STEPS Building a Path with Steps and Pavement BRIEF ABSTRACT This exercise uses the notions of arc length and slope to determine the cost of building a path up a slope that has steps when it's steep and is paved otherwise. GENERAL INFORMATION FileName: Steps Full title: Building a Path with Steps and Pavement Last Update: 5/30/96 Developers: Susan Clements, Mathematics Department, South Vigo High School, Terre Haute, IN 47802 USA Email: src@galileo.vigoco.k12.in.us Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Contact: Aaron 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 straight four-foot wide path going up a hill must be paved to a point and then continue with steps. The paving can only be done on portions of the path that have a slope of magnitude less than 1/3. The rest of the hill must be furnished with steps. Each step will rise exactly 8 inches high. Surveyors have gathered the following elevation data along the path. In[2]:= PathData = {{0,50},{20,47},{40,43},{60,38},{80,32}, {100,25},{120,17},{140,8},{160,6},{180,2},{200,0}}; In[8]:= TableForm[ PathData, TableHeadings -> {None,{"Dist from top of hill [feet]", "hill height [feet]"}}, TableAlignments -> Center] Out[8]//TableForm= Dist from top of hill [feet] hill height [feet] 0 50 20 47 40 43 60 38 80 32 100 25 120 17 140 8 160 6 180 2 200 0 In[10]:= p1 = ListPlot[PathData, PlotStyle -> PointSize[.02]] Out[10]= -Graphics- All relationships are in feet. For example, the ordered pair (60,38) represents a horizontal distance of 60 feet from the top of the hill, and 38 vertical feet from the bottom. 1) Fit the hill side data to a 3rd degree polynomial of the form y = A x^3 + B x^2 + C x + D. Does this offer a reasonable fit? You'll use this path equation to answer the rest of the questions. 2) How many steps are necessary, and what are their individual lengths, i.e. tread lengths? (You'll have to build a table since each step will likely have a unique tread length.) 3) How long is the paved section of the path? (Be sure to give the true length -- not the horizontal length.) 4) The cost of paving is $2.00 per square foot and the cost of putting in steps is $2.50 per square foot (horizontal area only). How much will the path cost to build? KEYWORDS Curve fitting, arc length, step functions, slope. TEACHER NOTES ISSUES RELATED TO THE PROBLEM In determining step length, the thought behind the process is that the length is going to change as the slope of the hill changes. The step length may therefore be calculated based on the derivative of the function at different points along the path. The student should be able to determine the point at which steps are to start and from this point determine the width of each individual step. There should be some discussion of the meaning of the term "length of tread", referring to the run of the step. This should not (but may) be confused with the width of the path. Prerequisites The first part of the problem, step lengths, may be completed with a knowledge of derivatives. The section concerning the length of the path which must be paved requires the use of integration to determine the arc length. Time allotment - time management Students, working in groups, should be able to complete the work in two hours. The first hour will be more of a discussion of what is necessary to complete the problem and the second will be used to solve it. Expectations Future payoffs This problem relates the use of the derivative to a practical problem involving changes in the slope of landforms. POSSIBLE SOLUTION(S) 1) Fit the hill side data to a 3rd degree polynomial of the form y = A x^3 + B x^2 + C x + D. Does this offer a reasonable fit? You'll use this path equation to answer the rest of the questions. In[11]:= Path[x_] = Fit[PathData, {1, x, x^2, x^3}, x] Out[11]= 2 3 49.4685 - 0.0274281 x - 0.00344406 x + 0.0000117521 x In[14]:= p2 = Plot[Path[x], {x, 0, 200}, DisplayFunction -> Identity] Out[14]= -Graphics- In[15]:= Show[p1, p2, DisplayFunction -> $DisplayFunction] Out[15]= -Graphics- By visualization, the fit looks pretty good! 2) How many steps are necessary, and what are their individual lengths, i.e. tread lengths? (You'll have to build a table since each step will likely have a unique tread length.) First we determine the sections to be stepped or paved. In[17]:= Bounds = Solve[Path'[x] == -1/3, x] Out[17]= {{x -> 68.2587}, {x -> 127.113}} Utilizing the graph in Problem 1 and the information we just got, it is clear that for x in the interval: 68.2587 < x < 127.113, the path is stepped; and otherwise, the path is paved. In[20]:= x1 = x /. Bounds[[1]] x2 = x /. Bounds[[2]] Out[20]= 68.2587 Out[21]= 127.113 Next, we determine the step lengths. In[26]:= sum = 0; x = x1; slps = {}; xs = {}; ys = {}; stlens = {}; lensums = {}; While[x < x2, {len = -2/(3 Path'[x]), sum = sum + len, AppendTo[slps, Path'[x]], AppendTo[xs, x], AppendTo[ys, Path[x]], AppendTo[stlens, len], AppendTo[lensums, sum], x = x + len}] In[30]:= StepData = Table[{slps[[i]], xs[[i]], ys[[i]], stlens[[i]], lensums[[i]]}, {i, 1, Length[xs]}]; In[35]:= TableForm[StepData, TableHeadings -> {None, {"Slope", "x", "y", "Step Length [ft]", "Cumm Length [ft]"}}, TableAlignments -> Center] Out[35]//TableForm= Slope x y Step Length [ft] Cumm Length [ft] -0.333333 68.2587 35.2872 2. 2. -0.337342 70.2587 34.6165 1.97623 3.97623 -0.341027 72.235 33.9461 1.95488 5.93111 -0.3444 74.1899 33.2761 1.93573 7.86685 -0.347475 76.1256 32.6064 1.9186 9.78545 -0.350262 78.0442 31.937 1.90334 11.6888 -0.352771 79.9475 31.2679 1.8898 13.5786 -0.355008 81.8373 30.5991 1.87789 15.4565 -0.356983 83.7152 29.9306 1.8675 17.324 -0.358699 85.5827 29.2622 1.85857 19.1826 -0.360164 87.4413 28.5942 1.85101 21.0336 -0.36138 89.2923 27.9263 1.84478 22.8783 -0.362352 91.1371 27.2587 1.83983 24.7182 -0.363082 92.9769 26.5914 1.83613 26.5543 -0.363573 94.813 25.9242 1.83365 28.388 -0.363826 96.6467 25.2573 1.83238 30.2203 -0.363842 98.4791 24.5906 1.8323 32.0526 -0.363621 100.311 23.9241 1.83341 33.886 -0.363163 102.145 23.2578 1.83572 35.7218 -0.362467 103.981 22.5917 1.83925 37.561 -0.361531 105.82 21.9259 1.84401 39.405 -0.360354 107.664 21.2603 1.85003 41.2551 -0.358932 109.514 20.5949 1.85736 43.1124 -0.357261 111.371 19.9297 1.86605 44.9785 -0.355338 113.237 19.2648 1.87615 46.8546 -0.353156 115.113 18.6001 1.88774 48.7424 -0.350711 117.001 17.9358 1.9009 50.6433 -0.347994 118.902 17.2716 1.91574 52.559 -0.344999 120.818 16.6078 1.93237 54.4914 -0.341716 122.75 15.9442 1.95094 56.4423 -0.338133 124.701 15.281 1.97161 58.4139 -0.334241 126.673 14.6182 1.99457 60.4085 We see from the table that the total of all stair lengths added together is (in feet) In[39]:= StepLength = lensums[[Length[lensums]]] Out[39]= 60.4085 3) How long is the paved section of the path? (Be sure to give the true length -- not the horizontal length.) In[40]:= PaveLength = NIntegrate[Sqrt[1+(Path'[x])^2], {x, 0, x1}] + NIntegrate[Sqrt[1+(Path'[x])^2], {x, x2, 200}] Out[40]= 144.572 4) The cost of paving is $2.00 per square foot and the cost of putting in steps is $2.50 per square foot (horizontal area only). How much will the path cost to build? In[41]:= PaveCost = 4 2 PaveLength Out[41]= 1156.58 In[42]:= StepCost = 4 2.5 StepLength Out[42]= 604.085 In[43]:= Cost = PaveCost + StepCost Out[43]= 1760.66 So, the total cost is $1760.66. ISSUES IN SOLUTION If there is a hill in close proximity to the student, they may benefit from a "field trip" to analyze the situation. This problem was developed after observing a path along a hill at Rose-Hulman Institute of Technology that was partially paved and partially steps. The steps were of varying lengths.