For the logarithmic spiral $\gamma(t) =(e^t \cos t, e^t \sin t)$ the angle between $\gamma(t)$ and the tangent vector at $\gamma(t)$ is independent of $t$. But is the angle value $\pi/4$ or $3\pi/4$?
Angle between tangent to a logarithmic spiral at a point and the curve
anglecurvesdifferential-geometry
Related Solutions
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.
Best Answer
With
$\gamma(t) = (e^t \cos t, e^t \sin t) = e^t(\cos t, \sin t), \tag 1$
we have
$\gamma'(t) = e^t(\cos t, \sin t) + e^t(-\sin t, \cos t), \tag 2$
and
$\gamma(t) \cdot \gamma'(t) = e^{2t}; \tag 3$
furthermore,
$\gamma(t) \cdot \gamma(t) = e^{2t}, \tag 4$
and
$\gamma'(t) \cdot \gamma'(t) = 2e^{2t}, \tag 5$
respectively; thus,
$\Vert \gamma(t) \Vert = (\gamma(t) \cdot \gamma(t))^{1/2} = e^t, \tag 6$
and
$\Vert \gamma'(t) \Vert = (\gamma'(t) \cdot \gamma'(t))^{1/2} = \sqrt 2 e^t; \tag 7$
therefore, the angle $\theta$ 'twixt $\gamma(t)$ and $\gamma'(t)$ satisfies
$\cos \theta = \dfrac{\gamma(t) \cdot \gamma'(t)}{\Vert \gamma(t) \Vert \Vert \gamma'(t) \Vert} = \dfrac{e^{2t}}{\sqrt 2 e^{2t}}= \dfrac{1}{\sqrt 2} = \dfrac{\sqrt 2}{2}, \tag 8$
whence
$\theta = \dfrac{\pi}{4}, \tag 9$
assuming that
$0 \le \theta \le \pi. \tag{10}$