Thursday, February 1, 2018

Lesson 10-1: Surface Areas of Prisms and Cylinders (Day 101)

Chapter 10 of the U of Chicago text is on surface areas and volumes. Measurement is usually the focus of the three-dimensional chapters in a Geometry text, not Euclid's propositions that we've been discussing the past two weeks.

Before we begin, let's return to the Queen of the MTBoS, Fawn Nguyen. Yesterday, she posted an interesting question about grading. I didn't mention it in yesterday's post since I was already tied up with the "super blue blood moon," traditionalists, and Sudbury Valley School. (Why would I mention Fawn's grading question in the same post as Sudbury, which doesn't even give grades?) And so I decided to save her grading question until today's post.

My 7th graders have a question on their exam that asks them to put eight numbers (integers and fractions) in order of their distance from 0 on the number line, starting with the smallest distance.
These types of questions are tricky for me to grade, and because there are eight numbers in this sequence, the task of grading it fairly suddenly becomes thorny and irksome.
Most of the time, I've seen ordering questions such as these only as multiple choice questions, so students can only choose the right answer or the wrong answer. But I suspect this sort of question is becoming more popular because they appear on the SBAC -- and on that test, this problem is not a multiple-choice question.

Nguyen continues:

Let’s change the question to this:
Put these numbers in order from least to greatest: 5, 7, 2, 3, 1, 4, 6, 8
The correct order is 1, 2, 3, 4, 5, 6, 7, 8 (you’re welcome) — for a possible score of 8 points. How many points would this response earn?

1,  2,  4,  5,  3,  7,  8,  6

This post has drawn many comments -- Nguyen's readers try to give their own scoring systems for how they would grade these.

Here's my take on this problem. One of the simplest possible scoring systems is what I call the "Hole in One" method. This is inspired by the game show The Price Is Right, where contestants are asked to order six grocery items from least to most expensive. A higher score entails that they get to make an easy golf putt from right next to the hole to win a car, while a lower score means they must putt from farther away.

Here is the Hole in One method defined -- the score is given as the quantity of increasing values before the first decrease. So in the example above, we have four increasing numbers (1, 2, 4, 5) before the first decrease (from 5 to 3). Therefore this student would score 4/8. Nguyen continues:

1, 2, 7, 8, 6, 4, 5, 3

This student would actually get the same 4/8 score, since there are likewise four increasing numbers (1, 2, 7, 8) before the first decrease (from 8 to 6).

Here are the rest of Nguyen's examples, scored using the Hole in One method:

Set A: 1, 2, 4, 5, 3, 7, 8, 6 (score 4/8)
Set B: 1, 2, 7, 8, 6, 4, 5, 3 (score 4/8)
Set C: 1, 2, 3, 5, 7, 8, 4, 6 (score 6/8)
Set D: 4, 1, 3, 2, 5, 7, 8, 6 (score 1/8)
Set E: 4, 7, 1, 8, 6, 3, 2, 5 (score 2/8)
Set F: 1, 2, 3, 4, 5, 7, 8, 6 (score 7/8)
Set G: 6, 8, 1, 2, 3, 4, 5, 7 (score 2/8)
Set H: 8, 7, 6, 1, 2, 3, 4, 5 (score 1/8)
Set I: 1, 2, 3, 7, 8, 5, 6, 4 (score 5/8)
Set J: 4, 5, 1, 2, 6, 8, 7, 3 (score 2/8)

The first thing we notice is that a score of 0/8 is impossible, even if the student begins with 8. On the actual game show, a player who starts with the most expensive item still gets to putt from the 1st hole, not the "0th hole" (whatever that would mean).

But here's a bigger flaw with the Hole in One method:

Set K: 2, 3, 4, 5, 6, 7, 8, 1 (score 7/8)
Set L: 2, 1, 3, 4, 5, 6, 7, 8 (score 1/8)

So apparently, the student who thinks the least number is the greatest number gets almost all the points, while the student who confuses only the two least numbers and gets everything else right scores only one point.

But believe it or not, this is how the actual Hole in One game on The Price Is Right works. Indeed, suppose a player doesn't know which of two items is the cheapest, yet knows the exact prices of all the remaining items. It's actually better strategy for that player to guess one of the cheap items first and then save the other for last (Set K). This guarantees the player 5/6 and a putt from the second closest line to the hole. But inevitably, the player will guess both cheap items first (Set L) -- a 50-50 shot at either 6/6 (a $500 bonus, which is nothing compared to the car) or 1/6 (the farthest line from the hole, where host Bob Barker or Drew Carey would make his "inspiration putt").

And using the Hole in One method has deeper implications for grading. Remember that this isn't a class of preschoolers ordering the numbers 1 to 8 -- these are seventh graders ordering rational numbers, with some positive and some negative. Student L makes an error ordering the two least (most likely negative) numbers -- an easy mistake to make. But Student K is claiming that a negative number is the largest -- a foolish error. Yet Student K scores 7/8 and Student L earns only 1/8. (Of course, on the real game show all prices are positive, so this isn't an issue.)

Perhaps a better scoring system is a reverse Hole in One method. Instead of starting with the first number and stopping at the first decrease, we start at the last number and go backwards until we reach the first increase. Forward Hole in One emphasizes getting the negative numbers right, while reverse Hole in One places the emphasis on the positive numbers. A student who can't even order the positive numbers correctly deserves a lower score than one who errs on the negatives.

Here are the examples again, using reverse Hole in One:

Set A: 1, 2, 4, 5, 3, 7, 8, 6 (score 1/8)
Set B: 1, 2, 7, 8, 6, 4, 5, 3 (score 1/8)
Set C: 1, 2, 3, 5, 7, 8, 4, 6 (score 2/8)
Set D: 4, 1, 3, 2, 5, 7, 8, 6 (score 1/8)
Set E: 4, 7, 1, 8, 6, 3, 2, 5 (score 2/8)
Set F: 1, 2, 3, 4, 5, 7, 8, 6 (score 1/8)
Set G: 6, 8, 1, 2, 3, 4, 5, 7 (score 6/8)
Set H: 8, 7, 6, 1, 2, 3, 4, 5 (score 5/8)
Set I: 1, 2, 3, 7, 8, 5, 6, 4 (score 1/8)
Set J: 4, 5, 1, 2, 6, 8, 7, 3 (score 1/8)
Set K: 2, 3, 4, 5, 6, 7, 8, 1 (score 1/8)
Set L: 2, 1, 3, 4, 5, 6, 7, 8 (score 7/8)

Now Sets K and L look better. The student who lists a negative number as the greatest earns only 1/8, while the student who gets all the positives right and confuses two negatives earns 7/8.

But still, something seems wrong here when we look at Nguyen's sequences. None of her examples correctly place the largest positive number, so the scores are low. Nonetheless, do Students G and H, who place the largest number near the beginning, really deserve 5/8 or 6/8? These could be students who know that the grading system is reverse Hole in One and are using a sort of strategy.

For example, suppose these are the actual numbers to be sorted:

-3, 36/7, 4, -5, 1, 17/6, 0, 2

A student sees two fractions in the list and balks, since fractions always confuse him. So he crosses out the fractions and is left with:

-3, 4, -5, 1, 0, 2

This student knows how to order integers properly:

-5, -3, 0, 1, 2, 4

And then, knowing that the scoring system is reverse Hole in One, he sticks the two fractions in front, even though they're positive:

17/6, 36/7, -5, -3, 0, 1, 2, 4

This student has exactly produced Nguyen's Set G from above. He knows that the scoring system is reverse Hole in One, and so he'd rather guarantee himself 6/8 rather than attempt to place the fractions correctly and risk getting 1/8 or 2/8. (If he'd known that the scoring system was forward Hole in One instead, he would have put the two fractions last.)

So does this student really deserve 6/8? On one hand, at least he knows how to order the integers, so he probably deserves more than 1/8 or 2/8. On the other, the scoring system rewards his bad behavior of not even attempting to place the fractions correctly. Indeed, students should be rewarded for at least knowing that the fractions are more than 0 (and indeed more than 1, since they're improper fractions), rather than punished with a 1/8 for thinking 36/7 < 4 while getting everything else correct (including the 17/6). Of course, all this assumes that the student knows what the grading system is -- I assume that Nguyen isn't going let her students know how she's grading them.

Some of Nguyen's commenters propose more sophisticated scoring systems, such as counting inversions, calculating deviations, and so on. These avert students who try to game the system as I did above with the -3, 36/7, ..., sequence. Of course, a teacher with five classes and 35 seventh graders in each one won't want to use such a complicated scoring system. One commenter coded a program that would calculate his formula given a student's answer sequence. I wonder whether the program would accept -3, 36/7, ..., as input, or must the teacher convert it to 1, 2, ..., as Nguyen does in her examples.

Last year, I had one group of about 25 seventh graders, so it was possible for me to try one of the more complicating grading formulas. But in the end, I didn't really reach comparing rational numbers before I left the class. The Hole in One methods are flawed -- but of course, game shows prefer simple rules to opaque formulas that confuse contestants and viewers -- imagine Bob or Drew explaining that due to some complicated formula, the player must putt from the second line instead of the fifth line. Such a player may sue the show after losing the car in this scenario!

Of course, there is one gold standard for how such a question should be graded -- since this is really an SBAC Prep question, how does the actual SBAC determine how many points to award? But of course, this is why many people criticize Common Core -- there is no transparency regarding how the tests are graded.

Lesson 10-1 of the U of Chicago text is called "Surface Areas of Prisms and Cylinders." In the modern Third Edition of the text, surface areas of prisms and cylinders appear in Lesson 9-9.

Two years ago, I actually borrowed a lesson from the (former) King of the MTBoS, Dan Meyer. (I already mentioned reigning Queen Fawn Nguyen earlier in this post.)

This is what I wrote two years ago about today's lesson:

Today's lesson is on surface areas. But recall that back at the end of the first semester, I mentioned Dan Meyer, the King of the MTBoS (Math Twitter Blogosphere), and his famous 3-act lessons. I pointed out how one of his lessons was based on surface area, and so I would wait until we reached surface area before doing his lesson. (In fact, Monday's lesson also comes from a MTBoS member, Lisa Winer. There were just so many ideas that happen to correspond to this week's lessons and I'm posting them all this week!)

Well, we've reached surface area. And so I present Dan Meyer's 3-act activity, "Dandy Candies," a lesson on surface area.

Meyer includes some additional questions for teachers to ask the students, but I only included what fits onto a single student page. Those who want the extra information can get it directly from Meyer:

Meanwhile, I also added some additional questions from U of Chicago. Notice that today's lesson is supposed to correspond to Dr. Franklin Mason's Lesson 12.2, and 12.3 is tomorrow, but notice how Dr. M divides the figures of surface areas into lessons:

Dr. M
Surface Area: Prisms and Pyramids, then Cylinders and Cones
Volume: Prisms and Cylinders, then Pyramids and Cones

U of Chicago:
Surface Area: Prisms and Cylinders, then Pyramids and Cones
Volume: Prisms and Cylinders, then Pyramids and Cones

So Dr. M combines the two polyhedra -- prisms and pyramids -- for surface area only, while for volume he follows the U of Chicago and combines prisms and cylinders. For getting through the formulas quickly, doing the prism and cylinder formulas together makes more sense, since they really are the same formula -- indeed, the U of Chicago calls them both cylindric solids. In other words, a cylinder is just a "circular prism."

Therefore this lesson incorporates Lesson 10-1 of the U of Chicago text, which is on surface areas of prisms and cylinders.

In last year's post, I spent a great deal of time discussing the PARCC PBA (Performance-Based Assessment) and the CAHSEE (California High School Exit Exam) tests. The idea was that students would have a tough trying to get through this week's material because they are so distracted by these big tests, depending on their state.

But this year, as it turns out, neither of those tests even exists anymore! The PARCC has been reduced to the EOY (End-of-Year) exam only, and the governor has signed a law dropping the CAHSEE as a graduation requirement in California. Hopefully, this will mean that students will be able to understand this week's material more clearly. Many Common Core opponents would agree -- less time for testing means more time for teaching!

Even though we're in Chapter 10 now, we might as well continue with Euclid. After all, David Joyce implies that he wouldn't mind teaching only "the basics of solid geometry" and throwing out surface area and volume altogether.

Proposition 14.
Planes to which the same straight line is at right angles are parallel.

This is another version of the Two Perpendiculars Theorem. Earlier, in Proposition 9, we had two line perpendicular to one plane, and now we have one line perpendicular to two planes. In all three theorems, two objects perpendicular to the same object are parallel.

Euclid's proof, once again, is indirect.

Indirect proof:
Point A lies in plane P and point B lies in plane Q, with AB perpendicular to both. Assume that planes P and Q intersect in point K. By the definition of a line perpendicular to a plane, AB is perp. to AK, and for the same reason, AB is perp. to BK. Then Triangle ABK would have two right angles, which is a contradiction since a triangle can have at most one right angle. (This is essentially Triangle Sum -- the two right angles add up to 180, so all three angles would be more than 180.) Therefore the planes P and Q can't intersect -- that is, they are parallel. QED

Euclid mentions a line GH where P and Q intersect, but Joyce tells us it's not necessary. Joyce also adds that Euclid forgot to mention the case where K, a point common to both planes, is actually on the original line that's perpendicular to them both. He doesn't tell us how to prove this case, but here's what I'm thinking -- consider a plane R that contains the original line. Planes P and R intersect in some line through K, and planes Q and R intersect in another such line, since the intersection of two planes is a line. By definition of a line perpendicular to a plane, these two new lines are both perp. to the original line. So in a single plane R, we have two lines perpendicular to a line through the same point K on the line, a contradiction.

Here is the worksheet for today's lesson/activity:

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