Recovering function from its gradient

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The discussion focuses on recovering a function ∅ from its gradient, given the vector field F = 3x^2y i + (x^3 + 2yz) j + y^2 k. The initial steps involve integrating ∂∅/∂x to find ∅ = x^3y + f(y,z) and determining f'(y,z) = 2yz. Participants emphasize the importance of using the correct notation for partial derivatives and suggest integrating with respect to y to find f(y,z). Additionally, a new problem is introduced regarding finding a vector field from its curl, prompting a discussion about setting up partial differential equations for each component of F. The conversation highlights the methods for solving these equations and the need for clarity in notation and approach.
Nero26
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Hi all,
I've to recover a function ∅ such that F=∇∅,F=3x^2y i +(x^3+2yz)j+y^2k.
So, ∂∅/∂x=3x^2y, Integrating w.r.t. x
∅=x^3y+f(y,z) ,Assuming there may be function of y and z.----------1
∴∂∅/∂y=x^3+f'(y,z)
now, ∂∅/∂y=(x^3+2yz)=x^3+f'(y,z)
∴f'(y,z)=2yz
To find ∅ from 1 I've to find f(y,z).
My problem is how to get f(y,z) from f'(y,z),should I integrate it w.r.t. y or z?
Thanks for your help.
 
Last edited:
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Nero26 said:
Hi all,
I've to recover a function ∅ such that F=∇∅,F=3x^2y i +(x^3+2yz)j+y^2k.
So, ∂∅/∂x=3x^2y, Integrating w.r.t. x
∅=x^3y+f(y,z)

OK so far

,Assuming there may be function of y and z.----------1
∴∂∅/∂y=x^3+f'(y,z)
now, ∂∅/∂y=(x^3+2yz)=x^3+f'(y,z)
∴f'(y,z)=2yz


You shouldn't use ' for a partial derivative. What you have is ##\phi_y = x^3 +f_y(y,z)## which must equal ##x^3+2yz##.
To find ∅ from 1 I've to find f(y,z).
My problem is how to get f(y,z) from f'(y,z),should I integrate it w.r.t. y or z?
Thanks for your help.

And that illustrates perfectly why you should use the partial symbol instead of the '. So what you actually have to solve at this step is ##f_y(y,z) = 2yz##. Can you finish it From there?
 
Last edited:
LCKurtz said:
OK so far




You shouldn't use ' for a partial derivative. What you have is ##\phi_y = x^3 +f_y(y,z)## which must equal ##x^3+2yz##.


And that illustrates perfectly why you should use the partial symbol instead of the '. So what you actually have to solve at this step is ##f_y(y,z) = 2yz##. Can you finish it From there?

Thanks a lot for your help.
Now I got it, I've to integrate it w.r.t. y to get ##f_y(y,z)## to get f(y,z),Right?
Btw, I've got a new problem,F=i(x3z-2xyz)+j(xy-3x2yz)+k(yz2-zx) is a solenoidal vector field.I've to find a V such that F=∇XV
So,now it comes to find a vector field from its curl.How can I find it? In my book only from gradient was explained but not from curl :cry: So please help me out.
 
Don't know if there's a smarter way, but the obvious is to write out three PDEs, one for each component of F.
 
haruspex said:
Don't know if there's a smarter way, but the obvious is to write out three PDEs, one for each component of F.
Sorry,I exactly didn't get you.
I think PDE=partial differential equations.If so with respect to which variable x,y,or z or all of them?
Please say something more,
Thanks a lot for your help.
 
Nero26 said:
Sorry,I exactly didn't get you.
I think PDE=partial differential equations.
Yes. You know the definition of and of cross-product, so given the vector that results, F, you can write down three PDEs, one for each component of ∇XV. That gives you three equations involving ∂V/∂x etc.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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