Show that $\bigtriangledown \vec{r} = \vec{1}$

Question

Show that $\bigtriangledown \vec{r} = \vec{1}$

My instructor put the r and 1 in bold. I am not sure what a bold one means. From my work I get $1ii + 1jj + 1zz$

Answer 1

Given a vector-valued function $\mathbf{r}:\mathbb{R}^3\rightarrow \mathbb{R}^3$, where $\mathbf{r}=\vec{r}=\langle x,y,z\rangle$, I wonder if your instructor meant to write the derivative of $\mathbf{r}$: $$ \begin{align*} D \mathbf{r} &= \begin{pmatrix} \partial_x(x) & \partial_y(x) & \partial_z(x) \\ \partial_x(y) & \partial_y(y) & \partial_z(y) \\ \partial_x(z) & \partial_y(z) & \partial_z(z) \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} = \mathbf{1}, \end{align*} $$ a matrix of partial derivatives of the component functions, where $\partial_x = \partial/\partial_x$.