[BUG] more on gradients

[Tevian Dray]

>>>>> Martin Jackson writes:

    MJ> In posing my question, I was curious about how you are setting up the
    MJ> contour diagrams (for linear functions) in which you are doing the
    MJ> counting.  Are you always using Delta z=1 in these?

The only example I did used Delta z = 1, but I don't think this matters so
long as Delta z is constant, as it must be for a linear function.

    MJ> In giving the specific example z=5x+3y+7, I used coordinates.  Is
    MJ> there some better initial exercise that doesn't use coordinates to
    MJ> specify a linear function?

I simply drew contour lines, without specifying the function.  I did draw axes
so that we could determine the partial derivatives by counting lines, but I
later discussed directional derivatives using a similar diagram without axes.

    MJ> I've defined gradient in the linear case to have direction
    MJ> perpendicular to the level curves.  Tevian defines the direction to be
    MJ> "that in which f increases the fastest."  I'm not sure if one approach
    MJ> is significantly better than the other.

>>>>> Matthias Kawski writes:

    MK> As long as the units in x and y direction are the same (meters or
    MK> dollars), there is no real issue. But as soon as the variables are
    MK> e.g. (pressure,volume) or (voltage, current) then perpendicular (and
    MK> dot product) make(s) no sense anymore.

So neither Martin's nor my phrasing works in this case; not only does
perpendicularity fail, but there's no notion of "fastest".  This is a clear
indication that the notion of gradient is not relevant to such problems,
although (as has already been pointed out) both differentials and local
linearity can still be used.

    MK> More generally, the perpendicularity issue appears as soon as anything
    MK> is graphed using different scales along the axes. E.g. plot the lines
    MK> y=10x and y=-10x on the window [-1,1] x [-10,10].

A similar Maple example is:
	> with(linalg):with(plots):
	> c:=contourplot(x^2+y^2,x=-3..3,y=-5..5):
	> f:=fieldplot(grad(x^2+y^2,[x,y]),x=-3..3,y=-5..5):
	> display(c,f,scaling=unconstrained);
which seems to imply the gradient is not perpendicular to the level curves.

>>>>> Alexander Smith writes:

    AS> It is convenient to quantify the condition that the two curves
    AS> meet tangentially by saying that their gradients are [parallel],
    AS> but this is just gravy.

Absolutely!

Tevian