Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 900: 9

\[ \vec{r}'(t)=\langle\sqrt{t-2},3,\frac{1}{t^2}\rangle \]

\( \vec{r}(t)=\langle\sqrt{t-2},2,\frac{1}{t^2}\rangle\) In order to get \(\vec{r}'(t)\) we simply take the derivate of each component. \(\vec{r}'(t)=\langle \frac{1}{2\sqrt{t-2}}, 0, \frac{-2}{t^3} \rangle \).