A regular curve $\textbf{$\gamma$}$ in $\mathbb{R}^3$ with curvature $> 0$ is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $\textbf{a}$. Show that the torsion $\tau$ and curvature $\kappa$ of $\textbf{$\gamma$}$ are related by $\tau = ±\kappa \cot \theta$. Show conversely that, if the torsion and curvature of a regular curve are related by $\tau = \lambda \kappa$ where $\lambda$ is a constant, then the curve is a generalized helix. Note that we have the Frenet equations $\textbf{t}'=\kappa \textbf{n}$ and $\textbf{b}'= -\tau \textbf{n}$.

I have done the first half of the proof. In the second half, I claim that $\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$ where $\textbf{t} $ and $\textbf{b}$ are the unit tangent and binormal vectors, respectively, satisfies the conditions for a general helix. I'm trying to show that the derivative of $\textbf{a}$ is $0$ if we assume that $\tau = \lambda \kappa$, proving that $\textbf{a}$ is constant. I tried showing that $\mathbf{a' \cdot a'}=0$, which would prove that $\mathbf{a'}=0$, but I was unable to make it work. Any suggestions or different ideas on how to approach the converse would be appreciated.

My attempt: $\textbf{a}'=\textbf{t}' \cos \theta ± \textbf{b}' \sin \theta =\kappa \textbf{n} \cos \theta ± \tau \textbf{n} \sin \theta = \kappa \textbf{n} \cos \theta ± \lambda \kappa \textbf{n} \sin \theta$

  • $\begingroup$ Don't over-complicate! Just show $\mathbf a’=0$ directly. $\endgroup$ Commented Mar 2, 2020 at 21:31
  • $\begingroup$ Any idea how? I took the derivative but couldn't get anywhere. $\endgroup$
    – A.B
    Commented Mar 3, 2020 at 2:36
  • $\begingroup$ Obviously, you need to use the Frenet equations. Please edit your question to include this. $\endgroup$ Commented Mar 3, 2020 at 2:37
  • $\begingroup$ @TedShifrin Added. $\endgroup$
    – A.B
    Commented Mar 3, 2020 at 2:43
  • $\begingroup$ No, I meant your computation of $\mathbf a'$. Don't forget to use the hypothesis! $\endgroup$ Commented Mar 3, 2020 at 2:44

2 Answers 2


Since $\mathbf a$ and $T$ are unit vectors we may write

$\mathbf a \cdot T = \Vert a \Vert \Vert T \Vert \cos \theta = \cos \theta, \tag 1$

where $\theta$ angle 'twixt $\mathbf a$ and $T$; we may differentiate this equation with respect to the arc-length $s$ along our curve $\gamma(s)$, yielding

$\dot {\mathbf a} \cdot T + \mathbf a \cdot \dot T = 0, \tag 2$

and since

$\dot {\mathbf a} = 0 \tag{2.5}$

and we have the first Frenet-Serret relation

$\dot T = \kappa N, \tag{2.6}$

(2) becomes

$\mathbf a \cdot \dot { \kappa N} = 0, \tag 3$

and with

$\kappa > 0 \tag{3.5}$

we find that

$\mathbf a \cdot N = 0, \tag 4$

which we may differentiate yet again with respect to $s$:

$\dot{\mathbf a} \cdot N + \mathbf a \cdot {\dot N} = 0, \tag 5$

and yet again via (2.5) we may write

$\mathbf a \cdot {\dot N} = 0; \tag{5.5}$

we now deploy the second Frenet-Serret equation

$\dot N = -\kappa T + \tau B \tag 6$

to obtain

$\mathbf a \cdot {(-\kappa T + \tau B)} = 0, \tag 7$


$-\kappa \mathbf a \cdot T + \tau \mathbf a \cdot B = 0, \tag 8$

whence, using (1),

$-\kappa \cos \theta + \tau \mathbf a \cdot B = 0; \tag 9$

we expand $\mathbf a$ in terms of $T$, $N$, $B$ using (1) and (4) as follows:

$\mathbf a = (\mathbf a \cdot T)T + (\mathbf a \cdot N)N + (\mathbf a \cdot B)B = (\cos \theta) T +(\mathbf a \cdot B)B; \tag{10}$


$\Vert \mathbf a \Vert = \Vert T \Vert = \Vert B \Vert = 1, \tag{11}$


$T \cdot B = \mathbf a \cdot N = 0, \tag{12}$

we infer from (10) that

$1 = \Vert \mathbf a \Vert^2 = \cos^2 \theta \Vert T \Vert^2 + (\mathbf a \cdot B)^2 \Vert B \Vert^2$ $= \cos^2 \theta + (\mathbf a \cdot B)^2, \tag{13}$

which implies that

$\mathbf a \cdot B = \pm \sin \theta; \tag{14}$

substituting this into (9) yields

$-\kappa \cos \theta \pm \tau \sin \theta = 0, \tag{15}$

and then

$\kappa \cos \theta = \pm \tau \sin \theta, \tag{16}$


$\tau = \pm \kappa \cot \theta, \tag{17}$

as per request.

Going the other way, given that

$\tau = \lambda \kappa \tag{18}$

for some constant

$\lambda \in \Bbb R, \tag{19}$

we may choose $\theta$ such that

$\lambda = \cot \theta = \dfrac{\cos \theta}{\sin \theta}; \tag{20}$

next, we set

$\mathbf a = (\cos \theta) T + (\sin \theta) B, \tag{21}$

and note this implies;

$\Vert a \Vert = \sqrt{\cos^2 \theta \Vert T \Vert^2 + \sin^2 \theta \Vert B \Vert^2}$ $= \sqrt{\cos^2 \theta + \sin^2 \theta} = \sqrt 1 = 1; \tag{21.1}$

and apply $d/ds$ to (21):

$\dot {\mathbf a} = (\cos \theta) \dot T + (\sin \theta) \dot B; \tag{22}$

we substitute (2.6) and the third Frenet-Serret equation

$\dot B = -\tau N, \tag{23}$

and obtain

$\dot{\mathbf a} = (\cos \theta)\kappa N - (\sin \theta)\tau N = (\kappa \cos \theta - \tau \sin \theta) N; \tag{24}$

in light of (18) and (20),

$\tau = \lambda \kappa = \dfrac{\cos \theta}{\sin \theta} \kappa, \tag{25}$

and thus

$\tau \sin \theta = \kappa \cos \theta, \tag{26}$

which in concert with (24) shows that

$\dot{\mathbf a} = 0. \tag{27}$

We have shown the existence of a constant vector $\mathbf a$ and a constant angle $\theta$ such that (1) binds; $\gamma(s)$ is a generalized helix.

Note Added in Edit, Monday 20 January 2020 6:31 PM PST: As we transit 'twixt (15) and (17), we have occasion to divide through by $\sin \theta$; thus we should address the question of just when

$\sin \theta = 0. \tag{28}$

Now (28) occurs precisely when

$\theta = 0, \pi, \tag{29}$

that is, when $T$ is aligned parallel or anti-parallel to $\mathbf a$. (We observe that

$0 \le \theta \le \pi \tag{30}$

since it is the angle between the vectors $\mathbf a$ and $T$.) But (29) implies

$T = \pm \mathbf a, \tag{31}$

which further implies that $\gamma(s)$ is a straight line; as such,

$\kappa = 0, \tag{32}$

which contradicts our hypothesis that the curvature of $\gamma(s)$ is positive. Therefore

$\sin \theta \ne 0, \tag{33}$

and the quotient

$\cot \theta = \dfrac{\cos \theta}{\sin \theta} \tag{34}$

may legitimately be formed. End of Note.

  • $\begingroup$ What does 'twixt mean? $\endgroup$ Commented May 11, 2021 at 1:20
  • $\begingroup$ @Ramanujan: between $\endgroup$ Commented May 11, 2021 at 6:29

OK, now that you added the relevant information. You never specified how you would choose $\theta$. I.e., you haven't yet related $\lambda$ and $\theta$. Everything you typed suggested that you should take $\theta$ so that $\lambda = \cot\theta$. If you do that, what is $\cos\theta - \lambda\sin\theta$? Now simplify. (Obviously, you need the correct sign choice.)

  • $\begingroup$ What do you mean by correct sign choice? Shouldn't both work? $\endgroup$
    – A.B
    Commented Mar 3, 2020 at 4:18
  • 2
    $\begingroup$ No, there can only be one direction that works. $\endgroup$ Commented Mar 3, 2020 at 5:23
  • $\begingroup$ So is it incorrect to claim $\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$? Should I have instead said that $\textbf{a}=\textbf{t} \cos \theta + \textbf{b}\sin \theta$? $\endgroup$
    – A.B
    Commented Mar 3, 2020 at 5:40
  • 1
    $\begingroup$ Figure out which sign makes it work and use that sign. Who told you to put $\pm$ in there? $\endgroup$ Commented Mar 3, 2020 at 5:42
  • $\begingroup$ I derived that $\textbf{a}=\textbf{t} \cos \theta ± \textbf{b}\sin \theta$ when showing the first part of the proof. See this post. They show $C = ± \sin \theta$. Is that wrong? math.stackexchange.com/questions/1478391/… $\endgroup$
    – A.B
    Commented Mar 3, 2020 at 5:47

You must log in to answer this question.

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