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I know that curvature for some curve $C$ defined parametrically is:

$$kappa=left|{dvec{T}over ds}right|$$

Which basically is the rate at which the tangent vector to the curve changes, as the arclength of the curve changes.

In another source, I saw the definition of curvature as the following:

If $P_1$ and $P_2$ are two points on the curve, $|P_1P_2|$ is the arclength between those two points, and $Phi$ is the limit of the angle between tangent vectors at the points $P_1$ and $P_2$ (as it goes to zero I assume), then the curvature is defined as:

$$kappa=lim_{|P_1P_2|to 0}{Phiover |P_1P_2|}$$

Which basically means, the rate at which the angle of tangent vectors in global frame of reference changes, as the arclength of the curve changes.

I assume that this second definition can be rewritten using the notation from the first example as:

$$kappa={dphiover ds}$$

Where $phi$ is the angle between the vector tangent to the curve, and some constant global axis of reference (which could be the x axis, but realy it could be any line or vector on the same plane).

Given the second (weird in my opinion) definition of curvature, I can't see how those two definitions can be equivalent. Maybe they are not, I don't know. May be they are; if yes, how?

Also, here's a picture of the section from the book where the second definition appears in (it's not in English):

Note that the text agrees with yet another definition of curvature, which I am aware of: $kappa=frac1r$, where $r$ is radius of curvature.

Let's work with the first definition. We have
begin{align} kappa (s) &= left| frac{dvec T}{ds}(s)right| \ &= lim_{delta sto 0}frac 1 {delta s}left| vec T(s + delta s) - vec T(s) right| \
&= lim_{delta s to 0} frac 1 {delta s} sqrt{(vec T(s + delta s) - vec T(s)).(vec T(s + delta s) - vec T(s))} \
&= lim_{delta s to 0} frac 1 {delta s} sqrt{| vec T(s + delta s)|^2 + | vec T(s) |^2 - 2vec T(s).vec T(s + delta s)}end{align}
But the curve is parameterised by arc length! So
$$ | vec T(s + delta s)|^2 = | vec T(s)|^2 = 1$$
and
$$ vec T(s).vec T(s + delta s) = cos Phi(s, s + delta s),$$
where $Phi(s, s + delta s)$ is the angle between $vec T(s)$ and $vec T(s + delta s)$.

Plugging this in, we get
begin{align}
kappa &= lim_{delta s to 0} frac 1 {delta s} sqrt{2 - 2 cos Phi(s, s + delta s)} \
&= lim_{delta s to 0} frac 1 {delta s} 2 sin left( frac { Phi(s, s + delta s) } {2}right) \
&= lim_{delta s to 0} frac {Phi(s, s + delta s)} {delta s} times frac{sin left( frac { Phi(s, s + delta s) } {2}right)}{frac{Phi(s, s + delta s)}{2}}
end{align}
Clearly, $lim_{delta s to 0} Phi(s, s + delta s) = 0$, so
begin{align}
kappa &= lim_{delta s to 0} frac {Phi(s, s + delta s)} {delta s} times lim_{Phi to 0} frac{sin left( frac { Phi } {2}right)}{frac{Phi}{2}} \
&= lim_{delta s to 0} frac {Phi(s, s + delta s)} {delta s} times 1 \ &= lim_{delta s to 0} frac {Phi(s, s + delta s)} {delta s}end{align}
which agrees with the second definition.

By your first statement, $T'(s)=kappa(s)N(s)$ where $kappa$ is the curvature and $N$ is the unit normal vector. Now, let's consider $T(s)$ and $T(s+Delta s)$, so that we can compare the angles between tangent vectors. Now, $T(s)$, $T(s+Delta s)$, and $N(s)$ are all unit vectors, so we can draw the following picture representing all of these vectors:

Here, $Delta theta$ is the angle between the two tangent vectors and $Delta T=T(s+Delta s)-T(s)$. From the diagram, we can find that $Delta Tcdot T(s)=1-cos Delta theta$ and $Delta Tcdot N(s)=sin dtheta$. From the x-component, we get the following equation:

$$frac{dT}{dtheta}cdot T(s)=lim_{Delta thetarightarrow 0}frac{Delta Tcdot T(s)}{Delta theta}=lim_{Delta thetarightarrow 0}frac{1-cos(Delta theta)}{Delta theta}=0$$

This was pretty obvious since $T'(s)=kappa(s)N(s)$ is orthogonal to $T(s)$, so nothing new there. However, from the y-component, we get:

$$frac{dT}{dtheta}cdot N(s)=lim_{Delta thetarightarrow 0}frac{Delta Tcdot N(s)}{Delta theta}=lim_{Delta thetarightarrow 0}frac{sin(Delta theta)}{Delta theta}=1$$

Now, let's use chain rule to figure out what $frac{dT}{dtheta}$ is:

$$frac{dT}{ds}=kappa(s)N(s)=frac{dT}{dtheta}frac{dtheta}{ds} rightarrow frac{dT}{dtheta}=frac{kappa(s)N(s)}{frac{dtheta}{ds}}$$

Finally, let's plug this value for $frac{dT}{dtheta}$ into the equation with the dot product:

$$frac{dT}{dtheta}cdot N(s)=1rightarrow frac{kappa(s)N(s)cdot N(s)}{frac{dtheta}{ds}}=1rightarrow frac{dtheta}{ds}=kappa(s)N(s)cdot N(s)=kappa(s)$$

(Note that the last step uses $N(s)cdot N(s)=1$ since $N(s)$ is a unit vector.)

At this point, we have shown $frac{dtheta}{ds}=kappa(s)$, which is what we originally set out to prove. Q.E.D.

Of course, this isn't exactly a formal proof since it assumes $T(s+Delta s)$ is in the plane spanned by $T(s)$ and $N(s)$, which isn't necessarily true. However, I think this is a fair approximation since $T'(s)=kappa(s)N(s)$, so $T(s+Delta s)approx T(s)+Delta sT'(s)=T(s)+[Delta skappa(s)]N(s)$. In any case, I feel like this geometric argument gives a better visual intuition for why $|T'(s)|=frac{dtheta}{ds}$.