Last night I tutored my geometry student. He is still working in Section 3-3 of the Glencoe text, which covers Slopes of Lines.
Well, I said that I'd reach slope on this blog during the second semester. Actually, we're not that far away from slope at all. I will include it as part of Coordinate Geometry in Chapter 11, when we'll finally jump back to Section 3-4 of the U of Chicago text, where slope is defined. But we have to get through Chapter 12 before we can cover slope.
I'll be teaching slope here on the blog during the second half of January. But my student is learning about slope right now. So I wasn't able to show him any material on slope on the blog, but there's one thing I want to say about the slope worksheet (from Glencoe) that the teacher gave him. One of the questions directed him to draw the line through point C that is parallel to line AB, where the coordinates of A, B, and C are given. I forget the exact coordinates, but they were something like, say, A(-2, -5), B(1, -3), and C(-2, 0). He immediately found point D(1, 2) and graphed line CD. I asked him whether he located point D by counting out the slope, and he told me that he had simply moved up five units from point B, since C was five units above A -- just as I suspected. In other words, he had performed a vertical translation, and I told him this. (Recall that I considered using translations to teach parallel lines, but I chose to follow Wu and use 180-degree rotations instead.)
I did what I did last time and showed him another worksheet on the Parallel Consequences, from Chapter 5 in U of Chicago and in Chapter 3 in Glencoe. He seemed to understand the worksheet well -- once I reminded him what the Parallel Consequences actually are. One of those Parallel Consequences will appear in one of the proofs that I will show you today.
Today was also my first day of subbing since winter break. I ended up covering two special-ed English classes in the morning, but in the afternoon I had two math classes -- one of them Algebra I, the other Integrated Math. You may be wondering, why would a school offer both? Well, recall that all sophomores and above are grandfathered into traditionalist classes. This includes those students who failed Algebra I as freshmen last year. So the current Algebra I class consists solely of sophomores and above who are repeating the class.
The Algebra I class was solving simple linear equations, but, since this is a Common Core blog, I want to discuss the Integrated Math class in more detail. The class was studying mathematical modeling -- specifically, applying a linear model to a time and distance problem. Here is the Common Core Standard to which this lesson refers:
Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
That's right -- once again, the freshmen are studying an eighth grade standard. Earlier, I wondered whether these integrated classes would move to ninth grade math later in the year, but I noticed that the teacher had packets for the whole school year in the back of the classroom, and all of them were eighth grade packets.
But the freshmen were nonetheless struggling with this assignment. The assignment was an interesting one -- it involves a student who is skateboarding to school. He starts 90 meters from the school, and every six seconds he skates another 18 meters closer to school. The students are supposed to notice not only that the function d(t) = -3t + 90 models the skateboarder's distance from the school, but that the slope -3 (Yes, there's that slope again!) is his rate of speed. Students were to fill in a table calculating the rate at three different intervals, first from 0 to 1 second, then from 1 to 3 seconds, and finally from 2 to 4 seconds, to discover that the rate was always -3 m/s.
Well, the students were having all sorts of trouble filling in the table. Part of the confusion is that to determine the distance from 0 to 1 second, students actually had to calculate d(1) - d(0) -- that is, the subtraction is done in the opposite direction from the order in which the times are mentioned. So many of the students found d(0) - d(1) instead -- assuming that they bothered to subtract at all. Some of the students tried to divide to find the distance or time, rather than the rate. And those who did find the rate from 0 to 1 second correctly did something different in trying to find the rate from 1 to 3 seconds and so got this part wrong. I doubt that anyone obtained -3 m/s for all three rates, and so the Aha! moment -- realizing that all three rates are in fact equal -- never arise.
It's lessons such as these that fuel the Common Core debates. This lesson has always appeared in Algebra I texts, but teachers typically skip over it and go straight to the lessons were slope and linear equations are explicitly taught. But Common Core emphasizes mathematical modeling. And I can see why -- we want students to see that slope isn't just something that math teachers make them calculate, but has a physical, real-world meaning. Yet if the students can't find the slope and realize that it's always the same, then they end up learning nothing at all.
Indeed, this demonstrates why, in the slope formula, students have to find y2 - y1 and x2 - x1, rather than vice versa. If students make both changes, then they are just multiplying by -1/-1 which is 1, so there's no effect on the slope. But x2 - x1 -- or for time, t2 - t1 -- is how we find intervals in reality. I ask, how long is it from 5:00 to 9:00? It's four hours -- that is, 9 minus 5, not 5 minus 9. Today is the 7th, so how long is it to our next holiday on the 19th. It's twelve days -- 19 minus 7. We always take the second time minus the first time.
I played my usual subbing game, but I ended up playing it only for some simple integer questions rather than the assigned work. My game actually works better for traditionalist math.
Section 12-3 of the U of Chicago text covers Properties of Size Changes. Once again, we're going to refer to size changes by their Common Core name -- dilations.
This section contains several theorems. The first is the Size Change Distance Theorem. Let's state the U of Chicago version of this theorem, with "size change" replaced with "dilation" and "magnitude" replaced with "scale factor":
Dilation Distance Theorem:
Under a dilation with scale factor k > 0, the distance between any two image points is k times the distance between their preimages.
As I alluded to yesterday, this is the same as Dr. Wu's Lemma 15:
Lemma 15. Let D be a dilation with scale factor r > 0. Then D changes distance
by a factor of r in the sense that, for any two points P and Q in the plane, if we
denote D(P) by P' and D(Q) by Q', then P'Q' = r * PQ.
Wu proves Lemma 15 by noting that if the center lies on line PQ, then the result follows by the definition of dilation -- otherwise, it follows from the Fundamental Theorem of Similarity (as its main hypothesis is that the center, P, and Q are not collinear.)
The next theorem is the main theorem of the lesson. It is the Size Change Theorem -- naturally, we'll call it the Dilation Theorem instead:
Under a dilation:
(a) angle measure is preserved;
(b) betweenness is preserved;
(c) collinearity is preserved;
(d) lines and their images are parallel
For Wu, this is Theorem 23:
Theorem 23. A dilation has the following properties:
(i) it maps lines to lines, rays to rays, and segments to segments,
(ii) it changes distance by a factor of r, where r is the scale factor of the dilation,
(iii) it maps every line passing through the center of dilation to itself, and
maps every line not passing through the center of the dilation to a parallel line,
(iv) it is degree-preserving.
We can see that these are basically the same theorem. Properties (a) and (iv) correspond, (b) and (c) are taken care of in (iii), and (ii) is just a restatement of Lemma 15/Dilation Distance Theorem. As it turns out, (iii) is a little more specific than (d). Since lines, for the U of Chicago, are parallel to themselves, (iii) implies (d). But (iii) tells us exactly which lines are invariant lines -- they are the lines passing through the center. In a way, dilations are just like 180-degree rotations -- every line passing through the center is an invariant line, but only the center itself is a fixed point. (In a way, a 180-degree rotation is actually a dilation with scale factor -1, but we won't discuss dilations with negative scale factors in this course.)
The specific Common Core Standards covered in Lemma 15 and Theorem 23 are:
The big difference between the U of Chicago and Wu is proof of their respective versions of the Dilation Theorem -- but most of the difference has already been taken care of yesterday with the proof of the FTS. In particular, parts (ii) and (iii) of Wu -- that is, part (d) and the Dilation Distance Theorem in U of Chicago -- are essentially restatements of the FTS. And for the only parts that need to be proved -- parts (i) and (iv) in Wu, parts (a) through (c) in the U of Chicago -- the proofs in the text are similar to Wu's. Since the U of Chicago proofs are already in two-column format, I will use those proofs -- as usual adding in an extra step for the Given section.
Dilations preserve betweenness and collinearity. U of Chicago (b) and (c), Wu (i)
Given: Let the images of P, Q, and R under a dilation be P', Q', and R',
Q is between P and R.
Prove: Q' is between P' and R'.
1. P', Q', R' dilation images, 1. Given
Q is between P and R
2. PQ + QR = PR 2. Definition of betweenness (meaning)
3. k(PQ + QR) = k * PR 3. Multiplication Property of Equality
4. k * PQ + k * QR = k * PR 4. Distributive Property
5. k * PQ = k * P'Q', 5. Dilation Distance Theorem -- Wu (ii)
k * QR = k * Q'R',
k * PR = k * P'R'
6. P'Q' + Q'R' = P'R' 6. Substitution (step 5 into step 4)
7. Q' is between P' and R' 7. Definition of betweenness (sufficient condition)
Note: I've seen college-level texts abbreviate "Q is between P and R" as P-Q-R. I wonder why this abbreviation has never caught on in high school texts.
Dilations preserve angle measure. U of Chicago (a), Wu (iv)
Given: Let the images of A, B, and C under a dilation centered at O be A', B', and C'.
Prove: Angle ABC = Angle A'B'C'.
1. A', B', C' dilation images. 1. Given
2. BC | | B'C', AB | | A'B' 2. A line and its image under a dilation are parallel.
U of Chicago (d), Wu (iii)
3. Angle ABO = A'B'O', 3. Corresponding Angles Parallel Consequence
Angle OBC = O'B'C' (I did warn you this was coming!)
4. Angle ABO + OBC = 4. Addition Property of Equality
Angle A'B'O' + O'B'C'
5. Angle ABO + OBC = ABC, 5. Angle Addition Property
Angle A'B'O' + O'B'C' = A'B'C'
6. Angle ABC = A'B'C' 6. Substitution (step 5 into step 4)
Notice that in these proofs, (ii) from Wu is used to prove (i), and later on (d) from U of Chicago is used to prove (c). Neither version of the Dilation Theorem has its parts written in the order in which one would prove them. In particular, the order of the U of Chicago version is supposed to remind the student of the A-B-C-D invariant properties of isometries, except that there is no D (distance preserving) property of dilations.
Finally, there is a Figure Size Change Theorem which states that one can perform the dilations one point at a time, just as for reflections and other isometries. This helps us determine the corresponding sides and angles for these transformations.