Exact differential of scalar field

[7thSon]

Suppose I have the scalar field f in the xy-plane and that it is smooth.

Its total derivative is given the normal way, i.e.

$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$

and the gradient of f is given the normal way as well.

I read in a paper that, due to the fact that the gradient and isocontours of f are perpendicular, the following relation holds:

$$\frac{dy}{dx} = \frac{f_y}{f_x}$$,

where $$f_x$$ is the partial derivative of f with respect to x.

My question is, how the heck do they get this and what does the quantity dy/dx mean? The motivation for writing this quantity in the first place is because there is a reason to find the quantities

$$\frac{dx}{dT}$$ and $$\frac{dy}{dT}$$

where those quantities are exact differentials and not partial differentials.

I guess my confusion lies in the fact that I am not used to seeing the exact differential dy/dx in the context of a function f of two variables, though I do understand that an increment df will, in general, vary dependently on x and y.

 

[Mute]

Suppose I have the scalar field f in the xy-plane and that it is smooth.

Its total derivative is given the normal way, i.e.

$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$

and the gradient of f is given the normal way as well.

I read in a paper that, due to the fact that the gradient and isocontours of f are perpendicular, the following relation holds:

$$\frac{dy}{dx} = \frac{f_y}{f_x}$$,

where $$f_x$$ is the partial derivative of f with respect to x.

Are you sure the formula you wrote is correct? I would expect $dy/dx = - f_x/f_y$? (e.g., see http://en.wikipedia.org/wiki/Implicit_function#Formula_for_two_variables) I would guess from the comment "gradient and isocontours of f are perpendicular" the derivative is the variation of y with respect to x along contours on which f is constant (an "isocontour").

 

[7thSon]

Are you sure you're not missing a minus sign? $dy/dx = - f_y/f_x$? I would guess from the comment "gradient and isocontours of f are perpendicular" the derivative is the variation of y with respect to x along contours on which f is constant (an "isocontour").

I too noticed that this just becomes the reciprocity relation when the sign is flipped, but wasn't sure whether or not that was a typo in the paper I was reading. I think you are probably right, and the lack of minus sign might have to do with the fact that the authors are more interested in the opposite of the gradient and were being sloppy.

Now that I think of it though, your comment makes a lot of sense, considering I remember a similar notation with exact differentials for phase lines in thermodynamics, i.e. there is a line dP/dT on some contour that represents the phase transition.

Your explanation of dy/dx thus makes sense, but I'm still wondering how to interpret the quantities

$$\frac{df}{dx}$$

vs.

$$\frac{\partial f}{\partial x}$$

The former looks/behaves kinda like a material derivative (having multiple contributions) but I still don't see the difference

 

[7thSon]

by the way, if anyone can explain this in terms of standard differential calculus, and then with the deeper interpretation of df as a one-form, that would be extremely useful to me.

 

[lavinia]

Suppose I have the scalar field f in the xy-plane and that it is smooth.

Its total derivative is given the normal way, i.e.

$$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$

and the gradient of f is given the normal way as well.

I read in a paper that, due to the fact that the gradient and isocontours of f are perpendicular, the following relation holds:

$$\frac{dy}{dx} = \frac{f_y}{f_x}$$,

where $$f_x$$ is the partial derivative of f with respect to x.

My question is, how the heck do they get this and what does the quantity dy/dx mean? The motivation for writing this quantity in the first place is because there is a reason to find the quantities

$$\frac{dx}{dT}$$ and $$\frac{dy}{dT}$$

where those quantities are exact differentials and not partial differentials.


I guess my confusion lies in the fact that I am not used to seeing the exact differential dy/dx in the context of a function f of two variables, though I do understand that an increment df will, in general, vary dependently on x and y.

If (dx,dy) is tangent to a level curve of f the df is zero on it.

 

[Rasalhague]

NOTATION. (Except for I and D, the particular letters used in this first paragraph are arbitrary, just for the sake of example, and have no significance to what follows. I'm introducing all this notation in the hope that it will make things clearer; if it doesn't work for you, or seems unnecessary, skip to the end where I've used a disambiguated form of the traditional Leibniz notation.) So, anyway... Let square brackets enclose the argument(s) of a function, thus a[x], b[x,y]. Let round brackets around items separated by commas denote a tuple, thus (p,q). Let I denote an identity function defined, for a given domain and codomain, by the rule I[t] = t. Let f_k denote the k'th component function of a vector-valued function, that is, a function with values in Rⁿ, for n > 1. Let D[g,x] denote the (total) derivative of a function, g, evaluated at x, which we'll identify with its Jacobian matrix. Let D_i[h,(x,y)] denote the i'th partial derivative of h, evaluated at (x,y).

Now, to your the particular example. Consider the case described in the Wikipedia section Mute linked to. I'll keep their letter F for this "outer" function from R² to R, or rather, its restriction to the subset of R² where its value, F[x,y], is identically equal to 0, but I'll introduce a new letter, f, which will stand for the "inner" function of the composition, thus

$$F:U \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}, \enspace F[x,y]=0,$$

$$f = (I,f_2):\mathbb{R} \rightarrow \mathbb{R}^2, \enspace f[t] = (I[t],f_2[t]) = (t,f_2[t]),$$

so that, letting the composition operator "o" make all necessary restrictions to ensure that the expression below is well defined (EDIT: I could have made this simpler by just defining the range of f to be U.),

$$F \circ f : \mathbb{R} \rightarrow \mathbb{R}, \enspace F \circ f[t] = F[I[t],f_2[t]]=F[t,f_2[t]].$$

Does this make sense so far? The rest is just application of the (multivariable) chain rule, and a little algebra. I'll present the whole thing in more traditional notation afterwards; if this is heavy going, you might want to look at the traditional versions first.

$$D[F \circ f,t] = D[F,f[t]] \circ D[f,t],$$

which, in matrix form (since composition of matrices, viewed as linear operators, amounts to matrix multiplication), is

$$\begin{pmatrix} D_1[F,(t,f_2[t])] & D_2[F,(t,f_2[t])] \end{pmatrix} \begin{pmatrix}D[I,t] \\ D[f_2,t]\end{pmatrix}$$

$$=D_1[F,(t,f_2[t])] \cdot 1 + D_2[F,(t,f_2[t])] \cdot D[f_2,t].$$

In more traditional notation:

$$\frac{\mathrm{d} }{\mathrm{d} t} F \circ f (t) \bigg|_{t_0} = \sum_{i=1}^{2} \partial_i F(x,y) \bigg|_{(t_0,f_2(t_0))} \cdot \frac{\mathrm{d} }{\mathrm{d} t} f_i(t) \bigg|_{t_0}$$

$$=\partial_1 F(x,y) \bigg|_{(t_0,f_2(t_0))} \cdot 1 + \partial_2 F(x,y) \bigg|_{(t_0,f_2(t_0))} \cdot \frac{\mathrm{d} }{\mathrm{d} t} f_2(t) \bigg|_{t_0}$$

$$=\partial_1 F(x,y) \bigg|_{(t_0,f_2(t_0))} + \partial_2 F(x,y) \bigg|_{(t_0,f_2(t_0))} \cdot \frac{\mathrm{d} }{\mathrm{d} t} f_2(t) \bigg|_{t_0}$$

$$=\frac{\partial }{\partial x} F(x,y) \bigg|_{(t_0,f_2(t_0))} + \frac{\partial }{\partial y} F(x,y) \bigg|_{(t_0,f_2(t_0))} \cdot \frac{\mathrm{d} }{\mathrm{d} t} f_2(t) \bigg|_{t_0}.$$

Now, a function identically equal to 0 is constant, so its derivative is also identically equal to 0:

$$\forall (x,y)(F[x,y] = 0) \Rightarrow \forall t(F[t,f_2[t]]=0) \Rightarrow \forall t (D[F \circ f, t] = 0).$$

Equating the above expression for the derivative of $F \circ f$ with zero, we can rearrange the equation by subtracting the first summand from both sides, then dividing by the partial derivative of F with respect to its second argument to get

$$D[f_2,t]=-\frac{D_1[F,(t,f_2[t]])}{D_2[F,(t,f_2[t])]}$$

or in good old Leibniz notation

$$\frac{\mathrm{d} }{\mathrm{d} t} f_2(t) \bigg|_{t_0}=-\frac{\partial_1F(x,y)}{\partial_2F(x,y)} \bigg|_{(t_0,f_2(t_0))} \equiv -\frac{\partial F (x,y)/\partial x}{\partial F(x,y)/\partial y} \bigg|_{(t_0,f_2(t_0))}.$$

Thanks to Arildno for helping me with this issue, in post #4 here; (if you look at the rest of that thread, best to ignore my confused ramblings at the beginning).

 

[7thSon]

If (dx,dy) is tangent to a level curve of f the df is zero on it.

Maybe Rasalhague's post will answer this but I can't read it until after work (Thanks for all of your repsonses).

I think part of my confusion is that I have been trying to reconcile all of this with trying to learn forms.

I see (dx,dy) and traditionally, I interpret it as the components of a vector, but in the context of forms it looks like a (dual) covector basis, not the components of a tangent vector.

I don't know what to properly call dx and dy... they are infinitesimal coordinate displacements? Am I correct to say they are a basis of the cotangent space at some point? Just to clarify, the notation for the basis of tangent vectors I would use is

$$\frac{\partial}{\partial x^i} = \mathbf{e_i}$$

Using this notation, I wouldn't think it's correct to write (dx,dy) as the components of a (tangent) vector.