= Solution
For a regular differentiable curve $\mathbf x(t)$, its speed is $ds/dt=|\mathbf x'(t)|>0$, where $s$ is <arc length>. The oriented <unit tangent vector> is $\widehat{\mathbf T}=\mathbf x'/|\mathbf x'|$. The <curvature of a space curve> is $\kappa=|d\widehat{\mathbf T}/ds|$. Equivalently, for a twice differentiable regular curve,
$$
\kappa=\frac{|\mathbf x'\times\mathbf x''|}{|\mathbf x'|^3}.
$$
For the <helix>, differentiation gives $\mathbf x'=(-a\sin t,a\cos t,b)$ and speed $v=\sqrt{a^2+b^2}$. Consequently
$$
\boxed{\widehat{\mathbf T}(t)=\frac{(-a\sin t,a\cos t,b)}{\sqrt{a^2+b^2}},\qquad\kappa=\frac{|a|}{a^2+b^2}.}
$$
Indeed $d\widehat{\mathbf T}/dt=(-a\cos t,-a\sin t,0)/v$ has magnitude $|a|/v$, and division by $ds/dt=v$ gives the result. This requires $(a,b)\ne(0,0)$; the excluded constant curve has no well-defined <unit tangent vector>. For $a=0,b\ne0$ the curve is a straight line with zero <curvature of a space curve>.
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