## The curvature and torsion: how to distinguish the shape of a curve

### Space curves

Given a curve $$f:\, I\rightarrow\mathbb{R}^{3}$$ parametrized by arc length, the vector $$T=\left(T_{1},T_{2},T_{3}\right)$$ given by $$f'(t)$$ is the tangent vector of $$f$$ at $$t$$ while $$N=\left(N_{1},N_{2},N_{3}\right)$$ given by $\frac{f''(t)}{\left|f''(t)\right|}$ is the normal vector.

In the 3D case, there is a third important vector: the binormal, defined as follows: $\begin{array}{ccl} B & = & \left(B_{1},B_{2},B_{3}\right)=T\times N=\\ & = & "\mbox{det}"\left(\begin{array}{ccc} e_{1} & e_{2} & e_{3}\\ T_{1} & T_{2} & T_{3}\\ N_{1} & N_{2} & N_{3} \end{array}\right)=\\ & = & \left(T_{2}N_{3}-T_{3}N_{2},T_{3}N_{1}-T_{1}N_{3},T_{1}N_{2}-T_{2}N_{1}\right) \end{array}$

It may be proved that also $$\left|B\right|=1$$. These three unitary vectors ($$T$$), ($$N$$) and ($$B$$) form the so called Frenet-Serret frame. They are orthogonal to each other; $$T$$ indicates the direction at which the curve moves; a $$N$$ indicates the direction at which the curve is turning while $$B$$ is the vector orthogonal to $$T$$ and $$N$$ such that the three vectors form an orthonormal basis with positive orientation.

The plane defined by tangent vector $$T$$ and normal vector $$N$$ is called the osculating plane. The osculating plane of the curve at a given point is the plane that better approximates the curve in that point. Note also that the binormal vector $$B$$ is orthogonal to the osculating plane.

A few examples: