If the position-vector description of a curve is given by $\mathbf{R}\left(t\right)\=x\left(t\right) \mathbf{i}\+y\left(t\right) \mathbf{j}\+z\left(t\right) \mathbf{k}$, then $\stackrel{\.}{\mathbf{R}}\=\stackrel{\.}{x} \mathbf{i}\+\stackrel{\.}{y} \mathbf{j}\+\stackrel{\.}{z} \mathbf{k}$, where the over-dot notation represents differentiation with respect to $t$. Hence, the integrand in the arc-length integral for R is $\sqrt{{\stackrel{\.}{x}}^2\+{\stackrel{\.}{y}}^2\+{\stackrel{\.}{z}}^2}$ = $∥\stackrel{\.}{\mathbf{R}}∥$.

For the given curve,

Hence, the arc-length function is $s\left(t\right)\=\frac{1}{2}{\int }_0^t\left(u^2\+2\right) \mathit{\ⅆ}u\=\frac{t^3}{6}\+t$. (Note that because the upper limit of integration is $t$, the integration variable itself must be some other variable, here chosen to be $u$.)

The arc-length function is therefore $s\left(t\right)\=t^3\/6\+t$. (Note that because the upper limit of integration is $t$, the integration variable itself must be some other variable, here chosen to be $p$.)

The arc-length function is therefore $s\left(t\right)\=t^3\/6\+t$. (Note that because the upper limit of integration is $t$, the integration variable itself must be some other variable, here chosen to be $p$.)