• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17

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.

...read more.

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

image09.png

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)

...read more.

Conclusion

th 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.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Extended Essay- Math

    48�����+��߿��_��?��[*�d��^Ǵ�K/1/2"�����!��L,y �]��kK�4� k��(c)k??�#F��YW8�ÅyÌ¿'��...�խ[7t1/41�1"��|�1�]C�+W�ͱc�ߣ :�"���W�^����Ø��x�Sh>[c9`�Q� #����c�-�L g1/4f�T'���a���I-9���[�l�l������0c-t��G� @hp"�`����o- -~o3/4���%�""0�)@��>?l Hl�O��(tm)�2h: (r)(tm)�2(r)(tm)�"�Õ�'''P��w��13�j�3/4 ��iÓ¦a~GÙ¡"� �`h---�G���>}��"�cV�>X��"�FL�/J1/2��{�&@Gjf� v"2� �L1/2N>!...��M���Mt9����1/2�7���... ,H"��`^"3/4Y1c"(tm)$��* �%�!1/2�O��P'_I�R)"'?:FLA"D��WÒ¬T�DG�äS% ��4+�"� ��c�D@IH~%�J��@tH��1P'_I�R)"'?:FLA"D��WÒ¬T�DG�äS% ��4+�"� ��c�D@I�� �J!N %IEND(r)B`�PK !�$ F�G�Gword/media/image7.png�PNG IHDR���k�+RiCCPICC Profilex�YwTͲ�(tm)�,KXr�9�s"�a�K�Q��"� H ���*��""(�PD@�PD$1/2A?�{�y��ys����v���� �JTT�@xD\��(tm)!��" ?n�:� ����F��Y#��\����5&�#�0�72�@��������k��~Q1q ~ �'ĸ(� ��(��-;8�7^�3/4�0���� ;xZ %&'0B�O� Bä�'#�(c)0�!X�/��G�#--���,�or�� S(3/4�ȤP����� 2(tm)ØFI����"��#���EFZÚ0�۰"��?��

  2. Ib math HL portfolio parabola investigation

    - (X4 - X3) |. + n D = | (X2 - X1) - (X4 - X3) |. D = | X2 - X1 - X4 + X3|. D =| | D = | D = | D= | This is the proof by deduction for the value of D.

  1. IB Mathematics Portfolio - Modeling the amount of a drug in the bloodstream

    Y: 3.7 +10.0 = 13.7 because it's left over amount of drug from previous dose plus 10 (new dose) for a total of 13.7 micrograms. New equation: 13.7= z e^(-.1863*6) Step 3: solve for z a.) -.1863*6=-1.1178; b.)e^-1.1178=.3260; c.)13.7/ .3260=41.896=z Step 4: graph the old equation replacing the 10.391 with z (41.896)

  2. Math Portfolio: trigonometry investigation (circle trig)

    0.5878=0.5878 When 13 is to represent the value of x and 77 is to represent the value of y in the conjecture sinx=cosy, sin(13) would equal to cos(77). Again this should add up to 90 degrees. sinx=cosy sin(13)=cos(77) sin ?=cos(90- ?)

  1. Math Portfolio Type II

    14 59994 5 47246 15 59997 6 53272 16 59999 7 56856 17 59999 8 58643 18 59999 9 59439 19 59999 20 59999 * When the harvest size = 2500 fishes We shall initiate the harvest of 2500 fishes after the 16th year when the population stabilizes at 59999.

  2. Stellar Numbers math portfolio

    Next the task considers stellar (star) shapes with p vertices, leading to p-stellar numbers. In this case, p, or the number of vertices has a value of 6. Images of the first six stages are shown below. S5 S6 Now here is a chart showing the number of dots (the stellar number)

  1. Mathematics Portfolio. In this portfolio project, the task at hand is to investigate the ...

    +Gz 0.01 18 0.03 14 0.1 11 0.3 14 1 7 3 6 10 4.5 30 3.5 In both these sets of data I will define the independent variable as the G-force represented along the y-axis, and the dependent variable as the time represented on the x-axis: � where +Gx represents the positive acceleration in the horizontal direction in grams.

  2. Artificial Intelligence & Math

    Such electronic criminals will now not be able to communicate and work freely without being caught by the surveillance system and dealt with by the judicial system. So not only does the system deter and scare people from partaking in illegal activities but it also helps the police find, catch, and punish such offenders.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work