The Tide is Turning BRIEF ABSTRACT Based on Coast Guard data of water levels in port over a 24 hour period we are asked to offer a mathematical model to "match" the data and to make intermediate predictions on height based on that model. Estimating with trigonometric functions proves to be very successful. GENERAL INFORMATION FileName: TIDE Full title: The Tide Is Turning Developer: Kimberly Foltz, Mathematics and Computer Science Division, Indiana Academy for Science, Mathematics, & Humanities, Muncie IN 47306 USA David Horn, Mathematics Department Rogers High School, Michigan City IN 46360 USA 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. STATEMENT OF THE PROBLEM Coast Guard instruments were used to take hourly readings of the height of the water in Oakland Bay on a given day. The findings are recorded in the table below. TIME WATER LEVEL (in feet) Midnight 15.63 1 a.m. 18.04 2 a.m. 19.53 3 a.m. 20.35 4 a.m. 19.74 5 a.m. 17.82 6 a.m. 15.63 7 a.m. 12.54 8 a.m. 11.19 9 a.m. 10.20 10 a.m. 10.96 11 a.m. 13.14 Noon 15.17 1. p.m. 18.04 2 p.m. 19.49 3 p.m 20.00 4 p.m. 19.87 5 p.m 17.77 6 p.m. 15.55 7 p.m 13.15 8 p.m. 10.82 9 p.m 10.38 10 p.m. 10.70 11 p.m 12.89 Midnight 15.21 PART A: Based on these data points, find a function that "best" models the height of the water at any time t. PART B: Approximate what the height of the water was at 1:30 p.m. on that day. Explain your conclusions. PART C: Was the water level rising or falling at 1:30 p.m.? PART D: Approximate what the height of the water was at 3:30 p.m. on that day. Explain your conclusions. PART E: Approximately how fast was the water level falling at 7 a.m.? PART F Approximate all the times within the first 12 hours at which the water level will be 18 feet. KEYWORDS Trigonometry, sine function, period, amplitude, mean, phase-shift, data, curve fitting, derivative, root finding. TEACHER NOTES ISSUES RELATED TO THE PROBLEM This problem requires to estimate the values of a, b, c, and d in the model h(t) = a sin b(t + c) + d where t is time in hours after midnight and h(t) is the height of the water in Oakland Bay. The approach is to eyeball estimate, not truly fit, and to compare the estimate model with the data and use the model to predict height and rate of change of height. With regard to the latter rate, one can take the formal derivative or use a difference quotient form of the derivative over a small interval to approximate the instantaneous change at time t. The problem is not much different or much more difficult than the standard simple harmonic motion problem. It does have the nice feature that the information is given to the students in the form of a table. This gives a nice introduction into the topics of data analysis and curve fitting. Prerequisites Students should have seen the basic function h(t) = a sin b(t + c) + d. They should have seen graphs and understand the significance of each of the terms a, b, c, and d. Time allotment - time management This activity can be done in one half hour or so with appropriate technology during class time. Expectations Future payoffs Extensions References and Sources POSSIBLE SOLUTION(S) PART A: We enter the information from the table as a data set. Orient the time values so that the day begins at t = 0 (i.e., t = 0 at midnight ). givendata = {{0, 15.63},{1, 18.04},{2, 19.53},{3, 20.35}, {4, 19.74},{5, 17.82},{6, 15.63},{7, 12.54},{8, 11.19}, {9, 10.20},{10, 10.96},{11, 13.14},{12, 15.17},{13, 18.04}, {14, 19.49},{15, 20.00},{16, 19.87},{17, 17.77},{18, 15.55}, {19, 13.15},{20, 10.82},{21, 10.38},{22, 10.70},{23, 12.89}, {24, 15.21}}; A graphical view of this data will assist the the investigation of the relationship of time and water level. points=ListPlot[ givendata, PlotStyle->{PointSize[.02]}]; joined=ListPlot[ givendata, PlotStyle->{PointSize[.02]}, PlotJoined->True]; Show[points,joined]; The data points seem to form a sinusoidal wave. Thus, an equation of the form h = a sin b(t + c) + d can be used to model this situation. We must determine values for the amplitude, period, phase shift, and vertical shift that will "best" fit this data. The water level seems to have a maximum value of 20.35 feet and a minimum value of 10.2 feet. Since the height varies by approximately 10 feet, we can estimate the amplitude of the sine function to be a = 10/2=5. The water level on this day begins at 15.63 feet. It returns to roughly this value at 6 a.m. and again at noon, showing a cyclic behavior. This can also be seen on the plot, where the curve completes one full cycle in a time period of approximately 12 hours. The period of the sine function is 2 Pi / b, where b is the coefficient of t. Therefore, the period is 2 Pi / b = 12, which gives that b = Pi / 6 . The curve that fits the points does not appear to have a horizontal phase shift from the graph of y = sin x, so c = 0 for our model. There is a vertical shift which moves the "mean" height from 0 to somewhere around 15, so we will start with d = 15 in our equation. Model[t_] =5 Sin[ Pi/6 t ] + 15.25; curve = Plot[ Model[t], {t,0,24}]; Show[curve,points]; This did fairly well, but maybe we can raise the graph just a little bit to get a better fit. Model2[t_] :=5 Sin[ Pi/6 t ] + 15.25; curve2 = Plot[ Model2[t], {t,0,24}]; Show[curve2,points]; PART B Using the time scale we have established, 1:30 p.m. is equivalent to 13.5 hours. Thus, we can simply use the model equation that we have developed with t = 13.5. Model2[13.5]//N 18.7855 This value seems reasonable because it is between the heights of 18.04 and 19.49 which were recorded on the table at 1 p.m. and 2 p.m. respectively. PART C The water clearly is rising. This question is somewhat trivial. Its value lies in that it can be answered graphically (by examining the plot), numerically (by examining the table), or symbolically (by examining the derivative). Here we illustrate the symbolic approach, which shows that the slope is positive for this value of time. Model2'[13.5]//N 1.8512 PART D At 3:30 p.m., our model time scale gives t = 15.5 hours. Model2[15.5]//N 20.0796 By the model, the height of the water would be 20.0796 feet. Is this reasonable? From 3 p.m. to 4 p.m., the water level decreases from 20.0 ft to 19.87 ft. Under the assumption that the water level is strictly decreasing during this time interval, the water could not have reached a height of more than 20 feet. Looking at the plot of the actual data points with the model function, it can be observed that the model curve does not fit the data points as neatly in this interval. This would explain the unusual result given by the model. Which is right, the model or the table? PART E This will again require using the derivative, this time evaluated when t= 7. The function appears to be decreasing there, so it is not surprising that the result is negative. The rate of decrease is 2.27 ft / hour Model2'[7]//N -2.26725 PART F To determine all the times within the first 12 hours at which the water level will be 18 feet we plot our Model2. Plot[Model2[t],{t,0, 12}] It appears as if there are two times at which the water level will be 18 feet in the first 12 hours - around 2 AM and around 4:30 AM. sol1 = FindRoot[Model2[t]==18,{t,2}] {t -> 1.11223} sol2 = FindRoot[Model2[t]==18,{t,4}] {t -> 4.88777} Thus we have the two times at which the water level will be 18 feet in the first 12 hours: t1 = t/.sol1[[1]] 1.11223 t2 = t/.sol2[[1]] 4.88777 And as a check we see that Model2[t1]//N 18. Model2[t2]//N 18. ISSUES IN SOLUTION