Why do we teach students about radians?

Throughout grades K-10, students are slowly introduced to the concept of angles. They are told that there are $90$ degrees in a right angle, $180$ degrees in a straight angle, and a circle has $60$ degrees. They are introduced to $30-60-90$ and $45-45-90$ right triangles. Fans of snowboarding even know the multiples of $180$ degrees up to $1440$ or even $1620$ degrees.

Then, in Precalculus, we make students get comfortable with $\pi$ , $\displaystyle \frac{\pi}{2}$ , $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{\pi}{4}$ , $\displaystyle \frac{\pi}{6}$ , and multiples thereof.

We tell students that radians and degrees are just two ways of measuring angles, just like inches and centimeters are two ways of measuring the length of a line segment.

Still, students are extremely comfortable with measuring angles in degrees. They can easily visualize an angle of $75^o$ , but to visualize an angle of $2$ radians, they inevitably need to convert to degrees first. In his book Surely You’re Joking, Mr. Feynman!, Nobel-Prize laureate Richard P. Feynman described himself as a boy:

I was never any good in sports. I was always terrified if a tennis ball would come over the fence and land near me, because I never could get it over the fence – it usually went about a radian off of where it was supposed to go.

Naturally, students wonder why we make them get comfortable with measuring angles with radians.

The short answer, appropriate for Precalculus students: Certain formulas are a little easier to write with radians as opposed to degrees, which in turn make certain formulas in calculus a lot easier.

The longer answer, which Precalculus students would not appreciate, is that radian measure is needed to make the derivatives of $\sin x$ and $\cos x$ look palatable.

Source: http://mathworld.wolfram.com/CircularSector.html

1. In Precalculus, the length of a circle arc with central angle $\theta$ in a circle with radius $r$ is

$s = r\theta$

Also, the area of a circular sector with central angle $\theta$ in a circle with radius $r$ is

$A = \displaystyle \frac{1}{2} r^2 \theta$

In both of these formulas, the angle $\theta$ must be measured in radians.

Students may complain that it’d be easy to make a formula of $\theta$ is measured in degrees, and they’d be right:

$s = \displaystyle \frac{180 r \theta}{\pi}$ and $A = \displaystyle \frac{180}{\pi} r^2 \theta$

However, getting rid of the $180/\pi$ makes the following computations from calculus a lot easier.

2a. Early in calculus, the limit

$\displaystyle \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

is derived using the Sandwich Theorem (or Pinching Theorem or Squeeze Theorem). I won’t reinvent the wheel by writing out the proof, but it can be found here. The first step of the proof uses the formula for the above formula for the area of a circular sector.

2b. Using the trigonometric identity $\cos 2x = 1 - 2 \sin^2 x$ , we replace $x$ by $\theta/2$ to find

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \frac{2\sin^2 \displaystyle \left( \frac{\theta}{2} \right)}{ \theta}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} = \displaystyle \lim_{\theta \to 0} \sin \left( \frac{\theta}{2} \right) \cdot \frac{\sin \displaystyle \left( \frac{\theta}{2} \right)}{ \displaystyle \frac{\theta}{2}}$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0 \cdot 1$

$\displaystyle \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta} =0$

3. Both of the above limits — as well as the formulas for $\sin(\alpha + \beta)$ and $\cos(\alpha + \beta)$ — are needed to prove that $\displaystyle \frac{d}{dx} \sin x = \cos x$ and $\displaystyle \frac{d}{dx} \cos x = -\sin x$ . Again, I won’t reinvent the wheel, but the proofs can be found here.

So, to make a long story short, radians are used to make the derivatives $y = \sin x$ and $y = \cos x$ easier to remember. It is logically possible to differentiate these functions using degrees instead of radians — see http://www.math.ubc.ca/~feldman/m100/sinUnits.pdf. However, possible is not the same thing as preferable, as calculus is a whole lot easier without these extra factors of $\pi/180$ floating around.

All I want to be is a high school math teacher. Why do I have to take Real Analysis?

In 2012, the Conference Board for the Mathematical Sciences published The Mathematical Education of Teachers II, providing recommendations for how universities prepare future teachers at all grade levels. From Chapter 6 of this report, here are the recommendations for future secondary teachers:

This report recommends that the mathematics courses taken by prospective high school teachers include at least a three-course calculus sequence, an introductory statistics course, an introductory linear algebra course, and 18 additional semester-hours of advanced mathematics, including 9 semester-hours explicitly focused on high school mathematics from an advanced standpoint. It is desirable to have a further 9 semester-hours of mathematics…

The report then goes on to describe how advanced mathematics courses that emphasize theorems and proofs, such as abstract algebra, real analysis, group theory, and number theory, are directly relevant for teaching the secondary mathematics curriculum.

In my experience, most math majors who want to be high school math teachers understand why they have to take calculus, statistics, linear algebra, and math courses specifically designed for their future career. But a perennial question that they often ask is, “Why do I have to take theorem-proof math classes if all I want to do is teach high school math?” Some years ago, I wrote the following document for my students to address this issue, which hopefully provide some good reasons for why these more abstract courses are required.

1. A central goal of theorem-proof courses (and subsequent courses) is to emphasize that mathematics is not only an exercise in quantitative skills; it is also an exercise in explaining why the rules of mathematics work the way they do. After you become a teacher, that’s also something that you can impress upon your students — not only the “how” of solving problems but also the “why.”

Through all of the “proof” classes, you will learn that the real purpose of math is not to be able to perform arithmetic, solve for $x$ , or even to be able to find the area under the curve. Rather, it is to be able to think logically, to reach a logical solution on well-defined terms, and to be able to problem-solve. So many of today’s youth do not encompass any of these qualities and/or abilities; when you think about it, that is a very scary thought! So when you become a teacher and are asked the inevitable questions “Why do I need to know (fill in a math topic)? How is that going to help me in life?”, you will be prepared to respond. (But don’t be surprised when they don’t like your response!)

2. Theorem-proof courses are specified by the National Council of Teachers of Mathematics (NCTM) and the National Council for Accreditation of Teacher Education (NCATE) as part of the national standard of best practices for preparing future high school math teachers. Furthermore, you will be responsible for the material in these classes when it’s time for you to take the TExES certification exam. All this to say, including these classes in the curriculum isn’t a requirement that somebody at our university thought was a bright idea; this is an “industry standard,” so to speak, for the preparation of highly qualified secondary math teachers.

3. Of course, simply saying “You have to do this because the ‘powers that be’ say that it’s good for you” may not be a terribly motivating reason for you to be in courses that emphasize mathematical abstraction. Another good reason to take these courses is because it’s always a good idea for teachers to be familiar with a few years’ worth of mathematics above the subject matter that they’re teaching. Right now, you probably expect (perhaps subconsciously) that your professor knows not only the content immediately pertinent to your class but also the content in subsequent classes, so that you’re prepared to take those subsequent classes if you elect to do so. The same logic will apply to you when you become a teacher yourself.

By analogy, UNT’s elementary teachers often complain, “Why do I have to take College Algebra if I’m only going to be teaching arithmetic?” Well, elementary students are learning algebra without the technical terms. For example: $3 + \fbox{~?~} = 8$ is a first-grade problem, and students will use the number line (or their fingers) to reach the conclusion that $\fbox{~?~} = 5$ . However, 6^th graders taking Pre-Algebra are replacing the box with the variable $x$ and now must show their work on why $x=5$ , which is a proof! Having taken theorem-proof courses will allow you, the teacher, to explain why we can add a $-3$ to both sides and it cancels out on one side and subtracts $3$ from the $8$ . Moreover, it explains that the minus sign just tells you which direction you are moving on the number line and why the word we came up with for that idea, subtraction, means “to take away, to go to the left.”

Ideally, elementary school teachers should teach their classes mindful of the fact that elementary school arithmetic is not a mere exercise in computation but part of a process of logical thinking that will be further developed in middle-school algebra. In the same way, when you become a secondary math teacher, you should be familiar with the topics that lie beyond algebra, geometry, trigonometry, and calculus.

4. When you’re in your future classroom, you should be equipped to answer just about any mathematical question that comes your way. Someday, a bright and inquisitive student will ask you an honest question about some of the deeper concepts covered in the course that you’re teaching. Such questions typically start with “why” instead of “how.” For example:

“Why is the number $e$ irrational?”
“Why is the Pythagorean theorem correct?”
“Why can’t complex numbers be defined using $\sqrt{-2}$ instead of $\sqrt{-1}$ ?”
“Why are all of the hypotheses of the intermediate value theorem needed?”
“Why do the rows in Pascal’s triangle add to powers of $2$ ?”
“Why do polynomials have a unique factorization using linear terms involving complex numbers?”

High school teachers (and, if they’re honest, college professors) will tell you that it’s quite embarrassing to be unable to provide an immediate answer to a student’s question, no matter how difficult. It’s even worse if the teacher is at a complete and utter loss as to who or what to consult to provide the answer. Theorem-proof classes will hopefully provide you the framework to answer such questions.

5. Many of the concepts in real analysis are directly related to concepts taught below the level of calculus. To give a few examples from the first few chapters of our textbook:

Unions and intersections of sets are important for developing the rules for computing $P(A \cup B)$ , the probability that event $A$ happens or event $B$ happens, and $P(A \cap B)$ , the chance that $A$ and $B$ both happen.
Important examples of equivalence classes are those derived from congruences, a notion which leads to the familiar grade-school rules for testing for divisibility by $2$ , $3$ , $4$ , $5$ , $6$ , $8$ , $9$ , $10$ or $11$ .
The notion of an injective function explains why the horizontal line test works.
The axioms of ordered fields are the logical framework behind high school algebra, and for why the “FOIL” method is correct but the “distributive” property $(a+b)^2 = a^2 + b^2$ is incorrect.
The density of both rational and irrational numbers is often taken for granted by high school students without explanation.

6. Many theorems in calculus, which are typically stated without proof when actually teaching a calculus course, rely on notions from real analysis. In fact, a major goal of theorem-proof courses is to “dot the i’s and cross the t’s” of familiar theorems that are often stated in a calculus class but not always completely proved for students. Here’s a sampling:

The order of quantifiers is important for distinguishing between the continuous functions and uniformly continuous functions. The latter notion is necessary to formally prove that every continuous function has a definite integral on a closed interval, often taken for granted in Calculus I.
The notion of supremum and the completeness property of $\mathbb{R}$ underlie some important concepts in calculus, including the proof of the intermediate value theorem, the rigorous definition of a definite integral, and the proof that the definite integral of the sum of two functions equals the sum of the two definite integrals.
Many optimization problems in calculus rely on the fact that continuous functions assume both an absolute minimum and absolute maximum value on any closed interval. The proof of this theorem relies on properties of closed sets and compact sets.
The proof that the composition of two continuous functions is continuous (and also the proof of the Chain Rule for differentiation) relies on properties of open sets.
Limit theorems that are often taken for granted in calculus may be proven using limit theorems about sequences.
The Mean Value Theorem (from Calculus I) is derived from the fact that limits preserve inequalities. Many important properties of calculus, including L’Hopital’s Rule, indefinite integration, curve sketching, and Taylor series, are direct consequences of the Mean Value Theorem.
The Root and Ratio Tests from Calculus II are derived using the notions of limsup and liminf.

Hopefully, while you’re still in college, you will begin the process of making connections between the topics that you will directly teach your students and the topics that your students will see after they graduate from high school. This may not fully sink in until you begin student teaching; only then will you realize the importance of being able to prove something (i.e. teaching a topic in a logical manner) when you are trying to explain it to inquiring minds.

As a student taking real analysis or abstract algebra, it’s easy to lose sight of the forest for all of the trees. That is, it’s easy to simply develop your skills in abstraction and theorem-proving without realizing that the topics you’re learning are indeed relevant to your future career as a mathematics educator. Teach North Texas and the UNT Math Department both wish you well as you continue through our degree program.

Area of a triangle: Pick’s theorem (Part 8)

The following is one of my all-time favorite paragraphs to ever appear in a professional mathematical journal.

Some years ago, the Northwest Mathematics Conference was held in Eugene, Oregon. To add a bit of local flavor, a forester was included on the program, and those who attended his session were introduced to a variety of nice examples which illustrated the important role that mathematics plays in the forest industry. One of his problems was concerned with the calculation of the area inside a polygonal region drawn to scale from field data obtained for a stand of timber by a timber cruiser. The standard method is to overlay a scale drawing with a transparency on which a square dot pattern is printed. Except for a factor dependent on the relative sizes of the drawing and the square grid, the area inside the polygon is computed by counting all of the dots fully inside the polygon, and then adding half of the number of dots which fall on the bounding edges of the polygon. Although the speaker was not aware that he was essentially using Pick’s formula, I was delighted to see that one of my favorite mathematical results was not only beautiful, but even useful.

D. DeTemple, cited in Branko Grunbaum and G. C. Shephard, “Pick’s Theorem,” American Mathematical Monthly, Vol. 100, pp. 150-161 (February 1993).

Suppose that the vertices of a triangle are $(1,1)$ , $(3,5)$ , and $(4,2)$ . What is the area of the triangle?

Because the vertices of the triangle have integer coordinates, Pick’s Theorem offers an exceedingly simple way of finding the area of this triangle.

There are $6$ points (marked white) that are inside the triangle.
There are $4$ points (marked red) that are on the boundary of the triangle, including the three corners.
Therefore, the area is $A = 6 + \frac{1}{2} (4) - 1 = 7$ .

You can confirm this area by drawing the rectangle with corners at $(1,1)$ , $(5,1)$ , $(5,5)$ , and $(1,5)$ and then taking away the three right triangles, leaving the triangle shown in the figure above.

Amazingly, this theorem is true for any polygonal figure — not just triangles — whose vertices have integer coordinates.

A decent classroom activity so that students can discover Pick’s theorem for themselves has been published by the National Council of Teachers of Mathematics. I modified this activity to teach my daughter and her friends last summer, so I say from first-hand experience that fourth-graders can use inductive reasoning to guess Pick’s theorem.

Additional references:

http://www.cut-the-knot.org/ctk/geoboard.shtml

http://www.cut-the-knot.org/ctk/Pick_proof.shtml

Area of a triangle: Incenter (Part 6)

Source: http://mathworld.wolfram.com/Incircle.html

The incenter $I$ of a triangle $\triangle ABC$ is defined by the intersection of the angle bisectors of its three angles. A circle can be inscribed within $\triangle ABC$ , as shown in the picture.

This incircle provides a different way of finding the area of $\triangle ABC$ commonly needed for high school math contests like the AMC 10/12. Suppose that the sides $a$ , $b$ , and $c$ are known and the inradius $r$ is also known. Then $\triangle ABI$ is a right triangle with base $c$ and height $r$ . So

$\hbox{Area of ~} \triangle ABI = \displaystyle \frac{1}{2} cr$

Similarly,

$\hbox{Area of ~} \triangle ACI = \displaystyle \frac{1}{2} br$

$\hbox{Area of ~} \triangle BCI = \displaystyle \frac{1}{2} ar$

Since the area of $\triangle ABC$ is the sum of the areas of these three smaller triangles, we conclude that

$\hbox{Area of ~} \triangle ABC = \displaystyle \frac{1}{2} r (a+b+c)$ ,

$\hbox{Area of ~} \triangle ABC = rs$ ,

where $s = (a+b+c)/2$ is the semiperimeter of $\triangle ABC$ .

This also permits the computation of $r$ itself. By Heron’s formula, we know that

$\hbox{Area of ~} \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$

Equating these two expressions for the area of $\triangle ABC$ , we can solve for the inradius $r$ :

$r = \displaystyle \sqrt{ \frac{(s-a)(s-b)(s-c)}{s} }$

For much more about the inradius and incircle, I’ll refer to the MathWorld website.

Area of a triangle: SSS (Part 5)

In yesterday’s post, we discussed how the area $K$ of a triangle can be found using SAS: two sides and the angle between the two sides. We found that

$K = \displaystyle \frac{1}{2} a b \sin C$

This can be used as the starting point for the derivation of Heron’s formula, which determines the area of a triangle using SSS (i.e., only the three sides). I won’t give the full derivation in this post — there’s no point in me retyping the details — but will refer to the Wikipedia page and the MathWorld page for the details. However, I will give the big ideas behind the derivation.

1. We begin by recalling that $\sin^2 C + \cos^2 C = 1$ . Since $0 < C < 180^o$ , we know that $\sin C$ must be positive, so that

$\sin C = \sqrt{1 - \cos^2 C}$

2. From the Law of Cosines, we know that

$c^2 = a^2 + b^2 - 2 a b \cos C$ ,

$\cos C = \displaystyle \frac{a^2 + b^2 - c^2}{2ab}$

3. Substituting, we see that

$K = \displaystyle \frac{1}{2} ab \sqrt{1 - \cos^2 C}$

$K = \displaystyle \frac{1}{2} ab \sqrt{1 - \frac{(a^2+b^2-c^2)^2}{4a^2b^2}}$

4. This last expression only contains the side lengths $a$ , $b$ , and $c$ . So the “only” work that’s left is simplifying this right-hand side and seeing what happens. After considerable work — requiring factoring the difference of two squares on two different steps — we end up with Heron’s formula:

$K = \sqrt{s (s-a) (s-b) (s-c)}$

where $s = \displaystyle \frac{a+b+c}{2}$ is the semiperimeter, or half the perimeter of the triangle.

A final note: If you actually are able to start with Step 3 and end with Heron’s formula on your own — without consulting a textbook or the Internet if you get stuck — feel free to cry out “More power!” and grunt like Tim “The Toolman” Taylor:

Area of a triangle: SAS, ASA, and the Law of Sines (Part 4)

The typical way students remember the area $K$ of a triangle is

$K = \displaystyle \frac{1}{2} \times \hbox{Base} \times \hbox{Height}$

However, there are other formulas for the area of a triangle which can be helpful if the height is not immediately known.

Case 1: SAS. Suppose that two sides and the angle between the sides — say, $b$ and $c$ and the measure of angle $A$ — are known.

If $\overline{CD}$ is an altitude for $\triangle ABC$ , then $\triangle ACD$ is a right triangle. Therefore,

$\sin A = \displaystyle \frac{\hbox{opposite}}{\hbox{hypotenuse}} = \displaystyle \frac{h}{b}$ , or $h = b \sin A$ .

Therefore,

$K = \displaystyle \frac{1}{2} ch = \displaystyle \frac{1}{2} bc \sin A$ .

Using the same picture, one can also show that

$K = \displaystyle \frac{1}{2} ac \sin B$

Also, with a different but similar picture, one can show that

$K = \displaystyle \frac{1}{2} ab \sin C$

An important consequence of the SAS area formula is the Law of Sines. Since all three formulas must give the same area $K$ , we have

$\displaystyle \frac{1}{2} bc \sin A = \displaystyle \frac{1}{2} ac \sin B = \displaystyle \frac{1}{2} ab \sin C$

Multiplying by $\displaystyle \frac{2}{abc}$ produces the Law of Sines:

$\displaystyle \frac{\sin A}{a} = \displaystyle \frac{\sin B}{b} = \displaystyle \frac{\sin C}{c}$

Case 2: ASA. Now suppose that we are given the measures of two angles and the length of the side in between them — say, angles $A$ and $B$ and side $c$ . Naturally, we can also get the measure of angle $C$ since the sum of the measures of the three angles must be $180^o$ .

From the SAS formula and the Law of Sines, we have

$K = \displaystyle \frac{1}{2} bc \sin A \quad \hbox{and} \quad \displaystyle b = \frac{c \sin B}{\sin C}$

Combining these, we find

$K = \displaystyle \frac{1}{2} \frac{c \sin B}{\sin C} \cdot c \sin A$

$K = \displaystyle \frac{c^2 \sin A \sin B}{2 \sin C}$

By similar reasoning, we can also find that

$K = \displaystyle \frac{a^2 \sin B \sin C}{2 \sin A} ~~$ and $~~ K = \displaystyle \frac{b^2 \sin A \sin C}{2 \sin B}$

Volume of pyramids, cones, and spheres (Part 3)

I’m in the middle of a series of posts concerning the area of a triangle. Today, however, I want to take a one-post detour using yesterday’s post as a springboard. In yesterday’s post, we discussed a two-dimensional version of Cavalieri’s principle. From Wikipedia:

Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.

In other words, if I have any kind of shape that has cross-sections that match those of the triangles above, then the shape has the same area as the triangles. Geometrically, we can think of each triangle a bunch of line segments joined together. So while the positioning of the line segments affects the shape of the region, the positioning does not affect the area of the region.

There is also a three-dimensional statement of Cavalieri’s principle, and this three-dimensional version is much more important than the above two-dimensional version. From MathWorld:

If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Pedagogically, I would recommend introducing Cavalieri’s principle with two-dimensional figures like those from yesterday’s post since cross-sections in triangles are much easier for students to visualize than cross-sections in three-dimensional regions.

This three-dimensional version of Cavalieri’s principle is needed to prove — without calculus — the volume formulas commonly taught in geometry class. Based on my interactions with students, they are commonly taught without proof, as my college students can use these formulas but have no recollection of ever seeing any kind of justification for why they are true. When I teach calculus, I show my students that the volume of a sphere can be found by integration using the volume of a solid of revolution:

$V = \displaystyle \int_{-R}^R \pi \left[ \sqrt{R^2 - x^2} \right]^2 \, dx = \frac{4}{3} \pi R^3$

Without fail, my students (1) already know this formula from Geometry but (2) do not recall ever being taught why this formula is correct. Curious students also wonder (3) how the volume of a sphere (or a pyramid or a cone) can be obtained only using geometric concepts and without using calculus.

For the sake of brevity, I only give the logical flow for how these volumes can be derived for students without using calculus. I’ll refer to this excellent site for more details about each step.

Using a simple foldable manipulative (see also this site), students can see that $V = \displaystyle \frac{1}{3} Bh$ for a certain pyramid — called a yangma — with a square base and a height that is equal to the base length.
Enlarging the yangma will not change the ratio of the volume of the pyramid to the volume of the prism.
Cavalieri’s principle then shows that $V = \displaystyle \frac{1}{3} Bh$ for any square pyramid.
Cavalieri’s principle then shows that $V = \displaystyle \frac{1}{3} Bh$ for any pyramid with a non-square base or even a cone with a circular base.
Finally, a clever use of Cavalieri’s principle — comparing a sphere to a cylinder with a cone-shaped region removed — can be used to show that the volume of a sphere is $V = \displaystyle \frac{4}{3} \pi R^3$ .

I note in closing that there are other ways for students to discover these formulas, like filling an empty pyramid with rice, pouring into an empty prism of equal base and height, and repeating until the prism is filled.

Area of a triangle: Equal cross-sections (Part 2)

Let’s take a second look at the familiar formula for the area of a triangle, $A = \displaystyle \frac{1}{2}bh$ .

The picture above shows three different triangles: one right, one obtuse, and one acute. The three triangles have bases of equal length and also have the same height. Therefore, even though the triangles have different shapes (i.e., they’re not congruent), they have the same height.

Let’s take a second look at these three triangles. In each triangle, I’ve drawn in three “cross-section” line segments which are parallel to the base. Notice that corresponding cross-sections have equal length. In other words, the red line segments have the same length, the light-blue line segments have the same length, and the purple line segments have the same length.

Why is this true? There are two ways of thinking about this (for the sake of brevity, I won’t write out the details).

Algebraically, the length of the cross-section increases linearly as they descend from the top vertex to the bottom base. This linear increase does not depend upon the shape of the triangle. Since the three triangles have bases of equal length, the cross-sections have to have the same length.
Geometrically, the length of the cross-sections can be found with similar triangles, comparing the big original triangle to the smaller triangle that has a cross-section as its base. Again, the scale factor between the similar triangles depends only on the height of the smaller triangle and not on the shape of the original triangle. So the cross-sections have to have the same length.

So, since the three triangles share the same height and base length, the three triangles have the same area, and the corresponding cross-sections have the same length.

The reverse principle is also true. This is called Cavalieri’s principle. From Wikipedia:

Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.

For example, here are three non-triangular regions whose cross-sections match those of the above triangles. The region on the right is especially complex since it has a curvy hole in the middle, so that the cross-sections shown are actually two distinct line segments. Nevertheless, we can say with confidence that, by Cavalieri’s principle, the area of each region matches those of the triangles above.

Though we wouldn’t expect geometry students to make this connection, Cavalieri’s principle may be viewed as a geometric version of integral calculus. In calculus, we teach that the area between the curves $x = f(y)$ and $x = F(y)$ is equal to

$A = \displaystyle \int_{y_1}^{y_2} [F(y) - f(y)] \, dy = \displaystyle \int_{y_1}^{y_2} d(y) \, dy$

where $d(y) = F(y) – f(y)$ is the difference in the two curves. In the above formula, I chose integration with respect to $y$ since the $y-$coordinates are constant in the above cross-sections. The difference $d(y)$ is precisely the length of the cross-sections. As with the triangles, the positioning of the cross-sections will affect the shape of the region, but the positioning of the cross-sections does not affect the length $d(y)$ and hence does not affect the area $A$ of the region.

Area of a triangle: Base and height (Part 1)

This begins a series of post concerning how the area of a triangle can be computed. This post concerns the formula that students most often remember:

$A = \displaystyle \frac{1}{2} b h$

Why is this formula true? Consider $\triangle ABC$ , and form the altitude from $B$ to line $AB$ . Suppose that the length of $AC$ is $b$ and that the altitude has length $h$ . Then one of three things could happen:

Case 1. The altitude intersects $AC$ at either $A$ or $C$ . Then $\triangle ABC$ is a right triangle, which is half of a rectangle. Since the area of a rectangle is $bh$ , the area of the triangle must be $\displaystyle \frac{1}{2} bh$ .

Knowing the area of a right triangle will be important for Cases 2 and 3, as we will act like a good MIT freshman and use this previous work.

Case 2. The altitude intersects $AC$ at a point $D$ between $A$ and $C$ . Then $\triangle ABD$ and $\triangle BCD$ are right triangles, and so

$\hbox{Area of~} \triangle ABC = \hbox{Area of ~} \triangle ABD + \hbox{~Area of~} \triangle BCD$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} b_1 h + \frac{1}{2} b_2 h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b_1 + b_2) h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} bh$

Case 3. The altitude intersects $AC$ at a point $D$ that is not in between $A$ and $C$ . Without loss of generality, suppose that $A$ is between $D$ and $C$ . Then $\triangle ABD$ and $\triangle BCD$ are right triangles, and so

$\hbox{Area of~} \triangle ABC = \hbox{Area of ~} \triangle BCD - \hbox{~Area of~} \triangle ACD$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b+t) h + \frac{1}{2} t h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b+t-t) h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} bh$

Engaging students: Completing the square

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Claire McMahon. Her topic, from Algebra: completing the square.

There were a lot of famous mathematicians that contributed to the notion of completing the square. The first of the mathematicians was that of the Babylonians. This culture started the notion of not only solving the quadratics but of arithmetic itself. The Babylonians started with the equations and then proceeded to solve them algebraically. Back then; they used pre-calculated tables to help them with solving for the roots. They were basically solving by the quadratic equation at this point. The man that came along has a very hard name to not only pronounce but to spell, and I will do my best. I will refer to him as Muhammad from here on out but his full name, or one of the common names to which he is referred is Muhammad ibn Musa al-Khwarizmi. He developed the term algorithm, which led to an algorithm for solving quadratic equations, namely completing the square.

The notion of completing the square has gone through a series of transformations throughout the history of mathematics. As mentioned before the Babylonians started with the notion and increased the knowledge by developing the quadratic formula to find the roots of a given quadratic equation. This spurred the thought that I can solve any equation and find its solution and roots by completing the square. Muhammad brought this notion to us, of which was mentioned before. More specifically the text that he developed was “The Compendious Book of Calculations by Completion and Balancing.” This book of course has been translated several times over but the general idea is laid out in the title. Modern mathematicians have developed a less compendious form that is now being taught in the math classes today. They take on many different forms and can be taught with manipulates as well.

The fabulous part to the story is there are a lot of resources that help the kids of today to deal with this “trick” of the math trade. There are numerous You Tube videos on the different methods of which show every step along the way with encouraging thoughts. Another great online resource is any of the math websites. I find it a little unfair that these resources were not readily available when I was struggling with such concepts. One of my personal favorites is the PurpleMath.com website. This website breaks everything down to basically that of a fourth grade level. They have pictures and fun problems to work out on your own. My favorite part is that you get your answers checked instantaneously to build the self-confidence and self-efficacy it takes to be a successful student. These particular websites are great tools for teachers as well, as they have a lot of great examples that can be used in the classroom and different ways that a student might present and calculate a problem.