Find the point where $\tau(t)$ is minimal.

Question

Question.

$\vec{r}(t)=\left[\frac{4}{9}(1+t)^{3/2},\frac{4}{9}(1-t)^{3/2},\frac{1}{3}t\right]$ is a parametrization of a curve $C$ ($t\in\left[-1,1\right]$)

find the point where $\tau(t)$ is minimal.

Attempt. $$ \tau(t) =\frac{(\vec{r}\,'(t)\times\vec{r}\,''(t))\cdot\vec{r}'''(t)}{||\vec{r}\,'(t)\times\vec{r}\,''(t)||^{2}}=\frac{\left[-\frac{1}{9}(1-t)^{-1/2},\frac{1}{9}(1+t)^{-1/2},\frac{4}{9}(1-t^{2})^{-1/2}\right]\cdot\left[-\frac{1}{6}(1+t)^{-3/2},\frac{1}{6}(1-t)^{-3/2},0\right]}{\left(\frac{\sqrt{2}}{3}(1-t^{2})^{-1/2}\right)^{2}} =\frac{\frac{1}{54}(1-t)^{-1/2}(1+t)^{-3/2}+\frac{1}{54}(1+t)^{-1/2}(1-t)^{-3/2}}{\frac{2}{9}(1-t^{2})^{-1}} =\frac{1}{12}(1-t^{2})\left((1-t)^{-1/2}(1+t)^{-3/2}+(1+t)^{-1/2}(1-t)^{-3/2}\right) =\frac{1}{12}(1-t)(1+t)\left((1-t)^{-1/2}(1+t)^{-3/2}+(1+t)^{-1/2}(1-t)^{-3/2}\right) =\frac{1}{12}\left((1-t)^{1/2}(1+t)^{-1/2}+(1+t)^{1/2}(1-t)^{-1/2}\right) =\frac{1}{12}\left(\frac{\sqrt{1-t}}{\sqrt{1+t}}+\frac{\sqrt{1+t}}{\sqrt{1-t}}\right)=\frac{1}{12}\left(\frac{1-t+1+t}{\sqrt{1-t^{2}}}\right)=\frac{1}{6}(1-t^{2})^{-1/2} $$

But I think my calculation is wrong, can you take a look?

Answer 1

The answer looks correct to me. Using Mathematica to perform the computation and simplify under the assumption that $-1 \leq t \leq 1$, I obtain the same thing, i.e. $\frac{1}{6\sqrt{1 - t^2}}$. Here was the Mathematica Code:

r[t_] := {4/9 (1 + t)^(3/2), 4/9 (1 - t)^(3/2), 1/3 t};
Cross[r'[t], r''[t]] . r'''[t]/Norm[Cross[r'[t], r''[t]]]^2;
FullSimplify[%, -1 < t < 1]

Answer 2

Your calculations are correct, as also checked by @ mathametics2X2life.

The $\pm$ plus and minus sign of parameterization admits both signs $(>0,<0) $

$$\vec{r_1}(t)=(+\frac49 (1+t)^{3/2},+\frac49 (1-t)^{3/2},\frac{1}{3}t)$$

$$\vec{r_2}(t)=(-\frac49 (1+t)^{3/2},-\frac49 (1-t)^{3/2},\frac{1}{3}t)$$

and acordingly we can have torsion $\tau$ at $t=0$ both as a minimum and as a maximum respectively. A rough plot of torsions: