Re: directional derivative,-

On Tue, 24 Apr 2007 01:43:40 +0100, JEMebius <jemebius@xxxxxxxxx>
wrote:

What about the well-known rectangular circular half-cone,
the graph of (x, y) -> z = sqrt(x^2 + y^2)?

That has no directional derivative in _any_ (non-zero) direction
(at the origin, which is presumably the point you're talking about.)

At least not according to what I've always thought was the
standard definition, as at

http://en.wikipedia.org/wiki/Directional_derivative

At first sight the problem seems underspecified? incompletely posed? badly posed?
provocative, i.e. aimed at putting in motion the reader's creativity?

Or did the author mean to ask:
"Prove that no real-valued function f ==that has a continuous derivative at a point C==
has a positive directional derivative at C for every possible direction u."

BTW, the paraboloid (x, y) -> z = x^2 + y^2 does not qualify because the slope at (0, 0)
is zero in every direction.
This example shows that zero local slope is not in contradiction to positive average slope.

Ciao: Johan E. Mebius


chrizm7@xxxxxxxxx wrote:
A problem asks me to show that no real-valued function f has a
positive directional derivative at a fixed point c for every possible
direction u.

But wouldn't a paraboloid work? Fix c corresponding to the vertex.
Then for any direction u, the slope will always be positive
(increasing).

To be clear, the exact problem is worded: Prove that there is no real-
valued function f such that f'(c;u) > 0 for a fixed point c in R^n and
every nonzero vector u in R^n.



************************

David C. Ullrich
.