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Given potential find F

  1. Mar 4, 2008 #1
    [SOLVED] given potential: find F

    1. The problem statement, all variables and given/known data

    Find the force for the following potential energy function:
    [tex]V=ce^{-(\alpha x+\beta y+\gamma z)}[/tex]

    2. Relevant equations

    [tex]\mathbf{F}=-\nabla V[/tex]

    [tex]F_x=- \frac{\partial V}{\partial x}[/tex]
    [tex]F_y=- \frac{\partial V}{\partial y}[/tex]
    [tex]F_z=- \frac{\partial V}{\partial z}[/tex]

    3. The attempt at a solution

    By the chain rule:

    [tex]F_x=-\frac{\partial V}{\partial x}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\alpha)][/tex]
    [tex]F_y=-\frac{\partial V}{\partial y}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\beta)][/tex]
    [tex]F_z=-\frac{\partial V}{\partial z}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\gamma)][/tex]

    Additionally,
    [tex]\mathbf{F}=F_x+F_y+F_z[/tex]

    So add it up and factor out the common factors and get
    [tex]\mathbf{F}=-(\alpha x+\beta y+\gamma z)ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})[/tex]

    And that would be my final answer, except that the back of the book says that
    [tex]\mathbf{F}=ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})[/tex]

    Can you help me find my error?

    Thanks!
     
  2. jcsd
  3. Mar 4, 2008 #2
    [SOLVED] given potential find F

    1. The problem statement, all variables and given/known data

    Find the force for the following potential energy function:
    [tex]V=ce^{-(\alpha x+\beta y+\gamma z)}[/tex]

    2. Relevant equations

    [tex]\mathbf{F}=-\nabla V[/tex]

    [tex]F_x=- \frac{\partial V}{\partial x}[/tex]

    [tex]F_y=- \frac{\partial V}{\partial y}[/tex]

    [tex]F_z=- \frac{\partial V}{\partial z}[/tex]

    3. The attempt at a solution

    By the chain rule:

    [tex]F_x=-\frac{\partial V}{\partial x}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\alpha)][/tex]

    [tex]F_y=-\frac{\partial V}{\partial y}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\beta)][/tex]

    [tex]F_z=-\frac{\partial V}{\partial z}=-[-(\alpha x+ \beta y + \gamma z)ce^{-(\alpha x+\beta y+\gamma z)}(-\gamma)][/tex]

    Additionally,
    [tex]\mathbf{F}=F_x+F_y+F_z[/tex]

    So add it up and factor out the common factors and get
    [tex]\mathbf{F}=-(\alpha x+\beta y+\gamma z)ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})[/tex]

    And that would be my final answer, except that the back of the book says that
    [tex]\mathbf{F}=ce^{-(\alpha x+\beta y +\gamma z)}(\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k})[/tex]

    Can you help me find my error?

    Thanks!
     
  4. Mar 4, 2008 #3
    Check the derivatives

    etc. Recall

    [tex]\frac{\partial}{\partial x}e^{f(x)} = e^{f(x)}\frac{\partial f}{\partial x}.[/tex]
     
  5. Mar 4, 2008 #4

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You used the chain rule incorrectly.

    [tex] \textrm{Let}\ \ \ u(x,y,z) = \alpha x+\beta y+\gamma z [/tex]

    Then

    [tex] \frac{\partial }{\partial x}(ce^{-u}) = \frac{d}{du}(ce^{-u}) \frac{\partial u}{\partial x} [/tex]

    [tex] = -\alpha ce^{-(\alpha x+\beta y+\gamma z)} [/tex]

    likewise for the partial derivatives wrt y and z.
     
  6. Mar 4, 2008 #5
    As I recall, chain rule goes like "Derivative of the inner function multiplied by the derivative of the outer function evaluated at the inner function".
     
  7. Mar 4, 2008 #6
    Right... it's the easy math that bites me in the butt.
    Thanks!
     
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