RAILROAD The Railroad Track Problem BRIEF ABSTRACT When a section of track buckles, the effects can be quite dramatic. Three models for such a scenario are considered. This can be used at the beginning of a calculus course. GENERAL INFORMATION FileName: RAILROAD Full title: The Railroad Track Problem Last Update: 6/5/96 Developer: Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Klebanoff@rose-hulman.edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM A Circular Bend: A railroad rail which is 1 mile long is firmly fixed at both ends. Unfortunately, the rail was installed during cool weather, and steel expands when it gets warm. Suppose that on a hot summer day the rail heats up by 30 degrees and expands one foot in length (a realistic value). The ends are fixed and the rail has no room to expand; let's suppose it buckles upward. Assuming the shape is that of a semi-circle, how far above ground would the middle of the rail be? A Triangular Bend: Assuming the shape is that of two lines meeting above the 1/2 mile mark (forming a triangular shape), how far above ground would the middle of the rail be? A Bell Shaped bend: Assuming the shape is bell shaped, how far above ground would the middle of the rail be? To be more precise, assume the curve is smooth, has zero slope at the end points, is symmetric about the center of the mile, and is positive. KEYWORDS Binary Search, geometry, trigonometry, Newton's Method, arc length, integration, error tolerance. TEACHER NOTES ISSUES RELATED TO THE PROBLEM This problem can be introduced at the precalculus level. Only geometry and trigonometry are necessary to do the first two problems, and the last can be done with the aide of a computer or graphing calculator. The last problem by itself is a nice application after discussing arc lengths in integral calculus, and in most cases would be inappropriate before then. Prerequisites Circular Case: With a computer algebra system to find roots of equations, high school algebra students with a geometry background can solve this problem. Without the computer algebra system, this problem is best left until after Newton's method is discussed in differential calculus. Triangular Case: Trigonometry. Bell Shaped Case: Arc length from an integral calculus course. Time allotment - time management Circular Case: 1 day in class or 1 week out of class Triangular Case: 10 minutes tops! Bell Shaped Case: 1-3 weeks out of class Expectations Students may have a difficult time determining a function that will work in the bell shaped case. Future payoffs These problems reinforce the need for numerical methods in solving more realistic problems. Expect your students to become more comfortable with one or more of these numerical methods. Extensions Try other shapes. Given the stretched length, can you determine how hot it was (need physical properties of the tracks as well as the bend shape.) References and Sources POSSIBLE SOLUTION(S) The circular case Let the origin be the center of the circle which contains the railroad track's arc. The circle has some radius which we will denote by the variable, r. Draw a secant line in the circle parallel to and above the x-axis to denote the railroad track. It's length is m = 5280 feet. Finally, label the angle, theta, between the x-axis and the radial segment from the origin to the right-hand edge of the secant line. Then 2 r cos(theta) = m (original track length) (Pi - 2 theta) r = m + 1 (circular buckled track length) (See the figure below, which has theta = 30 degrees only for sake of making an illustration.) In[44]:= x0 = Sqrt[3]/2; y0 = 1/2; In[45]:= TrackArc = ImplicitPlot[x^2 + y^2 == 1, {x, -x0, x0}, {y, y0, 1}, DisplayFunction -> Identity]; In[46]:= TrackFlat = Graphics[{AbsoluteDashing[{6, 3}], Line[{{-x0, y0}, {x0, y0}}]}]; In[47]:= Angle0 = Graphics[{AbsoluteDashing[{1, 4}], Line[{{0,0}, {1, 0}}]}]; In[48]:= Angle60 = Graphics[{AbsoluteDashing[{1, 4}], Line[{{0,0}, {x0, y0}}]}]; In[49]:= Angle90 = Graphics[{AbsoluteDashing[{1, 4}], Line[{{0,0}, {0, 1}}]}]; In[50]:= Angle120 = Graphics[{AbsoluteDashing[{1, 4}], Line[{{0,0}, {-x0, y0}}]}]; In[64]:= LengthLabels = Graphics[ {Text[FontForm["m + 1", {"Times-Italic", 18}], {0, 1.1}], Text[FontForm["m/2", {"Times-Italic", 18}], {0.45, 0.6}], Text[FontForm["h", {"Times-Italic", 18}], {-0.1, 0.7}], Text[FontForm["r", {"Times-Italic", 18}], {0.45, 0.3}], Text[FontForm["q", {"Symbol", 18}], {0.3, 0.07}]} ]; In[65]:= Show[TrackArc, TrackFlat, Angle0, Angle60, Angle90, Angle120, LengthLabels, Axes -> None, PlotRange -> All, AspectRatio -> Automatic, DisplayFunction -> $DisplayFunction] Out[65]= -Graphics- Since we have two equations in two unknowns we can solve for r and theta. Of course, we really want to know h, the height of the arc above the secant. This is simply h = r - r sin (theta) and thus can be immediately computed once we know r and theta. Rather than waste computation time on the computer, we immediately solve for r in the second equation and replace that in the first equation to have 1 equation with the unknown theta, m(Pi - 2 theta)/(m+1) = 2 cos(theta). Since this equation cannot be solved analytically, we must use the computer to estimate the solution. A good way to estimate the solution is to graph the left and right-hand sides of the equation above and see where they cross. In[66]:= m = 5280; In[67]:= Plot[{m(Pi - 2 theta)/(m+1), 2 Cos[theta]}, {theta, -1, 7}] Out[67]= -Graphics- It is clear from the graph that theta is about 1.5. We use that as a starting point for the Mathematica numerical solving command, FindRoot. In[69]:= sol = FindRoot[m(Pi - 2 theta)/(m+1) == 2 Cos[theta], {theta, 1.5}] Out[69]= {theta -> 1.53703} In[70]:= angle = theta /. sol[[1]]; In[71]:= r = (m + 1)/(Pi - 2 angle); h = N[r - r Sin[angle]] Out[72]= 44.5728 Thus, the railroad track is almost 45 feet off the ground at the center of the 1 mile strip! The triangular case Use the Pythagorean Theorem! In[73]:= height = N[Sqrt[(5281/2)^2 - (5280/2)^2]] Out[73]= 51.3834 The bell shaped case In[74]:= Clear[a, f, fun, x] Let f(x) = 1 + cos(2 Pi x) In[75]:= f[x_] = 1 + Cos[2 Pi x]; Plot[f[x], {x, -0.5, 0.5}] Out[76]= -Graphics- Note that f is satisfies the required conditions except for its length. We can find a function, g(x), which satisfies all conditions by scaling f(x) by a constant, i.e. g(x) = a f(x). We need the arc length of g(x) to be 5281. This is a bit tricky, since g(x) cannot be integrated symbolically, but instead must be done numerically, thus requiring a value for the parameter a. We'll use a binary search method to find a: And now we begin a binary search for a with the function... In[77]:= fun[a_] := N[NIntegrate[Sqrt[1 + a^2 (f'[x])^2], {x, -0.5, 0.5}] - (5281/5280)]; Define absolute error tolerance and max number of iterates. In[78]:= TOL = 0.000001; ITER = 20; Guess 2 points In[80]:= left = 0; right = 2; Evaluate the function to determine whether a sign change exists between 2 points thus indicating a root. In[82]:= atleft = fun[left]; In[83]:= atright = fun[right]; If there is no sign change over the interval, warn user that no root exists on the interval. In[84]:= If[atleft atright > 0, Print["Try Again -- Root not in this interval"]] Until the error tolerance is attained or its taking too many iterates, find the midpoint of the interval, compute the function at the left, mid, and right points; compare funleft funmid with funmid funright; If funleft funmid < 0, then amid becomes aright; otherwise amid becomes aleft. In[85]:= Do[ mid = 0.5 (left + right); atleft = fun[left]; atmid = fun[mid]; atright = fun[right]; leftprod = atleft atmid; rightprod = atmid atright; If[leftprod < 0, right = mid, left = mid]; If[Abs[left - right] < TOL, Break[]], {i, 1, ITER}] The root. In[86]:= root = 0.5(left + right) Out[86]= 0.00438023 Height is now found (in miles) In[87]:= height = a f[x] /. {a -> root, x -> 0} Out[87]= 0.00876045 Or in feet... In[88]:= 5280*height Out[88]= 46.2552 Thus this is higher than the semi-circular case which gave almost 45 ft. ISSUES IN SOLUTION The Circular Case: Some will find this to be a difficult problem. Give the students a week or so to work on this in groups and at least a day in class. The Triangular Case: This should be very easy for students and is not a sufficient problem on its own. The Bell Shaped Case: Students may find this quite difficult unless they are accustomed to having messy problems which are not always immediately solvable with a computer algebra system. Be prepared to remind the students of numerical methods, and help to convince them that they have set up the problem properly. There are many possible solutions to this problem since the form of the function is not specified. It will be quite difficult for most students to come up with any family of functions to consider, much less reasonable ones. Students who are used to studying families of functions (dependent on parameters) will more readily discover an appropriate function to work with here.