"When you flick the switch of your headlights from bright to dim, mathematics is at work. To be specific, the principles of a parabola do the trick."
This is the only page of the subsection "Parabolic Reflectors & Your Headlights." On this page, Pappas explains that the reflectors behind car headlights are parabolic in shape.
Here are some excerpts from this page:
"In fact, they are paraboloids (3-dimensional parabolas formed by rotating a parabola about its axis of symmetry. Thus, the light rays travel out parallel to the parabola's axis of symmetry. When the lights are dimmed, the light source changes location. The low beams now point down and up. The parabola is an ancient curve that was discovered by Menaechmus (circa 375-325 B.C.) while he was trying to duplicate the cube. For example, it was Galileo (1564-1642) who showed that a trajectile's path was parabolic. Today one can go into a hardware store and fight a highly energy efficient parabolic electric heater which uses only 1000 watts but produces the same number of BTU thermal units as a heater that operates on 1500 watts."
In math class, the first time students see the word "parabola" is in Algebra I. We know that the graph of a quadratic function is a parabola. But as we know, this is around the point in Algebra I when students start to ask, "What good is this for?" or "When will we ever use this?" While previously taught topics may appear useful in some cases, quadratic functions have no obvious purpose. This page tells us that parabolas do appear in real life after all, due to the curve's physical properties.
Furthermore, in Algebra II, students study parabolas along with the other conic sections -- and that's when they complain about other strange words that appear in their texts, "focus" and "directrix." The two pictures on this page depict parabolas with their respective foci identified. In one picture, Pappas shows the definition of parabola as the set (locus) of all points in a plane which are equidistant from a fixed point called its focus and a fixed line called its directrix. In the other picture, she shows us the difference between a light source at the focus and a light source at some other point. I was fascinated in learning that "dimming" a headlight is nothing more than moving the source away from the focus!
I know that I just wrote about traditionalists in my last post, but I guessed I already opened that can of worms in writing about students who complain about parabolas and quadratic functions. I want to continue writing about traditionalists because of today's post by Fawn Nguyen. Some things that both she and one of her commenters write are relevant to the traditionalist debate.
Let's start with a link to the post:
http://fawnnguyen.com/green-olives/
Fawn Nguyen:
My 7th graders are working on “percentages of” problems currently, and late last night, I saw this problem on one of Don Steward’s handouts.
Oh, that's right -- Nguyen teaches seventh grade math now. Meanwhile, Don Steward -- to whose blog Nguyen links above here -- is in fact a British teacher.
Steward's problem:
There are 75 olives, 40% of which are green. I eat some of the green olives until 10% of the olives that remain are green. How many green olives did I eat?
Nguyen:
How would you solve this?
I solved it using algebra. Then, immediately, I thought, Fawnzie, since when do you use algebra to solve stuff like this. C’mon, do your rectangles.
I think of 40% as 2 of 5 boxes [...]
At this point, you can see the rest of Nguyen's solution at her blog. The important thing is that she uses tape diagrams to solve the problem rather than algebra. We know that the Common Core encourages the use of tape diagrams to solve proportion problems, and so Nguyen is simply adhering to Common Core protocol. Of course, we can't be sure how Don Steward, the original poser of this problem, would solve it, considering that where he lives (England), there is no Common Core.At this point, it's easy to see why a traditionalist would object to this solution. Students should be learning to solve problems using algebra only, not tape diagrams -- especially seventh graders, who are just one year away from Algebra I (by the traditionalists' timetable). Yet Nguyen writes:
Nguyen:
Because if I tried to show my kids the [traditionalist algebra -- dw] work below, or versions thereof, a few might just [crap] in their pants.
Now let's read one of Nguyen's commenters, Michael Goldenberg:
Goldenberg:
Okay, the boxes make sense, though I tend not to think that way, particularly because I’m not always sure that the numbers I get out in the real world will behave nicely (that is, real-world data has a nasty habit of not being integral or even rational).
So, yes, in the end, if you have nice numbers, the boxes certainly work and the fact that I don’t think that way in no way would stop me from teaching this approach. But I really do jump to that algebra instinctively and I have a rule that no one is allowed to [crap] in my classroom. :^)
And so Goldenberg explains why algebra is superior to tape diagrams -- algebra generalizes, or transfers, to more problems, such as those involving fractional or irrational quantities. The green olive problem would never lead to irrational quantities, but Goldenberg gives an example involving area, and of course areas often involve irrational numbers. (What is the side length of a square of area 2, or the radius of a circle of area 2?)
Yes, a main concern of the traditionalists is that many simple methods don't transfer. But then again, let's return to quadratics again. Why do we teach students to solve quadratic equations by factoring, when this doesn't work when the solutions are -- here we go again -- irrational? Perhaps we should only teach the Quadratic Formula, which transfers to all quadratic equations.
I'd say that it's better to teach a method that sometimes works and which many of the students learn, rather than a method which always works, but which few of the students learn. A student who is taught the easier algorithm and learns it can solve some problems, while a student who is taught the harder algorithm and fails to learn it can solve no problems.
Of course, in her solution Nguyen converts 40% to the fraction 2/5 so easily. This means that even her solution requires students to know that 40% = 2/5, and so those who have trouble with this will be unable to solve it. But then again, someone who doesn't know that 40% = 2/5 would be unable to use the algebraic method either, since it begins by noting that 40% of 75 is 30.
This leads to another commenter, Frank Schorn:
Frank Schorn:
I like teaching problem solving to my 6th graders Singapore style, using boxes like you do, Fawn. It removes the mystery, and makes the process accessible to almost all the kids.
But I get soooo much push back from kids who don’t like trying this out.
And it’s frustrating when their “better” way doesn’t work.
BTW, I guffawed out loud when I read your true-but-profane comment!
There are two issues raised in this post. First of all, notice that Singapore Math, a text recommended by traditionalists, uses tape diagrams, a method not recommended by traditionalists. Then again, traditionalists tend not to use the sixth grade Singapore text, precisely for this reason. The Singapore text is only used by traditionalists for Grade 5 and below.The second issue is that some students don't like the tape diagram method and prefer solving it a "better" way, which doesn't work. So this "better" way probably isn't the algebraic method -- and in fact, the complainers are likely to be the students who don't know that 40% = 2/5 (and so they'll get wrong answers no matter what method they use). Still, the point I make is that no matter how many students complain when Schorn uses tape diagrams, still more would be objecting if he tried to teach them the algebraic method instead. (But that won't stop some traditionalists from claiming, "My own child complained because she wanted to use algebra instead of tape diagrams!")
As we prepare to read Chapter 6 of the U of Chicago text, it's time for our next side-along reading book, namely George G. Szpiro's Poincare's Prize: The Hundred-Year Quest to Solve One of Math's Greatest Puzzles.
And now you may be asking, "Already?" At one point, I was considering waiting until we reached Chapter 9 in January to read this book, since the Poincare Conjecture is related to dimensions, and Chapter 9 of the U of Chicago text is all about the third dimension. But I'm afraid that anything can happen between now and January. Perhaps I'll find another book to read by then, for example. I already purchased this book over a month ago, and now I want to read it. The new blog label for this book will be "Poincare." Yes, this post is already lengthy, but don't worry -- the first two chapters of this book are short.
Chapter 1 of George Szpiro's Poincare's Prize is called "Fit for a King." Here's how it begins:
"August is not the best time of year to visit Spain. Soaring temperatures during the late summer make even short sightseeing trips and brief walks through medieval cities a plight."
This book begins in medias res -- it is August 22nd, 2006, and mathematicians are gathering in Madrid, Spain, for the awarding of the prestigious Fields Medal. Szpiro explains:
"Considered the equivalent of the Nobel Prize, these medals are the highest honor the mathematics profession has to offer. Even though the monetary reward is only a tiny fraction of the Nobel Prize's, the Fields Medals are arguably even more select than the Nobel Prizes, with at most four winners chosen every four years."
One of these winners is Grigori Perelman, a Russian mathematician. He is to be honored for proving the famous Poincare Conjecture, which baffled the best math minds for over a century. It is one of the seven toughest math problems named by the Clay Mathematics Instititute. But, as Szpiro writes:
"But on this day in August, Perelman was nowhere to be found, In fact, he spent the festive day hidden away in the modest apartment that he shared with his mother in a drab neighborhood of St. Petersburg." (Yes, he's apparently another "mama's boy," just like Paul Erdos.)
Szpiro tells us about the history of the Fields Medal, named for a Canadian mathematician who died in 1932, and awarded every four years since 1950 (after World War II). There is an age limit -- medalists must be 40 years of age or younger. In 2006, Sir John Ball of Oxford is in charge of awarding the medals. One of the four Fields Medalists that year is Terence Tao of UCLA (Go Bruins!), a Chinese-Australian mathematician who is 31 years old at the time.
A few months before the ceremony, Ball meets Perelman to inform him of the award. But the Russian insists that he doesn't want the Fields Medal. Not only that, but he refuses to claim yet another award associated with the Poincare Conjecture. As Szpiro explains:
"The Clay Mathematics Institute in Boston, created in 1998 by the businessman Landon T. Clay and his wife, Lavinia D. Clay, to 'increase and disseminate mathematical knowledge,' offers a million-dollar prize for the solution of any of their seven Millennium Problems. As noted above, Poincare's Conjecture is one of them."
Here, by the way, is a link to the Clay Mathematics Institute:
http://www.claymath.org/millennium-problems
The author writes about another attempt to reach the reclusive Perelman:
"One person who wanted to know more about Perelman's intentions was the author Sylvia Nasar. Having previously written a best seller about the mathematician John Nash -- who had suffered from paranoid schizophrenia, been in and out of mental institutions, and then went on to win the Nobel Prize in economics -- she was curious about this strange mathematician."
But the Beautiful Mind author isn't able to make much more headway than the others. Szpiro tells us that the meeting is still a success, despite Perelman's absence. He concludes:
"ICM 2006 will be remembered as the event at which Poincare's Conjecture finally became a theorem. One of the seven hardest mathematical problems of our age had at last been solved."
Lesson 6-1 of the U of Chicago text is called "Transformations." There is no corresponding lesson in the modern Third Edition -- transformations are spread out in Chapters 3 through 6, with no separate introductory lesson.
This is what I wrote two years about this lesson:
Today we begin Chapter 6, which is on transformations -- the heart of Common Core Geometry. We see that the first lesson, Lesson 6-1, is simply a general introduction to transformations.
This lesson begins with a definition of transformation. Once again, I omit the function notation T(P) for transformations since I fear that they'll confuse students. But I do use prime notation. The best way to demonstrate transformations is on the coordinate plane, so I do use them.
The book uses N(S) to denote the number of elements of a set S, an example of function notation. I include this question in the review, since the number of elements in a set (cardinality) is such a basic concept for students to understand.
Interestingly enough, in college one learns about these special types of transformations:
- A transformation preserving only betweenness is called a homeomorphism. [2017 update: We will read more about homeomorphisms in Szpiro's Poincare's Prize.]
- Add in collinearity, and it becomes an affine transformation.
- Add in angle measure, and it becomes a similarity transformation.
- And finally, add in distance, and now we have an isometry.
The final question on my worksheet brings back shades of Jen Silverman -- the distance between lines is constant if and only if they are parallel in Euclidean geometry. I mentioned Silverman's page earlier when we were getting ready to learn about parallel lines.
One question on the worksheet involves the transformation mapping (x, y) to (x + 2, y - 3). The students can see that this transformation is clearly a translation. But so far, we haven't completed a proof that the transformation mapping (x, y) to (x + h, y + k) is a translation.
It's easy to show that the transformation mapping (x, y) to (x + h, y) is a translation -- where we define translation as the composite of two reflections in parallel mirrors, with the direction of the translation being a common perpendicular of the mirrors. In this case, we can let the y-axis and the line x = h/2 be the two mirrors, with their common perpendicular the x-axis. Likewise, it's trivial to show that the mapping (x, y) to (x, y + k) is a translation in the direction of the y-axis. What we want to show is that the composite of these two translations is itself a translation.
It seems as if it should be trivial to show that the composite of two translations is a translation, but in fact it isn't. After all, a translation is the composite of two reflections in parallel mirrors, so the composite oftwo translations is also the composite of four reflections. It's not obvious why the composite of these four reflections is also the composite of two reflections in parallel mirrors -- especially if these two new mirrors have nothing to do with the four original mirrors. Of course, at some point we'd like to say something about translation vectors -- for example, the composite of two translations is a new translation whose vector is thesum of the original two vectors. But this would need to be proved.
To understand the proof, let's use some notation that appears in the U of Chicago text. If m is a line, then we let r_m denote the reflection in mirror m. In the text, the letter m appears as a subscript, but this is hard to reproduce in ASCII, so we use r_m instead. Meanwhile a small circle o denotes the composite, and so we write r_n o r_m to denote the composite of two reflections -- first the reflection in mirror m, then the reflection in mirror n.
The first thing we know about reflections is the Flip-Flop Theorem, which tells us that if a reflection maps F to F', then it maps F' to F. That is, if r_m(F) = F', then r_m(F') = F. Therefore the composite of the reflection with itself must map every point to itself. This transformation is often called the identity transformation, and we can write it using the symbol I. So we write:
r_m o r_m = I
Since I is a transformation in its own right, we can compose it with other transformations. Of course, this is trivial -- the composite of I and any other transformation is the other transformation:
I o r_m = r_m
r_n o I = r_n
r_n o I o r_m = r_n o r_m
Notice that composition of transformations is associative, but not commutative -- so in general, we have that r_m o r_n is not the same as r_n o r_m. In fact, we can see how r_m o r_n and r_n o r_m are related by finding their composite:
r_m o r_n o r_n o r_m = r_m o I o r_m
= r_m o r_m
= I
So r_m o r_n is the transformation which, when composed with r_n o r_m, yields the identity. This transformation has a special name -- the inverse transformation. As it turns out, the inverse of any reflection is itself. But the inverse of a translation is a translation in the opposite direction, and the inverse of the counterclockwise rotation with magnitude theta is, quite obviously, the clockwise rotation with magnitude theta with the same center.
Notice that r_n o r_m, being the composite of two reflections, can't itself be a reflection -- it must be either a translation or rotation. So r_n o r_m can almost never equal its inverse r_m o r_n. But there is a special case -- notice that the inverse of a 180-degree rotation counterclockwise is the 180-degree rotation clockwise, but these are in fact that same rotation, since +180 and -180 differ by 360. So if r_n o r_m equals a 180-degree rotation, then r_m and r_n commute after all.
The last thing we need in our proof that the composite of two translations is a translation will be both of Two Reflections Theorems (one for Translations, the other for Rotations). The Two Reflections Theorem for Translations tells us that the direction and distance of a translation depend only on the common perpendicular and the distance between the mirrors. So if k, l, m, andn are all parallel mirrors, and the distance from k to l equals the distance from m to n, then r_n o r_m = r_l o r_k. And the same happens for rotations -- if k, l, m, and n are all concurrent mirrors, and the angle from k to l equals the angle from m to n, then r_n o r_m = r_l o r_k.
Notice that in some ways, we are actually performing a transformation on the mirrors -- that is, given two mirrors m and n, we wish to transform them to m' and n' such that r_n o r_m = r_n' o r_m'. Or if we want to get very formal, if we have some transformation T such that:
T = r_n o r_m
then we wish to find another transformation U such that:
T = r_U(n) o r_U(m)
and the point is that the transformation U has nothing to do with the transformation T. Indeed, if T is a translation, then U can be any translation. Likewise, if T is a rotation, then U can beany rotation with the same center as T.
So now let's prove a simple case, that the composite of two translations is a translation. Our simple case will be when the two translations are in the same direction (or in opposite directions). This means that the mirrors k, l, m, and n are all parallel, and we wish to prove that:
r_n o r_m o r_l o r_k
is a translation. Since all the mirrors are parallel, in particular m | | n. By the Two Reflections Theorem for Translations, we're allowed to slide the mirrors themselves. We can slide mirrorsm and n to two new mirrors, m' and n', such that m' | | n' with the same distance between them:
r_n o r_m o r_l o r_k = r_n' o r_m' o r_l o r_k
How does this even help us at all? That's easy -- we slide m until its image is exactly l! Then we have:
r_n o r_m o r_l o r_k = r_n' o r_m' o r_l o r_k
= r_n' o r_l o r_l o r_k
= r_n' o I o r_k
= r_n' o r_k
And so we've done it -- we've reduced four mirrors to two. And we know that n is parallel to its translation image n' and n is parallel to k as all the original mirrors were parallel. So k | | n' -- that is, r_n' o r_k is the composite of reflections in parallel mirrors. Therefore it is a translation. QED
So now we see our trick to reduce four mirrors to two -- we transform the mirrors a pair at a time, using the Two Reflections Theorems, until two of the mirrors coincide. Then the composite of a reflection with itself is the identity, which disappears, leaving two mirrors behind. But when we transform the mirrors, we must be careful to transform the correct mirrors. When we have:
r_n o r_m o r_l o r_k
we may transform k and l together, or l and m together, or m and n together. But we can't transform, say, k and m together, or l and n, or k and n, since these aren't listed consecutively in the composite.
Let's finally prove that the composite of a horizontal and a vertical translation is a translation. We begin by writing:
r_n o r_m o r_l o r_k
where k and l are horizontal mirrors (for the vertical translation), and m and n are vertical mirrors (for the horizontal translation). We notice that k, l, m, and n form the sides of a rectangle.
Now we begin transforming the mirrors. First, we notice that l and m are perpendicular (since l is horizontal, while m is vertical). As it turns out, reflections in perpendicular mirrors commute -- this is because by the Two Reflections Theorem for Rotations, the angle of rotation is double the angle between the mirrors, and since the angle between the mirrors is 90, the rotation angle is 180, which means that they commute. If you prefer, we could say that instead of commuting, we're actually rotating the two mirrors 90 degrees. The rotation image of l is m, and the rotation image of m is l. In either case, we obtain:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
Now we rotate the perpendicular pairs k and m together, and l and n together. We rotate the pairk and m until the image of m is concurrent with l and n, and then we rotate the pair l and n until the image of l is concurrent with k and m:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
= r_n' o r_l' o r_m' o r_k'
Now l' and m' are the same line -- the diagonal of the original rectangle. Let's call this new mirrord to emphasize that it's the diagonal of the rectangle:
r_n o r_m o r_l o r_k = r_n o r_l o r_m o r_k
= r_n' o r_d o r_d o r_k'
= r_n' o I o r_k'
= r_n' o r_k'
Even though d disappears, notice that d is still significant. Since both rotations preserved angle measures, we have that k' and n' are both perpendicular to the diagonal d. So, by Two Perpendiculars Theorem, k' | | n'. So we have the composite of reflections in parallel mirrors -- a translation. QED
But unfortunately, attempting to generalize this proof to the case when the directions of the two translations aren't perpendicular fails even in Euclidean geometry. Notice that the four mirrors now form merely a parallelogram, not a rectangle. But we know that in Euclidean geometry, the angles from k to m and from l to n are congruent because they're opposite angles of this parallelogram. After performing the rotations, the congruent angles have rotated into alternate interior angles formed by the lines k' and n' and the transversal d, which would mean that k' andn' are parallel. This may look appealing, but the problem is that the original proof commuted l andm, which is invalid unless l and m are perpendicular.
I'm not sure how to fix the proof that the composite of two translations is a translation, but luckily, we only need the perpendicular case to prove that mapping (x, y) to (x + h, y + k) is a translation.
Now here's the thing -- at some point this week, I wish to have students perform translations on a coordinate plane. I could give students this proof that mapping (x, y) to (x + h, y + k) is a translation, but look at how confusing this proof appears! It's highly symbolic and may be too abstract for students -- just look at lines like:
T = r_U(n) o r_U(m)
What in the world does this mean? It means that the lines which are the images of one transformation, U, are the mirrors (not the preimages, but the mirrors) of another transformation, T. If a student was to forget that U is a transformation, he or she might think that U is a line, and so r_U(m) would mean the image of the line m reflected in the mirror U. Notice that on a printed page rather than ASCII, it would be obvious that the subscript is all of U(m), not just the letter U.
As we've discussed so often recently, traditionalists like symbolic manipulation. [2017 update: Yes, like the symbolic, algebraic solution of the problem posed by Fawn Nguyen and Don Steward.] If they could see the symbolic manipulation involved with this, they might realize that transformational geometry isn't just hand-waving but is actually rigorous mathematics. But high school students will be confused if I were to give a proof as symbolic as this one, so I'd replace the symbols with words -- yet as soon as we did this, it will appear to the traditionalists that transformations are mere hand-waving yet again.
We might point out that verbal descriptions might be more understandable to high school students, while symbols for transformations might be better suited at the college-level. But this is when a traditionalist might say, fine, then let's save transformations for college-level math and teach high school students only math for which they can understand the symbols! In particular, a traditionalist would look at this proof and say one of two things. The first would be that students shouldn't be learning about translations anyway, so the mapping (x, y) to (x + h, y + k) is irrelevant. The other would be that if students really need to know that the mapping (x, y) to (x + h, y + k) is a translation, then they should prove it by using the Slope and Distance Formulas and not by playing with mirrors. (Our problem is that we can't use the Slope and Distance Formulas until we've reached dilations and similarity.)
Indeed, I've noticed that some college geometry texts use the notation A-B-C to denote a very simple concept -- namely that point B is between A and C. Often I wondered why no high school Geometry text ever uses this A-B-C notation, but now I realize why -- the A-B-C notation is considered too abstract for high school students to understand.
None of this has anything to do with today's lesson, so I'll make the decision regarding how to present the proof when we get there. Meanwhile, if traditionalists really want to "privilege the symbol," why don't they begin by enforcing A-B-C notation in high school Geometry and see how far they get?
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