Answer
$\mathbf{T}(2)=\langle\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\rangle$
Work Step by Step
1) Differentiate $\mathbf{r}(t)$ by differentiating each component with power rule
--> $\mathbf{r}'(t)=\frac{d(\langle{t^2-2t, 1+3t, \frac{1}{3}t^3+{1}{2}t^2}\rangle)}{dt}=\langle{2t-2, 3, t^2+t}\rangle$
2) Evaluate $\mathbf{r}'(t)$ at given parameter value $t=2$
-->$\mathbf{r}'(2)=\langle2, 3, 6\rangle$
3) Find magnitude of $\mathbf{r}'(2)$
--> $|\mathbf{r}'(2)|=\sqrt{2^2+3^2+6+2}=\sqrt{4+9+36}=\sqrt{49}=7$
4) By definition, $\mathbf{T}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$
5) Therefore, $\mathbf{T}(2)=\frac{\mathbf{r}'(2)}{|\mathbf{r}'(2)|}=\frac{\langle2, 3, 6\rangle}{7}=\langle\frac{2}{7}, \frac{3}{7}, \frac{6}{6}\rangle$
