show that $\phi'$ is zero

Question

let $\gamma: I \subset \mathbb{R} \to \mathbb{R^3} $ be a curve parameterized by arc length

We denote by $(\gamma(s),T(s),N(s),B(s))$ the frenet-frame.

suppose we assume the following :

$$\gamma''(s) \neq 0\;\;\forall s\in I$$
$$\tau(s) \neq 0\;\;\forall s\in I$$ here $\tau$ represents the torsion of $\gamma$
$$\rho'(s) \neq 0\;\;\forall s\in I $$ here $\rho = \frac{1}{\kappa}$ where $\kappa$ represents the curvature of $\gamma$
$$\rho^2+\sigma^2(\frac{d\rho}{d s})^2 = \text{a constant number}$$ here $\sigma = \frac{1}{\tau}$
$$\phi(s) = \gamma(s) +\rho(s)N(s) +\sigma(s)\frac{d\rho}{d s}(s)B(s)$$

show that : $\phi'(s) = 0 $ and conclude that there exists an $\Omega \in \mathbb{R^3}$ such that $||\gamma(s) - \Omega|| = \text{a constant number}$

my attempt : from 5

$$\phi'(s) = \gamma'(s) + \frac{d\rho}{d s}(s)N(s)+\rho(s)N'(s)+(\sigma(s)\frac{d\rho}{d s}(s))'B(s) + \sigma(s)\frac{d\rho}{d s}(s)B'(s) $$

by differentiating 4 you get :

$2\frac{d\rho}{d s}(s)\rho(s) + 2(\sigma(s)\frac{d\rho}{d s}(s))'\sigma(s)\frac{d\rho}{d s}(s) = 0 $

from now on I'm stuck I couldn't find a relation between the above equation and what I'm trying to prove any help/hints will be greatly appreciated. thanks !

First of all, have you identified that $\rho$ is the radius of curvature and that you are asked to prove that the curve is situated on a sphere ? — Jean Marie, Jun 06 '17 at 00:35
@JeanMarie I've never heard of the notion of 'radius of curvature' before I will google it — the_firehawk, Jun 06 '17 at 00:36
@JeanMarie a curve is situated on a sphere if it components $x,y,z$ satisfiy the equation $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = R^2$
I guess you're telling me that $\gamma$ satisfies such an equation where $R=\rho$ , right ? — the_firehawk, Jun 06 '17 at 00:42
I just translated in plain terms what $||\gamma(s) - \Omega|| = \text{a constant number}$ means. — Jean Marie, Jun 06 '17 at 00:46
Just after noting that $\gamma'$ has length 1, you change $T'(s)$ into $\gamma'(s)$ (following the first right-double-arrow) which is wrong. — John Hughes, Jun 06 '17 at 00:59
@JohnHughes thanks for noticing I shoudve changed it into $\kappa(s)N(s)$ — the_firehawk, Jun 06 '17 at 01:02
You can do so now, using the "edit" option at the bottom of your question. — John Hughes, Jun 06 '17 at 01:04

score 2 · Accepted Answer · answered Jun 06 '17 at 12:17

Hint: take your first equation $$ \phi'(s) = \gamma'(s) + \frac{d\rho}{d s}(s)N(s)+\rho(s)N'(s)+(\sigma(s)\frac{d\rho}{d s}(s))'B(s) + \sigma(s)\frac{d\rho}{d s}(s)B'(s) $$ and consider applying the Frenet formulas (which you've never used so far). For instance, you can replace $N'$ with $-\kappa T + \sigma B$ (I'm going to delete all the "s" arguments to make things simpler): \begin{align} \phi' &= \gamma' + \frac{d\rho}{d s}N+\rho N'+(\sigma\frac{d\rho}{d s})'B + \sigma\frac{d\rho}{d s}B'\\ \phi' &= \gamma' + \frac{d\rho}{d s}N+\rho (-\kappa T + \sigma B) +(\sigma\frac{d\rho}{d s})'B + \sigma\frac{d\rho}{d s}B'\\ \phi' &= \gamma' + \frac{d\rho}{d s}N- \rho \kappa T + \rho \sigma B +(\sigma\frac{d\rho}{d s})'B + \sigma\frac{d\rho}{d s}B' \end{align} Then you notice that $\rho \kappa = 1$, and things simplify a little. Do the same for $B'$, and maybe it'll get you somewhere. (Also: go ahead and write out the product rule for the second-to-last term.)

show that $\phi'$ is zero

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