I have a function as follows:
y= a*x^b
a=7e-5
b=-0.755
I attached a simple graph of the function. As it is apparent from the graph the CURVATURE of the function increases from ZERO to a finite value (around x=0.1-0.2) and then it decreases to reach a value of ZERO. I did my best to draw the CURVATURE of this function using the following formula:
K=f"/(1+f'^2)^1.5
However, using this formula I could not reach the predictable trend of the curvature. Do you have any idea what is the problem?
I can work with MATLAB and Excel.
Your help is appreciated in advance.
Kind regards,
Ebrahim
Hi, the curvature at a given point is given by the second derivative of your function.
Kappa(x)= y'' / (1+ y'^2) 3/2 Let: y(x) = a x-b
Then one has y'(x)= -a b x^(-b-1) and y''(x)= a b(b+1) x^(-b-2)
Kappa (x)= a b(b+1) x^(-b-2) / [1+(a b x^(-b-1))^2]^3/2
Substitute above last two expressions in the curvature formula given above you may obtain the curvature of any point on the curve as function of the abscissa denoted as x.
Hi Richard Epenoy and Tarık Ömer Oğurtanı and Peter Breuer
Thank you for your attention to my question.
I accept all answers and these are the things I achieved.
Let me rephrase my question. A quick look to the graph that I presented, one can say that the radius of the best fitted circle decreases to a point (I highlighted in the graph) and then increases again. I mean at that point the graph has minimum radius and thus the maximum curvature. I say that because it is looking like that two ends of the graph are line and in the aforementioned point it has a curvature due to transitions in the slopes!!
Now, using all points you mentioned the f" or K curve tend to decrease constantly while I expected to see a local maximum or minimum.
Is there any other way to locate the transition point?
kind regards,
Hello Peter Breuer
Thank you for your reply.
I am looking for a way to locate the coordinate of the point with smallest radius of curvature in the graph.
Cheers,
Thank you Peter Breuer
That is a good way to look at the problem. However, it does not solve the issue for all functions. I mean, for other complicated functions, like f= a*(b^x)*(x^c) and a-b*exp(-c*x^d) that I am working on , it is difficult to find a relationship as you mentioned.
Instead, given the f, f', and f", the function of curvature should be easily determined using the curvature formula (K=f"/(1+f'^2)^3/2). Then, it will be super easy to draw the graph for the curvature.
My problem is to find that graph for a function y=f(x).
Thank you for your time ...
Regards,
Best thing is to use vector analysis to define the curvature and even torsion of a curve given in 3D space by taking the position vector of an arbitrary point P on the curve as function of the arc length s with respect any arbitrary fixes point on the curve. Then curvature define as Kappa n = d t/ ds where t is unit tangent vector at P and n is called the Principal unit normal vector up there . One can easily show that the absolute value of above expression may be written as
Kappa= d teta /ds where teta is an angle which measures the inclination of the tangents at two success point P and P' on the curve. This derivative is the rate of turning of the tangent at point P and it is called the curvature of the curve at P. Actually I have used this primitive geometric definition of curvature to solve very complicated problem related to surfaces and interfaces in computer simulation to describe the capillary forces using discrete representations.
Similarly one can define the Binormal unit vector b = n ^ t Then torsion may be defined by db/ds = - Lambda n Where Lambda measures the arc-rate of turning of the unit vector b which is called torsion.
I say that because it is looking like that two ends of the graph are line and in the aforementioned point it has a curvature due to transitions in the slopes!! 5 days ago Ebrahim Sangsefidi mentioned
That is a wrong statement. As you might see from the derivative of the curve which is negative function and monotonously decreasing in magnitude and approaches asymptotically to (from the negative side !!!) ZERO.
Check the power index in the denominator of your expression for the curvature. It should be 3/2 rather than 1/2. Good luck
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