Partial Derivatives of Vectors and Gradients

The Head

I was reading a section on vector fields and realized I am confused about how to take partials of vector quantities. If V(x,y)= yi -xj, I don't understand why the [itex]\partialx[/itex]= y and the [itex]\partialy[/itex]= -x. The problem is showing why the previous equation is not a gradient vector field (because the second-order partials are not equal). 3 questions arise for me:

1) It seems to me that the only way you could obtain this answer for the [itex]\partialx[/itex] part of the gradient would be to only look at the i value and ignore the j value. But why can you ignore the j component? Is that because of the gradient's definition

2) Following off of the previous question, would[itex]\partialx[/itex] be something different (still using these vector components) if we were not interested in the gradient? And thus, is there something about taking gradients of vectors that can lead to an inequality among partials (because of only looking at certain components at certain times)?

3)And how is it that [itex]\partialx[/itex] yi = y? Does i serve as "x," so that it is like taking the partial of yx? If so why is that?

The Head

Sorry, every time I attempted to write the partial derivative of V with respect to x, it just ended up saying "partial" in red font. Just so you know what it was intended to be...

Some Pig

A gradient vector field means the vector field is the gradient of a scalar function.
For example, if ##\vec V## is a gradient vector field, then there is a scalar function ##\phi## such that ##\vec V=\nabla\phi.##
If ##\vec V=v_1\vec i+v_2\vec j##, then ##v_1=\frac{∂\phi}{∂x},v_2=\frac{∂\phi}{∂y},## so ##\frac{∂v_1}{∂y}=\frac{∂v_2}{∂x}=\frac{∂^2\phi}{∂x∂y}.##
For ##\vec V=y\vec i-x\vec j,\frac{∂}{∂y}(y)\ne\frac{∂}{∂x}(-x),## not a gradient vector field.

HallsofIvy

You have "\partialx" as a single "word" and it does not recognize that use "\partial x" instead with a space between \partial and x.

The Head

Thanks for your hep but I am still confused about a couple of things though. Why is [itex]\partial[/itex] / [itex]\partial x[/itex] of yi - xj = y? To me, it makes sense that when you are dealing with the i component and taking its partial with respect to x, you look at the value in front of the component, which is y. And [itex]\partial[/itex] / [itex]\partial x [/itex] (y) = zero. I just want to make sense of why the answer is y.

Also, I understand that equality of mixed partials must be equal is an implicit idea within the concept of a gradient, but is there a reason or proof for this, or just a definition?

Thank you again.

HallsofIvy

It isn't. Where did you get the idea that it was? The partial derivative, with respect to x, of yi- xj is -j.

As long as the derivatives are continuous,
[tex]\frac{\partial^2 f}{\partial x\partial y}= \frac{\partial^2 f}{\partial y\partial x}[/tex]

You should be able to find a proof of that in any Calculus book.

You can find a paper on how Euler derived it here:
http://www.maa.org/editorial/euler/H...erivatives.pdf

The Head

In Marsden and Tromba's Vector Calculus (5e, p.289), there is an example calculating the gradient of V(x,y)= yi -xj. In the next step it shows the gradient being equal to the product of [itex]\partial f [/itex] / [itex]\partial x [/itex] and yi, plus the product of [itex]\partial f [/itex] / [itex]\partial y [/itex] the and -xj. It then says that [itex]\partial f [/itex] / [itex]\partial x [/itex] = y and [itex]\partial f [/itex] / [itex]\partial y [/itex] = -x. I would have thought that the two partials would both be zero because the variable with which you are taking the partial is seemingly absent in each expression. Is this an error in the text, or am I misinterpreting something in the text?

Oh, and thanks for the interesting link to the paper!

HallsofIvy

I would appreciate it if you gave the full discussion. If you are taking "the gradient of V(x,y)" where did this "f" come from?

The Head

Here is the complete example:

Example 7: Show that the vector field V on R² defined by V= yi-xj is not a gradient vector field; that is, there is no C¹ function f such that V=[itex]\nabla[/itex] f(x,y)= [itex]\partial f [/itex] /[itex]\partial x [/itex] i + [itex]\partial f [/itex] / [itex]\partial y [/itex] j.

Solution Suppose that such an f exists. Then [itex]\partial f [/itex] /[itex]\partial x [/itex] = y and [itex]\partial f [/itex] /[itex]\partial y [/itex] = -x. Because these are C¹ functions, f itself must have continuous first and second-order partial derivatives. But... (goes on to show mixed partials are not equal.

Is this different because of the "f" you were talking about...somehow it changes the operation?

HallsofIvy

Thanks, that makes sense. But you are NOT taking the partial derivatives of yi and -xj. You want to determine if there is a scalar function f(x,y) so that the derivative of f with respect to x is equal to the coefficent of i (which in this problem is y) and the derivative of f with respect to y is equal tio the coefficient of j (which in this problem is -x).

In other words, rather than saying [itex]\partial/\partial x (yi- xj)= y[/itex] and that [itex]\partial/\partial y (yi- xj)= -x[/itex] they are asking if there exist a function f(x, y) such that [itex]\partial f/\partial x= y[/itex] and [itex]\partial f/\partial y= -x[/itex].

If it were true that [itex]\partial f/\partial x= y[/itex] then it would have to be true that [itex]\partial^2 f/\partial x\partial y= 1[/itex]. And if it were true that [itex]\partial f/\partial y= -x[/itex] then it would have to be true that [itex]\partial^2 f/\partial y\partial x= -1[/itex].

Here's a different way of thinking about it. If if were true that [itex]\partial f/\partial x= y[/itex], because taking the partial derivative with respect to x treats y like a constant, integrating, we must have f(x,y)= xy+ C- except that because we treating y like a constant, that constant of integration, C, might depend on y: f(x,y)= xy+ C(y). Differentiating with respect to y, we have [itex]\partial f/\partial y= x+ C'(y)[/itex]. But that must be equal to -x: x+ C'(y)= -x so that C'(y)= -2x- which is, of course, impossible since C only depends on y.

The Head

So much clearer now-- you have been a wonderful help!