Get Expert Tips for Proving the Product Rule Using Gradient Method

  • Thread starter Thread starter Fjolvar
  • Start date Start date
  • Tags Tags
    Class
Click For Summary

Homework Help Overview

The original poster is attempting to prove the product rule using the gradient of two scalar-valued functions, u and v, as part of a homework problem. The discussion revolves around the proper notation and setup for this proof.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the product rule for gradients and explore the necessary conditions for differentiability. There are questions about notation and setup, as well as the implications of certain assumptions on the solution.

Discussion Status

Some participants have provided guidance on how to approach the proof, including a breakdown of the product rule and considerations for the differentiability of functions involved. The original poster expresses gratitude and indicates they may have sufficient information to proceed.

Contextual Notes

There is mention of missing information regarding the differentiability of the functions, which is noted as necessary for the proof. The original poster is also working within the constraints of a specific homework problem.

Fjolvar
Messages
156
Reaction score
0
I've been up for hours trying to figure out the proper notation to prove problem #4 (1.6.5) on the attached files. I'm not sure how to set this up. It is basically proving the product rule using the gradient of two functions. The rest of the problem is on the backside file, but help getting started on the first part would be enough. Any help would be greatly appreciated. Thanks!
 

Attachments

Physics news on Phys.org
Any ideas?
 
It looks to me like a pretty standard "just plug it in and calculate"!

You are asked to show that [itex]\nabla(uv)= \nabla(u)v+ u\nabla(v)[/itex] where u and v are scalar valued functions of variables x and y. This is, basically, just the "product rule" for gradients.

Okay,
[tex]\nabla(uv)= \frac{\partial uv}{\partial x}\vec{i}+ \frac{\partial uv}{\partial y}\vec{j}[/tex]
Apply the product rule to both partial derivatives:
[tex]\nabla(uv)= \left(u\frac{\partial v}{\partial x}+ \frac{\partial u}{\partial x}v\right)\vec{i}+ \left(u\frac{\partial v}{\partial y}+ \frac{\partial u}{\partial y}v\right)\vec{j}[/tex]
and regroup.

For part (a), if f(u, v)= 0 (where f is differentiable- that is NOT stated in the text but is necessary), then
[tex]\frac{\partial f}{\partial x}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial x}= 0[/tex]
and
[tex]\frac{\partial f}{\partial y}= \frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+ \frac{\partial f}{\partial v}\frac{\partial v}{\partial y}= 0[/tex]
Think of those as two equations to solve for the "unknowns" [itex]\partial f/\partial u[/itex] and [itex]\partial f/\partial v[/itex]. What condition on the "coefficients", [itex]\partial u/\partial x[/itex], [itex]\partial u/\partial y[/itex], [itex]\partial v/\partial x[/itex], and [itex]\partial v/\partial y[/itex] is necessary in order that that have a solution (think about the determinant of the matrix of coefficients- that's the point of (b)).
 
Thank you very much, I think I've got it from here.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
14
Views
13K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
29
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K