{"id":151,"date":"2019-10-25T17:42:09","date_gmt":"2019-10-25T23:42:09","guid":{"rendered":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/?page_id=151"},"modified":"2019-10-25T17:42:09","modified_gmt":"2019-10-25T23:42:09","slug":"the-arc-length-problem-2","status":"publish","type":"page","link":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/the-arc-length-problem-2\/","title":{"rendered":"The Arc Length Problem"},"content":{"rendered":"\n

Problem<\/strong> Unpack Stillwell’s argument about the arc length of the circle starting on p. 96. What is an arctan? Where are $y$ and $x$?<\/em><\/p>\n\n\n\n

\"\"
The diagram accompanying Stillwell’s computation of \u03c0 in\u00a0Yearning for the Impossible<\/em> (p. 97)<\/figcaption><\/figure>\n\n\n\n

There are a few things about the way this is written that makes it harder for a reader to figure out what’s going on. First, $x$ and $y$ are not coordinate directions. There’s nothing illegal about this, but it messes with our usual assumptions.<\/p>\n\n\n\n

\"\"
All we know to start is $x$ and that we want to find $y$<\/figcaption><\/figure>\n\n\n\n

The assumption we do want to hold on to though is the one about starting with $x$ and looking for $y$. $x$ is distance up, and $y$ is distance along the curve. $dy$ is thus a little bit of distance along the curve. The basic idea of the problem is to use the idea of infinitesimals to add up all the little bits $dy$ to get $y$.<\/p>\n\n\n\n

Since we don’t yet know how to compute $y$, we can’t very well add up little bits of $dy$. But if we can write $dy$ in terms of $x$ and $dx$, we can actually get somewhere. That is our first goal.<\/p>\n\n\n\n

First Goal: Write $dy$ in Terms of $x$ and $dx$.<\/h2>\n\n\n\n

The next confusing thing that gets in the way is that Stillwell denotes $y$ as being $\\arctan{x}$. In the end, this isn’t wrong<\/em>, but it is<\/em> confusing. We don’t know what an $\\arctan$ is or why or how it is coming up. <\/p>\n\n\n\n

In fact, we don’t need $\\arctan{x}$ at all. If you don’t care, skip past the sidebar until later. If you want to know what the heck $\\arctan{x}$ is doing here, go ahead and read it.<\/p>\n\n\n\n

Sidebar\u2014Trying to See arctan<\/strong>
It is possible to do some sleuthing to see why something like the tangent is involved here. You probably remember the tangent as the ratio of the two legs of a right triangle, something like $\\tan{\\theta} = \\frac{\\text{opposite}}{\\text{adjacent}}$, where $\\theta$ is the angle of the triangle in question. In this drawing it looks like $\\angle BOA$ should be $\\theta$. In this case, $\\tan{\\theta} = \\frac{x}{1} = x$.

You may also remember the basic idea of $\\arctan$ to be the inverse<\/em> of $\\tan$. So if we have the equation $\\tan{\\theta}=x$, that can alternately be written (in algebra class you say “taking the $\\arctan$ of both sides” but higher level math teachers frown on this verbiage) $\\theta = \\arctan{x}$.

This is on the right track, but we are not there yet. In the drawing $y$ isn’t an angle, but a distance along a circle. If you’re asking the question about $\\arctan$ in the first place, the connection will be hard to see until we get much farther.
<\/p>\n\n\n\n

Finding $y$ in terms of $x$<\/h2>\n\n\n\n

Like a lot of math problems, we want to use $x$ to find $y$. Because this is calculus, we will need to first find $dy$ in terms of $x$ and $dx$ (with the hopes of adding up all the $dy$ later, i.e. integrating). To do this, we need to <\/p>\n\n\n\n

  1. Understand what is and what isn’t given in the drawing, and use a bit of algebra to find some basic, useful relations among its parts.<\/li>
  2. Make one big assumption using infinitesimals and then leverage this into some geometry that lets us actually get the equation we want relating $dy$ to $x$ and $dx$.<\/li><\/ol>\n\n\n\n

    1. Reading the Drawing<\/h3>\n\n\n\n

    Sooooo, forgetting about $\\arctan{x}$ for the moment and just thinking about finding out $y$ in terms of $x$, we need to identify some other parts of the drawing and understand their relations. The goal is to produce an equation for $dy$ in terms of $x$ and $dx$. If we get this, we can use the magical ability of Calculus to add up infinitesimals to find $y$. Remember, we want to know how long the curve is.<\/p>\n\n\n\n

    What can we see in the picture?<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n
    • There are two circles, one of radius 1 (little) and one of radius $\\sqrt{1+x^2}$ (big). Note that the big radius isn’t labeled by Stillwell directly.<\/em><\/li>
    • $dy$ is a little bit of arc length on the little circle.<\/li>
    • $BA’$ is a little bit of arc along the big circle. <\/li>
    • $BA’$ corresponds to $dy$ because it is the part of the circle marked out by the same two rays that mark out $dy$. The length of $BA’$ is thus related to the length of $dy$. Specifically, since the ratio of the radii of the two circles is $1:\\sqrt{1+x^2}$, that is also the ratio of the arcs. $BA’ = \\sqrt{1+x^2}dy$.<\/li><\/ul>\n\n\n\n

      To help you remember this last fact, I’m going to write it again:<\/p>\n\n\n\n

      $$BA’ = \\sqrt{1+x^2}dy.$$<\/p>\n\n\n\n

      Notice, this isn’t labeled in Stillwell’s drawing, and what he does label, $dx=dy(1+x^2)$\u2014we don’t know that yet<\/em>. Another reason the work is hard to follow.<\/p>\n\n\n\n

      2. Similar triangles give us $dx=dy(1+x^2)$<\/h3>\n\n\n\n

      Next in our quest to find $y$ in terms of $x$, which we will do by finding $dy$ in terms of $x$ and $dx$, we make an assumption using infinitesimals: $\\angle OBA = \\angle OB’A$.<\/p>\n\n\n\n

      \"\"
      $\\angle OBA = \\angle OB’A$, but why<\/em>?<\/figcaption><\/figure>\n\n\n\n

      This is a bit hard to swallow. You can see why the difference between them should be small (why?<\/em>), but it’s definitely not zero. Stillwell claims that the difference, being infinitesimal, can be called zero.<\/p>\n\n\n\n

      If you believe $\\angle OBA = \\angle OB’A$, then also note that $\\angle BAO$ and $\\angle B’A’B$ are both right angles and thus equal to one another. These two facts together tell us that the remains two angles of each triangle are equal (since all three must add up to two right angles), and so<\/p>\n\n\n\n

      $\\triangle B’A’B$ is similar to $\\triangle BAO$.<\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      This means corresponding sides of these two triangles form equal ratios. These ratios are what finally enable us to write an equation for $dy$ in terms of $x$ and $dx$. In particular,<\/p>\n\n\n\n

      $$\\frac{B’B}{BA’}=\\frac{BO}{AO}.$$<\/p>\n\n\n\n

      Notice, this isn’t labeled in Stillwell’s drawing, and what he does label, $dx=dy(1+x^2)$\u2014we don’t know that yet<\/em>. Another reason the work is hard to follow.<\/p>\n\n\n\n

      2. Similar triangles give us $dx=dy(1+x^2)$<\/h3>\n\n\n\n

      Next in our quest to find $y$ in terms of $x$, which we will do by finding $dy$ in terms of $x$ and $dx$, we make an assumption using infinitesimals: $\\angle OBA = \\angle OB’A$.<\/p>\n\n\n\n

      \"\"
      $\\angle OBA = \\angle OB’A$, but why<\/em>?<\/figcaption><\/figure>\n\n\n\n

      This is a bit hard to swallow. You can see why the difference between them should be small (why?<\/em>), but it’s definitely not zero. Stillwell claims that the difference, being infinitesimal, can be called zero.<\/p>\n\n\n\n

      If you believe $\\angle OBA = \\angle OB’A$, then also note that $\\angle BAO$ and $\\angle B’A’B$ are both right angles and thus equal to one another. These two facts together tell us that the remains two angles of each triangle are equal (since all three must add up to two right angles), and so<\/p>\n\n\n\n

      $\\triangle B’A’B$ is similar to $\\triangle BAO$.<\/p>\n\n\n\n

      \"\"<\/figure>\n\n\n\n

      This means corresponding sides of these two triangles form equal ratios. These ratios are what finally enable us to write an equation for $dy$ in terms of $x$ and $dx$. In particular,<\/p>\n\n\n\n

      $$\\frac{B’B}{BA’}= \\frac{BO}{AO}.$$<\/p>\n\n\n\n

      We also know $B’B=dx$, $BA’=\\sqrt{1+x^2}dy$, $BO=\\sqrt{1+x^2}$, and $AO=1$. By inserting these values into the equation above, we find out:<\/p>\n\n\n\n

      $$dy=\\frac{1}{1+x^2}dx.$$<\/p>\n\n\n\n

      We have thus completed our first goal. Having thus succeeded we are immediately depressed. Why was it so great to come up with this equation in the first place?<\/p>\n\n\n\n

      It had something to do with measuring arc length of a circle. Generically, we can add up all the little bits of $dy$ to get the full length of $y$. In symbols:<\/p>\n\n\n\n

      $$\\text{arc length} = y = \\int{dy} = \\int{\\frac{1}{1+x^2}dx}.$$<\/p>\n\n\n\n

      Second Goal: Actually Computing $\\pi$<\/h2>\n\n\n\n

      If we want to make the general idea of adding up little bits of $dy$ specific, we need to know<\/p>\n\n\n\n

      1. How much arc length we want to compute.<\/li>
      2. Where the specific numbers come in to compute the general relation above.<\/li>
      3. How to turn a general idea of adding up little bits<\/em> into an actual computation.<\/li><\/ol>\n\n\n\n

        1. How much arc length?<\/h3>\n\n\n\n

        Really, this is up to us. But in the book, a really specific amount is chosen: $\\frac{1}{8}$ of a circle. Recall, the section is about the computation of $\\pi$. We already know the circumference of a circle (and its fractions) in terms of $\\pi$. All the way around is $2\\pi$. $\\frac{1}{8}$ of the circle thus has arc length $\\frac{\\pi}{4}$.<\/p>\n\n\n\n

        Return of the Sidebar\u2014 arctan Revealed<\/strong>

        The insight that $2\\pi$ is all the way around the circle will help us see that in fact $\\theta = y$ in the sidebar above, even though they are defined differently. That the measure of the central angle sweeping out to make a circle (i.e. $\\theta$) is the same as the measure of the arc length of the circle(i.e $y$, when the radius is one) is a primary advantage of using radians for measuring angles. This makes me feel better about radians generally, but doesn’t help much with the problem at hand. Were I his editor, I would ask him to remove mention of $\\arctan$ altogether as it is a distraction.<\/p>\n\n\n\n

        This choice also tells us what $x$ varies between to give us $y$. Look at Figure 4.16 and think about $x$ as a growing quantity. $x$ starts at $0$ and grows upwards. How big does $x$ need to grow to produce $\\frac{1}{8}$ of a circle? The answer is $1$ (why?<\/em>). To compute $y$ (and thus $\\pi$), we need to add up the little bits of arc length from $x=0$ to $x=1$.
        <\/p>\n\n\n\n

        2. How do the specific values for $x$ allow us to compute arc length?<\/h3>\n\n\n\n

        The easy way to see this is to refer to the Fundamental Theorem of Calculus<\/em>. One of its claims is a method for computing integrals. Suppose you have a function $f(x)$ and you can compute its anti-derivative $F(x)$ (i.e. you can find $F(x)$ so that $\\frac{dF}{dx} = f(x)$). Then if you want to add up all the little bits<\/em> $f(x)dx$ from $x=a$ to $x=b$, you simply compute $F(b)-F(a)$. In symbols,<\/p>\n\n\n\n

        $$\\int_{a}^{b}{f(x)dx} = F(b) – F(a).$$<\/p>\n\n\n\n

        For our problem we thus want to compute<\/p>\n\n\n\n

        $$\\int_{0}^{1}{\\frac{1}{1+x^2}dx} = F(1)-F(0).$$<\/p>\n\n\n\n

        But if $f(x) = \\frac{1}{1+x^2}$, then how do we find $F(x)$?<\/p>\n\n\n\n

        3. How do we actually compute arc length?<\/h3>\n\n\n\n

        To find $F(x)$ in our case, we will need to see $f(x) = \\frac{1}{1+x^2}$ as an infinite polynomial. Earlier in the chapter, Stillwell brings up the geometric series<\/em>:<\/p>\n\n\n\n

        $$1+x+x^2+x^3+ \\ldots$$<\/p>\n\n\n\n

        and shows that its sum is $\\frac{1}{1-x}$. If we can think of $f(x)$ as having the same form as $\\frac{1}{1-x}$, then we can alternatively write it as an infinite polynomial. We just need to squint and see $(-x^2)$ in the place of the usual $x$. Hence, we can write<\/p>\n\n\n\n

        $$f(x) = \\frac{1}{1+x^2} = \\frac{1}{1-(-x^2)} = 1+(-x^2)+(-x^2)^2+(-x^2)^3+\\ldots = 1-x^2+x^4-x^6 +\\ldots$$<\/p>\n\n\n\n

        Although writing our function as an infinite polynomial complicates our understanding of it, from the perspective of integration, this makes things very simple indeed.<\/p>\n\n\n\n

        $$\\int_{0}^{1}{\\frac{1}{1+x^2}dx} = \\int_{0}^{1}{(1-x^2+x^4-x^6+\\ldots)dx} = \\int_{0}^{1}{dx}-\\int_{0}^{1}{x^2dx}+\\int_{0}^{1}{x^4dx}-\\int_{0}^{1}{x^6dx}+\\ldots.$$<\/p>\n\n\n\n

        And we know, also from earlier in the chapter, that $\\int{x^ndx}=\\frac{1}{n}x^{n+1}$. So <\/p>\n\n\n\n

        $$F(x) = x-\\frac{1}{3}x^3+\\frac{1}{5}x^5-\\frac{1}{7}x^7+\\ldots.$$<\/p>\n\n\n\n

        Finally, we see<\/p>\n\n\n\n

        $$\\frac{\\pi}{4} = F(1)-F(0)=1-\\frac{1}{3}+\\frac{1}{5}-\\frac{1}{7}+\\ldots$$<\/p>\n","protected":false},"excerpt":{"rendered":"

        Problem Unpack Stillwell’s argument about the arc length of the circle starting on p. 96. What is an arctan? Where are $y$ and $x$? There are a few things about the way this is written that makes it harder for a reader to figure out what’s going on. First, $x$ and $y$ are not coordinate … Continue reading “The Arc Length Problem”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-151","page","type-page","status-publish","hentry"],"jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/PbpbST-2r","_links":{"self":[{"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/pages\/151","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/comments?post=151"}],"version-history":[{"count":29,"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/pages\/151\/revisions"}],"predecessor-version":[{"id":184,"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/pages\/151\/revisions\/184"}],"wp:attachment":[{"href":"https:\/\/mathematicalimpossibilities.localgameslab.org\/2018\/wp-json\/wp\/v2\/media?parent=151"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}