For the curve defined by
\[
\vec{r}(t)=2 \cos (t) \vec{i}+2 \sin (t) \vec{j}+4 t \vec{k}
\]
evaluate
\[
s=\int_{0}^{t}\left\|\vec{r}^{\prime}(u)\right\| d u
\]
to find the arc length function.
\[
s=
\]
Solve the arc length function for $t$ :
\[
t=
\]
Parametrize $\vec{r}(t)$ by arc length by substituting in the solved arc length function.
\[
\vec{r}(s)=
\]
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Step 1 ：First, we need to find the derivative of the given vector function. The derivative of \(\vec{r}(t)\) is \(\vec{r}^\prime(t) = -2 \sin(t) \vec{i} + 2 \cos(t) \vec{j} + 4 \vec{k}\).

Step 2 ：Next, we calculate the magnitude of \(\vec{r}^\prime(t)\). The magnitude of a vector \(\vec{v} = a\vec{i} + b\vec{j} + c\vec{k}\) is given by \(\|\vec{v}\| = \sqrt{a^2 + b^2 + c^2}\). So, \(\|\vec{r}^\prime(t)\| = \sqrt{(-2 \sin(t))^2 + (2 \cos(t))^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).

Step 3 ：Now, we can evaluate the integral \(s = \int_{0}^{t} \|\vec{r}^\prime(u)\| du = \int_{0}^{t} 2\sqrt{5} du = 2\sqrt{5}t\). So, the arc length function is \(s = 2\sqrt{5}t\).

Step 4 ：To solve the arc length function for \(t\), we get \(t = \frac{s}{2\sqrt{5}}\).

Step 5 ：Finally, we substitute \(t\) into \(\vec{r}(t)\) to parametrize \(\vec{r}(t)\) by arc length. We get \(\vec{r}(s) = 2 \cos \left(\frac{s}{2\sqrt{5}}\right) \vec{i} + 2 \sin \left(\frac{s}{2\sqrt{5}}\right) \vec{j} + 4 \left(\frac{s}{2\sqrt{5}}\right) \vec{k}\).

Step 6 ：The final results are \(s = \boxed{2\sqrt{5}t}\), \(t = \boxed{\frac{s}{2\sqrt{5}}}\), and \(\vec{r}(s) = \boxed{2 \cos \left(\frac{s}{2\sqrt{5}}\right) \vec{i} + 2 \sin \left(\frac{s}{2\sqrt{5}}\right) \vec{j} + 4 \left(\frac{s}{2\sqrt{5}}\right) \vec{k}}\).