1. ## Curvature

Hello all,

I have a regular parametrized curve $\gamma: \mathbb{R} \rightarrow \mathbb{R}^{3}$ such that $\Vert\gamma''(t)\Vert =1$ for all $t \in \mathbb{R}$.
Assume that $\gamma(t)$ has constant curvature $k \neq 0$ and constant torsio $\tau=\frac{1}{\sqrt{2}}$. We also assume that:

$\gamma(0)=\left( \frac{1}{\sqrt{2}},0,0 \right)$

$\gamma'(0)=\left( 0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)$

$\mathbf{b}(t)=\frac{1}{\sqrt{2}} \left( \sin(t),-\cos(t),1 \right)$

where $\mathbf{b}(t)$ is the binormal of $\gamma(t)$

1) find the curvature K
2) find the $\gamma(t)$ explicitely

I have done the following:

1)

Using the Frenet equations I have that $\mathbf{n'}=-k\mathbf{t}+\tau\mathbf{b}$.
Differentiating on both sides yields $\mathbf{n''}=-k\mathbf{t'}+\tau\mathbf{b'}$ (*).

Using $\mathbf{b'}=-\tau\mathbf{n}$ and $\mathbf{t'}=k\mathbf{n}$ and substituting in (*) gives $k=\frac{\sqrt{2}}{2}$.

2)

I am still not done with this one but my suggestion is to use that $\mathbf{n}=\frac{\gamma''(t)}{\Vert \gamma''(t) \Vert}$ and integrate together with the fact that $\Vert\gamma''(t)t\Vert =1$.

Could someone verify 1) and 2)?

Thanks.

2. ## Re: Curvature

Sorry but 2) is wrong since $\Vert \gamma''(t) \Vert \neq 1$.

Suggestions would still be appreciated.