Lesson 10-8: The Volume of a Sphere (Day 108)

Lesson 10-8 of the U of Chicago text is called "The Volume of a Sphere." In the modern Third Edition of the text, the volume of a sphere appears in Lesson 10-6.

Today is Day 108 -- at least in my old subbing district. It can be considered as the end of the fifth quaver, which we determine by noting that one-quarter of the way from Day 85 (the end of the first semester) to Day 180 is 107.75. Of course, in my new district today is only Day 99. Thus it's not yet the end of the fifth quaver in my new district, seeing as it's only the second week of the semester.

Oh, and this is a good time to mention another difference between the two district calendars. In my old district, Lincoln's Birthday and President's Day are on two separate Mondays. But my new district does something different -- this one and several other districts observe Lincoln's Birthday on the Friday before President's Day, so that there can be a four-day break in February.

(Three years ago at the old charter, by the way, students observed a five-day weekend, from Thursday to Monday. But the first two days had nothing to do with Lincoln's Birthday. Instead they were PD days -- in other words, we teachers had only a three-day weekend. President's Day itself is the only school-closing February holiday in the LAUSD.)

What does this mean for the blog calendar? Well, I'm already committed to the calendar of the old district on the blog. This means that Lesson 10-9 won't be posted until Tuesday.

Meanwhile, today I subbed in a high school Chemistry class. Since it's a high school class that isn't math, I won't do "A Day in the Life" today.

Today is also 2/7 -- that's right, it's e Day. The digits of e, the base of the natural exponential function, is 2.718281828459045...., and so today is e Day. Last year on e Day, I subbed in a middle school classroom where I didn't wish to mention the number e. But since today is at a high school, I actually mentioned the number e today.

Almost as if to pay for having long breaks earlier this week, today's regular teacher is scheduled for all six periods -- five sections of regular Chem and one section of AP Chem.

I won't say much about the AP Chem class, which just has a worksheet today. I will point out that the number e is indirectly mentioned in the AP text. Exponential decay appears not just with radioactive materials, but ordinary chemical reactions as well. The text gives the "differential rate law":

Rate = -delta[A]/delta-t = k[A]

Thinking back to Starbird, we realize that in the limit, this becomes -d[A]/dt. To solve this differential equation, we must integrate it, obtaining the "integrated rate law":

ln[A]/ln[A_0] = -kt

The presence of ln -- the natural log to base e -- is because the integral of 1/A is ln A. There's also something called a "second-order" reaction, where we integrate 1/A^2 instead of 1/A. This time, the integral is just -1/A, with no need for natural logs or e.

I'll say a little about the general Chem classes -- once again, we can make some comparisons to science at the old charter, but again, high school science isn't appropriate for middle school.

Just like Wednesday's subbing, today's students are taking a "group test," although the regular teacher insists that it isn't really a test. The questions refer to a previous lab where a mysterious yellow liquid dissolves in water, but another blue liquid doesn't, and they must answer why.

The teacher wants students to distinguish (in color) between work that the group members come up with on their own vs. work that they must research on Chromebooks or in the text. And so I do what other teachers would do -- insist on closed (Chrome)books until the midpoint of the period, and then allow them to do research.

I suppose I could have had group tests at the old charter school for science (even as I maintain individual tests for math). I'm not sure how that would have worked out, though.

In all classes, I sing the "Number e Song." This song, a parody of "Sugar Sugar" by the Archies, was first written by Elizabeth Landau, aka Bizzie Lizzie. Apparently Landau has recently left JPL to work at NASA headquarters near Washington DC. I notice that Landau still writes song parodies, although these are more related to NASA and outer space, not numbers like e or pi:

http://www.lizlandau.com/music/

My lyrics for the "Number e Song" are incomplete, and I was hoping that Landau would post her original lyrics again, but there's no such luck.

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

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

Even though this isn't a math class, this rule fits the first part of the group assignment -- students must remember as much Chemistry as they can before looking at outside resources.

While the students are a bit talkative the entire day, one problem does occur in the last period of the day -- and it's indirectly related to the end-of-the-day clean-up yet again.

As in the earlier classes, I tell the students to work without Chromebooks until the midpoint of the period when they can use them. The problem is that so many students are talking that no one hears me at that time, and so no one takes out a Chromebook. This means that the groups are now at a disadvantage compared to the earlier periods. Earlier, all groups (of 3-5 students) finish except for one group of two, whom in hindsight I should have sent to other groups. But in the last period, several groups fail to finish, because none of them look up what they can't remember.

And of course, I exacerbate the problem by ending a few minutes earlier so that they'd have time to put away the Chromebooks -- the ones that this period never uses! It goes without saying that this is a struggle -- why should they put away laptops that they never use that day? (This problem also occurred last week at a middle school -- one class must charge another period's Chromebooks.)

What I probably should have done is at the midpoint of the period, walk around to make sure that groups are actually using the Chromebooks -- especially those groups that appear to be struggling to complete the assignment.

This is also part of a bigger issue in my classes -- I often need to give important instructions, but so many students are talking that no one hears me, and they all do the wrong thing. What I usually ended up doing was argue -- and we all know what problems that leads to.

Today I avoid arguing, so it's at least a step in the right direction. A student or two put Chromebooks away, and then I sing the e song, even though not as much work is completed as the other classes. As for what to do when I must give critical instructions and no one's listening, that's something for me to figure out another day.

Lecture 24 of Michael Starbird's Change and Motion is called "Calculus Everywhere." Here is an outline of this lecture:

I. A method using derivatives to solve equations is called the Newton-Raphson method.
II. We have taken a striking journey, exploring a bit of one of the greatest intellectual achievements in human history.
III. The potency of calculus is a testament to the power of abstraction.
IV. The development of calculus is a long story of human struggle and triumph. But the story is not over.
V. Not all functions are smooth.
VI. At this moment in history, computers and biology and biotechnology are having the biggest influence on human life.
VII. Newton said, "If I have seen farther than others, it is because I have stood on the shoulders of giants." In exploring calculus, we have stood on Newton's shoulders, on Leibniz's shoulders, on Archimedes's shoulders, and those of many other great thinkers. We have communed with some truly spectacular high points of human thought. All of us can share an appreciation for that giant conceptual development that shaped and continues to shape history -- calculus.

Since most of this lecture is just a summary of the course, I'll just write about how Starbird uses the Newton-Raphson method to find the value of an irrational number -- sorry, it's sqrt(5), not e:

f(x) = x^2 - 5 (to find sqrt(5), a zero of this function)
f'(x) = 2x

Guess: x = 2
f(2) = -1
f''(2) = 4
tangent line through (2, -1) crosses x-axis at 2.25

Guess: x = 2.25
f(2.25) = 0.0625
f' (2.25) = 4.5
tangent line through (2.25, 0.0625) crosses at 2.2361111...

Guess: x = 2.2361111...
next iteration gives 2.236067978...

checking for accuracy:
(2.25)^2 = 5.0625
(2.2361111)^2 = 5.000192852
(2.236067978)^2 = 5.000000002

The professor mentions that most curves aren't smooth -- yes, most curves are continuous everywhere but differentiable nowhere -- apologies to teacher Sam Shah and his blog name:

https://samjshah.com/

One thing I notice about Starbird's lectures on this e Day is that he says very little about e. I know he wanted to skip to the easier parts of Calculus, but what can be easier than the fact that the derivative of e^x is e^x? The only mention of e was in Lecture 22 on business and econ -- to make one of the integrals easier, he assumed that interest was compounded continuously, A = Pe^(rt). At no point does he explain what e is, or that it equals 2.718281828... approximately.

This is what I wrote last year about today's lesson. Here, I compared the treatment of the volume in two different texts, an old McDougal Littell California seventh grade text and the U of Chicago text:

The McDougal Littell text, in Lesson 10.7, demonstrates the sphere volume formula the same way that it does the cone volume formula. We take a cone whose height and radius are both equal to the radius of the sphere, and we find out how many conefuls of sand fill the sphere. The text states that two conefuls make up a hemisphere, and so four conefuls make up the entire sphere.

But of course, we want to derive the formula more rigorously. Recall that Dr. David Joyce states that a limiting argument is the best that can be done at this level -- but I disagree. Dr. Franklin Mason, meanwhile, enthusiastically gives another derivation of the sphere volume formula, and Dr. M's proof also appears in the U of Chicago text. Recall that Dr. M considers this day on which the sphere volume formula -- Lesson 12.6 of his text -- is derived to be one of the three best days of the year.

The U of Chicago text mentions that this proof uses Cavalieri's Principle. But it was hardly the mathematician Cavalieri who first proved the sphere volume formula. Indeed, according to Dr. M [and Michael Starbird -- 2020], this proof goes all the way back to Archimedes -- the ancient Greek mathematician who lived a few years after Euclid. (It's possible that their lives overlapped slightly.)

Here is a Square One TV video about Archimedes:

We mentioned earlier that Archimedes used polygons to determine the value of pi (also known as Archimedes' constant) -- hence the line in the song, "He was busy calculating pi." He was also famous for using the principle of buoyancy to determine whether the king's gold crown was a fake, and this is also mentioned in the song. Legend has it that the Greek mathematician was so excited when he discovered his principle -- he had been in a public bath at the time -- that he ran down the streets naked and shouted out "Eureka!" to announce his discovery. The Greek word eureka, meaning "I have found," is the motto of my home state of California.

But Archimedes himself actually considered the discovery of the sphere volume formula to be his crowning achievement -- to the extent that he requested it to be engraved on his tombstone. So let's finally derive that formula the way that Archimedes did over 2000 years ago. And no, he didn't simply drop a ball into water to determine the formula. Archimedes' sphere formula has nothing to do with Archimedes' principle of buoyancy.

We begin by considering three figures -- a cone, a cylinder, and a sphere. We will use the known volumes of the cone and cylinder to determine the unknown volume of the sphere -- thereby reducing the problem to a previously solved one.

Our cylinder will have the same radius as the sphere, while the height of the cylinder will equal the diameter (i.e., twice the radius) of the sphere. This way, the sphere will fit exactly in the cylinder.

Our cone, just like the cone mentioned in McDougal Littell, will have the its height and radius both equal to the radius of the sphere. Such a cone could fit exactly in a hemisphere. But we want there to be two cones, so that their combined height is the same as that of the cylinder. We set up the cones so that they have a common vertex (i.e., they are barely touching each other) and each base of a cone is also a base of the cylinder. The two touching cones are often referred to as a "double cone" -- Dr. M uses the term "bicone." (A bicone is also used to justify to Algebra II students why a hyperbola is a conic section with two branches. A hyperbola is the intersection of a bicone and a plane, such that the plane touches both cones.)

The focus is on the volume between the cylinder and the bicone. The surprising fact is that this volume is exactly equal to the volume of the sphere! Here is the proof as given by the U of Chicago:

[2020 update: Recall that this is slightly different from the proof given by Starbird in his Lecture 11.]

"...the purple sections are the plane sections resulting from a plane slicing these figures in their middles. These purple sections are congruent circles with area pi * r^2. At h units above each purple section is a section shaded in pink. In the sphere, by the Pythagorean Theorem, the pink section is a small circle with radius sqrt(r^2 - h^2). The area of this section is found using the familiar formula for the area of a circle.

"Area(small circle) = pi * sqrt(r^2 - h^2) = pi(r^2 - h^2)

"For the region between the cylinder and the cones, the section is the pink ring between circles of radius r and h. (The radius of that circle is h because the acute angle measures 45 degrees, so an isosceles triangle is formed.)

"Area(ring) = pi * r^2 - pi * h^2 = pi(r^2 - h^2)

"Thus the pink circles have equal area. Since this works for any height h, Cavalieri's Principle can be applied. This means that the volume of the sphere is the difference in the volume of the cylinder (B * 2r) and the volume of the two cones (each with volume 1/3 *B * r).

"Volume of sphere = (B * 2r) - 2 * (1/3 * B * r)
                            = 2Br - 2/3 * Br
                            = 4/3 * Br

"But here the bases of the cones and cylinder are circles with radius r. So B = pi * r^2. Substituting,

"Volume of sphere = 4/3 * pi * r^2 * r
                            = 4/3 * pi * r^3." QED

The sphere volume is indeed the crowning achievement of Chapter 10. We began the chapter with the volume of a box and end up with the volume of the least box-like figure of all, the sphere. The start of Lesson 10-8 summarizes how we did this:

"It began with a postulate in Lesson 10-3 (volume of a box). Cavalieri's Principle was then applied and the following formula was deduced in Lesson 10-5 (volume of a prism or cylinder). A prism can be split into 3 pyramids with congruent heights and bases. Using Cavalieri's Principle again, a formula was derived in Lesson 10-7 (volume of a pyramid or cone). In this lesson, still another application of Cavalieri's Principle results in a formula for the volume of a sphere."

So take that, Dr. Katharine Beals! After all, she was the one who derided Cavalieri's Principle as progressive fluff that the Common Core tests on instead of actual math. But without Cavalieri's Principle, we'd be stuck finding the volumes of only boxes and their unions. Well, I suppose if we simply declared the volume formulas by fiat (i.e., as postulates) rather than actually deriving them, then Cavalieri's Principle is not needed. But if we want to prove them, then the Principle gives us an elegant proof of the sphere volume formula that was discovered over 2000 years before there ever was a Common Core -- a proof that, if mastered, should permit one to date a mathematician's daughter (as Beals mentioned on her website regarding the Quadratic Formula proof).

Sadly, we don't know whether Archimedes ever dated anyone's daughter, or whether he ever had daughters of his own. His life ended tragically, being captured by an enemy army. Legend has it that he was busy working on a geometry problem when the Roman army captured him. His last words before he was killed are said to be, "Noli turbare circulos meos" -- Latin for "Do not disturb my circles!"

Today is an activity day. It's based on this lesson's Exploration question:

25. Unlike a cone or cylinder, it is impossible to make an accurate 2-dimensional net for a sphere. For this reason, maps of the earth on a sheet of paper must be distorted. The Mercator projection is one way to show the earth. How is this projection made?

Notice that the correct answer to this question is quite complex. Here's a link that describes both a misconception and the correct answer:

https://www.math.ubc.ca/~israel/m103/mercator/mercator.html

Let's get back to e Day. Here is a post from this year mentioning e Day:

https://www.ipwatchdog.com/2019/02/07/in-support-inaugurating-february-7-world-e-day/id=106100/

No math holiday is complete without music. Apparently e is a magic number. (Well, the singer does use e in the magic formula e^(i pi) + 1 = 0, so I guess that's magic.)

Here's another e song -- and this one actually explains why the magic formula works:

Here is Michael Blake's e song, based on the digits of e:

Here's how we can write our own song based on the digits of e in Mocha:

http://www.haplessgenius.com/mocha/

70 N=7
80 FOR X=1 TO 44
90 READ A
100 SOUND 261-N*(18-A),4
110 NEXT X
120 DATA 2,7,1,8,2,8,1,8,2,8
130 DATA 4,5,9,0,4,5,2,3,5,3
140 DATA 6,0,2,8,7,4,7,1,3,5
150 DATA 2,6,6,2,4,9,7,7,5,7
160 DATA 2,4,7,0

Notice that whereas we can avoid the zero problem in pi, zero appears early in e. Blake uses a rest for the zero in his song, but I decided to use 18EDL, with zero as the tonic. I included 44 digits of e so that the song ends on the tonic. Oh, and using N=7 places that tonic on red E. A song about the number e needs to be in the key of E. (The fundamental tonic for 16EDL would be E, but 16EDL contains only nine notes from tonic to tonic when we need ten.)

Since I lament how Starbird doesn't discuss the role of e in Calculus, let's make up for it with this 3 Blue 1 Brown video:

Numberphile, of course, has an e video.

Here's another new e video, by Zach Star:

Since I subbed in a Chemistry class today, here's an e video by "The Organic Chemistry Tutor":

Let me rewrite the lyrics to the Bizzie Lizzie e song that I perform today in class:

1st Verse:
I just can't believe the loveliness of graphing you.
I can't believe you're more than two.
I just can't believe the loveliness of graphing you.
I can't believe you're more than two. (to Refrain)

2nd Verse:
I just can't believe your digits go forever now.
As long as a number can be.
I just can't believe your digits go forever now.
As long as you're the number e. (to Bridge)

Bridge:
Put a little cash in the bank, money.
Put a little cash in the bank, baby.
I'll make more next year, yeah, yeah, yeah!
Put a little cash in the bank.

100% interest on my money.
Compound it continuously, baby.
I'm gonna take the limit now, yeah, yeah, yeah!
My cash is multiplied by you, e. (to Refrain)

The bridge is mostly mine -- Landau didn't mention anything about money in her song. I chose to include money since it rhymes with the original Archies lyrics ("honey") as well as retell the story of Jacob Bernoulli's discovery of this constant.

By the way, let's think back once more to e Day itself. Some people would suggest that in deference to Europeans and other little-endians, e Day should be 2nd July, not February 7th.

Meanwhile, e Approximation Day works out to be July 19th as 19/7 = 2.7142857... is approximately equal to e. So e Approximation Day would be just three days before Pi Approximation Day.

One last constant I wish to celebrate on the calendar is closely related to e -- eta. The number eta is used in tetration (which I explained in a June 2016 post). We define eta as e^(1/e), because if we define the sequence:

a_0 = 1
a_(n + 1) = eta^(a_n)

then the limit of a_n as n approaches infinity is e. But if we replace eta with any greater base -- even eta + epsilon for small epsilon -- the sequence diverges. The tetrater Gottfried Helms was the first to use the symbol eta for e^(1/e).

eta = e^(1/e) = 1.44466786...

Thus Eta Day would be January 4th -- yes, just two days before Phi Day. Little-endians might observe Eta Day in April -- on either the 1st (1/4) or 14th (14/4). As for Eta Approximation Day, we can clearly approximate eta by the rational number 1.444..., which is 13/9 for September 13th. We are more than two decades away from Eta Day of the Century.
Remember that Monday is Lincoln's Birthday in my old district (and on the blog calendar), and so my next post will be Tuesday.

END

Lesson 10-7: Volumes of Pyramids and Cones (Day 107)

Lecture 23 of Michael Starbird's Change and Motion is called "Palpitations, Populations, Perch, and Pachyderms," and here is an outline of this lecture:

I. Calculus can be applied to many aspects of biology, including personal health. Here is a problem of heartbeats; however, it is not medically accurate, so do not act on the advice of the answer.
II. Population change can be modeled mathematically.
III. The discrete version of the Verhulst model provides a cautionary tale about change. Change can be chaotic, in which case the underlying assumptions of calculus may not be appropriate.

Today's Starbird lecture is all about applying Calculus to Biology. He begins with the question of how much we need to exercise. Suppose we have a limited number of heartbeats. When we exercise our heart beats faster, but our resting heartrate is slower. (This immediately reminds me of the "Count the Ways" song about heartbeats.) So he writes:

u = unit of exercise = 3,600 heartbeats extra (This is about one hour of exercise -- our heartrate goes up from its usual 70 to 130.)
Resting heartrate = R(u) = 25/(u + 1) + 50

T(u) = 10,000R(u) + 3,600u (T = total heartbeats in 10,000 minutes, approx. one week)
T(u) = 10,000(25/(u + 1)) + 50 + 3,600u
T'(u) = (250,000)(-1/(u + 1)^2) + 3,600 + 0
250,000/3600 = (u + 1)^2
50/6 = u + 1
7 1/3 = u

Thus we should exercise about 7 1/3 hours per week. (Don't forget the professor's disclaimer -- he's not a doctor, so don't take this seriously.) But now he supposes, if we hate exercise, then how can we maximize our pleasurable weeks? We must ask, does our model reflect reality?

Total beats/beats per week = weeks

Want:
Max pleasurable wks = (wks)(happy heartbeats/wk)/(total heartbeats/wk)

Starbird's next example is about populations. He begins by showing a graph of population over time and assumes that a population doubles every year. This exponential growth can't last forever. So instead, his next graph shows that the growth rate should slow down as we approach the maximum:

dP/dt = KP(Max - P) (where P = population, K = some constant)

The professor now shows us the actual graph of an elephant population. It clearly increases and decreases, and so our differential equation model above doesn't reflect reality. He now writes the following recursive equation for the number of fish in a pond:

P_(n + 1) = P_n + KP_n(1 - P_n)

Starbird refers to this model is called the Verhulst model. He warns us that this model doesn't always reflect reality, as he shows us the following graphs:

y_(n + 1) = 2.6 * y_n * (1 - y_n) (the population stabilizes)
y_(n + 1) = 3.3 * y_n * (1 - y_n) (the population oscillates between two values)
y_(n + 1) = 3.6 * y_n * (1 - y_n) (the population follows a more complex pattern)
y_(n + 1) = 4.0 * y_n * (1 - y_n) (the population is completely chaotic)

The professor shows us all this to point out that now everything can be described using Calculus.

Today on her Daily Epsilon of Math 2020, Rebecca Rapoport writes:

What is the area (as a multiple of pi) of the largest semicircle that can be inscribed within a rhombus of edge length 6 and with one angle 60 degrees?

This is a very tricky question. We know what a rhombus is, as well as a semicircle, and we know how to find their areas. But it's not quite obvious how the semicircle should be drawn inside the rhombus in order to maximize its area (or its radius). Since one of the given angles is 60, we expect this problem to involve 30-60-90 triangles somehow -- but unless we know how the semicircle is oriented we can't even start to solve the problem.

Let's try drawing a rhombus first. We'll label this rhombus ABCD. All of its side lengths are 6, and so we know AB = BC = CD = DA = 6. We assume that Angle A = 60, which makes Angle C also 60, as well as Angle B = D = 120.

At this point, I had to guess where to draw the semicircle to maximize the radius. Here's my guess -- let O be a point somewhere in the interior of the rhombus. Then we choose points E, F, G, H on the rhombus such that A-E-B (that is, E is between A and B), B-F-C, C-G-D, and D-H-A. All four of these points lie on a circle centered at O. We'll let GH be a diameter of circle O, so that it forms a boundary of our desired semicircle. Also, to make it easier, we introduce some symmetry here -- O will lie on a symmetry diagonal BD of the rhombus, instead of on either side of this diagonal. (But O definitely does not lie on the other symmetry diagonal AC.)

Once again, I can't be sure that this really is the largest possible semicircle. But when I draw it, it certainly appears to be the largest.

Now we already know that we're looking for 30-60-90 triangles here, so let's start what math blogger Ben Leis calls an "angle chase" (and other posters call an "angle bash") -- we find as many angles in the diagram as we can and hope that 30-60-90 triangles appear.

I've drawn my semicircle such that sides AB and BC are tangent to (semi)circle O. Then by the Radius-Tangent Theorem (of Lesson 13-5), BE perp. OE and BF perp. OF.

Now quad. BEOF has three known angles -- Angle B = 120, Angle BEO = BFO = 90. Thus we find that the last angle of our quadrilateral (which turns out to be a kite) is EOF = 60.

Clearly Angles HOE, EOF, and FOG add up to 180, and EOF = 60. Now here's where symmetry comes in -- I know Angle HOE + FOG = 120, but we need to show HOE = FOG. We know that the symmetry diagonal of a rhombus bisects the angle (that is, DO bisects HDG) and that O is the midpoint of HG (since both OH and OG are radii).

The proof of this is trickier than it would seem at first:

Given: Ray DO bisects Angle HDG, O is the midpoint of HG
Prove: Triangles HDO and GDO are congruent.

Proof:
From the givens, we have Angle HDO = GDO and OH = OG. And of course, DO = DO. But unfortunately, this gives us only SSA, which doesn't guarantee congruence.

But SSA does give us some information -- if two triangles satisfy SSA, then the angles opposite the middle S must be either congruent or supplementary. (This makes since if we think about the ambiguous case of the Law of Sines -- the two angles must have the same sine.)

So assume that the angles opposite DO in each triangle -- DHO and DGO -- are supplementary. But then Triangle DHG would have three angles whose sum is greater than 180, a contradiction.

So Angles DHO and DGO are congruent. So Triangles HDO and GDO are congruent by AAS. QED

Notice that this makes Angles DOH and DOG a congruent linear pair, so each is 90. Then the angles BOH and BOG are right angles as well. And since we can easily show Angles BOE = BOF (as Triangles BOE and BOF are congruent by HL), we subtract to obtain HOE = FOG as desired. We already found their sum above to be 120, so each is 60.

Then we can draw in sides HE, EF, and FG. All three of these are the bases of isosceles triangles with radii as legs and vertex angle 60, so all of them are equilateral triangles. (In fact, H, E, F, G are four of the six vertices of a regular hexagon inscribed in Circle O.)

In particular, Angle HEO = 60, and since Angle AEO = 90, we conclude Angle AEH = 30. Then two of the three angles of Triangle AEH are known -- Angle A = 60 and AEH = 30. So we can conclude that Angle AHE = 90 -- there is our first 30-60-90 triangle. We know that EH = r, the radius of the circle, and so we can find the shortest leg AH = r sqrt(3)/3.

To find another triangle, we note that since Angle AHE + EHO + OHD = 180 and two of the angles are known, AHE = 90 and EHO = 60, we conclude DHO = 30. Then by symmetry again (the symmetry diagonal of a kite or rhombus bisects the angle), HDO = 60. And so Triangle HDO is another 30-60-90 triangle. Once again, the radius r is the side opposite the 60, but this time we want to find the hypotenuse HD = 2r sqrt(3)/3

But of course AH + HD = AD = 6. So now we finally have an equation for r:

r sqrt(3)/3 + 2r sqrt(3) = 6
r sqrt(3) = 6
r = 2sqrt(3)

And now we can find the area of the circle and semicircle:

r = 2sqrt(3)
r^2 = 12
pi r^2 = 12pi
pi r^2/2 = 6pi

Therefore the area of the semicircle is 6pi. Rapport already supplied us the factor of pi, so all we need to provide is the 6 -- and of course, today's date is the sixth.

Since the answer we found matches the date, it means that all of the assumptions we made earlier turn out to be correct. But how can we actually prove that this is the maximum? Since we're still in the middle of our Starbird lectures, we might think we can prove this using Calculus. But I don't see how this is possible -- Calculus allows us to find the maximum value of a function over some parameter, but there is no simple parameter that categorizes "set of all semicircles in rhombus ABCD."

Lesson 10-7 of the U of Chicago text is called "Volumes of Pyramids and Cones." In the modern Third Edition of the text, volumes of pyramids and cones appear in Lesson 10-4.

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

Lesson 10-7 of the U of Chicago text is on the volumes of pyramids and cones. And of course, the question on everyone's mind during this section is, where does the factor of 1/3 come from?

The U of Chicago text provides two ways to determine the factor of 1/3, and these appear in Exploration Questions 22 and 23. Notice that without the 1/3 factor, the volume formulas for pyramid and cone reduce to those of prism and cylinder, respectively -- so what we're actually saying is that the volume of a conic surface is one-third that of the corresponding cylindric surface. So Question 22 directs the students to create a cone and its corresponding cylinder and see how many conefuls of sand fill the cylinder. The hope, of course, is that the students obtain 3 as an answer. This is the technique used in Section 10.6 of the MacDougal Littell Grade 7 text that I mentioned in last week's post as well. (By the way, I just realized that I mentioned four different math texts in last week's post!)

But of course, here in High School Geometry, we expect a more rigorous derivation. In Question 23, students actually create three triangular pyramids of the same base area and height and join them to form the corresponding prism, thereby showing that each pyramid has 1/3 the prism's volume. But this only proves the volume formula for a specific case. We then use Cavalieri's Principle to show that therefore, any pyramid or cone must have volume one-third the base area times height -- just as we used Cavalieri a few weeks ago to show that the volume of any prism, not just a box, must be the base area times height.

I decided not to include either of the activities from Questions 22 or 23. After all, there will be an activity tomorrow and I wish to avoid posting activities on back-to-back days unless there is a specific reason to.

Lesson 10-6: Remembering Formulas (Day 106)

Today I subbed in a freshman Algebra I class. It's rare back-to-back visits to my old district (albeit at a different school), and thus it really is Day 106 here.

(As it turns out, today is the second anniversary of my first day in my newer district -- and I celebrate it by subbing in my older district!)

Since this is a math class, I will do "A Day in the Life" today:

8:50 -- Yes, this is the school that doesn't start until 8:50. And it's one of the schools with a block schedule, with odd classes meeting today. First period is the first of two Algebra I classes.

Today the students are taking the Chapter 6 Test. I so rarely sub for actual math classes in this district, so I don't know much about what the curriculum looks like here. But I clearly recognize the Big Ideas Algebra I text, since this is the same text used in my new district. Indeed, today's test is on the same unit the eighth graders were studying in my January 16th-17th posts, on exponents.

And since one of the questions from that test is on compound interest, of course this gives me an opportunity to sing the same "Compound Interest Rap" that I performed three weeks ago. I do this before passing out the test.

As it turns out, this is a group test -- the class of twenty students has been divided into five groups of four students each. The regular teacher has announced that he'll randomly choose one student from each group whose test to grade -- and then the entire group will receive that grade. Thus it behooves everyone in the group to discuss the questions and answers with each other.

This reduces the need for classroom management a little, since I don't need to keep reminding the students to be silent during the test. Instead, I make sure that students aren't speaking to members of other groups.

10:20 -- It is now embedded support time, which leads into nutrition. Most students have completed the test and are allowed to leave, but one group still needs a few more minutes.

As the students leave, I sing a second song. (I mentioned earlier that sometimes I might need to perform two songs on block days.) I choose Square One TV's "Ghost of a Chance" as the other song.

10:55 -- Third period arrives. This is the other Algebra I class.

During this period, some students ask for help. One problem that I've had on test days is figuring out how much help is too much help during the assessment. (I've written on the blog about the "special scholar" from the old charter school -- she learned how to "sweet-talk" me into giving her most of the answers, and then she just glanced at another student's test for the rest of them!)

What ends up happening is that the students ask the senior "academic mentors" for help. (Apparently, "academic mentors" are distinct from TA's, as this class has two mentors and one TA!) I hope that they know how much assistance they're allowed to give.

Unfortunately, this means that I, a math teacher, end up giving very little math advice today. After the test there is another worksheet for homework, but the students (understandably) work on the HW right after the test. Many teachers don't bother to assign HW after the test, but it's often trickier with block schedules where no HW means two HW-free nights. Still, this means that the worksheet is intended for tomorrow night. I help some students with the first few problems on the HW (for Chapter 7, on polynomials), and that's it.

12:25 -- Embedded support time leads into lunch. This time, all groups have completed the exam.

And that essentially completes my day of subbing. Fifth period isn't exactly a conference period, but it's time for the regular teacher to help out some special ed students. I'm not required to do anything this period, and so I can go home early. As I mentioned yesterday, this is the luck of subbing with the block schedule -- one day I have a long conference period (or other class I'm not expected to cover), and another day I have three full blocks of classes with no period off.

On the Eleven Calendar, today is Friday, the first day of the week:

Resolution #1: We are good at math. We just need to improve at other things.

And I definitely tell this to the students today during the test. In this case, the "other things" that they need to improve at are showing enough work on the test and knowing when to use each formula.

In fact, the only things I do today as a math sub (in other words, that a non-math sub wouldn't have done) are stating this math resolution and performing the two math songs! I hope that what I say and sing today has a positive influence on the students and their test scores, however slight it may be.

Lecture 22 of Michael Starbird's Change and Motion is called "Business and Economics -- Getting Rich and Going Broke." Here is an outline of this lecture:

I. Many aspects of business and economic conditions are described by functions that relate one feature of the situation to another. The rate at which a condition changes relative to another can be measured using derivatives, while the cumulative effect of a continuing process can be measured using integrals.
II. Here is an example of a business problem with the goal of choosing production and price levels that will maximize profit.
III. Maintaining inventory costs money. One the other hand, setting up production also costs money. How can we best balance the two?
IV. Calculus can help us estimate future values, which can help us decide whether we will be set up for retirement.

Today's Starbird lecture is all about the application of Calculus to Economics. He begins with some graphs -- the cost and revenue graphs (with quantity as the independent variable and dollars as the dependent variable). We see to find the point where revenue minus cost is the greatest. He explains that this point is where derivative of cost equals the derivative of revenue -- in Econ, the word "marginal" (marginal cost, marginal revenue) means derivative. Here's his example:

q = 20,000 - 50p (quantity of items sold at a certain price)
p = (20,000 - q)/50 = 400 - q/50
R(q) = (400 - q/50)q (revenue as a function of quantity)
R(q) = 400q - q^2/50
R'(q) = 400 - q/25

C(q) = 100,000 + 30q (cost as a function of quantity)
C'(q) = 30

R'(q) = C'(q)
400 - q/25 = 30
q = 9,250

p = 400 - 9,250/50
p = 400 - 185
p = $215

Therefore, the professor explains, we should produce 9,250 items and sell them for $215 each.

In the next example, the set-up cost = $2,000 and the storage cost = $3 per year, and we expect to sell 12,000 during the year. Starbird depicts this situation on a graph -- we start with x items, sell them until none are left, then make another production run:

y set-ups
costs for set-up = 2,000y
storage cost = (3)x/2 = 1.5x (on average there are half of x items waiting in storage)
Total cost = 2,000y + 1.5x
12,000 = xy
y = 12,000/x
T(x) = (2,000)(12,000/x) + 1.5x (total cost for producing and storing x items)

We wish to find the value of x for which the cost is minimized -- set the derivative equal to zero:

T'(x) = -24m/x^2 + 1.5 (where m = million)
24m/x^2 = 1.5
24m = 1.5x^2
16m = x^2
sqrt(16m) = x
4,000 = x

Therefore we should make three runs of 4,000 units each per year.

The professor's last example involves an integral. We're planning for retirement, and we do so by purchasing a vending machine that makes $100 per day. How much will we make in 20 years? This is a Calculus problem if we put the money in the bank and allow it to earn interest. (Cue the Compound Interest rap again!)

($36,500 delta-t)e^(.05(20 - t_1)) + ($36,500 delta-t)e^(.05(20 - t_2)) + ...

In other words, delta-t is a small interval (actually one day), but we let delta-t approach zero so that we can approximate the sum by an integral:

integral _0 ^20 $36,500e^(.05(20 - t)) dt

We will wind up with $1,254,354.74 in 20 years. (Yes, I know that the Algebra I students were able to find compound interest today, so why is this a Calculus problem? Well, here we're adding money throughout the whole 20 years, not just at the start as our Algebra I students calculate today. Also, we assume continuous compound interest A = Pe^(rt) here, likely because it's easier to find the integral.)

Lesson 10-6 of the U of Chicago text is called "Remembering Formulas." In the modern Third Edition of the text, remembering formulas appears in Lesson 10-5.

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

This isn't exactly a filler chapter, since students indeed must remember surface area and volume formulas. It's just that students can learn all of these formulas without the benefit of Lesson 10-6. The important part is that some formulas are clearly linked -- such as the formulas for a prism and a cylinder. It's no wonder then that the text calls both of these "cylindric solids."

...and that's all I wrote last year! That's going to be a problem on the blog now that we've reached the two-year anniversary of my first day in the new district. When I try to quote posts from last year, notice that on some days I wrote more about my subbing that day than the U of Chicago lesson -- and that includes the day I posted Lesson 10-6 last year.

And so let me fill the rest of this post in with some assorted topics. First, even though today's Rapoport problem isn't a Geometry question, I do want to mention it as it contains an error (likely just a typo):

x + y + z = 2
x - y + z = 4
x + y - z = 6

This is clearly a system of three variables in three equations. (Today's group test also contains a review question from the previous unit on systems -- but since this is Algebra I, there are only two equations in two variables.)

The error, though, is that Rapoport doesn't specify whether we're to solve for x, y, or z. In a real Algebra II class, of course we should find all three -- but on her calendar, the answer must be a single number (the date), not an ordered triple. Thus this counts as the first error I've spotted on Rapoport's new calendar this year.

We notice that the latter two equations are set up perfectly for elimination -- in fact, we get a very rare two-fer where adding them allows two variables to be eliminated:

2x = 10
x = 5

Therefore x equals five -- and of course, today's date is the fifth. And so the missing instruction that Rapoport forgets to give is "Find x." Fortunately, x is the easiest variable to solve for anyway. (It turns out that y and z are both negative, and there are no negative dates on the calendar.)

OK, that's enough for today's hodgepodge. Let's get to today's worksheet: