- Level: International Baccalaureate
- Subject: Maths
- Word count: 1839
Math Portfolio
Extracts from this document...
Introduction
Abhirath Singh Mathematics HL - Portfolio
IB – Year 12
Introduction
Different values measured can be represented on a scatter graph. To form a relation between the points that describes the value of each and every point may be a difficult task. In order to do this we take the Line or Curve Of Best Fit, that best describes the points. However, this line or curve of best fit may be a Linear Function, Polynomial Function To The nth Degree or an Exponential Function. In order to see which best describes the plotted points, we must see the r2 value of the lines or curves. The closer the r2 value is to the digit “1”, the more accurate the line or curve of best fit is.
In this portfolio, we are going to investigate the most accurate line or curve of best fit, that best describes the plotted points, which is the Flow Rate against Time of the Nolichucky River in Tennessee, between the 27th of October 2007, and 2nd November 2007.
Rivers carry a large Volume of water, at a particular Flow Rate at different moments of Time every day. All these details can be measured at a Weather Station.
Middle
0.7322
n = 5
0.6549
n = 6
0.9135
n = 7
0.9449
n = 8
0.9452
n = 9
0.9558
n = 10
0.9630
n = 11
0.9642
n = 12
0.9740
n = 13
0.9750
n = 14
0.9128
n = 15
0.9783
n = 16
0.9785
n = 17
0.9825
n = 18
0.9730
n = 19
0.9835
n = 20
0.9846
n = 21
0.9841
n = 22
0.9671
n = 23
0.9871
n = 24
0.9775
Graph 3
From Graph 3 we can see that we get the most accurate line of best fit in a Polynomial Function to the 23rd degree, as our r2 value is the highest. This means that the function of the plotted points is best described as:
f(x)=(1.9405965E-39)x23-(1.2327458E-36)x22+(2.5292048E-34)x21-(7.4544592E-33)x20-(2.470967E-30)x19-(2.370064E-28)x18+(9.0840112E-26)x17+(3.105029E-24)x16-(2.035099E-21)x15+(1.1461566E-19)x14-(1.7255951E-18)x13+(3.2613424E-15)x12-(7.9428277E-13)x11+(9.7793277E-11)x10-(8.7976202E-09)x9+(6.6934434E-07)x8-(4.1506499E-05)x7+(0.001891044)x6-(0.058446511)x5+(1.1527414)x4-(13.484413)x3+(82.955637)x2-(196.95227)x+439.95795
Therefore, we can derive the rate of change of this data, by taking the derivative of the above function.
d f(x) = f’(x)
d(x)
f’(x)= (4.463372E-380)x22-(2.7120407E-35)x21+(5.3113301E-33)x20-(1.4908918E-31)x19-(4.6948373E-29)x18-(4.2661153E-27)x17+(1.5442819E-24)x16+(4.9680464E-23)x15-(3.0526485E-20)x14+(1.6046193E-18)
Conclusion
The average Flow Rate of the River can be calculated by taking ∑ y
n
Where y is the values of the rate of flow every 6 hours, and “n” is the number of terms being added.
∑ y = 440+450+480+570+680+800+980+1090+1520+1920+1670+1440+1380+1300+1150+1060+970+900+850+800+780+740+710+680+660
= 24020
n = 25
Therefore Average Flow Rate = 960.8 cfs
Another way to find the Average Flow Rate is to take the Total Volume(Area Under The Graph) and divide it by the Total Time.
Total Volume = 7.6858E+4
Total Time = 144 Hours
Therefore Average Flow Rate = 7.6858E+4
144
= 533.7361 cfs
Difference between the two Average Flow Rates = 427.0639 cfs
Let The average Flow Rate = 960.8
Therefore, T when Flow Rate = 960.8 =
T = 37.9385 Hours, T = 93.2561 Hours
Conclusion
The difference between the Average Flow Rate derived by the formula ∑ y
n
and by dividing the area under the graph by the total time = 427.0639 cfs.
This shows us the inaccuracy of the function of the line of best fit, which best describes the points plotted, even though the r2 value is = 0.9871, which was the most close to the digit “1”.
From this I can conclude, that in order to calculate the right values for the Average Flow Rate or the Total Volume, we must have the exact line of best fit, where the r2 = 1. Even the smallest deviation, in this case the residual value that is 1 - r2 value = 1 – 0.9871 = 0.0129, should = 0 to give us the best results.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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