Confusion of derivation of radius of curvature

[noble cove]

Why is the angle of sector delta phi and why is the approximate arc length given as that expression

[indigo jungle]

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[gentle flare]

I'm confused by the question. The angle is delta phi because we have to call it something.

[stable flower]

The length of a circular arc with radius r and angle θ is L = rθ.
I assume this is an arbitrary curve in this case. In that case we should have L = rθ + o(θ), θ -> 0, so dL/dθ = r.

[noble cove]

I don’t see how delta phi in the circle is related to angle between the two different tangent vectors

[gentle flare]

Because the tangent is perpendicular to the radius.

[noble cove]

Where did u get the higher order terms from L from mathematically ? Is it an expansion

[noble cove]

Yep I get that but don’t see how I would use that fact

[gentle flare]

...because then if the angles of the radii differ by delta phi, then so do the angles of the tangents to those radii.

As depicted in the diagram.

Also, that Greek letter is actually psi.

[noble cove]

Hmm I still can’t see this inutively. How would I prove this mathematically

[gentle flare]

...it's basic circle geometry.

You have two radii of a circle whose angles differ by some amount. Call the angle of the first radius psi and the angle of the second psi + delta psi.

Now observe the tangent of the first radius. What angle does it have? Well, it's perpendicular to the radius, therefore its angle is psi + pi/2.

The tangent to the second radius is perpendicular to it, therefore its angle is psi + delta psi + pi/2.

Hence the tangents also differ by delta psi.

@noble cove Get it now?

[noble cove]

Yes I get it thanks

[noble cove]

Could u also explain the series approximation

[gentle flare]

Series approximation?

[gentle flare]

What series approximation?

[noble cove]

Of delta s . I don’t get where the higher order terms come from

[gentle flare]

To be honest, I don't either.

[stable flower]

That's because we have an arbitrary curve, not a circle.

[gentle flare]

Yes, but how do we know that the variance from a true circle is on the order of delta psi squared?

[stable flower]

Well, to be fair, (Δψ)^n = O((Δψ)^2) for any n > 2, too.

So, could be cubed or whatever.

But I do have a problem with that.

It could technically, maybe, be of the order (Δψ)^n where 1 < n < 2.

So, I'm rather proposing to write it as o(Δψ).

So, a term that goes to zero more quickly than just const*Δψ. Doesn't matter if (Δψ)^2, (Δψ)^1.0001 or whatever.

[gentle flare]

I don't really see why the error should have any particular relation to delta psi at all.

[stable flower]

Well, you can think of it as a polar function and expressing its Taylor series, I guess? Maybe...

Maybe not. To be honest, I don't quite remember the process here.

[hoary flower]

@noble cove

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[noble cove]

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