Answer
$$r'(t) = \lt 2, 2t, − 2t \gt, \quad T(t)= \frac{ \lt 2, 2t, − 2t \gt}{\sqrt{4+8t^2}}, \quad T(1)=\frac{ \lt 2, 2 , − 2 \gt}{\sqrt{12}}.$$
Work Step by Step
Since $ r(t) = \lt 1 + 2t, t2, 3 − t2\gt$, then $ r'(t) = \lt 2, 2t, − 2t \gt$ and $\|r'(t)\|=\sqrt{ 4+8t^2} $
Hence, we have
$$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt 2, 2t, − 2t \gt}{\sqrt{4+8t^2}}, \quad T(1)=\frac{ \lt 2, 2 , − 2 \gt}{\sqrt{12}}.$$
