Parabolas from String Art (Part 4)

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. 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:

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

At $x =4$ : $y = \displaystyle \frac{(4)^2}{16} - 4 + 8 = 5$

At $x =6$ : $y = \displaystyle \frac{(6)^2}{16} - 6 + 8 = 4.25$

At $x =10$ : $y = \displaystyle \frac{(10)^2}{16} - 10 + 8 = 4.25$

At $x =12$ : $y = \displaystyle \frac{(12)^2}{16} - 12 + 8 = 5$

At $x =14$ : $y = \displaystyle \frac{(14)^2}{16} - 14 + 8 = 6.25$

Therefore, the nine points in the above picture all lie on the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ .

In the next post, we’ll discuss a third way of convincing students that the points lie on this parabola.

Parabolas from String Art (Part 3)

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.

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. Let’s suppose that the curve is a parabola. The vertex form of a parabola is

$y = a(x-h)^2+k$ .

If the curve is a parabola, then clearly the vertex will be the lowest point on the axis of symmetry. By inspection, this point is $(8,4)$ , which is labeled $V$ in the above picture. So, if it’s a parabola, the equation has the form

$y = a(x-8)^2+4$ .

To find the value of $a$ , we note that the point $(x, y) = (16, 8)$ must be on the parabola, so that

$8 = a(x-8)^2 + 4$

$8 = 64a + 4$

$4 = 64a$

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

Therefore, the equation of the conjectured parabola is

$y = \displaystyle \frac{1}{16}(x-8)^2 + 4$

$= \displaystyle \frac{1}{16} (x^2 - 16x + 64) + 4$

$= \displaystyle \frac{x^2}{16} - x + 4 + 4$

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

So, if the curve is a parabola, it must follow the function this curve. By construction, this parabola passes through $(8,4)$ and $(16,8)$ . To show that this actually works, we can substitute the other seven values of $x$ :

At $x =0$ : $y = \displaystyle \frac{(0)^2}{16} - 0 + 8 = 8$

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

At $x =4$ : $y = \displaystyle \frac{(4)^2}{16} - 4 + 8 = 5$

At $x =6$ : $y = \displaystyle \frac{(6)^2}{16} - 6 + 8 = 4.25$

At $x =10$ : $y = \displaystyle \frac{(10)^2}{16} - 10 + 8 = 4.25$

At $x =12$ : $y = \displaystyle \frac{(12)^2}{16} - 12 + 8 = 5$

At $x =14$ : $y = \displaystyle \frac{(14)^2}{16} - 14 + 8 = 6.25$

Therefore, the nine points in the above picture all lie on the parabola $y = \displaystyle \frac{x^2}{16} - x + 8$ .

In the next couple of posts, we’ll discuss a couple of different ways of establishing that the points lie on this parabola.

Parabolas from String Art (Part 2)

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.

As discussed in the previous post, 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 now ask the following two questions:

For each of $x = 2, 4, 6, 8, 10, 12,$ and $14$ , which string has the largest $y-$ coordinate?
For each of these values of $x$ , what is the value of this largest $y-$ coordinate?

Evidently, for $x=4$ , the brown string that connects $(2,6)$ to $(10,2)$ has the largest $y-$ coordinate. This point is marked with the small brown circle. From the lines on the graph paper, it appears that this brown point is $(4,5)$ .

For $x=8$ , the horizontal green string appears to have the largest $y-$ coordinate, and clearly that point is $(8,4)$ .

For $x=12$ , the pink string that connects $(6,2)$ to $(14,6)$ has the largest $y-$ coordinate. From the lines on the graph paper, it appears that this point is $(12,5)$ .

Unfortunately, for $x=2$ , $x=6$ , $x=10$ , and $x=14$ , it’s evident which string has the largest $y-$ coordinate, but it’s not so easy to confidently read off its value. For this example, this could be solved by using finer graph paper with marks at each quarter (instead of at the integers). However, it’s far better to actually use the point-slope formula to find the equation of the colored line segments.

For example, for $x=2$ , the red string has the largest $y-$ coordinate. This string connects the points $(1,7)$ and $(9,1)$ , and so the slope of this string is $\displaystyle \frac{1-7}{9-1} = -\frac{6}{8} = -0.75$ . Using the point-slope form of a line, the equation of the red string is thus

$y - 7 = -0.75(x - 1)$

$y-7= -0.75x + 0.75$

$y = -0.75x + 7.75$

Substituting $x =2$ , the $y-$ coordinate of the highest string at $x=2$ is $y = -0.75(2) + 7.75 = 6.25$ .

Similarly, at $x=6$ , the equation of the orange string turns out to be $y=-0.25x+5.75$ , and the $y-$ coordinate of the highest string at $x=6$ is $y=-0.25(6)+5.75=4.25$ .

At $x=10$ , the equation of the blue string is $y=0.25x+1.75$ , and the $y-$ coordinate of the highest string at $x=10$ is $y=0.25(10)+1.75=4.25$ .

Finally, at $x=14$ , the equation of the purple string is $y=0.75x-4.25$ , and the $y-$ coordinate of the highest string at $x=14$ is $y=0.75(14)-4.25=6.25$ .

The interested student could confirm the values for $x=4$ , $x=8$ , and $x=12$ that were found earlier by just looking at the picture.

We now add the coordinates of these points to the picture.

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 next post in this series, I’ll discuss a way of convincing students that the curve is actually a parabola.

Parabolas from String Art (Part 1)

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.

To begin, we use graph paper to sketch to draw coordinate axes, the point $A(0,8)$ , the point $B(8,0)$ , the point $C(16,8)$ , line segment $\overline{AB}$ , and line segment $\overline{BC}$ .

Along $\overline{AB}$ , we mark the evenly spaced points $(1,7)$ , $(2,6)$ , $(3,5)$ , $(4,4)$ , $(5,3)$ , $(6,2)$ , and $(7,1)$ .

Along $\overline{BC}$ , we mark the evenly spaced points $(9,1)$ , $(10,2)$ , $(11,3)$ , $(12,4)$ , $(13,5)$ , $(14,6)$ , and $(15,7)$ .

Next, we draw line segments of different colors to connect:

$(1,7)$ and $(9,1)$
$(2,6)$ and $(10,2)$
$(3,5)$ and $(11,3)$
$(4,4)$ and $(12,4)$
$(5,3)$ and $(13,5)$
$(6,2)$ and $(14,6)$
$(7,1)$ and $(15,7)$

The result should look something like the picture below:

It looks like the string art is tracing a parabola. In this series of posts, I’ll discuss one way that talented algebra students can convince themselves that the curve is indeed a parabola.

Month: December 2022

Parabolas from String Art (Part 4)

Parabolas from String Art (Part 3)

Parabolas from String Art (Part 2)

Parabolas from String Art (Part 1)

Selection Bias