Express the velocity of $\gamma$ in terms of the TNB frame of $r$

Question

my question is:

Let $r(t)=<cos(t),sin(t),t>$ be a curve and let $N$ be the principal unit normal vector to the curve. Define the curve $\gamma (t)$ by $\gamma (t)=r(t)+N$. Express the velocity of $\gamma$ in terms of the TNB frame of r. ($\gamma '(t)=aT+bN+cB$ for some scalars $a,b,c$)

So far I have computed T,N and B for r(t) but i am lost on how to put these together to compute scalars $a,b,c$ for $\gamma '(t)$

Here is what I have:

$T = <-sin(t),cost(t),0>$

$N=<-cos(t),-sin(t),0>$

$B=<0,0,1>$

and $\gamma (t)=<0,0,t>$

thanks.

Answer 1

Recall that $\{\textbf{e}^1=T, \textbf{e}^2=N, \textbf{e}^3=B\}$ defines an orthonormal basis on $\gamma(t)$. Therefore, given $ \gamma'(t) \in T_{\gamma(t)} \textbf{Im}(\gamma)$ we have;

$$\gamma'(t) = \sum_{j=1}^3 (\gamma'(t) \cdot \textbf{e}^j) \ \textbf{e}^j$$

This is just due to the fact that $\textbf{e}^j \cdot \textbf{e}^i = \delta_{ij}$. Hence, $a = \gamma'\cdot T, b = \gamma' \cdot N, c = \gamma' \cdot B$.

Answer 2

First of all, it appears this problem may be solved in abstracto as follows:

Let $\rho(s)$ be a unit speed curve in $\Bbb R^3$, so that

$T(s) = \dot \rho(s) \tag{1}$

with

$\Vert T(s) \Vert = 1. \tag{2}$

We recall the Frenet-Serret equations for $\rho(s)$:

$\dot T(s) = \kappa(s) N(s), \tag{3}$

$\dot N(s) = -\kappa(s) T(s) + \tau(s)B(s), \tag{4}$

$\dot B(s) = -\tau(s) N(s); \tag{5}$

here

$B(s) = T(s) \times N(s), \tag{6}$

$\Vert N(s) \Vert = \Vert B(s) \Vert = 1, \tag{7}$

$\kappa(s)> 0 \tag{8}$

is the curvature, and $\tau(s)$ is the torsion of the curve $\rho(s)$. Setting

$\gamma(s) = \rho(s) + N(s), \tag{9}$

we find

$\dot \gamma(s) = \dot \rho(s) + \dot N(s); \tag{10}$

using (1) and (4) this becomes

$\dot \gamma(s) = T(s) - \kappa(s) T(s) + \tau(s) B(s) = (1 - \kappa(s))T(s) + \tau(s) B(s); \tag{11}$

taking $a(s) = 1 - \kappa(s)$, $b(s) = 0$ and $c(s) = \tau(s)$ we see that (11) expresses $\dot \gamma(s)$ in the requisite form. It should be observed however that $\gamma(s)$ is not, in general a unit-speed curve as is $\rho(s)$. Indeed,

$\Vert \dot \gamma(s) \Vert^2 = \langle \dot \gamma(s), \dot \gamma(s) \rangle$ $= \langle (1 - \kappa(s))T(s) + \tau(s) B(s), (1 - \kappa(s))T(s) + \tau(s) B(s) \rangle = (1 - \kappa(s))^2 + \tau^2(s), \tag{12}$

whence

$\Vert \dot \gamma(s) \Vert = \sqrt{(1 - \kappa(s))^2 + \tau^2(s)}. \tag{13}$

Note we have used (2), (7), and the orthogonality relation

$\langle T(s), B(s) \rangle = \langle T(s), T(s) \times B(s) \rangle = 0 \tag{14}$

in deriving (12)-(13).

We apply formula (11) to address the problem at hand. Given

$r(t) = (\cos t, \sin t, t), \tag{15}$

we find

$\dot r(t) = (-\sin t, \cos t, 1), \tag{16}$

whence

$\Vert \dot r(t) \Vert = \sqrt{(-\sin t)^2 + (\cos t)^2 + 1} = \sqrt{\sin^2 t + \cos^2 t + 1^2} = \sqrt{1 + 1} = \sqrt 2. \tag{17}$

(17) tells us that

$\dfrac{ds}{dt} = \sqrt 2, \tag{18}$

where $s$ is the arc-length along $r(t)$; thus we may take

$s = \sqrt 2 t \tag{19}$

as the unit-speed re-parametrization of the curve $r(t)$. With

$t = \dfrac{s}{\sqrt 2}, \tag{20}$

we may write

$r(s) = (\cos \dfrac{s}{\sqrt 2}, \sin \dfrac{s}{\sqrt 2}, \dfrac{s}{\sqrt 2}), \tag{21}$

whence

$\dot r(s) = \dfrac{1}{\sqrt 2} (-\sin \dfrac{s}{\sqrt 2}, \cos \dfrac{s}{\sqrt 2}, 1); \tag{22}$

it is easily seen that

$\Vert \dot r(s) \Vert = 1, \tag{23}$

that is,

$\dot r(s) = T(s), \tag{24}$

the unit tangent vector to $r(s)$. Then from (3),

$\kappa(s) N(s) = \dot T(s) = \ddot r(s) = \dfrac{1}{2}(-\cos \dfrac{s}{\sqrt 2} s, -\sin \dfrac{s}{\sqrt 2} s, 0); \tag{25}$

we conclude, using $\Vert N \Vert = 1$, that

$\kappa(s) = \dfrac{1}{2} \tag{26}$

and

$N(s) = (-\cos \dfrac{s}{\sqrt 2} s, -\sin \dfrac{s}{\sqrt 2} s, 0); \tag{27}$

from (9) we now see that

$\gamma(s) = r(s) + N(s) = (\cos \dfrac{s}{\sqrt 2}, \sin \dfrac{s}{\sqrt 2}, \dfrac{s}{\sqrt 2}) + (-\cos \dfrac{s}{\sqrt 2} s, -\sin \dfrac{s}{\sqrt 2} s, 0)$ $=(0, 0, \dfrac{s}{\sqrt 2}); \tag{28}$

we immediately have

$\dot \gamma(s) = (0, 0, \dfrac{1}{\sqrt 2}); \tag{29}$

in the spirit of the above we also compute $\dot \gamma(s)$ from (11);

$B(s) = T(s) \times N(s) = \dfrac{1}{\sqrt 2}(-\sin \dfrac{s}{\sqrt 2}, \cos \dfrac{s}{\sqrt 2}, 1)\times (-\cos \dfrac{s}{\sqrt 2} s, -\sin \dfrac{s}{\sqrt 2} s, 0)$ $= \dfrac{1}{\sqrt 2}(\sin \dfrac{s}{\sqrt 2}, -\cos \dfrac{s}{\sqrt 2}, 1); \tag{30}$

we see that

$\dot B(s) = \dfrac{1}{2}(\cos \dfrac{s}{\sqrt 2}, \sin \dfrac{s}{\sqrt 2}, 0) = -\dfrac{1}{2} N(s); \tag{31}$

following (5) we see that

$\tau(s) = \dfrac{1}{2}; \tag{32}$

finally,

$\dot \gamma(s) = (1 - \kappa(s))T(s) + \tau(s) B(s) = (1 - \dfrac{1}{2})T(s) + \dfrac{1}{2}B(s)$ $= \dfrac{1}{2}T(s) + \dfrac{1}{2}B(s) = \dfrac{1}{2}(T(s) + B(s)); \tag{33}$

this, (22), (24) and (30) yield

$\dot \gamma(s) = \dfrac{1}{2}(\dfrac{1}{\sqrt 2} (-\sin \dfrac{s}{\sqrt 2}, \cos \dfrac{s}{\sqrt 2}, 1) + \dfrac{1}{\sqrt 2}(\sin \dfrac{s}{\sqrt 2}, -\cos \dfrac{s}{\sqrt 2}, 1)) = \dfrac{1}{2}(0, 0, \dfrac{2}{\sqrt 2}) = (0, 0, \dfrac{1}{\sqrt 2}), \tag{34}$ in accord with (29). We can use the chain rule together with (17) to obtain $\dot \gamma(t) = d\gamma(t) / dt$, the "velocity" of $\gamma(t)$ in the original $t$-parametrization:

$\dot \gamma(t) = \dfrac{d \gamma(t)}{dt} = \dfrac{d \gamma(s)}{ds} \dfrac{ds}{dt} = \sqrt 2 (0, 0, \dfrac{1}{\sqrt 2}) = (0, 0, 1). \tag{35}$

In closing I point out that there appears to be an error in our OP sunsunsunsunsun's computation of $T(t)$, which he reports as

$T(t) = (-\sin t, \cos t, 0). \tag{36}$

With $r(t)$ as in (15) we have $\dot r(t)$ as in (16), whence

$T(t) = \dfrac{1}{\Vert \dot r(t) \Vert}\dot r(t) = \dfrac{1}{\sqrt 2}(-\sin t, \cos t, 1). \tag{37}$

Perhaps this discrepancy accounts for the difficulties sunsunsunsunsun encountered.