How much the curvature changing in a specific direction

[edgy smelt]

hey i wanted to calculate how much the curvature is changing on a certian point and i came across this which does work but it gives how much its changing both ways but i need to know how much its changing only in the right direction -> like for some function like cos(x^2) at -0.5 and 0.5 gives the same value for both since -0.5 is changing rapidly in the negative direction and 0.5 is changing rapidly in the right direction so is there anyway i can know how much its changing on specific direction?

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[civic pine]

I'm not sure what you mean "curvature in one direction". Curvature, at least from what I can tell, is scalar and pointwise.

[edgy smelt]

like at point -0.5 in the negative direction it changes rapidaly but in the right direction its more of flat

[edgy smelt]

this can work but both it only gets the result for both directions how

[civic pine]

If you're interested in the change in curvature, wouldn't you want the derivative of the curvature function?

[edgy smelt]

derivative of k(x)?

ill try one sec

[civic pine]

I mean, yeah, right? That's how you analyze how a function is changing.

[edgy smelt]

without the abs value tho right

[civic pine]

I dunno, frankly I'm not sure where this formula comes from. The derivative of absolute value is easy, though.

[edgy smelt]

https://tutorial.math.lamar.edu/classes/calciii/curvature.aspx

[regal ivy]

Maybe it's better to use a more general formula.
If we have a parametric curve x = x(t), y = y(t), then:
κ(t) = (x'y'' - x''y')/((x')^2 + (y')^2)^(3/2)
This gives the signed curvature.

[edgy smelt]

its exactly the same result

[regal ivy]

Yeah, I get that, but this formula is applicable to a wider range of curves.

[edgy smelt]

yeah but im looking to get the curvature change only in the positive direction 😭

[regal ivy]

Not sure what you mean by that.

[edgy smelt]

look what im trying to achieve is i want to know when the curve is being more of flat or rapidly changing but the thing is at point -0.5 as an example it does change rapidly in the negative direction but not in the positive direction but at point 0.5 it's the opposite so i want something that gives how much it's changing only to the positive direction on the x axis

[regal ivy]

Well, you can just look at the first derivative, then.

[edgy smelt]

that only tells weather it's increasing or decreasing it doesnt account for the curvature how rapidly the shape is changing beyond the immediate slope

[regal ivy]

I'm still not sure what you mean. Maybe try multiplying the second and first derivative?

[edgy smelt]

look im implementing an integral that makes a polygon having n the number of points across the curve the thing is im not having ∆x as js smth static like (b-a)/2 but it gives less points where the curvature is more of flat and more points where the curve shape is changing rapidly
so in my case if i want to get the integral from lets say -0.5 to 2 i would give the range from -0.5 to 0.5 less points than what i would give from0.5 to 2

so it's weighted based on how much the curve is changing

[civic pine]

Curvature at a point doesn't have direction. The curvature at -0.5 is constant. The curvature to either side of -0.5 may be behaving how you say, but to say the curvature at -0.5 is changing "differently" in "different directions" is nonsensical. -0.5 is one single point, it doesn't have "directions".

[edgy smelt]

okay but like i told u the behavior at -0.5 is different going positive direction than the negative direction so i only want to calculate how much it's changing that's what i need

[civic pine]

No, it's not, because that's not the behavior at -0.5, that's the behavior around -0.5.

[edgy smelt]

well sure that's still what i need to have a function to in order to have an optimal ∆x

[civic pine]

The point is that the curvature at a point doesn't change, and even insofar as it does "change", i.e. is differentiable, that's still not a directionally variable thing.

[edgy smelt]

yeah true what i need is how much the curvature change from -0.5 to the right

[civic pine]

Which is nonsensical.

[edgy smelt]

how come i mean if the eye can spot it so does math right 😭

[civic pine]

My eye can't spot it.

[edgy smelt]

can't u see that point -0.5 in the positive direction is more of flat but at like point -0.5 the shape ks rapidly changing

[civic pine]

No. What I can see is that at some point which is to the right of -0.5, the curvature is flat, and that at some point to the left of -0.5, the curvature is greater.

[edgy smelt]

yes that's what i mean

which is what i want to evaluate

how much of flat at -0.5 is

[civic pine]

Then you have to measure the curvature at those points.

[edgy smelt]

look it i used the curvature equation at let's say -0.1 I'll get a value showing it doesnt change much at it but at let's say 0.5 it DOES change much since going to the right of 0.5 the curve is changing much which would give a value for it BUT at -0.5 it has the same thing at -0.1 going to the right but it changes rapidly going to the left which is the same as 0.5 that's why im saying directionally

[edgy smelt]

Okay so what worked for me is

The second derivatives of the function from two points like -0.5 to 0 as an example so
sd(f(x))d (x1+x2)/2

Abs(NumericDerivative(D(Cos(x^2),x),x,(x*1+x*2)/2))

[formal pebble]

@edgy smelt

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[edgy smelt]

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