Tide Modelling

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IB Mathematics Standard Level Portfolio Assignment Task I
Tide Modeling
Kelvin Kwok

In this modeling assignment, I will develop a model function for the relationship between time of day and the height of the tide. I will first show the data in a table form copied from the assignment sheet. Then I will use the data to construct a scatter graph of time against height, then I will develop a function that models the behavior noted in the graph analytically, and describe any variables, parameters or constraints for the model. In the end I will modify the same model using the regression feature of graphing software to develop a best-fit function for the data. This modeling task will allow me to find out what is the best time that a good sailor should launch their boat. For all the diagrams, I am using autograph to plot my graphs and find the regression model.

Table 1 below shows the height of tide at different time from 27 December 2003 using Atlantic Standard Time(AST), the heights were taken at Grindstone Island.

Time(AST)| 00.00| 01.00| 02.00| 03.00| 04.00| 05.00| 06.00| 07.00| 08.00| 09.00| 10.00| 11.00| Height(m)| 7.5| 10.2| 11.8| 12.0| 10.9| 8.9| 6.3| 3.6| 1.6| 0.9| 1.8| 4.0|

Time(AST)| 12.00| 13.00| 14.00| 15.00| 16.00| 17.00| 18.00| 19.00| 20.00| 21.00| 22.00| 23.00| Height(m)| 6.9| 9.7| 11.6| 12.3| 11.6| 9.9| 7.3| 4.5| 2.1| 0.7| 0.8| 2.4|

Source: http://www.lau.chs-shc.dfo-mpo.gc.ca

Method 1: Using Autograph, plotted a scattered graph of time against height from the data collected on 27 December 2003

Diagram1 Scatter graph of hours after midnight against height of tide

The data points form a sinusoidal shape, high tide occurred at 03.00 and 15.00 which are 12.0m and 12.3m respectively. And the low tide occurred at 09.00 and 21.00 which are 0.9m and 0.7m respectively. If we look at the two crest(12.0m&12.3m) we can see a period of 12 hours from 03.00 to 15.00, which is 12 hours.

Method 2: Develop a function that models in the graph analytically

On the graph above, y is the height of tide and x is the hours after midnight. From the troughs of the sine curve above(09:00 and 21:00), we can see that the sine curve has a period of approximately 12 hours, and amplitude of approximately 6 ( Max. -Min.2≑12.3-0.72≈6). Standard sine curve has period of 2π(2π≈6.28) and the amplitude=1. Here I will attempt to model the data into y=AsinB(x)+D. A is the amplitude≈6 ( Max. -Min.2≑12.3-0.72≈6). As the period is 12months, so 2πB=12, B≈12. D is the principal axis, which is midway between max. and min. ∴D=12.3+0.72≈6. Hence we get the function y=6sinx2+6. So the function is dilating 6 times along y-axis, and 2 times along x-axis, and translate upwards for 6 units.

So the parameters are, amplitude=6 (dilation along y-axis), B=12 (dilation along x-axis), and principal axis at 6 unit.

The constraints are max. height at 12.3m and min. height at 0.7m, and a domain of from zero hours after midnight to 23 hours after midnight.

Method 3: Use Autograph to draw a graph of the function established in method 2.

Diagram2 Graph of the function y=6sinx2+6 established in method 2

On autograph, plotted the function y=6sin(x2)+6, it does not fit perfectly on the same set of axes as the graph in step1, the function passes through 3 data points, overall there are 24 data points, so 18 of the data points are passed through.

Method 4: Modify the function in method 3 to create a better fit

In method 1 I have attempted to model the data into y=AsinB(x)+D, so here I am going to modify the value of A, D, and I will add C as horizontal translation, following I am going to show how new parameters are established.

In method 2 when I am calculating amplitude, I have used 12.3 as max. point and 0.7 as min. point. Now I will use the average of the two crest which are 12.3 and 12.0, so the mean height at crest is...
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