Vector curl problem and potential

  • Thread starter navm1
  • Start date
  • #1
44
0

Homework Statement


Prove that the vector field F = (2xyz + 1, x^2 z, x^2 y) is irrotational. Find the potential φ associated with F (i.e. find the function φ for which ∇φ = F).

Homework Equations




The Attempt at a Solution



I figure for the first part I just calculate the curl, but for the second part, does this mean potential energy? perhaps it is irrelevant to me working out this question but if potential energy is mgx then taking the derivative with respect to x would leave us with a force mg. I am not sure how I would approach the second part. Thanks
 

Answers and Replies

  • #2
44
0
Am I to just integrate each component by its respective variable? Im not sure how to calculate ∇φ for a vector function

edit: just by looking at the components I got x^2yz+x because the rest have x^2 still and there would only be a +1 if there had been a x there. not sure if there was a mathematical way to calculate this too
 
Last edited:
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,368
1,035

Homework Statement


Prove that the vector field F = (2xyz + 1, x^2 z, x^2 y) is irrotational. Find the potential φ associated with F (i.e. find the function φ for which ∇φ = F).

Homework Equations




The Attempt at a Solution



I figure for the first part I just calculate the curl, but for the second part, does this mean potential energy? perhaps it is irrelevant to me working out this question but if potential energy is mgx then taking the derivative with respect to x would leave us with a force mg. I am not sure how I would approach the second part. Thanks
No. Not mgx.

mgx is a potential function for gravitational force near Earth's surface, provided that x is vertical distance.
Am I to just integrate each component by its respective variable? I'm not sure how to calculate ∇φ for a vector function

edit: just by looking at the components I got x^2yz+x because the rest have x^2 still and there would only be a +1 if there had been a x there. not sure if there was a mathematical way to calculate this too
Yes. Generally you integrate, but you need to be careful and/or clever regarding constants of integration.

This you can get pretty well by inspection.

Look at integration. Consider Fx first.
##\displaystyle\ \frac{\partial }{\partial x}\varphi(x,y,z)=F_x(x,y,z) \ ##

So that ##\displaystyle\ \varphi(x,y,z)=\int F_x(x,y,z)\,dx \ ## , treating y and z as constants.

Thus for this potential we get, ##\displaystyle\ \varphi(x,y,z)=\int (2xyz + 1)\,dx \ = x^2yz + x +C(y,z)##.

Notice that the constant of integration can be a function of y and z. However, if you compare the partial derivatives (w.r.t y and z) of this potential with Fy and Fz respectively, C(y,z) must be constant w.r.t. both y and z, since its partials w.r.t. each is zero.

In contrast to this, let us suppose that you chose to find the potential function by integrating w.r.t. y first. (or z if you wanted to: The choice is yours.)

##\displaystyle\ \varphi(x,y,z)=\int (x^2z)\,dy \ = x^2zy +C(x,z)##

Taking the partial w.r.t. z give Fz just fine so the constant of integration does not depend on z. So write C(x,z) as C(x).

However, ##\displaystyle\ \frac{\partial }{\partial x} (x^2zy +C(x))=2xyz+C'(x)## must be ##\ 2xyz + 1\ ## so ##\ C'(x) = 1\ ## thus ##\ C(x) = x\ + constant\,.##

Plug that back into the potential function.
 
  • #4
44
0
Thanks. That has definitely helped me build a little more intuition for what I was doing
 
  • #5
44
0
Also another quick related question, if I'm asked for the directional derivative in the positive x direction and calculated del-phi of a scalar function to be some (xi+yj) do I just plug numbers into the x component?

edit: I worked out that if I think of that as a vector with components (1,0) then it makes sense
 
Last edited:

Related Threads on Vector curl problem and potential

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
1
Views
661
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
2
Views
953
  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
3
Views
989
  • Last Post
Replies
2
Views
988
  • Last Post
Replies
1
Views
2K
Top