# Math Sl Fish Production

MODELING

The aim of this investigation is to consider commercial fishing in a particular country in two different environments, that is from the sea and a fish farm (aquaculture). The following data provided below was taken form the UN Statistics Division Common Database. The tables gives the total mass of fish caught in the sea, in thousands of tones (1 tone = 1000 kilograms). Year| 1980| 1981| 1982| 1983| 1984| 1985| 1986| 1987| 1988| Total Mass| 426.8| 470.2| 503.4| 557.3| 564.7| 575.4| 579.8| 624.7| 669.9|

Year| 1989| 1990| 1991| 1992| 1993| 1994| 1995| 1996| 1997| Total Mass| 450.5| 379.0| 356.9| 447.5| 548.8| 589.8| 634.0| 527.8| 459.1|

Year| 1998| 1999| 2000| 2001| 2002| 2003| 2004| 2005| 2006| Total Mass| 487.2| 573.8| 503.3| 527.7| 566.7| 507.8| 550.5| 426.5| 533.0| Table1. – The total mass of fish caught from 1980-2006

From the presented data, it is shown that in some years the amount of fish caught rises and on others the amount of fish caught in the sea decreases. The reason of the amount of fish to decrease can be influenced by ecological factors and weather phenomena’s or even the amount of fishermen on the particular area in the sea. Identifying variables: let y stand for the amount of fish caught from the sea and x stands for the years passed. It shows yx=amount of fish caught, so e.g. y1980=426.8. The set of possible values of the variable on the horizontal axis is called the domain. Therefore {x | 1980 < x ≤ 2006} is the domain of the amount of fish caught. The set which describes the possible y-values is called the range. Therefore the range of the fish caught is the total mass of fish from the sea.

From the data presented in Table1. it is possible to create a graph: A

B

(x)

Graph1. – The total amount of fish caught in the sea in years 1980 – 2006.

On the presented Graph1. it is easy to notice when the amount of fish caught from the sea increases and decreases. The most amount of fish 669.9 thousand tones, had been caught in the year 1988 as for the least amount of fish 356.9 thousand tones, had been caught in year 1991. As we can also see that the graph consists of a combination of functions: tangent and sine. The original graph will be divided into two parts to make it easier to read.

Graph2. – The tangent graph We attempt to model this data using the general tan function y=atanbx-c+d, or in this case M=atanbx-c+d. The period is in 19 years, so πb=19 and ∴b=π19.

The amplitude = max-min2≈ 669.9-426.82 ≈121.55, so a≈121.55. The principle axis is midway between the maximum and minimum, so d≈669.9+426.82≈548.35.

So, the model is M≈121.55tanπ19x-c+548.35, for some constant c. We can notice that the point B on the original graph, graph2, lies on the principles axis and so it is a point at which we are starting a new period. Since B is at (1983,557.3), c = 1983. The model is therefore M≈121.55tanπ19x-1983+548.35 and we can superimpose it on the original data as follows.

Graph3. Tan model

The graph shows what happens when we fit the original graph , Graph2. with its model graph. The slopes of the two graphs are presented to be nearly equal to each other.

Now considering the graph of sin function:

Graph4.- The sine graph

We attempt to model this data using the general sin function y=asinbx-c+d, or in...

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