Example 4.12.1.

The space curve \(x = a\cos t\hspace {0.2cm}, \hspace {0.2cm} y = a\sin t \hspace {0.2cm}, \hspace {0.2cm} z = t\) lies on the cylinder \(x^2 + y^2 = a^2\). Find the length of one turn of the curve of where \(t \in [0, 2\pi ]\). \[\overline {r}(t) = a\cos t \textbf {i} + a\sin t \textbf {j} + t\textbf {k}\] \[\overline {r}'(t) = -a\sin t \textbf {i} + a\cos t \textbf {j} + \textbf {k}\]

\begin {align*} \left |\overline {r}'(t)\right | & = \sqrt {a^2\sin ^2t + a^2\cos ^2t + 1}\\ & = \sqrt {a^2 + 1} \end {align*}

\[L = \int ^{2\pi }_0 \sqrt {a^2 + 1}\hspace {0.1cm}dt = 2\pi \sqrt {a^2 + 1}\]

The arclength function \(S\) of a curve \(C\) with vector function \(\overline {r}(t) = f(t)\textbf {i} + g(t) \textbf {j} + h(t)\textbf {k}\hspace {0.1cm}, \hspace {0.1cm} a\leq t \leq b\) where \(\overline {r}'(t)\) is continuous at \(C\) is transversed exactly once at \(t\) increases from \(a\) to \(b\) is given by \[S(t) = \int ^b_a\hspace {0.1cm}\left |\overline {r}'(t)\right |\hspace {0.1cm}dt = \int ^b_a \sqrt {(x')^2 + (y')^2 +(z')^2}\hspace {0.1cm}dt\]

Hence, \(S(t)\) is the length of the part of \(C\) between \(\overline {r}(a)\) and \(\overline {r}(b)\). From the formula above, we obtain \[\frac {dS}{dt} = \sqrt {\Bigg (\dfrac {dx}{dt}\Bigg )^2 + \Bigg (\dfrac {dy}{dt}\Bigg )^2 + \Bigg (\dfrac {dz}{dt}\Bigg )^2}\hspace {0.1cm}dt = \left |\overline {r}'(t)\right |\]