# Parabolic Properties from Pieces of String

I am pleased to announce that my latest paper, “Parabolic Properties from Pieces of String,” has now been published in Math Horizons. This was a really fun project for me. As I describe in the paper, I started wondering if it was possible to convince a student who hadn’t learned calculus yet that string art from two line segments traces a parabola. Not only was I able to come up with a way of demonstrating this without calculus, but I was also able to (1) prove that a quadratic polynomial satisfies the focus-directrix property of a parabola, which is the reverse of the usual logic when students learn conic sections, and (2) prove the reflective property of parabolas. I was really pleased with the final result, and am very happy that this was accepted for publication.

Due to copyright restrictions, I’m not permitted to freely distribute the final, published version of my article. However, I am able to share the following version of the article.

The above PDF file is an Accepted Manuscript of an article published by Taylor & Francis in College Mathematics Journal on February 24, 2022, available online: Full article: Parabolic Properties from Pieces of String (tandfonline.com)

# A New Derivation of Snell’s Law without Calculus

Last week, I posted that my latest paper, “A New Derivation of Snell’s Law without Calculus,” has now been published in College Mathematics Journal. In that previous post, I didn’t provide the complete exposition because of my understanding of copyright restrictions at that time.

I’ve since received requests for copies of my paper, which prompted me to carefully read the publisher’s copyright restrictions. In a nutshell, I was wrong: I am allowed to widely distribute preprints that did not go through peer review and, with extra restrictions, the accepted manuscript after peer review.

So, anyway, here it is.

The above PDF file is an Accepted Manuscript of an article published by Taylor & Francis in College Mathematics Journal on January 28, 2022, available online: Full article: A New Derivation of Snell’s Law Without Calculus (tandfonline.com).

# A New Derivation of Snell’s Law without Calculus

I’m pleased to say that my latest paper, “A New Derivation of Snell’s Law without Calculus,” has now been published in College Mathematics Journal. The article is now available for online access to anyone who has access to the journal — usually, that means members of the Mathematical Association of America or anyone whose employer (say, a university) has institutional access. I expect that it will be in the printed edition of the journal later this year; however, I’ve not been told yet the issue in which it will appear.

Because of copyright issues, I can’t reproduce my new derivation of Snell’s Law here on the blog, so let me instead summarize the main idea. Snell’s Law (see Wikipedia) dictates the angle at which light is refracted when it passes from one medium (say, air) into another (say, water). If the velocity of light through air is $v_1$ while its velocity in water is $v_2$, then Snell’s Law says that

$\displaystyle \frac{\sin \theta_1}{v_1} = \displaystyle \frac{\sin \theta_2}{v_2}$

I was asked by a bright student who was learning physics if there was a way to prove Snell’s Law without using calculus. At the time, I was blissfully unaware of Huygens’s Principle (see OpenStax) and I didn’t have a good answer. I had only seen derivations of Snell’s Law using the first-derivative test, which is a standard optimization problem found in most calculus books (again, see Wikipedia) based on Fermat’s Principle that light travels along a path that minimizes time.

Anyway, after a couple of days, I found an elementary proof that does not require proof. I should warn that the word “elementary” can be a loaded word when used by mathematicians. The proof uses only concepts found in Precalculus, especially rotating a certain hyperbola and careful examining the domain of two functions. So while the proof does not use calculus, I can’t say that the proof is particularly easy — especially compared to the classical proof using Huygens’s Principle.

That said, I’m pretty sure that my proof is original, and I’m pretty proud of it.

# Convexity and Orthogonality at Saddle Points

Today, the Texas Section of the Mathematical Association of America is holding its annual conference. Like many other professional conferences these days, this conference will be held virtually, and so my contribution to the conference is saved on YouTube and is available to the public.

Here’s the abstract of my talk: “At a saddle point (like the middle of a Pringles potato chip), the directions of maximum upward concavity and maximum downward concavity are perpendicular. The usual proof requires a fair amount of linear algebra: eigenvectors of different eigenvalues of a real symmetric matrix, like the Hessian, must be orthogonal. For this reason, the orthogonality of these two directions is not often stated in calculus textbooks, let alone proven, when the Second Partial Derivative Test for identifying local extrema and saddle points is discussed. In this talk, we present an elementary proof of the orthogonality of these two directions that requires only ideas from Calculus III and trigonometry. Not surprisingly, this proof can be connected to the usual proof from linear algebra.”

If you have 12 minutes to spare, here’s the talk.

# Adding by a Form of 0: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on adding by a form of 0 (analogous to multiplying by a form of 1).

Part 1: Introduction.

Part 2: The Product and Quotient Rules from calculus.

Part 3: A formal mathematical proof from discrete mathematics regarding equality of sets.

Part 4: Further thoughts on adding by a form of 0 in the above proof.

# A Clean Calculus Joke

The change of position over time is velocity.

The change of velocity over time is acceleration.

The change of acceleration over time is jerk.

And the change of jerk over time is an election.d

# Adding by a Form of 0 (Part 2)

Often intuitive appeals for the proof of the Product Rule rely on pictures like the following:

The above picture comes from https://mrchasemath.com/2017/04/02/the-product-rule/, which notes the intuitive appeal of the argument but also its lack of rigor.

My preferred technique is to use the above rectangle picture but make it more rigorous. Assuming that the functions $f$ and $g$ are increasing, the difference $f(x+h) g(x+h) - f(x) g(x)$ is exactly equal to the sum of the green and blue areas in the figure below.

In other words,

$f(x+h) g(x+h) - f(x) g(x) = f(x+h) [g(x+h) - g(x)] + [f(x+h) - f(x)] g(x)$,

or

$f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) - f(x) g(x)$.

This gives a geometrical way of explaining this otherwise counterintuitive step for students not used to adding by a form of 0. I make a point of noting that we took one term, $f(x+h)$, from the first product $f(x+h) g(x+h)$, while the second term, $g(x)$, came from the second product $f(x) g(x)$. From this, the usual proof of the Product Rule follows:

$[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}$

$\displaystyle = \lim_{h \to 0} f(x+h) \frac{g(x+h) - g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) - f(x) }{h} g(x)$

$= f(x)g'(x) + f'(x) g(x)$

For what it’s worth, a Google Images search for proofs of the Product Rule yielded plenty of pictures like the one at the top of this post but did not yield any pictures remotely similar to the green and blue rectangles above. This suggests to me that the above approach of motivating this critical step of this derivation might not be commonly known.

Once students have been introduced to the idea of adding by a form of 0, my experience is that the proof of the Quotient Rule is much more palatable. I’m unaware of a geometric proof that I would be willing to try with students (a description of the best attempt I’ve seen can be found here), and so adding by a form of 0 becomes unavoidable. The proof begins

$\left[\left( \displaystyle \frac{f}{g} \right)(x) \right]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h)}{ g(x+h)} - \frac{f(x)}{ g(x)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{ g(x) g(x+h)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x) - f(x) g(x+h)}{ h g(x) g(x+h)}$.

At this point, I ask my students what we should add and subtract this time to complete the derivation. Given the previous experience with the Product Rule, students are usually quick to chose one factor from the first term and another factor from the second term, usually picking $f(x) g(x)$. In fact, they usually find this step easier than the analogous step in the Product Rule because this expression is more palatable than the slightly more complicated $f(x+h) g(x)$. From here, the rest of the proof follows:

$[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x) + f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} + \frac{f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} - \frac{f(x) g(x+h) - f(x)g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{[f(x+h) - f(x)] g(x)}{h} - \frac{f(x) [g(x+h) - g(x)]}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) - f(x) }{h} g(x) - f(x) \frac{ g(x+h) - g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \frac{ f'(x) g(x) - f(x) g'(x)}{g(x)^2}$

P.S.

• The website https://mrchasemath.com/2017/04/02/the-product-rule/ also suggests an interesting pedagogical idea: before giving the formal proof of the Product Rule, use a particular function and the limit definition of a derivative so that students can intuitively guess the form of the rule. For example, if $g(x) = x^2$:

# Adding by a Form of 0 (Part 1)

Adding by a form of 0, or adding and subtracting the same quantity, is a common technique in mathematical proofs. For example, this technique is used in the second step of the standard proof of the Product Rule in calculus:

$[(fg)(x)]' = \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x+h) g(x) + f(x+h) g(x) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \left[ \frac{f(x+h) g(x+h) - f(x+h) g(x)}{h} + \frac{f(x+h) g(x) - f(x) g(x)}{h} \right]$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) g(x+h) - f(x+h) g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) g(x) - f(x) g(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{f(x+h) [g(x+h) - g(x)]}{h} + \lim_{h\ to 0} \frac{[f(x+h) - f(x)] g(x)}{h}$

$\displaystyle = \lim_{h \to 0} f(x+h) \frac{g(x+h) - g(x)}{h} + \lim_{h\ to 0} \frac{f(x+h) - f(x) }{h} g(x)$

$= f(x)g'(x) + f'(x) g(x)$

Or the proof of the Quotient Rule:

$\left[\left( \displaystyle \frac{f}{g} \right)(x) \right]' = \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h)}{ g(x+h)} - \frac{f(x)}{ g(x)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{ g(x) g(x+h)}}{h}$

$= \displaystyle \lim_{h \to 0} \frac{f(x+h) g(x) - f(x) g(x+h)}{ h g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x) + f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} + \frac{f(x)g(x) - f(x) g(x+h)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) g(x) - f(x)g(x)}{h} - \frac{f(x) g(x+h) - f(x)g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{[f(x+h) - f(x)] g(x)}{h} - \frac{f(x) [g(x+h) - g(x)]}{h }}{g(x) g(x+h)}$

$= \displaystyle \lim_{h \to 0} \frac{ \displaystyle \frac{f(x+h) - f(x) }{h} g(x) - f(x) \frac{ g(x+h) - g(x)}{h }}{g(x) g(x+h)}$

$= \displaystyle \frac{ f'(x) g(x) - f(x) g'(x)}{g(x)^2}$

This is a technique that we expect math majors to add to their repertoire of techniques as they progress through the curriculum. I forget the exact proof, but I remember that, when I was a student in honors calculus, we had some theorem that required an argument of the form

$|x - y| = |x - A + A - B + B - C + C - D + D - E + E - F + F - y|$

$\le |x - A| + |A - B| + |B - C| + |C - D| + |D - E| + |E - F| + |F - y|$

$\le \displaystyle \frac{\epsilon}{7} + \frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7} +\frac{\epsilon}{7}$

$= \epsilon$

But while this is a technique that expect students to master, there’s no doubt that this looks utterly foreign to a student first encountering this technique. After all, in high school algebra, students would simplify something like $x - A + A - B + B - C + C - D + D - E + E - F + F - y$ into $x-y$. If they were to convert $x-y$ into something more complicated like $x - A + A - B + B - C + C - D + D - E + E - F + F - y$, they would most definitely get points taken off.

In this brief series, I’d like to give some thoughts on getting students comfortable with this technique.

# My Favorite One-Liners: Part 103

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll use today’s one-liner to give students my expectations about simplifying incredibly complicated answers. For example,

Find $f'(x)$ if $f(x) = \displaystyle \frac{\sqrt{x} \csc^5 (\sqrt{x} )}{x^2+1}$.

Using the rules for differentiation,

$f(x) = \displaystyle \frac{[\sqrt{x} \csc^5 (\sqrt{x} )]'(x^2+1) -[\sqrt{x} \csc^5 (\sqrt{x} )](x^2+1)' }{(x^2+1)^2}$

$= \displaystyle \frac{[(\sqrt{x})' \csc^5 (\sqrt{x} ) + \sqrt{x} (\csc^5(\sqrt{x}))'](x^2+1) - \sqrt{x} \csc^5 (\sqrt{x} )](2x) }{(x^2+1)^2}$

$= \displaystyle \frac{[\frac{1}{2\sqrt{x}} \csc^5 (\sqrt{x} ) + 5 \sqrt{x} \csc^4(\sqrt{x}) [-\csc(\sqrt{x})\cot(\sqrt{x})]\frac{1}{2\sqrt{x}}(x^2+1) - \sqrt{x} \csc^5 (\sqrt{x} )](2x) }{(x^2+1)^2}$

With some effort, this simplifies somewhat:

$f'(x) = -\displaystyle \frac{\left(5 x^{5/2} \cot \left(\sqrt{x}\right)+3 x^2+5 \sqrt{x} \cot \left(\sqrt{x}\right)-1\right) \csc ^5\left(\sqrt{x}\right)}{2 \sqrt{x} \left(x^2+1\right)^2}$

Still, the answer is undeniably ugly, and students have been well-trained by their previous mathematical education to think the final answers are never that messy. So, if they want to try to simplify it further, I’ll give them this piece of wisdom:

You can lipstick on a pig, but it remains a pig.