# Calculus Identities

by bugatti79
Tags: calculus, identities
 P: 652 1. The problem statement, all variables and given/known data Suppose f is a continously differentiable real valued function on R^3 and F is a continously differentiable vector field Prove 1)##\oint (f \nabla g +g\nabla f) \cdot dr=0## 2) ##\oint(f \nabla f)\cdot dr=0## 2. Relevant equations ##\nabla f = f_z i+ f_y j+f_z k## Real valued function ##f(x,y,z)## and ##g(x,y,z)## 3. The attempt at a solution 1) ##f \nabla g =fg_x i +fg_y j+fg_z k## ##g \nabla f =gf_x i +gf_y j+gf_z k## ##\implies (f \nabla g + g \nabla f )\cdot dr## ##= (fg_x i +fg_y j+fg_z k+gf_x i +gf_y j+gf_z k)\cdot(dx i+dyj+dzk)## 2) ##(f \nabla f)\cdot dr= (ff_xi+ff_yj+ff_zk)\cdot(dxi+dyj+dzk)## How do these work out to be 0? Thanks
 P: 310 Any chance ## f \nabla g +g\nabla f ## and ## f \nabla f## might be conservative?
PF Gold
P: 836
In my opinion, there are some missing information in your problem.

Is ##g## also a continously differentiable real valued function in R^3?

 Quote by bugatti79 2. Relevant equations ##\nabla f = f_z i+ f_y j+f_z k##
Shouldn't this be: $\nabla f = f_x i+ f_y j+f_z k$?

P: 652

## Calculus Identities

 Quote by gopher_p Any chance ## f \nabla g +g\nabla f ## and ## f \nabla f## might be conservative?
 Quote by sharks In my opinion, there are some missing information in your problem. Is ##g## also a continously differentiable real valued function in R^3? Shouldn't this be: $\nabla f = f_x i+ f_y j+f_z k$?
Yes, you are correct sharks, that is a typo. It should be as you have stated.
##...f_xi...##

It does not mention anything about g but perhaps we take it that it is also a real valued function?

I believe I left out the following important information
C is a smooth, simple closed curve which lies on the surface of a paraboloid in R^3. I guess this means integrand is conservative, right?
But I still not sure how it goes to 0, there must be additional lines
Thanks
PF Gold
P: 836
 Quote by bugatti79 F is a continously differentiable vector field
Do you mean ##f## or did the question involve ##\vec F##, in which case, has any information been given about the latter?
P: 652
 Quote by sharks Do you mean ##f## or did the question involve ##\vec F##, in which case, has any information been given about the latter?
Here is the proper question asked in full. Apologies again.

Suppose that sigma and C satisfy the hypothesis of Stokes Theorem and that f and g have continous second order partial dervivatives. Prove each of the following

##\oint_C (f \nabla g) \cdot dr = \oint \oint_\sigma (\nabla f \times \nabla g)\cdot dS##

##\oint_C (f \nabla f) \cdot dr=0##

##\oint (f \nabla g +g \nabla f)\cdot dr=0##

I am interested in the last 2 but maybe the first one allows me to complete the last 2?

Thanks
Math
Emeritus
Thanks
PF Gold
P: 38,708
 Quote by bugatti79 1. The problem statement, all variables and given/known data Suppose f is a continously differentiable real valued function on R^3 and F is a continously differentiable vector field Prove 1)##\oint (f \nabla g +g\nabla f) \cdot dr=0## 2) ##\oint(f \nabla f)\cdot dr=0##
This makes no sense. You have "g" in the conclusion but not in the hypotheses and "F" in the hypotheses but not in the conclusion. What is the problem, really?

 2. Relevant equations ##\nabla f = f_z i+ f_y j+f_z k## Real valued function ##f(x,y,z)## and ##g(x,y,z)## 3. The attempt at a solution 1) ##f \nabla g =fg_x i +fg_y j+fg_z k## ##g \nabla f =gf_x i +gf_y j+gf_z k## ##\implies (f \nabla g + g \nabla f )\cdot dr## ##= (fg_x i +fg_y j+fg_z k+gf_x i +gf_y j+gf_z k)\cdot(dx i+dyj+dzk)## 2) ##(f \nabla f)\cdot dr= (ff_xi+ff_yj+ff_zk)\cdot(dxi+dyj+dzk)## How do these work out to be 0? Thanks
P: 652
 Quote by HallsofIvy This makes no sense. You have "g" in the conclusion but not in the hypotheses and "F" in the hypotheses but not in the conclusion. What is the problem, really?
The correct thread/question is post #6 and not #1. The is no 'F' involved, that was in another very similar question (#1 which I will ignore). Only f and g are involved.

THanks

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