## Calculus (3rd Edition)

$$r'(t) = \lt e^t, -e^{-t}, 2t \gt, \quad T(t)= \frac{ \lt e^t, -e^{-t}, 2t \gt}{\sqrt{e^{2t}+e^{-2t}+4t^2}}, \quad T(1)=\frac{ \lt e , -e^{-1}, 2 \gt}{\sqrt{e^{2}+e^{-2}+4}}.$$
Since $r(t) = \lt e^t, e^{-t}, t^2 \gt$, then $r'(t) = \lt e^t, -e^{-t}, 2t \gt$ and $\|r'(t)\|=\sqrt{ e^{2t}+e^{-2t}+4t^2}$ Hence, we have $$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt e^t, -e^{-t}, 2t \gt}{\sqrt{e^{2t}+e^{-2t}+4t^2}}, \quad T(1)=\frac{ \lt e , -e^{-1}, 2 \gt}{\sqrt{e^{2}+e^{-2}+4}}.$$