Monday, March 21, 2016

Lesson 10-2: Surface Areas of Pyramids and Cones (Day 127)

Chapter 5 of Rudy Rucker's The Fourth Dimension: Toward a Geometry of Higher Reality is called "Ghosts from Hyperspace." As the title implies, Rucker suggests that ghosts may actually live in the fourth dimension.

He writes:

One difficulty here was that if the spirits were so insubstantial, how could they do things like lifting a heavy seance table?

No such difficulty arises if we think of spirits as solid, substantial beings, but here of course we have the question of why one doesn't notice spirits very often if they really are so solid. The answer to this is to assert that the home of spirits is somehow outside our space entirely. How best to have spirits live outside space? One could have them be infinitely far away, but then one would have the problem of how they can get here so rapidly when summoned up by a medium. A much more satisfactory explanation is to say that spirits live in the fourth dimension.

As you can see, there is very little mathematics in this chapter, and so I don't want to spend much time on a math blog discussing it. But I will say this -- it's been almost one month since my grandmother's passing, and it's comforting to imagine that her spirit lives on in the fourth dimension, just slightly ana from where I am right now. I hadn't started reading Rucker's book yet at the time of her passing or funeral, but today's Chapter 5 would have been the perfect chapter to read at that time.

(Rucker also mention religion in Chapter 5. So it fits in well with all of the religious holidays taking place this week. I'll stop here, since I avoid religion on the blog except in Calendar-labeled posts.)

The little math that does appear in this chapter concerns knot theory. In particular, the one puzzle that Rucker includes in this chapter is all for "knot":

Puzzle 5.1:
A line or string can only be knotted in 3-D space: no string can be knotted in 2-D space, and no knot will stay tied in 4-D space. Why not? In 4-D space it is possible to knot a plane. Can you imagine how?

Answer 5.1:
A line can't be knotted in 2-D space, because there's no way to have a line in Flatland cross above itself. And a line won't stay knotted in 4-D space because, as is discussed in Chapter 5, the extra degree of freedom will cause any knot to slip through itself. Moving everything up a dimension, we expect that a plane can be knotted in 4-D space, but not in 3-D or 5-D. How do you get a knotted plane? The idea is to start with a knotted line, and then imagine moving the knotted line ana out of space. The trail the line traces out will be a knotted plane. The plane, it is important to realize, is knotted, but does not intersect itself. Of course, if we move the knot in 3-D space, the trail does intersect itself, but since ana is perpendicular to every space direction, the 4-D trail will not come back on itself anywhere.

As we've seen before, knots can be highly complex mathematical objects. I've mentioned one knot -- the trefoil knot -- in a previous post.

Meanwhile, today I subbed for a special ed class. The teacher I was covering observes two sections of English and then teaches two sections of math herself.

Officially, the class is labeled as Integrated Math I, but naturally these students were working well below grade level. They were working out of a McDougall Littell text -- Algebra Readiness. Now "Algebra Readiness" is the name for an old pre-Common Core math course in California. As its name implies, the class was for freshmen (and eighth graders, since the old California State Standards encouraged Algebra I in eighth grade) who were not yet ready to take Algebra I. (The text is dated 2008, just before the dawn of the Common Core.)

The students were working on Lesson 2.7 of the text, Divide Fractions. As we know, division of fractions is a sixth grade standard in the Common Core:

Apply and extend previous understandings of multiplication and division to divide fractions by fractions.
Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem. For example, create a story context for (2/3) ÷ (3/4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2/3) ÷ (3/4) = 8/9 because 3/4 of 8/9 is 2/3. (In general, (a/b) ÷ (c/d) = ad/bc.) How much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 3/4-cup servings are in 2/3 of a cup of yogurt? How wide is a rectangular strip of land with length 3/4 mi and area 1/2 square mi?.
Fraction division was also a sixth grade standard in the old California standards. The Algebra Readiness text mostly covers seventh grade standards, but sixth grade fraction division is repeated since the seventh grade standards (in both the old and Common Core Standards) require students to divide rational numbers.

As we expect at this level, many students are struggling. Just as the last time I subbed in a special ed class, I am not sure whether to have the students cross cancel or not. I avoid cross-cancelling, but the instructional aide tells me that these students actually have seen cross-cancelling before.

In fourth period, many students tried to use a calculator to solve these. I noticed that these were labelled "CASIO fx-45 FRACTION" -- that is, they could handle fractions! But none of the students knows how to enter fractions into the calculator -- and of course, I'm not about to tell them, since they are supposed to learn how to divide fractions without a calculator.

Because of this, there are some students who try to simplify 6/9 as 3/4.5 -- that is, they feel that they must always simplify by dividing by 2, and since they have a calculator, they try dividing 9 by 2 to obtain 4.5 in the denominator. And so for the sixth period class, I make sure that the students understand when to cancel, and that if they get a decimal, they're dividing by the wrong number!

Meanwhile yesterday, on her Mathematical Calendar 2016, Theoni Pappas asked this question:

(The diagram shows a circle centered at P, with A and B on the circle. Arc AB measures 45 degrees.)

Area of sector PAB = 12.5pi. The length of the diameter = ?

Notice that Chapter 8 of the U of Chicago text -- although it covers both arc length and circle area -- doesn't discuss the area of a sector. Many other Geometry texts do discuss the areas of sectors -- indeed the lesson where the formula A = pi r^2 appears is often called "Areas of Circles and Sectors."

But still, we can use the pattern established in the U of Chicago text for arclength -- the measure of the arc is 45 degrees, so the area of the sector is 45/360 times that of the entire circle:

Area(PAB) = 45/360 * A
12.5pi = 45/360 (pi r^2)
12.5pi = (pi r^2)/8
100pi = pi r^2
100 = r^2
r = +/-10

and of course, we only consider the positive solution. So the radius is 10 and the diameter is 20 -- and notice that yesterday's date was the 20th.

This might have been a great question for Pi Day. We may pose the question as, suppose we cut a pie (or pizza) into eight slices and we want each student to have 12.5pi (or about 40) square units, so how large must the pie be (in diameter)? Most pies don't measure 20 inches across, but 20 centimeters is reasonable for the diameter of a pie. Recall that the actual pie I brought to school last week had a diameter of 8 inches, which is just about 20 centimeters. Then the area of each slice would be about 40 square centimeters.

This Pappas question appears a full week behind my posting the corresponding lesson on the blog -- but I haven't given the test on it yet, so we're still in the same unit. Furthermore, this question is somewhat related to today's lesson.

This is what I wrote last year about today's lesson. Notice that I spent much of last year's post comparing the U of Chicago text to two other math texts (and I decided to preserve this discussion):

One of the texts was published by Merrill, the other by McDougal Littell. I ended up purchasing the latter, which is dated 2001. I actually recognize this text from when I spent one month in an advanced seventh grade math classroom back in 2012. Geometry is covered in Chapters 8 through 10. Chapter 8 covers points, lines, polygons, transformations, and similarity. The transformation section covers reflections and translations (but not dilations in the similarity section), but of course, this is an old pre-Common Core text, so transformations aren't used to define congruence. Chapter 9 is officially called "Real Numbers and Solving Inequalities," but the real numbers portion of the chapter segues from square roots to the Pythagorean Theorem and to the Distance Formula.

That takes us to Chapter 10. As it turns out, much of Chapter 10 of this seventh-grade text matches up with the same numbered chapter of the U of Chicago geometry text. Here are the sections:

Section 10.1: Circumference and Area of a Circle
Section 10.2: Three-Dimensional Figures
Section 10.3: Surface Areas of Prisms and Cylinders
Section 10.4: Volume of a Prism
Section 10.5: Volume of a Cylinder
Section 10.6: Volumes of Pyramids and Cones
Section 10.7: Volume of a Sphere
Section 10.8: Similar Solids

It's often interesting to see how much surface area and volume appears in pre-algebra texts. Wee see that this text gives all of the volume formulas, while only the cylindric solids have their surface areas included in the text. But let's keep in mind that this text was specifically written for the old California state standards that we had before the Common Core.

The final chapter, Chapter 12, of this text is on polynomials. This chapter actually goes a bit beyond the seventh grade standards -- most notably, Section 12.5 is "Multiplying Polynomials" and actually teaches the FOIL method of multiplying two binomials. I was only in the classroom that taught using this text for a month, but I was told that the honors class would cover Chapter 12 around the start of the second semester, with the rest of the chapters taught in numerical order. (Non-honors classes would not cover Chapter 12 at all.) The next section, Section 12.6, may also seem a bit advanced for a pre-algebra class -- "Graphing y = ax^2 and y = ax^3" -- but it appears in the 7th grade standards. 

If we compare this to the Common Core Standards, we see that much of Chapter 10 of the McDougal Littell text corresponds to an eighth grade standard in Common Core:

Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.

This is to be expected. The Common Core Standards are based on Algebra I in ninth grade, while the California Standards were based on Algebra I in eighth grade. So many eighth grade Common Core Standards must have been seventh grade standards in California.

Before we leave the McDougal Littell text, let me note that Section 4.3 is on "Solving Equations Involving Negative Coefficients." I point this out only because, if you recall, last month I subbed for a sixth grade teacher who unwittingly assigned a worksheet with a few negative coefficients, and the sixth graders were confused -- as they should have been, since this was a seventh grade standard.

2016 Update: Notice that the McDougal Littell text I mentioned in last year's post is not the same as the text I used when subbing today, despite both being based on the pre-Core California standards for seventh grade. Last year's text was for seventh graders who were on grade level -- today's text is for eighth and ninth graders who are below grade level. For comparison purposes, let's look at the McDougal Littell Algebra Readiness text in more detail:

1. Expressions, Unit Analysis, and Problem Solving
2. Fractions
3. Decimals and Percents
4. Integers
5. Rational Numbers and Their Properties
6. Exponents
7. Square Roots and the Pythagorean Theorem
8. Equations in One Variable
9. Inequalities in One Variable
10. Linear Equations in Two Variables

The purpose of Algebra Readiness was to prepare students for Algebra I. Therefore, as we can see, there is very little geometry in this text compared to the McDougal Littell Math 7 text. The only geometry that appears is in Chapter 7, with even less geometry content than Chapter 9 of the Math 7 text (as the Distance Formula doesn't appear in the Algebra Readiness text). Area and volume are nowhere to be seen in the Algebra Readiness text (except the appendix, "Skills Review Handbook").

In many ways, Algebra Readiness was more like a Common Core 7 text than Common Core 8, as Common Core 8 contains more geometry (and even a little more algebra) than the Readiness text.

Now leave MacDougal Littell and get back to last year's post. We continue with the Merrill text:

I didn't purchase the Merrill Pre-Algebra text, so I don't recall how old the text is. But I glanced at it and noticed that all of the equations that appear in Chapter 10 of the U of Chicago Geometry text also appear in this text, with the exception of the equations involving a sphere. That is, the surface area formulas of all cylindric and conic solids appear in this text. This is unusual since, as we've seen, neither the CAHSEE nor the Common Core Standard expect students to learn the more complex surface area formulas before high school Geometry. Since today's lesson is Lesson 10-2 of the U of Chicago text, which is on surface areas of pyramids and cones, I want to discuss what I remember about the Merrill lesson on these surface areas.

Both Merrill and the U of Chicago give the lateral area of the pyramid as the sum of the areas of its triangle lateral faces. But only the U of Chicago gives the formula for a regular pyramid, which it defines in Lesson 9-3 as a pyramid whose base is a regular polygon and the segment connecting the vertex to the center of this polygon is perpendicular to the plane of the base. The formula for the lateral area of a regular pyramid is LA = 1/2 * l * p.

But now we must consider the surface area of a cone. The Merrill text does something interesting here, as it considers the area of the net of the cone. We cut out the circular base and a slit in the lateral region, and then flatten this lateral region. What remains is a sector of a circle. Then the Merrill text simply gives the area of this sector as pi * r *s (where s, rather than l, is the slant height) without any further explanation.

The U of Chicago text, meanwhile, gives a limiting argument for the surface area of the cone, as its circular base is the limit of regular polygons as the number of sides approaches infinity. But there is Exploration Question 25, where the Merrill demonstration is done in reverse -- we begin with a sector of a disk and fold it into a cone.

But neither tells us why the area of the sector (and thus the lateral area of the cone) is pi *r * l. Let me give a demonstration of why the area of the sector is pi * r * l.

We begin with the area of a circle, pi * R^2. The reason why I used a capital R is to emphasize that the radius of the circle that appears in Question 25 is not the radius r of the base -- indeed, it's easy to see that the radius of the circle becomes the slant height l. So the area of the circle is pi * l^2 -- that is, before we cut out the sector. We want to fit the area after we cut it.

Let's recall another formula for the area of a circle given by Dr. Hung-Hsi Wu: A = 1/2 * C *R -- and once again, R = l, so we have A = 1/2 * C * l. But neither one of these gives us the circumference or area of a sector. If we let theta be the central angle of a sector, we obtain:

x = theta / 360 * C
L.A. = theta / 360 * A
        = theta / 360 * 1/2 * C * l

For lack of a better variable, I just let x be the arclength of our sector. But here I let L.A. be the area of the sector, since these equals the lateral area of the cone we seek. The big problem, of course, is that we don't know what angle theta is for the cone to have a particular shape. But we notice that we can simply substitute the first equation into the second:

L.A. = 1/2 * theta / 360 * C * l
        = 1/2 * x * l

And what exactly is the arclength x of our sector? Notice that once we fold the sector into a cone, the arclength of the sector becomes the circumference of the circular base of the cone! And this we know exactly what it is -- since the radius of the base is r, its circumference must be 2 * pi * r:

L.A. = 1/2 * (2 * pi * r) * l
        = pi * r * l

as desired. QED

I incorporate this demonstration into my lesson.

By the way, suppose we think of the sector mentioned in the Pappas question as one that we must cut out of the circle to form a cone. The circumference of the original circle is 20pi cm, and we've cut out one-eighth of the circumference, or 2.5pi cm, leaving 17.5pi cm, So the base of the cone must have diameter 17.5 cm -- that it, it has radius 8.75 cm. The slant height is the radius of the original circle, which is 10 cm. We may use the Pythagorean Theorem to find the vertical height of the cone -- which turns out to be approximately 5 cm. Yes, it is almost exactly half of the slant height because the numbers 4-7-8 are very close to a Pythagorean triple.

No comments:

Post a Comment