Measure of the "Sharpness of a curve"

Bavid

I have a set of curves that belong to the family of curves [itex]y=\frac{c}{x^m}[/itex], where [itex]m[/itex] and [itex]c[/itex] are parameters.
The attached picture (save.png) shows three such curves for different values of [itex]m[/itex] and [itex]c[/itex].
Now these curves have different 'sharpenss' of curvature (to see what I mean by sharpness, observe how 'sharp' a corner the lowermost curve forms compared to the uppermost).

I am trying to find a function F of [itex]m[/itex] and [itex]c[/itex] that can quantify this sharpness, i.e., larger value of F(m,c) indicates that the corresponding curve has a sharper corner or the vice versa.

Any ideas how to go about constructing the function F?

Attached Thumbnails

 

rudolfstr

I would recommend looking at how fast the derivative changes d²y/dx² and the higher value of this would mean a sharper curve.

HallsofIvy

It sounds to me like you are talkig about the curvature of a graph. That is the same 1 divided by the radius of curvature (the radius of the circle that best fits the curve at a given point). It can also be calculated as the length of the derivative of the unit tangent vector with respect to arclength or, if you are given x and y as functions of a parameter, t,
[tex]\kappa= \frac{x'y''- x''y'}{(x'^2+ y'^2)^{3/2}}[/tex]

As you can see, that will depend upon the second derivatives as rudolfstr suggests.

arildno

Basically (as HallsofIvy says), the idea of the "curvature" at a specific point, is logically related to answer the question:
"What is the radius of the circle that best approximates the curvature of the graph at that point?"
The smaller the (osculating) circle must be, the greater the curvature (the circle's radius is simply the absolute value of the reciprocal of the curvature!)

As has been said, in terms of derivatives, this is given by a disgusting formula involving second derivatives.

rcgldr

radius of curvature for a general function {x, y(x)}:

radius(x) = (1 + y'(x)²)^3/2 / | y''(x) |

where y' is first derivative of y, and y'' is second derivative of y.

For this case, to determine the minimum radius, you'd have to take the derivative of the radius of curvature and solve for x_minr :

radius'(x_minr) = 0

minimum radius = radius(x_minr)

Then F(c, m) = 1 / (minimum radius (c, m))

Although not needed for this case, the radius of curvature for a general function {x(t), y(t)}:

radius(t) = ( (x'(t))² + (y'(t))² )^3/2 / | (x'(t) y''(t)) - (y'(t) x''(t)) |

Bavid

Thanks everyone.

Bavid

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