{"id":2490,"date":"2018-10-13T15:47:06","date_gmt":"2018-10-13T20:47:06","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=2490"},"modified":"2018-10-13T16:48:58","modified_gmt":"2018-10-13T21:48:58","slug":"tether-your-goat","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/tether-your-goat\/","title":{"rendered":"Tether your goat!"},"content":{"rendered":"

A geometry problem from the Riddler blog<\/a>. Here it goes:<\/p>\n

\nA farmer owns a circular field with radius R. If he ties up his goat to the fence that runs along the edge of the field, how long does the goat’s tether need to be so that the goat can graze on exactly half of the field, by area?\n<\/p><\/blockquote>\n

Here is my solution:
\n[Show Solution]<\/a><\/p>\n

\n

There are many ways to solve this problem, so I figured I would pick an unorthodox way: using calculus! Consider the diagram below. <\/p>\n

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

The circular field is centered at $O$. The goat is tethered at $C$ and the tether has length $r$. Let $B$ be the point diametrically opposed to $C$ and let $A$ be the farthest point along the perimeter that the goat can reach. Let $\\theta$ be the angle $BOA$, as shown. If $r$ changes a little bit, by some amount $dr$, then $\\theta$ will change by some amount $d\\theta$. Likewise, the fraction of the total area reachable by the goat will change by $dA$. Let’s calculate how these quantities are related.<\/p>\n

First, note that $dA$ is simply the area between the two circular arcs centered at $C$ and passing through $A$ and $A’$, respectively. We must also divide by the total area of the field in order to get a ratio. As $dr\\to 0$, we have:
\n\\[
\ndA = \\frac{\\tfrac{1}{2} \\theta\\, d(r^2)}{\\pi R^2}
\n\\]Our next task is to find out how $d\\theta$ is related to $d(r^2)$. Look at the triangle $OAC$. By the law of cosines, we have:
\n$r^2 = 2R^2(1 + \\cos(\\theta))$. Taking differentials, we obtain:
\n\\[
\nd(r^2) = -2R^2 \\sin(\\theta)\\,d\\theta
\n\\]Substituting back into our expression for $dA$, we obtain:
\n\\[
\ndA = -\\tfrac{1}{\\pi} \\theta\\,\\sin(\\theta)\\,d\\theta
\n\\]We can now integrate! Note that $A=1$ when $\\theta=0$. Therefore, we have:
\n\\begin{align}
\nA(\\theta) &= 1 + \\int_{0}^\\theta \\tfrac{1}{\\pi} t\\, \\sin(t)\\,dt \\\\
\n&= \\frac{\\pi + \\theta\\,\\cos(\\theta)-\\sin(\\theta)}{\\pi}
\n\\end{align}This tells us the ratio of areas as a function of $\\theta$. If we want this as a function of $r$ instead, we can use the law of cosines again to eliminate $\\theta$ and write $A$ as a function of $r$. This yields:
\n\\[
\nA(\\rho) = 1+\\frac{\\left(\\frac{\\rho ^2}{2}-1\\right) \\cos ^{-1}\\left(\\frac{\\rho ^2}{2}-1\\right)}{\\pi }-\\frac{\\rho \\sqrt{4-\\rho ^2}}{2\\pi}
\n\\]where I made the substitution $\\rho := r\/R$. We can plot this quantity as a function of $\\rho$, and we get:<\/p>\n

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

As we might expect, when $r=0$, $A=0$. And the area ratio grows monotonically with $r$ until we reach $r=2R$, at which point the goat can reach the farthest point (which is point $B$) and therefore can reach the entire field. There is no closed-form expression for the exact tether length that leads to a particular area ratio, since that would require solving the above $A(\\rho)$ for $\\rho$. However, we can easily solve the equation numerically. Doing so, we obtain:<\/p>\n

$\\displaystyle
\n\\frac{\\text{length of tether}}{\\text{radius of field}} \\approx 1.15873
\n$<\/span><\/p>\n

Finally, here is a diagram of what the field and tether look like when the area ratio is exactly 1\/2.<\/p>\n

\"\"<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

A geometry problem from the Riddler blog. Here it goes: A farmer owns a circular field with radius R. If he ties up his goat to the fence that runs along the edge of the field, how long does the goat’s tether need to be so that the goat can graze on exactly half of … Continue reading “Tether your goat!”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":2494,"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":[7],"tags":[28,10,2],"class_list":["post-2490","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-calculus","tag-geometry","tag-riddler"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2490","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=2490"}],"version-history":[{"count":5,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2490\/revisions"}],"predecessor-version":[{"id":2498,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/2490\/revisions\/2498"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/2494"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=2490"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=2490"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=2490"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}