The Gradient Operator

The Gradient (also called the Hamilton operator) is a vector operator for any n-dimensional scalar function $f(x_1,\cdots, x_n)$ (temperature, concentration, pressure, etc.). The gradient vector represents (a) the direction in the n-D space along which the function increases most rapidly, and (b) the rate of the increment. Here we only consider 2D field:

$$\bigtriangledown\stackrel{\triangle}{=}\frac{\partial}{\partial x} \vec{i}+\frac{\partial}{\partial y} \vec{j}$$

When applied to a 2D function $f(x,y)$, this operator produces a vector function:

$$\vec{g}(x,y)\stackrel{\triangle}{=}\bigtriangledown f(x,y) =... ...artial}{\partial y} \vec{j}) f(x,y) =f_x \vec{i}+f_y \vec{j}$$

Now we show that $f(x,y)$ increases most rapidly along the direction of $\vec{g}(x,y)$ with the rate of increment rate is equal to the magnitude of $\vec{g}(x,y)$.

Consider the directional derivative of $f(x,y)$ along an arbitrary direction $r$:

$$\frac{d}{dr}f(x,y)= \frac{\partial f}{\partial x}\frac{dx}{d... ...l f}{\partial y}\frac{dy}{dr} =f_x cos \theta+f_y\sin\theta$$

This directional derivative is a function of $\theta$, defined as the angle between directions $r$ and $x$. To find the direction along which $df/dr$ is maximized, we let

$$\frac{d}{d\theta} \frac{df(x,y)}{dr}= \frac{d}{d\theta} (f_x cos \theta+f_y\sin\theta)= -f_x sin\theta +f_y cos (\theta)=0$$

Solving this for $\theta$, we get

$$f_x sin\theta=f_y cos \theta$$

i.e.,

$$\theta =tan^{-1} \frac{f_y}{f_x}$$

which is the direction of $\vec{g}(x,y)$.

From $tan \theta=f_y/f_x$, we can also get

$$sin\theta=\frac{f_y}{\sqrt{f_x^2+f_y^2}},\;\;\;\; cos \theta=\frac{f_x}{\sqrt{f_x^2+f_y^2}}$$

Substituting these into the expression of $df/dr$, we obtain its maximum magnitude,

$$\left. \frac{d}{dr}f(x,y) \right\vert _{max} =\frac{f_x^2+f_y^2}{\sqrt{f_x^2+f_y^2}}=\sqrt{f_x^2+f_y^2}$$

which is the magnitude of $\vec{g}(x,y)$.