The Algebra II classes are learning about the properties of logarithms. The worksheet is set up so that a calculator is not needed -- all the answers can be given either in exact form (such as log 6) or can be derived using the properties and a few given logs (for example, given both log 4 = 0.602 and log 7 = 0.845, derive log 28 = 0.602 + 0.845 = 1.447). Of course, a few students try to take the classroom TI-84 calculators anyway, so I have to stop them. This comes from Lesson 7-5 of the MacDougall-Littell Algebra II text.
Whenever I teach a log lesson, I always remind the students that the answer to every single question goes back to one of four rules -- Product, Quotient, Power, and Change-of-Base.
The Integrated Math II classes are learning about the division of complex numbers. This is, of course, not an easy topic. The students take notes on how to rationalize the denominator by multiplying by its complex conjugate. Afterwards, we move on to a much simpler topic -- the powers of the imaginary unit i. On the worksheet, we only have time to do the first five problems -- which were all on powers of i. This is based on Module 11.2 of the HMH Integrated Math II Text -- but I noticed that complex division doesn't actually appear anywhere in the text. So apparently, the teacher is completing complex arithmetic by going beyond the addition, subtraction, and multiplication that actually appear in the Math text.
To test for understanding, I had the students simplify the explanation i^2018. The reason I chose the exponent 2018 is that most Math II students are sophomores -- the class of 2018. Just as Theoni Pappas, on her Mathematical Calendar 2016, always uses the current date as the answer to every question, I try to incorporate the graduation date as much as possible. Hopefully, this will motivate the students and remind them what they're working for. One person gave 1 as the answer, but another student correctly answered that i^2018 = -1, not 1.
During the teacher's fourth period conference, I subbed for an Honors Pre-Calculus class. These students are working on Sequences and Series -- Lesson 8-1 of a Pre-Calc text which is also published by McDougal Littell.
Notice that most of the students in the Pre-Calc class were juniors. This is because this class is an Honors Pre-Calc class. Seniors don't take Honors Pre-Calc since this class is mainly to prepare students for AP Calculus next year. On the other hand, I notice that five of the students in the Honors Pre-Calc class are sophomores! So far, I see that at least the Class of 2018 is able to avoid Integrated Math II by moving into the grandfathered Algebra II and Honors Pre-Calc for the Class of 2017. But this year's freshmen, the Class of 2019, won't have this option next year.
All of this discussion reminds me of Governor Brown's proposed senior course from yesterday's post, the class that I'm tentatively calling Algebra III. Notice that none of the three topics I taught today -- complex numbers, sequences/series, and logarithms -- appear on the California State University placement test. Therefore none of them would be included in the Algebra III course! This is despite the fact that all three topics appear in Algebra II (with arithmetic and geometric sequences actually appearing in Algebra I in some Common Core texts).
Of course, this leads us to ask ourselves about the purpose of the governor's senior math course. The new class is supposed to prepare students for college-level math, but also to show students why certain math topics are useful.
I've mentioned recently that the question "When will we use this in real life?" is often asked during the unit on logarithms -- functions with a weird name and no apparent purpose. Sure enough, that question came up in today's class -- specifically when a student was asking what "ln" means. I ended up just explaining what the natural logarithm is and never actually gave a straight answer to what its real purpose is. It certainly isn't obvious where 2.718281829459045... appears in the real world.
If our goal is to design a senior year course where we tie the math to a particular purpose, it will be hard to come up with a real-world purpose for logarithms -- especially the natural log. And so we can see why logs might not appear in our Algebra III course.
I wonder whether this senior math course will be ready in time for the Class of 2018 to take it during their senior year. Our sophomores taking Algebra II and Honors Pre-Calc already know that they're headed for Calculus as seniors, but what about the sophs currently in Integrated Math II? They will take Integrated Math III next year -- and possible the governor's new course the following year.
Of course, how can we really answer that students question -- what exactly is the real-world purpose of the natural log? The natural log shows up in the real world only because of its properties in Calculus -- it is a solution to the differential equation dx = x dy. But when describing the natural log's effects to laypeople, we often use the Change-of-Base Formula to convert it to another base -- usually either two (half-life of an element, doubling time of an account) or ten. My old Chemistry text (the one with the deltas for slope) often converted the natural logs that appear in the solution of differential equations to common logs.
By the way, you may ask why natural logs appear in the doubling time of an account -- interest doesn't increase like the natural exponential function unless it is compounded continuously. Well, the answer is that if an account is compounded often enough (such as daily, which does occur in real life), it is close enough to the natural exponential function that natural logs can approximate its inverse.
Logarithms in general are often used to describe many measurements that can fluctuate by several orders of magnitude. The decibel scale, the pH scale, and the Richter scale are all logarthmic (all are based on the common log). You can imagine what the news reports about earthquakes would be like if numbers in the millions and billions were used to describe their strength, rather than their logs. I remember when I was tutoring an Algebra II student and I'd told her that anywhere that both large and small numbers appear, logs are inevitable. So then I read to her a word problem about decibels and I said to her that both large and small numbers are used. I asked her, "So what functions will we need to solve this problem" -- and by now, she knew that the answer was "logs."
So this tells us how logarithms are used in real-life applications. But of course, I didn't have time to say all of that to the student in that Algebra II class today.
Speaking of Pappas's Mathematical Calendar, today is the following question:
A floor can be tessellated using regular dodecagons and:
(1) squares
(2) regular pentagons
(3) regular rhombi
(4) equilateral triangles
On one hand, this seems like a good warm-up question to review the recently completed Lesson 8-2 of the U of Chicago text. On the other hand, if I were to assign this question to students following the text, they'd have several issues with it.
First of all, we have the word dodecagon. The U of Chicago text only emphasizes the names of polygons with up to ten sides and mentions the 12-sided dodecagon only in an Exploration question.
But here's the bigger issue -- all of the tessellations in the U of Chicago text consist of only a single polygon being tessellated! Even the Square One TV video I mentioned on tessellation day depicts the crew tessellating the beach with regular octagons and squares.
Also recall that the last thing I wrote in that text was a comment (originally found on another site) that tessellations can be made mathematically rigorous if the focus is on angle measures. The U of Chicago text briefly mentions how the angle measures around a point add up to 360. This is why the equilateral triangle, square, and regular hexagon all tessellate, but not the regular pentagon.
One common question on the Mathematical Calendar is to find the "gap angle" between two adjacent regular polygons. This equals twice the exterior angle. In Lesson 13-8 we learned that the exterior angle of a regular polygon is 360 divided by the numbers of side, so we conclude that the gap angle is 720 divided by the number of sides.
In some cases, the gap angle between two regular polygons is the angle of another regular polygon. So we see that the gap angle between two regular octagons is 720/8 = 90 -- which is the angle measure of the square. This is why regular octagons and squares tessellate the beach. Notice that the gap angle between two regular hexagons is 720/6 = 120, which is itself the angle of a regular hexagon. Thus regular hexagons tessellate without any help.
The gap angle between two dodecagons is 720/12 = 60 -- and this is the angle measure of the equilateral triangle. So regular dodecagons and equilateral triangles tessellate. The correct answer therefore is (4) -- and of course, today is the 4th.
By the way, here's one more thing I want to say about tessellations. Think about that age-old question "When will we use this in real life?" and ask yourself, when are you more likely to hear a student ask that question -- during a unit on tessellations or a unit on logarithms? Compare your answer to the thread I quoted in the earlier tessellation post.
This is what I wrote last year about today's activity:
One thing about Chapter 8 is that there are plenty of possible activities to do, since every lesson from 8-2 to 8-5 in the U of Chicago contains multiple questions in their Exploration sections. I've chosen to set up four activities this week -- the four exploration questions from the the most recent lessons, 8-3, and next week's lesson, 8-4.
Activity #1 (Lesson 8-3, Question 26):
Find a rectangular room where you live. Calculate its area a. to the nearest square foot b. to the nearest square meter.
This activity should be straightforward, except for the "where you live" part, since this isn't suitable for a classroom activity. Naturally, I ask the students to find the area of the classroom instead.
Activity #2 (Lesson 8-3, Question 27):
In 1860, the area of the United States was 3,021,295 square miles. By 1870, the area had become 3,612,299 square miles. What caused such a large change?
This is a good one for students who like history much more than math. This is a tricky one. At first glance, one might see the dates 1860 and 1870 and think "Civil War." Technically speaking, South Carolina did secede in December 1860, but the rest didn't secede until 1861. So if you're thinking about that war, the only Southern state that might be included in the 1870 area but not the 1860 area would be South Carolina, and SC would hardly have one-fifth the area of the whole United States.
A huge hint appears in the U of Chicago text itself, at the beginning of this section:
"An almanac gives the area of the 48 contiguous United States as about 3,026,000 square miles. (The other two states, Alaska and Hawaii, add about 593,000 square miles to the area.)"
And we see that the 1860 area is already close to the current area. But then we notice those important words "of the 48 contiguous United States" (emphasis mine). So we conclude that it's the territories of the last two states, Alaska and Hawaii, that made the difference -- we see that the difference between the 1860 and 1870 areas is right around 600,000 square miles, which is right around the extra area provided by these two states. Finally, we note that Hawaii's tiny islands can't make up the lion's share of that 600,000 square miles -- and besides, the annexation of Hawaii didn't occur until a few more decades after 1870. Therefore, the answer to this Exploration Question is that Alaska -- "Seward's Folly" -- caused the large change in area between 1860 and 1870.
I'm not sure how well our geometry students would fare on this history question. Here in California, American history is taught in the 5th, 8th, and 11th grades, and I'm not sure when Seward's Folly is taught, as it occurred on the cusp between the scopes of the 8th and 11th grade classes. Moreover, notice that most students take geometry in Grades 9-10 -- right between the U.S. history years. But still, this may be an interesting research activity.
Activity #3 (Lesson 8-4, Question 24)
Multiple choice. When the vertices of a polygon are on a lattice, there is a formula for its area. The formula is known as Pick's Theorem. Use the polygon [given] and test with other polygons, to answer this question. Let P be the number of lattice points on the polygon. LetI be the number of lattice points inside the polygon. Which is the polygon's area (in square units)?
[emphasis the U of Chicago's]
(a) 1/2 P + I - 1
(b) 1/2 P + I
(c) 1/2 P + I + 1
(d) 1/2 (P + I)
Notice that the given polygon in the text happens to be a pentagon. So far we only actually know the area of rectangles. The pentagon can be easily divided into triangles, but we won't know how to find the area of triangles until Lesson 8-5. Nonetheless, the text includes this question in Lesson 8-4, perhaps as a preview of Lesson 8-5.
Of course, students can experiment with rectangles, since these have known areas, in order to figure out which of the four formulas may be correct. In fact, we can quickly determine which of the four must be correct by considering the simplest possible rectangle -- the unit square. We see that the unit square contains four lattice points on the boundary, namely its four vertices (so P = 4), and there are no lattice points in the interior (so I = 0). When we plug in the values P = 4 and I = 0 into each of the four formulas above, we see that only formula (a) gives an area of 1 for the unit square. So we conclude that formula (a), that A = 1/2 P + I - 1, must be correct.
The text states that this result is known as Pick's Theorem. This theorem is named for Georg Alexander Pick, the Austrian mathematician who proved this theorem near the end of the 19th century (that is, right after Alaska and Hawaii became states). According to my favorite math bio page, Pick actually met Albert Einstein. Pick eventually died in the Holocaust, in his eighties:
http://www-history.mcs.st-and.ac.uk/Biographies/Pick.html
Pick's proof of his theorem is a bit complex. It essentially uses induction -- a technique that we've seen used for Polygon-Sum Theorem earlier on this blog. For both the Polygon-Sum Theorem and Pick's Theorem, we cut a polygon into triangles, show that the relevant formula holds for triangles, and conclude that the formula holds for all polygons.
Oh, and by the way, Pappas's Mathematical Calendar 2016 features a different mathematical idea on the page for each month. The page for the current month of March features equilateral triangles and -- you guessed it! -- Pick's Theorem.
Activity #4 (Lesson 8-4, Question 25)
[Here] is a pentagon that tessellates, of a type discovered by Marjorie Rice in 1975. Draw enough of a tessellation to show the pattern.
Even though most of the activities listed in Lessons 8-3 and 8-4 are about area, this one is about tessellations instead. Then again, tessellations naturally lend themselves to activity, so it would be unthinkable to have an activity day with none based on tessellations.
In Lesson 8-2, we learned that not every polygon tessellates, but every triangle does. In Question 14 (which I didn't include in the worksheet for that section), we find out that every quadrilateral tessellates, but performing the tessellation is tricky and requires several rotations. But not every pentagon tessellates -- in particular, the regular pentagon doesn't tessellate. According to the text, we don't know exactly which pentagons tessellate, and new discoveries are still being made today.
The tessellating pentagon in this activity was discovered by Marjorie Rice. According to the text, Rice was not a professional mathematician -- indeed, she was a homemaker. The text tells us that Rice is from right here in California. She was still alive in the 1990's, when the text was written -- and it appears that she is still alive today, now in her nineties. Here, in fact, is Rice's website:
https://sites.google.com/site/intriguingtessellations/home
The site made an announcement on Rice's 90th birthday that she was no longer updating the site, but a link to some of the tessellations she discovered is still there.
I tried to trace the tessellating pentagon as best I could, but this is the type of figure where if I'm off by just a few degrees, the pentagon won't tessellate. Of course, students can just take a ruler and straightedge and draw the pentagon themselves. Notice that four of the sides of the pentagon are congruent (all labeled by x in the diagram), and one side isn't congruent to the others (y). So students can just draw the four congruent sides of any length x, with the correct angles between them, then join the last vertex to the first.
A few months ago, in Chapter 13, we discussed how the LOGO programming language and turtle graphics work. The Rice pentagon is an excellent figure to draw in LOGO, if the students have access to a computer that has LOGO installed. One way to draw the Rice polygon is as follows -- the program takes a single parameter that corresponds to the valuex shown on the diagram. This one begins at the top and moves counterclockwise.
TO RICE :X
FORWARD :X
RIGHT 112
FORWARD :X
RIGHT 48
FORWARD :X
RIGHT 100
FORWARD :X
HOME
END
Recall that the angles in a LOGO program are exterior angles, so these are thesupplements of the interior angles shown on the diagram. If we begin from a clear screen, we may do RIGHT 110 before beginning the Rice polygon, so that the base of the polygon is horizontal such as it is on the diagram. Finally, the command HOME draws the missing side of the pentagon, since it returns the turtle to its "home" position at the center of the screen.
But using HOME in this situation is not elegant. It forces the turtle to go to the center of the screen, regardless of where we began drawing the polygon. We would want to draw the pentagon anywhere on the screen and at different angles, so that we can tessellate the screen.
What we need is to be able to calculate y, so that we can go FORWARD that many steps. Even if we're drawing this by hand and not by LOGO, we may still wish to find y, rather than follow the steps above and use a ruler to measure y.
One way to calculate y is just to use the vectors of Chapter 14. If we imagine the five sides of the polygon to be vectors, then notice that these five vectors must add up to the zero vector, since a polygon is a closed figure. So all we have to do is add up the first four vectors, and their sum must add to the fifth vector to give the zero vector -- that is, the opposite vector of their sum must be the fifth vector. We already know the magnitude of the first four vectors is x -- and for simplicity, we can simply let x be 1. The direction of one vector can be due east to represent the base, and then the direction of the second vector becomes 48 degrees north of east. That's right -- just as in LOGO, it's the exterior angles that give us the directions of the vectors.
Once we find the fifth vector, its magnitude must be y. Actually, it's the value of y when x = 1 -- but any pentagon that we wish to draw must be similar to the figure for x = 1, thus x and y must be directly proportional.
As you can see, I enjoy this activity because it brings together Chapters 13, 14, and 8. This is ironic since it occurs in Chapter 8 in the U of Chicago text -- a student reading the text in order wouldn't have reached Chapters 13 or 14 yet. But since we changed the order of the text, we can and should take advantage of our reshuffling of the chapters.
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