Differential Geometry Question

[Dahaka14]

Homework Statement

Find an explicit unit-speed non-degenerate space curve \vec{r}:(0,\infinity)\rightarrow\Re^{3} whose curvature and torsion \kappa,\tau:(0,\infinity)\rightarrow\Re are given by the functions \kappa(s)=\tau(s)=\frac{1}{s}.

Homework Equations

the only thing that I can think of that would help us here are the Frenet equations:
t'=\kappa n
n'=-\kappa t -\tau b
b'=\tau n

The Attempt at a Solution

If we are to have \kappa(s)=\tau(s)=\frac{1}{s}, then we must have
t'=\frac{1}{s} t and
b'=\frac{1}{s} t, thus
t'=b'. I'm not sure what to do after this point, as I messed with these equations for awhile to no avail.

 

[lanedance]

hi dahaka14

from your frenet equations you have
\textbf{t}'=\kappa \textbf{n}
\textbf{b}'=-\tau \textbf{n}

write down a vector a, with some constants c & d we will choose
\textbf{a}= c\textbf{t} + d \textbf{n}

differentiating
\textbf{a}'= c.\textbf{t}' + d .\textbf{b}'= c .\kappa .\textbf{n} - d.\tau .\textbf{n} = \frac{1}{s} (c-d) \textbf{n}

so choose c=d and the vector a is constant, might as well make a a unit vector so set:
c = d = \frac{1}{\sqrt{2}}

now think about the dot product of a with t and what this means...
hopefully this helps you get started...

 

[Dahaka14]

The dot product should give
\textbf{a}\cdot\textbf{t}=c\textbf{t}\cdot\textbf{t}=\frac{\textbf{t}\cdot\textbf{t}}{\sqrt{2}}

I'm not sure where to go from here. The only thing that I have been able to think of is that perhaps the curve should be a helix, since a helix is such that \frac{\tau}{\kappa} is constant.

Edit: that LaTeX image should have:
torsion/curvature=constant

 

[lanedance]

yeah i think you are on the right track, as i understand it a general helix is defined as when \frac{\tau}{\kappa} is constant, which is equivalent to the tangent vector making a constant angle with some vector, say a, which is what your dot product shows as t.t = 1

Not 100% where to go, but picking an aribtrary (a), then for s=0, a starting t which matches your dot product could be a good place to start