Confirming Einstein’s Theory of General Relativity With Calculus, Part 2d: Hyperbolas and Polar Coordinates

In this series, I’m discussing how ideas from calculus and precalculus (with a touch of differential equations) can predict the precession in Mercury’s orbit and thus confirm Einstein’s theory of general relativity. The origins of this series came from a class project that I assigned to my Differential Equations students maybe 20 years ago.

In a previous post, we showed that the polar equation

$r = \displaystyle \frac{a}{1 + e \cos \theta}$

is equivalent to the rectangular equation

$\displaystyle \frac{\left(x + \displaystyle \frac{\alpha e}{1-e^2} \right)^2}{\displaystyle \frac{\alpha^2}{(1-e^2)^2}} + \frac{y^2}{\displaystyle \frac{\alpha^2}{1-e^2}} = 1$

as long as $e \ne 0$ . Furthermore, if $0 < e < 1$ , then this represents an ellipse with eccentricity $e$ whose major axis lies on the $x-$ axis, with one focus located at the origin.

While not directly related to our discussion of precession, it turns out that this equation represents a hyperbola if $e > 1$ . Under this assumption, $1-e^2 < 0$ and $e^2-1>0$ , so let me rewrite the previous equation in terms of $e^2-1$ :

$\displaystyle \frac{\left(x - \displaystyle \frac{\alpha e}{e^2-1} \right)^2}{\displaystyle \frac{\alpha^2}{(e^2-1)^2}} - \frac{y^2}{\displaystyle \frac{\alpha^2}{e^2-1}} = 1$

This matches the form of a left-right hyperbola

$\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ ,

where the center of the hyperbola is located at

$(h,k) = \displaystyle \left( \frac{\alpha e}{e^2-1} , 0 \right)$

Also, for a hyperbola, the distance $c$ from the center to the foci satisfies

$c^2 = a^2 + b^2$ ,

so that

$c^2 = \displaystyle \frac{\alpha^2}{(e^2-1)^2} + \displaystyle \frac{\alpha^2}{e^2-1}$

$c^2 = \displaystyle \frac{\alpha^2 + \alpha^2 (e^2 - 1)}{(e^2-1)^2}$

$c^2 = \displaystyle \frac{\alpha^2 e^2}{(e^2-1)^2}$

$c = \displaystyle \frac{\alpha e}{e^2-1}$

The two foci are located a distance $c$ to the left of the right of the center. Since it happened to happen that $c = h$ , this means that the origin is, once again, one of the foci of the hyperbola.

Furthermore, the eccentricity $c/a$ of the hyperbola is easily computed as

$\displaystyle \frac{c}{a} = \frac{ \displaystyle \frac{\alpha e}{e^2-1} }{ \displaystyle \frac{\alpha}{e^2-1}} = e$ ,

so that, once again, the well-chosen parameter $e$ is the eccentricity.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 2c: Circles, Parabolas, and Polar Coordinates

In the previous post, we showed that the polar equation

$r = \displaystyle \frac{\alpha}{1 + e \cos \theta}$

converts to

$x^2 (1-e^2) + 2 \alpha e x + y^2 = \alpha^2$

in rectangular coordinates. Furthermore, if $0 < e < 1$ , then this represents an ellipse with eccentricity $e$ whose semi-major axis lies along the $x-$ axis with one focus at the origin.

It turns out that, for different non-negative values of $e$ , the same polar equation represents different conic sections. These are not particularly relevant for our study of precession, but I’m including this anyway in this series as a small tangential discussion.

Let’s take a look at the easy case of $e = 0$ . With this substitution, the equation in rectangular coordinates simplifies to

$x^2 + y^2 = \alpha^2$ .

Of course, this is the equation of a circle that is centered at the origin with radius $\alpha$ .

The other easy case is $e = 1$ , so that $1-e^2 = 0$ . Then the equation in rectangular coordinates simplifies to

$2 \alpha x + y^2 = \alpha^2$

$y^2 = -2\alpha x + \alpha^2$

$y^2 = -2 \alpha \displaystyle \left( x - \frac{\alpha}{2} \right)$

$y^2 = -4 \cdot \displaystyle \frac{\alpha}{2} \left( x - \frac{\alpha}{2} \right)$

This matches the form of a parabola that opens to the left with a horizontal axis of symmetry:

$(y-k)^2 = -4 p (x-h)$ .

In this case, the vertex of the parabola is located at

$(h,k) = \displaystyle \left( \frac{\alpha}{2} , 0 \right)$ ,

while the focus of the parabola is located a distance $p = \displaystyle \frac{\alpha}{2}$ to the left of the vertex. In other words, the origin is the focus of the parabola. (For what it’s worth, the directrix of the parabola would be the vertical line $y = \alpha$ , located $p$ to the right of the vertex.)

Confirming Einstein’s Theory of General Relativity With Calculus, Part 2b: Ellipses and Polar Coordinates

As part of our derivation, we’ll need to use the fact that, in polar coordinates, the graph of

$r = \displaystyle \frac{\alpha}{1 + e \cos \theta}$

turns out to be an ellipse if $0 < e < 1$ , with the origin at one focus.

We now prove this. Clearing the denominator, we obtain

$r + re \cos \theta = \alpha$ .

Switching to rectangular coordinates, this becomes

$\sqrt{x^2 + y^2} + e x = \alpha$

$\sqrt{x^2 + y^2} = \alpha - ex$

$x^2 + y^2 = (\alpha - ex)^2$

$x^2 + y^2 = \alpha^2 - 2\alpha ex + e^2 x^2$

$x^2 (1-e^2) + 2 \alpha e x + y^2 = \alpha^2$

$(1-e^2)\left(x^2 + 2 \displaystyle \frac{\alpha e}{1-e^2} x \right) + y^2 = \alpha^2$

$(1-e^2)\left(x^2 + 2 \displaystyle \frac{\alpha e}{1-e^2} x + \frac{\alpha^2 e^2}{(1-e^2)^2} \right) + y^2 = \alpha^2 + \displaystyle \frac{\alpha^2e^2}{1-e^2}$

$(1-e^2) \left(x + \displaystyle \frac{\alpha e}{1-e^2} \right)^2 + y^2 = \displaystyle \frac{\alpha^2(1-e^2)+\alpha^2 e^2}{1-e^2}$

$(1-e^2) \left(x + \displaystyle \frac{\alpha e}{1-e^2} \right)^2 + y^2 = \displaystyle \frac{\alpha^2}{1-e^2}$

$\displaystyle \frac{(1-e^2)^2}{\alpha^2} \left(x + \displaystyle \frac{\alpha e}{1-e^2} \right)^2 + \frac{1-e^2}{\alpha^2} y^2 = 1$

$\displaystyle \frac{\left(x + \displaystyle \frac{\alpha e}{1-e^2} \right)^2}{\displaystyle \frac{\alpha^2}{(1-e^2)^2}} + \frac{y^2}{\displaystyle \frac{\alpha^2}{1-e^2}} = 1$

Since we assumed that $0 < e < 1$ , we have $0 < 1 - e^2 < 1$ so that

$\displaystyle \frac{\alpha^2}{(1-e^2)^2} > \displaystyle \frac{\alpha^2}{1-e^2}$ .

Therefore, this matches the usual form of an ellipse in rectangular coordinates

$\displaystyle \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ ,

where the center of the ellipse is located at

$(h,k) = \displaystyle \left( -\displaystyle \frac{\alpha e}{1-e^2}, 0 \right)$ ,

the semi-major axis is horizontal with length

$a = \displaystyle \frac{\alpha}{1-e^2}$ ,

and the semi-minor axis is vertical with length

$b = \displaystyle \frac{\alpha}{\sqrt{1-e^2}}$ .

Furthermore, the distance $c$ of the foci from the center of the ellipse satisfies the equation

$b^2 + c^2 = a^2$ ,

so that

$\displaystyle \frac{\alpha^2}{1-e^2} + c^2 = \displaystyle \frac{\alpha^2}{(1-e^2)^2}$

$c^2 = \displaystyle \frac{\alpha^2}{(1-e^2)^2} - \displaystyle \frac{\alpha^2}{1-e^2}$

$c^2 = \displaystyle \frac{\alpha^2 - \alpha^2(1-e^2)}{(1-e^2)^2}$

$c^2 = \displaystyle \frac{\alpha^2 e^2}{(1-e^2)^2}$

$c = \displaystyle \frac{\alpha e}{1-e^2}$

From this, we derive two nice properties of the ellipse. First, looking back on previous work, we see that $c = -h$ . Therefore, since the foci of the ellipse are distance $c$ away from the center along the major axis, we conclude that one focus of the ellipse is located at $(-h+c,0)$ , or $(0,0)$ . That is, the origin is one focus of the ellipse. (For the little it’s worth, the other focus is located at $(-2h,0)$ .

Second, the eccentricity of the ellipse is defined to be the ratio $c/a$ . This is now easily computed:

$\displaystyle \frac{c}{a} = \displaystyle \frac{\displaystyle \frac{\alpha e}{1-e^2}}{\displaystyle \frac{\alpha}{1-e^2}} = e$ .

In other words, the letter $e$ was well-chosen to represent the eccentricity of the ellipse.

For what it’s worth, here’s an alternate derivation of the formulas for $a$ and $b$ . For this ellipse, the planet’s closest approach to the Sun occurs at $\theta = 0$ :

$r(0) = \displaystyle \frac{\alpha}{1 + e \cos 0} = \frac{\alpha}{1 + e}$ ,

and the planet’s further distance from the Sun occurs at $\theta = \pi$ :

$r(\pi) = \displaystyle \frac{\alpha}{1 + e \cos \pi} = \frac{\alpha}{1 - e}$ .

Therefore, the length $2a$ of the major axis of the ellipse is the sum of these two distances:

$2a = \displaystyle \frac{\alpha}{1 + e} + \frac{\alpha}{1 - e}$

$2a = \displaystyle \frac{\alpha(1-e) + \alpha(1+e)}{(1 + e)(1 -e)}$

$2a= \displaystyle \frac{2\alpha}{1 - e^2}$

$a = \displaystyle \frac{\alpha}{1 - e^2}$ .

Since $c = a\epsilon$ , we can also compute $b$ :

$b^2 = a^2 - c^2$

$b^2 = a^2 - a^2 e^2$

$b^2 = a^2 ( 1-e^2)$

$b^2 = \displaystyle \frac{\alpha^2}{(1 - e^2)^2} (1-e^2)$

$b^2 = \displaystyle \frac{\alpha^2}{1 - e^2}$

$b = \displaystyle \frac{\alpha}{\sqrt{1 - e^2}}$

Confirming Einstein’s Theory of General Relativity With Calculus, Part 2a: Graphically Exploring Precession

But what is precession? To explore this concept, let’s explore the graph of

$r = \displaystyle \frac{a}{1 + e \cos (1-k)\theta}$

for various values of $a$ , $e$ , and $k$ using Desmos. (Note that, in this context, the number $e$ does not mean Euler’s constant $2.718\dots$ . The reason for choosing the letter $e$ for this parameter will become clear shortly.) Naturally, this demonstration could also be done with other tools like a graphing calculator.

I suggest beginning by setting $e=0$ and $k=0$ and altering the value of $a$ . This is the easiest behavior to explain. From the equation, $a$ is directly proportional to the distance from the origin $r$ . So, not surprisingly, increasing $a$ produces a larger graph, and decreasing $a$ produces a smaller graph.

Second, I suggest setting $a=3$ and $k=0$ but altering the value of $e$ . Starting at $e=0$ , the graph is a circle. This makes complete sense: if $e=0$ , then the equation simply becomes $r=a$ , so the distance from the origin is the same for all angles. However, as $e$ increases, the original circle becomes more and more stretched out. We will prove this analytically in a later post, but it turns out that, for $0 < e < 1$ , the graph is an ellipse, and the origin is one of the foci of the ellipse. The number $e$ is called the eccentricity of the ellipse (hence the letter $e$ ).

Again, if the value of $e$ is fixed but $a$ varies, the graph becomes either larger or smaller as $a$ becomes larger or smaller.

We notice that if $0 < e < 1$ and $k=0$ , then the denominator of

$r = \displaystyle \frac{r}{1 + e \cos \theta}$

varies between $1-e$ and $1+e$ . In particular, the denominator is always positive. Therefore, the value of $r$ is least positive — the graph is closest to the origin — when the denominator is greatest. This happens when $\theta$ is a multiple of $2\pi$ . So, for example, when $\theta = 0$ , then $r = a/(1+e)$ is as close as the graph gets to the origin. Let’s call this closest distance $P$ ; in the context of a planet’s orbit around the sun, this represent perihelion. Then we have $P = \displaystyle \frac{a}{1+e}$ .

When $e=1$ , the graph switches from an ellipse to a parabola, where the origin is the focus of the parabola. For $e>1$ , the graph becomes a hyperbola. However, since we’re mostly going to be concerned with stable planetary orbits in this series, we won’t dwell too much on the case $e \ge 1$ .

Third, I suggest setting $a=3$ , $e=0.8$ , and then alter the value of $k$ . For $k=0$ , the graph is simply a single ellipse. However, by changing the value of $k$ , the graph changes into a spiral.

In the above figure, the spiral stopped “spiraling” because I had asked Desmos only to show the graph between $0 \le \theta \le 20 \pi$ . If I had changed the upper bound to something larger than $20\pi$ , the spiral would continue.

The precession in the spiral is defined to be the angular offset between each loop of the spiral. Clearly, this is a function of $k$ . To find this function, we again examine the function

$r = \displaystyle \frac{a}{1 + e \cos (1-k)\theta}$

Once again, if $0 < e < 1$ , then the denominator varies between $1-e$ and $1+e$ . In particular, the denominator is always positive. Therefore, the value of $r$ is least positive when the denominator is greatest, and the denominator is greatest when $(1-k)\theta$ is a multiple of $2\pi$ . So, for example, when $\theta = 0$ , then $r = a/(1+e)$ is as close as the graph gets to the origin.

When does the graph return to its closest point to the origin next? This would occur when $(1-k)\theta = 2\pi$ , or $\theta = \displaystyle \frac{2\pi}{1-k}$ . If $k =0$ , then the angle of closest approach to the origin would $\theta =2\pi$ , and the graph simply cycles over itself. However, if $k > 0$ , then this angle $\theta$ will be larger than $2\pi$ , thus producing a spiral. Indeed, the amount of precession would be equal to

$\displaystyle \frac{2\pi}{1-k} - 2\pi = \frac{2\pi k}{1-k}$ .

In the picture above, $k = 0.05$ . Therefore, the amount of precession would be $\displaystyle \frac{2\pi (0.05)}{1-0.05} = \frac{2\pi}{19}$ radians $\approx 18.95^\circ$ . Therefore, after 19 “leafs” of the spiral, the graph would begin to cycle on top of itself.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 1c: Outline of Argument

This is going to be a very long series, so I’d like to provide a tree-top view of how the argument will unfold.

We begin by using three principles from Newtonian physics — the Law of Conservation of Angular Momentum, Newton’s Second Law, and Newton’s Law of Gravitation — to show that the orbit of a planet, under Newtonian physics, satisfies the initial-value problem

$u''(\theta) + u(\theta) = \displaystyle \frac{GMm^2}{\ell^2}$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ .

In these equations:

The orbit of the planet is in polar coordinates $(r,\theta)$ , where the Sun is placed at the origin.
The planet’s perihelion — closest distance from the Sun — is a distance of $P$ at angle $\theta = 0$ .
The function $u(\theta)$ is equal to $\displaystyle \frac{1}{r(\theta)}$ .
$G$ is the gravitational constant of the universe.
$M$ is the mass of the Sun.
$m$ is the mass of the planet.
$\ell$ is the angular momentum of the planet.

The solution of this differential equation is

$u(\theta) = \displaystyle \frac{1 + \epsilon \cos \theta}{\alpha}$ ,

so that

$r(\theta) = \displaystyle \frac{\alpha}{1 + \epsilon \cos \theta}$ .

In polar coordinates, this is the graph of an ellipse. Substituting $\theta = 0$ , we see that

$P = \displaystyle \frac{1 + \epsilon}{\alpha}$ .

In the solution for $u(\theta)$ , we have $\alpha = \displaystyle \frac{\ell^2}{GMm^2}$ and $\epsilon = \displaystyle \frac{\alpha - P}{P}$ . The number $\epsilon$ is the eccentricity of the ellipse, while $\alpha = \displaystyle \frac{P}{1+\epsilon}$ is proportional to the size of the ellipse.

Under general relativity, the governing initial-value problem changes to

$u''(\theta) + u(\theta) = \displaystyle \frac{GMm^2}{\ell^2} + \frac{3GM}{c^2} [u(\theta)]^2$ ,

$u(0) = \displaystyle \frac{1}{P}$ ,

$u'(0) = 0$ ,

where $c$ is the speed of light. We will see that the solution of this new differential equation can be well approximated by

$u(\theta) = \displaystyle \frac{1 + \epsilon}{\alpha} + \frac{\delta}{\alpha^2} + \frac{\delta \epsilon^2}{2\alpha^2} + \frac{\delta\epsilon}{\alpha^2} \theta \sin \theta - \frac{\delta \epsilon^2}{6\alpha^2} \cos 2\theta - \frac{\delta(3+\epsilon^2)}{3\alpha^2} \cos \theta$

$\approx \displaystyle \frac{1}{\alpha} \left[1 + \epsilon \cos \left(\theta - \frac{\delta \theta}{\alpha} \right) \right]$ .

This last equation describes a spiral that precesses by approximately

$\displaystyle \frac{2\pi \delta}{\alpha} \quad$ radians per orbit

$\displaystyle \frac{6\pi G M}{a c^2 (1-\epsilon^2)} \quad$ radians per orbit,

where $a$ is the length of the semimajor axis of the orbit.

This matches the amount of precession in Mercury’s orbit that is not explained by Newtonian physics, thus confirming Einstein’s theory of general relativity.

To the extent possible, I will take the perspective of a good student who has taken Precalculus and Calculus I. However, I will have to break this perspective a couple of times when I discuss principles from physics and derive the solutions of the above differential equations.

Here we go…

Confirming Einstein’s Theory of General Relativity With Calculus, Part 1b: Precession of Mercury

The figure below shows the (greatly exaggerated) effect of precession on a planet’s otherwise elliptical orbit. In the figure, each perihelion is precessed by an angle of $40^circ$ . After nine orbits, the planet returns to its original position. Suppose, for the sake of argument, that each orbit of the planet depicted in the figure is four months, or one third of Earth’s year. Then the amount of precession would be $40^\circ$ per four months, or $120^\circ$ per year, or $12,000^\circ$ per century.

As I said, the figure above is greatly exaggerated. As we’ll see by the end of this series, Einstein’s general relativity predicts that, on top of the gravitational influences of the other planets, the orbit of Mercury should precess by 43″ of arc per century. That’s a really small angle, since 1 $^\circ$ is equal to 60′ (minutes) of arc and each 1′ is equal to 60″ (seconds) of arc, that means 1″ of arc is the same as $(1/3600)^\circ$ , so that 43″ of arc per century is about $0.012^\circ$ per century. That’s about a million times smaller than the precession of the fictitious planet in the above figure.

How small is $0.012^\circ$ , really?

Courtesy of Wikipedia, the pictures below are the Copernicus crater on the Moon as well as an indicator of its location on the Moon. It is visible with binoculars.

The diameter of the crater is 93 km. Since the Moon is 384,400 km from Earth, that means the angle subtended by the crater, as viewed from the Earth, is about

$\arctan \left( \frac{93}{384,400} \right) \approx 0.014^\circ$ .

So how much is 43″ of arc per century? That’s about the speed as, hypothetically, pointing at the left edge of this lunar crater (which cannot be seen by the naked eye) and then slowly moving your figure so that, about 115 years later, your finger is pointing at the right edge of the crater.

Said another way, the diameter of the Moon is about 3475 km, so that the angle subtended by the Moon, as viewed from the Earth, is about

$\arctan \left( \frac{3745}{384,400} \right) \approx 0.518^\circ$ .

So, at the rate of $0.012^\circ$ per century, it would take $0.518/0.012 \approx 43$ centuries, or about 43,000 years, to trace the angle subtended by the moon.

Needless to say, 43” of arc per century is really, really slow.

Nevertheless, and remarkably, this itty, bitty precession was observable by careful 19th century astronomers with the telescopes that were available then. At the time, this precession was the great unsolved mystery of Newtonian physics that was only answered after two generations later with the discovery of general relativity.

Confirming Einstein’s Theory of General Relativity With Calculus, Part 1a: Introduction

If the universe consisted of only Mercury and the sun, Mercury’s trajectory would trace the same ellipse over and over again. However, there are seven other planets in the solar system (not to mention the dwarf planets), and these planets tug and nudge the orbit of Mercury ever so slightly. (For what it’s worth, similar nudges in the orbit of Uranus led to the discovery of Neptune in 1846.)

The practical effect of these nudges is that the orbit of Mercury precesses, or rotates like a spiral. The figure below shows the (greatly exaggerated) effect of precession on a planet’s otherwise elliptical orbit. In the figure, each perihelion is precessed by an angle of $40^\circ$ . After nine orbits, the planet returns to its original position.

Since the planets are much smaller than the sun and are further away from the Sun than Mercury, this precession is very small. However, this effect can be measured. Every century, the perihelion of Mercury precesses by 574” of arc (roughly a sixth of a degree).

Newton’s Law of Gravitation can be used to calculate the amount of the precession of Mercury; however, they predict a precession of only 531” of arc per century. This discrepancy between observation and prediction was first observed in 1845 and was, for a long time, the outstanding unresolved difficulty in Newtonian physics.

Einstein’s general theory of relativity, which was published seventy years later in 1915, exactly accounts for the missing 43” per century (within the tolerances of observational error). This was the first physical confirmation of general relativity. Furthermore, general relativity predicted that the orbit of Venus also precesses, but by only about 9” of arc per century. This small discrepancy was unobservable in 1915 but was confirmed in 1960. (While not logically necessary, that’s certainly indicative of an accurate scientific theory… not that it merely explains the world but it makes a prediction that is currently unobservable.)

In this series, which might take me a few months to complete, I’m going to explore how to predict the precession in Mercury’s orbit — i.e., confirm Einstein’s theory of general relativity — using tools only from calculus and precalculus. I first introduced these ideas as a class project for my Differential Equations students maybe 20 years ago. As we’ll see, in a couple spots, ideas from first-semester differential equations can make the steps more rigorous. However, pretty much this whole series should be accessible to a good calculus student.

I should say at the outset that none of the mathematics in this series is particularly original with me. I gladly acknowledge that I first learned the ideas in this series as an undergraduate, when I took an upper-level physics course in mechanics. In particular, pretty much all of the ideas in this series can be found in the textbook Classical Dynamics of Particles and Systems, by S. T. Thornton and J. B. Marion (Brooks Cole, New York, 2003). If I’ve made any contribution, it’s the scaffolding of these ideas to make them accessible to students who won’t be taking (or haven’t yet taken) physics courses beyond the traditional first-year sequence.

Parabolas from String Art: 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: Introduction

Part 2: Identifying the highest points of the strings

Part 3: These nine points lie on a parabola: Method #1

Part 4: These nine points lie on a parabola: Method #2

Part 5: These nine points lie on a parabola: Method #3

Part 6: Proof that all of the highest points lie on a parabola without calculus, Part 1

Part 7: Proof that all of the highest points lie on a parabola without calculus, Part 2

Part 8: Proof that all of the highest points lie on a parabola with calculus

Part 9: Proof that the strings are indeed tangent to the parabola, with calculus

Part 10: Conclusion

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: