GEOTIME BRIEF ABSTRACT We determine the flow of soundwaves in an underground sonic boom through various media and we attempt to determine the shape of the underlying region given timings on soundwave propagation. GENERAL INFORMATION FileName: GEOTIME Full title: Using Sound to Determine Subsurface Geological Structure Last Revision Date: 3 June 1996. Developer: 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. 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. STATEMENT OF PROBLEM The speed of sound in various media can be used to confirm geological models near the surface of the earth. Consider the following diagram (see code to produce the figure below) which is the result of geological probing in western Kansas. Distances are in meters. We see three layers of material: clay/rock from the surface down to iron and on down to silver. The conjecture is that each layer is separated by a line (we keep the problem in two dimensions and ask the reader to consider the complexities of a three dimensional surface and interface) or function. y = f(x) separates iron and silver while y = g(x) separates clay/rock and iron. The speed of sound in silver is 2640 m/sec, in clay/rock it is 3480 m/sec, and in iron it is 5130 m/sec. A blast is set off at point O some 600 m below the surface. We set up an axes system and coordinatize our data. Point O is (0, -600) m. The conjecture is that silver is in the region below the function y = f(x) = 100(-6 + .5 + .3x/100 - sin(.5 x/100)) ; iron lies in the region bounded below by the function y = f(x) and above by the function y = g(x) = 100(-6 + 2.6 + .03(x/100-1)^2 + .4 cos(.2x/100)); and clay/rock lies above the function y = g(x) and below the surface. This conjecture comes from analysis of geophysical data such as several drill holes, gravitational changes, and surface geology. The code to produce the figure is below: f[x_] = 100(-6 + .5 + .3x/100 - Sin[ .5 x/100]); g[x_] = 100(-6 + 2.6 + .03(x/100-1)^2 + .4 Cos[.2x/100]); p = Plot[{f[x],g[x]},{x,-100,1000},PlotRange->{-700,50}, AxesLabel->{"length-y","depth-x"}, PlotLabel->{"Geology Conjecture"}] We label the regions and points. labs = Show[Graphics[{ Text["silver",{600,100-600}], Text["clay/rock",{250,430-600}], Text["iron",{300,190-600}], Text["U",{400,-25}], Text["V",{600,-25}], Text["W",{800,-25}], Text["O",{25,-600}], Text["surface",{700,25}] }]] Show[p,labs] A sound generator is placed in a hole, 600 m deep, at location O = (0, -600) m. Sounds are generated at O and the first reports of this sound are recorded at the points U, V, and W, each 50 m below the surface. U = (400, - 50) m, V = (600, -50) m, and W = (800, -50). With sensitive equipment it is determined that the first sounds reported at the respective points occur at the following times: 0.17 sec for the sound to reach point U; 0.20 sec for the sound to reach point V; and 0.22 sec for the sound to reach point W. Prepare a document which either supports or refutes the geological model of this underground region. Defend your decision with solid mathematical evidence. Make suggestions as to how the model could be better assessed. KEYWORDS Geology, speed of sound, optimization. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Be able to formulate and minimize time as a function of two variables . Realize that the first sound to arrive at the sites U, V, and W will travel along a path of least time through the various media. Time allotment - time management This problem can be introduced in class and given as a two night homework assignment so the students can ask clarifying questions on the second day. Expectations Expect the students to formulate a function of two variables to optimize and interpret the results in light of observed data. Future payoffs Students will get some idea as to the nagture of inverse problems if the teacher indicates how difficult the problem would be WITHOUT a proposed model for the separators. Extentions Teacher can make up several possible functions which divide the region and then, given the data, the teacher can ask which one of these surfaces is most likely to be the correct one. This will lead to a discssion of some measure of "best." The teacher can inquire as to how one might obtain the separator equations for f(x) and g(x) given just the times with perhaps of several more of the points. The problem can be extended to three dimensional surfaces. References and Sources POSSIBLE SOLUTION(S) We attempt to determine the path of sound which minimizes the time the sound will take to travel from O to U, from O to V, and from O to W. We enter the proposed separator functions. f[x_] = 100(-6 + .5 + .3x/100 - Sin[ .5 x/100]); g[x_] = 100(-6 + 2.6 + .03(x/100-1)^2 + .4 Cos[.2x/100]); And the data points Orig = {0, -600}; U = {400,-50}; V = {600,-50}; W = {800,-50}; We first determine the path and time the sound should take to go from O to A assuming the path of least time. Let (PA, f(PA)) be the point on the iron-silver interface which the sound hits on its least time path and let (QA, g(QA)) be the point on the clay/rock-iron interface which the sound hits on its least time path. Then we have the total time from O to A as a function of the two variables, PA and QA. We use Time = Distance/Rate and remind ourselves that the speed of sound in silver is 2640 m/sec, in clay/rock is 3480 m/sec, and in iron is 5130 m/sec. tA[PA_,QA_] = Sqrt[(Orig[[1]] - PA)^2 + (Orig[[2]] - f[PA])^2]/2640 + Sqrt[(PA - QA)^2 + (f[PA] - g[QA])^2]/5130 + Sqrt[(QA - 400)^2 + (g[QA] - (-50))^2]/3480; We ContourPlot this function to get a reasonable estimate of the PA and QA which give rise to the minimum value for tA(PA,QA). The contour plot will give a plot of the level curves of the surface t = tA(PA, QA). At the darkest points we find the lowest values of the function and at the brightest points we find the highest values of the function. ContourPlot[tA[PA,QA],{PA,0,100},{QA,0,400}] -ContourGraphics- And we obtain the minimum time of 0.17 sec which compares favorably to that reported. solA = FindMinimum[tA[PA,QA], {PA,30},{QA,300}] {0.169042, {PA -> 23.6034, QA -> 276.251}} Let (PB, f(PB)) be the point on the iron-silver interface which the sound hits on its least time path and let (QB, g(QB)) be the point on the clay/rock-iron interface which the sound hits on its least time path. Then we have the total time from O to B as a function of the two variables, PB and QB. We use Time = Distance/Rate and remind ourselves that the speed of sound in silver is 2640 m/sec, in clay/rock is 3480 m/sec, and in iron is 5130 m/sec. tB[PB_,QB_] = Sqrt[(Orig[[1]] - PB)^2 + (Orig[[2]] - f[PB])^2]/2640 + Sqrt[(PB - QB)^2 + (f[PB] - g[QB])^2]/5130 + Sqrt[(QB - 600)^2 + (g[QB] - (-50))^2]/3480; We ContourPlot this function to get a reasonable estimate of the PB and QB which give rise to the minimum value for tB(PB,QB). ContourPlot[tB[PB,QB],{PB,0,200},{QB,0,600}] -ContourGraphics- And we obtain the minimum time of 0.20 sec which compares favorably to that reported. solB = FindMinimum[tB[PB,QB], {PB,40},{QB,500}] {0.196665, {PB -> 28.7704, QB -> 472.648}} Let (PC, f(PC)) be the point on the iron-silver interface which the sound hits on its least time path and let (QC, g(QC)) be the point on the clay/rock-iron interface which the sound hits on its least time path. Then we have the total time from O to C as a function of the two variables, PC and QC. We use Time = Distance/Rate and remind ourselves that the speed of sound in silver is 2640 m/sec, in clay/rock is 3480 m/sec, and in iron is 5130 m/sec. tC[PC_,QC_] = Sqrt[(Orig[[1]] - PC)^2 + (Orig[[2]] - f[PC])^2]/2640 + Sqrt[(PC - QC)^2 + (f[PC] - g[QC])^2]/5130 + Sqrt[(QC - 800)^2 + (g[QC] - (-50))^2]/3480; We ContourPlot this function to get a reasonable estimate of the PC and QC which give rise to the minimum value for tC(PC,QC). ContourPlot[tC[PC,QC],{PC,0,400},{QC,0,800}] -ContourGraphics- And we obtain the minimum time of 0.22 sec which compares favorably to that reported. solC = FindMinimum[tC[PC,QC], {PC,40},{QC,600}] {0.223679, {PC -> 30.9791, QC -> 718.241}} Thus since the times of the travelling sound from O to all three points U, V, and W seem to agree with the fastest time sound travels from O to U, V, and W respectively we can say that the functions f and g seem reasonable as separators. ISSUES IN SOLUTION Once we set up the optimization function it is a routine to plot (to get some idea of a reasonable set of values for the respective points on the curves). Then it is a matter of minimizing a complex function of two variables, i.e. using FindMinimum or setting the partials equal to zero and checking to see if the times we get are close to the times offered and hence suggesting that the separator curves are reasonable.