tangents at unit circle - parametrization leads to strange result I'm thinking about the tangents at the unit circle in the upper right corner and I do so in terms of the parametrization
$$
\begin{eqnarray}
  v:&[0,1]& \rightarrow \mathbb{R}^2\\
  &t& \mapsto \left( t, \sqrt{1-t^2} \right)
\end{eqnarray}
$$
If I look at the derivative
$$
v'(t) = \left(1, \frac{-t}{\sqrt{1-t^2}}\right),
$$
I get $v'(0)=(1,0)$ and $v'\left(\frac{1}{2}\right)=(1,1)$ which is sensible. But for $v'(1)$, I get $(1,-\infty)$ instead of $(0,1)$.
My questions are: 


*

*Is this method correct in principle? I'm coming from a physics background and my formal training in calculus happened a long time ago.

*What happens at $t=1$? $v(1)$ itself gives a sensible value, only the derivative seems strange. Ultimately, my goal is to approximate the present segment of the unit cicle by a certain spline function. Can I just define $v'(t)=(1,0)$ for this?

 A: The method is correct, it's just that you've chosen an unsuitable parametrization of the curve. The tangent vector isn't parametrization invariant, and its magnitude reflects how quickly the parametrization moves along the curve. At $t=1$, the circle becomes vertical, so for a small change in $t = x$, you have arbitrarily large changes in $y$, which leads to the behavior that you see. 
To get rid of this behavior, you must reparametrize, and a standard way to do this is to parametrize using the arclength of the curve. This is called the arc-length or unit speed parametrization, and this parametrization guarentees that the tangent vector has norm $1$ everywhere. 
For our current curve, the arclength is just the angle from the $y$ axis. So the arclength $s$ as a function of $t$ is
$$s(t) = \pi/2 - \theta(t),$$
where $\theta(t)$ is the angle of the point at parameter $t$, as measured from the positive $x$-axis, as traditional. The angle is of course just given by
$$\theta(t) = \arccos(t),$$
so inverting $s$ for $t$ gives us
$$t(s) = \cos\left(\pi/2 - s\right) = \sin(s).$$
It follows that the unit speed parametrization is given by
$$s\in[0,\pi/2] \mapsto \left(\sin s, \sqrt{1-\sin^2 s}\right)=\left(\sin s,\cos s\right),$$
which is just the standard angle parametrization (it may look a little weird since we're starting at $(0,1)$ and moving clockwise rather than starting $(1,0)$ and moving counter-clockwise per usual). The tangent vector here is
$$\frac{d}{ds}\left(\sin s,\cos s\right) = \left(\cos s,-\sin s\right),$$
which is much better behaved.
