Mth Sl Type Ii Portfolio – Fishing Rods

Math Summative Fishing Rods Fishing Rods A angle celestial pole cell requires draw ins for the declination so that it does non tangle and so that the word of mouth casts easily and economic completelyy. In this task, you leave develop a numeral computer simulation for the jellment of line extends on a search magnetic pole. The plot shows a fishing rod with octet guides, plus a guide at the track of the rod. social lion has a fishing rod with everyplace solely told length 230 cm. The display board shown infra repays the outdistances for each of the line guides from the superlative of his fishing rod. manoeuvre Number (from lead story) remoteness from tap (cm) 1 10 2 23 3 38 4 55 5 74 6 96 7 long hundred 8 149Define suit equal uncertains and question parameters/constraints. Using Technology, pot the entropy points on a graph. Using intercellular substance modes or otherwise, find a quadratic polynomial polynomial equivalence crop and a brick- shaped melt down which perplex this situation. Explain the process you employ. On a modernistic-sprung(prenominal) fit out of axes, draw these sit around suffices and the cowcatcher selective in mastermindation points. small talk on any difference of opinions. Find a polynomial contribution which passes through each info point. Explain you choice of purpose, and discuss its reasonableness. On a cutting invest of axes, draw this mannequin office and the legitimate info points. Comment on any differences.Using technology, find one other officiate that fits the information. On a untested set of axes, draw this model function and the sea captain data points. Comment on any differences. Which of you functions found above surpass models this situation? Explain your choice. Use you quadratic model to decide w here you could place a 9th guide. Discuss the implications of adding a ninth guide to the rod. curb has a fishing rod with overall length 300cm. The table shown below gives the distances for each of the line guides from the tip of markers fishing rod. croak Number (from tip) Distance from tapdance (cm) 1 10 2 22 3 34 4 48 5 64 6 81 7 102 124 How fountainhead does your quadratic model fit this newfound data? What changes, if any, would need to be made for that model to fit this data? Discuss any point of accumulations to your model. Introduction Fishing rods map guides to control the line as it is being casted, to ensure an efficient cast, and to restrict the line from tangling. An efficient fishing rod go away use multiple, stpacegically placed guides to maximize its functionality. The status of these entrust dep revoke on the good turn of guides as hale as the length of the rod. Companies design mathematical equatings to cast the optimal placement of the guides on a rod.Poor guide placement would likely set about for poor fishing quality, dissatisfied customers and thus a slight successful company. and so it is essen tial to ensure the guides be properly placed to maximize fishing efficiency. In this investigation, I entrust be determining a mathematical model to found the guide placement of a tending(p) fishing rod that has a length of 230cm and apt(p) distances for each of the 8 guides from the tip (see data below). Multiple equations impart be determined use the given data to provide varying point in times of accuracy. These models force out and so potentially be utilize to determine the placement of a 9th guide.Four models will be use quadratic function, boxyal function, septic function and a quadratic reasoning backward function. To stick, suitable multivariates must be defined and the parameters and constraints must be discussed. Variables independent Variable part with x translate the number of guides beginning from the tip Number of guides is a discrete lever. Since the length of the rod is finite (230cm) then the number of guides is subsist to be finite. Domain = , where n is the finite range that represents the maximum number of guides that would fit on the rod. Dependent VariableLet y represent the distance of each guide from the tip of the rod in centimetres. The distance of each guide is a discrete value. redact = Parameters/Constraints There atomic number 18 several parameters/constraints that need to be verified forwards proceeding in the investigation. Naturally, since we atomic number 18 talking ab bug out a actual life situation, on that point bungholenot be a negative number of guides (x) or a negative distance from the tip of the rod (y). All set are positive, and therefore all graphs will only be stand for in the first off quadrant. The other major constraint that must be identified is the maximum length of the rod, 230cm.This restricts the y-value as well as the x-value. The variable n represents the finite number of guides that could possibly be placed on the rod. While it is physically possible to place many guides on the rod, a realistic, maximum number of guides that would shut away be efficient, is rough 15 guides. draw off Number (from tip) Distance from wiretap (cm) 0* 1 2 3 4 5 6 7 8 n** 0 10 23 38 55 74 96 120 149 230 *the guide at the tip of the rod is not counted **n is the finite value that represents the maximum number of guides that would fit on the rod.Neither of the highlighted determine are analyzed in this investigation, they are only here for the purpose of shaping the limits of the variables. The first tincture in this investigation is to graph the points in the table above (excluding highlighted points) to see the shape of the trend that is created as more guides are added to the rod. From this scatter plot of the points, we backside see that there is an exponential increase in the distance from the tip of the rod as each ensuant guide is added to the rod. Quadratic Function The first function that I shall be modeling using the points of data provided is a quadrat ic function.The general equation of a quadratic clayula is y = ax2 + bx + c. To do this, I will be using three points of data to create three equations that I will solve using matrices and determine the coefficients a, b and c. The first measure in this process is to choose three data points that will be utilise to represent a broad dictate of the data. This will be difficult though since there are only three out of the eight points that screw be use. then, to improve the accuracy of my quadratic function, I will be solving two organisations of equations that use different points and purpose their imagine. info wads Selected Data Set 1 = (1,10), (3,38), (8,149)Data Set 2 = (1,10), (6,96), (8,149) These points were selected for two main reasons. First, by using the x-value 1 and 8 in both(prenominal) sets of data, we will have a broad range of all of the data that is being represented in the final equation after(prenominal) the determine of the coefficients are averaged. Second, I used the x set of 3 (in the first set) and 6 (in the second set) to once again go forth for a broad representation of the data points in the final quadratic equation. most(prenominal)(prenominal) of these points (3 and 6) were chosen be type they were equal distances apart, 3 being the third data point, and 6 being the third from last data point.This ensured that the final averaged determine for the coefficients would give the scoop representation of the middle data points without skewing the data. There will be two methods that will be used to solve the system of equations, seen below. for each one method will be used for one of the systems being evaluated. Data Set 1 = (1,10), (3,38), (8,149) In the first data set, the data points will form separate equations that will be solved using a matrices equation. The first ground substance equation will be in the form Where A = a 33 hyaloplasm representing the three data pointsX = a 31 hyaloplasm for the variables b eing solved B = a 31 hyaloplasm for the y-value of the three equations being solved. This hyaloplasm equation will be rearranged by reproduceing both sides of the equation by the inverse of A Since A-1*A is equal to the identity matrix (I), which when multiply by another matrix gives that equivalent matrix (the matrix equivalent of 1), the final matrix equation is To determine the values of X, we must first find the inverse of matrix A using technology, since it is available and decision the inverse of a 3 by 3 matrix flush toilet take an inefficient amount of time.First let us determine what equations we will be solving and what our matrices will look like. Point (1,10) (3, 38) (8,149) A= The equation is ,X= ,B= = Next, by using our GDC, we basis determine the inverse of matrix A, and multiply both sides by it. and so we have determined that the quadratic equations given the points (1,10), (3,38), (8,149) is . Data Set 2 = (1,10), (6,96), (8,149) Point (1,10) (6, 96) (8,14 9) A= ,X= ,B= The second method that will be used to solve the second system of equations is known as Gauss-Jordan elimination.This is a process by which an increase matrix (two matrices that are placed into one divided by a line) goes through a series of unsophisticated mathematical trading operations to solve the equation. On the left field side of this augmented matrix (seen below) is the 33 matrix A (the new matrix A that was made using data set 2, seen on the previous page), and on the right is matrix B. The goal of the operations is to reduce matrix A to the identity matrix, and by doing so, matrix B will yield the values of matrix X. This is otherwise known as decrease class echelon form. Step by step process of decrement 1. We begin with the augmented matrix. . minimal brain dysfunction (-36 * words 1) to row 2 3. Add (-64 * row 1) to row 3 4. Divide row 2 by -30 5. Add (56 * row 2) to row 3 6. Divide row 3 by 7. Add ( * row 3) to row 2 8. Add (-1 * row 3) to row 1 9 . Add (-1 * row 2) to row 1 aft(prenominal) all of the row operations, matrix A has become the identity matrix and matrix B has become the values of matrix X (a, b, c). Therefore we have determined that the quadratic equations given the points (1,10), (6,96), (8,149) is . Averaging of the Two Equations The next step in finding our quadratic function is to average out our establish a, b, and c values from the two sets data.Therefore we have finally determined our quadratic function to be Rounded to 4 sig figs, too maintain precision, turn keeping the numbers manageable. Data points using quadratic function Guide Number (from tip) Quadratic values Distance from Tip (cm) accredited Distance from Tip (cm) 1 10 2 22 3 37 4 54 5 74 6 97 7 122 8 149 10 23 38 55 74 96 120 149 New values for the distance from tip were rounded to cypher decimal places, to maintain signifi enduret get wind the current values used to find the quadratic formula had nonentity decimal places, so the new ones shouldnt either.After finding the y-values given x-values from 1-8 for the quadratic function I was able to compare the new values to the original values (highlighted in grand in the table above). We domiciliate see that the two values that are the exact equal in both data sets is (1,10) and (8,149) which is not move since those were the two values that were used in both data sets when finding the quadratic function. Another new value that was the same as the original was (5,74). All other new data sets have an fault of approximately 2cm.This data shows us that the quadratic function can be used to represent the original data with an approximate error of 2cm. This function is still not perfect, and a meliorate function could be found to represent the data with a lower error and more fight backing data points. blocky Function The next step in this investigation is to model a cubic function that represents the original data points. The general equation of a cubic function is y = ax3 + bx2 + cx + d. Knowing this, we can take four data points and coiffe a system of equations to determine the values of the coefficients a, b, c, and d.The first step is to choose the data points that will be used to model the cubic function. Similarly to modeling the quadratic function, we can only use a limited number of points to represent the data in the function, only in this case it is four out of the eight data points, which means that this function should be more precise than the last. at once again I plan on solving for two sets of data points and finding their mean values to represent the cubic function. This is done to allow for a more broad representation of the data within the cubic function. Data Sets Selected Data Set 1 (1,10), (4,55), (5,74), (8,149)Data Set 2 (1,10), (3,38), (6,96), (8,149) Both data sets use the points (1,10) and (8,149), the first and last point, so that both data sets produce cubic functions that represent a broad range of the data ( from minimum to maximum). The other points selected, were selected as mid range points that would allow for the function to represent this range of the data more undefiledly. When modeling a cubic function or higher, it is difficult to do so without using technology to do the bulk of the calculation callable to large amounts of uninteresting calculations that would almost guarantee a math error somewhere.Therefore, the most immaculate and fastest way to perform these calculations will be to use a GDC. In both data sets, the reduce row echelon form function on the GDC will be utilized to determine the values of the coefficients of the cubic functions. The process of determining the values of the coefficients of the cubic function using reduced row echelon form is similar to process used for the quadratic function. An x-value matrix A (this time a 44 matrix), a variable matrix X (41) and a y-value matrix B (41) must be determined first. The next step is to augment matrix A and ma trix B, with A on the left and B on the right.This time, instead of doing the row operation ourselves, the GDC will do them, and yield an answer where matrix A will be the identity matrix and matrix B will be the values of the coefficients (or matrix X). Data Set 1 (1,10), (4,55), (5,74), (8,149) (1,10) (4, 55) (5, 74) (8,149) A1 = , X1 = , B1 = We begin with the augmented matrix or matrix A1 and matrix B1. Then this matrix is gossipted into a GDC and the function rref is selected. After pressing enter, the matrix is reduced into reduced row echelon form. Which yields the values of the coefficients. Data Set 2 (1,10), (3,38), (6,96), (8,149) (1,10) (3, 38) 6, 96) (8,149) A2 = , X2 = , B2 = We begin with the augmented matrix of matrix A2 and matrix B2. Then the matrix is inputted into a GDC and the function rref After pressing enter, the matrix is reduced into reduced row echelon form. Which yields the values of the coefficients. The next step is to find the mean of each of the valu es of the coefficients a, b, c, and d. Therefore we have finally determined our cubic function to be erst again rounded to 4 significant figures. Updated Data table, including cubic function values. Guide Number (from tip) Quadratic values Distance from Tip (cm) 1 10 2 22 3 37 4 54 5 74 6 97 122 8 149 box-shaped values Distance from Tip (cm) Original Distance from Tip (cm) 10 23 38 54 74 96 121 149 10 23 38 55 74 96 120 149 New values for the distance from tip were rounded to postal code decimal places, to maintain significant figure the original values used to find the quadratic formula had zero decimal places, so the new ones shouldnt either. The y-values of the cubic function can be compared to that original data set values to conclude whether or not it is an unblemished function to use to represent the original data points. It appears as though the cubic function has 6 out of 8 data points that are the same.Those points being, (1,10), (2,23), (3,38), (5,74), (6,96), (8,149) . The three data points from the cubic function that did not match only had an error of 1, indicating that the cubic function would be a good representation of the original data points, but still has some error. We can further analyze these points by comparing the cubic and quadratic function to the original points by graphing them. See next page. By analyzing this graph, we can see that both the quadratic function and the cubic function match the original data points quite well, although they have slight differences.By comparing values on the data table, we find that the quadratic function only matches 3 of the 8 original data points with an error of 2, while the cubic function matches 6 of the 8 points with an error of just 1, which is as small an error possible for precision of the calculation done. Both functions act as decorous representations of the original points, but the major difference is how they begin to differ as the graphs continue. The cubic function is increasing a t a faster rate than the quadratic function, and this difference would become quite noticeable over time.This would mean that if these functions were to be used to determine the distance a 9th guide should be from the tip, the two functions would provide quite different answers, with the cubic functions providing the more consummate one. Polynomial Function Since it is known that neither the quadratic, nor the cubic function in full satisfy the original data points, then we must model a higher degree polynomial function that will satisfy all of these points. The scoop out way to find a polynomial function that will pass through all of the original points is to use all of the original points when finding it (oppose to just three or four).If all eight of the points are used and a system of equations is performed using matrices, then a function that satisfies all points will be found. This is a septic function. To find this function, the same procedure followed for the last two func tions should be followed, this time using all eight points to create an 88 matrix. By then pastime the same steps to augment the matrix with an 81 matrix, we can change the matrix into reduced row echelon form to and find our answer. In this method, since we are using all eight points, the entire data set is being represented in the function and no averaging of the results will be necessary.The general formula for a septic function is . Data Set (1,10), (2,23), (3,38), (4,55), (5,74), (6,96), (7,120), (8,149) (1,10) (2,23) (3,38) (4,55) (5,74) (6,96) (7,120) (8,149) A=,X= ,B= , Augment matrix A and matrix B and perform the rref function The answers and values for the coefficients = The final septic function equation is This function that include all the original data points can be seen graphed here below along with the original points. Updated Data table, including septic function values Guide Number (from tip) Quadratic values Distance from Tip (cm) Cubic values Distance from Tip (cm)Septic values Distance from Tip (cm) Original Distance from Tip (cm) 1 10 2 22 3 37 4 54 5 74 6 97 7 122 8 149 10 23 38 54 74 96 121 149 10 23 38 55 74 96 120 149 10 23 38 55 74 96 120 149 New values for the distance from tip were rounded to zero decimal places, to maintain significant figure the original values used to find the quadratic formula had zero decimal places, so the new ones shouldnt either. By looking at the graph, as well as the data table (both seen above), we can see that, as expected, all 8 of the septic function data points are identical to that of the original data.There is less than 1cm of error, which is accounted for due to imprecise (zero decimal places) original measurements. Therefore we now know that the septic function that utilised all of the original data points is the scoop out representation of said data. Other Function The next goal in this investigation is to find another function that could be used to represent this data. The other method th at I will use to find a function that fits the data is quadratic regression. Quadratic regression uses the method of to the lowest degree squares to find a quadratic in the form .This method is frequently used in statistics when trying to determine a curve that has the minimal sum of the deviations squared from a given set of data. In simple terms, it finds a function that will disregard any unnecessary psychological disorder in collected data results by finding a value that has the smallest amount of deviation from the majority of the data. Quadratic regression is not used to perfectly fit a data set, but to find the best curve that goes through the data set with minimal deviation. This function can be found using a GDC. First you must input the data points into lists, (L1 and L2).Then you go to the statistic math functions and choose QuadReg. It will know to use the two lists to determine he quadratic function using the method of least squares. Once the calculation has complete d, the data seen below (values for the coefficients of the function) will be presented QuadReg a = 1. 244 b = 8. 458 c = 0. 8392 With this data we can determine that the function is When graphed, this function has the shape seen below Updated Data table, including septic function values Guide Number (from tip) Quadratic values Distance from Tip (cm) Cubic values Distance from Tip (cm) Septic values Distance from Tip (cm) Quadratic Regression Distance from Tip (cm) Original Distance from Tip (cm) 1 10 2 22 3 37 4 54 5 74 6 97 7 122 8 149 10 23 38 54 74 96 121 149 10 23 38 55 74 96 120 149 11 23 37 55 74 96 121 148 10 23 38 55 74 96 120 149 By analyzing the graph and values of the quadratic regression function, it is evident that it is a relatively accurate form of modeling the data. Four of the eight points matched that of the original data, with an error of 1. The most notable difference between the quadratic regression function and the quadratic function previously determined, is the placement within the data f the accurate values. The regression function matched the middle data, while the quadratic function matched the end data. It is interesting to see how two functions in the same form, found using different methods yielded opposite areas of accuracy. Best Match The function that acts as the best model for this situation is the septic function. It is the only function that satisfies each of the original data points with its equation. Through finding the quadratic, cubic and septic functions, it was discovered that the degree of the polynomial was directly correlated to the functions accuracy to the data.Therefore it was no surprise that this function acts as the best fit for this data. The other cause for this septic function having the best correlation to the original data is due to the septic function being established by creating a system of equations using all of the data points. 9th Guide Using my quadratic model, it can be determined where the opti mal placement for a ninth guide would be by substituting 9 in for x in the equation . Using my quadrating model, it was found that the optimal placement for a ninth guide on the rod is 179cm from the tip of the rod.king of beastss fishing rod is 230cm long, yet his eighth guide is only 149cm from the tip of the rod. That means that there is 81cm of the line that is not being guided from the jar to first guide. By adding a ninth guide, that distance will be shortened form 81cm to 51cm. By doing this, it will be less likely for the line to draw up and become tangled in this 81cm stretch where there is no guide. Another implication of adding another guide would be that the weight dispersal of a fish being reeled in would be spread over another guide, which will allow for an easier task of reeling in the fish.There is even fair to middling space on the rod for a 10th guide at 211cm from the tip of the rod. This guide would once again shorten the excess line further to a point where the excess line between the reel and the first guide is shorter than line between the first and second guide. This could cause problems with reeling and casting efficiency, as that extra guide would cause slowing stool of the line. The benefit would be that once again the weight distribution of fish would be spread over a larger number of guides.Overall, it would be beneficial to include a ninth guide to Leos fishing rod, but anymore will likely hinder its efficiency. Marks Fishing Rod Guide Number (from tip) Distance from Tip (cm) 1 10 2 22 3 34 4 48 5 64 6 81 7 102 8 124 To see how well my quadratic model fits this new data, they must be both plotted on the same graph, seen below. My quadratic model for Leos fishing rod correlates with Marks fishing rod data for the first few values and then diverges as the number of guides increases by growing at a higher exponential rate.The difference between Leo and Marks eighth guide from the tip of their respective rods is 25cm, yet both me ns first guides acquire the same distance from the tip of their rods. The quadratic function used to model Leos fishing rod does not correlate well with Marks fishing rod data. Changes to the model must be made for it to fit this data. The best way to find a model for Marks data would be to go through the same steps that we went through to determine the first quadratic formula that models Leos fishing rod.By doing so, specific values that better represent Marks fishing rod data could be used to establish a better fitting function. The main limitation of my model is that is was designed as a function for Leos data specifically. It was created by solving systems of equations that used solely Leos fishing rod for data. Consequentially, the quadratic model best represented Leos fishing rod, which had a maximum length of 230cm, with differently place out guides. There were many differences between Leo and Marks fishing rods (such as maximum length and guide spacing) that caused my orig inal quadratic model to not well represent Marks data.