$\def\vv#1{\mathbf{\vec{#1}}} \def\unitvv#1{\mathbf{\hat{#1}}} \def\derivative#1#2#3{\frac{d^{#3}#1}{{d#2^{#3}}}}$
I'm not sure if "domain scaling" is the correct word, but I had a problem:
Find the curvature of the curve $\vv r(t)= (9 + \cos 6t - \sin 6t)\unitvv i + (9 + \sin 6t + \cos 6t)\unitvv j+ 4 \unitvv k$
To do this, I used the formula that I had already derived for the curvature:
$\kappa := \frac{d\vv T}{ds} = \frac{\|{\vv{r'}\times\vv{r''}}\|}{{{\|\vv{r'}}\|}^3}$
It was an absolute mess with all sorts of factors of $6$ flying around the page, and I kept losing track of my work, so I decided to try and rescale by using $ u := \frac{t}{6}$ into:
$\vv r(u) = (9+\cos u - \sin u)\unitvv i + (9 + \sin u + \cos u) \unitvv j + 4 \unitvv k$.
I justified that the curvature would be the same by lemma:
Lemma 1: $ u(t) = \alpha t ; \alpha \in \mathbb{R \implies}\kappa (\vv r(t)) = \kappa(\vv r(u))$
Proof: $\kappa (\vv r(u)) = \frac{\|{{\derivative{\vv r}{u}{}}\times{\derivative{\vv r}{u}{2}}}\|}{{{\|\derivative{\vv r}{u}{}}\|}^3} = \frac{\|({\derivative{t}{u}{}\derivative{\vv r}{t}{} )\times((\derivative{t}{u}{})^2(\derivative{\vv r}{t}{2}) + (\derivative{t}{u}{2})(\derivative{\vv r}{t}{}))}\|}{{\|\derivative{t}{u}{}\derivative{\vv r}{t}{}\|}^3} = \frac{\|({\alpha^{-1}\derivative{\vv r}{t}{} )\times(\alpha^{-2}(\derivative{\vv r}{t}{2}) + 0(\derivative{\vv r}{t}{}))}\|}{{\|\alpha^{-1}\derivative{\vv r}{t}{}\|}^3} = \frac{\|\alpha^{-3}\|\|{{\derivative{\vv r}{t}{}}\times{\derivative{\vv r}{t}{2}}}\|}{{{\|\alpha^{-3}\|\|\derivative{\vv r}{t}{}}\|}^3} = \frac{\|{{\derivative{\vv r}{t}{}}\times{\derivative{\vv r}{t}{2}}}\|}{{{\|\derivative{\vv r}{t}{}}\|}^3} = \kappa(\vv r(t))$
It got me the right answer of $\frac{\sqrt2}{2}$ in my example problem, but I can't seem to wrap my head around it conceptually from what my idea of $\kappa$ represents. Perhaps I made an error in my lemma that made this only work because the curvature was a constant? If not, maybe my idea of $\kappa$ is wrong and someone can give be a better intuition of it.
"Scaling the domain"/reparametrizing the curve just changes the speed at which the curve is traced out. But the curvature at a point is a property of the shape that's traced out and doesn't care about the speed.