Prove Calculus Identities: f, g Real Valued Functions

In summary, the problem involves proving two statements using Stokes' theorem and the hypothesis that f and g are continuously differentiable real valued functions on R^3: 1) ##\oint_C (f \nabla f) \cdot dr = 0## and 2) ##\oint (f \nabla g + g \nabla f)\cdot dr = 0##. The question also includes the additional information that sigma and C satisfy the hypothesis of Stokes' theorem and that f and g have continuous second order partial derivatives.
  • #1
bugatti79
794
1

Homework Statement



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##

Homework Equations



##\nabla f = f_z i+ f_y j+f_z k##
Real valued function ##f(x,y,z)## and ##g(x,y,z)##

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
 
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  • #2
Any chance ## f \nabla g +g\nabla f ## and ## f \nabla f## might be conservative?
 
  • #3
In my opinion, there are some missing information in your problem.

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

bugatti79 said:

Homework Equations



##\nabla f = f_z i+ f_y j+f_z k##
Shouldn't this be: [itex]\nabla f = f_x i+ f_y j+f_z k[/itex]?
 
  • #4
gopher_p said:
Any chance ## f \nabla g +g\nabla f ## and ## f \nabla f## might be conservative?

sharks said:
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: [itex]\nabla f = f_x i+ f_y j+f_z k[/itex]?

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
 
  • #5
bugatti79 said:
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?
 
  • #6
sharks said:
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 continuous 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
 
  • #7
bugatti79 said:

Homework Statement



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?

Homework Equations



##\nabla f = f_z i+ f_y j+f_z k##
Real valued function ##f(x,y,z)## and ##g(x,y,z)##

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
 
  • #8
HallsofIvy said:
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
 

What is a "calculus identity"?

A calculus identity is an equation that is true for all values of the variables involved. In other words, it is a statement that is always true, regardless of the values of the functions or variables involved. In calculus, identities often involve functions and their derivatives.

What is the purpose of proving calculus identities?

The purpose of proving calculus identities is to demonstrate the validity of a statement or equation. By proving an identity, we can show that it holds true for all values of the variables involved, which allows us to use it in other mathematical calculations and proofs.

What is the process for proving a calculus identity?

The process for proving a calculus identity involves breaking down the equation into smaller, more manageable parts and using known mathematical principles and rules to manipulate and simplify these parts. This often involves using properties of derivatives, such as the product rule and chain rule, as well as basic algebraic techniques.

What are some common strategies for proving calculus identities?

Some common strategies for proving calculus identities include using common identities and properties, such as the power rule and quotient rule, to manipulate the equation. Other strategies may include simplifying using trigonometric identities or using known results from other areas of mathematics, such as the binomial theorem.

What are some tips for successfully proving calculus identities?

Some tips for successfully proving calculus identities include carefully examining the given equation and identifying any known identities or properties that can be used to manipulate it. It is also important to pay attention to the order of operations and to carefully keep track of any algebraic manipulations. Practice and familiarity with calculus concepts and techniques can also greatly aid in successfully proving identities.

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