I will continue to discuss some of the things I see in the science classrooms today, even though high school isn't quite comparable to the old charter middle school.
A few weeks ago, I wrote that this new district has replaced its spring Open House with the "Showcase," used to help sixth graders choose a middle school for seventh grade, as well as help eighth graders choose a high school. As it turns out, this school has its "Showcase" tonight.
For one thing, it means that today is a minimum day. For another, it means that the two classes that are in my own room today -- freshman Biology and sophomore Chemistry -- instead visited the art room, where student work has been put on display for tonight's Showcase.
How does the aide know that she should take the Bio class to the art room? That's easy -- she received a text from the regular teacher. So that's already a point of comparison to the old charter school -- the fact that I failed to communicate with the support aide in my own classroom. I should have had her number or email -- and likewise she mine -- so that we could regularly discuss the lesson plans.
The regular assignment (for both Bio and Chem) is on the human immunity system. Disease has been in the news a lot lately due to the Coronavirus outbreak in China.
The Chem class is my only solo class of the day. As I take them to the art room, some of the students start hitting and kicking each other. When they return to the classroom, I distract them from hitting with a few songs -- I start with my usual science go-to "Earth, Moon, and Sun," and then add a song that's related to the art (since they see art today) -- Square One TV's "Patterns" (Weird Al). I sang this at the old charter school around this time of year, and I ought to sing it whenever I sub for art (or do something related to art, like today).
Even though the students finally stop attacking each other, I still wonder whether there's something I could have done to prevent them from playing around -- as usual, the Bio freshman don't play around when the aide is watching them. I'm not sure whether a music incentive in advance would have worked, since I can't sing it in the art room (with the art teacher watching) and they might have forgotten the incentive by then.
The three co-teaching classes are next. Two of the classes are Biology. These students are learning about genetics and DNA.
Here are a few things I notice about the Bio classes. First, this teacher doesn't use interactive notebooks -- instead she has three-ring binders. These binders have six dividers -- Procedures, Notes, Journals, Classwork/Homework, Labs, and Tests. I keep writing that maybe I should have had interactive notebooks, but an argument could be made for having binders instead, even if they have just two dividers -- math and science. Just like interactive notebooks, today's binders have a table of contents and a strict page-numbering system.
The other thing I see is that the teacher is planning labs for tonight's Showcase -- and indeed, parents will have the opportunity to participate in the labs. One of them is DNA extraction from a strawberry, while the other is on evolution -- participants measure how long it takes for them to do ordinary tasks without opposable thumbs.
This is something that might have been interesting for me to do at the old charter school -- set up some sort of (Illinois State, of course) project for Back-to-School Night, which was the third Wednesday in September. There are a few ways this could have worked:
- During the actual Back-to-School Night week, I had the "Show Me the Numbers" STEM project on Tuesday and then a traditional lesson on Wednesday. This project was all about measuring different wheels, so it was more "math-y" than "science-y." (I did try to turn it into a physics lesson for eighth grade, but it went nowhere.)
- I could have had set up three real science labs, one for each grade.
- Since sixth grade should have had NGSS standards -- some of which might have overlapped with grandfathered lessons for Grades 7-8 -- it's possible that I might have needed to set up only two lessons. Perhaps sixth grade might have had the same lab as eighth grade (which would be convenient since seventh grade had music on Wednesdays), so that I'd only need to set up two labs.
- I could have just had one science lab for all three grades. I have might been able to justify this to parents if I could have found a lab that somehow tied to the previous week's trip to the fair.
- There's always the "Earth, Moon and Sun" unit that I mentioned above. That year, I taught that unit in advance of the Rosh Hashanah school closure (12 days after Back-to-School Night).
What's so frustrating whenever I think back to what I should have taught is that it's based on two curricula (Illinois State and Study Island) that I can't access without a password (and of course my old password expired after I left the school). It's easy for me to say that Grades 7-8 should have been based on the Illinois State Life and Physical Science texts, and that sixth grade should have been based on the Study Island NGSS curriculum, but I don't now know what those units are, or what labs they contain.
In the end, this is something I could have discussed with my support aide. (There I go again about keeping the lines of communication open.) I could have said, "I want to showcase some science projects for Back-to-School Night and even have the parents participate. Which labs do you recommend that we show them?"
The remaining class today is co-teaching Chemistry. Here the resident teacher is showing the students dimensional analysis and how calculators work (scientific notation).
Because one of his examples is 6.02 * 10^23, it's obvious that this lesson is in anticipation of Avogadro's number and the mole. I would have preferred this lesson to be taught closer to Mole Day in October. But the text for this class is Pearson -- known for having separate consumable texts for first and second semester. Stoichiometry ("moles") is the first topic in the second semester text. And so there's no way to teach this second semester topic in October.
Today is Sixday on the Eleven Calendar:
Decade Resolution #6: We ask, what would our heroes do?
There isn't much opportunity to discuss the 1955 heroes today. But not all heroes are born in 1955 -- in a way, today's student artists showcasing the work today can be considered heroes.
Lecture 18 of Michael Starbird's Change and Motion is called "Zeno, Calculators, and Infinite Series," and here is an outline of this lecture:
I. Recall that Zeno's arrow paradox considers an arrow flying through the air.
II. In the case of Zeno's arrow, we know in advance what the sum is, namely 1; however, the real significance of infinite series is that they are used to approximate quantities that we don't otherwise know.
III. Polynomials are easy to calculate because they involve only addition, subtraction, multiplication, and division. We often want to approximate a non-polynomial by a polynomial.
IV. While we are exploring the relationships between the volumes and surface areas of solids, let's consider an example of a solid with finite volume but infinite surface area. This apparently contradictory object involves the use of infinity.
V. Approximation techniques by infinite series are important in real-world applications.
Starbird begins the lecture with another diagram of Zeno's arrow paradox. This time, he shows how far the arrow goes after each step: 1 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ....
The professor now moves on to a clock problem -- if it is now 4:00, when will the minute hand catch up to the hour hand? In 20 minutes, the minute hand will be where the hour hand was, but the hour hand has since moved (just like Achilles and the Tortoise). We draw a chart:
Number Minutes Moved
Minute Hand Hour Hand
20 20/12
20/12 20/12^2
20/12^2 20/12^3
20/12^3 20/12^4
M = 20 + 20/12 + 20/12^2 + 20/12^3 + 20/12^4 + ...
M/12 = 20/12 + 20/12^2 + 20/12^3 + 20/12^4 + ...
11/12 M = 20
M = 20 * 12/11 = 21 9/11
We've mentioned this problem on the blog before, back when we were trying to add a clock to the Eleven Calendar.
Starbird now mentions some infinite series involving pi:
pi/4 = 1/1 - 1/3 + 1/5 - 1/7 + 1/9 - ...
As he points out, we need millions of terms of this sequence to get just a few digits of pi. He also mentions an infinite series for trig:
sin x = x - x^3/3! + x^5/5! - x^7/7! + ...
To show how well this formula works, he shows us how it approximates sin(pi/6) = 1/2. After just five terms, the approximation is already 0.5000000000 to ten decimal digits. This is how our calculators find the sine of a value. (Once again, we see how calculators work today -- except it's the sine button, not the scientific notation/exponent button.)
The professor's final example involves a "paint can" with finite volume and infinite surface area -- a horn whose cross-sections are circles of radius 1/x as measured on the x-axis starting from 1. (This is nothing like the pyramids/cones whose surface area we find in today's lesson!) He uses a clever method to prove it has finite volume and infinite area using diagrams. Instead of trying to draw the picture here, let me just show the link:
http://mathworld.wolfram.com/GabrielsHorn.html
Today on her Daily Epsilon of Math 2020, Rebecca Rapoport writes:
If the area of square ABCD is 36, what is the perimeter of ABCDE?
[Here is the given info from the diagram: E, lying outside square ABCD, is chosen so that ADE is an equilateral triangle.]
To answer this, we notice that if ABCD is a square of area 36, then its side length is 6. So ABCDE is an equilateral pentagon of side length 6. Therefore its perimeter is 30 -- and of course, today's date is the thirtieth.
I haven't done much Geometry from Rapoport's calendar yet. This is because there are so many more topics on her calendar compared to the Pappas calendar. There was an excellent Geometry problem back on MLK Day -- but that was a holiday, hence a non-posting day:
If the area is 24 and one diagonal is 1/4 longer than the other, what is the perimeter?
[Here is the given info from the diagram: the figure is a rhombus.]
One way to solve this is to let the diagonals be 3x and 4x -- or better yet, let the diagonals be 6x and 8x, so that 3x and 4x are the semidiagonals. Now we use an area formula that doesn't appear in the U of Chicago text -- the area of any kite (including a rhombus) is half the product of its diagonals:
(1/2)(6x)(8x) = 24
48x^2 = 48
x^2 = 1
x = 1 (since this is Geometry, we ignore x = -1)
Now the diagonals of any kite (including a rhombus) are also perpendicular. Thus 3x = 3 and 4x = 4 are the legs of a right triangle whose hypotenuse is the side length of the rhombus. We don't even waste time with the Pythagorean Theorem, as we recognize the 3-4-5 triangle.
So we have a rhombus of side length 5. Therefore its perimeter is 20 -- and of course, the date of that question was the twentieth.
I hope we'll see more Geometry in Rapoport's Daily Epsilon of Math soon. So far, the only epsilon we've see much of lately are the epsilons (and deltas) of Starbird's Calculus course.
Lesson 10-2 of the U of Chicago text is called "Surface Areas of Pyramids and Cones." In the modern Third Edition of the text, surface areas of pyramids and cones appear in Lesson 9-10.
This is what I wrote last year ago about today's lesson. Notice that I spent much of that post comparing the U of Chicago text to three other math texts (and I decided to preserve this discussion):
Let's continue with the next proposition in Euclid:
Proposition 15.
If two straight lines meeting one another are parallel to two straight lines meeting one another not in the same plane, then the planes through them are parallel.
This proof should be easy to modernize:
Given:AB, BC distinct lines in Plane P, DE, EF distinct lines in Plane Q, AB | | DE, BC | | EF
Prove: Plane P | | Plane Q
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. Point G in plane Q so that 2. Proposition 11 from Monday (construction)
BG perp. plane Q
3. H in Q so thatGH | | ED, 3. Existence of Parallels
K in Q so thatGK | | EF (sometimes called "Playfair," but uniqueness is not needed)
4.BG perp. GH, BG perp. GK 4. Definition of line perpendicular to a plane
5.BG perp. AB, BG perp. BC 5. Perpendicular to Parallels (planar)
6.BG perp. plane P 6. Proposition 4 from two weeks ago
7. Plane P | | Plane Q 7. Proposition 14 from yesterday (a form of Two Perpendiculars)
Euclid's original proof should be simple enough for high school students to understand without the need to convert it to two columns. All that's need is to replace phrases such as "therefore the sum of the two angles...is two right angles" with "Perpendicular to Parallels," for example.
Decade Resolution #6: We ask, what would our heroes do?
There isn't much opportunity to discuss the 1955 heroes today. But not all heroes are born in 1955 -- in a way, today's student artists showcasing the work today can be considered heroes.
Lecture 18 of Michael Starbird's Change and Motion is called "Zeno, Calculators, and Infinite Series," and here is an outline of this lecture:
I. Recall that Zeno's arrow paradox considers an arrow flying through the air.
II. In the case of Zeno's arrow, we know in advance what the sum is, namely 1; however, the real significance of infinite series is that they are used to approximate quantities that we don't otherwise know.
III. Polynomials are easy to calculate because they involve only addition, subtraction, multiplication, and division. We often want to approximate a non-polynomial by a polynomial.
IV. While we are exploring the relationships between the volumes and surface areas of solids, let's consider an example of a solid with finite volume but infinite surface area. This apparently contradictory object involves the use of infinity.
V. Approximation techniques by infinite series are important in real-world applications.
Starbird begins the lecture with another diagram of Zeno's arrow paradox. This time, he shows how far the arrow goes after each step: 1 = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ....
The professor now moves on to a clock problem -- if it is now 4:00, when will the minute hand catch up to the hour hand? In 20 minutes, the minute hand will be where the hour hand was, but the hour hand has since moved (just like Achilles and the Tortoise). We draw a chart:
Number Minutes Moved
Minute Hand Hour Hand
20 20/12
20/12 20/12^2
20/12^2 20/12^3
20/12^3 20/12^4
M = 20 + 20/12 + 20/12^2 + 20/12^3 + 20/12^4 + ...
M/12 = 20/12 + 20/12^2 + 20/12^3 + 20/12^4 + ...
11/12 M = 20
M = 20 * 12/11 = 21 9/11
We've mentioned this problem on the blog before, back when we were trying to add a clock to the Eleven Calendar.
Starbird now mentions some infinite series involving pi:
pi/4 = 1/1 - 1/3 + 1/5 - 1/7 + 1/9 - ...
As he points out, we need millions of terms of this sequence to get just a few digits of pi. He also mentions an infinite series for trig:
sin x = x - x^3/3! + x^5/5! - x^7/7! + ...
To show how well this formula works, he shows us how it approximates sin(pi/6) = 1/2. After just five terms, the approximation is already 0.5000000000 to ten decimal digits. This is how our calculators find the sine of a value. (Once again, we see how calculators work today -- except it's the sine button, not the scientific notation/exponent button.)
The professor's final example involves a "paint can" with finite volume and infinite surface area -- a horn whose cross-sections are circles of radius 1/x as measured on the x-axis starting from 1. (This is nothing like the pyramids/cones whose surface area we find in today's lesson!) He uses a clever method to prove it has finite volume and infinite area using diagrams. Instead of trying to draw the picture here, let me just show the link:
http://mathworld.wolfram.com/GabrielsHorn.html
Today on her Daily Epsilon of Math 2020, Rebecca Rapoport writes:
If the area of square ABCD is 36, what is the perimeter of ABCDE?
[Here is the given info from the diagram: E, lying outside square ABCD, is chosen so that ADE is an equilateral triangle.]
To answer this, we notice that if ABCD is a square of area 36, then its side length is 6. So ABCDE is an equilateral pentagon of side length 6. Therefore its perimeter is 30 -- and of course, today's date is the thirtieth.
I haven't done much Geometry from Rapoport's calendar yet. This is because there are so many more topics on her calendar compared to the Pappas calendar. There was an excellent Geometry problem back on MLK Day -- but that was a holiday, hence a non-posting day:
If the area is 24 and one diagonal is 1/4 longer than the other, what is the perimeter?
[Here is the given info from the diagram: the figure is a rhombus.]
One way to solve this is to let the diagonals be 3x and 4x -- or better yet, let the diagonals be 6x and 8x, so that 3x and 4x are the semidiagonals. Now we use an area formula that doesn't appear in the U of Chicago text -- the area of any kite (including a rhombus) is half the product of its diagonals:
(1/2)(6x)(8x) = 24
48x^2 = 48
x^2 = 1
x = 1 (since this is Geometry, we ignore x = -1)
Now the diagonals of any kite (including a rhombus) are also perpendicular. Thus 3x = 3 and 4x = 4 are the legs of a right triangle whose hypotenuse is the side length of the rhombus. We don't even waste time with the Pythagorean Theorem, as we recognize the 3-4-5 triangle.
So we have a rhombus of side length 5. Therefore its perimeter is 20 -- and of course, the date of that question was the twentieth.
I hope we'll see more Geometry in Rapoport's Daily Epsilon of Math soon. So far, the only epsilon we've see much of lately are the epsilons (and deltas) of Starbird's Calculus course.
Lesson 10-2 of the U of Chicago text is called "Surface Areas of Pyramids and Cones." In the modern Third Edition of the text, surface areas of pyramids and cones appear in Lesson 9-10.
This is what I wrote last year ago about today's lesson. Notice that I spent much of that post comparing the U of Chicago text to three other math texts (and I decided to preserve this discussion):
One of the texts was published by Merrill, the others 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:
CCSS.MATH.CONTENT.8.G.C.9
Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.
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." 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.
[2020 update: Recall that the traditionalists were recently complaining about how there is too much Geometry and too little Pre-Algebra in Common Core 8, even for students headed for freshman Algebra I. This text might be more acceptable, but it still lacks the rational equations that they also complained about.]
Now leave McDougal Littell and 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
Let's continue with the next proposition in Euclid:
This proof should be easy to modernize:
Given:
Prove: Plane P | | Plane Q
Proof:
Statements Reasons
1. bla, bla, bla 1. Given
2. Point G in plane Q so that 2. Proposition 11 from Monday (construction)
3. H in Q so that
K in Q so that
4.
5.
6.
7. Plane P | | Plane Q 7. Proposition 14 from yesterday (a form of Two Perpendiculars)
Euclid's original proof should be simple enough for high school students to understand without the need to convert it to two columns. All that's need is to replace phrases such as "therefore the sum of the two angles...is two right angles" with "Perpendicular to Parallels," for example.
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