Consider the space curve $\overrightarrow{r}(t)=⟨2\cos (t),2\sin (t),t⟩$.
a. Find the arc length function for $\overrightarrow{r}(t)$.
$$s(t)=$$
b. Find the arc length parameterization for $\overrightarrow{r}(t)$.
$$\overrightarrow{r}(s)=$$

Step 1 ：Consider the space curve $\overrightarrow{r}(t)=⟨2\cos (t),2\sin (t),t⟩$.

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

Step 3 ：The derivative of $\overrightarrow{r}(t)$ is ${\overrightarrow{r}}^{\prime }(t)=⟨-2\sin (t),2\cos (t),1⟩$.

Step 4 ：The magnitude of ${\overrightarrow{r}}^{\prime }(t)$ 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 $\overrightarrow{r}(t)$ 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 ${\overrightarrow{r}}^{\prime }(t)$ 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 $\overrightarrow{r}(t)$ is $\boxed{{\displaystyle s(t)=\sqrt{5}t-\sqrt{5}a}}$.