Consider the space curve $\vec{r}(t)=\langle 2 \cos (t), 2 \sin (t), t\rangle$.
a. Find the arc length function for $\vec{r}(t)$.
\[
s(t)=
\]
b. Find the arc length parameterization for $\vec{r}(t)$.
\[
\vec{r}(s)=
\]

Step 1 ：Consider the space curve \(\vec{r}(t)=\langle 2 \cos (t), 2 \sin (t), t\rangle\).

Step 2 ：To find the arc length function for \(\vec{r}(t)\), we need to compute the integral of the magnitude of the derivative of \(\vec{r}(t)\) from a to t.

Step 3 ：The derivative of \(\vec{r}(t)\) is \(\vec{r}'(t)=\langle -2 \sin (t), 2 \cos (t), 1\rangle\).

Step 4 ：The magnitude of \(\vec{r}'(t)\) is \(\sqrt{(-2 \sin (t))^2 + (2 \cos (t))^2 + 1^2}\).

Step 5 ：We then integrate this from a to t to get the arc length function.

Step 6 ：The integral of the magnitude of the derivative of \(\vec{r}(t)\) should simplify to a simpler expression.

Step 7 ：We realize that \(4\sin^2(t) + 4\cos^2(t)\) simplifies to \(4\), because \(\sin^2(t) + \cos^2(t) = 1\).

Step 8 ：So, the magnitude of \(\vec{r}'(t)\) simplifies to \(\sqrt{4 + 1} = \sqrt{5}\).

Step 9 ：The integral of a constant is just the constant times the variable, so the arc length function should be \(s(t) = \sqrt{5}t - \sqrt{5}a\).

Step 10 ：Final Answer: The arc length function for \(\vec{r}(t)\) is \(\boxed{s(t)=\sqrt{5}t - \sqrt{5}a}\).