\subsection{}
On sait que le vecteur tangante unitaire est donné par la formule \\
$\vec{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$\\
Calculons $\vec{r}'(t)$ soit: $\vec{r}'(t) = ((t\sin{t})', (t\cos{t})', (t)')$\\
\begin{math}
	(t\sin{t})' = \sin{t} + t\cos{t}\\
	(t\cos{t})' = \cos{t} - t\sin{t}\\
	(t)' = 1\\
	\vec{r}'(t) = (\sin{t} + t\cos{t}, \cos{t} - t\sin{t}, 1)\\
\end{math}

Déterminons à présent $\|\vec{r}'(t)\|$ soit:\\
\begin{math}
	\|\vec{r}'(t)\| = \sqrt{(\sin{t}+t\cos{t})^2 + (\cos{t}-t\sin{t})^2 + 1^2}\\
	\Leftrightarrow \|\vec{r}'(t)\|^2 = \sin^2{t}+\cos^2{t} + 2t\sin{t}\cos{t}-2t\sin{t}\cos{t} + t^2\cos^2{t}+t^2\sin^2{t}+1\\
	\Leftrightarrow 1 + 0 + t^2\cos^2{t}+t^2\sin^2{t}+1\\
	\Leftrightarrow 1 + 0 + t^2 + 1\\
	\Leftrightarrow t^2 + 2
\end{math}

On a donc:\\
\begin{math}
	\vec{T}(t) = \frac{(\sin{t}+t\cos{t}, \cos{t}-t\sin{t}, 1)}{\sqrt{t^2+2}}\\
	\boxed{\Leftrightarrow \frac{1}{\sqrt{t^2+2}}(\sin{t}+t\cos{t}, \cos{t}-t\sin{t}, 1)}
\end{math}

\subsection{}
On a 
\[
	\vec{r'}(t_0) \text{ -> vectur directeur}
\]
\[\vec{r}(t_0) \text{ -> points de passage}\]
Soit\\
\begin{math}
	\vec{r}(t_0) = (t_0\sin{t_0}, t_0\cos{t_0}, t_0) = (\frac{\pi}{2}, 0, \frac{\pi}{2})\\
	\Leftrightarrow (\frac{\pi}{2}\sin{\frac{\pi}{2}}, 0, \frac{\pi}{2})\\
	\Leftrightarrow (\frac{\pi}{2}, 0, \frac{\pi}{2}) \Rightarrow t_0 = \frac{\pi}{2}\\
\end{math}
On remplace dans \(\vec{R}'(t)\) soit\\
\begin{math}
	\vec{r}'(t_0) = (\sin{\frac{\pi}{2}} + \frac{\pi}{2}\cos{\frac{\pi}{2}}, \cos{\frac{\pi}{2}}-\frac{\pi}{2}\sin{\frac{\pi}{2}}, 1)\\
	\Leftrightarrow (1 + 0, -\frac{\pi}{2}, 1)
\end{math}

On a donc 
\[
\left\{
\begin{array}{l}
x = \frac{\pi}{2} + r \\
y = 0 - \frac{\pi}{2} r \\
z = \frac{\pi}{2} + r
\end{array}
\right.
\]

\subsection{}

On cherche \(\vec{r}(t)\) tel que \(\vec{r}'(t) = r(t)\) soit:\\
\begin{math}
	\vec{r}(t) = (t\sin{t_0}, t\cos{t_0}, t)\\
	\int{\vec{r}'(t)} = (\int{t\sin{t}}, \int{t\cos{t}}, \int{t})
\end{math}
\begin{itemize}
	\item \(\int{t\sin{t}dt} = -t\cos{t} - \sin{t} + C_1\)
	\item \(\int{t\cos{t}dt} = t\cos{t}-\cos{t} + C_2\)
	\item \(\int{tdt} = \frac{t^2}{2} + C_3\)
\end{itemize}
\begin{math}
	\int{\vec{r}(t) = (-t\cos{t}-\sin{t}, t\cos{t}-\cos{t}, \frac{t-2}{2}) + C \text{ (C = $C_1 + C_2 + C_3$)}}
\end{math}