Changing the Parameter to Arc Length

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*}