Torsion and curvature of the curve $X(t) = (at, bt^2, ct^3)$
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 $\tau$ :
$$
\tau(t) = \frac{3 a b c}{a^2b^2+9a^2c^2t^2+9t^4b^2c^2}
$$