Finding potential function for field F

In summary, the book says to find the potential function for the vector field by integrating each section respective to its component direction, but other wise. This gives the equation 2xy+2xz+2zy+C.
  • #1
skate_nerd
176
0
In doing a seemingly simple homework problem, I came across a snag...
It says to find the potential function \(f\) for the vector field \(\vec{F}\). The problem states
$$\vec{F}=(y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}$$
So I figured that just the simple integral of each section respective to its component direction would give the potential function
$$f=x(y+z)+y(x+z)+z(x+y)+C$$
However, the answer in the back of the book says other wise. It claims that
$$f=x(y+z)+zy+C$$
Now I can kind of recognize that distributing out my original answer would give something similar to this, but wouldn't it really be
$$f=2x(y+z)+2zy+C$$
Which is ultimately
$$f=2xy+2xz+2zy+C$$
Does anyone have any idea how they got to this answer? I'm a little stuck here.
 
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  • #2
skatenerd said:
In doing a seemingly simple homework problem, I came across a snag...
It says to find the potential function \(f\) for the vector field \(\vec{F}\). The problem states
$$\vec{F}=(y+z)\hat{i}+(x+z)\hat{j}+(x+y)\hat{k}$$
So I figured that just the simple integral of each section respective to its component direction would give the potential function
$$f=x(y+z)+y(x+z)+z(x+y)+C$$
However, the answer in the back of the book says other wise. It claims that
$$f=x(y+z)+zy+C$$
Now I can kind of recognize that distributing out my original answer would give something similar to this, but wouldn't it really be
$$f=2x(y+z)+2zy+C$$
Which is ultimately
$$f=2xy+2xz+2zy+C$$
Does anyone have any idea how they got to this answer? I'm a little stuck here.

I assume you have first shown that the field is irrotational, i.e. that \(\displaystyle \displaystyle \begin{align*} \nabla \times \mathbf{F} = \mathbf{0} \end{align*}\). Now, if it is irrotational, then there exists a scalar function \(\displaystyle \displaystyle \begin{align*} \phi \left( x, y, z \right) \end{align*}\) so that \(\displaystyle \displaystyle \begin{align*} \nabla \phi = \mathbf{F} \end{align*}\). This would mean that \(\displaystyle \displaystyle \begin{align*} \left( \frac{\partial \phi}{\partial x} , \frac{\partial \phi}{\partial y} , \frac{\partial \phi}{\partial z } \right) = \left( y + z, x + z, x + y \right) \end{align*}\). Equating each of these components gives

\(\displaystyle \displaystyle \begin{align*} \frac{\partial \phi}{\partial x} &= y + z \\ \phi &= \int{ \left( y + z \right) \, \partial x } \\ \phi &= x\,y + x\,z + f(y, z) \\ \\ \frac{\partial \phi}{\partial y} &= x + z \\ \phi &= \int{ \left( x + z \right) \, \partial y } \\ \phi &= x\,y + y\,z + g(x, z) \\ \\ \frac{\partial \phi}{\partial z} &= x + y \\ \phi &= \int{ \left( x + y\right) \, \partial z } \\ \phi &= x\,z + y\,z + h(x, y) \end{align*}\)

When we compare all the different components to \(\displaystyle \displaystyle \begin{align*} \phi \end{align*}\) we find that \(\displaystyle \displaystyle \begin{align*} \phi = x\,y + x\,z + y\,z + C \end{align*}\).
 
  • #3
Sorry to anybody who may have been trying to answer this at the moment, but I just realized how I was doing it wrong. I guess I didn't exactly understand the full process of finding a potential function. Nevermind!

- - - Updated - - -

Thanks ProveIt. That makes things pretty simple. Got it now
 

FAQ: Finding potential function for field F

What is a potential function?

A potential function is a mathematical function that is used to describe the behavior of a vector field. It helps to determine the potential energy of a system in a given field.

How do you find the potential function for a field?

To find the potential function for a field, you need to integrate the field with respect to one of the variables. This will result in a function that represents the potential energy of the system in that field.

Why is finding the potential function important?

Finding the potential function is important because it allows us to better understand the behavior of a vector field and determine the potential energy of a system in that field. It also allows us to simplify complex calculations and make predictions about the behavior of a system.

Can a potential function exist for all types of fields?

No, a potential function does not exist for all types of fields. It depends on the type of field and its properties. For example, conservative fields have a potential function, while non-conservative fields do not.

How do you know if a potential function is valid for a specific field?

To determine if a potential function is valid for a specific field, you can use the gradient test. If the gradient of the potential function is equal to the field, then the potential function is valid for that field.

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