Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises - Page 815: 41

$\textbf{r}''(t) = \langle 2, 0 , 0\rangle$ $\textbf{r}'''(t) = \langle 0, 0, 0\rangle$

To find the derivative, simply find the derivative of each component. $\textbf{r}(t) = \langle t^2+1, t+1 , 1\rangle$ $\textbf{r}'(t) = \langle 2t, 1 , 0\rangle$ $\textbf{r}''(t) = \langle 2, 0 , 0\rangle$ $\textbf{r}'''(t) = \langle 0, 0, 0\rangle$