- #1

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F(x,y,z)=-3x

**i**-2y

**j**+

**k**

f(x,y,z)=?

I'm not sure what the problem is asking. calcualting curl F needs integration and a boundary. I dont know why they ask for f(x,y,z). could someone explain what this problem as asking me to do?

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- Thread starter UrbanXrisis
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- #1

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F(x,y,z)=-3x

f(x,y,z)=?

I'm not sure what the problem is asking. calcualting curl F needs integration and a boundary. I dont know why they ask for f(x,y,z). could someone explain what this problem as asking me to do?

- #2

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- #3

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so to calculate f, i take the partial derivative of F?

- #4

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You differntiateUrbanXrisis said:so to calculate f, i take the partial derivative of F?

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So [itex] f(x,y,z) [/itex] will exist ONLY when [itex] \vec F [/itex] is conservative. So what does the gradient ask for?

So you must satisfy the following conditions:

[tex] \frac{\partial}{\partial x}f(x,y,z) = F_x [/tex]

[tex] \frac{\partial}{\partial y}f(x,y,z) = F_y [/tex]

[tex] \frac{\partial}{\partial z}f(x,y,z) = F_z [/tex]

Where [itex] F_x [/itex] is the x-component of the vector function.

- #6

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[tex] \nabla f(x,y,z) = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j +\frac{\partial f}{\partial z} k[/tex]

[tex] \nabla f(x,y,z) = -3 i-2j [/tex]

would this be f?

- #7

nrqed

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UrbanXrisis said:i-2yj+k

[tex] \nabla f(x,y,z) = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j +\frac{\partial f}{\partial z} k[/tex]

[tex] \nabla f(x,y,z) = -3 i-2j [/tex]

would this be f?

No. What you want is a potential f(x,y,z) such that [itex] \nabla f [/itex] gives you back the vector field [itex]{\vec F}[/itex] that you gave in your first post. This is what Frogpad was saying in his last post.

- #8

nrqed

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UrbanXrisis said:

F(x,y,z)=-3xi-2yj+k

f(x,y,z)=?

I'm not sure what the problem is asking. calcualting curl F needs integration and a boundary. I dont know why they ask for f(x,y,z). could someone explain what this problem as asking me to do?

First thinsg first. Have you computed the curl of F and checked if F was conservative?

- #9

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[tex] \int _C F dr=\int \int curlFdS[/tex]

- #10

nrqed

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UrbanXrisis said:

[tex] \int _C F dr=\int \int curlFdS[/tex]

Oh! So you haven't been taught the definition of the curl! I can see how things may be confusing! The above is an application of the curl (and could be taken as a definition, I suppose), but the curl of a vector field is defines to be [itex] \nabla \times {\vec F(x,y,z)} [/itex], that is, it is the cross product if you will of the nabla operator with the vector field. If your book does not define this it is because it assumes you are familiar with this. Look at a more basic math or E&M book and it will be defined there for sure.

The key point is that if the curl of a vector field is zero then the vector field can be written as a gradient of a scalar function (your small cap f). If the vector field represents a force field, then this implies that one can define a potential energy (because the force is conservative)

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- #12

nrqed

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The curl is a vector. Its x component is [itex] \partial_y F_z(x,y,z) -\partial_z F_y(x,y,z) [/itex], the y component is [itex] - \partial_x F_z(x,y,z) +\partial_z F_x(x,y,z) [/itex] and the z component is [itex] \partial_x F_y(x,y,z) -\partial_y F_x(x,y,z) [/itex]. Calculating this for your F you should get zero for all three components.nrqed said:Oh! So you haven't been taught the definition of the curl! I can see how things may be confusing! The above is an application of the curl (and could be taken as a definition, I suppose), but the curl of a vector field is defines to be [itex] \nabla \times {\vec F(x,y,z)} [/itex], that is, it is the cross product if you will of the nabla operator with the vector field. If your book does not define this it is because it assumes you are familiar with this. Look at a more basic math or E&M book and it will be defined there for sure.

The key point is that if the curl of a vector field is zero then the vector field can be written as a gradient of a scalar function (your small cap f). If the vector field represents a force field, then this implies that one can define a potential energy (because the force is conservative)

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