Potential function for conservative vector fields

In summary, you're integrating a function with respect to one variable, but the derivative of that function is also dependent on another variable.
  • #1
kasse
384
1
Task:

Find a potential function for the conservative vector field (y,x,1)

My work:

(Df/Dx, Df/Dy and Df/Dz are the partial derivatives)

Df/Dx=1, Df/Dy=x, Df/Dz=1

I take the first eq. and integrate, so that I get

f(x,y,z) = yx + C

I then derivate with respect to y:

Df/Dy= x + C '

I compare this to the other eq. Df/Dy=x, hence C ' is 0, and consequently C is also 0.

My answer is then

f(x,y,z)=yx

Which is wrong! Correct answer is yx+z. Where do I fail?
 
Physics news on Phys.org
  • #2
Hi,

I'm confused what you're doing. If you're looking for the potential function, you're looking for f such that it's gradient is the vector field you gave. Note also, that when integrating a function g(x,y,z) with respect to one variable say x, you won't just have a constant of integration, but a function dependent on y and z.
 
  • #3
Glass said:
Hi,

I'm confused what you're doing. If you're looking for the potential function, you're looking for f such that it's gradient is the vector field you gave. Note also, that when integrating a function g(x,y,z) with respect to one variable say x, you won't just have a constant of integration, but a function dependent on y and z.

Since the gradient of f is the vector field, we have that the partials are

Df/Dx=y, Df/Dy=x and Df/Dz=1, right?

I integrate Df/Dx=y with respect to x. This gives: f=yx+C(y,z). Then I derivate this with respect to y and get: Df/Dy=x + C ' (x,y). From before we have that Df/Dy=x, which means that C ' (x,y)= 0 and C(x,y)=0.

Southe potential function is

f(x,y,z)=xy

See what I've done?
 
  • #4
kasse said:
Since the gradient of f is the vector field, we have that the partials are

Df/Dx=y, Df/Dy=x and Df/Dz=1, right?

I integrate Df/Dx=y with respect to x. This gives: f=yx+C(y,z). Then I derivate this with respect to y and get: Df/Dy=x + C ' (x,y). From before we have that Df/Dy=x, which means that C ' (x,y)= 0 and C(x,y)=0.
Here, C should be a function of y and z. Also, the prime notation is ambiguous, since C is a function of two variables. You should write Df/Dy=x+DC/Dy, which tells us that C is not a function of y (on comparing this with your second equation above). You must then differentiate f with respect to z, which gives Df/Dz=DC/Dz, but from the third equation, this is equal to 1, which says that C=z; hence f=xy+z
 
Last edited:
  • #5
Its not a normal integral youre doing, its like a 'partial integral.' I remember this confusing me too.
 
  • #6
OK, I figured it our.
 
Last edited:

1. What is a potential function for conservative vector fields?

A potential function for conservative vector fields is a scalar function that represents the potential energy associated with a conservative vector field. It is also known as a conservative scalar field or simply a scalar potential.

2. How is a potential function related to conservative vector fields?

A potential function is related to conservative vector fields through the fundamental theorem of calculus. According to this theorem, if a vector field is conservative, then it can be written as the gradient of a potential function. This means that the potential function is the antiderivative of the vector field.

3. What are the properties of a potential function for conservative vector fields?

A potential function for conservative vector fields must satisfy two properties: it must be differentiable and its gradient must equal the vector field. In other words, the partial derivatives of the potential function with respect to each variable must equal the components of the vector field.

4. How is a potential function calculated for a conservative vector field?

To calculate a potential function for a conservative vector field, you can use the method of separation of variables. This involves solving a system of partial differential equations, where each equation corresponds to one of the components of the vector field.

5. What are some real-world applications of potential functions for conservative vector fields?

Potential functions for conservative vector fields have various applications in physics and engineering, such as in the study of electric and magnetic fields, fluid dynamics, and mechanics. They are also used in optimization problems, where finding the minimum or maximum value of a function involves finding its potential function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
947
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
706
  • Calculus and Beyond Homework Help
Replies
3
Views
828
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
849
  • Calculus and Beyond Homework Help
Replies
1
Views
683
  • Calculus and Beyond Homework Help
Replies
6
Views
695
Back
Top