Recent content by 7thSon

Reviving this dead thread. Thanks to those who posted. Suppose I have a contour f = f(t). In general, it is difficult to calculate an arc length parameterization of the contour f = f(s). We stated above df/ds = df/dt * dt/ds, by the chain rule. We cared about df/ds because it has unit...

What part of this terminology is outdated? And what are the more modern terms? Also, what is the proper, two sentence definition of the immersion?

Ok, I think 60% of my remaining doubts will be answered if someone can answer the following question: We know that the tangent vector \mathbf{v} = \nabla f = \frac{\partial f}{\partial x} \mathbf{e_1} + \frac{\partial f}{\partial y} \mathbf{e_2} points in the direction of maximum...

Thanks HallsOfIvy. So this makes sense that there is a restriction of y to be a function of x, or vice versa... given that the partial derivative identity i described looks like a consequence of the implicit function theorem (though strangely different for a reason that still eludes me). What...

This thread has gotten a lot of views but no responses, can someone at least say because they are perplexed or maybe my initial post is too long :P

Maybe someone could start out by explaining to me whether or not it is correct to write, as this author does, \frac{dx}{df} = \frac{\partial x}{\partial f} + \frac{\partial x}{\partial y} \frac{dy}{df} Isn't \frac{\partial x}{\partial y} always zero?

Suppose I have a smooth scalar function f defined on some region in the x-y plane. Its partial derivatives with respect to x and y are well-defined. Someone explain this "proof" to me that the quantity: \frac{dx}{df} = \frac{f_{,x}}{f_{,x}^2 + f_{,y}^2} (similar expression for dy/df) The...

Great, for future people reading this I think the reason it is so easy to be confused is simply notation (mathwonk would be glowing), but correct me if I'm wrong. For smooth f with the standard cartesian basis, \nabla f = \frac{\partial f}{\partial x^1} \mathbf{e_1} + \frac{\partial...

Woah wait a sec... I always used "one-form" and "covector" interchangeably. Were you humoring me with the first sentence? Is a gradient a one-form but not a covector, or are they in fact the same thing? Thanks for the clarification regarding gradient vs. differential being dual to each other.

Hate to revive this thread when you guys seem to have come to a conclusion, but I had another question about the cotangent space. The only example I see of a covector is the gradient of a scalar function, with the partial derivatives of some smooth scalar function f, being the components of a...

Thanks quasar, but if I might, allow me to try to make one final connection. Take an arbitrary one-form \omega and tangent vector \mathbf{v} at some point of interest \omega = f_1 \ dx^1 + f_2 \ dx^2 + f_3 \ dx^3 and \mathbf{v} = a^1 \frac{\partial}{\partial x_1} + a^2...

Bumping this old thread because I had a question that closely relates to it. I think this is an easy thing to confuse for someone just starting to learn these concepts. My question is, what exactly is meant by I understand that these differentials dx,dy, etc. map a vector to real number...

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...

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.

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...