Cartesian Definition of Gradient defines Gradient Operator

Theorem
Let $R$ be a region of Cartesian $3$-space $\R^3$.

Let $\map F {x, y, z}$ be a scalar field acting over $R$.

Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.

Let $\grad F$ be defined according to the Cartesian $3$-space definition of the gradient of $F$:

Then $\grad F$ is a gradient operator as defined by the geometrical representation:


 * $\grad F = \dfrac {\partial F} {\partial n} \mathbf {\hat n}$

where:
 * $\mathbf {\hat n}$ denotes the unit normal to the equal surface $S$ of $F$ at $A$
 * $n$ is the magnitude of the normal vector to $S$ at $A$.

Proof
The vector rates of increase of $F$ in the directions of the $3$ axes are:
 * $\dfrac {\partial F} {\partial x} \mathbf i$, $\dfrac {\partial F} {\partial y} \mathbf j$, $\dfrac {\partial F} {\partial z} \mathbf k$

Their sum will be a vector with the magnitude and direction of the most rapid rate of increase of $F$.

It remains to show that this expression:
 * $\dfrac {\partial F} {\partial x} \mathbf i + \dfrac {\partial F} {\partial y} \mathbf j + \dfrac {\partial F} {\partial z} \mathbf k$

is equivalent to:
 * $\dfrac {\partial F} {\partial n} \mathbf {\hat n}$

Let us take the dot product of both sides of the gradient equation with the position vector $\d \mathbf r$ of an arbitrary point $A$ on an equal surface $S$ of $F$.

Thus:

In Cartesian coordinates:

Thus the operations $\grad$ and $\nabla$ applied to a point in a scalar field are identical.