2 added 212 characters in body edited Aug 28 '16 at 20:23 sammy gerbil 22.5k55 gold badges2727 silver badges5959 bronze badges 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). TheYes, 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 $$\vec a$$ doescan 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. 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). The velocity $$\dot r$$ at each point is always tangential to the trajectory; there are no normal components of velocity. The Cartesian acceleration vector $$\vec a$$ does 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. 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. 1 answered Aug 28 '16 at 20:09 sammy gerbil 22.5k55 gold badges2727 silver badges5959 bronze badges 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). The velocity $$\dot r$$ at each point is always tangential to the trajectory; there are no normal components of velocity. The Cartesian acceleration vector $$\vec a$$ does 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.