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.

Thoughts on Numerical Integration: Index

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The links below show my series on numerical integration.

Part 1 and Part 2: Introduction

Part 3: Derivation of left, right, and midpoint rules

Part 4: Derivation of Trapezoid Rule

Part 5: Derivation of Simpson’s Rule

Part 6: Connection between the Midpoint Rule, the Trapezoid Rule, and Simpson’s Rule

Part 7: Implementation of numerical integration using Microsoft Excel

Part 8, Part 9, Part 10, Part 11: Numerical exploration of error analysis

Part 12 and Part 13: Left endpoint rule and rate of convergence

Part 14 and Part 15: Right endpoint rule and rate of convergence

Part 16 and Part 17: Midpoint Rule and rate of convergence

Part 18 and Part 19: Trapezoid Rule and rate of convergence

Part 20 and Part 21: Simpson’s Rule and rate of convergence

Part 22: Comparison of these results to theorems found in textbooks

Part 23: Return to Part 2 and accuracy of normalcdf function on TI calculators

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)

Square roots and logarithms without a calculator (Part 12)

I recently came across the following computational trick: to estimate $\sqrt{b}$ , use

$\sqrt{b} \approx \displaystyle \frac{b+a}{2\sqrt{a}}$ ,

where $a$ is the closest perfect square to $b$ . For example,

$\sqrt{26} \approx \displaystyle \frac{26+25}{2\sqrt{25}} = 5.1$ .

I had not seen this trick before — at least stated in these terms — and I’m definitely not a fan of computational tricks without an explanation. In this case, the approximation is a straightforward consequence of a technique we teach in calculus. If $f(x) = (1+x)^n$ , then $f'(x) = n (1+x)^{n-1}$ , so that $f'(0) = n$ . Since $f(0) = 1$ , the equation of the tangent line to $f(x)$ at $x = 0$ is

$L(x) = f(0) + f'(0) \cdot (x-0) = 1 + nx$ .

The key observation is that, for $x \approx 0$ , the graph of $L(x)$ will be very close indeed to the graph of $f(x)$ . In Calculus I, this is sometimes called the linearization of $f$ at $x =a$ . In Calculus II, we observe that these are the first two terms in the Taylor series expansion of $f$ about $x = a$ .

For the problem at hand, if $n = 1/2$ , then

$\sqrt{1+x} \approx 1 + \displaystyle \frac{x}{2}$

if $x$ is close to zero. Therefore, if $a$ is a perfect square close to $b$ so that the relative difference $(b-a)/a$ is small, then

$\sqrt{b} = \sqrt{a + b - a}$

$= \sqrt{a} \sqrt{1 + \displaystyle \frac{b-a}{a}}$

$\approx \sqrt{a} \displaystyle \left(1 + \frac{b-a}{2a} \right)$

$= \sqrt{a} \displaystyle \left( \frac{2a + b-a}{2a} \right)$

$= \sqrt{a} \displaystyle \left( \frac{b+a}{2a} \right)$

$= \displaystyle \frac{b+a}{2\sqrt{a}}$ .

One more thought: All of the above might be a bit much to swallow for a talented but young student who has not yet learned calculus. So here’s another heuristic explanation that does not require calculus: if $a \approx b$ , then the geometric mean $\sqrt{ab}$ will be approximately equal to the arithmetic mean $(a+b)/2$ . That is,

$\sqrt{ab} \approx \displaystyle \frac{a+b}{2}$ ,

so that

$\sqrt{b} \approx \displaystyle \frac{a+b}{2\sqrt{a}}$ .

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.

Thoughts on Numerical Integration (Part 23): The normalcdf function on TI calculators

I end this series about numerical integration by returning to the most common (if hidden) application of numerical integration in the secondary mathematics curriculum: finding the area under the normal curve. This is a critically important tool for problems in both probability and statistics; however, the antiderivative of $\displaystyle \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ cannot be expressed using finitely many elementary functions. Therefore, we must resort to numerical methods instead.

In days of old, of course, students relied on tables in the back of the textbook to find areas under the bell curve, and I suppose that such tables are still being printed. For students with access to modern scientific calculators, of course, there’s no need for tables because this is a built-in function on many calculators. For the line of TI calculators, the command is normalcdf.

Unfortunately, it’s a sad (but not well-known) fact of life that the TI-83 and TI-84 calculators are not terribly accurate at computing these areas. For example:

TI-84: $\displaystyle \int_0^1 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.3413447\underline{399}$

Correct answer, with Mathematica: $0.3413447\underline{467}\dots$

TI-84: $\displaystyle \int_1^2 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.1359051\underline{975}$

Correct answer, with Mathematica: $0.1359051\underline{219}\dots$

TI-84: $\displaystyle \int_2^3 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.021400\underline{0948}$

Correct answer, with Mathematica: $0.021400\underline{2339}\dots$

TI-84: $\displaystyle \int_3^4 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.0013182\underline{812}$

Correct answer, with Mathematica: $0.0013182\underline{267}\dots$

TI-84: $\displaystyle \int_4^5 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.0000313\underline{9892959}$

Correct answer, with Mathematica: $0.0000313\underline{84590261}\dots$

TI-84: $\displaystyle \int_5^6 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 2.8\underline{61148776} \times 10^{-7}$

Correct answer, with Mathematica: $2.8\underline{56649842}\dots \times 10^{-7}$

I don’t presume to know the proprietary algorithm used to implement normalcdf on TI-83 and TI-84 calculators. My honest if brutal assessment is that it’s probably not worth knowing: in the best case (when the endpoints are close to 0), the calculator provides an answer that is accurate to only 7 significant digits while presenting the illusion of a higher degree of accuracy. I can say that Simpson’s Rule with only $n = 26$ subintervals provides a better approximation to $\displaystyle \int_0^1 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx$ than the normalcdf function.

For what it’s worth, I also looked at the accuracy of the NORMSDIST function in Microsoft Excel. This is much better, almost always producing answers that are accurate to 11 or 12 significant digits, which is all that can be realistically expected in floating-point double-precision arithmetic (in which numbers are usually stored accurate to 13 significant digits prior to any computations).

Thoughts on Numerical Integration (Part 22): Comparison to theorems about magnitudes of errors

Numerical integration is a standard topic in first-semester calculus. From time to time, I have received questions from students on various aspects of this topic, including:

Why is numerical integration necessary in the first place?
Where do these formulas come from (especially Simpson’s Rule)?
How can I do all of these formulas quickly?
Is there a reason why the Midpoint Rule is better than the Trapezoid Rule?
Is there a reason why both the Midpoint Rule and the Trapezoid Rule converge quadratically?
Is there a reason why Simpson’s Rule converges like the fourth power of the number of subintervals?

In this series, I hope to answer these questions. While these are standard questions in a introductory college course in numerical analysis, and full and rigorous proofs can be found on Wikipedia and Mathworld, I will approach these questions from the point of view of a bright student who is currently enrolled in calculus and hasn’t yet taken real analysis or numerical analysis.

In this series, we have shown the following approximations of errors when using various numerical approximations for $\int_a^b x^k \, dx$ . We obtained these approximations using only techniques within the reach of a talented high school student who has mastered Precalculus — especially the Binomial Theorem — and elementary techniques of integration.

As we now present, the formulas that we derived are (of course) easily connected to known theorems for the convergence of these techniques. These proofs, however, require some fairly advanced techniques from calculus. So, while the formulas derived in this series of posts only apply to $f(x) = x^k$ (and, by an easy extension, any polynomial), the formulas that we do obtain easily foreshadow the actual formulas found on Wikipedia or Mathworld or calculus textbooks, thus (hopefully) taking some of the mystery out of these formulas.