Recent content by Syrus

Pardon me, I meant partial derivative fx of the implicit solution f(x,y) of F(x,y,z).

Homework Statement I understand the proof of the implicit function theorem up to the point in which I have included a photo. This portion serves to prove the familiar equation for the implicit solution f(x,y) of F(x,y,z)=c. My confusion arises between equations 8.1-4 and 8.1-5 when it is stated...

Ray- you're absolutely correct- the implicit function theorem is in fact discussed and I believe both your points apply here

Well put, I assume that is why they made that assumption- so the ensuing discussion applied to all points considered (as long as grad F is nonzero).

I'm also not finding any sources on the gradient being sufficient to have a well-defined surface. Any references for this?

Charles Link- I think I understand what you're saying, but let's consider the cone given by F(X,y,z) = z^2-x^2-y^2 = 0. Grad F = 0 at (0,0,0). But this seems to have nothing with the concept you're talking about- or does it?

I guess what I'm trying to ask is why is the assumption that grad(F) is nonzero necessary in the ensuing discussions? The text goes on to discuss topics such as using grad to find equations of tangent planes to surfaces and (eventually) to prove that a curve can result as the intersection of two...

It's just the gradient of an arbitrary surface F(,x,y,z). Not necessarily a function.

None of what you guys said attests as to why you couldn't have a "surface" with those properties. Link- I've never heard of "thickness" as a consideration of surfaces- but then again- how does this take away from the SURFACE aspect of it? Gilb- who cares if there's no local or absolute minima or...

Homework Statement I am working in "Intro to PDEs with Applications" on page 6. Gradients come up in discussions of surfaces expressed as F(x,y,z). In discussing such matters, the buildup includes the assumption that grad F is not equal to the zero vector. A later line reads, "Under the...

Anyone care to provide a meaningful response?

1. Homework Statement I am looking at problem 2.2 pictured above. I have solved all portions of the question except the last part, which asks for the angle between the normal vector to the surface and the z-axis. I am aware that the normal vector is simply equal to the gradient of the surface...

Yes, I see. But when, then, all the talk about "An eigenbasis for a linear operator that operates on a vector space is a basis for that consists entirely of eigenvectors of (possibly with different eigenvalues). Such a basis may not exist." See...

Homework Statement My quantum mechanics text (in an appendix on linear algebra) states, "f the eigenvectors span the space... we are free to use them as a basis..." and then states: T|f1> = λ1f1 . . . T|fn> = λnfn My question is: is it not true that fewer than n vectors might...

Precisely what I thought, thank you.