Arc Length Parametrization And Unit Tanget When looking at $r(t)=(\cos t,\sin t, t)$ what is the difference between
$$T(t)=\frac{r'(t)}{||r'(t)||}=\frac{2}{\sqrt{2}}(-\sin t, \cos t, 1)$$
And 
$$T(s)=(-\frac{1}{\sqrt{2}}\sin(\frac{s}{\sqrt{2}}),\frac{1}{\sqrt{2}}\cos(\frac{s}{\sqrt{2}}),\frac{1}{\sqrt{2}})$$
Where
$r(s)=(\cos(\frac{s}{\sqrt{2}},\sin(\frac{s}{\sqrt{2}}),(\frac{s}{\sqrt{2}}))$
[$s(t)=\int_{0}^{t}\|r'(t)\|\,du=\int_0^t\sqrt{\sin^2t+\cos^2t+1}du=\int_0^t\sqrt{2}=\sqrt{2}t\Rightarrow t=\frac{s}{\sqrt{2}}]$
The are both unit tangent? 
 A: When you have a regular curve $r\colon I \to \Bbb R^3$, you can write $\widetilde{r}(s) = r(s(t))$, where $\widetilde{r}$ has unit speed and $s = s(t)$ denotes arclength. When you write $T(t)$ and $T(s)$, you're supposed to think of "$T_r(t)$" and "$T_{\widetilde{r}}(s)$". They're related by $T_{r}(t) = T_{\widetilde{r}}(s(t))$, for all $t \in I$, point being that these vectors start at the same point in the curve (which is described by different parameters, depending on the parametrization you use).
A: We have cylindrical parametrization $r(t)=(\cos t,\sin t, t)$ which normally may define in $[0,2\pi]$, and unit tangent vector
$$T(t)=\dfrac{r'(t)}{||r'(t)||}=\dfrac{\sqrt{2}}{2}(-\sin t, \cos t, 1) ~~~~,~~~~ t\in[0,2\pi]$$
and also the arc-length 
$$s(t)=\int_{0}^{t}\|r'(t)\|\,dt=\int_0^t\sqrt{\sin^2t+\cos^2t+1}dt=\int_0^t\sqrt{2}=\sqrt{2}t$$
which varies respect to parameter $t$ and this defines another parametrization 
$$r(s)=\left(\cos\dfrac{s}{\sqrt{2}},\sin\dfrac{s}{\sqrt{2}},\dfrac{s}{\sqrt{2}}\right)$$
which is differ from $r(t)$ in spanning, where $s\in[0,2\sqrt{2}\pi]$, and in this case, the unit tangent vector is
$$T(s)=\left(-\dfrac{1}{\sqrt{2}}\sin\dfrac{s}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\cos\dfrac{s}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\right) ~~~~,~~~~ s\in[0,2\sqrt{2}\pi]$$
which shows they are the same with different spanning.
