You are free to derive a formula for curvature in any coordinate system that you want, and with respect to any parameter along the curve that you want. For example, you have probably also seen a formula, expressed in terms the $x$-coordinate parameterization $(x,f(x))$, for the curvature of the graph of a function $y=f(x)$:
$$\kappa = \frac{|f''(x)|}{(1+f'(x))^{3/2}}
$$
So the question is not why "all other alternatives are shut down", because they aren't (and by the way "shutting things down" is almost never how mathematics works).
Perhaps instead a better question might be
Why is the arc length parameterization the primary one used to express the formula for curvature?
I think the answer to this is simply that the arc length parameterization is so natural from a geometric viewpoint: it can be derived using nothing but Euclidean geometry and a limiting argument, as you learn in a real analysis course. So it might be the first thing a geometer would want to know about curvature: How do you write down a formula for curvature expressed in terms of the arc length parameter?
But let me suggest two still better questions:
Is there a definition of curvature independent of parameterization? And can one use that definition to derive to derive a formula in terms of the arc length parameteriation (or in terms of any other parameterization)?
There is indeed a nice definition which is independent of parameter, and it has three steps:
- The unit circle $S^1 = \{(x,y) \mid x^2+y^2=1\}$ has curvature $1$ at each point:
- Curvature varies inversely under similarity: Suppose $C$ and $C'$ are two curves such that $C$ is similar to $C'$. Let $f : \mathbb R^2 \to \mathbb R^2$ be a similarity map such that $f(C)=C'$. Let $r>0$ be the similarity factor, meaning that $d(f(p),f(q)) = r d(p,q)$ for all $p,q \in \mathbb R^2$. Then for all $x \in C$ with corresponding point $x' = f(x) \in C'$, the curvature of $C'$ at $x'$ is equal to $\frac{1}{r}$ times the the curvature of $C$ at $x$. (For example, by combining 1 and 2 one can prove easily that all radius $1$ circles have curvature $1$ at each point, and all radius $r$ circles have curature $1/r$ at each point.
- Curvature is a second order invariant: For any curve $C$ and $p \in C$, and for any circle $C' \subset \mathbb R$ which matches $C$ to second order at the point $p$, the curvatures of $C$ and $C'$ at $p$ are equal (this is the "osculating circle" condition referred to in the comment of @Kajelad).
Knowing this, one can prove the arc length parameterization formula for curvature, and any other formula you want such as the formula for the $x$-coordinate parameterizaiton given earlier.
Best Answer
HINT: To differentiate $|d\vec r/dt|$, you need to differentiate $$\left|\frac{d\vec r}{dt}\right|^2 = \frac{d\vec r}{dt}\cdot\frac{d\vec r}{dt}$$ by the product rule and then deduce what you want by the chain rule.