[BUG] A geometric approach to gradients

[Tevian Dray]

The traditional approach to the gradient simply defines the gradient of a
function in terms of its rectangular components, then derives the geometric
interpretation from the definition.

Modern treatments improve on this slightly, by starting with the notion of
directional derivative, then using local linearity to express directional
derivatives as a dot product, thus producing the gradient in rectangular
coordinates.  The geometric properties of the gradient now follow fairly
easily (since one already has the directional derivative).

One aspect of this treatment which I dislike, however, is that the gradient
is fundamentally defined in terms of its rectangular components.  I propose
turning the argument around.

1. DEFINE the gradient of a function to be the vector whose direction is that
   in which f increases the fastest, and whose magnitude is the rate of
   increase in that direction.

2. Use local linearity to argue that the rate of change of f in any direction
   is the dot product of the gradient with the given direction.  Therefore
   DEFINE this dot product to be the directional derivative.

3. Compute the rectangular components of the gradient by noting that they
   measure the rate of change of f along the axes, and must therefore be the
   (rectangular) partial derivatives of f.

I implemented this in my class this term as follows.  First, I discussed
contour diagrams of linear functions (of 2 variables).  After several
examples, I characterized such contour diagrams by a RESOLUTION (number of
contour lines per unit distance; call it M) and an ORIENTATION (oriented unit
vector orthogonal to the contour lines; call it g).  I then argued
geometrically that the number of contour lines crossed by any vector v would
be the resolution times the projection of v along g, that is M (v*g), or
equivalently (Mg)*v.  (I have written "*" for the dot product to avoid
confusion with "." or possible font problems with "·".)

How many lines are crossed by a vector of unit length in the x direction?
Mg*i, thus giving meaning to the rectangular components of Mg.

All that was left was to change the context to the tangent plane to the
graph of a function, with the gradient playing the role of the vector Mg
(by definition!), emphasizing that the number of contour lines crossed is just
the rate of change of the function, which we already knew to be the partial
derivatives in coordinate directions.

I chose to save the concept of directional derivative for a separate lecture,
reiterating that the number of contour lines crossed, and hence the rate of
change, in any direction is the dot product with "Mg", in this case the
gradient.  This gave me an excuse to relate the geometric properties of
the gradient to the notion of directional derivatives, effectively the
standard lecture normally used to derive those properties.  Not logically
necessary, but I think good for the students to see again.

It was also not until this later lecture that I expressed directional
derivatives in terms of limits, arguing by analogy with partial derivatives,
but not actually deriving the formula normally used to define the directional
derivative.  This may bother some folks, but I think it's the main point: The
dot product in the directional derivative is a geometric property of local
linearity, not an algebraic trick which somehow produces partial derivatives
in the limit expression.

I should also emphasize to this audience that I did not use differentials, and
in particular dr vector, anywhere in this argument, and that I therefore did
not mention the "master formula" df = grad(f)*dr or any of its derivative
versions.  To do so really requires discussing curves prior to partial
differentiation.  I think a good case can be made for this order (which is the
one we used at Mount Holyoke), but other constraints may make it difficult to
cover that material early enough.  Here at OSU, I will "review" gradient from
this point of view in the subsequent course, just after introducing dr vector.

Tevian