Answer
$\mathbf{r}'(t)$ is a vector that is tangent to the graph of a vector-valued function \(\mathbf{r}(t)\)
Work Step by Step
Step 1: A vector-valued function is differentiable if every component function of the vector \[ \mathbf{r}(t) = \langle \mathbf{f}(t), \mathbf{g}(t), \mathbf{h}(t) \rangle = \mathbf{f}(t)i + \mathbf{g}(t)j + \mathbf{h}(t)k \] is differentiable. In other words, \(\mathbf{r}(t)\) is differentiable if \(\mathbf{f}'(t)\), \(\mathbf{g}'(t)\), and \(\mathbf{h}'(t)\) exist.
Step 2: Geometrically interpreted, \(\mathbf{r}'(t)\) is a function whose output is a vector that is tangent to the graph of a vector-valued function \(\mathbf{r}(t)\).
