An algebra and trigonometry–based proof of Kepler’s First Law

The proofs of Kepler’s Three Laws are usually included in textbooks for multivariable calculus. So I was very intrigued when I saw, in the Media Reviews of College Mathematics Journal, that somebody had published a proof of Kepler’s First Law that only uses algebra and trigonometry. Let me quote from the review:

Kepler’s first law states that bounded planetary orbits are elliptical. This law is presented in introductory textbooks, but the proof typically requires intricate integrals or vector analysis involving an accidental degeneracy. Simha offers an elementary proof of Kepler’s first law using algebra and trigonometry at the high school level.
https://doi.org/10.1080/07468342.2022.2026089

Once upon a time, I taught Precalculus for precocious high school students. I wish I had known of this result back then, as it would have been a wonderful capstone to their studies of trigonometry and the conic sections.

The preprint of this result can be found on arXiv. (The proof only addresses Kepler’s First Law and not the Second and Third Laws.) The actual article, for those with institutional access, was published in American Journal of Physics Vol. 89 No. 11 (2021): 1009-1011.

Wonderful quote from College Mathematics Journal

As a member of the Mathematical Association of America, one of the journals that I subscribe to is College Mathematics Journal. Of late, there has been a pleasant uptick in the number of articles that have been co-written by undergraduate researchers under the mentorship of faculty advisers, which is a terrific development for the field.

In the latest issue, one such article on knot theory appeared. Truth in advertising: I know next to nothing about knot theory, I do not know the authors, and I’ve only driven through Colorado College while on a recent vacation to Colorado Springs. With all that said, I love the opening paragraphs of their recent article, which are shown below.

The first sentence of the second paragraph grabbed my attention:

The other author, undeterred by the challenge of a long standing open problem, decided she wanted to look at this question for her senior thesis.

I absolutely love the moxie behind this sentiment. For the little it’s worth, I offer my congratulations to both authors.

More truth in advertising: usually, when I see a journal article that’s outside my realm of expertise, I’ll make a half-hearted stab at scanning it; with rare exceptions, I then give up and move on to the next article. For this article, however, given this wonderful introduction, I’ll do my best to read and (try to) understand the full article.

Fermat’s First Theorem

Source: https://xkcd.com/2689/

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.

Parabolas from String Art (Part 7)

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.

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 $s + 8$ .

In the previous post, we established that the equation for string $s$ is

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

This has the appearance of a quadratic equation, but it’s actually a linear equation in $x$ for a fixed value of $s$ . For example, if s = 5, we find that the equation of string 5 is

$y = -\displaystyle \frac{25}{4} + \frac{5x}{4} - x +8 = 0.25x+1.75$ ,

matching the equation of the blue string we found in a previous post in this series.

To prove that the strings trace a parabola, we now determine which string $s$ maximizes the value of $y = -\displaystyle \frac{1}{4}s^2 + \frac{x}{4}s - x + 8$ for a given value of $x$ . Algebra students can determine this maximum by recalling that a quadratic function $y = as^2 + bs + c$ is maximized (for negative $a$ ) when $s = \displaystyle -\frac{b}{2a}$ . Therefore, the string with largest $y$ -coordinate for a given value of $x$ is

$s = \displaystyle -\frac{x/4}{2 \cdot (-1/4)} = \frac{x}{2}$ .

For example, if $x = 4$ , then string $s = 4 / 2 = 2$ has the largest $y$ -coordinate, matching our previous observations.
To complete the proof that the strings above trace a parabola, we substitute $s = \displaystyle \frac{x}{2}$ into $y = -\displaystyle \frac{s^2}{4} + \frac{xs}{4} - x + 8$ to find the value of this largest $y-$ coordinate:

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

$= -\displaystyle \frac{x^2}{16} + \frac{x^2}{8} - x + 8$

$= \displaystyle \frac{x^2}{16} - x + 8$ ,

matching the result that we found earlier in this series.

There’s also a bonus result. We further note that, for every $x$ , there is only one string $s = \frac{x}{2}$ that intersects the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ . Since each $x$ is associated with a unique string $s$ and vice versa, we conclude that each string intersects the parabola at exactly one point. In other words, string $s$ is tangent to the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ when $x=2s$ .

We note that all of the above calculations were entirely elementary, in the sense that calculus was not used and that only techniques from algebra were employed. That said, the word “elementary” in mathematics can be a bit loaded — this means that it is based on simple ideas that are perhaps used in a profound and surprising way. Perhaps my favorite quote along these lines was this understated gem from the book Three Pearls of Number Theory after the conclusion of a very complicated multi-page proof in Chapter 1:

You see how complicated an entirely elementary construction can sometimes be. And yet this is not an extreme case; in the next chapter you will encounter just as elementary a construction which is considerably more complicated.

In the next post, we take a second look at this derivation using techniques from calculus.

Parabolas from String Art (Part 6)

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.

As discussed previous posts, we begin our explorations with string art connecting evenly spaced points on line segments $\overline{AB}$ and $\overline{BC}$ with endpoints $A(0,8)$ , $B(8,0)$ , and $C(16,8)$ . We will call these colored line segments “strings.” We then found the string with the largest $y-$ coordinate at $x = 2, 4, 6, \dots, 14$ , resulting in the following picture:

In previous posts, we discussed three different ways of establishing that the colored points lie on the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ .

Unfortunately, checking that a statement is true for a few points (in our case, $x= 0, 2, 4, \dots, 14, 16$ ) does not constitute a complete proof for all points. Furthermore, it’s conceivable that “fuller” string art with additional strings, like the picture below, may identify a new string with a higher $y-$ coordinate than a colored point.

To prove that the string art indeed traces a parabola, we study 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 $s + 8$ .

Since the equations of $\overline{AB}$ and $\overline{BC}$ are $y=-x+8$ and $y=x-8$ , respectively, the $y-$ coordinates of $P$ and $Q$ are $-s+8$ and $(s+8)-8 = s$ , respectively. For example, if $s = 5$ , the coordinates of $P$ are $(s,8-s)=(5,3)$ and the coordinates of $Q$ are $(s + 8, s) = (13, 5)$ , matching the endpoints of the blue string in the first figure.
We now use standard algebraic techniques to find the equation of string $s$ . Its slope is

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

The coordinates of either $P$ or $Q$ can now be used to find the equation of string $s$ via the point-slope formula. As it turns out, the coordinates of $P$ are simpler to use:

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

$y-(8-s) = \displaystyle \frac{s-4}{4}(x-s)$

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

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

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

to finally arrive at the equation of string $s$ :

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

This has the appearance of a quadratic equation, but it’s actually a linear equation in $x$ for a fixed value of $s$ . For example, if s = 5, we find that the equation of string 5 is

$y = -\displaystyle \frac{25}{4} + \frac{5x}{4} - x +8 = 0.25x+1.75$ ,

matching the equation of the blue string we found in a previous post in this series.

We are now almost in position to prove that the string art traces a parabola. We demonstrate this in the next post.

Parabolas from String Art (Part 5)

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.

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. In the previous posts, we established algebraically that the curve is the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ .

A more modern way of convincing students that the points lie on a parabola is by using technology: specifically, quadratic regression. First, we can input the nine points into a scientific calculator.

Then we ask the calculator to perform quadratic regression on the data.

The calculator then returns the result:

The best quadratic fit to the data is $y = 0.0625x^2 - x + 8$ , or $y = \displaystyle \frac{x^2}{16} - x + 8$ as before. The line $R^2 = 1$ indicates a correlation coefficient of 1, meaning that the points lie perfectly on this parabola. A parenthetical note: If the $R^2$ line does not appear using a TI-83 or TI-84, this can be toggled by using DiagnosticOn:

I’ve presented three different ways that algebra students can convince themselves that the nine points generated by the above string art indeed lie on a parabola. I don’t suggest that all three methods should be used for any given student; as always, if one technique doesn’t appear to work pedagogically, then perhaps a different explanation might work.

However, our explorations aren’t done yet. Any of these three techniques may convince algebra students that the strings above trace a parabola. Unfortunately, checking that a statement is true for a few points (in our case, $x= 0, 2, 4, \dots, 14, 16$ ) does not constitute a complete proof for all points. Furthermore, it’s conceivable that “fuller” string art with additional strings, like the picture below, may identify a new string with a higher $y-$ coordinate than a colored point.

So we have more work to do to prove our assertion that string art traces a parabola. We begin this next phase of our investigations in the next post.

Parabolas from String Art (Part 4)

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.

However, perhaps it’s clearer to plot these points on a separate graph, without the clutter of the strings:

These points are definitely following some kind of curve. In the previous post, we established that the curve is a parabola by using the vertex form of a parabola $y = a(x-h)^2+k$ .

In this post, we use the other general form. If the curve is a parabola, then the equation of the curve must be $y = ax^2 + bx + c$ for some values of $a$ , $b$ , and $c$ . Since there are three unknowns, we need to have three equations to solve for them. This can be done by plugging in three $(x,y)$ pairs into this equation. While we can pick any three pairs that we wish, it seems convenient to use the points $(0,8)$ , $(8,4)$ and $(16,8)$ :

$a(0)^2+b(0)+c = 8$

$a(8)^2 + b(8) + c = 4$

$a(16)^2 + b(16) + c =8$

This simplifies to the $3\times 3$ system of linear equations

$c = 8$

$64a+8b+c=4$

$256a+16b+c=8$

In general $3\times 3$ systems of linear equations can be challenging for students to solve. However, while this is technically a $3\times 3$ system, it’s clear that $c =8$ , and so this reduces to a $2\times 2$ system

$64a+8b+8=4$

$256a+16b+8=8$

$64a+8b=-4$

$256a+16b=0$

$16a+2b=-1$

$16a+b=0$ .

In algebra, students are taught multiple ways of solving $2\times 2$ systems of linear equations, and any of these techniques can be used at this point to solve for $a$ and $b$ . Perhaps the easiest next step is subtracting the two equations:

$(16a + 2b) - (16a + b) = -1 - 0$

$b = -1$

Substituting into $16a+b=0$ , we see that

$16a - 1 = 0$

$16a = 1$

$a =\displaystyle \frac{1}{16}$ .

We conclude that $a = \displaystyle \frac{1}{16}$ , $b = -1$ , and $c = 8$ , so that, if the points lie on a parabola, the equation of the parabola must be

$y = \displaystyle \frac{x^2}{16} - x + 8$ .

By construction, this parabola passes through $(0,8)$ , $(8,4)$ , and $(16,8)$ . To show that this actually works, we can substitute the other six values of $x$ :

At $x =2$ : $y = \displaystyle \frac{(2)^2}{16} - 2 + 8 = 6.25$