CURVE FITTING | HOW TO FIND EQUATION OF STRAIGHT LINE, PARABOLA AND EXPONENTIAL CURVE USING fx82 MS CALCULATOR

CURVE FITTING



    WHAT IS CURVE FITTING?

    Curve fitting is the process of finding the best- fit curve for a given set of data. The relationship between two variables is represented by means of an algebraic equation.

    Suppose a set of n points of values (x1, y1), (x2, y2), ..., (xn , yn) of the two variables x and y are given.

    These values are plotted on the xy-plane. 


    Figure (i): Scatter Diagram

    The resulting set of points is known as a scatter diagram as shown in figure (i). 

    The scatter diagram exhibits the trend and it is possible to visualize a smooth curve approximating the data which is known as an approximating curve.

     

    LEAST SQUARE METHOD


    From a scatter diagram, generally, more than one curve may be seen to be appropriate to the given set of data. 

    The method of least squares is used to find a curve which passes through the maximum number of points.


    Let P (xi, yi) be a point on the scatter diagram as shown in figure (ii). 


    Figure (ii)

    Let the ordinate at P meet the curve y = f(x) at Q and the x-axis at M.

    Distance QP = MP-MQ
                           = yi - f(xi)

    The distance QP is known as deviation, error, or residual and is denoted by di

    It may be positive, negative, or zero depending upon whether P lies above, below, or on the curve. 

    Similar residuals or errors corresponding to the remaining (n-1) points may be obtained. 

    The sum of squares of residuals, denoted by E, is given as


    If E= 0 then all the n points will lie on y = f(x). If E
    0, f(x) is chosen such that E is minimum, i.e., the best fitting curve to the set of points is that for which E is minimum. 

    This method is known as the least square method. 

    This method does not attempt to determine the form of the curve y = f(x) but it determines the values of the parameters of the equation of the curve.


    FITTING OF LINEAR CURVES

    The normal equations for the straight line y = a + bx are

    y = na + bx

    ∑xy = a∑x + b∑x2

    These equations can be solved simultaneously to give the best values of a and b. 

    The best fitting straight line is obtained by substituting the values of a and b in the equation y = a + bx.

    Question:

    Fit a straight line to the following data:

    x

    100

    120

    140

    160

    180

    200

    y

    0.45

    0.55

    0.60

    0.70

    0.80

    0.85


    Solution:

    Let the straight line be fitted to the data be

                    y = a + bx

    The normal equations are

    y = na + bx

    ∑xy = a∑x + b∑x2

    X

    y

    x2

    xy

    100

    0.45

    10000

    45

    120

    0.55

    14400

    66

    140

    0.60

    19600

    84

    160

    0.70

    25600

    112

    180

    0.80

    32400

    144

    200

    0.85

    40000

    170

    x = 900

    ∑y = 3.95

    ∑x2 = 142000

    ∑xy = 621

     

    Substituting these values in normal equations

               3.95 = 6a + 900b                        …(i)

               621 = 900a + 14200b                …(ii)

    Solving equations (i) and (ii),

    a = 0.0476   and   b = 0.0041

    Hence, the required equation of the straight line is

                            y = 0.0476 + 0.0041x

    HOW TO FIND THE REQUIRED EQUATION OF STRAIGHT LINE USING fx82MS CALCULATOR:



     

    FITTING OF QUADRATIC CURVES

    The normal equations for the curve y = a+bx+cx² are

    ∑y = na + b∑x + c∑x2

    ∑xy = a∑x + b∑ x2 + c∑x3

    ∑x2y = a∑x2 + b∑x3 + c∑x4


    These equations can be solved simultaneously to give the best values of a, b, and c. 

    The best fitting parabola is obtained by substituting the values of a, b, and c in the equation y = a+bx+cx².


    Question:

    Fit a parabola to the following data:

    x

    1

    2

    3

    4

    5

    y

    5

    12

    26

    60

    97

    Solution:

    Let the equation of the parabola be a+bx+cx²

    The normal equations are

    ∑y = na + b∑x + c∑x2

    ∑xy = a∑x + b∑ x2 + c∑x3

    ∑x2y = a∑x2 + b∑x3 + c∑x4

    Here n = 5,

    X

    y

    x2

    x3

    x4

    xy

    x2y

    1

    5

    1

    1

    1

    5

    5

    2

    12

    4

    8

    16

    24

    48

    3

    26

    9

    27

    81

    78

    234

    4

    60

    16

    64

    256

    240

    960

    5

    97

    25

    125

    625

    485

    2425

    ∑x = 15

    ∑y = 200

    ∑x2 = 55

    ∑x3 = 225

    ∑x4 = 979

    ∑xy = 832

    ∑x2y = 3672

     

    Substituting these values in the normal equations,

          200 = 5a + 15b + 55c                       …(i)

          832 = 15a + 55b + 225c                  …(ii)

        3672 = 55a + 225b + 979c               …(iii)

    Solving equations (i), (ii) and (iii) we get

    a = 10.4,  b = -11.0857  and  c = 5.7143

    Hence, the required equation of the parabola is

               y = 10.4 – 11.0857x + 5.7143x2  

    HOW TO FIND THE REQUIRED EQUATION OF PARABOLA USING fx82MS CALCULATOR:



    FITTING OF EXPONENTIAL AND LOGARITHMIC CURVES

    For the curve y = abx

    Taking Logarithm on both sides of the equation y = abx,

    logey = logea + xlogeb

    Putting logey = Y, logea = A, logeb = B

    Y = A + BX

    The normal equations are

    Y = nA + BX

    ∑XY = a∑X + b∑X2

    Solving these equations, A and B, and, hence, a and b can be found. 

    The best fitting exponential curve is obtained by substituting the values of a and b in the equation y = abx.

    Question:

    Fit a curve of the form y = abx to the following data by the method of least squares:

    x

    1

    2

    3

    4

    5

    6

    7

    y

    87

    97

    113

    129

    202

    195

    193

    Solution:

                                         y = abx

    Taking Logarithm on both sides of the equation y = abx,

    logey = logea + xlogeb

    Putting logey = Y, logea = A, logeb = B

    Y = A + BX

    The normal equations are

    Y = nA + BX

    ∑XY = a∑X + b∑X2

    Here n = 7,

    x

    Y

    X

    Y

    X2

    XY

    1

    87

    1

    4.4659

    1

    4.4659

    2

    97

    2

    4.5747

    4

    9.1494

    3

    113

    3

    4.7274

    9

    14.1822

    4

    129

    4

    4.8598

    16

    19.4392

    5

    202

    5

    5.3083

    25

    26.5415

    6

    195

    6

    5.2730

    36

    31.6380

    7

    193

    7

    5.2627

    49

    36.8389

     

     

    ∑X = 28

    ∑Y = 34.4718

    ∑X2 = 140

    ∑XY = 142.2551

       
    Substituting these values in the normal equations,

              34.4718   = 7A + 28B                            … (i)

              142.2551 = 28A + 140B                       … (ii)

    Solving equations (i) and (ii),

               A = 4.3006   and   B = 0.156

                       logea = A

                       logea = 4.3006

                         a = 73.744

    logeb = B

    logeb = 0.156

     b = 1.1688

    Hence, the required curve is y = 73.744 (1.1688)

    HOW TO FIND THE REQUIRED EQUATION OF EXPONENTIAL CURVE USING fx82MS CALCULATOR:



    NOTES 

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