(*^ ::[ Information = "This is a Mathematica Notebook file. It contains ASCII text, and can be transferred by email, ftp, or other text-file transfer utility. It should be read or edited using a copy of Mathematica or MathReader. If you received this as email, use your mail application or copy/paste to save everything from the line containing (*^ down to the line containing ^*) into a plain text file. On some systems you may have to give the file a name ending with ".ma" to allow Mathematica to recognize it as a Notebook. The line below identifies what version of Mathematica created this file, but it can be opened using any other version as well."; FrontEndVersion = "NeXT Mathematica Notebook Front End Version 2.2"; NeXTStandardFontEncoding; fontset = title, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e8, 24, "Times"; ; fontset = subtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, bold, L1, e6, 18, "Times"; ; fontset = subsubtitle, inactive, noPageBreakBelow, noPageBreakInGroup, nohscroll, preserveAspect, groupLikeTitle, center, M7, italic, L1, e6, 14, "Times"; ; fontset = section, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, grayBox, M22, bold, L1, a20, 18, "Times"; ; fontset = subsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, blackBox, M19, bold, L1, a15, 14, "Times"; ; fontset = subsubsection, inactive, noPageBreakBelow, nohscroll, preserveAspect, groupLikeSection, whiteBox, M18, bold, L1, a12, 12, "Times"; ; fontset = text, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = smalltext, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 10, "Times"; ; fontset = input, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeInput, M42, N23, bold, L1, 12, "Courier"; ; fontset = output, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L-5, 12, "Courier"; ; fontset = message, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = print, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = info, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeOutput, M42, N23, L1, 12, "Courier"; ; fontset = postscript, PostScript, formatAsPostScript, output, inactive, noPageBreakInGroup, nowordwrap, preserveAspect, groupLikeGraphics, M7, l34, w282, h287, L1, 12, "Courier"; ; fontset = name, inactive, noPageBreakInGroup, nohscroll, preserveAspect, M7, italic, B65535, L1, 10, "Times"; ; fontset = header, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, italic, L1, 12, "Times"; ; fontset = leftheader, 12; fontset = footer, inactive, nohscroll, noKeepOnOnePage, preserveAspect, center, M7, italic, L1, 12, "Times"; ; fontset = leftfooter, 12; fontset = help, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = clipboard, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = completions, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12, "Courier"; ; fontset = special1, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special2, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special3, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special4, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; fontset = special5, inactive, nohscroll, noKeepOnOnePage, preserveAspect, M7, L1, 12; paletteColors = 128; automaticGrouping; magnification = 150; currentKernel; ] :[font = title; inactive; preserveAspect; startGroup] CORNER :[font = subtitle; inactive; preserveAspect] Cornering at Full Speed :[font = section; inactive; preserveAspect; startGroup] BRIEF ABSTRACT :[font = subsection; inactive; preserveAspect; endGroup] Curvature is a topic often left for vector calculus, but this curvature problem was designed for calculus students who have just learned about position, velocity, and acceleration. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect; endGroup] FileName: CORNER Full title: Cornering at Full Speed Last Update: 6/3/96 Developers: Lynn Kiaer, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA 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: Aaron.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. :[font = section; inactive; Cclosed; preserveAspect; startGroup] STATEMENT OF PROBLEM :[font = subsection; inactive; preserveAspect; startGroup] Curvature Preliminaries :[font = subsubsection; inactive; preserveAspect; endGroup] How would you measure the sharpness of a turn? In accordance with our notion that a small circle has greater curvature than a large one, the curvature of a circle is defined to be 1/radius. For noncircular curves, we can find the curvature at a point by finding a circle that is tangent to the curve and whose second derivative (remember, the second derivative gives us concavity) matches that of the circle. This circle is the circle closest approximation to the curve at that point. We then say that the curve has the same curvature as that circle at that point. :[font = subsection; inactive; preserveAspect; startGroup] Your Problem :[font = subsubsection; inactive; preserveAspect; startGroup] In many cases, a vehicle's speed is restricted by the curvature of the path it travels. If the curvature of the path is 1/r, with r in feet, then the maximum speed of our vehicle, in mph, is approximated by 30(1-1/(r+1)^(1/4)). The graph below shows that curvature has a substantial effect on the maximum speed of the vehicle. ;[s] 7:0,0;123,1;124,2;131,3;132,4;216,5;217,6;328,-1; 7:1,10,8,Times,1,12,0,0,0;1,10,8,Times,3,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,3,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,3,12,0,0,0;1,10,8,Times,1,12,0,0,0; :[font = input; preserveAspect; endGroup] speed[r_] = 30(1-1/(r+1)^(1/4)); Plot[speed[r],{r, 0, 12}, PlotLabel -> "speed vs. radius of curvature", AxesLabel -> {"radius [ft]", "speed [mph]"}]; :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] Find the minimum time needed to safely travel 1000 feet down a straight road, make a 90-degree turn of radius 30 feet, and then continue straight for another 1000 feet. Assume that the maximum safe acceleration and deceleration as 2 mph per second. :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsection; inactive; preserveAspect; endGroup] Curvature, optimization. :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; preserveAspect; startGroup] ISSUES RELATED TO THE PROBLEM :[font = subsubsection; inactive; preserveAspect; endGroup] Units play a large role in this problem. For example, the maximum safe acceleration and deceleration is given with mixed time units of mph/sec. :[font = subsection; inactive; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] While curvature is often a topic introduced in vector calculus, this problem can be given once the students are familiarized with the position, velocity, and acceleration functions in differential calculus. :[font = subsection; inactive; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect; endGroup] This problem could be completed in less than a single class period or as a homework problem. :[font = subsection; inactive; preserveAspect] Expectations :[font = subsection; inactive; preserveAspect; startGroup] Future payoffs :[font = subsubsection; inactive; preserveAspect; endGroup] This problem provides an introduction to curvature without having to burden students with the decomposition of the acceleration vector. :[font = subsection; inactive; preserveAspect] Extensions :[font = subsection; inactive; preserveAspect; endGroup] References and Sources :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION(S) :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] We'll first determine the top speeds on the Straight Path and the Curved Path. :[font = subsubsection; inactive; preserveAspect; startGroup] Our speed function given in the problem tells us that the maximum attainable speed is 30 mph since the radius of curvature of a straight line is infinite. :[font = input; preserveAspect; startGroup] MaxLineMPH = Limit[speed[r], r -> Infinity] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 30 ;[o] 30 :[font = subsubsection; inactive; preserveAspect; startGroup] On the turn we can go at most 17.286 mph. :[font = input; preserveAspect; startGroup] MaxTurnMPH = N[speed[30]] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 17.28604027754935 ;[o] 17.286 :[font = subsection; inactive; preserveAspect; startGroup] We must determine the length of time that we will travel at top speed along the line. Since our maximum deceleration is 2 mph per second, we will have to spend a little more than 6 seconds slowing down from our top speed of 30 mph to our curve speed of about 17 mph. :[font = input; preserveAspect; startGroup] AccTime = ((MaxLineMPH - MaxTurnMPH) mph)/ (2 mph/sec) :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 6.356979861225325*sec ;[o] 6.35698 sec :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] To determine how far we travel in that time, it will be helpful to get everything into feet per second rather than miles per hour. :[font = input; preserveAspect] fspeed[r_] = 5280 30(1-1/(r+1)^(1/4))/3600; :[font = subsubsection; inactive; preserveAspect; startGroup] Our top speed in feet per second is :[font = input; preserveAspect; startGroup] MaxLineFPS = Limit[fspeed[r], r -> Infinity] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 44 ;[o] 44 :[font = subsubsection; inactive; preserveAspect; startGroup] Our maximum deceleration in feet per second squared is the same as the maximum acceleration: :[font = input; preserveAspect; startGroup] MaxDecFPS2 = 5280 2/3600 :[font = output; output; inactive; preserveAspect; endGroup] 44/15 ;[o] 44 -- 15 :[font = input; preserveAspect; endGroup] MaxAccFPS2 = MaxDecFPS2; :[font = subsubsection; inactive; preserveAspect; startGroup] And our speed along the quarter circle must be :[font = input; preserveAspect; startGroup] MaxTurnFPS = fspeed[30] //N :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 25.35285907373905 ;[o] 25.3529 :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Next we find the acceleration and deceleration distances (in feet). :[font = subsubsection; inactive; preserveAspect; startGroup] We'll do deceleration here. Acceleration distance will be the same by symmetry. The speed starts at 44 fps and decreases to 25.35 fps at a rate of MaxDecFPS2 fps-squared over a time period of length AccTime. The distance travelled is given by integrating the speed over this time. :[font = input; preserveAspect; startGroup] DecFt = Integrate[MaxLineFPS - MaxDecFPS2 t, {t, 0, AccTime/sec}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 220.4373642250786 ;[o] 220.437 :[font = subsubsection; inactive; preserveAspect; startGroup] To check that this works, we'll do the Acceleration distance now... :[font = input; preserveAspect; startGroup] AccFt = Integrate[MaxTurnFPS + MaxAccFPS2 t, {t, 0, AccTime/sec}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 220.4373642250786 ;[o] 220.437 :[font = subsection; inactive; preserveAspect; startGroup] The Straight Line distances (in feet) travelled at full speed are given by subtracting the distance from 1000 on each straight away on which the vehicle is either slowing down or speeding up. Then, the time to travel over these linear stretches is found by dividing by the speed. :[font = input; preserveAspect; startGroup] LineFt = 2 (1000 - AccFt) :[font = output; output; inactive; preserveAspect; endGroup] 1559.125271549843 ;[o] 1559.13 :[font = input; preserveAspect; startGroup] LineTime = (LineFt ft)/(MaxLineFPS ft/sec) :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 35.43466526249643*sec ;[o] 35.4347 sec :[font = subsection; inactive; preserveAspect; startGroup] The time (in seconds) spent on the turn. :[font = input; preserveAspect; startGroup] TurnFt = N[(Pi/2) 30] :[font = output; output; inactive; preserveAspect; endGroup] 47.1238898038469 ;[o] 47.1239 :[font = input; preserveAspect; startGroup] TurnTime = (TurnFt ft)/(MaxTurnFPS ft/sec) :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 1.858720930321372*sec ;[o] 1.85872 sec :[font = subsection; inactive; preserveAspect; startGroup] So the total time spent was :[font = input; preserveAspect; startGroup] LineTime + 2 AccTime + TurnTime :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 50.00734591526845*sec ;[o] 50.0073 sec :[font = section; inactive; preserveAspect; endGroup] ISSUES IN SOLUTION ^*)