Fit a curve of the form v = ae^kt using Least Square Method.

Introduction:

A basic idea in statistics and mathematics, curve fitting is applied extensively in many disciplines, including biology, engineering, physics, and economics. The process entails determining which mathematical function best fits a given set of data points. The Least Square Method is one of the methods for curve fitting that is most frequently applied. With an emphasis on fitting curves of the type v=ae^kt, we will explore the different aspects of curve fitting with the Least Square Method in this blog article.

Understanding the Least Square Method:

The least square method is a mathematical approach to determining the best-fit curve for a set of data points. It operates by reducing the sum of the squares of the vertical deviations between each point and the curve. To put it another way, the goal is to minimize the total error and identify the curve that “best” fits the data.
The least squares method is based on the idea of creating an objective function that reflects the sum of squared disparities between observed data points and the associated model-predicted values. Minimizing this objective function allows us to identify the model parameters that best suit the data.

Fitting a Curve of the Form 𝑣=ae^kt:

The exponential growth or decay is shown by the curve of the type v=aekt, which varies based on the sign of the coefficient “k”. Numerous scientific and engineering applications, including population expansion, radioactive decay, and chemical kinetics, frequently deal with this curve.
We must figure out the values of the parameters a and k that best describe the data in order to fit a curve of this form to a set of data points (ti ,vi ), where ti represents the independent variable (for example, time) and vi represents the dependent variable (for example, velocity).

The general approach involves transforming the equation into a linear form by taking the natural logarithm of both sides:

ln⁡(v)=ln⁡(a)+kt

Now, we have a linear equation of the form y=mx+b, where:

The values of m and b that minimize the sum of the squares of the variations between the actual values of ln (𝑣) and the values predicted by the linear model can then be estimated using the Least Square Method.

Steps for Curve Fitting using Least Square Method:

Now, let’s focus on finding a parabola of the form y = a + bx + cx² using the least square method. This form of a parabola includes three parameters: a (the y-intercept), b (the coefficient of x), and c (the coefficient of x²).

To apply the least square method, we need a set of data points (x, y). Once we have the data, we can proceed with the following steps:

Advantages of Least Square Method:

Practical Applications:

The Least Squares Method is a widely used curve fitting technique in various areas.

Challenges and Considerations:

Although the Least Square Method is an effective technique for fitting curves, there are certain restrictions and things to keep in mind:

Conclusion:

The Least Squares Method for curve fitting is a strong data analysis and modeling approach applicable to a wide range of areas. We can estimate the parameters of mathematical models and forecast future trends by using this method, which minimizes the sum of squares of the differences between the observed and anticipated values. Whether you work as a scientist, engineer, or data analyst, becoming proficient in curve fitting will improve your capacity to draw conclusions from data that are relevant to your area and to make well-informed choices.

Question:

The voltage v across a capacitor at time 1 seconds is given by the following table. Use the principal of least squares to fit a curves of the form v = ae^kt to data: 

t

0

2

4

6

8

v

150

63

28

12

5.6

The voltage v across a capacitor at time 1 seconds is given by the following table. Use the principal of least squares to fit a curves of the form v = ae^kt to data: 

t

0

2

4

6

8

v

150

63

28

12

5.6

Solution:

Let         t = x , v = y

Constructing the table as shown below;

x

y

x2

Y = log (y)

xY

0

150

0

2.1760

0

2

63

4

1.7993

3.598

4

28

16

1.4471

5.788

6

12

36

1.0791

6.474

8

5.6

64

0.7481

5.984

∑x = 20

∑y = 258.6

∑x2 = 120

∑Y = 7.2496

∑xY = 21.844

 

 

                                                          V = aekt

It becomes                                       y = aekx       (Eq A) ꓼ v = y and t = x

Taking log on both sides

                                                         Log y = log a + log ekx

                                                           Y = log a + kx log e                                            ꓼ log mn = n log m

                                                                 Y = A + Bx

➡️ B = k log e

➡️ A = log a

The normal equations are

                                                                                  ∑Y = nA + B∑x            (Eq 1)

                                                                                 ∑xY = A∑x + B∑x2       (Eq 2)

From (Eq 1),

                                                                            7.2496 = 5A + 20B          (Eq 3)

From (Eq 2),

                                                                     21.844 = 20A + 120B            (Eq 4)

 

                                                                                 (Eq 4) – 4(Eq 3)

                                                                            21.844 = 20A + 120B

                                                                           28.9984 = 20A + 80 B

                                                                                 -7.1544 = 40B

                                                                                 B = -0.17886

Put in (Eq 3)

                                                                   7.2496 = 5A + 20(-0.17886)

                                                                               A = 2.16524

Since

                                                                                A = log a

                                                                          a = anti-log (A)

                                                                     a = anti-log (2.16524)

                                                                             a = 146.29

                                                                             a ≈ 146.3

Also,

                                                                             B = k log e

                                                                            k = B/log e

                                                                           k = -0.4118

Put in (Eq A) ,

                                                                              y = aekt

                                                                      y = 146.3 e-0.4118x

Reput,

                                                                     y = v and x = t

                                                                   v = 146.3 e-0.4118t

 

Picturized Solution:

Frequently Asked Questions (FAQs):

1. What is curve fitting, and why is it important?

With the use of curve fitting, a mathematical approach, we may characterize the relationship between variables and generate predictions based on observed data by determining which curve best fits a given set of data points. It is significant because it aids in the comprehension of data patterns, forecasting, and the extraction of insightful knowledge for decision-making.

2. What is the Least Square Method, and how does it work?

Through the mathematical process of reducing the sum of squares between the disparities between the observed and predicted values, the Least Square Method can be used to determine which curve best fits a given set of data points. It functions by reducing the residual sum of squares (RSS) between the values that the model predicts and the observed data points.

3. How does curve fitting using the Least Square Method differ from other curve fitting techniques?

The Least Square Method’s durability, adaptability, and simplicity make it a popular choice. In contrast to certain other methods, this strategy is applicable to a broad range of mathematical models, such as logarithmic, polynomial, exponential, and linear functions.

4. What are the advantages of fitting curves of the form v= ae^{kt} using the Least Square Method?

We can represent exponential growth or decay phenomena, which are prevalent in scientific, engineering, and financial applications, by fitting curves of this type. With the help of the Least Square Method, we may reliably assess patterns and make predictions by estimating the parameters a and k that best represent the data.

5. What are some practical applications of curve fitting using the Least Square Method?

The Least Square Method of curve fitting has many applications in science, engineering, economics, and medicine, among other domains. Among its many uses are the modeling of growth, decay, and diffusion processes; stock price prediction; market trend analysis; and medication kinetics modeling.

6. What are some challenges and considerations when using the Least Square Method for curve fitting?

Choosing the right mathematical model is crucial, handling outliers and missing values in data requires preprocessing, interpreting results while taking underlying assumptions and model parameter uncertainties into account, and there is a risk of overfitting complex models to sparse data.

7. How can I implement curve fitting using the Least Square Method in practice?

When applying the Least Square Method to curve fitting, the steps involved are usually data organization, equation transformation into a linear form, linear regression to estimate model parameters, fit evaluation using statistics like coefficient of determination, and interpretation of the findings to gain insights for decision-making.

8. Are there any recommended tools or resources for learning more about curve fitting and the Least Square Method?

Yes, there are a number of sites that offer implementations of curve fitting techniques, including software libraries like NumPy and SciPy for Python, online courses, and textbooks. Furthermore, research papers and scholarly journals frequently offer insightful analyses and practical examples of curve fitting applications in certain domains.