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# Tide Modeling

Extracts from this document...

Introduction

felpTide Modeling

 Using Microsoft Excel, plot the graph of time versus height.  Describe the result using the terminology you have acquired in the study of circular functions.

In order to come up with a plot graph for the bay of Fundy in Noca Scotia, Cana it was necessary to use Microsoft Excel and the data from www.lau.chs-shc.dfo-mpo.gc.ca.com. In order to plot the graph in excel all of the data had to be entered in the spreadsheet. Once that was done a scatter graph was created. Once it was done it was possible to analyze the graph. The first thing that is possible to notice is that this is a periodical graph. However, the graph is not completely periodical since the lines don’t always have the same height. Sometimes it is greater or lower. This can be attributed to the fact that it is a real life situation. Therefore, it would not be expected for it to be perfect. However, the graph is still periodical since it follows the same shape throughout. Use your understanding of circular functions and their transformations to develop a mathematical model for the behavior noted in the graph.  Explain your method and reasoning in detail.

In order to come up with the equation I had to come up with a series of averages. This was necessary since the graph is not completely periodical.

In order for me to find the vertical translation of the graph, meaning how much the graph was moved up I had to find the graph’s sinusoidal axis. In order to find it I had to find the graphs height for both crests and do an average of it. The highest point in the first crest was 12 meters. For the second crest it was 12.

Middle

0.087

0.325

0.412

0.29

0

0.26

0.438

𝑦=5.675,sin-,,𝜋-6.,𝜃...+6.475 and 𝑦=5.675,cos-,,𝜋-6.,𝜃−3...+ 6.475 Only one color is shown because graphs overlap each other since they are the same.

 Use the regression feature of your graphing calculator to develop a best-fit function for this data.  Compare this model with the one you developed analytically.

In order to come up with a best- fit line for the data, so that it is possible to see what the best model for the data would be it was used the regression feature of the TI 84. In order to use it  all the data values for the data had to be inserted in the table (Figure 3). Once that was done a regression had to be found and once again the TI 84 was used. In order to find it a feature called SinReg was used. Finally Graphing Package was used in order to graph the data and so that it could be compare to the sine model  Once that was done in order to better compare the model with the best- fit graph it was decided to plug in values for 𝜃 so that it the models could better be compared. When 3 was plug in for the sine model it was obtained 12.15 as the y values, and that would be expected since it was the average obtained in step 2.

However, when 3 was plug in the best- fit equation 12.24 was obtained. The difference between both was 0.091. In this case the sine model graph seems to have been better since according to the table used to create the excel graph it was supposed to be 12. In order to further prove, 17 was plug in for𝜃. However, this changed things. When it was plugged in for the sine model the value for y was 9.312. Finally, when it was plug in for the best- fit graph the value for y was 9.908. According to the data it was supposed to be 9.9. Therefore, this time the best- fit graph was better. After further Finally a conclusion can be arrived at. The best- fit graph is better than the sine model for all points but the through and the crest. This can be explained by the fact that the model was created mainly trying to make the through and the crest as close as possible, in the other hand the best fit model was created by the calculator in a way so that all the points would be as close as possible to the real one while the sine model sacrificed the other points in order to make sure the through and the crest were as accurate as possible. Therefore, that makes the best fit line graph better in the overall than the sine model. It is also possible to be sure that the best fit model is better than the sine model when looking at their percent error in relation to the original excel data. In the table below the points were not grouped together since for the best fit graph the values for 0 and 12 are not the same; therefore a point by point comparison was necessary. Using the same method as in number 3 with the percent error it is possible to determine that the sine models error point by point is %1.83 and for the best fit graph it was %0.25. Once again this shows that the model that the calculator came up with is better than the one that was developed analytically.

 Points 0 1 2 3 4 5 6 7 8 9 10 11 Model 6.475 9.313 11.39 12.15 11.39 9.313 6.475 3.638 1.56 0.8 1.16 3.638 Best fit 7.697 10.28 11.929 12.241 11.134 8.885 6.06 3.367 1.482 0.8782 1.708 3.762 Data 7.5 10.2 11.8 12 10.9 8.9 6.3 3.6 1.6 0.9 1.8 4 Points 12 13 14 15 16 17 18 19 20 21 22 23 Model 6.475 9.313 11.39 12.15 11.39 9.313 6.475 3.638 1.56 0.8 1.16 3.638 Best fit 6.525 9.303 11.4 12.29 11.75 9.908 7.237 4.402 2.116 0.9517 1.202 2.804 Data 6.9 9.7 11.6 12.3 11.6 9.9 7.3 4.5 2.1 0.7 0.8 2.4
 Apply the model you developed on the calculator to the following two situations:A sailor launched his boat at 7:45 on December 27, 2003.  Determine the height of the tide at that time using both analytical and graphical methods. Explain your methods and reasoning in detail.

Conclusion

-,0.5063958661,16.38624518..+6.586664794.

𝑦=5.713209557,sin-,8.297926821.+6.586664794.

𝑦=5.713209557∙0.9030642159+6.586664794

𝑦=5.159395109+6.586664794

𝑦=11.7460599

Finally, it is possible that the same answers were obtained both analytically and graphically. And so the height of the tides in which lazy sailors launched their boats was between 11.75 meters and 11.40 meters. In this problem only four significant figures were used because the problem gave the time with four significant figures.

 The table below lists the tide heights for December 28, 2003.  Does your previous model fit the data?  Why or why not?  What modifications are needed?  Confirm that your modified model fits the data.

The first model does not fit the new data as can be seen by the picture (Place picture). Mainly it does not fit because apparently the period of the graph is different. It is possible to conclude that because the first time the graph goes up it almost follows the same points. However, than the points are more to the right than the original model. That indicates that there is no or very little horizontal translation and rather a change in period. Further the graphs crest seems to be too high compared with the new data. However, the through seems to be good.

When actual points are analyzed it is better possible to come up with the problems of the first model and the modifications necessary. The first is the fact that the first data the graph starts at 7.5 and the second at 5. Further in the first data the highest point was 12.3 and in the second data it is 11.6 explaining why the model goes over the data. However, the lowest point in the graph is .7 and for the second data it is one. These were the main differences, however, modifications still need to be made to make the model fit the new data.

In order to come up with the sinusoidal axis of the same method described in two was used. (Put the equations).

For the amplitdude the method is once again the same again as 2.

For the period the method is also the same.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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