example, the path of a basketball is not really a parabola, because of air resistance and because gravity acts radially, not “constantly downward.” But again, as is usual in physics, parabolic motion is often a very, very good model.

@R. Wright

I completely agree with you that mathematical models are only very good approximations.

You’re right: the basketball is not *really* a parabola, because of all those physical and natural phenomena.

Why are the dolls not *really* a geometric sequence? Can we study it mathematically? (like we study air resistance and radial gravity?) I’m just afraid these dolls are off due to human craftsmanship, *not* something inherent to the dolls structure.

I guess my question is… if all I can see are the dolls A and B… why should I, as a student, use the ratio instead of the difference? Both of them lead to rather large errors off the actual measurements for Doll C.

Is there anything physically stopping the dolls from following a linear pattern? Remember, we know only the heights A and B.

I know our ideal set of stacking dolls would fit the geometric sequence, but how do we justify following that ideal? What is it about stacking dolls that makes them shrink geometrically?

Are they really similar?

Dan, could you provide measurements of the cross-sections? Does the wood’s thickness shrink? It should, if they were similar. If the heights of the dolls shrink by a ratio of B/A, so should the thickness.

But I would hypothesize that they don’t. This would mean the ratios of of each successive doll would have to shrink. Which the data shows that they do (after a bit). Look at the “actual height ratios” http://scottfarrar.com/algebra/data%20stacking%20dolls%20dan%20meyer.png

The first exception (the first ratio) is due to the pedestal on Doll A that is included in the height. No other dolls have this. **this could be something to have kids point out**

Then it holds steady at about 84%. So perhaps the thickness is shrinking here. But all of a sudden the height ratios make a break, and start rapidly descending. I hypothesize that this is where the thickness stops obeying similarity to satisfy artistry (and physics). My point is… the artist CHOSE to make more dolls, even after it was impossible physically to construct the dolls with thin enough wood.

Has no one else graphed this data? It provides awfully convincing evidence that Dan’s original task was phony (all due respect).

Scott’s model is much, much better. Although it poorly predicts the 15th doll.

Alright, this is getting interesting again. So, in light of the fact that the dolls do appear to run into a problem with their thickness near the end of the series, I propose a new model. Assume the dolls are each of width w, and attempt to get smaller by a factor of b at each step, until they run into the issue with their width.

I.e. H_n = min( H_{n-1} * b , H_{n-1} – w )

The height of the nth doll is the minimum of the height of the n-1th doll times b, and the height of the n-1th doll minus the thickness.

Doing this, we have a nice two stage behavior. With H_1 = 61.4, b =0.8, w = 1.5, I obtain the following: Model graph

This new model neatly incorporates an exponential decline, until the minimum becomes important, and then a linear decrease.

In light of the issue with the pedestal, I took the initial height to be 61.4. Now we are in a position to predict the number of dolls in terms of our two parameters, b and w, wherein this model neatly terminates the number of dolls when H_n attempts to go negative. Doing so, I obtain this plot , showing nicely the models behavior as the parameters vary.

Finally, to ease the eye, I’ve singled out the region of parameter space where 15 dolls are predicted here