Wednesday, May 18, 2016

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:
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.

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