Wednesday, February 5, 2020

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:


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