\subsection{}
On a:
\begin{itemize}
    \item \(x = e^t \cos t\)
    \item \(y = e^t\)
    \item \(z = e^t \sin t\)
\end{itemize}
Calculons les dérivées soit:
\begin{itemize}
    \item \(\frac{d}{dt} (e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t)\)
    \item \(\frac{d}{dt} (e^t) = e^t\)
    \item \(\frac{d}{dt} (e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t)\)
\end{itemize}
On a alors: \[
\vec{r'}(t) = \left( e^t (\cos t - \sin t), e^t, e^t (\sin t + \cos t) \right)
\]
Calculons la norme de \(\vec{r'}(t)\) soit:
\[\|\mathbf{r'}(t)\| = \sqrt{(e^t (\cos t - \sin t))^2 + (e^t)^2 + (e^t (\sin t + \cos t))^2}\]
\begin{itemize}
    \item \((e^t (\cos t - \sin t))^2 = e^{2t} (\cos t - \sin t)^2 = e^{2t} (\cos^2 t - 2 \sin t \cos t + \sin^2 t) = e^{2t} (1 - 2 \sin t \cos t)\)
    \item \(e^{2t}\)
    \item \((e^t (\sin t + \cos t))^2 = e^{2t} (\sin^2 t + 2 \sin t \cos t + \cos^2 t) = e^{2t} (1 + 2 \sin t \cos t)\)
\end{itemize}
On a donc\\
\begin{math}
    \|\mathbf{r'}(t)\| = \sqrt{e^{2t} (1 - 2 \sin t \cos t) + e^{2t} + e^{2t} (1 + 2 \sin t \cos t)}\\
    \|\mathbf{r'}(t)\| = \sqrt{e^{2t} (1 - 2 \sin t \cos t + 1 + 1 + 2 \sin t \cos t)}\\
    \|\mathbf{r'}(t)\| = \sqrt{3 e^{2t}} = \sqrt{3} e^t
\end{math}
L'abscice curviligne est donné par la formule \[s(t) = \int_{t_0}^{t} \sqrt{3} \, e^u \, du\] On a donc:\\
\begin{math}
    s(t) = \sqrt{3} \int_{t_0}^{t} e^u \, du \\
    \Leftrightarrow \sqrt{3} \left[ e^u \right]_{t_0}^{t} \\
    \Leftrightarrow \sqrt{3} \left( e^t - e^{t_0} \right)\\
\end{math}
Fixons \(t_0 = 0 \Rightarrow s(t) = \sqrt{3}(e^t-1)\)

\subsection{}
Sur l'intervale \(0 \leq t\leq 2\pi\) on a
\[\sqrt{3}(e^{2\pi}-e^0) = \sqrt{3}(e^{2\pi}) - 1\]