(*^ ::[ 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, L0, 12; fontset = footer, inactive, nohscroll, noKeepOnOnePage, preserveAspect, center, M7, italic, L1, 12, "Times"; ; fontset = leftfooter, L0, 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; currentKernel; ] :[font = title; inactive; preserveAspect; startGroup] HOW DEEP IS THE WELL :[font = section; inactive; preserveAspect; startGroup] BRIEF ABSTRACT :[font = subsection; inactive; preserveAspect; endGroup] We drop a small stone in a deep well. Given the time elapsed from release until we hear the splash determine the depth of the well. :[font = section; inactive; Cclosed; preserveAspect; startGroup] GENERAL INFORMATION :[font = subsection; inactive; preserveAspect; endGroup] FileName: DEEPWELL Full title: How deep is a well -- using sound of falling body striking the bottom of thewell? Last Revision Date: 27 May 1996. Developer: Dave Horn, Mathematics, Rogers High School, Michigan City IN 46363 USA. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. 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] 1. You drop a coin into a wishing well. Ten seconds later, you hear the sound of the coin hitting the water. How deep is the well? For what part of the 10 seconds was the coin falling? How long did it take the sound to travel back up the well? What assumptions did you make in solving the problem? :[font = subsection; inactive; preserveAspect] 2. Suppose another well is just that depth at which the time it takes your coin to hit bottom is the same as the time it takes the sound of it hitting to return to the top. How deep is this well? :[font = subsection; inactive; preserveAspect; endGroup] 3. Suppose another well is just that depth at which the time it takes your coin to hit bottom is 15 times greater than the time it takes the sound of it hitting to return to the top. How deep is this well? :[font = section; inactive; Cclosed; preserveAspect; startGroup] KEYWORDS :[font = subsection; inactive; preserveAspect; endGroup] Falling body, speed of sound, time, distance. :[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 problem combines two types of motion, each describe a different family of functions. The coin in free fall conforms to a quadratic equation, while the sound moving back up the well conforms to a linear equation. If you feel your students need to to review these concepts in isolation, there are some problems and solutions in the section ISSUES IN SOLUTION. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Prerequisites :[font = subsubsection; inactive; preserveAspect; endGroup] The students need to be able to use the formula for falling body (without resistance) and the distance = rate * time formula. They will also have to know how recognize that the total time elapsed consists of the time it takes for the rock to fall added to the time it takes the sound to reach the well top. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Time allotment - time management :[font = subsubsection; inactive; preserveAspect; endGroup] This problem can be done in one class period, even half a period if you let students work in groups and help each other realize the importance of adding up the times. D = R*T is where it is at for the proper start of this problem. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Expectations :[font = subsubsection; inactive; preserveAspect; endGroup] Students will sum the times and arrive at a complex function which when set to the observed time will need CAS to solve. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Future payoffs :[font = subsubsection; inactive; preserveAspect; endGroup] Students will be able to add functions from two different descriptions. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] Extensions :[font = subsubsection; inactive; preserveAspect; endGroup] The problem suggests several descriptions of wells which should give the teacher a number of further extensions. Indeed, collecting data on a given well might help. Of course, one could get into the differential equations describing the falling coin as it meets resistance in its fall as well. :[font = subsection; inactive; preserveAspect; endGroup] References and Sources :[font = section; inactive; Cclosed; preserveAspect; startGroup] POSSIBLE SOLUTION(S) :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] 1. All this work is based on the following simplifications, assumptions and approximations: We ignore air resistance. We assume the coin is dropped from ground level and its "echo" is heard at ground level. We approximate the speed of sound as 340 m/s. :[font = subsubsection; inactive; preserveAspect] Working in the metric system, the coin in freefall down conforms to the quadratic function d(t) = -4.9t^2 + v0 t + d0. Since the coin is dropped, not thrown, the initial velocity is zero, so the linear term, v0 t, drops out. Further, assume that the coin is dropped from ground level. This makes the constant term, d0, also drop out. Thus, the position of the the coin as a function of time is given, or at least well approximated, by d(t) = -4.9t^2. Note that this position function is negative. It should be, since the coin is going down a well. However, our solution will concern distances, not position, so it is fair to drop the negative sign. Thus, the distance the coin travels in t seconds is d(t) = 4.9t^2. :[font = subsubsection; inactive; preserveAspect] The motion of the sound back up the well conforms to the linear function d = 340t , where d is the distance in meters, t is in seconds, and the 340 represents the approximate speed of sound measured in meters per second. :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] However, this problem is about time, not distance, so solve each of the above for t. Then, add these two times and set that equal to 10 seconds. :[font = input; preserveAspect; endGroup] eq1 = Sqrt[d/4.9] + d/340 == 10; :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Further, Mathematica solves the equation with the following code :[font = input; preserveAspect] t[d_]=Sqrt[d/4.9]; :[font = input; preserveAspect; startGroup] distance=d/.Solve[eq1, d][[1]] :[font = output; output; inactive; preserveAspect; endGroup] 385.2487326858492 ;[o] 385.249 :[font = input; preserveAspect; startGroup] timedown=t[distance]//N :[font = output; output; inactive; preserveAspect; endGroup] 8.86691549210044 ;[o] 8.86692 :[font = input; preserveAspect; startGroup] timeup=distance/340//N :[font = output; output; inactive; preserveAspect; endGroup; endGroup] 1.133084507899556 ;[o] 1.13308 :[font = subsubsection; inactive; preserveAspect; endGroup] Thus, the well is approximately 385 meters deep. The 10 total seconds break down as 8.87 seconds spent falling down and 1.13 seconds for the sound to return. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] 2. The solution to this question uses the same time and distance formulas. In part one, we asked where the times added up to 10. This time, we wonder when the times are equal. :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Mathematica solves this equation with the following code :[font = input; preserveAspect; startGroup] Solve[Sqrt[d/4.9] == d/340] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {{d -> 0.}, {d -> 23591.83673469388}} ;[o] {{d -> 0.}, {d -> 23591.8}} :[font = subsubsection; inactive; preserveAspect; endGroup] We get two solutions. The first is trivial. If the well has no depth, the coin takes no time to fall, and the sound takes no time to return. The second is more interesting. A depth of 23, 591 meters is the other solution. This is a very deep well indeed, almost 15 miles!!! :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] 3. The solution to this question uses the same time and distance formulas and approach used in the previous two parts. This time, we wonder when the time it takes your coin to hit bottom is 15 times greater than the time it takes the sound of it hitting to return to the top. :[font = subsubsection; inactive; Cclosed; preserveAspect; startGroup] Mathematica solves this equation with the following code :[font = input; preserveAspect; startGroup] Solve[Sqrt[d/4.9] == 15 d/340] :[font = output; output; inactive; preserveAspect; endGroup; endGroup] {{d -> 0.}, {d -> 104.8526077097506}} ;[o] {{d -> 0.}, {d -> 104.853}} :[font = subsubsection; inactive; preserveAspect; endGroup; endGroup] Thus the well is 104.853 m deep. :[font = section; inactive; Cclosed; preserveAspect; startGroup] ISSUES IN SOLUTION :[font = subsection; inactive; preserveAspect] It might be useful to review some of the concepts required in isolation before combining them in the same problem. Here are some problems to illustrate each of the two main concepts. First, the motion of sound is the easier of the two. It is the well-known Distance = Rate * Time model. Sound travels through air at approximately 340 meters per second. This figure varies with the temperature. If your students would benefit from a review of this idea, here are two problems that you could use. :[font = subsection; inactive; Cclosed; preserveAspect; startGroup] First, Tommy was a timer at a track meet. He noticed he could see the smoke from the starter's gun before he heard the sound. He knew this happened because light travels much faster than sound. :[font = subsubsection; inactive; preserveAspect] a. After the meet, Tommy asked the timer to stand at the starting line for the 100m dash and fire his gun. Tommy stood at the finish line and used his stop watch to time the interval between seeing smoke and hearing sound. If Tommy has perfect reaction time, how long will this time interval be? (Restated: How long does it take sound to travel 100 meters?) (answer: 100m/340 m/s = .294 seconds) :[font = subsubsection; inactive; preserveAspect; endGroup] b. Next Tommy asked the starter to move to a far corner of the field and to fire his gun . Tommy found the interval between smoke and sound to be 1.6 seconds. How far was Tommy standing from the starter? (answer: 340 m/s * 1.6 s = 544 m) :[font = subsection; inactive; preserveAspect; endGroup; endGroup] Second, you may want to do some free falling body problems. They are readily available in math and physics books. Remind students to check which system they are working in. Using the wrong value for the acceleration of gravity is a common error and will wreck the solution. ^*)