The software also gives the r2 value of the graph that indicates the accuracy of the line best fit, in relation to the points that have been plotted. The closer the value to the digit “1”, the more accurate it is. Here we get the r2 value as 0.0009. This means that the line of best fit for this graph is very inaccurate.
In order to get the most accurate line of best fit, we must try different types of line of best fit, such as Linear, Polynomial to the nth degree, and exponential.
Graph 2(a)
Linear Function → r2 = 0.0009
Polynomial Function To The nth Degree
n = 2 → r2 = 0.6549
n = 3 → r2 = 0.7322
n = 4 → r2 = 0.8322
n = 5 → r2 = 0.9103
We can see that as the value of “n” increases, the r2 value approaches the digit “1”. We can then hypothesize that as we increase our nth degree, we will get a more accurate line or curve of best fit. In order to prove this, we consider more examples of increasing “n” values in a Polynomial Function to the nth degree for the line or curve of best fit.
Graph 2(b)
n = 6 → r2 = 0.9135
n = 7 → r2 = 0.9449
n = 8 → r2 = 0.9452
n = 9 → r2 = 0.9558
n = 10 → r2 = 0.963
Graph 2(c)
n = 11 → r2 = 0.9642
n = 12 → r2 = 0.9740
n = 13 → r2 = 0.9750
n = 14 → r2 = 0.9128
n = 15 → r2 =0.9783
Graph 2(d)
n = 16 → r2 = 0.9785
n = 17 → r2 = 0.9825
n = 18 → r2 = 0.973
n = 19 → r2 = 0.9835
n = 20 → r2 = 0.9846
Graph 2(e)
n = 21 → r2 = 0.9841
n = 22 → r2 = 0.9671
n = 23 → r2 = 0.9871
n = 24 → r2 = 0.9775
n = 25 → r2 = No Result due to Software Limitation – Not Enough Points For Calculation
Graph 2(f)
Exponential → r2 = 0.0492
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)x13-(2.2432736E-17)x12+(3.9136109E-14)x11-(8.7371105E-12)x10+(9.7793277E-10)x9-(7.9178582E-08)x8+(5.3547547E-06)x7-(0.00029054549)x6+(0.011346264)x5-(0.29223255)x4+(4.6109657)x3-(40.453239)x2+(165.91127)x-196.95227
We can also find the rate of change between 2 points by using the formula y2 – y1
x2 – x1
Graph 4
From Graph 4, we can see that between the time when x = 42 hours and x = 48 hours, the rate of change of the rate of flow of the river is the greatest.
The rate of flow when 42 ≤ x ≥ 48 =
y2 – y1
x2 – x1
= 1520 – 1090
48 – 42
= 71.67
From the data in Graph 1, we see that the rate of flow of the river increases exponentially, and then decreases exponentially. A weather pattern that could represent this kind of data would be of a Scanty Rainy Season, where there is an input of water in the form of rain to the river, increasing the rate of flow for a particular time period.
Between 28th October 2007 and 29th October 2007, a total volume of water flowed past the measuring station. In order to find this total volume, we can find the area under the graph, between the time 00:00 on the 28th and 00:00 on the 29th.
Area Under The Graph Between The Time 00:00 on the 28th (x = 24 hours), and 00:00 on the 29th (x = 48 hours) = 2.3947E+4 cubic feet (Calculated using Area Under Graph option on program “Graph” which uses the Simpson’s Formula, i.e:
The time when the flow rate is 1000cfs =
T = 38.8823 hours, T = 91.4791 hours
(Correct to 4 d.p.)
The amount of flowing water began to increase on the 27th of October, 2007, and kept increasing on each day until 6 am on the 29th of October, 2007.
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.
