Thursday, January 16, 2020

Lesson 9-3: Pyramids and Cones (Day 93)

Today I subbed in an eighth grade math class. It's the first day of a two-day assignment, and so I definitely need to do "A Day in the Life" today.

Before I start, let me point out that on the way to school today, Elle King's "Ex's and Oh's" happens to be playing on the radio. It's almost as if this is a sign -- my parody "x's and y's" is all about graphing linear equations, which the eighth graders figure to be studying today.

This is the school where all periods rotate, and today the rotation begins with sixth period. Normally, homeroom is the same as third period, but this teacher has third period conference. Thus he doesn't have a homeroom either.

8:25 -- So my day begins with sixth period. This is the first of two Algebra I classes.

The Algebra I students are studying exponential functions -- a fairly new Common Core topic. In fact, Dolciani's Algebra I text doesn't mention exponential functions at all. There is some discussion of exponential functions in Chapter 9 of the U of Chicago Algebra I text (which isn't surprising -- the common denominator of U of Chicago and Common Core is NCTM, as we've seen with Geometry).

Today's lesson is all about exponential growth and the equation y = a(1 + r)^t, where y is the final amount, a the initial amount, 1 for 100%, r the growth rate, and t the time. (In Lesson 9-2 of the U of Chicago Algebra I text, the only growth function given is b * g^x.) The regular teacher has left a bit of notes to take, and then the students can begin the homework assignment.

To establish a good list, I choose five students at random to help me with some HW questions. As you'd expect of an Algebra I (the higher of two possible eighth grade classes), there are no real behavior issues here.

9:20 -- Sixth period leaves and first period arrives. This is the second of two Algebra I classes.

This class goes even more smoothly than sixth period. I end up naming first period to be the best class of the day.

10:20 -- First period leaves. Ordinarily it would now be time for snack, except that the schedule for today is different. The reason for the change is the sixth grade orientation tour.

One thing that I've noticed about large districts with many schools is that the district is no longer divided into attendance zones for each school (especially the middle and high schools). Instead, students may attend any school in the district. Each middle and high school has a slightly different program (with regards to electives, etc.), and so the middle schools invite sixth graders from the K-6 elementary schools (and high schools invite eighth graders from the middle schools) so that they can learn more about the programs and make an informed decision. (Notice that magnet schools have done something similar for decades, but it's only recently that all secondary schools in a district are becoming schools of choice.)

In fact, last year the district changed Open House from its traditional springtime date to this month so that the secondary schools can showcase their distinctive programs to parents of sixth and eighth graders in the district. Some high schools have their Open House the week after finals, resulting in awkward schedules where the schools have almost two straight weeks of minimum or short days!

Today is Day 93 in my old district, but in this new district it's only Day 85. I don't necessarily like the idea of making sixth graders think about seventh grade when the year is less than half over (that is, they're closer to the last day of Grade 5 than the first day of Grade 7). But the deadline to submit first, second, etc., choices for middle school is the last school day before President's Day (with the admissions decisions to be revealed in the spring). Thus all of this hoopla is going on now.

(By the way, as bad as this is, the district where I graduated from as a young student is worse, at least for high schools. The deadline for eighth graders to apply was before Christmas, with the students receiving their decisions by President's Day. High schools held parent nights in November or December, but these are too early in the year to justify labeling them as "Open House.")

Anyway, the sixth grade tours today wreak havoc on the schedule. There are now three straight classes before snack break, with this last class (second period according to the rotation) being slightly longer since this is when the youngsters are visiting the classrooms. The sixth graders actually arrive just after the start of the day and begin with an assembly. The music classes begin with a performance and this is followed by an introduction to all the clubs. Thus musicians and club members don't attend their sixth or first period classes. Indeed, nearly half the students in my Algebra I classes today are absent due to these activities.

Oh, and my classroom is among those scheduled to be visited by the sixth grade tours.

Second period arrives. This is the first of three Common Core Math 8 classes.

These students are solving systems of linear equations. The class begins with going over last night's homework, followed by today's classwork. Both worksheets are preparation for tomorrow's quiz -- and in fact, the review sheets are in the exact same format as the quiz. Two of the ten systems are to be solved by graphing and four each by elimination and substitution.

Recently, I wrote that it's OK for opening material (Warm-Up and going over HW) to take a long time if the entire assignment is review. But the regular teacher has given a suggested schedule for the classwork, devoting ten minutes each of four tasks -- working alone without notes, working alone with notes, working with a partner, and going over these answers. With 58 minutes as the official length of the period, this leaves 18 minutes for going over the HW -- and it does take me 18 minutes (almost) to do so. I must cut this down to 12 minutes to accommodate shorter subsequent periods.

Just after I finish going over the HW, the first tour arrives. I decide that as sixth graders walk through the class, this is a great time to sing a song (as the eighth graders may be distracted anyway by the arrival of the youngsters). And I already mentioned above what song it is -- "x's and y's." It fits because the first two questions are solving systems by graphing -- and on each worksheet, one problem gives the equations in standard form, suggesting graphing them using intercepts (the topic of my parody song).

Many of the students enjoy my song, although one boy gets tired of hearing it over and over each time another tour arrives.

In addition to random names to help me with students on the board, I place one girl on the good list for spotting an error on the regular teacher's HW answer key. She's a Spanish speaker who speaks limited English, and so I'm impressed with her math skills.

11:20 -- Second period leaves, and it's finally time for snack break. Recall that third period is the teacher's conference period -- and these leads directly into lunch, since this is the only class today between the breaks. Thus I end up with a very long break today.

1:05 -- Fourth period arrives. This is the second of three Math 8 classes.

This class becomes a bit noisy. This worries me, since there is a quiz tomorrow. I can't help but think back to my last multi-day assignment (in my December 4th-6th posts). One class was loud all week, and continued to be so on quiz day. With all of the noise, some students take advantage to cheat on that quiz.

I do threaten to write some names on the bad list, particularly those who fail to attempt even one problem during the ten minutes of fully independent work (as if they're waiting for me just to tell them all the answers). I continue writing names on the good list for those who help me with some of the problems on the board. Once again, the loudest class has the most volunteers -- though unlike last Friday, the girls are the ones to volunteer today, and I revert to random choosing for one of the guys.

2:00 -- Fourth period leaves and fifth period arrives. This is the last of three Math 8 classes.

This is the only class with a co-teacher. He takes over the class, and I pay close attention, especially to his time management. He drops the HW time to around ten minutes -- he shows them some of the more difficult questions (graphing, elimination when multiplication is needed) and asks the students whether they need help on the others. He also ignores the "every ten minutes" plan suggested by the resident teacher and does the same with the classwork. He does the five more difficult problems on the board and then has them work on the other five independently.

I've seen this teacher co-teach in other rooms before, so I know that he's a strong classroom manager who sometimes helps student teachers and other novices when he co-teaches. Thus this class is the best of the day by far among the Math 8 classes. The students at least know to be quiet when he is speaking, although some fool around during the independent working time. In fact, very few of them complete the other five questions during the independent time.

2:55 -- Fifth period leaves, thus ending my day.

On the Eleven Calendar, today is Sunday, the third day of the week:

Decade Resolution #3: We remember math like riding a bicycle.

This is particular important during a lesson like today's. Students must remember how to add and multiply signed integers in order to be successful here -- and in fact, I mention this to both the second and fourth period Math 8 classes.

In second period, the repeated "x's and y's" song should also help establish the rules of graphing by intercepts ("When y = 0, that's the x-intercept," "Plus-y gets high, minus-y gets low"), since the song gets stuck in everyone's head. I also sing the song in fourth period, though not nearly as often.

Since this is a two-day assignment, I look ahead to tomorrow's lesson. In Algebra I, students will continue the exponential function chapter by learning about compound interest. And because today's song favors the Math 8 classes, I can prepare an Algebra I song for tomorrow. Obviously, I don't have many Algebra I songs having spent so much time teaching Math 8 and below, so I can either look one up on YouTube or create my own.

Actually, I'm able to find several good compound interest songs on YouTube:

"My Finance Song" by Donna Brecht (parody of Rachel Platten's "My Fight Song"):

I might have selected this one, but I don't necessarily want parody songs two days in a row (even if they're for different classes).

"Compound Interest Rap" by Richard Sanchez:

This last one is performed by students (just like "x's and y's"), which I like. But this song mentions continuous compound interest (Pe^rt), and I don't wish to mention e in an eighth grade class.

"Compound Interest Rap" by Scalar Learning:


This is the song that I've chosen. It's not really a parody, but just a basic rap. And the lyrics are right there in the video, so I can add them to my songbook and be ready for tomorrow.

Also an issue for Algebra I will be the return of half the class -- the students who miss class today due to the sixth grade tour activities. I should probably modify the lesson plan of the regular teacher (who likely knew about today's tour but not the extent of how many students would miss class today) and allow extra time to go over yesterday's lesson. I may not go over tonight's HW until late in the period.

But of course, my biggest concern for tomorrow is the Math 8 quiz. I'm considering not even singing a song to the Math 8 classes before the quiz. Even though my biggest concern is fourth period, I'll be keeping a close watch on second period -- after all, I don't really know whether this is a well-behaved group or not, since today's class is affected by the sixth grade tours.

I'm thinking about simply kicking out any students to try to make noise during the quiz -- the building has a central workroom, so they can just take this quiz there. This should send the message that I'm taking silence during the quiz seriously, so that there won't be another six cheaters as there were during the December multi-day.

Lecture 9 of Michael Starbird's Change and Motion is called "The Best of All Possible Worlds -- Optimization," and here is an outline of this lecture:

I. Suppose you have 600 feet of fencing with which you want to enclose a herd of camels in a rectangle field, one edge of which is bounded by a river. What dimensions should you make the fence to enclose the largest area?
II. The interesting strategy involved in analyzing this question is that, in a sense, we look at all possible answers at once.
III. Calculus to the rescue.
IV. Let's organize the possible widths and areas in a convenient form.
V. Here is an overview of our method [omitted here].
VI. The strategy of finding maxima and minima is extremely useful.
VII. Challenge problems were a feature of European mathematics. We will discuss one that involved looking at maxima and minima.
VIII. Virgil's Aeneid refers to an optimization problem now known as Dido's problem.

This problem reminds us of the Rapoport problem from January 9th post -- except because of the river, the answer isn't necessarily the square this time. But Starbird does mention the square as a possible guess -- the 200 ft. * 200 ft. square has 40,000 ft.^2 of area. But when he adjusts this slightly to a 199 ft. * 202 ft. square, the area increases to 40,198 ft.^2. Thus the square isn't the right answer.

The professor tells us that if two sides, the width, are w, then the length is 600 - 2w, and so the area function is A(w) = (600 - 2w)w = 600w - 2w^2. Then he graphs this function. This function equals zero at w = 0 and w = 300, and he tells us that at the maximum, the derivative is zero:

A(w) = 600w - 2w^2
A'(w) = 600 - 4w
0 = 600 - 4w
4w = 600
w = 150

So the maximum area occurs for a 150 ft. by 300 ft. rectangle. Starbird points out that we're able to find where the maximum occurs without calculating an area at all. Of course, the area is 45,000 ft.^2.

This generalizes. To find the maximum or minimum value of a function with a derivative, we should set the derivative equal to zero.

The professor now tells a story of Johann Bernoulli (1667-1748) and his brachystochrone problem -- down what shaped curve between two points shuld a rolling ball reach the lowest point fastest? He asks an audience member to help him draw the right solution to this problem, a cycloid, by using a rolling wheel. Then he demonstrates that the cycloid is faster than a linear path than a straight line by dropping two balls along two ramps, one straight and the other a cycloid.

I mention above that the lecture ends with a discussion of Dido's problem. I've blogged about Dido's problem before, in my discussion of Lesson 15-8.

Lesson 9-3 of the U of Chicago text is called "Pyramids and Cones." Our text refers to both pyramids and cones as "conic surfaces."

This is what I wrote last year about today's lesson:

This lesson is very similar to Lesson 9-2 in that the focus is on vocabulary. The difference is that today's lesson is not an activity, but a traditional worksheet.

Indeed, students are asked to calculate slant height in two of the problems on this worksheet. As we already know, this requires the Pythagorean Theorem. A right triangle can be formed with the slant height as the hypotenuse and the altitude height as one leg. The other leg is unnamed in the U of Chicago text, but notice that it's actually the apothem of the regular polygon base.

In fact, we discussed this in previous years -- many Geometry texts define "apothem," but not the U of Chicago text. Indeed, Lesson 8-6 explains the trapezoid area formula as follows:

"There is no known general formula for the area of a polygon even if you know all the lengths of its sides and the measures of its angles. But if a polygon can be split into triangles with altitudes or sides of the same length, then there can be a formula. One kind of polygon that can be split in this way is the trapezoid."

Well, another polygon that can be split in this way is the regular polygon. Indeed, all the altitudes and sides are of the same length, because the triangles are congruent. The common altitude of these triangles is the apothem of the regular polygon. (It has come to my attention that apothems do appear in the modern Third Edition of the text. This is in the new Lesson 8-7 on Special Right Triangles, since 30-60-90 and 45-45-90 triangles are used to find the apothems of equilateral triangles, squares, and regular hexagons.)

But let's get back on track with pyramids and cones. Again, we look at Euclid's definitions:





Definition 12.
pyramid is a solid figure contained by planes which is constructed from one plane to one point.
Definition 18.
When a right triangle with one side of those about the right angle remains fixed is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. And, if the straight line which remains fixed equals the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuse-angled; and if greater, acute-angled.
Definition 19.
The axis of the cone is the straight line which remains fixed and about which the triangle is turned.
Definition 20.
And the base is the circle described by the straight in which is carried round.


Notice that Euclid's definition of "pyramid" isn't that much different from the U of Chicago's. But Euclid's cones, like his cylinders, are solids of revolution. Therefore they are limited to right cones and right cylinders. The U of Chicago generalizes this with a definition of cone not unlike the definition of pyramid. This allows cones to be oblique as well as right, and pyramids and cones are examples of conic surfaces (the surfaces of conic solids).

Oh, and I also notice one more difference between the old Second and new Third Editions. The point at the top of the pyramid is called the vertex in the Second Edition and the apex in the Third. Euclid, meanwhile, never names the top point of the pyramid.

Let's look at the next proposition for us to prove:





Proposition 6.
If two straight lines are at right angles to the same plane, then the straight lines are parallel.


This is basically the Two Perpendiculars Theorem of Lesson 3-5, except that the two given lines are perpendicular to the same plane, not merely the same line. The line version of Two Perpendiculars appears as the last step of the proof.

Let's modernize Euclid's proof:

Given: Line AB perp. plane P, line CD perp. plane P (BD in plane P)
Prove: AB | | CD

Proof:
Statements                                    Reasons
1. bla, bla, bla                              1. Given
2. Draw DE in plane P such that 2. Point-Line-Plane, part b (Ruler/Protractor Postulates)
    DE perp BDDE = AB
3. AB perp. BDAB perp. BE,      3. Definition of line perpendicular to plane
    CD perp. BDCD perp. DE
4. BD = BD                                   4. Reflexive Property of Congruence
5. Triangle ABD = Triangle EDB 5. SAS Congruence Theorem [steps 2,3,4]
6. AD = BE, Angle ABD = EDB  6. CPCTC
7. AE = AE                                   7, Reflexive Property of Congruence
8. Triangle ABE = Triangle EDA 8. SSS Congruence Theorem [steps 2,6,7]
9. Angle ABE = Angle EDA         9. CPCTC
10. DE perp. DBDE perp. DA    10. Definition of perpendicular lines [steps 6,9]
11. Lines BDDADC coplanar   11. Proposition 5 from Friday (DE perp. to all, steps 2,10)
12. Lines BDABDC coplanar   12. Point-Line-Plane, part e (points A, B)
13. AB | | CD                                13. Two Perpendiculars Theorem [step 2]

Some steps look strange here, such as Step 12. Here, Euclid uses the "proposition" (actually a postulate) that all triangles lie in a plane, so the plane containing two sides of Triangle ABD (namely BD and DA) must contain the third (AB). Here, we use our Point-Line-Plane Postulate part e. This is because Step 11 establishes that a single plane (not plane P, by the way -- we can call it plane Q if we want) contains the lines BDDA, and DC -- that is, points A and B lie in plane Q. Thus by part e, the entire line AB must also lie in plane Q.

The reason the proof is so convoluted goes back to the original Two Perpendiculars Theorem:

If two coplanar lines l and m are each perpendicular to the same line, then they are parallel to each other. [emphasis mine]

Notice that key word coplanar. Most of the time we use the Two Perpendiculars Theorem, we take it for granted that the lines in question are coplanar. But in this proof, the bulk of the proof is just to establish that AB and CD are coplanar! After all, the two perpendicular statements (AB perp. BDCD perp. BD) were established all the way back in Step 2, so we could skip directly from Step 2 to Step 13 if we didn't have to prove that the lines are coplanar.

On the other hand, it is truly necessary to prove that all three lines (the two lines and the transversal) are perpendicular. Otherwise, we could prove, for example, that DB | | DA following Step 10, which is absurd. But since three lines (ABCD, and the transversal BD) were proved coplanar in Step 12 (they all lie in plane Q), we complete the proof that AB | | CD.

David Joyce points out that Euclid omitted a step here. It could be the case that the two lines meet the plane at the same point (i.e., B and D are the same point). But Euclid proves that this is impossible in a subsequent proposition.

Here are the worksheets:


No comments:

Post a Comment