# Measure of the Sharpness of a curve

1. Oct 25, 2012

### Bavid

Measure of the "Sharpness of a curve"

I have a set of curves that belong to the family of curves $y=\frac{c}{x^m}$, where $m$ and $c$ are parameters.
The attached picture (save.png) shows three such curves for different values of $m$ and $c$.
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 $m$ and $c$ 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?

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2. Oct 25, 2012

### rudolfstr

Re: Measure of the "Sharpness of a curve"

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

3. Oct 25, 2012

### HallsofIvy

Staff Emeritus
Re: Measure of the "Sharpness of a curve"

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,
$$\kappa= \frac{x'y''- x''y'}{(x'^2+ y'^2)^{3/2}}$$

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

4. Oct 25, 2012

### arildno

Re: Measure of the "Sharpness of a curve"

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.

5. Oct 25, 2012

### rcgldr

Re: Measure of the "Sharpness of a curve"

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 xminr :

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)) |

Last edited: Oct 25, 2012
6. Oct 26, 2012

### Bavid

Re: Measure of the "Sharpness of a curve"

Thanks everyone.

7. Oct 26, 2012

### Bavid

Re: Measure of the "Sharpness of a curve"

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Last edited: Oct 26, 2012