I'm using do Carmo's book of Differential Geometry of Curves and Surfaces. The literature tells us that the curvature $k(s)$ is defined by \begin{equation} k(s) = |\alpha''(s)| \end{equation} However, in one of the exercises (for reference: par 1.5, question 12b), we had the following:
Question 12 Let $\alpha : I \to \mathbb{R}^3$ be a regular parametrized curve and let $\beta : J \to \mathbb{R}^3$ be a reparametrization of $\alpha(I)$ by the arc length $s = s(t)$, measured from $t_0 \in I$. Let $t = t(s)$ be the inverse function of $s$ and set $d\alpha/dt = \alpha'$. Prove that \begin{equation} k(t) = \dfrac{|\alpha' \wedge \alpha''|}{|\alpha'|^3} \end{equation}
I need to calculate the curvature of \begin{equation} \alpha(t) = (t,t^2,t^3) \end{equation} The solution tells us that that we need to use the identity \begin{equation} k(t) = \dfrac{|\alpha' \wedge \alpha''|}{|\alpha'|^3} \end{equation} My question is: why can we not use the first identity? It yields a more, but different, answer. Thanks in advance!
The formula $k(s)=\lvert\alpha''(s)\rvert$ works for a regular curve $\alpha(s)$ parametrised by arclength $s$. It doesn't work if you parametrise your curve differently, as you discovered.
On the other hand, $k(t)=\dfrac{\lvert\alpha'(t)\times\alpha''(t)\rvert}{\lvert\alpha'(t)\rvert^3}$ works for all regular parametrisation of a space curve.