(*^ ::[ 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. 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Pahs Road Michigan City, IN 46360 Contact: Brian J. Winkel, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8412. Email: winkel@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] Consider a common bi-fold door. Such a door is pictured in cross section below, in both closed and partially opened position. :[font = subsection; inactive; preserveAspect] Both panels of the door are 50 cm wide. One edge of the door is fixed and pivots as the door opens. The other edge of the door is connected to a pin which slides along a track toward the fixed pivot as the door open. :[font = subsection; inactive; preserveAspect] 1. Describe the path of the sliding pin as the door moves from a closed to a fully open position. :[font = subsection; inactive; preserveAspect] 2. Describe the path of a point on the hinge as the door moves from a closed to a fully open position. :[font = subsection; inactive; preserveAspect] 3. Describe the path of the midpoint of panel A as the door moves from a closed to a fully open position. :[font = subsection; inactive; preserveAspect] 4. Describe the path of the midpoint of panel B as the door moves from a closed to a fully open position. :[font = subsection; inactive; preserveAspect] 5. Consider all points on panel A. Make a generalization concerning the path of all such points. :[font = subsection; inactive; preserveAspect] 6. Consider all points on panel B. Make a generalization concerning the path of all such points. :[font = subsection; inactive; preserveAspect; endGroup] 7. Consider the hinge and the sliding pin as points on panel B. Explain their motion in terms of your answer to problem 6. :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsubsection; inactive; preserveAspect; endGroup] Motion, trigonometry, parametric equations, conic sections. :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] ISSUES RELATED TO THE PROBLEM :[font = subsubsection; inactive; preserveAspect; endGroup] This represents a chance to model some motion problems with parametric equations. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] Standard precalculus topics--trig, conics, parametric equations :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect; endGroup] Good homework project--Have students go home and play with such a door so they can get a feel for what is happening. Some will be able to figure out the paths of points on the two panels is fundamentally different. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Expectations :[font = subsubsection; inactive; preserveAspect; endGroup] Fun! Students should quickly get a physical sense of what is being asked. If they don't have the opportunity to study an actual door, reasonable models can be built with a brief case or even a stiff folder. Half the problem (panel A) is deceptively easy. The other seems to be much harder, but order prevails over chaos without much complexity. The power of math to prove a conjecture is impressive. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Future payoffs :[font = subsubsection; inactive; preserveAspect; endGroup] Although no calculus is required, this problem is a good review of trigonometry, conic sections, and parametric equations . :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Extensions :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] This problem could be easily extended to include calculus topics by considering the velocity and acceleration of the points on the door. :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION(S) :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #1 :[font = subsubsection; inactive; preserveAspect; endGroup] This question is just to check to see if the students understand the situation. The sliding pin is stuck in a linear track. It must move along this line, or the door is broken. If one thought of the hinge being located at (0, 0) and the sliding pin at (100, 0) when the door is fully closed, then the sliding pin moves along the x-axis from 100 (fully closed) to 0 (fully opened). :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #2 :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] As the door opens, two congruent isosceles triangles are formed, each of which has a door panel as its hypotenuse. The hinge, located at (50,0) when the door is fully closed, may be considered as an endpoint of either or these panels. For this problem, think of the hinge as being the point on panel A farthest from the pivot. As the door rotates through an angle t radians from horizontal, the coordinates of the hinge are :[font = input; preserveAspect; endGroup] xhinge[t_] = 50 Cos[t]; yhinge[t_] = 50 Sin[t]; :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Here is a graph of the position of the hinge as the door opens. :[font = input; preserveAspect; endGroup] ParametricPlot[{xhinge[t],yhinge[t]},{t,0,Pi/2}, AspectRatio->Automatic] :[font = subsubsection; inactive; preserveAspect; endGroup] The path appears to be a circle, and indeed this is the case. It is easy to see this if you disregard panel B and focus on panel A as it rotates about the fixed point. The path is circular. Further, if you square both sides of the coordinate equations and then add them, you get: xhinge2 + yhinge2 = 50 cos2(t) +50 sin2(t) which quickly reduces to xhinge2 + yhinge2 = 50, a circle with a radius of 50. ;[s] 16:0,0;293,1;294,2;303,3;304,4;306,5;313,6;314,7;325,8;326,9;329,10;362,11;363,12;372,13;373,14;375,15;410,-1; 16:1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,12,9,Times,1,13,0,0,0;1,12,9,Times,33,13,0,0,0;1,12,9,Times,1,13,0,0,0;1,12,9,Times,33,13,0,0,0;1,12,9,Times,1,13,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,12,9,Times,1,13,0,0,0; :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #3 :[font = subsubsection; inactive; preserveAspect; endGroup] The motion of the midpoint of panel A, and indeed every point on panel A, is circular. The initial x-coordinate of the point when the door is closed gives the radius of circle. This point in particular moves from (25, 0) to (0, 25) as the door goes from fully closed to fully opened. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #4 :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] As panel A is rotated t radians, the midpoint of panel B moves according to the following parametric equations. These equations can be verified through right triangle trigonometry. :[font = input; Cclosed; preserveAspect; startGroup] xmidb[t_]=75*Cos[t] ymidb[t_]=25*Sin[t] ParametricPlot[{xmidb[t],ymidb[t]},{t,0,Pi/2}, AspectRatio->Automatic] :[font = output; output; inactive; preserveAspect] 75*Cos[t] ;[o] 75 Cos[t] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 25*Sin[t] ;[o] 25 Sin[t] :[font = subsubsection; inactive; preserveAspect] The midpoint of panel B begins its trip at (75, 0) [fully closed] and ends up at (50, 25) [fully opened]. Its path is elliptical. This can be verified from the coordinate equations as follows: x = 75 cos t ----> x/75 = cos t ----> x2/5625 = cos2 t y = 25 sint t ----> y/25 = sin t -----> y2/625 = sin2 t ;[s] 9:0,0;242,1;243,2;255,3;256,4;306,5;307,6;320,7;321,8;333,-1; 9:1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0; :[font = subsubsection; inactive; preserveAspect; endGroup] Adding the rightmost representation of these yields x2/5625 + y2/625 = cos2 t + sin2 t or x2/5625 + y2/625 = 1. This is the equation of an ellipse , centered at the origin, with a semimajor axis of length 75 and a semiminor axis of length 25. ;[s] 13:0,0;74,1;75,2;84,3;85,4;95,5;96,6;105,7;106,8;120,9;121,10;130,11;131,12;290,-1; 13:1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0; :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #5 :[font = subsubsection; inactive; preserveAspect; endGroup] Back on panel A, things are easier. Every point on panel A moves in a circular path. The radius of the circle is given by the x-coordinate of the point when the door is fully closed. In deriving this result, the basic trig identity cos2 t + sin2 t = 1 is used. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #6 :[font = subsubsection; inactive; preserveAspect] Every point on panel B moves in an elliptical path. For the point located at (a, 0) when the door is closed, the equation of this ellipse is :[font = subsubsection; inactive; preserveAspect; endGroup] x2/a2 + y2/(100 - a)2 = 1. The derivation of this is similar to that given in problem 4. ;[s] 9:0,0;4,1;5,2;7,3;8,4;12,5;13,6;23,7;24,8;96,-1; 9:1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0;1,10,8,Times,33,12,0,0,0;1,10,8,Times,1,12,0,0,0; :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Solution to #7 :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] In 6, we said every point on panel B moves in elliptical path. In 1, we said the sliding pin moves in a linear path. In 2, we said the hinge moves in a circular path. How can we consider the sliding pin and the hinge as points on panel B (its endpoints) and have these results be consistent? The answer is that both line segments and circles can be considered cases of degenerate ellipses. For the hinge, the circle is the ellipse with coincident foci. For the line segment, the foci coincide with the major vertices. :[font = section; inactive; preserveAspect; endGroup] ISSUES IN SOLUTION ^*)