Why curvature of a plane curve is k =d(phi)/ds?

[A330NEO]

Why curvature of a plane curve is k = [tex] \frac{d\phi }{ds} [/tex]?

I know that curvature of a plane is [tex] \frac{\left | r'(t) \times r''(t) \right |}{\left | r'(t) \right |^3} [/tex], and that led to this.

[tex]k = \frac{\left | \frac{d^2s}{dt^2} \right |sin\phi }{\left | \frac{ds}{dt}\right |^2}[/tex]

But I can't go any further. Any ideas?

 

[Fantini]

That is the definition of curvature. The expression you presented is the expression for the curvature given in terms of the derivatives of an arbitrary parametrization of a curve.

Consider a parametrization by arc length $\varphi: I \subset \mathbb{R} \to \mathbb{R}^3$. Then

$$\left\| \frac{d \varphi}{ds} \right\| = 1 \text{ and } \left\| \frac{d^2 \varphi}{ds^2} \right\| = \kappa,$$

which is the curvature of the curve.

 

[A330NEO]

That is the definition of curvature. The expression you presented is the expression for the curvature given in terms of the derivatives of an arbitrary parametrization of a curve.

Consider a parametrization by arc length $\varphi: I \subset \mathbb{R} \to \mathbb{R}^3$. Then

$$\left\| \frac{d \varphi}{ds} \right\| = 1 \text{ and } \left\| \frac{d^2 \varphi}{ds^2} \right\| = \kappa,$$

which is the curvature of the curve.

uhh... I still don't get it. [tex]k=\left|\d{T}{s} \right|[/tex](if length of curve is s and T is for its tangential vector), but how does that become [tex]\left| \d{\phi}{s} \right|[/tex]? why use [tex] \phi [/tex], which is the angle between T and i(1, 0, 0)? Is there an specific characteristic of 'plane' curve that separages it from other 3-dimentional curves?