$1$c$)(10\%)$ Justify that the scalar Laplacian operation $gra{d}^2-=$grad$*grad$ gives the distribution of "curvature". $($Note:

grad$*A=\frac{del{A}_x}{\text{d}\text{e}\text{l}\text{x}}+\frac{del{A}_y}{\text{d}\text{e}\text{l}\text{y}}+\frac{del{A}_z}{\text{d}\text{e}\text{l}\text{z}}$

in Cartesian coordinate.$)$