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Work and line integral

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! I would like to make sure I properly solved that problem because I find the result strange:

    Given a force field ##F_x = axy^3##, ##F_y = bx^2y^2##, ##F_z = cz^3##.
    Calculate the work with the line integral ##\int_{C} \vec{F} \cdot d\vec{r}## from point ##P_1(1,0,0)## to ##P_2(0,1,1)## along:
    a) a line;
    b) an helix by the z-axis.

    3. The attempt at a solution

    Okay so first I parametered the line:
    [tex]\vec{r}(\tau) = (1-\tau, \tau, \tau) \mbox{ with } \tau \in [0,1][/tex]
    Is the first component of my vector correct? I have a little doubt about it.
    [tex]\implies \dot{\vec{r}}(\tau) = (-1,1,1) \\
    \implies A = \int_{0}^{1} \vec{F} \cdot \vec{r} \cdot d\tau \\
    = \int_{0}^{1} (a (1 - \tau) \tau^3, b(1-\tau)^2\tau^2, c\tau^3)(-1,1,1)d\tau \\[/tex]
    Here again, I have a little doubt about what I did with the force field vector. Is that correct?
    [tex]\int_{0}^{1} (-a\tau^3 + a\tau^4 + b\tau^2 - 2b\tau^3 + b\tau^4 + c\tau^3) d\tau \\
    a[\frac{\tau^5}{5} - \frac{\tau^4}{4}]_0^1 + b[\frac{\tau^5}{5} + \frac{\tau^3}{3} - \frac{\tau^4}{2}]_0^1 + c[\frac{\tau^4}{4}]_0^1 \\
    A = -\frac{a}{20} + \frac{b}{30} + \frac{c}{4}[/tex]
    So yeah here is my result for (a). I'd like to know if I use the right method before I attempt to solve (b).

    What do you guys think?

    Thanks a lot in advance for your answers and suggestions, I appreciate it.


    Julien.
     
    Last edited: May 7, 2016
  2. jcsd
  3. May 7, 2016 #2

    TSny

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    Looks good. (In your integral for A you meant to type a dot over the r vector.)
     
  4. May 7, 2016 #3
    @TSny Ah yes, thanks a lot. I leave it as it is so that your message doesn't look strange in the feed :)

    Great that it is correct. I'm struggling to find the equation for (b) though, I don't even get how to search for it. There is a picture with the problem, I attached it.

    Thanks for all your help, I'm always making progress here.


    Julien.
     

    Attached Files:

  5. May 7, 2016 #4
    Is the equation ##x(\tau) = \cos \tau, y(\tau) = \sin \tau, z = \tau## actually?
     
  6. May 7, 2016 #5

    TSny

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    Close. Make sure the path goes through the final point (0, 1, 1).
     
  7. May 7, 2016 #6
    @TSny Oh yeah I said that kind of randomly but I think I get it now. What about that: ##x(\tau) = \cos( \tau), y(\tau) = \sin (\frac{\tau}{2}), z(\tau) = \sin(\frac{\tau}{2})## with ## \tau \in [0,\pi]##?
     
  8. May 7, 2016 #7

    TSny

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    This is no longer a helix. You were much closer before.
     
  9. May 7, 2016 #8
    @TSny Oops. The integral was also not that nice anymore. :biggrin:

    Okay another try: ##x(\tau) = \cos(\pi \tau), y(\tau) = sin(\frac{\pi}{2} \tau) , z(\tau) = \tau ## with this time ##\tau \in [0,1]##.
     
  10. May 7, 2016 #9

    TSny

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    This is not a helix either and it doesn't pass through P2. But it's pretty close!
     
  11. May 7, 2016 #10
    @Ah I think I was thinking too much about the path. This should be it: ##x(\tau) = cos(\frac{\pi}{2} \tau) , y(\tau) = sin(\frac{\pi}{2} \tau), z(\tau) = \tau ##, right?
     
  12. May 7, 2016 #11

    TSny

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    Right. Good luck with the integration. A little tedious. Can you use software?
     
  13. May 7, 2016 #12
    @TSny I guess I could, but I'll first give it a go. See you in 20 hours :DD
     
  14. May 7, 2016 #13

    TSny

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    :smile: (It's not THAT bad.)
     
  15. May 7, 2016 #14
    @TSny At the end I used a calculator because I must go to sleep. Did you also calculate it? I get ##A = \frac{-a}{5} + \frac{2b}{15} + \frac{c}{4}##.

    I have an extra question: in the next exercise, they ask for which values a, b, c does this force field have a potential, and what is it. I answered the first one like that:

    ##\vec{F} \mbox{ is conservative, if } \vec{\nabla} \times \vec{F} = 0\\
    \vec{\nabla} \times \vec{F} = (\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z})\vec{e_x} + (\frac{\partial F_x}{\partial x} - \frac{\partial F_z}{\partial x})\vec{e_y} + (\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y})\vec{e_z} \\
    = 0 \vec{e_x} + 0 \vec{e_y} + (2bxy^2 - 3axy^2) \vec{e_z} \\
    = xy^2 (2b - 3a) \vec{e_z} \\
    \implies \vec{F} \mbox{ is conservative, if } 2b - 3a = 0 \\
    \implies b = \frac{3}{2} a##
    Is that a correct answer? I'm very unexperienced with gradients. What does it mean physically that z doesn't play a role in the "conservativeness" of the field?

    Thanks a lot.

    Julien.
     
  16. May 7, 2016 #15

    TSny

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    Yes, that's what my computer got.
    Yes, very good.
    I'm not sure what you are asking here. In this particular problem, only the z-component of F depends on z. Thus, the curl of F does not involve z. But, I don't think that addresses your question.
     
  17. May 8, 2016 #16
    @TSny Thanks again for your answer. Nevermind my question, I think it was senseless. :)

    To calculate the potential, can I do the integral of the force field?
    [tex]
    \phi = - \int \vec{F} d\vec{r} = - \int (axy^3, \frac{3}{2} ax^2y^2, c z^3) d\vec{r} \\
    = - (\frac{a}{2} x^2 y^3, \frac{a}{2} x^2 y^3, \frac{c}{4} z^4) + \vec{c} [/tex]

    If that's correct, I guess I could also determine the constant with the help of the points ##P_1## and ##P_2##, right?

    Thanks a lot in advance for your answer.


    Julien.
     
  18. May 8, 2016 #17

    TSny

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    Note that you ended up with a vector expression for something that should be a scalar function. Suppose you integrated the force along a path from the origin to some arbitrary point (x, y, z)? What would the integral represent in terms of ##\phi(x, y, z)##?
     
  19. May 8, 2016 #18
    Okay, I am not sure to properly understand what you said but I give it a go using some resources from Internet too. It's kind of tedious, so after the 1st part I'll skip the steps:

    [tex]
    \vec{F} = -\nabla \phi = -(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}) \\
    \implies axy^3 = - \frac{\partial \phi}{\partial x} \implies d\phi = - \int axy^3 dx \\
    \implies \phi = -\frac{a}{2} x^2y^3 + g(y,z) \\
    \implies \frac{\partial \phi}{\partial y} = \frac{\partial g(y,z)}{\partial y} = -\frac{3}{2} a x^2 y^2 \\
    \implies g(y,z) = -\frac{3}{2}a \int x^2y^2 dy = -\frac{a}{2}x^2y^3 + h(z) \\
    \implies \frac{\partial h(z)}{\partial z} = -cz^3 \\
    \implies h(z) = -c \int z^3 dz = -\frac{c}{4} z^4 \\
    \implies \phi = -ax^2y^3 - \frac{c}{4} z^4
    [/tex]

    What do you think? Did I do that right now?


    Thanks a lot for your answers.

    Julien.
     
  20. May 8, 2016 #19

    TSny

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    Is the middle expression of the last line correct if you use the expression for ##\phi## at the end of the previous line?

    But your method is good.

    My earlier suggestion was to integrate F along a path from the origin to an arbitrary point (x, y, z). Up to a sign, this will give you the change in ##\phi##: ##\phi(x, y, z) - \phi(0,0,0)##. Thus, you will have an expression for ##\phi(x, y, z)##. The result is independent of path, so you can choose a path that makes it easy to do the integral. But this method is not very different than your method.
     
  21. May 8, 2016 #20
    @TSny Thanks for your answer (so quick!). I don't find the error though. I get it that the partial derivative of my final expression of Φ with respect to x doesn't match with the line you pointed out, but I didnt find yet why.
     
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