FISHING RODS

A fishing rod requires guides for the line so that it does not tangle and so that the line casts easily and efficiently. In this task, you will develop a mathematical model for the placement of line guides on a fishing rod.

Leo has a fishing rod with an overall length of 230 cm. The diagram shows a fishing rod with eight guides, plus a guide at the tip of the rod.

The table below shown below gives the distances for each of the line guides from the tip of his fishing rod.

The distance of each guide from the tip of the fishing rod is measured in cm. The x value, or the independent variable, is the guide number, and the y value, or the dependent variable, is the distance from tip. Only positive numbers can fit this scenario, because a fishing rod cannot have negative lengths. We can even further limit the domain, because the fishing rod only has 8 guides, so x

To find [X], we could use this formula:

So:

So, the Cubic Function is:

In order to graph accurately, I decided to only graph the quadratic and cubic functions only with the domain of the original data points. This domain is:  . Using this domain, I can determine the distances of the guides from the tip (cm) of the fishing rod.
To find the y values for this domain, on Excel, I created a formula that would allow me to plug in the x values and find the y values. The table loos like this:

The graph illustrating the original data points and the model functions looks like this:

The quadratic and cubic functions that I found were the same in the first quadrant. Because the guide number cannot be negative, I chose to only compare these two in the first quadrant. If we looked at the values in the other quadrants, they would be different.

Mark’s fishing rod:

My quadratic model does not quite fit this data set because there are some differences in the y value. Looking at the graph of the data, it seems that it fits the model but when looking at the equation from the graph, we can see that it is slightly different:

The r2 value for this graph is 0.998, while for my quadratic model it is 1. I would need to change the equation that I obtained. To do this I could use the matrix method, but instead use these three data points: (1, 10), (2, 22), (3, 34) Using these points in the matrix method, I could calculate the new quadratic model, and then compare it to the original data points. My equation would also change once I do this. The new values for a, b, c, would be 0, 12, -2 respectively. So the new quadratic model that would fit this data is: