Answer
$$ r'(t) = \lt 8t,9\gt, \quad T(t)= \frac{ \lt8t,9\gt}{\sqrt{64t^2+81}}, \quad T(1)=\frac{ \lt 8,9\gt}{\sqrt{145}}.$$
Work Step by Step
Since $ r(t) = \lt 4t^2, 9t\gt$, then $ r'(t) = \lt 8t,9\gt$ and $\|r'(t)\|=\sqrt{64t^2+81}=\sqrt{59}$.
Hence, we have
$$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt8t,9\gt}{\sqrt{64t^2+81}}, \quad T(1)=\frac{ \lt 8,9\gt}{\sqrt{145}}.$$
