Please note, this is not homework. This is a problem self-devised. Given a function f(x), find the points where it bends. We know that where f''(x) = 0, it is flat. f''(x) is the slope, is the rise/run. Where rise=run, this point is where it changes from more delta in the x direction to where there is more delta in the y direction, and vice versa. Therefore, I propose that where f''(x) = 1 is the bend x for a function, providing the 1 is in the domain of f''(x) I actually came up with this in calc class, then applied to towards optimization of car speeding over a distance(where does speedings efficency really become less effecient) I am pretty sure it has a programming use, I jsut havn`t thought up one yet. Bugle4d

Your problem statement is unclear... you should define what "bend" means.

Also, if you''re considering the trajectory of a car, it would be a bad idea to express it as y in terms of x; think about what happens when the car moves straight in the y direction, or when it curves back around.

Your math is fine, but you need to be thinking in terms of parametric equations, or vectors.

It isn''t clear what a bend means, but you may want to look at the second derivative, i.e. the slope of the slope.

Let f ℜ→ℜ² a parameterized function.

The curvature radius χ of the curve defined by f is
χ(t) = |f ′′(t)⊗f ′(t)|/|f ′(t)|³



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Stolen from Magmai Kai Holmlor, who held it from Oluseyi, who was inspired by Kylotan...

[edited by - Fruny on May 7, 2002 10:13:45 PM]

[edited by - Fruny on May 7, 2002 10:14:08 PM]

Well what I'm sayning is that you look at a graph and observe that it seems to bend around point(s).
I was interested in finding this point.
I'm not talking about the max/min, nor am I speaking of inflection points.


Its like with f(x)=x^2
It goes down, then seems to bend at a point, then goes to min, then mirrors the behavior on the other side.

Whats the bend point ?

Edit: algorithim correction; I forgot to put it in.
It should be where |f'(x)| = 1

Bugle4d

[edited by - Vlion on May 7, 2002 10:34:14 PM]

it still isn''t really clear what exactly you mean by bend. it seems to me that f(x)=x² is "bending" for all values of x. I don''t really see what is special about the point at which f''(x)=1, other than the fact that at that point f''(x)=1 .

as to the thing about the second deriv, I only see that being useful to tell which "way" the function is "bending" (concavity of course). is this what you meant?

ewen

So f(x) = x is always bending?

quote:

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Original post by Vlion
Its like with f(x)=x^2
It goes down, then seems to bend at a point, then goes to min, then mirrors the behavior on the other side.

Whats the bend point ?

---

That''s not true, the second derivative of f(x)=x² is a constant (2). That means that by your definition of "bend", f(x) = x² never bends.

For your car problem you''re actually looking for the point where the 2nd derivative is zero. This would be the point where the acceleration starts slowing down...

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codeka.com - Just click it.

WHAT IS A BEND POINT?

if you have some sort of sphere going tangential to the current curvepoint, and this sphere has the same "rotationspeed" (curvature?) as your curve, you mean where the radius has some minimum? (like it would have at x=0 in f(x)=x^2) you mean this midpoint of this sphere there? you wanna know this?

or what?

"take a look around" - limp bizkit
www.google.com

Hum, sure it isn''t just an artifact of raster graphics, i.e. the jaggies?