Determine times for three variables using Least squares powerfit

Hello, I'm having trouble figuring out the thinking time, braking car time and motorcycle braking time at speeds 20km/hr, 50km/hr, 80km/hr, 100km/hr, 120km/hr and 150km/hr using the least squares power fit method

[John D'Errico]

The crystal ball is so foggy. But for some reason, I cannot see into your notes, to then guess what model you would pose. And not knowing the model, is there anyway to know what model you are thinking about? Perhaps if you try really hard, and think about the model, get a picture of it in your mind, then you can telepatically tell us what it is.

Easier of course, if is you just edit your question, or add a comment (NOT an answer) that shows the model. But we can try the telepathy thing if you think it will work.

[Torsten]

In other words:

What is "least-squares powerfit" ?

[Juan Barragan]

yes, I've been having difficulty on figuring out the equations for each time

Answer 1

[Ravi]

Hi Juan Barragan,

I understand you are facing trouble in arriving at equations for the variables. I hope the following approach will be helpful to you.

The least squares power fit method is a technique used to find the best-fitting power function that minimizes the sum of the squared differences between the observed and predicted values.

A power-law relationship is represented by the equation,

y = a * x ^ b

Where a is a coefficient and b is the exponent.

The following function can help you in obtaining the equation.

speed = [30, 45, 60, 75, 90, 120];
thinking = [6, 9, 11, 15, 17, 22];
powerfit(speed, thinking);

Equation: y = 0.2436x^0.9432

y = 0.2436 * (20 ^ 0.9432)

y = 4.1097

function [] = powerfit(x, y) 
    % Perform least squares power fit 
    logx = log(x); 
    logy = log(y); 
    
    % Use polyfit for linear regression on transformed data 
    p = polyfit(logx, logy, 1); 
    
    % Extract coefficients 
    a = exp(p(2));  % Intercept term corresponds to log(a) 
    b = p(1);       % Slope corresponds to exponent b 
    
    % Display results 
    fprintf('Equation: y = %.4fx^%.4f\n', a, b); 
end 

The above function “equation” takes input two arguments “x” and “y” where “x” represents the Speed and “y” can take Thinking, Braking Car and Braking motorcycle.

Once after obtaining the equation, we can find the response variable by simply substituting the Speed value in place of “x” in the obtained equation.

To know more about the “polyfit” function, please refer to the following link.

polyfit - https://www.mathworks.com/help/matlab/ref/polyfit.html

Hope this explanation resolves the issue you are facing.

Thanks,

Ravi Chandra