= Solution
Use the <arc length> coordinate oriented in the direction of increasing $t$: $s(t)=\int_0^t2e^\tau\,d\tau=2(e^t-1)$, with $s>-2$. Also $\mathbf r''=e^t(\sqrt2,-2\cos t,-2\sin t)$, giving $|\mathbf r''|^2=6e^{2t}$ and $\mathbf r'\cdot\mathbf r''=4e^{2t}$. The <Lagrange identity for the cross product> yields
$$
|\mathbf r'\times\mathbf r''|^2=|\mathbf r'|^2|\mathbf r''|^2-(\mathbf r'\cdot\mathbf r'')^2=8e^{4t}.
$$
The <curvature of a space curve> is therefore
$$
\boxed{\kappa(t)=\frac{|\mathbf r'\times\mathbf r''|}{|\mathbf r'|^3}
=\frac{e^{-t}}{2\sqrt2},\qquad
\kappa(s)=\frac1{\sqrt2(s+2)}.}
$$
For the forward portion $t\ge0$, this has ordinary nonnegative <arc length> $s\ge0$. If instead nonnegative distance is measured backwards from $t=0$, put $s_b=2(1-e^t)$ for $t\le0$; that branch has $\kappa=1/[\sqrt2(2-s_b)]$, $0\le s_b<2$. Choosing orientation avoids confusing the two points at the same unoriented distance.
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