1
$\begingroup$

Let $\gamma:(a,b)\rightarrow\mathbb R^3$ be a unit speed space curve with non-vanishing curvature $\kappa(t)\neq0$. Define another parametrized curve $\beta:(a,b)\rightarrow\mathbb R^3$ by $$\beta(t)=\frac{d\gamma(t)}{dt}$$ Show that $\beta$ is regular, and that if $s$ is an arclength parameter along $\beta$, then $$\frac{ds}{dt}=\kappa$$ Show that the curvature $\tilde\kappa$ of $\beta$ is $$\tilde\kappa=(1+\frac{\tau^2} {\kappa^2})^\frac 12$$ where $\tau$ is the torsion of $\gamma$.

My attempt as follows: A point $\gamma(t)$ of a parametized curve $\gamma$ is a regular point if $\dot\gamma(t)\neq0$. A curve is regular if all points are regular. So, if $\dot\beta(0)\neq0$, then it is regular. $$\dot\beta(t)=\frac{d}{dt}\frac{d\gamma(t)}{dt}$$$$=\frac{d^2\gamma(t)}{dt^2}$$$$\dot\beta(0)=\frac{d^2\gamma(0)}{dt^2}$$ and as stated at the beginning, $\gamma$ has non-vanishing curvature, so $\gamma,\dot\gamma,\ddot\gamma$... $\neq0$ and, $\dot\beta(0)\neq0$ and $\beta$ is regular. Now, if $s$ is arclength of curve $\beta$ starting at $\beta(t_0)$, then we see that: $$\frac{ds}{dt}=\frac{d}{dt}\int_{t_0}^t\mid\mid\dot\beta(u)\mid\mid du = \mid\mid\dot\beta(t)\mid\mid=\kappa$$ For the next part, I'm completely lost. I know that the torsion, $\tau$, of a regular curve with non-vanishing curvature is found using the following $$\tau = \frac{(\dot\gamma \times\ddot\gamma)\cdot\dddot\gamma}{\mid\mid\dot\gamma \times\ddot\gamma\mid\mid^2}$$ and since $\gamma$ is unit speed, $\dot\gamma$ and $\ddot\gamma$ are perpendicular $$\mid\mid\dot\gamma\times\ddot\gamma\mid\mid = \mid\mid\dot\gamma\mid\mid\ddot\gamma\mid\mid = \mid\mid\ddot\gamma\mid\mid=\kappa$$ But I can't seem to get the required proof: curvature $\tilde\kappa$ of $\beta$ is $$\tilde\kappa=(1+\frac{\tau^2} {\kappa^2})^\frac 12$$

$\endgroup$

1 Answer 1

2
$\begingroup$

Using the formula $$\tilde{\kappa}=\frac{|\mathbf{\beta}'\times \mathbf{\beta}''|}{|\mathbf{\beta}'|^3}$$ and the Frenet formula: $\beta'=\kappa \mathbf{n}$, $\mathbf{n}'=-\kappa \mathbf{\beta}+\tau\mathbf{b}$. Thus $$\tilde{\kappa}=\frac{|\kappa \mathbf{n}\times(\kappa'\mathbf{n}+\kappa(-\kappa \mathbf{\beta}+\tau\mathbf{b}))|}{|\kappa \mathbf{n}|^3}=\frac{|\kappa^3 \mathbf{b}+\kappa^2\tau\mathbf{\beta}|}{\kappa^{3}}=\frac{\sqrt{\kappa^4(\kappa^2+\tau^2)}}{\kappa^3}=\sqrt{1+\frac{\tau^2}{\kappa^2}}$$

$\endgroup$
1
  • $\begingroup$ Could you explain how you simplified the expression for $\overline{\kappa}$ $\endgroup$
    – John
    Commented Feb 1, 2021 at 21:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .