What does the gradient mean geometrically? Along a particular path, $df$ tells us something about how $f$ is changing. But the Master Formula tells us that $df=\grad f\cdot d\rr$, which means that the dot product of $\grad f$ with a vector tells us something about how $f$ changes along that vector. So let $\Hat w$ be a unit vector, and consider \begin{equation} \grad{f} \cdot \Hat w = |\grad{f}| \> |\Hat w| \cos\theta = |\grad{f}| \cos\theta \end{equation} which is clearly maximized by $\theta=0$. Thus, the direction of $\grad{f}$ is just the direction in which $f$ increases the fastest, and the magnitude of $\grad{f}$ is the rate of increase of $f$ in that direction (per unit distance, since $\Hat w$ is a unit vector). If you visualize the value of the scalar field $f$ as represented by color, then the gradient points in the direction in which the rate of change of the color is greatest.

You can also visualize the gradient using the level surfaces on which $f(x,y,z)={\rm const}$. (In two dimensions there is the analogous concept of level curves, on which $f(x,y)={\rm const}$.) Consider a small displacement $d\rr$ that lies on the level surface, that is, start at a point on the level surface, and move along the surface. Then $f$ doesn't change in that direction, so $df=0$. But then \begin{equation} 0 = df = \grad{f} \cdot d\rr = 0 \label{fconst} \end{equation} so that $\grad{f}$ is perpendicular to $d\rr$. Since this argument works for any vector displacement $d\rr$ in the surface, $\grad{f}$ must be perpendicular to the level surface.

If you prefer working with derivatives instead of differentials, consider a curve $\rr(u)$ that lies in the level surface. Now simply divide ($\ref{fconst}$) by $du$, obtaining \begin{equation} 0 = \frac{df}{du} = \grad{f} \cdot \frac{d\rr}{du} = 0 \end{equation} so that $\grad{f}$ is perpendicular to the tangent vector $\frac{d\rr}{du}$ (which is just the velocity vector if the parameter $u$ represents time). Again, this argument applies to any curve in the level surface, so $\grad{f}$ must be perpendicular to every such curve. In other words, $\grad{f}$ is perpendicular to the level surfaces of $f$: \begin{equation} \grad{f} \perp \{f(x,y,z)={\rm const}\} \end{equation}

This orthogonality is shown for the case of level curves in Figure 1, which shows the gradient vector at several points along a particular level curve among several. You can think of such diagrams as topographic maps, showing the “height” at any location. The magnitude of the gradient vector is greatest where the level curves are close together, so that the “hill” is steepest.

An alternative way of seeing this orthogonality is to recognize that, since the gradient is a derivative operator, its value depends only on what is happening locally. If you zoom in close enough to a given point, the level surfaces are parallel, and the gradient points in the direction from one level surface to the next.

Like all derivative operators, the gradient is linear (the gradient of a sum is the sum of the gradients), and also satisfies a product rule \begin{equation} \grad(fg) = (\grad{f})\,g + f\,(\grad{g}) \end{equation} This formula can be obtained either by working out its components in, say, rectangular coordinates, and using the product rule for partial derivatives, or directly from the product rule in differential form, which is \begin{equation} d(fg) = (df)\,g + f\,(dg) \end{equation} as discussed in § {Using Differentials to Compute Derivatives}.