Hey all I am looking for help on a problem. I will post it, and than I will add what I have tried and my ideas etc. The question has been up now for a few days, I'm sure someone out there can help! I even put a bounty, I have spent a lot of time on this question!

I am interested in calculating the torsion ($\tau$) and curvature ($\kappa$) of the curve $$X(t)=(at,bt^2,ct^3), \quad t \ge 0 $$ and $a$, $b$, and $c$ are all positive constants.

So here is what I am having problems with. It seems like there are so many different formulas for curvature, and there are also the Frenet–Serret formulas so I am having issues deciding how to do it. I was thinking maybe I could reparametrize with respect to arc length, which would give me it in terms of unit length so I could use some of Frenet–Serret formulas, but I am not confident in that.

What I did so far was I calculated $X'(t)=(a,2bt,3ct^2)$ and $|X'(t)|=\sqrt{a^2+4b^2 t^2 +9c^2 t^4}$.

Then I calculated $$X''(t)=(0, 2b, 6ct)$$ and $$|X''(t)|=2\sqrt{b^{2}+9c^{2}t^{2}}.$$

I also know that the unit tangent, $$T(t)={X'(t)\over|X'(t)|}$$ and the unit normal is $$N(t)={T'(t) \over |T'(t)|}$$ and that the binormal is $B= T \times N$.

But I am really not sure how to take it further.

I know that in arc length parametrzation we would have $dT/ds= \kappa N$ and $dN/ds=-\kappa T + \tau B$ and $dB/ds= -\tau N$.

Should I keep it in the form it is and use the equation $$\kappa={|X'(t) \times X''(t)|\over |X'(t)|^3}?$$

A few of the other things I am thinking of is that maybe I could solve for torsion by the following:

I know that $T$, $B$ and $N$ form an orthonormal basis of $\mathbb R^{3}$ so can and when we write $N'=\alpha T + \tau B$ , is the coefficient, $\tau$ the definition of torsion?

Moreover, noting $B= T \times N$, we have $$B'= T'\times N + T \times N' = T \times N' =T \times (\alpha T+\tau B) = \tau T \times B= -\tau N $$ (because $N=B \times T$ so $T \times B=-N)$. I think this is how I understand the derivation for that equation.

Is it only in arc length parametrization that I can use the Frenet equations for example?

However, I think I should be able to do it just using the regular formulas, and not an arc length parametrization as the integral would be tough. I did conform this as well, so in terms of the basics of this question I'd like to do it without arc length parametrization.

I am also interested in seeing a intuitive derivation of the torsion formula (the one with the triple product).

I apologize if I didn't show enough work, this is all I could do but I am very happy to learn it. Thanks a lot to any help!


I am wondering about the validity of what I now have.

In addition to above I computed $X'''(t)=(0,0,6c)$

$X' \times X''= (6bct^2,-6act,2ab)$

$(X' \times X'') \cdot X''' = (12abc)$

$|X' \times X''| = 2 \sqrt {9b^2c^2t^4+9a^2c^2t^2+a^2b^2}$.

Now can I just apply the formulas

$$\tau = \frac{ (X' \times X'') \cdot X'''}{|X' \times X''|^2}.$$

  • $\begingroup$ The easiest way to calculate the curvature $\kappa$ is with $(dT/ds\cdot dT/ds)=\kappa^2(N\cdot N)=\kappa^2$, as has been done in my answer below (with IMO more common notation). The easiest way to calculate the torsion $\tau$ is with $\det(T,dT/ds,d^2T/ds^2)=\kappa^2\tau$, as has been done in the same answer. Are you familiar with the latter formula? $\endgroup$ Commented Mar 14, 2015 at 20:07
  • $\begingroup$ Im not familiar with the latter formula, but would be open to learning it. And sorry I am a little confused on the notation, of why you have (dT/ds $\dot$ dT/ds)=$\kappa(N \dot N) $? $\endgroup$
    – Quality
    Commented Mar 14, 2015 at 20:09
  • $\begingroup$ Yes, it is only in arc length parametrization that you can use the Frenet equations. But no, you don't need an integral to calculate the arc length, because it's sufficient to know that $dX/ds = (dX/dt)/(ds/dt)$ as has been done in my answer. $\endgroup$ Commented Mar 14, 2015 at 20:14
  • $\begingroup$ I've been trying to format an inner product. $\endgroup$ Commented Mar 14, 2015 at 20:16
  • $\begingroup$ The formula sequence in your update seems to be a more efficient method to calculate $\tau$ than the way I did it (with the $\det$ formula). $\endgroup$ Commented Mar 15, 2015 at 12:10

2 Answers 2


Those calculations with space curves tend to be tedious; therefore using a computer algebra system is not quite a luxury. Here goes: $$ \vec{r}(t) = \left[ \begin{matrix} a t \\ b t^2 \\ c t^3 \end{matrix} \right] \quad \Longrightarrow \\ \vec{\iota} = \, \stackrel{.}{\vec{r}} \, = \frac{d\vec{r}}{ds} = \frac{d\vec{r}}{dt}\frac{dt}{ds} = \left[ \begin{matrix} a \\ 2 b t \\ 3 c t^2 \end{matrix} \right] / \sqrt{a^2 + 4 b^2 t^2 + 9 c^2 t^4} \quad \Longrightarrow \\ \stackrel{..}{\vec{r}} \, = \frac{d^2 \vec{r}}{ds^2} = \, \stackrel{.}{\vec{\iota}} \, = \frac{d\vec{\iota}}{ds} = \frac{d\vec{\iota}}{dt}\frac{dt}{ds} = \left[ \begin{matrix} -2at(2b^2+9c^2t^2)\\2b(a^2-9c^2t^4)\\6ct(a^2+2b^2t^2) \end{matrix} \right]/(a^2+4b^2t^2+9c^2t^4)^2 \quad \Longrightarrow \\ \stackrel{...}{\vec{r}} \, = \, \stackrel{..}{\vec{\iota}} \, = \frac{d\stackrel{.}{\vec{\iota}}}{dt} \frac{dt}{ds} = \left[ \begin{matrix} 2a(2a^2b^2-24t^2b^4-162t^4b^2c^2+27a^2c^2t^2-405t^6c^4) \\ -8bt(27a^2c^2t^2-81t^6c^4+4a^2b^2) \\ 6c(a^4-6a^2b^2t^2-63a^2c^2t^4-8b^4t^4-90b^2t^6c^2) \end{matrix} \right] / (a^2+4b^2t^2+9c^2t^4)^{7/2} $$ With $\rho^2 = (\stackrel{.}{\vec{\iota}} \cdot \stackrel{.}{\vec{\iota}})$ - inner product - we find for the curvature $\rho$ : $$ \rho(t) = 2\sqrt{\frac{a^2b^2+9a^2c^2t^2+9t^4b^2c^2}{(a^2+4b^2t^2+9c^2t^4)^3}} $$ With the (hopefully) well-known formula $\;\det\left(\stackrel{.}{\vec{r}},\stackrel{..}{\vec{r}},\stackrel{...}{\vec{r}}\right) = \rho^2 \tau \;$ we find herefrom for the torsion $\tau$ : $$ \tau(t) = \frac{3 a b c}{a^2b^2+9a^2c^2t^2+9t^4b^2c^2} $$

  • $\begingroup$ I am a bit confused on why for example r' you include it over the sort and not just have (a 2bt, 3ct^2)? $\endgroup$
    – Quality
    Commented Mar 14, 2015 at 20:17
  • $\begingroup$ @LearningMath: The vector $r$ with the dot above it is the derivative to the arc length $s$. That is the derivative to $t$ divided by the derivative of the arc length $=ds/dt$ or multiplied by $dt/ds$. $\endgroup$ Commented Mar 14, 2015 at 20:22
  • $\begingroup$ Thanks, I think I am a little confused about deriving with respect to something else etc. Can the question just be solved by using X' X'', and X''' as a regular parameter, and applying the formulas $\endgroup$
    – Quality
    Commented Mar 14, 2015 at 20:23
  • $\begingroup$ @LearningMath: If I understand what you mean, then I think the answer is NO: you have to differentiate to the arc length in the end. $\endgroup$ Commented Mar 14, 2015 at 20:29
  • $\begingroup$ @LearningMath: Formula for the torsion of a regular curve paraemeterized by arc length $\endgroup$ Commented Mar 14, 2015 at 20:43

Let curvature $\kappa$ be defined by $\displaystyle \kappa = \frac{\big \|\mathbf T'(t) \big \|}{\big \| \mathbf r'(t)\big \|} $.

Since $\mathbf r(t) = (at, bt^2, ct^3)$,

$\mathbf r'(t)= (a, 2bt, 2ct^2)$ ,

$\|\mathbf r'(t) \| = \sqrt{a^2+4b^2t^2+9c^2t^4}$,

so $\mathbf T(t)= \frac{(a, 2bt, 2ct^2)}{\sqrt{a^2+4b^2t^2+9c^2t^4}}$

and $\mathbf T'(t)= \frac{(0, 2b, 4ct)}{\sqrt{a^2+4b^2t^2+9c^2t^4}}$.

and $\| \mathbf T'(t) \|= \frac{\sqrt{4b^2+16c^2t^2}}{\sqrt{a^2+4b^2t^2+9c^2t^4}}$

Now $\displaystyle \kappa = \frac{\big \|\mathbf T'(t) \big \|}{\big \| \mathbf r'(t)\big \|} = \frac{\sqrt{4b^2+16c^2t^2}}{a^2+4b^2t^2+9c^2t^4}$

For torsion do a similar thing but use the formula:

$\tau = \frac{\big | \mathbf r'(t) \ \mathbf r''(t) \ \mathbf r'''(t) \big| }{\kappa^2} $

where the numerator is the scalar triple product.


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