(*^ ::[ 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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Building on this question, PIPEPLUG (or even parts of it) could be used as a term project for multi-variate calculus. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect] FileName: PIPEPLUG :[font = subsection; inactive; preserveAspect] Full title: Pipes and Valves - Using a disk to cover or plug a pipe flow :[font = subsection; inactive; preserveAspect] Last Update: 6/3/96 :[font = subsection; inactive; preserveAspect; startGroup] Developers: :[font = subsubsection; inactive; preserveAspect] Aaron Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. :[font = subsubsection; inactive; preserveAspect; endGroup] Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. :[font = subsection; inactive; preserveAspect] 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. :[font = subsection; inactive; preserveAspect; endGroup] 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] (1) Consider a pipe of fixed diameter (an open circle measured inside the pipe) along with valves (covers) large enough to close the pipe. :[font = subsubsection; inactive; preserveAspect] In particular, consider the cases of a a) rectangle valve (large enough to plug the pipe) b) circular valve of the same radius as the pipe c) circular valve of a larger radius than the pipe :[font = subsubsection; inactive; preserveAspect; endGroup] The valve (either rectangular or circular) can be slid across the opening of the pipe to slow the flow of water or stop it all together. Let the center of the pipe be the origin, and let A(x) be the cross sectional area of the pipe through which water can flow with x labeling the location of the edge of the valve as it is slides across the opening of the pipe. For each case, set up the horizontal axis along which the valve moves and label x. Then, determine A(x). :[font = subsection; inactive; preserveAspect] (2) Find the rate at which the open area of pipe changes as the distance changes (constantly) for each of the three cases from (1). What is the maximum rate of change of open pipe area as the valve moves at constant rate across for each case? :[font = subsection; inactive; preserveAspect] (3) Consider each of the three valve coverings separately and answer this question. If a valve covering is closing at a constant rate (i.e, moved along the position axis at a constant rate), which valve allows the greatest flow and which allows the least flow of water? The water will flow greatest when the open pipe area is greatest. :[font = subsection; inactive; preserveAspect; endGroup] (4) An Inverse Problem: With a given shaped valve type (do this for each of the three cases considered), find the rule of motion for the valve so that rate of change of flow through the pipe is constant. :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsection; inactive; preserveAspect; endGroup] Circle geometry, trigonometry, coordinate systems, differential equations, differentiation, integration, cross sectional area, inverse problems, pipe, flow rate, valve (disk, cover, plug). :[font = section; inactive; Cclosed; preserveAspect; startGroup] TEACHER NOTES :[font = subsection; inactive; preserveAspect; startGroup] ISSUES RELATED TO THE PROBLEM :[font = subsubsection; inactive; preserveAspect] The simple idea of the problem, a valve shutting off water through a pipe, grows into a multifaceted and open ended investigation. The majority of the problem deals with a circular cylindrical pipe with valves (of varying shapes and sizes) that move straight through the cylinder (at varying rates) to stop flow through the pipe. The entire problem can be solved by the end of an integral calculus course, although some parts are appropriate earlier. Some instructors may wish to introduce the problem early, solving different parts as appropriate. :[font = subsubsection; inactive; preserveAspect] We see this problem as a list of questions in which the student can begin to sense how a simple idea can grow into a large and complicated problem which he/she can take as far as the student wants to. We hope you will encourage your students to go beyond what is asked of them here. :[font = subsubsection; inactive; preserveAspect] As an investigation, you can expect this to take up to 10 weeks if the students are working primarily out of class, although some in class time should be given. The problem is roughly outlined as follows. :[font = subsubsection; inactive; preserveAspect] 1. The main problem: Find the flow rate as a function of valve location. Main Math Tools: definite integrals as functions with parameters Suggested Time Allotment: 1--2 weeks out of class; 1 day in class :[font = subsubsection; inactive; preserveAspect] 2. Comparison of changing flow rates. A simple extension of 1. Main Math Tools: derivatives of definite integrals with parameters Suggested Time Allotment: 1--2 days out of class :[font = subsubsection; inactive; preserveAspect] 3. Comparison of total flow rates. Another simple, yet slightly harder extension of 1. Main Math Tools: setting up definite integrals with parameters Suggested Time Allotment: 1--2 days out of class :[font = subsubsection; inactive; preserveAspect] 4. An Inverse Problem. Quite difficult, yet a good problem for strong, motivated students with a good understanding integrals and a basic knowledge of differential equations. This requires a numerical differential equation solver for a thorough study. Main Math Tools: definite integrals and differential equations (not how to solve, but how to set up) Suggested Time Allotment: 2+ weeks out of class; 1 day in class. :[font = subsubsection; inactive; preserveAspect] 5. An open-ended extension of the original problem. This is enough for a full term project. Main Math Tools: Geometry and open minds as well as the concept of the definite integral Suggested Time Allotment: 3--7 weeks out of class; 1+ days in class. :[font = subsubsection; inactive; preserveAspect] 6. Similar to the original problem, but movement is by angle instead of along a straight line. This could be an extension of \#1 or an independent problem. Main Math Tools: Polar Coordinates and Integration in Polar Coordinates. Suggested Time Allotment: 1+ weeks out of class; 1 day in class. :[font = subsubsection; inactive; preserveAspect; endGroup] 7. Another inverse problem. Extremely difficult. Honest attempts deserve high praise. Give this only to motivated students. Main Math Tools: A willingness to try many things. A thorough understanding of the preceding problems. Suggested Time Allotment: 5--10 weeks out of class :[font = subsection; inactive; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] Integration, functions of two variables, differentiation, polar coordinates. Also see above: Issues Related to the Problem :[font = subsection; inactive; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect; endGroup] See above: Issues Related to the Problem :[font = subsection; inactive; preserveAspect] Expectations :[font = subsection; inactive; preserveAspect] Future payoffs :[font = subsection; inactive; preserveAspect; startGroup] Extensions :[font = subsubsection; inactive; preserveAspect] (5) Consider all or some of the previous problems with varying shaped valves and cross sections of pipe. For example, you might consider a pipe with a equilateral triangular cross section and a rectangular valve covering. :[font = subsubsection; inactive; preserveAspect] (6) Suppose the pipe opens by rotating the plug at a point of tangency on the pipe. Set up an appropriate coordinate system to determine how the cross sectional area varies as the valve is closed (at, say, a constant rate.) :[font = subsubsection; inactive; preserveAspect; endGroup] (7) A Much Harder Inverse Problem: If the plug moves at a constant rate, find the shape of the valve covering so that rate of change of flow (assuming a constant flow rate) through the pipe is constant. :[font = subsection; inactive; preserveAspect; endGroup] References and Sources :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION(S) :[font = subsection; inactive; preserveAspect; startGroup] Solution to (1) :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Rectangular Plug Case: :[font = text; inactive; preserveAspect] Let the origin be at the center of the pipe (circle), and let the leading edge of the rectangular plug be the x position. Let Rp = side length of the plug. The cross sectional area of the pipe through which water can flow as a function of x and Rp is given by Ac1[x, Rp] below. :[font = input; preserveAspect] Ac1[x_, Rp_] = 2 Integrate[Sqrt[Rp^2 - t^2], {t, x, Rp}]; :[font = text; inactive; preserveAspect] when x is in the interval [-Rp, Rp], and obvious otherwise. :[font = input; preserveAspect; startGroup] Rpval = 1; Plot[Ac1[x, Rpval], {x, -Rpval, Rpval}, AxesLabel -> {x, area}, AxesOrigin -> Automatic] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Circular Plug (of same radius) Case: :[font = text; inactive; preserveAspect] Let the origin be at the center of the pipe (circle), and let the center of the plug mark the x position. Then, :[font = input; preserveAspect] CrossSect1 = Sqrt[Rp^2 - t^2] - Sqrt[Rp^2 - (t - x)^2]; :[font = input; preserveAspect] CrossSect2 = Sqrt[Rp^2 - t^2]; :[font = input; preserveAspect] A1[x_, Rp_] = 2 (Integrate[CrossSect1, {t, x/2, Rp + x}]); :[font = input; preserveAspect] A2[x_, Rp_] = 2 (Integrate[CrossSect2, {t, Rp + x, Rp}]); :[font = input; preserveAspect] Ac2[x_, Rp_] = A1[x, Rp] + A2[x, Rp]; :[font = text; inactive; preserveAspect] when x is in the interval [-2 Rp, 0], and obvious otherwise. :[font = input; preserveAspect; startGroup] Rpval = 1; Plot[Ac2[x, Rpval], {x, -2 Rpval, 0}, AxesLabel -> {x, area}, AxesOrigin -> Automatic] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Circular Plug (of larger radius) Case: :[font = text; inactive; preserveAspect] Let the origin be at the center of the pipe (circle), and let the center of the plug mark the x position. Define Rv to be the radius of the valve cover. In this case , finding the intersection points of the two circle is a little tougher than when we used a rectangular cover. Then, :[font = input; preserveAspect] sol = Flatten[Simplify[ Solve[(t - x)^2 + y^2 - Rv^2 == t^2 + y^2 - Rp^2, t] ]]; :[font = input; preserveAspect] tint = t /. sol[[1]]; :[font = text; inactive; preserveAspect] Since x is in [-(Rv + Rp), Rp - Rv] (which is to the left of the origin, but :[font = input; preserveAspect] CrossSect1 = Sqrt[Rp^2 - t^2] - Sqrt[Rv^2 - (t - x)^2]; :[font = input; preserveAspect] CrossSect2 = Sqrt[Rp^2 - t^2]; :[font = input; preserveAspect] A1[x_, Rp_, Rv_] = 2 (Integrate[CrossSect1, {t, tint, Rv + x}]); :[font = input; preserveAspect] A2[x_, Rp_, Rv_] = 2 (Integrate[CrossSect2, {t, Rv + x, Rp}]); :[font = input; preserveAspect] Ac3[x_, Rp_, Rv_] = A1[x, Rp, Rv] + A2[x, Rp, Rv] ; :[font = input; preserveAspect; startGroup] Rpval = 1; Rvval = 2; xmin = -(Rvval + Rpval); xmax = Rpval - Rvval; Plot[Ac3[x, Rpval, Rvval], {x, xmin, xmax}, AxesLabel -> {x, area}, AxesOrigin -> Automatic] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = subsection; inactive; preserveAspect; startGroup] Solution to (2) :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Rectangular Plug Case: :[font = text; inactive; preserveAspect] The area changes at the following rate (as a function of x) :[font = input; preserveAspect] r1[x_, Rp_] = Together[D[Ac1[x, Rp], x]]; :[font = input; preserveAspect; startGroup] Rpval = 1; Plot[r1[x, Rpval], {x, -Rpval, Rpval}, AxesLabel -> {x, r1[x]}, AxesOrigin -> Automatic] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect; endGroup] The maximum rate of change of area occurs when x= 0. :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Circular Plug (of same radius) Case: :[font = text; inactive; preserveAspect] The area changes at the following rate (as a function of x) :[font = input; preserveAspect] r2[x_, Rp_] = Together[D[Ac2[x, Rp], x]]; :[font = input; preserveAspect; startGroup] Rpval = 1; Plot[r2[x, Rpval], {x, -2 Rpval, 0}, AxesLabel -> {x, r2[x]}, AxesOrigin -> Automatic] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect; endGroup] The maximum rate of change of area occurs when x= 0. :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Circular Plug (of larger radius) Case: :[font = text; inactive; preserveAspect] The area changes at the following rate (as a function of x) :[font = input; preserveAspect] r3[x_, Rp_, Rv_] = Factor[Together[D[Ac3[x, Rp, Rv], x]]]; :[font = input; preserveAspect; startGroup] Rpval = 1; Rvval = 2; xmin = -(Rvval + Rpval); xmax = Rpval - Rvval; Plot[r3[x, Rpval, Rvval], {x, xmin, xmax}, AxesLabel -> {x, r3[x]}, AxesOrigin -> Automatic] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] The Maximum rate of change of area. :[font = input; preserveAspect] der = Factor[Together[D[r3[x, Rp, Rv], x]]]; :[font = text; inactive; preserveAspect; endGroup; endGroup] Since Rv > Rp, the derivative can only be zero if x^2 = Rv^2 - Rp^2. For example, in the case shown above with Rv = 2 and Rp = 1, x = - Sqrt[3]. :[font = subsection; inactive; preserveAspect; startGroup] Solution to (3) :[font = subsubsection; inactive; preserveAspect; startGroup] Rectangular Plug Case: :[font = input; preserveAspect] valrp[Rp_] = Integrate[Ac1[x, Rp]/(2 Rp), {x, -Rp, Rp}]; :[font = input; preserveAspect; startGroup] N[valrp[Rp] /. Rp -> 1] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 1.570796326794897 ;[o] 1.5708 :[font = subsubsection; inactive; preserveAspect; startGroup] Circular Plug (of same radius) Case: :[font = input; preserveAspect] valcp[Rp_] = Integrate[Ac2[x, Rp]/(2 Rp), {x, -2 Rp, 0}]; :[font = input; preserveAspect; startGroup] N[valcp[Rp] /. Rp -> 1] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 1.80825932025646 ;[o] 1.80826 :[font = subsubsection; inactive; preserveAspect; startGroup] Circular Plug (of larger radius) Case: :[font = input; preserveAspect] Rpval = 1; Rvval = 2; xmin = -(Rvval + Rpval); xmax = Rpval - Rvval; :[font = input; preserveAspect; startGroup] NIntegrate[Ac3[x, Rpval, Rvval]/(xmax - xmin), {x, xmin, xmax}] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup] 1.672309002505973 ;[o] 1.67231 :[font = subsection; inactive; preserveAspect; startGroup] Solution to (4) :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Rectangular Plug Case: :[font = text; inactive; preserveAspect] Let y[t] be the x position at time t. We want to find y[t] so that the area changes at a constant rate ( with respect to time.) Solving for y'[t] instead of y[t] yields a single separable nonlinear ODE which can be easily solved after separating variables. However, since the implicit solution that we obtain will do little to add to the understanding of the solution, we'll opt to make our choices for the parameters here in an example and solve the ODE numerically. We choose k = 0.1, and Rp = 1. NOTE: Due to the singularities at plus and minus Rp, we start at the middle (x = y[0] = 0) and use symmetry to determine the full rule. :[font = input; preserveAspect] Remove[rhs,der] :[font = input; preserveAspect] ans = Flatten[Solve[D[Ac1[y[t], Rp], t]==k, y'[t]]]; rhs[t_] = y'[t] /. ans /. {k -> 0.1, Rp -> 1}; :[font = input; preserveAspect] soln = Flatten[NDSolve[ {y'[t] == rhs[t], y[0] == 0}, y[t], {t, 0, 50}]]; :[font = input; preserveAspect] rate1[t_] = y[t] /. soln; :[font = input; preserveAspect; startGroup] Plot[rate1[t], {t, 0, 50}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] The derivative of the rate... :[font = input; preserveAspect] der[t_] = rate1'[t]; :[font = input; preserveAspect; startGroup] Plot[der[t], {t, 0, 15.5}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] From the derivative of the rate, that near the beginning and end (don't forget the symmetry), the movement changes the fastest, while it changes at a constant rate in the middle. :[font = text; inactive; preserveAspect] A nice check of our work... :[font = input; preserveAspect; startGroup] Plot[Ac1[rate1[t], 1], {t, 0, 15.5}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect; endGroup] A line of slope 0.1! :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Circular Plug (of same radius) Case: :[font = text; inactive; preserveAspect] We'll handle this case in precisely the same way that we covered the rectangular case. We choose k = 0.1, and Rp = 1. :[font = input; preserveAspect] Clear[rhs] :[font = input; preserveAspect] ans = Flatten[Solve[D[Ac2[y[t], Rp], t]==k, y'[t]]]; rhs[t_] = y'[t] /. ans /. {k -> 0.1, Rp -> 1}; :[font = input; preserveAspect] soln = Flatten[NDSolve[ {y'[t] == rhs[t], y[0] == 0}, y[t], {t, 0, 50}]]; :[font = input; preserveAspect] rate2[t_] = y[t] /. soln; :[font = input; preserveAspect; startGroup] Plot[rate2[t], {t, 0, 50}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] The derivative of the rate... :[font = input; preserveAspect] der = rate2'[t]; :[font = input; preserveAspect; startGroup] Plot[der, {t, 0, 31.4}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] From the derivative of the rate, that near the beginning and end (don't forget the symmetry), the movement changes the fastest, while it changes at a constant rate in the middle. :[font = text; inactive; preserveAspect] A nice check of our work... :[font = input; preserveAspect; startGroup] Plot[Ac2[rate2[t], 1], {t, 0, 31.4}] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect; endGroup] A line of slope 0.1! :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Circular Plug (of larger radius) Case: :[font = text; inactive; preserveAspect] We must handle this case a bit differently due to the lack of symmetry in the solution. We choose k = 1, and Rp = 1, Rv = 2. :[font = input; preserveAspect] kval = 1; Rpval = 1; Rvval = 2; :[font = input; preserveAspect] Clear[rhs] :[font = input; preserveAspect] ans = Flatten[Solve[ D[Ac3[y[t], Rp, Rv], t]==k, y'[t]]]; rhs[t_] = Together[y'[t] /. ans /. {k -> kval, Rp -> Rpval, Rv -> Rvval}]; :[font = text; inactive; preserveAspect] Forward in time from initial conditions :[font = input; preserveAspect] soln = Flatten[NDSolve[ {y'[t] == rhs[t], y[0] == -Sqrt[Rvval^2 - Rpval^2]}, y[t], {t, 0, 50}]]; :[font = input; preserveAspect] rate3right[t_] = y[t] /. soln; :[font = input; preserveAspect] plotright = Plot[rate3right[t], {t, 0, 50}, DisplayFunction -> Identity]; :[font = text; inactive; preserveAspect] Backward in time from inital conditions :[font = input; preserveAspect] soln = Flatten[NDSolve[ {y'[t] == rhs[t], y[0] == -Sqrt[Rvval^2 - Rpval^2]}, y[t], {t, -10, 0}]]; :[font = input; preserveAspect] rate3left[t_] = y[t] /. soln; :[font = input; preserveAspect] plotleft = Plot[rate3left[t], {t, -10, 0}, DisplayFunction -> Identity]; :[font = text; inactive; preserveAspect] Full Solution Plot... :[font = input; preserveAspect; startGroup] Show[plotleft, plotright, DisplayFunction -> $DisplayFunction]; :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] The derivative of the rate... :[font = input; preserveAspect] derright = rate3right'[t]; derleft = rate3left'[t]; :[font = input; preserveAspect; startGroup] p1 = Plot[derright, {t, 0, 50}, DisplayFunction -> Identity]; p2 = Plot[derleft, {t, -50, 0}, DisplayFunction -> Identity]; Show[p1, p2, DisplayFunction -> $DisplayFunction] :[font = output; output; inactive; preserveAspect; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = text; inactive; preserveAspect] From the derivative of the rate, that near the beginning and end, the movement changes the fastest, while it changes at a constant rate in the middle. It should be no surprise that the slowest movement should occur when time is zero, since we chose the initial point in just this way. :[font = text; inactive; preserveAspect] A nice check of our work... :[font = input; preserveAspect; startGroup] p3 = Plot[Ac3[rate3right[t], 1, 2], {t, 0, 19}, DisplayFunction -> Identity]; p4 = Plot[Ac3[rate3left[t], 1, 2], {t, -45, 0}, DisplayFunction -> Identity]; Show[p3, p4, DisplayFunction -> $DisplayFunction] :[font = output; output; inactive; preserveAspect; endGroup; endGroup; endGroup; endGroup] Graphics["<<>>"] ;[o] -Graphics- :[font = section; inactive; Cclosed; preserveAspect; startGroup] ISSUES IN SOLUTION :[font = subsection; inactive; preserveAspect] Most of the integration is quite difficult and messy to do by hand (e.g., 1, part (a) requires trigonometric substitution), but can be handled nicely with a computer algebra system. The main emphasis should be on setting up the definite integrals, exploring the solutions' dependence on the parameters, and doing comparisons with different geometries. :[font = subsection; inactive; preserveAspect] Most students will guess that the derivative is the appropriate tool to use here. The instructor should emphasize why this makes sense, and encourage the students to play with parameters and consider other geometric shapes or other extensions not necessarily mentioned here. :[font = subsection; inactive; preserveAspect] Many students will not recognize to integrate here. Mentioning the concept of average area may help (this is what is placed in the solution.) :[font = subsection; inactive; preserveAspect; endGroup; endGroup] You may wish to give hints on setting up the differential equation. It is also important to get the students to verify that their work is correct as was demonstrated in the solution. Some students who are good technicians may be tempted to solve the differential equations outright. This is fine, as long as they become aware of the difficulties in studying the graphs of the solutions. (Actually, plotting the inverse of the solution is easy after solving the DEs explicitly.) Those who set out to solve explicitly should be encouraged to study the effects of varying the parameters. ^*)