Finding normal and tangential velocity and acceleration for motion along a 3D curve Basically, I have a major problem on my hands. 
As part of an assignment, I have been given a set of 3D points, as well as the time between them, and nothing else, and I'm supposed to find the tangential and normal velocities and accelerations between each point.
Now, I know that (tangential) velocity can be easily calculated simply by taking the differential of position over time, and that (tangential) acceleration can also be easily calculated by taking the differential of (tangential) velocity over time.
However, finding the normal velocities and accelerations is where I get stuck. I can find absolutely no information on how a normal velocity is calculated, except for the fact that any velocity vector can be broken down into its tangential and normal components.
However, since by calculating the (tangential) velocities by taking the differential of position over time, that means that the velocity vector IS the tangential velocity vector. Does that mean that the normal velocity for each and every point is just 0? That doesn't sound right to me.
Regarding normal acceleration, I have no idea how it's supposed to work in 3D, because we need the radius of a circle to be able to calculate that, but how do I calculate the radius of a sphere in 3D, given only two points?
Any help on this problem is greatly appreciated. This thing's driving me insane.
 A: Let's say you have a prescribed path parametrized as $\vec{r}(\alpha)$ for the position vector.

*

*The velocity vector is decomposed as $$ \vec{v}(\alpha,\dot{\alpha}) = \dfrac{\partial \vec{r}(\alpha)}{\partial \alpha} \dot{\alpha}$$

*The speed along the tangential direction is $$ v(\alpha,\dot{\alpha}) = \lvert d\dfrac{\partial \vec{r}(\alpha)}{\partial \alpha}\rvert \dot{\alpha} $$

*The tangential direction is thus $$\hat{e} = \dfrac{\vec{v}}{v} = \dfrac{ \dfrac{\partial \vec{r}(\alpha)}{\partial \alpha} }{\lvert\dfrac{\partial \vec{r}(\alpha)}{\partial \alpha}\rvert } $$

*The acceleration vector is decomposed as $$ \vec{a}(\alpha,\dot{\alpha},\ddot{\alpha}) = \dfrac{\partial \vec{v}}{\partial \alpha} \dot{\alpha}+\dfrac{\partial \vec{v}}{\partial \dot{\alpha}} \ddot{\alpha}$$

*The curvature is $$ {\kappa} = \dfrac{1}{\rho} = \dfrac{ \lvert \vec{v} \times \vec{a} \rvert}{ \lvert \vec{v} \rvert^3} $$

*The acceleration along the tangential direction is $$a_T = \hat{e} \cdot \vec{a} $$

*The acceleration towards the center of curvature is $$a_C = \dfrac{v^2}{\rho} = \vec{a} - a_T\; \hat{e}$$
I think this should be sufficient to deduce all the components you need.
Example
A particle is moving with constant speed along a helix with radius $R$ and pitch $h$.

*

*Position $\vec{r}(\alpha) = \begin{pmatrix} R \cos \alpha & R \sin \alpha & h \alpha \end{pmatrix}$

*Velocity $\vec{v}(\alpha,\dot{\alpha}) = \begin{pmatrix}-\dot{\alpha} R \sin \alpha & \dot{\alpha} R \cos\alpha & \dot{\alpha}\,h \end{pmatrix}$

*If the speed is $v$ then $\dot{\alpha} = \dfrac{v}{\sqrt{R^2+h^2}}$

*Similarly $\ddot{\alpha} = \dfrac{\dot{v}}{\sqrt{R^2+h^2}}$

*Acceleration $\vec{a} = \begin{pmatrix} -\dfrac{R v^2 \cos \alpha}{R^2+h^2}-\dfrac{R \dot{v} \sin \alpha}{\sqrt{R^2+h^2}} & -\dfrac{R v^2 \sin \alpha}{R^2+h^2}+\dfrac{R \dot{v} \cos \alpha}{\sqrt{R^2+h^2}}  & \dfrac{\dot{v} h}{\sqrt{R^2+h^2}}\end{pmatrix}$

*Curvature $\kappa = \dfrac{R}{R^2+h^2}$
A: The velocity $\vec{v}$ can be obtained by computing the derivate of each position component with respect to time.
The acceleration $\vec{a}$ can similarly be obtained as the second derivative.
The tangential component of the acceleration is then $\vec{a_T} = \vec{a} \cdot \frac{\vec{v}}{v}$, and the normal component $\vec{a_N}$ is such that $\vec{a} = \vec{a_T} + \vec{a_N}$.
A: As you note, the unit tangent vector to a 3D curve is $\hat T=\frac{\dot r}{|\dot r|}$ where $r(t)$ is the position vector and $\dot r=\frac{dr}{dt}$ (see MIT eqn 2.5 below).
There are two normals for a 3D curve : the principal normal and the bi-normal. 
The principal normal is in the osculating plane which contains the tangent vector and the local centre of curvature. The unit principal normal vector is $\hat N=\frac{\dot T}{|\dot T|}$ (see MIT eqn 2.18). 
The unit binormal is defined by $\hat B=\hat T \times \hat N$ (MIT eqn 2.40) which is equivalent to $\hat B=\frac{\dot r \times \ddot r}{|\dot r \times \ddot r|}$ (MIT eqn 2.41).
Yes, you are correct : the velocity $\dot r$ at each point is always tangential to the trajectory; there are no normal components of velocity. This happens because velocity is the time derivative of displacement, which is always along the curve. The motion curve or trajectory is the set of successive infinitesimal  displacements. 
The Cartesian acceleration vector $\vec a$ can have tangential and normal components. These can be found by taking the scalar product of $\vec a$ with the unit tangential and normal vectors - eg $\vec a_T=\vec a \bullet \hat T$. You can also use $\vec a = \vec a_T + \vec a_N + \vec a_B$.

The following may be of use :
Paul's Online Math Notes shows how to find the tangent, normal and bi-normal vectors.
The Differential Geometry of Curves is chapter 2 of a hyperbook from MIT for Computer Aided Design.
