Space curve torsion Hello I am looking for anyone to maybe look over my ideas and see if they think it is correct.
Say I am looking for the torsion $\tau$ of a space curve given by $r(t)=(cos(3t),sin(3t),4t)$
I know if it was in the form $(rcos(t),rsin(t),ct)$ I could use the formula $\tau = \frac{c^2}{c^2+r^2}$ so if I didnt have the 3t I could have just solved it using r as 1.
My question is is there any way to allow that formula to work for a helix with a constant infront of the t?
If not, would I just have to calculate using the formula $$\tau = \frac{ r''' \bullet ( r'' \times r')}{|r'' \times r' | ^2}$$ ? 
Thanks,
PS; when I calculated it regularly I got $\tau=\frac{12}{41}$ , but I am not sure, if anyone could conform/deny that be great,
 A: You can reparametrize it by arc-length instead of using that ugly formula, if you want. If ${\bf r}(t)=(\cos(3t),\sin(3t),4t)$, then ${\bf r}(t)=(-3\sin(3t),3\cos(3t),4)$, and from there: $$s(t) = \int_0^t \|{\bf r}(\xi)\|\,{\rm d}\xi = \int_0^t \sqrt{9\sin^2(3\xi)+9\cos^2(3\xi)+16}\,{\rm d}\xi = \int_0^t5\,{\rm dt} = 5t,$$ so that reparametrizing by $t(s) = s/5$ yields a unit speed curve. Now, with: $${\bf r}(t(s)) \equiv {\bf r}(s) = \left(\cos\left(\frac{3s}{5}\right),\sin\left(\frac{3s}{5}\right),\frac{4s}{5}\right)  $$ we can use the usual formulas. With this: $$\begin{align} {\bf T}(s) &= \left(-\frac{3}{5}\sin\left(\frac{3s}{5}\right),\frac{3}{5}\cos\left(\frac{3s}{5}\right),\frac{4}{5}\right)  \\ {\bf T}'(s) &=  \left(-\frac{9}{25}\cos\left(\frac{3s}{5}\right),-\frac{9}{25}\sin\left(\frac{3s}{5}\right),0\right) \implies \kappa(s) = \frac{9}{25} \\ {\bf N}(s) &=  \left(-\cos\left(\frac{3s}{5}\right),-\sin\left(\frac{3s}{5}\right),0\right) \\ {\bf B}(s) &= \left(\frac{4}{5}\sin\left(\frac{3s}{5}\right), -\frac{4}{5}\cos\left(\frac{3s}{5}\right),\frac{3}{5}\right) \\ {\bf B}'(s) &= \left(\frac{12}{25}\cos\left(\frac{3s}{5}\right), \frac{12}{25}\sin\left(\frac{3s}{5}\right),0\right) \end{align} $$
Now, apart from some sign (there doesn't seem to be a standard convention for it), we know by the Frenet-Serret Formulas that: $$\tau(s) = -\langle {\bf B}'(s), {\bf N}(s)\rangle = \frac{12}{25}.$$
Your direct computation probably gave a wrong result because your curve didn't have unit speed right off from the start.
