Explaination of Curvature

prace

1. The problem statement, all variables and given/known data
Find the curvature of y = x³


2. Relevant equations

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

3. The attempt at a solution

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

I got the answer numerically, but I am looking for an explaination 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:



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[sup]3[sup] 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!