# Thread: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1

1. ## 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.]

Let $V = y^2 U_1 - x U_3$ and let $f = xy$ [See attachment page 9 for definition of $U_1, U_2, U_3$, the natural frame field on $\mathbb{R}^3$

O'Neil uses the notation $v_p [f]$ for the derivative of f with respect to $v_p$ and derives the formula

$v_p [f] = \sum v_i \frac{\partial f}{\partial x_i} (p)$ [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 $V[f]$ 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:

$V(p) = (V_1, V-2, V_3) = V_1 U_1 + V_2 U_2 + V_3 U_3$

Also I assume that, analogous to $v_p [f] = \sum v_i \frac{\partial f}{\partial x_i} (p)$ we can write

$V_p [f] = \sum V_i \frac{\partial f}{\partial x_i} (p)$

Thus using the field and function in the exercise above i.e. $V = y^2 U_1 - x U_3$ and $f = xy$ we have

$V = y^2.U_1 + 0.U_2 + (-x).U_3$

i.e. $V_1 = y^2 , V_2 = 0 , V_3 = -x$

Also $\frac{\partial f}{\partial x} = y$ , $\frac{\partial f}{\partial y} = x$ and $\frac{\partial f}{\partial z} = 0$

Thus $V(p)[f] = V_1 \frac{\partial f}{\partial x} + V_2 \frac{\partial f}{\partial y} + V_3 \frac{\partial f}{\partial z}$

so

$V(p)[f] = y^2 . y + 0.x + (-x).0 = y^3$

However, at the back of the book, O'Neill gives the answer as $y^2$

This is really disconcerting ... can anyone locate my error? Could you also confirm that I am using the terminology properly.

Peter

2. ## Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1

I believe you are misunderstanding "V". If a curve is given by $y^2U_1- xU_3$ (which I would have called $y^2\vec{i}- x\vec{k}$) then V is its tangent vector at each point, $2y\vec{i}- \vec{k}$. Of course, that would still give $2y^2$ as the derivative, not $y^2$.

3. ## Re: Directional Derivatives - O'Neil - Elementary Differential Geonetry - Ch 1

Originally Posted by HallsofIvy
I believe you are misunderstanding "V". If a curve is given by $y^2U_1- xU_3$ (which I would have called $y^2\vec{i}- x\vec{k}$) then V is its tangent vector at each point, $2y\vec{i}- \vec{k}$. Of course, that would still give $2y^2$ as the derivative, not $y^2$.
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.

4. ## 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:

$V(p) = (V_1, V_2, V_3) = V_1 U_1 + V_2 U_2 + V_3 U_3 = y^2.U_1 + 0.U_2 + (-x).U_3$

I am seeking to understand what is wrong with breaking up a vector field in $\mathbb{R}^3$ into three coordinate functions?

[By the way, you are right, O'Neill has picked up the error - or someone has for him]

Peter

5. ## 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.

6. ## 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