The curve in terms of arc legth is \(r(s)=2\ \cos\ \left(\frac{s}{\sqrt{5}}\right)i\ +\ 2\ \sin\ \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{s}{\sqrt{5}}k\).
Given:
The function \(r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >\ s=(\sqrt{5t})\)
Calculate:
The given vector-function for the path is
\(r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >\)........(1)
The length of the curve is
\(s=(\sqrt{5t})\)

\(t=\ \frac{s}{\sqrt{5t}}\) Substituting this value of t in equation (1), we get \(r(s)=\ \langle2\ \cos\ \left(\frac{s}{\sqrt{5t}}\right),\ 2\ \sin\ \left(\frac{s}{\sqrt{5t}}\right),\ \frac{s}{\sqrt{5t}}\rangle\) Or \(r(s)=2\ \cos\ \left(\frac{s}{\sqrt{5}}\right)i\ +\ 2\ \sin\ \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{s}{\sqrt{5}}k\). Thus, the curve in terms of arc length is \(r(s)=2\ \cos\ \left(\frac{s}{\sqrt{5}}\right)i\ +\ 2\ \sin\ \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{s}{\sqrt{5}}k\).

\(t=\ \frac{s}{\sqrt{5t}}\) Substituting this value of t in equation (1), we get \(r(s)=\ \langle2\ \cos\ \left(\frac{s}{\sqrt{5t}}\right),\ 2\ \sin\ \left(\frac{s}{\sqrt{5t}}\right),\ \frac{s}{\sqrt{5t}}\rangle\) Or \(r(s)=2\ \cos\ \left(\frac{s}{\sqrt{5}}\right)i\ +\ 2\ \sin\ \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{s}{\sqrt{5}}k\). Thus, the curve in terms of arc length is \(r(s)=2\ \cos\ \left(\frac{s}{\sqrt{5}}\right)i\ +\ 2\ \sin\ \left(\frac{s}{\sqrt{5}}\right)j\ +\ \frac{s}{\sqrt{5}}k\).