Solving for a unit speed curve given curvature and torsion (diff. geo)

The answer is a little tricky:

Since the curve is torsionless, then the curve is planar, so that one can think that it lies completely on the xy-plane.

So, you have to find a curve a(s)=(x(s),y(s)). The tangent vector then is a'(s)=(x'(s),y'(s)), and we have the condition that x'^2(s)+y'^2(s)=1, since it is unitary. Here we do the trick: Since the ways to chose x and y are infinite, we assume that they are the antiderivative of a cosine and a sine, with the same angle:

x(s)=int(cos(f(s)))+a
y(s)=int(sin(f(s)))+b

So, the unitary condition on the tangent vector is fulfilled. We need now to find our function f.
To do this, we differentiate the tangent vector and then we calculate the norm of the new vector. Using the chain rule, we obtain that f'(s)=k(s). So, f(s)=int(k(s))+c, and you can then go back to find the curve you're looking for.

 

sorry for opening this old thread but i have to find a curve with given curvature and zero torsion. I read the answer but i couldn't understand the trick to solve the differential system. Is there anybody who could be more precise? Thanks

 

Hi

You don't need to solve a differential sistem.

The fact that your curve has zero torsion means that it is planar, so that you can study it as a curve on the xy-plane. This means that you can denote it as a(s)=(x(s),y(s)). But your curve has unit speed, which means that ||a'(s)||^2=x'(s)^2+y'(s)^2=1. A natural assumption then is to set x'(s)=cos(f(s)) and y'(s)=sin(f(s)). Integrating these expressions we obtain

x(s)=int(cos(f(s)))+a,
y(s)=int(sin(f(s)))+b. (1)

Now, if we differentiate the tangent vector a'(s), we obtain a''(s)=f'(s)(-sin(f(s)),cos(f(s))) using the chain rule. We know that the curvature k(s) is defined as

k(s)n(s)=a''(s),

where n(s) is the normal vector. This yields f'(s)=k(s). Integrating again, we find that f(s)=int(k(s)). Once we obtain f(s), we substitute in (1) and hence obtain a(s).