[BUG] A geometric approach to gradients

[Martin Jackson]

I like the spirit of the approach Tevian is 
proposing.  Not long after their first look at 
level curves, I have students think about zooming 
in at a point and looking at the local level 
curve structure.  Students generally have good 
intuition about this and can articulate that the 
"zoomed in" level curves will be lines.   I then 
point out that the difference in this local 
structure from one point to another can be 
characterized by an orientation and a density. 
In the past, I come back to this when we hit 
gradient (having gone through the traditional 
development).  I'm going to put thought into 
following Tevian's plan.

Tevian, can you give us more detail on how you 
want students to think about your RESOLUTION?  In 
particular, how do you want them to think about 
counting "the number of contour lines" in 
computing the number per unit distance?  Does 
Delta f enter into this explicitly?

Martin



>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
>
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