1 Attachment(s)
Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1
I am reading O'Neill: Elementary Differential Geonetry Ch 1. I am having a problem with Exercise 3 of Exercises 1.3 page 15, which I suspect may be due to me misuderstanding O'Neill's notation. [So that others may follow his terminology I have attached a copy of O'Neill Ch1 pages 6 to 15.]
Exercise 3 reads as follows:
Let
and let
[See attachment page 9 for definition of
, the natural frame field on 
O'Neil uses the notation
for the derivative of f with respect to
and derives the formula
[see the attachment pages 11 and 12]
O'Neill then defines the operation of a vector field V on a function f. The result is the real-valued function
whose value at each point p is the number V(p)[f], that is the derivative of f with respect to the tangent vector V(p) at p.
Thus following this (and the notation and definitions in O'Neil - see attachment) that we can (correctly) write the following:
 = (V_1, V-2, V_3) = V_1 U_1 + V_2 U_2 + V_3 U_3)
Also I assume that, analogous to
we can write
![V_p [f] = \sum V_i \frac{\partial f}{\partial x_i} (p)](http://latex.codecogs.com/png.latex? V_p [f] = \sum V_i \frac{\partial f}{\partial x_i} (p) )
Thus using the field and function in the exercise above i.e.
and
we have
.U_3 )
i.e. 
Also
,
and 
Thus ![V(p)[f] = V_1 \frac{\partial f}{\partial x} + V_2 \frac{\partial f}{\partial y} + V_3 \frac{\partial f}{\partial z}](http://latex.codecogs.com/png.latex? V(p)[f] = V_1 \frac{\partial f}{\partial x} + V_2 \frac{\partial f}{\partial y} + V_3 \frac{\partial f}{\partial z} )
so
![V(p)[f] = y^2 . y + 0.x + (-x).0 = y^3](http://latex.codecogs.com/png.latex?V(p)[f] = y^2 . y + 0.x + (-x).0 = y^3 )
However, at the back of the book, O'Neill gives the answer as 
This is really disconcerting ... can anyone locate my error? Could you also confirm that I am using the terminology properly.
Peter
Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1
I believe you are misunderstanding "V". If a curve is given by
(which I would have called
) then V is its tangent vector at each point,
. Of course, that would still give
as the derivative, not
.
Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1
Quote:
Originally Posted by
HallsofIvy
I believe you are misunderstanding "V". If a curve is given by

(which I would have called

) then V is its
tangent vector at each point,

. Of course, that would still give

as the derivative, not

.
i believe *you* are misunderstanding the question. V is a vector field, not a curve in R3.
i think calling the coordinate functions of V, "y2" and "x" is a bad idea...if i understand this correctly what we really have is the following:
V(x,y,z) = y2U1(x,y,z) - xU3(x,y,z)
= y2(1,0,0)(x,y,z) - x(0,0,1)(x,y,z)
= (y2,0,-x)(x,y,z).
in that case, i concur that V[f](x,y,z) = y2(y) + 0(x) + (-x)(0) = y3. the answer in the back appears to be incorrect.
EDIT: take a look here: http://www.calpoly.edu/~jborzell/Cou...ff_Geo_2ed.pdf
and see if the answer in question is on page 451.
Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1
Deveno,
Thanks for the help - and a thanks to HallsofIvy as well!
You write "i think calling the coordinate functions of V, "y2" and "x" is a bad idea"
I am trying to understand why you say this, but I think you are not happy with my statement:
 = (V_1, V_2, V_3) = V_1 U_1 + V_2 U_2 + V_3 U_3 = y^2.U_1 + 0.U_2 + (-x).U_3)
I am seeking to understand what is wrong with breaking up a vector field in
into three coordinate functions?
Can you please clarify?
[By the way, you are right, O'Neill has picked up the error - or someone has for him]
Peter
Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1
no, it's not your statement, you're fine. i think what is better is to say:
V1(x,y,z) = y2
V2(x,y,z) = 0
V3(x,y,z) = -x
notice is all 3 of these statements, x,y and z are "dummy variables" we get the exact same functions if we write:
V1(p1,p2,p3) = (p2)2
V2(p1,p2,p3) = 0
V3(p1,p2,p3) = -p1
understanding what function is INTENDED by V1,V2, and V3 depends on recognizing (x,y,z) as a "typical point", which is fine, but O'Neill uses "p"!
in other words: "-x" is isn't a function, it's the VALUE of a function (namely the function that takes the negative of the third coordinate of a triple of real numbers) when we take the first coordinate of p to be x. perhaps this seems like needless nit-picking, but i have to wonder: why be so careful about the definition of the "standard frame", and then so sloppy about defining the "coordinate functions"? especially since the point p in the definition of the vector field (for the tangent space) is just along for the ride, we actually USE the coordinate functions, and then we just tag the p as a subscript.
it's like if i say: "what is the derivative of a?" is "a" a number, the constant function f(x) = a, for all x in R, or the identity function f(a) = a, for all a in R? i haven't said, i've just written down a letter. it's little better if i say, "what is the derivative of y at a?" now there's two ambiguities, and you don't know whether you're trying to find y'(a) or evaluating the constant function 1 at a, or if (even *worse*) you have to use the chain rule:
y'(a(t))(a'(t))
in which case do i want the whole thing, or just the first factor?
perhaps i am way out in left field, here, but mathematical communication should possess clarity.
Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1
Thanks for that! Most helpful to me
I totally agree with you on the need for precision and clarity
Peter