Justification for Geometrical Representation of Gradient Operator

Proof
According to the geometrical representation:


 * Gradient-operator.png

Let $S$ be a constant value of $F$ giving rise to an equal surface of $F$.

Let $S + \d S$ be another constant value of $F$ giving rise to an equal surface of $F$ very close to $S$.

To economise on notation we refer to those equal surfaces themselves as $S$ and $S + \d S$ respectively.

Consider the point $A$ on $S$ whose position vector is $\mathbf r$.

Let $B$ be a point on $S + \d S$ whose position vector is $\mathbf r + \d \mathrm r$.

The smallest distance from $S$ to $S + \d S$ from $A$ is $AC$:
 * whose direction is that of the unit normal $\mathbf {\hat n}$ at $A$
 * whose length is $\norm {\d \mathbf n}$.

Let $\norm {\d \mathrm r}$ be the length of $AB$.

This is the magnitude of the rate of increase at $A$ of $S$ in the direction of $AB$ when $S$ and $S + \d S$ are infinitesimally close together.

The rate of increase is greatest in the direction is that of the unit normal, that is, along $AB$.

Hence at $A$ the rate of increase has value $\dfrac {\partial F} {\partial n}$.

Thus:
 * $\dfrac {\partial F} {\partial r} = \dfrac {\partial F} {\partial n} \cos \theta$

where $\theta$ is the angle between $AB$ and $AC$.

Hence if $\mathbf {\hat n}$ is the unit normal to $S$ at any point in a scalar field, the vector quantity $\mathbf {\hat n} \dfrac {\partial F} {\partial n}$ gives the greatest rate of increase of $F$ at that point in magnitude and direction.

This is the gradient of $F$ at that point and can be written:
 * $\grad F = \dfrac {\partial F} {\partial n} \mathbf {\hat n}$

Thus the gradient of $F$ is a vector field whose direction is the fastest rate of increase of $F$ and whose magnitude is that rate of increase.