{"id":3754,"date":"2023-09-10T09:18:32","date_gmt":"2023-09-10T14:18:32","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=3754"},"modified":"2023-09-10T14:57:48","modified_gmt":"2023-09-10T19:57:48","slug":"the-weaving-loom-problem","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/the-weaving-loom-problem\/","title":{"rendered":"The weaving loom problem"},"content":{"rendered":"

This week’s Fiddler<\/a> is a classic problem.<\/p>\n

\nA weaving loom<\/a> consists of equally spaced hooks along the x and y axes. A string connects the farthest hook on the x-axis to the nearest hook on the y-axis, and continues back and forth between the axes, always taking up the next available hook. This leads to a picture that looks like this:<\/p>\n

\n\"\"\n<\/div>\n

As the number of hooks goes to infinity, what does the shape trace out?<\/p>\n

Extra credit:<\/em> If four looms are rotated and superimposed as shown below, what is the area of the white region in the middle?<\/p>\n

\n\"\"\n<\/div>\n<\/blockquote>\n

My solution:
\n[Show Solution]<\/a><\/p>\n

\n

Suppose the weaving loom has size $1\\times 1$. If a given string starts a fraction $\\alpha$ away from the origin along the $y$-axis, so at the point $(0,\\alpha)$, it will connect to the point $(1-\\alpha,0)$ along the $x$-axis. Therefore, the set of all lines have equations:
\n\\[
\ny = \\alpha-\\frac{\\alpha}{1-\\alpha}x
\n\\]Suppose the limiting shape has equation $y = f(x)$. For a given $x$, $f(x)$ is the largest value of $y$ achievable for one of the lines above at that given $x$. This is illustrated in the diagram below:<\/p>\n

\"\"<\/a><\/p>\n

In other words, we have:
\n\\[
\nf(x) = \\max_{0\\leq\\alpha\\leq 1} \\left( \\alpha-\\frac{\\alpha}{1-\\alpha}x \\right)
\n\\]It’s clear that the maximum will not occur at a boundary point ($\\alpha=0$ or $\\alpha=1$), so it must occur at a point where the derivative with respect to $\\alpha$ is zero. In other words,
\n\\[
\n\\frac{\\mathrm{d}}{\\mathrm{d}\\alpha}\\left( \\alpha-\\frac{\\alpha}{1-\\alpha}x \\right)
\n= 1-\\frac{x}{(1-\\alpha)^2} = 0
\n\\]The optimal choice is $\\alpha_\\star = 1-\\sqrt{x}$, and this leads to
\n\\begin{align}
\nf(x) &= 1-\\sqrt{x}-\\frac{1-\\sqrt{x}}{\\sqrt{x}}x \\\\
\n&=1-2\\sqrt{x}+x \\\\
\n&= (1-\\sqrt{x})^2
\n\\end{align}A more symmetric way to express this function is to write it implicitly. If $y=f(x)$, the set of points $(x,y)$ that satisfy the equation above are precisely the points that satisfy the equation<\/p>\n

$\\displaystyle
\n\\sqrt{x} + \\sqrt{y} = 1
\n$<\/span><\/p>\n

Let’s dig a little deeper and see if we can learn more about the shape of this curve. Squaring both sides, rearranging, and squaring again, we obtain:
\n\\[
\nx^2+y^2-2xy-2x-2y+1 = 0
\n\\]This is a quadratic expression in $x$ and $y$, which means it is a conic section<\/a>! So either a circle, ellipse, hyperbola, or parabola. But which one? The easiest way to check is to examine the eigenvalues of the quadratic part (there will be two eigenvalues). Here is a handy table for reference:
\n\\[
\n\\begin{array}{l|l}
\n\\textbf{eigenvalue property} & \\textbf{conic section} \\\\ \\hline
\n\\text{equal eigenvalues} & \\text{circle} \\\\
\n\\text{same sign and nonzero} & \\text{ellipse} \\\\
\n\\text{different sign and nonzero} & \\text{hyperbola} \\\\
\n\\text{one zero eigenvalue} & \\text{parabola}
\n\\end{array}
\n\\]
\nIn our case, the quadratic terms are:
\n\\[
\nx^2+y^2-2xy = \\begin{bmatrix}x\\\\y\\end{bmatrix}^\\mathsf{T}
\n\\begin{bmatrix}1 & -1 \\\\ -1 & 1\\end{bmatrix}
\n\\begin{bmatrix}x\\\\y\\end{bmatrix}
\n\\]This matrix is singular (determinant is zero), so there is a zero eigenvalue, and the conic section must be a parabola! To make this more evident, we can rearrange the equation a bit more to obtain
\n\\[
\n\\left(\\frac{x+y}{\\sqrt{2}}\\right) = \\frac{1}{\\sqrt{2}}\\left(\\frac{x-y}{\\sqrt{2}}\\right)^2 + \\frac{1}{2\\sqrt{2}}
\n\\]If we imagine rotating our curve by $45$ degrees counterclockwise, this would correspond to the transformation $(x,y)\\mapsto\\left(\\tfrac{x-y}{\\sqrt{2}}, \\tfrac{x+y}{\\sqrt{2}} \\right)$, so the equation above in rotated coordinates $(\\hat x, \\hat y)$ is simply
\n\\[
\n\\hat y = \\frac{1}{\\sqrt{2}}\\hat x^2 + \\frac{1}{2\\sqrt{2}},
\n\\]which is clearly a parabola.<\/p>\n

Tangency.<\/strong> Based on the animation above, it looks like the line that crosses $(x,f(x))$ is also tangent to<\/em> the curve at that point. To verify this, we can compute the derivative directly and compare it to the slope of the line:
\n\\[
\nf'(x) = \\frac{\\mathrm{d}}{\\mathrm{d}x}(1-\\sqrt{x})^2 = 1-\\frac{1}{\\sqrt{x}}
\n\\]But the $\\alpha_\\star$ we found for a point $x$ produces:
\n\\[
\n\\text{(slope of line)} = \\frac{-\\alpha_\\star}{1-\\alpha_\\star} = \\frac{-(1-\\sqrt{x})}{1-(1-\\sqrt{x})} = 1-\\frac{1}{\\sqrt{x}},
\n\\]which is a perfect match!<\/p>\n

Extra credit<\/h3>\n

To find the area of the central region, we can break our shape into pieces as shown below.<\/p>\n

\"\"<\/a><\/p>\n

The point $x_0$ corresponds to the $x$-value of our curve where $y=\\frac{1}{2}$. This is given by:
\n\\[
\nx_0 = \\left(1-\\sqrt{\\frac{1}{2}}\\right)^2 = \\frac{3}{2}-\\sqrt{2} \\approx 0.0858
\n\\]The area of the central part is the total area of the square ($1$) minus four times the shaded area, which leads us to the integral:
\n\\begin{align}
\nA &= 1-4\\left( \\frac{1}{2}x_0 + \\int_{x_0}^{\\frac{1}{2}} (1-\\sqrt{x})^2\\,\\mathrm{d}x \\right) \\\\
\n&= \\frac{2}{3}\\left(4\\sqrt{2}-5\\right) \\\\[2mm]
\n&\\approx 0.4379
\n\\end{align}I decided to spare you the details of the integration because they’re not particularly interesting. So the white area in the middle occupies approximately $43.8\\%$ of the area of the square.\n<\/p><\/div>\n","protected":false},"excerpt":{"rendered":"

This week’s Fiddler is a classic problem. A weaving loom consists of equally spaced hooks along the x and y axes. A string connects the farthest hook on the x-axis to the nearest hook on the y-axis, and continues back and forth between the axes, always taking up the next available hook. This leads to … Continue reading “The weaving loom problem”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":3761,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[42],"tags":[28,11,43,10,4,26],"class_list":["post-3754","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-the-fiddler","tag-calculus","tag-conic-sections","tag-fiddler","tag-geometry","tag-integration","tag-optimization"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3754","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/comments?post=3754"}],"version-history":[{"count":8,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3754\/revisions"}],"predecessor-version":[{"id":3764,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/3754\/revisions\/3764"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/3761"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=3754"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=3754"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=3754"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}