Parabolas from String Art (Part 8)

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.

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.

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.

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