PARCC Practice Test Question 29 (Day 172)

Question 29 of the PARCC Practice Exam is our second straight trig problem:

29. A carpenter is constructing a triangular roof for a storage shed as shown in the figure:

[The triangle has a 45-foot base and two 15-degree base angles -- dw]

Part A
How high will the peak of the roof rise above the top of the shed?
Give your answer to the nearest foot.

Part B
After the roof is constructed, it will be covered with an asphalt roofing material. The carpenter needs to calculate the combined length of the two sloping sides. What will be the total length needed of the roof covering?
Give your answer to the nearest foot.

Once again, I lament that all the good stuff is showing up on the test too late. By "the good stuff," I mean the material I covered in the Algebra II and Honors Integrated Math I classes I covered earlier this month (trig and constructions), and by "too late," I mean that I reach the problems after the week I was in that classroom. Just think about how many trig and construction problems I've covered after Friday the 13th, when I left that class!

For Part A, notice that we can't use any of the three trig functions until we have a right triangle. We form right triangles by drawing in the altitude to the base, which as we know, divides the isosceles triangle (which it is, as it has two base angles) into two congruent right triangles. The base of each triangle -- the leg adjacent to the 15-degree angle -- is 22.5 feet, and the value we are trying to find is the leg opposite the given acute angle. This implies that we need the tangent ratio:

a/22.5 = tan 15
a = 22.5 tan 15
a = 6.03, which is about 6 feet.

For Part B, we need to find the sum of the two hypotenuses of the right triangles. This implies that we need the cosine ratio:

22.5/c = cos 15
c = 22.5/cos 15
c = 23.3
2c = 46.6, which is about 47 feet.

As usual, we must deal with rounding error. It's so easy to round 23.3 down to 23 feet first and then double it to 46 feet, when we see that 46.6 rounding up to 47 feet is more accurate. As usual, PARCC counts both 46 and 47 feet as correct answers. On the other hand, students really have no business rounding 6.03 to anything other than six feet. And we can already see what the common student error will be -- giving 23 feet as an answer to Part B instead of doubling it to 46 or 47 feet. Another error is to plug 45 feet directly into the formulas instead of halving it to 22.5 feet.

Also, my rule of thumb is to use exact values from the question to solve Part B -- not the rounded off value from Part A. So we use the cosine in 22.5/c = cos 15, not the sine in 6/c = sin 15 or even the Pythagorean Theorem in 6^2 + 22.5^2 = c^2. In this problem, it probably wouldn't make that much of a difference since 6 and 6.03 are so close -- about half a percent difference.

Then again, using 6/c = sin 15 gives c = 23.2 (instead of 23.3) and so 2c = 46.4, which rounds to 46 instead of 47 feet. (Using 6 in the Pythagorean Theorem still gives 46.6 feet.) This doesn't matter for PARCC, which counts both 46 and 47 feet as correct, but it could make a difference in a real class where 46 feet could be marked wrong.

And of course, under no circumstances should we round 22.5 off in these formulas. If we tried to round 22.5 to 23, we obtain c = 23/cos 15 = 23.8 and 2c = 47.6, which rounds off to 48 feet -- and this answer will be marked incorrect.

By the way, notice that the total length of the roof covering is about 47 feet -- just two feet longer than the base of the roof -- yet it sticks up six feet in the air. That is, adding two feet to the roof makes it stick up six feet in the air. This unexpected result occurs because of the Pythagorean Theorem -- the equation a^2 + b^2 = c^2 implies that c = sqrt(a^2 + b^2) -- and when a is much smaller than b, c works out to be very close to b.

A classic question exploits this property of square roots -- suppose there is a track of length one mile, but when it gets hot, the track expands one inch, causing it to buckle. How high in the air will the track buckle? This is equivalent to the original question where the "base" of the roof (cold track) is one mile and the length of the roof "covering" (hot track) is one mile and one inch, and we wish to find the peak of the roof "covering" (hot track). The answer works out to be 15 feet -- and we'd definitely notice a track rising 15 feet in the air!

PARCC Practice EOY Question 29
U of Chicago Correspondence: Lesson 14-4, The Sine and Cosine Ratios

Key Postulates: Definition of Sine and Cosine

Common Core Standard:
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

Commentary: Once again, this sort of question -- where we're given an isosceles triangle and we must cut it into two right triangles before we can apply trig -- is not in the U of Chicago text. I have run out of Pizzazz worksheets from the Algebra II class to apply, but I found an old worksheet from last year that covers another topic covered poorly in the U of Chicago text -- inverse trig functions. Both ideas -- applying trig to isosceles triangles and inverse trig -- are used to solve problems, so I decided to include last year's questions on today's worksheet.

PARCC Practice Test Question 28 (Day 171)

We have completed our reading Morris Kline's Mathematics and the Physical World. As I begin to turn my focus to the middle school math class that I will teach in the fall, books like Kline's can help motivate students to learn math as they learn about the connections to science.

For the sixth and seventh graders, I can show them ideas from the first six or seven chapters of the book, minus the trig parts. For the eighth graders, I can go a bit further in the book, since here in California eighth graders focus on the physical sciences. This may change under Next Generation Science standards, though, to a more integrated approach.

Today is a traditionalists post, but I think I'll do the PARCC question first. Question 28 of the PARCC Practice Exam is on trig:

28. The figure shows right triangle ABC.

[The triangle is labeled, with the right angle at A and a, b, and c opposite A, B, and C -- dw]

Which of the listed values are equal to the sine of B?

Select all that apply.

A. b/c
B. c/a
C. b/a
D. the cosine of B
E. the cosine of C
F. the cosine of (90 - B)
G. the sine of (90 - C)

And now we have a trig problem -- the other subject I taught during that full week I spent in a math classroom, this time the Algebra II class rather than Honors Integrated Math I. Again, all these great problems that match up with the worksheets are appearing one and two weeks after I subbed there.

The usual definition of sin B is choice (C). But we know the definition of cosine as complementary sine -- and this is a frequent PARCC question, as we've seen in the past. As it turns out, choices (E), (F), and (G) are all correct.

Frequent errors include marking only one answer instead of four, and getting confused with the relationship between sine and cosine. It may help to change 90 - B to C and 90 - C to B before attempting to answer -- then it becomes obvious that (E) and (F) both say cos C and (G) says sin B, which is the original question.

This is a great point to transition into another traditionalists post. I've said before that my classes will not be traditionalist, except when it comes to basic skills, where I agree with traditionalism. One traditionalist I haven't discussed on the blog yet, but whose ideas will be a strong influence in my own classes, goes only by the username "Bill." Bill often posts at the following link:

http://www.joannejacobs.com/

The creator of this website (Joanne Jacobs) regularly links to various education articles -- I often skip the middle woman and link directly to those articles, but if I do link to Jacobs herself, it's to highlight the comments of Bill and a few other traditionalists who post in the comments.

Bill strongly believes that students graduating these days lack basic skills. Here are a few of his most recent comments:

http://www.joannejacobs.com/2016/05/principal-teachers-dont-know-how-to-teach-reading/

Your 2nd grade teacher sounds like an idiot, if he or she couldn’t spot the properly spelled word and the improperly spelled one, but then again, in many elementary schools these days in the U.S., a student who fails all of their coursework in every subject isn’t held back, either, so they don’t have an issue with their self-esteem.

What happens in 6-7 years when they reach middle school and cannot read, write, or add/subtract/multiply/divide…

I'm actually confused with the first part of Bill's comment, since the "2nd grade teacher" mentioned here was trying to catch cheating. The important part of the comment is the second part. Bill complains that grade retention isn't used often enough, because students are allowed to move on to the next grade without having basic skills. This is a frequent traditionalist complaint.

http://www.joannejacobs.com/2016/05/math-excellence-is-it-just-for-asian-americans/#comments

That’s because in the 70’s and 80’s, persons who excelled academically or showed a strong interest in math/science got labeled as ‘geeks’, and the kudos always went to the football/baseball/basketball teams (or insert your sport here)…This is the way it has always been in high school/college/professional sports.

Given that it’s darn near impossible to hold back a student who has failed all of their courses in elementary school, is it any wonder why students who put the time and excel academically are in those ethnic groups which make education a priority?

I was looking at a article on using CRISPR-cas9 to edit DNA, and this was being discussed in a high school biotechnology classroom, I’d say all of the students were either caucasian, asian, or asian-american (go figure)…

By the way, most of Bill's comments are not about racial politics, but unfortunately, his most recent comment as of today is about racial politics (and I'd planned on writing this traditionalist post today about Bill long before I read this comment). Then again, I am burying this in a post just before a three-day weekend, which I often do in order to minimize the controversial subject of race. (The original article also mentioned gender -- not enough girls are going into STEM, but Bill and the other commenters chose to ignore gender and focus only on in this thread.)

Of course, the first part of the comment has nothing to do with race. Words like "geek" and "nerd" aren't specific to any race -- Bill laments that sports are considered high status and math is considered a low-status activity. We've seen before that students are willing to work on athletic drills, which are neither easy nor fun, if the drills would make them a champion athlete. But the same students, if they are asked to perform mathematical drills, will insist that they instead do only activities that are easy and fun, and ask questions like "Why?" and "When will we ever use this?" whenever asked to perform a task in math class that is not easy or fun.

Now it's the second part of Bill's comment where race places a factor. He repeats his complaint that there isn't enough retention -- a crude form of tracking -- in elementary school. But here's the thing -- Bill claims that this failure to track students, far from reducing the achievement gap, is actually contributing to the gap! Bill's post is implying that if elementary schools were to bring back retention, perhaps after repeating a grade, students will suddenly see the need to work hard in their classes in order to avoid being held back again. Then students of all races would suddenly have an incentive to work hard, and with everyone working hard, the achievement gap would be reduced.

This is a typical traditionalist argument, and I don't fully agree with it at all. I suspect that if an aggressive retention plan were adopted, students wouldn't start studying harder. Instead, the natural human reaction is to blame the system, rather than themselves, for their failure. And if we add the highly volatile factor of race, people are even more likely to blame the system -- and this is exactly why tracking no longer exists! Once again, the traditionalist plan won't give the results that the traditionalists desire.

And here's another, non-racial factor that the traditionalists fail to consider. Some students will refuse to learn, and so there will be huge age gaps in the classroom. Bill mentions self-esteem and the broken feelings of the students who must repeat grades, but I'm worried about the broken bones of the smarter students who are beaten up by kids in the same class who may be four or more years older than they are.

Last night was the Scripps National Spelling Bee, and every year after the competition, I watch one of my favorite movies, Akeelah and the Bee. It is about a girl who is branded as a nerd for being such an excellent speller, but becomes a local heroine as she advances to the National Spelling Bee. (Oh, and race definitely plays a factor among the characters in the movie as well.)

Now I said that I want to apply Bill's thoughts to my own classroom. I'm a middle school teacher, so there's obviously nothing I can do about elementary school retention (regardless of whether I agree with Bill's idea or not). But I do want to turn around the idea that students who are successful at math are "nerds" who are worthy of scorn.

Indeed, I turn around the word "nerd" to form the word "dren" -- an anti-nerd who is incapable of performing elementary-level math. Even though Bill doesn't use the word "dren" (of course not, since I made it up), it encapsulates Bill's opinion in a single syllable -- it's not the students who are good at math who have the problem. The students who can't do math are the ones with the problem!

In some of Bill's posts, he writes about college and career readiness. I will continue to write Bill's comments throughout the summer, and I may even repeat some of these in the classroom. The point is that the so-called "nerds" are actually popular with the only people who matter -- college admissions officers and employers -- and it's the "drens" who are unpopular with the people who matter. I'll have more to say about this idea as we read more of Bill's comments together on the blog.

My next post will be on May 31st, which is after Memorial Day.

PARCC Practice EOY Question 28
U of Chicago Correspondence: Lesson 14-4, The Sine and Cosine Ratios

Key Postulates: Definition of Sine and Cosine

Common Core Standard:
CCSS.MATH.CONTENT.HSG.SRT.C.7
Explain and use the relationship between the sine and cosine of complementary angles.

Commentary: The worksheet is based on the Pizzazz Worksheet from Algebra II. I changed it so that the hypotenuse is a, to match the PARCC problem on the previous page.

PARCC Practice Test Question 27 (Day 170)

Chapter 27 of Morris Kline's Mathematics and the Physical World is "Mathematics and Nature." Yet another very famous name is the author of today's quote:

"How can it be that mathematics, a product of human thought independent of experience, is so admirably adapted to the objects of reality?" -- Albert Einstein

Kline begins:

"There is a well-known story, largely apocryphal, concerning the visit of the French Encyclopedist, Denis Diderot, to the court of Catherine the Great...Diderot accepted [the invitation -- dw]. Euler appeared, rattled off some meaningless mathematical formulas, and concluded, 'So God exists.'"

If I remember correctly, the story was that Euler used his famous formula "e^(i*pi) = -1" in his proof of the existence of God. As we can see, this chapter of Kline is somewhat religious in nature. My earlier promise is that I would not discuss religion here on the blog unless it is in the context of calendars, which is why I added the "Calendar" label to this post.

Kline writes that many mathematicians, not just Euler, believed that the theorems of math were absolute truths, and so one could mathematically prove the existence of God. But he writes why this belief was flawed:

"The development of non-Euclidean geometry showed that man's mathematics did not speak for nature, much less lead to a proof of the existence of God."

So in a way, this is a continuation of the previous chapter. The fact that there exist geometries other than the familiar Euclidean geometry demonstrate that there is no absolute truth in mathematics -- truth is relative to the axioms and postulates that are chosen at a particular time. If we chose Euclid's Fifth Postulate, then the angles of a triangle add up to 180. If we choose a different postulate, then the angles of a triangle don't add up to 180. As Kline writes, even 2 + 2 = 4 is not absolutely true:

"But there are algebras, physically useful algebras, in which this statement does not hold."

Oh, and since I added the "Calendar" label to this post, I'd better say something about calendars and religion lest I break my own promise about writing about religion on the blog. Well, let's see:

-- On the Jewish Calendar, today is Lag B'Omer (33 days after Passover).
-- On the Christian Calendar, today is Corpus Christi (60 days after Easter). Yes, there's a Texas city named after this feast day.
-- On the Usher Calendar, Memorial Day weekend coincides with Pentecost (the 50th day after Usher Easter, which always falls April 5th-11th). I explained back on Leap Day that many American secular holidays and Christian holidays line up on the Usher Calendar.
-- On the Muslim Calendar, Ramadan will begin in just over a week (approx. June 6th).

Question 27 of the PARCC Practice Exam is about the diagonal of a screen:

27. A computer monitor is 20 inches wide. The aspect ratio, which is the ratio of the width of the screen to the height of the screen, is 16:9. What is the length of the diagonal of the screen, to the nearest whole inch?

This question involves similar rectangles and require us to set up a proportion. We'll let a be the altitude and b the base of the screen:

b/a = 16/9
20/a = 16/9
16a = 180
a = 11.25"

You may wonder, why did we use a for altitude and b for the base, rather than the more natural h for height and w for width? Oh, it's because we need to use the Pythagorean Theorem to find the diagonal:

a^2 + b^2 = c^2
(11.25)^2 + 20^2 = c^2
126.5625 + 400 = c^2
526.5625 = c^2
c = 22.9", which rounds to 23"

As is typical for measurement problems, PARCC will count either 22" or 23" as correct.

On one hand, we could bill this as a Pythagorean Theorem question from Lesson 8-7. But this question is very nearly like one that I included on the Chapter 12 Test (the lchapter on similarity) back in February. On that day, I alluded to the fact that TV's used to have a 4:3 aspect ratio, but now the standard is 16:9, just as in the PARCC problem.

But that question, as originally written in the SPUR section of the U of Chicago text, gave the ratio of a standard def TV as 9:7, referring to the diagonal-to-width ratio rather than width-to-length. This is approximately the same as a 4:3 width-to-length ratio. I converted this to an HD TV when I wrote the Chapter 12 Test, and I implied a diagonal-to width ratio of 8:7. This does correspond approximately to a 16:9 aspect ratio.

Notice that the 4:3 aspect ratio of the old TV sets should bring to mind the 3-4-5 Pythagorean triple and right triangle. Indeed, the ratio 9:7 is a poor excuse of an approximation to the diagonal-to-width ratio of the standard def TV's, considering that 5:4 is the exact value and is simpler than 9:7. (In music, we ordinarily don't play 9:7, a "septimal major third," instead of 5:4, the just major third.) On the other hand, the diagonal of a triangle with legs 9 and 16 is sqrt(337). I'd use 8:7 to approximate sqrt(337):16 well before using 9:7 to "approximate" 5:4.

Then again, the ratios 8:7 and 9:7 are easy to compare. The diagonal-to-width ratio is smaller for the HD TV. which makes sense, if you think about it. If we had a standard def TV of width 20", it's easy to calculate its diagonal as 25". Then if we played a widescreen video on this TV, black bars would appear near the top and bottom of the screen, so that the diagonal of the visible image is only 23" instead of 25". (This does not mean that the top and bottom bars are about 1" each. We must use the altitude, not the diagonal, to find the size of the black bars. We calculate the height of the 20" standard def TV as 15", and we already found the height of the widescreen image to be 11.25". This means that the top and bottom bars are closer to 2" each.)

PARCC Practice EOY Question 27
U of Chicago Correspondence: Lesson 8-7, The Pythagorean Theorem

Key Theorems: Pythagorean Theorem
In any right triangle with legs a and b and hypotenuse c, a^2 + b^2 = c^2.

Common Core Standard:
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

Commentary: The questions come from the old Pizzazz worksheet that I covered during the week I subbed in an Algebra II class. Oh why couldn't I have covered that class this week instead, when I'm discussing these questions on the blog?

PARCC Practice Test Question 26 (Day 169)

Chapter 26 of Morris Kline's Mathematics and the Physical World is "Non-Euclidean Geometry." I know that I've discussed non-Euclidean geometry extensively on the blog.

"One must regard nature reasonably and naturally as one would the truth, and be contented only with a representation of it which errs to the smallest possible extent." -- John Bolyai, 19th century Hungarian

Kline begins:

"Toward the ends of their lives, Euler, D'Alembert, and Lagrange agreed that the realm of mathematical ideas had been practically exhausted and that no new great minds were appearing on the mathematical horizons. Of course, these men had grown old and their vision was already dimmed, for Laplace, Legendre, and Fourier were already in young manhood."

But the hero of this chapter is Karl Friedrich Gauss, who needs no introduction. Here is a little of the great mathematician's bio as described by Kline:

"Karl Friedrich Gauss's brilliance was noted by his elementary school teachers, and they helped him to secure a good education. [No, Kline says nothing about adding up numbers 1 to 100 -- dw.] There is a story that at Gottingen he approached one of his university teachers, A.G. Kastner, with a proof that the 17-sided regular polygon is constructible with straightedge and compass -- at that time one of the outstanding construction problems. Of course, [Kastner] knew that the problem was of theoretical interest, but he did not believe Gauss could solve it. Then Gauss explained that he had reduced the solution of the seventeenth degree equation [that appears in a proof -- dw] to one of lower degree and then solved the latter. For this rebuff, Gauss repaid Kastner, who prided on being something of a poet, by lauding him as the best mathematician among poets and the best poet among mathematicians."

As we know now, Gauss was correct in that he really did construct a regular 17-gon. But no, neither the PARCC, nor any other Common Core test, will ever require high school students to know the steps to construct a regular heptadecagon, as it is very complicated. After this discovery, Gauss was convinced to spend the rest of his life studying mathematics.

And of course, one of his discoveries was non-Euclidean geometry -- one in which the Parallel Postulate is replaced with another axiom. Indeed, Gauss independently discovered the new geometry with two other mathematicians -- one of whom uttered the quote at the start of this chapter, Bolyai.

Kline writes:

"Their chief idea was the one we have already mentioned, namely, that Euclid's parallel axiom is an independent assumption about parallel lines and hence it is logically possible, whether or not it serves any scientific or practical purpose, to replace it by a contradictory axiom. What alternative axiom did these men adopt? One alternative would be to assume that every line through P met l; that is, there are no lines through P which are parallel to l. ...[T]hey had found that theorems deduced from this axiom and the other nine axioms of Euclid contradicted each other. This outcome meant that such an alternative to Euclid's parallel could not be entertained, since a body of inconsistent results certainly made no sense."

First of all, the total of ten axioms (a Parallel Postulate and "the other nine axioms") are mentioned in Lesson 13-6 of the U of Chicago text (five algebraic and five geometric postulates). All the talk about Euclid's "Fifth Postulate" refers to his fifth geometric postulate, not any algebraic postulates.

Anyway, we've discussed this contradiction that Gauss and others found before on the blog. We've seen that a geometry that satisfies all of Euclid's axioms except the Fifth Postulate is known as "neutral geometry." Spherical geometry is a well-known example of a non-Euclidean geometry in which there are no parallel lines. But, just as Kline implies above, spherical geometry is not a neutral geometry, since the Spherical Parallel Postulate contradicts the nine neutral axioms! Therefore the geometry that Gauss and Bolyai discovered is actually hyperbolic geometry -- the non-Euclidean geometry that is neutral, yet a bit harder to visualize than the non-neutral spherical geometry.

In fact, I myself had heard of non-Euclidean geometry, but I never knew that spherical geometry is not neutral until I researched the website of Dr. Franklin Mason for this blog. On an old version of his website. Dr. M used no version of a Parallel Postulate to prove the theorem "Through a point on a line, there is at least one line parallel to the given line." His proof was therefore neutral -- thus demonstrating that parallel lines exist in neutral, but not spherical, geometry.

Kline moves on to discuss the discovery of spherical geometry by Bernhard Riemann, another very famous mathematician. Kline writes:

"Hence, he proposed to replace the infiniteness of the Euclidean straight line by the condition that it is merely unbounded [like a circle -- dw]."

So Riemann had to replace Euclid's Second Postulate in order to make Spherical Geometry work. If Spherical Geometry were neutral, we'd have to keep the Second Postulate (and all others except the Fifth), but again, that's why spherical geometry is not a neutral geometry.

Last summer, I devoted some posts to Adrien-Marie Legendre's Geometry, which includes some work on spherical geometry. Notice that Legendre is one of the mathematicians that Kline mentions at the start of today's chapter as part of the "old" generation. We found out last year that Legendre's treatment has some gaps in it -- first he attempts to "prove" the Parallel Postulate, then much later on he discusses the properties of spherical lines without considering it to be a separate geometry. It wasn't until Riemann -- who was just six years old when Legendre died -- who put spherical geometry on a completely rigorous foundation.

Then last fall, I changed the order of the U of Chicago text so that congruence was taught earlier in the course, well before parallel lines. This meant that I had to pay more attention to which theorems required a Parallel Postulate and which ones didn't (i.e., were neutral). But this neutral geometry didn't connect to the spherical geometry I wrote over the summer, since spherical geometry is not a neutral geometry.

By the way, this summer I plan on returning to spherical geometry on the blog. I am adding the label "Spherical Geometry" to this post as a reminder that I want to finish Legendre this summer, starting at where I left off from last year.

Actually, Kline writes a little about spherical lines in the previous Chapter 25:

"If the surface of the earth is treated as a perfect sphere, and this may be assumed for navigation of the oceans, then the geodesic between two points on the surface is the shorter arc of the great circle passing through these points."

That chapter was all about the calculus of variations. I once took such a course in college, but I never did figure out the integral I was supposed to calculate to prove the geodesy of the great circle. (I couldn't figure out what to do with this strange "arctanh" term!)

In the current chapter, Kline writes more about the better-known results -- that the sum of the angles of a spherical triangle exceeds 180 (for a hyperbolic triangle, it's less than 180) and that in both spherical and hyperbolic geometry, all similar triangles are congruent. (In Common Core parlance, all similarity transformations are isometries.) Kline then describes the experiment in which Gauss tried to calculate the sum of the angles of a large triangle on earth to determine whether geometry was spherical, hyperbolic, or Euclidean. The U of Chicago mentions this experiment in Lesson 5-7, when the students first learn the Triangle-Sum Theorem.

After that long discussion on non-Euclidean geometry, fortunately today's PARCC question is short:

26. Which geometric figures have a measurable quantity?

Select each correct answer.
A. line
B. angle
C. point
D. line segment
E. ray

The correct answers are obviously (B) and (D). We even have two postulates that describe how these are measured -- the Protractor Postulate and the Ruler Postulate. The former is called the Angle Measure Postulate, and the latter part of the Point-Line-Plane Postulate, in the U of Chicago text. The common student errors include mistaking "line" for "line segment" and, as usual, marking only one correct answer instead of both.

PARCC Practice EOY Question 26
U of Chicago Correspondence: Lessons 1-7, Postulates and 3-1, Angles and Their Measures

Key Postulates: Point Line-Plane Postulate
(d) Distance assumption: On a number line, there is a unique distance between two points.

Angle Measure Postulate
(a) Unique measure assumption: Each angle has a unique measure from 0 to 180 degrees.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.A.1
Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, distance along a line, and distance around a circular arc.

Commentary: Since spherical geometry is on my mind today, let's try answering today's PARCC question in spherical geometry. Now a line is measurable since it has the same measure as a great circle. Some authors define a ray in spherical geometry to be half of a great circle, so it also has a measure. Only a point has zero measure in spherical geometry.

PARCC Practice Test Question 25 (Day 168)

Chapter 25 of Morris Kline's Mathematics and the Physical World is called "From Calculus to Cosmic Planning." In this chapter, Kline discusses the Calculus of Variations.

"The spectacle of the universe becomes so much the grander, so much the more beautiful, the worthier of its Author, when one knows that a small number of laws, most wisely established, suffice for all movements."

So we see that this chapter is all about minimizing things, including physical laws (which we've discussed in the previous chapter as well). Now Kline begins with a problem familiar to the readers of this blog:

"According to one of the legends of history, Dido, of the Phoenician city of Tyre, ran away from her family to settle on the Mediterranean coast of North Africa. There she bargained for some land and agreed to pay a fixed sum for as much land as could be encompassed by a bull's hide."

Ah yes, we are discussing the Isoperimetric Problem, Lesson 15-8 of the U of Chicago text. Actually, back when I was researching the Isoperimetric Problem for this blog, I saw references to the ancient Queen Dido, but I didn't mention her on the blog at all. Actually, what I knew then about the Dido problem was incomplete -- I didn't know about the following twist until I read it in Kline:

"Her second bright idea was to use this length to bound an area along the sea. Because no hide would be needed along the seashore she could thereby enclose more area."

We know that the solution to the Isometric Problem is the circle -- the curve that encloses the most area for its length. We've also seen questions in which we are to maximize area by building a fence along a river to enclose a rectangular area -- the answer is a rectangle whose width is exactly half of its length. Combining these two ideas, we can solve the Dido problem:

"According to the legend, Dido thought about the problem and discovered that the length of hide should form a semicircle."

So we see that without water, the largest area is a circle -- with water, it's a semicircle. If we restrict to rectangles, without water the largest area is a square -- with water, it's a semi-square (that is, half of a square, or a rectangle whose width is half of its length).

We found out earlier (Chapter 6 of Kline) that the reason for the semi-square involves reflections -- if we reflect the rectangle across the river, we have a larger rectangle whose area is maximized for the new fixed perimeter (double the original fence length). The solution to that problem is a square, so that of the original problem with the pre-image rectangle is a semi-square. Similarly, if we reflect Dido's land across the water, the figure should be a circle (the solution to the Isoperimetric Problem), so the pre-image is a semicircle.

Kline writes that the life of Dido -- perhaps the world's first female mathematician -- ended tragically:

"[Dido's new lover] Aeneas was a man on a mission, and he soon departed to found a new civilization in Rome. Dejected and distraught, Dido could do no more for Aeneas than to throw herself on a blazing pyre so as to help light his way to Italy...Rome made no contributions to mathematics whereas Dido might have."

Kline writes only three equations in this chapter -- all of them involving integrals. In this chapter, he proceeded to write an equation to describe an object falling on an arbitrary curve, which looks very ugly when we try to write the integral in ASCII:

T = 1/sqrt(2*32) int _0 ^x1 (sqrt((1 + y' ^2)/y) dx

This integral is used to solve the Brachistochrone Problem -- find the curve that minimizes the time it takes for an object to fall from P to Q not directly below P. Three Swiss mathematicians worked to solve this problem: the Bernoulli brothers (Johann and Jakob) and later on, a more famous name I've mentioned several times on the blog -- Leonhard Euler. Their work laid the foundation for a new branch of mathematics -- the Calculus of Variations.



Question 25 of the PARCC Practice Test is on constructions:

25. Using a compass and a straightedge, a student constructed a triangle in which XY is one of the sides.

The compass is opened to a set length and two intersecting arcs are drawn above XY using X and Y as the centers. The intersection of the two arcs is labeled as point Z.

Part A
What could be the set length of the compass so that triangle XYZ is isosceles but not equilateral?
Select all that apply.
A. less than 1/2 (XY)
B. equal to  1/2 (XY)
C. between 1/2 (XY) and (XY)
D. equal to (XY)
E. greater than (XY)

Part B
Select the correct phrase to complete the sentence.
If the opening of the compass is Choose... [from the same choices as A-E above -- dw], then triangle XYZ will be equilateral.

What's this? Here's yet another question that's on classical constructions,right after I spent an entire week in an Honors Integrated Math I class studying constructions!

Anyway, Part A of this question is similar to the earlier construction problem, Question 21. We recall how Jericho set his compass to be less than the length of the segment PQ. This prevented Choice (E), "an equilateral triangle," from being one of the correct answers. Since once again, we are asked not to construct an equilateral triangle, so it would appear that "less than XY" would be correct. There are three choices here that indicate less than XY -- (A), (B), and (C) -- and since this is a "select all that apply" question, all three could be correct.

At this point we should begin to wonder, what does 1/2 (XY) have to do with anything? Well, we can try an actual compass set to less than 1/2 (XY) and see what happens. The result is that the two arcs fail to intersect, and no triangle is formed. The problem here is the Triangle Inequality -- the longest side of the triangle can't be as long as or longer than the the sum of the other two sides. So if one side has length XY, the other two sides must be more than 1/2 (XY). Neither (A) nor (B) produce a triangle (in the latter case, the two arcs intersect right at the midpoint of XY).

Also, there's no reason that the compass setting can't be greater than XY (even though it's always less than the segment length in Question 21). This never violates the Triangle Inequality since the length of the long leg is always less than the sum of the other long leg and the base XY. Thus there are two correct answers to Part A, (C) and (E).

For Part B, we want the triangle to be equilateral. Of course we want all three sides to equal XY, so the fourth choice from the drop-down menu -- the same as choice (D) above -- is correct.

I suspect most student errors will occur in Part A. We begin with the usual problem with a multi-part answer, namely that students will only choose one answer even if two or more are correct. But even if the students realize that they must choose more than one answer, they might be influenced by Question 21 to choose answers like (A) or (B).

In fact, come to think of it, the earlier Question 21 is poorly worded. That question doesn't consider the possibility that the compass setting is less than half of PQ -- and if it were, none of the five choices from Question 21 would be constructed! And I admit that I probably would have ignored the less than half case -- except that just four questions later, we're forced to consider that case! Who knows -- perhaps we're supposed to know that the compass setting is more than half of PQ in Question 21 by looking at the diagram, which would be impossible otherwise. We're only given that the setting is less that PQ so that we know not to choose "an equilateral triangle" (or "a 60-degree angle") for that question.

PARCC Practice EOY Question 25
U of Chicago Correspondence: Lesson 3-6, Constructing Perpendiculars

Key Theorem: Construct the perpendicular bisector of a given segment AB.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.D.12
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.).Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Commentary: Inspired by the story of Dido, I decided to add two construction questions to construct a region over which she could rule. I changed "bull hide" to "fence" in order to avoid distracting the students who might wonder what the hide is for.

PARCC Practice Test Question 24 (Day 167)

Chapter 24 of Morris Kline's Mathematics and the Physical World is called "Differential Equations -- the Heart of Analysis." Differential equations are the next step beyond Calculus.

"Nature is pleased with simplicity, and affects not the pomp of superfluous causes." -- Isaac Newton.

Kline begins:

"Man's expectation that he can understand the multifarious workings of nature has rested upon the belief that nature is designed not only rationally but simply."

And so physicists wanted to derive as many facts from as few principles as possible. But to do this, they needed to solve differential equations.

Kline's first example of a differential equation is:

d = 32t

But this is clearly not a differential equation. Kline calls this an equation about velocity, so he clearly means to write d-dot here. In ASCII, we use the prime symbol instead:

d' = 32t

A solution of this differential equation is easily found:

d = 16t^2

So notice that every time we calculate an integral, we're actually solving a differential equation. Kline gives a more complicated differential equation:

v' = -96,000/(r^2)

From the solution of this, Kline calculates the escape velocity of a spacecraft to reach the moon. The solution of this equation gives 6.87 miles per second, and he writes that it would take a spacecraft about four days, 21 hours to reach the moon. (Kline wrote this book before the moon landing, but notice that Apollo 11 really did take a little over four days to reach the moon.)

Question 24 of the PARCC Practice Exam is on the circumference of a circle:

24. A landscaper is designing a display of flowers for an area in a public park. The flower seeds will be planted at points that lie on a circle that has a diameter of 8 feet. The point where any seed is planted must be at least 2 feet away from the seeds on either side of it.

Part A
What is the maximum number of flower seeds that can be planted using the design?

Part B
After planting the flower seeds, the landscaper has 20 seeds left over. The landscaper wants to plant all of the remaining seeds in another circle so that the seeds are 2 feet apart. To the nearest tenth of a foot, what is the diameter of the smallest circle that the landscaper can use to plant all of the remaining seeds?

This question clearly asks students to use the circle circumference formula, C = pi d. So for Part A, the circumference of the circle of diameter 8 feet is 8pi feet. Since we must allow for 2' per seed, we conclude that there must be at most 4pi seeds. Now 4pi is about 12.6, but there isn't quite enough room for the 13th seed (13 seeds would require a diameter of 8.3'), so there are only 12 seeds.

For part B, we see that 20 seeds require 40', so the diameter is 40/pi feet. We calculate that 40/pi is about 12.73', which we round to the nearest tenth as 12.7'.

For the second time in the last three questions, students may have some problem with rounding. Now according to the key provided by PARCC, 12.6', 12.7', and 12.8' are all acceptable answers. But for the number of seeds, only 12 is the correct answer. We see that PARCC is especially picky with rounding when it comes to real countable objects such as seeds, as opposed to measurements. There isn't enough room for the thirteenth seed, so the answer is twelve seeds. On the other hand, six bags of candy wasn't enough to fill the spherical candy jar -- we needed seven bags.

In each case, students need to read the question to know which way to round. For today's question, each seed needs at least two feet of clearance -- we're allowed to give each seed extra clearance, but we can't shortchange the amount of clearance by having too many seeds. Last week, we were asked to fill the sphere -- we're allowed to overfill it, but we couldn't leave any extra space. On the other hand, last week's question could have asked, "What's the most number of bags that can fit in the sphere?" To that question, the answer would have been six bags, because the seventh bag can't fit.

I also wonder whether students might be confused as to what to do to solve the problem. Some students might try to find the area of the circle or something like that, or they might find the circumference, but the fail to divide it by two since each seed need 2' clearance. Notice that once we knew what we had to find, the calculation was trivial.

Some readers might notice that the pre-rounded answers to Part A and Part B are almost the same, each about 12.6. But observe that we're calculating different things in each case -- the exact answer for Part A is 4pi, and for Part B, it is 40/pi. These are nearly equal because pi is so close to sqrt(10) -- remember our discussion about how sqrt(10) Day would be just two days after Pi Day.

I decided that since we never actually completed any questions from Lesson 8-8 of the U of Chicago text, I included some on this worksheet. But now I realize that some of the "review" questions from this section may be too difficult for the students. One of the questions asks for the area of an equilateral triangle -- this is a "review" question from Lesson 8-8, but in reality students can't answer until they learn about 30-60-90 questions in Lesson 14-1! (Of course, we have finished Lesson 14-1, so we have no such handicap.)

Another question asks to compare the perimeters of two triangles. Even though students might be able to eyeball the answer, a rigorous proof is quite tricky. Students must notice that one pair of sides form opposite sides of a parallelogram, and they have a second pair in common. The third pair of sides can be compared using the SAS Inequality (Hinge Theorem).

PARCC Practice EOY Question 24
U of Chicago Correspondence: Lesson 8-8. Arc Measure and Arc Length

Key Theorem: Circle Circumference Formula

If a circle has circumference C and diameter d, then C = pi d.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.GMD.A.1
Give an informal argument for the formulas for the circumference of a circle, area of a circle, volume of a cylinder, pyramid, and cone. Use dissection arguments, Cavalieri's principle, and informal limit arguments.

Commentary: Some questions from Lesson 8-8 are included.

PARCC Practice Test Question 23 (Day 166)

Today is a traditionalist-labeled post. I used last week's traditionalists posts to discuss the block schedule that my new school will use in the fall. Today's post will discuss the other aspect of my new school -- the fact that it is a charter school.

What exactly is a charter school? Well, why don't we ask the president? He has declared the first week of May each year as National Charter Schools week. This is what he said in this week's presidential proclamation about two weeks ago:

http://www.washingtonexaminer.com/what-obama-said-about-charter-school-week/article/2590212

"During National Charter Schools Week, we celebrate the role of high-quality public charter schools in helping to ensure students are prepared and able to seize their piece of the American dream, and we honor the dedicated professionals across America who make this calling their life's work by serving in charter schools," Obama said in a presidential proclamation. "With the flexibility to develop new methods for educating our youth, and to develop remedies that could help underperforming schools, these innovative and autonomous public schools often offer lessons that can be applied in other institutions of learning across our country, including in traditional public schools."

Charter schools are not without controversy, though. Once again, I want to avoid biting the hand that feeds me, and so I don't discuss this controversy on the blog. Instead, let's go to see what the traditionalists have to say about charter schools.

Before I was hired to work at one, I've only mentioned charter schools thrice on the blog. The first example I gave was a middle school that starts in fifth grade, and the second example I gave was a Waldorf-inspired school. The third example I gave was BASIS -- a charter school which prides itself on accelerating students as much as possible. We know that the traditionalists especially like BASIS, since traditionalists also emphasize acceleration.

No, the school I'll be working at in the fall isn't any of those three schools. My middle school begins at the more common sixth grade, not fifth (though it is a K-8 campus), it isn't a Waldorf school, and it certainly isn't BASIS (as there are no BASIS charters in California) -- otherwise I'd be preparing to teach Saxon Math 87, Algebra I, and Algebra II to my middle school students! But this does show that there are many different types of charters.

Charters are associated with choice. A quick glance at some traditionalist websites tells me that some traditionalists favor charters because they give teachers choices such as pedagogy -- meaning, of course, that they like charters that choose the traditionalist pedagogy over the more dominant progressive pedagogy. (I wonder whether they'd be as enthusiastic about charters if the dominant pedagogy were traditionalism and the progressives were the ones who wanted an alternative choice!)

Oh, and by the way, I subbed in at another class on the block schedule due to SBAC testing. This is the same school that I was at last week, with a Hybrid Block Schedule -- but as I pointed out last week, the blocks are now two full hours since school can't be out a half hour early everyday. This time it's a health class, but once again, the students watch a video (on the danger of STD's) and then study for a test coming up on Monday. Again, I can see traditionalists like Jeff Lindsay who oppose the block schedule point out that teachers use blocks just to play videos -- but once again, a counterargument is that teachers use blocks to have the sub play videos. Because I'm a sub, I can't be sure what these teachers do on block days when there's no sub. (And notice that the math teacher last week didn't play any videos on block days.)

Chapter 23 of Morris Kline's Mathematics and the Physical World is called "The Integral Calculus." I covered derivatives yesterday, so now it's time for integrals.

"As God calculates, so the world is made." -- Leibniz

Kline begins:

"One of the fascinating facts about the development of mathematics is that an idea that is created to solve one type of problem often solves another which on superficial examination seems to be totally unrelated. This turn of events occurred in the history of the calculus."

Notice that Kline quotes our other co-inventor, Leibniz. And the connection that he and Newton discovered is that the derivative and the integral are related -- the Fundamental Theorem of Calculus.

Kline's example to find the area of the curve y = x^2 between x = 3 and x = 5. We now think of this as the Riemann sum, but Newton and Leibniz originated the idea. Kline writes:

y_1 h + y_2 h + ... + y_n h

to approximate the area by n rectangles.

Actually, I won't write the rest of Kline's equations here, since the summation and integral signs look messy when converted into ASCII (and because I took up so much of this post with the discussion about traditionalists).

Question 23 of the PARCC Practice Exam is on translations on the coordinate plane:

23. Quadrilateral ABCD is shown graphed in the xy-coordinate plane.

Part A
Quadrilateral ABCD will be translated according to the rule (x, y) -> (x + 3, y - 4) to form A'B'C'D'.
Select the correct orientation of A'B'C'D' and place it correctly in the plane.

Part B
Quadrilateral ABCD maps onto A"B"C"D". It will undergo a different transformation that will map A(-6, 3) to A", B(-4, 5) to B", C(-1, 6) to C", and D(-3, 2) to D". The transformation will consist of a reflection over the y-axis followed by a translation. Point D" is shown plotted in the plane after the transformation.
Plot the point A" in the plane.

[The coordinates of D" are (6, 2) -- dw]

Part A is straightforward. The transformation is a translation, so we know that the image has the same orientation as the preimage. So we only need to translate a single point, say A(-6, 3) to A'(-3, -1), and place the quadrilateral so that A' is at the correct point.

Part B appears to be a glide reflection -- the composite of a reflection and a translation. Since we're given the image D", it's best just to reflect D first in the y-axis, to obtain (3, 2). The final point we need for D" is (6, 2), which is three units to the right of the mirror image. Thus the translation must be three units right. Transforming A in the same manner gives us (6, 3) as the mirror image and then (9, 3) as the final image.

Notice that for glide reflections, the translation is usually taken to be parallel to the mirror, but this time the translation is perpendicular to the mirror. We've discussed this in the past -- when the translation is perpendicular to the mirror, the composite turns out to be a simple reflection -- not a glide reflection at all! The mirror for this new reflection is found by taking the original mirror and sliding it half the distance of the translation. So we translate the y-axis 3/2 units to the right, giving us the line x = 3/2 as the new mirror. Reflecting A in this new mirror does indeed give us (9, 3), but it's actually easier to reflect in the y-axis and then translate.

PARCC Practice EOY Question 23
U of Chicago Correspondence: Lesson 6-2, Translations

Key Theorem: Two Reflection Theorem for Translations

If m | | l, the translation r_m o r_l slides figures two times the distance between l and m, in the direction from l to m perpendicular to those lines.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.A.5
Given a geometric figure and a rotation, reflection, or translation, draw the transformed figure using, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that will carry a given figure onto another.

Commentary: After not providing much of an activity last week, this week I decide to repeat an activity from last year that is all about compositions of transformations.

PARCC Practice Test Question 22 (Day 165)

Chapter 22 of Morris Kline's Mathematics and the Physical World is "The Differential Calculus." We apparently can't avoid Calculus forever!

"It's enough if you understand the Propositions with some of the Demonstrations which are easier than the rest." -- Newton's advice to prospective readers of his Principia Mathematica

Kline begins,

"Through the study of the motion of projectiles, planets, pendulums, sound, and light the scientific world of the seventeenth century became conscious of the pervasiveness of change. It had also become aware of the usefulness of a function to represent relationships between variables and to deduce new scientific laws."

Kline's opening quote is from one of the co-inventors of Calculus, Isaac Newton. As we can see so far, Calculus deals with functions and change. Kline writes:

"Next to the creation of Euclidean geometry the calculus has proved to be the most original and the most fruitful concept in all of mathematics."

I said that Newton was one of the co-inventors of Calculus. The other was Gottfried Wilhelm Leibniz, who created the subject independently from Newton. Here's a little of what Kline says about Leibniz:

"This man of universal gifts and interests, son and grandson of professors, was born in Leipzig [Germany] in 1646. At the age of fifteen he entered the University of Leipzig with the announced intention of studying law and the unannounced intention of studying everything. His brilliance so excited the jealousy and the envy of his teachers that they never granted him his doctor's degree. During the years in which he acquired his legal training Leibniz was also busily studying mathematics and physics."

Kline's first example of Calculus is to find the instantaneous velocity of a ball three seconds after it has been dropped. He writes the following calculation, starting with the formula Galileo had found for the distance traveled by an object in free fall:

d = 16t^2
d_3 = 144
d_3 + k = 16(3 + h)^2 (That is, in an additional h seconds the ball travels an extra k feet.)
d_3 + k = 16(9 + 6h + h^2)
d_3 + k = 144 + 96h + 16h^2
k = 96h + 16h^2 (Here Kline substitutes in the value of d_3 found earlier and subtracts.)
k/h = (96h + 16h^2)/h
k/h = 96 + 16h

Kline writes that to find the instantaneous velocity, we should plug in h = 0 into the penultimate equation in this list -- but this gives us 0/0. So we plug it in to the last equation instead, and we end up with k/h (distance divided by time, or velocity) equals 96 feet per second. According to Kline, the great insight by Newton and Leibniz is that this is justified using limits -- as h approaches 0, the value of 96 + 16h approaches 96, and so this is taken to be the instantaneous velocity.

The author goes on to demonstrate that at an arbitrary time t = x, we perform the same calculation to y = 16x^2 to obtain 32x as the instantaneous velocity. This is what we now know as the derivative of the expression 16x^2. Kline writes that Leibniz used the notation dy/dx to denote the derivative of y with respect to x, but he prefers Newton's notation, y-dot (that is, a dot above y). This is difficult for me to show in ASCII, so I'll use another commonly used symbol for derivative -- the apostrophe, often pronounced "prime" (just as we do with transformation images in Geometry). So we have:

y = 16x^2
y' = 32x

Kline also works out the velocity of a bob on a spring, which follows the formula y = sin x, at the initial time t = 0. He shows that this equals the limit of the expression (sin h)/h as h -> 0. He gives the following argument -- when h is a small central angle of a circle, we can let r be the radius of the circle, and then we draw a right triangle with hypotenuse r and opposite leg a (since sine equals opposite over hypotenuse, or a/r). Then s is the arc subtended by the central angle h, and so the measure of h in radians is the arclength s divided by the radius r, or s/r. Then (sin h)/h works out to be a/s, as the r's cancel. But as h approaches 0, a approaches the arclength s (which is why a regular polygon with many sides approaches a circle), so a/s is approaching 1. Nowadays we would write:

y = sin x
y' = cos x
y'_0 = cos 0
y'_0 = 1

but we must actually prove that the derivative of sine is cosine.

Here are a few more amusing examples. Kline begins with the area of a circle and differentiate it with respect to the radius:

A = pi r^2
A' = 2pi r

That's funny -- the derivative of the area is the circumference! Kline explains why:

"This result is intuitively clear, for as the radius increases, one might say that 'successive' circumferences are added to the area."

And of course, we do the same thing with the volume of a sphere:

V = (4/3) pi r^3
V' = 4pi r^2

which is the surface area of a sphere.

Actually, the volume of a sphere is a great place to segue into PARCC. Question 22 of the PARCC Practice Exam is on the volume of cylinders and -- you guessed it! -- spheres.

22. Hank is putting jelly candies into two containers. One container is a cylindrical jar with a height of 33.3 centimeters and a diameter of 8 centimeters. The other container is spherical. Hank determines that the candies are cylindrical in shape and that each candy has a height of 2 centimeters and a diameter of 1.5 centimeters. He also determines that air will take up 20% of the volume of the containers. The rest of the space will be taken up by the candies.

Part A
After Hank fills the cylindrical jar with candies, what will be the volume, in cubic centimeters, of the air in the cylindrical jar? Round your answer to the nearest cubic centimeter.

Part B
What is the maximum number of candies that will fit in the cylindrical jar?

Part C
The spherical container can hold a maximum of 260 candies. Approximate the length of the radius, in centimeters, of the spherical container. Round your answer to the nearest tenth.

Part D
Hank is filling the cylindrical candy container using bags of candy that have a volume of 150 cubic centimeters. Air takes up 10% of the volume of each bag, and the rest of the volume is taken up by candy. How many bags of candy are needed to fill the cylindrical container with 260 candies?

To answer this question, we obviously have to calculate volume. For Part A, we calculate the volume of the cylinder as pi(4)^2 (33.3) = 1673.8 cm^3. Only 20% of this is air, so this gives us an air volume of 334.8 cm^3, which rounds off to 335 cm^3.

For Part B, we see that 1339 cm^3 is left for the candy. Now the volume of a single candy is pi(0.75)^2 (2) = 3.53 cm^3, and so we divide the volumes to give a value of 378.88 candies. (This last value is exact -- the factors of pi cancel out to leave a rational number.) Notice that there isn't quite enough room for 379 candies -- so the answer is 378 candies.

For Part C, we begin by multiplying 3.53 cm^3 for each candy by 260 to obtain 918.92 cm^3. But notice that this doesn't include the air. The candy only takes us 80% of the space, so we must divide this volume by 0.8 to obtain the total volume of the sphere as 1148.64 cm^3. This is the volume of a sphere and we want to know the radius, so we use the sphere volume formula that we found in Kline:

(4/3) pi r^3 = 1148.64
r^3 = 274.21875 (exact rational value)
r = 6.497, which rounds to 6.5 cm

For Part D, first of all, I'm wondering, why does the question have Hank put 260 candies in the cylinder when we already know that exactly 260 candies fit in the sphere? Wouldn't it have been more logical to ask, "How many bags of candy are needed to fill the spherical container?" (Then there wouldn't have been any need to say "260 candies" again -- "fill the sphere" would've been sufficient.)

At this point, it's probably easier to notice that since air takes up only 10% of the 150 cm^3-bag, the remaining 135 cm^3 is candy. We already calculated the volume of 260 candies earlier (when trying to find the radius of the sphere) as 918.92 cm^3, so we only need to divide this by 135 to obtain an answer of 6.8 bags. Six bags aren't enough -- we need 7 bags.

This question has several places where student error can creep in. This includes how to handle the air factor in both cases. Students must know the formulas for both the cylinder and the sphere without mixing the two up, as well as plug in the radius (not the diameter) correctly and solve for it properly (for example, don't take the square root instead of the cube root of r^3).

As for rounding error, according to the PARCC answer key, both 334 and 335 are acceptable for Part A, and both 6.4 and 6.5 are acceptable for Part C. However, both Part B and Part D have only one acceptable answer, as there isn't enough room for the 379th candy, and six candy bags aren't enough to fill the sphere.

One problem with questions that include many values that need to be rounded off is that rounding error may accumulate throughout the problem. (I assume this is the reason that Parts A and C have two acceptable answers.) But notice that since both the cylinder and sphere have pi in their formulas, the irrational value of pi cancelled out. This means that for Parts A, B, and C, we could have simply ignored pi altogether (or used a crude value like 3.14) and obtained the correct answer. Only Part D requires a value for pi as it doesn't cancel out (since the volume of the bag is 150 cm^3, not anything in terms of pi). But even a crude value like 3.14 still produces seven bags as the correct answer. (In fact, the approximation pi = 3 gives an answer of exactly six and a half bags.)

With so many possibilities for error, I expect most students to get at least one part, if not several parts, of this question wrong.

PARCC Practice EOY Question 22
U of Chicago Correspondence: Lessons 10-5 and 10-8, Volumes of Cylinders and Spheres

Key Theorem: Volume Formulas

The volume V of any prism or cylinder is the product of its height h and the area B of its base.
V = Bh

The volume V of any sphere is (4/3)pi times the cube of its radius r.
V = (4/3)pi r^3

Common Core Standard:
CCSS.MATH.CONTENT.HSG.GMD.A.3
Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

Commentary: There are no problems just like this one in the U of Chicago text. Instead, I combine parts of two questions -- one from the SPUR section of Chapter 10 about the volume of a cylindrical cup, the other from Lesson 10-8 on the volume of a spherical tank -- and ask about how many cupfuls of water fit in the tank.

PARCC Practice Test Question 21 (Day 164)

Chapter 21 of Morris Kline's Mathematics and the Physical World is called "Mathematical Oscillations of the Ether." This is the chapter when electricity and magnetism join to be one force.

"For many parts of nature can neither be invented with sufficient subtlety, nor demonstrated with sufficient perspecuity, nor accommodated into use with sufficient dexterity without the aid and intervention of mathematics." -- Francis Bacon, 17th century English philosopher

Kline begins:

"The developments in electricity and magnetism that we have examined so far might be said to constitute the childhood of a new science."

Well, if the previous chapter was the childhood of electricity and magnetism, then the current chapter is their marriage. And the scientists who performed the marriage of electricity and magnetism were 19th century British physicists Michael Faraday and James Maxwell.

Maxwell's first equation is rather simple:

lambda f = c

where lambda is the wavelength, f is the frequency, and c is the wave velocity -- better known to us as the speed of light. (Kline doesn't point this out, but it's the speed of all electromagnetic waves, not just light). So wavelength and frequency are inversely proportional -- the longer the wavelength, the smaller the frequency.

The most interesting part of this chapter is when Kline explains how AM radio works. The letters AM stand for amplitude modulation -- that is the amplitude of the waves varies with time:

E = D(1 + .3 sin 2pi * 400t) sin 2pi * 1000000t.

So the wave has a frequency of one million cycles per second (or 1000 kilohertz), but the amplitude itself varies 400 cycles per second.

Now FM radio wasn't as common in Kline's day, but I know that the letters stand for frequency modulation -- so it's the frequency that must vary. Kline gives the formula for FM radio as:

E = D sin (2pi * 100000000t + a sin 2pi * 400t)

so that the factor that varies is placed with the frequency (which here is about 100 megahertz) rather than the amplitude. Notice that the radio station numbers refer to frequency -- 1000 AM means 1000 kilohertz, and 100.0 FM means 100 megahertz!

Afterward Kline moves on to light, and explains that different colors come from different frequencies, so that 4 * 10^14 hertz is red and 7 * 10^14 hertz is violet.

Question 21 of the PARCC Practice Test is on classical constructions:

21. Jericho is making several constructions based on the segment shown [PQ -- dw].

Part A

For his first construction, Jericho made the markings shown [two each with the tip at P and Q, one of which is above and the other is below PQ -- dw] with a compass open to a length less than the length of segment PQ. Jericho's markings are useful for the construction of which of the figures listed?

Select all that apply.

A. a 60-degree angle
B. a bisector of PQ
C. a line perpendicular to PQ
D. a rhombus with PQ as one diagonal
E. an equilateral triangle with side PQ

Part B

The first steps of Jericho's second construction are shown [with R not on PQ, S any point on PQ, and ray SR drawn in -- dw]. After drawing arcs from point S and point R, he adjusted the compass length using the intersection of the arc from point S with PQ and ray SR. Which figure is he constructing?

A. the bisector of PQ through point R
B. an angle congruent to angle RPQ with vertex R
C. a line through point R that is parallel to PQ
D. a circle containing points P, Q, and R

Of course, now we're wondering, why couldn't this question have shown up last week, back when I was covering that Honors Integrated Math I class? It would have been perfect to work with these constructions both on the PARCC and in the classroom at the same time!

I have a few things to say about this problem. First of all, I find it highly interesting that the name of the protagonist in this question is Jericho. You see, back on Sunday, after I watched my favorite TV show The Simpsons, an episode of Bob's Burgers aired. On that episode, the girl Tina Belcher has an imaginary horse named Jericho! And so, just three days after seeing the unusual name Jericho appear on TV, suddenly the same name appears on the PARCC as well.

Now in that Integrated Math I class, our constructions began with a circle and we were to inscribe an equilateral triangle or hexagon. On the PARCC, we are starting with just a segment, not a circle (even though the triangle and hexagon constructions appear in the Common Core Standards).

For Part A, it's clear that we're constructing the perpendicular bisector of PQ -- so we must choose choice (C) for "perpendicular" and choice (B) for "bisector." Less obvious is that we're also constructing a rhombus with diagonal PQ -- after all, a rhombus is just a quadrilateral with four congruent sides, and the four arcs drawn from P and Q all have the same radius, so the four sides are indeed congruent. So the correct answer is (B), (C), and (D). Common errors will include marking just one letter instead of all three, as will often occur with multiple-answer questions. Also, the rhombus will be easily missed by students. After all, we tell students to construct the perpendicular bisector of PQ, but we never ask them to construct a rhombus.

Notice that in this question, it is stated that to draw the four arcs, the compass setting is to be less than the length of PQ. In Lesson 3-6 of the U of Chicago text, students are told to set the compass to exactly the length of PQ. We see that by setting the compass to PQ, suddenly all five answer choices are correct. The four sides of the rhombus all have length PQ, and so PQ itself divides the rhombus into two equilateral triangles, each with side PQ. And of course each angle of these equilateral triangles must be 60 degrees.

For Part B, the correct answer is choice (C), the construction of parallel lines. I've discussed the construction of parallel lines several times on the blog -- a major weakness of the U of Chicago text is that this construction doesn't appear anywhere. I've written that it's possible to construct the parallel by constructing the perpendicular of PQ through R, labeling the intersection point S, and then constructing the perpendicular of RS through R. By the Two Perpendiculars Theorem, this last line must be parallel to PQ. But this isn't the construction that appears on the PARCC, which usually prefers the Corresponding Angles construction.

PARCC Practice EOY Question 21
U of Chicago Correspondence: Lesson 3-6, Constructing Perpendiculars

Key Theorem: Construct the perpendicular bisector of a given segment AB.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.CO.D.12
Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.).Copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Commentary: Lesson 3-6 of the U of Chicago text describes how to construct a perpendicular bisector of a segment -- the U of Chicago method is used for my first exercise. For my second exercise, students must construct a circle through three points. This is taught in Lesson 4-5 of the U of Chicago text.

PARCC Practice Test Question 20 (Day 163)

Today I subbed in a middle school history class. Normally I don't announce non-math subbing assignments here on the blog, but this one is significant because:

-- It's the first time I subbed at a middle school since the announcement that I will be teaching at a middle school full-time next year.
-- As SBAC testing continues, the middle school is on a block schedule. Wednesdays are Common Planning days at the elementary and middle schools in our district, and so the blocks are set up with odd periods on Mondays and Thursdays and even periods on Tuesdays and Fridays -- just like the middle school I'll be at in the fall.

So this is another chance to look critically at the block schedule and see how teachers plan for block periods, even in subjects other than my own. In this class, the eighth graders have four tasks to complete during the 1:50 block:

-- Finish the movie Glory. California eighth graders study U.S. History up to around the Civil War.
-- Take a 10-minute break.
-- Listen to a lecture about the Battle of Vicksburg.
-- Work on the Civil War project that is due next week.

Traditionalists often complain about watching movies when they occur in lieu of a task with more academic rigor. It's hard to know whether the students would have watched the video had there not been block periods, since the Civil War is always taught at the end of the year, right around the time of the SBAC. (Even before the SBAC, eighth graders always must take the NCLB science tests, which here in California are given to fifth, eighth, and tenth graders.)

But it's that 10-minute break that will annoy traditionalists the most. The savings in passing time are offset by the need to give the students a break during the period. The class is now divided essentially into two 50-minute "periods."

The project mentioned above may sound perfect for a block lesson, except that it's one of those multimedia projects that is best completed using technology at home. So if the students have no research to do, then it ultimately turns into study hall. So again, we see that there are many parts of this block period for traditionalists to dislike. Of course, I point out that even though I want to see what block periods look like, these block periods are biased by the fact that these are lesson plans written for a sub. For example, the "study hall" could occur because the teacher can't think about how to fill 1:50 for a sub, not because the teacher always has study time every block period. The same is true about the video -- there might be no video today had the regular teacher been present.

Chapter 20 of Morris Kline's Mathematics and the Physical World is "Old Foes With New Faces." In this chapter, the "old foes" are electricity and magnetism.

"The magnet's name the observing Grecians drew/From the magnetic region where it grew." -- Lucretius

Kline begins:

"The world of nature is vaster than what the hand can touch, the eye see, or the ear hear. Beyond the senses lies a world that has been effectively explored only within the last hundred years or so."

It is the world of electricity and magnetism. Kline's quote from the poet Lucretius tells us the origin of the word magnet, and Kline repeats this origin story:

"For example, Thales knew that iron ores containing lodestone, such as those found near Magnesia in Asia Minor, attract iron. He is also supposed to have known that after amber is rubbed it attracts light particles of matter."

And indeed, the word electricity comes from the Greek word for amber, although Kline does not point this out.

Now the key formula for this chapter was discovered by the 18th century French physicist Augustin Coulomb, and it describes the electrical force between two charged particles:

F = k q_1 q_2 / (r^2)

Kline points out that this is very similar to the equation for the gravitational force. Kline proceeds by pointing out that electricity and magnetism are closely related --the electromagnetic force.

Question 20 of the PARCC Practice Test is on inscribed angles:

20. In circle O, points A, B, C, and D lie on the circle; arc AD is congruent to BC; and the measure of arc AB is twice the measure of arc BC.

[Moving clockwise, the points lie on the circle in the order D, A, B, C.]

Part A
The measure of angle ACD is Choose... (a third, half, equal to, twice, three times) the measure of angle ADC.

Part B
The measure of angle ADC is Choose... (a third, half, equal to, twice, three times) the measure of angle BCD.

This question clearly involves the Inscribed Angle Theorem of Lesson 15-3, since all of the angles mentioned are inscribed angles. Of course, it may be tricky to look at the diagram and determine which angles subtend which arcs. An easy way to tell is to note that the endpoints of the arc are the two points named in the inscribed angle other than the vertex. Since the inscribed angle is half the measure of the arc, we substitute into the original problem:

Part A
Half the measure of arc AD is Choose... (a third, half, equal to, twice, three times) half the measure of arc AC.

Part B
Half the measure of arc AC is Choose... (a third, half, equal to, twice, three times) half the measure of arc BD.

Of course, we can multiply everything by two to get rid of the "half":

Part A
The measure of arc AD is Choose... (a third, half, equal to, twice, three times) the measure of arc AC.

Part B
The measure of arc AC is Choose... (a third, half, equal to, twice, three times) the measure of arc BD.

Both of these require us to calculate the measure of arc AC. We see that arc AC is the sum of arcs AB and BC, and the former is said to be twice the latter. Therefore AC has thrice the measure of BC, which in turn has the same measure as AD. So AD must have a third of the measure of AC -- that is, for Part A, the measure of angle ACD is a third of the measure of angle ADC.

For Part B, we see that arc BD is the sum of arcs AB and AD, and once again, the former is said to be twice the latter (as again, BC and AD have the same measure). So both AC and BD have thrice the measure of BC -- that is, AC and BD are congruent. Therefore the measure of angle ADC is equal to the measure of angle BCD.

Notice that we must be careful when just blindly substituting in the arc measures. If the angle is obtuse, then it must subtend a major arc, so we can't just substitute a minor arc measure (fortunately all the arcs needed for the problem are minor and all the angles are acute). Also, we can't generally ignore the factor of 1/2 -- here I did because all of the angles were inscribed. If we had a mix of inscribed and central angles then the factor of 1/2 becomes significant.

PARCC Practice EOY Question 20
U of Chicago Correspondence: Lesson 15-3, The Inscribed Angle Theorem

Key Theorem: Inscribed Angle Theorem

In a circle, the measure of an inscribed angle is one-half the measure of its intercepted arc.

Common Core Standard:
CCSS.MATH.CONTENT.HSG.C.A.2
Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

Commentary: This question may be slightly more abstract than a typical problem from the U of Chicago text. Question 9 from Lesson 15-3 and Question 33 from the SPUR Review for Chapter 15 are two questions where the arc lengths and angle measures are unknown.