If this sphere's radius is 3cbrt(6pi^2)/(2pi), what is its volume?
Well, Lesson 10-8 of the U of Chicago text gives us the formula for the volume of a sphere:
V = (4/3)pi r^3
V = (4/3)pi (3cbrt(6pi^2)/(2pi))^3
V = (4/3)pi 27(6pi^2)/(8pi^3)
V = (4/3)(6/8)(pi^3/pi^3)27
V = 27
Therefore the volume is 27 cubic units -- and of course, today's date is the 27th. This is another problem where the radius is some crazy irrational number just to make the volume work out to be a whole number.
Today, my old school hosts a Math Day, sponsored by the California Math Council. Here is a link describing such a festival in more detail:
Of course, while this is going on, I'm subbing in my new district, the second of five days in the Economics classroom. Today the seniors watch Boiler Room, a movie from the year 2000 whose main character becomes a stock broker. Meanwhile, the sophomore World History students watch a documentary, produced by the History Channel, about aircraft carriers during World War II.
The main classroom management issue involves a girl who asks for a restroom pass right at the start of third period. This is not the class right after snack (which would be fourth period), and so I allow her to go. But then she never returns. During the remaining three days of this assignment, I must watch out for this student. If she asks for another restroom pass, I might have to tell her that she can't go until the last 15 minutes of class, or just wait for snack.
Chapter 10 of Wickelgren's How to Solve Problems is "Topics in Mathematical Representation":
"As stated in Chapter 2, problems contain information concerning givens, actions, and goals. The first and most basic step in problem solving is to represent this information in either symbolic or diagrammatic form."
Wickelgren points out that many problems contain both forms -- for example, today's Pappas problem contains both symbolic and diagrammatic notation. I don't post the diagram part on the blog -- instead, only the symbolic form is given.
"The first step in solving such a problem is to translate from the representation given explicitly or implicitly in the original statement of the problem to a more adequate representation. This chapter is concerned with selected topics in the mathematical or precise representation of information in problems."
The author warns us that many of these problems are related to set theory, modern algebra, and combinatorial mathematics:
"Such students should consult regular mathematics books concerned with these topics, rather than try to master the material."
Even though I've written about these topics on the blog, they aren't the focus of this blog. Except for one example from Geometry (diagrammatic, of course), there isn't much of our favorite math course in this chapter. So let's keep this description mercifully brief -- a respite after yesterday's super-long post (where examples from Geometry abound).
Wickelgren lists several reasons why it's better to represent information about a problem on paper rather than in the head:
"Problems that involve tables or matrices of information are especially difficult to retain as a visual image purely in the mind."
Now let's get to the Geometry example. Wickelgren gives a diagram that we might see at the start of a Geometry problem. Since I can't post the diagram here, let's follow the author as he converts the given diagram to symbolic form:
"For example, the spatial information represented in [the figure] can be represented symbolically as follows: lines a, b, and h meet at common vertex B, lines a and d meet at vertex A, lines d, h, and c meet at vertex D, lines b and c meet at vertex C, lines c and d are collinear and line h is perpendicular to lines d and c."
I've had practice converting Pappas problems to symbolic form here on the blog. (By the way, Wickelgren's c and d are more like line segments than lines. Segment d can be also written
The authors says of postulates and other basic theorems:
"Since such prior knowledge is often assumed implicitly to be part of the givens in a problem, it is clearly important that you have access to your memory for such information."
Let's skip a few pages here. Wickelgren tells us that when converting a word problem to algebra, it's better to use as few variables as possible:
"However, it must be recognized that you are performing two steps at once: representing the concepts in the problem and expressing some relatively simple relations between concepts."
The author tells us there may be situations when subscripts are needed -- for example, if we're finding the volume of A, a complex figure consisting of a cylinder, two cubes, and a box:
"There are also complex containers B and C, each composed of subcontainers. How should you represent the volume of each subcontainer in container A, for example?"
Here's how I'd write subscripts in ASCII: V_c^A is the volume of the cylinder part of A. This means that c is a subscript and A is a superscript. It's also possible to have multiple subscripts:
"Such cases arise frequently in statistics, where x_ijk might represent the wheat yield on the kth plot of land, subjected to the ith value of one treatment dimension (for example, the amount of some kind of fertilizer), and subjected to the jth value on another treatment dimension (for example, the amount of water)."
Let's get to an example problem:
Tom, Dick, and Harry mow lawns in the summer to earn money. They each have a lawn mower, and one Saturday they decide to mow a 5,900 square foot lawn together, using all three lawn mowers. Tom mows 70 square feet per minute, Dick 50, and Harry 40. Dick and Harry start mowing the lawn at the same time, but Tom has trouble starting his mower and is delayed for 30 minutes. All three boys stop mowing at the same time, when the lawn is finished. How long does Tom mow?
Here Wickelgren lets t represent the time that Dick and Harry mow and t - 30 represent the time that Tom mows. This is better than using t_0, t_1, and t_2 for the times, or t_T, t_D, and t_H. Then he writes an equation, 70(t - 30) + 50t + 40t = 5900. I'd personally would prefer to let t represent the time that Tom mows and t + 30 for the others. Not only does this avoid negative values, but then I'm done as soon as I find t, whereas I might forget to subtract 30 from t = Dick/Harry's time. In either case, we find that Tom mows for 20 minutes.
A gym teacher wishes to put on a balancing demonstration in which one of the students will be to have four boys stand on each others' shoulders in a single tower. Out of the class of 20 boys, the gym teacher wishes to select the most stable tower of four boys once and time how long they are able to balance successfully on each others' shoulders without falling over. How many such towers of four boys must the gym teacher investigate? Stop reading and try to solve the problem.
The answer, found on a TI calculator, is 20 nPr 4 = 20!/(20 - 4)! = 20!/16!. Notice that it's nPr, not nCr, since the order matters -- one boy is on the bottom and other is on the top.
Wickelgren ends the chapter by discussing y = f(x) for function notation. One way to save ourselves a variable is to write y = y(x) -- so y is the name of the function as well as the dependent variable. (My first college math professor at UCLA wrote y = y(x) all the time.)
"This is a useful mnemonic trick in simplifying notation in problems where there is no possibility of confusing the concept of the function (f) with the concept of the dependent variable (y). However, when such confusion is possible, this trick should be avoided."
This is what I wrote last year about today's lesson:
Lesson 15-4 of the U of Chicago text is "Locating the Center of a Circle." According to the text, if we are given a circle, there are two ways to locate its center. The first is the perpendicular bisector method, which first appears in Lesson 3-6. (Recall that the perpendicular bisectors of a triangle are a concurrency required by Common Core.) This section gives the right angle method:
1. Draw a right angle at P (a point on the circle).
2. Draw a right angle at Q (another point on the circle).
3. The diameters
This method is based on the fact that a chord subtending a right angle is a diameter -- a fact learned in the previous lesson. Indeed, "an angle inscribed in a semicircle is a right angle" is a corollary of the Inscribed Angle Theorem.
Notice that unlike the perpendicular bisector method, this is not a classical construction. That's because the easiest way to construct a right angle is to construct -- a perpendicular bisector, which means that if we have a straightedge and compass, we might as well use the first method. The text writes that drafters might use a T-square or ell to produce the right angles, while students can just use the corner of a sheet of paper.
[2018 Update: Today is an activity day. I've decided to expand the Exploration Question from this lesson into a full activity page.]
"Each of the three circles below intersects the other two. The three chords common to each pair of circles are drawn. They seem to have a point in common. Experiment to decide whether this is always true."
As it turns out, these three chords are indeed concurrent, except for a few degenerate cases such as if the circles have the same center or if the centers are collinear. (The concurrency of perpendicular bisectors has the same exceptions.) The students are asked to experiment rather than attempt to prove the theorem that these three lines (called radical lines) intersect at a common point (radical center, or power center). The name "power center" refers to "power of a point" -- a dead giveaway that we must wait until Lesson 15-7 before we can attempt to prove the theorem.