[BUG] The geometry of directional derivatives?

[Tevian Dray]

One of our instructors suggested the following geometric description of
directional derivatives in 2 dimensions.

==============================================================================
The argument begins with the following facts:

Step 1: A normal vector to the plane ax+by+cz=d is ai+bj+ck.
Step 2: The equation of the tangent plane at to z=f(x,y) at (a,b) is
		z = f + df/dx (x-x0) + df/dy (y-y0)
	where of course f and its derivatives are evaluated at (a,b).

It now follows *algebraically* that:

Step 3: A normal vector to the tangent plane is
		n = - df/dx i - df/dy j + k

Now let m denote the slope in a particular direction given by the unit
vector u = ux i + uy j.

Step 4: If you proceed along u ("run"), then the vertical displacement
	("rise") must therefore be m.
Step 5: Connect the dots: Your actual displacement must be
		p = u + mk
Step 6: But p is a tangent vector!  Thus, n.p=0, or
		- df/dx ux - df/dy uy + m = 0

But m is the directional derivative of f along u, and we have shown that
		D_u f = grad(f) . u
==============================================================================

Every time I see this argument, I am attracted to it.  I like the
geometry of the normal vector (which is grad(z-f)!) and the orthogonal
tangent vector p; we use this notion in the hill lab in the special case
where u is parallel to grad(f).

On the other hand, there is a serious problem here: Except for hills,
the graph of z=f(x,y) is not an actual surface, so that neither the
tangent or normal vectors are vectors in the usual sense.  Yes, one
*can* make sense of vectors whose components have different dimensions,
but n and p turn out to live in different vector spaces (which are dual
to each other), and of course the "angle" between them is meaningless.
This is surely too sophisticated for a multivariable calculus course.

Unfortunately, the conclusion that I draw from this is that the normal
vector n above has no place in explaining the (2d) directional
derivative.  A cute argument, definitely.  Helpful in geometrically
reinforcing the various roles played by the gradient, probably.  But a
good way to *explain** the directional derivative, probably not.

And of course there is an alternative geometric description, which is
essentially the above argument without involving the normal vector
explcitly:

==============================================================================
Step 1: The definition of partial derivatives tells us that the equation
	of the tangent plane is 
		delta z = df/dx delta x + df/dy delta y
Step 2: If I move in the direction of a unit vector u = (delta x) i +
	(delta y) j, then delta z is both the slope and the directional
	derivative.
==============================================================================

Are we there yet?

So is it good or bad to use hills as examples?  They are not typical of
problems involving the gradient.  But perhaps useful?

I welcome further discussion.

Tevian