# What's the geometrical interpretation of the magnitude of gradient generally?

In the following picture, the author of the Field and Wave Electromagnetics shows the geometrical meaning of the direction of the gradient. That is, only by following the direction of the normal vector to the curve at that pointer could the rate of change be the maximum.

But what about the geometrical interpretation of the magnitude of gradient generally or maybe is there a geometrical interpreation of the magnitude of graident?

thanks.

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Here's a geometric way of thinking that might be helpful: Consider a family of level surfaces $f(x,y,z)=C$ for some evenly spaced values of $C$ (where the spacing should be fairly small). These level surfaces will lie closely stacked in space near points where $|\nabla f|$ is large, and farther apart near points where $|\nabla f|$ is small.

(The two-dimensional counterpart is curves of constant elevation on a map; they are densely packed where the slope of the terrain is steep.)

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If you look at the set of points satisfying f(x) = c, the gradient of f is normal to the surface and points in the direction of greatest increase of f. The magnitude of the gradient is proportional to the rate of increase.

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@Jichao: this is more or less by definition. –  Qiaochu Yuan Dec 2 '10 at 17:14
I think you mean the gradient of f is normal to the surface f(x) = c, yes? –  Rahul Dec 2 '10 at 17:43
Yes it should read normal to the surface. –  jennifer Dec 2 '10 at 17:45
Thanks, I corrected my answer. It's easy to see the gradient points in the normal direction: there's no change in f along the surface, so the direction of greatest change points away from the surface. –  John D. Cook Dec 2 '10 at 20:23
If you work out the algebra, the rate of change in a function $f$ in the direction of the unit vector $\eta$ is $\nabla f \cdot \eta$ (This is simply the chain rule.). This quantity reaches its maximum when $\eta$ shares the same direction as $\nabla f$ (to maximize the cosine of the angle that's part of the dot product formula). Thus the magnitude of $\nabla f$ gives the rate of change in that direction, as well.
The analogy with the hill might be a bit confusing here, since it refers to the gradient of a two-variable function $f(x,y)$, the hill being its graph $z=f(x,y)$. The picture in the question, on the other hand, seems to illustrate the gradient of a three-variable function $f(x,y,z)$. –  Hans Lundmark Dec 2 '10 at 19:07
If you are at a point ${\bf p}$ in the domain of $f$ then the rate of change of $f$ when starting from ${\bf p}$ to nearby points depends on the direction you take. The gradient $\nabla f({\bf p})$ is a vector attached at the point ${\bf p}$; it points into the direction of maximal positive rate of change of $f$. The actual rate of change in this direction is $|\nabla f({\bf p})|$, and in the opposite direction it is $-|\nabla f({\bf p})|$. For an arbitrary unit vector ${\bf e}$ attached at ${\bf p}$ the $directional\ derivative$ of $f$ in direction ${\bf e}$ is given by the scalar product $\nabla f({\bf p})\bullet {\bf e}$.