Represent the helix in Example 1(b). $$\begin{equation*} \mathbf{r}(t)=\cos t\mathbf{i}+\sin t\mathbf{j}+t\mathbf{k}\quad t\geq 0 \end{equation*}$$

so the parameter is arc length.

Solution The arc length function $s( t)$ along the curve $\mathbf{r}=\mathbf{r}( t)$, $a\leq t\leq b$, from $t=a$ to an arbitrary $t$ is given by the integral $$\begin{equation*} s( t)\;=\;\int_{a}^{t}\left\Vert \mathbf{r}^{\prime} (u)\right\Vert du \end{equation*}$$

Since the graph of the helix starts at $t=0,$ and $\left\Vert \mathbf{r}^{\prime}(t)\right\Vert\;=\;\sqrt{2}$ for all $t$ (Example 1(b)), we have $$\begin{equation*} s( t)\;=\;\int_{0}^{t}\sqrt{2}\kern.7ptdu=\Big[ \sqrt{2}\kern.7ptu\Big] _{0}^{t}= \sqrt{2}\kern.7pt t \end{equation*}$$

Then $t=\dfrac{s}{\sqrt{2}}.$ The helix using arc length $s$ as the parameter is expressed as $$\begin{equation*} \mathbf{r}( s)\;=\;\cos \dfrac{s}{\sqrt{2}}\kern.7pt\mathbf{i}+\sin \dfrac{s}{ \sqrt{2}}\kern.7pt\mathbf{j}+\dfrac{s}{\sqrt{2}}\kern.7pt\mathbf{k}\quad s\geq 0 \end{equation*}$$