In the following problems, we will work with spiral curves.
(a) Let
$$\gamma(t) = e^{at}(\cos t,\sin t)$$
be a logarithmic spiral. Show that the angle between the normal field $N(t)$ of $\gamma(t)$ and $\gamma(t)$ is constant. Conversely, show that if the angle between $N(t)$ and $\gamma(t)$ is constant, then $\gamma(t)$ is a logarithmic spiral.(b) Let
$$\gamma(t) = (r\cos t, r \sin t,at)$$
be a cylindrical spiral. Show that,
$$\kappa = \frac{r}{\sqrt{r^2 + a^2}}\tag{1},$$
$$aT + rB = (0,0,a^2 + r^2)\tag{2}.$$
Conversely, if $\kappa, T, B$ satisfying $(1)$ and $(2)$, then $\gamma$ is a cylindrical spiral of radius $r$ parallel to $z$-axis.
My attempts.
(a) Is easy to show $(\Longrightarrow)$. I could not to show the converse. I wrote
$$\frac{\langle N(t),\gamma(t) \rangle}{\|\gamma(t)\|} = c$$
and tried to get some ODE with coordinates of $\gamma(t)$, but I did not work.
(b) I showed $(\Longrightarrow)$ but at this moment I think that there is a mistake in the problem. I found
$$\kappa = \frac{r}{r^2 + a^2}$$
and
$$aT + rB = (0,0,\sqrt{r^2 + a^2}).$$
For the converse, defining
$$r = \frac{\kappa}{\kappa^2 + a^2}$$
and
$$a = -\frac{\tau}{\kappa^2 + \tau^2}$$
I think I can finish the problem, but I cannot use torsion and Frenet Equations so, I don't know how to solve the converse.
Notations. $\kappa$ – curvature, $T$ – tangent field, $N$ – normal field, $B$ – binormal field.
Best Answer
Part a. Write $\gamma(t)= r(t)(\cos \theta(t),\sin\theta(t))$. It's derivative is $$ \gamma'(t) = (r' \cos\theta - \theta'r\sin\theta, r'\sin\theta+r \theta \cos\theta) \tag{1} $$ (The argument $t$ is omitted for brevity.) Since $T$ is always perpendicular to $N$, the angle with $\gamma$ and $T$ is also constant, so $\langle \gamma,\gamma'\rangle = c \|\gamma\| \|\gamma'\|$. Squaring this equation and substituting $(1)$ gives $$ (r')^2 = c^2 ((r')^2 + (\theta')^2 r^2 ) $$ or $$ r' = a \theta' r \tag{2} $$ where $a = \pm \sqrt{c^2/(1-c^2)}$. Integrating gives $r = K e^{a\theta}$. If $r=1$ for $\theta = 0$, we have $K=1$. But $K$ can have any value; it just rescales the curve. Finally, substitute eq. $(2)$ into $(1)$. We see that $\gamma'(t)\neq 0$ iff $\theta'(t) \neq 0$. This means that $\theta(t)$ is a non-decreasing or non-increasing function, and has an inverse. Thus, after reparametrizing the curve, we can assume that $\theta(t) = t$ and we get the original parametrization.
Part b. Your expression for $\kappa$ and $aT + r B$ are indeed correct.
For the converse, take the derivative of $a T + r B$. Since the vector is constant, we get $$ 0 = (a \kappa - r \tau) N. $$ Therefore $$ \tau = \frac{a}{r}\kappa = \frac{a}{a^2 + r^2}. $$ We can check that $\kappa$ and $\tau$ are the same as the curvature and torsion of $(r\cos t, r\sin t, at)$, so by the existence and uniqueness theorem for curves, the curve $\gamma$ is also a cylindrical spiral with radius $r$.
For a cylindrical spiral $\gamma$ with radius $r$ the axis is parametrized by $\gamma - r N$. The derivative of this expression will thus give the direction vector of the axis: $$ \begin{align*} \left(\gamma + r N\right) ' &= T + \left(\frac{\kappa}{\kappa^2 + \tau^2}\right)(-\kappa T + \tau B) \\ &= \frac{\tau^2}{\kappa^2+\tau^2} T +\frac{\kappa \tau}{\kappa^2+\tau^2} B = \frac{\tau}{\kappa^2+\tau^2} D. \end{align*} $$
Note that this vector is a multiple of $aT + r B$, and we are done.
As a closing remark, I want to introduce you the so-called Darboux vector field $D=\tau T + \kappa B$. Along a cylindrical spiral, this vector field is constant and points in the direction of its axis.