Parabolas from String Art (Part 10)

Recently, I announced that my paper Parabolic Properties from Pieces of String had been published in the magazine Math Horizons. The article had multiple aims; in chronological order of when I first started thinking about them:

Prove that string art from two line segments traces a parabola.
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.
Prove the reflective property of parabolas.
Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. In this series of posts, I’d wanted to expand on the article with some pedagogical thoughts about connecting string art to parabolas for algebra students. After all, most mathematical studies of string art curves — formally known as “envelopes” — rely on differential equations or at least limits and calculus.

However, string art is simple enough for a young child to construct, and so this study was inspired by the quest of explaining this phenomenon using only simple mathematical tools.

The article linked above has further thoughts on this problem, including a calculus-free way of deriving the reflective property of parabolas. However, I think the article pretty much has all of my thoughts on this matter, and so I don’t think I need to elaborate upon them here.

This series of posts is dedicated to an inspired and inspiring Algebra I student who wanted to understand string art curves using tools that she could understand… even though she progressed much further into the mathematics curriculum by the time my article was published and this series of posts appeared on my blog.

Parabolas from String Art (Part 9)

Prove that string art from two line segments traces a parabola.
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.
Prove the reflective property of parabolas.
Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. While all the necessary mathematics is in the article, I think the article is somewhat lacking on how to sell the idea to students. So, in this series of posts, I’d like to expand on the article with some pedagogical thoughts about connecting string art to parabolas.

We have shown in the last couple of posts that if the three points that generate the Our explorations of string art led us to consider an arbitrary string $\overline{PQ}$ depicted below. For brevity, this string will be called “string $s$ ,” matching the (possibly non-integer) $x$ -coordinate of its left endpoint $P$ . Since $P$ is $s$ units to the right of $A$ , the right endpoint $Q$ must correspondingly be $s$ units to the right of $B$ . Therefore, the $x$ -coordinate of $Q$ is $s + 8$ .

Previously, we established that the equation for string $s$ is

$y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8$ .

We also obtained a bonus result that we obtained using only algebra: string $s$ is tangent to the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ , which is traced by the strings, when $x=2s$ . Of course, tangent lines are usually obtained using calculus, and so calculus should be able to confirm this result. The derivative of this function is

$y' = \displaystyle \frac{x}{8} - 1$ ,

so that the slope of the tangent line when $x=2s$ is $m = \displaystyle \frac{s}{4} - 1 = \frac{s-4}{4}$ . We observe that this matches the slope of line segment $\overline{PQ}$ in the above picture:

slope $= \displaystyle \frac{s - (s-8)}{(s+8) - 8} = \frac{2s-8}{8} = \frac{s-4}{4}$ .

Therefore, to show that $\overline{PQ}$ is the tangent line, it suffices to show that either $P$ or $Q$ is on the tangent line.

At $x = 2s$ , the $y-$ coordinate of where the tangent line intersects the curve is

$y = \displaystyle \frac{(2s)^2}{16} - 2s + 8 = \frac{s^2}{4} - 2s + 8$ .

Using the point-slope formula for a line, the equation of the tangent line is thus

$y-y_1 = m(x-x_1)$

$y-\displaystyle \left( \frac{s^2}{4} - 2s + 8 \right) = \frac{s-4}{4} (x-2s)$

$y = \displaystyle \frac{s-4}{4} (x-2s) + \frac{s^2}{4} - 2s + 8$ .

We now check to see if $P(s,8-s)$ is on the tangent line. Substituting $x =s$ , we find

$y = \displaystyle \frac{s-4}{4} (s-2s) + \frac{s^2}{4} - 2s + 8$

$= \displaystyle \frac{s-4}{4} (-s) + \frac{s^2}{4} - 2s + 8$

$= \displaystyle \frac{(s-4)(-s) + s^2}{4} - 2s + 8$

$= \displaystyle \frac{-s^2+4s + s^2}{4} - 2s + 8$

$= \displaystyle \frac{4s}{4} - 2s + 8$

$= s - 2s + 8$

$= -s + 8$

Therefore, the point $(s,8-s)$ is on the tangent line, thus confirming that $P$ is on the tangent line and that $\overline{PQ}$ is the tangent line.

Parabolas from String Art (Part 8)

Prove that string art from two line segments traces a parabola.
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.
Prove the reflective property of parabolas.
Accomplish all of the above without using calculus.

While I’m generally pleased with the final form of the article, the necessity of publication constraints somewhat abbreviated the original goal of this project: determining a pedagogically sound way of convincing a bright Algebra I student that string art unexpectedly produces a parabola. While all the necessary mathematics is in the article, I think the article is somewhat lacking on how to sell the idea to students. So, in this series of posts, I’d like to expand on the article with some pedagogical thoughts about connecting string art to parabolas.

Previously, we established that the equation for string $s$ is

$y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8$ .

Finding the curve traced by the strings is a two-step process:

For a fixed value of $x$ , find the value of $s$ that maximizes $y$ .
Find this optimal value of $y$ .

Previously, we showed using only algebra that the optimal value of $s$ is $s = \displaystyle \frac{x}{2}$ , corresponding to an optimal value of $y$ of $y = \displaystyle \frac{x^2}{16} - x + 8$ .

For a student who knows calculus, the optimal value of $s$ can be found by instead solving the equation $\displaystyle \frac{dy}{ds} = 0$ (or, more accurately, $\displaystyle \frac{\partial y}{\partial s} = 0$ ):

$\displaystyle \frac{dy}{ds} = -\frac{2s}{4} + \frac{x}{4}$

$0 = \displaystyle \frac{-2s+x}{4}$

$0 = -2s + x$

$2s = x$

$s = \displaystyle \frac{x}{2}$ ,

matching the result that we found by using only algebra.

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.

illuminating-illustration-parabolic-properties-from-pieces-of-string Download

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.

snell3 Download

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,

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

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

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:

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

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

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

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

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 $g(x) = x^2$ :

See also this previous post for another way of deriving the Product Rule without adding or subtracting the same quantity.