Still the curvature (II)

From a given curvature function it is also possible to determine a curve that has this function as its curvature function. Consider, for example, the following curvvature function:

Then, we have:

time \(t_{0}\)  \(t_{1}\)  \(t_{2}\)  \(t_{3}\)  \(t_{4}\)  \(t_{5}\)
curvature sign   0   +   -   +   0  
motion direction of the car   forward   left   right   left   forward  

The path of our car would be (in case it moves with constant speed \(1\)):

But is there another curve (i.e., another path for the car motion at constant speed \(1\)) whose curvature is given by the function above? Consider the following example:

Is the curvature of this curve different from the previous one?

Since the curvature at a point depends only on the tangent to the curve at this point, rotating the curve does not change it (since the tangents will also rotate accordingly). Therefore, the curve resulting from any rotation or any translation of the given curve will have the given curvature function.

In fact, the Fundamental Theorem of Plane Curves ensures that if two plane curves (assuming they are travelled at constant speed \(1\)) have the same curvature function, then there exists a rigid motion that transforms one curve in the other. This means that it is possible to take one curve precisely into the other with no deformation, only by translation and rotation.

Hence, any plane curve is uniquely defined by three things:

      - the curvature function;

      - the starting point of the curve (corresponds to the translation);

      - the tangent at the starting point (corresponds to the rotation).

See the following app and confirm experimentally that these three conditions are enough to define a plane curve:

2D app

Note that so far we have only studied the case of plane curves. Can this be generalized to any curve in space? Is there a similar notion of curvature for space curves? Is any three-dimensional curve also defined uniquely by the above three pieces of information?