Fitting Equations to Data

8/4/2019 Fitting Equations to Data

http://slidepdf.com/reader/full/fitting-equations-to-data 1/49

Reporters:Badilla, Harriet G.

Salva, Karen Frances Remey A.



� Spatial profile (Temp. vs. Time)

� Time history (Voltage vs. Time)

� Cause-and-effect (Force as a Function of Displacement)

� System output as a function of input (Yield of

a chemical reaction as a function of temperature)



� To determine the equation of a curve thatrepresents the aggregate of the data



� The overall strategy in fitting the curve to a

set of data points is to determine anappropriate set of value for the coefficients in

the equation of the chosen curve.



� Determine the value of the coefficients a and

b in the straight-line equation y= ax + b so

that the line will pass through n data points.



� Open a new worksheet and enter the x-data

(independent variable) in the leftmost

column� Enter the y-data (dependent variable) in the

next column

�

Plot the data as an x-y graph (i.e., an XYchart) with arithmetic coordinate. Do notinterconnect the individual data points

� Click on one of the plotted data points



� Choose Add Trendline from the Chart menu.

Then specify the type of curve and request

any pertinent options when the resultingdialog box appears.

� Press the OK button. The curve fitting willthen be carried out and the results displayed

automatically



� An engineer has measured the force exerted

by a spring as a function of its displacement

from its equilibrium position. The ff. datahave been obtained:

Data Point No. Distance (cm) Force (N)

2 2.0

2 4 3.5

3 7 4.5

4 11 8.0

5 17 9.5



y = 0.514x + 1.279

R² = 0.957

0.0

2.0

4.0

6.0

8.0

10.0

12.0

0.0 5.0 10.0 15.0 20.0

Force Exerted By A Spring



� Determine the values of a and b in the

equation y=ax + b in such manner that the

sum of the squares of the errors will beminimized

� Wrote each error term as ei = y i f( x i )e

i = y

i ( ax

i + b )

� Z = sum of the squares of the errors (SSE)z= ei

2 + e22 + e 3

2 + . . .



� A. Exponential functions

y = aebx

ln y = ln a + bx

Let u= ln y and c= ln a

u = bx + c



� The transient behavior of a capacitor has

been studied by measuring the voltage drop

across the device as a function of time. Theff. data obtained:

Time, sec Voltage Time, sec Voltage

0 10 6 0.5

1 6.1 7 0.3

2 3.7 8 0.2

3 2.2 9 0.1

4 1.4 10 0.07

5 0.8 12 0.03



y = 9.7534e-0.4923x

R² = 0.9988

0

2

4

6

8

10

12

0 2 4 6 8 10 12 14

Capacitor Discharge Rate



� B. Logarithmic Functions

y = a ln x + b

� C. Power Functions

y = ax b

ln y = ln a + b ln x

let u= ln x, v= ln y and c= ln a

v = bu + c



� A chemical engineer is studying at a rate at

which a reactant is consumed in a chemical

reaction involving the manufacture apolymer. The ff. data have been

obtained, showing the reaction rate (mol/sec)as function of the concentration of the

reactant (mol/cu.ft.)



Concentration, mol/cu.ft. Reaction Rate, mol/sec.

100 2.85

80 2.00

60 1.25

40 0.67

20 0.22

10 0.0725 0.024

1 0.0018



y = 0.0018x1.599

R² = 1

0

0.5

1

1.5

2

2.5

3

3.5

0 20 40 0 80 100 120

Reaction Rate vsConcentration



� D. Polynomials

y = c1 + c2x + c3x2 + . . . + ck+1xk



� The ff. table presents the time for a high-

performance sports car to reach the various

speeds. The times are given in seconds andthe speeds in miles per hour.

� Fit an equation to the data, thus obtaining anaccurate mathematical relationship bet.

Acceleration time and top speed.



Top speed, mph Time, sec.

30 1.9

40 2.8

50 3.8

60 5.2

70 6.5

80 8.3

90 10.4

100 12.7

110 15.6

120 19.0

130 23.2

140 31.2

150 45.1



y = 0.0000x5 - 0.0000x4 + 0.0008x3 - 0.0525x2 + 1.7770x - 20.9580

R² = 0.9998

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

AccelerationTime vsTop Speed



Some guideline that may prove helpful, asoutlined below:

1. Try to plot the data as a straight line. If this issuccessful, it will indicate the function thatshould be used to represent the data.

2. It a straight-line relationship cant beobtained, try fitting different types of curvesto the data.



3. If a satisfactory curve fit cant be obtained

using the first two guidelines, try plotting

the data differently.

4. In some cases, a better fit may be obtainedby scaling the data so that the magnitude of

the x-values is more-or-less the same as they-values.



� A. Plotting the Data as a Straight Line

EquationType Equation Coordinate System

Exponential y = aebx log y versus x (semi-log)

Logarithmic y = a ln x + b y versus log x (semi-log)

Power y = ax b

log y versus x (log-log)



� A. Plotting the Data as a Straight Line

Thus, if the data appear as a straight line whenplotted in one of these coordinate systems, wewill have a clear indication of what type of trendline to fit the data.



� In example 5.7, a power function was used torepresent a set of data. The selection of thepower function was based upon previousfamiliarity with the process from which the datahad been obtained. Verify the desirability of thepower function by plotting the data onarithmetic coordinates, semi-log coordinates(both log y versus x and x versus log y), and log-log coordinates. Seek a coordinate system thatwill result in a straight-line relationship.



Concentration, mol/cu.ft. Reaction Rate, mol/sec.

100 2.85

80 2.00

60 1.25

40 0.67

20 0.22

10 0.072

5 0.024

1 0.0018



Conc. Rxn Rate

100 2.85

80 260 1.25

40 0.67

20 0.22

10 0.072

5 0.024

1 0.0018

y = 0.028x - 0.228

R² = 0.970

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

0.00 20.00 40.00 60.00 80.00 100.00 120.00

Rxn Rate vsConc



y = 0.019e0.059x

R² = 0.763

0.00

0.01

0.10

1.00

10.00

0.00 20.00 40.00 60.00 80.00 100.00 120.00

RR vsConc

(semi-log plot on the y-axis)y = 0.533ln(x) - 0.691

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

1.00 10.00 100.00

RR vsConc

(semi-log plot on the x-axis)



y = 0.001x1.599

R² = 1

0.00

0.01

0.10

1.00

10.00

1.00 10.00 100.00

RR vsConc

(log-log plot)



� B. Fitting Multiple Functions to a Given Data

Set We may find that several different functions will

represent the data reasonably well. If severaldifferent functions appear to fit the data (using

visual inspection) more or less equally well, thesum of the squares of the errors (SSE) and the r2

may be helpful in deciding which function fitsbest.



� In example 5.8, we represented some

performance data for several popular sports

cars with a fifth-degree polynomial. Expandthis study by fitting several other commonly

used functions to the data. Determine whichfunction best fits the data by comparing their

respective SSE and r2 values.



R² = 0.8384

-10.0

0.0

10.0

20.0

30.0

40.0

50.0

0 20 40 60 80 100 120 140 160

Linear (Straight Line) R² = 0.9657

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

2nd-Deg Polynomial



R² = 0.9715

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

Power R² = 0.9909

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

3rd-Deg Polynomial



R² = 0.9922

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

Exponential R² = 0.9981

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

4th-Deg Polynomial



R² = 0.9998

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

5th-Deg Polynomial R² = 0.9999

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 20 40 60 80 100 120 140 160

6th-Deg Polynomial



Function Visual Assessment r2 SSE

Straight Line Poor 0.8384 312.0

2nd-Degree Polynomial Poor 0.965 66.237

Power Poor 0.9715 55.03

3rd-Degree Polynomial Fair 0.9909 17.57

Exponential Good 0.9922 15.06

4th-Degree Polynomial Good 0.9981 3.669

5th-Degree Polynomial Excellent 0.9998 0.3862

6th-Degree Polynomial Excellent 0.9999 0.1931



� C. Substituting Other Variables for y and x

It may be helpful to substitute

1/y for y

1/x for x

x1/2 for x

y1/2 for y



� The following data represent the rate at

which an oxygenation reaction occurs within

a water purification chamber as a function of temperature. The data were obtained by a

civil engineer, who must now plot the dataand fit an appropriate equation through the

data.



Temperature, K Reaction Rate, moles/sec

253 0.12

258 0.17

263 0.24

268 0.34

273 0.48

278 0.66

283 0.91

288 1.22

293 1.64

298 2.17

303 2.84

308 3.70



y = 2E-05x3 - 0.017x2 + 4.414x - 381.8

R² = 0.999

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

250 260 270 280 290 300 310 320

Reacti Rate vs Temperat reTemp Rxn Rate

253 0.12

258 0.17

263 0.24

268 0.34

273 0.48

278 0.66

283 0.91

288 1.22

293 1.64

298 2.17

303 2.84

308 3.70



1/Temp Rxn Rate

0.00395 0.12

0.00388 0.17

0.00380 0.24

0.00373 0.34

0.00366 0.48

0.00360 0.66

0.00353 0.91

0.00347 1.22

0.00341 1.64

0.00336 2.17

0.00330 2.84

0.00325 3.70

y = 28,666,742.1324e-4,887.2323x

R² = 0.9999

0.10

1.00

10.00

0.00300 0.00320 0.00340 0.00360 0.00380 0.00400 0.00420

Reaction Rate vs InverseTemp



� D. Scaling the Data

When fitting a curve to a data, it is sometimeshelpful to scale the data so that the magnitudes of the y-values do not differ significantly from themagnitudes of the x-values.

It is recommended to redefine one of the values of the variables so hat it is expressed in terms of somemultiple of the original unit of measurement.



� A group of environmental engineers andscientists have made a prediction of the damageto the ozone layer in the northern hemisphere asa function of time, based upon present trendsrelated to the use of fluorocarbon compounds.The ff. scenario has been suggested. (thethickness of the ozone layer is expressed in

terms of arbitrary scale ranging from 0-1). Fit anequation to the data, expressing the ozone layervalue as a function of time.



Year Ozone Layer

1995 1.00

1996 0.97

1997 0.881998 0.76

1999 0.63

2000 0.50

2001 0.39

2002 0.30

2003 0.22

2004 0.17

2006 0.11

2007 0.10



y = 8E-06x5 - 0.084x4 + 336.5x3 - 67460x2 + 7E+08x - 3E+11

R² = 1

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1994 1996 1998 2000 2002 2004 2006 2008

Change in the Ozone Layer



y = 8E-06x5 - 0.000x4 + 0.006x3 - 0.047x2 + 0.011x + 1

R² = 1

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 2 4 6 8 10 12 14

Change in the Ozone Layer

Fitting Equations to Data

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