Reparametrize the curve $\alpha(t)=(e^{t},e^{-t},\sqrt{2}t), \; \alpha: \mathbb{R} \rightarrow \mathbb{R}^{3}$, using $h(s)= \log(s)$ on $J:s>0$. Check the equation in Lemma in this case by calculating each side separately.
This Lemma : If $\beta$ is the reparametrization of $\alpha$ by $h$, then $$ \beta' (s)=\alpha'(h(s)) \frac{dh}{ds}(s) .$$
I don't know answer, so I need some help. Please thank you for reading.
After reparametrization you have a map $\beta: \mathbb R_{>0} \to \mathbb R^3$ defined by
$$\beta (s) = \alpha (h(s)) = (s, {1\over s}, \sqrt{2} \log s)$$
For the LHS of the lemma compute $\beta' (s)$ as
$$ \beta' (s) = {d \over ds} \beta (s) = (1, -{1 \over s^2}, {\sqrt{2}\over s})$$
For the RHS of the lemma compute
$$ {d \over ds} h(s) = {1 \over s}$$
and
$$ \alpha '(h(s)) = {d \over dh }\alpha (h) = (e^{h(s)}, -e^{-h(s)}, \sqrt{2})= (s, -{1\over s}, \sqrt{2})$$
and note that
$$ \alpha'(h(s)) \frac{dh(s)}{ds}= (s,-{1 \over s},{\sqrt{2}}) \cdot {1 \over s} = (1, -{1 \over s^2}, {\sqrt{2}\over s}) = \beta '(s)$$
as desired.
