Understanding Curvature in a Graph

[prace]

Homework Statement

Find the curvature of y = x³

Homework Equations

k(x) = \frac{f"(x)}{[1+(f'(x))²]^{3/2}

The Attempt at a Solution

k(x) = \frac{6x}{(1+9x^4)^{3/2}

I got the answer numerically, but I am looking for an explanation of the graph itself. I chose a relatively easy function in hopes that it would be easy to explain. Any help would be great. Please see the image below:

http://img100.imageshack.us/img100/8722/curvaturegraph0dg.jpg

The red curve represents the original function, and the blue curve represents the curvature. Could someone please explain how the blue curve represents the curvature? I just can't see how they are related.


** Sorry, I can't seem to get the LaTeX correct, but I will work on it and get the right equations on there. For the time being, I have them posted on the graph itself ** thanks

 

[HallsofIvy]

Compare the y value of the "blue" curve with the 'curviness' of the "red" curve (the colors don't show up on my reader). Right at x= 0, y= x³ is very "flat". What is its curvature? For x a little more than -1/2 y= x^{3 appears to be curving quite a lot, convex downward. What is the curvature there? Symmetrically, at x a little less than 1/2, y= x3 is curving a lot, convex upward. Do you see what happens to the curvature there? Finally, for x very "negative" or very "positive", at the two ends of the curve, y= x3 curves less and less and, sure enough the curvature graph is approaching 0. Those graphs are a very good idea.}

 

[slider142]

Also, in the plane, the curvature at a point is the reciprocal of the radius of the tangent circle to the graph at that point (See http://en.wikipedia.org/wiki/Curvature ). Thus, we define the curvature of a straight line to be zero (the radius of a tangent circle to a straight line increases without bound). Just a geometric aid to measure "curviness" a little more objectively; I'm not sure whether you've been exposed to this idea yet, but with this aid, you can see how the blue curve is generated.

 

[prace]

Awesome, thanks guys. You really cleared this up for me!